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t:o
}r;
+ (})- f(xo
E
L 1 ('JI') the
+B) I d(} = 0.
Multiplicatio n by the bounded function g preserves this continuity. The continuous image of a compact set is compact. Hence we can apply the uniform convergence on compacts to deduce that ( 1.2.11) holds uniformly on compact ¢ intervals.
1.2.2
Fourier Series of Finite Measures
The concept of a Fourier series can be easily extended from the class of integrable functions to the class of finite signed measures. Recall that a signed measure on 1I' is defined by a function of bounded variation, which can be represented as the difference of two monotone functions. The sum of two signed measures is the setwise sum: (J.L 1 + f.Lz)(A) = f.L 1 (A) + f.L2 (A). The convolution of two signed measures f.L J. f.L2 is, by definition, the signed measure f.L with the property that for every continuous function h E C('JI')
lx1f
h(x
+ y) dJ.L1 (x) dJ.Lz(y)
= 2n
J
h(z) dJ.L(Z).
The Fourier coefficients of a signed measure are defined by
{l(n) = _I_ { e-inO dJ.L(B). 2n lu Proposition 1.2.1 carries over with no essential change.
Proposition 1.2.9. The space of discrete Fourier transforms is an algebra, as expressed by (1.2.12)
fl1 (n)
+ flz(n)
= (J.LJ+f.Lz) (n),
(1.2.13)
The convolution product is commutative and associative: J.L 1 * J.Lz = f.Lz * J.L , 1 (J.L 1 * J.Lz) * f.L3 = f.L 1 * (J.Lz * f.L3). Furthermore the mapping f.L ~ fl is a contraction, meaning that ( 1.2.14)
~
IJ.L(n)i
0 and a > 1. Prove thatf is a constant, a.e. Hint: First show that Qp(h
+ k)
::: Op(h)
+ Qp(k); then iterate this to obtain a contradiction.
If we want to obtain a more precise estimation, we can assume that/ is absolutely
continuous and integrate by parts, as follows:
Proposition 1.2.14. Suppose that f
is absolutely continuous. Then f(n) = (1/in)f'(n); in particular f(n) = o(l/!nl), lnl - oo.A If in addition !'./"::.. ... ,J O,define G(x)
=
=
1""
e-"dS(t ),
X>
0.
0
On the one hand we may differen tiate under the integral for x > 0 to obtain G'(x) = -
"'1 o
e-xt sintdt = -
I I
,,
+ x~
X>
0.
Hence G(x) = C- arctanx for some constan t C. But the estimati on jG(x)l ::=: 1/x shows that 0 = G( oo) = C - rr /2, which shows that C = G(O) = rr /2. On the other hand we can transfor m the integral defining G(x) by a partial integrati on:
rr - arctanx
2
=
G(x)
= {oo xe-xtS(t )dt.
h
The latter integral tends toS(oo) whenx -
1.2.5
0. Hence Si(oo)
=
2/rr S(oo)
=
2/rr G(O)
=
I.
•
Pointw ise Conve rgence Criter ia
We are now in a positio n to prove some criteria for the converg ence of the partial sums of a Fourier series at a given point. All of the results describ ed below will be in the form of sufficie nt conditi ons. It is not possibl e to formula te any effectiv e necessa ry and sufficie nt conditi ons for the converg ence, as we will discuss below. The first step is to recall that the partial sum is express ed as a convolu tion with the Dirichl et kernel: (1.2.21 )
SN/(0) = (f
* DN )(0)
1 2rr
= -
1 1f
sin ( N + .!.2 ) ( ¢) . f(O - ¢) d¢. sm (¢/2)
This formula will be simplif ied in two ways. First, we will show that the factor sin (¢/2) in the denomi nator can be replace d by the simpler functio n ¢ ~ ¢/2. Second ly we will show that the integra l over the circle 1I' can be replace d by the integra l over a small interval about ¢ = 0. The details follow. The Dirichl et kernel is an even functio n and satisfie s the normal ization 1 - -
2rr
so that we can write (1.2.22 )
1
1rr
J1r{ DN(O)d O =
SNf(O) = -2 [/(0 rr o
+
¢)
1,
+ /(0-
¢)]
sin(N + 1 ) (¢) . 2 d¢, sm(¢/2 )
26
INTRODUCTION TO Hll'RIER ANALYSI'> >\ND W\VELiclS
and hence for any constant S (1.2.23)
s
SNf(())-
1 1~
=
2rr
[{(8
+ ¢)
+ j({)- ¢)- 25]
0
sin(N + ~) (¢) . (¢f2) d¢. sm
As a first reduction, we can replace the function l I sin(¢ /2) by the function 2/¢ with an error that tends to zero, uniformly in A. This comes from the fact that ¢~
sin(¢/2)
2 ¢
is bounded and continuous on the interval [ -rr. rr ]. The only possible difficulty is at ¢ = 0, where we can apply !'Hospital's rule to show that the difference tends to zero. Exercise 1.2.22. Show that JT
2 Hint: First show that
lx
~ sin xi S
s:xs:
JT
2
lxl' /6 and then simplify the fractions.
For each (), the function
~
¢
F&(¢) = [ .
l sm (¢/2)
-~]if(() +¢)+/(&-¢)) ¢
is an L 1 function, and the map () ~ F& is continuous from ( -rr, rr) to L 1 (1!'). Hence by the Riemann-Le besgue lemma the integral ofF&(¢) sin (N + 1/2)¢ tends to zero uniformly in() when n ~ oo. Thus we have reduced the problem of pointwise convergence to proving that (1.2.24)
lim N
j
rr
[f(B ()
+ ¢) + j(B- ¢ ) -
sin(N+l) ¢ 2 2S( d¢ = 0. 0 and some real numberS 8
1 ()
Proof. lemma.
Thi~
If(()+¢)+ f(()- ¢)- 2SI --=-------'----=---- ---- d ¢
2 (cos n8 flog n) is the Fourier series of an ~
Exercise 1.2.35. Suppose that g is a function of bounded variation on 1!'. Prove that for any f E L 1 (1!'),
lim { g(8)SNf(8) d(} = N~=}y
h{
g(8)f(8) d(}.
1.2.6.1 Convergence of Fourier series of measures The above ideas can also be used to give a quick treatment of the convergence of Fourier series of any finite signed measure on 1I'. The partial sum of its Fourier series is written n
S,J.-L(8) =
L ~n
[l(k)e'kH.
FOURIER SERIES ON THE CIRCLE
31
Proposition 1.2.36. Suppose that 1-L is a .finite signed measure on 'IT'. Then 1iffl
{b
la
SNt-t(B) d(}
=
1
t-L((a, b))+ 2~-t({a})
+
1 2~-t({b}).
Proof. The partial sum on the left is written in terms of the convolution with the Dirichlet kernel. Thus
Since 1 is of bounded variation, SN 1 converges boundedly to 1 (a.b> Applying the dominated convergence theorem completes the proof.
1.2.7
+
~ 1 fa)
+
~ 1 (h). •
Riemann Localization Principle
Fourier series in one dimension have the property that the limiting behavior of the partial sums at a point depends only on the values of the function in a neighborhood of the point, no matter how small. This is expressed as follows. L 1 ('If') is identically zero in an open interval (a, b). Then for any compact subinterval the Fourier partial sums tend uniformly to zero when n -+ oo.
Proposition 1.2.37. Suppose thatf
E
Proof. From formula ( 1.2.24) we have (1.2.29)
SN/(8) = o(l)
+ -I
rr
1" o
[{(8
+ ¢) + f(8
- ¢)]
sin
(N + 4>
!2 )
4>
d¢.
If [a 1, b.J is a subinterval of (a, b), let 28 = min(a 1 - a, b - b.). By hypothesis f(8 + ¢) + /(8 - ¢) = 0 if 8 E [a1, b.J, 4> < 8. Hence the integrand is identically zero when 8 E fa 1, b.], 4> s 8. On the other hand, the integral on [8, rr] tends to zero uniformly when e E [a I, bd, by the Riemann-Lebesgu e lemma. •
The Riemann localization principle allows us to infer that if two functions agree on an interval, then the Fourier series are equiconvergent meaning that limn (Snf1 -S11 f2) = 0 on that interval. This phenomenon is no longer present in higher dimensional Fourier analysis, as we shall see.
1.2.8
Gibbs-Wilbra ham Phenomenon
In the neighborhood of a discontinuity one cannot expect uniform convergence of the Fourier partial sums. The specific form of nonuniform convergence is best illustrated by the example (1.2.30)
f(x) = (rr - x)j2rr. f(x) = -(;r
+ x)j2JT,
0 <X< JT, -JT <X< 0.
32
INTRODUCTION TO FOURIER ANALYSIS AND WAVELI:TS
This is the simplest function of bounded variation that has a single jump of unit size. Now f is an odd function whose Fourier coefficients are
f
~ f(n) = -1
2n i
·~,
f(x)e~Jn'
1rr _n_ _x
sin nxdx
2n
Jr
dx
-i 2nn
so that the partial sum of the Fourier series is SNf(x)
~
=
[sinx
+ · · · + si~Nx].
This may be written as a definite integral by first computing the derivative I 1 (SNf) (x) = -(cosx
7r
+ · · · + cosNx]
1 = -(DN(X)- 1).
2n
Thus (1.2.31)
SNf(x) = -
Defining g(t) =
I
2n
1x
-x 1] dt = -
[DN(t) -
2n
0
1/ sin(t/2)- 2ft,
1-'
+ -1
2n
0
sin(N + 1/2)t . dt. sm(t/2)
we have by a single integration-b y-parts,
fox g(t) sin(N + !) tdt =
(~),
0
N
~
00
uniformly for 0 < x ::::; n, since g is a C 1 function on the interval [0, n]. Thus ( 1.2.32)
-x SNf(x) = -
2n
1x
sin(N + l2 )t dt no t
+ -1
If 0 < x < n, we see clearly that
-x 2n
limSNf(x) = N
To study the behavior when x sup [ SNf(x) O:ox:orr
+ ~ 2n
~
J : : ; _!_2
+-I n
1= - sin t t
0
+0 -x 2n
dt = -
( 1)
-
N
+-1 2
.
=f(x).
0, we note that on the one hand sup Si(x) o:ox f(x- 0) we can discuss the overshoot in a right neighborhoo d of x, defined as lim supN~co.xN~x SNf(xN) - f(x + 0), which will be proportional to the jumpf(x + 0)- f(x- 0). The numerical value of Si (n) = 1.18 ... can be computed by expanding sin t/t in a Taylor series and integrating term-by-term . This is carried out in Section 2.3.3. Exercise 1.2.39. Compute limN SNf(kn/N
+
t)fork = 2, 3, ...
Strichartz (2000) discovered a correspondin g behavior with respect to the arc length of the curve {(x, SNf(x)), -n < x < n}. The proof anticipates the behavior of the Lebesgue constants, to be studied in Section 1.5.
Proposition 1.2.40. Letf be defined by ( 1.2.30). Then when N-+ /_: y'1
where
cl
+
[(SNf)'(x)]2 dx = C 1 logN
OCJ
we have
+ 0(1),
is a positive constant.
Proof. From the computations following (1.2.30) we have (SN/)'(x) so that
(1.2.35)
=
(lj2rr)(DN(x ) -1),
::0 I
so that the error term in (1.2.35) is 0(1), N -+ oo. But Proposition 1.5.1 also shows that cl logN + 0(1) for a positive constant cl. •
r:Jr I cosx + ... + cosNxl dx =
Exercise 1.2.41. Suppose thatf(x), -n :::X::: 7r is a piecewise C 1function and that f is continuous on 1T'. Prove that when N -+ OCJ we have
f:
y'1
+
[(SNf)'(x)]2 dx-+
f:
y'1
+
[f'(x)]2 dx.
34
INTRODUCTION TO FOURIER ANALYSIS AND WAVELFT ¢and write 1 }.
L:::
M
SNf(B):::: (8j)(BI)SNJA, (B)+ :L: 0. The set of accumulation points of the partial sums SNf(xN) when N ~ oo and XN ~ ¢with¢ < XN < rr is described as follows:
( 1.2.37)
FOURIER SERIES ON THE CIRCLE
35
0.5
-6
2
2
6
4
FIGURE 1.2.3 The Gibbs-Wilbraham phenomenon for the functionj(/1) = sgn(&). FromM. Pinsky, Partial Differential Equations and Boundary-Value Problems with Applications. Used by the permission of ThP M(Graw-Hill Companies.
IfxN -+ ¢so that N(xN- ¢)-+ n. then ( 1.2.38)
!imSNf<xN) =f ¢.
1.3 1.3.1
FOURIER SERIES IN L 2 Mean Square Approximation-Pars eval's Theorem
Fourier series are well adapted to deal with the Euclidean geometry of the space L 2 (1I') where the inner product is defined by
<J, g) = _I { f(())g(()) d().
(1.3.1)
2n
Ir::
We can measure the degree of approximation by the mean square distance llf-gll~ = <J-g,f-g).
In particular, if g(()) = .L:~=-N bne 1118 is a trigonometric polynomial then
!If- gil~
=
II! II~ - N
This can be used to estimate the mean square error in terms of the smoothness off. If, for example,f E Ci (II'), thenf(n) = O(ini-J) and
L
IISNf- fllz < C
k- 2J = O(N 1 -
21
N ~ oo
).
k>N
which gives an upper bound for the mean square error when N ~ oo. In order to obtain more precise estimates, we introduce the L 2 -translation error, defined by (1.3.11)
llfh- !II~=
2~ lif(x +h)- f(x)i
2
dx,
This can be expressed directly in terms of the Fourier coefficients by using Parseval's identity to write
L
llfh- fll~ =
(1.3.12)
leinh- 11 2 1f(n)l 2 .
nEZ
The next theorem describes equivalent norms to measure the smoothness off. Theorem 1.3.9. Suppose thatf E L 2 (1I'), 0 0. Proof. From (1.3.12) we have for any M
llfi.-
/II~= (
L + L)
lni~M
~
L
2
lemh- ll 1f(n)l 2
lni>M
2
2
n h 1f(n)l 2 +4
lni~M
lf(n)l 2 •
lni>M
If IISnf- /II ~ en-", then the second sum is O(Mby parts, writing En := Likl~n lf(k)l 2 :
L
L
20
).
To estimate the first sum, we sum
M
n2 l](n)l 2
=
M 2 (EM- E 00 ) +
lni:OM
L(2n- 1)(£
00 -
En-d·
n=l
By hypothesis Eoo -En = O(n- 2"
),
so that both of these terms are 0(M 2 -
11/h- /II~ ~ C [h 2
2
M 2 - 2"'
20 ),
+4M- 2"'].
Choosing M = 1/h completes the proof. Conversely, iff satisfies an L 2 HOlder condition, we can write
2 L[1 -cos (nh)] lf(n)l 2 = 11/h -/II~ ~ K 2 h2"'. nEZ
Integrating this inequality over the interval [0, k] and dividing by k, we have " [1L...., nEZ
sin (nk) nk
J l/(n)l ~
2
~
c ea.
k > 0.
therefore
FOURIER SERIES ON THE CIRCLE
41
The terms on the left side are nonnegative. Restricting the sum to indices n for which lnlk ~ 2, we have
c ea ~
L: nEZ.Inlk~2
[t - sinn~k) Jll 0, IISNJ-!liz = O(N~a), N ~ oo. Prove that LnEZ lni 2 .Bi](n)l 2 < oojor any f3 ), N ~ oo, it is necessary and sufficient that f, f', ... , J be absolutely continuous and that J Then the Fourier series is absolutely convergent:
i·
LnEZ
if(n)i
0.
}1&1>8
Here C is a constant independent of r. In case k(r, (}) > 0, then ( 1.4.5) is superfluous and we can take C = 2rr.
By definition, a directed set is a set I together with a collection of subsets {A;} with the property that for each (i,j) there exists k with Ak C A; nAJ. In case I= [0, 1), the subsets can be taken in the form Ak = (1 1). A complex-valued function! on a directed set has a limit L, by definition, if for each E > 0, there exists a subset A so 1 that lf(x) - Ll < E for all x E A;. With this definition it is immediate that limits obey the usual laws for sums, products, and composition of functions.
t,
Remark. We choose the formulation with a general directed set in order to have maximum flexibility in the applications. For example, for the Fejer means we have the index set {1, 2, ... } with n --+ oo, whereas for the Poisson kernel associated with the Abel means we have the index set [0, 1) with r--+ 1. In the first case we may take Ak = (k, k + 1, ... ) whereas in the second case we take Ak as the open interval ( 1 1).
f,
Example 1.4.5. The Poisson kernel Pr((}) is an approximate identity.
JT
Indeed, we showed in (1.1.39) that Pr((}) dB = 2rr. Since Pr((}) > 0, the second property is automatically satisfied. To prove the third property, note that for IB I > 8 the denominator 1 + r 2 - 2r cos(} = (1- r) 2 + 2r(l- cos(}) > 2r(1- cos 8). Therefore in this interval we have Pr((}) < (1 - r 2 )/2r(l -cos 8), which tends to zero when r--+ 1. The fundamental use of approximate identities is described as follows:
Proposition 1.4.6. Suppose that k(r, (}) is an approximate identity. • If¢> E L 00 (1I') with limo--+O ¢> ( (}) = L, then
(1.4.7)
1 lim - - { k(r, B)¢>(B) dB = L. r 2rr }T
• If, in addition, for each 8 > 0 sup 101 ::: 8 1k(r, B)l--+ 0, then (1.4.7) holdsforall ¢> E L 1 (1I') with limo--+o ¢>((}) = L. Proof. For any 8 > 0, we have
-:}-- f 7r
J.r
1 k(r, 0)(0) de-L= ( 2 7r l101>o
f
+ f
Jllils.o
)
k(r, 0)((0)- L) dO+ o(l).
FOURIER SERIES ON THE CIRC~.E
47
The first integral tends to zero, for any o > 0. Given E > 0, the second integral can be made less than E by taking o sufficiently small, which proves the first statement. To prove the second statement, note that the first integral is bounded by sup 1111 >J lk(r, 8)1 x (L + 1111 d. which tends to zero by hypothesis. The second integral is bounded byE x j, lk(r, 0)1 de, which completes the proof. •
Exercise 1.4.7. Suppose that the approximate identity k(r, 8) has the additional property that k is even: k(r, 8) = k(r, -8)forall8 E 'Jr. Suppose that E L 00 (1r) with lim8 ..... 0 [(8) + (-8)] = 2L, for some complex number L. Prove that formula (1.4.7) holds. Exercise 1.4.8. Suppose that the approximate identity k(r, 8) is even and has the property that for each 8 > 0, sup 181 :o:.li lk(r, 8)1 ~ 0. Suppose that E L 1 (1r) with lim8 ..... o[(8) + ( -8)] = 2L, for some complex number L. Prove that formula (1.4. 7) holds.
In order to apply approximate identites to norm convergence, we recall the notation !¢>for the translate off E L 1 (11), defined by /¢>(8) =f(8 -¢>).The following definition is essential.
Definition 1.4.9. A subspace B c L 1 ('lr) with norm 1!·11 8 is called a homogeneous Banach subspace ifwe have II/III < 11/lla. the mapf ~ fo is B-normpreservi ng. and the map 8 ~ fo is continuous in the B norm. In detail we require that 11!8 1! 8 = llfllaforallf
E
Band al/8
E 'lr,
and that limo ..... o
lifo- fila~ Oforallf
E
B.
Example 1.4.1 0. The space C('lr) with the supremum norm is a homogeneous Banach subspace. The space LP (11) for 1 < p < (X) is also a homogeneous Banach subspace. Exercise 1.4.11. Prove these properties. Then prove that L 00 ('lr) with the supremum norm is not a homogeneous Banach space.
This notion is very effective for dealing with norm convergence, when we represent the convolution of two functions as a vector-valued integral. If K E L 1 (11'), we can write (K *f)(8)
= 2~
L
K(,P)f(8- ¢>) drp
= 2~
L
K(,P)f<J>(()) d,P.
The final integral is a vector-valued integral, defined as a limit in norm of Riemann sums. In particular, iff E B, then K *f is an element of Band we can estimate the B-norm of the vector-valued integral by the inequality
which follows from the triangle inequality for finite sums. Similarly
48
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
which can be analyzed by the more elementary techniques of Proposition 1.4.6. We formalize this as follows.
Theorem 1.4.12. If B is a homogeneou s Banach subspace and k(r, fJ) is an approximate identity, then
li~ II 2 ~
i
k(r,
~)fq, d~- f t
= 0.
Proof. The required norm is less than or equal to
2~ i
ik(r, cf>)lll/ot>- fiiBdcf>,
•
which tends to zero by Proposition 1.4.6.
The first application of approximate identities is to the sequence of Abel means of a Fourier series. To make the connection between Abel means and the Poisson kernel, we recall the basic identity of Fourier reciprocity, Proposition 1.2.3, which states in this case that for any f E L 1 ('lr) _1
2rr
rf(f}- ~)P,(~) d~ = L
J1r
Theorem 1.4.13. Iff
rlntj(n)einB.
nEZ
C('lr), then the Abel means converge uniformly to f. If f E U'('lr), 1 < p < oo then the Abel means converge to fin the norm of U'. If f E L 1 ('][') has right and left limits at f} E 'lr, then the Abel means converge to Hf I. To examine the second term, write (1.4.15)
(m
+
1)
i: (1- m+ __j__l) n+l
laJI :::0 M [<m
+
1)
log (m/n)-
(m- n)].
Now we let m, n ~ oo so that mjn ---+- (1 +a) and use the Taylor expansion of the logarithm; we see that the final term in (1.4.15) is less than (m + 1)8 2 , so that when we
60
INTRODUCTI ON TO FOURIER ANALYSIS AND WAVELETS
divide by (m- n) the sum is eventually less than 28. Thus we have limsuplsn -ani ::S 28.
•
But 8 was arbitrary, so that we conclude sn - O"n ~ 0, as desired.
How did Hardy think of the representa tion (1.4.14) for Sn in terms of O"n? Although we cannot say with certainty, we can surely provide a natural motivation in terms of the continuou s-paramet er analogue. Suppose that three functions a(t), s(t), a(t) are related by the formulas
1
11
1
s(t) =
a (t) = -
a(x) d.x,
t
1
s(x) d.x.
0
To study s(t) in terms of a(t), a(t), we can use Taylor's formula with remainder : t2a(t2)- tw(t,)
=
U2- t!)s(t,)
Solving for s(t 1 ), we obtain s(t 1 )
-
a (t,)
=
+
12
1,,
a(t2)- a(t,)
t2 - - - - t2- t,
(t2- x)a(x) d.x,
fr: (t2 2
x)a(x) d.x
t2- t,
which is the exact analogue of ( 1.4.14) in the continuou s-paramet er context.
Exercise 1.4.41. Suppose thatf(t), t > 0 is absolutely continuou s with an absolutely continuou s first derivative and that f"(t) = 0(1/t), f(t)/t --* s when t --* oo. Prove that f' (t) --* s when t --* oo. Hint: Takej(t)
=
ta(t)
above.
We can now reap some consequen ces of Hardy's theorem.
Corollary 1.4.42. Suppose that f is continuou s on 1I' and that its Fourier coefficients satisfy f(j) = 0(1/ljl), Ul --* oo. Then the Fourier series off converges uniformly on 1I' to f. Proof. We let ao = f(O) and ak = f(k)eik." + f( -k)e-ik1J fork = 1, 2, .... The Fejer means converge uniformly, so that O"n satisfies the hypotheses of Hardy's theorem, together with a, = O(lji)l) when)~ oo, uniformly in (J E 'lr. •
In particular, Hardy's theorem gives a new proof of the uniform convergen ce of the Fourier series of a continuou s function of bounded variation.
Corollary 1.4.43. Suppose that f is continuou s and of bounded variation on 'Jr. Then the Fourier series off converges uniformly on 1I' to f. Proof. We need to check the Fourier coefficients . By partial integration , we have f(n) = _21 { j((J)e-•nlld( J = __ 1_ { e-mBdj((J) = 0
(_!_).
J.r 2mrr J.r n Here we used the fact that the Fourier coefficients of a finite measure are bounded. 1T
•
FOURIER SERIES ON THE CIRCLE
1.5
61
IMPROVED TRIGONOMETRIC APPROXIMATION
1.5.1
Rates of Convergence in C('TI')
We now consider the Fourier approximation in the space C('JI'). In contrast with the space L 2 (1I'), the Fourier partial sum is not the closest trigonometric polynomial in this norm. In order to study the rate of convergence of the Fourier partial sum, we recall the representation of the Fourier partial sum in terms of the Dirichlet kernel: (1.5.1)
Suppose that gN is another trigonometric polynomial of degree N. Then clearly 8N = SNgN so that we have SNf- f = (SNf- SNgN) + (gN -f), from which we obtain (1.5.2)
Therefore the discrepancy ISNf - fl is measured in terms of the best trigonometric approximation and the L 1 norm of the Dirichlet kernel, which we now estimate. The Lebesgue constants are defined by the integrals (1.5.3)
Ln
= -1
2n
1
I sin (n + ~)tl frr . dt. sm (t/2) 2n -rr
= -1
IDn(¢)1 d¢
1r
The asymptotic behavior of the Lebesgue constants is provided as follows.
Proposition 1.5.1. Whenn--+ oo, Ln = (4lognjn 2 ) + 0(1). FurthermoreLn < 4 +log nfor all n ::: 1. Proof. Recall that on the interval (0, rr) the function t
~
-,--sin (t/2)
2 t
is bounded. Therefore we can write
1
1:rr I sin. (n + 4)tl dt
Ln = rr o
21:rr
=JT
0
sm (t /2)
I sin
(n
t
+ ~ )ti -
21(h)
=
-1
1!/m (sinmu)
m
~ I {'r/2 1 11 m
lf(x + 2u) + j(x- 2u) - 251 du,
o(h), h---+ 0. Referring to (1.5.5) and (1.5.6) we have
(1.5.7)
mn 2
1h
=
.
o
2
dct>(u) .:::: mct>(IIm) .::=: E,
SID U
I
(sin mu )2 dct>(u) < {rr/2 _1 dct>(u) sin u - 11/m mu 2 -4-0 ct>(rrl2)- mct>(llm)
(1.5.8)
mn-
+
!"
12
1~
2ct>(u) du. --,mu
1
The first term is O(m- ) and the second term is o(l). The final integral clearly tends to zero if we change the notation and write ct> ( u) I u = E (1 I v) ---+ 0, so that
f
1/m
I ~m
ct>(u)
rr/2
du = m
- -3
mu
•
E(llv) dv---+ 0.
2/rr
Corollary 1.5.7. The Fejer means of an L 1 function converge almost everywhere. Proof. The strong form of Lebesgue's differentiation theorem states that for almost all x E 1r, limh ..... o01h) foh lf(x + u) - j(x)l du ---+ 0. Therefore on this set we can take S =f(x) in the previous theorem. •
Corollary 1.5.8. The Abel means of an L 1 function converge almost everywhere. Proof. If a sequence {sn} is Cesaro-summ able, then it is also Abel-summab le to the same sum, from Proposition 1.4.25. Since CTn(f) converges almost everywhere, the same is true of the Abel means P, f. •
We close this section with a negative result, showing that the Fejer means have an inherent limitation in their ability to approximate functions to a higher order of approximati on. Proposition 1.5. 9. Suppose that f E C ('II') satisfies n ~ ex::>. Thenf is a constant, almost everywhere.
II an (f) - f II=
Proof. Recall that
Hence lklf(k) = _I_ n + I 2rr ~
lkll/(k)l .::::
n
+
r
J.K' 1
~
[f(e)- CT,;(j)]e-lkli de,
r lf(e)- an 1f
1
(f) I de,
lkl lkl
.::=: n,
.::=: n.
o(l/n),
65
FOURIER SERIES ON THE CIRCLE
For any fixed k, the right side tends to zero when In I -- oo, hence](__k) = 0 for all k By the uniqueness of Fourier coefficients, we conclude that a. e. f f (0).
=
Exercise 1.5.10. Suppose that f E L 1 (1I') satisfies n ---+ oo. Prove that f is a constant, almost everywhere.
¥:-
0. •
llan(j)- /Ill = o(I/n),
Finally, we note that the Fejer approximation holds with the rate O(n- 1 ) for wellbehaved functions. Exercise 1.5.11. Suppose that the Fourier coefficients of f E L 1 ('ll') satisfy Lnez. lnll](n)l < oo. Prove that llan(f)- /II= < C/nfor some constant C. Exercise 1.5.12. Suppose that the Fourier coefficients off E L 1 ('ll') satisfy Lnez lnll](n)l < oo. Prove that the uniform limit of n(an(f) - f ) exists and compute its Fourier series.
In the next two sections we introduce other approximate identities to obtain higherorder trigonometric polynomial approximations.
1.5.3
*Jackson's Theorem
Iff has additional smoothness properties, we can obtain quantitive estimates forf- gN by working with the Jackson means. The Jackson means of order four are defined by
JNf(x) = Zh1
1:rr/Z (sinNu)
N
. stnu
o
4
+ 2u) + f(x-
(f(x
2u)) du
where hN :=
l
2
:rr/ (sinNu)
0
4
du.
.
