Distributions: Theory and Applications

J.J. Duistermaat J.A.C. Kolk

Distributions: Theory and Applications

Springer Berlin Heidelberg New York Hong Kong London Milan Paris Tokyo

Preface

Not that much effort is needed, for it is such a smooth and simple theory (F. Tr`eves)

This book aims to be a thorough, yet concise and application-oriented introduction to the theory of distributions that can be covered in one semester. It evolved as a set of notes for lectures at Utrecht University over the last fifteen years, given mainly to bachelor’s students in their third year of theoretical physics, and to third-year mathematics students. In those courses, familiarity with measure theory, functional analysis or even some of the more theoretical aspects of real analysis, like compactness, could not be assumed. For that reason, the text was designed to be essentially self-contained: the reader is merely assumed to have a working knowledge of multidimensional real analysis (see [8], for instance), while a few of the problems also require some acquaintance with the residue calculus from complex analysis in one variable. Our aim is to make the reader familiar with the essentials of the theory in an efficient and fully rigorous way, while emphasizing applications to the theory of linear partial differential equations. In many aspects the text is really introductory; its sole ambition is to present theoretical physicists with an idea of what distributions are all about from the mathematical point of view, while it should give mathematicians a taste of the power of distributions as a natural method in analysis. Our preferred theory of integration is that of Riemann, and at a few occasions our arguments might be slightly shortened by the use of Lebesgue’s theory. Yet, in a very limited number of cases Lebesgue integration is essential and then we mention this explicitly. A large number of problems is included: some of these illustrate the theory itself while others explore its applications to other parts of mathematics. Solutions to many of the problems are provided, in substantial detail really to those problems included in take-home examinations. The theory of distributions was created by Laurent Schwartz in 1944. It gives a coherent presentation of many of the twentieth-century insights and techniques that

VIII

Preface

play a role in analysis; Schwartz’ book [23] made these a standard tool for analysts. Consequently, in the decennia since its publication great progress has been made in the theory of linear partial differential equations, with many innovations but also better understanding of classical results as an outcome. In particular, distributions were meant, right from the beginning, to provide the means of handling singular solutions of equations. They enable one to treat separately the questions of existence and uniqueness, on the one hand, and the questions of regularity, that is, of smoothness, on the other. Furthermore, both exposition and research have greatly gained by using the mechanisms of differentiation, convolution and Fourier transformation of distributions. For a comprehensive survey of this, see the standard treatise by H¨ormander [15]. Volume I of it, [16], gives a detailed treatment of distributions, supplemented with exercises, hints and answers. An unsurpassed summary of the theory of distributions is Chapter I in H¨ormander [14]. Books that devote much attention to examples of distributions are Gel’fand and Shilov [12] and Kanwal [17]. For a neat historical overview, consult G˚arding [11], and [24] for a retrospective view by Schwartz himself. Important predecessors of Schwartz were Heaviside (1893-94), Wiener (1926) [28], Dirac (1930) [5], Bochner (1932) [3] and Sobolev (1936) [26]. In the present text, the symbol  set against the right margin signifies the end of a proof, while  marks the end of a definition, example or remark. Each time the course was taught, the notes were corrected and refined, with the help of the students; we are grateful to them for their remarks. Also, we express our gratitude to our colleagues E.P. van den Ban, for making available the notes for his course in 1987 on Distributions and Fourier transformation and for very constructive criticism of the final draft, and R.W. Bruggeman, for the improvements and additional problems that he contributed over the past few years. Furthermore, we wish to acknowledge our debt to A.W. Knapp who played an essential role in the publication of this book. The original Dutch text has been translated with meticulous care by J.P. van Braam Houckgeest. In addition, his comments have led to considerable improvement in formulation. The responsibility for any imprecisions remains entirely ours, we would be grateful to be told of them, at [email protected]. Utrecht August, 2006

Hans Duistermaat Johan Kolk

Contents

1

Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 9

2

Test Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11 24

3

Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

27 36

4

Differentiation of Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

39 41

5

Convergence of Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43 45

6

Taylor Expansion in Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49 52

7

Localization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

55 59

8

Distributions with Compact Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

61 68

9

Multiplication by Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

71 75

10

Transposition. Pullback and Pushforward . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

79 92

11

Convolution of Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

X

Contents

12

Fundamental Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

13

Fractional Integration and Differentiation . . . . . . . . . . . . . . . . . . . . . . . . 13.1 The case of dimension one . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Wave family . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Appendix: Euler’s Gamma function . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

Fourier Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

15

Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

16

Fundamental Solutions and Fourier Transformation . . . . . . . . . . . . . . . 193 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

17

Supports and Fourier Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

18

Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

19

Appendix: Results from Measure Theory in the Context of Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

20

Solutions to Selected Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

131 131 134 140 145

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

List of Figures

Example 1.3. Graphs of φ and φ(x) x ............................. Equation (2.4). Scaling of function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Illustration for Problem 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Illustration for Problem 1.3. Graphs of g 00 = 2p , for  = 1/100, and of g 0 and g, for  = 1/1000 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 8 9

2.1 2.2 2.3 2.4 2.5 2.6

Equation (2.3). δ-neighborhood . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Lemma 2.7. Bump function in dimension one . . . . . . . . . . . . . . . . . . . . Lemma 2.9. Bump function in dimension two . . . . . . . . . . . . . . . . . . . . Definition 2.14. Partition of unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Formula (2.11). Gauss density in dimension two . . . . . . . . . . . . . . . . . . Problem 2.3. Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14 17 17 20 24 25

5.1 5.2

Problem 5.5. Heat kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problem 5.9. Nondifferentiable function . . . . . . . . . . . . . . . . . . . . . . . .

46 47

1.1 1.2 1.3 1.4

9

12.1 Problem 12.2. Equipotential curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 12.2 Problem 12.5. Wave operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 13.1 13.2 13.3 13.4 13.5

Formula (13.10). Nappes of cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Formula (13.28). Absolute value of Gamma function . . . . . . . . . . . . . . Lemma 13.4. Homotopic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Corollary 13.5. Reciprocal of Gamma function . . . . . . . . . . . . . . . . . . . Problem 13.9. Fundamental solution of wave operator . . . . . . . . . . . . .

135 141 143 144 150

14.1 Example 14.1. Graph of sinc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . P35 15.1 Example 15.7. Graph of x 7→ n=−35 einx . . . . . . . . . . . . . . . . . . . . . sin x 1 15.2 Problem 15.2. Graphs of x 7→ 2(1−cos x) and x 7→ 2 log(1 − cos x) . 15.3 Problem 15.3. Sawtooth function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

152 183 188 189

XII

List of Figures

18.1 Example 18.3. Cut-off function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Problem 18.5. 2 sgn(x)e−|x| . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3 Problem 18.5. C 3 function that is not C 4 . . . . . . . . . . . . . . . . . . . . . . . . P∞ kx 20.1 Problem 15.8. k=1 cos 1+k2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

215 219 221 278

1 Motivation

Distributions form a class of objects that contains the continuous functions as a subset. Conversely, every distribution can be approximated by infinitely differentiable functions and for that reason one also uses the term “generalized functions” instead of distributions. Even so, not every distribution is a function. In several aspects, the calculus of distributions can be developed more readily than the theory of continuous functions. For example, the derivative of any distribution is also a distribution (see Chap. 4). This may be compared with the extension of the system of integers to the system of rational numbers, where to any x and y with y 6= 0 corresponds the quotient x/y. We now discuss some more concrete types of problem that will be solved in the calculus of distributions. Example 1.1. The function f defined by f (x) = |x| for x ∈ R is continuous on R. It is differentiable on ] − ∞, 0 [ and on ] 0, ∞ [, with derivative equaling −1 and +1 on these intervals, respectively. f is not differentiable at 0. Yet it seems natural to say that the derivative f 0 (x) equals the sign sgn(x) of x, the value of f 0 (0) not being defined and, intuitively speaking, being of little importance. But beware, we obviously require that f 00 (x) = 0 for x < 0 and for x > 0, while f 00 (x) must have an essential contribution at x = 0. Indeed, if f 00 (x) ≡ 0, the conclusion is that f 0 (x) ≡ c for a constant c ∈ R; in other words, f (x) = c x for all x ∈ R, which is different from the function x 7→ |x|, whatever choice is made for c.  Example 1.2. Let v : R3 → R3 be a continuously differentiable mapping, interpreted as a vector field on R3 . Further, let 3 X ∂vj (x) x 7→ ρ(x) = div v(x) = ∂xj j=1 be the divergence of v; this is a continuous real-valued function on R3 . Suppose that X is a bounded and open set in R3 having a smooth boundary ∂X and lying at one

2

1 Motivation

side of ∂X; we write the outer normal to ∂X at the point y ∈ ∂X as ν(y). The Divergence Theorem then asserts Z Z ρ(x) dx = hv(y), ν(y)i dy, (1.1) X

∂X

see for instance Duistermaat–Kolk [8, Theorem 7.8.5]. Here the right-hand side is interpreted as an amount of volume that flows outwards across the boundary, while ρ(x) is rather like a local expansion (= source strength) in a motion whose velocity field equals v(x). Traditionally, one also wishes to allow point (mass) sources at a point p, for R R which X ρ(x) dx = c if p ∈ X and X ρ(x) dx = 0 if p ∈ / X, where X is the closure of X in R3 . (We make no statement for the case that p ∈ ∂X.) Here c is a positive constant, the strength of the point source in p. These conditions cannot be realized by a function ρ continuous everywhere on R3 . The divergence of the special vector field v(x) = kx − pk−3 (x − p)

(1.2)

vanishes at every point x 6= p; verify this. This implies that the left-hand side of (1.1) equals 0 if p ∈ / X and 4π if p ∈ X; the latter result is obtained when we replace the set X by X \ B, where B is a closed ball around p, having a radius sufficiently small that B ⊂ X (see [8, Example 7.9.4]). Thus we would like to conclude that the divergence of this vector field v equals the point source at the point p with strength 4π (see Problem 4.5 and its solution for the details).  Example 1.3. The function x 7→ x1 is not absolutely integrable over any bounded interval around 0, so it is not immediately clear what Z φ(x) dx R x is to mean if φ is a continuous function that vanishes outside a bounded interval. Even if φ(0) = 0, the integrand is not necessarily absolutely integrable. For example, let 0 <  < c < 1 and define (see Fig. 1.1)   if x ≤ 0, 0 φ(x) = 1  if 0 < x ≤ c.  | log x| This function φ is continuous on ] −∞, c ] and can be extended to a continuous function on R that vanishes outside a bounded interval. Then Z c φ(x) dx = log | log | − log | log c|, x  and the right-hand side converges to ∞ as  ↓ 0.

1 Motivation 1.4

3

50

0.5

Fig. 1.1. Graphs of φ and x 7→

0.5 φ(x) x

But if φ is continuously differentiable and vanishes outside a bounded interval, integration by parts and the estimate φ() − φ(−) = O() as  ↓ 0, which is a consequence of the Mean Value Theorem, give Z Z φ(x) φ0 (x) log |x| dx. (1.3) dx = − lim ↓0 R\[ −,  ] x R Note the importance of the excluded intervals being symmetric about the origin. The left-hand side is called the principal value of the integral of φ(x)/x and is also Z written as φ(x) PV dx. (1.4) R x This notion was introduced by Cauchy in the 19th century. Other, equally natural propositions can also be made. Indeed, always assuming φ to be continuously differentiable, Z φ(x) dx x + i R converges as  ↓ 0, or  ↑ 0, respectively, see Problem 1.1. The limit is denoted by Z φ(x) dx, (1.5) R x + i0 or

Z R

φ(x) dx, x − i0

(1.6)

1 1 respectively. Clearly, the “functions” PV x1 , x+i 0 and x−i 0 differ only at x = 0, that is, integration against a function φ will produce identical results if φ(0) = 0. In Problem 1.1 and its solution as well as in Example 3.2 one may find more information. 

Example 1.4. Another motivation, significant also for historical reasons, has its root in the calculus of variations. The starting point for the rather lengthy discussion is an obvious calculation. This is then followed by an existence theorem concerning minima of functions, the full proof of which is not allowed by the present context,

4

1 Motivation

however. The discussion ends in the statement of a problem that will be solved by means of distribution theory at a later stage. Z b Consider
0 2 2 1 1 F (v) = dx, 2 p(x) v (x) + 2 q(x) v(x) a

where p and q are given, nonnegative and sufficiently differentiable functions on the k interval [ a, b ]. Let Cα, β be the set of k times continuously differentiable functions v on [ a, b ] with v(a) = α and v(b) = β. For k ≥ 1, we consider F to be a real-valued k function on Cα, β. We now ask whether among these v a special u can be found for which F reaches k k its minimum, that is, u ∈ Cα, β and F (u) ≤ F (v) for all v ∈ Cα, β . If such u is k obtained, one finds that, for every φ ∈ C0, 0 , the function t 7→ F (u + tφ) attains its minimum at t = 0. This implies that its derivative with respect to t at t = 0 equals Z b 0, or
p(x) u0 (x) φ0 (x) + q(x) u(x) φ(x) dx = 0. a 2

If u ∈ C , integration by parts gives Z

b

− a


d (p(x) u0 (x)) + q(x) u(x) φ(x) dx = 0. dx

k As this must hold true for all φ ∈ C0, 0 , we conclude that u must satisfy the secondorder differential equation d (Lu)(x) := − (p(x) u0 ) + q(x) u = 0. (1.7) dx

This procedure may be applied to much more general functionals (functions on spaces of functions) F ; the differential equations that one obtains for the stationary point u of F are called the Euler–Lagrange equations. Until the middle of the 19th century the existence of a minimizing u ∈ C 2 was taken for granted. Weierstrass then brought up the seemingly innocuous example a = −1, b = 1, α = −1, β = 1, p(x) = x2 and q(x) = 0. One may then consider, for any  > 0, the function arctan(x/) v (x) = , arctan(1/) for x ∈ [ −1, 1 ]. The denominator has been included in order to guarantee that v (±1) = ±1. For x < 0, or x > 0, we have that arctan(x/) converges to −π, or +π, respectively, as  ↓ 0. Therefore, v (x) converges to the sign function sgn x as  ↓ 0. To study the behavior of F (v ) we write v0 (x) =

1 1 .  arctan(1/) (1 + (x/)2 )

The change of variables x =  y leads to

1 Motivation

F (v ) = 

Z

1 arctan(1/)

Note that in this expression the factor 2y 2 = φ0 (y) (1 + y 2 )2

if

1/

−1/

1 arctan(1/)

5

2

y dy. 2(1 + y 2 )2 converges to 1/π as  ↓ 0. One has

φ(y) = −

y + arctan y. 1 + y2

That makes the integral equal to (φ(1/) − φ(−1/))/4, from which we can see that the integral converges to π2 when  ↓ 0. The conclusion is that F (v ) =  ψ(), where ψ() converges to 12 as  ↓ 0. In particular, F (v ) converges to zero as  ↓ 0. 1 Thus we see that the infimum of F on C−1, 1 equals zero; indeed, even the in∞ fimum on the subspace C−1, 1 equals zero. However, if u is a C 1 function with F (u) = 0, we have du dx (x) ≡ 0, which means that u is constant. But then u cannot satisfy the boundary conditions u(−1) = −1 and u(1) = 1. In other words, the 1 restriction of F to the space C−1, 1 does not attain its minimum in this example. In the beginning of the twentieth century the following discovery was made. Let H(1) be the space of the square-integrable functions v on ] a, b [ whose derivatives v 0 are also square-integrable on ] a, b [ . Actually, this is not so easy to define. A correct definition is given in Chap. 18: v ∈ H(1) if and only if v is square-integrable and the distribution v 0 is also square-integrable. In order to understand this definition, we have to know how a square-integrable function can be interpreted as a distribution. Next, we use that the derivative of any distribution is another distribution, which may or may not equal a square-integrable function. If v is a continuously differentiable function on [ a, b ], application of the Cauchy– Schwarz inequality gives Z x Z x 1/2 Z x 1/2 0 0 2 |v(x) − v(y)| = v (z) dz ≤ v (z) dz dz y

y

y

≤ kv 0 kL2 |x − y|1/2 , where kv 0 kL2 is the L2 norm of v 0 . This can be used to prove that every v ∈ H(1) can be interpreted as a continuous function on [ a, b ] (also compare Example 18.3), with the same estimate |v(x) − v(y)| ≤ kv 0 kL2 |x − y|1/2 . The continuity of the v ∈ H(1) implies that one can meaningfully speak of the α,β subspace H(1) of the v ∈ H(1) for which v(a) = α and v(b) = β. Also, for every v ∈ H(1) the number F (v) is well-defined. Now assume that p(x) > 0 for every x ∈ [ a, b ]; this excludes the example of Weierstrass. The assumption implies the existence of a constant c with the property kv 0 k2L2 ≤ c F (v), for all v ∈ H(1) . In combination with the estimate for |v(x) − v(y)| this tells us α,β that every sequence vj in H(1) with bounded values F (vj ) is an equicontinuous and

6

1 Motivation

uniformly bounded sequence of continuous functions. By the Arzel`a–Ascoli Theorem (see Knapp [18, Theorem 10.48]), a subsequence vj(k) then converges uniformly to a continuous function u as k → ∞. A second fact here offered without proof is that α,β and that the values F (vj(k) ) converge to F (u) as k → ∞. This is now u ∈ H(1) applied to a sequence of vj for which F (vj ) converges to the infimum i of F on α,β α,β with F (u) = i. In other . Thus one can show the existence of a u ∈ H(1) H(1) α,β . words, F attains its minimum on H(1) This looked promising, but one then ran into the problem that initially all one could say about this minimizing u was that u0 is square-integrable. This does not even imply that u is differentiable under the classic definition that the limit of the difference quotients exists. Because so far we do not even know that u ∈ C 2 , the integration by parts is problematic and, as a consequence, so is the conclusion that u is a solution of the Euler–Lagrange equation. What we can do is to integrate by parts with the roles of u and φ interchanged and thereby conclude that Z b u(x) (Lφ)(x) dx = 0, (1.8) a

for every φ ∈ C ∞ that vanishes identically in a neighborhood of the boundary points a and b. For this statement to be meaningful, u need only be a locally integrable function on the interval I = ] a, b [. In that case the function u is said to satisfy the differential equation Lu = 0 in a distributional sense. Historically, a somewhat older term is in a weak sense, but this is not very specific. Assume that p and q are sufficiently differentiable and that p has no zeros in the interval I. In this text we will show by means of distribution theory that if u is a locally integrable function and satisfies the equation Lu = 0 in the distributional sense, u is in fact infinitely differentiable in I and satisfies the equation Lu = 0 on I in the usual sense. See Theorem 9.4. In this way, distribution theory makes a contribution to the calculus of variations: by application of the Arzel`a–Ascoli Theorem it demonstrates the existence of α,β α,β a minimizing function u ∈ H(1) . Every minimizing function u ∈ H(1) satisfies the differential equation Lu = 0 in the distributional sense; distribution theory yields the result that u is in fact infinitely differentiable and satisfies the differential equation Lu = 0 in the classical sense. This application may be extended to a very broad class of variational problems, also including functions of several variables, for which the Euler-Lagrange variation equation then becomes a partial differential equation.  The estimation result from the next Lemma 1.5 will play an important role in what follows. The functions in Examples 1.1, 1.2 and 1.3 are not continuous, or differentiable, respectively, at a special point. Singularities in functions can be mitigated by translating the function f back and forth and averaging the functions thus obtained with a weight function φ(y) that depends on the translation y applied to the original function. Let us assume that φ is sufficiently differentiable on R, that φ(x) ≥ 0 for all x ∈ R, that a constant c > 0 exists such that φ(x) = 0 if |x| ≥ c and, finally, that

1 Motivation

7

Z φ(x) dx = 1.

(1.9)

R

For the existence of such φ, see Problem 1.2. The averaging procedure is described by the formula Z Z (f ∗ φ)(x) = f (x − y) φ(y) dy = f (z) φ(x − z) dz. (1.10) R

R

The minus sign is used to obtain symmetric formulas; in particular, f ∗ φ = φ ∗ f . The function f ∗ φ is called the convolution of f and φ, because one of the functions is reflected and translated, then multiplied by the other one, following which the result is integrated. Another interpretation is that of a measuring device recording a signal f around the position x, where φ(y) represents the sensitivity of the device at displacement y. In practice, this φ is never completely concentrated in y = 0; because of built-in inertia, φ(y) will have one or more bounded derivatives. Yet another interpretation is obtained by defining Ty f , the function translated by y, via (Ty f )(x) := f (x − y). (1.11) Here we use the rule: the value of the translated function at the translated point equals the value of the function at the original point. In other words, under Ty the graph translates to the right if y > 0. If we now read the first equality in (1.10) as an identity between functions of x, we have Z f ∗φ= φ(y) Ty f dy. (1.12) R

Here the right-hand side is defined as the limit of Riemann sums in the space of continuous functions (of x), where the limit is taken with respect to the supremum norm. Thus, the functions f translated by y are superposed, with application of a weight function φ(y), similar to a photograph that becomes softer (blurred) if the camera is moved during the exposure. Indeed, differentiation with respect to x under the integral sign in the right-hand side in (1.10) yields, even in the case where f is merely continuous, that f ∗ φ is differentiable, with derivative (f ∗ φ)0 = f ∗ φ0 . In obtaining this result, we have not used the normalization (1.9). We can therefore repeat this and conclude that f ∗ φ is equally often continuously differentiable as φ. For more details, see the proof of Lemma 2.17 below. How closely does the smoothed signal f ∗ φ approximate the true signal f ? If a ≤ f (y) ≤ b for all y ∈ [ x − c, x + c ], we conclude from (1.10) and (1.9) that a ≤ (f ∗ φ)(x) ≤ b as well. This can be improved upon if we can bring the positive number c closer to 0. To achieve this, we replace the function φ by φ (see Fig. 1.2), for an arbitrary constant  > 0, with

8

1 Motivation

Fig. 1.2. Graph of φ as in (1.13) with  equal to 1 and 1/2, respectively

φ (x) =

1 x . φ  

(1.13)

Furthermore, φ is equally often continuously differentiable as φ. Lemma 1.5. If f is continuous on R, the function f ∗ φ converges uniformly to f on every bounded interval [ a, b ] as  ↓ 0. And for every  > 0, the function f ∗ φ on R is equally often continuously differentiable as φ. Proof. We have Z

Z φ (y) dy =

R

R

 y  dy Z φ = φ(z) dz = 1,   R

from which Z


f (x − y) − f (x) φ (y) dy

(f ∗ φ )(x) − f (x) = R

Z

c


f (x − y) − f (x) φ (y) dy,

= − c

where in the second identity we have used φ (y) = 0 if |y| >  c. This leads to the estimate Z c |(f ∗ φ )(x) − f (x)| ≤ |f (x − y) − f (x)| φ (y) dy − c

≤ sup |f (x − y) − f (x)|, |y|≤ c

where in the first inequality we have applied φ (y) ≥ 0 and in the second inequality we have once again used the fact that the integral of φ equals 1. The continuity of f gives that for every δ0 > 0 the function f is uniformly continuous on the bounded interval [ a − δ0 , b + δ0 ] (if necessary, see [8, Theorem 1.8.15 taken in conjunction with Theorem 2.2 below]). This implies that for every η > 0 there exists a 0 < δ ≤ δ0 with the property that |f (x − y) − f (x)| < η if x ∈ [ a, b ] and |y| < δ. From this we may conclude |(f ∗ φ )(x) − f (x)| ≤ η, if x ∈ [ a, b ] and 0 <  < δ/c.  Weyl [27] has called the operator f 7→ f ∗ φ a mollifier.

1 Motivation

9

Problems Problem 1.1. Determine the difference between (1.4) and (1.5), and between (1.5) and (1.6). Each is a complex multiple of φ(0). See Example 14.22 for a different approach.

1

1

-1

Fig. 1.3. Illustration for Problem 1.2

Problem 1.2. Determine a polynomial function p on R of degree six for which R1 p(a) = p0 (a) = p00 (a) = 0 for a = ±1, while in addition −1 p(x) dx = 1. Define φ(x) = p(x) for |x| ≤ 1 and φ(x) = 0 for |x| > 1. Prove that φ is twice continuously differentiable. Sketch the graph of φ (see Fig. 1.3). Problem 1.3. Let ψ be twice continuously differentiable on R and let ψ equal 0 outside a bounded interval. Set f (x) = |x|. Calculate the second-order derivative of g = f ∗ ψ by first differentiating under the integral sign, then splitting the integration at the singular point of f and finally eliminating the differentiations in every subintegral. 200

1

0.0015

-0.0015 -0.01

-1

0.01 0.003

0.003

-0.003 00

Fig. 1.4. Illustration for Problem 1.3. Graphs of g = 2p , for  = 1/100, and of g 0 and g, for  = 1/1000

10

1 Motivation

Now take ψ equal to φ with φ as per Problem 1.2 and φ as per (1.13). Draw a sketch of g 00 for small , and, by finding antiderivatives, of g 0 and g. Show the sketches of g, g 0 , g 00 next to those of f , f 0 and f 00 (?), respectively (see Fig. 1.4). Problem 1.4. We consider integrable functions f and g on R that vanish outside the interval [ −1, 1 ]. (i) Determine the interval outside which the convolution f ∗ g vanishes. (ii) Using simple examples of your own choice for f and g, calculate f ∗ g, and sketch the graphs of f , g and f ∗ g. (iii) Try to choose f and g such that f and g are not continuous while f ∗ g is. (iv) Try to choose f and g such that f ∗ g is not continuous. Hint: Let α > −1, f (x) = g(x) = xα if 0 < x < 1, f (x) = g(x) = 0 if x ≤ 0 or x ≥ 1. Prove that f and g are integrable. Show the existence of a constant c > 0 such that (f ∗ g)(x) = c x2α+1 if 0 < x ≤ 1. For what values of α is f ∗ g discontinuous at the point 0? R Problem 1.5. (i) Calculate PV R φ(x) x dx, for the following choices of φ: (a) φ1 is the characteristic function of the interval [ −2, 3 ], (b) φ2 (x) = x/(1 + x2 ), (c) φ3 (x) = 1/(1 + x2 ). Which of these three integrals converges absolutely as an improper integral? (ii) Also calculate the “integrals” over R of φ1 (x)/(x + i0) and φ1 (x)/(x − i0).

2 Test Functions

We will now introduce test functions and do so by specializing the testing of f as in (1.10). If we set x = 0 and replace φ(y) by φ(−y), the result of testing f by means of the weight function φ becomes equal Z to the “integral inner product” hf, φi =

f (x) φ(x) dx.

(2.1)

R

(For real-valued functions this is in fact an inner product; for complex-valued functions one uses the Hermitian inner product hf, φi.) In Chap. 1 we went on to vary φ, by translating and rescaling. The idea behind the definition of distributions is that we consider (2.1) as a function of all possible test functions φ, in other words, we will be considering the mapZ ping test f : φ 7→ f (x) φ(x) dx. R

Before we can do so, we first have to specify what functions will be allowed as test functions. The first requirement is that all these functions may be complexvalued. Definition 2.5 below, of test functions, refers to compact sets. In this text we will be frequently encountering such sets; therefore we begin by collecting some information on them. Definition 2.1. An open covering of a set K in Rn is a collection U of open sets in Rn such that their union contains K. That is, for every x ∈ K there exists a U ∈ U with x ∈ U . A subcovering is a subcollection E of U still covering K. In other words, E ⊂ U and K is contained in the union of the sets U with U ∈ E. The set K is said to be compact if every open covering of K has a finite subcovering. This concept is applicable in very general topological spaces. Next, recall the concept of a subsequence of an infinite sequence (x(j))∞ j=1 . This is a sequence having terms of the form y(j) = x(i(j)) where i(1) < i(2) < · · · ; in particular limj→∞ i(j) = ∞. Note that if the sequence (x(j))∞ j=1 converges to x, every subsequence of this sequence also converges to x.  For the sake of completeness we prove the following theorem, which is known from analysis (see [8, Section 1.8]).

12

2 Test Functions

Theorem 2.2. For a subset K of Rn the following properties (a) – (c) are equivalent. (a) K is bounded and closed. (b) Every infinite sequence in K has a subsequence that converges to a point of K. (c) K is compact. Qn Proof. (a) ⇒ (c). We begin by proving that a cube B = j=1 Ij is compact. Here Ij denotes a closed interval in R of length l, for every 1 ≤ j ≤ n. Let U be an open covering of B; we assume that it does not contain a finite covering of B and will show that this assumption leads to a contradiction. When we bisect a closed interval I of length l, we obtain I = I (l) ∪ I (r) where I (l) and I (r) are closed intervals of length l/2. Consider the cubes of the form B 0 = Qn (l) (r) 0 0 0 j=1 Ij , where for every 1 ≤ j ≤ n we have made a choice Ij = Ij or Ij = Ij . n 0 Then B equals the union of the 2 subcubes B . If it were possible to cover each of these by a finite subcollection E of U, the union of these E would be a finite subcollection of U covering B, in contradiction with the assumption. We conclude that there is a B 0 that is not covered by a finite subcollection of U. Applying mathematical induction we thus obtain a sequence B (t) of cubes with the following properties. (i) B (0) = B and B (t) ⊂ B (t−1) for every t > 0. Qn (t) (t) (ii) B (t) = j=1 Ij , where Ij denotes a closed interval of length 2−t l. (iii) B (t) is not covered by a finite subcollection of U. (t)

From (i) we now have, for every j, Ij

(t−1)

⊂ Ij

(t)

, that is, the left endpoints lj of

(t)

the Ij , considered as a function of t, form a monotonously nondecreasing sequence (t)

in R. This sequence is bounded; indeed, lj

(s)

∈ Ij

when t ≥ s. As t → ∞, the (s)

(s)

sequence therefore converges to an lj ∈ R; we have lj ∈ Ij because Ij is closed. Conclusion: the limit point l := (l1 , . . . , ln ) belongs to B (s) , for every s. Because U is a covering of B and l ∈ B, there exists a U ∈ U for which l ∈ U . Since U is open, there exists an  > 0 such that x ∈ Rn and |xj − lj | <  for all j, imply that x ∈ U . Choose s with 2−s < . Because l ∈ B (s) , the fact that x ∈ B (s) implies that |xj − lj | ≤ 2−s <  for all j; therefore x ∈ U . As a consequence, B (s) ⊂ U , in contradiction with the assumption that B (s) was not covered by a finite subcollection of U. Now let K be an arbitrary bounded and closed subset of Rn and U an open covering of K. Because K is bounded, there exists a closed cube B that contains K. Because K is closed, the complement C := Rn \ K of K is open. The collection Ue := U ∪ {C} covers K and C, and therefore Rn , and certainly B. In view of the e Removing C from E, e we foregoing, B is covered by a finite subcollection Ee of U. obtain a finite subcollection E of U; this covers K. Indeed, if x ∈ K, there exists U ∈ Ee with x ∈ U . Since U cannot equal C, we have U ∈ E. (c) ⇒ (b). Suppose that x(j) is an infinite sequence in K that has no subsequence converging in K. This means that for every x ∈ K there exist an (x) > 0 and an N (x) for which kx − x(j)k ≥ (x) whenever j > N (x). Let

2 Test Functions

13

U (x) = { y ∈ K | ky − xk < (x) }. The U (x) with x ∈ K form an open covering of K; condition (c) implies the existence of a finite subset F of K such that for every x ∈ K there is an f ∈ F with x ∈ U (f ). Let N be the maximum of the N (f ) with f ∈ F ; then N is well-defined because F is finite. For every j we find that an f ∈ F exists with x(j) ∈ U (x), and therefore j ≤ N (f ) ≤ N . This is in contradiction with the unboundedness of the indices j. (b) ⇒ (a). Suppose that K satisfies (b). If K is not bounded, we can find a sequence x(i) with kx(i)k ≥ i for all i. There is a subsequence x(i(j)) that converges and which is therefore bounded, in contradiction with kx(i(j))k > i(j) ≥ j for all j. In order to prove that K is closed, suppose limi→∞ x(i) = x for a sequence x(i) ∈ K. This contains a subsequence that converges to a point y ∈ K. But the subsequence also converges to x and in view of the uniqueness of limits we conclude that x = y ∈ K.  The preceding theorem contains the Bolzano–Weierstrass Theorem, which states that every bounded sequence in Rn has a convergent subsequence, see [8, Theorem 1.6.3]. The implication (a) ⇒ (c) is also referred to as the Heine–Borel Theorem, see [8, Theorem 1.8.18]. However, linear spaces consisting of functions are usually of infinite dimension. In normed linear spaces of infinite dimension, “compact” is a much stronger condition than “bounded and closed”, while in such spaces (b) and (c) are still equivalent. As a first application of compactness we obtain conditions which guarantee that disjoint closed sets in Rn possess disjoint open neighborhoods, see Lemma 2.3 below and its corollary. To do so, we need some definitions, which are of independent interest. Introduce the set of sums A + B of two subsets A and B of Rn by means of A + B := { a + b | a ∈ A, b ∈ B }.

(2.2)

It is clear that A + B is bounded if A and B are bounded. Also, A + B is closed whenever A is closed and B compact. Indeed, suppose that the sequence cj ∈ A + B converges in Rn to c. One then has cj = aj + bj for some aj ∈ A, bj ∈ B. By the compactness of B, a subsequence k 7→ bj(k) converges to a b ∈ B. Consequently, the sequence aj(k) = cj(k) − bj(k) converges to a := c − b as k → ∞. Because A is closed, a lies in A. The conclusion is that c ∈ A + B. In particular, A + B is compact whenever A and B are both compact. An example of two closed subsets A and B of R for which A + B is not closed, is the pair A = Z p and n ≥ 2, while p + 1/n converges to p as n → ∞. Furthermore, the distance d(x, U ) from a point x ∈ Rn to a set U ⊂ Rn is defined by

14

2 Test Functions

d(x, U ) = inf{ kx − uk | u ∈ U }. Note that d(x, U ) = 0 if and only if x ∈ U , the closure of U in Rn . The δneighborhood Uδ of U is given by (see Fig. 2.1) Uδ = { x ∈ Rn | d(x, U ) < δ }.

(2.3)

Fig. 2.1. Example of a δ-neighborhood

Observe that x ∈ Uδ if and only if a u ∈ U exists with kx − uk < δ. Using the notation B(u; δ) for the open ball of center u and radius δ this gives [ Uδ = B(u; δ), u∈U

which implies that Uδ is an open set. Also, B(u; δ) = {u} + B(0; δ), and therefore Uδ = U + B(0, δ). Finally, we define U−δ as the set of all x ∈ U for which the δ-neighborhood of x is contained in U . Note that U−δ equals the complement of (Rn \ U )δ and that, consequently, U−δ is a closed set. Now we are prepared enough to obtain the following two results on separation of sets. Lemma 2.3. Let K ⊂ Rn be compact and A ⊂ Rn closed, while K ∩ A = ∅. Then there exists δ > 0 such that Kδ ∩ Aδ = ∅.

Proof. Assume the negation of the conclusion. Then there exists an element x(i) ∈ K1/i ∩ A1/i , for every i ∈ Z>0 . Hence, one can select y(i) ∈ K and a(i) ∈ A satisfying ky(i) − x(i)k
0 and α(x) = 0 for x ≤ 0. Then α ∈ C ∞ (R) with α(x) > 0 for x > 0, and supp α = [ 0, ∞ [ . Proof. The only problem is the differentiability at 0. From the power series for the n exponential function one obtains, for every integer n > 0, the estimate ey ≥ yn! for all y ≥ 0. Hence, for x > 0, 1

α(x) = 1/e x ≤ n!/( x1 )n = n! xn . This tells us that α is differentiable at 0, with α0 (0) = 0. As regards the higher-order derivatives we note that for x > 0 the function α satisfies the differential equation α(x) α0 (x) = 2 . x By applying this in the induction step we obtain, with mathematical induction on k, 1 α(k) (x) = pk α(x), x where the pk are polynomial functions inductively determined by p0 (y) = 1

and


pk+1 (y) = pk (y) − pk 0 (y) y 2 .

In particular, pk is of degree 2k and therefore satisfies an estimate of the form |pk (y)| ≤ c(k) y 2k

(y ≥ 1).

From this we derive the estimate |α(k) (x)| ≤ c(k) n! xn−2k

(0 < x ≤ 1).

If we then choose n ≥ 2k + 2, we obtain, with mathematical induction on k, that α ∈ C k (R) and α(k) (0) = 0. 

2 Test Functions

17

Lemma 2.8. Let α ∈ C ∞ (R) be as in the preceding lemma. Let a and b ∈ R with a < b. Define the function β = βa,b by β(x) = βa,b (x) = α(x − a) α(b − x). One then has β ∈ C ∞ (R) with β > 0 on ] a, b [ and supp β = [ a, b ] (see Fig. 2.2). Z Furthermore, I(β) := β(x) dx > 0. R R 1 β has the same properties as β, while R γ(x) dx = The function γ = γa,b := I(β) 1.

Fig. 2.2. Graphs of α as in Lemma 2.7 on [ 0, 1/2 ] and of γ−1,2 as in Lemma 2.8, with the scales adjusted

Lemma 2.9. Let aj and bj ∈ R with aj < bj and define γaj ,bj ∈ C0∞ (R) be as in the preceding lemma, for 1 ≤ j ≤ n. Write x = (x1 , . . . , xn ) ∈ Rn . For a and b ∈ Rn , define the function Γa,b : Rn → R by (see Fig. 2.3) Γa,b (x) =

n Y

γaj ,bj (xj ).

j=1

Then we have Γa,b ∈ C ∞ (Rn ),

Γa,b > 0 on

n Y

] aj , bj [ ,

j=1

supp Γa,b =

n Y j=1

Z [ aj , bj ],

Γa,b (x) dx = 1. Rn

Fig. 2.3. Graph of Γ(−1,2),(2,3) as in Lemma 2.9

For a complex number c, the notation c ≥ 0 means that c is a nonnegative real number. For a complex-valued function f , f ≥ 0 means that f (x) ≥ 0 for every x in the domain space of f . If g is another function, one writes f ≥ g or g ≤ f if f − g ≥ 0.

18

2 Test Functions

Corollary 2.10. For every point p ∈ Rn and every neighborhood U of p in Rn there exists a φ ∈ C0∞ (Rn ) with the following properties: (a) φ ≥ 0 and φ(p) > 0. (b) supp φ ⊂ U. R (c) Rn φ(x) dx = 1. By superposition and taking limits of the test functions thus constructed we obtain a wealth of new test functions. For example, consider φ as in Corollary 2.10 and set 1 1  x . (2.4) φ (x) := n φ   Further let f be an arbitrary function in C0 (Rn ), the space of all continuous functions on Rn with compact support; these are easily constructed in abundance. By straightforward generalization of Lemma 1.5 to Rn , the functions f := f ∗ φ converge uniformly to f on compact subsets of Rn , as  ↓ 0. The f are test functions, in other words, f ∈ C0∞ (Rn ), (2.5) as one can see from Lemma 2.17 below. Consequently, C0∞ (Rn ) is dense in C0 (Rn ) in terms of uniform convergence on compact sets. We now review notations that will be needed for Definition 2.12 and Lemma 2.17 below, among other things. In this text we use the following notations for higherorder derivatives. A multi-index is a sequence α = (α1 , . . . , αn ) ∈ (Z≥0 )n of n nonnegative integers. The sum |α| :=

n X

αj

j=1

is called the order of the multi-index α. For every multi-index α we write ∂xα :=

∂α := ∂1α1 ◦ · · · ◦ ∂nαn ∂xα

where

∂j :=

∂ . ∂xj

(2.6)

Furthermore, we use the shorthand notation ∂α =

∂α ∂xα

when we only want to differentiate with respect to the variables xj . The crux is that the Theorem on the interchangeability of the order of differentiation (see for instance [8, Theorem 2.7.2]), which holds for functions sufficiently often differentiable, allows us to write every higher-order derivative in the form (2.6); also refer to the introduction to Chap. 6. Finally, in the case of n = 1, we define ∂ as ∂ (1) .

2 Test Functions

19

Remark 2.11. In (2.6) we defined the partial derivatives ∂ α f of arbitrary order of a function f depending on an arbitrary number of variables. For the k-th order derivatives of the product f g of two functions f and g that are k times continuously differentiable we have Leibniz’ formula: ∂ α (f g) =

X α
∂ α−β f ∂ β g, β

(2.7)

β≤α

for |α| = k. Here α = (α1 , . . . , αn ) and β = (β1 , . . . , βn ) are multi-indices, while β ≤ α means that for every 1 ≤ j ≤ n one has βj ≤ αj . The n-dimensional binomial coefficients in (2.7) are given by n Y α
αj
:= β βj

where

j=1

p! p
:= q (p − q)! q!

(p, q ∈ Z, 0 ≤ q ≤ p).

Formula (2.7) is obtained with mathematical induction on the order k = |α| of differentiation, using Leibniz’ rule ∂j (f g) = g ∂j f + f ∂j g

(2.8) 

in the induction step.

Definition 2.5 is supplemented by the following, which introduces a notion of convergence in the infinite-dimensional linear space C0∞ (X): Definition 2.12. Let φj and φ ∈ C0∞ (X), for j ∈ Z>0 and X an open subset of Rn . ∞ The sequence (φj )∞ j=1 is said to converge to φ in the space C0 (X) of test functions as j → ∞, notation lim φj = φ in C0∞ (X), j→∞

if the following two conditions are both met: (a) there exists a compact subset K of X with the property that supp φj ⊂ K for all j; (b) for every multi-index α the sequence (∂ α φj )∞ j=1 converges uniformly on K to ∂ α φ.  Observe that the data above imply that supp φ ⊂ K. The notion of convergence introduced in the definition above is very strong. The stronger the convergence, the fewer convergent sequences there are, and the more readily a function defined on C0∞ (X) will be continuous. Next we combine compactness and test functions in order to introduce the useful technical tool of a partition of unity over a compact set, in Definition 2.14.

20

2 Test Functions

Lemma 2.13. For every a ∈ Rn and r > 0 there exists φ ∈ C0∞ (Rn ) satisfying supp φ ⊂ B(a; 2r),

0 ≤ φ ≤ 1,

φ = 1 on

B(a; r).

Proof. By translation and rescaling we see that it is sufficient to prove the assertion for a = 0 and r = 1. By Lemma 2.8 we can find β ∈ C ∞ (R) such that β > 0 on R4 ] 1, 4 [ and supp β = [ 1, 4 ], while I = 1 β(x) dx > 0. Hence we may write Z 1 4 β(t) dt. η(x) := I x Then η ∈ C ∞ (R), 0 ≤ η ≤ 1, while η = 1 on ] −∞, 1 ] and η = 0 on [ 4, ∞ [ . Now set φ(x) = η(kxk2 ) = η(x21 + · · · + x2n ). 

0.6

2

-2 1

2

-2

Fig. 2.4. Example of a partition of unity

Definition 2.14. Let K be a compact subset of an open subset X of Rn and U an open covering of K. A C0∞ (X) partition of unity over K subordinate to U is a finite sequence ψ1 , . . . , ψl ∈ C0∞ (X) with the following properties (see Fig. 2.4): (i) ψj ≥ 0, for every 1 ≤ j ≤ l; Pl (ii) there exists a neighborhood V of K in X with j=1 ψj (x) = 1, for all x ∈ V ; (iii) for every j there is a U = U (j) ∈ U for which supp ψj ⊂ U .  Given a function f on X, write fj = ψj f in the notation above. Then we obtain Pl functions fj with compact support contained in U (j), while f = j=1 fj on V . Furthermore, all fj ∈ C k if f ∈ C k . In the applications, the U ∈ U are small neighborhoods of points of K with the property that we can reach certain desired conclusions for functions with support in U . For example, partitions of unity were used in this way in [8, Theorem 7.6.1], to prove the integral theorems for open sets X ⊂ Rn with C 1 boundary.

2 Test Functions

21

Theorem 2.15. For every compact set K contained in an open subset X of Rn and every open covering U of K there exists a C0∞ (X) partition of unity over K subordinate to U. Proof. For every a ∈ K there exists an open set Ua ∈ U such that a ∈ Ua . Select ra > 0 such that B(a; 2ra ) ⊂ Ua ∩ X. By criterion (c) in Theorem 2.2 for compactness, there exist finitely many a(1), . . . , a(l) such that K is contained in the union V of the B(a(j), ra(j) ), for 1 ≤ j ≤ l. Now select the corresponding φj ∈ C0∞ (X) as in Lemma 2.13 and set j Y ψ1 = φ1 ; ψj+1 = φj+1 (1 − φi ) (1 ≤ j < l). (2.9) i=1

Then the conditions (i) and (iii) for a C0∞ (X) partition of unity subordinate to U are satisfied by the ψ1 , . . . , ψl . The relation j X i=1

ψi = 1 −

j Y

(1 − φi )

(2.10)

i=1

is trivial for j = 1. If (2.10) is true for j < l, then summing (2.9) and (2.10) yields (2.10) for j + 1. Consequently (2.10) is valid for j = l and this implies that the ψ1 , . . . , ψl satisfy condition (ii) for a partition of unity with V as defined above.  Corollary 2.16. Let K be a compact subset in Rn . For every open neighborhood X of K in Rn there exists a χ ∈ C0∞ (Rn ) with 0 ≤ χ ≤ 1, supp χ ⊂ X and χ = 1 on an open neighborhood of K. In particular, for δ > 0 sufficiently small, we can find such a function χ with supp χ ⊂ Kδ . Proof. Consider the open covering {X} of K and let ψ1 , . . . , ψl be Pa subordinate partition of unity over K as in the preceding theorem. Then χ = j ψj satisfies all requirements. For the second assertion, apply Corollary 2.4 and use the open covering {Kδ } of K.  The function χ is said to be a cut-off function for the compact subset K of Rn . Through multiplication by χ we can replace a function f defined on X by a function g with compact support contained in X. Here g = f on a neighborhood of K and g ∈ C k if f ∈ C k . We still have to verify the claim in (2.5); it follows from Lemma 2.17 below. In the case of k equal to ∞, another proof will be given in Theorem 11.2. Later on, in proving Theorem 11.16, we will need an analog of Corollary 2.16 in the case of not necessarily compact sets. In preparation, we introduce two more concepts that are useful in their own right. The characteristic function 1U of a subset U of Rn is defined by

22

2 Test Functions

1U (x) = 1

if x ∈ U,

1U (x) = 0

if x ∈ Rn \ U.

Let X ⊂ Rn be an open subset. A function f : X → C is said to be locally integrable if, for every a ∈ X, there exists an open rectangle B ⊂ X with the properties that a belongs to the interior of B and that f is integrable over B. For the purposes of this book it will almost invariably be sufficient to interpret the concept of integrability, as we use it here, in the sense of Riemann. However, for distributions it is common usage to work with Lebesgue integration, see for example Stroock [25], which leads to a more comprehensive theory. Readers who are not familiar with Lebesgue integration can find a way around this by restricting themselves to locally integrable functions f with an absolute value whose improper Riemann integral exists, and otherwise taking our assertions about Lebesgue integration for granted. Some of these assertions do not apply to Riemann integration, but this need not be a reason for serious concern; after all, this is a text about distributions, not a course in Lebesgue integration. Lemma 2.17. Let f be locally integrable on Rn and g ∈ C0k (Rn ). Then f ∗ g ∈ C k (Rn ) and supp (f ∗ g) ⊂ (supp f ) + (supp g). Here (supp f )+(supp g) is a closed subset of Rn , compact if f , too, has a compact support; in that case f ∗ g ∈ C0k (Rn ). Proof. We study (f ∗ g)(x) for x ∈ U , where U ⊂ Rn is bounded and open. Define h(x, y) := f (y) g(x − y). Then the function x 7→ h(x, y) belongs to C k (U ) for every y ∈ Rn , because for every multi-index α ∈ (Z≥0 )n with |α| ≤ k ∂αh (x, y) = f (y) ∂ α g(x − y). ∂xα Let B(r) be a ball about 0 of radius r > 0 such that supp g ⊂ B(r). Then there exists an r0 > 0 with B(r) + U ⊂ B(r0 ); furthermore, the characteristic function α χ of B(r0 ) is integrable over Rn . For every x ∈ U the function ∂∂xαh (x, ·) vanishes outside B(r0 ); consequently, the latter function does not change upon multiplication by χ. In addition we have ∂αh ((x, y) ∈ U × Rn ), α (x, y) ≤ sup |∂ α g(x)| |f (y)| χ(y) ∂x x∈Rn where |f | χ is an absolutely integrable function on Rn . In view of a well-known Theorem on changing the order of differentiation and integration (inRthe context of Riemann integration, see [8, Theorem 6.12.4]) we then know that Rn h(x, y) dy is a C k function of x whose derivatives equal the integral with respect to y of the corresponding derivatives according to x of the integrand h(x, y). Furthermore, h(x, y) = 0 if x ∈ U and y ∈ / KU , where KU := (supp f ) ∩ (U + (− supp g)).

2 Test Functions

23

Now suppose u ∈ / (supp f ) + (supp g), then there exists a neighborhood U of u in Rn such that x ∈ / (supp f ) + (supp g) for all x ∈ U , for the complement of (supp f ) + (supp g) is open. But this means KU = ∅, which implies that (f ∗ g)(x) = 0 for all x ∈ U .  R Lemma 2.18. Let φ ∈ C0∞ (Rn ), φ ≥ 0, φ(x) dx = 1 and kxk ≤ 1 if x ∈ supp φ. Suppose that the subset U of Rn is measurable, that is, 1U is locally integrable. Select δ > 0 arbitrarily and define, for 0 <  < δ, 1 1  φ (x) := n φ x and χU,  := 1U ∗ φ .   Then χU,  ∈ C ∞ (Rn ),

0 ≤ χU, ≤ 1,

supp χU,  ⊂ Uδ .

Finally, χU,  = 1 on a neighborhood of U−δ . Proof. We have χU,  ∈ C ∞ (Rn ) by Lemma 2.17. Because φ ≥ 0, we obtain 0 = 0 ∗ φ ≤ 1U ∗ φ ≤ 1 ∗ φ = 1(φ ) = 1. Furthermore, if B denotes the -neighborhood of 0, the support of χU,  is contained in supp 1U + supp φ ⊂ U + B , and therefore also in the δ-neighborhood of U as  < δ. The latter conclusion is reached when we replace U by V = Rn \ U ; note that 1 − χU,  = 1 ∗ φ − 1U ∗ φ = (1 − 1U ) ∗ φ = χV,  .  Usually in applications of Lemma 2.18, the set U is either open or closed, but even then its characteristic function 1U will not always be locally integrable in the sense of Riemann (see [8, Exercise 6.1]), whereas it is in the sense of Lebesgue. This is an example where Lebesgue’s theory is superior. Finally, there are many situations where one prefers to use, instead of φ ∈ C0∞ (Rn ), functions like (see Fig. 2.5) n

2

γ(x) = γn (x) = π − 2 e−kxk .

(2.11)

The numerical factor is chosen such that the integral of γ over Rn equals 1; this γ is the Gaussian density or the probability density of the normal distribution, with expectation 0 and variance Z kxk2 γn (x) dx = 2n. Rn

For larger values of kxk the values γ(x) are so extremely small that in many situations we may just as well consider γ as having compact support. Naturally, this is 2 only relative: if we were to use γ to test a function that grows at least like ekxk as kxk → ∞, this would utterly fail.

24

2 Test Functions

Fig. 2.5. Graph of γ2 , with different horizontal and vertical scales

R For the sake ofQcompleteness we recall the calculation of In := γn (x) dx. n Because γn (x) = j=1 γ1 (xj ), we have In = (I1 )n . The change of variables x = r cos α and y = r sin α now yields Z π Z Z Z 2 2 2 1 1 e−(x +y ) dx dy = e−r r dα dr = 1, I2 = π R R π R>0 −π or

Z

2

e−x dx =

√

π.

(2.12)

R

We refer to [8, Exercises 2.73 and 6.15, or 6.41] for other proofs of this identity.

Problems Problem 2.1. Let φ ∈ C0∞ (R), φ 6= 0 and 0 ∈ / supp φ. Decide whether the se∞ quence (φj )∞ converges to 0 in C (R) as j → ∞ if: 0 j=1 (i) φj (x) = j −1 φ(x − j). (ii) φj (x) = j −p φ(j x). Here p is a given positive integer. (iii) φj (x) = e−j φ(j x). In each of these cases verify that, for every x ∈ R and every k ∈ Z≥0 , the se(k) quence (φj (x))∞ j=1 converges to 0 as j → ∞. And in addition, that in case (i) the convergence is even uniform on R. Problem 2.2. Let φ and φ be as in Lemma 2.18. Prove that, for every ψ ∈ C0∞ (X), the ψ ∗ φ converge to ψ in C0∞ (X) as  ↓ 0. Problem 2.3. For  > 0, define γ ∈ C ∞ (R) by 2 2 1 γ (x) = √ e−x / . π

2 Test Functions

25

25 0.5

-0.05

0.05

-0.5

0.5

Fig. 2.6. Illustration for Problem 2.3. Graphs of γ and | · | ∗ γ with  = 1/50

Calculate | · | ∗ γ . Prove that this function is analytic on R and examine how closely it approximates the function | · |. Also calculate its derivatives of first and second order. See Fig. 2.6. Problem 2.4. Let U be a measurable subset of Rn . Assume that both U and Rn \ U have positive measure. Let γ(x) be as in (2.11) and γ (x) = 1n γ( 1 x), for  > 0. Denote the “probability of distance to 0 larger than r” by Z ρ(r) = γ(x) dx. kxk>r

Give an estimate of the r for which ρ(r) < 10−6 . Prove that χ := 1U ∗ γ is analytic and that 0 < χ < 1. Further prove that χ (x) ≤ ρ(δ/) if d(x, U ) = δ > 0; finally show that χ (x) ≥ 1 − ρ(δ/) if d(x, Rn \ U ) = δ > 0. Problem 2.5. Consider β ∈ CR0∞ (R) with β ≥ 0 and β(x) = 0 if and only if |x| ≥ 1. Further assume that β(x) dx = 1. Let 0 <  < 1, β (x) = 1 β( x ), I = [ −1, 1 ] ⊂ R and let ψ = 1I ∗ β . Determine where one has ψ = 0, where 0 < ψ < 1 and where ψ = 1. And in addition, where ψ 0 = 0, ψ 0 > 0 and ψ 0 < 0, respectively. Now let φ(x) = β(x1 ) β(x2 ) for x ∈ R2 and let U = I ×I, a square in the plane. Consider χ = χU,  as in Lemma 2.18. Prove that χ(x) = ψ(x1 ) ψ(x2 ). Determine where one has χ = 0, or 0 < χ < 1, or χ = 1. And in addition, for j = 1 and 2, where ∂j χ = 0, ∂j χ > 0, ∂j χ < 0. Verify that if 0 < χ < 1, there is a j such that ∂j χ 6= 0. Prove by the Submersion Theorem (see [8, Theorem 4.5.2]) that for every 0 < c < 1 the level set N (c) := { x ∈ R2 | χ(x) = c } is a C ∞ curve in the plane. Is this also true for the boundary of the support of χ and of 1 − χ? Give a description, as detailed as possible, of the level curves of χ, including a sketch.

3 Distributions

For an arbitrary linear space E over C, a C-linear mapping : E → C is also called a linear form on E, or a linear functional on E. Definition 3.1. Let X be an open subset of Rn . A distribution on X is a linear form u on C0∞ (X) that is also continuous in the sense that lim u(φj ) = u(φ)

j→∞

lim φj = φ in C0∞ (X).

as

j→∞

Phrased differently, in this case continuity means preservation of convergence of sequences. The space of all distributions on X is denoted by D0 (X).  The notation derives from the notation D(X) used by Schwartz for the space C0∞ (X) of test functions equipped with the notion of convergence from Definition 2.12. (The letter D denotes differentiable.) It would in fact be more exact to use the notation φj → φ in D(X) instead of φj → φ in C0∞ (X), but this is not normally done. By the linearity of u, the assertion u(φj ) → u(φ) is equivalent to u(φj − φ) = u(φj ) − u(φ) → 0, while φj → φ is equivalent to φj − φ → 0. This implies that the continuity of a linear form u is equivalent to the assertion lim u(ψj ) = 0

j→∞

Example 3.2. We have PV of Example 1.3, by PV

1 x

1 : C0∞ (R) → C x

as

lim ψj = 0

j→∞

in C0∞ (X).

∈ D0 (R) if we define this linear form, in the notation

with C0∞ (R),



PV

1 (φ) = PV x

Z R

φ(x) dx. x

Indeed, consider arbitrary φ ∈ then there exists m > 0 with supp φ ⊂ [ −m, m ]. By the differentiability of φ we can write (see, for example, [8, Proposition 2.2.1.(ii)]), for all x ∈ R and with a suitable constant c > 0 and continuous function ψ on R,

28

3 Distributions

φ(x) = φ(0) + x ψ(x)

and

sup |ψ(x)| ≤ c sup |φ0 (x)|. |x|≤m

|x|≤m

The latter estimate follows from the Mean Value Theorem. We therefore have, for arbitrary  > 0, Z Z Z 1 φ(x) dx = φ(0) dx + ψ(x) dx. ≤|x|≤m x ≤|x|≤m ≤|x|≤m x Because of the continuity of ψ we get Z Z  φ(x) 1 PV (φ) = lim dx = ψ(x) dx. ↓0 ≤|x|≤m x x |x|≤m Consequently,  1  (φ) ≤ 2m sup |ψ(x)| ≤ 2cm sup |φ0 (x)|, PV x |x|≤m |x|≤m and this implies that PV

1 x

is a continuous linear form on C0∞ (R).



For every complex linear space E with a notion of convergence, it is customary to denote the space of continuous linear forms on E by E 0 ; this space is also referred to as the topological dual of E. If E is of finite dimension, every linear form on E is automatically continuous and E 0 is a complex linear space of the same dimension as E. For function spaces E of infinite dimension, this does not apply and it is therefore sensible also to require continuity of the linear forms. This will open up a multitude of conclusions that one could not obtain otherwise, while the condition remains sufficiently weak to allow a large space of linear forms. Remark 3.3. The study of general linear spaces E with a notion of convergence is referred to as functional analysis; this is outside the scope of this text. For the benefit of readers who (justifiably) find the preceding paragraph too vague, we add some additional clarification. We require that addition and scalar multiplication in E are continuous with respect to the notion of convergence in E. That is, xj + yj → x + y and cj xj → c x if xj , x, yj , y ∈ E, cj ∈ C and xj → x, yj → y and cj → c as j → ∞. Furthermore, we impose the condition that limits of convergent sequences are uniquely determined. In that case, E with its notion of convergence is also known as a topological linear space. A linear mapping u : E → C is said to be continuous if u (xj ) → u(x) as xj → x in E. If E is of finite dimension, there exists a basis (ei )1≤i≤n of E, where n is the dimension of E. ThePmapping that assigns to x ∈ E the coordinates (x1 , . . . , xn ) ∈ n Cn for which x = i=1 xi ei , is a linear isomorphism from E to Cn . The assertion is that, via this linear isomorphism, convergence in E is equivalent to the usual coordinatewise convergence in Cn .

3 Distributions

29

Proof. The property that E is a topological linear space immediately leads to the conclusion that coordinatewise convergence implies the convergence in E. We now prove the reverse by mathematical induction on n. Let xj → x in E as j → ∞. Select 1 ≤ i ≤ n and let cj , or c, be the i-th coordinate of xj , or x, respectively. Suppose that the complex numbers cj do not converge to c as j → ∞; we will show that this assumption leads to a contradiction. By changing over to a subsequence, if necessary, we can arrange for the existence of a δ > 0 with |cj − c| ≥ δ for all j. This implies that the sequence dj = 1/ (cj − c) is bounded. Changing over to a subsequence once again, if necessary, we can arrange that there is a d ∈ C for which dj → d in C; here we apply Theorem 2.2.(b). With respect to the E-convergence, this leads to yj := dj (xj − x) → d 0 = 0 as j → ∞. On the other hand, for every j the i-th coordinate of yj equals 1. This means that, for every j and k, the vector yj − yk lies in the (n − 1)-dimensional linear subspace E0 of E consisting of the elements of E whose i-th coordinates vanish. With respect to the E-convergence in E0 , the yj − yk converge to zero if both j and k go to infinity; therefore, by the induction hypothesis, the yj − yk converge coordinatewise to zero as j, k → ∞. In other words, the yj form a Cauchy sequence in E ' Cn with respect to the coordinatewise convergence. This implies that y ∈ E exists for which yj → y by coordinates as j → ∞, which in turn implies that yj → y in E. Because we had already obtained that yj → 0 in E, it follows from the uniqueness of limits that y = 0. This leads to a contradiction with the fact that the yj converge coordinatewise to y, for the i-th coordinate of yj equals 1 and the i-th coordinate of y = 0 equals 0. This negates the assumption that the complex numbers cj do not converge to c. The conclusion therefore is that, for every 1 ≤ i ≤ n, the i-th coordinate of xj converges to the i-th coordinate of x as j → ∞. In other words, xj → x in E implies the coordinatewise convergence of the xj to x as j → ∞. 

has

For an arbitrary linear function u on a linear space E with basis (ei )1≤i≤n , one X  X n n xi ei = u(x) = u xi u (ei ) . i=1

i=1

From this we can see that u is continuous with respect to coordinatewise convergence and is determined by its “matrix coefficients” ui = u (ei ), for 1 ≤ i ≤ n. The conclusion is that, for a topological linear space E of finite dimension, E 0 equals the space of all linear forms on E and this is a complex-linear space of the same dimension as E.  D0 (X) is a complex-linear space with respect to the addition and the scalar multiplication defined by (u + v)(φ) = u(φ) + v(φ)

and

(λ u)(φ) = λ u(φ),

for u and v ∈ D0 (X), φ ∈ C0∞ (X) and λ ∈ C. It is a mere matter of writing out the definitions to show that u + v and λ u ∈ D0 (X) if u, v ∈ D0 (X) and λ ∈ C. And,

30

3 Distributions

likewise, that with these definitions D0 (X) does indeed become a linear space over C. Many functions can be interpreted as distributions. To explain this, we introduce a new concept. In the theory of Lebesgue measure, a function f : X → C is said to be locally integrable if f is measurable and there exists, for every a ∈ X, an open rectangle B = Ba about a with the property Z |f (x)| dx < ∞. B

Theorem 3.4. For every locally integrable function f on X, Z f (x) φ(x) dx (φ ∈ C0∞ (X)) (test f )(φ) =

(3.1)

X

defines a distribution u = test f on X. Actually, let φj ∈ C0∞ (X), K ⊂ X compact and supp φj ⊂ K for all j. Then limj→∞ u(φj ) → 0, if limj→∞ φj = 0 uniformly on K. Proof. By the compactness of K := supp φ, there are finitely many a(i) ∈ X such that K is contained in the union of the rectangles Ba(i) . This means that the sum of the characteristic functionsR of the Ba(i) isR a majorant of the characteristic function of K. Using the identity K g(x) dx = 1K (x) g(x) dx, we deduce the absolute convergence of the integral in (3.1) from Z XZ |f (x) φ(x)| dx ≤ |f (x)| |φ(x)| dx ≤ C sup |φ(x)|, K

i

where C=

x∈K

Ba(i)

XZ i

|f (x)| dx.

Ba(i)

The continuity of u follows from the fact that, for all j, |u(φj )| can be estimated by a constant times the supremum norm of φj . Indeed, the supports of the φj are contained in a fixed compact K ⊂ X as φj → 0 in C0∞ (X). 

Lemma 3.5. The mapping f 7→ test f is linear and injective from C(X), the linear space of all continuous functions on X, to D0 (X), the space of distributions on X. Proof. By writing out the definitions one immediately verifies that the mapping is linear. A linear mapping is known to be injective if and only if its null space equals 0. In other words, assuming (test f )(φ) = 0 for all test functions φ, we have to prove that f (x) = 0 for all x ∈ X. With φ as in (2.4), we obtain, for x ∈ X and  > 0 sufficiently small,

3 Distributions

(f ∗ φ )(x) = (test f )(Tx Sφ ).

31

(3.2)

In the right-hand side the reflection S of a function ψ is defined by (Sψ)(z) = ψ(−z). We have already come across the translation Tx , in (1.11). This now yields (Tx Sφ )(y) = (Sφ )(y − x) = φ (x − y), which enables us to see that (3.2) is another way of writing (1.10), with φ replaced by φ . The assumption (test f )(φ) = 0, for all test functions φ, now implies (f ∗ φ )(x) = 0, for every  > 0 and for all x. But we have already seen that f ∗ φ converges uniformly on compact sets to f as  ↓ 0, where  ≤ 0 , while 0 depends on the compact set considered; and certainly, therefore, f (x) = lim(f ∗ φ )(x) = 0. ↓0

 Remark 3.6. This lemma justifies the usual identification of the continuous function f with the distribution test f . In other words, C(X) is identified with the linear subspace { test f | f ∈ C(X) } of D0 (X) and the premodifier “test” preceding the continuous function f is omitted when f is regarded as a distribution. A small amusement: 1 = test 1 is integration of test functions. In this case the left-hand side is identified with the function that constantly equals 1; when we are piling identifications on top of each other, we should not be surprised that the notations become ambiguous. For locally integrable functions f the situation is a little subtler. A function f is said to be equal to g almost everywhere if the set { x ∈ X | f (x) 6= g(x) } has Lebesgue measure equal to 0. In this case test f = test g, while f = g is not necessarily true in the strict sense that f (x) = g(x) for all x ∈ X. Conversely, if test f = test g, the functions f and g are equal almost everywhere. We outline the proof.  Proof. Write h = f − g. If φ is a continuous function with compact support in X, a sequence φj ∈ C0∞ (X) exists that converges uniformly to φ. This gives Z h(x) φ(x) dx = lim (test h)(φj ) = 0. j→∞

From the theory of Lebesgue integration it is known that the validity of this identity, for all continuous functions φ with compact support, implies that h vanishes almost everywhere.

32

3 Distributions

While not all readers may be familiar with the latter result, it will be realized that h vanishes almost everywhere if the integral of h over any rectangle U equals ∞ 0. R If h is tested with the functions χU, ∈ C0 (X) from Lemma 2.18, we obtain that h(x) dx = 0 by taking the limit as  ↓ 0.  U It is customary in the theory of Lebesgue integration to identify functions with each other when they are equal almost everywhere. We can therefore say that this custom amounts to interpreting locally integrable functions f as distributions, via the mapping f 7→ test f . Again, the word “test” is omitted. An example of a distribution not of the form test f for a locally integrable function f is the Dirac function or point measure δa , situated at the point a ∈ X. This is defined by δa (φ) := φ(a) (φ ∈ C0∞ (X)), in other words, by evaluating the test function φ at the point a. When a = 0, one simply uses the term Dirac function, without further specification, denoting it by δ. If δa = test f , for a locally integrable function f , the restriction of f to X \ {a} equals 0 almost everywhere. Because {a} has measure 0, we conclude that f equals 0 almost everywhere; thus, test f = 0. This leads to a contradiction, because there exists a test function φ with φ(a) 6= 0; consequently δa 6= 0. If we equate locally integrable functions with distributions, we should not be fussy about using function notations for distributions. In particular, for arbitrary u ∈ D0 (X) one encounters the notations Z u(φ) = u(x) φ(x) dx = hu, φi = hφ, ui (φ ∈ C0∞ (X)). (3.3) X

This will not lead to problems, provided one does not conclude that there is any meaning in the phrase “the value u(x) of the distribution u at the point x.” In the case of the Dirac function, for example, this certainly entails some problems at x = 0. So much for comments on the Dirac notation Z Z φ(a) = δ(x − a) φ(x) dx = δ(x) φ(x + a) dx. Rn

Rn

Normally, the continuity of a linear form u on C0∞ (X) is most easily proved by giving an estimate for u(φ) in terms of a so-called C k norm of φ ∈ C0∞ (X), see Example 3.2 for instance. In order to formulate this, we introduce, for every compact subset K of X, the space C0∞ (K) consisting of all φ ∈ C0∞ (X) with supp φ ⊂ K. On this space we have, for every k ∈ Z≥0 , the C k norm, defined by kφkC k =

sup

|∂ α φ(x)|.

x∈X, |α|≤k

In Chap. 8 we establish, in a general context, a relation between the continuity of a linear mapping and estimates in terms of (semi)norms. For linear forms on C0∞ (X) this looks as follows.

3 Distributions

33

Lemma 3.7. A linear form u on C0∞ (X) belongs to D0 (X) if and only if for every compact subset K of X there exist a constant c > 0 and an order of differentiation k ∈ Z≥0 with the property (φ ∈ C0∞ (K)).

|u(φ)| ≤ c kφkC k

(3.4)

Proof. φj → φ in C0∞ (X) means that a compact subset K of X exists for which φj and φ ∈ C0∞ (K) for all j while limj→∞ kφj − φkC k = 0 for all k. If u satisfies (3.4), it then follows that u(φj ) − u(φ) = u(φj − φ) converges to 0 as j → ∞. Now suppose that u does not satisfy the condition of the lemma. This means that a compact subset K of X exists such that, for every c > 0 and k ∈ Z≥0 , there is a φc,k ∈ C0∞ (K) for which |u(φc,k )| > c kφc,k kC k . This implies kψc,k kC k
0 such that K ⊂ [ −m,R m ]. If x ∈ R2 satisfies m kxk ∈ K, then |xj | ≤ kxk ≤ m, for 1 ≤ j ≤ 2. Since −m |x| dx = m2 we now obtain, for any φ ∈ C0∞ (K), |u(φ)| ≤ m4

sup t∈K, |j|≤2

|φ(j) (t)| = m4 kφkC 2 .

The assertion now follows from Lemma 3.7.



