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0, then g(x) = ljq. Investigate the differentiability of 9 at the number ex as above. 2. (a) Show that the function f(x) = Ixl is not differenliable at O. (b) Show Ihat the function g(x) = xix I is differentiable for all x. What is its derivative? 3. For a positive integer k, let flk) denote the k-th derivative of ao + a,x + ... + a.x' be a polynomial. Show that for all k, plkl(O)
f. Let P(x) =
= k! ak •
4. By induction, obtain a formula for the n-th derivative of a product, i.e. (fg),nl , in terms of lower derivatives /(Ii.). g(j)
70
DIFFERENTIATION
[III, §2]
III, §2. MEAN VALUE THEOREM Lemma 2.1. Let f be differentiable on the open interval a < x < b and let c be a number such that f(c) is a maximum, that is ftc) ~f(x)
for
a < x < b.
Then f'(c) = O. Proof For small h we have ftc
Ifh >
+ h) < ftc).
o then ftc
+ h)
- ftc) ~ h -
o.
If h < 0 then the Newton quotient is > O. By the theorem on limits of inequalities (Theorem 2.6 of Chapter II) we conclude that f' (c) = 0 as desired.
a
b
The conclusion of the lemma obviously holds if instead of a maximum we assume that ftc) is a minimum.
Lemma 2.2. Let [a, b] be an interval with a < b. Let f be continuous on [a, b] and differentiable on the open interval a < x < b. Assume f(a) = f(b). Then there exists c such that a < c < band f'(c) = o. Proof Suppose f is constant on the interval. Then any point c strictly between a and b wiu satisfy our requirements. If f is not constant, then suppose there exists some x E [a, b] such that f(x) > f(a). By a theorem on continuous functions, there exists c E [a, b] such that f(c) is a maximum value of f on the interval, and a < c < b. Then Lemma 21 concludes the proof. In case there exists XE [a, b] such that f(x) for a < x < b. Then f is strictly increasing on the interval [a, b].
°
Proof Let a ;;; x < y < b. By the mean value theorem, flY) - f(x) = j'(c)(y - x) for some c between x and y. Since y - x > 0, we conclude that f is strictly increasing. An analogous corollary holds for the three other cases when j'(x) < 0, j'(x) > 0, j'(x) ;;; 0 on the interval, in which cases the function is strictly decreasing, increasing, and decreasing respectively. Note especially the important special case: Corollary 2.5. Let f be continuous on Ell, b] and differentiable on a < x < b. Assume j'(x) = 0 for a < x < b. Then f is constant on the interval.
72
[III, §2]
DIFFERENTIA TION
Proof. Again, for a < x < b there exists a number c between a and x such that I(x) - I(a) = f'(c)(x - a) = O. Hence I(x) = I(a), and 1 is constant. The sign of the first derivative has been interpreted in terms of a geometric property of the function, whether it is increasing or decreasing. We shall now interpret the sign of the second derivative. Let 1 be a function defined on a closed interval [a, b]. The equation of the line passing through (a,f(a») and (b,f(b») is
y = I(a)
+ I(b) -
I(a) (x - a). b-a
The condition that every point on the curve y = I(x) lie below this line segment between x = a and x = b is that
I(x) < I(a)
(*)
for a
~
x
~
=
+ I(bi ~(a) (x -
b. Any point x between a and b can be written in the form x = a
with 0
a)
~ t :::
+ t(b -
a)
1. In fact, one sees that the map tf->
a
+ t(b - a)
is a strictly increasing map on [0, 1], which gives a bijection between the interval [0, 1] and the interval [a, b]. If we substitute the value for x in terms of t in our inequality (*), we find the equivalent condition (**)
1(1 - t)a + tb) < (l - t)/(a) + tf(b).
Suppose that 1 of points a < b in convex upward on < holds for 0 < t convex downward ~ and >.
is defined over some interval I, and that for every pair I the inequality (**) is satisfied. We then say that 1 is the interval. If the inequality (**) with < replaced by < I, we say that 1 is strictly convex upward. We define and strictly convex downward by using the signs
Theorem 2.6. Let 1 be continuous on [a, b]. Assume that the second derivative I" exists on the open interval a < x < b and that I"(x) > 0 on this interval. Then 1 is strictly convex upward on the interval [a, b].
[III, §2]
MEAN VALUE THEOREM
73
Proof If a ;;; c < d ;;; b, then the hypotheses of the theorem are satisfied for f viewed as a function on [c, d]. Hence it will suffice to prove (*) with ::: replaced by < for a < x < b. Let a < x < b and let g(x) = f(a)
=
+ f(bi ~(a) (x -
a) - f(x).
Then, using the mean value theorem on f, we get g'(x) = f'(c) - j'(x)
for some c with a < c < b. Using the mean value theorem on 1', we find g'(x) = j"(d)(c - x)
for some d between c and x. If a < x < c, then by Corollary 24 and the fact that j"(d) > 0 we conclude that 9 is strictly increasing on [a, c]. Similarly, if c < x < b, we conclude that 9 is strictly decreasing on [c, b]. Since g(a) = 0 and g(b) = 0, it follows that g(x) > 0 when a < x < b, and thus our theorem is proved. The theorem has the usual formulation when we assume that j"(x) ~O, < 0, ;5; 0 on the open interval. In these cases, the function is convex upward, strictly convex downward, convex downward, respectively. Let I be an interval, say a closed interval [a, b] with a < b. Let f be a function on [a, b]. We know the definition of differentiability on the open interval (a, b). We define f to be differentiable at a (or right differentiable at a) in the usual way, but taking h > O. Similarly, we define f to be (left) differentiable at b by taking h < O. Let p be an integer ;;:; O. If f is p-times differentiable, and if its first p derivatives are continuous, then we define f to be of class CPo It is clear that the CP functions on the given interval form a vector space. A similar definition can of course be made on an open interval, or on an interval which is half open. We also say that a function is COO if it is CP for l::very positive integer p. Thus a C" function is one all of whose derivatives exist (they are then automatically continuous I).
III, §2. EXERCISES 1. Let J(x) = a.x' + ... + ao be a polynomial with a. #0 O. Let c, < c, < ... < c, be numbers such that J(c,) = 0 for i = 1, ... ,r. Show that r ~ n. [Hint: Show that J' has at least r - 1 roots, continue to take the derivatives, and use induction.]
74
[III, §3]
DIFFERENTIATION
2. Let f be a function which is twice differentiable. Let c, < c, < ... < C, be numbers such that f(c,) = 0 for all i. Show that j' has at least r - I zeros (i.e. numbers b such that j'(b) = 0). 3. Let a" ... ,a. be numbers. Determine x so that
•
L: (a, -
x)' is a minimum.
i=J
4. Let j(x) = x' + ax' + bx + c, where a, b, c are numbers. Show that there is a number d such that f is convex downward if x ;;;; d and convex upward if x ~ d. 5. A function f on an interval is said to satisfy a Lipschitz condition with Upschitz constant C if for all x, y in the interval, we have If(x) - f(y)1 ;;;;
C1x - yl·
Prove that a function whose derivative is bounded on an interval is Lipschitz. In particular, a C' function on a closed interval is Lipschitz. Also note that a Lipschitz function is uniformly continuous. However, the converse is not necessarily true. See Exercise 5 of Chapter IV, §3. 6. Let f be a C' function on an open interval, but such that its derivative is oot bounded. Prove that f is not Lipschitz. Give an example of such a function. 7. Let f, g be functions defined on an interval [a, b], continuous on this ioterval, differentiable on a < x < b. Assume that f(a);;;; g(a), and j'(x) < g'(x) 00 a <x < b. Show that fIx) < g(x) if a < x ;;;; b.
III, §3. INVERSE FUNCTIONS
f be a function on [a, b], and assume that f is strictly increasing. Assume that f is continuous. We know from the intermediate value theorem that the image of f is an interval [IX, PJ. Furthermore, given IX ::: y < p, Lei
suppose that fIx) = y, and a ;;;; x then f(xt) and let (j = f(Xl) - f(c). The same argument works. If c = b, we let (j = f(c) - f(xl). This proves the theorem. Theorem 3.2. Let f be continuous on the interval [a, b] and assume a < b. Assume that f is differentiable on the open interval a < X < b, and that f'(x) > 0 on this interval. Then the inverse function g of f, defined on [lX, P], is differentiable on the intervallX < y < p, and , 1 1 g (y) = f'(x) = f'(g(y»" Proof Let lX < Yo
" X - Xo
76
[III, §3]
DIFFERENTIAnON
Since 9 is continuous, as y -+ Yo we know that x lows by taking the limit as x -+ Xo.
-+
Xo. The theorem fol-
Example. Let y = f(x) = x· for some positive integer n. Then j'(x) = nx·- I > 0
for all x > 0, whence f is strictly increasing. Its inverse function is the n-th root function. Since x· has arbitrarily large values when x becomes large, it follows by the intermediate value theorem that the inverse function is defined for all numbers > O. Let g(y) = denote the inverse function. Using Theorem 3.2, we verify at once that
y".
g'(y) =
~ yl,.-l. n
When studying the exponential and the logarithm, we shall give another proof that n-th roots of positive numbers exist. A function which is either increasing or decreasing is said to be monotone. If it is either strictly increasing Or strictly decreasing, it is said to be strictly monotone. For simplicity, Theorems 3.1 and 3.2 have been stated for increasing functions. Obviously their analogues hold for decreasing functions, and the proofs are the same, mutatis mutandis. We shall now systematize the notion of inverse for various kinds of maps. We have already met various categories of mappings, starting with just plain maps between sets, then continuous maps for functions defined on subsets of R, differentiable functions on intervals, Coo functions (i.e. infinitely differentiable functions). Later we shall deal with linear maps between vector spaces, or continuous linear maps. It has been found very valuable to use a certain terminology applicable to all these situations. Suppose we are given a category of mappings such as those listed above. Let f: S -+ T be a map in the given category. We call f an isomorphism in this category if f has an inverse g: T -+ S in the category. In other words, 9 0 f = ids, fog = idT> and 9 is in the same category as f Let S, T above be sets. The map f is a set-isomorphism if and only if f is bijective. Let S, T be subsets of R. The map f: S -+ T is a CO-isomorphism if and only if there exists a continuous function g: T -+ S such that fog = id T and go f = ids, i.e. f has a continuous inverse. Let S, T be open intervals in R, and let f: S -+ T be a Co function, i.e. p-times continuously differentiable function. Then f is a CO-isomorphism if and only if f has a CO-inverse g: T -+ S. Again if S, T are open intervals and f: S -+ T is differentiable, then f
[III. §3]
INVERSE FUNCTIONS
77
is a differentiable isomorphism if and only I has a differentiable inverse. Theorem 3.2 gives a criterion for I to be a differentiable isomorphism. Warning. A function I may be an isomorphism in one category but not in another. For example. consider the function I: R -+ R such that I(x) = Xl. Then I is a CO-isomorphism. whose continuous inverse function is given by g(x) = x 'fJ . However. I is not a differentiable isomorphism because its continuous inverse is not differentiable at O. Of course. I is bijective. i.e. I has a set-theoretic inverse. In the theory of vector spaces. it is proved that if L: E -+ F is a linear map between vector spaces. and L is bijective (i.e. there exists a map G: F -+ E such that Go L = idE and LoG = id F ). then G is linear. and hence L is a linear isomorphism. As we have done above, the category to which an inverse belongs is referred to by a prefix. as when we say set-isomorphism. or Co. isomorphism. or COO-isomorphism. or linear isomorphism in the case of linear maps between vector spaces.
III, §3. EXERCISES For each one of the following functions f restrict f to an interval so that the inverse function 9 is defined in an interval containing the indicated point. and find the derivative of the inverse function at that point. 1. J(x) = Xl
+ I; find g' (2).
2 J(x) = x 2
-
X
+ 5; find 9' (7).
3. f(x) = x· - 3x 2
+
I; find g' (-I).
+ 2x + I; find g' (2). + I; find g' (21).
4. f(x) = -x' 5. J{x)
= 2x'
6. Let J be a continuous function on the interval [a, b]. Assume that J is twice differentiable on the open interval a < x < b. and that f'{x) > 0 and j"{x) > 0 on this interval. Let 9 be the inverse function of f. (a) Find an expression for the second derivative of g. (b) Show that g"(y) < 0 on its interval of definition. Thus 9 is convex in the opposite direction to f. 7. In Theorem 3.2. prove that if J is of class CP with p ~ I, then its inverse function 9 is also of class CPo
CHAPTER IV
Elementary Functions
IV, §1. EXPONENTIAL We assume that there is a function f defined for all numbers such that f' = f and flO) = 1. The existence will be proved in Chapter IX, §7, by using a power series. We note that f(x) # 0 for all x. Indeed, differentiating the function f(x)f( - x) we find O. Hence there is a number c such that for all x,
f(x)f( -x) = c. Letting x
= 0 shows that c = 1. Thus for all x, f(x)f( -x) = 1.
In particular,f(x) # 0 and f( -x)
= f(X)-I. We can now prove the uniqueness of the function f satisfying the conditions f' = f and flO) = 1. Suppose 9 is any function such that g' = g. Differentiating gjf we find O. Hence gjf = K for some constant K, and thus 9 = Kf· If g(0) = 1, then g(O) = Kf(O) so that K = 1 and 9 = f. Since f(x) # 0 for all x, we see that f'(x) # 0 for all x. By the intermediate value theorem, it follows that f'(x) > 0 for all x and hence f is strictly increasing. Since f" = f' = f, the function is also strictly convex upward. We contend that for all x, y' we have
f(x + y) = f(x)f(y).
78
[IV, §I]
EXPONENTIAL
79
Fix a number a, and consider the function g(x) = f(a + x). Then g'(x) = r(a + x) = f(a + x) = g(x), so that g(x) = Kf(x) for some constant K. Letting x = 0 shows that K = g(O) = f(a). Hence f(a + x) = f(a)f(x) for all x, as contended. For every positive integer n we have f(na) = f(a)".
This is true when n = I, and assuming it for n, we have f«n
+ l)a) = f(na + a) = f(na)f(a) = f(arJ(a) = f(ar l ,
thus proving our assertion by induction. We define e = f(1). Then fen) = e" for any positive integer n. Since f is strictly increasing and f(O) = I, we note that 1 < e. Also, f( -n) = e-". In view of the fact that the values of f on positive and negative integers coincides with the ordinary exponentiation, from now on we write f(x) = e".
The addition formula then reads e"+' = eXe', and eO = 1. Since e > 1, it follows that e" -> 00 as n --+ 00. We already proved this in Chapter I, and it was easy: Write e = 1 + b with b > 0, so that
e" = (1 + b)"
~
I
+ nb.
The assertion is then obvious. Since eX is strictly increasing, it follows that x 00 as x -> 00. Finally, e- --+ 0 as x --+ 00. Hence the graph of e" looks like this:
e" ->
'f(x) -e'
Theorem 1.1. For every positive integer m we have . e" hm m= lI:-
coX
00.
80
[IV, §I]
ELEMENTARY FUNCTIONS
Proof Since e ~ I, this is a special case of Theorem 3.2 of Chapter II, whenever x is an integer n. Let f(x) = ex/x m • Then for x sufficiently large, f(x) is strictly increasing, because j'(x) > 0 if x> 111, as you verify at once by a direct computation. Since f is increasing for x > I and f(n) ..... 00 when n is an integer ..... 00, it follows that f(x) -+ 00 as x ..... 00, thus concluding the proof.
Remark 1. Exercise 2(a) will provide another proof, which you may find easier to remember. This other proof fits well with the fact, to be proved later, that eX is equal to the infinite series eX
x2
Xl
x"
2!
3!
n!
= I + x + - + - + ... + - + ...
that is
x'
n
eX = lim n .... (l()
L -kl. ' k=O
In particular, from this series expression, one gets n x' "L,-,=e. < X
1=0
n.
The direct argument of Exercise 2(a) shows how to get this inequality independently of showing that eX is equal to the series. Take your pick of all the possible proofs we are giving. Each one illustrates a different aspect of the exponential function, and some use less theory than others. You decide what you prefer. Remark 2. After you have the logarithm, one can also give another proof, based only on the fact that eX/x -+ 00 as x ..... 00. Namely, suppose we want to prove that ex/x m ..... 00 as x ..... 00. It suffices to prove that log(ex/x m ) -+ co. But log(ex/x m )
=x
- m log x
x = --(log x -
log x
m).
Putting x = e Y so log x = y, we see that x/log x ..... co as x ..... co because eY/y ..... 00 as y ..... co. Furthermore m is fixed, so log x - m-+
00
as
x~
00.
[IV, §1]
EXPONENTIAL
81
Hence log(ex/x rn ) -+ 00 as x -+ 00. as desired. Remember this technique of taking the logarithm, which can be used in other contexts. Bump functions. The exponential function can be used to construct bump functions. By this we mean a function 9 whose graph has the following shape:
a
b
Thus the function is 0 outside an interval [a, b], and g(x) > 0 if a < x < b. Furthermore. we require that 9 is COl, so it requires a bit of an argument to show the existence of such a function. Define f(x) =
Then
f
is also
fOl g(t) dt.
COl and the graph of f
looks as follows:
graph off
a
b
Thus f(x) = 0 if x < a. and between a and b. the function f climbs from 0 to a fixed number. This fixed number is actually the area under the bump. Multiplying f by some constant, one can obtain a function with a similar graph, but climbing from 0 to 1. You should now do Exercise 6 to write down all the details of the construction of the function g.
IV, §1. EXERCISES 1. Let f be a differentiable function such that f'(x) = - 2xf(x). Show that there is some constant C such that f(x)
= Ce- x '.
82
[IV, §1]
ELEMENTARY FUNCTIONS
2. (a) Prove by induction that for any positive integer n, and x i1; 0,
x2 it' x . x+-+···+-';;e l + n., 2 '. [Hint: Let J(x) = 1 + x + ... (b) Prove that for x i1; 0,
+ x'ln!
and g(x)
e- x > I - x =
x2
= eX.] x3
+ -2! - -. 3!
(c) Show that 27 < e < 3.
3. Sketch the graph of the following functions: (a) x" (b) xe- X 2 (c) x . . (d) x 2e- x 4. Sketch the graph of the following functions: (a) el/ x (b) e -I/x
:::; 0 and J(x) = e- I / x if x > O. Show that J is infinitely differentiable at 0, and that all its derivatives at 0 are equal to O. [Hint: Use induction to show that thc n-th derivative of J for x > 0 is of type P,(llx)e- I / x where P, is a polynomial.] (b) Sketch the graph of f.
5. (a) Let J be the function such that J(x)
= 0 if x
6. (a) Bump functions. Let a, b be numbers, a < b. Let
J
be the function such
that J(x) = 0 if x :::; a or x i1; b, and (a) J(x)
= e- 1/(x-0)(b-x)
if a < x < b. Sketch both a and b. (b) We assume you exists a em function strictly increasing on (c) Let li > 0 be so function 9 such that:
or
(b)
J(x) =
e-I/(x-o)e-I/(b-x)
the graph of f. Show that J is infinitely differentiable at know about the elementary integral. Show that there F such that F(x) = 0 if x :::; a, F(x) = 1 if x ~ b, and F is [a, b]. small that a + li < b - li. Show that there exists a em
g(x) = 0 if x :::; a and g(x) = 0 if x i1; b. g(x) = I on [a + li, b - li]. 9 is strictly increasing on [a, a + li] and strictly decreasing on [b -li, b]. Sketch the graphs of F and g. 7. Let J(x) = e- 1/ x ' if x "" 0 and J(O) = O. Show that J is infinitely differentiable and that P"(O) = 0 for all n. After you learn the terminology of Taylor's formula, you will see that the function provides an example of a em function which is not identically 0 but all its Taylor polynomials are identically O.
[IV, §2]
83
LOGARITHM
8. Let n be an integer
~ 1.
f.(x)e'"
Let 10' ... ,f. be polynomials such that
+ In_,(x)e(n-I)X + ... + lo(x) = 0
for arbitrarily large numbers x. Show that 10' ... .fn are identically O. [Hint; Divide by (f'x and let x -> 00.]
IV, §2. LOGARITHM The function f(x) = eX is strictly increasing for all x, and f(x) > 0 for all x. By Theorems 3.1 and 3.2 of the preceding chapter, its inverse function 9 exists, is defined for all numbers > 0 because f takes on all values > 0,
and
Thus we have found a function 9 such that g'(y) = I/y for all y > O. Furthermore, g(l) = 0 because f(O) = 1. The function 9 is strictly increasing, and satisfies g(xy)
= g(x) + g(y)
for all x, y > O. Indeed, fix a > 0 and consider the function g(ax) - g(x). Differentiating shows that this function is a constant, and setting x = I shows that this constant is equal to g(a). Thus g(ax) = g(x) + g(a), thus proving our formula. Let a > O. We see by induction that g(a n )
= ng(a)
for all positive integers n. If a > I, then g(a n ) becomes arbitrarily large as n becomes large. Since 9 is strictly increasing, we conclude that g(x) .... co as x .... co. The function 9 is denoted by log, and thus the preceding formulas read log(an) = n loga
and
log(xy) = log x
For x > 0 we have
o=
log I = log x
+ log(x- I ),
+ log y.
84
ELEMENTARY FUNCTIONS
[IV, §2]
whence log X-I = -log x. It follows that when x -+ 00, log IJx -+ - 00, i.e. becomes arbitrarily large negatively. Finally, the second derivative of the log of y is -lJy 2 < 0, so that the log is convex downward. Its graph therefore looks like this:
g(x)=logx
If a > 0 and x is any number, we define {i' = e 0, then al,n is an n-th root of a, because
(al'nr
= al = a.
Theorem 2.1. Let x> 0 and let f(x) = f'(x) = ax·- I .
Xo
for some number a. Then
Proof. This is an immediate consequence of the definition
and the chain rule. We now determine some classical limits.
[IV, §2]
85
LOGARITHM
Theorem 2.2. Let k be a positive integer. Then · {log I1m ""'co
Proof. Let x
xt
0
.
X
= e', that is z = log x.
Then
(log x)'
x
ez'
=
As x becomes large, so does z, and we can apply Theorem 1.1 to prove Theorem 2.2. Corollary 2.3. We have lim xlIx = I. Proof. Taking the log, we have, as x
-> 00,
I
log x
x
x
log{x l/") = -log x = - - ->
o.
Taking the exponential yields what we want. Finally, since the log is differentiable at I, we see that lim log{l h_O
+ h) =
h
lim log{l
+ h) -
log I = I.
h
h-O
But 10g{l + h) = log({l h
+ h)1/h).
Taking the exponential, we obtain lim {I
+ h)llh =
e.
h-O
The same limit applies of course when we take the limit over the set of h = Ijn for positive integers n, so that (I + Ijn)" -> e as n -> 00. In the exercises, we shall indicate how to prove some other inequalities using the log. In particular, we shall give an estimate for n l, namely:
86
[IV, §2]
ELEMENTAR Y FUNCTIONS
Later in this book. this estimate is made more precise. and one has
where 0 ~ 0 ~ 1. This is harder to prove, and in many applications, the first estimate given suffices.
IV, §2. EXERCISES
lex) = 2. Let f be as I. Let
x" for x > O. Sketch the graph of f.
in Exercise I, except that we restrict f to the infinite interval x > I/e. Show that the inverse function 9 exists. Show that one can write logy g(y) = log log y ljJ(y),
where lim ljJ(y) = I. 3. Sketch the graph of: (a) x log x; (b) x 2 log x.
4. Sketch the graph of: (a) (log xl/x; (b) (log x)/x 2 •
5. Let < > O. Show: (a) lim (log xl/x' = 0; (b) lim x' log x = O. (c) Let n be a x-co
positive integer, and let < > O. Show that
. (log x)' I1m , x- co
x-o
O.
X
Roughly speaking, this says that arbitrarily large powers of log x grow slower than arbitrarily small power of x. 6. Let lex) = x log x for x > 0, x # O. and f(O) = O. (a) Is f continuous on [0. I]? Is f uniformly continuous on [O,IJ? (b) If f right differentiable at O? Prove all your assertions. 7. Let f(x) = x 2 log x for x > O. x # 0, and f(O) = O. Is f right differentiable at O? Prove your assertion. Investigate the differentiability of lex) = x· log x for an integer k > 0, i.e. how many right derivatives does this function have at O. 8. Let n be an integer;;; I. Let fo' ... J, be polynomials such that .f.(x)(log xl"
+ .f._.(x)(log xr I + ... + fo(x) = 0
for all numbers x > O. Show that fo,'"
J, are identically O.
[Hillt: Let x = e'
[IV, §2)
87
LOGARITHM
and rewrite the above relation in the form
where
alj
are numbers. Use Exercise 8 of the preceding section.]
9. (a> Let a > I and x > O. Show that ~
x' - I
a(x - J).
(b) Let p, q be numbers ~ I such that I/p
+ I/q =
I. If x ~ I, show that
x I + -. -p q
xl'P S -
10. (a) Letll, v be positive numbers, and let p, q be as in Exercise 9. Show that
(b) Let u, v be positive numbers, and 0 < U'V I -
1
~ ttl
+
and that equality holds if and only if II
t
< I. Show that
(1 - t)v,
= v.
II. Let a be a number> O. Find the minimum and maximum of the function fIx) = x 2 /a x • Sketch the graph of fIx).
12. Using the mean value theorem, find the limit lim (nil' - (n
+
1)"').
Generalize by replacing! by 11k for any integer k ~ 2.
13. Find the limit .
11m
(I
+ 1r)'13 It
h-O
14. Show that for x
~
0 we have log(1
+ x)
15. Prove the following inequalities for x (a) log(1
(b) x -
+ x) x2
x2
;;; x -
x3
2' + 3
2' ;;; log(l + x)
- I
~
;;; x.
0:
.
88
[IV, §2]
ELEMENTAR Y FUNCTIONS
(c) Derive further inequalities of the same type. (d) Prove that for 0 ~ x ~ I, log (I
+ x) =
X
lim x - - 2 "-00 ( 2
+IX
+ ... + (- I)" -"). II
16. Show that for every positive integer k one has
1+;; (1+;;I)' <e< (1)"1 Taking the product for k = 1,2, ... ,II - I, conclude by induction that II"
< .,.-.,.-,.., (II - I)!
and consequently
For another way to get this inequality, see Exercise 20. 17. Show that lim n-co
(1+-x)"
= ...
II
18. Let {a,,}, {b,,} be sequences of positive numbers. Define these sequences to be equivalent. and write an ::::;: bn for '1 -+ co to mean that there exists a sequence of positive numbers {u,,} such that b" = u"a" and lim u~/" = 1. Alternatively, this amounts to the property that lim(a,,/b,,)'/" = l. (a) Prove that the above relation is an equivalence relation for sequences. (b) Show that II!:; lI"e-" for Give a similar equivalence for (31l)!. (e) Show that if an E a~ and hn ;; b~, then a"bn == a~b~ for 11 -+ co.
11-+ 00.
fo:~~wing limi(ts as ;' .)..../';;: (a) (311)!) (b) ~ 3ll n
(2)1 1"
19. Find the
n311
n e-
(c)
(II!)
n2n
2n
(d)
(
)1/.
(;,)!
For the next exercises, which concern the logarithm, we assume you know elementary integration and upper-lower sums associated with the integral. Some of the proofs are easiest using such sums.
20. We shall give here an alternate proof for the estimate of Exercise 16. Write down upper and lower sums for the integral of log x over the interval [I, II] for each positive integer II. Use the partition of the interval at the integers k such that I ~ k ~ II. Using the inequalities lower sum
~
integral
~
upper sum,
[IV, §2]
89
LOGARITHM
give a proof of the inequality n log n - n
+ I ~ log (n!) ~ (n + I) log n -
n
+ I.
Exponentiating, you have a proof of the inequality
21. (a) Using an upper and lower sum, prove that for every positive integer n, we have
I (I) I + - O. Let a(a+ l)···(a+n)
a. = --'---'-;---";---'n! n¢ Show that {a.} is monotonically decreasing for sufficiently large values of n, and hence approaches a limit. This limit is denoted by l/r(a), where r is called the gamma function.
IV, §3. SINE AND COSINE We assume given two functions f and 9 satisfying the conditions I' = 9 and g' = - f. Furthermore, f(O) = 0 and g(O) = I. Existence will be proved in Chapter IX, §7, with power series. We have the standard relation
f(X)2
+ g(X)2
=
1
for all x. This is proved by differentiating the left-hand side. We obtain 0, whence the sum f 2 + g2 is constant. Letting x = 0 shows that this constant is equal to I. We shall now prove that a pair of functions as the above is uniquely determined. Let f .. g, be functions such that
1'.
and
= g.
Differentiating the functions fg, - fig and ff. case. Hence there exist numbers a, b such that
+ gg.,
we find 0 in each
fg, - fig = a,
ff. + gg,
=
b.
We multiply the first equation by f, the second by g, and add. We multiply the second equation by f, the first equation by g, and subtract. Using f2 + g2 = 1, we find
g, = af + bg, (*)
f.
= bf -
ago
If we assume in addition that f,(O) = 0 and g.(O) = 1, then we find the
[IV, §3]
91
SINE AND COSINE
values a = 0 and b = I. This proves that f, = f and 9, = 9, thus proving the desired uniqueness. These functions f and 9 are called the sine and cosine respectively, abbreviated sin and cos. We have the following formulas for all numbers x, y: (1) (2)
sine-x)
(3)
=
-sinx,
cost - x) = cos x,
(4)
sin(x + y)
= sin x cos y + cos x sin y,
(5)
cos(x + y)
=
cos x cos y - sin x sin y.
The first formula has already been proved. To prove each pair of the succeeding formulas, we make a suitable choice of functions f" 9, and apply equations (*) above. For instance, to prove (2) and (3) we let f,(x)
=
cost -x)
and
9,(X) =
sine -x).
Then we find numbers a, b as before so that (*) is satisfied. Taking the values of these functions at 0, we now find that b = 0 and a = - I. This proves (2) and (3). To prove (4) and (5), we let y be a fixed number, and let f,ex) = sin(x
+ y)
and
91(X)
= cos(x + y).
We determine the constants a, b as before and find a = - sin y, b = cos y. Formulas (4) and (5) then drop out. Since the functions sin and cos are differentiable, and since their derivatives are expressed in terms of sin and cos, it follows that sin and cos are infinitely differentiable. In particular, they are continuous. Since sin 2 x + cos 2 X = 1, it follows that the values of sin and cos lie between -1 and I. Of course, we do not yet know that sin and cos take on all such values. This will be proved later. Since the derivative of sin x at 0 is equal to I, and since this derivative is continuous, it follows that the derivative of sin x (which is cos x) is > 0 for all numbers x in some open interval containing O. Hence sin is strictly increasing in such an interval, and is strictly positive for all x > 0 in such an interval. We shall prove that there is a number x > 0 such that sin x = I. In view of the relation between sin and cos, this amounts to proving that there is a number x > 0 such that cos x = O.
92
[IV, §3)
ELEMENTARY FUNCTIONS
Suppose that no such number exists. clude that cos x cannot be negative for mediate value theorem). Hence sin is and cos is strictly decreasing for all x >
o < cos 2a = cos 2 a -
Since cos is continuous, we conany value of x > 0 (by the interstrictly increasing for all x > 0, O. Let a > O. Then sin 2 a < cos 2 a.
By induction, we see that cos(2"a) < (cos a)2. for all positive integers n. Hence cos(2"a) approaches 0 as n becomes large, because 0 < cos a < 1. Since cos is strictly decreasing for x > 0, it follows that cos x approaches 0 as x becomes large, and hence sin x approaches 1. In particular, there exists a number b > 0 such that cosb
and
. b >,. I sm
Then cos 2b = cos 2 b - sin 2 b < 0, contradicting our assumption that the cosine is never negative. The set of numbers x > 0 such that cos x = 0 (or equivalently sin x = 1) is non-empty, bounded from below. Let e be its greatest lower bound. By continuity, we must have cos e = O. Furthermore, e > O. We define n to be the number 2e. Thus e = n/2. By the definition of greatest lower bound, there is no number x such that
and such that cos x = 0 or sin x = 1. By the intermediate value theorem, it follows that for 0 < x < n/2 we have 0 ~ sin x < 1 and 0 < cos x ~ 1. However, by definition,
n cos- = 0 2
and
. n sm
I 2= .
Using the addition formula, we can now find sinn=O,
cos n = -1,
sin 2n = 0,
cos 2n = 1.
For instance,
. . (nn) .n n smn=sm -+=2sm-cos-=0 2 2 2 2 . The others are proved similarly.
[IV, §3]
93
SINE AND COSINE
For all x, using the addition formulas (4) and (5), we find at once: sin(x
+
i)
= cos x,
cos(x
sin(x + 1t) = -sin x, sin(x
+ 21t) =
sin x,
+
cos(x
i)
=
-sin x,
+ 1t) =
-cos x,
cos(x + 21t) = cos x.
The derivative of the sine is positive for 0 < x < 1t{2. Hence sin x is strictly increasing for O:S x 1t{2. Similarly, the cosine is strictly decreasing in this interval, and the values of the sine range from 0 to I, while the values of the cosine range from 1 to O. For the interval1t{2 ~ x < 1t, we use the relation
:s
sin x
= cos(x -
i)
and thus find that the sine is strictly decreasing from 1 to 0, while the cosine is strictly decreasing from 0 to -1 because its derivative is - sin x < 0 in this interval. From 1t to 21t, we use the relations sin x = -sin(x - 1t) and similarly for the cosine. Finally, the signs of the derivatives in each interval give us the convexity behavior and allow us to see that the graphs of sine and cosine look like this:
cos x
sin x
..