SinU
By examining the transformations in (1.5.4), it is clear that JNf is the convolution of f with the square of the Fejer kernel, a trigonometric polynomial of degree 2N- 2. Therefore J Nf is a trigonometric polynomial of degree 2N - 2. Recall that/ satisfies a Lipschitz condition if there exists a constant K so that
1/(x)-f(y)l
< K!x-
Theorem 1.5.13. Iff has Lipschitz constant K
IJNf(x)- f(x)l ::::
yl. 1,
cl
then Kl
N '
where C1 is a universal constant. If, in addition, the derivative f' is Lipschitz continuous with Lipschitz constant K 2 , then Kz IJNf(x)- f(x)j :::=: Cz N 2
where
c2 is a
universal constant.
,
66
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Proof. We write 2hN
iJNf(x)- f(x)i :::0 - 1
4
2
JNf(x) - f(x) = - 1
2hN
1trl (sinNu) -.(f(x + 2u) smu
o
irr/ 0
2
+ f(x-
2u) - 2f(x)] du
4
(sinNu) -.4Ktlul du. smu
The denominator is estimated by hN
4
{tr/ 2 (sin Nu ) ~ Jo --u-
3
{Nrr/
2
du = N Jo
4
(sin v ) -v-
dv . . __ constN
3
•
To estimate the numerator, note that, on the interval [0, 7r /2), sin u is bounded below by 2u/7r. Making the change of variable v = Nu, the integral is no more than 4
7r
Kt 16
1rr 0
12
(sinNu)
4
4
1 2 7r -KN udu = 16
u
1N:n:f 0
2
4
(sin v) v dv, --
v
which proves the first statement and identifies the constant as 7r
4
Ct = 32
f 000 (sin vjv) 4 v dv J0 (sinvjv) 4 dv
=-'-';;=---'----00
If, in addition, f' is Lipschitz, then the mean-value theorem provides the estimate if(x + 2u) + f(x- 2u)- 2f(x)l :::: 8K2iul 2, which gives the improved estimate iJNf(x) - f(x)i :::0 - 1
hN
1rr; 0
2
4
(sinNu) --.8K2Iul 2 du. Sln U
Again we make the change of variable v = Nu and find the required estimate, with 7r 4
c2 = -
2
f 000 (sin vjv) 4 v 2
dv fo (sinvjv) 4 dv
::....::...,.,oo,--------
•
Exercise 1.5.14. Suppose thatf is Holder continuous with exponent et: lf(x) f(y)l ~ Klx- yla for some 0 < Ot < 1. Prove that IJNf(x)- f(x)l ~ C 01 KN- 01 for a universal constant C 01 • Exercise 1.5.15. Suppose that f is absolutely continuous and that f' is Holder continuous with exponent Ot, 0 < Ot < 1. Prove that IJNf (x) - f (x) I < c~ KN-I-Ot for a universal constant c~.
1.5.4
*Higher-Order Approximation
If r E z+, the space Cr (Ir') consists of functions whose rth derivativef(r) is a continuous function. If r E z+ and 0 < Ot ~ 1, the space cr,Ot ('Ir) consists of functions I E cr ('][') such thatj satisfies a Holder condition of order Ot. Since any differentiable function is Lipschitz continuous, we have the inclusion cr+ I ('Ir) c cr,l ('Jr). If 0 < Ot < 1, we often write, by abuse of notation, cr,Ot ('II') = cr+at ('Jr). It is natural to expect that iff E C('II') has derivatives of higher order, then we will obtain an improved rate of approximation by suitable trigonometric polynomials.
FOURIER SERIES ON THE CIRCLE
67
To make this concrete, consider for any even integer 2k, the difference operator 2k (2k) . ll.zkf(x;u )=L . (-1)'f(x+ u(k-j)) }=0
(1.5.9)
=f(x
J
+ ku)-
2kf(x
+
- 2kf(x- (k- l)u)
(k- l)u)
+ f(x-
+ ··· +
(-1) k(2k)r k (x)
ku).
The coefficients are those that occur in the binomial expansion of (ei 0 vanishes to order 2k at()= 0. Hence iff E C 2k(1r), the derivatives are
(::UY ll.zkf(x; u)lu=O (::U) 2k ll.zkf(x; u)lu=O
+ ···
= 0,
-
1 ) 2 k, which
0 < j < 2k- l,
= (2k)!f(2kl(x) ,
so that we have the bound
More generally, we can apply Taylor's theorem with remainder to prove that when u
~
0
u~o
==> ll.zkf(x; u) = O(u2k-l+a), f E C2k-1(1r) ==> Dozkf(x; u) = O(u2k-1), f E C2k-2,a ('][') ==> Dozkf(x; u) = O(u2k-2+a), f E C2k-l,a(1r)
u ~ 0,
0 n we have
S, T*n =
T*, so that n l
S,J =
T,:
+ SkR and hence
2n-l
l
2n-1
n
k=n
- L:s,J = r: +- L:skR. n
k=n
which can be written in terms of the delayed Fejer means:
But the Fejer kernel is a contraction in L 00 (1l'), thus
lr,(x)- T,:(x)l
=
Ia~.
(R)I .:::: IIRIIoo .:::: E,(j). In particular
12a2n-J(x)- a,_J(x)- T,:(x)l .:::: 2E,(j)
+
E,(j)
=
3E,(j)
so that
llr, -ill"".:::: II
,(x)l :::;
•~> 3Cr'"
For the remaining sum we use the mean value theorem and Bernstein's lemma (1.5.16) to write m(h)-1
L
m(h)-1
L
l;(x +h)- ;(x)l ::5 h
r=no+l
sup l;(x)l
r=no+l xET
m(h)-1
L
::5 h
C2' 2-cri
r=no+l
::5 hCa2m(h)(l-a) ::5 Cah". Hence lf(x +h)- f(x)l :5 IT2 no (x +h) - T 2 no (x)l
+ Cah" as required.
•
Exercise 1.5.24. Use Theorem 1.5.23 and the proof of Theorem 1.5.3 to show that iff satisfies a symmetric Holder condition: lf(x +h)+ f(x- h)- 2f(x)l < c ha for some 0 < a < 1, then f satisfies the usual one-sided Holder condition: lf(x +h)- f(x)l s Cha. (For a = 1 this is false; see Zygmund (1959).) If a = 1, the above estimates break down. Indeed, it is not generally true that En (f) < C In implies that f E Lip(1I'). The difficulty is in the estimate of m(h)-1
L
m(h)-1
l;(x +h)- ;(x)l < h
L
i=no+ I
i=no+ I
sup l;(x)l xe1'
m(h)-1
< h
L
cziz-i
i=no+1
shCm < Ch log (1/h).
Thus we have the general implication
c
En (f) < -
n
~
lf(x +h)- f(x)l < Kh log (1/h).
FOURIER SERIES ON THE CIRCLE
73
Bernstei n's inequali ty can also be used to characte rize the different iability off in terms of rates of converg ence of En(f). Indeed, suppose that En(f) < Cjnf3 for some f3 > 1. Then the triangle inequali ty gives l;(x)l < 3C2-f3i and Bernstei n's inequali ty (1.5.16) shows that for r E z+, r < {3, l~r\x)l < 2(x).
i=no+l
The first term is infinitely differenti able, hence Holder continuou s. The second term is handled exactly as in the proof of Theorem 1.5.23, replacing ; by ?>, to which the same
estimates apply.
Exercise 1.5.26. Suppose that En(f) < Cn-r where r E z+. f E c 2 for some constant C 1. Prove thatf satisfies the continui ty conditio n lf(x+h) -f(x)l .:::: C2flog (1/h) for 0 < h < Compar e with the result of Exercise 1.5.5 .
!.
1.6
DIVER GENCE OF FOURI ER SERIES
In this section we turn to some negative results, which have been instrume ntal in the developm ent of harmoni c analysis. In 1873 du Bois-Re ymond showed that there exists a continuo us function whose Fourier series diverges at a point. This was further develope d to show that any preassig ned set of Lebesgu e measure zero can be the set of divergen ce of the Fourier series of a continuo us function (Kahane and Katznels on, 1966). Meanwh ile, in 1915 Lusin had posed the problem of proving the almost-e verywhe re converg ence of the Fourier series of an arbitrary f E L 2 (1I'). This was proved by Carleson (1966) and extended by Hunt (1968) to the class U' ('JI') for p > 1. Another proof of Carleson 's theorem by C. Fefferm an (1973) has been useful in more recent developm ents of harmonic analysis. Many years earlier Kolmog orov (1926) had proved the existenc e of an L 1 function whose Fourier series diverges at every point of 'JI'. These results and counterexam ples are beyond the scope of this book. We will prove, by two different methods , the existenc e of continuo us function s with Fourier partial sums unbound ed at a point. We will also construc t L 1 function s whose Fourier series do not converge in the L 1 norm. In Chapter 3 we will prove the theorem of M. Riesz (1927) that for any function in ll'('JI'), 1 < p < oo the Fourier series converge s in the U' norm. The upshot of these results and countere xamples is that LP, p > 1 is a good space for one-dim ensional Fourier series, both in the a.e. sense and in the sense of norm
74
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
convergence. The space L 1 is bad in both senses. The space of continuous functions is bad for norm convergence and for convergence at a preassigned point, but good for almost-everywh ere convergence.
The Example of du Bois-Reymond
1.6.1
Proof. By suitable grouping of terms, we will construct a continuous function whose Fourier series diverges at a preassigned point. Without loss of generality, we will do this at () = 0. The desired function will be sought in the form (1.6.1)
where
aJ e'19
and where the integers mk. Nk will be chosen. Bk(()) is the partial sum of a Fourier series of a function of bounded variation with a jump discontinuity at the point () = 0 and otherwise smooth. For example, one may take f(O) = rr - () for 0 < () < rr, extended as an odd function. In this case ak = -i/k. Since/ is of bounded variation, the partial sums are uniformly bounded: IBk(O)i ::= M for a constant M, simultaneously for all() E T, k = 1, 2, .... Hence the series ( 1.6.1) is uniformly convergent to a continuous function, by the Weierstrass M test. In order that the different blocks Bk involve different frequencies, we will choose the integers mk> Nk so that
Although ( 1.6.1) is not written as a Fourier series, we claim that the nth Fourier coefficient of is given by the coefficient of em9 in the series. Indeed, since the series ( I.6.1) is uniformly convergent, when we multiply by e-mfJ we still obtain a uniformly convergent series that we can integrate term-by-term:
f
Since the terms of Bk contain different frequencies, all of the integrals will be zero, save for the value of k satisfying in - Nk I :5: mk> if there is one with an-N* :j:: 0. In that case the integral is 2rran-Nk and zero otherwise. Hence the nth Fourier coefficient is given by an-NJe or zero, which completes the required identification. Now we examine the partial sum at level Nk:
The first sum converges by the Weierstrass M test. The second sum can be evaluated exactly in the case ak = -ijk:
L-;-1I = mk
logmk
+ 0(1).
j=l
We now choose mk so that (log mk) j k 2 -+ oo, for example mk = 2k 3 will do. Having chosen mh we choose the sequence N"- so that Nk+ 1 - Nk > mk + m"-+l which is possible, for example by taking N1 = I and Nk+ 1 = r=::~ (m1 + m 1 + 1 + I) for k = I, 2, .... The proof is complete. •
FOURIER SERIES ON THE CIRCLE
75
The example of du Bois-Reymond depends ~ritically on the fact that the one-sided sums defined by L:J=o ajeiJ8 are divergent when k ---+ CX), whereas the corresponding two sided sums defined by Bk(()) are convergent when k ---+ CX).
Exercise 1.6.1 . Suppose that f is a function of bounded variation that has a jump discontinuity at () = 0 and is otherwise of class C 2 on the circle. Prove the asymptotic formula j(n) = C/n + O(l/n 2 ) and identify C in terms of the jump. 1.6.2
Analysis via Lebesgue Constants
In this section we re-examine the questions of convergence and divergence in a more general setting. A Banach space is a complete normed linear space. For example C('li'), LP (1I') for p ::: 1 are familiar Banach spaces. A mapping T : B 1 ---+ B2 is a bounded linear operator if it satisfies the conditions that (1.6.2)
T(f +g) = Tf
+
Tg,
T(cj) = cT(f),
IITflls2 < Kllflls,
Here cis any complex number and K is a positive real number. For example, if B 1 B2 = LP (1I') and g E L 1('II"), the convolution Tf = f * g defines a bounded linear operator. This is immediate from the B-valued norm computation
11Tflls2 = lllf(x- y)g(y) dyt = 2
0, choose 8 > 0 so that If(()+¢)+ j(B- ¢)- 2sl < 2E for 0 ~N -an = bN+l
there exists N(E) so that an < E, bn < E for n > N. Thus + · · · + bn 2: (n - N)bn, which proves that lim supn nbn .::5 E, which was to be proved. To prove the convergence of the series of second differences, use summation by parts to write E
N
L
n(6 2 a)n = ao- aN- NbN.
n=l
•
When N--+ oo, the right side tends to a 0 , as required.
We can use these techniques to construct trigonometri c series in L (1['). Let {an} 1
be a convex sequence and consider the sequence of functions N
SN((}) = ao
+
2
L
an
cos (nB).
n=l
Recall the Dirichlet kernel and Fejer kernels: Dn(B)
= l
+ 2cosB + · · · + 2cos (nB),
K (B) = Do(B) n
+ · · · + Dn(B). n+l
Inversely, 2cos(nB) = (f::t.D)n(B), Using summation by parts we can write N-1
(1.6.3)
SN(B) = L
n(!::t. 2 a)nKn-l (B) - NKN-1 (B)(f::t.a)N
+ aNDN(B).
n=l
For any B =I= 0, the last two terms tend to zero, while the first sum remains bounded in 1 • We define f e L 1 (II') by the L 1 convergent sum
L
00
(1.6.4)
f(B) = Ln(!::t. 2 a)nKn-I(B). n=l
From the above lemma and the normalizatio n of Kn we see that this series of nonnegative terms converges in L 1 , hence the sum is finite almost everywhere and defines an L 1 function. It remains to compute the Fourier coefficients. To do this we multiply j(B) by e-imB and integrate term-by-term , by dominated convergence . Thus we have 00
f(m) = L
n(!::t. 2 a)nKn-l (m)
n=O
the last step being the result of summing by parts twice. Having established that alml =f(m), we can investigate the L 1 norms.
Proposition 1.6.9. Suppose that {an} is a convex sequence and letf be defined by ( 1. 6.4 ). Then the partial sums s N remain bounded in L 1 if and only if {an log n} is a
80
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
bounded sequence. The partial sums are convergent in L 1 if and only if an log n ~ II 0 when n ~ <Xl. Proof. Returning to (1.6.3) we see that the first two terms are bounded in L 1 • Therefore if sN also remains bounded in L 1 the same must be true of the final term anDN. But its L 1 norm is asymptotic to ( 4 j rr 2 )aN log N, which proves the first statement. To prove the second statement, write SN(O)- /(0) =
L=
n(~ 2 a)nKn-l (0)- NKN-1 (O)(~a)N
+
aNDN(O).
n=N
=·
The first sum tends to zero in L 1 when N --+ as does the second term. Therefore the convergence to zero of llsN- /11 1 is equivalent to the convergence to zero of llaNDNII~> which is equivalent to aN log N --+ 0, which completes the proof. •
1.7
*APPENDIX: COMPLEMENTS ON LAPLACE'S METHOD
1.7.0.1
First variation on the theme-Gaussian approximation
The proof of Laplace's method can be modified at no expense to handle integrals of the form C(t) =
1b A(J-L)erB(iJ-)ei~-'-c
dJ-L.
Assume that A(J-L) and B"(J-L) are Lipschitz continuous. Without loss of generality, we may assume that the maximum of B is attained at f.-L = 0 E (a, b) and that B(O) = 0. Thus we apply Steps I, 2 and 3 of Section 1.1.5 to reduce to 2 C(t) = A(O) /_: e-tk11- ei11-c dJ-L
Applying Step 4 we replace the limits by so that C(t) = A(O)
i:
-<Xl
+
o( ~) .
< f.-L < <XJ and incur an exponential error
e-rk~-'- 2 eiw df.-L +
o( ~) .
But this is the Fourier transform of the Gaussian density, hence we have the asymptotic formula C(t) = A(O)Ifte-c2/4kr
+
o( ~).
All of the error estimates are independent of c, so that this can be used in cases when
c depends on t. Of course, to provide useful information, it is only interesting when c is restrained, for example c = 0(.../i); if cis too large the exponential term will be smaller than the error term when t
1.7.0.2
~ <XJ.
Second variation on the theme-improved error estimate
If we have additional information on the function B(J-L), we can refine the error estimate
to O(ljn 3 12 ), which comes up in many problems. Specifically, assume that A(J-L)
=
1,
FOURIER SERIES ON THE CIRCLE
and that B(JL) is four times differentiable with B' (0) for some b 3 real. We begin with C(t)
lb
=
=
1"
e 18 (J.L) df.L
+
-k < 0, B
111
(0)
= ib3
etB(J.L) df.L.
Assuming that the maximum is attained at f.L reduce to, C(t)
= 0, B" (0) =
81
=
0 and that B(O)
O(e-c
1
),
C >
=
0 we immediately
0.
-.5
Now we use the inequality (1.1.41) with
ZI
= B(JL), zz = -kf,L 2
+
if,L 3b3/3
-8 < f.L
1 the identity (1.9.2)
f. n=O
I
r2n
n.(n
I
+ m).
= Cm
2.n-
!7r sin2m ee2rcos0 de
for a suitable choice of the constant Cm.
-1r
== lm(2r) rm
86
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Exercise 1.9.1. Prove that Co= 1, Cm+t/Cm = 2/(2m + 1) and conclude that Cm = m!2 2 m/(2m)!form = 0, 1, ....
Equation (1.9.2) provides a definition of the Bessel function Im form = 0, 1, 2, ... as a power series convergent in the entire complex r-plane. If we interpret the factorial in terms of the gamma function, the definition can be extended to all complex values ofm.
Exercise 1.9.2. Prove that for r > 0 we have the inequality llm(2r)l < Cmrme 2r.
The higher-order Bessel functions can be recognized from the trigonometric series for the even functionf(B) = e 2rcosB. We begin with the power series expansion ~ (2r cos B)k e2rcos8 = ~ k=O k!
From the binomial theorem we have
leading to the absolutely convergent double series e2rcos8 =
f
r~
t
c.)ei(2j-k)B =
k=O k. j=O
f
t .,
r~
k=O j=O J .(k
. I ei(2j-k)B. j).
If m = 0, 1, 2, ... , the coefficient of e-imB is obtained by summing over those indices (j, k) for which 2j- k = -m. This is a line of slope +2 in the (j, k) plane, written as k = 2} + m. Thus oo r2j+m k Loa r2J = ~ = r = lm (2r) •t(k- .), ~ .,(j + )I .,(j )I (J.k):2J-k=ml· 1 · J=O 1· m · J=O J. + m ·
~ ~
rk
m = 0, 1, 2, ..
Noting thatf is even, if m = -1, -2, ... we obtain the same result with lml in place of m. Hence we obtain the absolutely convergent trigonometric series e2rcos8 = z=~lmln), for which we must prove that an --+ 0. Assume not; then there exists E > 0 and infinitely many indices n 1 < n 2 < · · · so that Ian, I 2:: E. Dividing by Ian, I. we conclude that cos (nkx+ l/>n•) --+ 0. Squaring this and using the double-angle formula, we have 1 + cos (2n* x)) ----+ 0 onE, with lEI > 0. This is a sequence of uniformly bounded functions, which we can integrate and take the limit. But if we apply the Riemann-Lebesgu e lemma to h, we conclude that the second integral tends to zero, thus lEI = 0, a contradiction. •
!(
Riemann introduces the function (1.10.2)
F(x) = Aox2 -
2
L
A~ em 0.
(1.10.4)
~-+0
We will prove that F, (x) is a convex function. Suppose not; then there is an interval (a, h) and a linear function g(x) = rx + s so that F, (a) = g(a), F, (b) = g(b) and F. (x0 ) > g(x0 ) for some x 0 E (a, b). The difference F, (x) - g(x) has a positive maximum at some point X max E (a, b). At this point we must have the inequality F. (x +h) +Fe (x - h) - 2F. (x) :::; 0 for small h. But this contradicts ( 1.1 0.4 ), so we have shown that F. is a convex function. But F is the limit of the sequence of convex fum:tions F., hence also convex. Applying the same reasoning to -F, we conclude that -F ~~also convex, hence F must be a linear function. To complete the proof of the uniquene~s theorem, from (1.10.3) for any h =1= 0, we may retrieve the Fourier coefficients as (sinnh/2)
A n
2
nh/2
_ _1_
2n
-
whereas
{ F(x +h)+ F(x- h)- 2F(x) h2
Ir
_ _I_ { F(x +h)+ F(x- h)- 2F(x)
A
o-
2n
J.r
h2
-in:c
e
_
dx -
o
(n =I= 0)
_ dx - 0.
•
The proof is complete.
Exercise 1.1 0.4. Suppose that LnEZ Aneinx converges to zero for all x {Xi, ... , xk}. Modify the above proof to show that An = Ofor all n E Z. Hint: F(x), defined by (1.10.2), will be piecewise linear.
E
"'I'\
CHAPTER
2 FOURIER TRANSFORM S ON THE LINE AND SPACE
2.1
MOTIVATION AND HEURISTICS
In parallel with Chapter 1, we can motivate the theory of the Fourier transform on the real line by considering an absolutely convergent trigonometric integral (2.1.1)
where
L1ccn1 d~
< oo.
In order to retrieve the coefficient function C (~) we multiply (2.1.1) by e-i 11·' and integrate over x E [ - T, T];
One cannot immediately take the limit T-+ oo without further hypotheses. If, for example, C(~) satisfies a Dini condition, then one can show that the right side converges to 2.nC(~) so that (2.1.2)
C(~) =
lim - 1 2rr
T-.oo
1T
.
f(x)e-'4 d~,
-T
which can be used to motivate the definition of the Fourier transform. The following exercise shows that C(~) can be retrieved under the hypothesis of continuity.
89
90
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Exercise 2.1.1. Show that
~loT (!~
1
Conclude that T-+ oo.
L
e-iqxf(x) dx) dt = 2
1
~;s!.~~ 2- ry) C(~) d~.
-
if C is continuous, then the right side converges to
Example 2.1.2. Let
C(~) = e-YI~I
j(x) =
for y > 0, x
E
2:rcC(ry) when
JR. Then
L
e-yl~le-lX~d~
= 2 Re
~a= e-Y~ e-ixl;
1 =2Re-y+ ix 2y
= x2
+
y2 ·
Apart from a constant, this is the Poisson kernel in the setting of the real line. The normalization is obtained from the elementary calculus integral for JTR dx I ( 1 + x 2) = :rc. Thus we have the normalized Poisson kernel · 1 Py(X) = 1T
v
y
2.
2
y > 0,
X E
+X
JR,
which has the properties of a positive approximate identity: (2.1.3)
Py(x) =::: 0,
L
Py(x)dx = 1,
{ Py(x) dx-+ 0 Jlxl>li
(y ..j... 0).
Exercise 2.1.3. Use the previous example andformula (2.1.2) to compute { e-ix~
lrrt
1 dx 1 +x 2
~ E
JR.
A more classical motivation, which goes back to Fourier, is to begin with a Fourier series on the interval [ -L, L]: (2.1.4)
f(x) ,.....,
L
Cnein:rcxfL'
nEZ
(2.1.5)
Cn =
2~ /_:_J(x)e-in:rcx/L.
These formulas will now be rewritten in terms of the functions
FOURIER TRANSFORMS ON THE LINE AND SPACE
91
Letting f.Ln = mrjL, we have (D..JL)n = rr/L so that we can write (2.1.4) as a formal Riemann sum (2.1.6) Formally taking L f(x)
---+ ~
oo, we find -1 2rr
1 lR
.
F(J-L)e' 1H dJL,
In order to have a more symmetrical theory, we let JL = 2rr I; in the first integral, to obtain (2.1.7)
f(x) ...__
i
F(2rr!;)e2rrtl;x dl;,
This symmetrical form will be in force in the systematic approach beginning in Section 2.2. In case f is real-valued, (2.1. 7) can also be written in terms of real-valued functions by writing F(JL) = HA(JL) - iB(J-L)] to obtain (2.1.8)
f(x)
'""1=
[A(2rrl;) cos(2rrl;x)
+ B(2rrl;) sin(2rr/;x)] dl;.
The above transformations are purely heuristic, with no pretense of rigor. We will show in the following sections that they can be systematically developed to obtain a powerful theory of Fourier analysis on the real line and in Euclidean space.
2.2
BASIC PROPERTIES OF THE FOURIER TRANSFORM
'
In order to formulate an unambiguous theory, we begin with a complex-valued , Lebesgue integrable function on Euclidean space, denoted by f (x), x E Rn. The Fourier transform is the complex-valued function](/;), I; E Rn defined by the integral (2.2.1)
f(l;) = (Ff)(!;) := {
f(x)e-2rril;·xdx.
jTP/'
The Euclidean dot product is I; · x = L.j= 1 t;1x1 . The basic elementary properties of the Fourier transform are summarized in the next statement. Proposition 2.2 .1 . Let f E L 1 (R n). Then
• Continuity: I; 1--4 f(l;) is a uniformly continuous function. • Contraction: The mapping f ---+ J is norm-decreasing from L 1 to L 00 , in the sense that
(2.2.2)
IFf(!;)!
M
Given E > 0, the second integral can be made less than The first integral is majorized by 27Tihl {
ftxt:sM
E
by taking M sufficiently large.
lxllf(x)l dx.
Therefore with the above choice of M, we have lim sup sup If(~ +h) -f
{
}Jil211
l/2(y)/J(Z)Id zdy= {
}fi(ll
=
l/2(y)! ( {
}IRn
1 ( { lfi an
lr.-t'
1
1/J(x-y)!dx )dy
(x- y).f2(y)l
dy) dx.
FOURIER TRANSFORMS ON THE LINE AND SPACE
93
The joint measurability of the product/1 (z)f2(y) is established by writing.fi (z) = limN Lkezkz--N1k2-NS"< z-N and similarly for fz(y). The product is then written as a pointwise limit of simple functions, especially jointly measurable. Therefore the convolution is finite almost everywhere and defines an L 1 function with II/I *h II ::::; 11/IIIIII/z III· Now we can compute Fz (5 )FI (5)
=
l
.. 2
=in =
ln
e-zrr.~-h ( y)/I (z) dz dy
e-2rrtl; Y!z(y)
(in
e-2rrtl; (x-y)II (x- y)
e-ZrrtH (LJI (x- y)fz(y) dy)
dx)
dy
dx.
The differentiation and multiplication properties will be proved below in a more amplified context. Finally, iff is a radial function, then we consider a rotation 'R in the ~-space, making ihe change-of-variabl e y = 'R' x with IYI = lxl, dy =
dx:
]CR5) = { ~(lxl)e-2rrtRI; x dx
Jan
= which shows that
{
}'Jtn
= { ~(lxl)e-2rrii;-R'x dx
Jan
~(lyl)e-zrr,~-y dy =f(~).
J is invariant by rotations, hence a function of 1~ 1.
•
We will also need a form of the Fourier reciprocity formula. Lemma 2.2.2. Suppose thatf E L 1 (IRn) and 1/r E L 1 (JRn) are integrable functions with Fourier transforms j and ~. Then we have the identity (2.2.3) Proof. We use the Fubini theorem to write { 1/f(~)f(5) d~ = { 1/f(~) ( { f(x)e-Zrril; xdx) d~ JlR, Jan }R.n
= =
which was to be proved.
1n (1.n 1/f(~)e-27ril;xd~) f(x)
{
}Rn
f(x).(f,(x)
dx
dx,
•
This applies in particular to a convolution operator with respect to a kernel which is written as an L 1 Fourier integral. In detail, if K(x) = JR.n k(ne2rri~·x d~ with k E L 1 (JR"), then (2.2.4)
94
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Exercise 2.2.3. If 1-L is a finite Borel measure on JR.", its Fourier transform is defined as jl(l;) := fl!?.n e- 2 7Ti~ x !-L(dx). Prove that if /-L. v are.finite Borel measures on JR.", then we have the Fourier reciprocity formula fii?.n v(x) !-L(dx) =fiR" fl(l;)v(dl;).
2.2.1
Riemann-Lebe sgue Lemma
The most basic analytic fact about the Fourier transform is the Riemann-Lebesg ue lemma, expressed as follows.
Theorem 2.2.4. For any f
E
L 1 (JR.")
lim f(l;) = 0,
I~ I~=
and the convergence is uniform on compact subsets of L 1 (JR."). Proof. From inequality (2.2.2), we need only prove this for a dense set of functions in the L 1 norm. From the dominated convergence theorem,
1
IJ(x)l dx-+ 0,
1-M
1
lf(x)l dx-+ 0
{x·IJ (x)l >M}
when M -+ =, so that we can approximate f in the L 1 norm by a bounded measurable function 1, which is supported on the cube [- M, M]". By rescaling we may suppose that M = 1. Now can be uniformly approximated by the simple function
1
2N
j=
L
k2
-N l{(k-1)2-N 4-sk2-N)
k=-2N
111-111
so that 1 :::: 2 x 2-N. Now we will prove that the theorem holds for any simple function,!= 2:~= 1 ck IE, where Ek are measurable subsets of [ -1, 1]" and Ck are arbitrary complex numbers. In case of an indicator function j(x) = n;=J l is a homogeneou s Banach space. This follows from the translation invariance of the Lebesgue integral and a density argument, beginning with continuous functions with compact support. Example 2.2.19. B = Buc(IR.n), the space of bounded and uniformly continuous functions on IR.n with the supremum norm, is a homogeneou s Banach space. Clearly the norm is translation invariant. The continuity of y -+ h is equivalent to the definition of uniform continuity of f.