Definition 3.10. If u ∈ D0 (X), the minimal k for which (3.4) holds true, for certain C, is known as the order of u on K. The supremum over all compact subsets K of X of the orders of u on K is called the order of the distribution u on X. 

34

3 Distributions

The order of a distribution u can be ∞, as for the distribution u on R defined by u(φ) =

∞ X

(−1)k φ(k) (k)

(φ ∈ C0∞ (R)).

(3.5)

k=0

To see why this is so, use Problem 3.2. In the next theorem we demonstrate that distributions of finite order k can be identified with the continuous linear forms on the space C0k (X) of C k functions with compact support in X. In this context, the convergence in C0k (X) is that of Definition 2.12 where condition (b) is only required for all multi-indices α with |α| ≤ k. Note that C0∞ (X) ⊂ C0k (X) ⊂ C0l (X) if k > l, in which there are no equalities. Furthermore, if ∞ ≥ k > l and the sequence φj ∈ C0k (X) converges in C0k (X) to φ, it also converges in C0l (X). This implies that the restriction to C0k (X) of a continuous linear form on C0l (X) defines a continuous linear form on C0k (X). In particular, the restriction to C0∞ (X) of a continuous linear form on C0l (X) is a distribution on X of order ≤ l. A converse assertion can also be proved: Theorem 3.11. Let X be an open subset of Rn , k ∈ Z≥0 and u a distribution on X of order ≤ k. Then u has a unique extension to a continuous linear form v on C0k (X). Proof. Let f ∈ C0k (X). With the functions φ as in (2.4) write f = f ∗ φ . We obtain, for every multi-index α with |α| ≤ k, that ∂ α f = (∂ α f ) ∗ φ converges uniformly to ∂ α f as  ↓ 0. From Lemma 2.17 it follows that, for sufficiently small  > 0, the supports of the f ∈ C0∞ (Rn ) are contained in a fixed compact subset of X. For these , therefore, f ∈ C0∞ (X) and these functions converge to f in C0k (X) as  ↓ 0. Thus, if v exists at all, it is given by v(f ) = lim u(f ). ↓0

This implies the uniqueness of a continuous linear extension of u to C0k (X). From this uniqueness it follows in turn that the right-hand side (provided it exists) does not depend on the choice of the family of functions φ as in (2.4). Now the right-hand side in |u(f ) − u(fη )| = |u(f − fη )| ≤ C kf − fη kC k converges to 0 as , η ↓ 0. This implies that  7→ u(f ) is a Cauchy sequence in C, and therefore converges to a complex number that we will denote by v(f ). Because the mapping f 7→ f ∗ φ is linear, we find that this defines a linear form v on C0k (X). Furthermore, taking the limit as  ↓ 0 in |v(f )| ≤ |v(f ) − u(f )| + |u(f )| ≤ |v(f ) − u(f )| + C kf kC k ≤ |v(f ) − u(f )| + C kf kC k + C kf − f kC k

3 Distributions

35

yields the result that |v(f )| ≤ C kf kC k for all f ∈ C0k (X), which implies that v is continuous on C0k (X). This also gives v(f ) = u(f ) if f ∈ C0∞ (X).  It is common to write v = u, in other words, to identify u with its continuous extension to C0k (X). A continuous linear form on C0 (X) = C00 (X), the space of continuous functions with compact support in X, is also known as a measure on X, or a Radon measure on X, if one wants to distinguish it from the general settheoretical concept of measure. Thus, the distributions of the order 0 are identified with measures. If f is a locally integrable function on X, the second assertion in Theorem 3.4 implies that test f is a distribution on X of order 0. We conclude that test f has a unique extension to a measure µ on X. This µ is called the measure with density function f ; but in view of the identifications made, we can also write f = test f = µ. An example of a measure without locally integrable density function is the Dirac function, which is therefore also called the Dirac measure. A distribution u ∈ D0 (X) is said to be positive if φ ∈ C0∞ (X) and φ ≥ 0 imply that u(φ) ≥ 0. In this case we write u ≥ 0. Here we use the convention that for a complex number c the notation c ≥ 0 means that c ∈ R and c ≥ 0. A measure u on X is said to be positive if u(f ) ≥ 0 for every nonnegative f ∈ C0 (X). Theorem 3.12. Every positive distribution is a positive measure. Proof. Suppose u ∈ D0 (X) and u ≥ 0. We begin by showing that u has order 0; on account of Theorem 3.11 it then has an extension to a measure on X. Let K be a compact subset of X. Corollary 2.16 yields a χ ∈ C0∞ (X) with 0 ≤ χ ≤ 1 and χ = 1 on K. This implies that u(χ) ≥ 0; in particular, u(χ) is real. Let φ ∈ C0∞ (K) be real-valued. With the notation c = kφk = supx |φ(x)| we then get c χ − φ ≥ 0, and therefore c u(χ) − u(φ) = u(c χ − φ) ≥ 0. This implies that c u(χ)−u(φ) is a nonnegative real number, so u(φ) is a real number and u(φ) ≤ u(χ) kφk. Now let φ ∈ C0∞ (K) be complex-valued. Then u(φ) = u(Re φ + i Im φ) = u(Re φ) + i u(Im φ) with u(Re φ) and u(Im φ) ∈ R. In particular, u(φ) = u(Re φ) if u(φ) ∈ R. Using the notations α = arg u(φ)

and

ψ = Re (e−iα φ)

we now obtain |u(φ)| = e−iα u(φ) = u(e−iα φ) = u(ψ) ≤ u(χ) kψk ≤ u(χ) kφk.

(3.6)

If f ∈ C0 (X) and f ≥ 0, then also f = f ∗ φ ≥ 0 because φ ≥ 0. Therefore u(f ) ≥ 0 and, consequently, also u(f ) = lim↓0 u(f ) ≥ 0. 

36

3 Distributions

Remark 3.13. In constructing the proof we have derived that u ≥ 0 implies that u(φ) ∈ R, for every continuous real-valued function φ with compact support. If both φ and ψ are real-valued and φ(x) ≤ ψ(x) for all x, then ψ − φ ≥ 0, whence u(ψ)−u(φ) = u(ψ−φ) ≥ 0, or u(φ) ≤ u(ψ). The conclusion is that the linear form u is monotone in the sense that u(φ) ≤ u(ψ), for φ and ψ real-valued and φ ≤ ψ, if and only if u ≥ 0.  Remark 3.14. One writes φj ↑ 1 as j → ∞ for a sequence of functions φj , if for every x we have φj (x) ↑ 1 as j → ∞. A positive measure µ is said to be a probability measure if a sequence φj of test functions exists with the property that φj ↑ 1 and µ(φj ) ↑ 1 as j → ∞. This implies that, for every sequence of test functions ψj with ψj ↑ 1 as j → ∞, one has µ(ψj ) ↑ 1 as j → ∞; one simply writes µ(1) = 1. For a (test) function φ, this µ(φ) is called the expectation of φ with R respect to the probability measure µ. If f is an integrable function with f ≥ 0 and f (x) dx = 1, then µ = test f is a probability measure. In this case f is called a probability density. l A finite linear combination X µ= pj δa(j) j=1

of point measures, where the a(j) are different points in Rn , is a probability measure Pl if and only if pj ≥ 0 for all j and j=1 pj = 1. The number pj is then called the probability of the alternative a(j). Thus we arrive at a distributional interpretation of the calculus of probabilities. One should remain aware, however, that many results for probability measures do not hold true for more general distributions. 

Problems Problem 3.1. Prove that u(φ) = φ(k) (0)

(φ ∈ C0∞ (R))

defines a distribution u ∈ D0 (R) of order ≤ k. Problem 3.2. Show that the distribution u from the preceding problem is not of order strictly lower than k in any neighborhood of 0. Hint: examine u(φ) and kφkC k−1 for φ = φ , where  > 0 and φ (x) = xk χ(x/) with χ ∈ C0∞ (R) and χ equals 1 on a neighborhood of 0. To calculate u(φ ) and kφ kC k−1 , use Leibniz’ formula (2.7) in the case n = 1. Problem 3.3. Demonstrate that for a continuous function f one has f ≥ 0 as a distribution if and only if f ≥ 0 as a function. 1 1 and x−i0 are distributions. Determine the orProblem 3.4. Show that PV( x1 ), x+i0 ders of these distributions. Prove that the mutual differences are scalar multiples of the Dirac function.

3 Distributions

37

Problem 3.5. Verify that u, v and w below are distributions on R2 : (i) u(φ) = R∂ α φ(x) where α is the multi-index (2, 3) and x the point (8, 5). (ii) v(φ) = R φ(t, 0) dt. R 2 (iii) w(φ) = R2 ekxk φ(x) dx. Problem 3.6. We consider functions and distributions on R2 . (i) r(x) = kxk defines a function on R2 . (a) Verify that log r and 1/r define distributions on R2 . What is the order of these distributions? (b) How about 1/r2 ? (ii) (a) Define Z π u(φ) = (∂1 φ(cos t, sin t) cos t + ∂2 φ(cos t, sin t) sin t) dt. 0

Show that this defines a distribution u on R2 . What can you say about the order of u? (b) The same questions for Z π v(φ) = (∂2 φ(cos t, sin t) cos t − ∂1 φ(cos t, sin t) sin t) dt. 0

4 Differentiation of Distributions

If f is continuously differentiable on an open set X in Rn , one obtains by means of integration by parts that for every test function φ the following holds: (test ∂j f )(φ) = −(test f )(∂j φ). Note that the boundary term on the right-hand side is absent because φ(x) = 0 for x sufficiently large, see [8, Corollary 7.6.2]. For an arbitrary distribution u on X we now define (∂j u)(φ) := −u(∂j φ)

(1 ≤ j ≤ n, φ ∈ C0∞ (X)).

Writing out the definition gives ∂j u ∈ D0 (X), which is called the distributional derivative of u with respect to the j-th variable. The form of the definition is such that ∂j (test f ) = test(∂j f ) for every continuously differentiable function f , so that we do not run into difficulties when we simply speak of “functions” when referring to test functions. By mathematical induction on the number of derivatives we find that partial derivatives of u of arbitrary order are also distributions. As in the case of C ∞ functions, the order of differentiation may be changed arbitrarily: Lemma 4.1. For every distribution u on the open subset X of Rn and for every pair of indices 1 ≤ j, k ≤ n, one has ∂j (∂k u) = ∂k (∂j u). Proof. For every φ ∈ C0∞ (X) one obtains (∂j ◦ ∂k u)(φ) = −∂k u(∂j φ) = u(∂k ◦ ∂j φ) = u(∂j ◦ ∂k φ) = −∂j u(∂k φ) = (∂k ◦ ∂j u)(φ). 

40

4 Differentiation of Distributions

With respect to the examples of continuous functions u for which ∂j ∂k u 6= ∂k ∂j u we may therefore note that the two sides are, in fact, distributionally equal. Lemma 4.1 makes it possible to write all higher-order derivatives in the form ∂ α u :=

∂αu . ∂xα

Note that this distribution is of order ≤ k + m when u is of order ≤ k and |α| = m. If U is open in R and u ∈ D0 (U ), we use the shorthand u0 for the distributional derivative ∂u, and u(k) for ∂ k u := ∂ (k) u, where k ∈ Z>1 . In particular, every continuous function considered as a distribution has partial derivatives of all orders. Conversely, we shall prove in Example 17.2 below that every distribution can locally be written as a linear combination of derivatives of some continuous function. If every continuous function is to be infinitely differentiable as a distribution, no proper subset of the space of distributions can therefore be adequate. In this sense, the distribution extension of the function concept is as economical as it possibly can be. Example 4.2. H := 1[ 0, ∞ [ , the characteristic function of the nonnegative x-axis in R, when interpreted as a distribution on R, is known as the Heaviside function. (At the end of the nineteenth century, Heaviside introduced a kind of distribution calculus for computations on electrical networks. H may be interpreted in terms of the sudden switching on of a current.) One has H 0 = δ. Indeed, for every φ ∈ C0∞ (R), Z 0 0 H (φ) = −H(φ ) = − φ0 (x) dx = φ(0). R>0

More generally, for a < b, the characteristic function 1[ a,b ] of the closed interval [ a, b ] satisfies in the notation (1.11) 1[ a,b ] = Ta H − Tb H,

and so

10[ a,b ] = δa − δb .

The next theorem asserts that every distribution on R has a distributional antiderivative and that this is uniquely determined up to an additive constant. Theorem 4.3. Let I be an open interval in R and f ∈ D0 (I). Then there exists u ∈ D0 (I) for which u0 = f . If, in addition, v ∈ D0 (I) and v 0 = f , then v = u + c for some constant c ∈ C. In particular, v ∈ D0 (I) and v 0 = 0 imply that v equals a constant function. R Proof. Choose χ ∈ C0∞ (I) such that 1(χ) = I χ(x) dx = 1. For φ ∈ C0∞ (I), define the function p(φ) on R by Z x Z x p(φ)(x) := φ(t) dt − 1(φ) χ(t) dt. −∞

−∞

4 Differentiation of Distributions

41

Then p(φ) ∈ C0∞ (I). Indeed, to see that p(φ) has compact support we note that there exist a and b ∈ I for which φ(t) = χ(t) = 0 if t < a or if t > b. This immediately implies that p(φ)(x) = 0 if x < a, while in the case where x > b one has p(φ)(x) = 1(φ) − 1(φ) 1 = 0. Denoting the derivative of φ by φ0 , we have 1(φ0 ) = 0 and therefore p(φ0 ) = φ. The mapping p : φ 7→ p(φ) from C0∞ (I) to C0∞ (I) is linear and sequentially continuous, that is, lim p(φj ) = 0

j→∞

in C0∞ (I) if

lim φj = 0 in C0∞ (I).

j→∞

This implies that u(φ) := −f (p(φ)), for φ ∈ C0∞ (I), defines a distribution u on I. That distribution satisfies u0 (φ) = −u(φ0 ) = f (p(φ0 )) = f (φ), for all φ ∈ C0∞ (I), in other words, u0 = f . To prove the second assertion observe that w := v − u satisfies w0 = 0, or w(φ0 ) = 0 for all φ ∈ C0∞ (I). In particular, 0 = w(p(φ)0 ) = w(φ − 1(φ) χ) = w(φ) − 1(φ) w(χ), for all φ ∈ C0∞ (I), which implies that w = w(χ) test 1 = test w(χ). Here we denote a constant function with value c by the same symbol c. 

Problems Problem 4.1. Prove that | · |0 = sgn and that | · |00 = 2 δ in D0 (R). Problem 4.2. Show that (log | · |)0 = PV( x1 ) in D0 (R). Problem 4.3. Let λ ∈ C and define f (x) = eλ x for x > 0 and f (x) = 0 for x ≤ 0. Prove that the derivatives of f satisfy f (k) = λk f +

k−1 X

λk−1−j δ (j)

(k ∈ Z≥0 ).

j=0

Now let p be a polynomial of degree m > 0 and p(λ) = 0. Does p(∂)f vanish? Calculate the order of p(∂)f . Problem 4.4. Determine a continuous function f on Rn and a multi-index α for which ∂ α f = δ. Establish how much smaller α may be chosen if f is merely required to be locally integrable.

42

4 Differentiation of Distributions

Problem 4.5. Let p ∈ Rn and vj (x) = kx − pk−n (xj − pj ), for 1 ≤ j ≤ n. For n = 3 this is the vector field v in (1.2). Verify that the vj are locally integrable on Rn and thus define distributions on Rn . Prove div v :=

n X

∂j vj = cn δp ,

j=1

where cn denotes the (n−1)-dimensional volume of the sphere S n−1 := { x ∈ Rn | kxk = 1 } in Rn . (See (13.31) for an explicit formula for cn .) Problem 4.6. (Sequel to Problem 4.5.) For x ∈ Rn \ {0}, define  1 2−n   if n 6= 2,  (2 − n) c kxk n E(x) =    1 log kxk if n = 2. 2π Prove that E is locally integrable on Rn and therefore defines a distribution on Rn . Demonstrate the existence of a constant c for which ∂j E = c vj as distributions. Then show ∆E = δ where n X ∆= ∂j 2 = div ◦ grad j=1

denotes the Laplace operator in Rn . Problem 4.7. Suppose Pqc1 , . . . , cq ∈ C and a1 < · · · < aq ∈ R. Find the solutions u ∈ D0 (I) of u0 = j=1 cj δaj . In which case does one find u = 1I as a solution, for an interval I in R?

5 Convergence of Distributions

Definition 5.1. We now introduce a notion of convergence in the linear space D0 (X) of distributions on an open set X in Rn . Let (uj )∞ j=1 be an infinite sequence in D0 (X) and let u ∈ D0 (X). One then writes lim uj = u

j→∞

in D0 (X),

(φ ∈ C0∞ (X)).

lim uj (φ) = u(φ)

if

j→∞

(5.1) This is also referred to as weak convergence or convergence of distributions. Instead of a sequence we can also take a family u of distributions that depend on one or more real-valued parameters . We then say that u converges in D0 (X) to u as  tends to a special value 0 , if for every test function φ the complex numbers u (φ) converge to u(φ) as  → 0 .  Example 5.2. Let fj be an infinite sequence of locally integrable functions on X and suppose that f is a locally integrable function on X with the property that, for every compact subset K of X, one has Z |fj (x) − f (x)| dx = 0. lim j→∞

K

0

Then fj → f in D (X), in the sense that test fj → test f in D0 (X) as j → ∞. Indeed, if φ ∈ C0∞ (X), application of standard inequalities for integrals yields Z | (test fj − test f ) (φ)| ≤ |fj (x) − f (x)| dx sup |φ(x)|. x

supp φ

Actually, this leads to convergence in the space of distributions of order zero, also referred to as weak convergence of measures.  Example 5.3. Set ut (x) = t eitx for x > 0 and ut (x) = 0, for x ≤ 0. Then Z Z ut (φ) = t eitx φ(x) dx = i φ(0) + i eitx φ0 (x) dx R>0

φ0 (0) 1 = i φ(0) − − t t

R>0

Z R>0

eitx φ00 (x) dx → i φ(0) as t → ∞,

44

5 Convergence of Distributions

for φ ∈ C0∞ (R). Hence limt→∞ ut = i δ in D0 (R).



The following principle of uniform boundedness will not be proved here. It is based on the Banach–Steinhaus Theorem, applied to the Fr´echet space C0∞ (K). See for example Rudin [22, Theorem 2.6] or Bourbaki [4, Livre V, Chap. III, §3, No 6 and §1, No 1, Corollaire]. Maybe the reader will see this principle explained as part of a course in functional analysis. Lemma 5.4. Let uj be a sequence of distributions on X with the property that, for every φ ∈ C0∞ (X), the sequence of complex numbers uj (φ) is bounded. Then, for every compact subset K of X, there exist constants c > 0 and k ∈ Z≥0 such that, for all j and all φ ∈ C0∞ (K), |uj (φ)| ≤ c kφkC k .

(5.2)

Note the analogy in formulation with Lemma 3.7. The lemma above leads to the following property of completeness of the space of distributions. Theorem 5.5. Let X be open in Rn and uj a sequence in D0 (X) with the property that for every φ ∈ C0∞ (X) the sequence uj (φ) converges in C as j → ∞; denote the limit by u(φ). Then u : φ 7→ u(φ) defines a distribution on X. Furthermore, limj→∞ uj = u in D0 (X). And what is more: if φj → φ in C0∞ (X), the sequence uj (φj ) converges to u(φ) as j → ∞. Proof. Writing out the definitions, we find that u defines a linear form on C0∞ (X). From the starting assumption it follows that the sequence uj (φ) is bounded for every φ ∈ C0∞ (X), and thus we obtain, for every compact K ⊂ X, an estimate of the form (5.2). Taking the limit in |u(φ)| ≤ |u(φ) − uj (φ)| + |uj (φ)| ≤ |u(φ) − uj (φ)| + c kφkC k as j → ∞, we find |u(φ)| ≤ c kφkC k for all φ ∈ C0∞ (K). According to Lemma 3.7 this proves that u ∈ D0 (X), and uj → u in D0 (X) now holds by definition. Regarding the last assertion we observe that if φj → φ in C0∞ (X), there exists a compact set K ⊂ X such that φj , φ ∈ C0∞ (K) for all j. Applying Lemma 5.4 once again, we obtain from this |uj (φj ) − u(φ)| ≤ |uj (φj − φ)| + |uj (φ) − u(φ)| ≤ c kφj − φkC k + |uj (φ) − u(φ)|, which converges to 0 as j → ∞.



Example 5.6. Let ut be a family of distributions on an open set X in Rn that depend on t ∈ R. Assume that the function t 7→ ut (φ) is differentiable on R, for every φ ∈ C0∞ (X). For every t ∈ R we then have the existence of d  ut+h (φ) − ut (φ) d 1 = (ut (φ)) =: ut (φ). lim (ut+h − ut )(φ) = lim h→0 h→0 h h dt dt By Theorem 5.5, then,

d dt ut

is a distribution on X.



5 Convergence of Distributions

45

The following lemma proves to be useful, for example in the theory of Fourier series from Chap. 15. Lemma 5.7. Let X be open in Rn and suppose that uk and u ∈ D0 (X) satisfy limk→∞ uk = u in D0 (X). Then, for every multi-index α, one also has limk→∞ ∂ α uk = ∂ α u in D0 (X). Proof. Let φ ∈ C0∞ (X) and 1 ≤ j ≤ n be arbitrary, then ∂j uk (φ) = −uk (∂j φ) → −u(∂j φ) = ∂j u(φ)

k → ∞.

as



Problems n Problem 5.1. Let (fj )∞ j=1 be a sequence of nonnegative integrable functions on R with the following properties: R (a) For every j one has fj (x) dx =R1. (b) For every r > 0 one has limj→∞ kxk≥r fj (x) dx = 0.

Prove that limj→∞ fj = δ in D0 (Rn ). Problem 5.2. Let f be an integrable function on Rn and define f (x) := −n f ( 1 x). Prove that Z lim f = c δ ↓0

in D0 (Rn )

f (x) dx ∈ C.

c=

with

Rn

Problem 5.3. Using the results from Problem 1.1, prove that lim

↓0 x2

x 1 = PV 2 + x

and

lim

↓0 x2

 = πδ + 2

in

D0 (R).

Problem 5.4. If e(j) denotes the j-th standard basis vector in Rn , show that lim

t→0

1 (δa−t e(j) − δa ) = ∂j δa t

in D0 (Rn ).

Problem 5.5. (Heat or diffusion equation). Consider, for t > 0, the following function on Rn : kxk2 n ut (x) = (4πt)− 2 e− 4t . Set n = 1, then make a sketch of this function and its first two derivatives, for a small positive value of t. Verify that ut (x) satisfies the partial differential equation on Rn d ut (x) = ∆ut (x), dt the heat equation or diffusion equation. Calculate the limit in D0 (Rn ) as t ↓ 0 of ut d and of dt ut .

46

5 Convergence of Distributions

Fig. 5.1. Graphs of ut and its first two derivatives, for n = 1 and t = 1/10

Problem 5.6. Let ut (x) as in Problem 5.5 and let n ≥ 3. Prove that for x 6= 0 the integral Z E(x) := − ut (x) dt R>0

converges and calculate E(x). Show that E is locally integrable on Rn and can therefore be interpreted as a distribution on Rn . Prove, for every φ ∈ C0∞ (Rn ), Z ∆E(φ) = −

T

Z

lim

s↓0, T ↑∞

ut (x) (∆φ)(x) dx dt = φ(0), s

Rn

and therefore ∆E = δ. Verify that E equals the E from Problem 4.6. Problem 5.7. Define ft (x) = ta eitx . For what a ∈ R does ft converge to 0 as t→∞ (i) with respect to the C k norm? (ii) in D0 (R)? Problem 5.8. Calculate the limit of ft (x) = t eitx log |x| in D0 (R), as t → ∞. Problem 5.9. (Riemann’s “nondifferentiable” function). Let f be the function on R defined by X sin(n2 x) . f (x) = n2 n∈Z>0

Weierstrass reported that Riemann suggested it as an example of a continuous function that is nowhere differentiable. (Actually, f is differentiable at points of the form pπ/q where p and q both are odd integers, with derivative equal to −1/2, but at no other points, see [6], for instance.) In the illustration below, the first ten thousand terms have been summed for ten thousand different values of 0 ≤ x ≤ π. Prove that lim

N →∞

N X

cos(n2 x) = f 0 (x) in D0 (R).

n=1

In the following problems the reader is asked to derive estimates of the form (5.2), so that we can use a fully proved variant of Theorem 5.5.

5 Convergence of Distributions

47

1

Π

Fig. 5.2. The second graph is an enlargement of the first one, in a neighborhood of x = π/2

Problem 5.10. Let cn ∈ C for all n ∈ Z and suppose that positive constants c and m exist such that |cn | ≤ c |n|m (n ∈ Z \ {0}). Let ω > 0 and write uj :=

j X

cn einωx

(j ∈ Z>0 ).

n=−j

Derive estimates of the form (5.2). Prove that a u ∈ D0 (R) exists for which uj → u in D0 (R). Finally, prove that the order of u is lower than or equal to m + 2. Problem 5.11. Let a(i) be a sequence of points in Rn with limi→∞ ka(i)k = ∞ and let ci ∈ C be an arbitrary sequence of complex coefficients. Give estimates of

48

5 Convergence of Distributions

the form (5.2) for uj :=

j X

ci δa(i) .

i=1

Show that there exists a u ∈ D0 (Rn ) with the property that uj → u in D0 (Rn ). Problem 5.12. Let f be a continuous function on R. Does 2

uN

1 := N

N X j=−N 2

f

j δj N N

converge in D0 (R) as N → ∞? If so, what is the limit? Problem 5.13. Show that for every φ ∈ C0∞ (R) the series in (3.5) converges; also, that (3.5) defines a distribution u on R of infinite order; and, finally, that lim

l→∞

l−1 X

∂ k δk = u

in D0 (R).

k=0

Problem 5.14. Let uj be a sequence of positive measures on X with the property that limj→∞ uj (φ) = u(φ), for every φ ∈ C0∞ (X). Use (3.6) to demonstrate that for every compact subset K of X there is a constant c > 0 with |uj (φ)| ≤ c sup |φ(x)| x

(j ∈ Z>0 , φ ∈ C0∞ (K)).

Prove that u is a positive measure. Furthermore, show that the uj converge weakly as measures to u. That is, limj→∞ uj (f ) = u(f ), for every continuous function f with compact support in X.

6 Taylor Expansion in Several Variables

Many classical asymptotic expansions imply interesting distributional limits. An example is the Taylor expansion of functions of several variables. Because we will be using this expansion elsewhere in this text, we begin by repeating its basic properties. A mapping f from an open subset U of Rn to Rp is given by p real-valued functions of n variables, that is f (x) = f (x1 , . . . , xn ) = (f1 (x1 , . . . , xn ), . . . , fp (x1 , . . . , xn )). Many properties of vector-valued functions can be derived from those of real-valued functions, by proving them for each of the coordinate functions fi (x1 , . . . , xn ), with 1 ≤ i ≤ p. For instance, the mapping f is continuously differentiable if, for every i and every j, the function fi is partially differentiable with respect to the j-th variable and if the function ∂j fi is continuous on U . In that case one writes f ∈ C 1 (U, Rp ), or f ∈ C 1 (U, V ) if f (U ) ⊂ V . The ∂j fi (x), for 1 ≤ j ≤ n and 1 ≤ i ≤ p, form the matrix of the linear mapping Df (x) : h 7→

n X

∂j f (x) hj : Rn → Rp .

j=1

This is the linear approximation at h = 0 of the increase h 7→ f (x + h) − f (x) of f and is also known as the total derivative of f at the point x. If p = 1, one normally writes df (x) instead of Df (x). The functions ∂j fi taken together can again be interpreted as a vector-valued function. This allows us to define, with mathematical induction on k, that f ∈ C k if Dk−1 f ∈ C 1 , where Dk f is inductively defined by Dk f (x) = D(Dk−1 f )(x). A theorem already mentioned is that ∂i (∂j f ) = ∂j (∂i f ) if f ∈ C 2 . For f ∈ C k this enables every higher-order derivative of f , where differentiation with respect to xj takes place αj times in total, to be rearranged in the form ∂ α f , as in (2.6). An important result that we will often be using is the chain rule. For describing this, consider f ∈ C 1 (U, Rp ) and g ∈ C 1 (V, Rq ) where U ⊂ Rn and V ⊂ Rp are

50

6 Taylor Expansion in Several Variables

open subsets. The composition g ◦ f of f and g is defined by (g ◦ f )(x) = g(f (x)) for all x ∈ U with f (x) ∈ V ; the continuity of f guarantees that the set of these x forms an open subset U ∩ f −1 (V ) of Rn . The chain rule asserts that g ◦ f ∈ C 1 (U ∩ f −1 (V ), Rq ), while D(g ◦ f )(x) = Dg(f (x)) ◦ Df (x);

D(g ◦ f ) = (Dg) ◦ f Df.

or

With mathematical induction on k one finds g ◦ f ∈ C k if f ∈ C k and g ∈ C k ; for the higher-order derivatives inductive formulas are obtained that rapidly become very cumbersome to write out in explicit form. Considering the matrix coefficients occurring in the chain rule, one obtains the following identity of functions on U ∩ f −1 (V ), for 1 ≤ j ≤ n and 1 ≤ i ≤ q, p X

∂j (gi ◦ f ) =

(∂k gi ) ◦ f ∂j fk =

k=1

p X

∂j fk (∂k gi ) ◦ f.

(6.1)

k=1

A practical way of studying f (x) for x near a point a ∈ U , is to consider the function on [ 0, 1 ] ⊂ R given by t 7→ g(t) := f (a + t (x − a)). We have g(0) = f (a) and g(1) = f (x); what we actually do is to consider f on the straight line from a to x. The difference vector h = x − a is usually small. In the following lemma we use the abbreviations xα =

n Y

α

xj j

and

j=1

α! =

n Y

αj !.

j=1

Lemma 6.1. If f ∈ C k (U, Rp ), we have, for 0 ≤ j ≤ k and a + th ∈ U , X hα 1 dj f (a + th) = ∂ α f (a + th). j j! dt α!

(6.2)

|α|=j

Proof. Using the chain rule we obtain n

X d f (a + th) = hj ∂j f (a + th). dt j=1 By mathematical induction on j this yields X dj f (a + th) = mα hα ∂ α f (a + th), dtj |α|=j

where the mα ∈ Z are determined by recurrence relations; in particular, they are independent of the choices made for f , a and t. To determine the mα , we take f (x) =

6 Taylor Expansion in Several Variables

51

xβ with |β| = j, a = 0 and differentiate at the point t = 0. Then f (a + th) = (t h)β = t|β| hβ = tj hβ ; the j-th order derivative with respect to t of this function equals j!hβ . But we also have, for every α with |α| = j, ( β! if α = β, α β ∂ x = (6.3) 0 if α 6= β. This proves that mβ = j!/β!.



k For f ∈ C k (U, Rp ), the Taylor polynomial T = Tf,a of order k at the point a ∈ U is defined by X (x − a)α ∂ α f (a). (6.4) T (x) = α! |α|≤k

k The remainder R = Rf,a of order k is defined as R(x) = f (x) − T (x). Then, by definition, f (x) = T (x) + R(x); we want to find out what information can be obtained about the remainder. The following formulation may be somewhat heavier than necessary for most applications, but it follows directly from the integral formula used for the remainder.

Theorem 6.2. Let f be a C l mapping defined on an open ball U in Rn . For every 0 ≤ k k ≤ l, the mapping (a, x) 7→ Rf,a (x) is a C l−k mapping on U × U . Furthermore, for every compact K ⊂ U and every  > 0, there exists a δ > 0 such that k kRf,a (x)k ≤  kx − akk

if

a, x ∈ K

and kx − ak < δ.

k Finally, R := Rf,a ∈ C l and ∂ α R(a) = 0 for all α with |α| ≤ k.

Proof. The Taylor expansion of a C k function g on an interval I containing 0 reads g(t) =

k−1 X j j=0

t (j) g (0) + j!

Z 0

t

g (k) (s)

(t − s)k−1 ds (k − 1)!

(t ∈ I);

the proof is obtained by mathematical induction on k. When we apply this formula, taking g(t) = f (a + t (x − a)) and substituting (6.2) for the derivatives, we obtain, for t = 1, the identity f (x) = T (x) + R(x), with X (x − a)α Z 1
R(x) = k (1 − s)k−1 ∂ α f (a + s (x − a)) − ∂ α f (a) ds. α! 0 |α|=k

On account of the locally uniform continuity of the k-th order derivatives of f , the integral converges locally uniformly to 0 as kx−ak → 0. The fact that the remainder is of class C l−k follows from the Theorem on differentiation under the integral sign. The function Rf,a is C l , being the difference of f and the Taylor polynomial T := Tf,a . From (6.3) we see that ∂ α T (a) = ∂ α f (a) for all multi-indices α with |α| ≤ k.  By way of application we give a “higher-order version” of Problem 5.2.

52

6 Taylor Expansion in Several Variables

Proposition 6.3. Let f be an integrable function of n variables with compact support. Let a ∈ Rn and  1 (x − a) , f (x) := −n f  for  > 0; we now define, for each multi-index α, the coefficient cα by Z cα := xα f (x) dx ∈ C. Rn

Next, we introduce, for every j ∈ Z≥0 , the distribution uj by uj :=

X cα ∂ α δa . α!

|α|=j

We then have, for all k ≥ 0, lim −k (f − ↓0

k X

(−)j uj ) = 0

in D0 (Rn ).

j=0

Proof. For every φ ∈ C0∞ (Rn ) we have Z Z  1 (x − a) φ(x) dx = f (h) φ(a + h) dh. f (φ) = −n f  Rn Rn The assertion now follows by substituting the Taylor expansion of φ at the point a and of order k.  The coefficients cα are known as the moments of the function f . A consequence of the proposition is that −k f converges distributionally to (−1)k uk as  ↓ 0, if cα = 0 for all multi-indices α with |α| < k.

Problems Problem 6.1. Formulate a variant of Proposition 6.3 in which f is not required to have compact support. Problem 6.2. Let ut (x) be as in Problem 5.5. Prove, for every k, lim t t↓0

−k

 ut −

k−1 X j j=0

t ∆j δ j!



Discuss the notation ut = et ∆ δ, for t > 0.

=

1 k ∆ δ k!

in D0 (Rn ).

6 Taylor Expansion in Several Variables

53

Problem 6.3. (Sequel to Problem 6.2.) Let φ ∈ C0∞ (Rn ) and define Z Z n Φ(r) := φ(x) dx = r φ(r x) dx. kxk0 , the function x 7→ x−k on R>0 has the extension

7 Localization k−1

57

k

d (−1) log |x| ∈ D0 (R) (k − 1)! dxk to a distribution on R. However, there are many cases where extension is not possible; for instance, the function x 7→ e1/x on R>0 has no distributional extension to any neighborhood of 0 in R. Proof. Suppose φ ∈ C0∞ (R>0 ) satisfies φ ≥ 0 and 1(φ) > 0 and set φ (x) = 1 x 1/x with φ , we see that for every N there  φ(  ). If a denotes the result of testing e −N exists a constant c > 0 such that a ≥ c  , for all 0 <  ≤ 1. (Hint: use the fact that e1/x ≥ x−N /N ! if x > 0 and apply the change of variables x =  y in the integral.) On the other hand, if u ∈ D0 ( ]−δ, ∞ [ ) with δ > 0, Lemma 3.7 asserts that there exist k and c0 , c00 > 0 with the property that |u(φ )| ≤ c0 kφ kC k ≤ c00 −k−1 for all 0 <  ≤ 1. Thus, the assumption that u = e1/x on R>0 leads to a contradiction.  There exists a counterpart of Theorem 7.1 in the form of an existence theorem: Theorem 7.4. Let U be a collection of open subsets of Rn with union X. Assume that for every U ∈ U a distribution uU ∈ D0 (U ) is given and that uU = uV on U ∩ V , for all U , V ∈ U. Then there exists a unique u ∈ D0 (X) with the property that u = uU on U , for every U ∈ U. Proof. Let K be a compact subset of X and ψj a partition of unity over K subordinate to the covering U. In particular, there exists U (j) ∈ U such that supp ψj ⊂ U (j) ∈ U. If φ ∈ C0∞ (X) and supp φ ⊂ K, then φ equals the sum of the finitely many ψj φ ∈ C0∞ (X). We therefore have to take u(φ) =

l X

uU (j) (ψj φ).

(7.3)

j=1

The uniqueness of u follows. Before continuing we first prove that the right-hand side of (7.3) is independent of the choice of the partition ψj of unity over K subordinate to U. Suppose χk is another partition of unity over K with supp χk ⊂ V (k) ∈ U. Then supp (ψj χk φ) ⊂ U (j) ∩ V (k). Consequently, uU (j) and uV (k) assume the same value on ψj χk φ. We therefore have l X j=1

uU (j) (ψj φ) =

l X j,k=1

uU (j) (ψj χk φ) =

l X j,k=1

uV (k) (ψj χk φ) =

l X

uV (k) (χk φ).

k=1

We now interpret the right-hand side in (7.3) as the definition of u. If α, β ∈ C0∞ (X) and a, b ∈ C, we take K = supp α ∪ supp β above and thus obtain that u(α), or u(β), or u(a α + b β), equal the right-hand side in (7.3), replacing φ by

58

7 Localization

α, or β, or a α + b β, respectively. From this it is immediate that u(a α + b β) = a u(α) + b u(β). In other words, u is a linear form on C0∞ (X). If φk → φ in C0∞ (X), there exists a compact subset K of X with the property that for every k one has that supp φk ⊂ K; this also implies that supp φ ⊂ K. Then, for every k, the u(φk ) equal the right-hand side of (7.3) with φ replaced by φk . From the continuity of the uU (j) it now follows that u(φk ) → u(φ) as k → ∞. We have now proved that (7.3) defines a continuous linear form u on C0∞ (X). Finally, if supp φ ⊂ U ∈ U, we have supp (ψj φ) ⊂ U (j) ∩ U , and therefore uU (j) (ψj φ) = uU (ψj φ). Summation over j gives u(φ) = uU (φ).  Remark 7.5. Thanks to their properties described in Theorem 7.4, the linear spaces D0 (U ), as functions of the open subsets U of X, form a system known in the literature as a (pre)sheaf. In many cases, sheaves consist of spaces of functions on U , with the usual restriction of functions to open subsets. One example is the sheaf U 7→ C ∞ (U ) of the infinitely differentiable functions. For these, the sheaf properties are a direct consequence of the local nature of the definition of the functions in the sheaf. Distributions form an example of a sheaf where the linear spaces contain elements that are not functions.  Definition 7.6. Let u ∈ D0 (X). The singular support of u, denoted by sing supp u, is the set of all x ∈ X that do not possess any open neighborhood U in X such that ρU X u ∈ C ∞ (U ).  The complement C of sing supp u has an open covering U such that ρU X u = uU ∈ C ∞ (U ), for every U ∈ U. This follows from the definition of sing supp . If U and V ∈ U, then uU = uV on U ∩ V , which implies that the uU possess a common extension to a C ∞ function f on C. In view of Theorem 7.1 we have u = f on C. In other words, u is of class C ∞ on C; every open subset of X on which u is of class C ∞ is contained in C. Obviously, sing supp u ⊂ supp u. In Problem 7.4 we use the following definition. In subsequent parts of this text we will repeatedly encounter linear partial differential operators with constant coefficients. Definition 7.7. Let

X

P (ξ) =

cα ξ α

|α|≤m

be a polynomial in n variables of degree m with coefficients cα ∈ C. Replacing ξ by ∂ = ∂x we obtain a differential operator P (∂) = P (∂x ) =

X |α|≤m

cα ∂ α =

X |α|≤m

cα

∂α , ∂xα

a linear partial differential operator of order m and with constant coefficients.

(7.4) 

Note that P (∂) maps D0 (X) into itself, for any subset X of Rn , see also Problem 7.4.

7 Localization

59

Problems Problem 7.1. Let U be a measurable subset of X. Prove that the support of ∂j 1U is contained in the boundary ∂U = X ∩ U \ U int of U in X. Here U int is the set of the interior points of U . Now, let U be an open subset of X with C 1 boundary ∂U . Prove that ∂j 1U = −νj δ∂U . Here νj denotes the j-th component of the outer normal to ∂U . Also discuss the case n = 1. Problem 7.2. Show that for a continuous function the support equals the support in the distributional sense. Problem 7.3. Determine the supports of the Dirac function, the Heaviside function and PV x1 , respectively. Problem 7.4. Prove that supp P (∂) u ⊂ supp u, for every u ∈ D0 (X) and every linear partial differential operator with constant coefficients P (∂). Problem 7.5. Let a and f ∈ C ∞ (X) and let f be real-valued. A point p ∈ X is said to be a stationary point of f if ∂j f (p) = 0, for all 1 ≤ j ≤ n; let Sf be the set of the stationary points of f in X. Prove that X \ Sf is an open subset of X and that for every m ∈ R lim tm a eitf = 0 in D0 (X \ Sf ). t→∞

Hint: use the formula eitf = ∂j (eitf )/(it∂j f ), wherever ∂j f 6= 0. Problem 7.6. Let a ∈ C and define xa+ = xa = ea log x if x > 0 and xa+ = 0 if x < 0. Prove the following assertions: (i) If Re a > −1, then xa+ is locally integrable in R and can therefore be interpreted as a distribution. (ii) If a is not a negative integer, xa+ =

1 dk a+k x (a + k) · · · (a + 1) dxk +

with

k > − Re a − 1,

defines a distribution on R that is an extension of xa+ on R\{0}. This distribution is independent of the choice of k. Determine the support and the order of this distribution. What happens if the real part of a is a negative integer (and the imaginary part does not vanish)? (iii) Let l+ (x) = log x if x > 0 and l+ (x) = 0 if x < 0. l+ is locally integrable in R and thus defines a distribution on R. For every k ∈ Z>0 (−1)k−1 dk l+ (k − 1)! dxk is a distribution on R that forms an extension of x−k + on R \ {0}. Again, determine its support and its order. (iv) Formulate similar results when starting from (−x)a+ or l+ (−x), respectively.

60

7 Localization

Problem 7.7. Show that for every C 1 function f on R>0 a real-valued C 1 function g on R>0 can be found, with the property that f eig has an extension to a distribution of order ≤ 1 on R. Hint: show that there exists a real-valued C 1 function g on R>0 such that g 0 (x) 6= 0, f (x)/g 0 (x) converges to zero as x ↓ 0 and f 0 /g 0 is absolutely integrable over a neighborhood of 0. Prove that Z d  f (x) φ(x)  dx (φ ∈ C∞ u(φ) = i ei g(x) 0 (R)) dx g 0 (x) R>0 defines a distribution u on R of order ≤ 1 and that u = f eig on R>0 . Problem 7.8. A subset D of X is said to be discrete if, for every a ∈ D, there is an open neighborhood Ua of a in X such that D ∩ Ua = {a}. Let X be an open subset of Rn and D a discrete subset of X. For every a ∈ D let there be a partial differential operator Pa (∂) of order ma . Write ua = Pa (∂) δa . Demonstrate the following: (i) There exists exactly one distribution u in X with supp u ⊂ D, such that for every a ∈ DPthere is an open neighborhood Ua of a in X on which u = ua . One writes u = a∈D ua . (ii) supp u = D, and the order of u equals the supremum of the ma , for a ∈ D. (iii) D is countable and for every enumeration j 7→ a(j) of D we have that u = Pk limk→∞ j=1 ua(j) . Problem 7.9. Let uj be a sequence of distributions in X. Suppose that for every x ∈ X there is an open neighborhood U of x in X with the property that the sequence of complex numbers uj (φ) converges for every φ ∈ C0∞ (U ) as j → ∞. Show that there exists a u ∈ D0 (X) such that limj→∞ uj = u in D0 (X). Problem 7.10. Determine the singular support of |x|, the Dirac function, the Heaviside function, PV x1 , 1/(x + i0) and 1/(x − i0), respectively. And also of the components vj of the vector field v in (1.2). Problem 7.11. Prove that sing supp (∂j u) ⊂ sing supp u for every u ∈ D0 (X), with j = 1, . . . , n.

8 Distributions with Compact Support

If u is locally integrable on an open set X in Rn and has compact support, the integral Z u(φ) = u(x) φ(x) dx X

is absolutely convergent for every φ ∈ C ∞ (X), as follows from a slight adaptation of the proof of Theorem 3.4. More generally, every distribution with compact support can be extended to a continuous linear form on C ∞ (X). Before we can give a precise formulation of this result, we have to define convergence in C ∞ (X); clearly, Definition 2.12 has to be modified. Definition 8.1. Let φj and φ ∈ C ∞ (X) and j ∈ Z>0 . We say lim φj = φ in C ∞ (X)

j→∞

if, for every multi-index α and for every compact subset K of X, the sequence ∂ α φj converges uniformly on K to ∂ α φ as j → ∞. A linear form u on C ∞ (X) is said to be continuous if u(φj ) → u(φ) whenever φj → φ in C ∞ (X), as j → ∞. The space of continuous linear forms on C ∞ (X) is denoted by E 0 (X). One says that, for uj and u ∈ E 0 (X), lim uj = u

j→∞

in E 0 (X)

if limj→∞ uj (φ) = u(φ) in C, for every φ ∈ C ∞ (X) (compare with Definition 5.1).  The notation E 0 (X) echoes the notation E(X) used by Schwartz to indicate the space C ∞ (X) endowed with the notion of convergence introduced in Definition 8.1. Note that for a linear form u on C ∞ (X) to be continuous it is sufficient that u(φj ) → 0 whenever φj → 0 in C ∞ (X). Furthermore, if φj → 0 in C0∞ (X), then certainly φj → 0 in C ∞ (X). It follows that, for every u ∈ E 0 (X), the restriction ρ u of u to C0∞ (X) belongs to D0 (X).

62

8 Distributions with Compact Support

A sequence of compact subsets Kj of X is said to absorb the set X if for every compact subset K of X there exists an index j with K ⊂ Kj . The sequence is said to be increasing if Kj ⊂ Kj+1 for all j. Lemma 8.2. Let X be an open subset of Rn . Then the following holds: (a) There exists an increasing sequence Kj of compact subsets of X that absorbs X. (b) There is a sequence χj ∈ C0∞ (X) with the property that, for every φ ∈ C ∞ (X), the sequence φj := χj φ ∈ C0∞ (X) converges in C ∞ (X) to φ, as j → ∞. (c) The restriction mapping ρ : E 0 (X) → D0 (X) is injective. Proof. (a). Using the notation C := Rn \ X, the complement of X in Rn , we define Kj = { x ∈ X | kxk ≤ j and d(x, C) ≥

1 } j

(j ∈ Z>0 ).

These are bounded and closed subsets of Rn , and therefore compact on account of Theorem 2.2. Furthermore, every Kj ⊂ X. If K is a compact subset of X, there exists an R with kxk ≤ R for all x ∈ K. In addition, the δ-neighborhood of K is contained in X for sufficiently small δ > 0 in view of Corollary 2.4; this implies d(x, C) ≥ δ, for all x ∈ K. If we choose j ∈ Z>0 sufficiently large that R ≤ j and δ ≥ 1j , then K ⊂ Kj . (b). On the strength of Corollary 2.16, we can find χj ∈ C0∞ (X) with χj = 1 on an open neighborhood of Kj . Let φ ∈ C ∞ (X). For every compact subset K of X there is a j(K) such that K ⊂ Kj(K) ; this implies K ⊂ Kj , if j ≥ j(K). But then one has φj = χj φ = φ on an open neighborhood of K, and therefore also ∂ α φj = ∂ α φ on K, for every multi-index α. It follows that certainly φj → φ in C ∞ (X) as j → ∞. (c). This is a consequence of (b). Indeed, consider u and v ∈ E 0 (X) with ρ u = ρ v. For every φ ∈ C ∞ (X) we have u(φ) = lim u(φj ) = lim v(φj ) = v(φ), j→∞

j→∞

where φj ∈ C0∞ (X) and φj → φ in C ∞ (X).



A subset D of a topological space E is said to be dense in E if every φ ∈ E is the limit in E of elements of D. Lemma 8.2.(b) implies that C0∞ (X) is dense in C ∞ (X). Lemma 8.2.(c) is a special case of the general principle that a continuous extension to the closure of a set is uniquely determined. Because of the injectivity of ρ, every u ∈ E 0 (X) can be identified with the distribution ρ u ∈ D0 (X); accordingly, we will write ρ u = u. Further note that uj → u in E 0 (X) implies that uj → u in D0 (X). In order to determine which elements in D0 (X) belong to E 0 (X), we first take a closer look at convergence in C ∞ (X).

8 Distributions with Compact Support

63

Definition 8.3. For every k ∈ Z≥0 and compact subset K of X, we define kφkC k , K =

sup

|∂ α φ(x)|

(φ ∈ C k (X)).

(8.1)

x∈K, |α|≤k

For φj and φ ∈ C k (X), we say that limj→∞ φj = φ in C k (X) if we have limj→∞ kφj − φkC k , K = 0, for every compact subset K of X. With this definition, φj → φ in C ∞ (X) is equivalent to the assertion that φj → φ in C k (X) for every k ∈ Z≥0 .  Because (8.1) is not subject to the condition supp φ ⊂ K, (8.1) does not define a norm on either C k (X) or C ∞ (X); for every k and K there exist φ ∈ C ∞ (X) with kφkC k , K = 0 and φ 6= 0. However, the nk,K : φ 7→ kφkC k , K do form a separating collection of seminorms. We now discuss the definition and properties of seminorms for arbitrary linear spaces. In Theorem 8.5 we will resume the characterization of the distributions in E 0 (X). For a linear space E over C, a seminorm on E is defined as a function n : E → R with the following properties. For every x and y ∈ E and c ∈ C, one has (a) n(x) ≥ 0, (b) n(x + y) ≤ n(x) + n(y), (c) n(c x) = |c| n(x). In other words, n has all properties of a norm, except that there is no requirement that n(x) = 0 implies x = 0. Instead of a norm on E, one may consider a collection N of seminorms on E that is separating in the following sense: (d) if n(x) = 0 for all n ∈ N , then x = 0, (e) if n and m ∈ N , there exists a p ∈ N such that n(x) ≤ p(x)

and

m(x) ≤ p(x)

(x ∈ E).

An example is E = C k (X) with the collection of seminorms N k := { nk,K | K ⊂ X, K compact }. S On C ∞ (X) we use N = k∈Z≥0 N k . A pair (E, N ) where E is a linear space and N a separating collection of seminorms on E, is said to be a locally convex topological linear space. A subset U of E is said to be a neighborhood of a in (E, N ) if there exist an n ∈ N and an  > 0 such that B(n, a, ) := { x ∈ E | n(x − a) <  } ⊂ U. The word “topological” relates to “neighborhoods”. Furthermore, a subset C of E is said to be convex if for every x and y ∈ C the line segment from x to y, the set of the x + t (y − x) with 0 ≤ t ≤ 1, is contained in C. For every n ∈ N and  > 0, one has that B(n, a, ) is convex; this is the origin of the term “locally convex”. The assertion about convexity follows from

64

8 Distributions with Compact Support

n(x + t(y − x) − a) = n((1 − t)(x − a) + t(y − a)) ≤ (1 − t) n(x − a) + t n(y − a). Readers familiar with topology will have noticed that the B(n, a, ) form a basis. A subset U of E is said to be open if U is a neighborhood of all of its elements. The open sets form a topology in E. For a deeper study of distribution theory we recommend that the reader peruse the general theory of locally convex topological linear spaces, as can be found in Bourbaki [4], for example. Let xj and x ∈ E. One then says that limj→∞ xj = x in (E, N ) if lim n(xj − x) = 0

j→∞

(n ∈ N ).

If (F, M) is another locally convex topological linear space, a linear mapping A from E to F is said to be sequentially continuous from (E, N ) to (F, M) if limj→∞ A(xj ) = A(x) in (F, M) whenever limj→∞ xj = x in (E, N ). A mapping A : E → F is said to be continuous from (E, N ) to (F, M) if for every a ∈ E and every neighborhood V of A(a) in (F, M) there exists a neighborhood U of a in (E, N ) such that A(U ) ⊂ V . For a linear mapping this is equivalent to the assertion that for every m ∈ M there exist a constant c > 0 and an n ∈ N with m(A(x)) ≤ c n(x) (x ∈ E). (8.2) For the sake of completeness we give a proof of this general result from functional analysis. Proof. Let a ∈ E and let V be a neighborhood of A(a) in (F, M), that is, there exist an m ∈ M and an  > 0 such that B(m, A(a), ) ⊂ V . Suppose that there are a constant c > 0 and an n ∈ N such that (8.2) holds true. In that case, x ∈ B(n, a, δ) implies m(A(x) − A(a)) = m(A(x − a)) ≤ c n(x − a) < c δ. Choosing δ = /c, we have δ > 0 and we see that A maps the neighborhood B(n, a, δ) of a into B(m, A(a), ), and therefore into V . From this we conclude that assertion (8.2) implies the continuity of A. Conversely, suppose that A is continuous and that m ∈ M. From the continuity of A we only use that there is some a ∈ E, that there are constants  > 0 and δ > 0, and that there exists an n ∈ N with the property that n(x − a) < δ implies m(A(x) − A(a)) < . (This is a weak form of the continuity of the mapping A at the point a.) Because A(x) − A(a) = A(x − a), we see that this is equivalent to the property that n(y) < δ implies m(A(y)) < . (This is a weak form of the continuity of the mapping A at the point 0.) Now define c = 2/δ; we are going to prove that for every x ∈ E one has m(A(x)) ≤ c n(x). If n(x) > 0, one has, for y = λ x where λ = δ/(2n(x)) n(y) = n(λ x) = λ n(x) = δ/2 < δ; consequently  > m(A(y)) = m(A(λ x)) = m(λ A(x)) = λ m(A(x)),

8 Distributions with Compact Support

65

which implies m(A(x)) < c n(x). If, on the other hand, n(x) = 0, then for every λ > 0 one has n(y) = n(λ x) = λ n(x) = 0 < δ; it follows that  > m(A(y)) = λ m(A(x)), which implies m(A(x)) ≤ 0 ≤ c n(x).  For F = C, with the absolute value as the only (semi)norm, we thus obtain the continuous linear forms on E, together forming the topological dual E 0 of E. The definition as given here is more exact than that in Chap. 3, because the definition there speaks rather vaguely about a “notion of convergence” in E. In E 0 we use the seminorms u 7→ |u(x)| for all x ∈ E. For these, limj→∞ uj = u in E 0 is equivalent to limj→∞ uj (x) = u(x), for every x ∈ E; this is the weak convergence in E 0 that we have introduced before, in Definitions 5.1 and 8.1. If the separating collection N of seminorms is finite, we may, as far as the notion of convergence is concerned, as well replace the collection N by the one norm given by the maximum of the n ∈ N . We then find ourselves in the familiar context of linear spaces endowed with a norm, where linear mappings that do satisfy estimates as in (8.2) are also said to be bounded linear mappings. For C k (X) and C ∞ (X), the collection N cannot be replaced by a single norm. It is possible, however, without changing the notion of convergence, to replace the collection of seminorms by the increasing countable sequence ni := ni,Ki , where Ki is an increasing sequence of compact subsets of X that absorbs X. Indeed, for every compact K ⊂ X there exists a j with K ⊂ Kj ; therefore nk,K ≤ ni if we choose i to be the maximum of k and j. The following lemma generalizes a familiar result about linear mappings between linear spaces endowed with a norm. Lemma 8.4. Let (E, N ) and (F, M) be locally convex topological linear spaces. Then every continuous linear mapping A : E → F is sequentially continuous. If N is countable, the converse is also true, that is, every sequentially continuous linear mapping A : E → F is continuous. Proof. Suppose that A is continuous and xj → x in E. For every m ∈ M we then have, on account of (8.2) m(A(xj ) − A(x)) = m(A(xj − x)) ≤ c n(xj − x) → 0

as

j → ∞.

Now suppose that k 7→ nk is an enumeration of N . By changing to νk := max nj , 1≤j≤k

we obtain an increasing sequence of seminorms νk that defines the same notion of neighborhood in E. Suppose A is not continuous. Then there exist an m ∈ M and, for every c and k, an x = xc,k ∈ E such that m(A(x)) > c νk (x). Next, take c = k and introduce yk = λ x with λ = 1/m(A(x)). We then find 1 = m(A(yk )) > k νk (yk ), from which we conclude that yk → 0 in (E, N ), without A(yk ) → 0 in (F, M). It follows that A is not sequentially continuous. 

66

8 Distributions with Compact Support

The proof of the second assertion was identical to that of Lemma 3.7. We have formulated Lemma 8.4 in this general form to avoid having to rewrite the same argument each time we work with another collection of seminorms. Theorem 8.5. Consider u ∈ D0 (X). One has u ∈ E 0 (X) if and only if supp u is a compact subset of X. If that is the case, u is of finite order and there exist constants c > 0 and k ∈ Z≥0 satisfying |u(φ)| ≤ c kχ φkC k ,

(8.3)

for every φ ∈ C ∞ (X) and every χ ∈ C0∞ (X) such that χ = 1 on an open neighborhood of supp u. Proof. Let u ∈ E 0 (X). By Lemma 8.4, with E = C ∞ (X) and F = C, it follows from the sequential continuity of u that u is continuous in the sense of (8.2). That is, there exist constants c > 0 and k ∈ Z≥0 and a compact subset K of X such that |u(φ)| ≤ c kφkC k , K

(φ ∈ C ∞ (X)).

(8.4)

From this it follows, in particular, that u(φ) = 0 if φ ∈ C0∞ (X \ K), in other words, supp u ⊂ K. The conclusion is that supp u is a compact subset of X and that u is of finite order, see Definition 3.10. The estimate (8.3) now follows from (8.4). Indeed, supp (φ − χ φ) ∩ supp u = ∅ implies u(φ − χ φ) = 0; in turn, this gives |u(φ)| = |u(χ φ)| ≤ c kχ φkC k , K ≤ c kχ φkC k . Next, suppose that supp u is compact. Corollary 2.16 yields a χ ∈ C0∞ (X) with χ = 1 on a neighborhood of supp u. Define v(φ) = u(χ φ) for every φ ∈ C ∞ (X); then v is a linear form on C ∞ (X). Now, φj → 0 in C ∞ (X) implies that χ φj → 0 in C0∞ (X) and therefore also v(φj ) → 0. It follows that v ∈ E 0 (X). Finally, if φ ∈ C0∞ (X), then supp (φ − χ φ) ∩ supp u = ∅, and so 0 = u(φ − χ φ) = u(φ) − v(φ). This means that u = ρ v.  In view of Theorem 8.5, E 0 (X) is said to be the space of distributions with compact support in X. Lemma 8.6. Let U and V be open subsets of Rn with U ⊂ V . For every u ∈ E 0 (U ) there exists exactly one v = i(u) ∈ E 0 (V ) such that v = u on U and supp v = supp u. This defines an injective continuous linear mapping i from E 0 (U ) to E 0 (V ). For v ∈ E 0 (V ) one has that v ∈ i(E 0 (U )) if and only if supp v ⊂ U . The mapping i is used to identify E 0 (U ) with the space of the v ∈ E 0 (V ) such that supp v ⊂ U , with the notation i(u) = u. Proof. The mapping ρU V from (7.1) defines a continuous linear mapping from C ∞ (V ) to C ∞ (U ). Thus, v := u ◦ ρU V ∈ E 0 (V ). It will be clear that v = u on U and that v = 0 on C := V \ supp u. Because V = U ∪ C, we conclude that

8 Distributions with Compact Support

67

w = v whenever w ∈ D0 (V ), w = u on U and w = 0 on C, which proves the uniqueness. It is also evident that u = 0 if v = 0, which yields the injectivity of the linear mapping i.  The estimate (8.4) is of the same form as (3.4); however, in the case of (3.4) the test functions were restricted to C0∞ (K). On C0∞ (K), sequential continuity of linear forms is equivalent to continuity, and Lemma 3.7 asserts that the distributions on X are precisely those linear forms u on C0∞ (X) for which the restriction of u to C0∞ (K), for every compact subset K of X, is continuous from C0∞ (K) to C. This is the reason why we took the liberty, in discussing the definition of distributions, to speak of continuous linear forms, although in fact we had introduced them as sequentially continuous linear forms. We may add that it is also possible to endow C0∞ (X) with an (uncountable) collection of seminorms N , in such a way that the distributions are precisely the continuous linear forms on C0∞ (X) with respect to N , see for example H¨ormander [16, Thm. 2.1.5]. Contrary to what might perhaps be expected, (8.4) does not generally hold true for K = supp u; a counterexample is given in H¨ormander [16, Example 2.3.2], see Problem 8.3. But, on the other hand: Theorem 8.7. Let a ∈ Rn and let U be an open neighborhood of a in Rn . Consider u ∈ D0 (U ) with supp u = {a}. Then u is of finite order, say, ≤ k, and one has X u(x 7→ (a − x)α ) . u= cα ∂ α δa with cα = α! |α|≤k

In particular, a measure on U with support in a is of the form c δa with c = u(1). Proof. Considering that the conclusion only relates to the behavior of u near a, we may assume that U is a ball of center a. Furthermore, we can reduce the problem to the case that a = 0, by means of a translation. Because u ∈ E 0 (U ), it follows from Theorem 8.5 that u has a uniquely determined continuous extension to C ∞ (U ), again denoted by u. Based on the estimate (8.3), we will first prove u(φ) = 0

if

φ ∈ C ∞ (U )

and

∂ α φ(0) = 0

(|α| ≤ k).

Applying Taylor expansion, with an estimate for the remainder according to Theorem 6.2, we obtain for such a function φ lim −k sup |φ(x)| = 0. ↓0

kxk≤

At 0, all derivatives to order k − |γ| of the function ∂ γ φ vanish, and so the preceding formula implies lim |γ|−k sup |∂ γ φ(x)| = 0. ↓0

kxk≤

Furthermore, we specify the functions χ to be used in (8.3). In fact, choose χ ∈ C0∞ (Rn ) with χ = 1 on a neighborhood of 0 and χ(x) = 0 if kxk > 1. Write

68

8 Distributions with Compact Support

χ (x) = χ( 1 x), for  > 0. We then have χ = 1 on a neighborhood of 0 and χ (x) = 0 if kxk > ; also, for every multi-index β there exists a constant cβ > 0 with the property |∂ β χ (x)| ≤ cβ −|β| (x ∈ Rn ). Next we substitute ψ = χ φ into the estimate (8.3). In doing so, we write ∂ α ψ(x) using Leibniz’ formula as a sum of a constant times terms of the form ∂ β χ (x) ∂ γ φ(x), with β + γ = α. Each of these terms can be estimated uniformly in x, as cβ −|β| sup |∂ γ φ(x)|, kxk≤

which converges to 0 as  ↓ 0, because −|β| = |γ| − |α| ≥ |γ| − k. The conclusion is that |u(φ)| converges to 0 as  ↓ 0. This is only possible if u(φ) = 0. For general φ ∈ C ∞ (U ), Taylor expansion to order k at 0 now gives φ(x) =

X ∂ α φ(0) xα + R(x), α!

|α|≤k

where R ∈ C ∞ (U ) and ∂ α R(0) = 0 whenever |α| ≤ k. The latter implies u(R) = 0, and therefore X ∂ α φ(0) u(φ) = u(x 7→ xα ). α! |α|≤k

Applying the definition of derivative of δ leads to the formula in the theorem.



The preceding theorem is definitely a local result. Still, its proof had to wait until the present chapter dealing with C ∞ , on account of the use of the Taylor expansion. Generally speaking, a remainder in the Taylor expansion of a function in C0∞ (X), with X open in Rn , does not belong to C0∞ (X), because Taylor polynomials do not belong to that space.

Problems Problem 8.1. Suppose that the singular support of u ∈ D0 (X) is a compact subset of X. Prove that u is of finite order. Problem 8.2. Let uj be a sequence of distributions with compact support in X that converges in D0 (X) to u ∈ D0 (X). If there exists a compact subset K of X with supp uj ⊂ K for all j, then supp u ⊂ K and uj → u in E 0 (X), in the sense that uj (φ) → u(φ) for every φ ∈ C ∞ (X). Prove this. Let a(j) be a sequence of points in X with the property that ka(j)k → ∞ or d(a(j), Rn \ X) → 0 as j → ∞. Prove that limj→∞ δa(j) = 0 in D0 (X), but not in E 0 (X).

8 Distributions with Compact Support

Problem 8.3. Let x(j)


j∈Z>0

69

be a sequence of points in Rn that converges to x ∈

Rn . In addition, assume that all x(j) and x differ from each other. Finally, let (j ) be a sequence of positive real numbers with the property X X j kx(j) − xk < ∞. j = ∞, while also j∈Z>0

j∈Z>0 (j)

(i) Prove the existence of x , x and j with these properties, for j ∈ Z>0 . (ii) Verify that X
j φ(x(j) ) − φ(x) u(φ) = (φ ∈ C0∞ (Rn )) j∈Z>0

defines a distribution u of order ≤ 1. Prove supp u = { x(j) | j ∈ Z>0 } ∪ {x} and show that this is a compact set. (iii) Verify that for every l ∈ Z>0 there exists a φl ∈ C0∞ (Rn ) with φl = 1 on a neighborhood of x(j) if 1 ≤ j ≤ l, while φl = 0 on a neighborhood of x(j) if j > l and φl = 0 on a neighborhood of x. Prove that for every k ∈ Z≥0 kφl kC k , supp u = 1,

while

u (φl ) =

l X

j .

j=1

(iv) Demonstrate that (8.4) does not hold for any k ∈ Z≥0 , with K = supp u. Problem 8.4. Prove that, for every r > 0, Z 2π ur (φ) = φ(r cos t, r sin t) dt

(φ ∈ C0∞ (R2 ))

0

defines a measure ur with compact support in R2 and determine its support. Is ur a locally integrable function? Calculate ur (x 7→ xα ) for all α with |α| ≤ 2. For what functions θ does one have limr↓0 θ(r) ur = δ? And what condition has to be met by the functions α and β so that
lim α(r) ur + β(r) δ = ∆δ? r↓0

Prove that, if α(r) ur + β(r) δ converges to a distribution v of order ≤ 2, it follows that v must be a linear combination of δ and ∆δ. Problem 8.5. Consider the function E of n+1 variables defined by E(x, t) = ut (x) for t > 0, with ut as in Problem 5.5, and E(x, t) = 0 for t ≤ 0. Prove that E is locally integrable, so that E can be interpreted as a distribution on Rn+1 . Determine sing supp E. Let v = ∂t E − ∆x E ∈ D0 (Rn+1 ), where ∆x denotes the Laplace operator only with respect to the x variables. Determine supp v and estimate the order of v. Finally, calculate v using integration by parts, as in Problem 5.6. Note that in this case there is no requirement that n ≥ 3.

9 Multiplication by Functions

If X is an open subset of Rn , the function ψ belongs to C ∞ (X) and f is locally integrable on X, one has, for every test function φ, (test(ψ f ))(φ) = (test f )(ψ φ). 0

For an arbitrary u ∈ D (X) we now define (ψ u)(φ) = u(ψ φ)

(φ ∈ C0∞ (X)).

This leads to a linear form ψ u on C0∞ (X). Using Leibniz’ formula (2.7) we see that the C k norm of ψ φ can be estimated by a constant times the C k norm of φ; using Lemma 3.7 we therefore conclude that ψ u ∈ D0 (X). The definition is formulated such that test(ψ f ) = ψ (test f ) for every locally integrable function f ; thus we do not immediately run into notation difficulties if we omit the premodifier “test”. From formula (9.2) below it follows that the mapping “multiplication by ψ”: u 7→ ψ u is linear and continuous from D0 (X) to D0 (X). One has u = ψ u on U if ψ(x) = 1 for all x in the open subset U of X. Because ψ φ = 0 if φ ∈ C0∞ (X) and supp ψ ∩ supp φ = ∅, we also obtain supp (ψ u) ⊂ supp ψ ∩ supp u. Example 9.1. Suppose X is an open set in Rn and let a ∈ X and ψ ∈ C ∞ (X). Then ψ δa = ψ(a) δa in D0 (X). Indeed, for every φ ∈ C0∞ (X), we have (ψ δa )(φ) = δa (ψ φ) = (ψ φ)(a) = ψ(a) φ(a) = ψ(a) δa (φ). In particular, x δ = 0 in D0 (X) if 0 ∈ X. In turn, this implies x δ 0 = −δ, because x δ 0 (φ) = δ 0 (x φ) = −δ(∂x (x φ)) = −δ(φ) − x δ(φ0 ) = −δ(φ). This leads to x2 δ 0 = 0 in D0 (X), on account of x2 δ 0 (φ) = x δ 0 (x φ) = −δ(x φ) = −x δ(φ) = 0.

72

9 Multiplication by Functions

= 1 in D0 (R). In fact, for φ ∈ C0∞ (R), Z Z 1 1 x PV (φ) = PV (x φ) = lim φ(x) dx = φ(x) dx = 1(φ). ↓0 R\[ −,  ] x x R

Example 9.2. We have x PV

1 x

Using Problem 1.1 and Example 9.1 we now obtain the following identities in D0 (R): 1 1 1 ± πi x δ = x . 1 = x PV = x x x ± i0 x ± i0 A first application of multiplication by functions is the following: Lemma 9.3. Let χj ∈ C0∞ (X) be a sequence as in Lemma 8.2.(b). For every u ∈ D0 (X), the sequence uj := χj u ∈ E 0 (X) then converges in D0 (X) to u, as j → ∞. In other words, E 0 (X) is dense in D0 (X). Proof. If φ ∈ C0∞ (X), then χj φ → φ in C0∞ (X) on account of Lemma 8.2.(b); therefore (χj u)(φ) = u(χj φ) → u(φ), as j → ∞.  By Theorem 3.11 we can similarly define the product ψ u if ψ ∈ C k (X) and the distribution u is of order ≤ k; in that case, ψ u is of order ≤ k. Thus ψ µ is a measure for every continuous function ψ and measure µ. The Leibniz rule (2.8), (2.7) holds true if f = ψ ∈ C ∞ (X) and g = u ∈ D0 (X). Indeed, for every φ ∈ C0∞ (X), (∂j (ψ u))(φ) = (ψ u)(−∂j φ) = u(−ψ ∂j φ) = u(∂j ψ φ − ∂j (ψ φ)) = (∂j ψ u + ψ ∂j u)(φ), which means that ∂j (ψ u) = ∂j ψ u + ψ ∂j u

(ψ ∈ C ∞ (X), u ∈ D0 (X)).