2. -I
-I
A function ffJ is called periodic, and a number s is called a period, if q:>(x + s) = q:>(x) for all x. We see that 21t is a period for sin and cos. U SI' S2 are periods, then
94
[IV, §3]
ELEMENTARY FUNCTIONS
so that s 1
+ S2 is a period. rp(x)
Furthermore, if s is a period, then
= rp(x -
s
+ s) = rp(x -
s),
so that - s is also a period. Since 2n is a period for sin and cos, it follows that 2nn is also a period for all integers n (positive or negative or zero). Let s be a period for the sine. Consider the set of integers m such that 2mn ~ s. Taking m sufficiently large negatively shows that this set is not empty. Furthermore it is bounded from above by s/2n. Let n be its maxi· mal element, so that 2nn < S but 2(n + I)n > S. Let t = S - 2nn. Then t is a period, and 0 ~ t < 2n. We must have sin(O
+ t) =
sin 0 = 0,
cos(O
+ t) =
cosO = 1.
From the known values of sin and cos between 0 and 2n we conclude that this is possible only if t = 0, and thus s = 2nn, as was to be shown. Theorem 3.1. Given a pair of numbers a, b such that a 2 + b 2 = I, there exists a unique number t such that 0 ~ t < 2n and such that a=cost,
b=sint.
Proof. We consider four different cases, according as a, bare :::: 0 or ~ O. In any case, both a and b are between -I and 1. Consider, for instance, the case where -I ~ a ::;; 0 and 0::;; b ~ 1. By the intermediate value theorem, there is exactly one value of t such that n/2 ::;; t ~ n and such that cos t = a. We have b 2 = 1 - a 2 = 1 - cos 2 t = sin 2 t.
Since for n/2 < t ::;; n the values of the sine are :::: 0, we see that b and sin t are both > O. Since their squares are equal, it follows that b = sin t, as desired. The other cases are proved similarly. Finally, we conclude this section with the same type of limit that we consider for the exponential and the logarithm. We contend that . sin h IIm--=I '-0 h .
This follows immediately from the definition of the derivative, because it is none other than the limit of the Newton quotient . sinh-sinO 11m h = sin'(O) = cos 0 = 1.
'-0
[IV, §4)
COMPLEX NUMBERS
95
IV, §3. EXERCISES I. Define tan x = sin x/cos x. Sketch the graph of tan x. Find tan h lim -,-.
h.... O
r
2. Restrict the sine function to the interval -1C/2 ~ x ~ 1C/2, on which it is continuous, and such that its derivative is > 0 on -1Cj2 < X < 1C/2. Define the inverse function, called the arcsine. Sketch the graph, and show that the derivative of arcsin x is 1/ JI - x 2• 3. Restrict the cosine function to the interval 0 ~ x ~ 1C. Show that the inverse function exists. It is called the arccosine. Sketch its graph, and show that the derivative of arccosine x is -I/JI - x 2 on 0 < x < 1C. 4. Restrict the tangent function to -1Cj2 < X < 1Cj2. Show that its inverse function exists. It is called the arctangent. Show that arctan is defined for all numbers, sketch its graph, and show that the derivative of arctan x is 1/(1 + x 2 ). 5. Sketch the graph of I(x) = x sin l/x, defined for x ~ O. (a) Show that I is continuous at 0 if we define 1(0) = O. Is I uniformly continuous on [0, I]? (b) Show that I is differentiable for x ~ 0, but not differentiable at x = O. (c) Show that I is not Lipschitz on [0, I]. 6. Let g(x) = x 2 sin I/x if x ~ 0 and g(O) = O. (a) Show that 9 is differentiable at 0, and is thus differentiable on the closed interval [0, I). (b) Show that 9 is Lipschitz on [0, I). (c) Show that g' is not continuous at 0, but is continuous for all x ~ O. Is g' bounded? Why? (d) Let g,(x) = x 2 sin (l/x2 ) for x ~ 0 and g,(O) = O. Show that g;(O) = 0 but g; is not bounded on (0, I). Is g, Lipschitz? 7. Show that if 0 < x < 1Cj2, then sin x < x and 2/1C < (sin xl/x. 8. Let 0 ~ x. (a) Show that sin x ~ x. (b) Show that cos x ;;; I - x 2 j2. (c) Show that sin x;;; x - x 3/3! (d) Give the general inequalities similar to the preceding ones, by induction.
IV, §4. COMPLEX NUMBERS The complex numbers are a set of objects which can be added and multiplied, the sum and product of two complex numbers being also complex numbers, and satisfying the following conditions: (1) Every real number is a complex number, and if ex,
P are real
numbers, then their sum and product as complex numbers are the same as their sum and product as real numbers.
96
[IV, §4]
ELEMENTARY FUNCTIONS
(2) There is a complex number denoted by i such that i Z = -1. (3) Every complex number can be written uniquely in the form a + bi, where a, b are real numbers. (4) The ordinary laws of arithmetic concerning addition and multiplication are satisfied. We list these laws: If a, 13, yare complex numbers, then
(a
+ 13) + y = a + (13 + y)
and
We have a(f3 + y) = af3 + ayand (13 + y)a We have af3 = f3a and a + 13 = 13 + a. If I is the real number one, then la = a. If 0 is the real number zero, then Oa = O. We have a + (-l)a = O.
(af3)y = a(f3y).
= f3a + ya.
We shall now draw consequences of these properties. If we write and then
If we call al the real part, or real component of ex, and az its imaginary part,
or imaginary component, then we see that addition is carried out componentwise. The real part and imaginary part of a are denoted by Re(a) and Im(a) respectively. We have
Let a = a + bi be a complex number with a, b real. We define IX = a - bi and call IX the complex conjugate, or simply conjugate, of a. Then
If a = a
+ bi is ,c 0, and if we let
then a.:l = Ace = I, as we see immediately. The number .:l above is called the inverse of 0< and is denoted by a-I, or l/a. We note that it is the only
[IV, §4]
97
COMPLEX NUMBERS
complex number z such that za = I, because if this equation is satisfied, we multiply it by Aon the right to find z = A. If a, Pare complex numbers, we often write Pia instead of a-Ip or pa-I. We see that we can divide by complex numbers "# O. We have the rules
ap = iiP,
a + P= ii + P,
ex
= a.
These follow at once from the definitions of addition and multiplication. We define the absolute value of a complex number a = a + bi to be lal =
Ja
2
+ b2 •
If we think of ex as a point in the plane (a, b), then lexl is the length of the line segment from the origin to ex. In terms of the absolute value, we can write
a
-1
ex
= lexl 2
provided a"# O. Indeed, we observe that Ia12 = aii. Note also that lexl = liil· The absolute value satisfies properties analogous to those satisfied by the absolute value of real numbers: lal > 0 and
= 0 if and only if ex = O.
lexPI = lailPI
la + PI ;;;; lexl + IPI· The first assertion is obvious. As to the second, we have lexpl'
= exPiiP = aiiPP = la1 2 1P1 2 •
Taking the square root, we conclude that lexilPI = laPI. Next, we have lex
because exP =
+ PI 2 = (ex + P}(ex + P) = (a + P}(ii + p> = aii + Pii + exP + PP 2 = lexl + 2Re(pii) + IW pii.
However, we have 2 Re(piij < 21PiXi
98
ELEMENTARY FUNCTIONS
because the real part of a complex number is
~
[IV, §4]
its absolute value. Hence
la + PI 2 ~ lal 2 + 21pal + IW 2 ~ lal + 21Pllal + IW = (Ial + IPW Taking the square root yields the final property. Let z = x + iy be a complex number #- O. Then zjlzl has absolute value 1. Let a + bi be a complex number of absolute value 1, so that a2 + b 2 = 1. We know that there is a unique real e such that 0 ~ e < 27t and a = cos e, b = sin If is any real number, we define
e. e
e'· =
cos e + i sin e.
Every complex number of absolute value 1 can be expressed in this form. If z is as above, and we let r = x 2 + y2, then
J
We call this the polar form of z, and we call (r, e) its polar coordinates. Thus
x=rcose
and
y = r sin e.
The justification for the notation ei• is contained in the next theorem. Theorem 4.1. Let
e, rp be real numbers.
Then
Proof. By definition, we have ei9+ i" = ei(" + "I = cos(e + rp) + i sin(e + rp). This is exactly the same expression as the one we obtain by multiplying out (cos e + i sin e)(cos rp
+ i sin rp)
using the addition theorem for sine and cosine. Our theorem is proved. We define e' = eXe" for any complex number z = x + iy. We obtain:
[IV, §4]
COMPLEX NUMBERS
99
Coronary 4.2. If a, Pare complex numbers, then
Proof· Let a
=
a.
+
iaz and
P= b 1 +
ib z . Then
Using the theorem, we see that this last expression is equal to
By definition, this is equal to eal!', thereby proving the corollary. Let S be a set. We denote the set of complex numbers by C. A map from S into C is called a complex valued function. For instance, the map
is a complex valued function, defined for all real O. Let F be a complex valued function defined on a set S. We can write F in the form F(x) = f(x)
+ ig(x),
where f, 9 are real valued functions on S. If F(O) = e;9, then f(O) = cos 0 and g(O) = sin O. We call f and 9 the real and imaginary parts of F respectively. If both the real and imaginary parts of F are continuous (resp. differentiable), we can say that F itself is continuous (resp. differentiable), whenever F is defined on a set of real numbers. Or we can give a definition using the complex absolute value in exactly the same way that we did for the real numbers. This will be discussed in detail in a more general context, in that of vector spaces and Euclidean n-space. For this section, we take the componentwise definition of differentiability. Thus we define F'(t) = f'(t)
+ ig'(t)
if F is differentiable on some interval of real numbers. We also write dF/dt instead of F'(t). Then the standard rules for the derivative hold:
100
ELEMENTARY FUNCTIONS
[IV, §4]
(a) Let F, G be complex valued functions defined on the same interval, and differentiable. Then F + G is differentiable, and (F
+ G)'
=
F'
+ G'.
If ex is a complex number, then
(exF)' = aF'. (b) Let F, G be as above. Then FG is differentiable, and (FG)'
= F'G + FG'.
(c) Let F, G be as above, and G(t) ,p 0 for all t. Then (FIG)'
= (GF' -
FG')/G z.
(d) Let qJ be a real valued differentiable function defined on some interval, and assume that the values of qJ are contained in the interval of definition of F. Then F 0 qJ is differentiable, and (F 0 qJ)'(t)
= F'(qJ(t»qJ'(t).
We shall leave the proofs as simple exercises.
IV, §4. EXERCISES I. Let a be a complex number #0 O. Show that there are two distinct complex num-
bers whose square is ex. 2 Let ex be complex, #0 O. Let tl be a positive integer. Show that there are exactly tl distinct complex numbers z such that z" = ex. Write these complex numbers in
polar form. 3. Let w be a complex number, and suppose that z is a complex number such that Ii' = w. Describe all complex numbers u such that t!' = w. 4. What are the complex numbers z such that Ii'
= I?
5. If 8 is real, show that and 6. Let F be a differentiable complex valued function defined on some interval. Show that
CHAPTER
V
The Elementary Real Integral
V, §1. CHARACTERIZATION OF THE INTEGRAL It is convenient to have the elementary integral available for examples, and exercises, and we need only know its properties, which can be conveniently summarized axiomatically. The proof of existence can be done either along the classical lines of Theorem 2.7, or as in Chapter X, which fits the larger perspective, applicable to more general integrals.
Theorem 1.1. Let a, d be two real numbers with a < d. Let f be a continuous function on [a, d]. Suppose that for each pair of numbers b ;:i; c in the interval we are able to associate a number denoted by I.(f) satisfying the following properties: (1) If M, m are numbers such thatm ;:i; f(x) ;:i; M fm' all x in the illlerval [b,c], then m(c - b) ;:i; lb(f) ;:i; M(c - b). (2) We have I~(f)
+ I.(f)
= I~(f).
Then the function x 1-+ I~(f) is differentiable in the illlerval [a, d], and its derivative is f(x). Proof. We have the Newton quotient, say for h > 0,
WV) h
I~(f)
-
I~(f)
+ I;+h(f) h
I~(f)
-
I;+h(f) h 101
102
[V, §1]
THE ELEMENTARY REAL INTEGRAL
Let s be a point between x and x + h such that f reaches a minimum at s on the interval [x, x + h], and let t be a point in this interval such that f reaches a maximum at t. Let m = f(s) and M = f(t). Then by the first property, f(sXx
+h-
x) ~ I~+h(f) ::; f(t)(x
+h-
x),
whence
Dividing by h shows that
As h -+ 0, we see that s, t -+ x, and since f is continuous, by the squeezing process, we conclude that the limit of the Newton quotient exists and is equal to f(x). If we take h < 0, then the argument proceeds entirely similarly. The Newton quotient is again squeezed between the maximum and the minimum values of f (there will be a double minus sign which makes it come out the same). We leave this part to the reader.
Corollary 1.2. An association as in Theorem 1.1 is uniquely determined. If F is any differentiable function on [a, d] such that F' = f, then I~(f) =
F(x) - F(a).
Proof· Both F and x 1-+ I~(f) have the same derivative, whence there is a constant C such that for all x we have F(x) =
Putting x
I~(f)
+ c.
= a shows that C = F(a) and concludes the proof.
For convenience, we define
whenever a ~ b. Then property (2) is easily seen to be valid for any position of a, b, c in an interval on which f is continuous.
[V, §1]
CHARACTERIZATION OF THE INTEGRAL
A function F on [a, b] (with a < b) such that F' = nite integral of f and is denoted by
f
103
is called an indefi-
f f(x)dx.
We use the usual notation. If c, d are any points on an interval [a, b] on which f is continuous, and if F is an indefinite integral for f, then d
f
d
f(x) dx = F(x)
!?; O. By Property (I), it follows that l:(q» !?; 0, whence the theorem follows by linearity.
Corollary 2.2. We have [f(X) dx < f!f(X)1 dx. Proof. Let g(x) = If(x) I in Theorem 2.1.
Corollary 2.3. Let M be a number such that I f(x) I ~ M for all x E [a, b]. Then for all c, d in the interval [a, b] we have
I
[f(X) dx < Mid - cl·
Proof. Clear if c < d, and also if d < c from the definitions.
Theorem 2.4. Let f be continuous on [a, b] with a < b and f!?; O. Assume that there is one point c E [a, b] such that f(c) > O. Then
Proof· Given f(c)f2, there exists b such that f(x) > f(c)f2 whenever x E [a, b] and Ix - cl < b. Suppose that c "# b. We take b small enough so that c + b < b. Then
as was to be proved. When c = b, we consider the interval [c - b, c] and proceed analogously.
[V, §2]
PROPERTIES OF THE INTEGRAL
105
Theorem 2.5. Let J I' J 2 be intervals each having more than one point, and let f: J, -+ J 2 and g: J 2 -+ R be continuous. Assume that f is differentiable, and that its derivative is continuous. Then for any a, bE J I we have
i
b g(f(x»)f'(x) dx =
f'(bl g(u) duo !(lJ)
G
Proof· Let G be an indefinite integral for g on J 2' Then by the chain rule, Go f is an indefinite integral for g(f(x»)f'(x) over J I' and our assertion folIows from the fact that both sides of the equation in the theorem are equal to G(f(b)} - G(f(a)}.
The next theorem is called integration by parts. Theorem 2.6. Let f, 9 be differentiable functions on an inter'Jal, and with continuous derivatives. Then
f
f(x)
~~ dx = f(x)g(x) -
f
g(x) ~ dx.
Proof. Differentiating the product fg makes this relation obvious.
For the definite integral, we have the analogous formula:
f
f(x)g'(x) dx = f(b)g(b) - f(a)g(a) -
f
g(x)f'{x) dx.
We end this section of basic properties with a discussion of Riemann sums, with which you should be acquainted from an earlier elementary course. Here, we indicate how certain properties which were probably left without proof can now be proved. By a partition P of [a, b] we mean a sequence of numbers denoted by (ao, . .. , an) such that 0, and consider an interval 0 < 8 ;;; x ;;; 2A - 8. Show that there exists a constant C, and for each positive integer n, there exists a polynomial Pn such that for all x in the interval, one has
Ilog(x) - p.(x)1 ;;; C/n. [Hint: Write x
= A + (x log x
A) so that
=
log A
+ log
X-A) .] (1 + -A-
We now suggest that you do Exercises 8, 9, 10 of Chapter VII, §3. These exercises have to do with approximating a function by polynomials. Suppose a function I is defined on an interval [a, b]. We say that I can be uniformly approximated by polynomials on [a, b] if given. there exists a polynomial P such that I/(x) - P(x)1 < •
for all x E [a, b].
The above-mentioned exercises will show you how to approximate the absolute value function uniformly on a given interval, say [-I, I]. They require nothing more than what you already know.
V, §4. ASYMPTOTIC ESTIMATES AND STIRLING'S FORMULA Functions defined on a set S containing arbitrarily large numbers can be ordered according to what is called their order of magnitude. More
[V, §4]
ASYMPTOTIC ESTIMATES AND STIRLING'S FORMULA
precisely, suppose given two functions
117
f and g defined on S, and suppose
g(x) > 0 for all XES, x sufficiently large. We say that f(x) is big oh of g(x) for x - co, and write f(x) = O(g(x», if there is a constant C > 0 such
that If(x) I ~ Cg(x)
for all x sufficiently large.
Instead of writing f(x) = O(g(x», we also use the notation or
f«g
f(x)« g(x)
for x - co.
For example, to say f(x) = 0(1), or f« 1, means that f is bounded. Thus we may write sin x = 0(1) for x _ co. For any polynomial P of degree d, we have P(x) = O(x d ),
P(x)« x d
also written
for x -+ co.
Similarly, we write f(x) = o(g(x» for x _ co if lim f(x)fg(x)
= O.
We then say that f is little oh of g, or that f has a strictly lower order of growth than g (for x - co). The standard orders of growth are given by the elementary functions log, polynomials, and exponentials. We can also iterate the logs and the exponentials, so by ascending order of growth, we can display functions as follows:
. .. I ogI og X, Iog X, To say that f(x)
= 0(1)
2
n
. .. ,X, X , ... ,X , . . . ,e
as x
-+
x2x
,e , ... ,e eX, ....
co means that lim f(x)
= O.
Each one of
the above functions is little oh of the next one on the right. Similarly, on the other side, we have in ascending order: ... ,e -2x, e-x, .. . ,x -n, ... ,x -2 ,x -I ,
Ifl og x, Iflog Iog x, . .. , 1, ....
Naturally, the above list does not include all possible orders of growth. For instance, we can have intermediate orders by multiplying two of the above, x log x, xOog X)2, x(log X)3, ••. , or in the other direction x x x log x' Oog X)2' Oog X)3'····
118
[V, §4]
THE ELEMENTARY REAL INTEGRAL
Or also, reflecting that powers of log x grow slower than powers of x: log x, (log X)2, (log x)3, ... ,(log x)", ... ,x', x, x 2, .. , . The following are basic properties of orders of growth, for x
+ f2
Property 1. If fl «g. and f2« g2' then f. Property 2. If f.(x)
+ g2'
= O(gl(x») and f2(X) = o(g(x»), then = o(glg(x)).
(fih)(x)
In particular, if 'I' is a bounded function, and Property 3. If f
«gl
-> 00.
= 0(1) and
f«
g, then
'l'f« g.
ip is a bounded function, then ipf = 0(1).
Property 4. If fl = o(g,) and f2 = o(gJ, then fl
+ f2 = 0(9. + g2)'
We leave the verification of these properties as exercises. One can make similar definitions when S contains numbers which are arbitrarily small in absolute value, in which case the order of growth is meant for x -> o. Example. If k, n are integers
0 with k < n, then
~
x" = o(lxl"> which means that x"flxl k
--+
0 as x
for x
->
--+
0,
O. On the other hand,
x k = o(x")
for x
--+
00.
Note the switch between k and n. Example. Taylor's formula expresses a function form
f(x) = Pn-I(X) + O(lxln ) where P"_I is a polynomial of degree
for x
~ n-
f of class
--+
C" in the
0,
1.
Example. We have sin x - x = o(lxl) and also sin x - x x --+ O. In fact sin x - x = O(lxI 3 ) for x --+ O.
= O(x i )
for
In general, say for x --+ 00, by an asymptotic expansion of a function f defined. on a set S containing arbitrarily large numbers, we mean an expression
f(x) = fl (x) where
1.+, (x) = o(.r.(x»)
+ f2(X). + ... + J.(x) + o(J.(x»),
for x
--+
00,
and k
= 1, ... ,n -
I. A similar defini-
[V, §4]
ASYMPTOTIC ESTIMATES AND STIRLING'S FORMULA
119
tion can be made for x -> O. We may then say that Taylor's expansion gives an asymptotic expansion of a function near 0, in terms of powers of x. In comparing orders of magnilude, we are led to an equivalence relation, two functions being equivalent if they have the same order of growth in the following precise sense. Let f, 9 be two functions defined on a set S contraining arbitrarily large numbers. We say that I(x) is asymptotic to g(x) for x --+ 00 (and XES) if I(x), g(x) "# 0 for all sufficiently large x, and lim l(x)/g(x) = I. We then write I(x) ~ g(x). A similar definition can be made for functions defined on a set S containing arbitrarily small numbers, replacing the limit to infinity by the limit as x approaches O. Observe that the relation I(x) ~ g(x) for x -> 00 can be formulated in different ways, as follows: There exists a function
II
There exists a function h such that
z-'"
and
lim hex) = 0 We have I(x)
=I
such that lim u(x)
= g(x)(1 + 0(1)
I(x)
for x ...
and I(x)
g(x)(1
=
= g(x)u(x).
+ hex)~.
00.
We have I(x) = g(x) + o(lg(x)l) for x --+
00.
Example. Let I be a function such that lim I(x) =
00.
Let C be a
z~'"
constant. Then I ~ I + C, as follows at once from the definitions. But also if 9 is a function such that g(x) = o(J(x)) for x --+ 00, then I(x) ~ I(x)
+ g(x)
for
In particular, if 9 is a bounded function, then
x --+
I
00.
~I
+ g.
Example. Let n
H(n)
I
= .f:, k'
Thus H(n) is the truncated harmonic series. Then H(n) ~ log n for n --+ 00. If we define H(x) = L 11k, then H(x) ~ log x for x --+ 00. Prove this as l:t"~x
an exercise, using Riemann sums. Example. We have sin x ~ x and log(l
+ x) ~ x
for x
--+
O.
We are now interested in giving an asymptotic expansion for log n!, with n --+ 00. In other words, we are interested in the order of growth of n! for positive integers n. The next theorem refines the rough estimate for n! of Chapter IV, §2.
120
THE ELEMENTARY REAL INTEGRAL
[V, §4]
Theorem 4.1 (Stirling's formula). Let n be a positive integer. Then there is a number between 0 and I such that
e
The steps in the proof are at the level of elementary calculus. The only difficulty lies in which steps to take and when, so we shall indicate the steps and leave the easy details to the reader. I. Let rp(x) =
I+x t log -I-x
x. Show that
rp'(x)
x2
= I -x 2 •
x3 2. Let l/J(x) = rp(x) - 3(1 _ x 2)' Show that l/J'(x)
=
-2x 4 3(1 _ X2)2 .
3. For 0 < x < I, conclude that rp(x) > 0 and l/J(x) < O.
4. Deduce that for 0 ;£ x < I we have 1
+X
I
o;£ -,log I _ 5. Let x
= 2
I n+1
x3
x - x ;£ 3(I _ x2)"
I+x n+1 Then - - ' = - - and I-x n x3 3(1 - x 2 )
+
12(2n
-
I 1)(n 2
+ n)'
6. Conclude that
n+1 o ~ -,I log --
n
I 2n
+I
I
n +- o ;£ (n + -,)1 log n
~
-
I =;::--=,...,,_...,. 12(2n + l)(n 2 + n)'
I (I I)
I ;£ - - - - - . 12 n n+1
7. Let and
[V, §4]
ASYMPTOTIC ESTIMATES AND STIRLING'S FORMULA
Then
bn . Show that
an ::;;
121
bn + I < 1 bn = .
and
Thus the an are increasing and the bn are decreasing. Hence there exists a unique number c such that
for all n.
8. Conclude that
for some number (J between 0 and 1. To get the value of the constant c, one has to use another argument. Our first aim is to obtain the following limit, known as the Wallis product. Theorem 4.2. We have
1t
-=
2
. 224466 2n 2n hm------·.. . n-oo 1 3 3 5 5 7 2n - 1 2n + 1
Proof. The proof will again be presented as an exercise. 1. Using the recurrence formulas for the integrals of powers of the sine, prove that
f.
"2
o sin /2
[o
2n
xdx
=
2n - 1 2n - 3 2n 2n - 2"
sin 2n + I x dx =
I 1t
'22'
2n-2 2 ... - . 2n+12n-l 3 2n
2. Using the fact that powers of the sine are decreasing as n = 1, 2, 3, ... and the second integral formula above, conclude that
Sn12 sin 2n +t x dx 1 . 2 d ::;; 1. 1 + 1/2n - SO~12 Sin nX X -
""---:-;:0-::;;
3. Taking the ratio of the integrals of sin 2n x and sin 2n + 1 x between 0 and 1t/2 deduce Wallis' product.
122
[V, §4]
THE ELEMENTARY REAL INTEGRAL
Corollary 4.3. We have (n !)22 2• _ hm (2)1 It-a> n. n 1/2 .
1/2 11
•
Proof. Rewrite the Wallis product into the form 224 2"'(2n)2 I -= hm 2 • 2 2 ._.,123 ···(2n-l) 2n 11
•
Tak¢ the square root and find the limit stated in the corollary. Fi:nally, show that the constant c in Stirling's formula is 1/.j2n, by arguing as follows. (Justify all the steps.)
.
(2n)2.+ 1/2e -2.
c = lim
(2 )1
n.
It-CO
_.
J2 [nn+1/2e- nJ2
(n !)22 2•
- lim (2)1 n. n 1/2 It-CO
Thus c
n.,
= 1/.j2n.
Remark. The above proof is tricky, and does not show the large structures behind the result. On the other hand, it has the advantage of using only very elementary means to give an asymptotic development for log n!, namely log n!
= n log n -
8
n + t log n + log .j2n + 12n
up to the term 8/12n which tends to 0 as n -+ 00. The preceding terms are ordered according to decreasing order of magnitude. The functions of positive integers given by
1
-, constant, log n, n, n log n, n
are in increasing order of growth, each one being little oh of the preceding one for n -+ 00. I don't like proofs like the above, but I found it worth including here to show how elementary calculus can be made to work. It takes a few more pages to establish the general techniques giving a full asymptotic expansion for the gamma function, with terms going beyond the term
[V, §4]
123
ASYMPTOTIC ESTIMATES AND STIRLING'S FORMULA
O(l/Il) for Il .... 00. This type of more structural proof is often done in courses in complex analysis, because it applies to the complex gamma function as well. Cf. my Complex Allalysis, Chapter XV, §2.
V, §4. EXERCISES I. Integrating by parts, prove the following formulas.
(a)
Jsin" x dx = -~ sin"-I x cos x + n : I Jsio"-2
(b)
Jcos"
x dx
n-IJ COS"-2
. x +-= -I COs"-l X sm n
X
X
dx
dx
II
2. Prove the formulas, where
II
is a positive integer:
J(log x)" dx x(log x)" - n J(log xr' dx (b) Jx"e dx = x"e n Jx"-le dx
(a)
=
X
x
x
-
3. By induction, find the value defined to be '"
J o
J;;' x"e-
fIx) dx
X
dx
= lim B-f»
= II!
J B
The integral to infinity is
fIx) dx.
0
4. Show that the relation of being asymptotic, i.e. fIx) - g(x) for x
equivalence relation.
S. Let r be a positive integer. Prove that for x --+ co. [Hint: Integrate between 2 and
-IX, and then between -IX and x.]
6. (a) Define x
Li(x) =
I
-I- dt. og t
J2
Prove that
x (x)
Lix =--+0 () log x (log xl'
so
(b) Let r be a positive integer, and let x
Lir(x) =
2
J
I
(log I)' dl.
x
Li(x) - -I- . ogx
--+ 00,
is an
124
[V, §4]
THE ELEMENTARY REAL INTEGRAL
Prove that Li,(x) - xj(log x)' for x -+
00.
Better, prove that
(c) Give an asymptotic expansion of Li(x) for x -+
00.
7. Let L(x)=
I
L 2~k;:!x log k
Show that L(x) = Li(x) + 0(1) for x -+
00,
so in particular, L(x) - Li(x).
8. More generally, let f be a positive function defined, say, for all x that f is decreasing, and let F(x)
=
~
2. Assume
r
f(t) dr.
2
Assume that F(x) is unbounded, i.e. lim F(x) = F(x) -
L
f(k)
00
for
for x -+
x --.
Show that
00.
00.
2~k~x
In fact, if we denote the sum on the right by S,(x), show that F(x)
= S,(x) + 0(1).
Exercises 6, 7, and 8 are especially significant because they are relevant to the probabilistic distribution of prime numbers. A prime number is an integer ~ 2 which is divisible only by itself and I. Thus the first few primes are 2, 3, 5, 7, II, 13, 17, 19,.... Roughly speaking, the probability that an integer n is prime is I/log n. What does this mean? It means that the number of primes in an interval [2, xl is given by the sum L(x) of Exercise 7, plus an error term wbich has lower order of growth. Thus if one denotes as usual by nIx) the number of primes :;; x, then one has the prime number theorem, nIx) = L(x)
+ E(x),
where E(x)
= o(L(x»)
for
x
-+
co.
According to Exercise 6, we then see that nIx) - x/log x for x -+ 00. This asymptotic relation was discovered experimentally by making tables of primes in the seventeenth century. It was proved only at the end of the nineteenth century (1896) by Hadamard and de la Vallee Poussin. Hadamard had developed a theory of complex analysis motivated precisely by the prime number counting problem. Perhaps the most famous unsolved problem in mathematics is the Riemann hypothesis, which says something much more precise about the error term, namely that E(x)
= O(x l /2 log x)
for x .....
00.
[V, §4]
ASYMPTOTIC ESTIMATES AND STIRLING'S FORMULA
125
Riemann was led 10 this conjecture (published in 1859) partly by experimentation, but mostly because of a much deeper investigation which could be described technically only after basic knowledge of complex analysis. For Exercises 9, 10, and II, we let F(x)
=(
J(I)
dl.
9. Let J and 11 be two positive continuous functions on R. lim l1(x) = O. Assume that F(x) --+ 00 as x --+ 00. Show that
r
J(I)I1(I)
2
dl =
for x
o(F(x)
-+ 00.
10. Assume that J, 11 are continuous positive, that J(x) x --+ 00. Show that
r
J(I)I1(I) d/
= o(F(x)
for x
Assume that
00
and l1(x) _ 0 as
-+ 00.
2
11. Suppose that
J is monotone positive (increasing or decreasing) and that J(x) = o(F(x)
for x
-+
00.
Prove that J(X)I12 =
0(( J(/)I12 dl)
for x .......
00.
PART TWO
Convergence
The notion of limit, and the standard properties of limits proved for real functions hold whenever we have a situation where we have something like I I, satisfying the basic properties of an absolute value. Such things are called norms (or seminorms). They occur in connection with vector spaces. It is no harder to deal with them than with real numbers, and they are very useful since they allow us to deal also with n-space and with function spaces. The chapters in this section essentially give criteria for convergence, in various contexts. We deal with convergence of maps, convergence of series, of sequences, uniform convergence. It is recommended that readers have understood Chapter II and have done the exercises in that chapter. We can then concentrate beller here on limits in the context of distances and normed vector spaces.
128
CHAPTER
VI
Normed Vector Spaces
VI, §1. VECTOR SPACES By a vector space (over the real numbers) we shall mean a set E, together with an association (v, w) 1-+ V + w of pairs of elements of E into E, and another association (x, v) 1-+ xv of R x E into E, satisfying the following properties : VS 1. For all u, v, weE we have associativity, namely (u
+ v) + w = u + (v + w).
VS 2. There exists an element 0 e E such that 0 veE.
+ v = v + 0 = V for
all
VS 3. If veE then there exists an element weE such that v+w=w+v=o.
VS 4. We have v + w = w + v for all v, weE. VS 5. If a, b e R and v, weE, then Iv = v, and (ab)v
= a(bv),.
(a + b)v
= av + bv,
a(v
+ w) =
av
+ aw.