99
FOURIER TRANSFORM S ON THE LINE AND SPACE
Example 2.2.20. B = C 0 (IR"), the space of continuou s functions vanishing at infinity, is a homogene ous Banach space. As a closed subspace of Buc(IR"), B is a homogene ous Banach space. In the following theorem we will assume, without loss of generality , that the limit in the directed index set is taken as t ~ 0.
Theorem 2.2.21. Suppose that B is a homogene ous Banach space and k, is an approxima te identity. Then f'J 0,
X
E
:R".
By different iation under the integral sign, it is immedia te that u = P(x, y) is a solution of the Laplace equation Uyy + L7= 1 Ux,x, = 0 in the half space {(x, y) : x E R", y > 0}.
Exercise 2.2.45. Suppose that FE L 00 (R"). Prove that
is a solution of Laplace 's equation
Uyy
+ L7= 1 Ux,x,
= 0 in the half space.
We have already shown, in case n = 1, that P(x, y) is an approxim ate identity, by an explicit computa tion. We now obtain an explicit formula for the n-dimen sional case. The general idea is called Bochner 's method of subordin ation, which allows us to obtain new kernels as suitable transform s of the heat kernel in the t-variabl e. In terms of operatio nal calculus , the heat kernel operator H, is the exponen tial of t times the Laplace operator , whereas the Poisson kernel operator Py is the negative exponen tial of y times the square root of the negative of the Laplace operator . Since the exponen tial is a simple and basic function , it is natural to expect that other kernels can be obtained by suitably transform ing the heat kernel. The method will also be applied in later sections to compute the Newtoni an kernel associat ed with Laplace 's equation and the more general Riesz kernels.
FOURIER TRANSFORM S ON THE LINE AND SPACE
107
To compute then-dime nsional Poisson kernel, we begin with the Laplace transform of the one-dimen sional heat kernel:
=
1
e-1~1 2 f4r
e-1~1...;;:
--==-- e -AI dt = - - = A > 0, l; E JR. o v'4iri 2--fi. This is proved by taking the one-dimen sional Fourier transform of both sides, which is justified by the Fubini theorem. Indeed, for the left side we have (2.2.23)
[
JTR
e2n:t~x
roo (e-1~12/4! e-At dt) dl; =
lo
v'4iri
{CX' e-Ale-4rr2rx2 dt =
lo
1 A+ 4rr 2x 2 '
whereas the Fourier transform of the right side is {
JTR
e2n:i~xe-llil...;;: 2--fi.
dl; =_I_ ( 1 ) +_I_ ( 2--fi. -.fi. - 2rr ix 2--fi. 1 A+ 4rr 2x 2 '
1
-.fi. + 2rr ix
which proves (2.2.23). Now we apply (2.2.23) with A 112 = 2rry and l; (2.2.24)
e-ll;l2n:y - --- = 4rr y
1oo
e-1~1 /4r
o
v'4iri
E
)
JR.n to obtain
2
e -4JT2},2! d t,
Finally, we compute the n-dimensi onal Fourier transform of both sides. In detail, we multiply (2.2.24) by e 2 rri~-x and integrate over l; E JR.n. On the right side we recognize the n-dimensi onal Fourier transform of the heat kernel, corrected by the factor (4rrt) e-4n:2ry2 dt
(4rr) 0 so that G(x) > 0 for 0 < x ::S 8. From this we can compute G on the rationals multiples of 8: G(8m/n) = G(8)"'1" and by continuity this formula extends to all real numbers in [0, 8] in the form G(x) = e- 8 ' , B := -8~ 1 log G(8). Now we can use the functional equation (2.2.31) to extend this to all x > 0. Since G is a bounded function, we must have B > 0, which completes the proof. •
In Chapter 5 we will see that this characterization of the Gaussian density is true in the wider context of probability measures on .!R.", not necessarily absolutely continuous with respect to Lebesgue measure.
Exercise 2.2.52. Suppose that G is a measurable and locally integrable function on lR and satisfies the functional equation G(x + y) = G(x)G(y) a. e. Prove that either G(x) = 0 or G(x) = e= for some a E .IR.. Hint: First show that G is a continuous function.
2.2.5
*Wiener's Density Theorem
The Fourier transform in one dimension can be effectively used to study the L 1 closure of the set of translates of a given L 1 function N
(2.2.32)
Lakf(x -xk) k=l
where ak are complex numbers andxk are real. A closely related set is formed by functions written as convolutions (2.2.33)
l
a(y)f(x- y) dy
where a E L 1 (IR). From Lebesgue's differentiation theorem it follows that any finite sum of the form (2.2.32) can be written as an L 1 limit of functions of the form (2.2.33); conversely, any convolution can be written as the limit of Riemann sums. Hence we see that the L 1 closure of (2.2.33) is identical to the L 1 closure of (2.2.32). Wiener's theorem characterizes this in terms of the Fourier transform. The following proof is adapted from Garding (1997).
FOURIER TRANSFORMS ON THE LINE AND SPACE
111
Theorem 2.2.53. Letf E L 1 (JR.). Then the L 1 closure o/(2.2.33) is the full space L 1 (JR.) if and only if the Fourier transform is never zero: f(~) =I= Ofor all~ E JR.. Proof. The necessity of the condition is immediate, since the Fourier transform of a * f is a(l;)](l;). If](~o) = 0, then the same is true for all convolutions and, by continuity, for all elements in the L 1 closure. Therefore the L 1 closure of {a *f: a E L 1 (1R)} is a proper subset of L 1 (IR). To prove the sufficiency, we first note that it's enough to prove that there exists a dense subset of L 1 (IR), all of whose elements can be written a *f for some a E L 1 . We let A 0 = {h Bo
=
L 1 (IR) :
E
h has compact support},
{h E Ao : his piecewise C 2 }.
Clearly A 0 is dense in L 1 (IR) since Fejer's theorem guarantees that we have the L 1 convergence h(x) = 1~
{M h(~) A
}_M
(
1-
~~~)
M
•
e27m/;
•
d~.
lf :
To proceed further, we introduce the notation A = f E L 1 (JR.)}, which consists of continuous functions vanishing at infinity, with the norm
llfiiA
:=
11/111·
We state and prove the following basic lemma. Lemma 2.2.54. For any f E L 1 (JR.) and g E B 0 , let G we have
lim
8~0
Proof.
]G8
lllf -f(O))GoiiA
= g,
G 0 (~)
=
G(~ 18). Then
= 0.
is the Fourier transform of t-+
i_f(s)8g(8(t- s)) ds,
whereas](O)G8 is the Fourier transform of t-+
8g(8t) (if(s) ds).
Therefore the required A norm is estimated as lllf- ](0))G8IIA
~i
( i lf(s)(g(t- 8s) - g(t))l ds) dt
where we have made the substitution t -+ 8t. The final integral tends to zero by the dominated convergence theorem. To complete the proof of the theorem, we let P be the piecewise linear function such that
P(~) = {~
ifl~l
2.
112
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Notice that P(~ 18) ::f. 0 implies that P(~ /28) = f(~) #- 0 to write
1, so that we can use the hypothesis
P8(~) . p(~ /8)f(~) p~~ / 8 ) /(~)
~
=
P(V8)
/(~) f(O) + P(~ /28)lf(~)
- f(O)).
From the lemma we see that for sufficiently sma118 > 0, the term P(~ /28)lf(~) -](0))/].(0) has norm less than 1/2, so that we can make the following convergent Taylor series expansion in the space A:
We have proved that P 8 = fQ 8 where Q 8 any translate of P 8 to obtain
for some Q8 E A. Note that L~N P(~ we can write
E
+ 3k)
A. In the same manner we can apply this to
= 1 for I~ I :5 3N
+
l. Hence for any h E Ao
which exhibits h as the convolution off and an L 1 function, which completes the proof of the theorem. •
2.3
FOURIER INVERSION IN ONE DIMENSION
In this section we give a self-contained treatment of convergence theorems for the Fourier integral in one dimension. Readers who have followed the treatment of Fourier series in Chapter 1 may wish to omit much of the current section, since many of the theorems are direct analogues of the corresponding theorems for Fourier series. An exception is the discussion of one-sided Fourier representations in Section 2.3.9, but this is not used in the sequej.
2.3.1
Dirichlet Kernel and Symmetric Partial Sums
The partial sum operator applied tof
E
L 1 (IR) is defined by
(2.3.1)
We now rewrite the integral defining the partial sum so that it makes no reference to the Fourier transform. This is called the explicit representation via the Dirichlet kernel.
FOURIER TRANSFORM S ON THE LINE AND SPACE
i: (/_: i: (/_:
113
In order to do this, we use Fubini to write
f:J(~)ezrr,l;x d~
SMf(x) =
= = =
! f
00
d~
e2rr1/;(x-y)dl; )j(y) dl;
sin 2n M(x- y) f(y) dy n(x- y)
-oo
=
e-2ml;}j(y ) dy) e2rril;x
sin 2n Mz f
oo
-oo
(x- z)
d
z.
1TZ
the required formula. The previous computati on is summariz ed by writing (2.3.2)
oo
sin 2nMz
-oo
JTZ
f
SMf(x) =
f(x- z) dz
or equivalent ly, since the kernel is an even function SMf(x) =
1
(X)
[f(x
0
+ z) + f(x-
z)]
sin 2n Mz
dz.
JTZ
The function z .._ (sin 2nMz)/n z is called the Dirichlet kernel and the integral operator is a convolutio n with the Dirichlet kernel. We recognize the Dirichlet kernel as the Fourier transform of the indicator function of the interval [- M, M]. As a first applicatio n, we use the Gaussian identity (2.2.9) to compute the (improper ) integral of the Dirichlet kernel. Applying (2.2.9) with x = 0 andf = 1r-I.IJ
11. m
f
t-+0
00
_
00
_si_n_2_n--'-~ _ 4 rr 2 1.; 2 e d~ = 1, JT~
or equivalen tly by changing variables to
f
.m 11
N-+oo
00
_ 00
z=
~
.../i, N
= 1j
_si_·n_2_n_N_z _ 4 rr 2 z2 e
JTZ
.../i we have
dz = 1.
This can be applied first to compute
f
1
-1
sin 2n Nz _ rr2z2 ----e 4 dz= JTZ
(f
oo
{
-oo-
Jlz!~l
)
sin 2rr Nz _ 4 rr2z2 d -1T-Z e
z.
The first integral tends to 1, while the second integral tends to zero, by the RiemannLebesgue lemma. Now
f
1
-1
sin 2rr Nz d z = rrz
f
1
-1
sin 2rr Nz _ rr2.2 (1- e 4 ~ ) dz rrz
+
f
1
sin 2rr Nz _ 4 rr2z2 e dz. _1 rrz
The first integral tends to zero by the Riemann- Lebesgue lemma and the final integral tends to 1, by the previous step. We have proved that 1 !ZNrr sin t lim - - dt =
N-+oo 1T
-2Nrr
t
lim N-.oo
f
1
_ 1
sin 2n Nz rr z
dz = 1.
114
INTRODUCTION TO HJURfER ANALYSiS AND WAVELETS
This is the famous sine integral, which is often computed from Cauchy's theorem on complex integration. In Chapter 1 this was computed from the properties of Fourier series. Now we have redone this using the Riemann-Le besgue lemma and the explicit Gaussian example. It is customary to write
21x
Si(x) = -
TC
o
sin t
- - dt,
t
so that limx->oo Si(x) = 1, lim_.->-= Si(x) = -1.
Exercise 2.3.1. Prove the inequality 11- Si(x)l Hint: Integrate-by-p arts
2.3.2
J,M (sin tft) dt and let M
< (4/rcx)fora llx > 0.
~ oo.
Example of the Indicator Function
We now consider in detail the case of the indicator functionf(x) = 1(a. b) (x). The Fourier transform is
for~ sum
=f: 0 andf(O) =
(b- a), by definition. Now we consider the nonsymrnetr ic partial
The first term is written
while the second term has an identical structure. When we take the real and imaginary part, we see that the real part may be written in terms of the sine integral Si(x) = (2/rc) (sin t/t) dt, hence convergent. But the imaginary part is written in terms of integrals involving J0M [cos ~(x- b)- cos ~(x- a)]/~ d~, which is convergent if x-# a, x-# b, but otherwise diverges logarithmica lly. Therefore the nonsymrnetr ic partial sum SM,Nf does not converge in general. Put otherwise, the improper Riemann integral will not suffice for the Fourier inversion of this function. This apparently anomalous behavior may be attributed to the generality of complex notation. Indeed, if we had begun with the basic trigonometri c form of the Fourier integral (2.1.8) this would not occur, since the correspondin g complex form will necessarily be the symmetric partial sum.
J;
Exercise 2.3.2. Showthatfor theaboveexa mple, So.Nf(a) ,...,_ ClogN andidentify the constant C.
FOURIER TRANSFORMS ON THE LINE AND SPACE
115
We now show explicitly that the symmetric partial sums converge. SMf(x) = !M f(l;)e2rrtl;r
d~
-M
= iM [sin =
2:rr~(x- b)- sin 2:rr~(x- a)]/n~ d~
t£Si(2Mrr(b- x)) - Si(2Mrr(a- x))].
It is immediate that if a < x < b this converges to 1, while if x < a or x > b it converges to zero. At the endpoints x = a, x = b it converges to ~. Furthermore these
approximating functions are uniformly bounded by 3.
Exercise 2.3.3. Check these statements. 2.3.3
Gibbs-Wilbra ham Phenomenon
The Fourier inversion of the indicator function provides the simplest occurence of the Gibbs- Wilbraham phenomenon. This is the detailed statement of nonuniform convergence that is present in the Fourier analysis of discontinuous functions. Indeed, if we had uniform convergence, then the above sequence of continuous functions would have a continuous limit. But the indicator fails to be continuous at its endpoints. In order to see this in more detail, we take the special case a = 0, b = 1. Applying the previous discussion, we see that SMf(x) = HSi(2M:rr(1 - x))
+ Si(2M:rrx)].
For any fixed x E (0, 1) this converges to 1, when M-H:JO. But if we take x = 1j2M ---+ 0, then fM(I /2M) ---+ (1/2)[1 + Si(:rr)], which is now shown to be larger than 1. Indeed Si(:rr) =
2 JT
rrr
lo
sinx 2 ~dx = :rr
=
~
rrr ( 1 -
lo
(:rr1l' 2
x2 3!
+
4
x
6
)
5!- 7!
dx+ . ..
~~ + ~- 35~~so) + · · · 1l' 4
1l' 6
300
17,640
=2-+-9 = 2 - 1.11
x
+ 0.33
- 0.04
+···
+ ...
= 1.18 to two decimal places
so that to two decimal places, limM SMf(l/2M) = 1.09, demonstrating the Gibbs overshoot.
2.3.4
Dini Convergence Theorem
Returning to the theory, we now develop a basic convergence theorem.
116
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Theorem 2.3.4. Suppose that f(x), -ex:>< x <ex:> is a complex-valued integrable function that satisfies a Dini condition at x: for someS E 0,
1 8
(2.3.3)
f_(_x_+_t_)_+_J_ _(x_ _ t)_ _2_S_I dt < ex:>.
--='
1
Then
!M
lim M-HXJ
f(f;)e2Jrii;x
-M
d~
= S.
(It is not asserted that S =f(x)). Proof. From the Dirichlet kernel representation, we have the Fourier partial sum SMf(x)
!
=
M
-M
~
f(t;)e 27"t;x d~
foe sin 2:rr Mz f(x +
=
-=
:rrz
z) dz.
Having proved Fourier inversion for the function e-rrx 2 , we can replacef(x) byf(x) -se-rrx 2 • The new choice off is also in L 1 and satisfies the Dini condition with S = 0. The function z -+ [f(x + z) + j(x - z}]jz is integrable, since the Dini condition takes care of ~z/:sa whereas
1
11
l:l>li
lf<x+z)+f(x-z )i - - - - - - - - dz ::5 z 8
lf(x
+ z) + f(x
- z)l dz < oo.
1:1>8
Using the Riemann-Lebesgu e lemma, it follows that the SMf(x)
-+
0, as required.
•
Corollary 2 .3.5. Suppose that f satisfies a local Holder condition with exponent a> 0: IY
-xl
< 8.
Then Fourier inversion holds with S = f(x). Proof. Taking S = lf(x
f
(x ),
+ t) + j(x- t)
But the integral
we have for 0 < t < 8 -
2/(x)l
=
!(f(x
+ t ) - f(x)) +
(f(x- t ) - f(x})l ::5 2Ct 01 •
J; ~-I dt is convergent, hence the Dini condition is satisfied.
Corollary 2.3.6. Suppose that f has right and left limits f(x a one-sided local Holder condition with exponent a > 0:
if(y)- f(x
+ 0)1
< Clx- Yla,
0) and satisfies
x
0.
Exercis e 2.3.20 . Check this directly from the Fourier inversio n theorem . The partial sums of the Fourier sine transfo rm satisfy (SMfooctHO) = 0, which suggest s the relation to the "bound ary conditi on" /(0) = 0. 2.3.9.3 Gener alized h-trans form It is also natural to conside r Fourier integra l represe ntation for functio ns that satisfy
a bounda ry conditi on of the form
/'(0) = hf(O). The case h = 0 corresp onds to the cosine transfo rm, while the limiting case h -+ oo corresp onds to the sine transfo rm. In order to motivat e the proper integra l transfo rm, we look for the combin ations of sin l;x, cos l;x that satisfy the bounda ry conditi ons. It is immed iately verified that the functio nf(x) = I; cos I; x + h sin l;x satisfie s the bounda ry conditi on. This functio n also has the propert y that hf - f' = (/; 2 + h 2 ) sin l;x, an odd functio n. This immed iately suggest s a new recipe for extensi on of an arbitrar y f to the half line x < 0, namely
126
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
to require thatj'(x) - hf(x) be an odd function. This leads to a first order differential equation, which is solved in detail by writing ](x) = f( -x) - 2h fo-x eh 0 Hk(ax) = Hk(O)
+ ak 1x Hk-1 (ay) dy,
1Hk(a.x)l.5 IHk(O)I +ak 1x iHk-1(ay)ldy.
Therefore 00
1
e-x
212
1Hk(ax)l
dx
.:5
~IHk(O)I + ak 1
.:5
~IHk(O)I +ak~ 1oo 1Hk-1(ay)le- h2 dy.
00
00
IHk-1(ay)l ( /
e-x
212
dx) dy
We perform the corresponding computation for x < 0. Combining the two, we see that the sequence At satisfies the system of inequalities. k ~ 1
136
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
which telescopes to
•
which was to be proved.
Hint: Use mathematical induction, applied to 2.4.4.2
H~
=
k.Hk-l·
Eigenfunctions of the Fourier transform
With this preparation: we can now list the eigenfunctions of the Fourier transform.
Proposition 2.4.17. (2.4.9)
Proof. To prove (2.4. 9), we can use the generating function (2.4. 7) with t real and x replaced by x../2 to write
We apply Lemma 2.4.15 with a =
../2, to see that the series
~ ~~ /_: e-x 12 1Hk (xJ2) I dx 2
converges for ltl < 1/2. Hence we can integrate term-by-term to find that
This integral can be evaluated by completing the square in the exponent and making the substitution y = x- t../2 to obtain e'2/zj"" e_Y212e-i~Cy+r-.12> dy =
.../2iie-t;2!2e-·~r../2er2Jz_
-oo
When we compare this with the original generating function, we see that the only difference is the replacement oft by -it. But the series defining the generating function converges for all complex t, from which we conclude that for ltl < ~.
•
FOURIER TRANSFORMS ON THE LINE AND SPACE
137
If we make the substitution x = y,./2ii, ~ = v,./2ii, this can be written in terms of the usual notations as (2.4.10) k = 0, 1, 2, ....
One can reinterpret the above result as providing a basis of functions in which the Fourier transform has a simple structure. For example, if a function is written as a finite sum:
then the Fourier transform is N
F(~) = 'L,C-i)kake-JT~ Hk(2~../ii). 2
k=O
2.4.4.3
Orthogonality properties
The orthogonality properties of the Hermite functions are obtained from a second-order differential equation which will be proved. Computing as above, we find
H;(x)-
H~(x) =
2 (!)k (terx-r ; 2)lr=O
H;(x) =
(!)
k
(t2erx-r2/2)1r=O
xH~(x) =
(!)k (xterx-r 212 )1r=O
xH~(x) =
(!)
2 k (t(t- x)erx-r ; 2)1r=O
2 = (!!__) (t!!__(erx-r /2)) dt
dt
I
t=O
We now prove the orthogonality of these functions with respect to the measure with 2 density e-x /Z. To do this we introduce the differential operator
138
INTRODUCTI ON TO FOURIER ANALYSIS AND WAVELETS
Thus iff, g are polynomia ls, we can integrate by parts as follows:
L
g(x)Lf(x) e-x 212 dx =
= =
L -l
2 g(x)[f' e-x 12 ]' dx
+
g'(x)[f'e-x 2/2]' dx
Lf(x)[g' e-x2/2]' dx
= Lf(x)Lg( x)e-x212 dx.
Applying this withf = Hm. g = Hn, we see that (n- m) Lf(x)g(x )e-x212 dx = 0,
which proves the orthogona lity. To obtain the normaliza tion, we write L
Hn(X) 2e-x 212 d.x = (-l)n L
= ( -l)n-l
Hn(x)D~(e-x 2 1 2 )dx
L
2 DxHn (x)D~-I (e-x 12) dx
= n(-l)n-1 L = n
l
Hn-!(x)D~-!(e-x2/2)
Hn-! (x)2e-x2/2 dx.
Proceedin g inductivel y, we see that fiR Hn(x) 2 e-x212 dx = n! fiR e-x212 = n!v'21r. The orthogona lity properties may be concisely written k =j k#j.
(2.4.11) We also introduce the normalize d Hermite functions ) -x2/4 (x ) -- (27r )-!/4Hk(X hk rr:~e ,
vk!
k = 0, 1, 2 ...
which satisfy k =J k #J
2.4.4.4
Complete ness
Finally we discuss the question of completen ess of the Hermite functions. We want to show that the closed linear span of finite linear combinati ons of the Hermite functions is the entire space L 2 (lR). If not, there would exist/ E L 2 (!R) which is orthogona l to all of the eigenfunc tions: f!Rf(x)hk( x) dx = 0 fork = 0, 1, 2, .... Since the Hermite
FOURIER TRANSFORMS ON THE LINE AND SPACE
139
polynomial = x" + lower-order terms, we conclude that JJRf(x)x"e-x 2 14 dx = 0 for n = 0, 1, 2, .... Now we can compute the Fourier transform off(x)e-x 2 14 by integrating term-by-term :
where the interchange of sum and integral is justified by noting that the modulus of the integrand is bounded by
which is an integrable function, for any~- Hence the Fourier transform ofj(x)e-x2 f 4 is zero, therefore f (x) = 0 a.e. This immediately shows that any L 2 function has an L 2 convergent Hermite series. Indeed, we define the Fourier-Herm ite coefficients off E L 2 (1R) by
Any finite linear combination fN = L::=o akhk is orthogonal tof- /N, thus
11/11 2
=II/- !Nil~+
II/NII~ 2: II/NII~ =
N
L lckl k=O
2
whichprove sBessel'sine quality: L:~o lckl 2 < 11/11~- Inparticular ] := L:~oakhdsan L 2 convergent series and the difference /- j is orthogonal to hk(x), fork = 0, 1, 2, ... , 00 hence by the above argument/ - f- = 0 a.e. thus/= Lk=O akhk. in the sense of L 2 .
2.5
SPHERICA L FOURIER INVERSIO N IN IR.n
Bochner ( 1931) studied the pointwise convergence of the spherical partial sums of the Fourier integral in Euclidean space. The purpose of this work is to determine the minimal smoothness assumptions necessary for pointwise Fourier inversion in lR". In this section we will give an up-to-date treatment of this material, based on Pinsky ( 1994) and Pinsky and Taylor ( 1997).
2.5.1
Bochner's Approach
The spherical partial sums are defined by (2.5.1)
140
INTRODUCTION TO FOURIER ANALYS~S AND WAVELETS
This integral may be rewritten as an integral transform onf by applying Fubini as follows: SMJ(x) =
1 ({ IH::N
)IR"
f(y)e-2rriy·l; dy)
= { f(y) ( { = =
e-2rri(x-y)·l;
}JR.n
jii;I5M
f
D'M(x- y)f(y) dy
f
D'M(z)f(x- z) dz
}JR.n
./.IR"
e2rrix·l;
dl;
dl;) dy
where the n-dimensional spherical Dirichlet kernel is defined by (2.5.2)
f
D'f.t (z) =
e-2rriz·l; dt; = D'f.t (lzl)
JII;/5M
where we abuse the notation and identify a radial function on lRn with a function on the positive real line. Noting that D'f.t is a radial function, we may further reduce SM.f in terms of the spherical mean value, defined by an integral over the surface of the unit sphere: (2.5.3)
fx(r) := -1Wn-1
1
sn-1
f(x
+ rw) dSw
resulting in (2.5.4) Lemma 2.5.1. The spherical Dirichlet kernel may be computed in terms ofBessel functions according to
(2.5.5) Proof. Taking a system of spherical polar coordinates with ~ 1 = f.J.. cos(}, we have D'J.t(r) = 9n
1M 1.,.
e2.,.irucos9f.J..n-l(sin9)"-2df.J ..d9.
From equation (2.6.6) in the appendix to this chapter, the(} integral is recognized as the Bessel function (rJ.J..)< 2-n>I 2J. The term IM(x) is estimated by repeated integration-by-parts, where we exploit the fact that the integrand has compact
142
INTRODU CTION TO FOURIER ANALYSI S AND WAVELET S
support to write
1
00
1
00
-
YJM(r)fx (r)D'J.t( r)r"- 1 dr =
o
00
=
1
-1 d 17M(r)r" - 1fx(r) - d D'J.t- 2 (r)dr 2 rrr r
d - [r"- 2 '7M(r)fx( r)]D'J.t- 2 (r)dr.
dr
We repeat the partial integrat ion k times to obtain IM(x)
=
Wn-1 --k
(2rr)
1
00
r
(
0
1 d
--d r r
)k
-
[r"- 2 '7M(r)fx (r)]
sin(2rrM r) rrr
dr.
Writing out the derivativ e by Leibnitz , we find that
r(~:rr [r"- 2YJM(r)fx(r)] =
t;C
1kr' (:r)J [YJM(r)] ..(r)]
for suitable constan ts c,k. For each) ~ 1 we have (djdr)'[' 7M(r)fx (r)]--+ (djdr)j] x(r) in L 1 (0, oo) when M --+ oo. Therefo re by the one-dim ensiona l Rieman n-Lebes gue lemma we have for M --+ oo 00
1 0
sin(2rrM r) - - - - r '. rrr
(d)' -
dr
-
['7M(r)fx (r)J dr--+ 0
j = 1, ... 'k.
For j = 0 we have the one-dim ensiona l Fourier inversio n of the absolute ly continuo us function '1M(r)].. (r) at r = 0, which gives
;)k Coifx(O + 0). 2
w, .J~oo IM(x) = (
1
-
The constan t is identifie d by choosin g a function for which we have already proved the Fourier inversio n, e.g.j(x) = e-,.lxl 2 • •
The sharpne ss of the conditi ons is reveale d by the followi ng basic exampl e in three dimens ions.
Examp le 2.5.6. Let n otherwi se.
3 and let j(x) -
1 for 0 < lxl < a and f(x) -
0
To comput e the spheric al mean value, we note that]x( r) is the fraction al area of the sphere S(x; r) which is contain ed in the ball B(O; a). This is zero if r > a + lxl and is one if r < a - lxl. Otherw ise it is comput ed as fx(r) = -1 47l'
1
{w:lx+rwl :5a}
4rlxl
dw
FOURIER TRANSFORM S ON THE LINE AND SPACE
143
Therefore if 0 < lxl < a, the spherical mean value]x(r) is ]x(r) = {
~2
if 0 < r < a - lx I lxl)2
(r -
-
if a - lx I < r < a
4rlxl 0
+
lx I
if r >a+ lxl.