(9.1)

In addition, the product ψ u is continuous as a function of ψ and u together. By this we mean ψj uj → ψ u in D0 (X) if ∞ ψj → ψ in C (X) and uj → u in D0 (X).

(9.2)

For the proof we write (ψj uj )(φ) = uj (ψj φ), for φ ∈ C0∞ (X). Next, we observe that ψj φ → ψ φ in C0∞ (X) and we use the latter assertion in Theorem 5.5. The mapping u 7→ ψ u, of multiplying by ψ ∈ C ∞ (X), is also written by means of the notation ψ. In quantum mechanics this is a familiar operation; multiplying by xj is said to be the j-th position operator. It will be necessary to recall the definition from time to time, in order to avoid confusion in the notations. Is ψ ∈ C0∞ (X), then ψ is a sequentially continuous linear mapping from D0 (X) to E 0 (X).

9 Multiplication by Functions

73

It is not very well possible to extend the product ψ u in a continuous manner to arbitrary distributions ψ and u. For example, let ut ∈ C ∞ (Rn ), for t > 0, be as in Problem 5.5. Then ut δ = ut (0) δ = (4πt)−n/2 δ according to Example 9.1, which does not converge in D0 (Rn ) as t ↓ 0. On the other hand, ut does converge to δ in D0 (Rn ) as t ↓ 0; this forms an obstacle to a well-behaved definition of the product δ δ ∈ D0 (Rn ). The product that maps both (ψ, u) ∈ C ∞ (X) × D0 (X) and (u, ψ) ∈ D0 (X) × 1 1 ∞ (x δ) = x+i0 0 = 0 by C (X) to ψ u ∈ D0 (X), is not associative. Indeed, x+i0 1 Example 9.1, while Example 9.2 implies ( x+i0 x) δ = 1 δ = δ 6= 0. Now that we have multiplication by functions available, we can study linear partial differential equations with variable coefficients, of the form X P (x, ∂)u = cα (x) ∂ α u = f, (9.3) |α|≤m

for given distributions f and desired distributions u. Here the coefficients cα are required to be C ∞ functions on the domain space X of the distributions. Variants involving C k coefficients with finite k are also possible, but the order of u must then be limited and one has to keep an administration of the degrees of differentiability and of the orders. By way of example we show how the theory of linear ordinary differential equations (n = 1) is developed in the distributional context. Theorem 9.4. Let I be an open interval in R, cj ∈ C ∞ (I) for 0 ≤ j ≤ m and cm (x) 6= 0 for all x ∈ I. Then there exists, for every f ∈ D0 (I), a solution u ∈ D0 (I) m of the equation X P u := cj (x) u(j) = f. (9.4) j=0 0

The subset in D (I) of all solutions consists of the u+h with h denoting any solution of the homogeneous equation, given by f = 0. Every such h is a classical C ∞ solution and together they form an m-dimensional linear space over C. If f ∈ C k (I), then u ∈ C m+k (I) and u is a solution in the classical sense. Proof. Because 1/cm ∈ C ∞ (I), we can multiply the equation by this factor; thus we obtain an equivalent equation for u, with cm replaced by 1, the cj by cj /cm and f by f /cm . Consequently, in what follows we may assume that cm = 1. We now perform the usual reduction to an m-dimensional first-order system. To achieve this, we introduce vj := u(j−1) , for 1 ≤ j ≤ m. Also, gj := 0 if 0 ≤ j < m and gm := f . Together these make (9.4) equivalent to a system of the form v 0 = L(x) v + g, where the m × m matrix L(x) can be obtained from the explicit form m X 0 vj0 = vj+1 (1 ≤ j < m), vm =− cj−1 (x) vj + gm . j=1

Let x 7→ Φ(x) be a classical fundamental matrix for the homogeneous system v 0 = L(x) v, that is, an m × m matrix whose columns form a basis for the linear

74

9 Multiplication by Functions

space of solutions of v 0 = L(x) v. In the theory of ordinary differential equations its existence is proved; the coefficients are shown to be C ∞ functions of x, and similarly those of the inverse matrix Φ(x)−1 . Furthermore, Φ satisfies the matrix differential equation Φ0 (x) = L(x) ◦ Φ(x). We now apply, in the distributional context, Lagrange’s method of variation of constants for solving the inhomogeneous equation. This consists of substituting v = Φ(x) w and then deriving the differential equation that w must satisfy to make v a solution of the inhomogeneous equation v 0 = L(x) v + g. Because a distribution may be multiplied by a C ∞ function, the substitution v = Φ(x) w yields a bijective relation between the vector distributions v and w on I. Using Leibniz’ rule we obtain for w L(x) ◦ Φ(x) w + g = ∂x (Φ(x) w) = L(x) ◦ Φ(x) w + Φ(x) w0 , in other words, w0 = Φ(x)−1 g. The notation used is shorthand for a calculation by components. But for every component this is an equation like the one we have already studied in Theorem 4.3. This yields the existence of a solution w ∈ (D0 (I))n and therefore of a solution v = Φ(x) w ∈ (D0 (I))n of v 0 = L(x) v+g and, consequently, of a solution u = v1 ∈ D0 (I) of (9.4). Owing to the linearity of the operator P , the solution space of the inhomogeneous equation consists of the u+h with h the solutions of the homogeneous equation. Now we have g = 0 if f = 0; this gives the equation w0 = 0, the only solutions of which are of the form w = c ∈ Rn according to the last assertion in Theorem 4.3. But this implies that v = Φ(x) c, which leads to a classical solution of the homogeneous equation. Finally, if f ∈ C k (I), there exists a classical solution u0 ∈ C k+m (I). Every solution u ∈ D0 (I) of the inhomogeneous equation is of the form u = u0 + h with h a classical C ∞ solution of the homogeneous equation. It follows that u ∈ C k+m (I) is a classical solution.  The condition that the highest-order coefficient cm should have no zeros, is essential. This is sufficiently demonstrated by the example of the equation x u = 0 in D0 (R), which has the nonclassical solution u = δ as follows from Example 9.1. In the next theorem we formulate a converse assertion, considering Rn straight away. Theorem 9.5. Let X be open in Rn and u ∈ D0 (X). If ψ ∈ C ∞ (X) and ψ u = 0, then supp u is contained in the zero-set of ψ. Consider real-valued ψj ∈ C ∞ (X), for 1 ≤ j ≤ n, and a ∈ X. Further suppose that ψj (a) = 0 and, additionally, that the total derivatives Dψj (a) of the ψj at a are linearly independent, with 1 ≤ j ≤ n. Then ψj u = 0, for all j, implies the existence of a c ∈ C and an open neighborhood U of a in X such that u = c δa on U . Proof. Let C = { x ∈ X | ψ(x) = 6 0 }. Then C is an open subset of X and ψ1 ∈ C ∞ (C); therefore u = ψ1 (ψ u) = ψ1 0 = 0 on C. This proves that supp u ⊂ X \ C, the zero-set of ψ.

9 Multiplication by Functions

75

Denote by ψ : X → Rn the C ∞ mapping having the ψj as component functions. Since ψ(a) = 0 we may write, for x sufficiently close to a, Z ψ(x) = ψ(x) − ψ(a) = 0

Z

1

d ψ(a + t(x − a)) dt dt

1

Dψ(a + t(x − a)) dt (x − a) =: Ψ (x) (x − a),

= 0

with Ψ an n × n matrix consisting of C ∞ functions. In particular, Ψ (a) = Dψ(a); and by assumption this matrix is invertible. By continuity (for instance, consider the determinant of Ψ (x)), there exists an open ball U about a in Rn such that Ψ (x) is invertible, for all x ∈ U . In view of Cramer’s rule for inverse matrices, the matrix coefficients of Ψ −1 belong to C ∞ (U ). We now have x − a = Ψ (x)−1 ψ(x). By combining this equality with the identity of vectors ψ(x) u = 0 we find (x − a) u = 0,

in other words

(xj − aj ) u = 0

(1 ≤ j ≤ n, x ∈ U ).

In view of the assertion proved above we may conclude that U ∩ supp u = {a}; on U , therefore, we may consider u as an element of E 0 (U ) on account of Theorem 8.5. Taylor expansion of an arbitrary φ ∈ C ∞ (U ) about a leads to φ(x) = φ(a) +

n X

(xj − aj ) φj (x)

(x ∈ U ),

j=1

for certain φj ∈ C ∞ (U ). We obtain n n X X u((xj −aj ) φj ) = φ(a) u(1)+ (xj −aj ) u(φj ) = u(1) δa (φ) u(φ) = u(φ(a))+ j=1

and from this we deduce u = u(1) δa on U .

j=1



Note that Example 9.1 shows that the nondegeneracy condition in the theorem, requiring that the derivatives be linearly independent, in general can not be omitted.

Problems Problem 9.1. Suppose p ∈ C 1 ( ] a, b [ ), p(x) > 0 for all x ∈ ] a, b [ and q and r continuous on ] a, b [. Prove that every measure u that is a solution of the variational equation (1.8) for all φ ∈ C0∞ ( ] a, b [ ), is in fact C 2 and satisfies the Euler–Lagrange equation (1.7). Hint: use v2 = p(x) u0 .

76

9 Multiplication by Functions

Problem 9.2. Let ψ ∈ C ∞ (Rn ). Describe the distribution ψ ∂ α δ, for every multiindex α, as a linear combination of the ∂ β δ with β ≤ α. Problem 9.3. Determine all solutions u ∈ D0 (R) of xk u = 0, for k ∈ Z>0 . Problem 9.4. Determine all solutions u ∈ D0 (R) of x u0 = 0. Show that these u form a linear space and determine its dimension. 1 is a solution of x u = 1, and prove that this Problem 9.5. Observe that u = x+i0 0 implies x u = −u. Let u ∈ D0 (R) be a solution of xk u = 1. Applying the differential operator x ∂x 0 k+1 v = 1. Finally determine, to this identity, show that v = −1 k u is a solution of x 0 for every k ∈ Z>0 , all solutions in D (R) of the equation xk u = 1.

Problem 9.6. Prove lim eitx PV

t→∞

Deduce lim

t→∞

1 = πi δ. x

(9.5)

sin tx = π δ. x

Next define ut, ∈ C ∞ (R) by ut, (x) = eitx /(x + i), where t and  ∈ R and  6= 0. Using Problem 1.1 prove the following equalities in D0 (R): (i)

limt→∞ (lim↓0 ut, ) = 0, (ii) limt→∞ (lim↑0 ut, ) = 2πi δ, (iii) lim→0 (limt→∞ ut, ).

In particular, limits in D0 (Rn ) can not be freely interchanged. Problem 9.7. Let u ∈ D0 (Rn ) and xn u = 0. Prove that there exists a uniquely determined v ∈ D0 (Rn−1 ) such that u(φ) = v(i∗0 φ) for every φ ∈ C0∞ (Rn ). Here (i∗0 φ)(x1 , . . . , xn−1 ) = φ(x1 , . . . , xn−1 , 0). Hint: use χ ∈ C0∞ (R) with χ(x) = 1 on a neighborhood of 0. For ψ ∈ C0∞ (Rn−1 ), define the function ψ ⊗ χ ∈ C0∞ (Rn ) by (ψ ⊗ χ)(y, z) = ψ(y) χ(z) and deduce that v must be defined by v(ψ) := u(ψ ⊗ χ). Problem 9.8. Prove that the differential equation (1 − x2 )2 u0 − 2x u = (1 − x2 )2 has no solution u ∈ D0 (I), if I is an open interval in R with [ −1, 1 ] ⊂ I. Problem 9.9. Let a ∈ C. Show that the solutions u ∈ D0 (R) of the differential equation x u0 = a u form a linear space Ha . Demonstrate that multiplication by x, and the differentiation ∂x , define linear mappings from Ha to Ha+1 , and Ha−1 , respectively. Determine Ha . Hint: use Problem 7.6 and deal with the case a ∈ Z 0 and an m ∈ M with l(B(y)) ≤ d m(y) for all y ∈ F . On account of the continuity of A, there exist for this m a constant c > 0 and an n ∈ N , such that m(A(x)) ≤ c n(x) for all x ∈ E. Combining these two estimates we obtain l(B ◦ A(x)) ≤ d m(A(x)) ≤ dc n(x) for all x ∈ E. Note that, in the case of linear spaces endowed with a norm, this was the proof of the theorem that the operator norm of B ◦ A is smaller than, or equals, the product of the operator norms of B and of A. For the proof of the second assertion, consider w ∈ G0 and x ∈ E. Then t

(B ◦ A)(w)(x) = w(B(A(x))) = t B(w)(A(x)) = t A(t B(w))(x). 

We have, in fact, applied the principle of transposition several times before, when introducing operations involving distributions. For example, we defined ∂j : D0 (X) → D0 (X) as the transpose of −∂j : C0∞ (X) → C0∞ (X). Also, for ψ ∈ C ∞ (X), we introduced the multiplication by ψ : D0 (X) → D0 (X) as the transpose of the multiplication by ψ : C0∞ (X) → C0∞ (X). If, in addition, ψ has compact support in X, multiplication by ψ is even continuous and linear from C ∞ (X) to C0∞ (X), which then immediately implies that multiplication by ψ, the transpose, defines a continuous linear mapping from D0 (X) to E 0 (X). Furthermore, the restriction mapping ρU V : D0 (V ) → D0 (U ) from Chap. 7 also is a transpose, namely of the identical embedding ιV U : C0∞ (U ) → C0∞ (V ). We summarize the situation by the following diagram, where v ∈ D0 (V ) and φ ∈ C0∞ (U ): ιV U : C0∞ (U ) −−−−→ C0∞ (V ) D0 (U ) ←−−−− D0 (V ) : ρU V

with

(ρU V v)(φ) = v(ιV U φ).

For general continuous linear A : C0∞ (U ) → C0∞ (V ), it is customary to denote A : D0 (V ) → D0 (U ) by B if B is a previously introduced operator acting on functions and t A = B on the subspace C0∞ (V ) of D0 (V ). This is what we have done in the case of differentiation and multiplication by functions and this will also be our guiding principle in the subsequent naming of transpose operators. In Corollary 11.7 below we will see that C0∞ (V ) is dense in D0 (V ), which makes this procedure almost inevitable. Sometimes, however, the transpose operator turns out to be truly novel, for example when t A(φ) ∈ D0 (U ) is not a function for some φ ∈ C0∞ (V ). In later parts of this chapter we will encounter further examples of this, for instance in Proposition 10.16. t

10 Transposition. Pullback and Pushforward

81

Let X be an open subset of Rn , let Y be an open subset of Rp , and Φ : X → Y a C ∞ mapping. Then ψ ◦ Φ : x 7→ ψ(Φ(x)) belongs to C ∞ (X), for every ψ ∈ C ∞ (Y ). The mapping Φ∗ : ψ 7→ ψ ◦ Φ

C ∞ (Y ) → C ∞ (X)

:

is linear (albeit between infinite-dimensional spaces) and by means of the chain rule for differentiation it can be shown also to be continuous. Note that (Φ∗ ψ)(x) = ψ(y) if y = Φ(x); thus, Φ∗ is something like “the change of variables y = Φ(x) in functions of y”. The operator Φ∗ is said to be the pullback of functions on Y to functions on X, under the mapping Φ : X → Y . Actually, transposition is the pullback of continuous linear forms under continuous linear mappings, although the latter may have been defined on infinite-dimensional spaces. Furthermore, observe that, for X = R and Y = {0} ⊂ R, we get (Φ∗ ψ)(x) = ψ(0), for ψ ∈ C0∞ (Y ) and x ∈ R. In this case, therefore, we do not have Ψ ∗ : C0∞ (Y ) → C0∞ (X). Example 10.2. Let Φ : X → Y and Ψ : Y → Z be C ∞ mappings where X ⊂ Rn , Y ⊂ Rp and Z ⊂ Rq are open sets. Then we have, on account of the chain rule p (6.1) X (∂j ◦ Φ∗ )Ψi = ∂j (Ψi ◦ Φ) = ∂j Φk (Φ∗ ◦ ∂k )Ψi , k=1

for 1 ≤ j ≤ n and 1 ≤ i ≤ q. In particular, if q = 1 this equality is valid for all Ψ = Ψ1 ∈ C ∞ (Y ); therefore we obtain the following identity of continuous linear mappings, which describes composition of partial differentiation and pullback, for 1 ≤ j ≤ n: p X ∂j ◦ Φ∗ = ∂j Φk ◦ Φ∗ ◦ ∂k : C ∞ (Y ) → C ∞ (X). (10.1) k=1

The transpose of the continuous linear mapping Φ∗ : C ∞ (Y ) → C ∞ (X) is a continuous linear mapping t (Φ∗ ) : E 0 (X) → E 0 (Y ). This is written as t (Φ∗ ) = Φ∗ and is said to be the pushforward of distributions on X with compact support to distributions on Y with compact support, under the mapping Φ : X → Y . The defining equation is (Φ∗ u)(ψ) = u(Φ∗ ψ) = u(ψ ◦ Φ)

(u ∈ E 0 (X), ψ ∈ C ∞ (Y )).

(10.2)

We describe the preceding results by means of the following diagram: X

Φ ∗

/Y

C ∞ (X) o

Φ

C ∞ (Y )

E 0 (X)

Φ∗

/ E 0 (Y )

In mathematics, Φ∗ is said to be covariant with respect to Φ, because Φ∗ acts in the same direction as Φ itself, while Φ∗ is said to be contravariant with respect to Φ, seeing that the sense of the arrow is reversed.

82

10 Transposition. Pullback and Pushforward

If Ψ is a C ∞ mapping from Y to an open subset Z of Rq , then Ψ ◦ Φ is a C ∞ mapping from X to Z. As in the proof of the second assertion in Lemma 10.1, we see that (Ψ ◦ Φ)∗ = Φ∗ ◦ Ψ ∗ : C ∞ (Z) → C ∞ (X). Transposing this in turn we obtain (Ψ ◦ Φ)∗ = Ψ∗ ◦ Φ∗

:

E 0 (X) → E 0 (Z).

(10.3)

Example 10.3. For every x ∈ X one has Φ∗ δx = δΦ(x) . Indeed, for every ψ ∈ C ∞ (Y ) we obtain (Φ∗ δx )(ψ) = δx (ψ ◦ Φ) = ψ(Φ(x)) = δΦ(x) (ψ).  Example 10.4. Set V = { x ∈ R2 | kxk = 1 } and consider ∆ δV ∈ E 0 (R2 ) where ∆ denotes the Laplace operator as in Problem 4.6, thus, for φ ∈ C ∞ (R2 ), Z Z π (∆ δV )(φ) = ∆φ(x) dx = ∆φ(cos α, sin α) dα. V

−π

Define Φ : R2 → R by Φ(x) = kxk2 . Then we have Φ∗ ∆ δV = 8π(δ1 00 − δ1 0 ) in E 0 (R). Indeed, using the chain rule repeatedly we obtain, for ψ ∈ C ∞ (R), ∂1 2 (ψ ◦ Φ)(x) = 4x1 2 ψ 00 (kxk2 ) + 2ψ 0 (kxk2 ), and an analogous formula for differentiation with respect to x2 . It follows that (Φ∗ ∆ δV )(ψ) = (∆ δV )(Φ∗ ψ) = δV (∆(ψ ◦ Ψ )) Z π =4 ((cos2 α + sin2 α) ψ 00 (1) + ψ 0 (1)) dα = 8π(δ1 00 − δ1 0 )(ψ). −π



See Problem 10.13.(vi) for another proof.

In the following theorem Φ∗ is extended, subject to additional conditions on the mapping Φ, to distributions that may not have compact support. For a closed subset A of X, we denote the space of the u ∈ D0 (X) with supp u ⊂ A by D0 (X)A . We then say that uj → u in D0 (X)A if uj ∈ D0 (X)A for all j and uj → u in D0 (X) as j → ∞. This implies that u ∈ D0 (X)A . The mapping Φ : X → Y is said to be proper on A if, for every compact subset L of Y , the set Φ−1 (L) ∩ A is a compact subset of X. Theorem 10.5. Let Φ be a C ∞ mapping from the open subset X of Rn to the open subset Y of Rp . Let A be a closed subset of X and suppose that Φ is proper on A. Then there exists a uniquely determined extension of Φ∗ : E 0 (X) ∩ D0 (X)A → 0 E (Y ) to a linear mapping Φ∗ : D0 (X)A → D0 (Y ) with the property that Φ∗ uj → Φ∗ u in D0 (Y ) if uj → u in D0 (X)A . One has supp (Φ∗ u) ⊂ Φ(supp u)

(u ∈ D0 (X)A ).

(10.4)

In particular, if Φ is proper from X to Y , then Φ∗ : E 0 (X) → E 0 (Y ) possesses an extension to a sequentially continuous linear mapping Φ∗ : D0 (X) → D0 (Y ) and one has (10.4) for all u ∈ D0 (X).

10 Transposition. Pullback and Pushforward

83

Proof. If u ∈ D0 (X)A , the sequence uj = χj u ∈ E 0 (X) from Lemma 9.3 converges to u in D0 (X)A . This leads to the uniqueness: if the extension exists, we have to take Φ∗ u = lim Φ∗ uj . j→∞

Let L be an arbitrary compact subset of Y and χ ∈ C0∞ (X) with χ = 1 on a neighborhood of the compact set K = Φ−1 (L) ∩ A, see Corollary 2.16. For all φ ∈ C0∞ (L), we define v(φ) = (Φ∗ (χ u))(φ) = u(χ Φ∗ φ). Because the linear mapping Φ∗ is continuous from C0∞ (L) to C ∞ (X) and multiplication by χ is continuous from C ∞ (X) to C0∞ (X), v is a continuous linear form on C0∞ (L). Furthermore, if uj → u in D0 (X)A , then (Φ∗ uj )(φ) = uj (Φ∗ φ) = (χ uj )(Φ∗ φ) = uj (χ Φ∗ φ) converges to v(φ) as j → ∞. From the resulting uniqueness it follows that this v has an extension to C0∞ (Y ); this is the Φ∗ u with the desired continuity properties. With respect to (10.4) we observe that (supp φ) ∩ Φ(supp u) = ∅ implies that supp (Φ∗ φ) = Φ−1 (supp φ) is disjunct from supp u, and so (Φ∗ u)(φ) = u(Φ∗ φ) = 0.  We will now discuss the pushforward of distributions under C ∞ mappings having various additional properties. We begin by considering diffeomorphisms and then, in Theorem 10.14 and Proposition 10.16, we study mappings between open sets of different dimensions. A C ∞ diffeomorphism Φ : X → Y is a C ∞ mapping from X to Y that is bijective and whose inverse mapping Ψ := Φ−1 is a C ∞ mapping from Y to X. First, we recall some basic facts about diffeomorphisms. Applying the chain rule to Ψ ◦ Φ(x) = x and to Φ ◦ Ψ (y) = y, we find, with the notation y = Φ(x), that DΨ (y) ◦ DΦ(x) = I and DΦ(x) ◦ DΨ (y) = I. In other words, DΦ(x) : Rn → Rp is a bijective linear mapping, with inverse DΨ (y); this also implies that n = p. Conversely, if Φ is an injective C ∞ mapping from an open subset X of Rn to Rn , such that DΦ(x) is invertible for every x ∈ X, the Inverse Function Theorem asserts that Y := Φ(X) is an open subset of Rn and that Φ is a C ∞ diffeomorphism from X to Y . See, for example, [8, Theorem 3.2.8]. In the following theorem we use the notation jΦ (x) = | det DΦ(x)|

(x ∈ X),

for a differentiable mapping Φ from an open subset X of Rn to Rn , The matrix of derivatives DΦ(x) is also referred to as the Jacobi matrix of Φ at the point x and det DΦ(x) as the Jacobi determinant or Jacobian. Note that jΦ ∈ C ∞ (X) if Φ : X → Rn is a C ∞ mapping.

84

10 Transposition. Pullback and Pushforward

Theorem 10.6. Suppose that X and Y are open subsets of Rn and Φ : X → Y a C ∞ diffeomorphism. Then Φ is proper and the pushforward Φ∗ : D0 (X) → D0 (Y ) is a sequentially continuous linear mapping. In particular, for a locally integrable function f on X, Φ∗ f is the locally integrable function on Y given by Φ∗ f = jΨ Ψ ∗ f = Ψ ∗

f  jΦ

where

Ψ := Φ−1 .

In particular, we have jΦ Φ∗ ◦ Φ∗ = I on C0∞ (X). Proof. The properness of Φ follows from the continuity of Ψ ; accordingly, the existence of the pushforward Φ∗ : D0 (X) → D0 (Y ) is a consequence of Theorem 10.5. For every ψ ∈ C0∞ (Y ) we obtain by means of the change of variables x = Ψ (y) Z (Φ∗ f )(ψ) = f (ψ ◦ Φ) = f (x) ψ(Φ(x)) dx X Z (10.5) f (Ψ (y)) ψ(y) jΨ (y) dy = (jΨ Ψ ∗ f )(ψ). = Y

See, for example, [8, Theorem 6.6.1] for the change of variables in an n-dimensional integral.  As a consequence of the theorem, the restriction of Φ∗ to C0∞ (X) is a continuous linear mapping A from C0∞ (X) to C0∞ (Y ). Its transpose t A is a continuous linear mapping from D0 (Y ) to D0 (X). Let ψ ∈ C ∞ (Y ). For every φ ∈ C0∞ (X) one has (t A ψ)(φ) = ψ(Aφ) = ψ(Φ∗ φ) = (Φ∗ φ)(ψ) = φ(Φ∗ ψ) = (Φ∗ ψ)(φ), and therefore t A ψ = Φ∗ ψ. In other words, t A : D0 (Y ) → D0 (X) is an extension of Φ∗ : C ∞ (Y ) → C ∞ (X). This extension is also denoted by Φ∗ and is said to be the pullback under Φ of distributions on Y to distributions on X. Because C0∞ (Y ) is dense in D0 (Y ), see Corollary 11.7 below, this continuous extension is uniquely determined and the naming is entirely natural. Summarizing these results we obtain: Theorem 10.7. Under the conditions of Theorem 10.6 the pullback Φ∗ : D0 (Y ) → D0 (X) is a continuous linear mapping. For any v ∈ D0 (Y ) and φ ∈ C0∞ (X), it satisfies   φ  (Φ∗ v)(φ) = jΦ−1 v((Φ−1 )∗ φ) = v (Φ−1 )∗ . jΦ Example 10.8. It is a direct application of Theorem 10.6 that jΦ Φ∗ ◦ Φ∗ = I on D0 (X). In particular, Example 10.3 therefore implies Φ∗ δΦ(x) =

1 δx jΦ(x)

(x ∈ X).

10 Transposition. Pullback and Pushforward

85

From Example 10.8 it follows that, for a diffeomorphism Φ, the pushforward under Φ and the pullback under the inverse Φ−1 of Φ are closely related to each other. Indeed, we have Φ∗ ◦ jΦ = (Φ−1 )∗

:

D0 (X) → D0 (Y ).

(10.6)

In particular, Φ∗ and (Φ−1 )∗ are identical if and only if jΦ equals 1, that is, if Φ is volume-preserving. For u ∈ D0 (X), the distribution Φ∗ u ∈ D0 (Y ) is also said to be the transform of u under Φ as a distributional density, while (Φ−1 )∗ u is said to be the transform of u as a generalized function. Remark 10.9. The pullback of distributions under the diffeomorphisms that act as coordinate transformations enables one to define distributions on manifolds. See, for example, H¨ormander [16, Section 6.3].  Example 10.10. The reflection S : x 7→ −x in Rn about the origin satisfies S −1 = S and jS = 1, and therefore, in this case S∗ = (S −1 )∗ = S ∗ . This transformation of distributions on Rn is denoted by the same letter S; for every u ∈ D0 (Rn ), Su is said to be the reflected distribution. Somewhat more generally, if A is an invertible linear transformation of Rn , then jA equals the constant | det A| so that (10.6) leads to the following identity of continuous linear mappings on D0 (Rn ): | det A| A∗ = (A−1 )∗ .

Example 10.11. For every h ∈ Rn , let Th : x 7→ x + h : Rn → Rn be the translation in Rn by the vector h. This is evidently a diffeomorphism, whose Jacobi matrix equals the identity and whose Jacobi determinant therefore equals 1. Consequently, the mapping Th ∗ : C ∞ (Rn ) → C ∞ (Rn ) has an extension to a continuous linear mapping Th ∗ = (Th −1 )∗ = (T−h )∗ : D0 (Rn ) → D0 (Rn ). (10.7) If e(j) denotes the j-th basis vector in Rn , one has, for every φ ∈ C ∞ (Rn )
Tt e(j) ∗ φ − φ (x) = φ(x + te(j)) − φ(x) =

Z 0

Z =t

1

d φ(x + s te(j)) ds ds

1

∂j φ(x + s te(j)) ds. 0

From this we find
1 Tt e(j) ∗ φ − φ = ∂j φ t→0 t lim

in

C0∞ (Rn ).

(10.8)

86

10 Transposition. Pullback and Pushforward

In conjunction with (10.7) this implies that, for every u ∈ D0 (Rn ) and φ ∈ C0∞ (Rn ), 1


 1 1 Tt e(j) ∗ u − u (φ) = (T−t e(j) )∗ u − u (φ) = u T−t e(j) ∗ φ − φ t t t converges to u(−∂j φ) = ∂j u(φ) as t → 0. In other words
1 Tt e(j) ∗ u − u = ∂j u t→0 t lim

in

D0 (Rn ).

(10.9)

For example, if u is a layer on a surface V , as introduced in Example 7.3, one may regard ∂j u as having been obtained by translating 1/t times the charge distribution on V by −t e(j), adding the opposite on V of this and then taking the limit as t → 0. So long as one does not pass to the limit, this is similar to a capacitor from the theory of electricity. For this reason the derivative ∂j u of a layer u is also said to be a double layer and, more generally, the distribution ∂ α u, with u a layer and |α| > 0, is said to be a multiple layer. (The question whether a k-th order derivative should be called a 2k -fold layer or a (k + 1)-fold layer is undecided.) In (1.11) translation of functions was written more concisely as Th (f ) := f ◦ T−h = (T−h )∗ (f ), in other words Th = (T−h )∗ = (Th )∗ . For example, using this notation we obtain Th (δa ) = δa+h . The disadvantage of this notation is that now ∂j u is the derivative of T−t e(j) u with respect to t at t = 0. But that is the way it is: when the graph of an increasing function is translated to the right, it will lie lower.  Equation (10.8) can be generalized as follows. Let I be an open interval in R and let As be a continuous linear mapping from (E, N ) to (F, M), for every s ∈ I. This one-parameter family of linear mappings is said to be differentiable at the point a if there exists a continuous linear mapping B from (E, N ) to (F, M) such that for every x ∈ E
1 (As (x) − Aa (x) = B(x) in F. lim s→a s − a d In this case one writes B = ds As |s=a . Writing out the definitions one immediately sees that this implies that the one-parameter family s 7→ t (As ), of continuous linear mappings from F 0 to E 0 , is also differentiable at s = a, while  d dt (As ) =t As . ds ds s=a s=a

The generalization to families of mappings that depend on more than one parameter is obvious. Let X and Y be open subsets of Rn and Rp , respectively. Let I be an open interval in R and let Φ : X × I → Y be a C ∞ mapping. Then Φt (x) = Φ(x, t)

10 Transposition. Pullback and Pushforward

87

defines, for every t ∈ I, a C ∞ mapping from X to Y . Denote by Φt,k the k-th component function of Φt , for 1 ≤ k ≤ p. Applying (10.1) with j = n + 1 to the Φ under consideration, one deduces that the one-parameter family t 7→ Φt ∗ , of continuous linear mappings from C ∞ (Y ) to C ∞ (X), is differentiable and that one has the identity of continuous linear mappings p

d ∗ X d Φt = Φt,k ◦ Φt ∗ ◦ ∂k dt dt

C ∞ (Y ) → C ∞ (X)

:

(10.10)

k=1

for the derivative with respect to t of the pullback. In particular, if we restrict the operators in this equality to C0∞ (Y ) we obtain an identity of continuous linear mappings sending C0∞ (Y ) to C0∞ (X). Transposition and Lemma 10.1 as well as the fact that the transpose of ∂k is −∂k then immediately lead to the formula p

X d d (Φt )∗ = − ∂k ◦ (Φt )∗ ◦ Φt,k dt dt

:

D0 (X) → D0 (Y )

k=1

for the derivative with respect to t of the pushforward. If we restrict the operators in this formula to C0∞ (X) and transpose once more, we find that (10.10) also holds when we have the continuous linear mappings act on D0 (Y ). These formulas become simpler if X = Y , 0 ∈ I and Φ0 (x) = x, for all x ∈ X; that is, Φ0 = I, the identity on X. In this case v(x) =

d Φt (x) dt t=0

is said to be the velocity vector field of Φ at time t = 0. This defines a C ∞ vector field on X with components vj ∈ C ∞ (X); we now find for the pullback n X d ∗ Φt = vj ◦ ∂j dt t=0 j=1

:

D0 (X) → D0 (X),

(10.11)

and for the pushforward n X d (Φt )∗ =− ∂j ◦ vj dt t=0 j=1

:

D0 (X) → D0 (X).

(10.12)

For an arbitrary C ∞ vector field v on X, let Φ(x, t) be the solution x(t) at time t of the system of differential equations x0 (t) = v(x(t))

with initial condition

x(0) = x.

Suppose, for simplicity, that the solutions for all (x, t) ∈ X × R do exist. This is equivalent to the condition that none of the solutions will leave every compact subset of X in a finite period of time. According to the theory of ordinary differential equations, Φ in that case is a C ∞ mapping from X × R to X and Φt : x 7→ Φt (x) =

88

10 Transposition. Pullback and Pushforward

Φ(x, t) is said to be the flow over time t with velocity vector field v. The latter satisfies the so-called group law Φt+s = Φt ◦ Φs ,

which also implies

Φt+s ∗ = Φs ∗ ◦ Φt ∗

(t, s ∈ R).

By differentiation with respect to t at t = 0 we find, in view of (10.11)

Transposition gives

n X d ∗ Φs = Φs ∗ ◦ vj ◦ ∂j . ds j=1

(10.13)

(Φt+s )∗ = (Φt )∗ ◦ (Φs )∗ , from which we obtain, by differentiation with respect to s at s = 0, and in view of (10.12) n X d (Φt )∗ = −(Φt )∗ ◦ ∂j ◦ vj . (10.14) dt j=1 The following theorem gives an application of this. The distribution u ∈ D0 (X) is said to be invariant under the diffeomorphism Φ : X → X as a generalized function if (Φ−1 )∗ u = u. Applying Φ∗ , we see that this is equivalent to Φ∗ u = u. This u is said to be invariant under Φ as a distributional density, if Φ∗ u = u. Theorem 10.12. Let v be a C ∞ vector field on X whose solutions are defined for all (x, t) ∈ X × R. Then u ∈ D0 (X) is invariant as a generalized function, and as a distributional density, under the flow with velocity vector field v if and only if n X

vj ◦ ∂j u = 0,

n X

and

j=1

∂j ◦ vj u = 0,

respectively.

j=1

Proof. Because Φ0 = I, the identity in X, we have Φ0 ∗ u = I ∗ u = u. Therefore, Φt ∗ u = u for all t ∈ R is equivalent to the assertion that Φt ∗ u is constant as a function of t. By testing with some φ ∈ C0∞ (X) and applying the theorem that a real-valued function of a real variable is constant if and only if its derivative vanishes, we find that the first assertion follows from (10.13). Likewise, the second assertion follows from (10.14), on account of (Φ0 )∗ u = I∗ u = u.  Homogeneity of functions and homogeneity of distributions can be discussed in analogous ways. For c > 0, we denote the mapping x 7→ c x : Rn → Rn by c. Then u ∈ D0 (Rn ) is said to be homogeneous of degree a ∈ C if, for every c > 0, c∗ u = ca u. Theorem 10.13. u ∈ D0 (Rn ) is homogeneous of degree a ∈ C if and only if it satisfies Euler’s differential equation: n X j=1

xj ∂j u = a u.

10 Transposition. Pullback and Pushforward

89

Proof. x 7→ et x is the flow of the vector field v(x) = x. Define At (u) = e−at (et )∗ u. Then At ◦ As = At+s and we obtain: u is homogeneous of degree d At (u)|t=0 = 0. P By Leibniz’ rule and a ⇐⇒ At (u) = u for all t ∈ R ⇐⇒ dt (10.11), the left-hand side in the latter equality equals −a u + j xj ∂j u.  Thus, Problem 9.9 amounted to determining all homogeneous distributions on R. As we have already established, pushforward under Φ of distributions with compact support by application of formula (10.2), is possible for arbitrary C ∞ mappings Φ from an open subset X of Rn to an open subset Y of Rp , even if n > p or n < p. We conclude this chapter with a few remarks concerning these two cases under the assumption that Φ be generic, i.e., that all of its corresponding tangent mappings are of maximal rank. If n > p, and under the further assumption that Φ be a submersion, we can define pullback under Φ of distributions; this result requires the preceding theory for diffeomorphisms. In the case where n < p, and under the further assumption of Φ being a proper immersion, we encounter a new phenomenon: the pushforward under Φ of a function is not necessarily a function anymore. This is related to Φ(X) not being an open subset of Y . n > p in case of a submersion. We first derive an intermediate result. Write n = p + q with q > 0 and x = (y, z) with y ∈ Rp and z ∈ Rq . Let Π : (y, z) 7→ y : Rn → Rp be the projection onto the first p variables. This mapping is not proper and therefore Π∗ is only defined on E 0 (Rn ). For a continuous function f on Rn with compact support, (10.2) now gives Z Z Z Z  f (y, z) dz ψ(y) dy, (Π∗ f )(ψ) = f (y, z) ψ(y) dz dy = Rp

Rq

Rp

Rq

for all ψ ∈ C ∞ (Rp ). This means that the distribution Π∗ f is given by the continuous function of y obtained from f (x) = f (y, z) by integration over the fiber of the mapping Π over the value y, that is, by “integrating out” the z-variable: Z (Π∗ f )(y) = f (y, z) dz. (10.15) Rq

On the strength of the Theorem about differentiation under the integral sign, we have Π∗ f ∈ C0∞ (Rp ) if f ∈ C0∞ (Rn ). We conclude that the restriction of Π∗ to C0∞ (Rn ) is a continuous linear mapping from C0∞ (Rn ) to C0∞ (Rp ). Its transpose is a continuous linear mapping from D0 (Rp ) to D0 (Rn ). Since u(Π∗ φ) = (Π∗ φ)(u) = φ(Π ∗ u) = (Π ∗ u)(φ), if u ∈ C ∞ (Rp ) and φ ∈ C0∞ (Rn ), this transpose corresponds on C ∞ (Rp ) to Π ∗ . For this reason we denote the transpose by Π ∗ . In other words, Π ∗ : C ∞ (Rp ) → C ∞ (Rn ) has an extension to a continuous linear mapping Π ∗ : D0 (Rp ) → D0 (Rn ), defined as the transpose of the restriction to C0∞ (Rn ) of Π∗ .

90

10 Transposition. Pullback and Pushforward

Recall that n > p and let Φ be a C ∞ mapping from an open X ⊂ Rn to an open Y ⊂ Rp , with the property that its derivative DΦ(x) ∈ Lin(Rn , Rp ) at x is surjective, for every x ∈ X. One says that such a mapping is a submersion, see for example [8, Definition 4.2.6]. Theorem 10.14. Let Φ be a C ∞ submersion from an open X ⊂ Rn to an open Y ⊂ Rp . Then the restriction of Φ∗ to C0∞ (X) is a continuous linear mapping from C0∞ (X) to C0∞ (Y ). Its transpose defines an extension of Φ∗ : C ∞ (Y ) → C ∞ (X) to a continuous linear mapping from D0 (Y ) to D0 (X); this is also denoted by Φ∗ and is said to be the pullback of distributions under Φ. Proof. By means of the Submersion Theorem (see [8, Theorem 4.5.2], for example) one can find, for every x ∈ X, a neighborhood U = Ux of x in X and a C ∞ diffeomorphism K from U onto an open subset V of Rn , such that Φ = Π ◦ K on U . (Since K and Π map open sets to open sets, it now follows that this holds for Φ as well.) Applying this we obtain in E 0 (U ) Φ∗ = Π∗ ◦ K∗ = Π∗ ◦ jΛ ◦ Λ∗

with

Λ := K −1 .

(10.16)

Here we have used (10.3) and Theorem 10.6. In particular, it now follows that the restriction of Φ∗ to C0∞ (U ) is a continuous linear mapping from C0∞ (U ) to C0∞ (Y ). If, for every compact subset K of X, we use a partition of unity over K subordinate to the covering by the Ux , for x ∈ X, we conclude that the restriction of Φ∗ to C0∞ (K) is a continuous linear mapping from C0∞ (K) to C0∞ (Y ).  Remark 10.15. As we have seen, (Π∗ g)(y) is obtained by integrating g over the linear manifold Π −1 ({y}), for a continuous function g with compact support. If f is a continuous function with compact support contained in U , one can see, by application of (10.16), that (Φ∗ f )(y) is obtained by integrating g := jΛ (f ◦ Λ) over Π −1 ({y}). But this amounts to an integration of f over the C ∞ submanifold Λ(Π −1 ({y}) = K −1 (Π −1 ({y}) = Φ−1 ({y}) ⊂ X, the fiber of the mapping Φ over the value y. If one chooses Euclidean (n − p)dimensional integration over the inverse image Φ−1 ({y}), then f must first be multiplied by a C ∞ factor that does not depend on f but only on the mapping Φ. We state here (see Problem 10.14 and its solution for a proof, or Problem 10.15 in the special case of p = 1) that this factor equals gr Φ (with gr associated with gradient and Grammian), where p (x ∈ X), (gr Φ)(x) := det(DΦ(x) ◦ tDΦ(x)) which is the p-dimensional Euclidean volume of the parallelepiped in Rn spanned by the vectors grad Φj (x), with 1 ≤ j ≤ p. Here tDΦ(x) ∈ Lin(Rp , Rn ) denotes the adjoint of DΦ(x), the matrix of which is given by the transpose of the matrix of

10 Transposition. Pullback and Pushforward

91

DΦ(x). Observe that gr Φ = k grad Φk if p = 1. Summarizing, in the notation of Example 7.3 we have Z 1 f (x) (Φ∗ f )(y) = dx = δ −1 (f ) (y ∈ Y ). (10.17) (gr Φ)(x) gr Φ Φ ({y}) −1 Φ ({y}) Pushforward under a submersion Φ is not well-defined for all distributions, whereas pullback under Φ of distributions is. The latter will play an important role in Chap 13. n < p in case of a proper embedding. Suppose, for convenience, that Φ is proper. If, moreover, Φ is injective and DΦ(x) : Rn → Rp is injective for every x ∈ X, then Φ is a proper C ∞ embedding of X onto the n-dimensional locally closed C ∞ submanifold V = Φ(X) of Y . The n-dimensional Euclidean integral of a continuous function f with compact support in V then equals the integral of jΦ Φ∗ f over X, where jΦ (x) now denotes the Euclidean n-dimensional volume in Rp of the parallelepiped spanned by the DΦ(x)(e(j)) ∈ Rp , 1 ≤ j ≤ n. Here e(j) denotes the j-th standard basis vector in Rn . See [8, Theorem 1.8.6, Exercise 4.20 and Section 7.3] among other references, for more on these facts concerning proper embeddings and integration over submanifolds. We further recall the distribution g δV of Euclidean integration over V with weight function g, as introduced in Example 7.3. Proposition 10.16. Let n ≤ p, suppose X is open in Rn and let Φ be a proper C ∞ embedding of X into the open subset Y of Rp , having as its image the manifold V = Φ(X). Using the notation Ψ = Φ−1 : V → X we obtain Φ∗ f = Ψ ∗

f  δ ∈ D0 (Y ), jΦ V

for every continuous function f on X. Note that δV is a distribution of order 0, multiplied by the continuous function Ψ ∗ (f /jΦ ) on V . Proof. This follows from a variant of (10.5). For ψ ∈ C0∞ (Y ) one has Z Z (Φ∗ f )(ψ) = f (x) ψ(Φ(x)) dx = (f /jΦ )(x) ψ(Φ(x)) jΦ (x) dx X ZX = (f /jΦ )(Ψ (y)) ψ(y) dy = Ψ ∗ (f /jΦ ) δV (ψ). V

 Observe that in this case of a proper embedding Φ, the pushforward Φ∗ of distributions is well-defined on account of Φ being proper; however, Φ∗ does not map functions to functions. Owing to the latter phenomenon, it is not possible to define the pullback Φ∗ of distributions in a manner similar to the case of diffeomorphisms.

92

10 Transposition. Pullback and Pushforward

Problems Problem 10.1. (Composition of pushforward and partial differentiation). Consider a C ∞ mapping Φ from an open subset X of Rn to an open subset Y of Rp . Prove, for every 1 ≤ j ≤ n, the following identity of continuous linear mappings: Φ∗ ◦ ∂j =

p X

∂k ◦ Φ∗ ◦ ∂j Φk

:

E 0 (X) → E 0 (Y ).

k=1

Problem 10.2. Let u be a probability measure on X and Φ : X → Y a continuous mapping. Discuss how to define Φ∗ u so as to make it a probability measure on Y . Problem 10.3. Let Φ : X → Y be a C ∞ mapping that is not proper on the closed subset A of X. Demonstrate the existence of a sequence a(j) in A with the following properties: (i) Φ(a(j)) converges to a point b ∈ Y as j → ∞; (ii) for every compact subset K of X there exists a j0 with a(j) ∈ / K, whenever j ≥ j0 . Prove that δa(j) → 0 in D0 (X)A , while Φ∗ δa(j) → δy in D0 (Y ), as j → ∞. Prove that the condition in Theorem 10.5, requiring Φ to be proper on A, is necessary for the conclusion in Theorem 10.5. Problem 10.4. (Interpretation of divergence). Suppose Φ0 = I. Verify, by combining Theorem 10.6 with (10.11) and (10.12), that n X d ∂j vj = div v. det(DΦt ) = dt t=0 j=1

Problem 10.5. Consider a C ∞ vector field v on X with globally defined flow (Φt )t∈R and let x ∈ X. Then prove the following: δx is invariant under Φt , for all t ∈ R, as a distributional density ⇐⇒ Φt (x) = x for all t ∈ R ⇐⇒ v(x) = 0. And also: δx is invariant under Φt , for all t ∈ R, as a generalized function if and only if v(x) = 0 and div v(x) = 0. Problem 10.6. The rotation in the plane R2 about the origin by the angle t is the linear mapping (x1 , x2 ) 7→ (x1 cos t − x2 sin t, x1 sin t + x2 cos t). Demonstrate that u ∈ D0 (R2 ) is invariant under all rotations if and only if x1 ∂2 u = x2 ∂1 u. Problem 10.7. A distribution u in R is said to be periodic with period a > 0 if u is invariant under translation by a. Prove that every distributional Fourier series from Problem 5.10 is periodic with period a = 2π ω .

10 Transposition. Pullback and Pushforward

93

Problem 10.8. Let ω > 0 and suppose u ∈ D0 (R) is the distributional Fourier series given by X u= einω where einω (x) = einωx . Prove:

n∈Z

(i) eiωx u = u, for all x ∈ R. (ii) u is periodic with period 2π/ω. P (iii) There exists a constant c ∈ C such that u = c k∈Z δk 2π/ω . For the determination of c = 2π/ω, see Example 15.7. Problem 10.9. (Distributions homogeneous of fixed degree). Let Ha be the set of homogeneous distributions of degree a ∈ C in Rn . Verify the following assertions. (i) Ha is a linear subspace of D0 (Rn ). Furthermore, Ha is a closed subspace of D0 (Rn ), that is, if u ∈ D0 (Rn ), uj ∈ Ha and uj → u in D0 (Rn ) as j → ∞, then u ∈ Ha . (ii) If u ∈ Ha , one has ∂j u ∈ Ha−1 and xk u ∈ Ha+1 . (iii) If ψ ∈ Ha ∩ C ∞ (Rn ), with ψ 6= 0, then a ∈ Z≥0 and ψ is a homogeneous polynomial function of degree a. Hint: first prove that if the function t 7→ ta : R>0 → C possesses a C ∞ extension to R, then a ∈ Z≥0 ; for example, by studying the behavior of the derivatives of this function as t ↓ 0. (iv) Every continuous function f on the sphere kxk = 1 possesses a uniquely determined extension to a homogeneous function g of degree a on Rn \ {0}. This g is continuous. If Re a > −n, then g is locally integrable on Rn and g ∈ Ha . (v) Ha is infinite-dimensional if n > 1 and a ∈ C. Hint: use (iv) when Re a > −n, and combine this with (ii) when Re a ≤ −n. Problem 10.10. For all α ∈ (Z≥0 )n , prove that ∂ α δ ∈ D0 (Rn ) is homogeneous of degree −n − |α| Problem 10.11. Let Φ : X → Y be a C ∞ submersion and let K : X → X be a C ∞ diffeomorphism such that Φ ◦ K = Φ. Prove that for every v ∈ D0 (Y ) the distribution u = Φ∗ v ∈ D0 (X) is invariant under K (as a generalized function). Problem 10.12. (Composition of partial differentiation and pullback under submersion). Let X be open in Rn , Y open in Rp and Φ : X → Y a C ∞ submersion. Prove that (10.1) can be extended to an identity of continuous linear mappings: D0 (Y ) → D0 (X). Problem 10.13. (Second-order differential operator composed with pullback under quadratic form). Consider the second-order linear partial differential operator P =

n X

Bij ∂i ∂j

i,j=1

in Rn , where (Bjk ) is an invertible symmetric real n × n matrix. Let (Akl ) be the inverse matrix and

94

10 Transposition. Pullback and Pushforward

Φ : x 7→

n X

Akl xk xl ∈ R

k,l=1

the corresponding quadratic form on Rn . Verify the following assertions. (i) The restriction of Φ to Rn \ {0} is a submersion from Rn \ {0} to R. (ii) One has the following identity of distributions in D0 (Rn \ {0}) (P ◦ Φ∗ )v = Φ∗ (4y ∂y 2 + 2n ∂y )v

(v ∈ D0 (R)).

Here y = Φ(x) denotes the variable in R. (iii) If v ∈ D0 (R) is homogeneous of degree a, then Φ∗ v is homogeneous of degree 2a on Rn \ {0}. (iv) If v 0 is homogeneous of degree − n2 , then u = Φ∗ v defines a solution on Rn \{0} of the partial differential equation P u = 0. When is u homogeneous? Of what degree? (v) From part (ii) deduce the following two identities of continuous linear mappings: P ◦ Φ∗ = Φ∗ ◦ (4y ∂y 2 + 2n ∂y ) Φ∗ ◦ P = (4∂y 2 ◦ y − 2n ∂y ) ◦ Φ∗

: :

C ∞ (R) → C ∞ (Rn ), E 0 (Rn ) → E 0 (R).

(vi) Apply the latter equality in part (v) to derive the result in Example 10.4. Problem 10.14. Verify formula (10.17) by using linear algebra to rewrite jΛ with jΛ as in (10.16). To this end, parametrize U ∩ Φ−1 ({y}) using Λ and show that the tangent space T of Φ−1 ({y}) at x = Λ(x0 ) equals ker φ where φ = DΦ(x). Apply the chain rule to Φ ◦ Λ = Π and deduce that T is spanned by certain column vectors λk that occur in the matrix of λ := DΛ(x0 ). Next, project the remaining column vectors in λ along T onto T ⊥ = im tφ and express (det λ)2 = det(tλ λ) in terms of these projections and the λk spanning T . Finally, write the contribution coming from the projections in terms of the row vectors of φ. Problem 10.15. (Pushforward of function under submersion to R as consequence of integration of a derivative). Let X be an open subset of Rn and Φ : X → R a C ∞ submersion. Denote by H the Heaviside function on R. (i) Show, for all φ ∈ C0∞ (X) and y ∈ R, (Φ∗ φ)(y) = (Φ∗ δy )(φ) = Φ∗ (∂ Ty H)(φ) and apply Problem 10.11 to derive ∂j Φ Φ∗ δy = ∂j Φ ◦ Φ∗ ◦ ∂(Ty H) = ∂j (Φ∗ (Ty H)). (ii) By approximating H by C ∞ functions, verify Z (Φ∗ (Ty H))(φ) = Φ−1 ( ] y, ∞ [ )

φ(x) dx.

10 Transposition. Pullback and Pushforward

95

(iii) Apply the Theorem on Integration of a Total Derivative (see [8, Theorem 7.6.1]) to deduce Z ∂j Φ(x) ∂j (Φ∗ (Ty H))(φ) = φ(x) dx. k grad Φ(x)k Φ−1 ({y}) Here ∂j Φ/k grad Φk denotes the j-th component of the normalized gradient vector field of Φ; the value of the vector field at x is the inner normal to Φ−1 ({y}) at x. Furthermore, dx denotes Euclidean (n − 1)-dimensional integration over Φ−1 ({y}). (iv) Combine the results from parts (i) and (iii), replace φ by φ ∂j Φ/k grad Φk and sum over 1 ≤ j ≤ n to obtain formula (10.17): Z φ(x) dx (y ∈ Y ). (Φ∗ φ)(y) = Φ−1 ({y}) k grad Φ(x)k Problem 10.16. (Pullback of Dirac measure under submersion). Let X be an open subset of Rn and consider a C ∞ submersion Φ : X → R. (i) Let φ ∈ C0∞ (X). Prove (Φ∗ φ)(y) =

1 δ −1 (φ) k grad Φk Φ ({y})

(y ∈ R).

Verify the following identity in D0 (X) (compare with Example 10.8): Φ∗ δy =

1 δ −1 k grad Φk Φ ({y})

(y ∈ R).

(ii) On the basis of part (i) conclude (compare with [8, Exercise 7.36]) 1] −∞, y [ (Φ∗ φ) = 1Φ−1 ( ] −∞, y [ ) (φ) and deduce from this, for all y ∈ R, Z (Φ∗ φ)(y) = ∂y φ(x) dx, Φ−1 ( ] −∞, y

(y ∈ R);

Φ∗ δy = ∂y ◦ (1Φ−1 ( ] −∞, y [ ) ).

[)

(iii) Give an independent proof of the second identity in part (ii) by means of successive application of the Fundamental Theorem of Integral Calculus on R (see [8, Theorem 2.10.1]), interchange of the order of integration and integration by parts to ψ(Φ∗ φ), where ψ ∈ C0∞ (R). Problem 10.17. (Integration of total derivative). Let X be an open subset of Rn and Φ : X → R a C ∞ submersion. Define the open set Ω = Φ−1 (R>0 ), and suppose that Ω is nonempty and bounded. Note that the boundary ∂Ω of Ω equals Φ−1 ({0}) and that this set is a C ∞ submanifold in Rn of dimension n − 1. Derive from Problem 10.16.(i)

96

10 Transposition. Pullback and Pushforward

Φ∗ δ =

1 δ∂Ω k grad Φk

in D0 (X).

With H the Heaviside function on R, prove on the basis of Problem 10.12 ∂j (Φ∗ H) =

∂j Φ δ∂Ω . k grad Φk

Deduce that, for all φ ∈ C0∞ (X), Z Z ∂j φ(x) dx = Ω

φ(y) νj (y) dy,

∂Ω

where ν(y) denotes the outer normal to ∂Ω at y and dy denotes Euclidean (n − 1)dimensional integration over ∂Ω. Conclude that by this method we have proved the following Theorem on Integration of a Total Derivative, see [8, Theorem 7.6.1]: Z Z Dφ(x) dx = φ(y) tν(y) dy. Ω

∂Ω

Problem 10.18. (Wave equation). Denote the points in R2 by (x, t) and consider the mappings plus : (x, t) 7→ x + t

and

minus : (x, t) 7→ x − t

from R2 to R. Verify that these are submersions. Prove, for every pair of distributions a ∈ D0 (R) and b ∈ D0 (R), that the distribution u := (plus)∗ a + (minus)∗ b on R2 satisfies the wave equation ∂t 2 u = ∂x 2 u. This gives a distributional answer to the classical question whether (x, t) 7→ a(x + t) + b(x − t) is acceptable as a solution of the wave equation also when a and b are not C 2 functions, although being continuous, for example. Describe (plus)∗ δ and (minus)∗ δ. Problem 10.19. Let Φ be a C ∞ mapping from an open subset X of Rn to an open subset Y of Rp . Let C be the closure of Φ(X) in Y . Prove supp (Φ∗ u) ⊂ C

(u ∈ E 0 (X)).

Now assume that the p-dimensional measure of C equals 0, so that Φ∗ u = 0 if Φ∗ u is locally integrable. Prove the existence of φ ∈ C0∞ (X) such that Φ∗ φ is not locally integrable. Hint: approximate δx by φj ∈ C0∞ (X).

10 Transposition. Pullback and Pushforward

97

Problem 10.20. Let Φ be a C ∞ mapping from an open subset X of Rn to an open subset Y of Rp . Suppose that Ψ : Y → X is a C ∞ mapping such that Ψ ◦ Φ(x) = x, for all x ∈ X. Prove that Φ is a proper embedding. Write V := Φ(X) and define E 0 (V ) := { v ∈ E 0 (Y ) | ψ v = 0 if ψ ∈ C ∞ (Y ) and ψ = 0 on V }. Prove that Φ∗ is a bijective linear mapping from E 0 (X) to E 0 (V ), with the restriction of Ψ∗ to E 0 (V ) as its inverse. Hint: use the fact that, for every φ ∈ C ∞ (Y ), the function ψ = φ − Ψ ∗ ◦ Φ∗ (φ) equals 0 on V . Finally, let f be a real-valued C ∞ function on the open subset X of Rn . Denote the points of Rn+1 by (x, y), where x ∈ Rn , y ∈ R. Define Φ : X → X × R by Φ(x) = (x, f (x)), for x ∈ X. Let v ∈ E 0 (X × R). Prove that (y − f (x)) v = 0 if and only if there exists a u ∈ E 0 (X) with v = Φ∗ u.

11 Convolution of Distributions

Convolution involves translation; that makes it difficult to define the former operation for functions or distributions supported by arbitrary open subsets in Rn . Therefore we initially consider objects defined on all of Rn . In (3.2) we described the convolution (f ∗ φ)(x) of a continuous function f on Rn and φ ∈ C0∞ (Rn ) as the testing of f with the function Tx ◦ Sφ : y 7→ φ(x − y). Here S and Tx are the reflection, and the translation by the vector x, of functions, as described in Example 10.10, and Example 10.11, respectively. This suggests defining the convolution product u ∗ φ, for u ∈ D0 (Rn ) and φ ∈ C0∞ (Rn ), as the function given by (u ∗ φ)(x) = u(Tx ◦ Sφ) (x ∈ Rn ). (11.1) In Theorem 11.2 we will show that this function belongs to C ∞ (Rn ). Furthermore, with a suitable choice of φ = φ we can ensure that u ∗ φ converges in D0 (Rn ) to u for  ↓ 0, see Lemma 11.6. Example 11.1. For a ∈ Rn we have (δa ∗ φ)(x) = (Tx ◦ Sφ)(a) = (Sφ)(a − x) = φ(x − a). Consequently, δa ∗ φ = Ta φ for every φ ∈ C ∞ (Rn ), and in particular δ∗φ=φ

(φ ∈ C0∞ (Rn )).

In other words, δ acts as a unit element for the operation of convolution.

(11.2) 

Theorem 11.2. If u ∈ D0 (Rn ) and φ ∈ C0∞ (Rn ), then u ∗ φ ∈ C ∞ (Rn ). One has Ta (u ∗ φ) = (Ta u) ∗ φ = u ∗ (Ta φ),

(11.3)

∂ α (u ∗ φ) = (∂ α u) ∗ φ = u ∗ (∂ α φ),

(11.4)

for every a ∈ Rn , and

for every multi-index α. Furthermore,

100

11 Convolution of Distributions

supp (u ∗ φ) ⊂ supp u + supp φ.

(11.5)

Proof. Because x 7→ Tx ◦ Sφ is continuous from Rn to C0∞ (Rn ), the function u ∗ φ is continuous on Rn . Formula (11.3) follows when we compare (Ta (u ∗ φ))(x) = (u ∗ φ)(x − a) = u(Tx−a ◦ Sφ), ((Ta u) ∗ φ)(x) = Ta u(Tx ◦ Sφ) = u(T−a ◦ Tx ◦ Sφ), (u ∗ (Ta φ))(x) = u(Tx ◦ S ◦ Ta φ) = u(Tx ◦ T−a ◦ Sφ). Substituting a = −t e(j) in (11.3), we find
1
1 u ∗ φ)(x + t e(j)) − (u ∗ φ)(x) = Tt e(j) ∗ u − u (Tx ◦ Sφ) t t

1 = u Tx ◦ S ◦ Tt e(j) ∗ φ − φ . t In view of (10.9) and (10.8) this converges as t → 0. We see that u ∗ φ is partially differentiable with respect to the j-th variable, with derivative ∂j (u ∗ φ) = (∂j u) ∗ φ = u ∗ (∂j φ). As the partial derivatives are continuous, the conclusion is that u ∗ φ is of class C 1 . By mathematical induction on k one can then show that u ∗ φ is of class C k and that (11.4) holds for all multi-indices α with |α| = k. x∈ / supp u + supp φ means that supp u has an empty intersection with x + (− supp φ) = x + supp Sφ = supp (Tx ◦ Sφ), which implies (u ∗ φ)(x) = u(Tx ◦ Sφ) = 0. Because supp u is closed and supp φ compact, C = supp u + supp φ is a closed subset of Rn , see the text following (2.2). Because u ∗ φ = 0 on the open subset Rn \ C of Rn , the conclusion is that supp (u ∗ φ) ⊂ C.  An operator U from functions on Rn to functions on Rn commutes with all translations if U ◦ Ta = Ta ◦ U , for all a ∈ Rn . Convolution with u can be characterized in terms of such operators: Theorem 11.3. Let u ∈ D0 (Rn ). Then u ∗ : φ 7→ u ∗ φ is a continuous linear mapping from C0∞ (Rn ) to C ∞ (Rn ) that commutes with all translations. Conversely, if U is a continuous linear mapping from C0∞ (Rn ) to C ∞ (Rn ) that commutes with all translations, then there exists a uniquely determined u ∈ D0 (Rn ) with U (φ) = u∗φ, for all φ ∈ C0∞ (Rn ).

11 Convolution of Distributions

101

Proof. It is evident that u ∗ is linear and, as (11.3) shows, u ∗ commutes with all translations. The proof of the continuity is more involved. Let A and B be compact subsets of Rn . Then K := A + (−B) is compact and according to Lemma 3.7 there exist a constant c > 0 and k ∈ Z≥0 such that (ψ ∈ C0∞ (K)).

|u(ψ)| ≤ c kψkC k

Applying this to ψ = Tx ◦ Sφ, with x ∈ A and φ ∈ C0∞ (B), we find (φ ∈ C0∞ (B)).

ku ∗ φkC 0 , A ≤ c kφkC k

Because ∂ α (u ∗ φ) = u ∗ (∂ α φ), it now follows that for every m ∈ Z≥0 there exists a constant c0 > 0 with ku ∗ φkC m , A ≤ c0 kφkC k+m

(φ ∈ C0∞ (B)).

Because this holds for all A, B and m, the conclusion is that u ∗ is continuous from C0∞ (Rn ) to C ∞ (Rn ). Now consider U as described above. If U is of the form u ∗, then (U ψ)(0) = u(Sψ); thus, u is determined by u(ψ) = (U ◦ Sψ)(0), for ψ ∈ C0∞ (Rn ). If we now interpret this as the definition of u, we see that u ∈ D0 (Rn ) and (u ∗ φ)(x) = u(Tx ◦ Sφ) = (U ◦ S ◦ Tx ◦ Sφ)(0) = (U ◦ T−x φ)(0) = (T−x ◦ U φ)(0) = U (φ)(x), for every x ∈ Rn ; that is, u ∗ φ = U (φ).



In proving Theorem 11.5 we will use the following principle of integration under the distribution sign. Lemma 11.4. Let X be an open subset of Rn and suppose that the mapping A : Rm → C0∞ (X) has the following properties: (a) There exists a compact subset T of Rm such that A(t) = 0 whenever t ∈ / T. (b) There is a compact subset K of X such that supp A(t) ⊂ K for all t ∈ Rm . (c) For every k ∈ Z≥0 and every  > 0 there exists a δ > 0 such that ks − tk < δ implies kA(s) − A(t)kC k < . Then

Z

Z  A(t) dt (x) =

Rm

A(t)(x) dt

Rm

R defines a function A(t) dt belonging to C0∞ (X). For every u ∈ D0 (X), one has that u ◦ A : Rm → C is a continuous function with compact support while Z  Z u A(t) dt = (u ◦ A)(t) dt. Rm

Rm

102

11 Convolution of Distributions

Proof. It follows from the assumptions made that the Riemann sums X Sh := hm A(h t) (h ∈ R>0 ) t∈Zm

R converge in C0∞ (X) to A(t) dt as h ↓ 0. One begins by proving this in C00 (X) and then uses the Theorem about differentiation under the integral sign to obtain the result in C0∞ (X). From the continuity of A and of u it follows that u ◦ A is a continuous function. The complex number X X u(Sh ) = hm u(A(h t)) = hm (u ◦ A)(h t) t∈Zm

t∈Zm

is an approximating Riemann sum for the integral of u ◦ A. Using the continuity of u once again we now find Z Z 
u A(t) dt = u lim Sh = lim u(Sh ) = (u ◦ A)(t) dt. h↓0

Rm

h↓0

Rm

 Theorem 11.5. For u ∈ D0 (Rn ) and φ and ψ in C0∞ (Rn ) one has (u ∗ φ)(ψ) = u(Sφ ∗ ψ).

(11.6)

Proof. The mapping A : x 7→ ψ(x) (Tx ◦ Sφ)

:

Rn → C0∞ (Rn )

satisfies the conditions of Lemma 11.4. Applying (1.12) we find Z Z (u ∗ φ)(ψ) = (u ∗ φ)(x) ψ(x) dx = u(Tx ◦ Sφ) ψ(x) dx Z =

u(ψ(x) (Tx ◦ Sφ)) dx = u

Z

 ψ(x) (Tx ◦ Sφ) dx = u(Sφ ∗ ψ). 

Lemma 11.6. Let φ ∈ C0∞ (Rn ) and 1(φ) = 1. Write φ (x) = −n φ( 1 x), for  > 0. For every u ∈ D0 (Rn ) we then have that u ∗ φ ∈ C ∞ (Rn ) converges in D0 (Rn ) to u as  ↓ 0.

11 Convolution of Distributions

103

Proof. Let ψ ∈ C0∞ (Rn ). Because 1(Sφ) = (S1)(φ) = 1(φ) = 1 and S(φ ) = (Sφ) , we find, in view of Lemma 1.5, which can be extended to Rn in a straightforward manner, that Sφ ∗ ψ → ψ in C0∞ (Rn ). In combination with Theorem 11.5 this implies (u ∗ φ )(ψ) = u(Sφ ∗ ψ) → u(ψ) as  ↓ 0. Since this holds for every ψ ∈ C0∞ (Rn ), we conclude that u ∗ φ → u in D0 (Rn ).  Next we extend a sharpening of this result to open subsets of Rn by “cutting and smoothing”. Corollary 11.7. Let X be an open subset of Rn . For every u ∈ D0 (X) there exists a sequence uj ∈ C0∞ (X) with limj→∞ uj = u in D0 (X). Proof. Choose an increasing sequence of compact sets Kj in X that absorbs X, see Lemma 8.2.(a). Further, choose cut-off functions χj ∈ C0∞ (X) such that χj = 1 on an open neighborhood of Kj . Finally, choose (j) > 0 such that limj→∞ (j) = 0 and supp χj + supp φj ⊂ X for every j. Here φj := φ(j) with φ as in Lemma 11.6. The third assumption is satisfied if (j) c < δj , where c is the supremum of the kxk with x ∈ supp φ and where δj > 0 has been chosen such that the δj neighborhood of supp χj is contained in X.

∆ j -nbhd of supp Χ j

supp Χ j

Kj

X

Because χj u ∈ E 0 (X), we can interpret it as an element of E 0 (Rn ) ⊂ D0 (Rn ), see Lemma 8.6. We now take uj := (χj u) ∗ φj ∈ C ∞ (Rn ). In view of (11.5) we find that supp uj is a compact subset of X, and so uj ∈ C0∞ (X). Let ψ ∈ C0∞ (X) ⊂ C0∞ (Rn ). There exists a j0 such that, for all j ≥ j0 , the δj neighborhood of supp ψ is contained in Kj ; consequently, χj (Sφj ∗ ψ) = Sφj ∗ ψ. Thus, on the strength of Theorem 11.5 we get, for j ≥ j0 , uj (ψ) = (χj u)(Sφj ∗ ψ) = u(χj (Sφj ∗ ψ)) = u(Sφj ∗ ψ).