As with numbers, we note that the element w of VS 3 is uniquely determined. We denote it by - v. Furthermore, Ov = 0 (where 0 denotes the zero number and zero vector respectively), because Ov
+ v = Ov + Iv = (0 + I)v = Iv = v. 129
130
NORMED VECTOR SPACES
[VI, §I]
Adding - v to both sides shows that Ov = o. We now see that -v = (-I)v because v + (-I)v = (1 + (-I»v = Ov = O. An element of a vector space is often called a vector. Example 1. Let E = R' be the set of k-tuples of real numbers. If X = (x" ... ,x,) is such a k-tuple and Y = (y" . .. ,y,), define X
+
Y
= (Xl + y" ... ,X, + Yo);
and if a E R, define
aX = (ax" ... ,ax,). Then the axioms are easily verified. Example 2. Let E be the set of all real valued functions on a non-empty set S. If f, 9 are functions, we can define f + 9 in the usual way, and af in the usual way. We now see that E is a vector space. Let E be a vector space and let F be a subset such that 0 E F, if v, WE F then v + W E F, and if v E F and a E R then av E F. We then call F a subspace. It is clear that F is itself a vector space, the addition of vectors and multiplication by numbers being the same as those operations in E. Example 3. Let k > 1 and let j be a fixed integer, 1 ~ j ~ k. Let E = R' and let F be the set of all elements (x" ... ,x,) of R' such that Xj = 0, that is all elements whose j-th component is O. Then F is a subspace, which is sometimes identified with R'-' since it essentially consists of (k - I)-tuples. Example 4. Let E be a vector space, and let v" ... ,Vr be elements of E. Consider the subset F consisting of all expressions
with XI E R. Then one verifies at once that F is a subspace, which is said to be generated by v" ... ,vr • As a special case of Example 4, we may consider the set of all polynomials of degree ~ d as a vector space, generated by the functions I, x, ... ,xd• One can also generate a vector space with an infinite number of elements. An expression like X,v, + ... + XrVr above is called a linear combination of V,' ... 'Vr. Given any set of elements in a vector space, we may consider the subset consisting of linear combinations of a finite number of them. This subset is a subspace. For instance, the set of all polynomials is a subspace of the space of all functions (defined on R). It is generated by the infinite number of functions I, x, x 2 , ••••
[VI, §2]
131
NORMED VECTOR SPACES
Example S. Let S be a subset of the real numbers. The set of continuous functions on S is a subspace of the space of all functions on S. This is merely a rephrasing of properties of continuous functions (the sum of two continuous functions is continuous, and the product of two continuous functions is continuous, so a constant times a continuous function is continuous). Example 6. One of the most important subspaces of the space of functions is the following. Let S be a non-empty set, and let !?4(S, R) be the set of bounded functions on S. We recall that a function f on S is said to be bounded if there exists C > 0 such that I f(x) I ::: C for all XES. If f, 9 are bounded, say by constants C 1 and C 2> respectively, then
If(x)
+ g(x)1
S;
If(x)1
+ Ig(x) I ;:;; C, + C z
so f + 9 is bounded. Also, if a is a number, then laf(x)1 ;:;; laiC so af is bounded. Thus the set of bounded functions is a subspace of the set of all functions on S. Example 7. The complex numbers form a vector space over the real numbers. Example 8. Let S be a non-empty set and E a vector space. Let "'t(S, E) denote the set of all mappings of S into E. Then .,(t(S, E) is a vector space, namely we define the sum of two maps f, 9 by (f
+ gXx) =
f(x)
+ g(x)
and the product cf of a map by a number to be
(cf)(x) = cf(x). The conditions for a vector space are then verified without difficulty. The zero map is the constant map whose value is 0 for all XES. The map - f is the map whose value at x is - f(x).
VI, §2. NORMED VECTOR SPACES Let E be a vector space. A norm on E is a function satisfying the following axioms:
v...... lvl from
= Oifandonlyifv = o. If a E R and vEE, then lavl = lallvl· For all v, WEE we have Iv + wi < Ivl + Iwi.
NI. We have Ivl;;; Oandlvl N 2. N 3.
The inequality of N 3 is called the triangle inequality.
E into R
132
[VI, §2]
NORMED VECTOR SPACES
A vector space together with a norm is called a normed vector space. Avector space may of course have many norms on it. We shall see examples of this. Example I. The complex numbers form a normed vector space, the norm being the absolute value of complex numbers. If
I I is a norm on a vector space E and jf F is a subspace, then the
restriction of the norm to F is a norm on F, which is thus also a normed vector space. Indeed, the properties N I, N 2, N 3 are a fortiori satisfied by elements of F if they are satisfied by elements of E. As with absolute values, if VI' ... 'Vm are elements of a normed vector space, then
This is true for m = I, and by induction:
lv, + ... + Vm-I + vml
+ Iv,1 +
+ vm-II + Ivml + Ivml·
S; IVI ~
We shall deal with normed vector spaces of functions, and in these cases, it is useful to denote the norm by II II to avoid confusion with the absolute value of a function. Example 2. Let S be a non-empty set, and let !!I(S, R) be the vector space of bounded functions on S. If J is a bounded function on S, we define
1IJ11
O. The open ball of radius r and center W in E consists of all those elements x E E such that Ix - wi < r. The closed ball of radius r and center w in E is the set of all x E E such that Ix - wi :;; r. The sphere of radius r and center w in E is the set of all x E E such that Ix - wi = r. We shall see the justification for this terminology in the next section. We use the notation B,(w),
B,(w),
S,(w)
for the open ball, closed ball, and sphere of radius r, centered at w. We shall now discuss in greater detail the standard norms used throughout the book. We shall see that a vector space may have two distinct useful norms on it. It is therefore important to have some notion concerning these norms which describes when they will affect the notion of limit, discussed later. Let I I. and I b be norms on a vector space E. We shall say that they are equivalent if there exist numbers C h C 2 > 0 such that for all vEE we have
If I I" I Iz, I 13 are norms on E such that I I. is equivalent to I 12 and I b is equivalent to I 13, then I I. is equivalent to I b. Also if I I. is equivalent to I b, then I b is equivalent to I I•. We leave the easy proofs to the reader. We define a subset S of a normed vector space to be bounded if there exists a number C > 0 such that Ix I ~ C for all XES. It is clear that if a set is bounded with respect to one norm, it is bounded with respect to any equivalent norm. Spheres and balls are bounded.
VI, §2. EXERCISES I. Let S be a set. By a distance function on S one means a function d(x, y) of pairs of
elements of S, with values in the real numbers, satisfying the following conditions: d(x, y) ;;; 0 for all x, YES, and d(x, y)
= 0 if and only if x = y.
= d(y, x) for all x, yES.
d(x, y) ;;; d(x, z)
+ d(z, y) for all x, y, z E S.
Let E be a normed vector space. Define d(x, y) this is a distance function.
= Ix -
yl for x, y
E
E. Show that
136
[VI, §2]
NORMED VECTOR SPACES
2. (a) A set S with a distance function is called a metric space. We say that it is a bounded metric if there exists a number C > 0 such that d(x, y) ~ C for all x, yES. Let S be a metric space with an arbitrary distance function. Let Xo E S. Let r > O. Let S, consist of all XES such that d(x, xo) < r. Show that the distance function of S defines a bounded metric on S,. (b) Let S be a set with a distance function d. Define another function d' on S by d'(x, y) = min(l, d(x, y)). Show that d' is a distance function, whicb is a bounded metric. (c) Define d(x. y)
d"(x, y)
= 1 + d(x. y)
Show that d" is a bounded metric.
3. Let S be a metric space. For each XES, define the function formula f.(y)
f.: S -+ R
by lhe
= d(x, y).
(a) Given two points x, a in S show that f. - f. is a bounded function on S. (b) Show that d(x, y) = IIf. - frll. (c) Fix an element a of S. Let g. = f. - f •. Show that the map
is a distance preserving embedding (i.e. injective map) of S into tbe normed vector space of bounded functions on S, with the sup norm. [If the metric on S originally was bounded, you can use f. instead of g••] Tbis exercise sbows that the generality of metric spaces is illusory. In applications, metric spaces usually arise naturally as subsets of narmed vector spaces. 4. Let I I be a norm on a vector space E. Let a be a number> O. Show that the function x ....... a Ixl is also a norm on E.
5. Let I 11 and I I, be nOrms on E. Show that the functions x .... Ixl,
+ Ixh
and x .... max(lxl,.lxh) are norms on E.
6. Let E be a vector space. By a seminorm on E one means a function u: E -+ R such that u(x) ~ 0 for all x E E. u(x + y) ~ u(x) + u(y), and u(ex) = Icl"(x) for all eER, x, YEE. (a) Let "I. u2 be seminorms. Show that ", + "2 is a seminorm. If).1.).2 are numbers ~ 0, show that ).. + ).2"2 is a seminorm. By induction show that if 0"1' ... ,G are seminorms and It . ...• J... are numbers ~ 0 then Alai + ... + ).,.uII is a seminorm. (b) Let " = max(u, • ( 2 ). Show that" is a seminorm.
"1
II
0'.
be a norm and (]2 a seminorm on a vector space. Show that and max(al' 0'2) are norms.
7. Let
(11
+ 112
[VI, §3]
n-SPACE AND FUNCTION SPACES
8. LeI a be a seminorm on a vector space E. Show that the set of all x that a(x) = 0 is a subspace.
137 E
E such
9. The cP seminorms. Let p be an integer !1; O. Let E = 0([0, 1]) be the space of p-times continuously differentiable functions on [0, 1]. Define ap and Np by ap(J)
= sup IPp)(x)! x
and
Np(J)
=
max a,(J), O:!,.:;;p
where the maximum is taken for r = O•... •p. (a) Show that ap is a seminorm and N p is a norm. Note that ao = No is just the ordinary sup norm. The norm Np is called the cP-norm. (b) Describe the subspace of E consisting of those functions f such that ap(f) = 0 for p = 0 and also for p> O. This is the subspace of Exercise 8. 10. Consider a scalar product on a vector space E which instead of satisfying SP 4 (that is positive definiteness) satisfies the weaker condition that we only have (v, v) !1; 0 for all VEE. Let IV E E be such that (IV. IV) = O. Show that (IV. v) = 0 for all vEE. [Hint: Consider (v + tlV, V + tlV) !1; 0 for large positive or negative values of r.] 11. Notation as in the preceding exercise. show that the function IV I-> I/IVI/ = J(IV,IV)
is a seminorm, by proving the Schwarz inequality just as was done in the text. 12. Let E be a vector space with a positive definite scalar product, and the corresponding norm I/vl/ =~. Prove the parallelogram law for all v, IV E E:
I/v + 1V1/2 + I/v - 1V1/2 = 21/vl/ 2 + 21/1V1/2. Draw a picture illustrating the law. For a follow up. see §4, Exercises 5, 6, and 7.
VI, §3. n-SPACE AND FUNCTION SPACES The euclidean norm. Let E = Rn be the space of n-tuples of real numbers. If A = (ai' ... ,an) and B = (b., ... ,bn) and n-tuples of numbers, we define
The four properties of a scalar product are then immediately verified. The last one holds because if A# 0, so that A . A > O. The others are left to the reader. We therefore obtain a norm on Rn given by
IAI = Ja~ + ... + a;.
138
[VI, §3]
NORMED VECTOR SPACES
This will be called the euclidean norm, because it is a generalization of the usual norm in the plane, such that the norm of a vector (a, b) is a 2 + b2 • Consider the euclidean norm, in R 2 • Then the open ball of radius r centered at the origin will be called the open disc (of radius r centered at the origin), and it corresponds geometrically to such a disc. Similarly we define the closed disc. The sphere of radius r centered at the origin is nothing but the circle of radius r centered at the origin. In R3 , with the euclidean norm, the ball and sphere have the usual interpretation of these words. This is the reason for adopting the same terminology for R", n > 3, or for normed vector spaces in general. Ordinary intuition of euclidean geometry can be used to justify the definition that A is perpendicular (or orthogonal) to B if and only if
J
A·B=O.
This is done as follows. By euclidean geometry, A is perpendicular to B if and only IA -
BI
= jA
+ BI,
as shown on the figure.
A
A IA -BI ""--+---~B
IA - BI B
-B -B
But this condition is equivalent with (A
+ B)2
= (A - B)2
or in other words, A
2
+ 2A . B + B 2 =
A 2 - 2A . B
+ B 2.
This is equivalent with 4A· B = 0, whence equivalent with A· B = 0, as desired. Similarly, in an arbitrary vector space E with a scalar product, one defines two vectors v, W to be perpendicular or orthogonal if and only if v·w = O.
[VI, §3]
139
n-SPACE AND FUNCTION SPACES
The sup norm. We can define another norm on Rn which will be denoted by II II or II 110' We let IIAII = max la,l, ;
the maximum being taken over all i = 1, ... ,no Thus IIAII is the maximum of the absolute values of the components of A. We contend that this is a norm. Clearly, if II A II = 0 then A = 0 because all ai = O. Furthermore, if A # 0, then IIAII > 0 because some la,l > O. Let
Then
+ BII
IIA
= max la,
+M
i
We have laj
+ bj !
0 for x in some neighborhood of c, so the above integral is actually > O. This proves that the sequence {f.} has no V-limit in Co([O, I]), which is therefore not complete. In my Real and Functional Analysis, Chapter VI, you can see how one finds a complete normed vector space of functions in a natural way,
[VI, §4]
149
COMPLETENESS
extending the L '-norm above. This involves a study relating pointwise convergence, sup-norm convergence, and L'-convergence simultaneously.
VI,
§4_ EXERCISES
I. Give an example of a sequence in CO([O, I]) which is L'-Cauchy but not sup
norm Cauchy. Is this sequence L'·Cauchy? If it is, can you construct a sequence which is L'-Cauchy but not L'-Cauchy? Why? 2. Let I,,(x) = x", and view (f.,l as a sequence in CO([O, 1]). Show that approaches 0 in the L'-norm and the L'·norm. but not in the sup norm.
{J.l
3. Let I I, and I b be two equivalent norms on a vector space E. Prove that limits, Cauchy sequences, convergent sequences are the same for both norms. For instance, a sequence {x,,} ha~ the limit" for one norm if and only if it has the limit" for the other norm. 4. Show that the space CO([O, I]) is not complete for the L '-norm. Almost all the time it has been appropriate to deal with subsets of normed vector spaces. However, for the following exercises, it is clearer to formulate them in terms of metric spaces, as in Exercises I and 2 of §2. The notion of Cauchy sequence can be defined just as we did in the text, and a metric space X is said to be complete if every Cauchy sequence in X converges. We denote the distance between two points by d(x" x,). 5. The semiparallelogram law. Let X be a complete metric space. We say that X satisfies the semiparallelogram law if for any two points x, in X, there is a point z such that for all x E X we have
x,.
d(x" x,)'
+ 4d(x, z)';:;; 2d(x. x,)' + 2d(x, x,)'.
(a) Prove that tl(z, x,) = d(z, x,) = tl(x" x,)/2. [Him: Substitute x = x, and x = x, in the law for one inequality. Use the triangle inequality for the other.] Draw a picture of the law when there is equality instead of an inequality. (b) Prove that the point z is uniquely determined, i.e. if =' is' another point satisfying the semiparallelogram law then z = z'. In light of (a) and (b), one calls z the midpoint of x" x,. 6. (Bruhat-Tits-Serre.) Let X be a complete metric space and let S be a bounded subset. Then S is contained in some closed ball ii.(x) of some radius Rand center x E X. Define I' (depending on S) to be the inf of all such radii R with all possible centers x. By definition, there exists a sequence {r,,} of numbers;;; I' such that lim 1'" = T, together with a sequence of balls B,.(x,,) of centers x,,, such II-X
that ii, (x,,) contains S. In general, it is nO/ true that there exists a ball ii,(x) with radius precisely I' and some center x, containing S. If such a ball exists, it is called a ball of minimal radius containing S. Prove the following theorem: LeI X be a complele melric space salisfying fhe semiparallelogram law. LeI S be a houllded suhsel. Theil Ihere exisls a ullique closed ball ii,(x,) of millimal radius conw;n;119 s.
150
[VI, §4]
NORMED VECTOR SPACES
[Hint: You have to prove two things: existence and uniqueness. Use the semiparallelogram law to prove each one. For existence, let {x n } be a se'!.uence of points which are centers of balls of radius rn approaching r, and B,.(xn ) contains S. Prove that {x n } is a Cauchy sequence. Let c be its limit. Show that Ii,(c) contains S. For uniqueness, again use the semiparallelogram law. Let ii,(x l ) and ii,(X2) be balls of minimal radius centered at x,, X2' Let z be the midpoint, and use the fact that given X
such that 9 preserves distances. In other words, for all x" X2 E X we have d(9(X,), 9(X2»
= d(x"
x 2).
If plane geometry was properly taught in high school, you should know that translations, rotations and reflections are isometries of the euclidean pJane, and that all isometries of the plane can be obtained by composition of these special ones. In any case, these mappings provide examples of isometries. Note that if 9,,92 are isometries, so is the composite 91 092' Also if 9 is an isometry, then 9 has an inverse mapping (because 9 is a bijection), and the isometry condition immediately shows that 9-': X -> X is also an isometry. Note that the identity mapping id: X -> X is an isometry. Let G be a set of isometries. We say that G is a group of isometries if G contains the identity mapping, G is closed under composition (that is, if 9" 92 E G then 9,092 E G), and is closed under inverse (that is, if 9 E G, then 9-' E G). One often writes 9192 instead of 9,092' Note that the set of all isometries is itself a group of isometries. Let x' E X. The subset Gx' consisting of all elements 9(X') with 9 EGis called the orbit of X' under G. Let S denote this orbit. Then for all 9 E G and all elements XES it follows that 9x E S. Indeed, we can write x = 91 x' for some 9, E G, and then 9(91 x') = 9(9,(X'» = (9 0 9, )(x') E S, In fact, 9(S)
= S because G contains the
and
9 0 9, E G
by assumption.
identity mapping.
After these preliminaries, prove the following major result. 7. Bruhat-Tits fixed point theorem. Let X be a complete metric space satisfyin9 the semiparallel09ram law. Let G be a group of isometries. Suppose that an orbit is bounded in X. Let x, be the circumcenrer of this orbit. Then x, is a fixed point of G, that is, 9(X,) = x, for all 9 E G. Historical comments. The last three exercises resulted from a century of research in rather fancy mathematics. We can't go into it, but it involves research
[VI, §5]
OPEN AND CLOSED SETS
151
on surfaces of negative curvature by von Mangoldt and Hadamard at the end of the nineteenth century, and by Cartan in the 1920s, when Cartan formulated a fixed point theorem under conditions of compactness and negative curvature. In 1972, Bruhat-Tits set up the semiparallelogram law as a fundamental hypothesis, and proved the fixed point theorem given here in Exercise 7. [" Oroupes r a is open. It closure is the infinite closed interval [a, (0). (c) The rational numbers are dense in R. (d) In R", the open ball of radius r > 0 is dense in the closed ball of radius r. (e) In R", the set of all points (al"" ,a") whose coordinates are rational numbers is dense in R". Thus R, and R", have denumerable dense subsets, even though neither R nor R" are denumerable. Let S be a subset of a normed vector space E. Let X be a subset of S. We say that X is open in S if there exists an open set U in E such that X = S n U. Alternatively, we see that X is open in S if and only if, for each point x e X, there exists E > 0 such that B,(x) n S is contained in X. We define a subset X to be closed in S if there is a closed set Z in E such thatX=SnZ. Remark. Let S be open in E. Then a subset X of S is open in S if and only if it is open in E. Similarly, let S be closed in E. Then a subset of S is closed in S if and only if it is closed in E. It is an exercise to give the proofs of these statements. See Exercise 9. Identify R with the x-axis in R 2 • Then R is open in R, but is not open in R 2 • Similarly, an open interval on R is open in R, but is not open in R 2 • If the interval is, say, (a, b) with a < b, then this interval is the intersection of an open set in R 2 with R. For instance, we can take the open set in R 2 to consist of all numbers (x, y) with a < x < b, and -1 0 such that
if v E S and Iv - xl < 0, then If(v) - yl
0 such that
This means precisely that f is continuous at x, and concludes the proof of the first statement. We leave the proof of the statement about closed sets to the reader. Example 5. A polynomial f is continuous. Thus the set of numbers x such that f(x) < 3 is open in R because the set of numbers y such that y < 3 is an open set V in R, and the set of numbers x such thatf(x) < 3 is equal torl(V). The norm function on a normed vector space E is continuous (with a vengeance). Therefore: Let r be a number > O. Then the r-sphere S,(O) and the r-ball 8,(0) are closed. Proof For the sphere, we have S,(O) = g-l(r), where 9 is the norm, g(x) = Ixi. Since a single real number is closed, it follows that g-1 (r) = S,(O) is closed. For the ball, we have 8,(0) = g-I([O, r]), and the closed interval [0, r] is closed. Hence so is the ball 8,(0).
Warning. The previous statements are made concerning certain closed sets with respect to a given norm. Of course these sets will also be closed with respect to any equivalent norm. However, they may not be closed with respect to a norm which is not equivalent to the given one.
VI, §5. EXERCISES I. Let S be a subset of a normed vector space E. and let S denote the set of all points of E which are adherent to S. (a) Prove that S is closed. We call S the closure of S. (b) If S. T are subsets of E. and SeT. show that SeT. (c) If S, T are subsets of E, show that S v T = S v T.
158
NORMED VECTOR SPACES
[VI, §5]
(d) Show that S = S. (e) If S c:: T c:: S, prove that 't = S. ([) Let E, F be normed vector spaces, S a subset of E and T a subset of F. Take the sup norm on E x F. Show that (S x T) = S x f. 2. A boundary point of S is a point VEE such that every open set U which contains v also contains an element of S and an element of E which is not in S. The set of boundary points is called the boundary of S, and is denoted by as. (a) Show that as is closed. (b) Show that S is closed if and only if S contains all its boundary points. (c) Show that the boundary of S is equal to the boundary of its complement. 3. An element u of S is called an interior point of S if there exists an open ball B centered at u such that B is contained in S. The set of interior points of S is denoted by Int(S). It is obviously open. It is immediate that the intersection of Int(S) and as is empty. Prove the formula
S = Int(S) u as. In particular, a closed set is the union of its interior points and its boundary points. If S, T are subsets of normed vector spaces, then also show that Int(S x T)
= Int(S) x
Int(T).
4. Let S, T be subsets of a normed vector space. Prove the following: (a) iJ(S u T) c:: as u or. (b)
a(s" T) c:: as u or.
(c) Let S - T denote the set of elements XES such that x ¢ T. Then iJ(S - T) c:: as u aT. [Nole: You may save yourself some work if you use the fact that iJf{fS = iJS, where '6S is the complement of S, and use properties like S - T = S " '6T, as well as '6(S " T) = '6S u '6T.] (d) iJ(S x T) = (as x T) u (S x aT). 5. Let S be a subset of a normed vector space E. An element v of S is called isolated (in S) if there exists an open ball centered at v such that v is the only element of S in this open ball. An element x of E is called an accumulation point (or point of accumulation) of S if x belongs to the closure of the set S - {xl. (a) Show that x is adherent to S if and only if x is either an accumulation point of S or an isolated point of S. (b) Show that the closure of S is the union of S and its set of accumulation points. 6. Let V be an open subset of a normed vector space E, and let VEE. Let V v be the set of all elements x + v where x E V. Show that V v is open. Prove a similar statement about closed sets. 7. Let V be open in E. Let I be a number> O. Let IV be the set of all elements IX with x E V. Show that IV is open. Prove a similar statement about closed sets. 8. Show that the projection R x R --> R given by (x, Y)H X is continuous. Find an example of a closed subset A of R x R such that the projection of A on the
[VI, §5]
159
OPEN AND CLOSED SETS
first factor is not closed. Find an example of an open set U projection is closed, and U .;. R2 •
In
R2 whose
9. Prove the remark before Theorem 5.6. 10. Let U be open in a normed vector space E and let V be open in a normed vector space F. Let g: V_ U
and
f:U - V
be continuous maps which are inverse to each other, that is fog
= id v
go f = id u ,
and
where id means the identity map. Show that if U, is open in U then flU ,) is open in V, and that the open subsets of U and V are in bijection under the association and
V,
1-+
g(V ,).
II. Let B be the closed ball of radius r > 0 centered at the origin in a normed vector space. Show that there exists an infinite sequence of open sets U. whose intersection is B. 12. Prove in detail that the following notions are the same for two equivalent norms on a vector space: (a) open set, (b) closed set, (c) point of accumulation of a sequence, (d) continuous function, (e) boundary of a set, and (f) closure of a set.
13. Let I I, and I It be two norms on a vector space E, and suppose that there exists a constant C > 0 such that for all x E E we have Ixl, ;;; C1xlz. Let fi (x) = Ixl,. Prove in detail; Given e, there exists 0 such that if x, y E E and Ix - ylz < 0, then If, (x) - f, (y)1 < e. [Remark. Since f, is real valued, the last occurrence of the signs I I denotes the absolute value on R.] In particular, f, is continuous for the norm liz. 14. Let BS be the set of all sequences of numbers
which are bounded, i.e. there exists C > 0 (depending on Xl such that Ix.1 ;;; C for all II E Z·. Then BS is a special case of Example 2 of §l, namely the space of bounded maps ~(Z·, R). For X E BS, the sup norm is IIXII
= sup Ix,'!' •
i.e. IIXII is the least upper bound of all absolute values of the components. (a) Let Eo be the set of all sequences X such that x. = 0 for all but a finite number of II. Show that Eo is a subspace of BS. (b) Is Eo dense in BS? Prove your assertion. [Note: In Theorem 3.1 of Chapter VII, it will be shown that BS is complete.]
CHAPTER
VII
Limits
VII, §1. BASIC PROPERTIES A number of notions developed in the case of the real numbers will now be generalized to normed vector spaces systematically. Let S be a subset of a normed vector space. Let f: S -+ F be a mapping of S into some normed vector space F, whose norm will also be denoted by I I. Let v be adherent to S. We say that the limit of f(x) as x approaches v exists, if there exists an element WE F having the following property. Given E, there exists {) such that for all XES satisfying
Ix - vi < {) we have If(x) -
wi < E.
This being the case, we write lim f(x) = w. x-v XES
Proposition 1.1. Let S be a subset of a normed vector space, and let v be adherent to S. Let S' be a subset of S, and assume that v is also adherent to S. Let f be a mapping of S into some normed vector space F. If lim f(x) X-V XES
160
[VII, §IJ
161
BASIC PROPERTIES
exists, then
lim f(x) x-v xeS'
also exists, and these limits are equal. In particular, is unique. Proof· Let w be the first limit. Given everxeSand Ix - vi < /5 we have
f,
If(x) - wi
'"
limf(x) x-v
,w.).
Given f, there exists 0 such that if Ix - v I < 0 then If(x) -
Let f(x)
= y = (Y"
wi
0 and approaching 0 have to be closer and closer together to make If(x) - f(y)1 < E. For instance, let x = lin and y = I/(n + 1). Then 1 Ix - yl = n(n
+ 1)
but
If(x) - f(y)1 = 1.
Another way to see the non-uniform continuity is to use the mean value theorem. We have If(x) - f(y) I = 1f'(c)llx - yl,
where c is between x and y. To get 1f'(c)llx - yl < E, we need Ix - yl < E/lf'(c)I, and 1f'(c)l-+ 0 as x, y -+ O. Thus the choice of b continuity cannot be made independently of x, y.
In
the definition of
Example 2. If f satisfies a Lipschitz condition on an interval, with Lipschitz constant C i' 0, then f is uniformly continuous on the interval. Given E, we can select b = E/C, independently of x, y in the interval, because by hypothesis, If(x) - f(y)1
~
Clx - yl·
If you have not yet done it, you should now do Exercises 5 and 6 of Chapter lV, §3, to get a feeling for uniform or non-uniform continuity.
Example 3. In Chapter II, Theorem 4.6, we proved that a continuous function on a bounded closed interval is uniformly continuous. This theorem wilI be generalized considerably in Chapter VIII. For instance, let S be a closed bounded set in R". Let f be a continuous function on S. Then f is uniformly continuous.
[VII, §2]
CONTINUOUS MAPS
175
Example 4. Let E = CO([a, b]) be the vector space of continuous functions on [a, b], with the sup norm, and a < b. Let I~
= I: E--.R,
be the integral, that is, for fEE, I(f) =
r
f(x) dx. Then I is a linear
map. Let C = b - a. Then by Corollary 2.2 of Chapter V, we have
II(f)1 where IIfll is the sup norm of f actually Lipschitz, because
~
Cllfll,
Then I is uniformly continuous, and is
II(f) - I(g)1 = II(f - g)1 ~ C1Jf - gil. So if IIf - gil < E/C then II(f) - I(g)1 < E. Let {f.,} be a sequence in E, converging to a function f in E, with respect to the sup norm. Then by continuity, we have
lim I(f.,)
=
I (f)
or with the integral sign,
b
b
r f.,(x) dx = Jro f(x) dx. n-oo Ja lim
Thus the integral of the limit is the limit of the integral. This limit property of the integral often occurs in the context of products. If {f.,}, {gn} are sequences of continuous functions on [a, b] converging to f, g, respectively for the sup norm, then the product f.,gn converges to fg (cf. Example 4 of §1), and so
!~~
r
f,,(x)gn(x) dx
=
r
f(x)g(x) dx.
In the next section, we shall discuss more systematically limits with respect to the sup norm, which are called uniform limits. See also Exercises 10-14 at the end of the present section for a continuation of some of the above ideas. Example S. For this example, we define a fundamental notion, that of distance not only between points, but between a point and a set. Let S be a non-empty subset of a normed vector space E, and let vEE. The set of numbers Ix - vi for XES is bounded from below by o. We call its
176
LIMITS
[VII, §2]
greatest lower bound the distance from S to v, or v to S, and denote it by d(S, v) or d(v, S). Theorem 2.4. (a) (b)
We have d(S, v) = 0 if and only if v lies in the closure of S. The map u H d(S, u) is uniformly continuous from E into R.
The proofs, which are easy, will be left for Exercise 1. Counterexample. There may not be any point w of S such that d(S, v) = Iw - vi. In the first place, if S is not closed, then a boundary point v of S but not in S is at distance 0, and there is no point of S at distance 0 from v. But much more seriously, even if S is closed, and v ¢ S, this phenomenon may still occur in the case of infinite dimensional spaces. We now give an example. Let E be the space of bounded sequences of numbers, i.e. bounded maps of Z+ into R. Let en be the n-th unit vector, with component 1 in the n-th place, and 0 otherwise. Let X n = (1 + l/n)e n • Let S consist of all points X n with n ;;; 2. Then you can verify that S is closed. Furthermore, d(e l , S) = 1, but there is no point v E S such that d(e l , v) = 1. However, see Exercise 16. Such an example does not exist is finite dimensional spaces. In Chapter VIII, §2, Exercise 4, you will prove that if S is a closed subset of a finite dimensional space, then there always exists some WE S such that d(S, v) = Iw- vi· The distance function can be used to give an easy proof of the following important theorem. Theorem 2.5. Let S, T be two non-empty disjoint closed sets in a normed vector space E. Then there exists a continuous function f on E such that o ~ f ~ 1, f(S) = 0 and f(T) = 1. You can prove this in Exercise 1. The next theorem shows how the condition of uniform continuity allows one to extend a continuous mapping to the closure of a set. Theorem 2.6. Let f: S --> F be a mapping of a subset S of a normed vector space E into a complete normed vector space F. Assume that f is uniformly continuous on_ S. Then there exists an extension J of f to a continuous mapping of S into F.
Proof See Exercise 8. For an important application of Theorem 2.6, see Exercises 10-14, as well as Theorem 1.2 of Chapter X.
[VII, §2]
177
CONTINUOUS MAPS
VII, §2. EXERCISES I. (a) Prove Theorem 2.4(a). (b) Prove Theorem 2.4(b). (Hint: Show that for all v, IV E E, we have the inequality diS, v) - d(S, IV) ;:;; d(v, IV), or equivalently d(S, v) ;:;; d(S, IV) + d(v, IV). First note that for all XES, d(x, v) ;:;; d(x, 11') + d(v, 11'). Then take the g.l.b. for XES on the left, and follow that by taking the g.l.b. of d(x, IV) + d(v, 11') for XES on the right.] (c) Let S, T be two non·empty closed subsets of E, and assume that they are disjoint, i.e. have no point in common. Show that the function f(v)
=
d(S, v) d(S, v) + d(T, v)
is a continuous function, with values between 0 and I, taking the value 0 on S and 1 on T. (For a continuation of this exercise, cf. the next chapter, §2.) 2. (a) Show that a function f which is differentiable on an interval and has bounded derivative is uniformly continuous on the interval. (b) Let [(x) = x 2 sin(l/x 2 ) for 0 < x ;:;; 1 and [(0) = O. Is [ uniformly continuous on [0, I]? Is the derivative of f bounded on (0, I)? Is [ uniformly continl10us on the open interval (0, I)? Proofs?
3. (a) Show that for every c > 0, the function f(x) = l/x is uniformly continuous forx;;;;c. (h) Show that the function [(x) = e- X is uniformly continuous for x ;;;; 0, but not on R. (c) Show that the function sin x is uniformly continuous on R.
= sin(l/x) is not uniformly continuous on the interval 0 < x ;:;; n, even though it is continuous.
4. Show that the function [(x)
5. (a) Define for numbers t, x: [(t, x)
sin ex
= -t-
if t,. 0, [(0, x)
= x.
Show that [ is continuous on R x R. [Hint: The only problem is continuity at a point (0, b). If you bound x, show precisely how sin IX = tx + O(IX).] (b) Let (y2 _ X)2
[(x, y)
=
2 if (x, :4 +x {
y) ,. (0, 0),
if (x, y) = (0,0),
Is [ continuous at (O,O)? Explain.
178
[VII, §2J
LIMITS
6. (a) Let E be a nonned vector space. Let 0 < r, < r,. Let vEE. Show that there exists a continuous function [ on E, such that:
[(x) [(x)
= I if x is in the ball of radius r, centered at v, = 0 if x is outside the ball of radius r, centered at v.
We have 0
~
[(x)
~
1 for all x.
IHilll: Solve first the problem on R, then for the special case v = 0.) (b) Let v, IVE E and v #' IV. Show that there exists a continuous function [ on E such that [(v) = I and [(lV) = 0, and 0 ~ [(x) ~ I for all XE E.
7. Let S be a subset of nonned vector space E, and let [: S --> F be a map of S into a nonned vector space. Let S' consist of all points vEE such that v is adherent to S and lim [(x) exists. Define f(v) to be this limit. If v E S, then .l"-l'
_
xeS
_
_
[(v) = [(v) by definition, so [is an extension of [to S'. Show that [is continuous on S'. S'. We show continuity of f at vo. Given e there exists 0 such that if XES and Ix - vol < 0 then I[(x) - f(vo)1 < e. Let v E S' and Iv - vol < 0/2. Since lim [(x) = f(v) (x E S.X --> v), there exists XES such that Ix - vi < 0/2 and I[(x) - f(v)1 < F be a map into some nonned vector space. Show that [ is continuous on A if and only if its restrictions to S to T are continuous. Continuous linear maps 10. Let E, F be normed vector spaces, and let L: E --> F be a linear map. (a) Assume that there is a number C > 0 such that IL(x)1 ~ C1xl for all x E E. ShQw that L is continuous. (b) Conversely, assume that L is continuous at O. Show that there exists such a number C. IHiIll: See §l of Chapter X.I k
II. Let L: R --> F be a linear map of Rk into a nonned vector space. Show that L is continuous. 12. Show that a continuous linear map is uniformly continuous.