The function r ---+ ]x(r) is Lipschitz continuou s, in particular absolutely continuou s, hence Theorem 2.5.5 applies to prove convergen ce at x. However if x = 0 we have ]x(r) = 1 for r < a and zero otherwise , a discontinu ous function. The Fourier inversion fails in a very simple way in this case, since we can use (2.5.4) to write SMf(O) = 4.rr
faa D~(r)r 2 dr a
= -2
1 0
d r-D1(r) dr dr
= -2aDl,(a )
+
zfoa D,{,(r) dr
2 = - - sin(2M.rra ) .rr
+2
1a
sin(2.rrMr ) dr . o .rr r
The second term tends to 1 when M ---+ oo, whereas the first term oscillates between ±2/.rr; in detail 2
lim inf SMf(O) = I - - , M
7r
limsupSM f(O) = 1 + M
~7r
This example provides a concrete illustratio n of the nonlocal dependen ce of Fourier inversion in three dimension s. The function is smooth in a neighborh ood of x = 0 but has a jump at Jxl = a. The jump effects the impossibi lity of Fourier inversion at x = 0. Figure 2.5.1 gives the profile of the spherical partial sum for this example. Kahane ( 1995) generalize d this example to the setting of a bounded region in JR 3 bounded by a smooth surface. If the surface is analytic and if the spherical Fourier inversion fails at a single point, then the surface must be a sphere and the point must be the center. For more general smooth surfaces, one may have divergenc e of the spherical Fourier partial sum at any preassigne d finite set of points. This example can be modified to provide a concrete example of nonlocaliz ation.
Example 2.5.7. Let f(x) = I for 0 < a < lxl < band zero otherwise. Then by applying the previous example twice and subtractin g, we have 2 SMf(O) = -[sin(2M .rra)- sin(2M.rrb )] + o(l) . .rr Clearlyf( x) = Ofor lxl (a;)
-;. dr
)k ( n- 2
r
-
TJM(r)fx(r )
)
sin(27rM r) 1rr
dr.
!f each of the jump terms is zero, then we obtain the desired converge nce: limM SMf(x) = fx(O + 0). Converse ly, suppose that limM SMf(x) exists. Then it must be equal to]x(O + 0), by Gaussian summabil ity. The final integral also converge s to]x(O+O ), by one-dime nsional Fourier inversion . Therefore the sum involving the jump terms must converge to zero. If we now divide by Moo L~ 1 C; cos(xa; - (}) = 0, then C; = 0.
E lR
is such that
146
INTROL "CTION TO FOURIER ANALYSIS AND WAVELETS
Proof. Multiply the sum by cos(xa1 -(}),integrate over [0, M] and divide by M. Each term in the sum tends to z
\
lim '""'- 8fx0 ' ~ ... ) cos(Ma, - (}n_z) = 0.
M-+oo~ l=l
-
'!
Applying the lemma once more we see that o]}nx(a;) inductively proves the result.
=
0 fori
=
I, ... , K. Continuin~ •
The results of this section can be reformulated in terms of the smoothness index, which is defined as follows: If the spherical mean value r -+ fx (r) is discontinuous, we set J(f; x) = -1. Otherwise r-+ jx(r) is continuous with a certain number of continuous derivatives, denoted J (f; x). The convergence theorem for piecewise smooth functions can be rephrased as follows, where [ ] denotes the integral part.
Theorem 2.5.16. Suppose that f E L 1 (lR.n) is piecewise smooth with respect to x E Rn with smoothness index J(f; x). Then the spherical partial sum SM.f(x) converges when M-+ oo if and only if J(f; x) > [(n- 3)/2], in which case the limit is]x(O + 0). If J(f; x) < [(n- 3)/2], then we have -00< lim inf M-k(SMf(x) -}x(O + 0))< lim supM-k(SM.f(x) -}x(O + O)) 0
2.5.3
Relations with the Wave Equation
We have already seen the close relation between Fourier analysis and the partial differential equation of heat flow, whose steady-state limit is the Laplace equation. In each case the solution is defined by integration of an approximate identity applied to the initial-boundary data, as we have seen in detail. When we come to the wave equation the situation is different, since the solution is no longer expressed as an integral with respect to a positive kernel, but rather a Schwartz distribution on the surface or interior of a sphere. To see this in detail, consider the initial-value problem for the wave equation
(2.5.7)
a2 u -a2 t
(2.5.8)
L n
=
f).u
:=
u(x; 0) =f(x),
a2 u
--2 i=l axi
au
-(x; 0) = 0
at
where/ E Sis a rapidly decreasing function. This initial-value problem can be solved in terms of the Fourier transform] by the formula
(2.5.9)
u(x; t) =
{
]JR.
cos(2rrti~l)j(~)e 21ri~·x d~.
FOURIER TRANSFORMS ON THE LINE All..!5 SPACE
147
It is immediately verified that u solves the wave equation with the given initial conditions, in the sense that lim 1 ~o u(x; t) = f(x) and lim 1 ~o au;at(A:·t) = 0. A corresponding formula can also be developed for the more g~eral initia' ~onditions, which is left as an • • > exercise. · ~ In order to proceed further, we write (2 :. 'J) as ,, .
u(x; ,., ................
~) =
1 c~~(2rrtJ.L)S'(J.L) 00
dJ.L
4
where
e2rril;·xj(~) d~'
S(J.L) = SJ(x) := {
S'(J.L) =
11/;I<JL
{
e2rril;·xj(~)wJL(d~),
11/;I=JL
where wJL (d~) is the surface measure on the indicated sphere. Since/ E Sit follows that j E Sand that both J.L -+ S'(J.L) and t-+ u(x, t) are rapidly decreasing functions. Hence we can apply one-dimensional Fourier inversion to obtain
1
00
S'(J.L)
=
cos(2rrtJ.L)u(x; t) dt.
Integrating once more on [0, M] and applying Fubini, we obtain the following proposition. Proposition 2.5.17. Suppose that f E S and that u(x; t) is the solution of the initial-value problem (2.5.7) and (2.5.8)/or the wave equation. Then the Fourier partial sum can be retrieved through the formula 00
(2.5.10)
SMJ(x) =
1 0
sin(2rr Mt) ]"( t
u(x; t) dt.
Formula (2.5.1 0) was developed and applied by Pinsky and Taylor ( 1997) to study pointwise Fourier inversion oniEuclidean space and other classical spaces on which the wave equation has a known solution. Proposition 2.5.17 will now be used to find an explicit representation for u(x; t), not involving the Fourier transform. To do this, recall the representation of SMf(x) in terms of the Dirichlet kernel
1
00
SMf(x) = Wn-l
D'M(r)]x(r)rn-l dr.
Since/ E S. the function r -+ ]Ar) is also smooth and rapidly decreasing, so that we can integrate-by-parts and use the properties of the Dirichlet kernel to write D'M(r) _ -(Ij2rrr)(8j8r)D'M- 2(r) and obtain if n = 2k + 1 Wn-l { SMJ(x) = (2n-)k
1o
00
sin 2rrMr ( a rr r r rar
)k [rn-2-fx(r)] dr.
But Proposition 2.5.17 also provides a representation in terms of the same onedimensional Dirichlet kernel. Therefore we obtain the following proposition.
148
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Proposition 2.5.18. The solution of the initial-value problem (2.5. 7) and (2.5.8) is given by the explicit formula (2.5.11)
u(x; t) =
Wn-I k
(27r)
t
0 ( tot )
k
-
[tn- 2fx(t)].
Example 2.5.19. lfn = 3, then the solution is u(x; t) = (d/dt)[t]x(t)] = t];(t) + jx(t). The second term corresponds to a measure on the sphere {y: ly-xl = t}, while the first term corresponds to a (dipole) distribution, which is the first derivative of a measure-in the sense of Schwartz distributions. The representation formula (2.5 .11) can be used to exhibit the finite speed of propagation of the wave equation, as contr~ted with the "instantaneous speed of propagation" of the heat equation.
Proposition 2.5.20. Suppose that f(y) = 0 in a ball of radius a centered at x. Then u(x; t) = 0 fort < a. Proof. If f derivatives.
=0
in the ball,
then
]x €, for some € > 0. For example we can take an iterated Fejer kernel (2.5.12)
1/J"(t) = CN8 ( 1
-COS
2:rr8t)N
(:rr8t)2
where eN is a positive constant. Explicit computation shows that {/J-(t) = 0 for ltl > N8, so that we can take 8 = € IN. Without loss of generality, we take 8 = 1 in what follows.
150
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Now lro.MJ is the restriction to the positive real axis of the even function 1r-M.MJ- Writing l[-M,M] = 1[-M,M] * 1/F + 0[-M,M) - 1[-M,M) * 1/F), we have SMf(x) = SlJ(x)
+ S1f(x)
where
r
S1f(x) : =
}JRn
lr-M.M]
0[-M.M) -
* 1/F)(I~ l)](~)e2Jri~-x d~.
The success of the method depends on two fundamental points: (i) S}.,J(x) depends only on the values off in a ball of radius E about x. (ii) S1f(x) is essentially bounded by a constant multiple of fM:siH:sM+l lf(~)l d~. To see this, first write the convolution Or-M.MJ
* 1/F)(t)
=
l_
2
e rristj[-M.MJ(s)..,lr( s) ds 00
=2 SlJ(x) = 2
1
cos(2rrst)
~
1/F(s)ds
Jl"S
foo lfr(s) sin 2 rr Ms ( { ](I.; l)e 2rri~-x cos(2rr sl.; I) Jo rrs Jan 00
= 2
sin 2rrMs
1
1/F(s)
sin2rrMs
0
d.;)
ds
u(s, x) ds
Jl"S
where u is the solution of the wave equation uu = du with u(O, x) = j(x), u 1 (0, x) = 0. Indeed, this equivalence has been demonstrated for the class S, and the solution formula (2.5 .11) easily extends to f E L 1 + L 2 . This formula shows two important properties.
=
=
• Ifj 0 in a ball of radius E about x, then S1f(x) 0. • If s ---+ u(s, x) satisfies a Dini condition at s = 0, then limM S1f(x)
= j(x).
To estimate S1f(x), we note that lr-M.MJ * 1/F(t) = 1,~: lfr(y) dy. Assuming the iterated Fejer kernel (2.5.12) with 8 = 1, we see that forM> 1
1
/ti>M
11/F(t)j dt
1
1/F(t) dt = 1
which can be combined into the overall bound
1
/t/>M
.l.(t) dt < o/
-
(1
eN
+ M)2N-l,
0 < M < oo.
FOURIER TRANSFORM S ON THE LINE AND SPACE
We use this to estimate the difference Or-M.MJ
* 1/f)(t)
151
- lr-M.MJ(t) separately for ltl >
M and ltl < M. In the first case 1[-M.MJ(t) = 0 and we have
1[-M.M]
* 1/f(t)
=
f
t+M
1/f(y) dy
0
1
1->l>li
which proves that
K'M
iK:,(x)ldx =
1
Kf(y)dy--+- 0,
Jy/>o/M
is an approximate identity.
M--+- co,
•
FOURIER TRANSFOR MS ON THE LINE AND SPACE
153
One can also prove a.e. summab ility, as follows. Proposi tion 2.5.25. /fa> (n- 1)/2 andf E IJ'(lRn) with 1 < p < oo, then for every Lebesgu e point, limM Uf.,f(x) = f(x). In particul ar this holds at almost every x E lRn. Proof. Without loss of generality we can suppose that x = 0. Furtherm ore we may replace 2 - j(O)e-1x1 to reduce to the casej(O) = 0. Define
f by j(x)
(r) E(r) := - -
lf(y)l dy,
(r) := {
r"
Jly\9
From the hypothes is of Lebesgue point we have E (r) --+ 0 when r --+ 0. Now if 1 for larger we can use Holder's inequality to write (r) ~ Cn,p
E(r) < C
r" 1P',
II fliP
-
n,p
r-n(l-lfp')
'
~
p < oo,
p' =pf(p- 1)
so that E is a bounded function. In case p = oo it is immediat e that E is a bounded function. Now we can write UXJ(O) = jUXJ(O)I
{
JIR"
K'f-t(x)j(x ) dx
~ c,.M" Ln =
+ Mlx~)a+ -1.
FOURIER TRANSFORMS ON THE LINE AND SPACE
155
This power series converges in the entire complex plane. If m is not an integer, we define the factorial as the Gamma function m!
=~a= tme- dt 1
= r(m
+
1).
Proposition 2.6.1. lfm, v > 0, the Bessel functions satisfy the relations d -[tmJm(t)] = tmJm-J(t) dt
(2.6.2)
t > 0
d
-[t-m Jm(t)] = -t-m Jm+l (t) dt
(2.6.3)
t > 0
(2.6.4) J:;,(t)
+
1 tJ:r,(t)
+
( 1-
m2) Jm(t) fZ
= 0
(' = : , )
t>O
(2.6.5) (2.6.6) The first three are obtained by termwise differentiation of the power series. For the fourth, we substitute the definition of J m into the integral and integrate term-by-term:
1
1
o
Jm(Rs)sm+l(l- s2)vds =
1' ( o
1
00
~~)-lY (Rsj2)m+2j) ., . I sm+l(l- s2)vds j=O J.(j + m).
oo
.
(R/2)m+2j
1' .
= - L(-1)' . . rm+1 (1- r)vdr 1 + m).1 o 2 j=O J.(j 2vv! oo . (R/2)m+v+l+2j = Rv+l L j=O( - l ) ' J."I(m + v + J· + 1)1. 2vv! = Rv+l Jm+v+l (R) The final integral formula is obtained by making the substitution s = sin 8 and recognizing the power series coefficients from Chapter 1.
Exercise 2.6.2. Prove (2.6.2) by termwise differentiation of the power series definition (2.6.1). Exercise 2.6.3. Prove (2.6.3) by termwise differentiation of the power series definition (2.6.1 ).
Exercise 2.6.4. Prove (2.6.4) by termwise differentiation of the power series definition (2.6.1 ). Exercise 2.6.5. Complete the details ofthe proofof(2.6.6).
156
INTRODUcriON TO FOURIER ANALYSIS AND WAVELETS
The asymptotic be.havior of J m(t), t --+ oo is most efficiently deduced from a differential equation. Let y(t) = v'fJm(t). From (2.6.4). we have by successive differentiatio n Cy 2 1 (2.6.7) C . ·-m 0 < t < 00, -y" +Y = 2• 4" t
This implies that y and Y 1 remain bounded when t --+ oo. since d -(y2 + 12) = 2 I+ 2 I fl = 2Cyy < _ICI _(y2 +y12). dt y yy y y t2 t2 I
Hence fort > to. y(t) 2 + y (t) 2 < [y(to) 2 + Y 1 (to)] exp 1
(1c1 i' ~:)
< [y(to)
2 + y'(to)]e 1 CI/to,
which proves the required boundedness . From this we have the representatio n (2 .6. 8)
y(t) =
l
~
r
.
0W
.
sui(t-s)-2--ds+A 1 cost+A2sm t. S
Indeed. the first term on the right of (2.6.8) is a solution of the differential equation (2.6. 7). so that it differs from y(t) by a solution of the homogeneou s equation z" + z = 0. whose general solution is A 1 cost+ A 2 sin t. From this follows an asymptotic formula for y(t). since the integral term is bounded in the form
~~~ sin(t- s) c~;s) dsl ~ ~~ IC~;s)l ds =
o( ~),
t--+
00.
We summarize the above work as follows.
Proposition 2.6.6. The Bessel function Jm(t) satisfies the asymptotic relation .JiJm(t) = A1 cost+ A2 sin t
+
o( ~),
t--+
00
for suitable constants A1, A2. Equivalently , we may write Jm(t) =
~ cos(t- &) + o(t31/2)'
t--+
00
for suitable constants A, &.
The constants A, e can be explicitly identified in the case of integer m by the method of stationary phase. described in the next section. In the case of half integer m. it is often possible to identify the constants from elementary formulas. beginning with J1;2(x) = ...;'21irX sinx.
Exercise 2.6.7. Prove that the derivative of the Bessel function satisfies the
asymptoticf ormula (d/dt)(v'tJm (t)) = -A 1 sint +A 2 cost+ 0(1/t), t--+ oo. Hint: Compute Y 1 (t) from (2.6.8).
FOURIER TRANSFORMS ON THE LINE AND SPACE
2.6.1
157
Fourier Transforms of Radial Functions
Iff e L 1 (.!Rn), the Fourier transform j is
This follows from the representation of Bessel functions, specifically (2.6.6) with = .rrrl~l- For details, see Stein and Weiss, 1971, p. 154. We can use the representation (2.6.9) to prove additional properties of Fourier transforms of radial functions for restricted values of p e (1, 2). m
=
(n/2)- 1, t
Proposition 2.6.9. Suppose that 1 < p < 2nj(n+1). ThentheFouriertransfor m (2.6.9) is a continuous function for~ =I= 0, which vanishes at infinity and satisfies
Proof. Recalling the bounds on the Bessel function and applying Holder's inequality, we have
J(n-2)/2(2rrrf~f) ~-1 d f
{M
I
Jo
cp(r)
(rl~ i) 1.
Corollary 2.6.13. Suppose that f
E LP(JRn)
Then J;I;I=J(r~)w(d~) ---. 0 when r---.
where 1 < p < (2n
+
2)/(n
+ 3).
<X>.
One may note that the results on radial transform s are valid on a wider set of l.J' spaces than the L 2 -restrictio n theorem, since 2n + 2 2n n> 1 ==> n+3 "(x) = cosx. The only stationary point is x = 0, where (/> 11 (0) = +1, g(O) = 3. Applying (2.7.2), we have
t-+
2.7.2
00
Application to Bessel Functions
As a primary application of the method of stationary phase, we propose to identify the constants in the asymptotic behavior when t-+ oo of the Bessel function Jm(t), which is represented by the integral (2.7.5)
m = 0, I, 2, ....
Proposition 2.7.3. The Besselfunction has the asymptotic behavior
(2.7.6)
Jm(t)
=If cos (t- rr/4- mrr/2) + 0 ( ~).
t-+
00.
Proof. From the integral representation (2. 7 .5) we have (2. 7.1 ), where rp(x) = cos x, g(x) = (lj2rr)e-imxe-im:n:f 2 . Since rp'(x) -sinx, rp"(x) -cosx, there are three stationary points: x 0, x 1r, x -1r, with rp"(O) -1, rp"(rr) 1 rp"(-rr); also g(O) (1/27r)e-im7r/ 2 ,g(7r) (l/2rr)e- 3 im,./ 2 g(-1r). Weapplythemeth odofstationaryph ase,
=
=
=
=
=
=
=
=
=
=
=
FOURIER TRANSFORM S ON THE LINE AND SPACE
165
noting that the endpoints contribute with a factor of ~. Hence e''e-i:n:/ 4
J (t) = m
{2ir
27r
+
(
1 1) Ff-7r . . 14 e-31m:n:/2 ( 1) - +-e-"e':n: + 0 2 2 t 27r t
+ {2ir e-it ei:n:/4 eim:n:/2 +
Vt
H,
(.!.) t
v~
=
0
2rr
f l (ei(t-:n:/4-mrr/2) + e-•(x)d x
=
c
ld .g~x) c
=
d (eitv>(xl) dx
ttcp (x)
.!..id
g(x) eitv>(x) lx=d _ itcp'(x) x=c it
c
eitv>(x) !!._ dx
(
g(x) ) dx.
cp'(x)
Both terms are 0(1/t), t-+- oo, and can therefore be included in the remainder term. Therefore we can restrict attention to contributio ns from intervals containing the stationary points. Assume that x 1 is a stationary point for which cp" (x 1 ) > 0, and let ~ > 0 be chosen so that cp(x)- cp(x 1) > 0 in the 'interval x 1 - ~ < x < x 1 +~.We introduce a new variable of integration v through the equation v = (x- x 1 )
cp(x) - cp(x1) (x- x 1 ) 2
X1 -
~
0, is a locally integrable function and lim 1 ~ 00 f(t) = L. 00 Then limpJ.O j 0 f(t)pe-P 1 dt = L. • Suppose that g(s), s > 0 is a locally integrable function and that the improper 00 integral ] 0 g(s) ds converges to some L. Then
lim pJ.O
roo g(s)e-psds =
lo
L.
168
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Proof. The proof consists of writing
1""
f(t)pe-P' dt- L
=
1""
(f(t)- L)pe-P' dt.
Given E > 0, we split the region of integration into the two regions 0 < t < T and T < t < oo, where Tis chosen such that 1/(t) - Ll < E fort > T, so that the second integral is less than E. The first integral is less than p J:J" 1/(t) - Ll dt which tends to zero when p ,J.. 0 and the first statement follows. To prove the second statement, we setf(t) = J~ g(s)ds. By hypothesisf(t) - 4 L when t - 4 oo. Interchanging the orders of integration yields
1""
1 = 1"" 1"" 00
f(t)pe-P'dt
=
= Applying the first limpJ.O fo"" g(s)e-P-'ds
statement L.
=
gives
the
pe-P'
(1'
g(s) ds) dt
g(s) ([""pe-P' dt) ds g(s)e-ps ds.
result:
if
lim,_ 00 J~ g(s) ds = L,
then
•
CHAPTE R
3 FOURIER ANALYSI S IN LP SPACES
3.1
MOTIVATION AND HEURISTICS
Much of modem Fourier analysis is concerned with bounded linear operators on the Lebesgue spaces LP('JI') and LP(lRn). This chapter is devoted to the development of systematic methods for proving the boundedness of relevant operators by the method of interpolation. Following M. Riesz, if we can first prove that an operator is bounded on two different pairs of Lebesgue spaces, then we can often deduce boundedness on the intermediate spaces. A more general concept, that of weak boundedness, can often be used in place of strict boundedness, following the work of Marcinkiewicz. These techniques are applied to prove the LP boundedness of the classical Hilbert transform, both on the circle and on the real line. This yields theM. Riesz theorem on LP convergence of Fourier series and integrals in one dimension. This chapter also includes the HardyLittlewood maximal inequality. which proves the weak L 1-boundedness of a fundamental operator that underlies the Lebesgue differentiation theorem and many other almosteverywhere convergence results in Fourier analysis.
3.2
THEM. RIESZ-THORI N INTERPOLAT ION THEOREM
In order to introduce the ideas, we first develop some elementary properties of LP spaces. Lemma3.2.1. Supposethatf
E
LP"(lRn)nL 00 (1Rn). Thenf E LP'(lR")forany
PI> PO· Proof. Letting M
= llfll""' we write {
}R''
lf(x)IP' dx
~
Mr,-ro {
~n
lf(x)IPo dr: < oo.
• 169
170
INTRODUCTION TO f'OURIER ANALYSIS AND WAVELETS
We say that/ lives on a set offinite measure if l{x : f(x) =I= 0} I < Lemma 3.2.2. Suppose that f E If' (ll~.n) lives on a set Then f E LP" (JR.") for any Po < p 1.
CXJ.
B of finite measure IBI.
Proof. From Holder's inequality we can write
{ if(x)i"" dx = ~
{
1.
if(x)i"0 1s(x) dx _:: : : ( {
1.
Lemma 3.2.3. Let 0 < p Thenf E LP(JR..").
1
< p < p
1
if(x)lf' 1 dx)POIP• (IBI)(pl-Pol/Pi .
•
and suppose that f
E
LP0 (lR")
n
LP' (lR").
Proof. We write/= jl 1111 sll + jl 11 fi>IJ = / 1 + /2. Both/1 and/2 are dominated by f, in particular fi E L"" andf2 E L"'. Butj1 is bounded and/2 lives on a set of finite measure, since l{x :f2(x)
¥=
0}1 = l{x: if(x)i > 1}1 ::0 {
Therefore by the preceding lemmas,j1 space, hencej E L"(!Rn).
I~.n
E
L"(IRn) andh
E
if(x)l"0 dx . L"(!Rn). But LP(!Rn) is a linear •
Lemma 3.2.3 has a sort of converse, as follows. Lemma 3.2.4. Let 0 < Po < p < p 1 and suppose that f E LP(IR"). Then there existfo E LPo and/1 E LP' such thatf =fo + j,.
•
Proof. It suffices to take fo = f lu1:s 1 and /1 = f llfl > 1.
The last two lemmas can be restated as follows: if Po < p < p
1,
then
(3.2.1) The above lemmas show that for any measurable function/, the set {p : II/ lip < CXJ} is a connected subset of the real line. The theory of M. Riesz-Thori n quantifies this by showing, for example, that the mappingp--- + log II fliP is a convexfunct ion of 1/p. At the same time we deal with linear operators that are simultaneou sly defined on two different LP spaces, developing the interpolation properties of these linear operators. Exercise 3.2.5. Use Holder's inequality to show directly that ifl :::=Po < Pt < andO (-.z) := La:'IPe;a,IA, = La:'/p(Z)ej. j=l
]=I
N
N
1/F(·,z) := Lb'J'/q'e' 13'1o1 = L b f 1q'cpj• J=l
J=l
where we have set
It is immediate that r/> ( ·, z) Therefore the integral
(3.2.2)
F(z)
=
E
LP1 (M), 1/f ( ·, z)
1
Ar/>(·. z)l/f(-, z) dv
E L
=
q; (N),
t
in particular A 1/f ( ·, z)
dj 1Pbeq'
j,k=l
N
E
U1 (N).
1
(AE>j)<J>k dv
N
is a finite sum of exponential functions, in particular an analytic function in the open strip 0 < Re(z) < I and is bounded and continuous on the closed strip 0 ~ Re(z) < I. On the boundary we have by direct computation
llr/>(·,
iy)llp0
llr/> (-. 1 +
iy) lip,
111/F(·, iy)llq~ 111/F(·, 1 +
iy)llq'
I
= 111/lp/poiiPO = IIJII;1Po = 1, = II IJIPIPI lip, = llfii;1PI = I. = II lglq'!q~llq~ = llgll!jqb = I, = lllglq'/q', llq' = llgllq/q', = 1. q I
1
By the definition of F(z) above and Holder's inequality, we have from (3.2.2)
IF(iy)l ::::; liAr/>(·, IF(l
iy)!lq0
111/f(iy)llqb ::::; leo
+ iy)l:::::: liAr/>(·, 1 + iy)llq,lll/F(·, 1 + iy)llq;
:::::: kl.
On the other hand, fort E (0, 1), r/> (x, t) = f(x), 1/f(y, t) = g(y), so that F(t) = JN Afg dv. Applying Lemma 3.2.11, we conclude IF(t)l::::; k~-~k~. which completes the proof. •
Now we examine some important applications of theM. Riesz-Thorin interpolation theorem.
174
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
3.2.0.1
Generalize d Young's inequality
The interpolation theorem of M. Riesz-Thorin can be used to prove the generalized Young's inequality for convolutions : (3.2.3)
+ .!.r
( 1
=
.!. + .!.) q
p
where 1 < p < ex:>, I < q < ex:> with 11p + I I q :::: I. Here f, g are measurable functions on JR.n and the integrals are taken over all of JR.n. Proof. To prove (3.2.3), we begin with the elementary estimates from Lebesgue integration theory:
f
II/* gll1 ::: ll/ll1 llgll1
E
L 1, g E L 1
/EL=, gEL 1 •
11/ *gil= ::: 11/11"'-llgiii
This shows that, for a fixed g E L 1 , the map f ---+ f * g defines a bounded operator on L 1 and L 00 , with norm less than or equal to one. Therefore, by the M. Riesz-Thorin theorem, this map can be extended to LP, with the same operator norm, to yield (3.2.4)
f
In other words, the map g ---+ f inequality, iff E LP, g E U',
E
LP, g
E
L 1•
* g is bounded from L 1 to LP. On the other hand, by HOlder's
(3.2.5)
p' = pj(p- 1).
Therefore the map g ---+ f * g is bounded from LP' to L oc. Hence we can apply theM. RieszThorin theorem with Po= 1, qo = p, p 1 = p', ql = oo. In detail, I 1 1-t t -=-=--+-
(3.2.6)
q
I I - =r q,
(3.2.7) Solving (3.2.6) for t have
E
[0, I], we have t I
I
p,
p'
I - t
= --+0. p
= pI q', which is possible since p
I
I-t I I -r = -q, = - = p- - p q'
which proves (3.2.3).
I = -
p
I
+-q -
::;: q'. Finally, we
I,
•
Beckner ( 197 5) has shown that the value of the "best constant" in the generalized Young's inequality (3.2.3) has the precise value Mpq = (p 11P lq 11q)nl 2 • The proof is beyond the scope of this book.
3.2.0.2
The Hausdorff- Young inequality
The M. Riesz-Thorin theorem can be immediately applied to the setting of Fourier transforms on lRn. Let fl.! = N = lRn with Lebesgue measure and p 0 = 1, q = ex:>. The 0 Fourier transform! ---+ f is a bounded operator from L 1 (M) to L 00 (N) with norm 1. Also from the Plancherel theorem the Fourier transform is a bounded operator from L 2 (M) to
FOURIER ANALYSIS IN U SPACES
175
L2(N) with norm 1. Therefore we conclude that
Hausdorff- Young: If 1 < p < 2, the Fourier transform is defined and is a bounded operator from LP(lRn) to LP' (lRn) where p' = pj(p- 1) is the conjugate exponent.
Theorem 3.2.12.
Beckner ( 1975) has shown that the best constant in the Hausdorff-Young iw•quality has the value (p 11P jq 11q)nl 2 where q = p'. 2
Exercise 3.2.13. Letf(x) = e-rrlx1 , so that](~) = e-rrl~ 12 . Show that llfllq/IIJIIP
attains the Beckner bound, where 1 < p < 2.