104

11 Convolution of Distributions

Because Sφj ∗ ψ → ψ in C0∞ (X) we find uj (ψ) → u(ψ) as j → ∞ and conclude that uj → u in D0 (X).  By Corollary 11.7, many identities that hold for test functions can be extended to distributions. The principle is that if A and B are sequentially continuous operators on D0 (X) with A(φ) = B(φ) for all φ ∈ C0∞ (X), then A(u) = B(u) for all u ∈ D0 (X). Indeed, with the sequence uj in C0∞ (X) as in Corollary 11.7 we obtain A(u) = lim A(uj ) = lim B(uj ) = B(u). j→∞

j→∞

(The principle as introduced here is the same as the “principle of uniqueness of continuous extensions to the closure”). This was the basis for conjecturing (9.1), (11.3), (11.4) and (11.6), for instance, and even (11.5). The proof by transposition was then a matter of writing out the definitions. It is a very fruitful principle, if only for generating conjectures, but it should not be overlooked that its application presupposes that A and B have already been shown to be sequentially continuous extensions to D0 (X) of A = B on C0∞ (X). Because these extensions are often composed of operators defined by means of transposition, the method of transposition has not become redundant. The fact that C0∞ (X) is dense in D0 (X) supports the concept of distributions as generalized functions. One can even go a step further, observing that, for example, the linear subspace in C ∞ (X) consisting of the polynomial functions is dense in C ∞ (X) with respect to uniform convergence on compacta. This is Weierstrass’ Approximation Theorem, see Example 14.23 below. Because C ∞ (X) is dense in D0 (X), the polynomial functions are also dense in D0 (X). Our next aim is to define the convolution u ∗ v of the most general distributions u and v in D0 (Rn ) possible. If u and v are continuous functions and if at least one of them has compact support, we obtain the following symmetric expression for testing u ∗ v with φ ∈ C0∞ (Rn ): Z Z Z Z (u ∗ v)(φ) = u(x) v(z − x) dx φ(z) dz = u(x) v(y) φ(x + y) dx dy. (11.7) Actually, the left-hand side should read test(u ∗ v), instead of u ∗ v. In the integration over z we have applied the change of variables z = x + y. In the right-hand side of (11.7), the function u ⊗ v : (x, y) 7→ u(x) v(y) on Rn × Rn is tested against the function Σ ∗ φ : (x, y) 7→ φ(x + y).

(11.8)

The function u ⊗ v is said to be the direct product or the tensor product of u and v. In (11.8) we have used the sum mapping

11 Convolution of Distributions

Σ : (x, y) 7→ x + y

:

105

Rn × Rn → Rn ,

(11.9)

by which φ ∈ C ∞ (Rn ) is pulled back to a function in C ∞ (Rn × Rn ). Using the transpose Σ∗ of Σ ∗ in a formal way, see (10.2), we can write (u ∗ v)(φ) = (u ⊗ v)(Σ ∗ φ) = (Σ∗ (u ⊗ v))(φ)

(φ ∈ C0∞ (Rn )).

This suggests the following definition for the convolution product u ∗ v of the distributions u and v in Rn : u ∗ v = Σ∗ (u ⊗ v), (11.10) the direct product of u and v pushed forward under the sum mapping. In explicit form this becomes (u ∗ v)(φ) = u(x 7→ v(y 7→ φ(x + y)))

(φ ∈ C0∞ (Rn )).

(11.11)

To justify this, we consider the extension of the definition of the direct product u ⊗ v to arbitrary distributions u and v. Theorem 11.8. Let X and Y be open subsets of Rn and Rp , respectively, and let u ∈ D0 (X) and v ∈ D0 (Y ). Then there exists exactly one distribution on X × Y , denoted by u ⊗ v and said to be the distributional direct product or tensor product of u and v, with the property (u ⊗ v)(φ ⊗ ψ) = u(φ) v(ψ),

(11.12)

for all φ ∈ C0∞ (X) and ψ ∈ C0∞ (Y ). One has, for θ ∈ C0∞ (X × Y ), (u ⊗ v)(θ) = u(x 7→ v(y 7→ θ(x, y))) = v(y 7→ u(x 7→ θ(x, y))).

(11.13)

The mapping (u, v) 7→ u ⊗ v

:

D0 (X) × D0 (Y ) → D0 (X × Y )

is sequentially continuous with respect to each variable separately. Proof. We begin by verifying that the right-hand side in (11.13) is well-defined and is continuously dependent on the test function θ. Given x ∈ X, introduce the partial function θ(x, ·) : y 7→ θ(x, y) in C0∞ (Y ); then x 7→ θ(x, ·) is a continuous mapping: X → C0∞ (Y ). Hence composition with the continuous v : C0∞ (Y ) → C implies that (V θ)(x) := v ◦ θ(x, ·) defines a continuous function V θ on X with compact support. Using the linearity, and once again using the continuity of v, one can show
 ∂θ  1 (V θ)(x + t e(j)) − (V θ)(x) = V (x). t→0 t ∂xj lim

106

11 Convolution of Distributions

By mathematical induction on k we find that V θ is of class C k and that ∂ α (V θ) = V

∂αθ , ∂xα

for all multi-indices α with |α| ≤ k. Thus we see that V θ ∈ C0∞ (X); moreover, making use of the estimates for v of the form (3.4), we conclude that V : θ 7→ V θ

:

C0∞ (X × Y ) → C0∞ (X)

defines a continuous linear mapping. Consequently, u ⊗ v ∈ D0 (X × Y ) if we set u⊗v =u◦V

:

C0∞ (X × Y ) → C.

(11.14)

Because (V (φ ⊗ ψ))(x) = v(φ(x) ψ) = φ(x) v(ψ)

(x ∈ X),

it follows that V (φ ⊗ ψ) = v(ψ) φ. If we then apply u to this, we come to the conclusion that (11.12) holds true. This proves the existence part of the theorem. Now suppose that w ∈ D0 (X × Y ) and w(φ ⊗ ψ) = u(φ) v(ψ) whenever φ ∈ C0∞ (X) and ψ ∈ C0∞ (Y ). Then ω := w − (u ⊗ v) has the property that ω(φ ⊗ ψ) = 0 for all φ ∈ C0∞ (X) and ψ ∈ C0∞ (Y ). We are going to demonstrate that this implies ω = 0. In that case, w = u ⊗ v and thus we will also have proved the uniqueness. Let (x0 , y0 ) ∈ X × Y . We can find a corresponding open neighborhood U of (x0 , y0 ) whose closure is a compact subset K of X × Y . If χ ∈ C0∞ (X × Y ) equals 1 on an open neighborhood of K, one has χω ∈ E 0 (X × Y ) ⊂ D0 (Rn × Rp ). Then we get in the notation of Lemma 11.6, if  > 0 is sufficiently small, ((χω)∗(φ ⊗ψ ))(x, y) = (χω)((Tx Sφ )⊗(Ty Sψ )) = ω((Tx Sφ )⊗(Ty Sψ )) = 0, for all (x, y) ∈ K, in view of supp Tz Sζ = z +(− supp ζ). In particular, we have (χω) ∗ (φ ⊗ ψ ) = 0 on U . On account of Lemma 11.6, the limit of the left-hand side, as  ↓ 0, equals χω; thus we obtain χω = 0 on U , and so ω = 0 on U . By applying Theorem 7.1 we see that this implies ω = 0. Furthermore, the right-hand side of the second identity in (11.13) defines a distribution that also satisfies (11.12); in view of the uniqueness just proved, this distribution also equals u ⊗ v. To prove the last assertion we observe that, on the strength of (11.14), it follows from limj→∞ uj = 0 in D0 (X) that, for every θ ∈ C0∞ (X × Y ), lim uj ⊗ v(θ) = lim uj (V θ) = 0,

j→∞

j→∞

because V θ ∈ C0∞ (X). The analogous result holds for u ⊗ vj .



The previously defined direct product (tensor product) of continuous functions u and v satisfies (11.12), and in view of the uniqueness it therefore equals the distributional direct product (tensor product) of u and v. In other words, the direct product of distributions is an extension of the direct product of continuous functions.

11 Convolution of Distributions

107

Write C0∞ (X) ⊗ C0∞ (Y ) for the linear subspace of C0∞ (X × Y ) consisting of the finite linear combinations of the functions φ ⊗ ψ, where φ ∈ C0∞ (X) and ψ ∈ C0∞ (Y ). The Hahn–Banach Theorem is a basic result in functional analysis; using one of its corollaries, see [22, Theorem 3.5], one now deduces from the proof of Theorem 11.8 that the linear subspace C0∞ (X)⊗C0∞ (Y ) is dense in C0∞ (X ×Y ). In other words, test functions of both the variables x and y can be approximated by linear combinations of products of test functions of x and y, respectively. Example 11.9. If we denote the variable in X or Y by x, or y, respectively, then u ∈ D0 (X) and v ∈ D0 (Y ) can also be denoted by the function symbols u(x) and v(y), respectively. In that case it is usual to use the function notation (u ⊗ v)(x, y) = u(x) v(y) also for distributions u ∈ D0 (X) and v ∈ D0 (Y ). If a ∈ X and b ∈ Y are given points, then (δa ⊗ δb ) (φ ⊗ ψ) = δa (φ) δb (ψ) = φ(a) ψ(b) = δ(a, b) (φ ⊗ ψ), from which we see that δa ⊗ δb = δ(a, b) . In function notation: δa (x) δb (y) = δ(a, b) (x, y). For example, if X = Rn and Y = Rp , then δ(x) δ(y) = δ(x, y), a formula popular in physics literature. Now let X = Y = Rn . Then δ(x + y) δ(x − y) = 21 δ(x, y) in D0 (Rn × Rn ). Indeed, δ(x+y) δ(x−y) corresponds to Φ∗ (δ⊗δ) ∈ D0 (Rn ×Rn ), where Φ(x, y) = (x + y, x − y). Now we have, for φ ∈ C0∞ (Rn × Rn ), Φ∗ (δ ⊗ δ)(φ) = (δ ⊗ δ)(Φ∗ φ) = jΦ−1 (δ ⊗ δ)((Φ−1 )∗ φ) =

1 1 φ(0, 0) = δ(φ). 2 2

where, in order to establish the second identity, we have used Theorem 10.6 and  where we have observed that Φ−1 = 12 Φ and jΦ−1 = 12 for the third. We now want to define the convolution u ∗ v of the distributions u and v belonging to D0 (Rn ) as the direct product, pushed forward under the sum mapping, as in (11.10). In the pushing forward we intend to use Theorem 10.5. We are, however, faced with the problem that the sum mapping is not a proper mapping from Rn ×Rn to Rn , unless n = 0. Indeed, the inverse image of z ∈ Rn consists of the set of all (x, z − x) with x ∈ Rn . In the following lemma we use sets of sums as defined in (2.2). Lemma 11.10. Let A and B be closed subsets of Rn . Then the following assertions are equivalent. (a) The sum mapping (11.9) is proper on A × B. (b) For every compact subset L of Rn , A ∩ (L + (−B)) is bounded in Rn . (c) For every compact subset L of Rn , B ∩ (L + (−A)) is bounded in Rn . Assertions (a) – (c) obtain if either A or B is compact. Proof. (a) means that for every compact subset L of Rn the set V of the (x, y) ∈ A × B, with x + y ∈ L, is compact. Because this set is closed, we may replace

108

11 Convolution of Distributions

the word “compact” by “bounded”. Furthermore, the projection of V onto the first component equals A ∩ (L + (−B)) and the projection onto the second component is B ∩ (L + (−A)). This proves (a) ⇒ (b) and (a) ⇒ (c). On the other hand,



Σ −1 (L) ⊂ A ∩ (L + (−B)) × L + − A ∩ (L + (−B)) . This implies (b) ⇒ (a) and by an analogous argument one shows that (c) ⇒ (a).  Definition 11.11. For every u and v in D0 (Rn ) such that the sum mapping (11.9) is proper on supp u × supp v, (11.10) defines a distribution u ∗ v ∈ D0 (Rn ), said to be the convolution of u and v. In particular, u ∗ v is defined if at least one of the two distributions u and v has compact support.  Example 11.12. By substituting a Dirac measure into (11.11) we obtain u ∗ δa = δa ∗ u = Ta u

(u ∈ D0 (Rn ), a ∈ Rn ).

(11.15)

In particular (compare with (11.2)), u∗δ =δ∗u=u

(u ∈ D0 (Rn )).

(11.16)

Let A be a closed subset of Rn . In the following theorem, as in Theorem 10.5, we use the notation D0 (Rn )A for the set of the u ∈ D0 (Rn ) such that supp u ⊂ A. We say that uj → u in D0 (Rn )A if uj ∈ D0 (Rn )A for all j and uj → u in D0 (Rn ). In that case, it automatically follows that u ∈ D0 (Rn )A . For the properness of the sum mapping, see Lemma 11.10. Theorem 11.13. Let A and B be closed subsets of Rn such that the sum mapping (11.9) is proper on A × B. The convolution product (u, v) 7→ u ∗ v is the uniquely determined extension of the convolution product (u, φ) 7→ u ∗ φ

:

D0 (Rn )A × C0∞ (Rn )B → C ∞ (Rn )

as introduced in (11.1), to a mapping from D0 (Rn )A × D0 (Rn )B to D0 (Rn ) that is sequentially continuous with respect to each variable separately. The convolution of distributions satisfies the following computational rules: supp (u ∗ v) ⊂ supp u + supp v, u∗v = v∗u (commutative rule), (u ∗ v) ∗ w = u ∗ (v ∗ w) (associative rule), (u ∗ v)(φ) = u(Sv ∗ φ) (φ ∈ C0∞ (Rn )), Ta (u ∗ v) = (Ta u) ∗ v = u ∗ (Ta v) (a ∈ Rn ), ∂ α (u ∗ v) = (∂ α u) ∗ v = u ∗ (∂ α v) (α ∈ (Z≥0 )n ).

(11.17) (11.18) (11.19) (11.20) (11.21) (11.22)

Rule (11.19) holds if the sum mapping Σ : (x, y, z) 7→ x + y + z is proper on supp u × supp v × supp w. This certainly is the case if at least two out of the three distributions u, v and w have compact support.

11 Convolution of Distributions

109

Proof. Regarding the extension of the convolution product to D0 (Rn )A ×D0 (Rn )B , we first note the following. For P and Q closed subsets of Rn with P ∩ Q compact, there exists χ ∈ C0∞ (Rn ) with χ = 1 on a neighborhood of P ∩ Q. Given any u ∈ D0 (Rn )P and ψ ∈ C ∞ (Rn )Q , then u(χ ψ) is well-defined and independent of the choice of χ. In this manner, we extend u to a continuous linear functional on C ∞ (Rn )Q and the extension is uniquely determined, because C0∞ (Rn )Q is dense in the latter space. Now consider v ∈ D0 (Rn )B , then Sv ∈ D0 (Rn )−B ; so we obtain, for any compact L ⊂ Rn and φ ∈ C0∞ (Rn )L , Sv ∗ φ ∈ C ∞ (Rn )L+(−B) . Next apply the first result in the proof choosing P = A and Q = L + (−B), then P ∩ Q is compact by Lemma 11.10.(b) and accordingly we may define, for u ∈ D0 (Rn )A , (u e ∗ v)(φ) = u(Sv ∗ φ). If v ∈ C0∞ (Rn ), then (11.6) implies that the right-hand side is equal to (u ∗ v)(φ). This being the case for every φ ∈ C0∞ (Rn ), we conclude u e ∗ v = u ∗ v if v ∈ C0∞ (Rn ): the convolution product of distributions is an extension of the convolution product of a distribution and a test function as introduced in (11.1). Furthermore, (11.20) now follows on identifying e ∗ with ∗. Using (10.4), we see supp (u ∗ v) = supp Σ∗ (u ⊗ v) ⊂ Σ(supp (u ⊗ v)) = Σ(supp u × supp v) = supp u + supp v. Let τ : (x, y) 7→ (y, x) be the interchanging of variables. The second identity in (11.13) can be written as (u ⊗ v)(θ) = (v ⊗ u)(τ ∗ (θ)) for every test function θ, that is, u ⊗ v = τ∗ (v ⊗ u). The commutativity of addition means that Σ ◦ τ = Σ. Application of (10.3) leads to u ∗ v = Σ∗ ◦ τ∗ (v ⊗ u) = (Σ ◦ τ )∗ (v ⊗ u) = Σ∗ (v ⊗ u) = v ∗ u. Thus, the commutativity of convolution is seen to be a direct consequence of the commutativity of addition. By writing out the definitions we find that (u ∗ (v ∗ w))(θ) equals u(x 7→ v(y 7→ w(z 7→ θ(x + (y + z))))), while ((u ∗ v) ∗ w)(θ) is equal to u(x 7→ v(y 7→ w(z 7→ θ((x + y) + z))))). It follows that (11.19) is a direct consequence of the associativity of addition.

110

11 Convolution of Distributions

By virtue of Theorem 11.8 we conclude that u 7→ u ∗ v = Σ∗ (u ⊗ v), being a composition of two such mappings, is sequentially continuous; using the commutativity of convolution, we find that v 7→ u ∗ v is also sequentially continuous. Finally, (11.21) and (11.22) can be proved by writing out the definitions and applying transposition. These rules can also be derived from the corresponding formulas for v ∈ C0∞ (Rn ), approximating v ∈ D0 (Rn ) by a sequence of test functions and using the sequential continuity of the convolution product.  Remark 11.14. The associativity of convolution implies that there is no need to use parentheses in the case of multiple convolution. The convolution u1 ∗ u2 ∗ · · · ∗ uk of k distributions u1 , u2 , . . . , uk is well-defined if the sum mapping Σ : (x(1) , x(2) , . . . , x(k) ) 7→

k X

x(j)

j=1

is proper on supp u1 × supp u2 × · · · × supp uk . One then has u1 ∗ u2 ∗ · · · ∗ uk = Σ∗ (u1 ⊗ u2 ⊗ · · · ⊗ uk ).

(11.23)

The notation for the repeated direct product in the right-hand side of this formula suggests that the direct product is associative as well. In fact, for test functions this is evident. Using Corollary 11.7 and the continuity of the direct product, we then also find the associativity of the direct product for arbitrary distributions. The proof of (11.23) can be given by mathematical induction on k.  Remark 11.15. Assume that u and v ∈ D0 (Rn ) and that the sum mapping is proper on supp u × supp v. If in this situation v ∈ C ∞ (Rn ), then also u ∗ v ∈ C ∞ (Rn ), while (u ∗ v)(x) = u(Tx ◦ Sv) (x ∈ Rn ). (11.24) Although this is the same formula as (11.1), we first of all need to justify the right-hand side in (11.24), because we have not assumed u or v to have compact support. Consider a compact subset L ⊂ Rn . On account of Lemma 11.10.(b) there exists a compact K ⊂ Rn with the property that supp u ∩ (L + (− supp v)) ⊂ K. Next, write A = L + (− supp v), then A is a closed subset of Rn while supp u ∩ A is compact. Therefore there exists a uniquely determined sequentially continuous extension of u to the space of φ ∈ C ∞ (Rn ) with supp φ ⊂ A. This extension is also denoted by u. To verify its existence, we write u(φ) := u(χ φ), where χ ∈ C0∞ (Rn ) and χ = 1 on a neighborhood of K (see Corollary 2.16). Now consider any x ∈ L and write φ = Tx ◦ Sv, then supp φ = {x} + (− supp v) ⊂ A, and it follows that u(Tx ◦ Sv) = u(φ) is well-defined. We will now show that u ∗ v ∈ C ∞ (Rn ) and that (11.24) holds true. Let U be an arbitrary bounded open subset of Rn with compact closure L. According to the assumption made, K := Σ −1 (L) ∩ (supp u × supp v)

11 Convolution of Distributions

111

is a compact subset of Rn × Rn . Let χ ∈ C0∞ (Rn × Rn ) with χ = 1 on a neighborhood of K. The definition of distributions pushed forward, given in the proof of Theorem 10.5, leads to the formula (u ∗ v)(θ) = (u ⊗ v)(χΣ ∗ (θ)), for every θ ∈ C0∞ (U ). In addition, we can take χ = α⊗β, with α and β ∈ C0∞ (Rn ) equal to 1 on a large bounded subset of Rn . But this means that on U u ∗ v = (αu) ∗ (βv). Here βv ∈ C0∞ (Rn ); in Theorem 11.2 we have concluded that the right-hand side is a C ∞ function given by ((αu) ∗ (βv))(x) = (αu)(Tx ◦ S(βv)). For x ∈ U this is equal to the right-hand side of (11.24), with the interpretation given above.  In general, we have the following estimate for the singular support of the convolution of two distributions. Theorem 11.16. Consider u and v ∈ D0 (Rn ) and suppose that the sum mapping (x, y) 7→ x + y is proper on supp u × supp v. Then sing supp (u ∗ v) ⊂ sing supp u + sing supp v.

(11.25)

Proof. Let A = sing supp u and B = sing supp v. For every δ > 0 we can find a cut-off function α ∈ C ∞ (Rn ) with α = 1 on an open neighborhood of A and such that supp α is contained in the δ-neighborhood Aδ of A. For this purpose we can choose the convolution of the characteristic function of the 21 δ-neighborhood of A and φ , for sufficiently small  > 0, see Lemma 2.18. Let β be an analogous cut-off function, having the same properties with respect to B instead of A. Write u1 := α u and u2 := (1 − α) u, and analogously v1 := β v and v2 := (1 − β) v. Then u = u1 + u2 with supp u1 ⊂ Aδ , u2 ∈ C ∞ (Rn ). Furthermore, v = v1 + v2 with supp v1 ⊂ Bδ and v2 ∈ C ∞ (Rn ). Out of the four terms in u ∗ v = u1 ∗ v1 + u1 ∗ v2 + u2 ∗ v1 + u2 ∗ v2 the last three are of class C ∞ on account of Remark 11.15. This implies sing supp u∗v = sing supp u1 ∗v1 ⊂ supp u1 ∗v1 ⊂ supp u1 +supp v1 ⊂ Aδ +Bδ . If we can demonstrate that for every x ∈ / A + B there exists a δ > 0 with x ∈ / Aδ + Bδ , that is, x ∈ / sing supp (u ∗ v), then we will have proved that sing supp (u ∗ v) ⊂ A + B. If this is not the case, then there exists an x ∈ / A + B such that

112

11 Convolution of Distributions

x = a(j) + b(j), where d(a(j), A) → 0 and d(b(j), B) → 0 as j → ∞. This means that there are a ˜(j) ∈ A and ˜b(j) ∈ B such that ka(j) − a ˜(j)k → 0 and ˜ kb(j) − b(j)k → 0 as j → ∞. Because the sum mapping is proper on A × B, we can ensure, by changing to a subsequence, that a ˜(j) → a ∈ A and ˜b(j) → b ∈ B. But then x = a + b ∈ A + B, a contradiction.  Remark 11.17. It may also happen that the convolution u ∗ v can be formed without making any assumptions about the supports of u and of v. The space of all integrable functions on Rn is denoted by L1 (Rn ), or simply L1 , and provided with the integral norm Z kvkL1 := |v(x)| dx. Rn

One of the main theorems from the theory of Lebesgue integration asserts that L1 is complete (a Banach space, that is, a normed linear space that is complete). Let E be a linear subspace of D0 (Rn ) endowed with a norm u 7→ kuk, with the following four properties: 1

(i) For every a ∈ Rn and u ∈ E, one has Ta u ∈ E and kTa uk = kuk. (ii) E is complete. (iii) C0∞ (Rn ) ⊂ E and convergence in C0∞ (Rn ) implies convergence in E. (iv) C0∞ (Rn ) is dense in E. An example of such a space is E = L1 . In that case, y 7→ ψ(y) Ty φ is continuous from Rn to E, for every φ and ψ ∈ ∞ C0 (Rn ). In view of (1.12), the integral of this function is equal to φ ∗ ψ. Applying the triangle inequality to the approximating Riemann sums, we find the inequality kφ ∗ ψk ≤ kψkL1 kφk. Now let u ∈ E and v ∈ L1 . Then there exist sequences uj and vj in C0∞ (Rn ) with ku − uj k → 0 and kv − vj kL1 → 0 as j → ∞. The sequence uj ∗ vj is a Cauchy sequence in E. The limit, denoted by u ∗ v, is independent of the choice of the sequences uj and vj and satisfies ku ∗ vk ≤ kvkL1 kuk. The mapping (u, v) 7→ u ∗ v thus defined is continuous from E × L1 to E and is the only continuous extension to E × L1 of the convolution product on C0∞ (Rn ) × C0∞ (Rn ). It is left to the reader as a problem to prove these assertions concerning u ∗ v. Applying this for E = L1 , one can prove by means R of the theory of Lebesgue integration that for every u and v ∈ L1 the integral u(x − y) v(y) dy converges to (u ∗ v)(x) for almost all x. The example u = v = |x|−a 1[ −1, 1 ] in L1 (R), for 1  2 ≤ a < 1, shows that the integral need not converge for all x. 1

See Appendix 19.

11 Convolution of Distributions

113

Problems Problem 11.1. For each of the following cases, find a combination of a distribution u and a test function φ on R that solves the equation. (i) u ∗ φ = 0. (ii) u ∗ φ = 1. (iii) u ∗ φ = x. (iv) u ∗ φ = sin. Problem 11.2. If limj→∞ uj = u in D0 (Rn ) and limj→∞ φj = φ in C0∞ (Rn ), then lim uj ∗ φj = u ∗ φ in C ∞ (Rn ).

j→∞

Prove this by using the principle of uniform boundedness from Chap. 5. Problem 11.3. Given a mapping A : C0∞ (Rn ) → C ∞ (Rn ), verify that the following assertions are equivalent. (i) A is a continuous linear mapping and commutes with the partial differentiations ∂j , for all 1 ≤ j ≤ n. (ii) There exists a u ∈ D0 (Rn ) such that A(φ) = u ∗ φ, for every φ ∈ C0∞ (Rn ). Hint: differentiate a 7→ Ta ◦ A ◦ T−a . Problem 11.4. If t 7→ ut is a continuous mapping from Rm to D0R(Rn ) with ut = 0 for all t outside a bounded subset of Rm , we define the integral ut dt ∈ D0 (Rn ) by Z Z  ut dt (φ) := ut (φ) dt (φ ∈ C0∞ (Rn )). Rm

Rm

0

Now prove that, for u ∈ D (R ) and φ ∈ C0∞ (Rn ), Z u∗φ= φ(x) Tx u dx. n

Rn

Problem 11.5. Let u ∈ D0 (Rn ) with ∂j u = 0, for all 1 ≤ j ≤ n. Prove the existence of a constant c ∈ C such that u = c. Problem 11.6. Calculate δa ∗ δb , for a and b ∈ Rn . Problem 11.7. Let P be a linear partial differential operator in Rn with constant coefficients. Prove that P u = (P δ) ∗ u for every u ∈ D0 (Rn ). Also prove that P (u ∗ v) = (P u) ∗ v = u ∗ (P v) if u and v ∈ D0 (Rn ) and if the sum mapping is proper on supp u × supp v.

114

11 Convolution of Distributions

Problem 11.8. Let E be the linear subspace of D0 (Rn ) consisting of the finite linear combinations of the δa with a ∈ Rn . Prove that for every continuous function f in Rn there exists a sequence uj ∈ E such that uj → f in D0 (Rn ) as j → ∞. Hint: see Problem 5.12. Now go on to show that for every u ∈ D0 (Rn ) there exists a sequence uj ∈ E with uj → u in D0 (Rn ) as j → ∞. Given that the convolution of distributions is continuous in each of the variables, use this to prove, once again, the properties (11.21), (11.22), (11.20), (11.19) and (11.18). Problem 11.9. Define D0 (R)+ as the union of the D0 (R)[ l, ∞ [ over all l ∈ R. We say that uj → u in D0 (R)+ if there exists an l ∈ R with uj → u in D0 (R)[ l, ∞ [ . Prove that for every u and v ∈ D0 (R)+ the sum mapping is proper on supp u × supp v. And, additionally, that the convolution product (u, v) 7→ u ∗ v

:

D0 (R)+ × D0 (R)+ → D0 (R)+

satisfies all computational rules for the convolution product. In assertions concerning sequential continuity, one only needs to replace convergence in D0 (R) by convergence in D0 (R)+ . Problem 11.10. Define, by mathematical induction on k, χk+ by χ1+ = H, the Heavif k > 1. Calculate all χk+ and all derivatives iside function, and χk+ = H ∗ χk−1 + k (l) (χ+ ) . Prove that, for every k ∈ Z>0 and f ∈ D0 (R)+ , there exists exactly one u ∈ D0 (R)+ with u(k) = f . For φ ∈ C k (R), verify the formula (φ H)(k) =

k−1 X

φ(k−j−1) (0) δ (j) + φ(k) H.

j=0

Write out the formula resulting from the convolution of the left-hand side and the right-hand side, respectively, and χk+ . Do you recognize the result? Problem 11.11. Calculate (1 ∗ δ 0 ) ∗ H and 1 ∗ (δ 0 ∗ H). Problem 11.12. Let u ∈ E 0 (Rn ). Prove that φ 7→ u ∗ φ is a continuous linear mapping from C ∞ (Rn ) to C ∞ (Rn ). Problem 11.13. Prove that the direct product of two probability measures is also a probability measure. Verify that for two probability measures µ and ν on Rn the convolution µ ∗ ν is well-defined and defines a probability measure on Rn . The convolution µ ∗ ν is said to be the independent sum of the probability measures µ and ν. Do you recognize this from probability theory? Finally, calculate the probability measure µ∗µ∗· · ·∗µ, the independent sum of N copies of µ, if µ = p δ1 +(1−p) δ0 and 0 ≤ p ≤ 1. Problem 11.14. Calculate the distribution δ 0 (x + y) δ(x − y) on R2 . Express the result in terms of the Dirac measure at 0 ∈ R2 and its derivatives.

11 Convolution of Distributions

115

Problem 11.15. (Young’s inequality). In this problem we use the notation from Example 14.16. Let p, q and r ≥ 1 and 1/p + 1/q + 1/r = 2. Then we have Young’s inequality which asserts, for all f , g and h ∈ C0 (Rn ), Z f (x) (g ∗ h)(x) dx ≤ kf kLp kgkLq khkLr . Rn

Indeed, we are free to assume that f , g and h are real and nonnegative. Introduce p0 ≥ 1 by 1/p + 1/p0 = 1, and similarly q 0 and r0 . Write the integral on the left-hand side as Z Z Z Z I= f (x) g(x − y) h(y) dy dx = a(x, y) b(x, y) c(x, y) dx dy Rn

Rn

Rn

Rn

p/r 0

a(x, y) = f (x)

0

g(x − y)q/r , 0

0

b(x, y) = g(x − y)q/p h(y)r/p ,

with

0

0

c(x, y) = f (x)p/q h(y)r/q . Noting that 1/p0 +1/q 0 +1/r0 = 1, we can use H¨older’s inequality for three functions to obtain |I| ≤ kakLr0 kbkLp0 kckLq0 . But kakLr0 =

Z Rn

Z

f (x)p g(x − y)q dx dy

Rn

1/r0

p/r 0

q/r 0

= kf kLp kgkLq ,

and similarly for b and c. The second equality above is a consequence of changing variables from y to x − y and integrating first with respect to y. This leads to the desired inequality. Now prove the following. For f and g ∈ C0 (Rn ) and p, q and r ≥ 1 satisfying 1 + 1/p = 1/q + 1/r, one has the inequality kf ∗ gkLp ≤ kf kLq kgkLr ,

in particular

kf ∗ gkLp ≤ kf kLp kgkL1 . 0

In order to obtain this estimate, apply Young’s inequality with f = s |g ∗ h|p /p , where s is the function defined by s (g ∗ h) = |g ∗ h|. This implies kg ∗ hkLp0 ≤ kgkLq khkLr ,

where

1+

1 1 1 = + . p0 q r

Then replace p0 by p. Finally, prove the validity of the estimates if the functions belong to the appropriate spaces of type Lp (Rn ) .

12 Fundamental Solutions

P α Definition 12.1. Let P = P (∂) = |α|≤m cα ∂ be a linear partial differential n operator in R with constant coefficients, as introduced in (7.4). A fundamental solution of P is a distribution E ∈ D0 (Rn ) such that P E = δ, the Dirac measure at the origin.  Every linear partial differential operator with constant coefficients (not all of them equal to 0) has a fundamental solution. This was first proved by Ehrenpreis [9] and Malgrange [20]. For more on this, see Remark 16.10 below. For linear partial differential operators with variable coefficients the existence of fundamental solutions need not be the case. The importance of fundamental solutions lies in the following: Theorem 12.2. Suppose E is a fundamental solution of P . Then we have P (E ∗ f ) = f

(f ∈ E 0 (Rn )),

(12.1)

u = E ∗ (P u)

(u ∈ E 0 (Rn )).

(12.2)

Proof. Using (11.22) and (11.16), we find P (E∗f ) = (P E)∗f = δ∗f , and therefore (12.1). Combination of (11.22) and (12.1) yields E ∗ (P u) = (P E) ∗ u = u, and therefore (12.2).  For every distribution f on Rn with compact support, formula (12.1) implies the existence of a distributional solution u = E ∗ f ∈ D0 (Rn ) of the inhomogeneous linear partial differential equation P u = f . Additionally, under the assumption that the solution u of P u = f has compact support, (12.2) means that the solution is uniquely determined and given by u = E ∗ f . A word of warning: it may seem as if E ∗ is a two-sided inverse of P , which would imply that P is bijective. But that is not actually the case, because the domain spaces do not correspond: generally speaking, we do not know more about E ∗ than that it is a mapping from E 0 (Rn ) to D0 (Rn ). The differential operator P maps

118

12 Fundamental Solutions

D0 (Rn ) to D0 (Rn ) and E 0 (Rn ) to E 0 (Rn ), but not D0 (Rn ) to E 0 (Rn ). In general, partial differential operators are far from injective, as we will see in the next example. Another consequence of this is that fundamental solutions are not uniquely dee is a fundamental solution of termined. If E is a fundamental solution of P , then E 0 n e P if and only if E = E + u, with u ∈ D (R ) a solution of P u = 0. Pn Example 12.3. Let P = ∆ = j=1 ∂j 2 be the Laplace operator in Rn . A C 2 function u is said to be harmonic on the open subset U of Rn if ∆u = 0 on U ; this terminology is carried over to distributions u ∈ D0 (U ). In other words, the kernel of ∆ consists of the harmonic functions, or distributions, respectively. If n > 1, then the harmonic polynomials on Rn form a linear space of infinite dimension. Indeed, the functions x 7→ (x1 + i x2 )k on Rn are harmonic, for all k ∈ Z≥0 . (If n = 1 and U is an interval, then u ∈ D0 (U ) and ∆u = 0 if and only if u on U is equal to a polynomial function of degree ≤ 1.) In Problem 4.6 a fundamental solution of the Laplace operator was found to be the locally integrable function E on Rn \ {0} with  1   if n 6= 2,  (2 − n) c kxkn−2 n (12.3) E(x) =    1 log kxk if n = 2. 2π For cn , see also (13.31). For f ∈ E 0 (Rn ), the distribution u = E ∗ f is said to be the potential of the distribution f , a terminology that has its origin in the situation where n = 3 and where f denotes a mass or charge density. We conclude that the potential u of f satisfies Poisson’s equation ∆u = f

in D0 (Rn );

and in particular the potential u is harmonic on the complement of supp f , the largest open set on which f = 0. Furthermore, if u is a solution of Poisson’s equation that has compact support, then u necessarily equals the potential of f . The general fundamental solution of the Laplace operator is equal to the sum of the δ-potential and a harmonic function on Rn . If n ≥ 3, then E can be characterized as the fundamental solution of ∆ that converges to 0 as kxk → ∞. This can be proved by means of Fourier transformation, see Problem 16.2.  As far as the study of the singular supports (see Definition 7.6) of solutions is concerned, addition of C ∞ functions is irrelevant. This can also be expressed by saying that calculations are performed modulo C ∞ when summands of class C ∞ are neglected. A distribution E on Rn is said to be a parametrix of P if there exists an ω ∈ C ∞ (Rn ) such that P E = δ + ω. In other words, if E modulo C ∞ satisfies the equation for a fundamental solution. For certain P a parametrix can be obtained by iterative methods.

12 Fundamental Solutions

119

Theorem 12.4. Suppose that P possesses a parametrix E with sing supp E = {0}. Then, for every open subset X of Rn , sing supp u = sing supp P u

(u ∈ D0 (X)).

(12.4)

Proof. We see immediately that sing supp P u ⊂ sing supp u, this is the case for every linear partial differential operator P in X with C ∞ coefficients. We will now prove the converse inclusion. If u has compact support, we can interpret u as an element of E 0 (Rn ) according to Lemma 8.6. We have E ∗ (P u) = (P E) ∗ u = δ ∗ u + ω ∗ u. Since δ ∗ u = u and ω ∗ u ∈ C ∞ (Rn ) by Remark 11.15, we find, using (11.25), sing supp u = sing supp E ∗ (P u) ⊂ sing supp E + sing supp P u = {0} + sing supp P u = sing supp P u. Now let x ∈ X \ sing supp P u. Choose χ ∈ C0∞ (X) with χ = 1 on an open neighborhood U of x. Then P (χ u) = P u on U , therefore x ∈ / sing supp P (χ u). The above yields sing supp P (χ u) = sing supp (χ u), because χ u has compact support. Therefore x ∈ / sing supp (χ u), which in turn implies that x ∈ / sing supp u, because χ = 1 on a neighborhood of x.  The theorem means that, if u ∈ D0 (X) is a solution of the partial differential equation P u = f , then u ∈ C ∞ on every open subset U of X where f ∈ C ∞ . A linear partial differential operator P with C ∞ coefficients is said to be hypoelliptic if P has this property, that is, if (12.4) holds. If E is a parametrix of a hypoelliptic operator P with constant coefficients, then necessarily sing supp E = sing supp P E = sing supp δ = {0}. The term “hypoelliptic” conveys the fact that this condition is weaker than the condition that P be elliptic, see Theorem 16.6. For the definition and explanation of the term “elliptic”, see Definition 16.2 and the observation following Lemma 16.4. Remark 12.5. If all derivatives can be estimated such that convergence of each of the Taylor series is guaranteed, the proof of Theorem 12.4 can be modified so as to yield results on the analyticity of solutions. Thus one has that u is analytic wherever P u = 0, if there exist E ∈ D0 (Rn ) and an open neighborhood U of 0 in Rn with the following properties: (a) E is analytic on U \ {0}, (b) P E − δ is analytic on U . Estimation of all derivatives can be circumvented by means of complex analysis, see Remark 12.13; this is then followed by the proof of this assertion. 

120

12 Fundamental Solutions

Remark 12.6. For linear partial differential operators P = P (x, ∂) with variable coefficients there is a well-developed theory in which the convolution operator E ∗, defined by Z (E ∗ f )(x) =

E(x − y) f (y) dy,

is replaced by a singular integral operator K of the form Z (Kf )(x) = k(x, y) f (y) dy. Here the integral kernel k(x, y) is a distribution on X × X, if the open set X in Rn denotes the domain space of the functions f and Kf . The operator K is then said to be a parametrix of P if P ◦ K = I + R and K ◦ P = I + S, where R and S are integral operators with a C ∞ kernel on X × X. It is not difficult to prove that P is hypoelliptic when P has a parametrix K with sing supp k ⊂ { (x, y) ∈ X × X | x = y }. However, the construction of parametrices, for sufficiently general operators with variable coefficients, involves too much work to be dealt with in this text.  Example 12.7. Every ordinary differential operator (the case n = 1) is hypoelliptic in the open set where the coefficient of the highest-order term does not vanish. See Theorem 9.4.  Example 12.8. The fundamental solution E of the Laplace operator given in (12.3) satisfies sing supp E = {0}; consequently, the Laplace operator is hypoelliptic. In particular, every harmonic distribution is a harmonic C ∞ function. E is not only C ∞ , but even analytic on the complement of the origin. Remark 12.5 therefore implies that u is analytic wherever P u = 0. In particular, every harmonic function, or distribution, respectively, is analytic. A note on history:RWeyl [27, Lemma 2] proved that every quadratically integrable function u such that u(x) ∆φ(x) dx = 0 for every φ ∈ C02 , is in fact C ∞ . This assertion is known as Weyl’s Lemma and has become the prototype of Regularity Theorems like Theorem 12.4. Weyl’s proof follows the same lines as the proof of Theorem 12.4 given here.  Example 12.9. In Problem 8.5 a fundamental solution E was found for the heat operator P = ∂t − ∆x in (n + 1)-dimensional (x, t)-space. This, too, satisfies sing supp u ⊂ {0}; consequently, the heat operator, too, is hypoelliptic. If the distribution u is a solution of the heat equation P u = 0 on an open subset U of Rn+1 , then u is a C ∞ function on U . However, in the present case E is not analytic on the entire plane of the (x, t) with t = 0, so here it is not true that u is analytic wherever P u is analytic. P is called the n-dimensional heat operator, although this operator is defined in Rn+1 . This derives from the interpretation of x = (x1 , . . . , xn ) as the position coordinates and t as the time coordinate, which are often treated as playing different roles. 

12 Fundamental Solutions

121

Example 12.10. Problems 10.18 and 12.5 describe u ∈ D0 (R2 ) that satisfy the equation (∂1 2 −∂2 2 )u = 0 in R2 but are far from C ∞ functions. This shows that the wave operator ∂1 2 − ∂2 2 is not hypoelliptic. Actually, for every n ≥ 1 the n-dimensional ∂2 n wave operator  := ∂t 2 − ∆x , with x ∈ R , is not hypoelliptic, see Theorem 13.2.  Example 12.11. We now consider complex analysis from a distributional point of view. A function f , defined on an open subset V of C, is said to be complex(z) converges as h → 0 in C. differentiable at the point z ∈ V when f (z+h)−f h 0 The limit is said to be the complex derivative f (z) ∈ C of f at z. If we apply the identification C ' R2 , by writing z ∈ C as z = x + iy in the usual way, with x and y ∈ R, then this condition is stronger than the condition that f be differentiable, as a complex-valued function of two real variables (x, y). Indeed, when f depends on the combination z = x + iy, its partial derivatives satisfy ∂x f (z) = f 0 (z) ∂x z = f 0 (z)

∂y f (z) = f 0 (z) ∂y z = i f 0 (z).

and

This means that f is complex-differentiable if and only if f is real-differentiable and if, moreover, ∂y f (z) = i ∂x f (z)

(Cauchy–Riemann equation).

(12.5)

It is then natural to call a distribution u ∈ D0 (V ) complex-differentiable if u satisfies (12.5). In other words, if P u = 0, with P equal to the first-order linear partial differential operator with constant coefficients P = i ∂x − ∂y = i(∂x + i ∂y ). The formula for integration by parts in Rn reads, with 1 ≤ j ≤ n, Z Z Z f (x) ∂j g(x) dx = − g(x) ∂j f (x) dx + f (y) g(y) νj (y) dEucl y. (12.6) U

U

∂U

Here U is an open subset of Rn with C 1 boundary ∂U and with the property that U lies at one side of ∂U . Furthermore, f and g ∈ C 1 (U ) and (supp f ) ∩ U , for example, is compact. Finally, νj (y) is the j-th component of the outer normal to ∂U at y. See, for example, [8, Corollary 7.6.2]. In R2 ' C we can write i ν1 − ν2 = i(ν1 + iν2 ) = iν, where the vector ν(y) is now interpreted as a complex number. The vector iν(y) is equal to the tangent vector of length 1 to ∂U at y, oriented to have U to the left. If ∂U is locally parametrized by a C 1 curve γ : [ a, b ] → C, where γ 0 (t) has the same orientation as iν(γ(t)), one has, for a continuous function f with support in γ([ a, b ]) Z

Z f (z) iν(z) dEucl z =

∂U

a

b

f (γ(t)) γ 0 (t) dt.

122

12 Fundamental Solutions

Here the vector γ 0 (t) is also interpreted as a complex number. This is said to be the complex line integral of f over ∂U with respect to the orientation of ∂U described above, and is denoted as Z f (z) dz with dz = iν(z) dEucl z. ∂U

Thus we obtain for the operator P the following transposition formula: Z Z Z f (z) P g(z) dz = − g(z) P f (z) dz + f (z) g(z) dz. U

U

(12.7)

∂U

In this formula, U , f and g are as above, with n = 2. The integrals over U are Euclidean 2-dimensional integrals, while that over ∂U is a complex line integral. For a first application of (12.7) we assume that U is bounded, that f is also a complex-analytic function on U and that g equals the constant function 1. We then immediately find the following version of Cauchy’s Integral Theorem: Z f (z) dz = 0. (12.8) ∂U

In order to find a fundamental solution of P , we consider the function z 7→ z1 on C \ {0}. This function is locally integrable on C = R2 (use polar coordinates) and can therefore be interpreted as a distribution. From (12.5) it is evident that P z1 = 0 on R2 \ {0}. For φ ∈ C0∞ (R2 ) we obtain, applying (12.7), Z P φ(z) 1 1 P (φ) = − (P φ) = − lim dz ↓0 z z z |z|> Z = − lim ↓0

0

2π

φ( e−it ) −it  e (−i) dz = 2πi φ(0).  e−it

Here we use that the outer normal to { z ∈ C | |z| =  } points towards the origin; therefore this circle has to be traversed clockwise, which is the case under the 1 mapping t 7→  e−it . We conclude that P z1 = 2πi δ, in other words, E(z) = 2πi z defines a fundamental solution of P . Because this fundamental solution is C ∞ , and even complex-analytic, outside the origin, we conclude on the strength of Theorem 12.4 that P is hypoelliptic. It follows that, in particular, every complex-differentiable distribution on an open set V equals a C ∞ function on V and as such is complex-differentiable in the classical sense. Now let U be an open subset of V with C 1 boundary ∂U . Assume that U is a compact subset of V , f ∈ C 1 (V ) and g ∈ C0∞ (V ). Interpreting the function g in (12.7) as a test function and writing the left-hand side as (f 1U )(P g) = −P (f 1U )(g), we can rewrite (12.7) as the identity compl P (f 1U ) = (P f ) 1U − f δ∂U

12 Fundamental Solutions

123

compl in the space of distributions on V of order ≤ 1 with compact support, where δ∂U denotes the complex line integration over the boundary. If we now apply the convolution operator E ∗ to this identity, and recall (12.2), the left-hand side becomes equal to f 1U ; and thus we obtain Pompeiu’s integral formula Z Z P f (z) 1 f (z) 1 dz + dz (ζ ∈ U ). (12.9) f (ζ) = 2πi U ζ − z 2πi ∂U z − ζ

In particular, if f is complex-differentiable on V , that is, P f = 0 and f ∈ C ∞ (V ), one finds the well-known Cauchy integral formula Z 1 f (z) f (ζ) = dz (ζ ∈ U ). (12.10) 2πi ∂U z − ζ Thus, f is expressed on U in terms of the restriction f |∂U of f to the boundary ∂U . Not every analytic function g on ∂U is of the form g = f |∂U for a complex-analytic function f on U , see Problem 15.6. Cauchy’s integral formula gives an arbitrary complex-differentiable function f as a “continuous linear combination” of the very simple complex-differentiable func1 , where the variable z runs over the boundary ∂U . The singularity at tions z 7→ z−ζ z = ζ need not bother us here, provided we ensure ζ ∈ U steers clear of ∂U . In particular, for a given point a ∈ U and with ζ in a sufficiently small neighborhood U (a) of a, we can substitute the power series 1 1 1 1 = = ζ−a z−ζ (z − a) − (ζ − a) z − a 1 − z−a X (ζ − a)k = (z − a)k+1 k∈Z≥0

into (12.10). This implies that the complex-differentiable f can be expressed by a convergent complex power series in a neighborhood of a. More precisely,  X  1 Z f (z) dz (ζ − a)k (ζ ∈ U (a)). f (ζ) = 2πi ∂U (z − a)k+1 k∈Z≥0

Conversely, it is known that every convergent complex power series is complexdifferentiable and even infinitely differentiable, and the power series is equal to the Taylor series. See any textbook about analysis in one variable. It follows that the conditions “complex-differentiable distribution” and “is locally equal to a convergent complex power series” are equivalent. In this case, f is said to be complex-analytic. Instead of working with the operator P = i ∂x − ∂y , it is more usual in the literature to use 1 1 1 ∂z := (∂x − i ∂y ) and ∂z¯ := (∂x + i ∂y ) = P. 2 2 2i df That makes (12.5) equivalent to ∂z¯f = 0. If this is the case, then dz = ∂z f . The operator ∂z¯ is said to be the Cauchy–Riemann operator; it has the fundamental solution 1 πz and in the literature (12.9) is mostly written in the form

124

12 Fundamental Solutions

f (ζ) = −

1 π

Z U

∂z¯f (z) 1 dz + z−ζ 2πi

Z ∂U

f (z) dz z−ζ

(ζ ∈ U ).

(12.11)

In several variables one has an analogous theory: a function z 7→ f (z) on Cn is complex-differentiable if f is complex-differentiable as a function of each of the variables zj , that is, if ∂z¯j f = 0, for 1 ≤ j ≤ n. These are referred to as the Cauchy–Riemann equations for f . It can be proved that a distributional solution of these equations is of class C ∞ and can locally be developed into power series, where the proof makes use of a higher-dimensional version of Cauchy’s integral formula. These functions are said to be complex-analytic in several variables.  Remark 12.12. The notation becomes clear when we write f (x, y) = F (z, z¯), with z = x + iy and z¯ = x − iy. This yields ∂x f = ∂z F + ∂z¯F

and

∂y f = i ∂z F − i ∂z¯F.

If we now solve this system for ∂z F and ∂z¯F in terms of ∂x f and ∂y f , we obtain formulas that we recognize as the definitions of ∂z and ∂z¯.  Remark 12.13. In Lemma 2.6, a function f on an open subset U of Rn was said to be analytic on U if f has local power series representations at all points of U . To distinguish it from the above, such f is also said to be real-analytic on U . By substituting complex values for the variables into the power series, we see that f has a complex-analytic extension to an open neighborhood V of U in Cn ' R2n . Conversely, the restriction to U := V ∩ Rn of a complex-analytic function on V is a real-analytic function on U . It follows that a function f is real-analytic if f can be extended to a function on a complex neighborhood that satisfies the Cauchy–Riemann equations on that neighborhood. In many cases, this involves less effort than working out the estimates for all derivatives, required to establish the convergence of the Taylor series.  Finally, we come back to the proof of the assertion in Remark 12.5. We observe that the analytic singular support sing supp anal (u) may be defined in the obvious manner, for any u ∈ D0 (Rn ). The proof of the next lemma follows the method of H¨ormander [16, Thm. 4.4.3]. Lemma 12.14. Let u ∈ D0 (Rn ) and v ∈ E 0 (Rn ). Then we have sing supp anal (u ∗ v) ⊂ sing supp anal u + supp v.

Proof. Set U0 = Rn \ sing supp anal u and denote by u0 the restriction of u to U0 . The function u0 has a complex-analytic extension to an open subset U ⊂ Cn satisfying U ∩ Rn = U0 . Selecting a ∈ Rn with a ∈ / sing supp anal u + supp v, we have a + (− supp v) ⊂ U0 . Let V be an open neighborhood of a in Cn such that V + (− supp v) ⊂ U . For z ∈ V , write

12 Fundamental Solutions

125

Z u0 ∗ v(z) =

u0 (z − x) v(x) dx. Rn

This equality should be interpreted in the sense of distributions and defines a complex-analytic function on V . We note that, on V ∩ Rn , the function u0 ∗ v coincides with the usual convolution product u ∗ v. Thus, the latter is analytic on the open neighborhood V ∩ Rn of a in Rn ; this implies a ∈ / sing supp anal (u ∗ v).  Theorem 12.15. Let P be a linear partial differential operator in Rn with constant coefficients. Suppose there exist E ∈ D0 (Rn ) and an open neighborhood U of 0 in Rn with the following properties: (a) E is analytic on U \ {0}, (b) F := P E − δ is analytic on U . We then have, for every open subset X ⊂ Rn and u ∈ D0 (X), that u is analytic on X if P u = 0 on X. Proof. We fix a point x ∈ X and will prove that u is analytic on a neighborhood of x in X. To this end, we select  > 0 such that the open ball B(0; ) ⊂ U while B(x; ) ⊂ X. Furthermore, we consider χ ∈ C0∞ (B(x; )) satisfying χ = 1 on an open neighborhood of x in X. Then we have P (χu) = f with f ∈ E 0 (Rn ) and ˚ ) := B(x; ) \ {x}. supp f ⊂ supp Dχ ⊂ B(x;

(12.12)

We obtain E ∗ f = E ∗ P (χu) = P E ∗ (χu) = F ∗ (χu) + χu, which shows that it is sufficient to verify (i) x ∈ / sing supp anal (E ∗ f ), (ii) x ∈ / sing supp anal (F ∗ (χu)). (i). On account of Lemma 12.14 and (12.12) we have ˚ ). sing supp anal (E ∗ f ) ⊂ sing supp anal E + supp f ⊂ sing supp anal E + B(x; ˚ ) = ∅ on the strength of condition (a); therefore 0 ∈ Now sing supp anal E ∩ B(0; / ˚ ), which implies x ∈ ˚ ). In other sing supp anal E + B(0; / sing supp anal E + B(x; words, (i) is valid. (ii). We note that, again on account of Lemma 12.14, sing supp anal (F ∗(χu)) ⊂ sing supp anal F +supp χ ⊂ sing supp anal F +B(x; ). Furthermore, condition (b) gives sing supp anal F ∩ B(0; ) = ∅, which leads to 0 ∈ / sing supp anal F + B(0; ); hence x ∈ / sing supp anal F + B(x; ) and we conclude that (ii) holds too. 

126

12 Fundamental Solutions

Problems Problem 12.1. Determine all fundamental solutions Ek ∈ D0 (R) of P = ∂ k , for k ∈ Z>0 . Which of these are homogeneous? Of what degree? P Problem 12.2. (Dipoles). The distribution j vj ∂j δa is said to be a dipole at a point a with dipole vector v. Calculate the potential in R3 of the dipole at the origin, with dipole vector equal to the first basis vector. Sketch the corresponding level curves in the (x1 , x2 )-plane. What do the level curves of the potential of δ look like? Also work out this problem for the potential in R2 .

Fig. 12.1. Equipotential curves in (x1 , x2 )-plane in R3 and in R2 , respectively

Problem 12.3. (Potential of bar). Calculate the potential u in R3 of the distribution f defined by Z a f (φ) = φ(x1 , 0, 0) dx1 (φ ∈ C0∞ (R3 )), −a

a bar of length 2a. Indicate where the distribution u is harmonic. Is u a (locally integrable) function? Determine sing supp u. How does u behave as a → ∞? Problem 12.4. Suppose E ∈ D0 (Rn ) is a fundamental solution of the Laplace operator ∆ and consider a harmonic function u on the open subset X of Rn . Let U be an open and U a compact subset of X. Then prove that v := ∆(u 1U ) is a distribution on X with supp v ⊂ ∂U , the boundary of U . Also prove that v ∈ E 0 (X) and u 1U = E ∗ v. Now assume that the boundary ∂U is of class C 1 and denote the outer normal to the boundary at the point y ∈ ∂U by ν(y). For a C 1 function f on a neighborhood of ∂U , the normal derivative ∂ν f of f is defined by n X ∂ν f (y) := νj (y) ∂j f (y) (y ∈ ∂U ). j=1

This is a continuous function on ∂U . Now prove Green’s formula Z
u(x) = u(y) ∂ν (y 7→ E(x − y)) − E(x − y) ∂ν u(y) dy ∂U

(x ∈ U );

12 Fundamental Solutions

127

and furthermore, that the right-hand side vanishes for x ∈ X \ U . Observe that this formula expresses u in U in terms of u and ∂ν u on the boundary ∂U . However, that is not the whole story, for it is known that u is completely determined on U by u|∂U ; this is the so-called Dirichlet problem (not discussed in the present book). Problem 12.5. (One-dimensional wave operator). Define the open set V = { (x, t) ∈ R2 | |x| < t }. Compute a constant a ∈ C such that E = a 1V is a fundamental solution of the wave operator  = ∂t 2 − ∂x 2 . Hint: for the computation one has to evaluate an integral. The evaluation of this integral can be simplified in several ways. One possibility is to apply the change of variables (x, t) = Ψ (y), where Ψ is the rotation in R2 about the origin by the angle π/4. Alternatively, one can apply Green’s Integral Theorem (see [8, Theorem 8.3.5]). In this case, given a test function φ, determine a vector field v such that  φ = curl v = ∂x v2 − ∂t v1 . t

V

x

Fig. 12.2. The set V

Determine supp E and sing supp E. Show that u = E ∗ f is a well-defined distribution on R2 if the support of f ∈ D0 (R2 ) is contained in a half-space of the form H = { (x, t) | t ≥ t0 }, for some t0 ∈ R. Prove that u is a solution of the inhomogeneous wave equation  u = f . Problem 12.6. (Three-dimensional wave operator). Denote the coordinates in R4 by (x, t), where x ∈ R3 and t ∈ R, and the wave operator by  = ∂t 2 − ∆x . Let u be the characteristic function of V = { (x, t) | kxk < t }, the interior of the forward light cone. Further, let v =  u and w =  v. Write q(x, t) = t2 − kxk2 and P = { (x, t) | t > 0 }, the positive half-space. Verify that u = q ∗ H on P . Use Problem 10.13 to determine v and w in P ; describe v. Prove that v and w are homogeneous distributions on Rn and determine their degrees. Prove that w = c δ for some constant c ∈ C. Determine c by testing with the function t 7→ e−t , and justify this procedure. Calculate a fundamental solution E of  . If σ : (x, t) 7→ (x, −t), then prove that σ ∗ (E) is also a fundamental solution of  and that U = E −σ ∗ (E) satisfies the wave equation  U = 0. Determine supp U and sing supp U . Problem 12.7. Assume that P and Q are hypoelliptic operators and show in that case that the composition P ◦ Q is also a hypoelliptic operator. Calculate the composition

128

12 Fundamental Solutions

of ∂z and ∂z¯. Use this to determine, from the fundamental solution (12.3) of ∆, a fundamental solution of ∂z , and also of ∂z¯. Problem 12.8. (Determining an antiderivative if integrability conditions are satisfied). Let E be a fundamental solution of ∆ in Rn and let gj , for 1 ≤ j ≤ n, be distributions with compact support in Rn , with the property that ∂j gk = ∂k gj for all 1 ≤ j, k ≤ n. Prove that the distribution f=

n X

∂j E ∗ gj

j=1

satisfies the system of differential equations ∂j f = gj

(1 ≤ j ≤ n).

(The problem with the usual method to obtain f , namely, by integration of the gj along lines, is that distributions on Rn cannot in all cases be restricted to curves.) Now prove that, for arbitrary distributions gj in Rn , there exists a distribution f in Rn with ∂j f = gj if and only if ∂j gk = ∂k gj , for all 1 ≤ j, k ≤ n. Hint: in the formula above for f , replace the distributions gj by φ gj , where φ denotes a cut-off function. Problem 12.9. (Helmholtz’ equation). Let k ∈ [ 0, ∞ [ and define E± ∈ C ∞ (R3 \ {0})

by

E± (x) =

e±ikkxk . kxk

(i) Prove that E± ∈ D0 (R3 ). Begin by writing, for x ∈ R3 \ {0}, E± (x) = cos kkxk

1 sin kkxk 1 ±i =: g(x) ± i h(x) kxk kxk kxk

and show that both functions g and h belong to C ∞ (R3 ). 1 Next one is asked to show, by two different methods, that − 4π E± is a fundamental solution of Helmholtz’ differential equation with parameter k, or, differently phrased, that in D0 (R3 ) one has

(?)

(∆ + k 2 )E± = −4πδ.

The first approach is a verification that the distribution E± does in fact satisfy equation (?), the second one is an actual construction of the solution. (ii) Verify the following equalities in D0 (R3 ): grad g(x) = −k h(x) x,

grad

 1  1 (x) = − x, k·k kxk3

∆g = −k(k g + 2h).

12 Fundamental Solutions

129

Prove that, in addition, one has in D0 (R3 )  1  D  1 E  1  1 = g∆ + 2 grad g, grad + ∆g ∆ g k·k k·k k·k k·k = −k 2 g

1 − 4π δ, k·k

and use these results to prove (?). Now the second method. Suppose f ∈ C ∞ (R3 \ {0}) to be a radial function, that is, there exists f0 ∈ C ∞ (R>0 ) having the property f (x) = f0 (kxk). (iii) Evaluate ∆f in terms of derivatives of f0 . (iv) Further assume that (∆+k 2 )f = 0. Determine the differential equation satisfied by fe0 ∈ C ∞ (R>0 ), where fe0 (r) = rf0 (r), and use it to prove, for a and b ∈ C, f (x) = a

sin kkxk cos kkxk +b kxk kxk

(x ∈ R3 \ {0}).

(v) Use Green’s second identity (see for instance [8, Example 7.9.6]) to show that, in D0 (R3 ), (∆ + k 2 )f = −4πa δ. (vi) Conclude that every rotation-invariant fundamental solution of Helmholtz’ equation is given by X X c± E± where c± ∈ C, c± = 1. ±

±

13 Fractional Integration and Differentiation

d In this chapter we deal with “complex powers of the operator dx ”, an idea already found in the posthumous works of Riemann and elaborated by Marcel Riesz in the 1930s and 1940s. The relevant article [21] is lengthy, but with the help of a little distribution theory and complex analysis all results can readily be proved. We will also deal with Riesz’ treatment of the wave operator  = ∂t 2 − ∆x in arbitrary dimension; thus we will obtain, among other things, a fundamental solution of .

13.1 The case of dimension one Let D0 (R)+ be the space of distributions u ∈ D0 (R) such that there exists an l ∈ R with supp u ⊂ [ l, ∞ [. For u and v ∈ D0 (R)+ the convolution product u ∗ v ∈ D0 (R)+ is well-defined, see Problem 11.9. Every u ∈ D0 (R)+ possesses a uniquely determined k-th order antiderivative in D0 (R)+ , that is, a v ∈ D0 (R)+ with ∂ k v = u. This is obtained via the definition (compare with [8, Exercise 2.75]) v = χk+ ∗ u

where

χk+ (x) =

xk−1 H(x) (k − 1)!

(x ∈ R).

Indeed, by mathematical induction on k ∈ Z>0 one finds ∂ k χk+ = δ, since on account of the Leibniz rule and Example 9.1 ∂χk+1 + (x) = ∂

 xk k!

 H(x) =

xk−1 xk xk−1 H(x) + δ= H(x) = χk+ (x). (k − 1)! k! (k − 1)!

In other words, finding k-th order antiderivatives is equal to convolution with the function χk+ . Conversely, k-th order differentiation can also be regarded as a convolution operator, namely, as convolution with the distribution δ (k) . It was Riesz who discovered that these operators can be embedded in an analytic a way into a family of convolution operators I+ = χa+ ∗ that depend on an arbitrary a ∈ C. Here the χa+ have the following properties, for all a and b ∈ C:

132

13 Fractional Integration and Differentiation

χa+ ∈ D0 (R)

and

supp χa+ ⊂ R≥0 ,

(13.1)

(k ∈ Z>0 ),

(13.2)

k−1

χk+ (x) =

x H(x) (k − 1)!

(k) χ−k + =δ

χa+

∗

χb+

=

(k ∈ Z≥0 ), χa+b + .

(13.3) (13.4)

a By (13.1), the operator I+ of convolution with χa+ is a continuous linear operator 0 0 k from D (R)+ to D (R)+ . On the strength of (13.2), I+ is equal to finding a k-th −k order antiderivative in D0 (R)+ , while according to (13.3), I+ is equal to k-th order a+b a b differentiation. Finally, (13.4) implies the validity of the group law I+ ◦ I+ = I+ . 0 0 0 Because χ+ = δ, one has I+ = I, the identity in D (R)+ . On account of the group a law with b = −a we also deduce from this that I+ is bijective from D0 (R)+ to −a 0 a D (R)+ , with inverse I+ . If one describes I+ as “finding antiderivatives of order −a a ∈ C”, then I+ could be called “differentiation of order a”. The family (χa+ )a∈C forms a complex-analytic family of distributions, in the sense that a 7→ χa+ (φ), for every φ ∈ C0∞ (R), is a complex-analytic function on C. In this case one also uses the term distribution-valued complex-analytic function . This property will be used to deduce the validity for all a ∈ C of numerous identities involving χa+ from the validity on an arbitrary nonempty open subset U of C. Thus, we are free to choose U such that the desired identity can be verified on U by direct calculation. This procedure involves the following principle. If f and g are analytic functions on a connected open subset V of C, one has f = g on V whenever f = g on a nonempty open subset U of V . This follows by applying Lemma 2.6 to φ = f − g. The functions f and g may be taken as the left-hand side and the right-hand side or vice versa, of the identity involving χa+ , after testing with an arbitrary test function. This is called the principle of analytic continuation of identities.

We now give the definition of the family χa+ , followed by a summary of the properties. The starting point is the definition  a−1  x if x > 0, a χ+ (x) = Γ (a) (13.5)   0 if x < 0. Here xc = ec log x for x > 0 and c ∈ C. The function χa+ (x) is locally integrable on R if and only if Re a > 0; for these values of a we interpret χa+ as an element of D0 (R). The factor Γ (a) in the denominator is in terms of Euler’s Gamma function, see the Appendix in Sect. 13.3 below and, in particular, Corollary 13.5. This is a complex-analytic function on C \ Z≤0 without zeros. For every k ∈ Z≥0 , Γ has a simple pole at −k, with residue (−1)k /k!. Consequently, 1/Γ possesses an extension to an entire analytic function on C, with zeros only at the points −k, with k ∈ Z≥0 , and derivative (−1)k k! at those points. Further, Γ (k) = (k − 1)! for k ∈ Z>0 , which means that (13.2) is satisfied.

13.1 The case of dimension one

133

Integration by parts and the formula Γ (a + 1) = a Γ (a) imply the following identity in D0 (R): ∂χa+1 = χa+ (Re a > 0). (13.6) + This enables us to define χa+ ∈ D0 (R) for every a ∈ C, by means of χa+ = ∂ k χa+k +

(k ∈ Z≥0 , Re (a + k) > 0).

(13.7)

The right-hand side does not depend on the choice of k, while for Re a > 0 the definition is identical to the χa+ that we have defined above. This also implies that (13.6) holds true for all a ∈ C. Furthermore, for every φ ∈ C0∞ (R), the following function is complex-analytic on C: (k) a 7→ χa+ (φ) = (−1)k χa+k ). + (φ

For every φ ∈ C0∞ (R) with supp φ ⊂ ] −∞, 0 [ one has χa+ (φ) = 0 if Re a > 0; it follows by analytic continuation that this identity holds for all a ∈ C; this proves (13.1). In the same way, analytic continuation yields the validity of (13.5) for all a ∈ C. From the analytic continuation of (13.6) we obtain χ0+ = ∂χ1+ = ∂H = δ. By repeated differentiation of this we now find (13.3). From (13.5) we see sing supp χa+ = {0}

(13.8) χa+

⊂ {0}; in this case we if a ∈ / Z≤0 . For a ∈ Z≤0 , (13.5) implies that supp conclude again that (13.8) is valid, by using (13.3). Because χa+ has order 0 if and only if Re a > 0, we see that the order of χa+ equals k if −k < Re a ≤ −k + 1, with k > 0. This can also be formulated by saying that the singularity of χa+ at 0 becomes worse and worse as Re a → −∞. If Re a > 0 and Re b > 0, we have, for x > 0, Z x a b Γ (a) Γ (b) (χ+ ∗ χ+ )(x) = y a−1 (x − y)b−1 dy 0

= xa+b−1

Z

1

ta−1 (1 − t)b−1 dt = Γ (a + b) B(a, b) χa+b + (x).

0

Here B is Euler’s Beta function, see (13.30) below. In view of (13.29), this implies that (13.4) holds if Re a > 0 and Re b > 0. The identity (13.4) now follows by analytic continuation (after testing) for all a ∈ C, for given b ∈ C with Re b > 0. And then also, by analytic continuation with respect to the variable b, for all a ∈ C and b ∈ C. By analytic continuation one also proves the following identity in D0 (R): x χa+ = a χa+1 +

(a ∈ C).

134

13 Fractional Integration and Differentiation

Combining this with (13.6) we now find x ∂χa+ = (a − 1) χa+ .

(13.9)

In other words, for every a ∈ C we conclude, on the strength of Theorem 10.13, that χa+ is a distribution homogeneous of degree a − 1. a Marcel Riesz called I+ (u) = χa+ ∗ u a Riemann–Liouville integral of u; he had encountered it in his work on so-called Ces`aro means.

13.2 Wave family We now describe the version for the wave operator  = ∂t 2 − ∆x = ∂n+1 2 −

n X

∂j 2

j=1

in Rn+1 , whose points we denote by y = (x, t). Here (x1 , . . . , xn ) ∈ Rn are the position coordinates and t ∈ R is the time coordinate. (For generalization to multidimensional time, see Kolk–Varadarajan [19].) All definitions will be given in terms of the corresponding quadratic form q on Rn+1 , of Lorentz type, defined by q(y) = q(x, t) = t2 − kxk2 = t2 −

n X

xj 2 .

j=1

Further, we denote the interior C+ and C− of the (solid) forward and the backward cone, respectively, by  C± = y = (x, t) ∈ Rn+1 | q(y) > 0, t ≷ 0 . (13.10) Then the boundary ∂C+ of C+ is the nappe of the cone ∂C+ = { (x, t) ∈ Rn+1 | kxk = t }.