[VII, §3]
LIMITS IN FUNCTION SPACES
179
13. Let L: E --+ F be a continuous linear map. Show that the values of L on the closed ball of radius I are bounded. If r is a number > 0, show that the values of L on any closed ball of radius r are bounded. (The closed balls are centered at the origin.) Show that the image under L of a bounded set is bounded. Because of Exercise 10. a continuous linear map L is also called bounded. If C is a number such that IL(x)l;:;; C1xl for all x E E. then we call C a bound for L. 14. Let L be a continuous linear map. and let ILI denote the greatest lower bound or all numbers C such that lL(xll;:;; C1xl for all XE E. Show that the continuous linear maps of E into F form a vector space. and that the function L ...... ILj is a norm on this vector space. 15. Let a < b be numbers. and let E = CO([a. b]) be the space of continuous functions on [a, b]. Let I~: E --+ R be the integral. Is I~ continuous: (a) for the L'·norm; and (b) for the L2 -norm on E1 Prove your assertion. 16. Let X be a complete metric space satisfying the semiparallelogram law. (Cr. Exercise 5 of Chapter VI, §4.) Let S be a closed subset. and aSsume that given Xl> X 2 E S the midpoint between x, and X 2 is also in S. Let VEX. Prove that there exists an element IV E S such that dIu, IV) = dIu, S). Remark. The condition that S contains the midpoint between any two of its points amounts to a generalized convexity condition. In a linear context. see Exercises II, 12, and 13 of Chapter XII, §1.
VII, §3. LIMITS IN FUNCTION SPACES Let S be a set, and F a normed vector space. Let if,,} be a sequence of maps from S into F. For each XES we may then consider the sequence of elements of F given by (f,(x), fix), ...}. Thus for each x, we may speak of the convergence of the sequence (fn(x)}. If if,,} is a sequence of maps such that for each XES the sequence (fn(X)} converges, then we say that {f,,} converges pointwise. On the other hand, suppose eachf" E ~(S, F) is an element of the vector space of bounded maps from S into F, with its sup norm. Then we may speak of the convergence of the sequence {f,,} in this space. If the sequence {fn} converges for the sup norm, we say that it converges uniformly. Con· vergence in ~(S, F) is called uniform convergence. We shall denote the sup norm by II II"" and usually just II II. to shorten the notation. Observe that in defining a convergent sequence fn --. f, we take the difference fn - f. All we really need is that we can take the sup norm of this difference. Hence we shall say that an arbitrary sequence of maps {f,,} from S into F converges uniformly to a map f if given " there exists N such that for all n ~ N the difference f - f" is bounded, and such that IIf - .f.,11 < E. Similarly, we can define a sequence of maps {.f.,} to be uniformly Cauchy, wilhout each f" being bounded. All we need is Ihat
180
LIMITS
[VII, §3]
each difference I. - 1m is bounded. In Theorem 3.1 we shall prove that a uniformly Cauchy sequence is uniformly convergent (assuming F is complete, of course). Its limit is called the uniform limit. Let T be a subset of S. If I is a map on S, into F, and I is bounded on T, we write
IIf11T = sup II(x)l. xeT
If {I.} is a sequence of maps from S into F and if this sequence converges uniformly for the sup norm with respect to T, then we say that it converges uniformly on T.
Example 1. For each n, let In(x) = l/nx be defined for x > O. Then for each x, the sequence of numbers {l/nx} converges to O. Thus the pointwise limit of {f.{x)} is O. We also say that {f.} converges pointwise to the function O. However, this convergence is not uniform. Indeed, for any given x the N needed to make l/nx < E for all n ~ N depends on x. We could write N = N(x). As x approaches 0, this N(x) becomes larger and larger. However, let c be a number > O. View each I. as defined on the set T consisting of all numbers x ~ c. Then {fn} converges uniformly to 0 on this set T. Indeed, given E, select N such that l/N < EC. Then for all x ;;; C and all n ;;; N we have 1 --0 nx Hence
IIf.lIT
0, the limit lim In(x) = 0,
[VII, §3]
LIMITS IN FUNCTION SPACES
181
because for each x there exists N such that fn(x) = 0 if n ::: N. Thus again the sequence {fn} converges pointwise to O. However, it does not converge uniformly to O. If the peaks all go up to I, then
=
IIfnll
I
= IIfn - 011,
so the sequence of functions cannot converge uniformly. Note that each fn is bounded, and continuous, and the limit function (pointwise) is also continuous. Example 3. Let fn(x) = (I - x)" be defined for 0 ;;i; x ;;i; 1. For each x 'f. 0 we have lim (I - x)"
= O.
However, lim fn(O)
=
lim (I - 0)" = 1.
Each fn is continuous, but the limit function (pointwise) is not continuous on [0, I]. We shall now prove two basic theorems concerning uniform limits of functions. Theorem 3.1. Let F be a complete normed vector space, and S a nonempty set. Let {In} be a sequence of maps fn: S -+ F which is uniformly Cauchy. Then the sequence is uniformly convergent to a map f: S -+ F. If (fn} is bounded for each n, then so is f Thus the space !J4(S, F) with the sup norm is complete. Proof Let (fn} be a Cauchy sequence of maps of S into F. Given there exists N such that if n, m ~ N then
E,
IIfn - fmll < E. In particular, for each x, Ifn(x) - fm(x) I < E, and thus for each x, {!.(x)} (k = 1,2, ...) is a Cauchy sequence in F, which converges since F is complete. We denote its limit by f(x), i.e. f(x) = lim fix). k-oo
Let n ~ N. Given
XES,
select m ~ N sufficiently large (depending on x)
182
[VII, §3]
LIMITS
such that If(x) -
fm(x)1
we have II.(x) - I.(v) I < £. Then If(x) - f(v) 1 ~ If(x) -
I.(x)l +
Ifn(x) - fn(v)l
+ II.(v)
The first and third term on the right are bounded by IIf middle term is < E. Hence
- f(v)l·
I.II
(lV, v) in the product
+ I/w(x) - LI·
If I is close to 11', then (1) shows that the first term on the right is small by (ii). If x is close to v then f".(x) is close to L by the preceding step. This proves our last limit, and concludes the proof of the theorem.
VII, §3. \ EXERCISES 1. Let I.(x) = x'/(I + x') for x ~ O. (a) Show that I. is bounded. (b) Show that the sequence {1.1 converges uniformly on any interval [0, c] for any number 0 < C < 1. (c) Show that this sequence converges uniformly on the interval x ~ b if b is a number > I. but not on the inlerval x ~ I. 2. Let 9 be a funclion defined on a set S, and let a be a number > 0 such that g(x) ~ a for all XES. Show that Ihe sequence
ng g,
= 1 + ng
converges uniformly to the constant function I. Prove the same thing if the assumption is that Ig(x)1 ~ a for all XES. 3. lei I.(x) = x/(I + nx 2 ). Show that that each function I. is bounded.
{1.1 converges uniformly for
x
E
R, and
4. Lei S be Ihe interval 0 ~ x < I. Let J be the funclion defined on S by J(x) = 1/(1 - x). (a) Determine whether J is uniformly continuous. (b) Let p,(x) = I + x + ... + x'. Does the sequence {p,l converge uniformly to J on S? (c) Let 0 < C < 1. Show that J is uniformly continuous on the interval [0, c], and that the sequence {p,l converges uniformly 10 J on this interval.
5. Let I.(x)
= x 2/(1 + nx 2 ) for
all real x. Show that the sequence
{1.1 converges
uniformly on R. 6. Consider the function defined by J(x) = lim lim (cos In!
1IX)2'.
the values of J at rational and irrational numbers. 7. As in Exercise 5 of §2, lei (y2 - x)2
J(x, y)
i
4
= : +x
2
if (x, y) # (0, 0), jf (x, y)
= (0, 0).
Find explicitly
188
[VII, §4]
LIMITS
Is f continuous on R 2 ? Explain, and determine all points where f is continuous. Determine the limits lim lim fIx, y)
x-o)'-o
and
lim lim fIx, y). )'-0 x-O
8. Let S, T be subsets of normed vector spaces. Let f: S --> T and g: T --> F be mappings, with F a normed vector space. Assume that 9 is uniformly continuous. Prove: Given c, there exists lJ such that if f.: S --> T is a map such that IIf-f.1I F is a map such that IIg-g,1I 0 such that IIxll ;;; C for all XES, where II II is the sup norm on RO. Let I be the closed interval - C ;;; x ;;; C. Then I is compact (by the Weierstrass':-Bolzano theorem I), and S is contained in the product
Ixlx···xI which is compact by Theorem 1.4. Since S is closed, we conclude from Theorem 1.2 that S is compact, as was to be shown. Remark. The notions of being closed or bounded depend only on the equivalence class of the given norm, and hence in fact apply to any norm on R" since all norms on RO are equivalent. Example. Let r be a number > 0 and consider the sphere of radius r centered at the origin in RO. We may take this sphere with respect to the Euclidean norm, for instance. Let f:Ro -+ R be the norm, Le. f(x) = Ixi. Then f is continuous, and the sphere is f - I(r). Since the point r is closed on R, it follows that the sphere is closed in RO. On the other hand, it is obviously bounded, and hence the sphere is compact. It is not true in an arbitrary normed vector space that the sphere is compact. For instance, let E be the set of all infinite sequences (XI' x" ...) with Xi E R, and such that we have Xi = 0 for all but a finite number of integers i. We define addition componentwise, and also multiplication by numbers. We can then take the sup norm as before. Then the unit vectors
e,
=
(0, ... ,1, 0, ...)
having components 0 except I in the i-th place form a sequence which has no point of accumulation in E. In fact, the distance between any two elements of this sequence is equal to I.
[VIII, §2]
CONTINUOUS MAPS ON COMPACT SETS
197
VIII, §1. EXERCISES 1. Let S be a compact set. Show that every Cauchy sequence of elements of S has a
limit in S. 2. (a) Let SI' ... ,Sm be a finite number of compact sets in E. Show that the union S. U ... U Sm is compact. (b) Let (S'},E/ be a family of compact sets. Show that the intersection () S, is compact. Of course, it may be empty. '.1 3. Show that a denumerable union of compact sets need not be compact. 4. Let (x.) be a sequence in a normed vector space E such that (x.) converges to v. Let S be the set consisting of all x. and v. Show that S is compact.
VIII, §2. CONTINUOUS MAPS ON COMPACT SETS Theorem 2.1. Let S be a compact subset of a normed vector space E, and let f: S -> F be a continuous map of S into a normed vector space F. Then the image of f is compact. Proof. Let {Yn} be a sequence in the image of f. Thus we can find X n E S such that Yn = f(x n). The sequence (x n) has a convergent subsequence, say {xn.l, with a limit v E S. Since f is continuous, we have
lim Yn. = lim f(x n.) = f(v). k-co
"-co
Hence the given sequence (Yn) has a subsequence which converges in f(S). This proves that f(S) is compact. Theorem 2.2. Let S be a compact set in a normed vector space, and let f: S .... R be a continuous function. Then f has a maximum on S (there exists v E S such that fIx) ;::; f(v) for all XES), and also a minimum. Proof By Theorem 2.1 the image f(S) is closed and bounded. Let b be its least upper bound. Then b is adherent to f(S). Since f(S) is closed, it follows that bE f(S), that is there exists v E S such that b = f(v). This prove the theorem for a maximum. The minimum case follows by considering - f instead of f
We can now prove for a compact set what is not true in general for closed sets. Corollary 2.3. Let K be a compact set in a normed vector space E, and let vEE. Then there exists an element WE K such that d(K, v) = Iv - wi.
198
[VIII, §2]
COMPACTNESS
Proof The function x...... Ix - vi for x e K has a minimum, at an element we K which satisfies the property asserted in the coroIlary.
Corollary 2.4. Let S be a closed subset of Rk with the sup norm. Let ve Rk • Then there exists we S such that d(S, v) = Iw - vi. Proof Exercise 4.
For infinite dimensional spaces, we gave a counterexample to the analogous statement in Chapter VII, foIlowing Theorem 2.4. Example 1. Using Theorem 2.2, in Exercise 3, you will give a much shorter proof that all norms on a finite dimensional vector space are equivalent. It is easier to remember this proof rather than the inductive proof given in Chapter VI. §4. Let S be a subset of a normed vector space, and let f: S ..... F be a mapping into some normed vector space. We recaIl that f is uniformly continuous if given £ there exists D such that whenever x, yeS and Ix - YI < Dthen If(x) - f(Y)1 < £. Theorem 2.5. Let S be compact in a normed vector space E, and let f:S ..... F be a continuous map into a normed vector space F. Then f is uniformly continuous.
E,
Proof Suppose the assertion of the theorem is false. Then there exists and for each n there exists a pair of elements xn, Yn e S such that
IXn - Ynl < I/n
but
I f(x n) -
f(yn)1 >
£.
There are an infinite subset I I of Z + and some v e S such that Xn ..... v for n 00, nell' There are an infinite subset 1 2 of I" and weS, such that Yn w for n ..... 00 and ne1 2 . Then, taking the limit for n ..... 00 and ne1 2 , we obtain Iv - wi = 0 and v = w. On the other hand, by continuity of J, we have f(x n) ..... f(v) and f(Yn) ..... f(w) = f(v) as xn ..... v and Yn ..... w. n e J2 • Hence taking the limit of the right side, we find If(v) - f(w)1 ~
E,
a contradiction which proves the theorem. Remark. It is sometimes possible, in proving theorems about functions or mappings, to consider their restrictions to compact subsets of the set on which they are defined. Thus a continuous function on R is uniformly continuous on every compact interval. Furthermore, if f is a continuous
[VIII, §2J
CONTINUOUS MAPS ON COMPACT SETS
199
map on a set S, and if I is uniformly continuous on S, then f is uniformly continuous on every subset of S. Thus a continuous function on R is uniformly continuous on every bounded interval. Theorem 2.6. Let S. T be subsets of normed vector spaces, and let f: S --+ T be continuous. Assume tilat S is compact und I is bijective, so f Iras lin inverse g: T --+ S. Theil 9 is contimwlIs.
Proof. Let {y,,} be a sequence in T converging to an clement IV E T. We have to show that g(y,,) converges to g(IV). Let v = g(IV). If {g(y,,)} does not converge to I', then there exists 0 and .r E T. Let I = g-I, so I(O) = 0 and fix) = x - I if x E (I, 2]. Then j: !I are bijective, f is continuous but 9 is not.
VIII, §2. EXERCISES
J: T - F be a mapping into some normed vector space. We ~ay that J is relatively uniformly continuous on S if given f there exists b such that whenever XES, yET and Ix - YI < b, then IJ(x) - J(y)1 < f. Assume that S is compact and J is continuous at every point of S. Verify that the proof of Theorem 2.5 yields that J is relatively uniformly continuous on S.
I. Let S c: T be subsets of a normed vector space E. Let
2. Let S be a subset of a normed vector space. Let J: S - F be"'a map of S into a normed vector space. Show that J is continuous on S if and only if thc restriction of J to every compact subsct of S is continuous. [Hillt: Givcn v E S, consider sequences or clements of S converging
(0
v.]
3. Prove that two norms on R" are equivalent by the following method. By the
200
[VIII, §2]
COMPACTNESS
first part of the proof of Chapter VI, Theorem 4.3, if a is a norm, then there exists C, > 0 such that a ~ C,II II (the sup norm). Thus a is continuous. Then a has a minimum on the unit sphere, so there exists 0> 0 such that a(v) ~ 0 for all vERn with IIvll = I. For any'" we have a(IV/IIIVII) ~ 0, so we can take C, = 1/0. 4. (Continuation of Exercise I, Chapter VII, §2.) Let E = R' and let S be a closed subset of R', Let v E R'. Show that there exists a point W E S such that dIS, v) = IIV - vi· [Hint: Let B be a closed ball of some suitable radius, centered at v, and consider the function x ..... Ix - vi for x E B,.., S.)
5. Let K be a compact set in R' and let S be a closed subset of R'. Define d(K, S) = glb Ix -
,.s
yi.
)CEil.
Show that there exist elements Xo
E
K and Yo
E
S such that d(K, S) = Ixo - Yol·
[Hint: Consider the continuous map x ..... dIS, x) for x E K.] Note. In Exercise 5, if K is not compact, then the conclusion does not necessarily hold. For instance, consider the two sets Sl and 52:
There is no pair of points xo, Yo whose distance is the distance between the sets, namely O. (The sets are supposed to approach each other.) 6. Let K be a compact set, and let I: K I is expanding, in the sense that II(x) -
-+
K be a continuous map. Suppose that
I(Yll
~
Ix - yl
for all x, yE K. (a) Show that I is injective and that the inverse map I-I: I(K) -+ K is continuous. (b) Sho,:, that I(K) ~ K. [Hint: Given Xo E K, consider the sequence Un(x o)}, where I IS the n-th Iterate of f. You might use Corollary 2.3.] 7. Let U be an open subset of R". Show that there exists a sequence of compact subsets Kj of U such ~at Kj c Int(Kj +.) for all j, and such that the union of all K j is U. [~int:.. Let BJ be the closed ball of radius j, and let K j be the set of points x E U ,.., Bj such that d(x, OU) ~ l/j.)
[VIII, §3]
ALGEBRAIC CLOSURE Of THE COMPLEX NUMBERS
201
VIII, §3. ALGEBRAIC CLOSURE OF THE COMPLEX NUMBERS A polynomial with complex coefficients is simply a complex valued function f of complex numbers which can be written in the form a; E C.
We call ao, ... ,an the coefficients of f, and these coefficients are uniquely determined, just as in the real case. H an ¢ 0, we call n the degree of f. A root of f is a complex number Zo such that f(zo) = O. To say that the complex numbers are algebraically closed is, by definition, to say that every polynomial of degree ~ 1 has a root in C. We sha/l now prove that this is the case. We write f(t) = antn +
... + ao
with an ¢ O. For every real number R, the function
If I such that
t ...... lf(t)1
is continuous on the closed disc of radius R, which is compact. Hence this function (real valued!) has a minimum value on this disc. On the other hand, from the expression
+ ... + -ao) f (t ) = a" t n( 1 + -an-I an t an t n we see that when It I becomes large, If(t)! also becomes large, i.e. given C> 0, there exists R > 0 such that if It I > R then I f(t) I > C. Consequently, there exists a positive number R o such that, if Zo is a minimum point of Ifl on the closed disc of radius R o, then If(t)1 ~ If(zo)1
for all complex numbers t. In other words, Zo is an absolute minimum of If I· We shall prove thatf(zo) = o. We express f in the form
with constants c;. If f(zo) ¢ 0, then Co = f(zo) ¢ O. Let z = t - Zo and let m be the smallest integer > 0 such that Cm ¢ O. This integer m
202
[VIII, §3]
COMPACTNESS
exists because I is assumed to have degree ~ 1. Then we can write
1(/) = I,(z) =
+ cmf' +
Co
zm+ 'g(z)
for some polynomial g, and some polynomial I, (obtained from I by changing the variable). Let z, be a complex number such that
and consider values of z of the type z = We have
1(/)
= 1,(Az,) = Co = co[1
Az" where A is real, 0
~
A ~ 1.
+ Am+'zi+1g(Az,) Am + Am+'zi+'co'g(AZ,)].
AmCo -
There exists a number C > 0 such that for all A with 0 ~ A < I we have
(continuous function on a compact set), and hence
If we can now prove that for sufficiently small A with 0 < A < I we have
o
O. There exist a finite number of open balls of radius r lVith cenlers at elements of S, whose IInion contains S. Proof Suppose this is false. Let x I E S and let B(x 1) be the open ball of radius r centered at x,, Then B(x l ) does not contain S, and there is some X2 E S, X2 ¢ B(x,). Proceeding inductively, suppose we have found open balls B(x,), ... ,B(xn ) of radius r, we select x n E S such that X n +1 does not lie in the union B(x l ) U .. · U B(xn ), and we let B(xn+d be the open ball of radius r centered at x n + l . By hypothesis, the sequence {x n } has a point of accumulation v E S. By definition, there exist positive integers m, k with k > m such that
+'
lx, Then
r 2
vi N, we can express the partial sums Sn and s~ for n ;:: N in the form n
Sn
= V, + ... + Vn = V, + ... + VN + I
Vb
k=N+ 1 n
s~ = V~
+ ... +
V~
=
V'n
L
+ ... + UN +
Uk-
k=N+1
The last sums from N + I to n on the right are equal to each other. Hence {sn} has a limit if and only if {s~} has a limit, as n -+ 00. Finally, we observe that if I Vn converges, then lim
Vn
= 0,
because in particular ISn+' - snl = Ivn+,1 must be less than £ for n sufficiently large. However, there are plenty of series whose n-th term approaches 0 which do not converge, e.g. I lin, as we shall see in a moment.
IX, §2. SERIES OF POSITIVE NUMBERS We consider first the simplest case of series, that numbers.
IS
series of positive
Theorem 2.1. Let {an} be a sequence of numbers;;:; O. The series
""
Ian n=1
converges if and only if the partial sums are bounded.
[IX, §2]
209
SERIES OF POSITIVE NUMBERS
Proof Let s"
= at
+ ... + an
be the n-th partial sum. Then {sn} is an increasing sequence of numbers. If it is not bounded, then certainly the series does not converge. If it is bounded, then its least upper bound is a limit, and hence the series converges, as was to be shown. In dealing with series of numbers;;; 0, one sometimes says that the series diverges if the partial sums are not bounded. Theorem 2.2 (Comparison test). Let L an and L bn be series of numbers with an' bn ;;; 0 for all n. Assume that L bn converges, and that there is a number C > 0 such that 0 ~ an ~ Cbn for all sufficiently large n. Then L an converges. Proof Replacing a finite number of the terms an by 0, we may assume that an ~ Cb n for all n. Then
'" al + ... + an ~ C(b, + ... + bn ) ~ C Lb•. k=l
This is true for alln. Hence the partial sums of the series L an are bounded, and this series converges by Theorem 2.1, as was to be shown. The comparison test is the test used most frequently to prove that a series converges. Most are compared for convergence either with the geometric series
L'" c", n=1
where 0 < c < I, or with the series
'"
L n=1
n1+("
The geometric series converges because 1 c"+' 1 +c+···+c n = - - - - I-c
I-c
and taking the limit as n -+ 00, we find that its sum is 1/(1 - c). The other series will later be proved to converge. Theorem 2.3 (Ratio test). Let L an be a series of numbers ;;; 0, and let c be a number, 0 < c < I, such that a,,+1 ~ ca" for all n slifficienlly large. Then L an converges.
210
[IX. §2]
SERIES
Proof We shall compare the series with the geometric series. Let N be such that 0.+ I ~ ca. for all n ~ N. We have
and in general by induction.
Hence m
I
aNH ~ aN(l
+ e + ... + em).
k= 1
and our series converges, by comparison with the geometric series. The next theorem concludes the list of criteria for the convergence of series with terms ~ O. Theorem 2.4 (Integral test). Let f be a function defined for 0/1 numbers ~ I. Assume that f(x) ~ 0 for a/l x. that f is decreasing. and that
f"' f(x) dx =
J
lim B--lXJ
1
fB f(x) dx
J
1
exists. Then the series
I"'
f(lI)
n=J
callverges. If tile imegl'al diverges. thell the series diverges. Proof For all n
~
2 we have f(lI) ;;;
~
I
2
f-,
f(x) dx.
--3
n
I
n
[IX, §2]
211
SERIES OF POSITIVE NUMBERS
Hence if the integral converges, J(2)
f
+ ... + J(n) ~
J(x) dx
~
r
J(x) dx.
Hence the partial sums of the series are bounded, and the series converges. Suppose conversely that the integral diverges. Then for n ~ I we have
" f.
+1
J(n)::::"
J(x) dx,
and consequently J(1)
+ ... + J(n) >
"+l
f
1
J(x) dx
and the right-hand side becomes arbitrarily large as n --+ co. Consequently the series diverges, and our theorem is proved. The integral test shows us immediately that the series because we compare it with the integral a>
f " f
1
which diverges. Indeed,
L lin diverges,
1 -dx
x
1 -dx = log n,
1
x
which tends to infinity as n --+ co. Observe that the integral test is essentially a comparison test. We compare the series with another series whose terms are the integrals of J from n - 1 to n. Example 1. Using the integral test, we can now prove the convergence of a>
L
n=1
1 n1 +£'
We compare this series with the integral
f.
a>
1 ---r+. dx.
1 X
212
[IX, §2]
SERIES
We have 8
J I
I xl+' dx =
x-' 18 I
-€
=
I
I
€B'
€
--+-.
Here of course, the £ is fixed. Taking the limit as B --+ 00, we see that the first term on the right approaches 0, and so our integral approaches 1/£; so it converges. For instance, I
L n3/2 converges. Note that the ratio test does not apply to the series
Indeed,
a
n+ ,
l
n +
I to prove convergence by means of a standard trick, as follows. We wish to show that
converges for s > I. Write s large,
= I + £ + (j with (j
> O. For alln sufficiently
log n -n. - ~ I, and so the comparison works, with the series I I/n' + '. Example 2. One can also use a comparison test to show that a series diverges. For instance, there is 110 integer d such that the series I I/(Iog nld converges. Indeed, given d, for all but a finite number of n, we have (log nld ~ n, so 1/11 ~ I/(Iog nld and since I I/n diverges, i.e. the partial sums become arbitrarily large as n --+ 00, it follows that the partial sums of I I/(Iog nld also become arbitrarily large as n --+ 00.
[IX, §2]
213
SERIES OF POSITIVE NUMBERS
One reason for having gone through systematically series with positive numbers is that they provide a test for arbitrary series. Theorem 2.S. Let L: an be a series of numbers. Assume that converges. Then the series L: an itself converges.
L: lanl
Proof Let
be the partial sum of the series. It will suffice to prove that the sequence {sn} is a Cauchy sequence. Indeed, for n > m we have n
ISn - sml =
L: ak k=m+l
n
;;;
L: lakl. k=m+l
By assumption, given € there exists N such that if m, n > N, then the right side is < €. This proves the theorem. A series L: an such that L: lanl converges is said to converge absolutely. We shall analyze this situation more systematically in §4, even for normed vector spaces.
IX, §2. EXERCISES 1. (a) Prove the convergence of the series L I/n(log 11)1 +. for every (b) Does the series L 1/11 log 11 converge? Proof? (c) Does the series L I/n(log n)(log log n) converge? Proof? What if you stick an exponent of I + E to the (log log n)?
€
> O.
2. Let L a. be a series of terms ~ O. Assume that there exist infinitely many integers n such that a. > I/n. Assume that the sequence (a.) is decreasing. Show that La. diverges.
3. Let L a. be a convergent series of numbers ~ 0, and let (bl> b2 , b3 ,· •• ) be a bounded sequence of numbers. Show that L a.b. converges. 4. Show that L(log 11)/11 2 converges. If s > I, does LOog n)3/n' converge? Given a positive integer d, does L(log nt/n' converge? 5. (a) Let II! = n(n - 1)(11 - 2)··· I be the product of the first 11 integers. Using the ratio test, show that L 1/1I! converges. (b) Show that L l/n· converges. For any number x, show that L x"/II! converges, and so does L X·/II·. 6. Let k be a integer ~ 2. Show that ~
L l/n 2 < I/(k n=k
I).
214
[IX, §2]
SERIES
L a: and L b; converge, assuming a, ;;; 0 and b, ;;; 0 for La,b, converges. [Hint: Use the Schwarz inequality but
all n. Show that be careful: The Schwarz inequality has so far been proved only for finite sequences.]
7. LeI
8. Let {a,} be a sequence of numbers;;; 0, and assume that the series L a,/n' con· verges for some number 5 = So' Show that the series converges for 5;;; So' 9. Let {a,} be a sequence of numbers ;;; 0 such that L a, diverges. Show that: (a)
L ~ diverges.
(b)
L 1+a'z converges. n a,
(c)
L
(d)
L~ sometimes converges and sometimes diverges. I + an
I
+ a,
an sometimes converges and sometimes diverges. 1 + nan
10. Let {a,} be a sequence of real numbers ;;; 0 and assume that lim a, = O. Let
,
n (I + a,} = (I + a,)(1 + aZ)"'(1 + a,).
11:::1
We say that the product converges as n exists, in which case it is denoted by
00
if the limit of the preceding product
'"
n (I + a,). ,-, L
Assume that a, converges. Show that the product converges. [Hint: Take the log of the finite product, and compare 10g(1 + a,) with a,. Then take exp.] II. Decimal expansions. (a) Let a be a real number with O:S; a :s; I. Show that there exist integers a, with 0 ~ a, ~ 9 such that '" a
a= ,_, L~' 10"
L
The sequence (a" a z , ... ) or the series a,/IO" is called a decimal expansion of a. [Hint: Cut [0, I] into 10 pieces, then into IOz, etc.] (b} Let a =
L'"
k=m
a,/IO' with numbers a, such that la,l;;;; 9. Show that lal ;;;;
1/10"'-'. (c) Conversely, let {a,) be integers with
a=
la,l
~ 9,
a,
¢
±I
for all k. Let
'" a L-"
'=1 10'
Suppose that lal ~ 1/10" for SOme positive integer N. Show that k = I, .. . ,N - I.
a. =
0 for
[IX, §2]
(d) Let
SERIES OF POSITIVE NUMBERS
{J.
., ., = 11=1 L a.llf1' = 11=1 L b,/IO' with
215
integers a.. b, such that 0 ~ a, ~ 9 and
o ~ b, ~ 9.
Assume that there exist arbitrarily large k such that a, ¢- 9 and similarly b, ¢- 9. Show that a, = b, for all k.
12. Let S be a subset of R. We say that S has measure zero if given ( there exists a sequence of intervals (J.J such that
.,
L length(J.) < C, n= I
and such that S is contained in the union of these intervals. (a) If Sand T are sets of measure O. show that their union has measure O. (b) If S,. S, •... is a sequence of sets of measure O. show that union of all S,(i = 1,2•...)
has measure O. 13. The space {'. Let {, be the set of sequences of numbers
such that
converges. (a) Show that {, is a vector space. (b) Using Exercise 7. show that one can define a product between two elements X and Y = (y,. y, • ... ) by
.,
(X,
Y> = L
X.Y.·
n=1
Show that this product satisfies all the conditions of a positive definite scalar product, whose associated norm is given by
(c) Let Eo be the space of all sequences of numbers such that all but a finite number of components are equal to O. i.e. sequences
Then Eo is a subspace of {'. Show that Eo is dense in {'. (d) Let {Xi} be an {'-Cauchy sequence in {I. Show that {X;) is {'-convergent to some element in {'. So {, is complete. 14. Let S be the set of elements e. in the space {, of Exercise 13 such that e. has
216
[IX, §2]
SERIES
component I in the n-th coordinate and 0 for all other coordinates. Show that S is a bounded set in E but is not compact. 15. The spa« ('. Let (, be the set of all sequences of numbers X that the series
= {x.}
such
m
L Ix.1
"=1
converges. Define IIXII 1 to be the value of this series. (al Show that (' is a vector space. (b) Show that X f--> II XII , defines a norm on this space. (c) Let Eo be the same space as in Exercise 13(c). Show that Eo is dense in ('. (d) Let {Xi} be an (I-Cauchy sequence in {'. Show that {Xi} is ('-convergent to an element of {I. SO (, is complete.
La.
16. Let {a.1 be a sequence of positive numbers such that converges. Let (a.1 be a sequence of seminorms on a vector space E. Assume that for each x E E there exists C(x) > 0 such that a.(x) :;; C(x) for all n. Show that L a.a. defines a seminorm on E.
17. (Khintchine) Let f be a positive function, and assume that m
L
f(q)
q= I
converges. Let S be the set of numbers x such that 0 :;; x :;; I. and such that there exist infinitely many integers q, P > 0 such that
Show that S has measure O. [Hint: Given c, let qo be such that
L f(q)
0 such that for all integers q > 0 and integers p we have
m
Let IjJ be a positive decreasing function such that the sum
.-,
L ljJ(n)
converges.
[IX, §3]
217
NON-ABSOLUTE CONVERGENCE
Show that the inequality
has only a finite number of solutions. [Hine: Otherwise, t/J(q) > Cfq for infinitely many q. Cf. Exercise 2.] 19. (Schanuel) Prove the converse of Exercise 18. That is, let a. be a real number. Assume that for every positive decreasing function t/J with convergent sum L t/J(n), the inequality Ia. - p/q I < t/J(q) has only a finite number of solutions. Show that there is a number C > 0 such that Ia. - p/q I > Cfq for all integers p, q, with q > O. [Hine: If not, there exists a sequence I < q 1 < q 2 < ... such that
Ia. -
pJq;j < (1/2')q,.