3.2.1
Stein's Complex Interpolation Theorem
The M. Riesz-Thorin theorem deals with a single operator A, initially defined and bounded on yo n LP• and subsequently extended to LP, p 0 < p < PI. E.M. Stein discovered a remarkable extension to a family of operators Az which depend analytically on a complex parameter z. where 0 < Re(z) < 1. For the complete theory, see Stein and Weiss ( 1971 ), Chapter 4. The following is a special case of the general theory. Theorem 3.2.14. Let (M, J.L) and (N, v) be measure spaces with a family oflinear
operators Az defined on the class ofsimple functions S(M) so that z ~ JN (AJ)g dv is analytic and bounded for 0 < Re(z) < 1 whenever f e S(M) and g e S(N). Furthermore suppose thatforsome 1 < po, qo,pi, qi < CXJ we have IIA;y/llqo < MollfiiPo
f
E
S(M), y
E
lR
~Mill/lip,
f
E
S(M), y
E
JR.
IIAI+iy/llq, Then for all t e (0, 1)
IIA,fllq, < MJ-'M~IIfllp,
f
E
S(M),
where 1
1- t
t
Pt 1
Po 1- t
Po t
-=--+-
-=--+-. q, Qo QI
Proof. We can follow the steps of the proof of the M. Riesz-Thorin theorem, defining g E S(N) with 11/IIPo = 1, llgllqb = 1 we set
N n-N n
+
.!.](N)e iNfJ 2
+
.!.](-N )e-iNf J 2
and are bound ed by the
If
norm off for
FOURIER ANALYSIS IN V SPACES
177
The representatio n (3.3.2) shows that H cannot be bounded on L 1 (1I").
Proposition 3.3.2.
IIH/III
(3.3.3)
sup II/III
= +oo
where the supremum is taken over all trigonometri c polynomials f =I= 0. Proof. Suppose not; then we would have the estimate IIH/IIt ::: Cllfllt for some constant C and all trigonometric polynomials f. Referring to (3.3.2), this implies that IISN/IIt :S (1 + C)ll/11 1 • which implies that the (1,1) operator norm of SN is bounded by a constant, independent of N. But we showed in Section 1.6.3 of Chapter 1 that IISNIII,l = LN ,.....,
(4 log N) j rr 2 when N -+ oo. Thus we have a contradiction, which proves that the supremum in (3.3.3) is infinite. •
3.3.1
LP Theory of the Conjugate Function
In order to prove uniform boundedness of SN on the space LP(1r) for 1 < p < oo, it suffices to prove that the operator H can be extended to a bounded operator on the space LP(1r). This will be accomplishe d by interpolation , as follows: Lemma 3.3.3. The operator His bounded on L 2 (1r). Proof. From Parseval's identity, for any trigonometric polynomial/
IIH/11~ =
L
L
lcnf :S
0#neZ
lcnl = 11/11~2
nEZ
•
This allows us to extend the definition of H to the space L 2 ('][') as a bounded operator with norm at most 1. But this bound is attained, since IIH(e;8 )112 = lle;8 112 = 1. Since LP ('][') C L 2 ('][') for p > 2, we also obtain the existence of Hf when f e LP (']['), p > 2. We will now show by several steps that H is a bounded operator on any LP space for p > 2. We first prove that H is bounded on LP (1r) if p = 2k is an even integer. The following lemma is attributed to M. Riesz ( 1927). Lemma 3.3.4. For any k = 2, 3, ... there exists a constant C 2 k, so that iff is any trigonometri c polynomial, IIHflb < C2kll/ll2k·
~roof. It suffices to first prove this for real-valued functions. If/ is real-valued (f( -n) = /(n)), then so is Hf and we have Pf = ~ (/ + iHf). Expanding this by the binomial theorem we note that (Pfi has no constant term, so that 0
t ( ~) 2
= { (2P/) 2* =
J.r
j=O
1
{ /j(iH/) 2*-;. JT
Taking the real part and writing j = 2r we have
0
=
t (2k) (
-1 )k-r { J2r (Hj)2k-2r.
r=O
2r
JT
178
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
We isolate the term with r
=
0 and apply Holder's inequality to the remaining terms:
t (2k) {
{ (H/)2k :::::
J2r (H/)2k-2r
J.rr:
r=l
J.rr:
2r
Dividing both sides by llfii~Z. we have the polynomial i~equality x2k :::::
(3.3.4)
t
(2k)x2k-2r 2r
r=l
where X:=
2
If X ::::: 1 there is nothing to prove. If X >
(;~) xzk-z, so that the sum is bounded by x2k :::::
x2k-2
L k
r=l
equivalently X 2
:::::
2 2k
-
I
IIHflb = (f-r;(Hf) k) llflb !-r;J2k
:rr
1, then each term in (3.3.4) is bounded by
(2k) =
x2k-2
2r
1. The lemma is proved with C 2k
=
.../2 2k
-
1.
•
Since the space of trigonometric polynomials is dense in lP('li'), we immediately obtain the extension of H as follows. · Corollary 3.3.5. The map f k = 2, 3, ....
-+ Hf is a bounded operator on L 2 k('li') for any
Proof. In order to prove the boundedness of H on the intermediate LP spaces, we can apply the M. Riesz-Thorin theorem to conclude that H can be extended to a bounded operator from LP('][') to LP('][') for any I < p < 2k. But k was arbitrary, so we conclude boundedness for any p > 2. In order to prove boundedness for 1 < p < 2 we use the duality of the norms, namely IIHJIIP =
sup O,CgeLP' ("IT:)
where the supremum is taken over all trigonometric polynomials. We can apply HOlder's inequality and the boundedness result on LP', p' > 2 to see that the numerator is bounded by
and thus conclude that
IIHfllp ::::: which completes the proof.
We list the result as follows.
Cp' II fliP
f
E
U' (']['), I < p < 2
•
FOURIER ANALYSIS IN LP SPACES
179
Propositi on 3.3.6. The mapping f---+ Hf is a bounded operator on LP('Jr) whenever 1 < p
I I!SNfllp.p :':: l + CP < oo . Therefore we have norm convergenc e on the entire space U' (1r). •
Exercise 3.3.8. Let Qr be the conjugate Poisson kernel, defined on trigonome tric polynomia ls by (3.3.5)
Qrf(B) = - i
L
sgn (n)f(n)rinl einiJ.
O¥nEZ
Prove that for any k E z+, there exists a constant c2k such that 11Qrfll2k :::; Ckilfll2k for all f E P and 0 < r < 1. Hint: Prf + iQrf = 2 Ln>.J(n)r 1"1em 11 . Now copy the proof of Lemma 3.3.4 and use the U' boundedne ss of the Poisson kernel from Chapter 1.
Exercise 3.3.9. Prove that for any 1 < p < IIQrfliP < Cpllfllpfo r allf
E
CXJ
there exists a constant Cp so that
P and 0 ::S r < 1.
Hint: Use the result of exercise 3.3.8 and theM. Riesz-Thor in theorem for 2 < p < oo. The duality argument takes care of 1 < p < 2.
3.3.2
L 1 Theory of the Conjuga te Function
An alternative method for proving LP boundedn ess of the conjugate function is to first prove the following inequality of Kolmogor ov: (3.3.6)
1{8: lHf(B)l
>
a}l
Cilfllt.
l> ~ ¢ 2 /:rr 2 for let> I ~ :rr to write
1z = 1/zl
~
-(1 -
r) 2
1
l4>l >< (1
(1- r)2 [ 2 (11 1l4>l>< r
~ :rr4(1 _ rf {
SID
~
:n¢1
[f(e-
- cos c/>)
[
f(O - c/>) -
/(O)j
de/>
¢)- f(O)]dc/>
lf(8 - c/>) - /(8)1 de/>
= :rr4(1 - r)21 14>1 >
1>•
where we have set F(¢) =
c/>
+ r- - 2r cos c/>) (1
J;}'
dF(c/>) 3 let> 1
1/(8- u)- /(8)1 du. Integration-by-parts shows that
~ :rr4(1- r)z
( iFjlll4>1=rr
let> I
+ 3 [
111 ~•
ll=•
IF(~)I
de/>).
ct>
But F(lj))jlj) := '10/v) --+ 0 a.e. when lj)--+ 0 and 1- r = E shows that the first term tends to zero. Similarly, the second term = u17(u) du = o(E- 2 ) when E --+ 0, which shows that h --+ 0. Recalling that Hf(8) = limr__. 1 Qrf(8) a.e. completes the proof of (3.3.8). •
J/j;
Exercise 3.3.14. Prove that 1- cos¢~ ¢ 2 j1r 2 for 1¢1 <Jr. Exercise 3.3.15. Let '7 limx-+oo x- 2
3.4
E ~~(lR)
with 'f'/(X) ~ 0 when x ~ oo. Prove that
J; U'fl(u) du ~ 0 when x
~ oo.
THE HILBERT TRANSFORM ON IR
On the circle we developed the conjuga,te function beginning with its Fourier representation Hf(n) = - i sgn(n)](n) for trigonometric polynomials f. eventually leading to the singular integral representation (3.3.8). When we pass to the corresponding problem on the real line, the relevant operator is the Hilbert transfonn, defined formally as the singular integral (3.4.1)
1 H'.1r+( X ) = -
1'lffi
1f "__.Q,M__.oo
Proposition 3.4.1. Iff
E
[
E
dy
Y
= - fI<M f(x; y) dy
2:rrH,,M f(x) =
1 f(x- y)- f(x Y
+
y) dy.
S
FOURIER ANALYSIS IN LP SPACES
If/
E
S, this family of integrals converge when l Hf(x) = 2rr
(3.4.2)
Example 3.4.2. Let f x =I= a, b.
=
1
E
--+ 0. M--+ oo and we have
f(x - y) - f(x
Then Hf(x)
l[a,bJ·
+ y)
d
Y
JR
185
=
•
Y·
(lj:rr) log (ix - alflx - bi) for
Exercise 3.4.3. Prove this. Explicit calculation reveals that in this example Hf(x) ~ (lj:rr)(b- a)jx when x ~ oo, showing that Hf fj. L 1 (lR). The same behavior is generically true whenever frrtf =I= 0.
Exercise 3.4.4. Suppose thatf
E
S(lR). Prove that limx-+oc xHf(x) = 1/:rr J'Rf.
Exercise 3.4.5. Suppose that frrt lf(x)l/(1 + lxl)dx < oo and thatf satisfies a Dini condition at x. Prove that the integral in (3.4.2) is absolutely convergent.
3.4.1
L2 Theory of the Hilbert Transform
In order to define H on L 2 (JR), we let K".M(x) = (lj:rrx)l"x(t)
y
+y if(x
+ u) + j(x- u)
- 2/(x)l du.
From Lebesgue's theorem, we have for almost every x, ct>,(t)/t-+ 0 when t-+ 0. On the other hand,/ E B 1 implies that ct>, (t) ~ Ct for all t ~ 0. Now we integrate-by-parts:
=
2ty
1
IPJ(x)- j(x)l ~
(t
0
where the estimate lct>,(t)i in the integration gives
~
+
2
y2
)
2
ct>x(t) dt,
Ct allows one to discard the term at the limits. Setting t = yz
IP,/(x) - /(x)l ~
=
2z
1
" "
ct>,(zy)
O+z-)-
y
dz.
But the integrand is bounded by an L 1 function and tends to zero pointwise when y -+ 0, hence P y f (x) -+ f (x) as required. •
We now introduce a norm on the space B 1 , by defining
11/IIB, =
(3.4.6)
_.!._ {
rr
]TR.
lxf(x)l dx. 1 + x2
Theorem 3.4.15. Suppose that fEB,. Then Qvf(x) converges when y .!- 0 almost everywhere to a limiting function j (x) and we have for almost every x E JR, 1 f(x) = Hf(x) := -lim
(3.4.7)
1
f(x- y)
Proof. Any complex-valued function can be written as f = ~ 0. We begin with the conjugate Poisson kernel operator
J;
(3.4.8)
Qyf(x)
= -1 rr
1 IR
tf(x- !) t2
+y2
dt,
Clearly IQJ(O)I ~ 11/!ls,. Then (3.4.9)
PJ(x)
. + 1Q,./(x) = -i Tr
For any f mapping (3.4.10)
E B~o
y > 0,
1 R
dy.
Y
IYI>€
7l' €-->O
/(t)
.
X+ lY- t
/1 X
-
E
h +
i(/J - / 4 ) where
JR.
dt.
(3.4.9) defines an analytic function in the upper half plane y > 0. The (x, y) --+ exp [ -(P,/(x)
+ iQJ(x))]
is a bounded analytic function in the upper half plane. By the Fatou theorem, it possesses a.e. limits when y ,j.. 0. But Py/(x) converges to a finite limit a.e. whenever f E B 1 c B2. Hence we deduce the existence of the a. e. limit of exp ( -iQ,.f(x)] when y ,j, 0. From this it follows that Qyj(x) can have only one accumlation point ~hen y ,j.. 0. hence the existence of Hf(x) = lim,.'- 0 QJ(x). •
FOURIER ANALYSIS IN lJ' SPACES
191
It remains to identify the Hilbert transform as defined in (3.4. 7), with the boundary values of Qy/. namely to show that for a.e. x E IR, (3.4.11)
i
1
Qyj(x) = -
tf(x- t) dt t 2 +y 2
lR
7r
Lemma 3.4.16. Suppose that/
~
B 1 • Then
E
lim ( { tf(x- t) dt- {
.v~o for almost every x
E
J'R
+ y2
t2
y ~ 0.
Hj(x),
llrl>y
f(x- t) dt) t
= 0
JR.
Proof. We write the above difference as 1 1
+ h where
tf(x- t) dt 12
r (
h =
Jlti?.Y
The function t-+ t/(t
+y
2
2
)
t
2
+
t
y
.!.)t<x- t) dt.
2 -
t
is odd and increasing for ltl < y, so that we can write
1 + 1
=
/1
+ y2
ltl_v
r
1/zl ::::
y2)
[f(x- t) -f(x)]dt
2
Jlti>y
= y2
t(t
1
y 1 if(x- t)- f(x)i dt itidF(t)
lti>y~
where F(t) = J~ if(x- s) -f(x)l ds. Clearly F(t)/t-+ 0 at every Lebesgue point when t-+ 0, whereas F(t) ::;: Ct when for all t. Therefore we can integrate-by-p arts to obtain
1
dF(t) = F(y)
I l>y
itl 3
IYI 3
The term at the limits is clearly o(yF(t)/t = 17(1), v = l/t to obtain
1
ltl>v
F(t) -
t
4
-
dt =
which completes the proof that h
2
)
+3
1
ltl>y
when y
-+
11/v V1'/(ljv)dv =
F(t) dt 4
1
•
0. To analyze the new integral, write
o(y- 2 ),
y-+ 0,
0
-+
0 when y
-+
0 for almost every x e JR.
•
As a special case, we can deduce the properties of the Hilbert transform on LP(R), 1 ::; p < oo.
}92
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Corollary 3.4.17. Suppose that f E £P(JR), 1 ::S p < oo. Then Pyj(x)
--+ f for almost every x E lR and the convergence takes place in LP (lR). Furthermore Qy/(x) converges almost everywhere to a limitingfunction ](x) and the convergence takes place in LP ifp > 1.
The convergence properties of Py/ follow from the properties of the Poisson kernel: It is readily verified that t --+ t/(t 2 + y 2 ) satisfies the properties of an approximate identity. Therefore if B is any homogeneous Banach space, we have Py/--+ fin the Bnorm when y --+ 0. The almost-everywh ere convergence of Qy/ follows from Theorem 3.4.15. • 3.4.3. 1 Kolmogorov's inequality for the Hilbert transform Following the discussion of the conjugate function on the circle, we can establish a corresponding inequality for the distribution function of the Hilbert transform whenever f E L, {lR). This takes the form
l{x: IH/(x)l ::: a}l < c { lf(x)l dx
(3.4.12)
J'R.
a
where Cis an absolute constant. This will be deduced as a limiting case of a corresponding inequality for functions in the space B 1 • Define a weighted measure by
11
(3.4.13)
J.L(A) = -
7r
Theorem 3.4.18. Suppose that f
E
A
dx
1+
.
x2
B I· Iff > 0, then we have the weak inequality
(3.4.14) JL{X:
iHf(x)l::: a}
llfll,.
For any complex-valued / E B 1 , (3.4.14) holds with four terms on the right side and with a replaced by aj4. Proof. We consider the harmonic function J,.(w), defined for Re(w) > 0 as the harmonic measure of the two rays {w = iv, v ::: a} and {w = iv, v :5 -a}. This is the harmonic function that takes the value 1 on these rays and takes the value zero on the segment {w = iv, -a :5 v :5 a}. Equivalently, it can be obtained as the imaginary part of (ljrr) log [(wia)j(w + ia)] for a suitable branch of the logarithm. The set {w : la(w) ::: is the exterior of the semicircle described by {w : Re(w) > 0, lwl = a}. On the strip llm(w)l < a, we have
1}
(3.4.15)
la(u +
iv)
=
_!._ [arctan rr
(-u-) + a-v
artctan
(-u-)], a+v
We now consider the harmonic function Uu(X, y)
= la[P,.j(x) + iQJ(x)].
We first recall a basic fact on harmonic functions.
u > 0, lvl 0, thenfora ny y 1 , y 2 > 0, we have
(3.4.16) Applying (3.4.I6) with X= 0, Y2 (3.4.17)
=
I,
u = u," we have
. Ja {PI+vf(O ) + tQI+y/(0 ))
=
I
{ J 01 (Pyj(t)
+
iQy/(t))
I+ t2
1T }R
dt.
The right side of (3.4.17) is underesti mated by
_!_ { JT }R
J 01 (Pyj(t)
+ iQy/(t))
I +
t
dt > _I_l.
2
2rr
-
(t
~-
IQ,.f(t)l~l 1 + t
Using the inequality Iarctan (x)l :5 Jxl, applied to (3.4.I5), we can overestim ate the left side of (3.4.17) by writing . I ( PI+J(O) PI+y/(0) ) Ja(PI+J( O) + IQI+.vf(O )) :5 1T a - H2I+.vf(O)I + a+ IQI+y/(0) 1 . Therefore we have
(3.4.I8) Recall that Hf(x) = limy ... o Qyj(x) a.e., in particula r we have converge nce in measure. Now from (3.4.6), IQI/(0)1 :5 11/IIB" PI/(0) = II/IIB and the right side of(3.4.18 ) is only 2 increased when we replace Qd(O) by its upper bound 11/IIB • Hence 1
J.dx: IH/(x)l >a}
a}
0. Thus the right side of (3.4.18) becomes (4/rr)PI+ J(O) ~ 411/11 82 /rr when y ~ 0. The inequality (3.4.I4) contains the classical Kolmogo rov inequality (3.4.I2) as a limiting case, when we introduce a scaling paramete r Y.
194
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
In detail, define (3.4.19)
11
y
= -
t.ty(A)
2 2 ll"At+Y
dt.
Then we have the following scaled replacement for (3.4.14) when a > IQyj(O)I: (3.4.20)
t.Ly{x: IHf(x)i :=:::a} :S
2 ( 7l"
Pyf(O) Qyf(O)
a_
Pyf(O)
+a+ Qyf(O)
) ·
Now multiply (3.4.20) by Y and take Y -+ CXl. For the left side, we note that for any Borel set of finite Lebesgue measure we have from the dominated convergence theorem 0
hm Yt.ty(A)
Y-oo
For the right side, we see that when Y -+ that for any f E L 1 (JR.)
=
1 -IAI. 1f
the dominated convergence theorem shows
CXl,
hm YPyf(O) = -1 0
Y--~>oo
Jr
1
j(x) dx
IR
whereas IQyj(O)i ::: 11/IIL'<JR> x sup tEIR
iti
1
2
+
Hence when we multiply (3.4.20) by Y and take Y -+ form of Kolmogorov's inequality.
3.4.4
y-+
y 2 -+ 0, CXl,
CXl
we obtain (3.4.12), the original •
Application to Singular Integrals with Odd Kernels
The theory of the Hilbert transform can be transplanted to study n-dimensional singular integrals of the form (3.4.21)
K".Mf(x)
J
k(y)f(x- y) dy.
L,'=,iJ, lfl
>a
L
IB;I =a
t=l
This proves (3.5.2) and hence part (i). Taking a ---+
CXJ
shows that Mf
a}l =
{
'R"
Ica.x>(lf(x)l) dx.
The l.J' norm can be written in terms of the distribution function by writing
a.e.
FOURIER ANALYSIS IN LP SPACES
199
and applying Fubini's theorem to obtain {
lrrt"
lf(x)IP dx
(roc
= {
Jrx, Jo
=
1=
=
1
puP-
1
puP-ll(O.If(x) l)(u)) dx
(in
l(u.oo)(f(x))d x) du
00
pup- I AJ(u) du.
We will apply this identity to estimate liM/lip· To do this, we first decomposef into a bounded part and an unbounded part (without loss of generality, we can assume thatf 2: 0). (3.5.4) Since the operator M is sublinear, we have Mf(x) ::::: Mfc,(x)
+ Mf"' (x)
::::: ex
+ Mf"' (x).
Hence if Mf > 2ex, then Mfa > ex. In terms of the distribution function, we have AMJ(2ex) = l{x: Mf(x) > 2ex}l::::: l{x: Mf"'(x) > ex}j.
Butf"' lives on a set of finite measure, since l(x :f"(x) > 0}1 = l{x :f(x) > ex}l ::::: llfiiP exP
while l/0 1 Sex+ lfl so that/a E U(lRn) to which we can apply the Hardy-Littlew ood maximal inequality; we change ex to 2ex and write
IlM/II~ =
1
00
p
(2ex)p-I A.M1 (2ex)d(2cx)
1 (i. 1 p2~'C l, (1f 0, there exists a continuous function g such that II f g + h with llh III < E, we have
- g III
0, l{x: rlf(x) > o}l ::::; l{x: Mlhl(x) But the left side does not depend on
E,
~
8/2}1 ::0
2E
8
hence we conclude that for every 8 > 0,
l{x: Qf(x) > 8}1 = 0 which means that Qf(x) = 0 almost everywhere, which was to be proved.
•
The above reasoning can be strenthened and clarified in terms of the Lebesgue set of the locally integrable function f. This is defined as (3.5.6)
Leb{f) =
{
X E
n
.
JR : hm
fa(x;r)
r---+0
lf(y)- f(x)jm(dy) } = 0 . IB(x; r)l
Proposition 3.5.4. For any locally integrable f, i (Leb(f) Y! = 0. Proof. For each real number c we apply the previous proof to the function I f - c!. Thus we obtain for almost every x E !Rn,
(3.5.7)
.
fBix r)
hm r-o
·
lf(y) - clm(dy) IB(x; r)l
=
lf(x)- cl.
If Cj is an enumeration of the rational numbers, we obtain a countable collection of excep· tional sets Ej· Then E := U1 E1 also has Lebesgue measure zero. But the right side and left side of (3.5.7) are continuous functions of c; the right side is obvious, but so is the left since the triangle inequality gives llf(y)- c,l- if(y) -· c2!1 ::::; lei - c 2 1. Hence if x ¢Ewe let
c1
~
f(x) to conclude that
lim r-o which was to be proved.
JB\x r)
lf(y)-:- /(~/m(d):!._ IB(x; r)/
=
O
x ¢E.
•
202
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
The Hardy-Littlewood maximal function can also be used to investigate certain questions of nontangential convergence. Normally this is considered in the framework of convergence in the unit disk or a higher-dimensional space. But the basic ideas are already present in the above framework, in the context of convergence of the type when Xm--+ x. We do not expect that this will be true unrestrictedly, since x --+ f 1;m(x) is a continuous function whereas the limit f is not continuous in general-hence we cannot expect uniform convergence. Nevertheless we have the following proposition.
Proposition 3.5.5. Suppose that Xm f
E L 1 (1Rn),
--+ x so that lxm -xi < lfm. Then for any limm ft;m(Xm) =f(x) atalmosteveryx.
Proof. Repeating the above steps, we write f Define
=
g
+
h where g is continuous and II h 11 1
8}1 ~ l{x: Mh(x) =:::: 8/2}1 ~ - . €
But the left side does not depend on
E,
hence we conclude that for every 8 > 0,
l{x: Qj(x) > 8}1 = 0.
3.5.2
•
Application to Radial Convolution Operators
The Hardy-Littlewood maximal function can be used to estimate more general convolution operators of the form (3.5.8)
(k, *f)(x) =
{
' R"
k,(y)f(x- y) dy,
where the radial kernel k, E L 1 (lRn) is obtained from a monotone decreasing nonnegative function K : JR+ --+ JR+ by setting k,(y) = t-nK(yft). The example of the HardyLittlewood operator is obtained when we set K(x) = l[O.I](Ixl), the indicator function of the closed unit ball. The following lemma shows that the Hardy-Littlewood maximal function can be used as a universal bound for any radial convolution. The original form is attributed to K.T. Smith ( 1956), and appears in Stein ( 1970). The proof below is attributed to S. Saeki, reproduced by Banuelos and Moore ( 1999).
FOURIER ANALYSIS IN U SPACES
203
Lemma 3.5.6. For any radial kernel, we have the estimate lkJ(x) I < (Mj)(x) Ilk !II 1,
f
E
1
L (lR.n).
Proof. Let J-1.-x be the measure defined by J-1.-,(A) =fA lf<x- y)l dy, for any Borel setA. In particular if A is a ball, we have J-1.-x{A) ::::0 IAI(Mj)(x). Applying this to the ball defined by B>. = {y E Rn : k,(y) > ).}, we have
l(k, *f)(x)l
::::0 .(., k,(y)lf(x-
= {
}fi>.n
=
k,(y)f.lx(dy)
r=
Jo
y)l dy
px(B;,_) d).
::::0 (Mj)(x) {""
Jo
IB"I d).
= <Mf)(x) Ilk, ll1, where we have used the definition of Mf and twice used the representation of the L 1-norm as the integral of the distribution function. • A typical application is to the n-dimensional heat kernel, where K(x) = j(4rr)nf2. Lemma 3.5.6 can then be combined with the Hardy-Littlewood maximal inequality to give a new proof of Proposition 2.2.35, that for any f E L 1 (JR.n), and for almost every x E IR.n, (k 1 *f)(x) --+ f(x) when t --+ 0. e-lxl
2
/
4
Exercise 3.5.7. Complete the details of this argument. Hint: Define Qj(x) =lim supt- 0 (k, *f)(x)- lim inft-o(k, *f)(x) and argue as in the proof of the Lebesgue differentiation theorem to prove that lim,~o(k, * f)(x) exists a.e. Then identify the limit by a density argument.
As a second application of Lemma 3.5.6, consider the case of then-dimensional Poisson kemel-K(x) = Cn/0 + lxl 2 )0
Ill.vl=l { f(x- ty)dw(y)l
where dw is the normalized surface measure on the sphere sn-l c IR.n. Results of this type can be used to prove Fatou-type theorems for solutions of the wave equation u 11 = l::!..u for suitable values of n, p. In the treatment below we restrict attention to the simplest
204
INTROpUCTION TO FOURIER ANALYSIS AND WAVELETS
case, where p = 2, n > 4. A complete account of the subject for more n,p can be found in Stein (1993), Chapter XI. We begin with the spherical averaging opeator f
~
jvl=l
(AJ)(x) :=
f(x- ty) dw(y) = (f
genera~
values of
* dwr)(x).
We will also make use of the square function, defined by (3.5.10) which is well-defined whenever f Lemma 3.5.9. For any f
E
E
C 1 has compact support.
C 1 (lRn) with compact support, we have
IArf(x)l < (Mf)(x) +Sf::; ...;2n where Mf is the Hardy-Littlewood maxima/function. Proof. We write
1
=
t-n
d -(snAJ(x)) ds p ds ·
1
=I, +/z where
1 1
I,= t-n
nsn- 1 (AJ)(x)ds
d
]z = sn ds (Asf)(x) ds.
I, is majorized by the Hardy-Littlewood maximal function, since for any Cn I, = -;;t
1
t >
0,
f(y) dy .::; (Mf)(x).
ly-xi9
Meanwhile, [z is estimated in terms of Sf by using Cauchy-Schwarz: 1
liz I = tn
1 o
.:S tn .::;
(
d sn-o; 2 >Js- (Asf)(x) ds ds
Jo{ ' s 2
n-J
ds
) 1/2 (
s
[
d ds (A_,f)(x)
J ds 2
)
l/2
1
r,:o-(Sj)(x),
v2n
which completes the proof.
•
FOURIER ANALYSIS IN LP SPACES
205
Lemma 3.5.1 0. /fn > 4 andf is C 1 with compact support, then
where the L 2 nonn is taken over all of~" and the constant Cn depends only upon the dimension. Proof. The Fourier representation of the spherical average is
where w(~) = c,.J12 (I~ I)/I~ l1 2 is the Fourier transform of the normalized measure w. Now by the chain rule
(3.5.11) where we have set n
J.L(~) := L:~j j=l
aw -. a~,
Now from the asymptotic behavior of the Bessel function, we have
aw a~j On the other hand, when we can write
I~ I
--. 0, we have
= O(I~I(J-n)/ 2 ).
=
aw;a~j
0(1~1).
1~1 __. 00.
Combining these estimates,
(3.5.12) Applying Plancherel's theorem to (3.5.11), we have
Integrating both sides with respect to the measure s ds and applying Fubini, we have (3.5.13)
{ ISJ(x)l 2 dx = { lf(~)l 2 }JRn }JRn
(
{co IJ.L(s~W ds).