(13.11)

a The point of departure this time is the function R+ on an open dense subset of R , defined by  a−n−1  c(a) q(y) 2 if y ∈ C+ , a R+ (y) = (13.12)  0 if y ∈ Rn+1 \ C+ . n+1

Here c(a) = cn (a) is a constant to be determined, see Lemma 13.1 below. Note that a for Re a < n + 1 the function R+ is unbounded along ∂C+ . It is locally integrable n+1 on R if and only if Re a > n − 1; if that condition is satisfied, we obtain an element of D0 (Rn+1 ).

13.2 Wave family

135

Fig. 13.1. C+ and C− is bounded by the upper and the lower nappe of the cone, respectively

Lemma 13.1. If we choose c(a) =

Γ ( a+1 2 ) π Γ (a) Γ ( a−n+1 ) 2 n 2

(13.13)

we obtain, for Re a > n − 1 and Re b > n − 1, a+b a b R+ ∗ R+ = R+ .

(13.14)

b a at y = (x, t) we have to ∗ R+ Proof. For the calculation of the function R := R+ integrate a b R+ (x − ξ, t − τ ) R+ (ξ, τ )

over the η = (ξ, τ ) ∈ C+ with y − η ∈ C+ . The latter two relations imply that kξk < τ < t, which means that this set of integration is uniformly bounded whenever t varies in a bounded set; thus, there are no problems with the convergence of the integral. Next, on account of y − η ∈ C+ and η ∈ C+ , we see that the sum y belongs to C+ , and therefore supp R ⊂ C+ . And, finally, we observe that a change of variables η = ρ ζ, with ρ > 0, results in R(ρ y) = ρ2m R(y) where 2m = (a − n − 1) + (b − n − 1) + (n + 1) = a + b − n − 1. That is, R is homogeneous of degree a + b − n − 1. For the further determination of R we use the Lorentz group L of all linear transformations A in Rn+1 with A∗ q = q, det A = 1 and A(C+ ) = C+ (see [8, Exercise 5.70]). (The name Lorentz group is normally used for the somewhat larger group of

136

13 Fractional Integration and Differentiation

all transformations A such that A∗ q = q, but our terminology is also found in the a literature.) It is clear that R+ is invariant under all A ∈ L. Conversely: if f is a function on C+ that is invariant under the Lorentz group and homogeneous of degree 2m, then there exists a constant c ∈ C with f = c q m on C+ . Indeed, f is constant on all orbits { Ay | A ∈ L } of points y ∈ C+ under the action of L. It is known that the orbits in C+ under the action of L are equal to the level surfaces { y ∈ C+ | q(y) = constant } of q in C+ . This implies that there exists a function g on R>0 such that f (y) = g(q(y)) for all y ∈ C+ . From the homogeneity of f we deduce that g must be homogeneous of degree m on R>0 . For every A ∈ L, the change of variables η = A ζ in the integral for R = a b R+ ∗ R+ yields that R(A y) = R(y); in other words, R is invariant under all A ∈ L. Applying the preceding characterization, we find a constant c(a, b) ∈ C such that a+b a b . R+ ∗ R+ = c(a, b) R+

(13.15)

For the calculation of c(a, b) we test both sides of this identity in D0 (Rn+1 ) with the function (x, t) 7→ e−t in C ∞ (Rn+1 ). On C+ this function decreases fast enough to ensure convergence of the integrals. If we now write Z Z a T (a) = e−t R+ (x, t) dx dt, R

Rn

and note that et = et−τ eτ , we find that testing of the left-hand side of (13.15) equals T (a) T (b); therefore, c(a, b) is given by T (a) T (b) = c(a, b) T (a + b). √ On the other hand, with the changes of variables kxk = r = t s and the notation cn for the Euclidean (n − 1)-dimensional volume of the unit sphere in Rn we obtain Z Z a−n−1 T (a) = e−t (t2 − kxk2 ) 2 dx dt c(a) R>0 kxk0

0

1 a−n−1 n−2 cn e−t ta−1 (1 − s) 2 s 2 ds dt 2 R>0 0 n a − n + 1 cn = Γ (a) B , 2 2 2  n   a − n + 1 .  a + 1  cn = Γ (a) Γ Γ Γ . 2 2 2 2

Z

Z

=

Here B is Euler’s Beta function, see (13.30) and (13.29). Now substituting formula (13.31) for cn , we see that the choice (13.13) means that T (a) = T (b) = T (a + b) = 1; therefore, c(a, b) = 1. 

13.2 Wave family

137

If a+1 = −k with k ∈ Z≥0 , then a = −2k − 1 ∈ Z n − 1).

(13.17)

The specification “on Rn+1 \ C− ” is necessary because q ∗ χc+ behaves identically on C− and C+ . Indeed, the reflection y 7→ −y leaves q ∗ χc+ invariant and carries C+ a into C− . However, R+ = 0 everywhere on Rn+1 \C+ , which contains C− \{0}. On −1 q ( ] − ∞, 0 [ ), that is, the complement of C+ ∪ C− , equation (13.16) is correct, since both sides vanish there. If the target variable of q in R is also denoted by q, we obtain from Problem 10.13.(ii) the following identity in D0 (Rn+1 \ {0}): (q ∗ v) = q ∗ (2(n + 1) v 0 + 4q v 00 ). a−n+1

a−n−1

Apply this identity to v = χ+ 2 . Note that v 0 = χ+ 2 , see (13.6). This means that v 0 is homogeneous of degree a−n−3 , see (13.9), so that Theorem 10.13 2 implies a−n−3 0 q v 00 = v; therefore 2(n + 1) v 0 + 4q v 00 = (2a − 4) v 0 . 2 On Rn+1 \ C− this yields a  R+ = d(a) (2a − 4) q ∗ v 0 =

Because Γ ( a+1 2 )= is

a−1 a−1 2 Γ( 2 )

d(a) (2a − 4) a−2 R+ . d(a − 2)

and Γ (a) = (a − 1)(a − 2) Γ (a − 2), the end result

a−2 a  R+ = R+

(Re a > n + 1).

This now enables us to define, for all a ∈ C, as we did for the a tions R+ by a+2k a R+ =  k R+

(13.18) χa+ ,

the distribu-

(k ∈ Z≥0 , Re a + 2k > n − 1).

a The right-hand side is independent of k and a 7→ R+ (φ) is a complex-analytic ∞ n+1 ). By analytic continuation, the identities function on C, for every φ ∈ C0 (R (13.12), (13.14), (13.16) and (13.18) are now seen to be valid for all a ∈ C, in the a distributional sense. With respect to the distribution R+ , (13.12) here means

138

13 Fractional Integration and Differentiation a supp R+ ⊂ C+ ,

(13.19)

a while on the open set C+ the distribution R+ equals the analytic function c(a) q This also implies in the notation of (13.11) a sing supp R+ ⊂ ∂C+ .

a−n−1 2

(13.20)

We further mention the equation n+1 X

a a yj ∂j R+ = (a − n − 1) R+ ,

(13.21)

j=1 a which says that, for every a ∈ C, the distribution R+ is homogeneous of degree a − n − 1. Furthermore, by analytic continuation of identities this is seen to imply a that, for every a ∈ C, the distribution R+ is invariant under the Lorentz group and satisfies a+2 a q R+ = a (a − n + 1) R+ . (13.22) 0 2 Theorem 13.2. R+ = δ and E+ := R+ is a fundamental solution of  . For n = 1 or n even, one has supp E+ = C+ , while supp E+ = ∂C+ if n > 1 is odd. In all cases, sing supp E+ = ∂C+ .

Proof. On account of the pole of Γ at 0, which is not compensated by a pole in the numerator in (13.17), it follows by analytic continuation of (13.16) to a = 0 0 = 0 on Rn+1 \ C− . By combining this with (13.19), we can conclude that that R+ 0 supp R+ ⊂ {0}. But on the strength of Theorem 8.7 this leads to X 0 R+ = cα ∂ α δ, |α|≤m 0 for some m ∈ Z≥0 and cα ∈ C. Now (13.21) implies that R+ is homogeneous of α degree −n − 1, while ∂ δ is homogeneous of degree −n − 1 − |α| according to Problem 10.10. As a consequence we obtain

0 = (n + 1 +

n+1 X j=1

0 yj ∂j ) R+ =−

X

cα |α| ∂ α δ.

(13.23)

|α|≤m

Further, testing with y 7→ y β shows that the distributions ∂ α δ are linearly independent in D0 (Rn+1 ). This implies that (13.23) can only be true if cα = 0, for all α 6= 0. 0 This now proves that R+ = c δ, for some c ∈ C. But then (13.14) yields 0+a a 0 a a a R+ = R+ = R+ ∗ R+ = c δ ∗ R+ = c R+ , a which implies that c = 1, because there certainly exists an a ∈ C such that R+ 6= 0. This proves the first assertion; the second one follows from (13.18) for a = 2.

.

13.2 Wave family

139

As regards the description of supp E+ and sing supp E+ we observe that (13.16) leads to 3−n
1 ∗ on Rn+1 \ C− . E+ = χ+2 n−1 q 2 2π If n = 1, we have that χ1+ = H and therefore E+ = 21 1C+ , a fundamental solution 1 ∗ that we have encountered before, in Problem 12.5. For n = 3 we find E+ = 2π q δ 4 on R \C− ; compare this with Problem 12.6. More generally we find, for n = 2k+3 with k ∈ Z≥0 E+ =

1 q ∗ δ (k) 2 π k+1

on

Rn+1 \ C− .

From this we can see that supp E+ = ∂C+ . If, however, n is even, then we have (3−n)/2 supp χ+ = R≥0 , and so supp E+ = C+ . In (13.20) we already established that sing supp E+ ⊂ ∂C+ . For n odd, the equality of these sets follows from the foregoing, while if n is even, the equality follows from the observation that the 3−n 2 -th power of q is not differentiable coming from within C+ at points y satisfying q(y) = 0, that is, at points belonging to ∂C+ .  Hadamard referred to the assertion supp E+ = ∂C+ for n > 1 odd as the Huygens principle. In view of what Huygens himself wrote about wave propagation, a case can also be made for calling the assertion sing supp E+ = ∂C+ , which holds in every dimension, the Huygens principle. Remark 13.3. The fact that supp E+ ⊂ C+ enables us to form E+ ∗ u for every u ∈ D0 (Rn+1 ) such that supp u is contained in the half-space { (x, t) ∈ Rn+1 | t ≥ 0 }; see the first part of the proof of Lemma 13.1. If, moreover, the distribution u satisfies the wave equation  u = 0, then u = 0. Indeed, u = δ ∗ u = (  E+ ) ∗ u =  (E+ ∗ u) = E+ ∗  u = 0. This implies that E+ is the only fundamental solution E with support in the halfspace t ≥ 0. Indeed, in that case u := E+ − E also has support in t ≥ 0, while  u =  E+ −  E = δ − δ = 0, and therefore E+ − E = u = 0. By elaborating this further we can also obtain the uniqueness of solutions u = u(x, t) for t > 0, of the Cauchy problem  u = f (x, t),

lim u(x, t) = a(x), t↓0

lim ∂t u(x, t) = b(x), t↓0

(13.24)

for a given inhomogeneous term f and given initial functions a and b. To prove the uniqueness, we have to demonstrate that u(x, t) = 0 for t > 0, in the case where  u = 0 and u(x, t) and ∂t u(x, t) both converge to 0 as t ↓ 0. On the strength of the foregoing we can indeed draw that conclusion, if we can show that extension of u by u(x, t) = 0 for t < 0 yields a distribution u on Rn+1 , while, additionally, this distribution satisfies  u = 0 on Rn+1 .

140

13 Fractional Integration and Differentiation

We now consider this Cauchy problem for the partial differential operator  as d2 an initial-value problem for the ordinary differential operator dt 2 −∆x in the variable 0 n t, with integral curves of the form t 7→ ut ∈ D (R ). If we write ut (x) = u(x, t), then ut is a function on Rn for every t > 0 and we obtain, for every test function φ Z with support in t > 0, u(φ) = ut (φt ) dt. (13.25) R>0

This leads to the idea of defining a distributional solution of the Cauchy problem (13.24) as a family (ut )t∈R>0 of distributions on Rn with the following properties: (a) t 7→ ut (ψ) is of class C 2 , for every ψ ∈ C0∞ (Rn ), d2 (b) dt 2 ut − ∆x ut = ft , for every t > 0, d (c) ut → a and dt ut → b in D0 (Rn ), as t ↓ 0. Here we assume that (ft )t∈R>0 is a continuous family of distributions on Rn , and that a and b ∈ D0 (Rn ). We obtain uniqueness for this Cauchy problem if ft ≡ 0 and a = b = 0 imply that ut ≡ 0. In the first instance one may think of C 2 functions u, instead of distributions. The identity (13.25), now for all φ ∈ C0∞ (Rn+1 ) and without the restriction t > 0 for the support of φ, defines a distribution u on Rn+1 , with support in the half-space t ≥ 0. This can be demonstrated by means of the principle of uniform boundedness, while also using the continuity of t 7→ ut : R≥0 → D0 (Rn ). All we have to do is show that  u = 0. But this follows from Z ∞  2  d ( u)(φ) = u( φ) = lim ut φt − ∆φt dt 2 s↓0 s dt Z ∞ 2    d  d d φs + us (φs ) + u − ∆u = lim −us (φ ) dt = 0. t t t s↓0 ds ds dt2 s For C 2 functions this is clear; for the justification in the distributional context, the principle of uniform boundedness is invoked once more. An existence theorem for the Cauchy problem is given in Example 17.4 below, by means of Fourier transformation. 

13.3 Appendix: Euler’s Gamma function Euler’s Gamma function is defined by Z Γ (a) :=

e−t ta−1 dt.

(13.26)

R>0

This is an absolutely convergent integral if a ∈ C and Re a > 0. Differentiation with respect to a under the integral sign leads to the conclusion that Γ satisfies the Cauchy–Riemann equation in the complex right half-plane, and therefore defines a complex-analytic function on that open set.

13.3 Appendix: Euler’s Gamma function

141

d −t By writing e−t = − dt e and integrating by parts, we find

Γ (a) = (a − 1) Γ (a − 1)

(Re a > 1).

(13.27)

Because Γ (1) = 1, it follows by mathematical induction on k that Γ (k) = (k − 1)!

(k ∈ Z>0 ).

Euler’s brilliant idea was now to define, for every a ∈ C with a ∈ / Z≤0 Γ (a) :=

Γ (a + k) a(a + 1) · · · (a + k − 1)

(k ∈ Z>0 , Re a + k > 0).

(13.28)

Using (13.27), one sees that the right-hand side does not depend on the choice of k, while in the complex right half-plane the definition is identical to the Γ (a) introduced before. In this way we obtain a complex-analytic extension of the Gamma function to C \ Z≤0 . Equation (13.27) holds for all a ∈ C where the left-hand side and the right-hand side are both defined.

Fig. 13.2. Graph of |Γ |

For Re a > 0 and Re b > 0 we can write

142

13 Fractional Integration and Differentiation

Z

Z

e−(t+u) ta−1 ub−1 dt du

Γ (a) Γ (b) = R>0

Z

R>0

Z

= R>0

Z

s

e−s ta−1 (s − t)b−1 dt ds

0

e−s sa+b−1 ds

= R>0

Z

1

ra−1 (1 − r)b−1 dr.

0

This has been obtained by first substituting u = s − t and then t = r s. In other words, Γ (a) Γ (b) = Γ (a + b) B(a, b)

(Re a > 0, Re b > 0),

where B is Euler’s Beta function, defined by Z 1 ta−1 (1 − t)b−1 dt. B(a, b) :=

(13.29)

(13.30)

0

Lemma 13.4. For 0 < Re a < 1, one has Γ (a) Γ (1 − a) = B(a, 1 − a) =

π . sin πa

Proof. The first identity follows from Γ (a + b) = Γ (1) = 1 if a + b = 1. We will now try to calculate (13.30) by replacing the integral of f (t) := ta−1 (1− b−1 t) over the real interval [ 0, 1 ] by a complex line integral over a suitable closed contour γ in V := C \ ({0} ∪ {1}). The problem then arising is that f is not a single-valued complex-analytic function on V . More exactly, in view of the formula tc = ec log t = |t|c ec i arg t , we see that after traversing γ, the value f (t) has been multiplied by the factor e2πi((a−1) w(γ,0)+(b−1) w(γ,1)) , if w(γ, z) denotes the net winding number of γ with respect to the point z. This means that f is single-valued along γ if and only if a w(γ, 0) + b w(γ, 1) ∈ Z. When we have R made choices for arg t and arg(1 − t) at the initial point of γ, the integral Iγ := γ f (t) dt is well-defined and, in view of Cauchy’s Integral Theorem, invariant under homotopy of γ. (For the Integral Theorem, see for example formula (12.8), or else [8, Theorem 8.3.12 or Exercise 8.11]; homotopy = continuous deformation in which the deformed contour stays away from the singular points t = 0 and t = 1.) As a next step, we may try to find a homotopy of γ in which the contour lies alongside [ 0, 1 ] in such a way that B(a, b) can be expressed in terms of Iγ . If we can also find another homotopy with a limit position such that Iγ can be calculated, our attempt will have succeeded.

13.3 Appendix: Euler’s Gamma function

143

Let us try to achieve the latter, by choosing γ homotopic to the contour γR (τ ) = + R eiτ , where τ from 0 to 2π, for large values of R. Then w(γ, 0) = w(γ, 1) = 1, so we need a + b ∈ Z. If, for τ = 0, we now choose the argument of t = 12 + R equal to 0 and that of 1 − t = 21 − R equal to π, we obtain 1 2

Z

Z

γR

2π 

a−1  1 b−1 1 R eiτ i dτ + R eiτ − R eiτ 2 2 0 Z 2π a+b−1 πi(b−1) eiτ (a+b−1) (1 + φ(τ ))a−1 (1 − φ(τ ))b−1 dτ, = iR e

f (t) dt =

0

with φ(τ ) := 1/(2Reiτ ). If a + b = 1, this has the limit 2πi eπi(b−1) as R → ∞.

ΓR

∆Ε 0

Ε

1-Ε

1

1 €€€€€ +R 2

Fig. 13.3. γR and δ are homotopic

Now γ is also homotopic to the contour δ which is obtained by first going from  to 1 −  along the real axis, then traversing the circle c1 (τ ) := 1 −  eiτ with τ from 0 to 2π, then returning along the real axis from 1 −  to  and, finally, traversing the circle c0 (τ ) :=  eiτ with τ from 0 to 2π. Under the homotopy from δ to γR , the choice we have made, i.e. taking the arguments of t and 1 − t at the initial point both equal to 0, corresponds to the choice we made before. The integrand has been multiplied by the factor e2πi(b−1) after c1 has been traversed, so that we obtain Z lim f (t) dt = (1 − e2πi(b−1) ) B(a, b). ↓0

δ

Thus, the conclusion is 2πi eπi(b−1) = (1 − e2πi(b−1) ) B(a, b)

(a + b = 1),

which is equivalent to the second identity in the lemma. For other proofs, see [8, Exercises 6.58.(iv) or 6.59]. 

144

13 Fractional Integration and Differentiation

in the formula of the lemma, we obtain Γ ( 12 )2 = π. Here Z Z 1 Z 2 −t − 12 −x2 = Γ e t dt = 2 e dx = e−x dx. 2 R>0 R>0 R

Substituting a =

1 2

This indirectly reconfirms (2.12). A formula for the Euclidean (n − 1)-dimensional volume cn of the unit sphere { x ∈ Rn | kxk = 1 } in Rn follows from Z Z 2 n −kxk2 2 e dx = cn e−r rn−1 dr π = Rn

R>0

Z = cn

e−t t

n−1 2

R>0

1 −1 1 n
t 2 dt = cn Γ . 2 2 2

This yields (see [8, Example 7.9.1 or Exercise 7.21.(viii)] for alternative proofs) n

cn =

2π2 . Γ ( n2 )

(13.31)

With n = 2m even, one has Γ ( n2 ) = Γ (m) = (m − 1)!. With n = 2m + 1 odd, one finds m  n  1 Y 1 1 Γ + m = π2 j− =Γ . 2 2 2 j=1 π , for all a ∈ C \ Z. For every Corollary 13.5. We have Γ (a) Γ (1 − a) = sin(πa) a ∈ C with a ∈ / Z≤0 we have Γ (a) 6= 0. If k ∈ Z≥0 , then lima→−k (a + k) Γ (a) = (−1)k /k!. The function Γ1 possesses an extension to an entire analytic function on C, with zeros only at −k, for k ∈ Z≥0 , and with derivative at those points equal to (−1)k k!.

Fig. 13.4. Graph of the restriction of 1/Γ to a segment in R

Proof. The first assertion follows by analytic continuation. The assertion that Γ has simple poles at the points −k for k ≥ 0, with residue (−1)k /k! at those points, follows from this formula, or directly from (13.28). 

13.3 Appendix: Euler’s Gamma function

145

Problems Problem 13.1. For a ∈ C with Re a > 0, define χa− ∈ D0 (R) by χa− (φ) =

Z

0

−∞

(−x)a−1 φ(x) dx Γ (a)

(φ ∈ C0∞ (R)).

P Furthermore, set |χ|a = ± χa± . Copy the theory about χa+ for the case of χa− , by proving χa− (φ) = χa+ (Sφ), and also for the case of |χ|a . In particular, show ( (k) 2δ if k ∈ Z≥0 even, −k |χ| = 0 if k ∈ Z>0 odd. Problem 13.2. For −1 < Re a < 0 and φ ∈ C0∞ (R), prove that the function x 7→ xa−1 (φ(x) − φ(0)) is integrable over [ 0, ∞ [ and that Z xa−1 a (φ(x) − φ(0)) dx. χ+ (φ) = R>0 Γ (a) a−1

In particular, for these values of a it is not true that χa+ = test xΓ (a) H. More generally, show that, for n ∈ Z≥0 and −n − 1 < Re a < −n, χa+ (φ)

Z = R>0

=

n X xa−1  xk (k)  φ (0) dx φ(x) − Γ (a) k!

1 lim Γ (a) ↓0

k=0

Z

∞

xa−1 φ(x) dx +



n X φ(k) (0) a+k   . k! (a + k)

k=0

Conclude that upon restriction to the linear subspace L ⊂ C0∞ (R) consisting of the φ ∈ C0∞ (R) with 0 ∈ / supp φ, one actually has Z  xa−1 xa−1  φ(x) dx, that is, χa+ L = test H . χa+ (φ) = Γ (a) L R>0 Γ (a) Verify that the last identity holds for all a ∈ C. Problem 13.3. Similarly to Problem 13.2, introduce xa± ∈ D0 (R), for −n − 1 < Re a < −n with n ∈ Z≥0 , by setting xa± (φ) =

n−1   X φ(k) (0) (±x)k dx xa φ(±x) − k! R>0

Z

(φ ∈ C0∞ (R)).

k=0

P Then xa± is well-defined. Next, let |x|a = ± xa± ∈ D0 (R). Replace n by 2n in the formulas above, then demonstrate that |x|a is well-defined for −2n − 1 < Re a < −2n + 1 and satisfies

146

13 Fractional Integration and Differentiation

|x|a (φ) =

Z

xa

X

R>0

φ(±x) − 2

±

n−1 X k=0

φ(2k) (0) 2k  x dx (2k)!

(φ ∈ C0∞ (R)).

In particular, for n ∈ Z≥0 even, deduce that −n

|x|

Z (φ) =

−n

x R>0

X

n 2 −1

φ(±x) − 2

±

Problem 13.4. Prove ∂x PV

 X φ(2k) (0) x2k dx (2k)!

(φ ∈ C0∞ (R)).

k=0 1 x

= −|x|−2 in D0 (R).

Problem 13.5. Let a ∈ C, k ∈ Z and 0 < Re a < k. Prove that for every f ∈ C k (R) with supp f ⊂ [ l, ∞ [ there exists exactly one continuous function u in R such that supp u ⊂ [ l, ∞ [ and Z x (x − y)a−1 u(y) dy = f (x) (x ≥ 0). l

Derive a formula for u(x) in terms of f , for every x ≥ 0. How does u = uj behave as j → ∞, if f = fj converges to the Heaviside function in D0 (R)+ ? Problem 13.6. Prove −2k R+ =  kδ

(k ∈ Z≥0 ).

a ⊂ ∂C+ ? For what a ∈ C does one have that supp R+

Problem 13.7. Consider the distribution ρ :=

d a R . da + a=0

Prove E+ =

1 q ρ if n > 1, (1 − n)

and

n+1 X

yj ∂j ρ = −(n + 1) ρ + δ.

j=1

Is ρ homogeneous? Problem 13.8. (Wave operator). In this problem, we compute a fundamental solution in D0 (R4 ) of the wave operator  = ∂t 2 − ∆x , where (x, t) ∈ R3 × R ' R4 . (i) Let X ⊂ Rn be open. Consider a C ∞ function Φ : X → R with the property grad Φ(x) 6= 0, for every x ∈ X. Write N (y) = { x ∈ X | Φ(x) = y }, for every y ∈ R; from the Submersion Theorem (see, for instance, [8, Theorem 4.5.2.(ii)]) it is known that N (y) is a closed C ∞ submanifold in X of dimension n − 1. Let φ ∈ C0∞ (X). For every y ∈ R, prove that Z φ(x) dx. Φ∗ φ ∈ C0∞ (R) satisfies (Φ∗ φ)(y) = N (y) k grad Φ(x)k (Compare with [8, Exercise 7.36] or with (20.10).) Here Euclidean (n − 1)dimensional integration over N (y) has been used.

13.3 Appendix: Euler’s Gamma function

147

(ii) On the basis of part (i), conclude that, for t ∈ R and φ ∈ C0∞ (X), Z

t

Z (Φ∗ φ)(y) dy =

−∞

φ(x) dx; { x∈X|Φ(x)0 B(t) t2 − kxk2 First we consider the case of n = 3. (i) Prove q∗ φ(0) =

1X 2 ±

Z R3

X φ(x, ±kxk) dx =: δ± (φ). kxk ±

Hint: use that q : R4 \ {0} is a submersion and note that q −1 ({0}) = while ∂C± = graph(h± ) with h± : R 3 → R

defined by

S

±

∂C±

h± (x) = ±kxk.

Next √ apply [8, Special case in Section 7.4.III]. Finally, note k grad q(x, t)k = 2 2 kxk, for (x, t) ∈ q −1 ({0}). (ii) Show that δ± ∈ D0 (R4 ) is homogeneous of degree −2 by verification of the definition, by application of a result in the theory and in a problem, respectively. Prove that δ± is a Lorentz-invariant measure supported by ∂C± . (iii) Verify (13.32) by introducing spherical coordinates in R3 .

13.3 Appendix: Euler’s Gamma function

149

Next we derive (13.33) from (13.32) by means of Hadamard’s method of descent. In this method solutions of a partial differential equation are obtained by considering them as special solutions of another equation which involves more variables and can be solved. Accordingly, consider φ ∈ C0∞ (R3 ) and define φe ∈ C0∞ (R4 ) by (x ∈ R2 , x3 ∈ R, t ∈ R).

e x3 , t) = φ(x, t) φ(x,

(iv) The sphere S(t) minus its equator is the union of the graphs of the functions p h± : B(t) → R given by h± (x) = ± t2 − kxk2 . Imitate the method from [8, Example 7.4.10] to prove Z Z φ(x, t) e t) dx = 2t p dx φ(x, t2 − kxk2 S(t) B(t)

(t ∈ R>0 ).

Now prove (13.33). Formula (13.33) can also be proved directly as follows. (v) For p > 0, show F (p) := q −1 ({p2 }) ∩ C+ = { (x, t) ∈ R3 | t2 − kxk2 = p2 , t > 0 }, which is one sheet of a two-sheeted hyperboloid. Deduce F (p) = graph(h(p, ·)), where p h : R × R2 → R is given by h(p, x) = p2 + kxk2 . Derive, for φ ∈ C0∞ (R3 ), 1 q∗ φ(p ) = 2 2

Z R2

φ(x, h(p, x)) dx. h(p, x)

(vi) Using a change of variables in R>0 , verify that, for ψ ∈ C0∞ (R), Z 1 1 1 2 √ χ (ψ) = ψ(p2 ) dp. π R>0 2 π + (vii) Combining parts (v) and (vi) deduce, for φ ∈ C0∞ (R3 ), Z Z 1
1 1 φ(x, h(p, x)) √ q ∗ χ+2 (φ) = dp dx. 2π R2 R>0 h(p, x) 2 π (viii) p Next introduce the change of variables h(p, x) = t in R>0 , that is, p = t2 − kxk2 . Then p > 0 implies t > kxk. Deduce Z Z ∞ 1 φ(x, t) p E+ (φ) = dt dx. 2π R2 kxk t2 − kxk2 Furthermore, interchange the order of integration to obtain (13.33).

150

13 Fractional Integration and Differentiation ÈÈ x ÈÈ

t

Fig. 13.5. Domain of integration in Problem 13.9.(viii)

Finally we give an application of the results obtained. (ix) Given f ∈ C0∞ (R4 ), verify that a solution u ∈ C ∞ (R4 ) of the inhomogeneous wave equation  u = f is provided by the retarded potential Z 1 f (y, t − kx − yk) u(x, t) = dy. 4π R3 kx − yk

14 Fourier Transformation

Let λ ∈ Cn . The function eλ : x 7→ ehx, λi = ex1 λ1 +···+xn λn in C ∞ (Rn ) has the remarkable property that ∂j eλ = λj eλ

(1 ≤ j ≤ n).

As a consequence, for every linear partial differential operator with constant coefficients X P = P (∂) = cα ∂ α : C ∞ (Rn ) → C ∞ (Rn ) one has

|α|≤m

P eλ = P (λ) eλ

with

P (λ) =

X

cα λα .

|α|≤m

In other words, eλ is an eigenvector of all of the operators P simultaneously. We note in passing that eλ is also eigenvector of all translations Ta , for a ∈ Rn , with eigenvalue eλ (−a). The idea behind Fourier transformation is that the operators P can better be understood if the vectors on which they act can be written as linear combinations of eigenvectors; indeed, P then acts as a scalar multiplication on each summand, the scalar being the eigenvalue corresponding to the eigenvector in question. A particularity is that here the eigenvalues do not form a discrete set but an entire continuum, namely Cn . For this reason it is natural to replace the linear combinations by integrals over λ, where a weight function plays the role of the coefficients in the integration. The approximating Riemann sums (in the λ-space) then become finite linear combinations. As will appear later on, for a large class S 0 of distributions, every u ∈ S 0 can be written as an integral of this type over the set i Rn = { iξ ∈ Cn | ξ ∈ Rn }. The function eiξ is constant on the planes { x ∈ Rn | hx, ξi = c }, for c ∈ R, and behaves as a harmonic oscillation transversely to them; accordingly, this is called a

152

14 Fourier Transformation

plane harmonic wave. These are precisely the eλ , with λ ∈ Cn , that are bounded on Rn , a great help in obtaining estimates. For a Fourier series on R of the form X φ(x) = cn einωx n∈Z

the Fourier coefficients cn are given in terms of the function φ by Z 2π/ω ω e−inωx φ(x) dx, cn = 2π 0 the average of e−inω φ over one period of φ. If we cast off the shackles of periodic functions, the numerical factor and the interval of integration are no longer determined. For an integrable function u on Rn one now defines the Fourier transform Fu (often written as u b in the literature) as Z Fu(ξ) = e−ihx, ξi u(x) dx (ξ ∈ Rn ). (14.1) Rn

Several other conventions are current in the literature. Example 14.1. If 1[ a, b ] denotes the characteristic function of the interval [ a, b ], then F1[ a, b ] (ξ) = i

e−ibξ − e−iaξ . ξ

(14.2)

In particular, if −a = b = 1, the right-hand side takes the form eiξ − e−iξ 2 sin ξ = =: 2 sinc ξ, iξ ξ 

where the name sinc originates from sinus cardinalis. 1

20 Π

-20 Π

Fig. 14.1. Graph of sinc

The following result is known as the Riemann–Lebesgue Theorem. Theorem 14.2. If u is a Lebesgue integrable function on Rn , then Fu is continuous on Rn and Fu(ξ) converges to 0 if ξ ∈ Rn and kξk → ∞. In particular, Fu is bounded on Rn and satisfies Z |Fu(ξ)| ≤ |u(x)| dx (ξ ∈ Rn ). (14.3) Rn

14 Fourier Transformation

Proof. The estimate (14.3) follows from Z Z Z −ihx, ξi −ihx, ξi ≤ dx = u(x) e u(x) dx e

|u(x)| dx,

Rn

Rn

Rn

153

because the absolute value of the product equals the product of the absolute values, and the absolute value of e−ihx, ξi equals 1. Let C(0) be the space of continuous functions f on Rn with the property that f (ξ) → 0 as kξk → ∞. If v is the characteristic function of an n-dimensional rectangle { x ∈ Rn | aj ≤ xj ≤ bj , 1 ≤ j ≤ n } , we see from (14.2) that Fv ∈ C(0). If v is a finite linear combination of characteristic functions of rectangles, then again Fv ∈ C(0). Furthermore, it is known from the theory of Lebesgue integration that for every Lebesgue integrable function u on Rn and every  > 0 there R exists a finite linear combination v of characteristic functions of rectangles with Rn |u(x) − v(x)| dx <  2 . On account of (14.3), with u replaced by u − v, this implies that sup |Fu(ξ) − Fv(ξ)| < ξ∈Rn

 . 2

Because Fv ∈ C(0), there exists an R > 0 with the property |Fv(ξ)| < kξk > R. From this it follows that |Fu(ξ)| ≤ |Fu(ξ) − Fv(ξ)| + |Fv(ξ)|
R).

Thus we have shown that Fu(ξ) → 0 as kξk → ∞. The limit of a uniformly convergent sequence of continuous functions is continuous; see, for example, a text on analysis in one variable. Because Fu is the uniform limit of the above continuous functions Fv, we conclude that Fu is continuous. This completes the proof that Fu ∈ C(0).  If u has compact support, the definition of Fourier transformation can be extended to distributions and the Fourier transform is a complex-analytic function: Lemma 14.3. If u is an integrable function with compact support, then Fu(ξ) = u(e−iξ ). For every u ∈ E 0 (Rn ) the function ζ 7→ Fu(ζ) := u(e−iζ ) is complexanalytic on Cn . Proof. The first assertion is evident. Let u ∈ E 0 (Rn ). Then, by the continuity of ζ 7→ e−iζ : Cn → C ∞ (Rn ), it follows that Fu is continuous on Cn . The difference quotient 1 (e−i(ζ+t e(j)) − e−iζ ) (t ∈ C) t converges to the function x 7→ −ixj e−iζ (x) in C ∞ (Rn ) as t → 0. From this we see that Fu is complex-differentiable with respect to every variable, with partial derivative equal to the continuous function

154

14 Fourier Transformation

∂j Fu(ζ) = u(−ixj e−iζ )

(ζ ∈ Cn ).

(14.4)

On the strength of Cauchy’s integral formula (12.10), applied to each variable, it now follows that v = Fu is a complex-analytic function on Cn . That is, for every z ∈ Cn the Taylor series of v X ∂ α v(z) (ζ − z)α α! α converges for ζ in a suitable neighborhood in Cn of the point z, equaling v(ζ).



Example 14.4. The function sinc is complex-analytic on C. This can also be seen from the power series expansion X (−1)k z 2k sinc z = (z ∈ C). (2k + 1)! k∈Z≥0

For η ∈ Rn the function u(e−η ) = Fu(−iη) is said to be the Laplace transform of u. The classic case is that where n = 1, where u is a probability measure with support in R≥0 , and η > 0. For this reason the complex-analytic function Fu on Cn , for u ∈ E 0 (Rn ), is said to be the Fourier–Laplace transform of u. The analog of the Completeness Theorem for Fourier series is that “arbitrary” functions u can be written as continuous superpositions of the eiξ with weight factors c Fu(ξ), where c is a constant to be determined. Here we run into a problem, because Fu is not necessarily integrable over Rn for every integrable function u. For example, if 1[ a, b ] denotes the characteristic function of the interval [ a, b ], then, by (14.2), F1[ a, b ] is a real-analytic function on R that is not absolutely integrable over R. The following space S of test functions, which lies in between C0∞ (Rn ) and C ∞ (Rn ), proves to be very useful for constructing the theory. Definition 14.5. A function φ on Rn is said to be rapidly decreasing if, for every multi-index β, the function x 7→ xβ φ(x) is bounded on Rn . One defines S = S(Rn ) as the space of all φ ∈ C ∞ (Rn ) such that ∂ α φ is rapidly decreasing for every multi-index α. If φj is a sequence in S and φ ∈ S, then φj is said to converge to φ in S, notation limj→∞ φj = φ in S, if, for all multi-indices α and β, the sequence of functions xβ ∂ α φj converges uniformly on Rn to xβ ∂ α φ.  Two alternative formulations can be given: a function φ on Rn
is rapidly decreasing if and only if for every N > 0 one has φ(x) = O kxk−N as kxk → ∞. We have φ ∈ S if and only if φ ∈ C ∞ and P (x, ∂)φ is bounded on Rn , for every linear partial differential operator P (x, ∂) with polynomial coefficients. S = S(Rn ) is a linear subspace of C ∞ = C ∞ (Rn ). When provided with the norms φ 7→ kφkS(k,N ) := sup |xβ ∂ α φ(x)| (14.5) |α|≤k, |β|≤N, x∈Rn

it becomes a locally convex topological linear space in which convergence of sequences corresponds to that in Definition 14.5. Because it would be sufficient to

14 Fourier Transformation

155

consider the countable nondecreasing sequence of norms that is obtained by taking N = k, sequential continuity of linear operators on S is equivalent to continuity with respect to these norms, see Lemma 8.4. We have C0∞ ⊂ S ⊂ C ∞ . Furthermore, Definitions 2.12, 14.5 and 8.3 imply φj → φ in C0∞ ⇒ φj → φ in S ⇒ φj → φ in C ∞ . This is tantamount to the assertion that the identity, interpreted as a mapping from C0∞ to S, is continuous, and the same applies to the identity from S to C ∞ . Phrased differently, we have continuous inclusions in the following: C0∞ (Rn ) ⊂ S(Rn ) ⊂ C ∞ (Rn ).

(14.6)

By Lemma 8.2.(b), C0∞ (Rn ) is dense in C ∞ (Rn ) and therefore, a fortiori, S(Rn ) is dense in C ∞ (Rn ). This implies that a continuous linear operator on C ∞ (Rn ) is uniquely determined by its restriction to S(Rn ). The following lemma shows that, likewise, every continuous linear operator on S(Rn ) is uniquely determined by its restriction to C0∞ (Rn ). Lemma 14.6. Let χ ∈ C0∞ (Rn ) with χ(x) = 1 for kxk < 1. Write (∗ χ)(x) = χ( x). For every φ ∈ S(Rn ) and  > 0, one has that φ := (∗ χ) φ belongs to C0∞ (Rn ) and converges in S(Rn ) to φ as  ↓ 0. Proof. By application of Leibniz’ formula we find that the value xβ ∂ α (φ − φ)(x) equals (χ( x) − 1) xβ ∂ α φ(x) plus terms that can be estimated by a constant times |γ| |xβ ∂ α−γ φ(x)|, for |γ| 6= 0 and γ ≤ α. But the first term can be estimated by the supremum over the x with kxk ≥ 1/ of |xβ ∂ α φ(x)|. From this we see that lim↓0 φ → φ in S(Rn ).  Theorem 14.7. If P is a linear partial differential operator with polynomial coefficients, then P is a continuous linear mapping from S to S. Proof. For the norms defined in (14.5) and φ ∈ S we have the estimates k∂j φkS(k,N ) ≤ kφkS(k+1,N ) ,

kxj φkS(k,N ) ≤ kφkS(k,N +1) + k kφkS(k−1,N ) ,

where we have used that ∂ α (xj φ) = xj ∂ α φ + αj ∂ α−δj φ

(1 ≤ j ≤ n).

Here δj is the multi-index having a 1 in the j-th position and zeros everywhere else. From this follows the assertion for P = xj and for P = ∂j . The general assertion then follows by mathematical induction on the order of the operator and the degree of the polynomials in the coefficients.  For the next lemma the following notation proves useful:

156

14 Fourier Transformation

Dj :=

1 ∂j i

(1 ≤ j ≤ n).

The operator ~ Dj = ~i ∂j is known in quantum mechanics as the j-th momentum oph erator. In this expression ~ = 2π , where h is Planck’s constant, a quantity describing the atomic scale. Lemma 14.8. The Fourier transformation F : u 7→ Fu defines a continuous linear mapping from S to S. For every 1 ≤ j ≤ n, φ ∈ S and ξ ∈ Rn , one has F(Dj φ)(ξ) =

ξj Fφ(ξ),

(14.7)

F(xj φ)(ξ) = −Dj Fφ(ξ).

(14.8)

Proof. Let φ ∈ S. Then the estimate |φ(x)| ≤ c (1 + kxk)−N for all x ∈ Rn and some c > 0 and N > n, implies that φ is integrable over Rn . On account of the Riemann-Lebesgue Theorem 14.2, it follows that Fφ is a bounded continuous function on Rn . But xj φ is also integrable and Dj = 1i ∂ξj acting on the integrand in (14.1) yields −e−ihx, ξi xj φ(x). Therefore, the Theorem on differentiation under the integral sign implies that Fφ is differentiable with respect to ξj , with continuous derivative given by (14.8). By mathematical induction on k we find that Fφ ∈ C k (Rn ), for all k ∈ Z≥0 , and that all derivatives of Fφ are bounded. We obtain (14.7) when we apply integration by parts to the approximating integral over a large rectangle. For every pair of multi-indices α and β the function ξ β Dα (Fφ) = ξ β F((−x)α φ) = F(Dβ ((−x)α φ)) =: Fψ

(14.9)

is bounded, because ψ ∈ S according to Theorem 14.7. Here we have used the following self-explanatory notation: Dβ =

n Y

Dj βj = (−i)|β| ∂ β .

j=1

We conclude that F(S) ⊂ S. Finally we prove the continuity of F on the basis of (8.2). To this end we first apply (14.3) to ψ in (14.9), then use the estimate Z Z |ψ(x)| dx = (1 + kxk)−(n+1) (1 + kxk)n+1 |ψ(x)| dx Rn

Rn

≤ C sup (1 + kxk)n+1 |ψ(x)|, x∈Rn

where C =

R Rn

(1 + kxk)−(n+1) dx < ∞.



14 Fourier Transformation

157

Example 14.9. For every a ∈ C with Re a > 0, 2

ua (x) = e−a x

/2

(x ∈ R)

defines a function ua ∈ S(R); the proof is analogous to that of Lemma 2.7. Therefore, Fua ∈ S. Now Dua = iax ua , and we obtain by Fourier transformation of both sides, also taking into account (14.7) and (14.8), 1 DFua = i ξ Fua . a This yields Z Fua = c(a) u a1

with

2

e−ax

c(a) = (Fua )(0) =

/2

dx.

R

For a > 0, the change of variables x = (2/a)1/2 y leads to r 2π c(a) = , (14.10) a R √ 2 where we have also used R e−x dx = π, see (2.12). Now a 7→ c(a) is a complex-analytic function on H = { a ∈ C | Re a > 0 }; differentiation under the integral sign shows, for example, that c satisfies the Cauchy– Riemann equation. With the definition 1

1

1

a− 2 = |a|− 2 e− 2 i arg a

where

−

π π < arg a < , 2 2

the right-hand side of (14.10) is a complex-analytic function on H as well. The difference v between the two sides equals 0 on the positive real axis. Therefore, the Taylor series of v vanishes at every point on that axis, which implies that v = 0 on an open neighborhood in C of R>0 . On account of Lemma 2.6 we conclude that v = 0 on H. In other words, (14.10) holds for all a ∈ C with Re a > 0. The preceding results can be generalized to Rn by ua (x) = e−

Pn

Since e−ihx, ξi = we conclude

j=1

n Y

aj xj 2 /2

(a ∈ Cn , Re aj > 0, x ∈ Rn ).

e−ixj ξj

and

j=1

Fua = (2π)

n 2

ua (x) =

n Y 1 √ u a1 . aj j j=1

n Y

uaj (xj ),

j=1

The following lemma prepares the theorem that every φ ∈ S can be written as a continuous linear combination of the eiξ , with ξ ∈ Rn and c Fφ(ξ) as weight function, giving the “Fourier coefficients”.

158

14 Fourier Transformation

Lemma 14.10. Let L : S → S be a linear mapping that commutes with the position and momentum operators. That is, L(xj φ) = xj (Lφ) and L(Dj φ) = Dj (Lφ), for all 1 ≤ j ≤ n and every φ ∈ S. Then there exists a constant c ∈ C such that L = c I, with I the identity on S. Proof. We first show that (Lψ)(a) = 0 if ψ ∈ S, a ∈ Rn and ψ(a) = 0. Note that there exist ψ1 , . . . , ψn ∈ S such that ψ(x) =

n X

(xj − aj ) ψj (x)

(x ∈ Rn ).

(14.11)

j=1 d ψ(a + t (x − a)), Indeed, by writing ψ(x) − ψ(a) as the integral from 0 to 1 of dt ∞ e we obtain (14.11), with ψj ∈ C instead of the ψj ∈ S. Now choose χ ∈ C0∞ with χ = 1 on a neighborhood of a. Then xj − aj (1 ≤ j ≤ n) ψj (x) := χ(x) ψej (x) + (1 − χ(x)) ψ(x) kx − ak2 satisfy (14.11). This implies X  X n n n X (xj − aj ) ψj = Lψ = L L(xj ψj ) − L(aj ψj ) = xj Lψj − aj Lψj , j=1

j=1

j=1

from which we conclude that (Lψ)(a) = 0 if ψ ∈ S and ψ(a) = 0. Now let γ ∈ S be chosen such that γ is nowhere zero on Rn . For arbitrary φ ∈ S, define ψ = φ − φ(a) γ(a) γ ∈ S; then ψ has a zero at a, and so 0 = (Lψ)(a) = (Lφ)(a) −

φ(a) (Lγ)(a). γ(a)

Because this holds true for all a ∈ Rn , the conclusion is that Lφ = c φ

with

c :=

Lγ ∈ C ∞ (Rn ). γ

Finally, we see that c is constant, because application of this identity with φ successively replaced by γ and Dj γ implies γ Dj c + c Dj γ = Dj (c γ) = Dj (Lγ) = L(Dj γ) = c Dj γ, and so γ Dj c = 0; in other words, Dj c = 0, for every 1 ≤ j ≤ n.



Theorem 14.11. The Fourier transformation F : S(Rn ) → S(Rn ) is bijective, with inverse F −1 = (2π)−n S ◦ F = (2π)−n F ◦ S. This is expressed by the formula Z 1 φ(x) = eihx, ξi Fφ(ξ) dξ (φ ∈ S(Rn ), x ∈ Rn ). (14.12) (2π)n Rn

14 Fourier Transformation

159

Proof. From (14.8) and (14.7) we deduce F ◦ F ◦ xj = F ◦ (−Dj ) ◦ F = −xj ◦ F ◦ F, F ◦ F ◦ Dj = F ◦ xj ◦ F = −Dj ◦ F ◦ F. For the reflection Sφ(x) = φ(−x) we have S ◦ (−xj ) = xj ◦ S

S ◦ (−Dj ) = Dj ◦ S;

and

thus we conclude that L := S ◦ F ◦ F commutes with the position and momentum operators. That is, L equals multiplication by a constant c ∈ C. 2 To determine c, we can apply L to φ(x) = e−kxk /2 , for example. Then we see from Example 14.9, with aj = 1, that Fφ = (2π)n/2 φ. Therefore L(φ) = (2π)n φ, that is, c = (2π)n . We can now conclude that ((2π)−n S ◦ F) ◦ F = I. This is formula (14.12); in other words, F has left-inverse (2π)−n S ◦ F. It then follows that F is injective. A more symmetric form is obtained if we left-multiply by (2π)n S: F ◦ F = (2π)n S. If we now right-multiply this in turn by (2π)−n S, we see that (2π)−n F ◦ S is a right-inverse of F; its existence implies that F is surjective. The end result is that F is bijective, with inverse as per the theorem.  Identity (14.12) is known as the Fourier inversion formula. In the case n = 1 the constant in the formula can also be determined by means of Fourier series, see Example 15.7 below. In the following theorem we use the notations Z hφ, ψi = φ(x) ψ(x) dx, Rn Z φ(x) ψ(x) dx. (φ, ψ) = hφ, ψi =

(14.13)

Rn

Here the overline denotes complex conjugation; thus, (14.13) defines a Hermitian inner product in the space of functions. These expressions are meaningful if φ ψ is integrable. Theorem 14.12. For all φ, ψ ∈ S we have hFφ, ψi = hφ, Fψi, (φ, ψ) = (2π)−n (Fφ, Fψ) (Parseval’s formula), φ ∗ ψ ∈ S and F(φ ∗ ψ) = Fφ Fψ, F(φ ψ) = (2π)−n Fφ ∗ Fψ.

(14.14) (14.15) (14.16) (14.17)

160

14 Fourier Transformation

Proof. Both sides in (14.14) are equal to the double integral Z Z φ(x) e−ihx,ξi ψ(ξ) dx dξ, Rn

Rn

where we also use hx, ξi = hξ, xi. The proof of (14.15) begins with the observation that the complex conjugate of e−ihx, ξi equals eihx, ξi , which implies that Fψ = S ◦ Fψ. Combining (14.14) and Theorem 14.11 we then obtain hFφ, Fψi = hFφ, S ◦ Fψi = hφ, F ◦ S ◦ Fψi = (2π)n hφ, ψi. The convolution φ ∗ ψ is bounded; this merely requires one of the factors to be integrable and the other one to be bounded. Using xβ = (x − y + y)β =

X β
(x − y)β−γ y γ γ

γ≤β

we see that xβ (φ ∗ ψ) =

X β
(xβ−γ φ) ∗ (xγ ψ). γ

γ≤β

Combining this with (11.4), we find that xβ ∂ α (φ∗ψ) is bounded for all multi-indices α and β, which implies φ ∗ ψ ∈ S. The formula F(φ ∗ ψ) = Fφ Fψ is obtained by substituting e−ihx, ξi = e−ihx−y, ξi e−ihy,ξi in the double integral that represents the left-hand side. For (14.17) we replace φ and ψ in (14.16) by Fφ and Fψ, respectively. Further, we apply F −1 to both sides and use F 2 = (2π)n S. Thus we get Fφ ∗ Fψ = F −1 (F 2 φ F 2 ψ) = (2π)n F ◦ S(Sφ Sψ) = (2π)n F(φ ψ).  Definition 14.13. A tempered distribution on Rn is a (sequentially) continuous linear form on S = S(Rn ). The space of tempered distributions is denoted by S 0 or S 0 (Rn ). The term originates from the French “distribution temp´er´ee”. One says that, for uj and u ∈ S 0 , lim uj = u in S 0 j→∞

if limj→∞ uj (φ) = u(φ) in C, for every φ ∈ S (compare with Definitions 5.1 and 8.1).  Restriction of u ∈ E 0 = E 0 (Rn ) to S yields a ρ u ∈ S 0 , restriction of u ∈ S 0 to gives a ρ u ∈ D0 = D0 (Rn ). The restriction mappings ρ from E 0 to S 0 and from S to D0 are continuous and linear. These mappings are also injective, on account C0∞ 0

14 Fourier Transformation

161

of Lemmas 8.2.(c) and 14.6, respectively. In both cases ρ u is identified with u; this leads to the continuous inclusions (compare with (14.6)) E 0 (Rn ) ⊂ S 0 (Rn ) ⊂ D0 (Rn ). We further note that Lemma 14.6 implies that for every u ∈ S 0 the (∗ χ) u ∈ E 0 converge in S 0 to u as  ↓ 0. Consequently, E 0 is dense in S 0 . Because C0∞ is dense in E 0 , see Corollary 11.7, C0∞ is dense even in S 0 , in the sense that for every u ∈ S 0 there exists a sequence uj ∈ C0∞ such that uj → u in S 0 as j → ∞. For example, take uj = φ ∗ ((∗ χ) u) , with φ as in Lemma 11.6, χ as in Lemma 14.6 and  = 1/j, and apply (11.17). Consequently, every continuous operator on S 0 is uniquely determined by its restriction to C0∞ . For u ∈ D0 , one has that u ∈ S 0 if and only if there exist a constant c > 0 and an S norm n such that |u(φ)| ≤ c n(φ), for all φ ∈ C0∞ . The continuous extension of u to S is obtained by taking u(φ) = lim↓0 u(φ ), for φ ∈ S and φ as in Lemma 14.6. Example 14.14. Define Z u(φ) =

φ(x) ex cos ex dx

(φ ∈ S(R)).

R

Note that the integral converges, because integration by parts and limx→±∞ |φ(x)| = 0 imply Z Z |u(φ)| = φ0 (x) sin ex dx ≤ |φ0 (x)| dx R Z R 1 dx sup (1 + x2 )|φ(j) (x)|. ≤ 2 0≤j≤1, x∈R R 1+x Actually, this estimate shows that u ∈ S 0 (R). Observe that there exists no polynomial p on R such that |ex cos ex | ≤ |p(x)|, for all x ∈ R.  A direct consequence of Theorem 14.7 is: Theorem 14.15. If P is a linear partial differential operator with polynomial coefficients, P is a continuous linear mapping from S 0 (Rn ) to S 0 (Rn ). Example 14.16. For u integrable over Rn , we are going to prove that u ∈ S 0 . Denoting the space of Lebesgue integrable functions on Rn by L1 = L1 (Rn ) as usual, we have the continuous inclusions C0∞ (Rn ) ⊂ S(Rn ) ⊂ L1 (Rn ) ⊂ S 0 (Rn ). More generally: if 1 ≤ p < ∞, then Lp is defined as the space of the measurable functions u on Rn , modulo functions that are equal to zero almost everywhere, such that |u|p is integrable. In Lp

162

14 Fourier Transformation

kukLp :=

Z

1/p |u(x)|p dx

Rn

defines a norm; L is complete with respect to this norm. L∞ denotes the space of the essentially bounded measurable functions on Rn (modulo functions that are equal to zero almost everywhere); this is a Banach space with respect to the norm p

kukL∞ := ess sup |u|. Here ess sup f , the “essential supremum” of a function f , is defined as the infimum of all sup g, if g runs through the set of all functions that are equal to f almost everywhere. Let q be the real number such that p1 + 1q = 1, on the understanding that q = ∞ if p = 1 and q = 1 if p = ∞. H¨older’s inequality (see, for example, [8, Exercise 6.73]) then implies, for u ∈ Lp and φ ∈ S, |u(φ)| ≤ kukLp kφkLq . Next, choose N ∈ Z≥0 such that N q > n, with N = 0 if q = ∞, that is, p = 1. If we then write φ(x) = (1 + kxk)−N (1 + kxk)N φ(x) and use the integrability of (1 + kxk)−qN , we conclude that there exists a constant cN > 0 with the property that, for every u ∈ Lp and φ ∈ S, |u(φ)| ≤ kukLp cN kφkS(0,N ) . This proves that we have the following continuous inclusions: C0∞ (Rn ) ⊂ S(Rn ) ⊂ Lp (Rn ) ⊂ S 0 (Rn ). In other words, convergence with respect to the Lp norm implies convergence in S 0 . Combining this with Theorem 14.15, we find that the space of tempered distributions is really quite large: for every 1 ≤ p ≤ ∞ we can start with u ∈ Lp , then apply to it an arbitrary linear partial differential operator P with polynomial coefficients, and conclude that the result P u is a tempered distribution.  Definition 14.17. If u is a tempered distribution in Rn , its Fourier transform Fu ∈ S 0 (Rn ) is defined by Fu(φ) = u(Fφ)

(φ ∈ S(Rn )). 

The slightly frivolous notation (3.3) is continued to Z Fu(ξ) = e−ihx, ξi u(x) dx

(14.18)

Rn

where we are not allowed to give ξ a value, as the expression is only meaningful after integration over ξ, with a test function as weight function.

14 Fourier Transformation

163

Theorem 14.18. For every u ∈ S 0 (Rn ) we have Fu ∈ S 0 (Rn ). The mapping F : u 7→ Fu is a continuous linear mapping from S 0 (Rn ) to S 0 (Rn ) that is an common extension of the Fourier transformation on S(Rn ), on the space L1 (Rn ) of Lebesgue integrable functions, or on the space E 0 (Rn ) of distributions with compact support, respectively. If u ∈ E 0 (Rn ), then Fu ∈ S 0 (Rn ) corresponds to the analytic function Fu from Lemma 14.3, and therefore certainly Fu ∈ C ∞ (Rn ). For every u ∈ S 0 (Rn ) and 1 ≤ j ≤ n we have F(Dj u) = ξj Fu

and

F(xj u) = −Dj Fu.

(14.19)

Finally, F is bijective from S 0 (Rn ) to S 0 (Rn ), with inverse equal to (2π)−n S ◦F = (2π)−n F ◦ S. Proof. These assertions follow from Lemma 14.8 and Theorem 14.11, where we recognize F : S 0 → S 0 as the transpose of F : S → S. The assertion that F is an extension of the Fourier transformation on S follows from (14.14). On L1 , F corresponds to the Fourier transformation, because the Fourier transformation on L1 is also a continuous extension of that on S. Now let u ∈ E 0 (Rn ). As we have seen above, E 0 (Rn ) ⊂ S 0 (Rn ); therefore u ∈ S 0 (Rn ), and thus Fu ∈ S 0 (Rn ) is defined. In order to prove that this distributional Fourier transform Fu is equal to the analytic function u b = Fu from Lemma 14.3, we write, for every φ ∈ C0∞ (Rn ), Z  Z (Fu)(φ) = u(Fφ) = u e−iξ φ(ξ) dξ = u(e−iξ ) φ(ξ) dξ Rn Rn Z = u b(ξ) φ(ξ) dξ = (test u b)(φ), Rn

where for the third equality we have used Lemma 11.4 with A(ξ) = φ(ξ) e−iξ as well as the linearity of u. So as to satisfy condition (b) from that lemma, i.e. that the support of A(ξ) be contained in a compact set for all ξ ∈ Rn , we have, in fact, to perform the calculation above with u replaced by χ u, where χ ∈ C0∞ (Rn ) is a test function that equals 1 on a neighborhood of supp u. Thus, we have shown that Fu = test u b. The formulas in (14.19) follow from those in S by continuous extension or by transposition. The same holds for the formula F ◦ F = (2π)n S in S 0 .  Example 14.19. We have, for all a ∈ Rn , Ta ◦ F = F ◦ eia

and

F ◦ Ta = e−ia ◦ F

in S 0 (Rn ).

(14.20)

Indeed, the substitution of variables x = y−a shows, for all φ ∈ S(Rn ) and ξ ∈ Rn , Z Z −ihx, ξi iha, ξi (F ◦ T−a )φ(ξ) = e φ(x + a) dx = e e−ihy, ξi φ(y) dy Rn

= (eia ◦ F)φ(ξ).

Rn

164

14 Fourier Transformation

Hence, for any u ∈ S 0 (Rn ), (Ta ◦ F)u(φ) = u(F ◦ T−a φ) = u(eia Fφ) = (F ◦ eia )u(φ). The second identity in (14.20) follows from the first by means of transposition. Example 14.20. Because Fδ(φ) = δ(Fφ) = Fφ(0) = 1(φ), for every φ ∈ S(Rn ), we see that Fδ = 1 in S 0 (Rn ). (14.21) Since S1 = 1, we find on account of Theorem 14.18 F1 = (2π)n δ. In the notation (14.18) this looks rather spectacular: Z 1 eihx,ξi dξ. δ(x) = (2π)n Rn More generally, if P is a polynomial function in n variables, then P (x) = P (x) 1 for all x ∈ Rn , and so (14.19) leads to F(P (∂) δ) = P (i ξ)

and

FP = (2π)n P (i ∂) δ.

In particular, F(∆ δ) = −k · k2 . Furthermore, application of (14.20) implies, for every a ∈ Rn , S ◦ F δa = S ◦ F ◦ Ta δ = S ◦ e−ia ◦ F δ = Se−ia = eia .

(14.22) 

Example 14.21. According to (1.3) we obtain, for PV x1 ∈ D0 (R) and any φ ∈ Z C0∞ (R), 1 log |x| PV (φ) = − (1 + x2 )φ0 (x) dx, x 1 + x2 R and so Z log |x| 1 dx sup (1 + x2 )|φ0 (x)|. PV (φ) ≤ 2 x 1 + x x∈R R This estimate proves that PV x1 ∈ S 0 (R). On account of Example 9.2 we have x PV x1 = 1 in D0 (R), hence (14.19) implies   1 1 i ∂x F PV = F x PV = F1 = 2π δ x x in view of the preceding example. Example 4.2 and Theorem 4.3 then give F PV x1 = −2πi H + c, for some c ∈ C, where H denotes the Heaviside function, which is a tempered distribution. Furthermore, PV x1 is an odd distribution and so is F PV x1 in view of SF = FS; this implies −2πi + c = −c, which leads to c = πi. Therefore, with sgn denoting the sign function on R,

14 Fourier Transformation

F PV

1 = −πi sgn; x

1 F sgn = −2i PV , ξ

and so

165

(14.23)

because 2π PV x1 = SFF PV x1 = −πi SF sgn = πi F sgn. For any φ ∈ S(R), the latter identity in (14.23) entails Z Z Z φ(x) dx. φ(x) sin ξx dx dξ = lim ↓0 R\[ −,  ] x R>0 R Indeed, Z

Z

F sgn(φ) = sgn(Fφ) = sgn(ξ) e−ixξ φ(x) dx dξ R R Z Z Z Z iξx =− e φ(x) dx dξ + e−iξx φ(x) dx dξ R>0 R R>0 R Z Z = −2i φ(x) sin ξx dx dξ. R>0

Finally, note that H =

1 2

R

+

1 2

sgn. Applying (14.23) we now see

1 FH = π δ − i PV . x

(14.24)

Example 14.22. According to Example 4.2 one has δ = ∂ H, and on account of (14.21) this implies 1 = Fδ = F(iDH) = iξ FH. But this does not uniquely determine FH, because for every constant c ∈ C, the equation above is also satisfied by FH + c δ. (All solution are described in Problem 9.5.) For every  > 0, however, H (x) = e− x H(x) is a Lebesgue integrable function on R, and therefore FH is a continuous function. Furthermore, ∂ H = − H + δ, and so (iξ + ) FH = 1;

that is,

(FH )(ξ) =

1 , iξ + 

because FH is continuous. Since lim↓0 H = H in S 0 we now conclude, on the strength of Theorem 14.18, FH = lim ↓0

1 1 1 = iξ +  i ξ − i0

(14.25)

with convergence in S 0 (R). Combination of (14.24) and (14.25) now leads to the following Plemelj–Sokhotsky jump relations (compare with Problem 1.1): 1 1 ± πi δ = PV . x ± i0 x

166

14 Fourier Transformation

Example 14.23. Consider u ∈ E 0 (Rn ) satisfying u(x 7→ xα ) = 0, for every multiindex α. From (14.4) we derive ∂ α Fu(ζ) = (−i)|α| u(x 7→ xα e−iζ ). Since Fu is complex-analytic on Cn on account of Lemma 14.3, we obtain by power series expansion about 0 X (−i)|α| u(x 7→ xα ) ζ α = 0 (ζ ∈ Cn ). Fu(ζ) = α! α In other words, Fu = 0, but then Theorem 14.18 implies u = 0. The Hahn–Banach Theorem is a basic result in functional analysis; using one of its corollaries, see [22, Theorem 3.5], one now deduces that the linear subspace in C ∞ (Rn ) consisting of the polynomial functions is dense in C ∞ (Rn ). In other words, given a function φ ∈ C ∞ (Rn ), there exists a sequence of polynomial functions pj on Rn such that, for every multi-index α and compact set K in Rn , the sequence ∂ α pj converges to ∂ α φ uniformly on K as j → ∞. This is Weierstrass’ Approximation Theorem, see [8, Exercise 6.103] for a proof along classical lines; conversely, the vanishing of u is a direct consequence of this theorem.  n Next, we generalize Parseval’s formula (14.15). A function R f on R 2 is said to be square-integrable if f is a measurable function such that Rn |f (x)| dx < ∞. The space of square-integrable functions is denoted by L2 = L2 (Rn ). According to the Cauchy–Schwarz inequality the so-called L2 inner product (φ, ψ) from (14.13) is well-defined for all φ and ψ ∈ L2 . The corresponding norm Z 1/2 1/2 kf k = kf kL2 := (f, f ) = |f (x)|2 dx Rn

is said to be the L2 norm. L2 (Rn ) is complete with respect to this norm. In other words: L2 (Rn ) is a Hilbert space, an inner product space that is complete with respect to the corresponding norm. Additionally, the space of continuous functions with compact support is dense in L2 (Rn ); also refer to the proof of Theorem 14.2. These facts are known from the theory of Lebesgue integration; they also hold when, for example, the integration over Rn is replaced by integration over an open subset X of Rn . See, for example, [25, §6.3]. From Example 14.16 we obtain the continuous inclusion L2 ⊂ S 0 . For a general Hilbert space H, a linear mapping U : H → H is said to be a unitary isomorphism if U (H) = H and (U f, U g) = (f, g)

(f, g ∈ H).

This implies that U is bijective and that U −1 from H to H is unitary as well. (Observe that in an infinite-dimensional Hilbert space an injective linear mapping is not necessarily surjective.) Theorem 14.24. If u belongs to L2 (Rn ), this is also true of its Fourier transform Fu. Parseval’s formula (14.15) applies to all φ and ψ ∈ L2 (Rn ). It follows that n the restriction of Fe := (2π)− 2 F to L2 (Rn ) defines a unitary isomorphism from L2 (Rn ) to L2 (Rn ).

14 Fourier Transformation

167

Proof. As in the proof of Theorem 14.2, we obtain a sequence φj ∈ C0∞ (Rn ) with limj→∞ ku − φj k = 0. On account of (14.15) this leads to e j − Fφ e k k = kφj − φk k ≤ kφj − uk + ku − φk k → 0, kFφ e j therefore form a Cauchy sequence in L2 and in view of the for j, k → ∞. The Fφ e j k = 0. This implies completeness of L2 there exists a v ∈ L2 with limj→∞ kv − Fφ 0 0 e j → v in S . Because, in addition, φj → u in S , we find Fφ e j → Fu e in that Fφ 0 e S , and on account of the uniqueness of limits we conclude that Fu = v. Because e ∈ L2 . v ∈ L2 , the conclusion is that Fu Combining the estimates e ≤ kFu e − Fφ e j k + kFφ e jk kFuk

and

e j k = kφj k ≤ kuk + ku − φj k kFφ

and taking the limit as j → ∞, we also find e ≤ kuk. kFuk This implies that Fe is continuous with respect to the L2 norm. It follows that the right-hand side in (14.15) is continuous with respect to both variables φ and ψ in L2 , as is the left-hand side. If we now approximate φ and ψ by φj ∈ C0∞ and ψj ∈ C0∞ , e j , Fψ e j ), we obtain respectively, relative to the L2 norm and use (φj , ψj ) = (Fφ (14.15) for φ and ψ in L2 . e According to Theorem 14.11, we have Finally we prove the surjectivity of F. e F(S) = S, while S is dense in L2 . It follows that the image of L2 under Fe is dense in L2 , and because Fe is unitary the image is closed; therefore, the image is equal to e j → v in L2 , then kFu e j − Fu e k k = kuj − uk k shows that the L2 (Rn ). Indeed, if Fu 2 uj form a Cauchy sequence in L . Hence there exists u ∈ L2 with uj → u in L2 , e j → Fu e in L2 . But this leads to v = Fu. e and therefore Fu  We give an extension of (14.16) to distributions. Note that the product Fu Fv is well-defined as an element of D0 (Rn ) if u ∈ S 0 (Rn ) and v ∈ E 0 (Rn ), because then Fv ∈ C ∞ (Rn ) according to Theorem 14.18. Theorem 14.25. If u ∈ S 0 (Rn ) and v ∈ E 0 (Rn ), then u ∗ v ∈ S 0 (Rn ) while F(u ∗ v) = Fu Fv. In particular, Fu Fv ∈ S 0 (Rn ). Proof. On the strength of (11.20), we recognize that to prove u ∗ v ∈ S 0 we only have to demonstrate that φ 7→ Sv ∗ φ : C0∞ → C ∞ possesses an extension to a continuous mapping from S to S. This means that, for every pair of multi-indices α and β, we have to estimate the number xβ ∂ α (Sv ∗ φ)(x) = xβ (Sv ∗ ∂ α φ)(x) uniformly in x by an S norm of φ; this only requires a uniform estimate by an S norm of ψ = ∂ α φ. In view of (11.1),

168

14 Fourier Transformation

(Sv ∗ ψ)(x) = Sv(Tx ◦ Sψ) = v(S ◦ Tx ◦ Sψ) = v(T−x ψ). Using the continuity of v on C ∞ (Rn ), we deduce from (8.4) that it is sufficient to show that, for every multi-index γ and compact K ⊂ Rn , the number
xβ ∂y γ (ψ(y + x)) = xβ ∂ γ ψ (y + x) can be estimated uniformly in x ∈ Rn and in y ∈ K by an S norm of ψ. The change of variables x = z − y and the uniform estimate |(z − y)β | ≤ C (1 + kzk)|β| , for all z ∈ Rn and y ∈ K, then yield the desired result. We now prove F(u ∗ v) = Fu Fv by continuous extension of this formula for u and v ∈ C0∞ . We write, for φ ∈ C0∞ , F(u ∗ v)(φ) = (u ∗ v)(Fφ) = u(Sv ∗ Fφ). Because Fφ ∈ S, we obtain from the above that Sv ∗ Fφ ∈ S. Following Definition 14.13 we observed that for every u ∈ S 0 there exists a sequence uj ∈ C0∞ with the property that uj → u in S 0 as j → ∞. From this we now conclude that F(uj ∗ v) → F(u ∗ v) in D0 , because uj → u in S 0 as j → ∞. But we also have Fv φ ∈ C0∞ , and therefore we deduce from (Fu Fv)(φ) = Fu(Fv φ) = u(F(Fv φ)) that one also has Fuj Fv → Fu Fv in D0 . Thus we conclude that F(u ∗ v) = Fu Fv for all u ∈ S 0 , if it holds for all u ∈ C0∞ , for given v ∈ E 0 . The commutativity of the convolution in turn implies that F(u ∗ v) = Fu Fv for all v ∈ E 0 , because it holds for all v ∈ C0∞ , for given u ∈ C0∞ .  In specific examples of homogeneous distributions it is often straightforward that the distributions are tempered. More generally one has H¨ormander’s Theorem [16, Thm. 7.1.18]: Theorem 14.26. If u ∈ D0 (Rn ) and the restriction of u to Rn \{0} is homogeneous (of arbitrary degree a ∈ C), then u ∈ S 0 (Rn ). Proof. We begin with a ρ ∈ C0∞ ( ] 0, ∞ [ ) such that ρ ≥ 0 and ρ not identically zero. Through multiplication by a suitable positive number we can ensure that R ρ(1/s) s−1 ds = 1. Let r > 0. By the change of variables s = t/r we find Z r
dt =1 (r > 0). ρ t t R>0 Z ∞ Now define kxk
dt ψ(x) := 1 − ρ . t t 1 If x is bounded, the interval of integration can be replaced by a bounded interval, and the Theorem on differentiation under the integral sign gives ψ ∈ C ∞ (Rn ). On

14 Fourier Transformation

169

the other hand, if kxk is large, the interval of integration can be replaced by ] 0, ∞ [, which implies that ψ(x) = 0. In conclusion, ψ ∈ C0∞ (Rn ). Using the notations χ(x) := ρ(kxk) and t : x 7→ t x, we now find, for every φ ∈ C0∞ (Rn ), Z ∞ Z ∞  ∗ −1 u(φ) = u(ψ φ) + u (1/t) χ t dt φ = u(ψ φ) + u ((1/t)∗ χ φ) t−1 dt 1 1 Z ∞ = u(ψ φ) + u (χ t∗ φ) ta+n−1 dt, 1

where we have used (1/t)∗ χ φ = (1/t)∗ (χ t∗ φ) and (1/t)∗ = tn t∗ (see Theorem 10.6) and t∗ u = ta u on Rn \ {0}. Because both ψ u and χ u are distributions with compact support, we deduce the existence of a constant C > 0, a k ∈ Z≥0 and compact subsets K of Rn and L of Rn \ {0}, such that Z ∞ |u(φ)| ≤ C kφkC k , K + C kt∗ φkC k , L tRe a+n−1 dt, 1

see (8.4). Now kt∗ φkC k , L = tk kφkC k , t L and for every N there exists a constant C 0 such that kφkC k , t L ≤ C 0 t−N kφkS(k,N ) ,

(14.26)

with the notation of (14.5). Indeed, it is sufficient to prove the estimate for k = 0. Because 0 ∈ / L, there is a constant m > 0 such that m ≤ max1≤j≤n |xj | for all x ∈ L. Consequently, for x ∈ t L there exists a j such that xj ≥ t m, and therefore −N −N |φ(x)| = |xN ≤ |xN ≤ m−N t−N kφkS(k,N ) . j ||φ(x)||xj | j ||φ(x)|(tm)

Choosing N > k + Re a + n in (14.26), we find a constant C 00 with the property |u(φ)| ≤ C 00 kφkS(k,N )

(φ ∈ C0∞ (Rn )).