Let t/J(I)
= L~. -,- '-"". I 2 qj
]
j""
IX, §3. NON·ABSOLUTE CONVERGENCE We first consider alternating series. Theorem 3.1. Let {an} be a sequence of numbers ~ 0, monotone decreasing to O. Then the series L (_I)n an converges, and
.,
L (-I)n an
~ al
·
n=1
Remark. Theorem 3.1 is an immediate corollary of the much more powerful Theorems 3.2 and 3.3. See the example after Theorem 3.3. However, we give an ad hoc proof first, because the special case has concrete features which deserve separate emphasis, at the cost of some inefficiency. Proof. Let us assume say that a, > 0, so that we can write the series in
the form
with bn ,
Cn ~
0 and b l
= al'
Let
sn = bl
-
CI
+ b2
-
C2
+ ... + bn ,
tn = bl
-
CI
+ b2
-
C2
+ ... + bn -
Cn •
(IX, §3]
SERIES
218
Then
Since 0 ~ bn + 1
N then
and if m ;;; N also, and say m ;;; n, then
Hence the series converges, the limit being viewed as either the greatest lower bound of the sequence {sn}' or the least upper bound of the sequence {tn }. Finally, observe that this limit lies between S 1 and
This proves our last assertion. Example 1. The series I (-l)"fn converges, by a direct application of Theorem 3.1. However, the series I lin does not converge, as we saw from the integral test. More generally, let 0;;;; x ~ 1. You should have done Exercise 15 of Chapter IV, §2, from which it follows that
The series on left are truncated at terms with a minus sign, and the series
[IX, §3]
219
NON-ABSOLUTE CONVERGENCE
on the right are truncated at terms with a plus sign. The inequalities are valid also at x = I, and for this value, let 5,. =
truncated alternating series at even terms,
52>+1 = truncated alternating series at odd terms,
i.e. put x = I in the above inequality, and let 52> be the left side, while 52k+1 is the right side of the inequality. Then we find 52> ~ log 2 ::;;
52k+1.
It is an exercise to see that the sequence {52.} is increasing, while the sequence (52k+d is decreasing. Furthermore, 52k+1 - 52> = 1/(2k + 1).
From this it immediately follows that the least upper bound of the increasing sequence {52>} is equal to the greatest lower bound of the decreasing sequence (52k+d, and both are equal to log 2. Example 2. There is another example similar to the alternating series. Let f(x) = (sin xl/x. The graph of f looks like this:
graph of sin x x
,
--- - -1
N and say In
IVml + ... + Iv,,1
N so large that if 11 > N I then U(I1) > N. This can be done because U is injective, and there is only a finite number of integers II such that U(II) ~ N. Then if k, I > N I we have
(2) because the terms in this sum are among the terms in the sum (1). This proves that the partial sums of the series IVol") I form a Cauchy sequence, and hence that the rearranged series is also absolutely convergent. We must see that it has the same limit. We want to estimate
L
m
(3)
m
N
k=1
11=1
00
00
L V.(k) - L v" = L V.1k) - L v" - L
11=1
11=1
V"
n=N+l
for III sufficiently large. Select M > N such that every integer /1 with 1 ~ II ~ N can be written in the form u(k) for some k ~ M. Such M exists because u is surjective. Consider those III > M. Then by (I), m L:
V.(k) -
1k=1
N L:
1/=1
I
v" ~
L: Iv,,1 ~ £, 00
,,=N+I
because the difference on the left contains only terms v" such that I12:N+l. Consequently we obtain the estimate for (3), namely m
.,
L Va(k) - I
1 lie, so that for n ~ m ~ N the above expression is < e. Note how the estimate on the right is independent of x E [0, I], so the convergence is uniform. Example 2. The series L (-I)·(x + n}/n 2 converges uniformly for every interval - C ;::; x ;::; C. Indeed, for all n sufficiently large, (x + n)ln 2 is positive, and for such n, x+n C I 2 0 O. One distinguishes the three cases in which for a given x E A we have x E An' or x ¢ An but x ¢ Bn, or x E Bn. The desired inequality of I. is then obvious in each case, using the inductive hypothesis. From our inequalities (*), we then conclude that the series go
+ g. + ... + gn + ...
converges pointwise, and furthermore converges to f on A. The uniform bounds imply at once that the limit function is continuous, thus proving Theorem 5.2, when f has values in [0, I]. The restriction to the interval [0, IJ is of course unnecessary, and the theorem extends at once to any other closed bounded interval, for instance by mapping such an interval linearly on [0, I]. Now suppose that f is unbounded. Using the arctangent map we reduce the theorem to the case when f takes values in the open interval (- I, I) and we must then know that the extension can be so chosen that its values also lie in the open interval (-I, I). Let B be the closed set where the extension f* (which we have constructed with values in [ -I, IJ) takes on the values I or -I. Then A and B are disjoint, so there exists a continuous function 11 on E with values in [0, IJ such that h is I on A and 0 on B. Then hf* has values in the open interval (-I, I), as desired. This concludes the proof of Theorem 5.2.
IX, §5. EXERCISES I. Show that the following series converge uniformly and absolutely in the stated
interval for x. I
(a)
L It 2 + X 2 for 0 ;;; x
(cl
L x"e-n., for x f; O.
(b)
sin
IlX
L ---",for all x /I
2. Show that the series x"
L I + x"
[IX, §5]
ABSOLUTE AND UNIFORM CONVERGENCE
233
converges uniformly and absolutely for 0 ~ Ixl ~ C, where C is any number with 0 < C < 1. Show that the convergence is not uniform in 0 ~ x < 1. 3. Let "
fIx) =
I
.~1 I + n2 x·
Show that the series converges uniformly for x i1; C > O. Determine all points x where f is defined, and also where f is continuous. 4. Show that the series
converges absolutely and uniformly on any closed interval which does not contain an integer. 5. (a) Show that
converges uniformly on any interval [0, C] with C > O. (b) Show that the series x2+n
co
L(-I)" n"" I
2
n
converges unifonnly in every bounded interval, but does not converge absolutely for any value of x. 6. Show that the series L eM'/n is uniformly convergent in every interval [~, 2n - ~J for every ~ such that O 0, whence r ~ l/t. Conversely, given £ there exist infinitely many n such that 10.1'/· ~ t - £, and therefore
[IX, §6]
237
POWER SERIES
Hence Ihe series L 0,,2" does not converge if Izi = 1/(1 - E), because its /I-th term does not even tend to O. Therefore the radius of convergence r satisfies r ;;; 1/(1 - E) for every E > 0, whence r;;; 1/1. This concludes the proof in case I oF 0, 00. The case when I = 0 or 00 will be left to the reader. The above arguments apply, even with some simplifications. Corollary 6.5. If lim 10,,1 1/" =
I
exisls, Ihen r = 1/t.
Proof If the limit exists, then I is the only point of accumulation of the sequence ja"I"", and the theorem states that 1= l/r.
Let {a,,}, {b,,} be two sequences of positive numbers. We shall write for n -->
00
if for each n there exists a positive real number u" such that lim u~/" = 1, b l' , eXists ,· . and an -- bnUll. If I·1m an11"· eXists, an d an =- b then I·Im,1 and IS equal to lim a~/". We can use this result in the following examples. II'
Example. The radius of convergence of the series Ln! z" is O. Indeed, we have n! ;;: n"e-" and (n !)I/" is unbounded as n --> 00. Example. The radius of convergence of L(I/n !)2" is infinity, because l/n! ;;: e"/n" so (I/n!)I/" --> 0 as n --> 00. Observe that our results apply to complex series, because they involve only taking absolute values and using the standard properties of absolute values. The discussion of absolute convergence in normed vector spaces applies in this case, and the comparison is always with series having real lerms ~ O. The next lemma is for the next section. Lemma 6.6. Let s be Ihe radius of convergence of the power series
L a"z".
Then the derived series
L na"z"-I
converges absolutely for
121 < s. Proof Recall that lim nil" = 1 as n --> 00. Without loss of generality, we may assume that a" 2: O. Let 0 < C < S and let c < CI < s. We know that L a"c'1 converges. For alln sufficiently large, nl/llc < c. and hence
converges. This proves that the derived series converges absolutely for 121 ;;; c. This is true for every c such that 0 < c < s, and consequently the derived series converges absolutely for Izl < s, as was to be shown.
238
[IX, §6]
SERIES
Of course, the integrated series
a z"
lJ+ I
I: a"z = zL: -"1l+1 "+1 also converges absolutely for 1=1 < S, but this is even more trivial since its terms are bounded in absolute value by the terms of the original series, and thus the integrated series can be compared with the original one. In the next section, we shall prove that the term-by-term derivative of the series actually yields the derivative of the function represented by the series.
IX, §6. EXERCISES 1. Determine Ihe radii of convergence of the following power series.
(a)
L nx"
(b)
L n'x"
(c)
X' Ln
(d)
L =--II"x"
(c)
L 2"X'
(f)
L 2"
(g)
L (II' x"+ 211)
(h)
L (sin 1Il')x"
x"
2. Determine Ihe radii of convergence of the following series. (a)
L (log II)X'
(b)L,ognX'
(c)
1 " L II10'" X
(d)
L
(e)
L(411 _ I)!
(f)
L (21i + 7)! x"
II
1
1i(log Ii)'
X'
2"
x"
3. Suppose that L a"z" has a radius of convergence r > O. Show that given A > I/r there exists C > 0 such that
la,,1 4. Let
~ CA"
for all Ii.
{a,,} be a sequence of positive numbers, and assume that lim a"+t/a,, =
A ;;; O. Show that lim a~/"
= A.
5. Determine the radius of convergence of the following series. II! (a) "-z" tl
'- n
b
(1i!)3"
( ) L(311)!z
6. Let {a,,} be a decreasing sequence of positive numbers approaching O. Prove
[IX, §7]
DIFFERENTIATION AND INTEGRATION OF SERIES
239
L a.z· is uniformly convergent on the domain of complex
that the power series z such that
Izi ~ I
Iz -
and
II
~
15,
where 15 > O. Remember summation by parts, and Theorem 3.3. ~
L a.z· be a
7. Abel's theorem. Let
power series with radius of convergence ~ I.
It""l ~
Assume that the series
L a. converges.
.-1
Let 0 ~ x < I. Prove that
~
lim
~
L a.x· = L a•.
x-I It=l
n=1
(Hint: Use Theorem 3.3 to show that the conditions of Chapter VII, Theorem 3.5 are satisfied. Actually it falls out that the partial sums (s.(x)} converge uniformly on the closed interval [0,1].
IX, §7. DIFFERENTIATION AND INTEGRATION OF SERIES We first deal with sequences.
Theorem 7.1. Let {f,,} be a sequence of continuous functions on an interval [a, b], converging uniformly to a jimction f (necessarily continuous). Then
Proof We have ff. Given
E,
ff
~
f(f" -
f)
~ (b -
a)II.r.. - fll·
we select n so large that
IIf" -!II
O. Hence on the open interval (0, 21t) we have g'(x) = If (x). In the present example, the convergence is more delicate than in Example 2. In the chapter on Fourier series, you will be able to see that the series of the present example represents a very simple function. Cf. Exercises 1 and 9 of Chapter XII, §4. Next we deal with power series. Corollary 7.4. Lee I a.x· be a power series wieh radius oj convergence s > O. Lee J(x) = I a.x·. Then J'(x) =
I
na.x·- 1
Jorlxl < s.
Proof Let 0 < c < s. Then the power series converges uniformly for
Ix I ~ c, and so does the derived series by the lemma of the preceding section. Hence J'(x) = I na.x"- I for Ixl < c. This is true for every c such that 0 < c < s, and hence our result holds for Ix I < s. Corollary 7.4 shows that even though a series may not converge uniformly on a certain domain. nevertheless this domain may be the union of subintervals on which the series does converge uniformly. Thus on each such interval we can differentiate the series term by term. The result is then valid over the whole domain. In particular, uniform convergence is usually easier to determine on compact subsets, as we did in Corollary 7.4, selecting c such that 0 < c < s and investigating the convergence on 0< Ixl ~ c. Corollary 7.5. Let I a.x· have radius oj convergence s, and lee J(x) =
I
a.x·
Jor
Ixl < s.
[IX, §7]
DIFFERENTIATION AND INTEGRATION OF SERIES
243
Let •.n+ 1
F(x)
= 2: --"a"_A--,n +1
Then F'(x) = f(x).
Proof Differentiate the series for F term by term and apply Corollary 7.4. We have now derived the theorems giving a proof of the existence of functions f, 9 such that f'
=9
and
f(O)
=0
and
g'=-j,
g(O)
= 1.
Indeed, we put X2n-1
00
f(x) = J,(_1)"-1 (2n - I)!' x2n
00
g(x)
= J}-I)" (2n)!"
This fills in the missing existence of Chapter IV, §3, for the sine and cosine. Similarly, we have proved the existence of a function f(x) such that and
f'(x) = f(x)
f(O) = I,
namely f(x) =
x" 2:-. n!
This fills in the missing existence of Chapter IV, §l, for the exponential function.
IX, §7. EXERCISES I. Show Ihal if f(x) = L 1/(n 2 Ihis series term by term.
+ x 2)
then f'(x) can be obtained by differentiating
[IX, §7]
SERIES
244 2. Same problem if f(x) =
L 1/(n '
- x'), defined when x is not equal to an integer.
3. Let F be the vector space of continuous functions on [0,2,,] with the sup norm. On F define the scalar product
f'·
~o. If f
~
g, then
The first statement follows from the fact that we can find a convergent
[X, §4]
PROPERTIES OF THE INTEGRAL
257
sequence of step functions {fn} converging to f such that fn > 0 for all n. The second follows from the first by considering the integral of g - f.
Theorem 4.5. Let f be regulated on [a, b]. Let c E [a, b]. Then for x E [a, b] we have
1'f ;;;Ix-cillfll· The map x I-t
LX f
is continuous.
Proof. To prove the inequality if x < c we reverse the limits of integration. Otherwise, the inequality follows by the limiting process from the integral of step maps. As for the continuity statement, it is clear, because if we let F(x) =
1'f
then IF(x
+ h) -
F(x)1
; c ;;; b. Assume that f is continuous at c and that f(c) > 0, and also that f(t) ~ 0 for all t E [a, b]. Then
ff>O. Proof. Given f(c), there exists some {) such that f(t) > f(c)/2 if It - cl < {) and t E [a, b]. If c #' a, let 0 < ,t < {) be such that the interval
[c - J., c] is contained in [a, b]. Then
thereby proving our assertion. If c = a, we take a small interval [a, a and argue in a similar way.
+ ,t]
258
THE INTEGRAL IN ONE VARIABLE
[X, §4]
Consider now the special case when f takes its values in k-space R'. Then f can be represented by coordinate functions, f(t) = (Ji(t), ... J,(t). It is easily verified that f is a step map if and only if each f; is a step function, and that if f is a step map, then
In other words, the integral can be taken componentwise. Thus by taking limits of step maps, we obtain the same statement for integrals of regulated maps: Theorem 4.7. If f: [a, b] -+ R' is regulated, then each coordinate function fl> ... J, of f is regulated, and
Thus the integral of a map into R' can be viewed as a k-tuple of integrals of functions. However, it is useful not to break up a vector into its components for three reasons. One, the geometry of k-space can be easily visualized without components, and the formalism of analysis should follow this geometrical intuition. Second, it is sometimes necessary to take values of f in some space where coordinates have not yet been chosen, and not to introduce irrelevant coordinates, since the pattern of proofs if we don't introduce the coordinates follows the pattern of proofs in the case of functions. Third, in more advanced applications, one has to integrate maps whose values are in function spaces, where there is no question of introducing coordinates, at least in the above form. However, in some computational questions, it is useful to have the coordinates, so one must also know Theorem 4.7. Note that in the case of the complex numbers, if f is a complex valued function, f =
0
and
CE
[a, b).
lim J(c + h)
'-0
. 0, respectively, the left limit with h < O. [Hint: First prove the result if J is an increasing function.]
X, §4. APPENDIX. THE LEBESGUE INTEGRAL For the present course, it is not necessary to deal with integrals of functions other than those considered in this chapter, or those for which a rather naive notion of improper integrals can be defined as in Chapter XIII. These integrals can be handled in very short space, and are serviceable for the numerous applications we give in this book. However, it may be illuminating to see how these integrals can be subsumed under a larger class, which is covered in a subsequent course. The definition of this class may be based on what I call the fundamental lemma of Lebesgue integration, which is presented in this appendix. Instead of approximating certain functions uniformly by step maps, we approximate them in the L 1_ seminorm by step maps. In other words, instead of considering Cauchy sequences of step maps for the sup norm, we consider Cauchy sequences of step maps for the L'-seminorm. We then want to see how such a sequence converges pointwise. The answer is given in the present appendix. The proof is an e/2" proof. Both the result and its proof provide examples of notions which have been defined so far, so this appendix can be used both to illuminate past results in a new light, and to serve as an introduction for future happenings. Note that our approach to integration, starting in §I, continuing in §2, and also in the present appendix, sets things up along the following general pattern. To prove something in integration theory at the basic level, first prove it for a subspace of functions for which the result is obvious or easy, then extend by linearity and continuity to the largest possible subspace. The very definition of the integral, whether in §2 or in this appendix, follows the general pattern. Nothing will be used about the space of values of functions or mappings except linearity, a norm, and completeness, so there is no reason to assume anything more. Thus we deal with mappings f: R .... E, where E is a complete normed vector space. The length of an interval A will be called the measure of A and will be denoted by ptA). In this definition, the interval may be open, closed, or half closed. Thus the interval may consist of a single point. A subset S of R will be said to have measure 0 if given e > 0 there exists a sequence of
[X. §4. ApP.]
263
THE LEBESGUE INTEGRAL
intervals {An} such that and
I'"
J.l(A n )
0, we have
Let r = (I - ( 2 ). Then 0 < r < I, and (11 + I)r" approaches 0 as 11 -> 00. This proves condition DIR 3. (The integral on the other side has the same value because of the symmetry of K n .) Thus {K n } is a Dirac sequence. There remains to show only that
=
fn(x)
is a polynomial. But
f
f:
Kn(x - t)f(t) dt
is equal to 0 outside [0, I]. Hence fn(x)
=(
Kn(x - t)f(t) dt.
Observe that Kn(x - t) is a polynomial in t and x, and thus can be written in the form
where 90' ... ,g 2n are polynomials in t. Then
where the coefficients Q, are expressed as integrals Q,
=(
9,(t)f(t) dt.
This concludes the proof of the Weierstrass theorem. The functions K n used in this proof are called the Landau kernels.
XI, §2. EXERCISES I. Let! be continuous on [0, I]. Assume that
fa' !(x)x" dx
=
0
290
APPROXIMATION WITH CONVOLUTIONS
[XI, §2]
for every integer n = 0, I, 2, .... Show that J = O. [Hinr: Use the Weierstrass theorem to approximate J by a polynomial and show that the integral of J2 is equal to 0.] 2. Prove that if J is a continuous function, then lim h-O '>0
f
h
l
2
_Jh +x
2
J(x}dx
= nJ(O}.
3. An integral operator. Let K = K(x, y) be a continuous function on the rectangle defined by inequalities and
e~y~d.
For J E CO([e, d]), define the function TJ = TK! by the formula
J, d
TK!(x) =
K(x, y}J(y} dy.
(a) Prove that TK is a continuous linear map CO([e, d]) -+ CO([a, b]), with the sup norms on both spaces. (b) Prove that Tx is a continuous linear map with the L 2- norm on both spaces. Remark. One often denotes TKJ by K. J. When we take a convolution in the text, we use addition implicitly to get a function of two variables out a function of one variable, that is, given the K. in the text, we could define L.(x, y} by L.(x, y) = K.(x - y).
CHAPTER
XII
Fourier Series
XII, §1. HERMITIAN PRODUCTS AND ORTHOGONALITY We shall consider vector spaces over the complex numbers. These satisfy the same axioms as vector spaces over the reals, except that the scalars are now taken from C. Let E be a vector space over C. By a hermitian product on E we mean a map E x E -+ C denoted by (v, w) t-+ 0 whenever v # 0, we say that the product is positive definite. However, we don't assume that. Example I. This is the example with which we are concerned throughout the chapter. Let E be the vector space of complex valued functions on R which are piecewise continuous (on every finite interval) and periodic of period 2n. Thus these are essentially the piecewise continuous functions on the circle, as we say. Iff, geE, we define
(/. g)
= f/(X)g(X) dx.
The standard properties of the integral show that this is a hennitian product satisfying the four conditions. The complex conjugate which appears in the definition is to guarantee HP 4. In the case of real valued functions, it is of course not needed. For complex valued functions, we have ff = Ifl 2 and we know that the integral of a function ::=: 0 is also ~ o. We let E R be the space of real valued functions in E. Thus E R is the space of real valued piecewise continuous functions of period 2n. If f is complex valued, and f = f. + if2 is its decomposition into a real part and an imaginary part, then fe E if and only if fl.!2 e E R. This is obvious. We return to an arbitrary vector space E over C, with a hermitian product. We define v to be perpendicular or orthogonal to w if (v, w) = O. Let S be a subset of E. The set of elements veE such that (v, w) = 0 for all w e S is a subspace of E. This is easily seen and will be left as an exercise. We denote this set by SJ.. In particular, we may take S to consist of E itself. Thus EJ. is the subspace of E consisting of those elements veE such that (v, w) = 0 for all weE. We denote EJ. by Eo, and call Eo the null space of the hermitian product. Example 2. If E is the space of periodic piecewise continuous functions as before, and feE is such that (/.f) = 0, this means that
We know that if 9 is continuous at a point and # 0 at that point, and if 9 is otherwise ~ O. then its integral is > O. Hence we conclude that If 12 is equal to 0 except at a finite number of points. It follows that f is equal to 0 except at a finite number of points. Conversely, if f has this property, then (f, g) = 0 for all geE. Hence Eo consists of all functions which are equal to 0 except at a finite number of points.
[XII, §l]
HERMITIAN PRODUCTS AND ORTHOGONALITY
293
The next theorem shows in general Ihat the subspace Eo can be characterized by a weaker property. Indeed, we show that if weE and w is orthogonal to itself, then w is orthogonal to all elements of E. Formally stated: Theorem 1.1. If weE is such that (w, w) (w, v) = 0 for all ve E.
= 0,
then weEo , that is
Proof Let t be real, and consider
o:;; (v + tw, v + tw) =
(v, v)
+ 2t Re(v, w) + t 2 (w, w)
= (v, v) + 2t Re(v, w). If Re(v, w) '" 0 then we take t very large of opposite sign to Re(v, w). Then (v, v) + 2t Re( v, w) is negative, a contradiction. Hence
Re(v, w)
= O.
This is true for all ve E. Hence Re(iv, w) = 0 for all ve E, whence Im(v, w) = O. Hence (v, w) = 0, as was to be shown.
J
We define IIvll = (v, v), and call it the length or norm of v. By definition and Theorem 1.1, we have Ilvll = 0 if and only ifveEo . We note another property of the norm, which we use in a moment, namely for any number a e C we have (av, av) = cxa(v, v), and so lIavll = lalllvll, where lal is the absolute value of a, and IIvll is the norm of v. Pythagoras theorem. If u, weE are perpendicular, then Ilu
+ wll 2 = lIull 2 +
IIw1l
2
•
The proof is immediate from the definitions. Next let v, weE. We want to orthogonalize v from w; in other words we seek a number c such that v - cw is perpendicular to w. Suppose (w, w) '" O. Then: v - cw.l w ..,. (v - cw, w) = 0, ..,.
(v, w) - (cw, w) = 0,
..,.
(v, w) - c(w, w)
(v, w) ..,. c = (w, w)'
= 0,
294
FOURIER SERIES
[XII, §I]
Thus there exists a unique number c such that v - cw 1- v, and this number c is given by c=
(v, w) . (w, w)
We call c the component of v along w, or the Fourier coefficient of v with respect to w. It is natural to define the projection of v along w to be the vector cw, because of the following picture where c is the Fourier coefficient.
We shall give examples of such Fourier coefficients later. For the moment, we give an application of Pythagoras and the orthogonalization process to prove: Theorem 1.2 (Schwarz inequality). For all v, wEE we have
I(v, w)1 Proof If
IIvll or IIwll
~
IIvllllwll
0, then both sides of the inequality are equal to Suppose w ¢ O. Let c be the component of v along w, so v - cw 1- w. Write =
oso the theorem is proved.
v=v-cw+cw.
Then v - cw and cw are perpendicular, so by Pythagoras,
because IIv - cwll 2 ~ O. We substitute the value for c found above, and cross multiply to get
I(v, w>l'
~
IIv1l 2 11w1l 2 •
Taking the square root proves the theorem. Just as in the real case, we see that the hermitian product satisfies the continuity condition which allows us to conclude that the product of a limit is the limit of a product. Also as in the real case, we conclude:
[XII, §I]
HERMITIAN PRODUCTS AND ORTHOGONALITY
295
The function v ...... IIvll is a seminorm on E, that is: SN 1. We have
II vii
~ 0, and
IIvll
= 0
SN 2. For every complex ex, we have SN 3. For v,
WE
if and only if VE Eo.
lIexvll
E we have IIv + wll
~
Iex III vII. IIvll + IIwll. =
Proof. The first assertion follows from Theorem 1.1. The second has
already been mentioned. The third is proved with the Schwarz inequality. If suffices to prove that
To do this, we have
+ wl1 2 = (v + w,v + w) = (v, v) + (w, v) + (v, w) + (w, w). (w, v) + (v, w) = 2 Re(v, w) ~ 21(v, w)l. Hence by Schwarz, IIv + wll 2 ~ IIvll 2 + 21(v, w) I + IIwll 2 ~ IIvll 2 + 211vllllwll + IIwll 2 = (IIvll + IIwll)2.
IIv But
Taking the square root of each side yields what we want. An element v of E is said to be a unit vector if II vII = I. If II vII # 0, then "III vII is a unit vector. Examples in the function space Example 3. We return to the space E which consists of piecewise continuous functions, periodic of period 2n. We shall determine Fourier coefficients with respect to certain important functions, defined as follows. We let Xn be the function
Xn(x) = i nx for each integer n (positive, negative, or zero). Then
(Xn' Xn) Thus we view
= 2n,
(Xn' Xm) = 0 if m # n.
xnl.J2ir. as a unit vector.
If f is a function, then its Fourier coefficient with respect to Xn is
Cn
1 = -2
n
In f(x)e-,nX. dx. -n
296
[XII, §I]
FOURIER SERIES
We let lfJo = I, and for every positive integer n we let lfJn(x) = cos nx and t/Jn(x) = sin nx. We agree to the notation that for a positive integer n, we put lfJ-n(x)
= sin nx = t/Jn(x).
We thus have a unified notation {lfJn}n.Z' Elementary integrations show that for all nEZ, me Z, (lfJn, lfJm) = 0 if m # n.
Observe that we have lfJo lfJn' t/Jn and Xn' X-n' namely: einx
= Xo'
+ e- inx = 2 cos nx
Furthermore, we have relations between
and
einx
e- inx = 2i sin nx.
_
Of course, einx
= cos nx + i sin nx.
We shall usually use the letters ao, an, bn = a_ n to denote the Fourier coefficients of f with respect to I, cos nx, and sin nx respectively, so that: ao = -2 1
J'
n _,
J'
an = -1 f(x) cos nx dx, n _,
f(x) dx,
J'
a_ n = bn = -1 f(x) sin nx dx. n _,
Having the Fourier coefficients, one may then form what is called the Fourier series of f, which can be expressed either in tenns of the functions X. or the functions I, cos nx, sin nx. Let us first deal with the partial sums. Consider first the case of the functions {X,}, with n E Z, so that n ranges over all positive or negative integers and O. We agree to the convention that the partial sums of the Fourier series are or
s.(x) =
•
L
Ck eikx
k=-n
At the moment, nothing is said about the convergence of this series. We shall study various types of convergence later. The reader will immediately verify that if {a.}n. z are the Fourier co-
[XII, §1]
HERMITIAN PRODUCTS AND ORTHOGONALITY
297
efficients of f with respect to the family {lI'n}neZ, then n
n
-II
-II
L akll'k = L CkXk, in other words ao
n
n
k=1
k=-n
+ L (ak cos kx + bk sin kx) = L
Cke'''''
Let ERas before be the real vector space of real valued piecewise continuous periodic functions. Then the functions I, cos nx, sin nx (n E Z, n ~ 1) may be viewed as generating a subspace of E R as well as a subspace of E = Ec . The Fourier coefficients of a real function with respect to the functions ({in are real, so using the functions lI'n has the advantage of never leaving the real category of functions in Fourier theory. All the above results could have been carried out for a real vector space, as could subsequent results, for instance, Theorems 1.4 and 1.5. Readers should think this through. We now return to the complex case for definiteness. The Fourier series itself is defined to be Sf
'" =L
-0:0
a.lI'.
= -co L'" C.X.,
or simply L a. lI'n , it being undersJood that n e Z, and that the sum is taken in the prescribed order, lim
•
L
"-IX) 11:=-,.
akll'k = lim
•
L
CkXk'
n... c:ok=-n
The Fourier series mayor may not converge. One writes the series Sf(x) = ao
+
..
L (a. cos nx + bn sin nx) n=1
independently of whether the series converges or not. It is then a problem to determine when the series converges and gives the value of the function f(x) at a given point x. Example 4. Let f(x) = x on the interval (-n, n), and extend f by periodicity to R. It doesn't matter what the values are at the end points.
298
[XII, §I]
FOURIER SERIES
However, note that there is no way to define a value at the end points so that the extended periodic function is continuous on all of R. The extended function is the sawtooth function whose graph is shown on the figure.
graph of I(x) - x, -][ < x
R be a continuous linear map. Show that there exists y E E such that .l(x) = (x, y) for all x E E. [Hilll: Let F be the subspace of all x E E such that .l(x) = O. Show that F is closed. If F # E, use Exercise 13 to get an element Z E E, z ¢ F, Z # 0, such that z is perpendicular to F. Dividing by its norm, one may assume IIzll = 1. Let y = .l(z)z.) Verify that Exercises 11-14 hold for a hermitian positive definite product over C.
15. Let E be a vector space over C with a hermitian product which is positive definite. Let v" ... ,v" be elements of E, and assume that they are linearly independent. This means: if c,v, + ... + C"V" = 0 with C;EC, then C, = 0 for all i. Prove that for each k = I, ... ,11 there exist elements IV" ... ,IV, which are of length I, mutually perpendicular (that is (w;. Wi) = 0 if i # j), and generate the same
306
[XII, §2]
FOURIER SERIES
subspace as v" ... ,v.. These elements are unique up to multiplication by complex numbers of absolute value I. [Hint: For the existence, use the usual orthogonalization process: Let
where c; are chosen to orthogonalize. Divide each in all the details and complete this proof.]
U;
by its length to get w;. Put
16. In this exercise, take all functions to be real valued, and all vector spaces over the reals. Let K(x, y) be a continuous function of two variables, defined on the square 0 ;:;; x ;:;; band 0 ;:;; y ;:;; b. A continuous function f on [0, b] is said to be an eigenfunction for K, with respect to a real number A, if
f
b
K(x, y)f(y) dy
= Af(x).
Q
Use the L2 ·norm on the space E of continuous functions on [0, b]. Prove that if f" ... ,f. are in E, mutually orthogonal, and of L 2·norm equal to I, and are eigenfunctions with respect to the same number A #< 0, then n is bounded by a number depending only on K and A. [Hint: Use Theorem 1.5.]
XII, §2. TRIGONOMETRIC POLYNOMIALS AS A TOTAL FAMILY By a trigonometric polynomial, we mean a finite linear combination n
P
= k:::-n L c.X.,
with coefficients Ck
E
n
that is
P(x)
n
= k=-n L CkXk(X) = k=-n L cke ikx,
C. Equivalently, we may write P in the fonn
P(x) = ao +
n
L (a k=1
k
cos kx
+ bk sin kx),
and the coefficients ao, ak , b,. are also in C. Usually, we use this second representation when dealing with real functions, so that ao , ak , bk are taken to be real. In this section, we assume the following theorem.
[XII, §2]
TRIGONOMETRIC POLYNOMIALS AS A TOTAL FAMILY
307
Theorem 2.1. Let f be a continuous function, periodic of period 27 0 and an open interval containing x such that for y in this interval, we have
If(x) - f(y)1 Or alternatively. there exists
a Clx -
yl.
°
If(x
such that for all
+ e) -
f(x) I
e with lei < 0,
we have
a Clel·
Theorem 4.5. Let f be a periodic, uniform limit of step functions on [ - n. n]. Suppose f satisfies a Lipschiez condieion ae a given poine x.
Then ehe Fourier series Sf(x) converges eo fIx). Proof With x fixed, also fix Dn* fIx) - fIx) =
=
r
£
> O. Using D 2 we get
Dn(e) [fIx - e) - fIx)] de
f' + f-' + f. _,
-n
n
Dn(e) [fIx - e) - fIx)] de.
By the Lipschitz hypothesis we can pick 0, such that for
If(x - t) - f(x)1 for
lei
a 0,.
a Clel
Then using D 3, the integral over [-0,0] with 0
a). Let a < c. We define
ra> f
Ja
if b
=
lim [
".. . 0
a+d
f + lim b..... co
c
Finally, if f is piecewise continuous on every bounded interval, we define
f
a>
f = lim [ f + lim
-«I
a... ao
-0
b ..... co
if b
c
These are independent of the choice of c. Example 2. The integral
converges, being equal to
From the properties of limits, it is clear that the improper integrals are again linear in f For instance, with co as a limit, and with any constant ex,
and
[XIII, §1]
DEFINITION
329
whenever each improper integral for f and g converges. Furthermore, if f is real valued and f > 0 for x > a, then
provided the integral converges, and thus if f ~ g then
again provided each one of these integrals converges. Finally, we define an integral
to be absolutely convergent if
f"f(X)' dx converges. We say that f is absolutely integrable (on R) jf
converges. Equivalently, we also say that f is in L 1•
XIII, §1. EXERCISES I. Let f be complex valued, f continuous. (a) Show that
= f, + if2
where fl' f2 are real valued, and piecewise
1 a we have
J8 I9 I for all B is a limit for this inte-
The least upper bound of all values gral, which therefore converges. •
Q
Example 1. We shall give an example with an improper integral over a finite interval. We wish to show that the integral
,
log r d
f.o r
1'2
t
converges. Write logr _
---'--/2 - (r r
114
log t)
I
r
3/4'
We know that lim (t l14 log t) = 0
'-0 (why?) and hence the function (t ' /4log t) is continuous on [0, I], hence bounded. On the other hand let f(t) = I/t 3 / 4 • Then the improper integral of f exists:
f.