Jo
S
But the inner integral can be estimated using (3.5.12). For any M, we have
:S
Cni~I 2 1M sds + Cni~IJ-n Leo s:~2
= Cni~I 2 M 2 + Cn
~~~ -n 3
(n -
2)M"- 3
206
INTRODUC TION TO FOURIER ANALYSIS AND WAVELETS
The two terms are comparab le to one another by taking M
1= iJ.t(s~)l ~s ~ 2
Cn ( 1 + n
~ 2)
=
I /I~ I, leading to the estimate
= C,.:
=~ ·
Referring to (3.5.13) and applying Planchere l 's theorem once again, the proof is complete .
•
Combin ing Lemmas 3.5.9 and 3.5.10, with the L 2 bound for the Hardy-L ittlewoo d maxima l function , we have proved the followin g result.
Theorem 3.5.11. Suppose that n :?: 4. Then we have the followin g estimate for any f E C 1 (IR") with compact support:
II sup(AJ) (x) liz
(3.5.14)
< Cllfllz
t>O
where the constant depends only on the dimensio n. The estimate (3.5.14) can be extended to any f function s with compact support is dense in L 2 .
E L 2 (IR:")
by noting that the set of C 1
Example 3.5.12. Taking n = 1 and f an unbound ed function shows that sup, AJ(x) = +(X) for every x E JR. 1, hence the estimate (3.5.14) cannot hold for general ! E L 2 (1R.). Example 3.5.13. To obtain an example in higher dimensio ns, take f(x) = lxl 1-n[log( l/lxl)r 1 I[O.IJ(Ixi ), which fails to be in L 2 (1Rn) andfor which it is verified that sup, AJ (x) = +(X) for every x E JR.n.
3.6
THE MARCI NKIEW ICZ INTERP OLATIO N THEOR EM
In this section we develop the notion of weakly bounded operator s and apply it to prove the theorem of Marcink iewicz. To orient the thinking , we first define the notion of weak Lebesgu e space.
Definiti on 3.6.1.
for some C > 0, where 1 < p
1
1/1>0'
if(x)i" d.x:?: cxPi{x : if(x) :?: cx}j.
FOURIER ANALYSIS IN U' SPACES
207
But the converse is not true. For example f (x) = 1 I ( 1 + lx I) is wkL 1 (IR 1 ) but is not integrable . In parallel with the discussion preceding the M. Riesz-Tho rin theorem 0, (3.6.2)
aqJ.L{X:
ITf(x)l
>a}< CPIIfll~-
The converse is not true. This leads us to the notion of weak type (p, q).
Definition 3.6.7. An operator for which (3.6.2) holds is said to be ofweak type (p, q).
For example, the Hardy-Little wood maximal operator is of weak type (1, 1) but does not satisfy (3.6.1) with (p, q) = (1, 1). In case q = oo, condition (3.6.2) implies that Tf E L <Xl so that the definition can be taken in the strong sense.
Theorem 3.6.8. LetT be a sublinearop erator that is defined on the space LP0 +LP' so that it is weakly bounded of type (po, Po) and weakly bounded of type (Pt, Pt). where 1 ~Po l~amlf 0. Then there so that E
(i) lf(x)l < aforx E F. (ii) Q = UkQk, where Qk are cubes with disjoint interiors so that
-1
1
a < IQkl (iii) ~n =
n u F, IQI
oan un(x,y). lflani < MandR < 1,thenth isseries converge s uniform ly in the closed disk x 2 + y 2 < R 2 together with the different iated series, thus defining a harmoni c function in the open disk x 2 + y 2 < 1. In particula r, a useful class of harmoni c function s are provided by the Poisson integrals of integrab le function s (or more generall y of finite measure s):
( 3 ·9 · 1)
u(x, y)
=
1 [ 2JT Jlf' R2
2
+
r
2
R r 2 - 2rR cos (8- ¢/(¢) d¢
Proposi tion 3.9.1. Suppose thatf
"""" (
=~
r) lnl f(n)ern A
•
R
0
·
L 1 (1f') and u(x, y) is defined by ( 3.9.1 ). Then u is a harmoni c function in the disk x 2 + y 2 < R 2 and limr....,.R u(rei0 ) = f(8) for almost every 8 E T. Iff E C('JI'), the converge nce is uniform in 'JI'. Iff E LP(1f'), 1 < p < ex::>, then the converge nce is in the norm of LP('JI'). E
Proof. We simply note that lf(k)l::; 11/11 1 so that we can infer that the sum of the series in (3.9.1) is a harmonic function. The converge nce to the boundary values was already proved in Chapter 1 as part of the Abel summabi lity of Fourier series in the spaces L 1 ('lr), C('lr), and LP('lr). •
Exercise 3.9.2. Suppose that f E ux'(1I') and that u is defined by ( 3.9.1). Prove that iu(x, y)l :::::: llfll 00 in the disk x 2 + y 2 < R 2 . Exercise 3.9.3. Suppose that m(d8) is a finite Borel measure on 'JI'. Defining 1
u(x, y) := 2JT
1 1f'
R2
+
R2- r2 m(d
The first tenn on the right is zero since u is harmonic. Applying Green's theorem transfonns the double integral of the divergence into a line integral on the boundary. Thus (
lao
u gradu dS
= ( (u; + u~) dx dy.
Jn
But the left side is also zero, since u is zero on the boundary. Hence we conclude that the continuous function u7 + u: = 0 in D, meaning that u is constant on D. But the boundary value is zero, hence u 0 in D. •
=
Corollary 3.9.5. Any harmonic function in the disk x 2 + y 2 < 1 can be represented on the closed disk x 2 form
+
y 2 :S R 2
(~)" n~l 1+ r
provided that R > (1 + r)/2. For any fixed r < 1 the last sum is finite, which proves the uniform convergence of the Poisson kernel. Hence we have established (3.9.4). The convergence of u(re' 0 ) follows from the almost-everywher e Abel summability of the Fourier series. •
The next result concerns representation of nonnegative harmonic functions.
Theorem 3.9.8. Suppose that u is a nonnegative harmonic function in the disk x 2 + y 2 < 1. Then there exists a nonnegative Borel measure m on '['so that
1
u(x, y) = - 1 2n 'lf 1 + r 2 For any f E C('JI'), we have (3.9.7)
(3.9.8)
lim r--.1
f
}']f
2
-
1- r m(df/J), 2r cos (B- f/J)
u(rei0 )f(B) d() =
f j(B)m(dB).
}']f
Proof. Again we consider the linear functional (3.9.5), now defined on the space C('li') whose dual space is the set of finite Borel measures on 'JI'. We have ILRfl ::::0 llflloc
2~ 1u(rei
0
)
d().
But from the Poisson integral representation of u in the disk x 2 + y 2 :::; R 2 , we have u(O) = 1/21r 111: u(reiiJ) dO, hence ILJI :::; u(O)IIflloo· Applying the weak compactness argument once again yields a measure m on 1I' and a weak* convergent subsequence LR· so that LRJ --+ J..Jf'f(O)m(d()). As in the previous proof we have (3.9.6) on which w~ can take the limit R, --+ 1 to conclude (3.9.7). To prove the convergence, we multiply (3.9.7) by a continuousf(O) and integrate on the circle of radius r. The left side is Lrf = 1-r u(re;0 )f(8) dO. The right side can be written as 1"J PJ(8)m(d8). But Prf converges uniformly tof, so that the right side converges to 11ff(8)m(d8), from which it follows that limr-.1 Lrf exists and is given by (3.9.8). •
216
INTRODUCTI ON TO FOURIER ANALYSIS AND WAVELETS
To complete the picture, we state and prove the theorem on LP boundedn ess. Theorem 3.9.9. Suppose that u is a harmonic function in the disk x 2 with the property that
where 1 < p < ex:>. Then there exists u 1
(3.9.9)
1
u(x, y) = -2 n
i1r
E
+ y2
< 1
l!'("JJ') so that
2
1
+
r
2
l-r -
2
cos
r
(8
-
¢)
UJ
(¢) d¢,
In particular
lim [ iu(rei0 r-+ I
J1r
andlimr-+ l u(re' 8 ) = Ut(8)fora .e. 8
) - UJ
E
(8)1P d8
=
0
"JJ'.
Proof. Again we consider the linear functional (3.9.5), now defined on the space U' (1r) which is the dual space of £P (1r) where p' = pI ( p - 1). We have for any f E u' (1r),
ILRfl :5 Mil/lip· so that the linear functionals LR have bounded norms. Applying the weak compactne ss argument once again yields an U function/ on 1r and a weak* convergent subsequenc e LR so that LRif--+ f..rf(O)u 1 ((})dO. As in the previous proof we have (3.9.6) to which we can1 apply take the limit Rj --+ 1 to conclude (3.9.9). The convergenc e follows from the £P and a. e. Abel summabilit y of the Fourier series proved in Chapter I. •
3.9.3
Represe ntation Theorem s in the Upper Half Plane
The results in the previous section can be transform ed to obtain representa tion theorems for harmonic functions in the upper half plane JR 2 + = { (x, y) : -ex:> < x < ex:>, y > 0}. To see this, we write z = x + iy and introduce the fractional linear transform ation (3.9.10)
z-i z+i
x+i(y-1 ) x+i(y+ l)
w= ---- = --------- --
which maps ito 0 and maps the real axis -ex:> < x < oo to the unit circle lwl = 1, deleted by the point w = 1, which correspon ds to the point z = ex:>. The upper half plane JR 2 + is mapped 1: 1 conformal ly onto the unit disk D = {w: lw) < 1}. If U(z) is a harmonic function defined for z = x + iy E JR 2 +, we obtain a harmonic function on D by setting u(w) = U(z). This can be seen directly by computing the partial derivative s by the chain rule or by observing that u is the real part of the holomorp hic function obtained by compositi on of a holomorp hic function with the fractional linear transform ation (3.9.10). If U is a bounded harmonic function on JR 2 +, then u is a bounded harmonic function in D. If U is a nonnegati ve harmonic function in JR 2 +, then u is a nonnegati ve harmonic function in D. We state and prove the correspon ding Fatou theorems.
217
FOURIER ANALYSIS IN LP SPACES
Theorem 3.9.1 0. Suppose that U is a bounded harmonic function in IR 2 +. Then there exists U 1 E L 00 (IR) so that U(x, y) =
(3.9.11)
y 7r
f
JJR
Ur (t) (t - x)2
Proof. From the Fatou theorem in the disk, we set w (3.9.12) 1 u(w) = 2rr
1 1f
+ y2
=
re' 11 and
2
1
+ r2
-
dt.
1- r 1 u 1 (¢J)d¢J = 2r cos (0 - ¢J) 2rr
1 1f
2
1. - lwl UJ(¢J)d¢J. je•
0,
sup
Y>y>O
1lR
1U(x, y)!P d + x 2 t.::: M < oo. 1
Then there exists U 1 E If(lR; dt/(1
(3.9.16)
U(x,y) =
~ 7r
+ {
JlR
t 2 ))
(
so that
\ t - x)
+ y2
U 1 (t)dt. .
Proof. We have t 2 + (y + 1) 2 > 1 + t 2 , thus
1
1
---------- < -2 - t2+(y+l)2 t +1
so that 2(y + 1)
L. t2 ~~~; ~~:)2 dt :s 1
2(y + 1)
L. ~?.::IP dt :s 4M(Y 1
Transforming the integrals as in (3.9.15), sup Y>y>O
1"' I ( u
-n:
-yy
+}
+
1
· ) ----e'"'
y
+
1
lp d'JJ
0, •
CHAPTER
4 POISSON SUMMATION FORMULA AND MULTIPLE FOURIER SERIES
4.1
MOTIVATION AND HEURISTICS
Up until now we have treated Fourier series and Fourier integrals independently of one another. Given the strong parallels between the one-dimensional theories, we may ask if there is a systematic link for passing back and forth. This is supplied by the notion of periodization and the closely related Poisson summation formula. For the sake of clarity, we will first pursue these ideas in one dimension where the formulas are simpler. As applications, we will obtain the Shannon sampling formula for band-limited signals and the transformation formulas for Gaussian sums from number theory using these ideas. The higher-dimensional Poisson summation formula allows us to study multiple Fourier series. As a by-product we can derive the famous Landau asymptotic formula for the number of lattice points in a large sphere in Euclidean space. In order to simplify the notations, we will consider Fourier series for functions of period 1; these are defined by their restriction to the interval (0, 1). In this setting a Fourier series is written
f(x) ,...._
L nEZ
222
1
1
f(n)e27rinx,
f(n) :=
j(x)e- 21rinx dx,
223
POISSON SUMMATION FORMULA AND MULTIPLE FOURIER SERIES
and the Dirichlet kernel of Fourier series is written DMFS (x) =
sin((2M + l)rrx) sinrrx
M
"""' e2:rik.x = - - - - - - ~
k=-M
THE POISSON SUMMATION FORMULA IN 1R 1
4.2 4.2.1
Periodization of a Function
Given a function on the line, one can construct a periodic function by summing over the integer translates, defining (4.2.1)
](x) = L f ( x
+ n)
nEZ
where the series is supposed to converge in the sense of symmetric partial sums. This is called the periodization off. Example 4.2.1. Iff E L 1 (0, 1) and is defined to be zero elsewhere, then simply the periodic extension off to the entire real line.
j is
Example 4.2.2. Iff is the heat kernel, defined by f(x) = (4rrt)- 112 e-x 2 / 41 for t > 0, x E lR, thenj(x) = (4rrt)- 112 LnEZ e-<x+n) 2 / 41 is the periodic heat kernel. Example 4.2.3. Let f be the Dirichlet kernel relative to the Fourier transform, studied in Chapter 2:
f(x) = D[{ (x) :=
sin(2Mrrx) Jl"X
=
1M
.
e 2 :r•~x d~.
-M
In this case the series defining j is not absolutely convergent, but we can compute the periodization] directly as follows: (4.2.2)
t
sin 2rrM(x rr(x
n=-N
+
n)
+ n)
=1M
e2:ri~x (sin(2~ + l)rr~) d~. smrr~
-M
We apply the pointwise inversion of Fourier series on each interval (k- 1/2, k each time obtaining a contribution from the center. If M ¢ Z, we obtain (4.2.3)
lim N-+oo
t
n=-N
f
+
sin 2rrM(x + n) = ez:r,k.x = sin((2[M] + l)rrx) rr(x + n) k=-!Ml sinrrx
1/2),
224
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Furthermore the partial sums (4.2.2) converge boundedly for each M < oo, since the right side consists of a finite number of Fourier partial sums for the function ~ ---+ e 2 :n:ixl; at~= 0, ±1, ... , ±[M]. We have shown that the periodization of the Dirichlet kernel D';J is the periodic Dirichlet kernel D[~ 1 • Exercise 4.2.4. /fM E Z, showthatwemu staddanadditio naltermofcos2M rrx to the right side of(4.2.3).
We can further exploit the bounded convergence to represent the Fourier partial sum of an arbitrary f E L 1 (0, 1):
i
N_
f(x + t)Df.J (t) dt =
1'o
-N
~ sin2rrM(t+n) dt---+ f(x + t) ~ n=-N rr(t + n)
1' 0
f(x + t)Df~ 1 (t) dz
WhenN ---+ oo, the left side converges to the Cauchy principal value, or symmetric improper integral, which can be summarized by writing M¢Z.
As a final example of periodization, we consider the kernel of the Hilbert transform, which was studied in Chapter 3. Example 4.2.5. Letf(x) = 1/xfor x
=I= 0.
The periodization is determined from the identity 1 " " - - - = rr cot rrx, ~x-n nEZ
(4.2.4)
X¢
Z.
This can be proved by residue calculus, which the reader is invited to supply. We offer a Fourier-analytic proof as follows: Proof. Let j(x) = Ln€Z I I (x - n) - rc cot rc x for x ¢ Z. Then x ~ j(x) is an odd function of period I. It is also a continuous function, since Lo;o 0, x E R Show that the periodization of P,_. is the Poisson kernel of Fourier series. Hint: Begin with the representation
4.2.2
of~·
as a Fourier integral: P,(x)
= JJR e 2rrixl; e-21TiHr d~.
Statement and Proof
The Poisson summation formula allows us to compute the Fourier series of] in terms of the Fourier transform off at the integers.
!'fleorem 4.~8. Suppose that f E L 1 (R). Then f(x) is finite a.e., satisfies f(x + 1) = f(x) a.e. and is an integrable function on any period, e.g., [0, 1]. The Fourier coefficient is obtained as t](x)e-2rrimx dx =](m) =
lo
r= f(x)e-2rrimx
}_=
dx,
mE
Z.
/fin addition L~-= i](n)l < oo, then the Fourier series of] converges and we have the a.e. equality
(4.2.8)
L nEZ
f(x
+ n)
=f(x) =
L nlf:-_'?..
j(m)e2rrim~.
226
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
In particular j is a. e. equal to a continuous function on 1r = ~/Z. Redefining f necessary, then the equality holds everywhere and we have the Poisson identity (4.2.9)
Lf(n) =](0) =
L:J<m).
nEZ
mEZ
if
Proof. We have
t (n~oo lf(x + n)l oo
Jo
)
dx =
n~oo oo
fn+
I
n
1/(x)l dx =
~oo
-oo
lf(x)l dx
~we have the estimate
i·
(4.2.13)
~
f(t)-L-f(n) nEZ
sin rr (t - n) ( ) Tf t - n
!
IF(~)Id~.
POISSON SUMMATION FORMULA AND MULTIPLE FOURIER SERIES
229
Proof. Let F(~) = EnEZ F(~ + n) be the periodization ofF E L 1 (IR). The Fourier transform ofF is x -+ f( -x) so that by the Poisson summation formula, we have the Fourier series F(~) ~
(4.2.14)
L
L
j( -n)e2rrml; =
neZ
f(n)e-21rml;.
nEZ
From the results of Chapter 1, any L 1 Fourier series may be integrated term-by-term after by a function g of bounded variation. Applying this with g(~) = e 2rrul;, we have
multiplicat~on
1/2
F(~)e2rrul;
=
1 -1/2
L
.f(n)
11/2
nEZ
=
e2rrrl;(t-n) d~
-1/2
"'"'
sin rr(t - n)
nEZ
Jr(t-n)
L., .f(n)
.
If A.::; ~.then F is the periodic extension ofF to JR, so that F(~) = F(~) for I~ I < ~.and the left side reduces to ~~~~ 2 F(~)e 2 ""~; =.f(t), which proves (4.2.12). Otherwise, we can rewrite the left side as
1
1/2
F(~)e2rru/; d~
=
-1/2
L
11/2
nEZ
= =
L
F(~
nEZ
l
L
e-2rrmt
F(u)e21ru(u-n) du
n-1/2
.f(t) -11/2
F(~)e2rru/; dl;l
-1/2
L n;o"O
=
but
F(u)e2rriru du;
thus
n-1/2
,,..0
::: 2
F(u)e2rruu du;
n+l/2
L
nEZ
l
l
n+l/2
n-1/2
l = j.L(l=
n)e2rrul; d~
n+l/2
nEZ
.f(t)
+
-1/2
i
e-2rrmr) in+l/2 F(u)e2rritu du) n-1/2
n+l/2
IF(u)l du
n-1/2
2 {
J/u/~l/2
IF(u)l du,
•
which completes the proof.
Band-limited signals have the further property that the total signal strength
JJR f(t) dt can be computed by sampling at the integers.
4
Proposition 4.2.14. Suppose that A < and that F e L 1 (-A. A) satisfies a Dini condition at~ = 0 with valueS. Then the series LneZ f(n) converges and we have the identity
(4.2.15)
lim
T-+oo
JT -
T
f(t) dt = S
= ""'f(n). L._,;
neZ
230
INTRODUCTI ON TO FOURIER ANALYSIS AND WAVELETS
Proof. From ( 4.2. 11 ), the left side of (4.2.15) is computed as
I
T
-T
f(t) dt
f).
= -). F(l;)
sin 2rri;T 7r
I;
dl;,
which converges to S by one-dimen sional Fourier inversion. The convergenc e of the series on the right side of ( 4.2.15) can be seen directly by computing from (4.2.11 ): ~
~f(n) -N
=
f). F(l;) sin(2N. + t:l)rrl; dt:s· -).
Sin Jrs
Since J... _:::: ~, this is an integral on a single copy of the basic period interval (- ~, ~) and converges to S by applying one-dimen sional Fourier inversion. •
Exercise 4.2.15. Suppose that f(t) = J~). e 2 rrit~ J-L(dl;), where 11 is a finite Borel measure and A. < Prove the identity
1·
1 lim - T--+oo 2T
(4.2.16)
!T
f(t) dt =lim
-T
N
1
2N
+
1
t
f(n).
n=-N
Hint: Suitably apply the dominated convergenc e theorem and identify both sides with J.L({O}).
!
Remark. The formula (4.2.12) for a band-limi ted signal with A. < is not canonical. If instead we integrate (4.2.14) on the interval [-A., A.], we obtain the alternative representa tion (4.2.17)
f
(t) =
"'"'
L.; f
sin 7r J...(t - n) (n) _7r_(_t--'- -n)-
t E JR.
nEu...
Equivalen t formulas can be obtained for any v
E
[A.,
1J.
Exercise 4.2.16. Suppose that f(t) is any band-limi ted signal. Show that we can reconstru ct! from its values at the points nj2A., n E Z by means of the formula
f
"'"'f ( - n
(t) = 2 A. L..., "" nEu...
2A.
) sin ;r(2At- n) . ;r(2A.t- n)
Hint: Apply the Shannon sampling formula to F"(l;) := F(2J...I;).
4.3
MULTIPL E FOURIE R SERIES
Multiple Fourier series are naturally associated with functions on the torus ']f'd = (0, 1)d = JRd ;zd. There is a natural 1:1 correspon dence between functions on ']f'd and functions on JRd, which are periodic in each coordinate : f (x 1 , ••• , x; + 1, ... , xd) = f(x,, ... ,x;, ... ,xd) for I< i < d, (x 1 , ... ,xd) E !Rd. We begin with an integrable function on the d-dimensi onal torus ']f'd = (0, l)d. The L 1 norm is denoted 11!11 1 = f1fd lf(x)l dx and the Fourier coefficien ts of
POISSON SUMMATION FORMULA AND MULTIPLE FOURIER SERIES
f(n) := {
lu"
j(x)e~2n:m x
231
dx,
and the Fourier series is written f(x) ~
(4.3.1)
L
f(n)e2n:m x.
nEZd
Basic L1 Theory
4.3.1
The elementary properties of multiple Fourier series may be obtained from the periodic heat kernel. We first develop the lemma of Fourier reciprocity in the following form.
Proposition 4.3.1. Suppose that K(x) := LnEZd K(n)e 2:n:mx is an absolutely convergent trigonometric series: LnEZd
(4.3.2)
1
IK(n)l
f(y)K(x- y) dy =
L
0, X E JRd.
232
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Exercise 4.3.2. Prove that the series LnEZd
J1fd
2141
if(y)ie-lx-y-nl dy converges 1 whenever f E L (1rd) and use this to show that the left side of (4.3.5) is equal to the left side of(4.3.6).
From this we conclude the following.
Proposition 4.3.3. (i) Jff(n)
(ii)
= 0, thenf =
0 a.e.
If
LnEZ" lf(n)l 0.
lxl>li
Exercise 4.3.4. Prove the three properties (4.3. 7) of K 1 • A multiple Fourier series is said to be Gauss-summable if (4.3.5) converges tof when t -+ 0. Since the periodic heat kernel is an approximate identity, we immediately obtain the following properties of Gauss summability.
Proposition 4.3.5. Suppose that f E L 1 (1rd). Then the Fourier series is Gausssummable in L 1 (1rd) to f. If, in addition, f is continuous at x E 'JI"G', then the Fourier series is Gauss-summable to f(x).
The Gauss-summability of Fourier series has the further consequence that the set of trigonometric polynomials is dense in L 1(1fd). A direct proof can be obtained by applying properties of one-dimensional Fourier series, as follows. Lemma 4.3.6. The trigonometric polynomials Llnl:;;N ane 2rrin·x are dense in L' (1rd). Proof. From Chapter I, we know that the Fourier series of an indicator function converges boundedly and in the L 1 norm. Now if we have a product of indicator functions l(a,.b,) each approximated by a Fourier partial sumS~, then we can write
n s~- n d
t=l
d
t=l
d
t(a,.b,)
=
L t=l
[s~-
t(a,.b,)J
c,
POISSON SUMMATJ 0:-.1 FORML1LA AND MULTIPL E FOURIER SERIES
233
where C, contains i - I factors of S~ and d - i factors of I
2
2
lc(J-t)l 1c(v)l .
JJ.,V 0 0 and for almost every x E ']['d, the number of integer lattice points in a ball centered at x satisfies the estimate
z+ Exercise 4.5.3. Show that if R
3 R -.. oo.
oo along the sequence of squares R = j we have the improved estimate: VE > 0 and almost every x E ']['d, ---+
N(x· R)- (rrR2)df21 < R 3. Then the spherica l partial sums SMf(O) =
'"""'
L- a
k Jd/2
d
lkajJ
lkl:s:M
are unbound ed when M--+
CXJ,
(2.rr lkla)
wheref( x) = l[O.aJ(Ixi ).
Proof. We can repeat the asymptot ic analysis done above in case d function A(JL) =ad (Jd 12 (27r JLa))j(aJL )dl 2 has the asymptot ic behavior A(JL) =
A'(JL)
=
JL/ 2
[cos(27r aJL-
~~:~>~d
3. The Bessel
(d-/)Jr) +0(~)]. -/)Jr) + o(~) J.
[sin( 2JraJL- (d
Letting JLk be the consecuti ve zeros of the above cosine function, we estimate as before:
When we integrate- by-parts, we find that the terms at the limits yield A(JLk) [N<JLk) _ CdJLn = O(k-(d+l) /2 X kd-2+2/(d +l)) = O(k(d-3)/ 2-(d-I)j(d +l)) while the new integral is
Thus we have when k --+ oo,
On the other hand, the explicit form of A (JL) shows that the integral has the explicit asymptot ic expressio n
1:*+'
JLk-IA(JL )
$
=
1:•+'
=
const k1 2(-l)k [1
JL)
=
0, ' e C 00 (1l'), hence Lis •
The Fourier representatio n can also be used to define new distributions .
249
POISSON SUMMATION FORMULA AND MULTIPLE FOURIER SERIES
Exercise 4.6.4. Suppose that Ln is a bilateral sequence of complex numbers of
polynomial growth: ILnl < C(l + lni)N for some C > 0, N > 0. Prove that there exists a periodic distribution L such that L (n) = L, for every n E Z. Hint: For every cf> for every n E Z.
E
C""('JI'), and every k > 0, there exists Ckn > 0 so that lcP(n)l .:S CknO
+
lnl)-k
Exercise 4.6.5. Define the convolution of a periodic distribution L with ¢
C 00 (1I') by L * ¢(x) = L(¢(x - ·)). Show that L have the convergent Fourier series L
* ¢(x)
=
L
*¢
E
E
C 00 ('Jf) and that we
L(n)rP(n)e 2rrmx.
nEZ
An important class of distributions are those that are sums of a finite number of delta measures at equally spaced points; in detail we write L = L:;: 1 cJfJJ/N where 0
The Fourier coefficients are N-1 L(k) =
L
cie-2rrtkJ/N.
j=O
This sequence is periodic with period N, since L(k
+ N) =
N-1
L
Cje-2rri(k+N)j/N
J=O
=
N-1
L
Cje-2rrikJ/N
=
L(k).
j=O
Conversely, suppose that we are given a bilateral sequence Lk with the property that for some N E z+, Lk+N = Lk for all k E Z. The smallest such value N is called the period. Then we can uniquely solve the system of linear equations c·e-2rrijk/N J
O=:::_ko,lt-rrl >o
Thus md/21
f(t)2m dt = 2md/2 {
(-rr/2.1rrj2) d
j(t)2m dt
+ O(md/2(1
- T/1 )2m}
Jltl
o.
;.k=l
3. If m 1 , m 2 are two measures, the Fourier transform of the convolution is the product of their Fourier transforms, where the convolution is defined by (m1
* mz)(B)
1
=
m1 (dx)mz(dy).
{ (x ,y):x+yEB)
Equivalently , for any bounded continuous function g
f
~d
g(z)(m 1 * m 2 )(dz) =
{
~u
g(x
+ y)m1 (dx)mz(dy).
Proof. The continuity of m follows from the dominated convergence theorem: If l;n ~ l;, then the complex-valu ed functions e''" x are bounded by 1, and converge to e'< x when n ~ oo. The positive definite property is a direct computation:
::::0.
To prove the convolution property, multiply the two transforms to obtain ml (l;)m2(l;)
= {
}fftu
ei< (x+y)ml (dx)m2(dy)
= {
}fftd
e'l; (x+y)(m,
* m2)(dz),
•
which was to be proved.
Example 5.2.2. The centered Gaussian distribution with variance parameter 0 is the measure with density e-lxl 2 12a 2 • Its Fourier transform can be computed
a >
in terms of a product of one-dimensi onal transforms as
m(~) =
=
f
J"R,d
eiS"·xe-lx1 2 /2a 2 dx
n(
f
;=I
j'R
eiS"}x}e-x}/2a 2 dxj)
n( d
=
v'2JTa2e-.;}a 2)
;=I
= (2JTa2)dl2e- a 2 fl;i 2 /2.