This implies that u can be extended to a continuous linear form on S (Rn ); on account of Lemma 14.6 this extension is uniquely determined. 

Problems Problem 14.1. (Characterization of exponentials). Suppose u ∈ D0 (Rn ) and ∂j u = λj u, for some λ ∈ C and all 1 ≤ j ≤ n. Demonstrate that there exists a constant c such that u = c eλ .

170

14 Fourier Transformation

Problem Q 14.2. Assume that φ1 , . . . , φn are integrable functions on R. Prove that n φ(x) := j=1 φj (xj ), for x ∈ Rn , defines an integrable function on Rn with the n property Y Fφ(ξ) = Fφj (ξj ). j=1

Problem 14.3. (Constant in Fourier inversion formula). Prove that, in the context of Problem 14.2, one has φ ∈ S(Rn ) if φj ∈ S(R), for every j = 1, . . . , n. Use this to prove thatR cn = (c1 )n , if cn is the constant in (14.12). Compute c1 by calculating L(φ)(0) = Fφ(ξ) dξ for φ(x) = e−|x| . A problem with this method is that φ does not belong to S(R). Argue that this can be overcome through approximating φ by a suitable sequence of functions in S(R). Problem 14.4. Suppose that u and v := Fu are Lebesgue integrable on Rn . Prove that u equals the continuous function (2π)−n S ◦ Fv almost everywhere on Rn . Problem 14.5. (Eigenvalues of Fourier transformation). What are the possible eigenvalues of the Fourier transformation acting on S(R)? And when acting on L2 (R)? For three eigenvalues, try to find corresponding eigenfunctions. Problem 14.6. Consider the following functions on R: 2

a(x) = e−x

+2x

,

b(x) = e−x H(x),

c(x) = e−|x| ,

d(x) =

1 . 1 + x2

(i) For each of these, sketch its graph. (ii) Verify that these functions belong to L1 = L1 (R). Which of them belong to S = S(R) and which to L2 = L2 (R)? (iii) Calculate the Fourier transforms of these functions. Sketch the graphs of the Fourier transforms of the functions b, c and d; make separate sketches of the real and imaginary parts. Hint: in the case of the function d, both Laplace’s integral and complex analysis may be used for the calculation. For the latter method of calculating Fd(ξ), the choice of contour depends on whether ξ > 0 or ξ < 0. (iv) For each of the Fourier transforms found, determine whether it is a function in S, in L1 , or in L2 . (v) Suppose that, for functions that belong to L1 ∩ L2 , the Fourier transform in the sense of L2 is given by the Fourier integral (almost everywhere). Then calculate the Fourier transform of the function d once again. Problem 14.7. Differentiate x 7→ e−|x| twice and use this to calculate the Fourier transform of e−|x| . Problem 14.8. Let N be the space of complex-analytic solutions u on Cn of the equation P u = 0. Prove that N is infinite-dimensional whenever n > 1 and the linear partial differential operator P with constant coefficients is not of order 0.

14 Fourier Transformation

171

Problem 14.9. Let u be a locally integrable function on Rn such that there exists an N ∈ R with u(x) = O(kxkN ) as kxk → ∞. Prove that u ∈ S 0 . Problem 14.10. Prove that every tempered distribution is of finite order. Problem 14.11. (Characterization of tempered exponentials). Suppose u ∈ S 0 (Rn ) and φ ∈ S(Rn ). Prove that there exist constants C and N such that |u(Ta φ)| ≤ C (1 + kak)N

(a ∈ Rn ).

Now let ξ ∈ Cn . Prove that the function eiξ : x 7→ eihx,ξi is a tempered distribution on Rn if and only if ξ ∈ Rn . p

Problem 14.12. For what combinations of p and q ∈ Z>0 does f (x) = ex ei e define a tempered distribution on R?

xq

Problem 14.13. Calculate the Fourier transform of the following distributions: (i) δa ∈ E 0 (Rn ), for a ∈ Rn . (ii) A dipole in Rn . (iii) e−ax H on R, for a > 0. (iv) x 7→ e−a|x| for a constant a > 0. (v) x 7→ 1/(x2 + a2 ) for a constant a > 0. Discuss possible complex-analytic extensions of the Fourier transforms to open subsets of C. Problem 14.14. Let ξ ∈ Rn be given. Determine the Fourier transform of x 7→ eihx, ξi . Problem 14.15. Determine the Fourier transforms of the functions: R → R given by the following formulas: (i) cos, sin, x 7→ x sin x, cos2 , cosk for k ∈ Z>0 . (ii) sinc. (iii) x 7→ x H(x) (use Problem 13.4), | · |, sin | · |. Note that only in case (ii) the Fourier transform is a function. In the notation of Problem 13.3 deduce from part (iii) that F| · |−2 = −π | · |. Qm Problem 14.16. Let P (D) = j=1 (D − λj ) be a differential operator on R, where λj ∈ C for 1 ≤ j ≤ n. Prove that all fundamental solutions of P (D) are tempered distributions if and only if λj ∈ R for all 1 ≤ j ≤ n. Now let P (D) be a linear partial differential operator with constant coefficients on Rn , with n > 1. Prove that there are always solutions u ∈ D0 (Rn ) of P (D)u = 0 such that u ∈ / S 0 (Rn ). Conclude that not all fundamental solutions of P (D) are tempered distributions. Problem 14.17. Prove the first identity in (14.19) by means of (10.9) and the second one in (14.20).

172

14 Fourier Transformation

Problem 14.18. (Fourier transform of homogeneous distribution). Let u ∈ S 0 (Rn ) be homogeneous of degree a ∈ C. Show that Fu is homogeneous of degree −a − n. Problem 14.19. Let A be an invertible linear transformation of Rn . Write the inverse −1 of the transpose of A as B = t A . Show that we have the following identity in S 0 (Rn ): F ◦ A∗ = | det B| B ∗ ◦ F. Verify that u ∈ S 0 (Rn ) is invariant under all orthogonal transformations if and only if Fu has that property. Problem 14.20. Consider the distributions χa+ ∈ D0 (R), defined in Chap. 13. Prove that χa+ ∈ S 0 (R) for every a ∈ C, and that, for every φ ∈ S(R), a 7→ χa+ (φ) is a complex-analytic function on C. Gather as much information as possible on the Fourier transforms F(χa+ ). Problem 14.21. Try to establish whether the functions from Problem 14.13.(iii), (iv) and (v) converge in S 0 , as a ↓ 0. If so, determine the Fourier transforms of the limits. Calculate F −1 H. 2

Problem 14.22. Consider ua (x) = e−ax /2 as in Example 14.9, but now for Re a ≥ 0. Let t ∈ R. Prove that uit ∈ S 0 and that lima→it, Re a>0 ua = uit in S 0 . Calculate Fuit . Problem 14.23. For what λ ∈ C does the difference equation T1 u = λ u have a solution u ∈ S 0 (R) with u 6= 0? Describe the solution space for every such λ. The same questions for the difference equation 12 (T1 u + T−1 u) = λ u. Problem 14.24. Let u ∈ S 0 (Rn ) with u 6= 0 be a solution of the differential equation ∆u = λ u, with λ ∈ C. Prove that λ ≤ 0 and that u has an extension to an entirely analytic function on Cn . Show that u is a polynomial if λ = 0. Discuss the rotationally symmetric solutions u, for every λ ≤ 0. Problem 14.25. Let a > 0. Compute F sinc and show Z Z π sinc a x dx = sinc2 a x dx = . a R R Determine F sinc2 and obtain the value of the second integral once more. Problem 14.26. Use Parseval’s formula to calculate the integrals Z Z x2 1 dx and dx. 2 2 2 2 R (x + 1) R (x + 1) Problem 14.27. Let µ be a probability measure on Rn . Prove that µ ∈ S 0 and that Fµ is a continuous function on Rn , with |Fµ(ξ)| ≤ 1 for all ξ ∈ Rn and Fµ(0) = 1. Hint: take χr ∈ C0∞ , 0 ≤ χr ≤ 1 and χr (x) = 1 whenever kxk ≤ r. Verify that F(χr µ) converges uniformly on Rn as r → ∞. Also prove F(µ ∗ ν) = Fµ Fν if µ and ν are probability measures.

14 Fourier Transformation

173

Problem 14.28. (Hilbert transformation). The Hilbert transform Hφ ∈ C ∞ (R) of φ ∈ C0∞ (R) is defined as Z 1 1 1 φ(x − y) φ ∗ PV (x) = lim (Hφ)(x) = dy. ↓0 π y π |y|> y Show that Hφ ∈ S 0 (R) and that (F ◦ H)φ(ξ) = −i sgn(ξ) Fφ(ξ)

(ξ ∈ Rn ).

Prove that the L2 norm of Hφ equals the L2 norm of φ. Verify that H possesses an extension to a continuous linear mapping H from L2 (R) to L2 (R) that satisfies H2 = −I. Problem 14.29. If u ∈ L1 , v ∈ E 0 and u ∗ v = 0, then prove that u = 0 or v = 0. Calculate 1 ∗ ∂j δ, for 1 ≤ j ≤ n. Problem 14.30. Prove the following assertions, in which the convolution product is as defined in Remark 11.17: (i) If u, v ∈ L1 then u ∗ v ∈ L1 and F(u ∗ v) = Fu Fv. (ii) If u ∈ L1 and u ∗ u = u, then u = 0. (iii) If u ∈ L2 and v ∈ L1 , then u ∗ v ∈ L2 and F(u ∗ v) = Fu Fv. (iv) If v ∈ L1 , there exists an f ∈ L2 such that the equation u∗v = f has no solution u ∈ L2 . Problem 14.31. Let u0 ∈ S 0 (Rn ). Prove that there exists exactly one differentiable family t 7→ ut : R>0 → S 0 (Rn ) such that

d dt ut

= ∆x ut and limt↓0 ut = u0 in S 0 (Rn ). Calculate Fut and ut .

Problem 14.32. Let v be the distribution on R determined by the function that equals
−1/2 1 − x2 for |x| < 1 and vanishes for |x| ≥ 1. (i) Demonstrate that v is a tempered distribution. (ii) Demonstrate that the Fourier transform of v is given by an analytic function h. (iii) Calculate h(0) and h0 (0). (iv) p is the tempered distribution denoted by x 7→ x. Calculate p ∗ v. Problem 14.33. (Homogeneous fundamental solution of Laplace operator). Let n ≥ 3. Prove the following assertions: (i) v(ξ) = −kξk−2 , for ξ ∈ Rn \ {0}, defines a locally integrable function on Rn , and therefore a distribution on Rn that we denote by v as well. (ii) v is homogeneous of degree −2 and v ∈ S 0 (Rn ). (iii) Define E := F −1 v. Then ∆E = δ, E ∈ S 0 (Rn ) and E is homogeneous of degree 2 − n. (iv) Every homogeneous fundamental solution of the Laplace operator is equal to E.

174

14 Fourier Transformation

(v) E is C ∞ on Rn \ {0} and invariant under all rotations. (vi) There exists a constant c such that E is equal to the locally integrable function x 7→ c kxk2−n . (vii) c = 1/((2 − n)cn ) with cn as in (13.31). Problem 14.34. (Cauchy-Riemann operator). In this problem the notation is as in Example 12.11. Our goal is to obtain a fundamental solution E of the Cauchy– ∂ by means of Fourier transformation. Differently phrased, we Riemann operator ∂z use that transformation to solve the following partial differential equation for E on R2 : ∂E 1 ∂E (x, y) − (x, y) = 2δ(x, y) = 2δ(x)δ(y). ∂x i ∂y To this end, suppose that x 7→ E(x, ·) is a C 1 family in S 0 (R) and deduce on the basis of Fourier transformation with respect to the other variable, that under this assumption dFE (x, η) − ηFE(x, η) = 2δ(x). dx Here FE denotes the partial Fourier transform. Write H for the Heaviside function on R and prove that, for a function η 7→ c(η) still to be determined, FE(x, η) = 2(c(η) + H(x))exη

((x, η) ∈ R2 ).

Observe that η 7→ exη does not define a tempered distribution on R if xη > 0. Obviate this problem by means of the choice c(η) = −H(η) and show that in this case FE(x, η) = −2 sgn(η) H(−x sgn(η)) exη ((x, η) ∈ R2 ). Now prove (compare with the assertion preceding (12.11)) E(x, y) =

1 1 , π x + iy

in other words,

E(z) =

1 πz

and show that E is indeed a tempered distribution on C. Problem 14.35. (Laplace operator). In this problem we construct a fundamental solution of the Laplace operator ∆ on Rn by a method analogous to the one in Chap. 13. To this end we define, for a ∈ C with Re a > 0, the function Ra : Rn → C by Γ ( n2 ) 2 Ra (x) = c(a) kxka−n where c(a) = . n a = cn Γ ( 2 ) π 2 Γ ( a2 ) (i) Prove that Ra is a locally integrable function on Rn and in consequence defines a distribution in D0 (Rn ). Prove that, in fact, Ra ∈ S 0 (Rn ) and verify 2

Ra (e−k · k ) = 1.

14 Fourier Transformation

175

(ii) Let q : Rn → R be the quadratic form x 7→ kxk2 ; then verify that q : Rn \ {0} → R is a C ∞ submersion. Now prove that on Rn \ {0} and for Re a > 0 we have a−n+2
2Γ ( a−n+2 ) ) Γ ( n2 )Γ ( a−n+2 2 2 with d(a) = Ra = d(a) q ∗ χ+ 2 . = n a a cn Γ ( 2 ) π 2 Γ(2) Show that a 7→ c(a) admits an extension to a complex-analytic function on C and recall that a 7→ χa+ is a family of distributions on R that is complex-analytic on C. Use this to verify that the formula above is valid on Rn \{0} for all a ∈ C. (iii) Prove in two different ways that, on Rn \ {0} and for Re a > 0, ∆ Ra = 2(a − n) Ra−2 .

(14.27)

Now suppose n ≥ 3. Show that the definition of Ra ∈ S 0 (Rn ) may be extended to all a ∈ C by setting, for k ∈ Z≥0 and 0 < Re a + 2k ≤ 2, Ra =

2k

Q

1 ∆k Ra+2k . (a + 2j − n) 1≤j≤k

(14.28)

(iv) Next prove that R0 = 0 on Rn \ {0} and use positivity to deduce R0 = δ. Conclude that  k · k2−n  ∆ = δ. (2 − n)cn This is the way in which the fundamental solution of ∆ from Problem 4.6 was obtained. More generally, demonstrate R−2k =

2k

Q

(−1)k ∆k δ 0≤j0 ).

(v) For all a ∈ C, verify that Ra is homogeneous of degree a − n and invariant under the orthogonal group of Rn . Background. The occurrence of numerical factors in (14.27) and (14.28) can be avoided by multiplying Ra by Γ ( n−a 2 ) , 2a Γ ( n2 )

in other words, by introducing

n−a ea = Γ (n 2 ) k · ka−n . R 2a π 2 Γ ( a2 )

ea ∗f = f , where f (x) = eix1 . Observe This normalization results if we require that R ea is not even that we do not get a complex-analytic family on all of C, because R well-defined if a ∈ n + 2Z≥0 . On the other hand, we obtain, for all k ∈ Z≥0 and for a ∈ C satisfying Re a + k < n, ea = (−1)k ∆k R ea+2k R

and

e−2k = (−1)k ∆k δ. R

ea ∗ R eb = R ea+b is valid, be it with restrictions on a and Furthermore, the group law R b. This can be proved from the convolution identity above, or by means of Fourier transformation.

176

14 Fourier Transformation

Problem 14.36. (Radon transformation). For ω ∈ S n−1 and t ∈ R, define Φω : Rn → R

by

Φω (x) = h ω, x i

and

N (ω, t) = Φ−1 ω ({t}).

Note that N (ω, t) equals the hyperplane { x ∈ Rn | h ω, x i = t }. (i) Prove that Φω satisfies k grad Φω (x)k = 1, for all x ∈ Rn . Next, show (see Theorem 10.14) that, for any φ ∈ C0∞ (Rn ) and t ∈ R, Z (Φω )∗ φ(t) = φ(x) dn−1 x =: Rφ(ω, t), N (ω,t)

where dn−1 x denotes integration with respect to the Euclidean hyperarea on N (ω, t) and Rφ : S n−1 × R → R is called the Radon transform of φ. (ii) Prove R(Th ∗ φ)(ω, t) = Rφ(ω, t + h ω, h i) (h ∈ Rn ) and deduce, for 1 ≤ j ≤ n and with ∆ as usual denoting the Laplacian acting on Rn , R(∂j φ)(ω, t) = ωj ∂t Rφ(ω, t)

and

R(∆φ)(ω, t) = ∂t 2 Rφ(ω, t).

(iii) Suppose n ∈ Z>0 is odd. In the notation of Problem 13.1, prove n−1 (φ ∈ C0∞ (Rn )). Φω ∗ (|χ|−n+1 )(φ) = 2∂t n−1 Rφ t=0 = 2R(∆ 2 φ) t=0 Let a ∈ C and let dω denote integration with respect to the Euclidean hyperarea on S n−1 . Define Aa ∈ D0 (Rn ) by Z Aa (φ) = Φω ∗ (|χ|a+1 )(φ) dω (φ ∈ C0∞ (Rn )). S n−1

(iv) Prove that (Aa )a∈C is a complex-analytic family of distributions and that, for Z Z Re a > 0, |h ω, x i|a Aa (φ) = dω φ(x) dx. Rn S n−1 Γ (a + 1) In particular, Aa is a continuous function on Rn for Re a > 0. Also show Z −n A (φ) = 2 ∂t n−1 Rφ(ω, t) t=0 dω. S n−1

(v) Prove that the function Aa is homogeneous of degree a and is invariant under rotations acting on Rn , for all Re a > 0. Deduce the existence of d(a) ∈ C such that Aa = d(a) Ra+n ∈ D0 (Rn ) for all a ∈ C, where (Ra )a∈C is the complex-analytic family from Problem 14.35. Prove, for a ∈ C \ (−2Z>0 ), d(a) =

21−a π n + 1)

Γ ( n2 )Γ ( a2

and conclude

Ra+n =

1 Aa . d(a)

Note that it is sufficient to evaluate Aa (en ) for Re a > 0 owing to the invariance under rotations, where en denotes the n-th standard basis vector in Rn . Furthermore, use Legendre’s duplication formula Γ (2z) = 22z−1 π −1/2 Γ (z)Γ (z + 21 ) (see [8, Exercises 6.50.(vi) or 6.53]).

14 Fourier Transformation

177

(vi) Now assume that n ∈ Z>0 is odd. Apply Problem 14.35.(iv), part (ii) and the reflection formula for Γ from Lemma 13.4 to obtain, for x ∈ Rn and φ ∈ C0∞ (Rn )), n−1 Z (−1) 2 φ(x) = ∂t n−1 Rφ(ω, t) t=h ω, x i dω. n−1 2(2π) S n−1 For n odd, the formula above is called Radon’s inversion formula, that is, it recovers φ from its Radon transform Rφ. Differently phrased: φ is known when its integrals over all hyperplanes in Rn are known. Note that { N (ω, h ω, x i) | ω ∈ S n−1 } is the collection of all hyperplanes in Rn passing through x. Hence, this inversion formula is local, in the sense that only the integrals over hyperplanes close to x are needed. If n is even, the constant 1/d(−n) is undefined, while the definition of A−n involves the constant 1/Γ (1 − n) = 0. In this case, proceed as follows. (vii) Assume n to be even. Apply Legendre’s duplication formula twice, then apply the reflection formula in order to show n

(−1) 2 (n − 1)! 1 = . d(−n)Γ (1 − n) (2π)n In the notation of Problem 13.3, deduce Z n (−1) 2 (n − 1)! δ= Φω ∗ (| · |−n ) dω; (2π)n n−1 S that is, for φ ∈ C0∞ (Rn ) and x ∈ Rn , Z Z n X (−1) 2 (n − 1)! t−n Rφ(ω, h ω, x i ± t) φ(x) = n (2π) S n−1 R>0 ± n 2 −1

−2

X k=0

t2k  dt dω. ∂s 2k Rφ(ω, s) s=h ω, x i (2k)!

Note that in the case where n is even, the inversion formula is not local. Consider the special case of n = 2 and define, for x ∈ R2 and t ≥ 0, Z X 1 Rφ(x, t) = Rφ(ω, h ω, x i ± t) dω. 4π S 1 ± Then Rφ(x, t) is the average of Rφ over the set of all lines in R2 at a distance t from the point x (note that N (ω, h ω, x i + t) = N (−ω, h −ω, x i − t)). The inversion formula now takes the form Z Z 1 Rφ(x, t) − Rφ(x, 0) 1 ∂t Rφ(x, t) φ(x) = − dt = − dt 2 π R>0 t π R>0 t Z 1 d Rφx (t) =− . π R>0 t

178

14 Fourier Transformation

Here the second identity is obtained by means of a formal integration by parts. The last expression is a Riemann-Stieltjes integral and equals, apart from differences in notation, the formula given by Radon in his 1917 paper.

15 Fourier Series

The theory of Fourier series asserts that a sufficiently smooth function u (of class C 2 is sufficient) on R which is periodic with period a > 0, can be written as a uniformly convergent series X u(x) = cn einωx . (15.1) n∈Z

Here ω =

2π a

> 0 and cn , the n-th Fourier coefficient of u, is given by 1 cn = a

Z

s+a

e−inωx u(x) dx

(n ∈ Z, s ∈ R).

s

For a locally integrable periodic function f with period a the integral Z 1 s+a M (f ) := f (x) dx a s

(15.2)

is said to be the mean of f . Note that the periodicity of f implies that f (x + na) = f (x), for all n ∈ Z, and consequently Z s+na 1 f (x) dx (n ∈ Z>0 ). M (f ) = na s From this it follows in turn that 1 M (f ) = lim t−s→∞ t − s

Z

t

f (x) dx. s

This result can be verified by choosing an n ∈ Z>0 such that t ≤ s + n a < t + a, then writing Z t 1 M (f ) − f (x) dx t−s s Z s+a Z s+n a  1 1  1 = − n f (x) dx + f (x) dx, na t − s t−s t s

180

15 Fourier Series

n a 1 and then using | a1 − t−s | = | t−s−n a (t−s) | ≤ t−s . This further shows that (15.2) is independent of the choice of s, as one also verifies directly. Our aim now is to reconstruct the theory of Fourier series from the theory of Fourier transformation of Chap. 14. In doing so, we will obtain (15.1), with convergence in the sense of distributions, even for arbitrary periodic distributions with period a. We begin by formulating an alternative definition of the mean M (u) of a periodic distribution u. This definition is required because the formula

Z

s+a

u(x) dx = u(1[ s, s+a ] ) s

cannot be generalized: the characteristic function 1[ s, s+a ] of the interval [ s, s + a ] is not a C ∞ function. P Lemma 15.1. For every a > 0 there exists a µ ∈ C0∞ (R) such that n∈Z Tna µ = 0 1. If u ∈ D (R) and Ta u = u, then u(ν) = u(µ) if in addition ν ∈ C0∞ (R) and P n∈Z Tna ν = 1. If u is locally integrable, then u(µ) = a M (u). Proof. Begin byPchoosing a χ ∈ C0∞ (R) such that χ ≥ 0 and χ > 0 on [ s, s + a [ . ∞ Then θ(x) := function on R. Indeed, note that n∈Z χ(x − na) defines a C for every bounded open interval I in R there exists a finiteP subset E of Z with χ(x − na) = 0 whenever x ∈ I and n ∈ / E. This gives n∈Z χ(x − na) = P χ(x − na), for x ∈ I, and from this immediately follow both the convergence n∈E and the fact that θ ∈ C ∞ (R). Furthermore, θ is periodic with period a. For every x ∈ R there exists an n ∈ Z with x − na ∈ [ s, s + a [ , and therefore θ(x) = θ(x − na) ≥ χ(x − na) > 0. This gives µ := χ/θ ∈ C0∞ (R) and X

Tna µ =

n∈Z

X n∈Z

Tna

χ θ

=

X Tna χ X Tna χ θ = = = 1. Tna θ θ θ

n∈Z

n∈Z

Using the periodicity of u we obtain, by a summation that is in fact finite,  X  X X u(ν) = u ν Tna µ = T−na u(µ T−na ν) = u(µ T−na ν) = u(µ). n∈Z

n∈Z

n∈Z

(15.3) To obtain the latter assertion we apply the foregoing to µ = 1[ s, s+a ] , which is permitted if u is locally integrable. Naturally, in this case u(µ) = a M (u).  Definition 15.2. Let a > 0. For an arbitrary u ∈ D0 (R) satisfying Ta u = u, the number M (u) := a1 u(µ), with µ as in Lemma 15.1, is said to be the mean of u.  Theorem 15.3. Every periodic distribution on R is tempered.

15 Fourier Series

181

Proof. For u ∈ D0 (R) we have an estimate of the form (3.4), with K = supp µ and µ as in Lemma 15.1. Formula (15.3), with φ instead of ν, implies, for every φ ∈ C0∞ (R), X X sup |∂ α φ(x + na)|. u(µ T−na φ) ≤ c |u(φ)| = n∈Z |α|≤k, x∈K

n∈Z

There exists a constant b > 0 with 1 + |n| P≤ b (1 + |x + na|), for all n ∈ Z and x ∈ K. Now choose d > 1 such that s := n∈Z (1 + |n|)−d < ∞. This yields |u(φ)| ≤ c bd s

(1 + |x + na|)d |∂ α φ(x + na)| ≤ c0 kφkS(k,b) ,

sup |α|≤k, x∈K, n∈Z

for a constant c0 > 0, see (14.5).



The sequence of coefficients (cn )n∈Z is said to be of moderate growth if positive constants c and N exist such that |cn | ≤ c |n|N

(n ∈ Z \ {0}).

(15.4)

Lemma 15.4. Let ω > 0. The distribution X v= cn δnω n∈Z

on R is tempered if and only if the sequence (cn )n∈Z is of moderate growth. In that k case X lim cn δnω = v j,k→∞

and lim

k X

j,k→∞

n=−j

cn einω = u := S ◦ Fv,

(15.5)

n=−j

both with convergence in S 0 . Proof. We have v ∈ S 0 if and only if there exist c > 0, k ∈ Z≥0 and N ∈ Z≥0 such that |v(φ)| ≤ c kφkS(k,N ) , for all φ ∈ C0∞ (R). See (14.5) and Definition 14.13. First, suppose that v ∈ S 0 . Let χ ∈ C0∞ (R), supp χ ⊂ ] −ω, ω [ and χ(0) = 1. For n ∈ Z and φ = Tnω χ we obtain v(φ) = cn , while on the other hand kφkS(k,N ) ≤ d (1+|n|)N , where d is a constant that is independent of n but contains the C k norm of χ. This implies that the sequence cn is of moderate growth if v ∈ S 0 . Conversely, if the sequence cn satisfies (15.4), we obtain, for every b > 1, X X |v(φ)| = cn φ(nω) ≤ c (1 + |n|)N |φ(nω)| n∈Z

≤c

X n∈Z

n∈Z

−b (1 + |n| ) sup (1 + |n|)N +b |φ(nω)|, n∈Z

182

15 Fourier Series

for all φ ∈ C0∞ (R), which implies that v ∈ S 0 . The convergence in S 0 follows from analogous estimates, with v replaced by the difference of v and the finite sum. The latter assertion follows from (14.22) and the continuity of S ◦ F from S 0 to S 0 , see Theorem 14.18.  When (15.5) holds, u is said to be the distributional Fourier series with coefficients (cn )n∈Z . One uses the notation (see also Problem 5.10) X u= cn einω in S 0 . n∈Z

Lemma 15.5. The derivative of a distributional Fourier series can be obtained by termwise differentiation; more exactly, if the sequence (cn )n∈Z is of moderate growth, then X X ∂ cn einω = inω cn einω in S 0 . n∈Z

n∈Z

Proof. Application of Lemma 5.7 yields ∂

X

cn einω = lim ∂ j,k→∞

n∈Z

k X

cn einω

in S 0 .

n=−j

But from Lemma 15.4 we find, considering that the sequence (inω cn )n∈Z is of moderate growth, k X X cn einω = lim ∂ inω cn einω in S 0 . j,k→∞

n=−j

n∈Z

 Theorem 15.6. Let a and ω > 0 and a ω = 2π.PThe mapping S ◦ F then is a linear bijection from the linear space of distributions n∈Z cn δnω , where cn is a sequence of moderate growth, onto the linear space of periodic distributions with period a. If X u= cn einω in S 0 n∈Z

then cn = M (e−inω u)

(n ∈ Z).

(15.6)

Proof. If u ∈ D0 (R) and Ta u = u, then u ∈ S 0 on the strength of Theorem 15.3. We 1 Fu ∈ S 0 and according to (14.20) the equation Ta u = u is equivalent find v := 2π to e−ia v = v, that is, (eia − 1) v = 0. On account of Theorem 9.5 we conclude that

15 Fourier Series

183

P v = n∈Z cn δnω , for a sequence cn ∈ C that can at most be of moderate growth, because v ∈ S 0 . In reaching this conclusion P we have also seen that, conversely, u := S ◦ Fv is 0 periodic with period a if v = n∈Z cn δnω ∈ S , where we have used Theorem 14.18. The assertion in (15.5) concerning the convergence follows from Lemma 15.4 and the fact that F is continuous from S 0 to S 0 , see Theorem 14.18. For (15.6) we write X  X M (e−inω u) = M ck e−inω eikω = ck M (ei(k−n)ω ) = cn . k∈Z

k∈Z

The convergence does not pose a problem, because M is a continuous linear form on S 0 , as one straightforwardly reads from the definition of the mean.  Example 15.7. Let a and ω be positive numbers satisfying a ω = 2π and consider X uω = δnω ∈ S 0 (R). n∈Z

P

Then Fuω = S ◦ Fuω = n∈Z einω on the basis of (14.22). A substitution of the index of summation implies Tω uω = uω , hence (14.20) leads to (eiω − 1) Fuω = 0. As in the proof of Theorem 15.6 wePdeduce the existence of a sequence cn ∈ C of moderate growth such that Fuω = n∈Z cn δna . But eia δnω = δnω , for all n ∈ Z, implies eia uω = uω . Applying (14.20) once again, we obtain that Fuω is a measure invariant Punder Ta ; therefore P Fuω has the same mass, say c, at every point. In other words, n∈Z einω = c n∈Z δna . We now find from (15.6)

60

-Π

Fig. 15.1. Graph of x 7→

1 = M (c ua ) =

Π

P35

n=−35

einx

X  c c c ua (µ) = δ Tka µ = , a a a k∈Z

and therefore c = a = 2π/ω. In explicit form the formula becomes, for φ ∈ S(R),

184

15 Fourier Series

X

einω = a

n∈Z

X

δka ,

that is,

X

X

Fφ(n ω) = a

n∈Z

k∈Z

φ(k a).

(15.7)

k∈Z

This result is known as Poisson’s summation formula (see Exercise 15.8 for another proof). Next, apply it with T−x φ instead of φ and note that F(T−x φ) = eix Fφ on account of (14.20); this leads to X X eixn ω Fφ(n ω) = a φ(x + k a) (x ∈ R, φ ∈ S(R)). (15.8) n∈Z

k∈Z

The distribution uω can also be used to determine the constant in the Fourier inversion formula (14.12) in an alternative way. In the proof of (15.6), and therefore also of (15.7), formula (14.12) is not used. We now obtain, with a ω = 2π, F ◦ Fuω = a Fua = a ω uω ; thus, S ◦ F ◦ F = c I is seen to act on uω as the scalar multiplication by a ω = 2π. Because uω 6= 0, it follows that c = 2π.  P A distribution of the form cn δnω reminds one of the functions on a lattice Ω := { nω ∈ R | n ∈ Z } used in numerical mathematics. By taking the limit as ω ↓ 0 one can use these to approximate arbitrary distributions on R, compare Problem 5.12. Conversely, by application of the convergence in S one derives the Fourier transformation in R from the theory of Fourier series. There also exists a Fourier series variant of Parseval’s formula. To formulate this on the natural maximal domain space, we introduce the following Hermitian inner product in the space of periodic functions with period a: (f, g)R/aZ = M (f g). The corresponding L2 norm is kf k = kf kL2 (R/aZ) = (f, f )R/aZ


1/2

.

The linear space of measurable periodic functions f on R with period a such that kf k < ∞ is denoted by L2 (R/aZ). This space is complete (a Hilbert space) and contains the space C ∞ (R/aZ) of periodic C ∞ functions with period a as a dense linear subspace. Furthermore one defines the Hermitian inner product in the space of sequences n 7→ cn : Z → C X (c, d) := cn dn . The corresponding norm is kck = kckl2 =

n∈Z

X

2

1/2

|cn |

n∈Z

and the space of sequences c with kck < ∞ is denoted by l2 = l2 (Z). This, too, is a Hilbert space; the sequences cn of which only finitely many cn differ from zero (the sequences with compact support) lie dense in this space.

15 Fourier Series

185

Theorem 15.8. Consider a and ω > 0 with a ω = 2π. TheP mapping Fω that assigns to (c)n∈Z ∈ l2 (Z) the distributional Fourier series u = n∈Z cn einω , is a linear bijection from l2 (Z) to L2 (R/aZ). We have Parseval’s formula X k(cn )n∈Z kl2 (Z) = |cn |2 = M (|u|2 ) = kuk2L2 (R/aZ) . n∈Z

Proof. For a finite Fourier series, Parseval’s formula holds on account of X X cm cm M (ei(n−m)ω ) = cn cn . M (|u|2 ) = M (u u) = n,m

n

By continuous extension we see that Fω is a linear isometry from l2 (Z) to L2 (R/aZ); in particular, it is injective. Furthermore, this implies that the image is complete with respect to the L2 (R/aZ) norm, and therefore constitutes a closed subset of L2 (R/aZ). For every φ ∈ C ∞ (R/aZ) and every N ∈ Z≥0 , repeated integration by parts yields cn = M (e−inω φ) = O((1 + |n|)−N ) as |n| → ∞. From this we deduce c ∈ l2 (Z), and so φ ∈ Fω (l2 (Z)). Approximating an arbitrary u ∈ L2 (R/aZ) with respect to the L2 norm by means of such φ, and using the fact that Fω (l2 (Z)) is closed, we see that u ∈ Fω (l2 (Z)). The conclusion is that the mapping Fω is also surjective from l2 (Z) to L2 (R/aZ).  Remark 15.9. The extension of the theory of Fourier series to the n-dimensional variant below does not involve any real difficulties; our only reason for starting with the one-dimensional theory was to avoid being confronted with too many different aspects at the same time. Let a(j), for 1 ≤ j ≤ n, be a basis for Rn . A distribution u ∈ D0 (Rn ) is said to be n-fold periodic with periods a(j) if Ta(j) u = u, for all 1 ≤ j ≤ n. From this one immediately derives by mathematical induction that Ta u = u, for all a ∈ A, where A :=

n nX

o kj a(j) | k ∈ Zn ,

j=1

is the lattice in Rn generated by the vectors a(j). The distribution u is also said to 0 n be A-invariant and the space of A-invariant distributions is denoted P by D (R /A). ∞ n As in Lemma 15.1, one finds aPµ ∈ C0 (R ) such that a∈A Ta µ = 1. If, furthermore, ν ∈ C0∞ (Rn ) with a∈A Ta ν = 1, then u(µ) = u(ν), for every u ∈ D0 (Rn /A). The mean of u ∈ D0 (Rn /A) is defined as M (u) =

1 u(µ), jA

186

15 Fourier Series

where jA equals the n-dimensional volume of the parallelepiped spanned by the vectors a(j), a fundamental domain for the lattice A. This corresponds to the usual mean if u is locally integrable. As in Theorem 15.3, we can use this function µ to show that every A-invariant distribution is tempered. To study the Fourier transforms we need the dual lattice Ω in Rn , generated by vectors ω(k), for 1 ≤ k ≤ n, that are determined by the equations ha(j), ω(k)i = 2π δjk . 1 ω(k) form the so-called dual basis of the The ω(k) also form a basis for Rn ; the 2π a(j), see Problem 9.11. A multisequence of coefficients ck ∈ C, with k ∈ Zn , is said to be of moderate growth if positive constants c and N exist such that |ck | ≤ c (1 + kkk)N , for all k ∈ Zn . As in Lemma 15.4, we obtain X v := ck δω(k) ∈ S 0 (Rn ) k∈Zn

if and only if the multisequence ck is of moderate growth; the ω(k) :=

n X

kj ω(j)

j=1

run through the points of the lattice Ω. The n-dimensional version of Theorem 15.6 asserts that the mapping S ◦ F is a bijection from the space of these v ∈ S 0 (Rn ) to the space of A-invariant distributions in Rn . The relevant coefficients ck in the distributional Fourier series X u= ck eiω(k) k∈Zn

are given by the formula ck = M (e−iω(k) u)

(k ∈ Zn ).

The n-dimensional Poisson summation formula takes the following form: X X F(δΩ ) = jA δA if δΩ = δω and δA := δa . ω∈Ω

a∈A

This also gives F 2 (δΩ ) = jA jΩ δΩ = (2π)n δΩ , an alternative calculation of the factor in the Fourier inversion formula. Finally, we have Parseval’s formula. It asserts that the mapping that assigns the corresponding distributional Fourier series to a multisequence c ∈ l2 (Zn ), is a bijective isometry from l2 (Zn ) to L2 (Rn /A).  Remark 15.10. The notations D0 (Rn /A) and L2 (Rn /A) derive from the following. The quotient space Rn /A is defined as the space of cosets x + A with x ∈ Rn . A

15 Fourier Series

187

function f is A-invariant if and only if f is constant on every coset relative to A; that is, if f (x) = g(x + A), for all x ∈ Rn , where g denotes a uniquely determined function on Rn /A. To use yet another formulation, if π : x 7→ x + A is the canonical projection from Rn to Rn /A, the mapping π ∗ is bijective from the space of functions on Rn /A to the space of A-invariant functions on Rn . It is customary to identify f and g with each other. Rn /A can be equipped with the structure of an n-dimensional C ∞ manifold in such a way that π ∗ creates a correspondence between the C ∞ functions on this manifold and the A-invariant C ∞ functions on Rn , which explains the notation C ∞ (Rn /A) for the space of these functions. This manifold can in fact be obtained as a submanifold of R2n . Let C := { z ∈ C | |z| = 1 } be the unit circle in the complex plane C ' R2 ; this is a compact, C ∞ , onedimensional submanifold of R2 . The n-fold Cartesian product T := C × C × · · · × C is a compact, C ∞ , n-dimensional submanifold of R2n , known as an n-dimensional torus. With respect to the complex multiplication we can treat C, and therefore also T , as a group. This implies that Φ : x 7→ (eihx, ω(1)i , . . . , eihx, ω(n)i ) is a homomorphism from the additive group Rn to the multiplicative group T , with kernel equal to the period lattice A. This means that Φ induces a bijective mapping from Rn /A to T , by which we can transfer the manifold structure of T to Rn /A. On the basis of this identification, Rn /A may also be referred to as an n-dimensional torus. It is possible on manifolds to integrate with respect to densities; considering, on T = Rn /A, the density that locally corresponds to the standard density on Rn , we have Z Z 1 u(x) dx. u(µ) = u(x) dx and M (u) = jA T T In Remark 10.9 we have noted that it is possible to define distributions on arbitrary manifolds. Thus we now see that distributions on T correspond to the Ainvariant distributions on Rn . The point of Remark 15.9 is that there also exists a Fourier transformation defined for distributions on the torus that leads to the space of functions on Zn , the multisequences. In the case of compact manifolds X, like the torus, we have the simplification that every distribution on X automatically has compact support; in this case, therefore, C ∞ (X) = C0∞ (X) and D0 (X) = E 0 (X). By way of example, for the torus this yields a convolution product u∗v of arbitrary u and v ∈ D0 (T ), such that F(u∗v) = Fu Fv.

188

15 Fourier Series

For Lie groups G, groups that also possess a manifold structure, there exists a convolution product, which, however, is not commutative when G is not commutative. For compact Lie groups there is a theory of Fourier series that generalizes the theory outlined above for tori. For noncompact, noncommutative Lie groups, Fourier analysis is much more difficult. Here, Harish-Chandra has done groundbreaking work. 

Problems Problem 15.1. Demonstrate that Poisson’s summation formula may also be applied to φ(x) = 1/(1 + x2 ). Calculate X 1 , Rh = h 1 + h2 n 2 n∈Z

a Riemann sum approximation of the integral of φ. Determine the limit, as h ↓ 0, of e2π/h (Rh − π). PN Problem 15.2. (Cancellation in sum of exponentials). Calculate n=−N einx , for x ∈ R, and discuss the convergence in S 0 (R) as N → ∞. Also deal with PN PN n=0 cos(nx) and n=1 sin(nx) in the same way. Determine the limits in all of these cases (in the last case, the Fourier transform of the limit will suffice). Remark: it is also possible to prove, without using Fourier P transformation (see, for example, [8, Exercise 0.18]), that the distribution u := n∈Z>0 sin(nx) on R \ 1 x sin x 2πZ equals the analytic function x 7→ 2(1−cos x) = 2 cot 2 . As x → 0, this function asymptotically equals x 7→ 1/x. On account of the 2π-periodicity of u, it has similar, not absolutely integrable behavior at all integer multiples of 2π. For a test function φ with support in the interval ] −2π, 2π [ one has Z Z (φ(x) − φ(−x)) sin x 1 u(φ) = dx = − log(1 − cos x) φ0 (x) dx, 4(1 − cos x) 2 where both integrals are absolutely convergent. Also, u equals the distributional derivative of the locally integrable function x 7→ 21 log(1 − cos x). 3 -Π

-Π

Π

Π

-3 -3

Fig. 15.2. Graphs of x 7→

sin x 2(1−cos x)

and x 7→

1 2

log(1 − cos x)

15 Fourier Series

189

Problem 15.3. Draw a sketch of the sawtooth function s(x) = x − [x]. Here [x] is the integer k such that x − 1 < k ≤ x, the integer part of x. Calculate the Fourier transform of s. Also compute s0 and its Fourier transform.

1

-2

1

-1

Fig. 15.3. Problem 15.3: Graph of x 7→

1 2

−

P100

n=1

2

sin(2πnx) πn

Problem 15.4. Calculate the Fourier transform of u (x) := tan(x + i) for  > 0. Investigate the limiting behavior of u and Fu as  ↓ 0. Problem 15.5. Let u0 be a periodic distribution with period a. Prove that there exists exactly one differentiable family t 7→ ut : R>0 → D0 (R) with the following properties: (i) For every t ∈ R>0 , the distribution ut is periodic with period a, 2 d ut = ddxu2t , (ii) dt (iii) limt↓0 ut = u0 in D0 (R). Calculate Fut and ut . Problem 15.6. Write D± = { z ∈ C | |z| ≶ 1 }

and

C = ∂D+ = ∂D− = { z ∈ C | |z| = 1 }.

Let f be complex-analytic on D+ and write ur (x) := f (r eix ), for 0 ≤ r < 1 and x ∈ R. Verify that, as r ↑ 1, ur converges in D0 (R) if and only if the sequence cn := f (n) (0)/n!, for n ∈ Z≥0 , is of moderate growth. If this is the case, the limit f+ is said to be the distributional boundary value of f , via the parametrization x 7→ eix : R → C. Prove that f+ is periodic with period 2π. Does every complex-analytic function on D+ have a distributional boundary value? Does every 2π-periodic distribution u on R equal the distributional boundary value of an analytic function on D+ ? Let g be complex-analytic on D− . It is known that g(z) → 0 as |z| → ∞ if and only if g(z) = h( z1 ) for an analytic function h on D+ with h(0) = 0. Write vr (x) := g(r eix ), for r > 1 and x ∈ R. Prove that, as r ↓ 1, vr converges in D0 (R) if and only if h possesses a distributional boundary value h+ . If this is the case, the limit g− is said to be the distributional boundary value of g. Prove that g− = Sh+ . Show that every 2π-periodic distribution u on R can uniquely be written as f+ + g− , with f+ distributional boundary value of a complex-analytic function f on D+

190

15 Fourier Series

and g− distributional boundary value ofPa complex-analytic function g on D− with g(∞) = 0. Determine f and g for u = n∈Z δ2πn . Problem F sinc2 and by using Poisson’s transformation formula deP 15.7. Compute 2 duce k∈Z sinc (x + k π) = 1, for all x ∈ R. Compare with [8, Exercise 0.14]. Derive the following partial-fraction decomposition: X 1 π2 = sin (πx) k∈Z (x − k)2

(x ∈ R \ Z).

2

See [8, Exercises 0.12 and 0.13] for many interesting applications of this identity, for instance, values of Riemann’s zeta function at points 2n for n ∈ Z>0 , MacLaurin’s series of the tangent, partial-fraction decompositions of various trigonometric functions and Wallis’ product. Problem 15.8. (Poisson summation formula). Define u ∈ D0 (R) by X 1 δ2πn . u= δ+ 2 n∈Z>0

(i) Verify that u ∈ S 0 (R). Deduce the following equality of distributions in S 0 (R): X 1 Fu = + e−2πik . 2 k∈Z>0

−x

(ii) Prove that u = lim e ↓0

u in S 0 (R) and furthermore that

Fu = lim ↓0

1 2

X

+

 e−2πk e−2πik .

k∈Z>0

Show by means of summation of a geometric series Fu(ξ) = lim ↓0

1 1 cot π(ξ − i) =: cot π(ξ − i0). 2i 2i

Now use addition of the resulting identities to obtain X 2i e2πikξ = cot π(ξ − i0) − cot π(ξ + i0). k∈Z

(iii) In addition, show that the complex-analytic function z 7→ π cot πz on C has a simple pole of residue 1 at every point z = n, for n ∈ Z. Conclude by means 1 1 − ξ+i0 = 2πiδ (see Problem 1.1) of the Plemelj–Sokhotsky jump relation ξ−i0 that one has X X X cot π(ξ −i0)−cot π(ξ +i0) = 2i δn ; deduce e2πik = δn , n∈Z

k∈Z

n∈Z

where the latter identity of tempered distributions is Poisson’s summation formula known from (15.7) and Problem 10.8.

15 Fourier Series

(iv) Prove

X

X

δn0 (x) = −4π

n∈Z

Now define s(x) =

191

k sin 2πkx.

k∈Z>0

X cos kx 1 + k2

(x ∈ R).

k∈Z>0

(v) Show  1  cosh(x − π) π −1 (0 ≤ x ≤ 2π). 2 sinh π To obtain this result, note that Poisson’s summation formula implies the following identities in S 0 (R) X X 1 cos kx = − + π s(x) − s00 (x) = δ2πn . 2 s(x) =

k∈Z>0

n∈Z

Now solve the differential equation for s using the fact that s is even and periodic. Conclude (compare with [8, Exercise 6.91.(iii)]) X k∈Z>0

1 1 = (−1 + π coth π) = 1.076 674 047 468 581 · · · . 1 + k2 2

16 Fundamental Solutions and Fourier Transformation

Fourier transformation in S 0 is a very useful tool in the analysis of linear partial differential operators X P (D) = cα Dα , |α|≤m

with constant coefficients cα . If u ∈ S 0 , then (14.7) implies F(P (D)u) = P (ξ) Fu. Here P (ξ) =

X

(16.1)

cα ξ α

|α|≤m

is the polynomial in ξ obtained by substituting ξj for Dj everywhere. P (ξ) is said to be the symbol of P (D). In other words, on the Fourier transform side the differential operator P (D) changes over into multiplication by the polynomial P (ξ). Clearly, this is the expansion in eigenvectors that we discussed in the introduction to Chap. 14. Remark 16.1. A tempered distribution u ∈ S 0 is a solution of the equation P (D)u = 0 if and only if the tempered distribution Fu satisfies P (ξ) Fu = 0. If P (ξ) 6= 0, for all ξ ∈ Rn , this leads to Fu = 0 and therefore u = 0. In general, one has supp Fu ⊂ N := { ξ ∈ Rn | P (ξ) = 0 }. If DP (ξ) 6= 0 for all ξ ∈ N , then N is a C ∞ (even analytic) manifold in Rn , of dimension n − 1. In this case Fu is given by a “distribution on N ”; but for a proper formulation we have to know what distributions on submanifolds are. This therefore represents a natural occasion to introduce these. An example of an entirely different nature was given by the distributions on the torus, in connection with Fourier series, see Remark 15.10.  Suppose that E is a fundamental solution of P (D) and, in addition, that E ∈ S 0 . Using (16.1), we conclude from (14.21)

194

16 Fundamental Solutions and Fourier Transformation

1 = F(P (D)E) = P (ξ) FE.

(16.2)

Conversely, if Q ∈ S 0 is a solution of the equation P (ξ) Q = 1, then E := F −1 Q ∈ S 0 is a fundamental solution of P (D). Indeed, in that case one has F(P (D)E) = P (ξ) Q = 1, therefore P (D)E = δ. It goes without saying that this result makes use of Theorem 14.18. Equation (16.2) implies that, on the complement C = { ξ ∈ Rn | P (ξ) 6= 0 } in Rn of the zero-set N of the polynomial P (ξ), the distribution Q = FE equals the C ∞ function 1/P . In particular, a possible tempered fundamental solution of P (D) is uniquely determined if P does not have real zeros. Even in this case it is not obvious, a priori, whether 1/P is a tempered distribution. This might lead to problems in cases where P (ξ) were to converge to 0 too fast in too large subsets of Rn , as kξk → ∞. At the present stage we do not have sufficient background on polynomials in n variables to reach general conclusions on this point. Instead, we now concentrate on a subclass of operators. Definition 16.2. The operator P (D) of order m is said to be elliptic if the homogeneous part X Pm (ξ) := cα ξ α |α|=m

of degree m of the symbol P (ξ) of P (D) does not have any real zeros ξ except ξ = 0. That is, ξ ∈ Rn

and ξ 6= 0

⇒

Pm (ξ) 6= 0. 

Example 16.3. The Laplace operator ∆ is elliptic. The heat operator ∂t − ∆x and the wave operator ∂t 2 − ∆x are not elliptic.  Note that the condition of ellipticity applies only to the highest-order part Pm (D) of P (D). Thus, if P (D) is elliptic, the operator P (D) + Q(D) is elliptic for every Q(D) of order lower than m. The condition of ellipticity becomes effective with the estimate in the following lemma. Lemma 16.4. P (D) is elliptic of order m if and only if there exist constants c > 0 and R ≥ 0 such that |P (ξ)| ≥ c kξkm

(ξ ∈ Rn , kξk ≥ R).

(16.3)

Proof. Let P (D) be elliptic. The continuous function ξ 7→ |Pm (ξ)| attains its minimum on the compact subset S = { ξ ∈ Rn | kξk = 1 } of Rn . That is, there exists

16 Fundamental Solutions and Fourier Transformation

195

an η ∈ S such that |Pm (ξ)| ≥ |Pm (η)|, for all ξ ∈ S. Now η 6= 0, and therefore µ := |Pm (η)| > 0. For arbitrary ξ ∈ Rn \ {0} one has kξk−1 ξ ∈ S, and therefore  1  ξ | = kξk−m |Pm (ξ)|. µ ≤ |Pm kξk Furthermore, Q(ξ) = P (ξ) − Pm (ξ) is a polynomial of degree ≤ m − 1, which implies the existence of a constant C such that |Q(ξ)| ≤ C kξkm−1

(kξk ≥ 1).

Combining the estimates we find, for kξk ≥ 1,  C  kξkm . |P (ξ)| ≥ |Pm (ξ)| − |Q(ξ)| ≥ µ kξkm − C kξkm−1 = µ − kξk This yields the desired estimate, with c = µ − C/R, whenever R > C/µ and R ≥ 1. Conversely, if P (D) is not elliptic, there exists a ξ ∈ Rn such that ξ 6= 0 and Pm (ξ) = 0. This implies Pm (t ξ) = tm Pm (ξ) = 0 and consequently P (t ξ) = Pm (t ξ) + Q(t ξ) = Q(t ξ) is a polynomial in t of degree < m. This is in contradiction with (16.3).  The estimate (16.3) implies that, for any real number w, the set of the ξ ∈ Rn with |P (ξ)| ≤ w is bounded. This is the origin of the name “elliptic”. Lemma 16.5. If P (D) is elliptic, it has a parametrix E with sing supp E ⊂ {0}. Proof. Let R and c > 0, as in (16.3). Choose a χ ∈ C0∞ (Rn ) with χ(ξ) = 1 for kξk ≤ R. The function 1/P is C ∞ on an open neighborhood U of supp (1 − χ), therefore v ∈ C ∞ (Rn ) if ( (1 − χ(ξ))/P (ξ) if ξ ∈ U, v(ξ) := 0 on interior of set where χ = 1. Furthermore, v(ξ) = 1/P (ξ) if ξ ∈ / supp χ, that is, if kξk is sufficiently large. On account of (16.3), one there has |v(ξ)| ≤ 1c kξk−m . It follows that v is certainly bounded and defines a tempered distribution on Rn . If we now choose E := F −1 v, then F(P (D)E) = P (ξ)FE = P (ξ)v = 1 − χ. Applying F −1 , we obtain P (D)E = δ − F −1 χ. Now F −1 χ = (2π)−n S ◦ Fχ is a C ∞ function on Rn , even with a complex-analytic extension to Cn ; it follows that E is a parametrix of P . To prove that sing supp E ⊂ {0}, we begin by deriving an auxiliary estimate, guided by the knowledge that the way in which v decreases at infinity improves under differentiation. We observe that, by mathematical induction on l, Dj l (P (ξ)−1 ) = Ql (ξ) P (ξ)−1−l ,

196

16 Fundamental Solutions and Fourier Transformation

if P (ξ) 6= 0. Here Q0 = 1, and Ql (ξ) = Dj Ql−1 (ξ) P (ξ) − l Ql−1 (ξ) Dj P (ξ) is a polynomial in ξ of degree ≤ l(m − 1). Because l(m − 1) − (1 + l)m = −l − m this yields, in combination with (16.3),
ξ α Dj l v = O kξk|α|−m−l , kξk → ∞. For given k ∈ Z>0 we choose l > k − m + n, or k − m − l < −n; then ξ α Dj l v ∈ L1 , for all multi-indices α with |α| ≤ k. But this means that in the integral representation of F −1 (Dj l v) = (−xj )l E we may differentiate to order k under the integral sign, which implies that xj l E ∈ C k , that is, E ∈ C k on the complement Cj of the hyperplane xj = 0. Because this holds for all k, we have ∞ ∞ E Sn∈ C on Cnj and because this in turn is true for all j, it follows that E ∈ C on  j=1 Cj = R \ {0}. From the proof it will be clear that we could have chosen E = F −1 (1/P (ξ)) if P (ξ) has no zeros in Rn . This applies, for example, if P = ∆ − λ I, with λ ∈ C and not λ ≤ 0. In that case, E also is a fundamental solution of P , with the additional properties that E ∈ S 0 and sing supp E ⊂ {0}. By combining Lemma 16.5 and Theorem 12.4 we obtain: Theorem 16.6. Every elliptic operator P (D) is hypoelliptic. Remark 16.7. For the parametrix E in the proof of Lemma 16.5, P (D)E − δ equals the Fourier transform of a function with compact support, and is therefore analytic. By replacing the integration over Rn in the formula for E by an integral over a suitable submanifold in Cn , which, on the strength of Cauchy’s Integral Theorem (see (12.8)), leaves the result unchanged, one also proves that E is analytic on Rn \ {0}. Thus, it follows from Theorem 12.15 that, for every elliptic operator P , the distribution u is analytic wherever P u = 0.  Remark 16.8. An elliptic operator P with variable (C ∞ ) coefficients has a parametrix K as discussed in Remark 12.6. Its construction is a generalization of the proof of Lemma 16.5 and the result is a so-called pseudo-differential operator K. This has a kernel k(x, y) ∈ C ∞ for x 6= y and the conclusion is that elliptic operators with C ∞ coefficients, too, are hypoelliptic. This theory can be found in H¨ormander [15, Ch. 18], for example.  Theorem 16.9. Every elliptic operator P (D) has a fundamental solution E ∈ D0 . Proof. Let R be given and choose η ∈ Cn such that Pm (η) 6= 0; for example, η ∈ Rn \ {0}. For z ∈ C we write P (ξ + zη) as a polynomial in z: P (ξ + zη) = Pm (η) z m +

m−1 X l=0

Rl (ξ) z l ,

16 Fundamental Solutions and Fourier Transformation

197

here the coefficients Rl (ξ) are polynomials in ξ. With Cl = supkξk≤R |Rl (ξ)| this yields the estimate m−1   X |P (ξ + zη)| ≥ rm |Pm (η)| − Cl rl−m

(|z| = r, kξk ≤ R).

l=0

Because the entity within parentheses converges to |Pm (η)| > 0 as r → ∞, we conclude that there exist  > 0 and r > 0, such that |P (ξ + zη)| ≥ 

(z ∈ C, |z| = r, ξ ∈ Rn , kξk ≤ R).

Write B := { ξ ∈ Rn | kξk ≤ R }

and

C := Rn \ B.

We now define E := F + G, with F := F −1 (1C /P ) and G the analytic function given by Z Z eihx, ξ+zηi 1 1 dz dξ. G(x) := (2π)−n B 2πi |z|=r P (ξ + zη) z Here the integral over z is the complex line integral, whose value can be regarded as the sum of all residues in the disk { z ∈ C | |z| < r } of the meromorphic function z 7→ eihx,ξ+zηi /z P (ξ + zη). This yields P (D)F = F −1 (1C ), while Z Z 1 1 eihx, ξ+zηi dz dξ P (D)G(x) = (2π)−n z B 2πi |z|=r Z = (2π)−n eihx, ξi dξ = (F −1 (1B ))(x), B

where in deriving the latter identity we have used Cauchy’s integral formula (12.10). The conclusion is P E = F −1 (1C ) + F −1 (1B ) = F −1 (1) = δ.  Note that sing supp E ⊂ {0} for every fundamental solution of an elliptic operator, this follows from Theorem 16.6. Remark 16.10. The construction given here is a simple variant of H¨ormander’s construction [16, Thm. 7.3.10], of a fundamental solution for an arbitrary linear partial differential operator P (D) 6= 0 with constant coefficients. The simplification is made possible by the ellipticity of P (D), which ensures that no problems arise due to P (ξ) becoming small for large kξk. By combining Fourier transformation with deep results from algebraic geometry, Bernstein and Gel’fand [2] and Atiyah [1] were able to prove that every P (D) 6= 0 has a fundamental solution E ∈ S 0 . However, in the nonelliptic case these solutions often have less favorable local properties than those of H¨ormander.

198

16 Fundamental Solutions and Fourier Transformation

Problems Problem 16.1. For what linear partial differential operators P in Rn , with constant coefficients and of order > 0, does the equation P u = 0 possess a solution u 6= 0 in (i) D0 , (ii) S 0 , (iii) E 0 , (iv) L1 , (v) C ∞ and (vi) S, respectively? Problem 16.2. Prove that every harmonic and tempered distribution u on Rn is a polynomial. Prove that u is constant if u is a bounded harmonic function. Prove that for the Laplace operator ∆ in Rn , with n ≥ 3, the potential of δ is the only fundamental solution E with the property that E(x) → 0 as kxk → ∞. Problem 16.3. (Liouville’s Theorem). Suppose that u is a tempered distribution on C ' R2 and is complex-differentiable in the sense that ∂u/∂ z¯ = 0; then prove that u is a complex polynomial. Prove Liouville’s Theorem, which asserts that any bounded complex-analytic function u on C is constant. Problem 16.4. Let P (D) be an elliptic operator of order m. Prove the following assertions. (i) If P (D) possesses a homogeneous fundamental solution E, then P (D) = Pm (D), that is, P is homogeneous of degree m. Furthermore, E is homogeneous of degree m − n. (ii) Suppose that P is homogeneous and that E is a homogeneous fundamental soe lution E of P (D). On account of Theorem 14.26, E is tempered. Prove that E is a tempered fundamental solution of P if and only if there exists a polynomial e = E + f and P f = 0. Furthermore, if m ≥ n, then function f such that E e E is a homogeneous fundamental solution for P if and only if there exists a hoe = E + f and mogeneous polynomial function f of degree m − n such that E e P f = 0. If m < n and if E is a homogeneous fundamental solution of P , then e = E. E (iii) If P is homogeneous and m < n, then 1/P is locally integrable on Rn and P has exactly one homogeneous fundamental solution. (iv) The Laplace operator in R2 has no homogeneous fundamental solution. d − λ has exactly one Problem 16.5. Let λ ∈ C and λ ∈ / R. Show that P = 1i dx tempered fundamental solution E0 . Calculate the fundamental solution E from the proof of Theorem 16.9. Show that E is not a tempered distribution on R whenever r > (|λ| + R)/|η|.

17 Supports and Fourier Transformation

In Lemma 14.3 we have proved that the Fourier transform Fu of a distribution with compact support can be extended to a complex-analytic function U on Cn , called the Fourier-Laplace transform of u. In this chapter we deal with questions like: What complex-analytic functions U occur as Fu for some u ∈ E 0 ? and: How can U provide information on the support of u? For a compact subset K of Rn we define sK : Rn → R

by

sK (η) := sup hx, ηi.

(17.1)

x∈K

Theorem 17.1. Let K ⊂ Rn be compact and u ∈ E 0 (Rn ) with supp u ⊂ K. If u is of order k, there exists a constant C > 0 such that |Fu(ζ)| ≤ C (1 + kζk)k esK (Im ζ)

(ζ ∈ Cn ).

(17.2)

(ζ ∈ Cn ).

(17.3)

If u ∈ C k , there exists a constant Ck > 0 such that |Fu(ζ)| ≤ Ck (1 + kζk)−k esK (Im ζ)

Proof. We are going to apply Theorem 8.4; this requires some preparation. For given  > 0 we can, on the strength of Corollary 2.4, cover the compact set K by a finite number of open balls whose union U is contained in the /3-neighborhood K/3 . Applying Lemma 2.18 with this set U and with  and δ replaced by /3 and 2/3, respectively, we conclude that it is possible to find χ ∈ C0∞ (Rn ) with χ = 1 on an open neighborhood of K, while supp χ ⊂ K . Then there exists, for every multi-index β, a constant Cβ > 0 such that sup |∂ β χ (x)| ≤ Cβ −|β|

( > 0).

x

Writing ζ = ξ + iη with ξ, η ∈ Rn , we have, for every multi-index γ,

200

17 Supports and Fourier Transformation

|∂x γ e−ihx, ζi | ≤ |ζ γ | ehx, ηi . On account of hy, ηi ≤ kyk kηk, the Cauchy–Schwarz inequality, we get ehx, ηi ≤ esK (η)+ kηk

(x ∈ K ).

If we now apply Theorem 8.5 with χ = χ , and work out an explicit form of the C k norm of χ e−iζ by means of Leibniz’ rule, we obtain an estimate of the form |Fu(ζ)| = |u(e−iζ )| ≤ C 0

sup

−|β| |ζ γ | esK (η)+ kηk .

|β|+|γ|≤k

With the choice  = 1/kηk we now find (17.2), where we have used −|β| |ζ γ | = kηk|β| |ζ γ | ≤ (1 + kζk)|β|+|γ| , while e kηk = e may be included with the constant. As regards (17.3), we observe that Z Z ehx, ηi |u(x)| dx ≤ esK (η) kukL1 , |Fu(ζ)| = e−ihx, ζi u(x) dx ≤ x∈supp u

0

n

whenever u ∈ E (R ) is integrable. On account of (14.19) this yields, for all multiindices α with |α| ≤ k and for all u ∈ C0k , |ζ α Fu(ζ)| ≤ esK (η) kDα ukL1 . 

This proves (17.3).

For u ∈ C0∞ (Rn ) we thus find that an estimate of the form (17.3) obtains for every k ∈ Z≥0 . To account for the fact that the constant C will depend on k, it is written with subscript k. Example 17.2. Let u ∈ E 0 (Rn ). Then there exist a continuous function f on Rn and an N ∈ Z≥0 such that u = (1 − ∆)N f . Indeed, every distribution with compact support is of finite order k and Theorem 17.1 therefore yields an estimate of the form |Fu(ξ)| ≤ C (1 + kξk)k

(ξ ∈ Rn ).

Now choose N sufficiently large to ensure that 2N − k > n. One then has g ∈ L1 (Rn ) if
−N g(ξ) := 1 + kξk2 Fu(ξ) (ξ ∈ Rn ). On account of Theorem 14.2, the Riemann-Lebesgue Theorem, f = F −1 g is a continuous function on Rn with f (x) → 0 as kxk → ∞. Furthermore,
N

N F (1 − ∆ f (ξ) = 1 + kξk2 g(ξ) = Fu(ξ), It follows that (1 − ∆)N f = u. By using a smooth cut-off function we can deduce, for every distribution u on an open subset X of Rn and for every compact subset K of X, the existence of a continuous function f on X and an N ∈ Z≥0 such that u = (1 − ∆)N f on a neighborhood of K. This implies that every distribution can locally be written as a linear combination of derivatives of a continuous function. 

17 Supports and Fourier Transformation

201

We now give a converse of Theorem 17.1. For a real-valued function s on a subset Y of Rn \ {0} we define the subset Ks of Rn by \ Ks := { x ∈ Rn | hx, ηi ≤ s(η) }. (17.4) η∈Y

See Theorem 17.8 below for a characterization of the sets of the form Ks . Theorem 17.3. Let Y ⊂ Rn \ {0} and s : Y → R such that Ks is a bounded (and therefore compact) subset of Rn . Let N ∈ R. Suppose that U is a complexanalytic function on Cn with the property that for every η ∈ Y there exists a constant C(η) > 0 such that |U (ξ + iτ η)| ≤ C(η) (1 + kξk)N eτ s(η)

(ξ ∈ Rn , τ ≥ 0).