1
o
I
-3/d t -- 4r l / 4 4 t
1
= 4,
0
whence our integral of (log l}/t'/2 converges. Note that we don't evaluate it.
332
[XIII, §2]
IMPROPER INTEGRALS
Example 2. Let A > O. We claim that
lim x A x- 0 as x --> co, thus proving our claim. The example iUustrates the general principle that to estimate, we can replace the expression by some other which can be handled much more easily. ~
Theorem 2.4. Let f be defined for x
1, say, and let
·+l
a. = If the integral
f
f
•
r f converges, then so does the series I a•.
Proof Obvious from the definition. The next theorem is a converse of Theorem 2.4. Theorem 2.S. Let f be defined for x
~
1, say, and let
·+l
a. =
f
•
f
Assume that
lim f(x)
= O.
If the series L a. converges, then so does the integral
r
f
. Prf Select N so large that if x> N then If(x)1 < £, and such that If s. IS the n-th partial sum of the series, then Is. - Sm I < £ whenever n,
[XIII, §2]
CRITERIA FOR CONVERGENCE
333
m> N. If N < B) < B 2 , then
fJ'"
+
+
fB''" f·
We select n;;;; B. < n + 1 and m < B 2 ;;;; m + 1. The length of each interval [n, B.] and [m, B 2 ] is at most 1. Using the hypothesis on f, we conclude that each one of the terms on the right of the inequality is < £; hence we have a 3£-proof of the theorem. Theorem 2.6. Let f, 9 be continuous for x ;;; a. Assume that f(x) is real monotone decreasing to 0 as x -+ aJ and that Jb 9 is bounded for all b ;;; a. Then the integral
roo fg
Ja
0
converges. In fact, let
G(x)
=
r
g.
Then we have the estimate
Proof In general, the stated estimate is a direct consequence of the
Bonnet mean value theorem, Exercise 1 of Chapter V, §2, and the Cauchy criterion. Since this theorem is delicate in general, we reprove the current theorem assuming that f is C·, which is all that matters in the applications. Integrating by parts, just as we summed by parts for the analogous theorem about series, and noting that G(a) = 0, we find
f
fg = f(b)G(b) -
1:
Gf'·
The assumption f decreasing implies f' :,; 0 so - f' ~ 0, whence we get the estimate
r
I
fgl :,; f(b)IIGII",
+ II Gil",
f(-
f') = f(b) II Gil",
+ II Gil", (f(a) -
f(b»)
= f(a) II Gil"" thus proving the estimate stated in the theorem. Then given e, for all A ;;;; B are sufficiently large, we get
334
[XIII, §2)
IMPROPER INTEGRALS
which proves the Cauchy criterion and also the convergence of the integral J:" fg, thus finishing the proof of the theorem. Example 3. The integral
f'" sin x dx
J. converges. We take f(x)
= I/x
x
and g(x)
= sin x.
Example 4. We can sometimes prove the convergence of an integral by integrating by parts. Suppose f, g, 1', g' are continuous functions. Then for B ~ a, we have
f
fg' = f(B)g(B) - f(a)g(a) -
f
gf'.
If f(B)g(B) -+ 0 as B -+ 00 and the integral on the right has a limit as B ..... 00, then so does the integral on the left. For example, the integral
can be handled in this way. It can also be handled, for instance, by writing
and observing that for x sufficiently large, xe- x / 2 ~ I. In that case, we can compare the integral with
f~-X'2 dx which is seen to converge by a direct integration between a, B and letting B tend to infinity.
XIII, §2. EXERCISES I. Show that the following integrals converge absolutely. We take a > 0, and P is a
polynomial. (a)
1~ p(x)e-' ,Ix
(b) 1"'p(X)e- O 'dX
(c)
1'"
(d)
(e)
{"(I +
P(x)e-'" dx
J~",p(x)e-OI'1 dx
Ixl)°e-'" dx for every positive integer /I
[XIII, §2]
CRITERIA FOR CONVERGENCE
335
2. Show that the integrals converge. "/2
(a)
f
0
1 Isin xll/2 dx
(b)
" f
1 I' 1"2 dx rr/2 sm x
3. Interpret the fOllowing integral as a sum of integrals between
nt,
and (n
+ l)n,
and then show that it converges.
1 2 o (x + l)lsin xl,,2 dx
f
a>
4. Show that the following integrals converge: a>
(a)
f
0
1
fi e-
x
dx
(b)
'f " r o
1 -e-xdx fors
converges uniformly for
f Proof. Given we have
£,
f(t, x) dt = i<X> f(t, x) dt Q
X E
Q
J. Then
L<X> f(t, x) dt dx
=
L<X>
f
f(t, x) dx dt.
there exists B o such that for all B ~ B o and all x E J
f
£
f(t, x) dt - L<X> f(t, x) dt O. Let sin e t
f( t,x ) =--e Then
f
-f%
.
is continuous for t ~ 0 and for all x. We shall consider first Then
x~a>O.
is absolutely integrable for x ~ a, that is
EO Isin tie-IX dt
$;
fa'" e- Ix de
converges. The other conditions of Theorem 3.4 are clearly satisfied, so that the function 9
() f. '" x
=
o
sin t -IX d --e t t
is differentiable, and
g'(x)
= -
fa"'e-f%Sin
tdt.
Since this formula for the derivative is true for each a > 0 and x :::: a, it is true for x > O. An estimate as in Theorem 2.6 can be used to show that the integrals above converge uniformly for x ~ O. We shall leave this to the reader. Example 2. Let tp(t) be a continuous function for t
converges. Then
~
0 such that
342
[XIII, §3]
IMPROPER INTEGRALS
converges, and g'(x)
= L"'it=
V 2n
f'"
l e -),'1/2 e -iyx dy=e -x l2 .
-'0
We are left with one theorem to prove for the case when both integrals are taken from 0 to co, in Theorem 3.3. In that case, we must put some supplementary condition, as shown in the example given in Exercise 8. Theorem 3.5. Let f be a continuous function of two variables, defined for t ;;:; a and x ;;:; c. Assume that: (I) The integrals
J.'' If(t, x)1 dt
and
i"'lf(t,X)ldX
converge uniformly for x lit every finite interval, and for finite interval respectively. (2) One of the integrals
i'" i'"
I f(t,
x) I dt dx
i'" i'"
or
I f(t,
converges. Then the following integrals exist, and they are equal.
J'" J% f(t,x) dt dx = J'" J'" f(l,x) dx dt c
a
a
c
t
in every
x) I dx dt
[XIII, §3]
343
INTERCHANGING DERIVATIVES AND INTEGRALS
Proof In condition (2), assume for instance that the first repeated integral converges. Assume first that f ~ 0, so that we may omit absolute value signs. Then by Theorem 3.3,
s: r rs: ~ rr· =
This is true for all b ~ a, and since all our integrals are ~ 0, the least upper bound of the integral on the left for b ;;:>: a is a limit of that integral, which therefore converges. Thus we have proved that the second integral also converges, and is less than or equal to the first. We can now use symmetry to conclude that they are equal. To deduce the general case from the special case just considered, we split f into its imaginary and real parts. The assumptions (1) and (2) apply to these, so that we may assume that f is real. Finally, we write f = 9t - 92 where 9. = max(O, f) and 92 = max(O, -f). Then 91> 92 are both ~ 0, and 9. ~ If I, 92 ~ If I· The hypotheses of the theorem apply separately to 9t and 92, and by linearity we see that our theorem is proved for f
XIII, §3. EXERCISES 1. Show that the integral
f
eo
g(x) =
sin
o
converges uniformly for x
~
I
-
e- ox dl
I
0 bUI does not converge absolutely for x
= o.
2. Let 9 be as in Exercise 1. (a) Show that you can differentiate under the integral sign with respect to x. Integrating by parts and justifying all the steps, show that for x> 0, g(x)
= -arctan x + const.
(b) Taking the limit as x .... 00, show that the above constant is n12. (c) Justifying taking the limit for x .... 0, conclude that
:f " o
sin I dl = ~. I
2
3. Show that for any number b > 0 we have
~ sin bl dl
fo
I
= ~. 2
344
[XIII, §3]
IMPROPER INTEGRALS
4. Show that there exists a constant C such that
[Hin/: Let fIx) be the integral. Show that f'(x) = -xf(x)/2. See the proof of Theorem 1.3 of the next chapter. Using the value
one sees that C = ,J;f2. The preceding value is best computed by reduction to polar coordinates as in elementary calculus. We deal with this later in the book.] 5. Determine the following functions in terms of elementary functions:
(a) fIx)
= f:~eO"~ sin Ix dl
6. Determine whether the following integrals converge:
fo xJI + x dx (b) J: sin (I/x) dx 7. Show that $; sin (x dx converges. [Hin/: Use the substitution x ~
I
(a)
2
2
)
2
= I.]
8. Evaluate the integrals
f
~ f~ I - X
I
1
(x
+ I)
3
dx dr f ~ f ~ .,..:./_--=X' (x + I) 3
and
dl dx
I
I
to see that they are not equal. Some sort of assumption has to be made to make the interchange of Theorem 3.5 possible. 9. For x
~
0 let g(x)
=f
~
log (U 2X 2 2
o
u
+ I)
+I
du
so that g(O) = O. Show that 9 is continuous for x ~ O. Show that 9 is differentiable for x > O. Differentiate under the integral sign and use a partial fraction decomposition to show that 1t 9 '(x ) = I +x
and thus prove that g(x) =
1t
log (I
+ x).
forx>O, (lowe this proof to Seeley.)
[XIII, §3]
INTERCHANGING DERIVATIVES AND INTEGRALS
345
10. (a) For y > 0 let
O. 12. Show that the formula in Exercise lI(d) is valid when x is replaoed by any com-
346
[XIII, §3]
IMPROPER INTEGRALS
plex number z not equal to a real number;'; O. Show that lim
fa>
)' ..... «1
0
P,(e) de iy + t
= O.
13. The gamma fUllCtion. Define
for x> O. (a) Show that f is continuous. (b) Integrate by parts to show that f(x + I} = xf(x). Show that f(l} = I. and hence that f(n + I} = n! for n = 0, 1,2, .... (c) Show that for any a > 0 we have
(d) Sketch the graph of f for x > O. showing that f has one minimum point, and tends to infinity as x - co, and as x - O. (e) Evaluate = [Hinl: Substitute 1= u' and you are allowed to use the value of the integral in the hint of Exercise 4.] (f) Evaluate f(3f2}, f(5f2}, ... ,j(n + t}· (g) Show that
fm fi.
fif(2n)
= 2"-'f(n)f(n + !}.
(h) Show that f is infinitely differentiable, and that
For any complex number s with Re(s} > 0 one defines the gamma funcrion
I(s) =
a>
f
o
dl e-'c'-. I
Show that the gamma function is continuous as a function of s. If you know about complex differentiability, your proof that it is differentiable should also apply for the complex variable s. 14. Show that
" f_a>(u' +
du =
I
I)'
r,; ....(s - t}
v"
I(s}
for Re(s} >
[Hint: Multiply the desired integral by I(s} and lettH (u'
t.
+ I}I.]
[XIII, §4]
THE HEAT KERNEL
347
15. A Bessel function. Let a, b be real numbers > O. For any complex number s define K t.
Show that the integral converges absolutely. For c > 0 define K2
Show that K 3.
K,(a, b) =
(~y K,(ab).
Show that
K 4.
= K _. 0 the function IxlmI(x) is bounded. Since Ixlm + lj(x) is bounded, it follows that lim IxlmI(x) = 0 Ixl- co
for every positive integer In. We let S be the set of all infinitely differentiable functions I such that I and everyone of its derivatives decrease rapidly at infinity. There are such functions, for instance e- x '. It is clear that S is a vector space over C. (We take all functions to be complex valued.) Every function in S is bounded. If I E S, then its derivative DI is also in S. and hence so is the p-th derivative D'f for every integer p 2': O. We call S the Schwartz space. Since
J OD
-OD
1 """-'---'2 dx
1+x
converges, it follows that every function in S can be integrated over R, i.e. the integral
fJ(X)dX 353
354
[XIV, §I]
THE FOURIER INTEGRAL
converges absolutely. For simplicity, from now on we write
since we don't deal with any other integrals. If P is a polynomial, say of degree m, then there is a number C > 0 such that for alllxl sufficiently large, we have
Hence if f E S, then Pf also lies in S. If f, g E S Ihen fg E S. (Obvious.) We see that S is an algebra under ordinary multiplication of functions. We shall have to consider the function -ixf(x), i.e. multiply by -ix. To avoid the x, we may use the notation (Mf)(x) = -ixf(x), and iterate, MPf(x) = (- ixY'f(x) for every integer p E; O. In order to preserve a certain symmetry in subsequent results, it is convenient 10 normalize integrals over R by multiplication by a conslant factor, namely 1/~. For this purpose, we introduce a notation. We write
We now define the Fourier transform of a function f
E
S by the integral
The integral obviously converges absolutely. But much more: Theorem 1.1. Iff E S, then J E S. We have
D"!= (M"lY
and
(D"lY
= (-I)"MPj.
[XIV, §l]
355
THE SCHWARTZ SPACE
Proof The function J is continuous and bounded since
The partial derivative
B.
.
- (f(x)e- ox") = -ixf(x)e- OX'
By
is bounded by Ixllf(x)1, and so we can differentiate the Fourier transform under the integral sign. We obtain
whence the first formula by induction, for D"f. As to the second, we integrate by parts the integral
using u = e-;x, and dv = Df(x) dx. Taking the integral over a finite interval [ - B, B] and then taking the limit (obviously converging), we find that (Df)'(y) = iy!(y).
By induction, we obtain the second formula From it, we conclude that! lies in S, because DPf E S, hence (DPf)' is bounded, and thus lyIPI!(y)1 is bounded. This proves Theorem 1.1. We now introduce another multiplication between elements of S. For f, 9 E S the convolution integral f
* g(x) =
J
f(e)g(x - e) dIe
is absolutely convergent. In fact, if C is a bound for g, then
356
[XIV, §I]
THE FOURIER INTEGRAL
* 9 E S. We have f an algebra under the product (f, g) ...... f * g. We have
Theorem 1.2. If f, 9 E S, then f
* 9 = 9 * f,
and S is
and (f*g)" =l{j. Proof Changing variables in the convolution integral, letting u = x - t, du = -dt, between finite bounds and letting the bounds tend to infinity, we see that f * 9 = 9 *f. The product is obviously linear in each variable. Since Dg is in S, we have a uniform bound If(t)Dg(x - t)1 ~ C1f(t)1
for some constant C, whence we can differentiate under the integral sign and find that D(f * g) = f * Dg. Iterating by induction gives the first formula, DP(f * g) = f * DPg. We now show that f * 9 is in S. Fix a positive integer m. For any x, t we have
with fixed numbers c,.. Then
is bounded, so f * 9 is in S. There remains but to prove the last formula. We have
Since f * 9 is absolutely integrable, being in S, we can interchange the orders of the integrals, and find (f * g)"(y) = JJf(t)g(X - t)e- ixy d.x d.t = Jf(t>[J g(x - t)e-i(x-,)y d.x}-i'Y d.t =
j(y){j(y)
after a change of variables u
=x-
t. This proves Theorem 1.2.
[XIV, §I]
THE SCHWARTZ SPACE
We conelude with a useful example of a function f such that f Theorem 1.3. Let f(x)
= e- x '/2.
Then f
357
= /.
= /.
Proof We know that
We integrate by parts, using u = e- ixy and dv = _xe- x '/2 dx. We first integrate between - B and B and let B -> 00. The term with uv will vanish because e- B'/2 will pull it to O. The other term shows that Dj(y)
=-
yj(y).
We differentiate the quotient j(y)
e -)'2/2 and find O. Hence there is a constant C such that
On the other hand 1(0)
=
f
e- x '/2 d,x
=I
(this is where the normalized integral is useful i). Hence the constant C is equal to I, thus proving the theorem. [As already mentioned, evaluation of fe-x' dx is best done with polar coordinates as in elementary calculus. We shall redo it later in the book.]
i,
and so forth. We shall From Theorem 1.3 we co.nelude that I = l\.eneralize the equality f = i to arbitrary functions in S, and find that i(.>:) = I( -x). In the special case of Theorem 1.3, the minus sign disappears because of the evenness of the function.
XIV, §1. EXERCISES I. Let 9 E S and define g.(x)
= g(ax) for a > O. O.(y)
[n particular, if g(x) = e- x ', find {J.{x).
Show that
= ~1_(Y) 9 ~ .
3)8
[XIV, §1]
THE FOURIER INTEGRAL
2. Normalize the Fourier series differently, for the interval [0, 1]. That is, define the scalar product for two functions J, 9 periodic of period I to be
J:
J(t)g(t) dt.
The total orthogonal family that corresponds to the one studied in Chapter 12 is then the family of functions
neZ. These are already unit vectors, that is these functions form an orthonormal family, which is often convenient because one does not have to divide by 2n. The theorems of Chapter XII go over to this situation, of course. In particular, if we deal with a very smooth function g, its Fourier series is uniformly convergent to the function. That's the application we are going to consider now. Let J be in the Schwartz space. Define a different normalization of the Fourier transform for the present purposes, namely define the Poisson dual
r
(x) =
f
J(t)e- 2 •
, <x
dc.
Prove the Poisson summation formula:
L nE
L
J(n) =
Z
ne
r(n).
Z
[Hint: Let g(x)
= L J(x + n). •• z
Then 9 is periodic of period 1 and infinitely differentiable. Let
Cm
=
f
1
o
,
g(x)e- 2• = dx
=
f
1
0
L nE
J(x
+ n)e- 2 • ,=
dx,
Z
Then
L
me Z
Cm
= g(O) = neLZ J(n).
On the other hand, using the integral for Cm, insert the factor 1 = e- 2"'mn, change variables, and show that Cm = JV (m). The formula drops out.] 3. Functional equation of the theta function. Let Ii be the function defined for x > 0 by ~
Ii(x) =
L e-""'. -~
[XIV, §2]
359
THE FOURIER INVERSION FORMULA
Prove tbe functional equation, namely
4. Functional equation of the zeta function (Riemann). Let s be a complex number. s = G + it witb G, t real. If G > I, and a > I, sbow tbat tbe series
L'" ...I
{Is) =
n= 1
n
converges absolutely, and uniformly in every region tion of s defined for G > I by
Let g(x) =
L'" .-n''', so tbat 2o(x) = It::::
G ;;;
a > I. Let F be tbe func-
O(x) - I. Sbow tbat
I
F(s) =
'f " o
dx x'/2 g(X) X
(I)
= f '"x'l2g(x) dx _ + f'" X-,/2 g I
X
I
_ dx -. X
X
Use tbe functional equation of tbe tbeta function to sbow tbat F(s)
=-I- - -I + s-I
s
f'" (X"2 + X"- m2 )g(x) dx- . x
I
Sbow tbat tbe integral on the rigbt converges absolutely for all complex s. and uniformly for s in a bounded region of the complex plane. The expression on tbe right tben defines F for all values of s oF 0, I, and we see tbat F(s)
= F(I
- s).
XIV, §2. THE FOURIER INVERSION FORMULA If f is a function, we denote by f- the function such that f-(x) = f( -x). The reader will immediately verify that the minus operation commutes with all the other operations we have introduced so far. For instance:
Note that
O. Then
and hence f !(x)e-iXVh(ox) d,x = f J(t)
= f J(au
~ f{ : y) d1t - Y)h(u) diu
after a change of variables, t+y
u = -a-' Both integrals depend on a parameter a --> 0 and find
0,
and are continuous in a. We let
h(O)j(y) = f( - y) f h(u) du = J( -
y)~(O).
Let h be the function of Theorem 1.3. Then Theorem 2.1 follows.
Theorem 2.2. For every J J = ')
for every E > O. It is known that the error term cannot be O(R'I2(log R)') for any positive integer k (result of Hardy and Landau). Prove the following best known result of Sierpinski-Van der Corput-Vinogradov-Hua: N(R)
[Hint: Let
tp
= nR' + O(R'/3).
be the characteristic function ofthe unit disc, and put
Let '" be a C" function with compact support, positive, and such that
J "'(x) dx = I,
and let
R'
Then ("',) is a Dirac family for
l _
0, and we can apply the Poisson summation
[XIV, §3]
AN EXAMPLE OF FOURIER TRANSFORM
367
formula to the convolution 'P. - 1/1, to get
L 'P. - I/I,(m) = L
mEZ 2
fiJ.(m)~,(m)
meZ 2
+
= nR'
L nR'fiJ(Rm)~(Em). mOO
We shall choose
E
depending on R to make the error term best possible.]
Note that 'P. -I/I,(x) = 'P.(x) if dist(x, S.) > R. Therefore we get an estimate
E,
where S. is the circle of radius
Ileft-hand side - N(R)I
~ lR.
Splilling off the term with m=O on the right-hand side, we find by Theorem 3.4:
L R'fiJ(Rm)~(lm) ~ R'-)f2 L Iml-)f2~(Em). m~O
m~O
But we can compare this last sum with the integral
Therefore we find N(R) = nR'
We choose l = R-
I/)
+ O(lR) + O(R ",, - 112 ).
to make the error term O(R "'), as desired.
2. Verify that the proof of Theorem 2.1 applies under the following more general conditions: The function 1 is in L 1, bounded and continuous. Here you can take L I to mean that
roo I/(x)\ elx exists, Le. converges. 3. Let
o. Under conditions of convergence, we define the Mellin transform M
O. Such a product expression is called an elementary partial differential operator. The sum
is called its degree or order. For instance DfD) has degree 2
+I=
3.
Remark. Some smoothness assumption has to be made in order to have the commutativity of the partial derivatives. See Exercise 12 for a counterexample. In the exercises, we deal with polar coordinates. Let x = r cos 9 and y = r sin 9. Let
f(x, y) = g(r, 9). We wish to express og/or and og/o9 in terms of of/ox and of/oy. We have
g(r, 9) = f(r cos 9, r sin 9). Hence
og ox or = D.!(x, y) or
oy
+ D,f(x, y) or
so
og or
(*)
.
= D I f(x, y) cos 9 + D2 f(x, y) sm 9.
Similarly,
og
ox
iJ6 = D.!(x, y) 09
oy
+ D,f(x, y) 09'
so (**)
og
.
iJ6 = D.J(x, yX -r sm 9)
+ D2 f(x, y)r cos 9.
Instead of writing Dtf(x, y), D2 f(x, y), we may write of/ox and of/oy respectively.
[XV, §l]
375
PARTIAL DERIVATIVES
In 2-space, the operator or
02
02
+ox 2 oy2
is called the Laplace operator. In Exercises 6 and 7 you will be asked to express it in terms of polar coordinates. A function / such that /t,f = 0 is called harmonic. Important functions which are harmonic are given in polar coordinates by r" cos n8
and
r" sin n8,
where n is a positive integer. You can prove easily that these functions are harmonic by using the formula of Exercise 6. They are fundamental in the theory of harmonic functions because other harmonic functions are expressed in terms of these, as infinite series
g(r,8) =
L'" anr" cos n8 + L'" bnr" sin n8, n=O
n= I
with appropriate constant coefficients an and bn. In general, a C 2 -function / on an open set of R n is called harmonic if (D~ + ... + D;)/ = O.
XV, §1. EXERCISES In Ihe exercises, assume Ihat all repeated partial derivatives exist and are continuous as needed. I. Let f, 9 be two runctions or two variables with continuous partial derivatives or order ~ 2 in an open set U. Assume Ihat
and
of oy
og
= ox'
Show that 02f ox
02f oy2
-+-=0. 2 2. Lei f be a runclion or Ihree variables, defined ror X that 02f 02f 02f +-+-=0 ox 2 oy2 OZ2
¢ 0
by f{X)
= I/IXI· Show
ir the three variables are (x, y, z). (The norm is the euclidean norm.)
[XV, §l]
FUNCTIONS ON n-SPACE
376 3. Let J(x, y)
= arctan(Ylx) for x > O.
Show that
iJ'J iJx'
iJ'J iJ}"
-+-=0
.
4. Let 8 be a fixed number, and let Y = u sin II
x = u cos 8 - v sin fJ,
LetJbe a function of two variables, and letJ(x, y)
(~:r + (~~r =
(rxr
+ v cos II.
= g(u, v).
+
Show that
(~r
5. Assume that J is a function satisfying
J(lx,ly) for all numbers x, y, and
I.
=
t"'J(x, y)
Show that
o'J iJ'J iJ'J x' - , + 2xy - - + i -, iJx iJx iJy iJy
= m(m -
l)J(x, y).
[Hinl: Differentiate twice with respect to I. Then put I = I.]
6. Let x = r cos II and y = r sin II. Let J(x, y) = g(r, II). Show that
iJJ iJx
= cos II iJg iJr
iJJ = sin 8 iJg iJy iJr
_ sin II iJg , r iJ8
+ cos II iJg • r
iJlI
[Hinl: Solve the simultaneous system of linear equations (.) and (••) given in the example of the text.]
7. Let x = r cos II and y = r sin II. Let J(x, y) = g(r, II). Show that
iJ'g 1 iJg 1 iJ'g iJ'J iJ'J -+--+--=-+iJr' r iJr r' oil' ox' iJy" This exercise gives the polar coordinate form of the Laplace operator, and we can write symbolically:
iJ)' (iJ)' (iJ)' 1 iJ 1 (iJ)' ( ox + iJy = or + ;: or + r' iJlI [Hint Jor Ihe prooJ: Start with (.) and (••) and take further derivatives as needed. Then take the sum. Lots of things will cancel out leaving you with
Dl!+DIfJ
[XV, §I]
377
PARTIAL DERIVATIVES
8. With the same notation as in the preceding exercise, show that
+..!.. (Og)'
( Og)' or
9. In R', suppose that f(x. y)
r'
00
= (Of)'
ox
+ (Of)'. oy
= g(r) where r = Jx' + y'. o'f
o'f
d'g
Show that
I dg
+ - - - + -r dr -. ox' oy' - dr' 10. (a) In R 3 , suppose that f(x, y, z) o'f
= g(r) where r = J
o'f
o'f
d'g
x'
+ y' + z'.
Show that
2 dg
+ - + - - - + -r dr' ox' oy' oz' - dr' (b) Assume that f is harmonic except possibly at the origin on R", and that there is a C' function 9 such that f(X) = g(r) where r = X . X. Let n ;E: 3. Show that there exist constants C, K such that g(r) = Kr'-" + C. What if n = 2?
J
J
11. Let r = x' + y' and leI r, 0 be the polar coordinates in the plane. Using the formula for the Laplace operator in Exercise 7 verify that the following functions are harmonic: (a) r" cos nO = g(r,O) (b) r" sin nO = g(r.O) As usual, n denotes a positive integer. So you are supposed to prove that the
expression o'g 1 og 1 o'g + + -or' r or r' 00' is equal to 0 for the above functions g. 12. For x
E
R" leI x' = xj
+ ... + x:.
For r real >0, let
f(x, t)
= t-""e- x "".
L
If l!. is the Laplace operator, l!. = 0'/ox!, show that l!.f = of/ot. A function satisfying this differential equation is said to be a solution of the beat equation.
13. This exercise gives an example of a function whose repeated partials exist but such that D,D,[ #< D,D,f. Let
",' - y'
f(x, y)
=
xY,
{
x
o
,
+Y
if (x, y) #< (0, 0), if (x, y)
= (0, 0).
Prove: (a) The partial derivatives o'f/ox oy and o'f/oyox exist for all (x, y) and are continuous except at (0, 0). (b) D,D,f(O,O) #< D, D,[(O, 0).
378
FUNCTIONS ON n-SPACE
[XV, §I]
Green's function 14. Let (a, b) be an open interval, which may be (a, (0). Let
=-
My
(~r + P(Y),
where p is an infinitely differentiable function. We view My as a differential operator. Iff is a function of the variable y, then we use Ihe notation Myf(y)
= -j"(y) + P(Y)f(y)·
A Green's function for the differential operator M is a suitably smooth function g(y, y) defined for y, y in (a, b) such that My
f
g(y,y')f(y') dy
= feY)
for all infinitely differentiable functions f on (a, b) with compact support (meaning f is 0 outside a closed interval contained in (a, b». Now let g(y, y) be any continuous function satisfying the following additional conditions: GF I. 9 is infinitely differentiable in each variable except on the diagonal, that is when y = y. GF 2. If Y ¢ y, then Myg(y, y)
= O.
Prove: Let 9 be a function satisfying GF 1 and GF 2. Then 9 is a Green's function for the operator M if and only if 9 also satisfies the jump condition
GF 3. D,g(y, y+) - D,g(y, y-)
= I.
As usual, one defines D,g(y, y+)
= lim D.g(y, y'). }"-y »'>),
and similarly for y- instead of Y+, we take the limit with y' < y. [Hint: Write the integral
(=( + (J IS. Assume now that the differential equation f" - pf = 0 has two linearly independent solutions J and K. (You will be able to prove this after reading the chapter on the existence and uniqueness of solutions of differential equations. See Chapter XIX, §3, Exercise 2.) Let W = J K' - J' K. (a) Show that W is constant ¢ O.
[XV, §2]
DIFFERENTIABILITY AND THE CHAIN RULE
379
(b) Show that there exists a unique Green's function of the form A(y')J(y) if y' < y, g(y, y') = { B(y')K(y) if y' > y,
and that the functions A, B necessarily have the values A
= K/W, B = J /W.
16. On the interval (-co, co) let M, = -(d/dy)2 + c2 where c is a positive number, so take p = c > 0 constant. Show that e" and e- 0 if lim ((J(h) = • -0
o.
Ihl
Observe that here, h = (hi"" ,h.) is a vector with components hi which are numbers. We use any norm I I on R· (usually in practice the euclidean norm or the sup norm). Of course we cannot divide a function by a vector, so we divide by the norm of the vector. If a function ((J(h) is o(h), then we can write it in the form ((J(h)
= Ih It/J(h),
where lim t/J(h) = O. • -0
All we have to do is to let t/J(h) = ((J(h)/Ihl for h # O. Thus at first t/J is defined for sufficiently small h # O. However, we may extend the function t/J by continuity so that it is defined at 0 by t/J(O) = o.
380
[XV, §2]
FUNCTIONS ON n-SPACE
We say a function f: U a vector A E R" such that f(x
R is differentiable at a point.x if there exists
-+
+
h)
= f(x) +
A .h
+
o(h).
By this we mean that there is a function qJ defined for all sufficiently small values of h '" 0 such that X2 + h 2,··· 'Xn + hn)h, + ... + Dnf(x., .. . ,xn-I> cn)hn, where C., ... 'Cn lie between Xi + hi and Xi' respectively. By continuity, for each i there exists a function IjJ i such that lim ljJ,(h)
=0
h-O
and such that DJ(XI>'" ,xi-I> Ci"" ,Xn + hn) = DJ(x)
+ ljJi(h).
Hence f(x
+ h) -
n
f(x) =
L (DJ(x) + ljJi(h»)h, i= 1
n
n
i= 1
i= I
= L DJ(x)h i + L ljJi(h)h i· It is now clear that the first term on the right is nothing but grad f(x) , h and the second term is o(h), as was to be shown. Remark. Some sort of condition on the partial derivatives has to be placed so that a function is differentiable. For instance, let f( x, ) Y
= x 2 xy + y2
f(O,O)
= O.
,'f (X, Y) ¥ (0, 0),
You can verify that Dt/(x, y) and D2 f(x, y) are defined for all (x, y), including at (0, 0), but f is not differentiable at (0, 0). It is not even continuous at the origin.
382
FUNCTIONS ON n-SPACE
[XV, §2]
When the function f is differentiable, we see from Theorem 2.1 that the gradient of f takes the place of a derivative. We say that f isdifferentiable on U if it is differentiable at every point of U. The rules for derivative of a sum hold as usual: Iff, 9 are differentiable, then
= grad f + grad g;
grad(f + g) and if c is a number,
grad(cf)
= c grad f.
One could formulate a rule for the product of two functions as usual, but we leave this to the reader. At the moment, we do not have an interpretation for the gradient We shall derive one later. We shall use a technique reducing certain questions in several variables to questions in one variable as follows. Suppose f is defined on an open set U, and let q>:
[a, b]
-+
U
be a differentiable curve. Then we may form the composite function f given by
0
q>
(f 0 q»(t) = f(q>(t». We may think of q> as parametrizing a curve, or we may think of q>(t) as representing the position of a particle at time t. If f represents, say, the temperature function, then f(q>(t» is the temperature of the particle at time 1. The rate of change of temperature of the particle along the curve is then given by the derivative df(q>(t»/dt. The chain rule which follows gives an expression for this derivative in terms of the gradient, and generalizes the usual chain rule to n variables. Theorem 2.2. Let q>: J -+ RO be a differentiable function defined on some interval, and with values in an open set U of RO. Let f: U -+ R be a differentiable function. Then f 0 q>: J -+ R is differentiable, and
if q>)'(t) 0
=
grad f(q>(t»· q>'(t).
Proof By the definition of differentiability, say at a point t E J, there is a function'" such that
lim t/J(k)
>-0
= 0,
and
f(q>(t + h» - f(q>(t»
= gradf(q>(t». (q>(t + h) - q>(t» + Iq>(t + h) - q>(t)I"'(k(h)~
[XV, §2)
DIFFERENTIABILITY AND THE CHAIN RULE
where k(h) = q>(t
f(q>(t
+ h) -
q>{t). Divide by Ihe number h to get
+ h») - f(q>(t») h
=
grad f(q>(t») . q>{t
± Take the limit as h
-+
383
q>{t
+
h-
+ h) - q>{t) h
q>(t) I/J(k(h»).