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INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Example 5.2.3. The Fourier transform of the uniform measure on the rectangle
nf=l (aj, bj)
is computed as
If ~J = 0 for some j, then the correspondi ng factor is replaced by bJ - aJ. We now prove that the mapping m
----+
mis 1 : 1.
Proposition 5.2.4. The measure m can be retrieved from its Fourier transform by the inversionfor mula
provided that them-measu re of the boundary ofthis rectangle is zero. In particular. m is uniquely determined by m. Proof. Multiply the defining equation (5.2.1) by e-cr 2i 0, which complete s the proof. •
Remark . We can render more transpar ent the computa tions in Proposit ion 5.2.4 and Corollar y 5.2.5 followin g by using the notation s of Chapter 2, beginnin g with the Fourier represen tation of the heat kernel H 1 , a bounded function with H E L 1 (JR. d): 1
260
INTRODUCTION TO FOURIER ANALYSIS ANO WAVELETS
Multiply both sides by lA ( y)m(dx) and integrate over JR.d x JR.d: { (Hr
}Rd
*
IA)(x)m(dx) =
{
Jocd
m( -0 iA
m(A) =lim { t---+0
= =
}Rd
(~)Hr(~) d~
m( -~)
L (~)Hr(~) d~
if m(aA) = 0
f m(-~)LC~)d~
if ,n
}Rd
1(ld e rrtl;·xm(~) d~) 2
E
L'CJRd)
by Fubini.
dx
The above methods can also be used to prove the following continuity theorem for Fourier transforms of measures. Proposition 5.2.7. Suppose that mn, n = 0, 1, 2, ... is a sequence of finite measures whose Fourier transforms converge:
lim mn(~) = n
mo(~).
Then the measures converge on every rectangle whose boundary has mo measure zero. Proof. For any ¢
E
S, let 1fr (/;)
= ¢. (-~), so that ¢ = Vr.
Then by Fourier reciprocity we
have
Letting n -----+ oo, the dominated convergence theorem implies that
li~ If R
i.
¢(x) mn(dx)
= li~
i . 1/f(~)mn(~) d~ = i . lfr(~)mo(~) d~ = i .
= n7=1 (a;, b;) is any rectangle, let¢± lim supmn(R) .:::lim sup { n
n
JR."
E
s
cp(x)mo(dx).
so that¢- .::: IR.::: ¢+.Thus
¢+(x)mn(dx) =
{
}TR.n
cp+(x)mo(dx)
liminfmn(R):;:: liminf {
{ ¢-(x)mo(dx). n hn ¢-(x)mn(dx) = hn
n
Now let¢+ ..!- lR, ¢-
t
lR to conclude that
m 0 (R").::: liminfmn(R).::: limsupmn(R).::: mo(R). n n
If m 0 (aR) = 0, then the extreme values are equal, so that the limit exists as required.
5.2.1
•
The Central Limit Theorem
Fourier analysis of measures is particularly well suited to study the convolutions of a single probability measure. In the case of the measure p8 1 + (1- p)8 0 this was effectively studied in Chapter 1 in connection with the DeMoivre-Lapla ce local limit theorem. The central limit theorem extends this to an arbitrary probability measure with a finite second moment, which we now describe.
261
APPLICATI ONS TO PROBABILI TY THEORY
Theorem 5.2.8. Suppose that m is a probabil ity measure on the rea/line with
L
i
x m(dx) = 0
2
x m(dx) = u
Then for any interval A,
* m * · · · * m)(A-vCn) =
. hm(m "
1
~2nu 2
2
=
.J2ir
1 '
e __,·; 2 dx.
A
Hint: It suffices to show that the Fourier transform satisfies the limiting relation lim,~ 1 n~ 0 m(trny' I- r 2 ) = e-r'1 2 . Use the fact that m(~) = e-t.'1 2 when~~ 0.
5.2. 1. 1
Restate ment in terms of indepen dent random varia/;Jies
The central limit theorem is presente d as a result on the convolut ion powers of a single probabil ity measure . This can also be recast as a result about the measure induced by a sum of independ ent random variable s, as follows:
Definiti on 5.2.1 0. A set of real-valu ed function s X 1 (t), ... , X 11 (t) on a proba-
bility measure space (Q, ~) is mutually independ ent if for every choice of real numbers x 1 , ••. , X 11 , we have
262
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Let m; be the distribution of X;, namely the measure induced on IR by the equation m;(A) = J.L{t : X;(t) E A}. Then the distribution of the sum X, (t) + · · · + Xn(t) is the convolution m 1 * · · · * mn, so that (m * · · · * m)(A.y'n) is the distribution of the sum (X 1 (t) + ... + Xn(t))/Jn. In probability theory, the term "random variaple" is synonymous with "real-valued measurable function." The central limit theorem can now be recast as follows: Theorem 5.2.11. Suppose that {X,.(t)},.= 1 , 2 .... is a sequence ofmutually indepen-
dent random variables with distribution m, where fiT?. x m(dx) = 0, JR x 2m(dx) = u 2 < oo. Then the distribution of the normalized sum [X 1 (t) + · · · + Xn(t)]/u Jn converges to a standard normal distribution when n --* oo.
Independent random variables may be constructed on the unit interval Q = [0, 1] as follows. Let cj> : N --* N 2 be a biject~ve mapping. For example, this may be constructed by listing all of the integers in a doubly infinite array as follows: 1
3
6
10
15
2
5
9
14
20 ...
4
8
13
7
12
11
21 ...
19 ...
18 ...
17 ...
16 In this example we have, for example c/>(8) = (2, 3), c/>(18) = (3, 4) and so forth. Now we expand t = L:~ 1 wk /2k and define
and so forth. Then for every n, {X1 (t),X2(t), ... ,Xn(t)} are independent random variables, each of which is distributed according to Lebesgue measure on [0, 1]. To achieve more general distributions, it suffices to form Borel functions in the form Y;(t) = c/>;(X;(t)) fori= 1, 2, ....
5.3
EXTENSION TO GAP SERIES
The asymptotic normal distribution is not restricted to sums of independent random variables. In this section we consider a class of trigonometric series that are asymptotically normal. More general results are found in the book of Zygmund (1959, Volume 2,
APPLICATION S TO PROBABILITY THEORY
263
Chapter XVI). Here we consider sums of the form k
(5.3.1)
Sk(t)
=
L
a1 cos n1 t
}=1
where (a1 ) are real numbers and n 1 < n2 < · · · are integers which satisfy (5.3.2)
(k
=
1, 2, ... )
for some q > 1. The growth of the sum is measured by the L 2 norm, which is
(5.3.3) where we assume that
(5.3.4)
(k -----* 00).
As preparatio n for the theorem, we first prove a simple lemma.
Lemma 5.3.1. Under the conditions ( 5.3.4 ), we have (k-+ oo). Proof. Given E > 0, let K, be such that (Ak) is increasing, we have for k > K,
-
1
la,I!Ak
max la1 1 ::;
Ak K, ~J~k
I -E Ak
< E fork> K,. On the one hand, since
max IAjl =
K, ~j~k
E,
while (k-+ CXi).
Hence lim
supk~= (l flAk
I) max,~ 1 ~k la1 I ::;
E.
But E was arbitrary, so the proof is complete .
•
This lemma allows one to conclude, for example, that for any p > 2
Theorem 5.3.2. Suppose that the integers (nk) and the real numbers (a ) satisfy 1 (5.3.2) and (5.3.4) with q > 3. Then we have fork-+ oo (5.3.5)
-I
2rr
I{
Sk(t) t E 1I' : YI ::::: - :::=: Y2
Ak
}I
--+ -1-
../iii
1Y2 e-u y1
2
12
du.
264
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
Proof. We compute the Fourier transform, defining
It suffices to prove that k(~) --+ e~< 1 2 when k --+ oo. From the power series of the exponential function, we have for small lzl, 2
(I
+ z)e~o
=
(I -
+ z)
= (I
z
+ ~ + O(ld))
z2
3 2 + 0(1zl )
I -
so that (z--+ 0)
and we can write
where the o(l) term is uniform in t E 'lf'. Now we write
noting that I
-2
1
rr
2
Tk(t) dt
1f
I
=8
L k
J=l
a4
~-+ 0
(k-+ 00),
Ak
in particular Tdt) --+ 0 in measure. Meanwhile
n(t
I}=I
2
+
i~ Aka} cos njr) 1
:s
Ii }=I
2
(1
+
~Ak~J)
so that we can write
k(~) = 2 ~ 1Il (1 + i~ ~j 1f J=l
k
2
cosnjt) exp [ - x (1 2
+ ~k(t) + o(l))] dt
APPLICATION S TO PROBABILITY THEORY
265
It remains to analyze the final integral. For this purpose we expand the cosine products using repeatedly the identity 2 cos a cos b = cos (a +b) +cos (a -b) to obtain a finite sum (5.3.6) where the sum is over those indices of the form v = n,, ± n, ± · · · with n,, > n, > · · ·. 2 2
Lemma 5.3.3. Suppose that nk+I > qnk with q > 3. Suppose that an integer vis represente d in two (possibly) different ways
Proof. If all of the subscripts are equal, there is nothing to prove. Otherwise there is a first subscript that differs in the two representat ions. By relabeling the subscripts, we may assume without loss of generality that i 1 > j 1 • By a further relabeling and moving all of the terms to one side, we may write
where the coefficients a,
E
{0, ±1, ±2} and i 1 > k 1 > k 2 > ···.Hence
+···) which is a contradicti on. To complete the proof of the theorem, we note that in the product (5.3.6) the only contributio n to the term ao occurs when all of the frequencies are zero, hence a = 1. 0 Applying the orthogonal ity of cos vt, we conclude that
Il (1
1 { 2 rr }.,. = 1 1
+
which completes the proof of the theorem.
i~ Ak a 1
cos n1
t)
dt = l,
•
Exercise 5.3.4. Suppose that the coefficient s (an) satisfy (5.3.4). Prove that 1
n- log lanl--+ 0 when n--+ oo.
Hint: Write a~/A~ =En ~ 0 and solve for a~ = a~(En/EN )n:Z,+ (1/(l -Ed) for n > N, where 1 0 and Ek < E for n > N.
EN >
Exercise 5.3.5. Suppose that the coefficien ts (an) satisfy ( 5.3.4) and the integers (nk) satisfy (5.3.2). Let Sk(t) = L~=I a cos(n t - (} ) where (} E JR.. Prove 1 1 1 1 that ( 5.3.5) holds for this wider class of series.
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INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
5.3.1
Extension to Abel Sums
The central limit theorem for gap series can naturally be extended to obtain the limiting distribution of the harmonic function oc
(5.3.7)
u(r, t)
=
2:a1rn1 cosn1 t j=l
under the same conditions as in Theorem 5.3.2. We define (5.3.8)
Then Exercise 5.3.5 implies that A(r) < oo for 0 < r < I. The following result was first proved -by Kac ( 1939) and later extended by Salem and Zygmund (1948). Theorem 5.3.6. Suppose that the integers (nk) and the real numbers (a ) satisfy 1 (5.3.2) and (5.3.4) with q > 3. Thenfor any interval C C "JJ', we have for r ~ 1
I{t
(5.3.9)
E '][' :
u(r t) '
A(r)
E
C
}
I~
1
2
e-Y 12
c
-./2]i dy. 2n
We first develop the Abelian counterpart of Lemma 5.3 .1. Lemma 5.3.7. Let bk ::: 0 for k ::: 1 and set Bo = 0, Bn = bt + · · · + bn for n ::: 1. Suppose that Bn ~ oo and bn/Bn ~ 0 when n ~ 00. Furthermore let B(r) = Ln::: 1 bnrn for 0 < r < 1. Then 1
- - supbkr B(r) k:;o:l when r
~
k
~
0
I.
Proof. We set e,. = b,.J B,. for n ~ 2. Without loss of generality we may assume that b = I 1 and e,. < 1 for all n. Then B,_ 1 JB, = 1 - e,., so that we can write for n ~ 2
APPLICATIONS TO PROBABILITY THEORY
Since Bn
---+
oo, we have B(r)
---+
oo when r
---+
267
1. On the other hand we can write for any
NEz+,
B(r)
"' = Lb,r" n=l N
=::: L
(B,- Bn-1 )r"
n=l N
= r'\ B"
+ (1 -
r)
L
r"Bn
n=l
= r
Given then
E
N( 1 - I
> 0, let K, be such that e,
K,. If sup, bkrk is attained at some k > K_,
On the other hand, if the supremum is attained at some k
But
E
~
K,, then
•
was arbitrary, which completes the proof.
We can apply this lemma by taking b 11 , =a~ and b 11 = 0 if n ~ {nk,, nk , 2 conclude that r"k akf A (r) ~ 0 when r ~ 1, unifonnly in k E z+.
•• • },
to
Proof of the Theorem. We compute the Fourier transform
We proceed as in the proof of Theorem 5.3.2, beginning with the estimate e 0 = (1 + z)exp[z 2 /2 + O(lzi 3 )] applied to z1 = (ii;/A(r))ajr"' cosn t, noting that z ---+ 0 1 1 when r ---+ 1, uniformly in}. Then
Write
268
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
The L 2 norm of \llr is estimated as before r-+1
where we have used Lemma 5.3.7 to replace one factor by the supremum, the remaining sum being equal to 1. Meanwhile, we have the uniform bound
n1 ""
,~
2 I
1
t;uk + i- cosnkt 1
A(r)
n
::: = ( 1
t; u~ ) + --)-~
•=I
1
A(r -
so that we can apply the dominated convergence theorem to conclude that
But this infinite product can be expanded as a sum:
n ""
(
1+
1'/;
= a,r nk cosnkt ) =L:a.,(r)cosvt .
A(r)
k=l
v=O
From Lemma 5.3.3 and the condition (5.3.4) we have cx 0 (r)
=
o(l), so that we conclude
•
which completes the proof.
5.4
WEAK CONVERGEN CE OF MEASURES
To probe the deeper aspects of the convergence question, we develop the following notion of weak convergence of measures.
Definition 5.4.1. A sequence of finite Borel measures (m,) is said to converge weakly to a limit measure m iffor every bounded continuous function g
lim { n
}Rd
g(x)m,(dx) =
{ g(x)m(dx). }'R,d
If A is any set in JR.d, the interior A o is the set of points x E A such that A contains an open ball about x. The closure A is the complement of the interior of the complement; in symbols (A)c = (N') 0 and we have A o c A c A. The boundary is defined as aA = A \A o. The portmanteau theorem gives equivalent conditions for weak convergence.
Theorem 5.4.2. The following conditions are equivalent 1. m, converges weakly to m.
2. For every closed set A, lim sup, m,(A)
::::=:
m(A).
APPLICATI ONS TO PROBABILI TY THEORY
269
3. For every open set A, lim infn mn(A) > m(A). 4. For every Borel set A with m(3A) = 0, limn mn(A) = m(A). The proof of this theorem , which has nothing to do with Fourier analysis, can be found in Billingsl ey ( 1999). It is also helpful to develop the appropri ate notions of compact ness in the context of weak converg ence of measure s. In this setting we refer to a tight family of measure s, formaliz ed as follows:
Definiti on 5.4.3. A sequenc e offinite measure s (mn) is tight if lim supnmn( {x : lxl >A}) = 0.
A--><XJ
Theorem 5.4.4. Suppose that mn is a tight sequenc e of finite measure s with supnmn( Rd) < oo. Then there exists a weakly converge nt subseque nce. Again we refer to Billingsl ey (1999) for the details.
Exercise 5 .4.5. Prove that any weakly converge nt sequenc e of finite Borel mea-
sures is tight.
= 1, let mn = 8n, a point mass at the point n. Then mn is not a tight sequence , since for any n, mn {x : lxl > n/2} = l. Exampl e 5.4.6. With d
5.4.1
An Improv ed Contin uity Theore m
The theory of tightness can be used to formulat e an improve d version of the continui ty theorem for sequenc es of characte ristic function s of probabil ity measure s. In the previous version, Proposit ion 5.2.7, we required that the Fourier transform s mn(~) converge to a limit M(~). which is assumed to be a characte ristic function . We now have the followin g improve d version.
Theorem 5.4.7. Suppose that mn is a sequenc e of probabil ity measure s on Rd with Fourier transfor ms mn(~). with the property that there exists (5.4.1)
M(~)
=lim mn(~) n
and that M is continuo us at ~ = 0. Then there exists a probabil ity measure m so that mn converge s weakly tom and m(~) = M(~). Proof. From the hypothes es, we can write
1-
mn(~)
=
{
JTR.d
{1 - e'" ') mn(d.x).
270
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
This equation is integrated over the cube Cu
1
[I - m,(~)] d~
=
Ca
Id --
(2a)
1 c,
= n;= 1 { 1~1 I s
1( Rd
[1-m,(~)] d~ =
a} to obtain
d 2sinax·) 1 (2a)d - n m,(dx)
1(
J=l
XJ
1 sin ax I - nd - ) m,(dx)
JRd
J= 1
ax1
since the integrand on the right side is nonnegative. If (x 1 •••• , xd) lies outside the cube C 2 a-' , then the indicated product contains at least one factor that is less than ~, the remaining factors are less than I . Hence - -I d
(2a)
1
~
m,(~)] d~
[I -
C,
:::: -I 2
1
m,(dx).
(C , - 1 )' 2
From the dominated convergence theorem we conclude that lim sup {
" J' la
1
m,(dx) < ~ { - (2a)'
Jc
[1 -
M(~)] d~.
u
From the continuity of M, the right side can be made arbitrarily small by taking a sufficiently small. This proves that the measures m, are tight, hence we can extract a weakly convergent subsequence. If we had two different subsequential limits m 0 and no, then both of these measures must have the same characteristic function, hence they must be the same measure by the uniqueness of Fourier transforms, Proposition 5.2.4. Hence every subsequence has a subsubsequen ce that converges weakly to the same measure m 0 • From this it follows that the original sequence converges weakly to mo. •
Exercise 5.4.8. Show that the hypothesis of continuity at~ = 0 in Theorem 5.4. 7 can be weakened to the hypothesis that ~ = 0 is a Lebesgue point for M, in the sense that lima ....... o a-d Jc, II - M(~)l d~ = 0. 5.4. 1. 1 Another proof of Bochner's theorem In Chapter 3 we introduced the concept of positive-defi nite function and used a Fatou theorem for harmonic functions in the upper half plane to prove Bochner's theorem, which affirms that any continuous positive-defi nite function on lR is the Fourier transform of a nonnegative measure. In this section we will give an independent proof of Bochner's theorem using the theory of weak convergence applied to a Gaussian convolution. Proof. We begin with/(~), a complex-valu ed positive-defin ite function that is assumed to be continuous. Define
(5.4.2)
m, (x)
= - 1-
By hypothesis, we have for every 1fr
2n
E
r e-uue-'"
]IR
L 1 (lR),
2
/(u) du.
APPLICATIONS TO PROBABILITY THEORY
271
2 Taking the choice 1/F(~) = e'-"<e-•< 12 we infer that
0
~I12 f(~- T/)e•x<J;-ry)e-• 0 2
(5.4.3) Taking~
= 0 and using Fatou's lemma and the continuity at u = 0 shows that
1
m,(x)dx
~
liminf8-o
1
8 2 2
m,(x)e- x 1 dx =f(O)
A) < r::xJfor each A > 0 and F(A) tends
to zero when A --+
r::xJ.
Proof. We first prove this for the symmetrized distribution F!(dx) whose characteristic function is fl~n
=
(/;)
=
Fn(dx)
* Fn(-dx),
e-(2/n)Y,(/;}.
We apply the technique of (5.5.6) with the interval [ -1, 1] replaced by the interval [-2/A, 2/A] and use the inequality I - e-' ::5:. y to write
F:(l/;1
>A) ::5:. 2A { 111;1A)
f
I<J~A I
+X"
and we have shown above that the supremum over n tends to zero when A --+ oo.
•
Proof of Theorem 5.5.4. Now we prove the represent ation theorem. From the definition
1/r(~) =lim /r(~) r~o
=lim n
-
I
t
{(en~j,R
I)nF,(dx ).
For any n we can write (5.5.7) where J.l-, = JD?. sin xF, (dx). The integrand in (5.5.7) is defined by continuit y at x = 0. Note that Gn({O}) = 0 for all n. From Lemma 5.5.5, we may take a subseque nce for which the total masses G,(IR) converge to a limit. Since the measures Gn are tight from Lemma 5.5.7, we may take a further subseque nce that converge s weakly to a limit measure G. All of the limits below will be taken through this new subseque nce. The integral term on the right side of (5.5.7) converge s to
1
e•x~ -
I - il; sin x
- - - -- - - ( 1 2 D?. X
+
,
x") G(dx)
while the left side converge s to 1/r (i;). Letting ~ = I and taking the imaginar y part shows that J.l-n converge s to Im 1/r(l ). We have obtained the Levy-Khi ntchine represent ation in Feller's form: (5.5.8)
1/r(i;) =
1
e"' - I - ii; sinx
L
0, n < oo the integrand is an integrable function, so that we can apply the Riemann-Leb esgue lemma to let a--+ -oo and obtain (5.6.6) (F,
* KT-
1T e-'~"'
, ') dl;. - . - ( 1 -\/;-\) ( F~ ( - 1; )" - e-~-;~
1 * Kr)((-oo, x]) = 2rr
_ 7 -tl;
T
..jii
It remains to estimate the integrand and apply Letnma 5.6.2. From the definition of the Fourier transform,
~~= 2
IF(!;)- 1
+
IL
: :; i
(e'~x- I - il;x + \/;:3\ F(dx)
\I; 13 =m36·
2 l;:x ) F(dx)l
278
INTRODUC TION TO FOURIER ANALYSIS AND WAVELETS
We apply the same estimates to e-< 1 12 and use Holder's inequality to obtain 13 le-< 212 -1 +e/21:::: : 1 ~ 6
f
Jfil:.
lxl 3e-x 212
__:!!__ ~
21~1'
- 3./2ii
1 (g) = e-~ a}/ 2 is the Fourier transform of a normal distribution with mean zero and variance so that we can write 2
2
2
aJ,
L (fr,j
Arguing as before, we have for sufficiently large n,
so that
IFt(;/An) · · ·Fn(l;/An)- e-1;
2 /
2
1 :S:
2
e-1; /
n
3
L j=l
Now
IFJ(;/An)- ,(;/An)l.
280
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
so that
Applying Lemma (5.6.2) we have
AfT( 1 - 1~1) -
sup IG(x)- (x)l ::: xEIR 3An
T
-T
1~1 2 e-~ 2 13 d~
C + -.
T
Choosing T = An balances the two error terms and leads to the desired result.
5.7
THE LAW OF THE ITERATED LOGARITHM
The central limit theorem, which gives the limiting distribution of normalized sums of independent random variables, has an almost-everywhere counterpart, as follows: (5.7.1)
. hmsup k-.oo
Sk(t)
J2A~ log logAk
= +1
a.e. t.
The proof of (5.7.1) depends on careful bounds for the distribution of Sk(t)/Ak and a simple estimate for the distribution of the maximum of S 1 (t), ... , Sk(t). We will prove the following result: Theorem 5.7 .1. Suppose that {Xk (t) hEz+ are independent functions on 'II' = [0, 1] with f1fXk(t)dt = 0, f1fXk(t) 2 dt =a~, f1r 1Xk(t)l 3 dt < oo and that the conditions (5.6.8) are satisfied. Suppose further thatfor each x E Rand k E z+
(5.7.2)
l{t : Xk(t) < -x}l = l{t: Xk(t) > x}l.
Then ( 5. 7.1) holds. The distribution of the maximum is estimated as follows. Lemma 5.7.2. Let SZ(t) = max{SI (t), ... , Sk(t)} where (5.7.2) is satisfied. Then
(5.7.3)
j{t: s;(t)
> x}l < 21{t: Sk(t) > x}l.
Proof. Let A} = {t : sl (t) .:::X, ... 'sj-1 (t) .:::X, SJ(t) > x}. Then the event {t : s;(t) > x} is written as the disjoint union U~=IAr From the symmetry hypothesis (5.7.2) we have for any·n1 _::: n2, i{t: I:;,!n 1 Xj(l) 3::- 0}1 ::=:: Now
4·
SJ(t) >X, XJ+l (t) + ... + Xk(t) :::: 0 ==> Sk(t) >X hence A 1 n {X1 +1
+ · · · + Xk
:::: 0} C A1
n {Sk
> x}.
APPLICATIONS TO PROBABILITY THEORY
281
Then k
l{t: Sk(t) > x}l ::=::
L lA, n {t: Sk(t) >
x}l
j=l
k
: : :. L IA
1
n {t: X 1 +1(t)
+ ··· +
Xk(r):::::.
Oll
j=l
k
=
L
lA, II{!:
x,+1(t)
+ · · · +Xk(t)::::: Oll
;=I
1
k
: : : ;zLIA1
1
j=l
•
1 S*k(t) > x}l. = ;zl{t:
Lemma 5.7.3. lfxk--+ oo so that xU2 -logAk--+ -oo, then t: I{
I
Sk(t) > x } = exp[-xf(l +o(l))]. ~
Proof. From (5.6.9) we have
ll{t: s~~) > x}l- (1- (x))l ~ ~ where C is a constant. When x -+ oo we have 1 - (x) = exp [ -x /2( 1 + o( 1))]. The 2
hypothesis on xk is equivalent to 1/Ak = o(e-xif ) and therefore the error term can be absorbed into the Gaussian term when this is satisfied. • 2
We also need the first and second Borel-Cantelli lemmas, as follows. Lemma 5.7.4. Suppose that (Bd are measurable sets with L~! IBkl
1 -28. This proves that for any 8 > 0 there is a subsequence j --+ so that S1 (t)/A1 j2Iog IogA1 :::-_ (1 - 28). Hence the limsup of this ratio is greater than or equal to 1, which was to be proved. •
=
CHAPTER
6 INTRODUCTION TO WAVELETS
6.1
MOTIVATION AND HEURISTICS
Classical Fourier analysis may be viewed as the problem of reconstructing a function f from dilations of a fixed sinusoidal function x ~ e 2rrix by writing f(x) = JR e 2rcit;xj(l:,) dl;. The Fourier transform}(!;.) may be thought of as the amount of the sinusoidal oscillation e 2rrit;x present in the function f. The Fourier representation is instrumental in analyzing translation-invariant operators such as convolution operators and linear differential operators with constant coefficients, where we can write l j ( x - y)K(y) dy = l P
(~)f(x)
K(l;)e2nit;xj(l;) dl;,
= l.p(2JTil;)e2nit;xj(l;,)dl;.
However classical Fourier analysis suffers from the defect of nonlocality: The behavior of a function in an open set, no matter how small, influences the global behavior of the Fourier transform. We have also remarked on the simultaneous nonlocalizability in connection with the uncertainty principle. The theory of wavelets is concemed with the representation of a function in terms of a two-parameter family of dilates and translates of a fixed function that, in general, is not sinusoidal, for example: f(x) =
l.
\a\-11/1 (x a b) W..pf(a, b)dadb 2
where W ..pf is a suitably defined transform off. Altematively one may envision a series expansion f(x)
=
L j,k
284
c1 .k2j12 1/1(2jx- k)
INTRODUCTION TO WAVELETS
285
where we sum over the dilates in geometric progression. The factors of lai-I/ 2 and 2j 12 are inserted to preserve the L 2 -norm of the basic wavelet 1jr. In this chapter we will describe the properties of wavelets in one dimension, making full use of the tools of Fourier analysis.
6.1.1
Heuristic Treatment of the Wavelet Transform
The wavelet transform off with respect to 1/f is defined by the integral W>!ff(a,
b)= J~[
j(y){/f
(Y-a b) ~lal %.
It is straightforward to compute this transform and the inverse transform on the Fourier
exponentialsj(x) = e 2rrit;x; from the definition of the Fourier transform, we have W>!ff(a,
b) = =
b) -JjQf
{ e2rr'b{/J ( y -
JR
l
viJQT
dy
a
e2rnt;(h+a~l{/J(z) dz
= Me2rrit;bVr(a~).
Now we form the adjoint operator
w; W>!ff(x) = = =
{ (W>!ff)(a, b)ljr
J~
(x- b) vial ~ (X- b) db a
MVr(a~)
{ e2rrit;bVr
JR
MVr(a~)M
L
a
-JjQf
e 2rrd;(x-az)1/r(Z) dz
= lall~(a.;)/2e2rrit;x { W* W
}JR.
1/f
A
f( ) da = 1/1
x a2
2rrit;x {
JTR.
e
11/r(a.;)i Ia!
2
d a.
The final integral is independent of~, which is seen by making the substitution v = a~, from which we obtain the inversion formula ·~ jlR(W.~, W>!ff ja 2 )da f(x) = e 2 rr,.,x = A"' . 11/r(v)l 2 /I vi dv
JR
This leads us to impose the normalization fiR I~ ( v) 12 j 1v 1d v = 1, in order to obtain the wavelet representation f =
[
JR
w; W"'fda a 2
valid when f(x) = e 2rr,t;x. It now remains to investigate this inversion procedure for arbitrary f E L 2 (1R).