(17.5)

Then U = Fu for a u ∈ E 0 (Rn ) with supp u ⊂ Ks . If, in addition, N < −n − k, then u ∈ C k . Proof. Denote the restriction of U to Rn by v. First assume that N < −n − k. In that case, the functions ξ α v are integrable on Rn if |α| ≤ k and, consequently, u := F −1 v is a bounded C k function on Rn . Now let η ∈ Y . Because η 6= 0, there exists a linear mapping A : Rn−1 → Rn such that B : (σ, µ) 7→ σ η + Aµ a is bijective linear mapping from Rn to Rn ; the latter has a complex-linear extension B : Cn → Cn . The change of variables ξ = B(σ, µ) yields Z Z u(x) = | det B| F ◦ B(σ, µ) dσ dµ Rn−1

where

R

F (ξ) = (2π)−n eihx, ξi U (ξ).

On account of Cauchy’s Integral Theorem (see (12.8)), applied to the complexanalytic function F ◦ B(ρ, µ) of ρ = σ + iτ ∈ C, the integration over the real σ-axis may be replaced by integration over σ 7→ σ + iτ . Here the estimate (17.5) guarantees that the contributions over t 7→ σ + itτ , for t ∈ [ 0, 1 ], converge to 0 as |σ| → ∞. Returning to the old variables, we get Z u(x) = (2π)−n eihx, ξ+iτ ηi U (ξ + iτ η) dξ. Rn

Since |eihx, ξ+iτ ηi | = e−τ hx, ηi , substitution of the estimate (17.5) gives |u(x)| ≤ C e−τ hx, ηi eτ s(η) , with a constant C that is independent of τ ≥ 0. If hx, ηi > s(η), the right-hand side converges to 0 as τ → ∞; in that case, therefore, |u(x)| ≤ 0, implying u(x) = 0. The conclusion is that hx, ηi ≤ s(η), for all x ∈ supp u. Because this is true for all η ∈ Y , we have proved supp u ⊂ Ks .

202

17 Supports and Fourier Transformation

If we drop the assumption N < −n − k, we still have v ∈ S 0 ; so v = Fu, for some u ∈ S 0 . Choose φ ∈ C0∞ (Rn ) with φ(x) = 0 if kxk > 1 and 1(φ) = 1. For φ (x) := −n φ(−1 x) we find that u := u ∗ φ → u in S 0 as  → 0. Furthermore one has, on the strength of Theorem 14.25, v (ξ) := F(u )(ξ) = Fu(ξ) Fφ (ξ) = U (ξ) Fφ( ξ). The right-hand side has an extension to a complex-analytic function on Cn , which we denote by U . Using the Cauchy–Schwarz inequality we find sK (η) = kηk, if K = {x ∈ Rn | kxk ≤ 1}. Combining the estimate (17.3) for φ with (17.5) we get, for ζ = ξ + iτ η, |U (ζ)| ≤ C(η, k) (1 + kζk)−k eτ kηk (1 + kξk)N eτ s(η) . Choosing k > N + n we may apply the first part of the theorem with k set to zero, to conclude that supp u is contained in the half-space H (η) := { x ∈ Rn | hx, ηi ≤ s(η) + kηk }. On account of supp uδ ⊂ Hδ (η) ⊂ H (η), whenever 0 < δ ≤  and u = limδ↓0 uδ , this implies supp u ⊂ H (η). Because this is true for every  > 0, we conclude that supp u ⊂ H0 (η). Finally, as this applies for all η ∈ Y , it follows that supp u ⊂ Ks .  Theorems of this kind are called Paley–Wiener Theorems and the estimates for the function U that they contain are the Paley–Wiener estimates. Theorems 17.1 and 17.3 in this form are due to Schwartz; Paley and Wiener studied the Fourier transformation of integrable functions on R with support in a half-line. Example 17.4. It is a classical idea, due to Cauchy, to solve the wave equation d2 ut = ∆ut dt2 in (x, t)-space by treating t as a parameter in the Fourier transformation with respect to the variable x. We will use the notations from Remark 13.3. To initiate this treatment, we assume that t 7→ ut is a C 2 family in S 0 (Rn ). Because F : S 0 → S 0 is a continuous linear mapping, t 7→ Fut in that case also is a C 2 family in S 0 and  d2  d2 Fu = F ut = F(∆ut ) = −kξk2 Fut , t dt2 dt2 in view of Theorem 14.18. But this implies Fut (ξ) = cos(t kξk) p + with

sin(t kξk) q, kξk

17 Supports and Fourier Transformation

203

d  d =F , Fut ut dt dt t=0 t=0 or p = Fa and q = Fb in the notation of Remark 13.3. This causes us to turn to the functions p = Fu0

q=

and

At (ξ) := cos(t kξk)

and

Bt (ξ) :=

sin(t kξk) , kξk

(17.6)

the solutions for p = 1 and q = 0, and for p = 0 and q = 1, respectively. Both functions possess an extension to a complex-analytic function on Cn , owing to the fact that sin z = S(z 2 ), cos z = C(z 2 ) and z where C and S are complex-analytic functions on C; this can be seen from the power series, for example. Accordingly, we may write At (ξ) = C(t2 kξk2 )

and

Bt (ξ) = t S(t2 kξk2 );

here ξ 7→ kξk2 has the complex-analytic extension ξ + iη = ζ 7→

n X

ζj 2 = kξk2 − kηk2 + 2ihξ, ηi.

j=1

In order to estimate At (ζ), we observe that | cos z| ≤ e|y| , if z = x + iy and x, y ∈ R. Solving y 2 from x2 − y 2 + 2ixy = z 2 = t2 (kξk2 − kηk2 + 2ihξ, ηi), we find y2 =

p t2 (kηk2 − kξk2 ± (kξk2 − kηk2 )2 + 4hξ, ηi2 ) ≤ t2 kηk2 . 2

Here we have used the Cauchy–Schwarz inequality. This leads to |At (ξ + iη)| ≤ e|t| kηk

(ξ, η ∈ Rn ).

Now hx, ηi ≤ |t| kηk, for all η ∈ Rn , if and only if kxk ≤ |t|. Similar Paley– d Wiener estimates hold for Bt , if it is observed that dt sin tz = z cos tz, which imR1 |y| plies | sin z| = |z 0 cos tz dt| ≤ |z| e . On the basis of Theorem 17.3 we thus find that At = Fat and Bt = Fbt , for distributions at and bt in Rn with supp at and supp bt both contained in the ball about 0 of radius |t|. Moreover, these are C 2 families in E 0 (Rn ) with d2 at = ∆at dt2 and initial conditions

and

d2 bt = ∆bt dt2

204

17 Supports and Fourier Transformation

a0 = δ,

d =0 at dt t=0

and

b0 = 0,

d = δ. bt dt t=0

(17.7)

The fact that at and bt are families of distributions with compact support enables us to convolve them with arbitrary distributions. The conclusion is that, for every a and b ∈ D0 (Rn ), the distributions ut := at ∗ a + bt ∗ b,

(t ∈ R),

give a C 2 family in D0 (Rn ) that forms a solution of the Cauchy problem d2 ut = ∆ut dt2

with

u0 = a

and

d = b. ut dt t=0

(17.8)

Combining this with the uniqueness from Remark (13.3), we conclude that the Cauchy problem for the wave equation has a unique solution, even for arbitrary distributional initial values. Next, we consider the wave equation with an inhomogeneous term f ∈ D0 (R × n R ) as in (13.24). Once again, we assume f toRbe associated with a family ft of distributions in D0 (Rn ) via the formula f (φ) = R ft (φt ) dt, for all φ ∈ C0∞ (Rn+1 ), with the notation φt : x 7→ φ(t, x). Furthermore, we suppose that the ft ∈ D0 (Rn ) depend continuously on the parameter t ∈ R. Recall that this means that the function t 7→ ft (φ) is continuous, for every φ ∈ C0∞ (Rn ). (The principle of uniform boundedness is required in showing that one really defines a distribution in this manner.) In order to get an idea about the solution, we initially consider the situation where t 7→ ft is continuous with values in S 0 (Rn ). In that case, t 7→ Fft is also continuous with values in S 0 (Rn ). Upon Fourier transformation the inhomogeneous wave equation takes the form d2 Fvt = −kξk2 Fvt + Fft . dt2 Moreover, let us suppose that Fft (ξ) depends continuously on t and ξ. Under that assumption we may, for fixed ξ, consider the equation as an inhomogeneous secondorder equation with constant coefficients and continuous inhomogeneous term. As in the proof of Theorem 9.4, we write it as a first-order inhomogeneous system d Vt (ξ) = L(ξ) Vt (ξ) + Ft (ξ), where dt  0   0   Fv  1 L(ξ) = and F = . Vt = d t , t −kξk2 0 Fft dt Fvt Let Φt (ξ) be the fundamental matrix determined by At (ξ) and Bt (ξ) for the homogeneous system, more precisely,   A Bt Φt (ξ) = d t = exp t L(ξ) and Φ0 (ξ) = I. d dt At dt Bt Next, we use Lagrange’s method of variation of constants to find a solution of the inhomogeneous system having the form Φt (ξ) Wt (ξ). This leads to

17 Supports and Fourier Transformation

Z

t

Vt (ξ) = Φt (ξ)

Z

t

Φ−s (ξ) Fs (ξ) ds = t0

205

Φt−s (ξ) Fs (ξ) ds, t0

for some arbitrary t0 ∈ R. Thus, Z t Z t Z t Fbt−s Ffs ds Bt−s Ffs ds = Φt−s Fs ds = Fvt = (1, 0) Vt (ξ) = (1, 0) =F

t

t0

t0

t0

Z

Z



bt−s ∗ fs ds ,

t0

and hence

t

bt−s ∗ fs ds.

vt =

(17.9)

t0

Recall that supp bt−s is contained in the closed ball around 0 of radius |t−s|. Therefore (17.9) also makes sense for a continuous family ft of distributions in D0 (Rn ). The family vt thus defined is a C 2 family in D0 (Rn ). A direct verification that the family defined in (17.9) provides a solution to the inhomogeneous equation is also feasible. Indeed, Z t Z t d d d vt = bt−s ∗ fs |s=t + bs−t ∗ fs ds = bs−t ∗ fs ds dt t0 dt t0 dt in view of (17.7). Using these conditions once more, we obtain Z t 2 Z t d2 d d v = b ∗ f + b ∗ f ds = δ ∗ f + ∆bs−t ∗ fs ds t t−s s s−t s t 2 dt2 dt t0 dt t0 s=t = ft + ∆vt . Furthermore, the initial values can be adjusted by adding to vt a solution of the Cauchy problem (17.8) with suitably chosen constants a and b. If ft is a family of distributions with ft = 0 for t < t0 , then the lower limit t0 may be replaced by −∞. For the corresponding distribution v we thus obtain the following formula, where φ ∈ C0∞ (Rn+1 ): Z Z Z Z Z v(φ) = vt (φt ) dt = bt−s ∗fs (φt ) ds dt = bt−s (Sfs ∗φt ) dt ds. R

R

s≤t

R

t≥s

If we take f = δRn+1 , the Dirac measure on Rn+1 , then certainly t 7→ ft is not a continuous family. Yet, we have obtained a formula that makes sense, viz., Z Z n v(φ) = bt−0 (SδR ∗ φt ) dt = bt (φt ) dt. t≥0

R>0

This formula really defines a fundamental solution. In fact, Z Z  d2  φ dt − bt (∆φt ) dt  v(φ) = v(  φ) = bt t dt2 R>0 R>0 Z Z Z  d2    d2   d2 φ dt − ∆b (φ ) dt = bt φt − 2 bt (φt ) dt. = bt t t t 2 2 dt dt dt R>0 R>0 R>0

206

17 Supports and Fourier Transformation

Using integration by parts we now deduce h d  i∞ d  v(φ) = bt φt − bt (φt ) = δRn (φ0 ) = φ(0) = δRn+1 (φ). dt dt 0 Denoting the fundamental solution thus obtained by E+ , we may write Z E+ (φ) = bt (φt ) dt (φ ∈ C0∞ (Rn+1 )). R>0

supp bt is contained in the closed ball around 0 of radius |t|. Therefore E+ = 0 on the union of the set in R × Rn where t < 0 and the set where kxk > t. In other words, supp E+ is contained in the solid cone C+ = { (t, x) ∈ R × Rn | kxk ≤ t }. On account of the uniqueness result from Remark 13.3, this fundamental solution equals the one from Theorem 13.2. This implies that bt is analytic on { x ∈ Rn | kxk < t } and even vanishes on that set for odd n > 1. More precisely, we obtain the formula  3−n  n−1 1 (t > 0). bt = π − 2 qt ∗ χ+2 2 Here qt = q ◦ it : x 7→ t2 − kxk2 : Rn → R, in other words, the composition of q by the mapping it : x 7→ (x, t) from Rn to Rn+1 . Furthermore, qt is a submersion from Rn \ {0} to R; there is not a real 3−n

problem at x = 0, however, since bt ∈ C ∞ at x = 0 and χ+2 ∈ C ∞ at q = t. One hesitates to write bt = it ∗ (E+ ), because it is not a submersion. For bt with t < 0 one uses b−t = −bt , while at can be expressed in terms of bt by means of the formula at =

d bt . dt

(17.10)

The analyticity of at and bt in kxk < t can also be demonstrated directly, if the paths of integration in the integral formulas for at = F −1 At and bt = F −1 Bt , respectively, are slightly rotated in suitably chosen complex directions.  Remark 17.5. It is natural to try and derive explicit formulas for the distributions at and bt , whose respective Fourier transforms At and Bt are given by (17.6). In view of (17.10), only bt needs to be determined. A uniform formula for all dimensions n is given by bt (x) = pn lim Im kxk2 − (t + i)2 ↓0


−(n−1)/2

with pn :=

Γ ((n + 1)/2) . π (n+1)/2 (n − 1) (17.11)

17 Supports and Fourier Transformation

207

If n = 2, then bt equals the locally integrable function (

sgn t (2π)−1 (t2 − kxk2 )−1/2

if kxk < |t|,

0

if kxk > |t|,

bt = as can be seen from (17.11) for n = 2, for which some extra work is required, however. If n = 3, then bt (φ) equals t times the mean of φ over the sphere of radius |t| about the origin. For n = 3, the derivation of this classical description of bt from (17.11) involves further complications. For n ≥ 4 it is useful to observe that kxk2 − (t + i)2


−(n−1)/2

= (n−3)−1 (t+i)−1


−(n−3)/2 d , kxk2 − (t + i)2 dt

which shows that the right-hand side of (17.11) can be identified with a constant d p−1 times the differential operator (t−1 dt ) of order p − 1 applied to the distributions lim Im kxk2 − (t + i)2


−1/2

lim Im kxk2 − (t + i)2

or

↓0


−1

↓0

,

depending on whether n = 2p or n = 2p + 1, respectively. To within a constant
−1/2 factor, the latter distributions equal the integral of t2 − kxk2 φ(x) over the n-dimensional ball kxk2 < t2 , and tn−2 times the mean of the test function φ over the sphere of radius |t| about 0, respectively. (17.11) may be verified as follows. The function 2

q(x, t) = (πt)−1/2 e−x satisfies the diffusion equation

∂q ∂t

=

∂2q ∂x2

/4t

(x ∈ R, t > 0),

(see Problem 5.5), and furthermore

Z q(x, t) dx = 1. R>0 k

Under the assumption x > 0, all derivatives ∂∂tkq (x, t) converge to 0 as t ↓ 0 and also as t → ∞. This enables us to conclude that the integral Z I(x) := e−t q(x, t) dt R>0

converges and that I 00 (x) =

Z

e−t

R>0

∂2q (x, t) dt = ∂x2

q(x, t) R>0

e−t

R>0

∂q (x, t) dt ∂t

−t

Z =−

Z

∂e dt = I(x). ∂t

Indeed, the first identity follows by differentiating twice under the integral sign, the second from the diffusion equation and the third through integration by parts. By

208

17 Supports and Fourier Transformation

solving the differential equation we obtain the existence of a constant c such that I(x) = c e−x , because I(x) → 0 as x → ∞. As a result we have Z Z Z c= I(x) dx = e−t q(x, t) dt dx R>0

R>0

Z

e−t

= R>0

R>0

Z

Z

R>0

This proves the formula Z e−x =

e−t dt = 1.

q(x, t) dx dt = R>0

2

e−t (π t)−1/2 e−x

/4t

dt

(x > 0).

R>0

Replacing x by kξk with ξ ∈ Rn , then multiplying by (2π)−n eihx, ξi , integrating over ξ ∈ Rn , changing the order of integration and finally using Example 14.9, we obtain the following formula for the Fourier inverse of the function f (ξ) = e−kξk : Z 2 −1 −n F f (x) = (2π) e−t (π t)−1/2 (4π t)n/2 e−t kxk dt R>0

= π −(n+1)/2 Γ ((n + 1)/2) 1 + kxk2


−(n+1)/2

.

If we now define fτ (ξ) := e−τ kξk , then (F −1 fτ )(x) = π −(n+1)/2 Γ ((n + 1)/2) τ kxk2 + τ 2


−(n+1)/2

.

If n > 1, which we assume from now on, Fτ (ξ) := e−τ kξk /kξk defines a locally integrable function of ξ whose derivative with respect to τ equals −fτ (ξ). This leads to
−(n−1)/2 Gτ (x) := (F −1 Fτ )(x) = pn kxk2 + τ 2 , (17.12) where we have used the fact that both sides converge to zero as τ → ∞, and where the constant pn is as in (17.11). Now Fτ is a complex-differentiable function of τ with values in S 0 (Rn ), defined on the complex half-plane Re τ > 0. Furthermore, this S 0 (Rn )-valued function is continuous on { τ ∈ C | Re τ ≥ 0 }. Because F −1 is a continuous operator, we arrive at the same conclusions, with Gτ := F −1 Fτ instead of Fτ , where formula (17.12) remains valid for Re τ > 0. If we write τ =  − it = −i(t + i), with t ∈ R and  > 0, and let  ↓ 0, this leads to F −1

 eitkξk  kξk

(x) = pn lim kxk2 − (t + i)2 ↓0


−(n−1)/2

.

(17.13)

Because Fτ (−ξ) = Fτ (ξ), the complex conjugate of F −1 Fτ equals the Fourier inverse of the complex conjugate of F−τ ; therefore, we also have Im F −1 Fτ = F −1 (Im Fτ ). On account of Bt = Im Fit , (17.11) now follows from (17.13). 

17 Supports and Fourier Transformation

209

Remark 17.6. The theory of the wave operator  has a generalization to a wide class of linear partial differential operators P = P (D) with constant coefficients. Let P (ξ) be the symbol of P and Pm (ξ) the homogeneous part of degree m of P (ξ), where m is the order of P . Let η ∈ Rn \ {0}. The operator P is said to be hyperbolic with respect to η if, for every ξ ∈ Rn , the polynomial σ 7→ Pm (ξ + ση) of degree m has only real zeros. For n > 1, this excludes the possibility that P is elliptic, because in that case σ 7→ Pm (ξ + ση) does not have any real zeros if ξ and η ∈ Rn and ξ is not a multiple of η. Hyperbolicity is equivalent to the assertion that Pm (ξ + iτ η) 6= 0 whenever ξ ∈ Rn , τ ∈ R and τ 6= 0. Using the homogeneity of Pm , we obtain |Pm (ξ + iτ η)| ≥ c (kξk + |τ |)m

(ξ ∈ Rn , τ ∈ R),

where c is the infimum of the left-hand side for kξk + |τ | = 1; that is, c > 0. Because the other terms in P (ξ) are of degree < m, there exist a τ0 ≥ 0 and a constant C > 0 such that |P (ξ + iτ η)−1 | ≤ C (kξk + |τ |)−m

(ξ ∈ Rn , τ > τ0 ).

This allows the definition of a distribution E on Rn by means of the formula Z F −1 φ(ξ + iτ η) dξ (φ ∈ C0∞ (Rn )), E(φ) = P (ξ + iτ η) Rn where we have chosen τ > τ0 . For F −1 φ(ζ) = (2π)−n Fφ(−ζ) we use Paley– Wiener estimates of the form (17.3). On the basis of F −1 (t P φ)(ζ) = P (ζ) F −1 φ(ζ), translation of the path of integration leads to Z Z P E(φ) = E(t P φ) = F −1 φ(ξ + iτ η) dξ = F −1 φ(ξ) dξ = φ(0). Rn

Rn

That is, P E = δ; in other words, E is a fundamental solution of P . In addition, by Cauchy’s Integral Theorem (see (12.8)), the definition of E is independent of the choice of τ ≥ τ0 . Using the Paley–Wiener estimates for Fφ(−ζ) we see that the limit as τ → ∞ equals 0 if hx, ηi > 0 for all x ∈ supp φ. That is, supp E ⊂ { x ∈ Rn | hx, ηi ≤ 0 }. We obtain a fundamental solution with support in the opposite half-space by substituting −η for η. Moreover, the Cauchy problem with initial values on the hyperplane hx, ηi = 0 can be shown to have a unique solution. In this text, we will not elaborate the theory any further; the main purpose of the remarks above was to demonstrate that the ideas underlying the Paley–Wiener Theorems can find application in quite a few situations.  Remark 17.7. The constructions of parametrices and solutions of the Cauchy problem of hyperbolic differential equations with variable coefficients find their natural

210

17 Supports and Fourier Transformation

context in the class of the so-called Fourier integral operators. These form an extension of the class of pseudo-differential operators mentioned in Remark 16.8. See, for example, H¨ormander [15, Chap. 25] or Duistermaat [7].  Theorem 17.8. For every Y ⊂ Rn \ {0} and s : Y → R ∪ {∞}, Ks as defined by (17.4) is a closed and convex subset of Rn . Conversely, if K is a closed and convex subset of Rn , then K = Ks if s = sK is defined by (17.1). Proof. A subset K of a linear space is said to be convex if x, y ∈ K implies that l(x, y) ⊂ K; here l(x, y) is the line segment l(x, y) := { x + t (y − x) | 0 ≤ t ≤ 1 } from x to y. It is evident that a half-space H(η, c) := { x ∈ Rn | hx, ηi ≤ c } is convex for η ∈ Rn \ {0} and c ∈ R. Because the intersection of a collection of convex sets is also convex, we deduce the convexity of Ks , the intersection of all H(η, s(η)), where η ∈ Y . Furthermore, Ks is closed, being the intersection of the closed subsets H(η, s(η)) of Rn . With respect to the converse assertion we begin by observing that K ⊂ Ks , for every K ⊂ Rn , if s = sK ; this immediately follows from the definitions. What remains to be shown, therefore, is that if K is convex and closed, and if y ∈ / K, it follows that y ∈ / Ks . That is, there exists an η ∈ Rn such that sK (η) < hy, ηi. Because K is closed, there exists an a ∈ K such that ky − ak = d(y, K) := inf ky − xk. x∈K

Indeed, there exists a sequence xj ∈ K such that ky−xj k → d(x, K). This sequence is bounded, it contains a subsequence that converges to some a; then ky − ak = d(y, K). Because K is closed, a ∈ K. Now let x ∈ K. Because K is convex, we see, for every t ∈ [ 0, 1 ], that a + t (x − a) ∈ K; as a result ky − ak2 ≤ ky − (a + t (x − a))k2 = ky − ak2 − 2t hy − a, x − ai + t2 kx − ak2 . This implies that the derivative of the right-hand side with respect to t cannot be negative at t = 0, that is, hx, y − ai ≤ ha, y − ai

(x ∈ K).

In other words, sK (y − a) = ha, y − ai < hy, y − ai, where the inequality is a consequence of the assumption that y ∈ / K; therefore, y 6= a. This is the desired inequality, with η = y − a. 

17 Supports and Fourier Transformation

211

Problems Problem 17.1. (Sequel to Problem 14.24.) Prove that there exist constants C and k such that √ |u(z)| ≤ C (1 + kzk)k e −λ k Im zk (z ∈ Cn ). n Problem 17.2. Suppose that Y ⊂ Rn consists Pn of a basis η(j) of R , where 1 ≤ j ≤ n, supplemented by a vector η(n + 1) = j=1 cj η(j), with cj < 0 for all j. Further suppose that U is a complex-analytic function on Cn and that for every  > 0 there exist constants C and N with

|U (ξ + iτ η)| ≤ C (1 + kξk)N e τ

(ξ ∈ Rn , η ∈ Y, τ ≥ 0).

Prove that U is a polynomial. Problem 17.3. (Klein–Gordon operator). Discuss fundamental solutions and the Cauchy problem for P =  + λ, where λ ∈ C is a given constant. (For λ = m2 > 0 this is the Klein–Gordon operator.) Problem 17.4. Let u be an integrable function on R. Let H = { ζ ∈ C | Im ζ < 0 } and write H = { ζ ∈ C | Im ζ ≤ 0 }, the closure of H in C. Prove that supp u ⊂ [ 0, ∞ [ if and only if Fu has an extension to a bounded continuous function U on H that is complex-analytic on H.

18 Sobolev Spaces

The degree of differentiability of a function u with compact support cannot easily be seen from the Fourier transform Fu. For instance, there is a difference of more than n between the exponents in the estimates for Fu in Theorems 17.1 and 17.3, which are necessary and sufficient, respectively, for u ∈ C k . The question how to decide by inspection whether a function is the Fourier transform of a bounded continuous function is very difficult to answer in general. In contrast, the situation for the space L2 (Rn ) and the corresponding L2 -norm is n much simpler, because the mapping Fe = (2π)− 2 F is a unitary isomorphism from L2 (Rn ) to L2 (Rn ), see Theorem 14.24. Now FDα u = ξ α Fu, see (14.19), implying that u is an L2 function whose derivatives of order ≤ k all belong to L2 (Rn ) if and only if (1 + kξk)k Fu ∈ L2 (Rn ). Hence, in this case the degree of differentiability can be derived from the Fourier transform. The space of these functions u is denoted by H(k) = H(k) (Rn ). The following extension of this definition proves very flexible. Definition 18.1. For every real number s we define the Sobolev space H(s) = H(s) (Rn ) of order s as the space of all u ∈ S 0 (Rn ) such that (1 + kξk2 )s/2 Fu ∈ L2 (Rn ). This space is provided with the norm 1/2 Z 2 e |Fu(ξ)| (1 + kξk2 )s dξ . (18.1) kuk(s) := Rn

 The norm has been chosen such that H(0) = L2 . The convention to use (1 + kξk ) , instead of 1 + kξk, derives from the fact that 1 + kξk2 is the symbol of 1 − ∆, where ∆ is the Laplace operator. The operators 2 1/2

(1 − ∆)z := F −1 ◦ (1 + kξk2 )z ◦ F can be defined for all z ∈ C; they form a complex-analytic family of operators, analogous to the families of Riesz from Chap. 13. Thus, under this convention H(s)

214

18 Sobolev Spaces

can also be characterized as the space of u ∈ S 0 such that v := (1 − ∆)s/2 u ∈ L2 , and the H(s) norm of u as the L2 norm of v. For estimates the norm convention is hardly of importance, considering that 1 + kξk2 ≤ (1 + kξk)2 ≤ 2 (1 + kξk2 ). Part (a) of the following result is a version of Sobolev’s Embedding Theorem. Theorem 18.2. (a) If u ∈ H(s) and s > k + n2 , one has u ∈ C k and ∂ α u(x) → 0 as kxk → ∞, for every multi-index α with |α| ≤ k. (b) For every u ∈ E 0 , we have that u is of finite order, say k, and u ∈ H(s) if s < −k − n2 . Proof. (a). s > k + n2 means 2(k − s) < −n, that is, (1 + kξk)k−s ∈ L2 . The Cauchy–Schwarz inequality for the integral inner product yields the integrability of the product of two L2 functions. Thus we see, for every multi-index α with |α| ≤ k, that the function ξ 7→ (F(Dα u))(ξ) = ξ α (1 + kξk)−s (1 + kξk)s Fu(ξ) is integrable. In view of Theorem 14.2 this proves (a). (b). Let u ∈ E 0 . From Theorem 8.5 we know that u is of finite order. From (17.2) we read that there exists a C > 0 such that |Fu(ξ)| ≤ C (1 + kξk)k

(ξ ∈ Rn ).

This implies that the square of (1 + kξk)s |Fu(ξ)| can be estimated as (1 + kξk)2(s+k) , which is integrable if 2(s + k) < −n, that is, s < −k − n2 .  Corollary:

\

H(s) ⊂ C ∞

s∈R

and

E0 ⊂

[

H(s) ,

s∈R

but the assertions in Theorem 18.2 are more detailed, of course. Example 18.3. In Example 1.4 we discussed functions v defined on an interval ] a, b [ , such that v and its derivative v 0 are quadratically integrable. If φ belongs to C0∞ ( ] a, b [ ), we find by means of Leibniz’ rule that φ v ∈ H(1) ; therefore, on account of Theorem 18.2, φ v is continuous on R. Because this holds for every φ ∈ C0∞ ( ] a, b [ ), we conclude that v is continuous on ] a, b [ . Furthermore, one has theRestimate (1.8), for every a < x < y < b. Indeed, let φ ∈ C0∞ (R) with φ ≥ 0 and φ(z) dz = 1. As usual, we introduce φ (z) = 1 φ( z ) and define the function ψ in terms of φ (z) by Z z
ψ (z) = φ (ζ − x) − φ (ζ − y) dζ. −∞

18 Sobolev Spaces

215

1

-1

-0.5

1

1.5

2

Fig. 18.1. Graph of ψ with  = 1/5, a = −1, x = −1/2, y = 1 and b = 2

If  is sufficiently small, one has ψ ∈ C0∞ ( ] a, b [ ). Furthermore, v(x) − v(y) equals the limit of the v(ψ0 ) as  ↓ 0, that is, of the −v 0 (ψ ). On the strength of the Cauchy–Schwarz inequality, the absolute value of −v 0 (ψ ) is smaller than or equal to the product of the L2 norm of v 0 and the L2 norm of ψ . Now, ψ converges, in the L2 -sense, to the characteristic function of [ x, y ] and so the L2 norm of ψ converges to (y − x)1/2 as  ↓ 0. This proves (1.8). Applying the Cauchy criterion for convergence we now also see that v(x) converges as x ↓ a or x ↑ b, respectively; write the corresponding limits as v(a) and v(b). Thus we have extended v to a continuous function on [ a, b ]. It now makes α,β sense to speak of the space H(1) of all v ∈ L2 ( ] a, b [ ) such that v 0 ∈ L2 ( ]a, b[ ) and for which v(a) = α and v(b) = β. What is more, v is uniformly H¨older continuous with exponent 1/2, because (1.8) now holds for all x and y ∈ [ a, b ].  Theorem 18.4. Let P (D) be a linear partial differential operator with constant coefficients, of order m. One then has, for every s ∈ R, (a) P (D) is a continuous linear mapping from H(s) to H(s−m) . (b) If P (D) is elliptic, u ∈ E 0 and P (D)u ∈ H(s−m) , then u ∈ H(s) . Proof. As regards (a), one only needs to observe that ξ 7→ (1 + kξk)−m P (ξ) is e bounded, so that the L2 norm of (1 + kξk)s−m P (ξ) Fu(ξ) can be estimated as a 2 s e constant times the L norm of (1 + kξk) Fu(ξ). To prove (b) we use (16.3). If kξk ≥ R we have U (ξ) := (1 + kξk)s Fu(ξ) =

(1 + kξk)m (1 + kξk)s−m F(P (D)u)(ξ). P (ξ)

Write B = { ξ ∈ Rn | kξk < R }. The function |U (ξ)|2 is integrable over Rn \ B, because the factor (1 + kξk)m /P (ξ) is bounded on Rn \ B Rand the other factor in U (ξ) is quadratically integrable over Rn . The conclusion B |U (ξ)|2 dξ < ∞ follows from the fact that U is continuous (and even analytic).  The definition of H(s) (Rn ) makes use of the global operation of Fourier transformation, which precludes a similar definition of a Sobolev space H(s) (X), for arbitrary open X ⊂ Rn . Still, there exists a localized version of the Sobolev spaces. Definition 18.5. Let X be an open subset of Rn and u ∈ D0 (X) a distribution on X. The distribution u is said locally to belong to H(s) if φ u ∈ H(s) (Rn ) for every loc φ ∈ C0∞ (X). The space of all such u is denoted by H(s) (X). 

216

18 Sobolev Spaces

Theorem 18.6. Let k ∈ Z≥0 and s ∈ R. loc loc (a) H(s) (X) ⊂ C k (X) ⊂ H(k) (X) for s > k + n2 . loc (b) If u is a distribution of order ≤ k in X and s < −k − n2 , then u ∈ H(s) (X). If loc u ∈ H(−k) (X), it follows that u is a distribution of order k.

loc Proof. (a). Let u ∈ H(s) (X) and s > k + n2 . For every φ ∈ C0∞ (X) one then n has φ u ∈ H(s) (R ), and therefore φ u ∈ C k (Rn ) on account of Theorem 18.2. This implies u = φ1 φ u ∈ C k , where φ(x) 6= 0. For every a ∈ X there exists a φ ∈ C0∞ (X), with φ(x) 6= 0 for all x in an open neighborhood of a; this implies that u ∈ C k (X). The second inclusion follows from the fact that a continuous function with compact support is quadratically integrable. As regards the first assertion in (b) we note that φ u ∈ E 0 (X) is of order ≤ k, hence Theorem 18.2 implies φ u ∈ H(s) (Rn ). Because this holds for every φ ∈ loc C0∞ (X), we conclude that u ∈ H(s) X. The proof of the second assertion is left to the reader; it may be seen in relation to the continuous inclusion of C k (X) in loc H(k) (X). 

Theorem 18.7. (a) Let X be an open subset of Rn and let P (x, D) be a linear partial differenloc tial operator of order m with C ∞ coefficients on X. If u ∈ H(s) (X), one has loc P (x, D)u ∈ H(s−m) (X). (b) In addition, suppose that P (D) is an elliptic operator with constant coefficients loc loc and let P (D)u ∈ H(s−m) (X). Then u ∈ H(s) (X). Proof. (a) follows if the following two assertions are combined: loc loc (X), one has Dj u ∈ H(s−1) (X), for every 1 ≤ j ≤ n. (i) If u ∈ H(s) loc ∞ loc (ii) If u ∈ H(s) (X) and ψ ∈ C (X), one has ψ u ∈ H(s) (X).

In the arguments below, φ is an arbitrary element of C0∞ (X). As regards (i) we note that φ Dj u = Dj (φ u) − (Dj φ) u. Here φ u ∈ H(s) (Rn ), hence Dj (φ u) ∈ H(s−1) (Rn ) on the strength of Theorem 18.4. Because the second term lies in H(s) ⊂ H(s−1) , the conclusion is that φ Dj u ∈ H(s−1) , for all φ ∈ loc (X). With respect to (ii) we note that φ ψ ∈ C ∞ (X), C0∞ (X); that is, Dj u ∈ H(s−1) 0 and therefore φ (ψ u) = (φ ψ) u ∈ H(s) . (b). Let K be a compact subset of X. On account of Theorem 8.5, there exists a k ∈ Z≥0 such that ψ u is of order ≤ k whenever ψ ∈ C0∞ (X) and supp ψ ⊂ K. Choose l ∈ Z such that s − l < −k − n2 . In view of Theorem 18.2 we then have ψ u ∈ H(s−l) . Now write

18 Sobolev Spaces

P (D)(φ u) = φ P (D)u +

X

P(α) (D)(Dα φ u),

217

(18.2)

α6=0, |α|≤m

where P(α) (D) are operators of order < m. This is true for every operator P (D) of order ≤ m. The proof proceeds by first considering the case of P (D) = Dα , by mathematical induction on |α|. Now suppose that φ ∈ C0∞ (X) and supp φ ⊂ K. In that case, the support of ψ := Dα φ is contained in K, and so ψ u ∈ H(s−l) . Using Theorem 18.4.(a), we find that the terms following the summation sign in (18.2) belong to H(s−l−m+1) . Because it is given that φ P (D)u ∈ H(s−m) , we can conclude that P (D)(φ u) ∈ H(s−l−m+1) , that is, so long as l ≥ 1. If we now apply Theorem 18.4.(b), we find that φ u ∈ H(s−l+1) . By descending mathematical induction on l we arrive at the result that φ u ∈ H(s) , whenever φ ∈ C0∞ (X) and supp φ ⊂ K. The fact that this loc holds true for every compact subset K of X implies u ∈ H(s) (X).  loc Remark 18.8. An important fact that justifies the name H(s) , is that loc H(s) (Rn ) ⊂ H(s) (Rn ).

This means that, for every φ ∈ C0∞ (Rn ), the operator A := (1 − ∆)s/2 ◦ φ ◦ (1 − ∆)−s/2 maps the space L2 (Rn ) in itself. The most transparent proof of this is obtained in the framework of pseudo-differential operators: A is a pseudo-differential operator of order 0, and every pseudo-differential operator of order ≤ 0 is a continuous linear mapping from L2 (Rn ) to L2 (Rn ). By means of pseudo-differential operators one also proves that the second assertion in Theorem 18.7 remains valid when an elliptic operator P (x, D) with C ∞ coefficients on X is substituted for P (D).  Remark 18.9. The fact that Sobolev norms are better suited to partial differential operators than C k norms is also illustrated by the fact that there exist continuous functions f with compact support in Rn whose potential is not C 2 ; at least, not if n > 1. For example, let χ(x) x1 2 , f (x) = − kxk2 log kxk if 0 < kxk < 1 and f (x) = 0 in all other cases. Here χ ∈ C0∞ (Rn ), χ(0) > 0 and χ(x) = 0 if kxk ≥ r, where 0 < r < 1. Then f is continuous if we take f (0) = 0. Furthermore, f has compact support and sing supp f ⊂ {0}. For the potential u = E ∗ f of f we obtain ∆u = f , and therefore sing supp u ⊂ {0}. However, further computation yields lim

x→0, x6=0

∂1 2 u(x) = ∞,

218

18 Sobolev Spaces

in other words, ∂1 2 u is certainly not continuous at 0. The function f is said to be H¨older continuous if there exist a constant C > 0 and an α > 0 such that |f (x) − f (y)| ≤ C kx − ykα , for all x and y in the compact support of f . If f is H¨older continuous, the potential u of f is actually C 2 and all second-order derivatives of u are also H¨older continuous. Furthermore, Sobolev norms are meaningful for general distributions. However, the notion of “C k plus H¨older” is preferred if uniform estimates for function values are required. 

Problems Problem 18.1. Let P (D) be an elliptic operator of order m and E the parametrix of P (D) constructed in Lemma 16.5. Prove that E ∗ : u 7→ E ∗ u is a continuous linear mapping from H(s−m) to H(s) . Problem 18.2. Let E be the potential of δ in Rn , for n ≥ 3. Is E ∗ u ∈ H(s) for every u ∈ H(s−2) ? Problem 18.3. Let at and bt be the solutions of the Cauchy problem for the wave equation from Example 17.4. Prove, for arbitrary s ∈ R, d at ∗ are continuous linear mappings from H(s) to H(s+1) , to (i) bt ∗, at ∗ and dt H(s) and to H(s−1) , respectively. (ii) If ut is a solution of the homogeneous wave equation with u0 ∈ H(s) and d d dt ut |t=0 ∈ H(s−1) , one has, for every t ∈ R, that ut ∈ H(s) and dt ut ∈ H(s−1) .

Problem 18.4. Examine the following functions on R and decide in which Sobolev spaces they are contained, and in which they are not (parts (iii) and (iv) are difficult): (i) the characteristic function of the interval ] 2, 5 ], (ii) x 7→ e−x H(x), (iii) x 7→ H(x)/(x + 1), (iv) x 7→ x H(x), (v) the function given by x 7→ 1 − x2 for |x| ≤ 1, and zero elsewhere. Problem 18.5. (Fourth-order ordinary differential operator). Define the differential operator P (∂) acting in D0 (R) by P (∂)u = ∂ 4 u + u. (i) Prove that P (∂)u ∈ C ∞ (R) implies u ∈ C ∞ (R). Conclude that N ⊂ C ∞ (R) if N is the solution space of the homogeneous equation P (∂)u = 0. (ii) Verify that N is spanned by the following four functions: x 7→ ew(±1±i)x

with

w=

1√ 2 2

and where all combinations of the + and − signs occur. Conclude N ∩ S 0 (R) = {0}.

18 Sobolev Spaces

219

(iii) Consider a fundamental solution E of P (∂). Show that E does not belong to E 0 (R), but that there exists a unique E such that E ∈ S 0 (R). Suppose u ∈ S 0 (R). We now study the properties of u in the case where this distribution satisfies the inhomogeneous equation P (∂)u = f ∈ D0 (R). (iv) Show that f ∈ S 0 (R) is a necessary condition under this assumption. Next, consider the special case of f ∈ L2 (R). (v) Prove that ∂ j u ∈ L2 (R), for 0 ≤ j ≤ 4. Show that in addition u ∈ C 3 (R), but that u ∈ / C 4 (R) if f ∈ / C(R). f (ξ) 2 2 (vi) Verify that g ∈ L (R) if g(ξ) = F1+ξ 4 , and prove that a solution u ∈ L (R) of −1 P (∂)u = f is uniquely determined and is given by u = F g.

f 2

-6

3

-3

6

-2

Fig. 18.2. Graph of f

Finally, define f ∈ L2 (R) by f (x) = 2 sgn(x)e−|x| ; observe that f is discontinuous at 0. We now prove that the solution u ∈ L2 (R) of P (∂)u = f is given by (?)

u(x) = sgn(x)e−|x| + e−w|x| (sin wx − sgn(x) cos wx).

(vii) By an a priori argument show that sing supp u ⊂ {0}. (viii) Determine the solution u from (?) by the method of part (vi). To do so, prove (if necessary, by means of [8, Exercise 0.8]) Ff (ξ) = −4i

ξ ξ2 + 1

and

Fu(ξ) =

1 ξ − ξ3 Ff (ξ) − 2i 4 . 2 ξ +1

Using the theory of complex functions, compute Z ξ eixξ 4 dξ = πie−w|x| sin wx =: h(x) ξ +1 R

(x ∈ R).

Note that different contours are required depending on whether x > 0 or x < 0, and that the contour is followed in the clockwise direction in the latter case; or use the fact that the integral is an odd function of x. Next, conclude by means of Fourier theory Z ξ3 (??) eixξ 4 dξ = −∂ 2 h(x) = πi sgn(x)e−w|x| cos wx. ξ +1 R

220

18 Sobolev Spaces

(Note that the integral is not absolutely convergent and should, in fact, be interpreted as the Fourier transform of a square-integrable function.) Conclude that u is given by (?). Background. The following is an example of the situation described in part (v). The function u from (?) belongs to C 3 (R) but not to C 4 (R) (that is, u is not a classical solution). Indeed (see Fig. 18.3 below), √ ∂ 2 u(0) = 0, ∂ 3 u(0) = −1, u(0) = 0, ∂u(0) = −1 + 2, limx↑0 ∂ 4 u(x) = −2 6= 2 = limx↓0 ∂ 4 u(x). Alternatively, the solution u in (?) may be obtained by requiring that the L2 solutions u± on ±R>0 given by u+ (x) =

e−x + a+ ew(−1+i)x + b+ ew(−1−i)x

u− (x) = −ex + a− ew(

1+i)x

+ b− ew(

1−i)x

for x ∈

R>0 ,

for x ∈ −R>0 ,

have coinciding derivatives of order 3 or lower at 0, and solving the resulting system of linear equations for the constants a± and b± . This amounts to a computation of the following integrals without complex analysis: Z ξ cos xξ π dξ = e−w|x| sin wx, 4+1 ξ 2 R>0 Z ξ 3 cos xξ π dξ = sgn(x)e−w|x| cos wx. 4+1 ξ 2 R>0

18 Sobolev Spaces

u 0.4

-10

10

-0.4

0.4145

¶u

-0.1

0.1

0.1

-0.1

0.1 ¶2 u

-0.1 -0.8

¶3 u

-0.1

0.1 2 ¶4 u

-0.1

0.1

-2

Fig. 18.3. Graphs of ∂ j u, for 0 ≤ j ≤ 4

221

19 Appendix: Results from Measure Theory in the Context of Distributions

For the benefit of readers not familiar with measure theory and the theory of Lebesgue integration, we discuss some important results from these theories of particular relevance to the theory of distributions. Completeness of a linear space E provided with a norm k · k means that every Cauchy sequence with respect to k · k in E converge to an element of E. That is, if E is not complete, then some Cauchy sequences in E do not converge. By emulating the Cantor process for getting the reals from the rationals, one successively adjoins ideal elements to E that act as limits for the nonconvergent Cauchy sequences. Thus b in which E is dense, and an extension of one ultimately obtains an enlarged space E b b is then k · k to a norm on E such that the latter is a complete normed linear space. E said to be a completion of E. Any two completions of a normed linear space E are linearly isometric. Indeed, suppose i : E → F and j : E → G are norm-preserving linear injections from E into the normed linear spaces F and G with dense images, respectively. Then j ◦ i−1 : i(E) → G is uniformly continuous while i(E) is a dense linear subspace of F . Hence, this mapping can be uniquely extended as a normpreserving linear mapping that sends all of F to G; a similar statement is true for i ◦ j −1 . One of the main results in the theory of Lebesgue integration is the explicit description of a completion of the linear space C0 (X) provided with the Lp norm. Here X denotes an open subset of Rn and C0 (X) is the linear space of all continuous functions on X with compact support, which contains C0∞ (X); furthermore, C0 (X) is provided with the Lp norm, for 1 ≤ p < ∞, Z 1/p kf kLp = |f (x)|p dx (f ∈ C0∞ (X)). X

Minkowski’s inequality (see, for instance [8, Exercise 6.73].(iii)) implies that one does indeed obtain a norm. The completion of C0 (X) is usually given by the Banach space Lp (X) of all Lebesgue measurable functions u on X modulo functions that vanish almost everywhere, such that |u|p is integrable, and provided with the norm kukLp . The case where p = 1 has already been discussed.

224

19 Appendix: Results from Measure Theory in the Context of Distributions

Remark 19.1. With a view to the description in Proposition 19.4 below of the spaces Lp (X), and of other generalizations as much as possible in the context of distribution theory we now list some properties of the spaces Lp (X). The inclusion mapping Lp (X) → D0 (X) given by u 7→ test u is injective, as a consequence of Remark 3.6. On account of H¨older’s inequality (see, for instance [8, Exercise 6.73].(i)) we obtain, for arbitrary u ∈ Lp (X) and φ ∈ C0∞ (X), Z |u(φ)| = u(x)φ(x) dx ≤ kukLp kφkLq . (19.1) X

Here q is the real number such that 1/p + 1/q = 1, with the proviso that q = ∞ if p = 1 and that L∞ denotes the supremum norm. This implies that the injection Lp (X) → D0 (X) is continuous. Furthermore, consider a sequence (uj )j∈Z>0 in C0∞ (X) that is a Cauchy sequence with respect to the Lp norm and satisfies limj→∞ uj = 0 in D0 (X). Then we claim that limj→∞ kuj kLp = 0. Indeed, on account of the completeness of Lp (X), there exists u ∈ Lp (X) ⊂ D0 (X) such that limj→∞ kuj − ukLp = 0. In view of (19.1) this implies u(φ) = limj→∞ uj (φ) = 0, for every φ ∈ C0∞ (X); and therefore u = 0 in Lp (X). This in turn leads to the desired conclusion.  It is our next goal to show that the arguments above can be put in a more general form, as follows: (i) E is a normed linear subspace of D0 (X) provided with a norm k · k. (ii) The linear injection E → D0 (X) is continuous, in other words, the norm topology on E is stronger than the restriction to E of the topology on D0 (X). (iii) E possess the Radon–Riesz property, which by definition means limj→∞ kuj k = 0 if (uj )j∈Z>0 is a Cauchy sequence in E satisfying limj→∞ uj = 0 in D0 (X). The term Radon–Riesz property used in (iii) is justified by the fact that in the case of E = Lp (X), Theorem 19.3 below is related to a special case of the Radon– Riesz Theorem, see Hewitt and Stromberg [13, Exercise 15.17]. The present proof, however, is different; in particular, contrary to the one loc. cit., it does not use local uniform convexity. Example 19.2. Select φ ∈ C0∞ (R) with supp φ ⊂ [ −1/4, 1/4 ] and define ej ∈ C0∞ (R) by ej = Tj φ, for j ∈ Z>0 (see Example 10.11). Clearly, the ej are linearly independent in C0∞ (R); let E be the linear subspace of C0∞ (R) consisting of all finite linear combinations of the ej . We now construct an inner product on E such that the associated normSdoes not satisfy the Radon–Riesz condition. Indeed, let En = ⊕nj=1 Rej ; then E = n∈Z>0 En . We provide each En with an inner product by mathematical induction on n as follows. Suppose an inner product has been fixed for En ; then extend this to an inner product on En+1 by the following two requirements: en+1 − en ∈ En+1 is perpendicular to En and ken+1 − en k = n > 0 where P 0 n n∈Z>0 n < ∞. Obviously, we obtain limn→∞ en = 0 in D (R ) whereas, for j < k,

19 Appendix: Results from Measure Theory in the Context of Distributions

kej − ek k ≤

k−1 X

ken − en+1 k =

n=j

k−1 X

225

n ,

n=j

ken+1 k2 = k(en+1 − en ) + en k2 = ken+1 − en k2 + ken k2 > ken k2 . The former inequality expresses that (en ) is a Cauchy sequence with respect to the norm on E, whereas the latter says limn→∞ ken k 6= 0. Phrased differently, E does not satisfy condition (iii) above.  Theorem 19.3. Given E possessing the properties (i) – (iii) above, there exists a b of D0 (X) whose norm is also denoted by k · k. complete normed linear subspace E b with respect to the norm k · k and E b is uniquely determined up to E is dense in E b linear isometries. To be more precise, E may be taken as { u ∈ D0 (X) | ∃ Cauchy sequence (uj )j∈Z>0 in E with lim uj = u in D0 (X) } j→∞

and

kuk := limj→∞ kuj k

b (u ∈ E).

Proof. According to (8.2), condition (ii) means that for every φ ∈ C0∞ (X) there exists a constant c(φ) > 0 such that |u(φ)| ≤ c(φ) kuk

(u ∈ E).

Now consider a Cauchy sequence (uj ) in E. Then, for arbitrary φ ∈ C0∞ (X), |uj (φ) − uk (φ)| ≤ c(φ) kuj − uk k

(j, k ∈ Z>0 );

hence, in view of the completeness of C there exists u(φ) ∈ C such that u(φ) = limj→∞ uj (φ). Clearly, the mapping φ 7→ u(φ) defines a linear functional, say u, on C0∞ (X). Next fix a compact set K ⊂ X and apply Lemma 5.4, a consequence of the principle of uniform boundedness, to find a constant C > 0 and a k ∈ Z≥0 such that |uj (φ)| ≤ C kφkC k (j ∈ Z>0 , φ ∈ C0∞ (K)). By taking the limit for j → ∞ in |u(φ)| ≤ |(u − uj )(φ)| + |uj (φ)| ≤ |(u − uj )(φ)| + C kφkC k , we obtain |u(φ)| ≤ C kφkC k , for all φ ∈ C0∞ (K). But Lemma 3.7 then implies u ∈ D0 (X), while limj→∞ uj = u in D0 (X) by definition. b as in the theorem. If u ∈ E b and (uj ) is an approximating Cauchy Next, define E sequence in E, then (kuj k) is a Cauchy sequence in R; accordingly kuk := lim kuj k j→∞

226

19 Appendix: Results from Measure Theory in the Context of Distributions

exists and is actually independent of the particular choice of the Cauchy sequence (uj ). The proof of the latter assertion requires the Radon–Riesz condition (iii). It is b k · k) is a normed linear space. Furthermore, now a standard verification that (E, b b In fact, if u ∈ E, b then there E is dense in E with respect to the norm k · k on E. 0 exists a Cauchy sequence (uj ) in E with limj→∞ uj = u in D (X). In that case, b for every (uk − uj )k∈Z>0 is a Cauchy sequence in E determining u − uj ∈ E, j ∈ Z>0 . Hence, by definition, ku − uj k = lim kuk − uj k, k→∞

and so, (uj ) being a Cauchy sequence in E, lim ku − uj k = lim ( lim kuk − uj k) = 0.

j→∞

j→∞ k→∞

b is complete with respect to its norm. To this end, Finally, we prove that E b Given that E is dense in E, b there exist consider a Cauchy sequence (uj ) in E. uj,k(j) ∈ E such that kuj,k(j) − uj k
0 ).

(19.2)

Accordingly, for every  > 0, there exists N ∈ Z>0 such that, for all j and j 0 > N , kuj,k(j) − uj 0 ,k(j 0 ) k ≤ kuj,k(j) − uj k + kuj − uj 0 k + kuj 0 − uj 0 ,k(j 0 ) k 1 1 < +  + 0. (19.3) j j Hence (uj,k(j) )j∈Z>0 is a Cauchy sequence in E and therefore in D0 (X) as well. By the completeness of D0 (X) (see Theorem 5.5), there exists u ∈ D0 (X) satisfying b by definition of E. b In addition, limj→∞ uj,k(j) = u in D0 (X); but then u ∈ E 0 b In fact, by taking limits in (19.3) as j → ∞, we obtain u = limj→∞ uj in E. kuj,k(j) − uk < 2, if j sufficiently large. Using (19.2) we then get limj→∞ kuj − uk = 0.  Applying Theorem 19.3 to the space E = C0 (X) provided with the Lp norm b∼ and to its completion E = Lp (X) (here we use uniqueness of completions), we may deduce the following characterization of Lp (X), which does not involve the notion of Lebesgue measurability. Proposition 19.4. Lp (X) considered as a linear subspace of D0 (X) equals the collection of all u ∈ D0 (X) for which there exists a Cauchy sequence (uj )j∈Z>0 in C0 (X) with respect to the Lp norm that satisfies limj→∞ uj = u in D0 (X). Furthermore, kukLp = limj→∞ kuj kLp , independently of the particular choice of the sequence (uj )j∈Z>0 .

19 Appendix: Results from Measure Theory in the Context of Distributions

227

Observe that measure theory has been applied in obtaining this proposition. For the benefit of readers not familiar with that theory, we construct, in Corollary 19.9 below, a completion contained in D0 (X), of the space C0 (X) provided with the Lp norm, without making use of measure theory. Again, we study the problem in a more general setting, seeking to obtain sufficient conditions to ensure the Radon–Riesz property is met; Lp norms on C0 (X) will clearly satisfy such conditions. First we note that many norms on C0∞ (X) actually are restrictions to that space of translation-invariant norms on C0∞ (Rn ). This often allows the use of convolution. Invariance under translations is not guaranteed. For instance, consider the case of integral norms involving a weight function, like R1 kuk = ( −1 |u(x)|2 (1 − x2 )−1/2 dx)1/2 , which prescribe a certain behavior of u near the boundary of X. A remarkable property of the Lp norm k · k, is the following locality property: ku + vkp = kukp + kvkp

if

supp u ∩ supp v = ∅.

Actually, a weak form of such a property is necessary for the validity of the conclusion of Theorem 19.3. Indeed, when that theorem applies, then for every Cauchy sequence (φj ) in C0∞ (X) there exists u ∈ D0 (X) so that limj→∞ φj = u in D0 (X) and limj→∞ kφj − uk = 0. Under the additional assumption that the supports of the (φj ) move away towards infinity, limj→∞ φj = u in D0 (X) implies that u = 0. Thus, the conclusion is limj→∞ kφj k = 0; and this in turn gives kφk = limj→∞ kφ − φj k, for every φ ∈ C0∞ (X). This argument is formalized in the following Definition 19.5. We say that a sequence (φj ) in C0∞ (Rn ) localizes at infinity if for every compact subset K of Rn there exists N having the property that supp φj is contained in the complement of K, for all j ≥ N . The norm k · k on C0∞ (Rn ) is said to possess the weak locality property if, for any φ ∈ C0∞ (Rn ) and any Cauchy sequence (φj ) that localizes at infinity, one has kφk ≤ lim supj→∞ kφ − φj k.  Example 19.6. For a suitable choice of the n in Example 19.2, the norm in the example does not have the weak locality property. Indeed, select the n so that P 2 2 n∈Z>0 n < ke1 k . Then (en ) is a Cauchy sequence that localizes at infinity, whereas n−1 n−1

2 X X

kej − ej+1 k2 k2e1 − en k2 = e1 + (ej − ej+1 ) = ke1 k2 + j=1 2

≤ ke1 k +

X

j=1

2j

2

< 4ke1 k .

j∈Z>0

Hence, k2e1 k > lim supn→∞ k2e1 − en k. Note that the norm is not defined on all of C0∞ (Rn ) and neither is it invariant under translation.  We are now sufficiently equipped to formulate Theorem 19.7. Suppose

228

19 Appendix: Results from Measure Theory in the Context of Distributions

(i) C0∞ (Rn ) is provided with a norm k · k that is translation-invariant. (ii) The norm topology on C0∞ (Rn ) is weaker than the topology of C0∞ (Rn ), but stronger than the restriction to C0∞ (Rn ) of the topology of D0 (Rn ). (iii) The weak locality property is satisfied by k · k. Then k·k has the Radon–Riesz property. In particular, D0 (Rn ) contains a completion of C0∞ (Rn ) with respect to this norm. The first step in the proof of this theorem is an auxiliary result on convolution, which is of interest in its own right. The formulation of the lemma below is slightly more general than would be necessary for the proof of the theorem. Lemma 19.8. Let (uj ) be a sequence in D0 (Rn ) such that limj→∞ uj = 0 in D0 (Rn ). Let φ ∈ C0∞ (Rn ) be arbitrary and write vj = φ ∗ uj ∈ C ∞ (Rn ). Then we have the following assertions. (a) limj→∞ vj (x) = 0, for all x ∈ Rn . (b) Stronger even, limj→∞ vj = 0 in C ∞ (Rn ). Proof. (a). This follows from vj (x) = uj (Tx ◦ Sφ) and limj→∞ uj = 0 in D0 (Rn ). (b). We may write, for x, y and z ∈ Rn , Z 1 d φ(z − y − t(x − y)) dt (Tx − Ty )φ(z) = φ(z − x) − φ(z − y) = 0 dt Z 1X n = ∂l φ(z − y − t(x − y)) dt (yl − xl ) 0

=

l=1

n Z 1 X l=1

Ty+t(x−y) (∂l φ)(z) dt (yl − xl ).

0

Therefore (Tx − Ty )φ =

n Z X

1

Ty+t(x−y) ◦ ∂l φ dt (yl − xl );

0

l=1

thus, for any j ∈ Z>0 , vj (x) − vj (y) = uj ((Tx − Ty )Sφ) X  n Z 1 = uj Ty+t(x−y) ◦ ∂l (Sφ) dt (yl − xl ) l=1

=

n Z X l=1

(19.4)

0

1


uj (Ty+t(x−y) ◦ ∂l ◦ S)φ dt (yl − xl )

0

Let K be an arbitrary compact subset of Rn . The mapping [ 0, 1 ] × supp φ × K 2 → Rn

with

(t, z, x, y) 7→ −z + y + t(x − y)

19 Appendix: Results from Measure Theory in the Context of Distributions

229

is continuous with a compact domain space; hence, its image L is compact. Note supp ((Ty+t(x−y) ◦ ∂l ◦ S)φ) ⊂ L

(0 ≤ t ≤ 1, x, y ∈ K, 1 ≤ l ≤ n).

Since (uj ) satisfies the conditions of Lemma 5.4, we may apply the lemma to the compact set L, to find c > 0 and k ∈ Z≥0 such that, for all j ∈ Z>0 , 0 ≤ t ≤ 1, x, y ∈ K and 1 ≤ l ≤ n, |uj ((Ty+t(x−y) ◦ ∂l ◦ S)φ)| ≤ c k(Ty+t(x−y) ◦ ∂l ◦ S)φkC k , L ≤ kφkC k+1 . It follows from (19.4) that there exists d > 0 such that, for all j ∈ Z>0 and x, y ∈ K, |vj (x) − vj (y)| ≤ d kx − yk. Next, let  > 0 be arbitrary. Then we can find finitely many points xm ∈ K, with 1 ≤ m ≤ M , such that K is contained in the union of the balls in Rn of center xm and radius /(2d), for 1 ≤ m ≤ M , see Theorem 2.2. Since limj→∞ vj (xm ) = 0, for every 1 ≤ m ≤ M , there exists N ∈ Z>0 such that, for all j ≥ N and 1 ≤ m ≤ M,  |vj (xm )| < . 2 Furthermore, for arbitrary x ∈ K, we can find xm ∈ K with kx − xm k < /(2d). Thus we obtain, for all j ≥ N , |vj (x)| ≤ |vj (x) − vj (xm )| + |vj (xm )| < d

  + = . 2d 2

In other words, limj→∞ vj = 0 uniformly on compact subsets of Rn , that is, in C 0 (Rn ). Finally, we obtain limj→∞ vj = 0 in C ∞ (Rn ) by application of the preceding argument to ∂ α (φ ∗ uj ) = ∂ α φ ∗ uj .  Next, we take up the proof of Theorem 19.7. Proof. Let (uj ) be a sequence in C0∞ (Rn ) that is a Cauchy sequence with respect to k · k and satisfies limj→∞ uj = 0 in D0 (Rn ). Let φ and vj be as in Lemma 19.8. First, observe that (vj ) is a Cauchy sequence with respect to k · k. Indeed, on account of assumption (i) in the theorem we have, for u ∈ C0∞ (Rn ),

Z

Z

kφ ∗ uk = φ(x) Tx u dx ≤ |φ(x)| kTx uk dx ≤ kφkL1 kuk. Rn

Rn

Next, select a sequence (Ki )i∈Z>0 of compact subsets of Rn that absorbs Rn while supp ui ⊂ Ki , and let (χi )i be a corresponding sequence of functions in C0∞ (Rn ) such that χi = 1 on Ki . Then χi = 1 on supp vj , for all j ≤ i. For fixed i, we have on account of Lemma 19.8.(b) that limk→∞ χi vk = 0 in C0∞ (Rn ). Accordingly, limk→∞ kχi vk k = 0 in view of the norm topology being weaker than the topology of C0∞ (Rn ) (assumption (ii)). It follows that we can find a sequence (k(i)) converging to ∞ and having the property that η(i) := kχi vk(i) k converges

230

19 Appendix: Results from Measure Theory in the Context of Distributions

to 0 as i → ∞. Furthermore, denote by (j) the supremum over all k ≥ j of the kvj −vk k; then (vj ) being a Cauchy sequence implies limj→∞ (j) = 0. Combining these results we find, for given j and taking i so large that k(i) ≥ j, kvj − (1 − χi ) vk(i) k ≤ kvj − vk(i) k + kχi vk(i) k ≤ (j) + η(i). However, ((1 − χi )vk(i) )i is a sequence that localizes at infinity and therefore the weak locality property from assumption (iii), in conjunction with the preceding estimate, implies kvj k ≤ lim sup kvj − (1 − χi ) vk(i) k ≤ (j) + lim sup η(i) = (j). i→∞

i→∞

Thus we may conclude lim kvj k ≤ lim j = 0.

j→∞

(19.5)

j→∞

Finally, we return to the (uj )j . For all j and k ∈ Z>0 , we have kuj k ≤ kuj − uk k + kuk − vk k + kvk k. The three terms at the right can each be made arbitrarily small by taking j and k sufficiently large: the first term because (uj ) is a Cauchy sequence and the third on account of (19.5); as regards the second term, consider the following estimate, for k fixed and sufficiently large. For any δ > 0, select φ = φδ ∈ C0∞ (Rn ) as in Lemma 2.18 and observe

Z

Z

kφδ ∗ uk − uk k = φδ (x)(Tx uk − uk ) dx ≤ φδ (x) kTx uk − uk k dx Rn

≤

sup

Rn

kTx uk − uk k.

{ x∈Rn |kxk 0 be arbitrarily selected, then the Cauchy condition implies the existence of N ∈ Z>0 such that

19 Appendix: Results from Measure Theory in the Context of Distributions

 2

kuj − uk k
0 ). For k · k we now take the norm k · k(−1) on the Sobolev space H(−1) (R) from Definition 18.1. The identity F(Th u)(ξ) = e−ih h, ξ i Fu(ξ) shows that the norms on the Sobolev spaces are invariant under translations. Now δ ∈ H(−1) (R), Z because Fδ = 1 implies dξ kδk2 = = π. 1 + ξ2 R Furthermore, since Fuj (ξ) = Fu(ξ/j), we have 2

Z

kδ − uj k = R

|1 − Fu( ξj )|2 1 + ξ2

dξ.

The Paley-Wiener estimate from Theorem 17.1 implies the existence of a constant c > 0 such that |1 − Fu(ξ)|2 ≤ c, for all ξ ∈ R. Given arbitrary  > 0, we can find C ∈ R>0 such that

232

19 Appendix: Results from Measure Theory in the Context of Distributions

Z { |ξ|≥C }

 dξ < . 2 1+ξ 2c

The continuity of Fu on [ −C, C ] gives the existence of N ∈ Z>0 such that, for all |ξ| ≤ C and j ≥ N , we have |1 − Fu(ξ/j)|2 < /(2π). Hence, for all j ≥ N , kδ − uj k2 =

Z

Z +

{ |ξ|≤C }

 ≤ 2π

Z R



|1 − Fu( ξj )|2

{ |ξ|≥C }

1 + ξ2

dξ

 dξ +c = . 2 1+ξ 2c

In other words, limj→∞ uj = δ in H(−1) (R); in particular, (uj )j is a Cauchy sequence in C0∞ (X) with respect to k · k. 

20 Solutions to Selected Problems

1.1 Let φ be continuously differentiable and suppose φ(x) = 0 if |x| ≥ m > 0. We introduce some notation. For  ≥ 0 and x ∈ R, set φ(x) − φ(0) ; and Ia = ] −a, a [ (a ∈ R). x + i Note that Φ is continuous on R (see [8, Proposition 2.2.1.(ii)], for all  ≥ 0. Now we get, writing φ(x) = φ(0) + (φ(x) − φ(0)), Z Z Z Z φ(x) φ(x) x − i dx = dx = φ(0) dx + Φ (x) dx. 2 2 R x + i Im x + i Im x +  Im Φ (x) =

As x 7→ x/(x2 + 2 ) is an odd function on R and Im is symmetric about the origin, we obtain by means of the change of variables x =  y Z Z Z x − i  1 dx = −i dx = −i dy 2 2 2 2 2 Im x +  Im x +  Im y + 1 

m = −2i arctan → −πi as  ↓ 0.  Furthermore, we have Z

Z lim ↓0

Φ (x) dx = Im

Φ0 (x) dx.

(20.1)

Im

Postponing for the moment the proof of this identity, we apply it to obtain Z Z m  Z −δ   1 Φ0 (x) dx + Φ0 (x) dx = PV lim Φ (x) dx = lim (φ). ↓0 I δ↓0 x −m δ m With the notation δ(φ) = φ(0), combination of these identitites leads to the following Plemelj–Sokhotsky jump relations: 1 1 1  1 1  ± πi δ = PV , δ= − . x ± i0 x 2πi x − i 0 x + i 0

234

20 Solutions to Selected Problems

Formula (20.1) follows from Arzel`a’s Dominated Convergence Theorem, see [8, Theorem 6.12.3]. Indeed, we have x |x| |Φ (x)| = Φ0 (x) = √ |Φ0 (x)| ≤ |Φ0 (x)|. x + i x2 + 2 A more elementary proof of (20.1) is as follows. The integrand on the left-hand side is a continuous function that admits a majorant which is independent of . Accordingly, for arbitrary η > 0 there exists m > δ > 0 such that, for every  ≥ 0, Z η Φ (x) dx < . 3 Iδ Next, observe that, for all  > 0 and x ∈ Iδ0 := [ −m, m ] \ Iδ , |Φ (x) − Φ0 (x)| =

|Φ0 (x)| 1 1 |Φ0 (x)| − |φ(x) − φ(0)| =  ≤ . x + i x |x + i| δ

In other words, the convergence of Φ to Φ0 is uniform on Iδ0 as  ↓ 0. This implies the existence of 0 > 0 such that, for all 0 <  ≤ 0 , Z
η Φ (x) − Φ0 (x) dx < . 3 Iδ0 As a consequence we obtain, for all 0 <  ≤ 0 , Z Z Φ (x) dx − Φ0 (x) dx Im

Im

Z Z Z ≤ Φ (x) dx + Φ0 (x) dx + Iδ


η Φ (x) − Φ0 (x) dx < 3 = η. 3

Iδ0

Iδ

√ Finally note that, for x =  = 1/n with n ∈ Z>0 , we have 1/|x + i| = n/ 2; this implies that the convergence of Φ to Φ0 is not uniform on all of Im . R1 1.2 Consider pe(x) = −(x − 1)3 (x + 1)3 = 1 − 3x2 + 3x4 − x6 . Then −1 pe(x) dx = 35 32/35 and therefore p(x) = 32 (1 − x2 )3 . That φ is twice continuously differentiable on some neighborhood of ±1, respectively, is an easy verification. 1.3 Integration by parts leads to (| · | ∗ ψ)00 = 2ψ. Indeed, Z 00 00 (| · | ∗ ψ) (x) = (| · | ∗ ψ )(x) = |y| ψ 00 (x − y) dy R

Z

0

Z

00

=−

y ψ (x − y) dy + −∞

Z

0

Z

ψ (x − y) dy + −∞

y ψ 00 (x − y) dy

0

0

=−

∞

0

∞

ψ 0 (x − y) dy = 2ψ(x).