0 to obtain the statement of the chain rule.
Application: interpretation of the gradient From the chain rule we get a simple example giving a geometric interpretation for the gradient. Let x be a point of U and let v be a fixed vector of norm 1. We define the directional derivative of f at x in the direction of v to be
d
Dvf(x)
= dJ(x + tv)
,=0
This means that if we let g(t) = f(x + tv), then
Dvf(x) = g'(O). By the chain rule, g'(/) = grad f(x
I
+ IV) . v, whence
D.f(x) = grad f(x)· v.
I
From this formula we obtain an interpretation for the gradient. We use the standard expression for the dot product, namely
D.f(x)
= Igradf(x)Jlvl cos 0,
where 0 is the angle between v and grad f(x). Depending on the direction of the unit vector v, the number cos 0 ranges from -1 to + 1. The maximal value occurs when v has the same direction as grad f(x), in which case for such unit vector v we obtain
D.f(x) = Jgradf(x)l·
384
FUNCTIONS ON n-SPACE
[XV, §2]
Therefore we get an interpretation for the direction and norm of the gradient:
The direction of grad f(x) is the direction of maximal increase of the function f at x. The norm Igrad f(x) I is equal to the rate of change of f in its direction of maximal increase. Example. Find the directional derivative of the function f(x, y) = X 2 y 3 at (I, -2) in the direction of (3, I). Let A = (3, 1). Direction is meant from the origin to A. Note that A is not a unit vector, so we have to use a unit vector in the direction of A, namely
v
1
= r.n (3, 1). ",10
We have grad f(x, y) = (2xy 3, 3X 2y 2) and grad f(l, - 2) Hence the desired directional derivative is D v f(l, -2)
= (-16, 12).
1 = (-16,12)· r.n (3,1)
",10
1 =-(-36).
flO
Consider the set of all x e U such that f(x) = 0; or given a number c, the set of all x e U such that f(x) = c. This set, which we denote by Sc> is called the level hypersurface of level c. Let x e Sc and assume again that grad f(x) '# O. It wiII be shown as a consequence of the implicit function theorem that given any direction perpendicular to the gradient, there exists a differentiable curve
(I.:J .... U defined on some interval J containing 0 such that (1.(0) = x, (1.'(0) has the given direction, and f((I.(t» = c for all t e J. In other words, the curve is contained in the level hypersurface. Without proving the existence of such a curve, we see from the chain rule that if we have a curve (I. lying in the hypersurface such that (1.(0) = x, then
d 0= dt f((I.(t» = grad f«(I.(t»· «(t).
[XV, §2]
DIFFERENTIABILITY AND THE CHAIN RULE
385
In particular, for t = 0,
o=
grad f(a(O»)· a'(O) = grad f(x) . a'(O).
Hence the velocity vector a'(O) of the curve at t = 0 is perpendicular to grad f(x). From this we make the geometric conclusion that grad f(x) is perpendicular to the level hypersurface at x. Thus geometrically the situation looks like this: gradf(x)
level surface f(x)
= O.
Application: the tangent plane We want to apply the chain rule to motivate a definition of the tangent plane to a surface. For this we need to recall a little more explicitly some properties of linear algebra. We denote n-tuples in R" by capital letters. If A = (a" . .. ,a")
and
are elements of R", we have already seen that A is perpendicular to B if and only if A . B = O. Let A E R", A '" 0 and let P be a point in R". We define the hyperplane through P perpendicular to A to be the set of all points X such that (X - P)· A
= 0,
or also X . A = P . A. This corresponds to the figure as shown. The set of points Y such that y. A = 0 is the hyperplane passing through the origin, perpendicular to A, and the hyperplane through P, perpendicular to A, is a translation by P.
386
FUNCTIONS ON n-SPACE
[XV, §2]
P+A
~i
A
P
x-p/ __- - - -
Example. The plane in 3-space passing through (I, - 2, 3), perpendicular to A = (- 2, 4, 1) has the equation
-2x
+ 4y + z = -2 - 8 + 3 = -7.
Let f be a differentiable function on some open set U in R". Let c be a number, and let S be the set of points X such that f(X) = c,
but
grad f(X) ¢ O.
The set S is called a hypersurface in RO. Let P be a point of S. We define the tangent hyperplane of S at P to be the hyperplane passing through P perpendicular to grad f(P). Example. Let f(x, y, z) = x 2 + y2 + Z2. The surface S of points X such that f(X) = 4 is the sphere of radius 2 centered at the origin. Let
P We have grad f(x, y, z)
= (1, 1, .j'i).
= (2x, 2y, 2z) and so grad f(P)
= (2, 2, 2.j2).
Hence the tangent plane at P is given by the equation 2x
+ 2y + 2.j2z =
8.
Functions depending only on the distance. Let f be a differentiable function on R" - {O}, depending only on the distance from the origin, that is, there exists a differentiable function g of one variable r > 0 such that
f(X) = g(r)
where
[XV, §2]
DIFFERENTIABILITY AND THE CHAIN RULE
387
Then a routine differentiation using the chain rule shows that
(grad f)(X) = g'(r) X. r
Carry out the differentiation as Exercise 3.
XV, §2. EXERCISES I. Show that any two points on the sphere of radius 1 (or any radius) in n-space centered at the origin can be joined by a differentiable curve. If the points are not
antipodal, divide the straight line between them by its length at each point. Or use another method: taking the plane containing the two points, and using two perpendicular vectors of lengths I in this plane, say A, B, consider the unit circle <X(I) = (cos I)A
+ (sin I)B.
2. Let I be a differentiable function on Rn, and assume that there is a differentiable function h such that
(grad I)(X) = h(X)X. Show that I is constant on the sphere of radius r centered at the origin in Rn. [Hint: Use Exercise I.] 3. Prove the converse of Exercise 2, which is the last statement preceding the exercises, namely if I(X) = gIrl, then grad I(X) = g'(r)X/r. 4. Let I be a differentiable function on Rn and assume that there is a positive integer m such that I(IX) = Im/(X) for all numbers I >F 0 and all points X in Rn. Prove Euler's relation: Xl -
01 + ... + X 01 n -
ox.
ox"
= ml(X).
5. Let I be a differentiable function defined on all of space. Assume that I(IP)
= t/(P)
for all numbers I and all points P. Show that I(P)
= grad 1(0)' P.
6. Find the equation of the tangent plane to each of the following surfaces at the specified point.
388 (a) (b) (c) (d)
FUNCTIONS ON n-SPACE
[XV, §3]
+ y2 + Z2 = 49 at (6, 2, 3) x 2 + xl + y3 + Z + 1 = 0 at (2, -3,4) x 2l + xz - 2y' = 10 at (2, 1,4) sin xy + sin yz + sin xz = 1 at (I, n/2, 0) x2
7. Find the directional derivative of the following functions at the specified points in the specified directions. (a) Iog(x 2 + y2)112 at (I, I), direction (2, 1) (b) xy + yz + xz at (-1, 1,7), direction (3, 4, -12)
+ y)2 + (y + z? + (z + X)2. What is the direction of greatest increase of the function at the point (2, -I, 2)? What is the directional derivative off in this direction at that point?
8. Let f(x, y, z) = (x
9. Let f be a differentiable function defined on an open set U. Suppose that P is a point of U such that f(P) is a maximum, that is suppose we have f(P) Show that grad f(P)
~f(X)
for all X in U.
= O.
10. Let f he a function on an open set U in 3-space. Let g he another function, and let S be the surface consisting of all points X such that
g(X) = 0
grad g(X) #< O.
but
Suppose that P is a point of the surface S such that f(P) is a maximum for f on S, that is
f(P)
~f(X)
for all X on S.
Prove that there is a number Asuch that grad f(P) 11. Let f: R2
--+
= Agrad g(P).
R be the function such that f(O, 0)
f(x, y) =
x3 x
2
+Y
2
= 0 and
if (x, y) #< (0,0).
Show that .f is not differentiable at (0, 0). However. show that for any differenti2 able curve 0 around the origin is parametrized by P(t)
= (a cos t, a sin t).
The integral along a curve is independent of the parametrization. This is essentially proved in the next theorem. Theorem 4.1. Let J 1 = [ah b,] and J 2 = [a2, b2] be two intervals, and let g:J 1 -J 2 be a C 1 map such that g(a l } = a2 and g(b l ) = b 2. Let 1 IX: J 2 U be a C path into an open set U of R". Let F be a continuous vector field on U. Then
[Xv, §4]
397
CURVE INTEGRALS
Proof This is nothing more than the chain rule. By definition, (
F = (b'F(a(g(t»). da(g(t») dt
J'I
Ja ' 9 =
dt
L b')
da(u) F(a(u») . du
du
9(41)
=
=
(b'F(a(U»)' a'(u) du
Ja2
(F.
Jex
This proves our theorem. Suppose the vector field is on R 2 , say F(x, y) = (f(x, y), g(x, y»).
Then one denotes the integral of F along a curve a formaUy by the expression IF = I I dx
+ gdy.
The curve a can be represented by coordinates, a(t) = (x(t), Y(t»)
with a < t
~ b,
and therefore in terms of the parameter t the integral is given by (
(b[
J/ = J.
dx I dt
dY]
+ 9 dt
dt.
Note that the expression inside the integral sign is precisely the dot product: dx I dt
dy
da
+ 9 dt = F(a(t») . dt .
Example. Let us go back to the vector field F(x, y) = (x 2 y, y3) to be integrated along the line segment between (0,0) and (1, 1). Then we can write the integral in the form IF =
I
x 2 y dx 3
= L't dt _.I.
-
2'
+ y3 dy
+ t 3 dt
[with x
= t, Y = t]
398
[XV. §4)
FUNCTIONS ON n-SPACE
Example. We want to find the integral of the vector field F(x
y)
x
)_(
,y -
X2
+ y2 ' X2 + y2
around the circle of radius 2, counterclockwise. We parametrize the circle by x = 2 cos 6, y = 2 sin 6, so the integral is equal to
f F = f 2 X 2 dx + 2 Y 2 dy Jex + y x +Y
Je
2.
=
f
0 COS
6(-sin 0) dO + sin O(cos 0) dO
=0.
If a = {a" ... ,cx,.} is a path such that each a, is C', we define
I I F =
F
+ ... +
«1ll1
I
F
G: r
a,.
to be the sum of the integrals of F taken over each i = 1•... ,r. We shall say that the path a is closed if its beginning point is equal to its end point. The next theorem is concerned with closed paths, and with the dependence of an integral on the path between two points. For this we make a remark. Let a: J -+ U be a C' path between two points of U. Say a is defined on J = [a, b) and P = a(a), Q = cx(b). We can define a path going in reverse direction by letting a-(t) = cx(a
+b
- t).
When t = a we have a-(a) = a(b), and when t = b we have a-(b) = cx(a). Also, a- is defined on the interval [a. b). A simple change of variables in the integral shows that
We leave this to the reader. We call a- the opposite path of a, or inverse path. The piecewise C I path consisting of the pair {a, a-} is a closed path, which comes back to the beginning point of a. More general closed paths look like this:
[XV, §4]
399
CURVE INTEGRALS
Theorem 4.2. Let U be a connected open set in Rn. Let F be a continuous vector field on U. Then the following conditions are equivalent:
(I) F has a potential function on U. (2) The integral of F between any two points of U is independent of the path. (3) The integral of F along any closed path in U is equal to O. Proof. Assume condition (I), and let
(a(t))) . a'(t) dt d
• dt 0, we obtain G'(O) = g(O) = f.{Q), thus showing that the i-th partial derivative of
O. 5. Let r = (x' + y')'" and F(X) = r-' X for X = (x, y). Find the integral of F over the circle of radius 2, centered at the origin, taken in the counterclockwise direction. 6. Let C be a circle of radius 20 with center at the origin. Let F(X) be a vector field on R' such that F(X) has the same direction as X (that is there exists a differentiable function g(X) such that F(X) = g(X)X, and g(X) > 0 for all X). What is the integral of F around C, taken counterclockwise? 7. Let P, Q be points in 3-spaces. Show that the integral of the vector field given by
F(x, y, z)
=
(z', 2y, 2xz)
from P to Q is independent of the curve selected between P and Q. 8. Let F(x, y)
= (xlr', Ylr') where r = (x' + y')'''.
Find the integral of F along the
curve
"(I) = (e' cos I, e' sin I)
from the point (I, 0) to the point (e", 0).
404
FUNCTIONS ON n-SPACE
[XV, §4]
9. Let F(x, y) = (x'y. xy'). (a) Does this vector field admit a potential function? (b) Compute the integral of this vector field from (0, 0) to the point P
= (lj,fi. Ij,fi)
along the line segment from (0. 0) to P. (c) Compute the integral of this vector field from (0,0) to P along the path which consists of the segment from (0.0) to (1.0). and the arc of circle from (1.0) to P. Compare with the value found in (b).
10. Let F(x,y)
r)
r y cos = ( X cos . -r - . r
r Jx
2 where = + y'. Find the value of the integral of this vector field: (a) Counterclockwise along the circle of radius I, from (I. 0) to (0. I). (b) Counterclockwise around the entire circle. (c) Does this vector field admit a potential function? Why?
I I. Let
F(x, y)
xe' Ye') = ( ----;' ----; .
Find the value of the integral of this vector field: (a) Counterclockwise along the circle of radius I centered at the origin. (b) Counterclockwise along the circle of radius 5 centered at the point (14, (c) Does this vector field admit a potential function? Why?
-17~
12. Let -y
X)
G(x. y) = ( , ", ,. x +y x +y
(a) Find the integral of G along the line x + y (b) From the point (2, 0) to the point (-I. figure.
= I from (0, I) to (I, 0).
J3) along the path shown on the
2
(2.2)
(-I.j3)
-2
-I
2
[XV, §5]
405
TAYLOR'S FORMULA
13. Let F be a smooth vector field on R 2 from which the origin has been deleted, so F is not defined at the origin. Let F = (f, g). Assume that D 2 f = D,g and let
k ~iF. =
2" c
where C is the circle of radius I centered at the origin. Let G be the vector field
x)
-y
G(x, y) = ( 2 2' 2 2' x +y x +y
Show that there exists a function tp defined on R2 from which the origin has been deleted such that F=gradtp+kG
[Hint: Follow the same method as in the proof of Theorem 4.2 in the text, but define rp(P) by integrating F - kG from (1,0) to P as shown on the figure.] p
XV, §S. TAYLOR'S FORMULA Let f be a function on an open set U of Rn. We may take iterated partial derivatives (if they exist) of the form
D"I ... Dij n where i" ... ,in are integers ~ O. It does not matter in which order we take the partials (provided they exist and are continuous) according to Theorem 1.1. If cio""n are numbers, we may form finite sums ~ Ci,,··jn Di1, ..• l.J
Din n
406
FUNCTIONS ON n-SPACE
[XV, §5]
which we view as applicable to functions which have enough partial derivatives. More precisely, we say that a function f on U is of class CP (for some integer p ;;; 0) if all partial derivatives
exist for i, + ... + in ~ P and are continuous. It is clear that the functions of class CP form a vector spaoe. Let i" ... , in be integers ;;; 0 such that i, + ... + in = r ~ p. Let Fp be the vector spaoe of functions of class CP. (For p = 0, this is the vector space of continuous functions on U.) Then any monomial D~' ... D~n may be viewed as a linear map F p .... F p_, given by
f
1--+
I Dj'. ... D;,.
We say that f is of class C'" if it is of class CP for every positive integer p. If f is of class C"', then D~' ... D~"f is also of class C"'. We can take the sum of linear maps in the usual way, and thus I I -- "" ("" i..JCll···i" D1, ••• Dln)f n LJCil,··in D1, ••• Din,! n'
if the sum is taken over all n-tuples of integers (i" ... ,in) such that
A linear map such as the above, expressed as a sum of monomials of partial derivatives with constant coefficients, will be called a partial differential operator with constant coefficients. We multiply such operators in the obvious way using distributivity. For example,
In terms of two variables (x, y), say, we write this also in the form
(OX0 + oy0)2 (0ox)2 + 2 ox0oy0+ (0)2 oy . =
Similarly, we write in terms of n variables Xl>"
• ,)Cn:
In Taylor's formula, we shall use especially the expansion
(h,D, + ... + hn D)' = "" . hi,1 ... hinDi, ... Din n lJ c.11"'1" n I n
[XV, §5]
TAYLOR'S FORMULA
407
if hI> . .. ,h. are numbers. In the special case where n = 2 we have
In the general case, the coefficients are generalizations of the binomial coefficients, which we don't need to write down explicitly. It will be convenient to use a vector symbol
v
= (DI> ... ,D.) =
(a:, ,... 'a~J
its form in terms of XI> ... ,)C. being used when the variables are called XI> •.. ,)C•• If H = (hl> ... ,h.) is an n-tuple of numbers, then we agree to let
We view H . Vas a linear map, applicable to functions. Observe that (H· V)f = h,Dd +
... + h.D.f = H· gradf,
or in terms of a vector X = (XI> .•. ,)C.), (H· V)f(X) = h,Dd(X) =
+ ... + h.D.f(X)
H . (grad fXX).
This last dot product is the old dot product between the vectors Hand grad f(X). Of course one should write V)f)(X), but as usual we omit the extra parentheses. This notation will be useful in the following application. Let f be a C' function on an open set U in R·. Let P E U, and let H be a vector. For some open interval of values of t, the vectors P + tH lie in U. Consider the function g of t defined by
«H·
g(t) = f(P
+ tH)o
By a trivial application of the chain rule, we find that dg(t) dt
= g'(t) = grad f(P + tH) . H =
=
+ tH) + ... + h.D.f(P + tH) (H· V)f(P + tH)o h,Dd(P
We can generalize this to higher derivatives:
408
[XV, §5]
FUNCTIONS ON n-SPACE
Theorem 5.1. Let r be a positive integer. Let f be a function of class C' on an open set U in n-space. Let P E U. Let H be a vector. Then
(:J(J(P +
tH») = (H· V)'f(P
+ tH)o
Proof For r = 1 we have just proved our formula. By induction, assume it proved for 1 ~ k < r. Let qJ = (H . V)'1 and apply the derivative dJdt to the function t ...... 0 and a constant C such that
411 ~ s-
I.
IJ(X) - g(X)I :£ CI X I' for all X such that IXI ~ Q. Show that Taylor's formula is uniquely determined.
J = g. In particular, lIie polynomial of
8. Let U be open in R' and let J: U - R be a function of class CPo Let g: R - R be a function of class CPo Prove by induction that 9 0 J is of class CPo Furthermore, assume that at a certain point P E U all partial derivatives
D" ... D,.f(P)
=0
for all choices of i ,_ ... ,i, and r ~ k. In other words, assume that all partials of J up to order k vanish at P. Prove that the same thing is true for go J. [Hint: Induction.]
XV, §6. MAXIMA AND THE DERIVATIVE In this section, we assume that the reader knows something about the dimension of vector spaces. Furthermore, if we have a subspace F of R' and if we denote by Flo the set of all vectors w e R' which are perpendicular to all elements of F, then Flo is a subspace, and
dim F
+ dim Flo
= n.
In particular, suppose that dim F = n - I. Then dim Flo = I, and hence Flo consists of all scalar multiples of a single vector w, which forms a basis for Fl.. Let U be an open set of R' and let f: U -> R be a function of class C' on U. Let S be the subset of U consisting of all x e U such that f(x) = 0 and gradf(x) # O. We call S the hypersurface determined by f The next lemma will follow from the inverse function theorem, proved later. Lemma 6.1. Given xeS and given a vector we R" perpendicular to grad f(x), there exists a curve a: J -> U defined on an open interval J containing 0 such that a(O) = x, a(t) E S for all t e J (so the curve is contained in the hypersurface), and a'(t) = W. Theorem 6.2. Let f: U -> R be a function of class C', and let S be the subset of U consisting of all x e U such that f(x) = 0 and grad f(x) # O. Let PeS. Let g be a differentiable function on U and assume that P is a maximum for g on S, that is g(P) ;;; g(x) for all XES. Then there exists
412
[XV, §6]
FUNCTIONS ON n-SPACE
a number f.l such that
grad g(P) = f.l grad f(P).
Proof. Let ex: J -+ S be a differentiable curve defined on an open interval J containing 0 such that ex(O) = P, and such that the curve is contained in S. We have a maximum at t = 0, namely g(ex(O») = g(P) ~ g(cx(t»)
for all t have
E
J. By an old theorem concerning functions of one variable, we
o=
(g 0 ex)'(O) = grad g(ex(O») . «(0) = grad g(P) . ex'(O).
By the lemma, we conclude that grad g(P) is perpendicular to every vector w which is perpendicular to grad f(P), and hence that there exists a number f.l such that grad g(P)
= f.l grad f(P)
since the dimension of the orthogonal space to grad f(P) is equal to n - 1. This concludes the proof. The number f.l in Theorem 6.2 is called a Lagrange multiplier. We shall give an example how Lagrange multipliers can be used to solve effectively a maximum problem. Example. Find the minimum of the function f(x, y, z)
= x 2 + y2 + Z2
subject to the constraint x 2 + 2y2 - Z2 - 1 = O. The function is the square of the distance from the origin, and the constraint defines a surface, so at a minimum for f, we are finding the point on the surface which is at minimum distance from the origin. Computing the partial derivatives of the functions f and g(X)
= x 2 + 2y2 -
Z2 - 1
we find that we must solve the system of equations g(X) (a) 2x = f.l2x. (b) 2y = f.l4y. (c) 2z = f.l( - 2z).
= 0, and:
[XV, §6]
MAXIMA AND THE DERIVATIVE
413
Let (xo, Yo, zo) be a solution. If Zo # 0 then from (c) we conclude that Jl = -1. The only way to solve (a) and (b) with Jl = -1 is that x = y = O. In that case, from the equation g(X) = 0 we must have
zij = -1, which is impossible. Hence any solution must have Zo = O. If Xo # 0 then from (a) we conclude that Jl = 1. From (b) and (c) we then conclude that Yo = Zo = O. From the equation g(X) = 0 we must have Xo = ± 1. In this manner we have obtained two solutions satisfying our conditions, namely (1,0,0)
and
(-1,0,0).
Similarly if Yo # 0, we find two more solutions, namely (0,
A, 0)
and
(0,
-A,o).
These four solutions are therefore the extrema of the function f subject to the constraint 9 (or on the surface 9 = 0). If we ask for the minimum of f, then a direct computation of f(P) for P anyone of the above four points shows that the two points P
= (0, +A, 0)
are the only possible solutions because 1 > l Next we give a more theoretical application of the Lagrange multipliers to the minimum of a quadratic form. Let A = (aij) be a symmetric n x n matrix of real numbers. "Symmetric" means that a'j = a j,. If (x, y) denotes the ordinary dot product between elements x, y of R", then we have (Ax, y) = (Ay, x). The function f(x)
= (Ax, x)
is called a quadratic form. If one expresses x in terms of coordinates Xl>"',x" then f(x) has the usual shape f(x) =
L" a'jx,xj' i,i= 1
But this expression in terms of coordinates is not needed for the statement and proof of the next theorem. A vector v E R", v # 0 is cal1ed an eigenvector of A if there exists a number c such that Au = cv.
414
[XV, §6]
FUNCTIONS ON n-SPACE
Theorem 6.3. Let A be a symmetric matrix and let f(x) = (Ax, x). Let v be a point of the sphere of radius 1 centered at the origin such that v is a maximum for f, that is f(v)
~f(x)
for all x on the sphere.
Then v is an eigenvector for A. Proof Let IX be a differentiable curve passing through v (that is a(0) = v) and contained in the sphere S. Using the rules for the derivative of a product, and composition with a linear map, we know that
d
- f(lX(t» dt
d
= -dt (AIX(t), lX(t» = (Acx'(t), a(t»
+ (AIX(t), 1X'(t»
= 2(AIX(t), 1X'(t» using the fact that A is symmetric. Since IX(O) we conclude that
o=
= v is
a maximum for
f,
(f 0 IX)'(O) = 2(AIX(0), cx'(0» = 2(Av, IX'(O».
Now by the lemma, we see that Av is perpendicular to IX'(O) for every differentiable curve IX as above, and hence that Av = cv for some number c. The theorem is proved.
XV, §6. EXERCISES 1. Find the maximum of 6x 2 (x, y) such that
+
17y' on the subset of R2 consisting of those points
(x - 1)3 -
2. Find the maximum value of x 2 centered at the origin.
I = o.
+ xy + y2 + y= + 2 2
on the sphere of radius 1
3. Let! be a differentiable function on an open set U in R", and suppose that P is a minimum for! on U, that is !(P) ;i;; !(X) for all X in U. Show that all partial derivatives DJ(P) = O. 4. Let A, B, C be three distinct points in R". Let !(X)
= (X
- A)2
+ (X
- B)2
+ (X
- C)2.
Find the point where! reaches its minimum and find Ihe minimum value.
[XV, §6]
415
MAXIMA AND THE DERIVATIVE
5. Find Ihe maximum of the function f(x, y, z) x ~ 0, y ~ 0, z ~ 0, and xy + yz + xz = 2.
= xyz
subject to the constraints
6. Find the shortest distance from a point on the ellipse x 2 x + y = 4.
+ 4y'
=
4 to the line
7. Let S be the set of points (x" ... "'.) in R' such that and Show that the maximum of g(x) =
X, ...
for all i. x. occurs at (I/n, ... ,I/n) and that for all XES.
[Hint: Consider log g.] Use the result to prove that the geometric mean of n positive numbers is less than or equal to the arithmetic mean. 8. Find the point nearest the origin on the intersection of the two surfaces
x2
_
xy
+ y2
_
Z2
=I
and
X2 +y2=1.
9. Find the maximum and minimum of the function f(x, y, z) (a) on the ball x 2 + y2 + Z2 :i! I; (b) on the plane triangle X + y + z = 4, X ?; I, y ~ I, z ~ 1.
= xyz:
10. Find the maxima and minima of the function
if a, b are numbers with 0 < a < b. II. Let A, B, C denote the intercepts which the tangent plane at (x, y, z) (x>O,y>O,z>O)
on the ellipsoid
makes on the coordinate axes. Find the point on the ellipsoid such that the following functions are a minimum: (a) A + B + C. (b) JA 2
+ B2 +
C2•
12 Find the maximum of the expression
+ 6xy + 3y 2 x 2 _xy+y2·
x2
Because there are only two variables, the following method will work: let y = tx, and reduce the question to the single variable t.
416
[XV, §6]
FUNCTIONS ON n-SPACE
Exercise 12 can be generalized to more variables, in which case the above method has to be replaced by a different conceptual approach, as follows. 13. Let A be a symmetric n x /I matrix. Denote column vectors in R" by X, Y, etc. For X E R" let fIX) = (AX. X), so f is a quadratic form. Prove that the maximum of f on the sphere of radius I is the largest eigenvalue of A.
Remark. If you know some linear algebra. you should know that the roots of the characteristic polynomial of A are precisely the eigenvalues of A. 14. Let e be a symmetric n x n matrix, and assume that X ...... (ex. X) defines a symmetric positive definite scalar product on R". Such a matrix is called positive definite. From linear algebra. prove that there exists a symmetric positive definite matrix B such that for all X E R" we have
(ex, X) = (BX, BX) = IIBXII 2 • Thus B is a square root of e, denoted by e l12• [Hint: The vector space V = R" has a basis consisting of eigenvectors of e, so one can define the square root of e by the linear map operating diagonally by the square roots of the eigenvalues of C.]
15. Let A, e be symmetric n x n matrices, and assume that e is positive definite. Let Q.(X) = (AX, X) and Qclx) = (ex. X) = (BX, BX) with B = e l12 • Let fIX) = Q.(X)/QclX)
for
X#< o.
Show that the maximum of f (for X #< 0) is the maximal eigenvalue of B-1AB- 1 • [Hint: Change variables. write X = BY.] 16. Let a, b, C, e, f, g be real numbers. Show that the maximum value of the expression
ax 2 + 2bxy + cy2 ex 2 + 2fxy + gy2
(eg - f2 > 0)
is equal to the greater of the roots of the equation (ac - b 2) - T(ag - 2bf
+ ec) + T'(eg
- f')
= O.
For Exercise 16, both methods, that of Exercise 12 and the one coming from quadratic forms in several variables, work. If you don't mind computations, check that they give the same answer.
CHAPTER
XVI
The Winding Number and Global Potential Functions
Theorem 4.2 of Chapter XV gave us a significant criterion for the existence of a potential function, but falls short of describing completely the nature of global obstructions for its existence if we know that the vector field is locally integrable. The present chapter deals systematically with the obstruction, which will be seen to depend on a single vector field. The same considerations are used in subsequent courses on complex analysis and Cauchy's theorem. The fundamental result proved in the present chapter is valid more generally, but will constitute perfect preparation for those who will subsequently deal with Cauchy's theorem. In fact, Emil Artin in the 1940s gave a proof of Cauchy's theorem basing the topological considerations (called homology) on the winding number (cf. his collected works). I have followed here Artin's idea, and applied it to locally integrable vector fields in an open set U of R 2 • A fundamental result, quite independent of analysis, is that if the winding number of a closed rectangular path in U is 0 with respect to every point outside U, then the path is a sum of boundaries of rectangles completely contained in U. See Theorem 3.2. If one knows that for certain vector fields their integrals around boundaries of rectangles are 0, then it immediately follows that their integrals along paths satisfying the above condition is also O. This is the heart of the proof of the global integrability theorem, and may be viewed as a general theorem on circuits in the plane. Some readers may be only interested in this aspect of locally integrable vector fields, and they may omit the entire subsequent discussion leading to homotopy. However, since I have found that homotopy is not satisfactorily treated from the point of view of an undergraduate analysis course anywhere else, I have stilI included three sections which deal with arbitrary continuous curves and homotopy. 417
418
WINDING NUMBER AND GLOBAL POTENTIAL FUNCTIONS
[XVI, §1]
In §1 we establish some technically convenient results about integrals along paths. In §2 we state the global integrability theorem, and mention some applications. We shall see that the vector field G mentioned in Chapter 15 is (up to translations) essentially the only obstruction for a locally integrable vector field to have a global potential function. In §3 we prove the global integrability theorem. In §4 we define the integral along an arbitrary continuous path. This is useful to deal with the homotopy form of the integrability theorem, which we give in §5. We discuss homotopies more extensively in §6.
XVI, §1. ANOTHER DESCRIPTION OF THE INTEGRAL ALONG A PATH Let V be a connected open set in R 2. Let F = (f. ,[2) be a vector field on U. For simplicity, we assume F is of class C'. In §4 we shall indicate a generalization which allows more flexibility in dealing with certain questions. For our purposes here, we define F to be locally integrable if D,fl = D.f2· Exactly the same proof given for Theorem 3.3 of Chapter XV shows that if D is a disc contained in V, then F has a potential function on D. All we needed was to be able to integrate along certain line segments from one point to another, and such integration is possible within a disc as well as within a rectangle. By a path throughout until §4, we shall mean a piecewise C' path. If F has a potential function
i is a potential function for G on Di , and so '1'1 represents a choice of angle on each disc D, . Let us select '1'0 in some fashion on Do. Then select '1', on D, such that q:>,(Pd = q:>o(P,). Continuing inductively, choose q:>i+1 on Di+l such that
Since p. = Po because y is assumed to be a closed path, it follows that q:>.(P.) is an angle for p. and also for Po, but it may differ from q:>o(Po) by an integral multiple of 211. Thus there is an integer k such that
J, G = 211k. We define the winding number W(y, 0) to be W(y, 0) =
2~
ff
G = k.
[XVI, §2]
THE WINDING NUMBER AND HOMOLOGY
423
Similarly, let P be an arbitrary point of R 2 such that P does not lie on y. We define the winding number of y with respect to P to be W(y, P) =
2~ J, G
p •
Applying the chain rule to a translation, we see that W(y, P) is an integer. For example, the winding number of the curve in Figure 4 with respect to P is equal to 2. y
Figure 4
Lemma 2.1. Let y be a path. Then the function of P defined by
for P not on the path, is a cominuous function of P. Proof The path consists of a finite number of C 1 curves, so without loss of generality, we may suppose y is C 1 , so the integral is of the form
f
G(y(t) -
p). /(t) dt.
The mapping P ...... G(y(t) - P) is continuous, being composed of continuous maps, and all other operations involved in the above expression are continuous, for instance taking the dot product with /(t) and integrating. Hence P ......
f
G(y(t) - p). y'(t) dt
is a continuous function of P, as asserted.
424
WINDING NUMBER AND GLOBAL POTENTIAL FUNCTIONS
[XVI, §2]
Lemma 2.2. Let y be a closed path. Let S be a connected set not intersecting y. Then the function
is constant for P in S. If S is not bounded, then this constant is O. Proo! The integral is the winding number, and is therefore an integer.