286
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
6.2
WAVELET TRANSFO RM
Let 1fr
E
L 2 (lR). The dilated-trans lated function is defined by
(6.2.1)
Vra,b(x) = lal -1/2 Vr
(X- b) -a-
,
0=/=aElR ,bER
This function is obtained from 1fr by first dilating by the factor a and then translating by b. Clearly lll/Fa,bll2 = llo/112·
Definition 6.2.1. 1fr
E L 2 (lR)
is a continuum wavelet
if
(6.2.2) The wavelet transform off E L 2 (1R) by 1/r is defined by
(6.2.3)
WlfJf(a, b) =
l
{/fa,b(x)f(x) dx.
From the Cauchy-Sch warz inequality, we see that W lfJf is a bounded function with iWlfJf(a, b)i < ll1/rll2llfll2· The intuitive meaning of WlfJf(a, b) is the amount of the dilated-trans lated waveform VFa.b that is present in the functionf.
Remark. If, in addition, 1/r
L 1 (lR), then the integrability condition (6.2.2) implies that JR lfr(x) dx = 0. Indeed, -(/;- is continuous at I; = 0 with -(/;-(0) = Jff?. lfr. If this is nonzero, then the integral (6.2.2) is divergent. The following is a form of Parseval's theorem for the wavelet transform: E
Proposition 6.2.2. Suppose that 1/r is a continuum wavelet with (1/r, 1/r)w Thenforan yf, g E L 2 (1R), we have
(6.2.4)
1
f(x)g(x)dx =
lR
11
IRIR
1.
-
dadb WlfJf(a, b)WlfJg(a, b ) -- .
a2
Proof. Let lfr(x) lfr(-~). Then Wy,f(a, b) is the convolution off with lfra.O• ~hose Fourier transform is v'JQT~(a~). Hence the Fourier transform of Wy,f is](/;)v'JQT~(a/;), and similarly for Wy,g. Therefore from Parseval's theorem for the Fourier transform we have (6.2.5)
We integrate both sides with respect to dafiai 2 • apply the Fubini theorem to the right side, and use the definition of (1/F, 1/F)w to remove this constant factor. The remaining integral is transformed by another application ofParseval's theorem in the form fr~. fg = JJR which completes the proof. •
Jg.
INTRODU CTION TO WAVELET S
287
This proposi tion can be interpre ted as the stateme nt that/ -+ W y,f is an isometr y from L2(JR; dx) to L 2 (JR 2 ; da dbfiai 2 ), where the inner produc t is defined as ((F, G)) :=
1 JR2
-
dadb F(a, b)G(a, b ) -2- . a
Theore m 6.2.3. Suppos e that 1/F is a continu um wavele t with (1/r, 1/r)w = 1. Then 2
forany f E L (1R), we have the L 2 inversi onform ula f(x) =
1 JR2
(6.2.6) =
dadb Wy,f(a, b)1/ra ,b(x)-a2
lim
€ a
But this is the Fourier transform of the convolution/* g€, where g€ is the Gaussian density 2 2 g€ (x) = e-rrx / 2 € 1.../2Ei, which is an approximate identity in the sense of Chapter 2, from which we conclude that for any homogeneous Banach space B, we have S€ f ~fin norm. In particular, iff is bounded and uniformly continuous we have liS€/- flloo ~ 0 when E ~ 0. Iff E LP(IR), 1 < p < oo, then liS€/ -fliP~ OwhenE ~ OandS€f(x) ~ j(x) for almost every x E JR.. This follows from the results on Gaussian summability in Chapter 2. These desirable properties are not shared by the partial inversion of the Fourier transform, for example.
Exercise 6.2.8. For the Mexican hat wavelet, define the partial inversion by (6.2.7) and explicitly compute S€f as the convolution with an integrable function, in particular verify that liS€!- flip ~ 0 in case off E LP(JR.), 1 < p < oo or f E BucOR). Exercise 6.2.9. Formulate the wavelet transform in n dimensions, beginning
with 1fr E L 2 (1Rn) satisfying fn~.· 1~(~)1 2 d~ /I~ In < oo and defining 'tfra.b(x) = lfr((x- b)ja)flaln/ 2 forb E lR.n and 0 =I= a E JR.
6.2.0. 1
Wavelet characterizatio n of smoothness
We can use the wavelet transform to characterize the smoothness of measured by the Sobolev norm
f
E
L 2 (JR.) as
Exercise 6.2.10. lfllfll2.s < oo,provethatfha skcontinuousde rivatives, where k < s- ~· Hint: Apply Cauchy-Schwarz to the Fourier integral representation of J.
The next proposition applies to continuum wavelets that possess a certain number of vanishing moments: JR xklfr(x) dx = 0. The result states that for this class of wavelet expansions, the Sobolev norm is equivalent to a weighted L 2 norm of the wavelet transform.
INTRODUCTION TO WAVELETS
291
Proposition 6.2 .11 . Suppose that 1/F is a continuum wavelet with ( 1/F, 1/F) w = 1 and C
·- { 1/f.s . -
}JR
1~(~)12 ~~~1+2s
de < oo 5
for some s > 0. Then [
(6.2.8)
[
JJR JJR
2
=
Proof. Returning to (6.2.5) with/ respect to a E .IR. Thus {
{ IW /(
JrR JL'ii
"'
a,
dadb
2
IWwf(a, b)l lal 2 + 2 , = Cw.sll/llz..,·
lal2+ 2s and integrate with
g, we divide both sides by
h)l7 dadb = { { 1](.;)121~(a.;)l2 dad.; - lal2+ 2 ' JR JR lal 1+ 2s
= { lf(.;)l2 ( { l~l(v)l
JrR
=
JrR
c1/f.s
i
2
lvll+ls
dv) l~l2s d.;
2
1.; 1 ' 11 l. Then Holder's inequality gives X
E
hn ==> IPnf(x) I ::0 2n IPnf<xW ::: 2np
(fn 1/( (f" If
IP
mg(A)I
n I' lc--mg(x)l
= 2 "'I {
Jo
=
2
mp
g(y)dyl
iK lg(y)l~'dy
(2K)f'lp'
0
{""' IP-mg(x)ll' dx
Jo
which tends to zero when m the same fashion.
= 2"'2-"'~'ljK lg(y)li' dy! (2K)PIP', -K
-+ <XJ.
The contribution from the negative axis is estimated in •
Hence we conclude the following. Proposition 6.3.12. Let 1 < p < converges in the norm of U'(.IR).
CXJ.
Foranyf
E U'(lR),
theHaarseries(6 .3.15)
In the case p = 1 there is a simple example to show that this proposition is sharp. Exercise 6.3.13. Letf = 1[O.I J· Prove that the Haar series (6.3. 1 5) is not convergent in L 1 (lR).
However this anomaly is not present for the one-sided Haar series. Exercise 6.3.14. Let 1 ::: p < CXJ and let f E U' (lR). Prove that the one-sided Haar series (6.3.11) converges in the norm of LP(JR). 6.3.3.2 Pointwise convergence of Haar series Since the projection operator P, agrees with the average over dyadic intervals, it follows from Lebesgue's theorem that P,f(t) -+ f(t) for almost every t E JR. In particular if/ is continuous at t, then we have lim,_.= ~J(t) = f(t). Iff has a jump discontinuity at a
INTRODUCTION TO WAVEUoTS
299
dyadic rational t, then we note that K, (t. y) = 0 for y < t and sufficiently large n. Then we can write P,,j(t) - f(t
+
0) =
lex. lf(
y) - f(t
+
0)]K11 ( t, y) dy-----+ 0,
n----"' CXJ
to conclude that P,J(t) -
f(t
+
t =
0)
k/2".
n----"' CXJ.
One can also con finn the absence of a possible Gibbs phenomenon for Haar series. Indeed, the kernel K 11 (x. y) ;::: 0 with fo. K 11 (.>.:. y) dy = I. Therefore iff E U'"(lR)
-llflk ::C P,,f(x) ::C llflk. which implies that for any sequence
X
11
----+
x, we must have
-II f lk ::=: lim inf P,,f (x,) ::=: lim sup P,,f (xn) ::=: II f II x. II
II
Exercise 6.3.15. Suppose that f(t) = I for 0 ::=: t < I /3 and that f(t) = 0 for 1/3 < t < I. Show that we have lim inf, P,,f(l/3) < lim sup, Pnf0/3), so that the Haar series diverges at t = I j3.
6.3.4
*Construction of Standard Brownian Motion
The Haar wavelet expan~ion can be used to make an effective construction of the standard Brownian motion process. By definition, this is an indexed family of real-valued functions X 1 (w) where 0 ::=: t ::=: I and cv E Q, where (Q, :F, P) is a measure space of total measure I. In this context, the functions w-----+ X 1 (w) are called random variables. They are assumed to have the following properties:
1. For each 0 S variance
t -
s < t S s: in detail
I, X 1
-
X, has a normal distribution with mean zero and 1
P[w: X,(w)- X,(w) < v] = · --/2rr(t- s)
~-'
2
e-u
/
2 lr-,)
du.
-oo
2. For any subdivision 0 = to < t 1 < · · · < t,v S I, the random variables X,, Xro, ... , x,N - Xr, are independent. 3. For a.e. w, the function t - X 1 (w) is continuous, with Xo = 0. I
From properties (I), (2), it follows that the random variables X 1 ,, . . • , X 1, have a joint normal distribution with mean values zero and covariance matrix defined by E[X1,X1) = min(t,, t 1 ).
Exercise 6.3.16. Prove this. The Haar functions !fr,k are not continuous, so we would not expect to be able to construct the Brownian motion as a Haar series. But the functions t - J~ 1/r k(s) ds are 1 continuous and can be used to construct the Brownian motion. In order to prove the
300
INTRODU CTION TO FOURIER ANALYSI S AND WAVELET S
distribu tional propert ies ( 1) and (2), we will first conside r a general orthono rmal basis of the space L 2 [0, I]. The Browni an motion will be constru cted as the infinite series
(6.3.I6 ) Here ( n) is an orthono rmal basis of the space L 2 [0, I] and (Zn) is a sequen ce of independen t standar d normal random variable s; in detail P[Zn
,(u) du)
s
00
= Z:::::(lr,.rJ·¢,>2 n=O
=t-s where we have used Parseva l's identity for the orthono rmal basis (¢,).Th is proves that the series (6.3.16) converg es in L 2 (Q); the partial sums of the series are normall y distribu ted with mean zero, so that the limit is also a normall y distribu ted random with mean zero and the asserted variance , proving (1 ). To prove (2), we note that the partial sums of the series define a Gaussia n distribut ion on JRN so that the indepen dence can be inferred from the covarian ce function by showing that the increme nts are orthogo nal in pairs. Now if s < t ::S u < v, we have
00
=
L
(1 r-.rl' ¢,) (I [u.v]' ¢,)
n=O
= Ors.t]• lru.vJ)
,o where we have used the bilinear version of Parseva l's identity and the disjointn ess of the interval [s, t] from the interval [u, v]. This proves the pairwise orthogon ality. Since the vector is multivar iate normal, the indepen dence is thereby proved. •
301
INTRODUCTION TO WAVELETS
6.3.5
*Haar Function Represent ation of Brownian Motion
The normalized Haar functions l/f1k(t) together with the constant function provide a convenient orthonormal basis of the Hilbert space L 2 (0, I). We find it convenient to relabel them as follows: 1/fo(t) = I,
If n :::: I, then we can write n we set
=
1/f,(t)
2!
=
+
k for j
l/l"jk(t)
=
=
0, 1, 2, ... and k
=
0, I, ... , 21 -
I and
2112 l/f(21 t - k).
From the one-sided Haar series representatio n (6.3.1 1), we see that {1/f, (t) }n=O.I.2 .... is an orthonormal basis of £ 2 (0, 1). To display the Haar series representatio n of Brownian motion, we introduce a sequence of independent standard normal random variables Z,, n :::: 0 with (6.3.17)
P[Z, .:5 x] =
Jx e- u v 2n -·= ~
2
1
2
du.
These may be defined on a probability space, denoted (Q, .F, P). The Brownian motion is sought in the form
(6.3.I8)
It is immediate from the orthonormal basis properties of 1/f, that for each t E [0, 1], the series (6.3.18) converges in L 2 (Q, .F, P). From this it is immediate from the proofs of (I), (2) above that X 1 has the required distributiona l properties of Brownian motion.
6.3.6
*Proof of Continuity
We will now prove property (3) of Brownian motion, by showing that the series (6.3. I 8) converges uniformly for almost all w E n. Lemma 6.3.18. There exists M = M(w)
P
[
w: sup n
I, and we have the general term of a convergent series, and by the first Borel-Cantell i lemma the series In;::,,Az < <Xl for almost all w. Therefore for n sufficiently large ni:,A~ fails to occur, in particular for some k, Xk; 2 n - Xck-l >/ 2" ::: c ./ii72", which proves that the Levy modulus is a sharp lower bound also. •
I:;:,
Exercise 6.3.21. Prove that the elementary estimates (6.3.19) hold with the constant
c, = c2 =
1/(1 - 2- 112 ).
Hint: Compare a sum with an integral, which can be estimated by partial integration.
6.4
MULTIRES OLUTION ANALYSIS
In this section we return to the construction of general wavelets. The main features of the Haar wavelet expansion can be abstracted as follows.
Definition 6.4.1. An orthonorma l wavelet is a function \II doubly indexed set
{2il 2
\lf(2 1 t -
L 2 (lR) such that the k)}J,k.eZ is an orthonorma l basis ofL 2 (.1R). E
304
INTRODlJCTI ON TO FOURIER ANALYSIS AND WAVELETS
We have already seen that the Haar function provides an example of an orthonormal wavelet. To develop a systematic method for producing orthonorm al wavelets, we introduce another notion. which generalize s the Haar constructi on.
Definitio n 6.4.2. A multiresol ution analysis ( MRA) is an increasing sequence of
subspaces { V,}
C
L 2 (JR.) defined for n
··· c together with a function
E
V~1
E
Z with
C Vo
c
V,
c ···
L 2 (JR.) such that
(i) U~~oo V, is dense in L 2 (1R.), n~~= V, (ii) f E V, if and only iff(2~"-) E Vo
=
{0}
(iii) {(x- k)}kE:Z is an orthonorm al basis of Vo. is called the scaling function of the MRA. Clearly Vo is uniquely defined by through (iii), and V, is further uniquely determine d through (ii). However we do not require that be unique; a given family {Vn} may have several different possible choices of . The job of the theory is to show that there exist other nontrivial examples of multiresol ution analyses, to construct the correspon ding orthonorm al wavelet bases and to discuss their properties .
Example 6.4.3. Let V, be the set off intervals hn = [(k- 1)/2", k/2").
E L 2 (1R.),
which are constant on the dyadic
Clearly all of the properties are satisfied, with the Haar scaling function (x) 1ro.l) (x).
Example 6.4.4. Let V, be the setoff on each dyadic interval hn-
E L 2 (JR.), which are continuou s and linear
It is straightfo rward to see that properties (i) and (ii) of Definition 6.4.2 are satisfied. The choice of a scaling function is less obvious and will be obtained in this section. This example is related to piecewise linear spline approxima tion. In order to develop scaling functions for more general MRA systems, we first develop the necessary properties of orthonorm al systems and Riesz systems.
6.4. 1
Orthono rmal Systems and Riesz Systems
Let H be a Hilbert space with inner product(, ). A set of vectors (xn) is an orthononn al system, by definition , if (Xn, Xm) = Omn·
INTRODU CTION TO WAVELET S
Lemma 6.4.5. The set (x,) is orthono rmal comple x number s (a,), we have
305
if and only if for every finite set of
(6.4.1) Proof. If (x,) is orthono rmal, then the left side of (6.4.1) is the finite sum
Lama, (x",x,) = La,a, = L la,l 2 . tn,n
n
n
Convers ely, if (6.4.1) holds, first we choose a, = 8,N to obtain (xN, XN) = I. Then choosin g a,= 8,M- 8,N with M =1= M gives 2 llxN- xMII 2 2 - (xN,xM )- (xM,XN) hence 0 (XN, XM) (XM, XN). Replaci ng XM by iXM We Obtain 0 (XN, XM) - (XM, XN ), from
=
+
=
=
=
which the result follows.
•
This leads us to formula te a more general concep t.
Definit ion 6.4.6. Let H be a Hilbert space. A set of vectors (x,) is, by definiti on, a Riesz system, if there exist constan ts 0 < c < C < oo such that for any finite set of comple x number s (a,) (6.4.2)
c
L
la,l
2
:::
n
II
L:a,x,f < cL: ja,j n
2
.
n
Clearly any orthono rmal system is a Riesz system , where c = C = 1. If (xn) is a Riesz system , then the vectors (xn) are linearly indepen dent: Ln anXn = 0, implies that an = 0 for all n. Examp le 6.4.7. Let H
=e
(ll~.) and Xn (t)
=
A (t- n) where A is the tent functio n
i\(t) = (1- itl)l[-I. I)(t).
To verify the Riesz propert y, we note that the linear combin ation A(t) Ln ani\(t - n) is piecew ise linear with A(n) =a, for all n. Hence
Jm{ IA(t)i 2 dt = IR
ln+l i(n + 1 -
L nEZ
We use the Cauchy inequal ity i2abl < lai 2
i
2
IA(t)i dt:::
IR
+
t)a,
n
+
( t - n)an+ti 2 dt
ibi 2 to obtain the upper bound
~ L(lan l 2 + lan+tl 2 )
L
=
nEZ
ianl 2
nEZ
and the lower bound
1IA(t )l 2 dt
>
~ L(ian l 2 +
R
Therefo re (6.4.2) is satisfie d with c
lan+Ii
nEZ
=
~, C
=
1.
2
)
=
!_ 3
L nEZ
ja,j 2 .
306
INTRODUCTI ON TO FOL;RIER ANALYSIS AND WAVELET 1, with equality if and only if = 1K for some measurable set K with iKi = 1. Proof. From Corollary 6.4.9 we have Lta lei>(~ From Parseval's identity
+ /)1 2
-
I a.e., hence lci>(~)l < I a.e.
which proves that 1supp ci> I ~ l. If equality holds, then the middle terms give
But the integrand is nonnegative a.e., hence I -lci>(~)l 2 means that ci> = IK a.e., where IKI = lsupp 1 = l.
=
0 a.e. on the support of ci>, which •
Proposition 6.4.8 allows us to obtain the following orthogonalizatio n procedure to generate scaling functions from a Riesz sequence. Proposition 6.4.15. Let EL 2 (JR) be such that {(t - m)}mEZ is a Riesz sequence. Then there exist complex numbers bn with LnEZ ibn 12 < oo such that {r (t- m)}mEZ is an orthonormal sequence, where 1 (t) := LnE:Z bn(t- n). Furthermore, the span of {r ( t - n)}nEZ equals the span of {(t- n)}nEZ· Proof. From Proposition 6.4.8, it suffices to find bn such that LtEZ 1 ci> 1 (~ From the definition of 1 , we have
+ I) 2 = 1
1 a.e.
INTRODUC TION TO WAVELETS
:= L lcPI (~ +
309
B (!;) cP (!;)
/)1 2 =LIB(~+ 01 2 lcP(~
feZ.
+ 01
2
IE:...
=
IB(l;)l 2
L
lcP(~
+
1)1
2
.
IEZ
Therefore we must choose the constants bn so that
1
IB(~)I2:=1I;:bne-hm 0, Ke-fllnl and obtain an estimate for {3.
6.4.2
f3
>
0 so that Ibn I
k(t) 1 and let P1 be the orthogonal projection on the space V,. In detail
PJ =
=
2JI 2 ct>(21t- k)
L (/' ct>jk) ct>jk. kEZ
These projection operators satisfy the bounds IIP1 /II ::=:: 11/11. To prove the MRA property, it suffices to show that limj~oo P1 / = f and lim1 ~-oc P1 f = 0 for allf E L 2 (~). This is done in two separate lemmas.
314
INTRODUCTION TO FOURIER ANALYSIS AND\,, ' :' : ·.,.,
Proof. Since IIP1 11 = 1, it suffices to prove the result on a dense set, e.g., L 2 functions with compact support. Iff has support in [ -R, R], then
fu 2 (f: = 11/112 fu f_~::J2~R
=
11111:
1
14>(2
1
s- k)l ds) 2
j 0, there exists g whose Fourier transform is bounded and supported in ( -R, R] for some R > 0 and so that II!- gil < E. Hence IIP1 gll = IIPJ(g- f) II < E. Applying Lemma 6.4.29 yields the estimate E
2
>
IIPJgll 2 = /_: lg(l;)l 2 1(2-1 01 2
~ l(O)I 2 11gll 2 2: l(O)I\11!1 1- £) 2 •
This holds for every
E
> 0, which is a contradictio n if E is sufficiently small.
• The continuity condition (iii) in Theorem 6.4.27 can be weakened to lim l(2-1 l;)l = 1 }---"> 00
a.e.
.;
E JR.,
and this condition is also necessary. Indeed, the sufficiency is apparent from application of Lemma 6.4.29 to f, whose Fourier transform is bounded with compact support. To see the necessity, we anticipate a result from the next section, that lmo(l;)l ::::: 1 a.e. From this it follows that
In addition 1 = LiE:Z l(l; + 1)1 2 > l(l;)l 2 , so that we have the existence of the limit g(l;) = 1im1 -">oo 1(2-1 !;)1 and g(l;) < 1. Applying Lemma 6.4.29 toj with]= lr-I.IJ. we see that
where we have applied the Lebesgue dominated convergence theorem, thanks to the bound lg(l;)/ < 1. Hence f~ 1 (1 - /g(l;)l) dl; = 0 where the integrand is nonnegative , hence g(l;) = 1 a.e. • If the scaling function E L 1 (JR.) n L 2 (JR.), then the conditions of Theorem 6.4.27 are necessary as well as sufficient. To see this, note that E L 1 (JR.) implies that is continuous, especially at t; = 0. If generates an MRA, then we can apply Lemma 6.4.29 to f =1= 0 whose
INTRODUCTION TO WAVELETS
317
Fourier transform is bounded with compact support. Taking j ~- oo in (6.4.17), we obtain 11/11 = lcP(O)IIIfll. hence lcP(O)I = 1, as promised. • If the scaling function 1:1> E L 1 (ffi.) n L 2 (ffi.), then cP (l) = 0 for 0 =I= l E Z and LkEZ t:l>(t - k) = cP(O) a.e. Indeed, from the orthonormali ty relation, we have a.e. 1 > IcP (~) 12 + IcP (~ + l) l2 . But cP is continuous, hence we can take ~ ~ 0 avoiding the exceptional set to obtam lcP(l)l < 0, which was to be proved. From this it also follows from the Poisson summation formula that the periodized scaling function LkEZ t:l>(t- k) = cP(O) a.e., since its Fourier coefficients are all zero except for one term. • If the scaling function has compact support with JJ(x) dx = 1, then the Fourier transform argument of Lemma 6.4.29 can be avoided. This is formulated as follows.
Proposition 6.4.33. Suppose that 11> is the scaling function for a compact MRA with t:l>(x) dx = 1. Then u}EZ vi is dense in L 2 (ffi.).
JR
Proof. The orthogonal projection P1 onto 1-j is given by PJ = 2 1
L
(2 1 x - y) (
yEZ
f
J'R
f(y)(2 1 y - y)dy)
and satisfies (6.4.18) Therefore, to show that P1 f --+ f in L 2 , it suffices to prove that (6.4.19)
IIP;/11
--+
llfll-
Since the operators P1 have norm 1, it suffices to prove (6.4.19) on a dense set, e.g., linear combinations off= /A, where A= [a, b]. If supp E [ -M, M]
= 2-;
L y(l;) is of class Ck, then itkU>(t)i < Ckt for all real t and I E z+. In particular if.; ---+ ../8 (/;) is of class C 00 , then E S. Proof. We first check the conditions of Theorem 6.4.27. Since e is supported in an interval of length 2, the sum Lta I¢(~ + I) 12 = LtE:: 8(~ + /) consists of at most two nonzero terms, of the form 8
+
8=
ec-~>
+eo+ n
=
I,
which proves the orthonormality of {(t- m)LnEZ· To prove the scaling equation, define
mo(~) = /8(2~) m 0 (~ + 1) = m 0 (~), especially =I= 0, so that the equation ¢(2~) =
and extend m 0 to the real line as a !-periodic function: moE L 2 (~/Z). In addition 8(~) = 1 whenever 8(2~)
mo(~)cP(~) holds for J~l _:: : ~- In addjtion m 0 (~) is zero for ~ _:::: ~ _::::j, so tha~ we have the scaling equation (2~) = m 0 (~)(~) for all~ E ~. since both (~) and (2~) are zero when I~ I > ~, while m 0 is zero when < I~ I < ~. Finally, the condition (6.4.23) guarantees that is continuous at~ = 0 with_ I¢(0) I = 1. Hence by Theorem 6.4.27 there exists an MRA corresponding to . Since¢ has compact support, is of class c= and we have the bounds I1
2
= 1,
IEZ
(6.4.28) (6.4.29)
Vk
E
Z,
(6.4.30)
The !-periodic function mr is called the wavelet filter. It allows us to pass directly from the scaling function to the wavelet via (6.4.25). It remains to periodize these relations. At the same time we formulate the periodized version of (6.4.15) from the previous discussion. The idea is that the scaling equations allow us to rewrite the orthogonality relations as identities on the circle IR/Z in terms of the scaling filter and wavelet filter.
Proposition 6.4.36. Suppose that is a scaling function of an MRA with scaling .filtermo defined by (6.4.16). Then m 0 (0 satisfies (6.4.31) a.e.
320
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
( i)
if \11 is an orthonormal wavelet with respect to . then the 1 -periodic functions m 0 (~). m 1 (~)satisfy the relations (6.4.32)-(6.4.33 ), a.e. lmo(01 2 + Jmo (.;
(6.4.31)
+ ~) J2
= 1,
lm,(~)l 2 + Jm, (~ + ~)J = mo(~)m, (~) + mo (~ + ~) m, (~ + 2
(6.4.32) (6.4.33)
1,
n
= 0.
(ii) Conversely, given m 1 E L 2 ("'1R./Z) satisfying (6.4.32) and (6.4.33), if we define \II by (6.4.28), then {\11(2lt- k)}kE;z is an orthonormal system in Vt e Vo.
The equations (6.4.31) and (6.4.32) suggest the term quadrature mirror filter for the functions m 0 (s), m 1 (~).since the point~+~ is the mirror image of~ in the circle of unit circumference, with respect to which the quadratic functional equations are satisfied. Proof. We apply Corollary 6.4.9 to formula (6.4.15) summing separately over the odd and even IE Z.
=
L:: 1
(~)
. ~ . = ze ml; sm :rr~.
Returning to the theory, we now solve for the wavelet \II by means of the function
m 1 • We give the first row of the matrix M and must find the second row. The orthogonality condition (6.4.33) requires that we have (mt (~). mt (~
~)) =a(~)
+
(mo
(~+~),-rna(~))
for some 1-periodic complex-valued function a(~). The normalizations (6.4.31) and (6.4.32) further require that Ia(~) I = 1. Finally, the substitution~ --+ ~ + ~ shows that a must satisfy the half period condition a(~ + ~) = -a(~). Thus we find the general solution (6.4.35)
m1
(t - k)dt = 0 for all k E Z. In terms of Fourier transforms, we have
'Vk
E
Z.
Now we can apply the same computations as those following (6.4.34) to conclude that
mo(~)C(~) + mo (~ +
4) C (~ + 4) =
0.
But we have already seen from the proof of Proposition 6.4.36 that the general solution of this equation is obtained as (6.4.35). Finally, this can be written as a multiple of (~ /2) defined in (6.4.38 ), expressed as
/ [ -1, 1]d is a cube centered at 0 with 0 , I\IJ I < ~I \IJ I > 0. Then
J
..JN _:: : 2v'12QT/I JR.d
then
Jfii.d
J'Rd
\llf. In particular, if {\IJ1d;EZ•.kE:Zd is an orthonormal set,
\IJ = 0.
Proof. Let fJFid \II = Re'H with R > 0. Replacing \II by e -IIJ \II, we preserve the orthonormality of \111k while achieving fJRd \II = R > 0. Now let Q ::;, [- 1, I ]d be a cube centered at 0 so that 1f0 , \II I < R/2. Then for any set A with Q c A, we have I fA' \II I < R/2 and IRe fA \II - Re fJFid \II I .:::: R/2, so that Re fA \II :::: R - R/2 = R/2. Now let (6.4.40)
'YJ := L\11(2 1 x-k) ==> 1111 11~ = kES1
L:r
1
d
= l.
kES1
On the other hand,
But if Q = [ -M, M]" and x Q C 2 1 + 1 Q- k fork E S, and x
E E
(6.4.40
Q and j :::: 0, then l(x, Q. This means that Re
f
]2Q
11
::::
+ k,)j2i+ 1 I
J (0 =
o(.L 2J ) mo (£) 2 ···mo(.L 2J ).
We claim that 1 EM for all}~ 0. To see this, we will use mathem atical inductio n to prove that { (t- k) ha is ortho1 normal. For j = 0 we have L: .,z lo(~ + /)1 2 = 1, so that o(t- khez is 1 an orthono rmal set, i.e., o E M. Assumi ng that J--l EM we write J(~) = m 0 (~ /2) _, (~ /2) and compute 1
.L;: 11