20 Solutions to Selected Problems

Now select ψ = φ , then integration implies Z x/ Z 0 g (x) = −1 + 2 p(t) dt and g(x) = −x + 2 −1

x/

−1

Z

235

ξ/

p(t) dt dξ. −1

1.4 R (i) f ∗ g equals 0 outside [ −2, 2 ]. Indeed, R f (x − y) g(y) dy 6= 0 implies that there is y ∈ [ −1, 1 ] with −1 ≤ x − y ≤ 1, which leads to −2 ≤ y − 1 ≤ x ≤ y + 1 ≤ 2. (ii) Take both f and g equal to the characteristic function 1[ −1,1 ] . Then f ∗ g(x) = R1 1 (x−y) dy, where the interval of integration may actually be restricted −1 [ −1,1 ] to [ x − 1, x + 1 ]. Hence, for −2 ≤ x ≤ 0 and 0 ≤ x ≤ 2 we find that f ∗ g(x) is equal to Z x+1 Z 1 dy = 2 + x and dy = 2 − x, respectively. −1

x−1

It follows that f ∗ g(x) = max{ 0, 2 − |x| }, for all x ∈ R. (iii) The functions from part (ii) satisfy all demands in (iii). (iv) Following the hint, we find by means of the substitution y = x t Z x Z 1 f ∗ g(x) = (x − y)α y α dy = x2α+1 (1 − t)α tα dt, 0

0

where the latter integral converges. f ∗ g is discontinuous at 0 if 2α + 1 < 0, that is, if −1 < α < −1/2. 2.1 Note that supp φ(k) ⊂ supp φ, for all k ∈ Z≥0 . (i) Suppose there exists a compact set K ⊂ R such that {j}+supp φ = supp φj ⊂ K, for all j ∈ Z>0 . Then j ∈ K + (− supp φ), which is a compact set; hence no convergence in C0∞ (R). On the other hand, for fixed k and arbitrary j and x, (k)

|φj (x)| =

1 1 (k) | φ (x − j)| ≤ sup |φ(k) (x)|, j j x∈R (k)

which shows that limj→∞ φj = 0 converges uniformly on R. (ii) φ cannot be a polynomial, owing to the compactness of supp φ; consequently, there exists x0 ∈ R with φ(p+1) (x0 ) 6= 0. Now set xj = x0 /j. Then we have, as j → ∞, x0 (p+1) |φj (xj )| = j p+1−p |φ(p+1) (j )| = j |φ(p+1) (x0 )| → ∞. j (p+1)

This implies that (φj )j is not uniformly convergent on R, and therefore (φj )j is not convergent in C0∞ (R). Now consider x ∈ R fixed. Since 0 does not belong to the compact set supp φ, there is N = N (x) such that jx ∈ / supp φ, (k) for all j ≥ N . Accordingly, for such j and fixed k ∈ Z≥0 , we have φj (x) = (k)

j k−p φ(k) (jx) = 0 and thus limj→∞ φj (x) = 0.

236

20 Solutions to Selected Problems

(iii) Select a < 0 < b such that supp φ ⊂ [ a, b ]. Then supp φj ⊂ [ a, b ], for all j. Fix k arbitrarily, then we have, for all j and x,

1 j

supp φ ⊂

(k)

|φj (x)| = j k e−j |φ(k) (jx)| ≤ j k e−j sup |φ(k) (x)|, x∈R

(k)

whereas limj→∞ j k e−j = 0. It follows that limj→∞ φj = 0 uniformly on R, and this holds for every k. As a consequence, we have convergence to 0 in C0∞ (R). 2.2 supp φ is contained in the closed ball B(0; ), while supp ψ is a compact subset of the open set X. According to Corollary 2.4, there exist 0 > 0 and a compact set K ⊂ X such that the -neighborhood of supp ψ is contained in K, for 0 <  ≤ 0 . Now Lemma 2.17 implies, for 0 <  < 0 , supp (ψ ∗ φ ) ⊂ supp ψ + B(0; ) ⊂ K ⊂ X, which shows that condition (a) in Definition 2.12 is satisfied. According to the proof of Lemma 2.17 one has ∂ α (ψ ∗ φ ) = ∂ α ψ ∗ φ , for arbitrary multi-index α. On account of Lemma 1.5, it follows that ∂ α ψ ∗ φ converges uniformly to ∂ α ψ on K as  ↓ 0; this leads to the uniform convergence of ∂ α (ψ ∗ φ ) to ∂ α ψ on K. Therefore condition (b) in Definition 2.12 is satisfied too. R −x R∞ 2 2.3 We have −y γ (y) = 2 γ0 (y) and furthermore −∞ γ (y) dy = x γ (y) dy since γ is an even function. Hence, for any x ∈ Rn , Z Z ∞ Z x (| · | ∗ γ )(x) = |x − y| γ (y) dy = − (x − y) γ (y) dy −∞

R

Z =x

x

∞

Z

x 2

 ([γ ]x−∞ − [γ ]∞ x ) 2 −∞ x Z x Z x/ 2 x 2 e−y dy + 2 γ (x), =x γ (y) dy +  γ (x) = √ π −x/ −x −



γ (y) dy +

where we have obtained the last integral by a change of variables. In view of (2.12)  Z x/ x > 0;  1, 2 1 0, x = 0; e−y dy = lim 2 γ (x) = 0 and lim √  ↓0 ↓0 π −x/ −1, x < 0. Accordingly lim(| · | ∗ γ )(x) = |x|. ↓0

A standard computation now gives 1 (| · | ∗ γ ) (x) = √ π 0

Z

x/

−x/

2

e−y dy

and

(| · | ∗ γ )00 (x) = 2γ (x).

20 Solutions to Selected Problems

237

3.2 Suppose there exists a neighborhood U of 0 in R such that u is of order < k on U . Then, for every compact subset K ⊂ U , there exists c > 0 such that |φ(k) (0)| = |u(φ)| ≤ c

|φ(j) (x)|

sup 0≤j 0. For the corresponding φ , we obtain by means of Leibniz’ rule (2.7) |φ(k) (0)| = |k! χ(0)| = k!

( > 0).

On the other hand, also by Leibniz, for j < k, φ(j)  (x) =

X j
x
X xk−l (j−l) x
xk−l = cl j−l χ . k(k−1) · · · (k−l+1) j−l χ(j−l) l     l≤j

l≤j

In view of the properties of supp χ, we see that χ(j−l) ( x ) 6= 0 implies |x| ≤ a . For l ≤ j, this gives the existence of a constant dl > 0 such that |xk−l χ(j−l) ( x )| ≤ dl (a )k−l , for all x ∈ R. Hence, for K as above we can find C > 0 such that X sup |φ(j) k−j ,  (x)| ≤ C 0≤j0 × S Ψ (r, y) = r y; one then has | det DΨ (r, y)| = rn−1 ω(y), where ω denotes the Euclidean (n − 1)-dimensional density on S n−1 . The local integrability of the vj over Rn , for 1 ≤ j ≤ n, then follows from Z Z  Z |vj (x)| dx = r−n+1+n−1 dr |yj | ω(y) dy B(0;) 0 S n−1 Z = |yj | ω(y) dy < ∞ ( > 0). S n−1

For any φ ∈

C0∞ (Rn ),

we have

(div v)(φ) = −

n X

Z vj (∂j φ) = −

n X

vj (x) ∂j φ(x) dx

Rn j=1

j=1

Z

Z

=−

h v(x), grad φ(x) i dx = − Rn

Dφ(x)v(x) dx. Rn

Upon the introduction of spherical coordinates in Rn this leads to Z Z (div v)(φ) = − rn−1 Dφ(r y)v(r y) dr ω(y) dy S n−1

Z

R>0

Z

=−

Z

Z

Dφ(r y)y dr ω(y) dy = − S n−1

R>0

= cn φ(0) = cn δ(φ).

S n−1

R>0

d φ(r y) dr ω(y) dy dr

20 Solutions to Selected Problems

239

4.6 The local integrability of E over Rn follows as in the preceding problem. For any φ ∈ C0∞ (Rn ) and n 6= 2 we obtain, through integration by parts Z (∂j E)(φ) = −E(∂j φ) = − E(x) ∂j φ(x) dx Rn

Z

Z 1 kxk2−n ∂j φ(x) dxj dxbj (n − 2)cn Rn−1 R Z Z Z 1 1 xj 1−n xj = φ(x) dx φ(x) dxj dxbj = kxk cn Rn−1 R kxk cn Rn kxkn 1 = vj (φ). cn =

Finally, ∆E = div(grad E) = in a similar way.

1 cn

1 cn

div v =

cn δ = δ. The case of n = 2 is treated

5.1 For any 0 6= φ ∈ C0∞ (Rn ) and r > 0, we have in view of condition (a) Z (test fj )(φ) − δ(φ) = fj (x) (φ(x) − φ(0)) dx Rn

Z

Z fj (x)(φ(x) − φ(0)) dx +

= kxk≤r

fj (x) (φ(x) − φ(0)) dx =: kxk≥r

2 X

Ii .

i=1

Now select  > 0 arbitrarily. In view of the continuity of φ there exists r > 0 such that |φ(x) − φ(0)| < /2, for all kxk ≤ r. Since fj ≥ 0, condition (a) implies for all j ∈ Z>0 Z Z   |I1 | ≤ fj (x)|φ(x) − φ(0)| dx < fj (x) dx ≤ . 2 kxk≤r 2 kxk≤r Next, write m = supx∈Rn |φ(x)| > 0. On account of condition (b) there exists N ∈ Z>0 such that for all j ≥ N Z Z   0≤ fj (x) dx < , so |I2 | ≤ 2m fj (x) dx < . 4m 2 kxk≥r kxk≥r Combination of the estimates leads to the desired conclusion. P 5.2 By decomposing f = ± (|f | ± f )/2, we see that f may be assumed to be nonnegative integrable on Rn . Write f◦ = 1c f , then (f◦ ) satisfies conditions (a) and (b) from Problem 5.1. Indeed, Z Z r f ◦ (x) dx with lim = ∞. f◦ (x) dx = r ↓0  kxk≥r kxk≥  Thus the conclusion follows from the same problem. Alternatively, one can proceed as follows. For any φ ∈ C0∞ (Rn ), one has

240

20 Solutions to Selected Problems

Z (test f )(φ) =

Z f (x)φ(x) dx =

Rn

f (x)φ( x) dx. Rn

Now |f (x)φ( x)| ≤ (supx∈Rn |φ(x)|) |f (x)|, for all x ∈ Rn and all  > 0, with a dominating function that is integrable over Rn . Furthermore, lim↓0 f (x)φ( x) = φ(0) f (x). As one can see by applying either the Dominated Convergence Theorem of R Arzel`a (see [8, Theorem 6.12.3]) or that of Lebesgue, one has lim↓0 (test f )(φ) = f (x) dx δ(φ). Rn 5.5 Note that ut (x) = f√t (x) in the notation of Problem 5.2, where Z kxk2 n and f (x) dx = 1. f (x) = (4π)− 2 e− 4 Rn

Application of that same problem therefore implies limt↓0 ut (x) = δ in D0 (Rn ). The following formulas show that ut satisfies the heat equation:  kxk2 d n ut (x) ut (x) = x, − ut (x), gradx ut (x) = − 2 dt 4t 2t 2t  kxk2 ut (x) 1 n ∆ut (x) = − divx x − h gradx ut (x), x i = − ut (x), 2t 2t 4t2 2t where we have used that ∆ = div grad. Consequently, limt↓0 D0 (Rn ). Indeed, for any φ ∈ C0∞ (Rn ), lim t↓0

d dt ut

= ∆δ in

d ut (φ) = lim ∆ut (φ) = lim ut (∆φ) = δ(∆φ) = ∆δ(φ). t↓0 t↓0 dt

6.2 According to Problem 6.1 we may apply Proposition 6.3 to the function ut . 2 Indeed, if 2t = 2 , then ut (x) = 1n f ( 1 x) where f (x) = (2π)−n/2 e−kxk /2 . In the notation of the proposition and on account of [8, Exercise 6.51] we have, for all α ∈ (Z≥0 )n , Z cα =

xα f (x) dx =

Rn

=

   

n Z 1 2 1 Y α x j e− 2 xj dxj n (2π) 2 j=1 R j

0, n 2

n n Y

2 π    (2π) n2 2 n2

(αj )! αj αj j=1 ( 2 )! 2 2

α!

= 2

|α| 2

( α2 )!

,

if there exists j with αj odd; for α ∈ (2Z≥0 )n .

Therefore we have, with j ∈ Z≥0 ,  0,   X cα 1 X 1 uj = ∂αδ = ∂ α δ,  α! j  ( α2 )! |α|=j 22

j odd; j even.

|α|=j

Now apply Proposition 6.3 with k replaced by 2k. This leads to

20 Solutions to Selected Problems

0 = lim ↓0

241

2k k   X X X 1 1 1  1 j j 2α f − lim u − (−) u = t ∂ δ .  t j 2k 2k t↓0 tk α! j=0 j=0 |α|=j

The desired formula now follows by application of the following Multinomial Theorem (see [8, Exercise 2.52.(ii)]): Pn X 1 X x2α ( i=1 x2i )j 1 = , that is, ∂ 2α = ∆j . α! j! α! j! |α|=j

|α|=j

7.1 See [8, Formula (7.44)]. 7.5 See [8, Exercise 6.95]. 7.6 (i) and (ii). For these results, see the beginning of Chap. 13. (iii) Obviously l+ is of order 0. Suppose k ≥ 1 and select φ ∈ C0∞ (R) with supp φ ⊂ [ −1, 1 ] and ∂ k−1 φ(0) = 1. Now define φj (x) =

1 φ(j x) j k−1 log j

(j ∈ Z>0 ).

Then Z j (−1) ∂ l+ (φj ) = ∂ φj (x) log x dx = ∂ k φ(j x) log x dx log j R>0 R>0 Z Z 1 x 1 k = ∂ φ(x) log dx = ∂ k φ(x) log x dx + ∂ k−1 φ(0). log j R>0 j log j R>0 k

k

Z

k

As a consequence, limj→∞ (−1)k ∂ k l+ (φj ) = 1. On the other hand, for 0 ≤ l < k we find   1 k∂ l φj kC 0 = O k−l−1 = o(1), j → ∞. j log j It follows that ∂ k l+ is of order k on R, by arguments similar to those in the solution to Problem 3.2. 8.2 If C = X \ K, then C open in Rn . For any x ∈ C, there exists an open neighborhood U of x with K ∩ U = ∅. Now consider φ ∈ C0∞ (U ). As supp uj ∩ supp φ ⊂ K ∩ U , formula (7.2) says uj (φ) = 0 and thus u(φ) = limj→∞ uj (φ) = 0. In other words, u = 0 on U and then Theorem 7.1 asserts u = 0 on C; that is, supp u ⊂ K. This implies u ∈ E 0 (X), in view of Theorem 8.5. Next, select χ ∈ C0∞ (X) with χ = 1 on a neighborhood of K and let ψ ∈ C ∞ (X) be arbitrarily chosen. Then supp (ψ − χ ψ) ∩ K = ∅, and so uj (ψ − χ ψ) = 0

(j ∈ Z>0 )

and

u(ψ − χ ψ) = 0.

Because limj→∞ uj = u in D0 (X) and χ ψ ∈ C0∞ (X), we obtain

242

20 Solutions to Selected Problems

lim uj (ψ) = lim uj (χ ψ) = u(χ ψ) = u(ψ),

j→∞

j→∞

which implies limj→∞ uj = u in E 0 (X). The final assertion is a direct consequence of the preceding results. n+1 8.5 The because it equals 0 if t ≤ 0, R function E(x, t) is locally integrable over R while Rn E(x, t) dx = 1, if t > 0, which defines a locally integrable function on R. Further, sing supp E = {0}. In order to see this, we only have to consider a neighborhood of a point (x, 0) with x 6= 0. In fact, we have to show that all partial derivatives of E(x, t) tend to 0 as t ↓ 0. We have, for φ ∈ C0∞ (Rn+1 ),

v(φ) = (∂t E − ∆x E)(φ) = −E(∂t φ + ∆x φ) Z Z =− E(x, t)(∂t φ + ∆x φ)(x, t) dt dx =: lim I + J , Rn

Z

↓0

R>0

Z

I = −

∞

Z (E ∂t φ)(x, t) dt dx,

Rn

∞

Z

J = −



with

(E ∆x φ)(x, t) dx dt. 

Rn

Integration by parts with respect to the variable t gives Z Z Z ∞ I = (E φ)(x, ) dx + (φ ∂t E)(x, t) dt dx. Rn

Rn



Owing to the fact that φ has compact support, ∂Rn does not make a contribution; application of Green’s second identity (see [8, Example 7.9.6]) therefore leads to Z ∞Z J = − (φ ∆x E)(x, t) dx dt Rn



n Since E satisfies the heat equation √ on R × ] , ∞ [ we therefore obtain by means of the change of variables x = 2  y Z Z √ 2 −n 2 e−kyk φ(2  y, ) dy. I + J = (E φ)(x, ) dx = π Rn

Rn

On account of the Mean Value Theorem and the compactness of supp φ we can find c > 0 such that for all 0 <  ≤ 1 and y ∈ Rn √ √ |φ(2  y, ) − φ(0)| ≤ c  kyk. √ √ Hence the equality φ(2  y, ) = φ(0) + φ(2  y, ) − φ(0) leads to Z 2 −n 2 e−kyk dy + R , I + J = φ(0) π Rn

where

|R | ≤

√

n

 cπ − 2

Z

√ 2 kyk e−kyk dy = O( ),

Rn

It follows that v = δ ∈ D0 (Rn+1 ), in view of

 ↓ 0.

20 Solutions to Selected Problems

243

v(φ) = lim I + J = φ(0) + lim R = φ(0) = δ(φ). ↓0

↓0

9.2 For every φ ∈ C0∞ (Rn ), we have on account of Leibniz’ rule (2.7) (ψ ∂ α δ)(φ) = (∂ α δ)(ψ φ) = (−1)|α| ∂ α (ψ φ)(0) X α
X α
= (−1)|α| ∂ α−β ψ(0) ∂ β φ(0) = ∂ α−β ψ(0) (−1)|α|−|β| ∂ β δ φ β β β≤α

=

X

(−1)|α−β|

β≤α

β≤α

α
α−β ∂ ψ(0) ∂ β δ φ. β

9.3 Application of Problem 9.2 implies xk ∂ m δ =

m X m
(−1)m−l (∂ m−l xk )(0) ∂ l δ l

 m l=0   X m
(−1)m−l k(k − 1) · · · (k − m + l + 1)xk−m+l ∂ l δ, m − l ≤ k; x=0 l =  l=0  0, m − l > k. Therefore xk ∂ m δ 6= 0 only if there exists an l such that l ≥ 0 and k − m + l = 0, that is, l = m − k ≥ 0. If this is the case, we obtain xk ∂ m δ = (−1)k

m! ∂ m−k δ. (m − k)!

Now consider u ∈ D0 (R) satisfying xk u = 0. On account of Theorem 9.5 we have P supp u ⊂ { x ∈ R | xk = 0 } = {0}, while Theorem 8.7 then implies u = m cm ∂ m δ, with cm ∈ C. Now 0 = xk u =

X m

cm xk ∂ m δ =

X

cm (−1)k

m

Accordingly cm = 0, for m ≥ k, and so u = for xk u = 0.

Pk−1

m! ∂ m−k δ. (m − k)!

m=0 cm ∂

m

δ is the general solution

9.4 According to Problem 9.3 we have u0 = c δ with c ∈ C arbitrary. In view of Example 4.2 the latter equation has the particular solution u = c H, while Theorem 4.3 says that u = c H + d is the general solution, where d ∈ C. 9.5 Since x δ = 0, the desired formula may be easily deduced from the formula 1 1 x+i 0 = −πi δ + PV x , which occurs in the solution of Problem 1.1. The problem k of solving x u = 1 can be reduced to finding all solutions u0 to the homogeneous problem xk u0 = 0, which is done in Problem 9.3, plus a particular solution to the inhomogeneous problem xk u = 1. In the case where k = 1, we already know that PV x1 solves the latter problem. Next, suppose that uk satisfies xk uk = 1. This implies

244

20 Solutions to Selected Problems

0 = x (xk uk )0 = kxk uk + xk+1 u0k = k + xk+1 u0k ,  1  so xk+1 uk+1 := xk+1 − u0k = 1. k Therefore, the general solution to xk u = 1 is given by, see Problem 4.2, u=

k−1 k−1 X X (−1)k−1 k−1 1 (−1)k−1 ∂ log(k) |x| + PV + cm ∂ m δ = cm ∂ m δ. (k − 1)! x m=0 (k − 1)! m=0

9.6 Let φ ∈ C0∞ (R) and suppose φ(x) = 0 if |x| ≥ m > 0. We have 

eitx PV

1 (φ) = lim J , ↓0 x

where, by means of a change of variables, Z Z m  itx eitx φ(x) e φ(x) − e−itx φ(−x)  J := dx = dx. x x ≤|x|≤m  Writing φ(x) = φ(0) + xψ(x) for all x ∈ R, with ψ continuously differentiable on R, we get Z m itx Z m e − e−itx J = φ(0) dx + (eitx ψ(x) + e−itx ψ(−x)) dx x   Z m Z tm sin x dx + = 2i δ(φ) (eitx ψ(x) + e−itx ψ(−x)) dx. x   Both integrands on the right-hand side are continuous, which implies Z tm sin x dx + Kt , lim J = 2i δ(φ) ↓0 x 0 where

Z Kt =

m

(eitx ψ(x) + e−itx ψ(−x)) dx.

0

R Now we have the well-known evaluation R>0 sinx x dx = π2 , see Problem 14.25 or [8, Example 2.10.14 or Exercises 0.14, 6.60 or 8.19]. Furthermore, the identity eitx = it1 ∂x eitx and integration by parts lead to, for t > 0, i Z m 1Z m |Kt | = (eitx ψ 0 (x) + e−itx ψ(−x)) dx ≤ |ψ 0 (x)| dx. (20.2) t 0 t −m We now immediately obtain (9.5). The second identity follows by adding together (9.5) and the analogous identity obtained by replacing x by −x. By combination of Problem 1.1 and (9.5) it follows that  1 1 lim eitx = lim ∓πi eitx δ + eitx PV = ∓πi δ + πi δ, t→∞ x ± i 0 t→∞ x

20 Solutions to Selected Problems

245

and this proves (i) and (ii). Assertion (iii) follows from the fact that the inner limit is already equal to 0, as can be demonstrated in a way similar to the proof of (20.2). 9.7 Theorem 9.5 implies that supp u is contained in the hyperplane H = { x ∈ Rn | xn = 0 }. Select χ ∈ C0∞ (R) with χ = 1 on an open neighborhood of 0 and define χ e ∈ C ∞ (Rn ) by χ e(x1 , . . . , xn ) = χ(xn ). Then 1 − χ e = 0 on an open neighborhood of H, hence u = χ e u. Now consider φ ∈ C0∞ (Rn ), then Z 1 φ(x1 , . . . , xn ) = φ(x1 , . . . , xn−1 , 0) + xn ∂n φ(x1 , . . . , xn−1 , txn ) dt 0

= φ(π(x)) + xn η(x), where π : x 7→ (x1 , . . . , xn−1 , 0) and η ∈ C0∞ (Rn ). This leads to u(φ) = u(e χ φ) = u(e χ (φ ◦ π)) + u(xn χ e η) = u(e χ (φ ◦ π)). Now χ e(x) (φ ◦ π)(x) = φ(x1 , . . . , xn−1 , 0) χ(xn ) = i∗0 φ ⊗ χ(x), so u(φ) = ∗ u(i0 φ ⊗ χ). Finally, we note that ψ 7→ ψ ⊗ χ defines a continuous linear mapping C0∞ (Rn−1 ) → C0∞ (Rn ). Hence, v : ψ 7→ u(ψ ⊗ χ) belongs to D0 (Rn−1 ) and satisfies u(φ) = v(i∗0 φ). 10.1 Taking the transpose of (10.1) and using Lemma 10.1 as well as the fact that the transpose of ∂k is −∂k , leads to the desired identity. 10.4 The mapping t 7→ det DΦt is continuous and nonvanishing on the connected set R, while det DΦ0 = det I = 1. Therefore, det DΦt > 0, for all t ∈ R. On account of (10.6) and the fact that (Φ−t )−1 = Φt , one obtains on D0 (X)
d d d (Φ−t )∗ = ((Φ−t )−1 )∗ (det D(Φ−t ))−1 = (Φt ∗ det DΦt ) . dt dt dt t=0 t=0 t=0 According to (10.12) and (10.11), this implies n X

∂j ◦ vj =

j=1

=

d ∗ d Φt det DI + I ∗ det DΦt dt dt t=0 t=0 n X j=1

vj ◦ ∂j +

d det DΦt . dt t=0

Hence, n

n

X X d det DΦt = (∂j ◦ vj − vj ◦ ∂j ) = ∂j vj = div v. dt t=0 j=1 j=1 Actually, one can prove that d det DΦt = (div v) ◦ Φt det DΦt dt

(t ∈ R).

In view of det DΦ0 = 1, solving the differential equation (compare with [8, Formula (5.32)]) yields

246

20 Solutions to Selected Problems Rt

det DΦt (x) = e

0

div v(Φτ (x)) dτ

((t, x) ∈ R × X).

10.5 δx is invariant under (Φt )t∈R as a distributional density iff (Φt )∗ δx = δx for all t ∈ R, and in view of Example 10.3 this is the case iff δΦt (x) = δx , that is, iff Φt (x) = x, which is equivalent to v(x) = 0. Furthermore, assume δx is invariant under (Φt )t∈R as a generalized function. On account of Example 10.8, we have Φt ∗ δx =

1 δΦ (x) | det DΦt (Φ−t (x))| −t

(t ∈ R).

(20.3)

By considering the supports of the distributions occurring in (20.3), it follows that the assumption implies x = Φ−t (x), or equivalently Φt (x) = x, for all t ∈ R. In turn, (20.3) then leads to det DΦt (x) = 1, for all t ∈ R. Upon differentiation with respect to t at t = 0 and application of Problem 10.4 we obtain the desired conclusion. d Φt (x) t=0 = 10.6 Writing Φt for the rotation by the angle t, we find v(x) = dt (−x2 , x1 ). On account of Theorem 10.12, the distributional density u ∈ D0 (R2 ) is invariant under all of the Φt iff 2 X

∂j (vj u) = −∂1 (x2 u) + ∂2 (x1 u) = −x2 ∂1 u + x1 ∂2 u = 0.

j=1

Furthermore, u ∈ D0 (R2 ) is invariant as a generalized function iff −x2 ∂1 u + x1 ∂2 u = 0.

P2

j=1

vj ∂j u =

10.9 Pn (i) According to Theorem 10.13, we have k=1 xk ∂k uj = a uj , for all j. Since ∂k : D0 (Rn ) → D0 (Rn ) is a continuous linear mapping, on account of (8.2), it followsP that ∂k uj → ∂k u in D0 (Rn ); by taking the limit as j → ∞, we therefore n obtain k=1 xk ∂k u = a u, which implies u ∈ Ha . (ii) Applying ∂j to the last identity in (i), we see X X xk ∂j ∂k u + ∂j u + xj ∂j2 u = xk ∂k ∂j u + ∂j u = a ∂j u; k6=j

k

in other words, ∂j u ∈ Ha−1 . The assertion for xk u follows by a similar argument. (iii) If a ∈ R, then t 7→ ta belongs to C ∞ (R) if and only if a ∈ Z≥0 . For any a ∈ C \ R,
ta = tRe a ei(Im a) log t = tRe a cos((Im a) log t) + i sin((Im a) log t) , where Im a 6= 0. Hence, t 7→ ta does not belong to C ∞ (R) in this case. Now suppose x ∈ Rn and a ∈ Z≥0 . Then ψ(t x) = ta ψ(x) is an identity of functions in C ∞ (R). Now differentiate this identity a times with respect to t at t = 0; the right-hand side then gives a! ψ(x), while on account of the chain rule the lefthand side leads to

20 Solutions to Selected Problems

247

Da ψ(tx)(x, . . . , x)|t=0 = Da ψ(0)(x, . . . , x) =: p(x), with a copies of x, and where p is a polynomial function on Rn . Accordingly, 1 ψ = a! p. 1 x). The local integrability of g can be proved using (iv) Set g(x) = kxka f ( kxk spherical coordinates. And g ∈ Ha , because according to Theorem 10.6 we have, for any φ ∈ C0∞ Rn , Z c∗ g(φ) = g(c∗ φ) = g(c−n (c−1 )∗ φ) = g(x) φ(c−1 x) c−n dx n R Z a = g(cx) φ(x) dx = c g(φ). Rn

10.10 On the basis of Theorem 10.6 we see, for all c > 0 and φ ∈ C0∞ (Rn ), c∗ δ(φ) = δ(c∗ φ) = δ(c−n (c−1 )∗ φ) = c−n δ(φ). This proves that δ is homogeneous of degree −n. Using Problem 10.9.(ii) and mathematical induction over |α| we now find that ∂ α δ is homogeneous of degree −n − |α|. 10.12 Problem 10.1 implies that Φ∗ ◦ ∂j =

p X

∂k ◦ Φ∗ ◦ ∂j Φk

:

C0∞ (X) → C0∞ (Y )

k=1

is an identity of continuous linear mappings. On account of Theorem 10.14, the pullback Φ∗ : D0 (Y ) → D0 (X) is well-defined as the transpose of Φ∗ : C0∞ (X) → C0∞ (Y ), the mapping Φ being a C ∞ submersion. Therefore the desired identity can be obtained by transposition. 10.13 (i) From Φ(x) = h Ax, x i one obtains grad Φ(x) = 2Ax. Because A is invertible, its kernel consists of 0 only. This proves the assertion. (ii) Apply Problem 10.12 with p = 1 to find, as Φ(x) = y and y denotes the variable in R, (∂j ◦ Φ∗ )v = (∂j Φ) Φ∗ v 0 in D0 (Rn \ {0}). Using this identity leads to ((∂i ∂j ) ◦ Φ∗ )v = ∂i (∂j ◦ Φ∗ )v = (∂i ∂j Φ) Φ∗ v 0 + (∂j Φ)(∂i ◦ Φ∗ )v 0 = (∂i ∂j Φ) Φ∗ v 0 + (∂j Φ)(∂i Φ) Φ∗ v 00 . Pn Part (i) now implies ∂j Φ(x) = 2 k=1 Ajk xk and ∂i ∂j Φ = 2Aji . Hence, in D0 (Rn \ {0}),

248

20 Solutions to Selected Problems

(P ◦ Φ∗ )v =

=

n X i,j=1 n X

Bij ((∂i ∂j ) ◦ Φ∗ )v ∗ 0

Bij 2Aji Φ v +

i,j=1

=2

n X

n X

Bij 4Ajk Ail xk xl Φ∗ v 00

i,j,k,l=1

δii Φ∗ v 0 + 4y Φ∗ v 00 = Φ∗ (2n v 0 + 4y v 00 ),

i=1

because n X

n X

Bij Ajk Ail xk xl =

i,j,k,l=1

δik Ail xk xl =

i,k,l=1

n X

Akl xk xl = y.

k,l=1

(iii) Application of the first identity in part (ii) and of part (i) leads to n X

xj (∂j ◦ Φ∗ )v =

j=1

n X

xj (∂j Φ) Φ∗ v 0 = 2

j=1

n X

xj (Ax)j Φ∗ v 0

j=1

= 2y Φ∗ v 0 = 2Φ∗ (y v 0 ) = 2a Φ∗ v. Thus, the claim is a consequence of Theorem 10.13. (iv) The first assertion also follows from Theorem 10.13. If v is homogeneous of degree 1 − n2 , then u is homogeneous of degree 2 − n. (v) The former identity follows upon restriction of the identity in part (ii) from D0 (R) to C ∞ (R), while the latter is obtained by means of transposition. (vi) In this case, the matrix B is the identity matrix on R2 and so is A. Hence Φ as per this problem coincides with Φ as per Example 10.4. Furthermore Z π (Φ∗ δV )(φ) = φ(1) dα = 2π δ1 (φ). −π

Since y δ1 = δ1 according to Example 9.1, one derives Φ∗ ∆ δV = 4(∂y 2 ◦ y − ∂y )2π δ1 = 8π(δ1 00 − δ1 0 ). 10.14 We use the notation of the proof of Theorem 10.14 and Remark 10.15. Given y ∈ Y , write V (y) = Π −1 ({y}) ∩ V = { z ∈ Rq | (y, z) ∈ V }. In view of formulas (10.15) and (10.16), we obtain Z (Φ∗ f )(y) = jΛ (y, z) f (Λ(y, z)) dz (f ∈ C0 (U ), y ∈ Y ).

(20.4)

V (y)

We will rewrite this integral in a more intrinsic manner, as an integral with respect to Euclidean integration over the fiber of Φ over the value y ∈ Y , which possesses the following parametrization:

20 Solutions to Selected Problems

Λy : V (y) → U ∩ Φ−1 ({y})

given by

249

Λy (z) = Λ(y, z).

To this end, we derive the identity (20.9) below for jΛ (y, z). Identify x0 ∈ Rn with (y, z) ∈ Rp × Rq and write x = Λ(x0 ) = Λ(y, z), for (y, z) ∈ V . Suppose x = Λ(x0 ) ∈ U , then Φ(x) = Π ◦ K ◦ Λ(x0 ) = Π(x0 ) = y; in other words x ∈ Φ−1 ({y}). On account of the Submersion Theorem, near x the set Φ−1 ({y}) is a C ∞ submanifold of Rn of dimension n − p = q. According to [8, Theorem 5.1.2] the tangent space of Φ−1 ({y}) at x is the linear subspace of Rn of dimension q T := Tx Φ−1 ({y}) = ker φ

where

φ = DΦ(x) ∈ Lin(Rn , Rp ).

(20.5)

Application of the chain rule to the identity Φ ◦ Λ = Π on V gives the following identity in Lin(Rn , Rp ): DΦ(x) ◦ DΛ(x0 ) = DΠ(x0 ) = Π

(x0 ∈ V ).

For the moment we keep x and x0 fixed and rewrite this equality more concisely as φ λ = φ(λ` λk ) = (φ λ` φ λk ) = Π.

(20.6)

Here λ = DΛ(x0 ) ∈ Aut(Rn ), while, for 1 ≤ j ≤ n, λj = Dj Λ(x0 ) ∈ Rn ,

λ` = (λ1 · · · λp ),

λk = (λp+1 · · · λn ) = DΛy (z).

As customary, we identify a linear mapping and its matrix with respect to the standard basis vectors. It is a consequence of (20.6) and (20.5) that T is spanned by the vectors that occur in λk ; these are the linearly independent vectors λk , for p < k ≤ n. Furthermore, (20.6) implies that φ λj = ej , for 1 ≤ j ≤ p. In particular, the vectors that occur in λ` are transversal to T in view of (20.5). Now consider the parallelepiped in Rn spanned by all λj , for 1 ≤ j ≤ n. Its volume equals J := | det DΛ(x0 )| = jΛ (y, z) = | det(λ1 · · · λn )| = | det λ |. Denote the orthocomplement in Rn of T by T ⊥ , then dim T ⊥ = p. The value of J does not change if we replace λj ∈ Rn by its projection πj ∈ T ⊥ along T , for all 1 ≤ j ≤ p. Using the notation π = (π1 · · · πp ) this leads to J 2 = (det λ)2 = (det(λ1 · · · λn ))2 = (det(π1 · · · πp λp+1 · · · λn ))2 = (det(π λk ))2 = det(t(π λk ) (π λk )). The vectors in π belong to T ⊥ while those in λk belong to T , hence (π λk ) (π λk ) = tπ π ⊗ tλk λk ,

t

where the former factor on the right-hand side belongs to Aut(Rp ) and the latter one to Aut(Rq ). This implies

250

20 Solutions to Selected Problems

J 2 = det(tπ π) det(tλk λk ).

(20.7)

Next, we evaluate det(tπ π). From linear algebra it is well-known that T ⊥ may be described by T ⊥ = (ker φ)⊥ = im tφ,

in particular,

Rn = ker φ ⊕ im tφ.

(20.8)

(Observe that (20.8) proves the identity rank tφ = rank φ, see [8, Rank Lemma 4.2.7].) Indeed, for p ∈ Rn , p ∈ ker φ

⇐⇒

hφ p, qi = 0 for all q ∈ Rp

⇐⇒

φp = 0

⇐⇒

hp, t φ qi = 0 for all q ∈ Rp

⇐⇒

p ∈ (im tφ)⊥ ,

which shows that (ker φ)⊥ = (im(tφ))⊥⊥ = im(tφ). Now (20.8) yields the existence of vectors η1 , . . . , ηp ∈ Rp such that π = tφ η. Since λj − πj ∈ T , for 1 ≤ j ≤ p, it follows from (20.5) that φ(λ` − π) = 0. Hence, we obtain from (20.6) Ip = φ λ` = φ π = φ tφ η = γ η

where

γ = φ tφ.

Note that γ ∈ Aut(Rp ) because it is Gram’s matrix associated with the linearly independent row vectors of φ. As a consequence η = γ −1 and thus π = tφ γ −1 . In turn, this leads to π π = (tγ)−1 (φ tφ)γ −1 = γ −1 γ γ −1 = γ −1 .

t

On account of (20.7) we therefore find 1 and det(tπ π) = det γ

J2 =

det(tλk λk ) . det(φ tφ)

More explicitly, this means p p det(tDΛy (z) ◦ DΛy (z)) det(tDΛy (z) ◦ DΛy (z)) = . jΛ (y, z) = q
(gr Φ)(Λ(y, z)) det DΦ(Λ(y, z)) ◦ tDΦ(Λ(y, z)) (20.9) On account of the definition of Euclidean q-dimensional integration (see [8, Section 7.3]) one now deduces from (20.4), for f ∈ C0 (U ) and y ∈ Y , Z q f (Λy (z)) (Φ∗ f )(y) = det(tDΛy (z) ◦ DΛy (z)) dz V (y) (gr Φ)(Λy (z)) (20.10) Z f (x) = dx. U ∩Φ−1 ({y}) (gr Φ)(x) 10.15 (i) The first identity follows from (Φ∗ φ)(y) = δy (Φ∗ φ) = (Φ∗ δy )(φ) and δy = ∂ Ty H, which is a consequence of Z ∞ φ(y) = − ∂φ(x + y) dx = −H(T−y ∂φ) = ∂ Ty H(φ). 0

The second may be obtained from (10.1).

20 Solutions to Selected Problems

251

(ii) Select χ ∈ C ∞ (R) satisfying 0 ≤ χ ≤ 1, χ = 0 on ] −∞, −1 [ and χ = 1 on ] 1, ∞ [ . For t > 0, write (t∗ χ)(y) = χ(ty). Then, for x ∈ X, Φ∗ (Ty t∗ χ)(x) = (Ty t∗ χ)(Φ(x)) = χ(t(Φ(x) − y)) ∗

and so

∗

lim Φ (Ty t χ)(x) = H(Φ(x) − y) = 1Φ−1 ( ] y, ∞ [ ) (x).

t→∞

Arzel`a’s Dominated Convergence Theorem, see [8, Theorem 6.12.3], now leads to Z Φ∗ (Ty H)(φ) = lim Φ∗ (Ty t∗ χ)(φ) = lim Φ∗ (Ty t∗ χ)(x) φ(x) dx t→∞ t→∞ X Z Z = 1Φ−1 ( ] y, ∞ [ ) (x) φ(x) dx = φ(x) dx. Φ−1 ( ] y, ∞ [ )

X

(iii) Part (ii) and the Theorem on Integration of a Total Derivative imply Z ∂j (Φ∗ (Ty H))(φ) = −(Φ∗ (Ty H))(∂j φ) = − ∂j φ(x) dx Φ−1 ( ] y, ∞ [ )

Z =

φ(x) Φ−1 ({y})

∂j Φ(x) dx. k grad Φ(x)k

Here we have used that grad Φ(x) points inward in Φ−1 ( ] y, ∞ [ ). (iv) Parts (i) and (iii) lead to   ∂j Φ (∂j Φ)2 ∗ ∗ Φ δ (φ) = ∂ (Φ (T H)) φ y j y k grad Φk2 k grad Φk2 Z (∂j Φ(x))2 = φ(x) dx, k grad Φ(x)k3 Φ−1 ({y}) and so Pn Z 2 φ(x) j=1 (∂j Φ) ∗ (Φ∗ φ)(y) = Φ δ (φ) = dx. y 2 k grad Φk Φ−1 ({y}) k grad Φ(x)k 10.16 (i) The first identity is a direct consequence of (10.17). The second follows from the first by noting that (Φ∗ δy )(φ) = δy (Φ∗ φ) = (Φ∗ φ)(y). (ii) From part (i) we obtain, for any y ∈ R, Z y Z y Z φ(x) (Φ∗ φ)(t) dt = dx dt k grad Φ(x)k −1 −∞ −∞ Φ ({t}) Z = φ(x) dx. Φ−1 ( ] −∞, y [ )

In view of the Fundamental Theorem of Integral Calculus on R this implies Z y Z (Φ∗ δy )(φ) = (Φ∗ φ)(y) = ∂y (Φ∗ φ)(t) dt = ∂y φ(x) dx. −∞

Φ−1 ( ] −∞, y [ )

252

20 Solutions to Selected Problems

(iii) We have Z ψ(Φ∗ φ) = (Φ∗ ψ)(φ) = φ(x) ψ(Φ(x)) dx Rn Z Z ∞ Z Z =− φ(x) ∂ψ(y) dy dx = − ∂ψ(y) φ(x) dx dy Rn Φ(x) R Φ−1 ( ] −∞, y [ ) Z Z = ψ(y) ∂y φ(x) dx dy. Φ−1 ( ] −∞, y [ )

R

10.17 Combination of Problem 10.12 and Φ∗ δ =

1 k grad Φk

∂j (Φ∗ H) = ∂j Φ Φ∗ (∂H) = ∂j Φ Φ∗ δ =

δ∂Ω leads to

∂j Φ δ∂Ω . k grad Φk

For arbitrary φ ∈ C0∞ (X) we now find Z

∗

Z

∂j (Φ H)(φ) = −(H ◦ Φ)(∂j φ) = −

∂j φ(x) dx = − Φ−1 (R>0 )

∂j φ(x) dx, Ω

while ∂j Φ δ∂Ω (φ) = k grad Φk

Z φ(y) ∂Ω

∂j Φ(y) dy = − k grad Φ(y)k

Z φ(y) νj (y) dy. ∂Ω

Here we have used that ∂j Φ(y)/k grad Φ(y)k equals the j-th component of the normalized gradient vector of Φ at y, which points inward in Ω since the latter is the inverse image Φ−1 (R>0 ); phrased differently, it equals −νj (y) where νj (y) is the outer normal to ∂Ω at y. The desired equality is now obvious. 11.1 (i) u = φ = 0. R (ii) Select u = 1 ∈ C ∞ (R) and φ ∈ C0∞ (R) satisfying R φ(y) dy = 1. Then Z Z (u ∗ φ)(x) = u(x − y)φ(y) dy = φ(y) dy = 1. R

R

∞

(iii) Select u ∈ C (R) with u(x) = x and φ as in part (ii) and, in addition, assume that φ is an even function. Then Z Z Z (u ∗ φ)(x) = (x − y)φ(y) dy = x φ(y) dy − y φ(y) dy = x. R

R

C0∞ (R)

R

R

(iv) Select u(x) = sin x and an even φ ∈ with R φ(y) cos y dy = 1. Then Z (u ∗ φ)(x) = sin(x − y) φ(y) dy R Z Z = sin x φ(y) cos y dy − cos x φ(y) sin y dy = sin x. R

R

20 Solutions to Selected Problems

253

11.3 The implication (ii) ⇒ (i) follows from (11.4). For the reverse implication, note that, for every φ ∈ C0∞ (Rn ) and x ∈ Rn , ∂aj Ta φ(x) = ∂aj (a 7→ φ(x − a)) = −∂j φ(x − a) = −Ta ∂j φ(x) = −∂j Ta φ(x), because ∂j commutes with Ta . Application of the chain rule to a 7→ (a, a) 7→ Ta ◦ A ◦ T−a φ(x)

:

Rn → Rn × Rn → R,

taken in conjunction with assumption (i), implies ∂aj (Ta ◦ A ◦ T−a )φ = (∂aj Ta ) ◦ A ◦ T−a φ + Ta ◦ A ◦ (∂aj T−a )φ = −Ta ◦ ∂j ◦ A ◦ T−a φ + Ta ◦ A ◦ ∂j ◦ T−a φ = 0. Apparently a 7→ Ta ◦ A ◦ T−a is a constant mapping, with value A, for a = 0; or Ta ◦ A = A ◦ Ta . The desired conclusion now follows from Theorem 11.3. 11.5 Consider φ ∈ C0∞ (Rn ) with 1(φ) = 1. On the basis of Problem 5.2 we then have lim↓0 φ = δ. Accordingly, u ∗ φ ∈ C ∞ (Rn ) satisfies ∂j (u ∗ φ ) = ∂j u ∗ φ = 0, for 1 ≤ j ≤ n. Thus there exists c ∈ C with u ∗ φ = c and this leads to u = lim↓0 u ∗ φ = lim↓0 c in D0 (Rn ). This in turn implies c := u(φ) = lim↓0 c 1(φ) = lim↓0 c in C and therefore u = c ∈ C. 11.6 In view of (11.10) and Example 11.9 we obtain (δa ∗ δb )(φ) = (δa ⊗ δb )(Σ ∗ φ) = δ(a,b) (φ ◦ Σ) = φ(a + b) = δa+b (φ). 11.7 From (11.16) and (11.22) we deduce P v = δ ∗ P v = (P δ) ∗ v. The second assertion is a direct consequence of (11.22). 11.11 On account of (11.22), 1 ∗ δ 0 = 10 ∗ δ = 0 and so (1 ∗ δ 0 ) ∗ H = 0. On the other hand, δ 0 ∗ H = δ ∗ H 0 = δ ∗ δ = δ and so 1 ∗ (δ 0 ∗ H) = 1. This result does not violate (11.19), because neither of the distributions 1 and H has compact support. 12.2 Successively using (11.4), (11.15) and (11.3), we obtain for the potential u ∈ D0 (R3 ) of the dipole f ∈ E 0 (R3 ) X   X X vj ∂j δa ∗ E = u= vj Ta δ ∗ ∂j E = Ta vj δ ∗ ∂j E j

= Ta

j

X

vj ∂j E



j

where

j

1 1 E(x) = − . 4π kxk

Now ∂j E(x) =

xj , 4πkxk3

so

 hv, xi  1 hv, x − ai u(x) = Ta x 7→ = , 4πkxk3 4π kx − ak3

for x 6= a. In particular, for a = 0, v = e1 and x in the (x1 , x2 )-plane provided with polar coordinates (r, α), we see

254

20 Solutions to Selected Problems

u(x) =

1 x1 1 cos α . = 4π (x21 + x22 )3/2 4π r2

Note that the equipotential lines satisfy a polynomial equation of degree 6. In the case of n = 2, E(x) =

1 log kxk, 2π

∂j E(x) =

1 xj , 2π kxk2

so

u(x) =

1 x1 . 2π kxk2

The equipotential lines now become circles centered on the x1 -axis and tangent to the x2 -axis; they satisfy (x1 − c)2 + x22 = c2 , where c ∈ R. 12.3 In order to obtain concise formulas, we introduce some notation. Denote by S ⊂ R2 the strip parallel to the main diagonal, given by S = { (x1 , ξ1 ) ∈ R2 | x1 ∈ R, x1 − a ≤ ξ1 ≤ x1 + a } = { (x1 , ξ1 ) ∈ R2 | ξ1 ∈ R, ξ1 − a ≤ x1 ≤ ξ1 + a }; x1 + a furthermore, ρ(x) = k(0, x2 , x3 )k and β(a, x) = (x ∈ R3 ). ρ(x) √ R dt Moreover, recall the antiderivative √1+t = log(t + 1 + t2 ). Then, for any φ ∈ 2 C0∞ (R3 ), we obtain by changing the order of integration over S, Z −4π ua (φ) = −4π (E ∗ f )(φ) = −4π E(x)f (y) φ(x + y) d(x, y) R3 ×R3 Z a Z 1 = φ(x1 + y1 , x2 , x3 ) dy1 dx R3 kxk −a Z Z x1 +a 1 = φ(ξ1 , x2 , x3 ) dξ1 dx 3 kxk x1 −a ZR Z 1 φ(ξ1 , x2 , x3 ) d(x1 , ξ1 ) d(x2 , x3 ) = kxk 2 R S Z Z Z ξ1 +a 1 p = dx1 φ(ξ1 , x2 , x3 ) dξ1 d(x2 , x3 ) 2 x1 + ρ(x)2 R2 R ξ1 −a Z p β(a,x) = [log(t + 1 + t2 )]β(−a,x) φ(x) dx. R3

Hence p 1 + β(a, x)2 p exp(−4π ua (x)) = β(−a, x) + 1 + β(−a, x)2 p 1 + x1 a−1 + ρ(x)2 a−2 + (1 + x1 a−1 )2 p = . −1 + x1 a−1 + ρ(x)2 a−2 + (−1 + x1 a−1 )2 √ For the purpose of studying the behavior of ua as a → ∞, we note that 1 + h = 2 1 + h2 − h8 + O(h3 ) as h → 0. This implies, as a → ∞, β(a, x) +

20 Solutions to Selected Problems

255

p p ρ(x)2 a−2 + (±1 + x1 a−1 )2 = 1 ± 2x1 a−1 + kxk2 a−2 1 = 1 ± x1 a−1 + ρ(x)2 a−2 + O(a−3 ). 2 Therefore, as a → ∞, exp(−4π ua (x)) =

4a2 2 + 2x1 a−1 + O(a−2 ) = (1 + x1 a−1 + O(a−2 )). 1 2 −2 + O(a−3 ) ρ(x)2 2 ρ(x) a

The unnormalized potential ua becomes unbounded as a → ∞. To remedy this, consider va = ua − ua (e3 ). Then we have, for x ∈ R3 with ρ(x) > 0 and as a → ∞, 4a2 (1 + x1 a−1 + O(a−2 )) exp(−4π ua (x)) = exp(−4π va (x)) = exp(−4π ua (e3 )) 4a2 ρ(x)2 (1 + O(a−1 )) 1 = (1 + O(a−1 )). ρ(x)2 This leads to 1 1 1 va (x) = log ρ(x) − log(1 + O(a−1 )) = log ρ(x) + O(a−1 ), 2π 4π 2π and accordingly 1 1 lim va (x) = log ρ(x) = log k(0, x2 , x3 )k. a→∞ 2π 2π Observe that in this manner we have obtained the fundamental solution as in (12.3) of the Laplacian ∆ acting on R2 . 12.4 The first assertions follow in a straightforward manner. Green’s second identity (see [8, Example 7.9.6]) asserts that, for u as given and φ ∈ C0∞ (Rn ) and in the notation of Example 7.3, Z Z (1U u ∆φ − 1U φ ∆u)(x) dx = (δ∂U u ∂ν φ − δ∂U φ ∂ν u)(x) dx. Rn

Rn

This implies the following identity in E 0 (Rn ): ∆(u 1U ) − (∆u) 1U = −∂ν (u δ∂U ) − (∂ν u) δ∂U . u being a harmonic function on X, it satisfies ∆u = 0; therefore one finds the identity ∆(u 1U ) = −∂ν (u δ∂U )−(∂ν u) δ∂U in E 0 (Rn ). Convolution of this identity with a fundamental solution E ∈ D0 (Rn ) of ∆ gives, for any x ∈ U , u(x) = ∆E ∗ (u 1U )(x) = E ∗ ∆(u 1U )(x) = −E ∗ ∂ν (u δ∂U )(x) − E ∗ (∂ν u δ∂U )(x) = −∂ν E ∗ (u δ∂U )(x) − E ∗ (∂ν u δ∂U )(x) Z   = ∂ν (y 7→ E(x − y)) u(y) − E(x − y) ∂ν u(y) dy. ∂U

256

20 Solutions to Selected Problems

12.5 The direct approach to prove that a = 21 is as follows. Write, for arbitrary φ ∈ C0∞ (R2 ), Z  1V (φ) = 1V (  φ) = (∂t 2 − ∂x 2 ) φ(x, t) d(x, t) V Z Z t Z Z ∞ ∂x 2 φ(x, t) dx dt. ∂t 2 φ(x, t) dt dx − = R

|x|

R>0

−t

Continuing this evaluation is straightforward but leads to tedious calculations. Instead, we discuss two alternative, more conceptual, approaches. First, note that the rotation Ψ as described in the problem satisfies V = Ψ (R2>0 ) and is given by 1  1 −1  the matrix Ψ = √ . In particular, DΨ (y) = Ψ and so det DΨ (y) = 1, for 2 1 1 all y ∈ R2 . For arbitrary φ ∈ C0∞ (R2 ), write φe = φ ◦ Ψ : R2 → R. Successively applying the chain rule, the transposition of a matrix and the orthogonality of Ψ , we then find D(φ ◦ Ψ ) = (Dφ) ◦ Ψ DΨ,

=⇒

grad φe = (DΨ )t (grad φ) ◦ Ψ,

e ^φ = (grad φ) ◦ Ψ = ((DΨ )t )−1 grad φe = Ψ grad φ. grad

=⇒

As a consequence we obtain, for 1 ≤ j ≤ 2, 1 1 j 2 e j g 2 g D j φ = √ (D1 + (−1) D2 )Dj φ = (D1 + (−1) D2 ) φ, 2 2
1 e =⇒ ( φ) ◦ Ψ = (D1 + D2 )2 − (D1 − D2 )2 φe = 2D1 D2 φ. 2 The Change of Variables Theorem (see [8, Theorem 6.6.1]) implies Z Z  1V (φ) = 1V ( φ) =  φ(x) dx =  φ(x) dx Ψ (R2>0 )

V

Z ( φ) ◦ Ψ (y)| det DΨ (y)| dy

= R2>0

Z

Z

Z e 1 , y2 ) dy1 dy2 = −2 D1 (D2 φ)(y

=2 R>0

R>0

e y2 ) dy2 D2 φ(0, R>0

e = 2φ(Ψ (0)) = 2φ(0) = 2δ(φ). = 2φ(0) Next, we describe the second method. Consider the vector field   01 v = −S grad φ : R2 → R2 with S=− ∈ Mat(2, R). 10 Introducing the notation J = Ψ 2 , the rotation in R2 by π/2, and (g1 , g2 ) = (g1 , −g2 ) for a vector field g, we have S=

 0 −1  1 1

0

0 =J◦ 0 −1

.

20 Solutions to Selected Problems

257

Furthermore, J ∈ SO(2, R) implies curl v = div(J t v) = − div(J t J grad φ) = − div( grad φ ) = (D22 − D12 )φ =  φ. It is a straightforward verification that a positive parametrization y : R → ∂V is given by y(s) = (s, sgn(s) s), for s ∈ R. Differentiation then leads to the following formula for Dy(s), where we note that sgn is a locally constant function: Dy(s) =



1  sgn(s)

and

Dy(s) = sgn(s)S Dy(s)

(s ∈ R \ {0}).

Next we obtain, on account of the chain rule and S t = S, for s ∈ R \ {0}, (φ ◦ y)0 (s) = Dφ(y(s)) Dy(s) = sgn(s)hgrad φ(y(s)), SDy(s) i = sgn(s)h (S grad φ) ◦ y(s), Dy(s) i = sgn(−s)h v ◦ y, Dy i(s). On the basis of Green’s Integral Theorem and the compact support of φ we find Z Z Z Z  φ(x) dx = curl v(x) dx = h v(y), d1 y i = h v ◦ y, Dy i(s) ds V

V

∂V

Z =

0

R

Z

0

sgn(−s) (φ ◦ y) (s) ds = (φ ◦ y)0 (s) ds R −∞ Z ∞ 0 − (φ ◦ y) (s) ds = φ(y(0)) − (−φ(y(0))) = 2φ(0). 0

It is straightforward that supp E = V and sing supp E = ∂V . For the last two assertions, we verify that the condition of Theorem 11.13 is satisfied with A = V and B = H or, differently phrased, that the sum mapping Σ : V × H → R2 is proper. Indeed, consider arbitrary (x, t) ∈ V and (x0 , t0 ) ∈ H and suppose (x + x0 , t + t0 ) belongs to a compact subset in R2 . Then x + x0 and t + t0 belong to compact subsets in R. From t ≥ 0 and t0 ≥ t0 we get that both t and t0 belong to a compact subset of R. But now |x| ≤ t implies that x belongs to a compact subset of R, which finally implies that x0 belongs to a compact subset of R. This proves the claim. The theorem says that u is a well-defined element of D0 (R2 ), while (11.22) leads to  u = ( E) ∗ f = f . 12.7 For the first assertion, iterate (12.4). Note that on D0 (R2 ) ∂z ∂z¯ =

1 1 1 (∂x − i ∂y )(∂x + i ∂y ) = (∂x 2 + ∂y 2 ) = ∆. 4 4 4

Hence 1 4 1 ∆(log k(x, y)k) = ∂z¯∂z log(x2 + y 2 ) 2π 2π 2   1   1  1 1 1  x − iy  = ∂z¯ ∂x − i ∂y log(x2 + y 2 ) = ∂z¯ 2 = ∂ = ∂ . z ¯ z π 2 π x + y2 πz π¯ z

δ=

258

20 Solutions to Selected Problems

12.8 The idea underlying the solution is that, formally speaking, the equation grad f = g in (D0 (Rn ))n leads to ∆f = div g, which in turn implies f = E ∗∆f = E ∗ div g ∈ D0 (Rn ). Select φ ∈ C0∞ (Rn ) such that φ = 1 on an open neighborhood of the closed ball around 0 of radius r. Now introduce n X gej = φ gj and fe = ∂j E ∗ gej . j=1

Note that fe is well-defined since the gej have compact support. On account of the identity ∆E = δ as well as the integrability conditions satisfied by the gj , this leads to n n n X X X ∂k fe = ∂k ∂j E ∗ gej = ∂j2 E ∗ gek + (∂k ∂j E ∗ gej − ∂j2 E ∗ gek ) j=1

= gek +

j=1 n X

j=1

∂j E ∗ (∂k gej − ∂j gek ) = gek +

j=1

n X

∂j E ∗ ((∂k φ) gj − (∂j φ) gk )

j=1

=: gek + hk . Take E as in (12.3), note that supp ((∂k φ) gj − (∂j φ) gk ) ⊂ Rn \ B(0; r) and apply Theorem 11.25. One obtains sing supp hk ⊂ Rn \ B(0; r). Furthermore, in view of hk = ∂k fe − gek one has on B(0; r), for all j and k, ∂j hk − ∂k hj = ∂k gej − ∂j gek = ∂k gj − ∂j gk = 0. Applying Poincar´e’s Lemma in the classical setting (see [8, Lemma 8.2.6]), one finds H ∈ C ∞ (B(0; r)) such that in C ∞ (B(0; r)) ∂k H = hk

(1 ≤ k ≤ n).

Any two such solutions H differ by an additive constant. Hence, one may assume that solutions H and H 0 associated with balls B(0; r) and B(0; r0 ), for r < r0 , coincide on B(0, r). It follows that f := fe − H defines a distribution on B(0; r) satisfying ∂fk = gk on B(0; r), for 1 ≤ k ≤ n, while any two such solutions f agree on the intersection of open balls centered at 0. Therefore Theorem 7.4 implies the existence of f ∈ D0 (Rn ) with the desired properties; actually, it is of the form given in the problem. 12.9 (i) Because the power series expansions of g and h only involve powers of kxk2 = P3 2 ∞ 3 j=1 xj , both g and h belong to C (R ). According to Problem 4.6, the func1 3 tion x 7→ kxk on R \ {0} defines an element of D0 (R3 ), and therefore E± does too. (ii) By direct computation or by using a computer algebra system like Mathematica one directly verifies that h is a solution of the homogeneous equation and

20 Solutions to Selected Problems

259


1 furthermore that the identities for grad g, grad k·k on R3 \ {0} and ∆g are
1 valid. On the basis of Problem 4.6 we have ∆ kxk = −4π δ. Furthermore, Leibniz’ rule (9.1) may now be used to obtain the second identity (compare with [8, Exercise 2.40.(i)]), which then leads to  1  h(x) g(x) h(x) 2 g(x) (x) = −4π g(x) δ+2k −k −2k = −k 2 −4π δ. ∆ g k·k kxk kxk kxk kxk (iii) According to the well-known formula for the Laplacian in spherical coordinates in R3 (see [8, Exercise 3.8.(vi)]), one has (∆f )0 (r) = f000 (r) +

2 0 f (r). r 0

(iv) The function fe0 : r 7→ r f0 (r) satisfies 0 fe0 (r) = r f00 (r) + f0 (r),

  00 2 fe0 (r) = r f000 (r) + f00 (r) ; r

and this implies 00 fe0 (r) = r(∆f )0 (r) = −k 2 r f0 (r) = −k 2 fe0 (r).

Accordingly, there exist constants a and b ∈ C such that fe0 (r) = r f0 (r) = a cos kr + b sin kr,

so

f (x) = a

g(x) + b h(x). kxk

(v) h ∈ C ∞ (R3 ) is a classical solution of Helmholtz’ equation, as has been obg served in part (i). Therefore, one only needs to compute (∆ + k 2 ) k·k in D0 (R3 ). ∞ 3 For arbitrary φ ∈ C0 (R ) one finds Z  g g(x) g  (φ) = (∆ + k 2 )(φ) = (∆ + k 2 )φ(x) dx (∆ + k 2 ) k·k k·k R3 kxk Z g(x) = lim (∆ + k 2 )φ(x) dx =: lim I . ↓0 kxk> kxk ↓0
g Since (∆ + k 2 ) k·k = 0 on the open set { x ∈ R3 | kxk >  }, application of Green’s second identity leads to Z  g(y) g(y)  ∂ν φ(y) − φ(y) ∂ν d2 y. I = kyk kyk= kyk For the evaluation of the integral, introduce spherical coordinates (r, ω) ∈ R>0 × S 2 for x ∈ R3 and use dx = r2 dr dω where dω indicates twodimensional integration over the unit sphere S 2 . On S 2 , considered as the boundary of the complement of the unit ball, the direction of the outer normal is the opposite of the radial direction; hence,

260

20 Solutions to Selected Problems

g(y) cos kr kr sin kr + cos kr ∂ν = −∂r . = kyk y=rω r r2 Conclude Z

Z

S2

=:

3 X

Z

∂ν φ(ω) dω − k sin k

I =  cos k

φ(ω) dω − cos k S2

φ(ω) dω S2

I(i) .

i=1 (1)

To estimate I , observe that, for any y ∈ R3 , |∂ν φ(y)| = |h grad φ(y), ν(y) i| ≤ sup k grad φ(x)k =: m. x∈R3

Hence |I(1) | ≤ | cos k| m

Z

dω → 0, for  ↓ 0, Z (2) |I | ≤ | sin k| | sup |φ(x)| dω → 0, for  ↓ 0, x∈R3 S2 Z Z (3) I = − cos k φ(0) dω − cos k (φ(ω) − φ(0)) dω. S2

S2

S2

On account of the Mean Value Theorem one has |φ(ω) − φ(0)| ≤  sup kDφ(x)k =:  M, x∈R3

which implies Z Z (φ(ω) − φ(0)) dω ≤ S2

|φ(ω) − φ(0)| dω ≤  4πM.

S2

Accordingly  g  (φ) = lim I = lim I(3) = −4πφ(0) = −4π δ(φ). (∆ + k 2 ) ↓0 ↓0 k·k (vi) The results above imply that rotation-invariant fundamental solutions are only obtained by taking P a = 1 and b ∈ C arbitrarily in part (iv). This amounts to c± = 1∓bi , so that ± c± = 1. 2 13.1 For Re a > 0 we have Z 0 a−1 Z ∞ a−1 x x φ(−x) dx = Sφ(x) dx = χa+ (Sφ). χa− (φ) = − Γ (a) 0 ∞ Γ (a) The identity now follows for all a ∈ C by analytic continuation. In particular, it −k (k) follows from (13.3) that χ−k (Sφ) = (−1)k δ (k) (φ). − (φ) = χ+ (Sφ) = δ 13.2 Taylor expansion of φ around 0 leads to the existence of ψ ∈ C ∞ (R), so that φ(x) − φ(0) = xψ(x). Hence

20 Solutions to Selected Problems

261

xa−1 (φ(x) − φ(0)) = xa ψ(x). This function is integrable over ] 0, ∞ [ ; indeed, the right-hand side is locally integrable at 0 since −1 < Re a, and the left-hand side is locally integrable at ∞ because d χa+1 Re a − 1 < −1. On account of (13.7) one has χa+ = dx + , where 0 < Re a + 1. Therefore, integration by parts gives Z h xa i∞ xa a+1 0 a χ+ (φ) = −χ+ (φ ) = − φ0 (x) dx = − (φ(x) − φ(0)) Γ (a + 1) 0 R>0 Γ (a + 1) Z Z a−1 a−1 x x + (φ(x) − φ(0)) dx = (φ(x) − φ(0)) dx. R>0 Γ (a) R>0 Γ (a) More generally, for n ∈ Z≥0 and 0 < Re(a + n + 1) < 1, define φn+1 = φ(n+1) and also Z x φk (x) =

φk+1 (t) dt 0

by means of downward mathematical induction on 0 ≤ k ≤ n. Using mathematical Rx induction and 0 φ(k) (t) dt = φ(k−1) (x) − φ(k−1) (0) one proves n−k X xl φ(k+l) (0). φk (x) = φ(k) (x) − l! l=0 Then ( a+k x O(xn−k ) = O(xa+n ) = o(1), x → ∞, xa+k φk (x) = xa+k O(xn−k+1 ) = O(xa+n+1 ) = o(1), x ↓ 0. In the particular case of k = 0, the O-estimates also imply that x 7→ xa−1 φ0 (x) is integrable over [ 0, ∞ [ . On the basis of these estimates, (n + 1)-fold integration by parts leads to dn+1 a+n+1 χ (φ) = (−1)n+1 χa+n+1 (φ(n+1) ) + dxn+1 + Z xa+n = (−1)n+1 φn+1 (x) dx R>0 Γ (a + n + 1) n h i∞ Z X xa+k xa−1 = (−1)k+1 φk (x) φ0 (x) dx + Γ (a + k + 1) 0 R>0 Γ (a) k=0 Z n X xk (k)  xa−1  φ(x) − φ (0) dx = k! R>0 Γ (a) k=0 n Z ∞ X 1 φ(k) (0) a+k  = lim xa−1 φ(x) dx +  . Γ (a) ↓0 k! (a + k) 

χa+ (φ) =

k=0

The last equality follows by means of a simple integration. Finally, (13.3) implies that χk+ L = 0 if k ∈ Z≤0 , which is also the case for k−1 (test x H) because of the zero of 1/Γ at k, see Corollary 13.5. Γ (k)

L

262

20 Solutions to Selected Problems

13.4 Indeed, for φ ∈ C0∞ (R), 1 1 ∂x PV (φ) = − PV (φ0 ) = − lim ↓0 x x

Z R\[ −,  ]

φ0 (x) dx =: − lim I . ↓0 x

Integration by parts gives − ∞ Z φ(x) φ(x) φ(x) + + dx 2 x −∞ x  R\[ −,  ] x Z ∞ 1X 1 X =− φ(±) + φ(±x) dx.  ± x2 ± 

I =

0 2 Next, Taylor expansion where Pof φ about 0 implies φ(x) P= φ(0)+x φ (0)+x Rψ(x), ∞ 1 ∞ 2 ψ ∈ C (R). Hence ± φ(±) = 2φ(0) +  ± ψ(±). In view of  x2 dx = 1  , one finds  Z ∞ X X 1 I = φ(±x) − 2φ(0) dx −  ψ(±), 2 x  ± ±

and therefore Z

x−2

lim I = ↓0

R>0

X

 φ(±x) − 2φ(0) dx = |x|−2 (φ).

±

13.7 Using (13.16) and (13.17), we obtain outside C− ρ=

a−n+1 d a d Γ ( a+1 2 ) ∗ q (χ+ 2 ) . R+ = n da da π 2 Γ (a) a=0 a=0

Next, we use Leibniz’ rule and Corollary 13.5 to conclude that outside C− ρ=

−n+1 −n+1 Γ ( 12 ) d 1 1−n q ∗ (χ+ 2 ) = π 2 q ∗ (χ+ 2 ). n π 2 da Γ (a) a=0 1−n

n−1

In the particular case of n − 1 even, (13.3) implies ρ = π 2 q ∗ (δ ( 2 ) ) outside C− , which is a distribution with support contained in ∂C+ . In view of (13.22) we have qρ = q

d a 2 R = (−n + 1) R+ = (1 − n)E+ . da + a=0

Finally, the last equality is immediate from (13.21). This equality in turn implies that ρ is not homogeneous. 13.8 (i) Because Φ is a submersion, Φ∗ φ belongs to C0∞ (R) in consequence of Theorem 10.14. Further, the validity of the formula for Φ∗ φ results from the following, in which ψ ∈ C0∞ (R) is arbitrary:

20 Solutions to Selected Problems

Z

ψ(y)(Φ∗ φ)(y) dy = (Φ∗ φ)(ψ) = φ(Φ∗ ψ) =

Z

263

φ(x)(Φ∗ ψ)(x) dx

Rn

R

Z

Z Z

=

φ(x)ψ(Φ(x)) dx = Rn

R

Z

Z

=

ψ(y) R

N (y)

N (y)

φ(x)ψ(Φ(x)) dn−1 x dy k grad Φ(x)k

φ(x) dn−1 x dy. k grad Φ(x)k

(ii) From part (i) we obtain, for any t ∈ R, Z t Z t Z (Φ∗ φ)(y) dy = −∞

−∞

N (y)

φ(x) dn−1 x dy k grad Φ(x)k

Z =

φ(x) dx. { x∈X | Φ(x)