If a function takes its values in the integers, and is continuous, then it is constant on any curve. More formally, let J be a continuous function on S, integer valued, and let 0 S be a continuous curve. Then J 0 ex: [0, b] --> Z is a continuous Z-valued function on the interval [0, b], and must be constant (give the formal details proving this assertion). The assumption that S is connected (pathwise) then implies that J is constant on S. Suppose next that S is not bounded. The path y consists of a finite number of C' curves. We shall prove that for each C' curve II the integral of Gp over II approaches 0 as IFI --> 00. The following figure illustrates Lemma 2.2.
p
Figure 5 We can compute the integral in the usual way, i.e. letting P = (P" pz), we compute
by substituting the parametrization x = X(II(t) and Y = Y(II(t) and use the derivative II'(t) which is bounded. Then the integrand is bounded by a constant times
IX -PI'
[XVI, §2]
THE WINDING NUMBER AND HOMOLOGY
425
for X on the curve (which is compact), and IPI tending to infinity. Thus the integrand is arbitrarily small. Since the integral over )' giving the winding number is an integer, and has arbitrarily small value, it must be 0, as was to be shown. Let U be an open set. Let)' be a closed path in U. We want to give conditions that
for every locally integrable vector field F on U. We already know from the example of a winding circle that if the path winds around some point outside of U (in this example, the center of the circle), then definitely we can find a vector field whose integral is not equal to 0, and even with the special vector field Gp , where P is a point not in U. The remarkable fact about the global integrability theorem is that it will tell us this is the only obstruction possible to having
for all possible F. In other words, the vector fields Gp (for P '" U) suffice to determine the behavior of f, F for all possible F. With this in mind, we want to give a name to those closed paths in U having the property that they do not wind around points in the complement of U. The name we choose is homologous to 0, for historical reasons. Thus formally, we say that a closed path )' in U is homologous to 0 in U if Wi)', P) = 0
for every point P not in U, or in other words, { Gp=O
for every such point. Similarly, let )', 'I be closed paths in U. We say that they are homologous in U if Wi)', P)
= W('1, P)
for every point P in the complement of U. It will also follow that if)' and
426
WINDING NUMBER AND GLOBAL POTENTIAL FUNCTIONS
[XVI, §2]
'1 are homologous, then
for all locally integrable F on U. Next we draw some examples of homologous paths. In Figure 6, the curves y and '1 are homologous. Indeed, jf P is a point inside the curves, then the winding number is 1, and if P is a point in the connected part going to infinity, then the winding number is O.
Figure 6
In Figure 7 the path indicated is supposed to go around the top hole counterclockwise once, then around the bottom hole counterclockwise once, then around the top in the opposite direction, and then around the bottom in the opposite direction. This path is homologous to O.
Figure 7
[XVI, §2]
THE WINDING NUMBER AND HOMOLOGY
427
In Figure 8 we are dealing with a simple closed curve, whose inside is contained in U, and the figure is intended to show that)' can be deformed to a point, so that )' is homologous to O. This will be proved formally in Theorem 5.4.
Figure 8
Given an open set U, we wish to determine in a simple way those closed paths which are not homologous to O. For instance, the open set U might be as in Figure 9, with three holes in it, at points P" P2 , P3 , so these points are assumed not to be in U.
•
P2
•
P,
Figure 9
Let )' be a closed path in U, and let F be locally integrable on U. We illustrate)' in Figure to.
Figure 10
428
WINDING NUMBER AND GLOBAL POTENTIAL FUNCTIONS
[XVI, §2]
In that figure. we see that l' winds around the three points, and winds once. Let 1',. 1'2. 1'3 be small circles centered at p,. P2 • P3 , respectively, and oriented counterclockwise. as shown on Figure 10. Then it is reasonable to expect that
f f +f +f F=
y
F
F
)',
)'z
F.
)'3
This will in fact be proved after the integrability theorem. We observe that taking 1'., 1'2' 1'3 together does not constitute a .. path" in the sense we have used that word. because. for instance, they form a disconnected set. However, it is convenient to have a terminology for a formal sum like 1'. + 1'2 + 1'3. and to give it a name '1. so that we can write
The name that is standard is the name chain. Thus let, in general, 1'" .. ·,l'n be curves. and let m••... ,mn be integers which need not be positive. A formal sum l' = m.l'.
+ ... + mnl'n =
n
L mil'i i=1
will be called a chain. If each curve l'i is a curve in an open set U, we call l' a chain in U. We say that the chain is closed if it is a finite sum of closed paths. If l' is a chain as above, we define
If l' = Lmil'; is a closed chain, where each 1'1 is a closed path, then its winding number with respect to a point P not on the chain is defined as before, W(l'. P) =
2~
f.
Gp •
If 1', '1 are closed chains in U. then we have W(l'
+ '1, P) = W(l'. P) + W('1, Pl.
Therefore, if l' is a closed chain as above, l' = L mil'" 1=1
[XVI, §2]
THE WINDING NUMBER AND HOMOLOGY
429
then for P not on any y;, we have W(y, P) =
Ii=l" m W(y" P). i
We say that y is homologous to '1 in U, and write y ~ '1. if W(y, P) = W(lj, P)
for every point P
U be a continuous curve. Let
be a partition of [a, b] such that the image y([a" a i+ I ]) is contained in a disc D" and D, is contained in U. Let F be locally integrable on U and let g, be a potential of F on Di . Let p, = yea;). Then the sum 0-1
L [g,(Pi+l) -
g,(P;)]
i=O
is independent of the choices of partition, discs D i , and potentials g, on D, subject to the stated conditions.
Proof. First let us work with the given partition, but let Bi be another disc containing the image y([a i , ai+ I ]), and B, contained in U. Let hi be a potential of f on B,. Then both gi' hi are potentials of F on the intersection B, n D" which is open and connected. Hence there exists a constant C, such that gi = h, + Ci on Bi n D,. Therdore the differences are equal:
Thus we have proved that given the partition, the value of the sum is independent of the choices of potentials and choices of discs. Given two partitions, we can always find a common refinement, as in elementary calculus. Recall that a partition
is called a refinement of the partition {!} if every point of {!} is among the points of fi2, that is if each aj is equal to some bi • Two partitions always have a common refinement, which we obtain by inserting all the points of one partition into the other. Furthermore, we can obtain a refinement
[XVI, §4]
THE INTEGRAL OVER CONTINUOUS PATHS
of a partition by inserting one point at a time. that if the partition 12 is a refinement of the inserting one point, then Lemma 4.2 is valid suppose that 12 is obtained by inserting some [a., ak+l] for some k, that is 12 is the partition
441
Thus it suffices to prove partition i?J' obtained by in this case. So we can point c in some interval
We have already shown that given a partition, the value of the sum as in the statement of the lemma is independent of the choice of discs and potentials as described in the lemma. Hence for this new partition !2, we can take the same discs Di for alI the old intervals [ai' ai+'] when i # k, and we take the disc Dk for the intervals [a.. c] and [c, ak+l]. Similarly, we take the potential gi on Di as before, and gk on Dk • Then the sum with respect to the new partition is the same as for the old one, except that the single term
is now replaced by two terms
This does not change the value, and concludes the proof of Lemma 4.2. For any continuous path 1': [a, b]
-->
U we may thus define
for any partition [ao,a" ... ,an ] of [a,b] such that 1'([a"ai +l]) is contained in a disc D" Di C U, and gi is a potential of F on Di • We have just proved that the expression on the right-hand side is independent of the choices made, and we had seen previously that if l' is piecewise C' then the expression on the right-hand side gives the same value as the definition used in Chapter XV, Theorem 4.2. It is often convenient to have the additional flexibility provided by arbitrary continuous paths. As an application, we shalI now see that if two paths lie "close together," and have the same beginning point and the same end point, then the integrals of F along the two paths have the same value. We must define precisely what we mean by "close together." After a reparametrization, we may assume that the two paths are defined over the same interval [a, b]. We say that they are close together if there exists a partition
442
WINDING NUMBER AND GLOBAL POTENTIAL FUNCTIONS
[XVI, §4]
and for each i = 0, ... ,n - I there exists a disc D, contained in U such that the images of each segment [a" ai+.J under the two paths y, '1 are contained in D" that is, and Given the locally integrable vector field F, we say that the paths are F-c1ose together if they satisfy the above conditions, and also if F has a potential function on each disc D,. Lemma 4.3. Let y, '1 be two paths in an open set U, and assume that they have the same beginning pOint and the same end point. Let F be a locally integrable vector field on U, and assume that paths are F-c1ose together. Then
Proof. We suppose that the paths are defined on the same interval [a, b], and we choose a partition and discs D, as above. Let g, be a potential of F on D,. Let
Po = yea,)
and
We illustrate the paths and their partition in Figure 18.
Figure 18
The functions gi+l an.d g, are potentials of F on the connected open set Di+l n D" so g'+1 - g, IS constant on. Di+l n Di . But Di+1 n D, contains
[XVI. §4] PI+ I and
THE INTEGRAL OVER CONTINUOUS PATHS
Qi+I'
443
Consequently
Then we find
.-1
=
I
[(g,(P,+d - g,(Qi+I») - (g,(P,) - g,(Q,»)]
i=O
=0. because the two paths have the same beginning point Po = Qo. and the same end point p. = Q.. This proves the lemma. One can also formulate an analogous lemma for closed paths. Lemma 4.4. Let Y. '7 be closed paths in the open set U, say defined on the same interval [a, b]. Assume that they are F-close together. Then
Proof The proof is the same as above, except that the reason why we lind 0 in the last step is now slightly different. Since the paths are closed, we have Po = p.
and
Qo
= Q••
y
Figure 19 as illustrated in Figure 19. The two potentials g.-I and go differ by a constant on some disc contained in U and containing po. Qo· Hence the
444
WINDING NUMBER AND GLOBAL POTENTIAL FUNCTIONS
[XVI, §5]
last expression obtained in the proof of Lemma 4.3 is again equal to 0, as was to be shown.
XVI, §S. THE HOMOTOPY FORM OF THE INTEGRABILITY THEOREM Let Y, fl be two paths in an open set U. After a reparametrization if necessary, we assume that they are defined over the same interval [a, b]. We shall say that Y is homotopic to fl if there exists a continuous function
!/J: [a, b] x [c, d]
-+
U
defined on a rectangle [a, b] x [c, d], such that
!/J(t, c)
= y(t)
and
!/J(t, d) = flit)
for all t E [a, b]. For each number s in the interval [c, d], we may view the function !/J, such that
!/J,(t) = !/J(t, s) as a continuous curve, defined on [a, b], and we may view the family of continuous curves !/J, as a deformation of the path y to the path fl. The picture is drawn on Figure 20. The paths have been drawn with the same end points because that's what we are going to use in practice. Formally, we say that the homotopy !/J leaves the end points fixed if we have
!/J(a, s) = y(a)
and
!/J(b, s) = y(b)
for all values of s in [c, d]. In the sequel it will be always understood that
when we speak of a homotopy of paths having the same end points, Ihen the homotopy leaves the end points fixed, unless othenvise specified. _--=~,...,(bl
.., (al
Figure 20
[XVI, §5]
THE HOMOTOPY FORM OF THE INTEGRABILITY THEOREM
445
If )' is homotopic to '1 (by a homotopy leaving the end points fixed), we denote this property by )' "" '1 (relative to end points). In line with our convention, we might omit the reference to the end points. Similarly, when we speak of a homotopy of closed paths, we assume always that each path t/J, is a closed path. These additional requirements are now regarded as part of the definition of homotopy and will not be repeated each time. Theorem 5.1. Let )', 'I be paths ill an opell set U having the same begillllillg point and the same end point. Assume that they are homotopic ill U relative to the end poillts. Let F be locally integrable on U. Theil
Theorem 5.2. Let)','1 be closed paths ill U, and assume that they are homotopic ill U. Let F be locally integrable on U. Theil
In par'licular,
if )'
is homotopic to a point in U, then
If )', 'I are closed paths in U alld are homotopic, then they are homologous. We prove Theorem 5.2 in detail, and leave Theorem 5.1 to the reader; the proof is entirely similar using Lemma 4.3 instead of Lemma 4.4. The idea is that the homotopy gives us a finite sequence of paths close to each other in the sense of these lemmas, so that the integral of F over each successive path is unchanged. The formal proof runs as follows. Let
t/J: [a, b]
x [c, d] -+ U
be the homotopy. The image of t/J is compact, and hence has distance > 0 from the complement of U. By uniform continuity we can therefore find partitions a
= Qo ~
Q1
~ .,. ;;; an =
c = Co ;;1 c, ;;1 .•. ;;1 Cm
b,
= d,
446 WINDING NUMBER AND GLOBAL POTENTIAL fUNCTIONS [XVI, §5) of these intervals, such that if
Sij
= small
rectangle [ai' ai+') x [Cj' cj+,)
then the image if/(Sij) is contained in a disc D'j which is itself contained in U and such that F has a potential Oij on Dij • Let if/j be the continuous curve defined by j = 0, ... ,m.
Then the continuous curves if/j, if/j+1 are F-c1ose together, and we can apply Lemma 4.4 to conclude that
f =f F
!/IJ
Since
if/o = y and if/m = I},
F.
"'J+I
we see that the theorem is proved.
Remark. It is usually not difficult, although sometimes it is tedious, to exhibit a homotopy between continuous curves. Most of the time, one can achieve this homotopy by simple formulas when the curves are given explicitly. Example. Let P, Q be two points in R 2 • The segment between P, Q, denoted by [P, Q), is the set of points
P +!(Q - P),
o ~! ~
I,
or equivalently, (I - !)P
+ !Q,
O~!~l.
A set S in R 2 is called convex, if whenever P, Q E S, then the segment [P, Q) is also contained in S. We observe that a disc and a rectangle are convex. Lemma 5.3. Le! S be a convex set, and let Y, I} be continuous closed curves in S. Then y, I} are homotopic in S.
Proof We define if/(!, s) = sy(t)
+ (I
It is immediately verified that each curve
- s)I}(!).
if/,
defined by if/,(t)
= if/(!, s)
is a
[XVI, §5]
THE HOMOTOPY FORM OF THE INTEGRADILITY THEOREM
447
closed curve, and t/J is continuous. Also t/J(t,O) = tI(t)
and
"'(t, 1) = j,(t),
so the curves are homotopic. Note that thc homotopy is given by a linear function, so if )', /1 are smooth curves, that is C' curves, then each curve t/J., is also of class C'. We say that an open set V is simply connected if it is connected and if every closed path in V is homotopic to a point. By Lemma 5.3 a convex open set is simply connected. Other examples of simply connected open sets will be given in the exercises. From Theorem 5.2, we conclude at once: Theorem 5.4. Let F be a locally illlegrable vector field cOllllected opell set U. Theil F has a potemial fUllctioll
011
011
a simply
V.
Proof. Theorem 5.2 shows that the third condition of Theorem 4.2 in
Chapter XV is satisfied, and so the potential function may be defined by the integral of F from a fixed point Po to a variable point P in U, independently of the path in U from Po to P. Thus we have derived one useful sufficient condition on an open set V for the global existence of a potential for F, namely simply connectedness. Corollary 5.5. Let F be a locally imegrable vector field
all opell set U. Theil F admits a potemial fUIIClioll 011 every disc alld every rectallgle cOlltaille,1 ill U. ~r R is a closed rectallgle cOlltailled ill U, thell
f
011
F =0.
c'R
Pl'Oof. The first assertion comes from the fact that a disc or a rectangle
is convex. As to the second, since a closed rectangle is compact, there exists an open rectangle W containing R and contained in V (take W with sides parallel to those of R, and only slightly bigger). Then W is simply connected, and we can apply Theorem 5.4 to conclude the proof. Corollary 5.6. If /lVO paThs are close togelher ill V, thell they are F-close together for every locally illlegrable vector field F all V
We can also give a proof of Lemma 2.2 based on a different principle. Indeed, the closed path y being compact, it is contained in some disc D. It is therefore homotopic to a point Q in D. If P lies outside D (which is the case when P is at sufficiently large distance from the curve), it follows
448
WINDING NUMBER AND GLOBAL POTENTIAL FUNCTIONS
[XVI, §5]
from Theorem 5.2 that
because that integral is the same as the trivial integral over the constant curve with value Q. Although in this chapter we are principally concerned with open sets in the plane, it is useful to develop a general formalism of homotopy for more general spaces. A metric space S is called pathwise connected if any two points in the space can be joined by a contin uous curve in the space. Given such a space S, let p. Q E S. Let Path(P, Q) be the set of all con· tinuous curves 0:: [0.11- S such that 0:(0) = P and 0:(1) = Q. We define a homotopy between two such curves 0:, p relative to the end points (i.e. relative to P, Q) to be a continuous map
h: [0, I] x [0, I]
-->
S
such that h(t,O) = <x(t) and h(t. I) = P(t). We denote the property that <X is homotopic to P relative to the end points by 0: ~ p. For homotopies, the interval of definition of a curve. and the interval for the parameter are chosen to· be [0, I] for convenience. One may use other intervals also, and then use the fact that given two intervals, there is a polynomial of degree I which maps one on the other. One can also use the following remark.
Lemma 5.8. (a)
Let [a, b) and [c, d) be tlVO intervals, and let f: [c. d)
-->
[a, b)
be continuous such that f(c) = a and f(d) = b. Let 0:, p: [a, b) -+ S be continuous curves in a metric space S, from P to Q. If <X ~ P. then o:of~pof. (b) Let <X: [0, I] - S be a continuous curve in a metric space S. Let f: [0. I] -+ [0. I] be a continuous function such that f(O) = 0 and f(l) = I. Then 0: ~ 0: 0 f
Proof We leave the first assertion to the reader. As to the second, a homotopy leaving the end points fixed is given by h(t, u) = O:«(i - u)t
+ uf(t)}.
[XVI, §5]
THE HOMOTOPY FORM OF THE INTEGRABILITY THEORE:I1
449
Theorem 5.9. Let U be open in R 2 and let F be a locally i/llegrable vector field on U. Let ex: [a, b] -+ U be a cO/llilillOus curve. and let
I: [c, d] -+ [a, b] be continuous such that ftc) = a and I(d) = b. Then
J F=J
fJ.o!
Q'
F
Proof This is an immediate consequence of what we have already done. First, after a translation and a linear function, one reduces the proposition to the case when both intervals are [0, I]. Then one applies Lemma 5.8 as well as Theorem 5.1, which tells us that the integrals of F over two homotopic paths from P to Q have the same value. Note that in Theorem 5.9 the map I is a very general kind of reparametrization of the interval. We put no condition of any kind on I except continuity, and the value at the end points. We have now finished our discussion of the integrability theorem in the context of homotopies. The next, and final section of this chapter continues with properties of homotopies.
XVI, §S. EXERCISES I. Let A be a closed annulus bounded by two circles IXI = r, and IXI = r, with o < r, < r,. Let F be a locally integrable vector field on an open set containing the annulus. Let V, and V, be the two circles, oriented counterclockwise. Show
that
J J F=
)'1
F.
)'1
2. A set S is called star-shaped if there exists a point Po in S such that the line segment between Po and any point P in S is contained in S. Prove that a star-shaped set is simply connected, that is. every closed path is homotopic to a point. 3. Let U be the open set obtained from R' by deleting the set of real numbers ~ O. Prove that U is simply connected. 4. Let V be the open set obtained from R' by deleting the set of real numbers ~ O. Prove that V is simply connected.
450
WINDING NUMBER AND GLOBAL POTENTIAL FUNCTIONS
[XVI, §6]
XVI, §6. MORE ON HOMOTOPIES In this section we deal with homotopies for their own sake, to complement §4, and give more criteria for paths to be homotopic in various ways. The section can be used for further study, but it will not be used in the rest of this book. Let T be the triangle with vertices (0,0), (0, 1), (1,0), as shown on the figure, and let h: T -+ S be a continuous map of the triangle into a metric space S. (0, I) h
-_.::....-_. S
T (0, 0)
Yo
(I, 0)
Figure 21 Then the map
o ~ t ,s; 1,
1'0: tt-+ h(t, 0),
is a curve 1'0 in S. If we look at the restriction of h to the diagonal, from (0, I) to (1,0), then we may also view the image of this diagonal as a curve in S, but the domain of definition is not the interval [0, I]. By a simple device, we can change the parametrization to a standardized one defined on a square. Indeed, let cp: [0, 1] x [0, 1]
-+
T
be a continuous map of the unit square onto the triangle which keeps the left vertical side fixed, and also the bottom horizontal side fixed, and maps the top side of the square on the diagonal, namely (t, 1) t-+ (t, 1 - t). (0, I)
(I, I)
(0, I) qJ
(0,0)
Yo
•
(I, 0)
T (I, 0)
Figure 22
Yo
(I, 0)
[XVI, §6]
451
MORE ON HOMOTOPIES
For instance, one could take qJ(t, u) = (t, (1 - t)u). Then the composite h 0 qJ gives a homotopy of the image of the bottom curve with the image of the diagonal under h. The homotopy h 0 qJ is defined on the unit square. For each fixed u with 0 S; u ;;; I, the map tI-+h(qJ(t, u»)
is one of the curves in the homotopy. Since h 0 qJ(t, 1) = h(t, 1 - t), we see that the image of the top line of the square under h 0 qJ is precisely the image of the diagonal under h. In practice, up to a point, one gets away with not writing down the homotopy by a formula but just drawing pictures which convince people that the formulas can be written down. The rest of this section consists of exercises, which we state as propositions, although we give occasional hints. Throughout we let S be a pathwise connected metric space.
Proposition 6.1. Let P, Q E S. If If
IX, P, l' E Path(P, Q) IX :::< P, then P :::< IX.
and
IX :::
F by the condition
A(v)
= lim
An(v).
If v, v' E E, then
A(V
+ v') =
lim A.(v
+ v') =
+ An(v'»)
lim (An(v)
= lim
An(v)
+
"- co
lim An(v') n-co
= A(v) + A(v'). If c is a number, then
A(CV) = lim An(cv) = lim cAn(v) = c lim An(v) = c.I.(v).
Hence Ais linear. Furthermore, for each n we have
458
DERIVATIVES IN VECTOR SPACES
[XVII, §1]
whence taking limits and using the properties of limits of inequalities together with the fact that the norm is a continuous function, we find that IA(V) I ;§;
Clvi
= lim
IAnl.
where C
Finally, the sequence {An} converges to A in the norm prescribed on L(E, F). Indeed, given £. there exists N such that for m, n ~ N and all v with Ivl = 1 we have
Since we have seen that An(V) -. A(v) as n -+ co, we take n sufficiently large so that IAn(V) - A(v)1
XW
for x E R is obviously a linear map of R into F. We may denote it by A"" so that A",(X) = XW. Conversely, suppose that A: R _ F is a linear map. Then for all x E R we have
A(X)
= A(x· I) = XA(l).
Now A(l) is a vector in F. Let Wo = A(l). We see that A = A",o' In this way we have described all linear maps of R into F by the elements of F itself. To each such element corresponds a linear map, and conversely; namely to the element w corresponds the linear A",: R _ F such that
for all x E R.
[XVII, §I]
THE SPACE OF CONTINUOUS LINEAR MAPS
461
Observe that a linear map into R'" can be viewed in terms of its coordinate functions. Theorem 1.2. Let E be a normed vector space, and A: E -+ R"'. Let A = (1.1>'" ,A",) be its expression in terms of coordinate functions AI' Then A is a continuous linear map if and only if each Ai is continuous linear for i = I, ... ,m. Proof This is obvious from the definitions.
Remark. One need not restrict consideration to maps into R"'. More generally, if FI>"',F mare normed vector spaces, we can consider maps A: E -+ FIx ... x F", into the product space consisting of all m-tuples of elements (XI> .•. ,x",) with XI E Fl' We take the sup norm on this space, and Theorem 1.2 applies as well. Let us reconsider the case of Rn -+ R'" as a special case of Theorem 1.2. Let A: Rn -+ R'" be a linear map, and A = AA for some matrix A = (ai;)' Let (AI> ••• ,J.",) be the coordinate functions of,t By what we have seen concerning the product Ax of A and a column vector x, we now conclude that if A I> .•. ,A", are the row vectors of A, then Ai(X)
= Ai' X
is the ordinary dot product with Ai' Thus we may write
Finally, let E, F, G be normed vector spaces and let w:E-+ F
and
kF-+G
be continuous linear maps. Then the composite map 1.0 W is a linear map.
Indeed, for v, VI> V2 E E and c E R we have
A(W(V 1
+ V2»
= A(w(vI)
+ w(V2»
= A(w(v 1 »
+ A(w(v2»
and
A(W(CV» = A(cw(v» = d(w(v». A composite of continuous maps is continuous, so 1.0 W is continuous. In terms of matrices, if E = Rn, F = R"', and G = R', then we can represent wand A by matrices A and B respectively. The matrix A is m x n and the matrix B is s x m. Then A W is represented by BA. One verifies this directly from the definitions. 0
462
DERIVATIVES IN VECTOR SPACES
[XVII, §1]
XVII, §1. EXERCISES I. Let E be a vector space and let v, . ... .v. E E. Assume that every element of E has a unique expression as a linear combination XJVI + ... + XnVn with XjE R. That is, given vEE, there exist unique numbers Xi E R such that
Show that any linear map A: E .... F into a normed vector space is continuous. 2. Let Mat m •• be the vector space of all m x n matrices with components in R. Show that Matm •• has elements e'j (i = I, ... ,m and j = I, ... •n) such that every element A of Mat m •• can be wrillen in the form m
A =
•
L LIaijeij' i= 1 j=
with number a'j uniquely determined by A. 3. Let E. F be normed vector spaces. Show that the association L(E, F) x E .... F
given by (.l, y) .-.,l(y)
is a product in the sense of Chapter VII, §l. 4. Let E, F, G be normed vector spaces. A map A: Ex F .... G
is said to be bilinear if it satisfies the conditions .l(v, w, A(V,
+ w2 )
+ V2' w) = .l(v" .l(cv, w)
for all v,
Vi E
E,
W, Wi E
= .l(v, w,)
F, and
CE
w)
+
A(v, w 2 ),
+ A(v2' w),
= cA(v. w) = A(v, cw)
R.
(a) Show that a bilinear map A is continuous if and only if there exists C> 0 such that for all (v, w) E E x F we have IA(v, IV) 1 ~
ClvlllVl.
(b) Let vEE be fixed. Show that if A is continuous, then the map A,: F .... G given by w.... .l(v, w) is a continuous linear map.
[XVII, §2]
THE DERIVATIVE AS A LINEAR MAP
463
For the rest of this chapter, we let E, F, G be euclidean spaces, that is R" or Rm• The reader will notice however that in the statements and proofs of theorems. vectors occur independently of coordinates, and that these proofs apply to the more general situation of complete normed vector spaces. We shall always accompany the theorems with an explicit determination of the statement involving the coordinates, which are useful for computations. The theory which is independent of the coordinates gives, however, a more faithful rendition of the geometric .flavor of the objects involved.
XVII, §2. THE DERIVATIVE AS A LINEAR MAP Lei U be open in E, and let x E U. Let f; U -+ F be a map. We shall say that f is differentiable at x if there exists a continuous linear map A: E -+ F and a map'" defined for all sufficiently small h in E. with values in F, such that lim .-0
"'(h) = 0,
and such that (*)
f(x
+ h) = f(x) + A(h) + Ihl"'(h).
Setting h = 0 shows that we may assume that '" is defined at 0 and that 0/1(0) = O. The preceding formula still holds. Equivalently, we could replace the term Ihlo/l(h) by a term qJ(h) where qJ is a map such that lim qJ(h) = O. Ihl
• -0
The limit is taken of course for h ;f 0, otherwise the quotient does not make sense. A mapping qJ having the preceding limiting property is said to be o(h) for h -+ O. (One reads this "little oh of h.") We view the definition of the derivative as stating that near x, the values of f can be approximated by a linear map A, except for the additive term f(x), of course, with an error term described by the limiting properties of'" or qJ described above. It is clear that if f is differentiable at x, then it is continuous at x. We contend that if the continuous linear map A exists satisfying (*), then it is uniquely determined by f and x. To prove this, let AI> A2 be con-
464
DERIVATIVES IN VECTOR SPACES
[XVII, §2]
tinuous linear maps having property (*). Let veE. Let I have real values > 0 and so small that x + tv lies in u. Let h = tv. We have
f(x
+ h) -
f(x) = A1(h)
+ Ihlr/ll(h)
= Aih) + Ihlr/l2(h) with lim r/lJ{h)
=0
0-0
for j = 1,2. Subtracting the two expressions for
f(x
+ tv)
- f(x),
we find
and setting h =
IV,
using the linearity of A,
We divide by I and find
Take the limit as I - O. The limit of the right side is equal to O. Hence A1(V) - A2(V) = 0 and A,(V) = Aiv). This is true for every ve E, whence A, = A2' as was to be shown. In view of the uniqueness of the continuous linear map A, we call it the derivative of f at x and denote it by f'(x) or Df(x). Thus f'(x) is a continuous linear map, and we can write
f(x + h) - f(x)
= f'(x)h + Ih Ir/I(h)
with lim ljJ(h)
= O.
0-0
We have written f'(x)h instead of f'(x)(h) for simplicity, omitting a set of
[XVII, §2]
THE DERIVATIVE AS A LINEAR MAP
465
parentheses. In general we shall often write }'h
instead of }'(h) when}. is a linear map. If f is differentiable at every point x of V, then we say Ihat f is differentiable on U. In that case, the derivative I' is a map
1': V -+ L(E, F) from V into the space of continuous linear maps L(E, F), and thus to each x E V, we have associated the linear map I'(x) E L(E, F). We shall now see systematically how the definition of the derivative as a linear map actually includes the cases which we have studied previously. We have three cases: Case 1. We consider a map f: J -+ R from an open interval J into R. This is the first case ever studied. Suppose f is differentiable at a number x E J in the present sense, so that there is a linear map }.; R -+ R such that
f(x
+ h) -
f(x) = }'(h)
+ IhlojJ(h)
with lim ojJ(h) = O. • -0
We know that there is a number a such that A(h) = ah for all h, that is A = }.•. Hence
f(x
+ h) - f(x) = ah + Ih IojJ(h).
We can divide by h because h is a number, and we find
f(x
+ h) h
f(x) = a
+ !!:! ojJ(h)
h'
But Ihl/h = I or -1. The limit of (lhl/h)ojJ(h) exists as h -+ 0 and is equal to O. Hence we see that f is differentiable in the old sense, and that its derivative in the old sense is a. In this special case, the number a in the old definition corresponds to the linear map "multiplication by a" in the new definition. (For differentiable maps over closed intervals, cf. the exercises.)
466
DERIVATIVES IN VECTOR SPACES
[XVII, §2]
Case 2. Let U be open in R" and let f: U -+ R be a map, differentiable at a point x E U. This is the case studied in Chapter XV, §1. There is a linear map A.: R· -+ R such that f(x
+ h) -
+ IhlI/J(h)
f(x) = A.(h)
with lim I/J(h)
= O.
h-O
We know that A. corresponds to a vector A, that is A. = A... , where A..(h) = A . h. Thus f(x
+ h) - f(x)
= A· h
+ IhlI/J(h).
This is precisely the notion of differentiability studied in Chapter XV, and we proved there that A = grad f(X) = (of/ox l' ... ,of/ox.). In the present case, the old "derivative" A corresponds to the new derivative, the linear map" dot product with A." Case 3. Let J be an interval in R, and let f J -+ F be a map into any normed vector space. This case was studied in Chapter X, §5. Again suppose that f is differentiable at the number x E J, so that f(x
+ h) -
f(x)
= A.(h) + Ih II/J(h)
for some linear map A.: R -+ F. We know that A. corresponds to a vector WE F, that is A. = A.,. is such that A...(h) = hw. Hence f(x
+ h) - f(x)
= hw
+ IhlI/J(h).
In the present case, h is a number and we can divide by h, so that f(x
+ h~ -
f(x) = w
+
':1
I/J(h).
The right-hand side has a limit as h ..... 0, namely w. Thus in the present case, the old derivative, which was the vector w, corresponds to the new derivative, the linear map Aw , which is "multiplication by w on the right" We have now identified our new derivative with all the old derivatives, and we shall go through the differential calculus for the fourth and last time, in the most general context. Let us consider mappings into Rm.
[XVII, §2]
THE DERIVATIVE AS A LINEAR MAP
467
Theorem 2.1. Let U be all opell set of R", alld let f: U - R'" be a map which is differelltiable at x. Theil the cOlltilluous lillear map f'(x) is represented by the matrix
where j; is the i-th coordinate function of f Proof Essentially this comes from pUlling together Case 2 discussed above, and Theorem 1.2. We go through the proof once morc from scratch. We have using Case 2:
f(x
+ h) -
+ h) -
fl(x f(x) =
=
(
fm(x
(AI'
+ h
A", . h
fl(X»)
: h) - fm(x)
~ ipl(h») + ip",(h)
= (AI:' h)
+
(ipl~h»)
A",· h
ip",(h)
oj;
oj; ) 'ox" '
where
A;
= grad j;(x) = ( ox l ""
and ip!..h) = o(h). It is clear that the vector ip(h) = (ipl(h), ... ,ip",(h») is o(h), and hence by definition of f'(x), we see that it is represented by the matrix of partial derivatives, as was to be shown. The matrix
is called the Jacobian matrix of f at x. We see that if f is differentiable at every point of U, then x ...... J I(x) is a map from U into the space of matrices, which may be viewed as a space of dimension mil.
468
DERIVATIVES IN VECTOR SPACES
[XVII, §3]
We defined f to be differentiable on U if f is differentiable at every point of U. We shall say that f is of class CIon U, or is a C 1 map, iff is differentiable on U and if in addition the derivative
1': U -+ L(E, F) is continuous. From the fact that a map into a product is continuous if and only if its coordinate maps are continuous, we conclude from Theorem 2.1: Corollary. The map f: U -+ Rm is of class CI if and only if the partial derivatives iJfJiJxj exist and are continuous functions, or put another way, if and only if the partial derivatives Dj /;: U -+ R exist and are continuous.
XVII, §2. EXERCISES I.
Find explicitly the Jacobian matrix of the polar coordinate map
x
= rcosO
y = r sin O.
and
2. Find the Jacobian matrix of the map (u, v) = F(x, y) where u = eX cos y,
v = eX sin y.
Compute the determinants of these 2 x 2 matrices. The determinant of the matrix
is by definition ad - be. 3. Let A: R" - Rm be a linear map. Show that A is differentiable at every point, and that A'(x) = A for all x E R".
XVII, §3. PROPERTIES OF THE DERIVATIVE Sum. Let U be open in E. Let f, g: U -+ F be maps which are differenti-
able at x E U. Then f
+ 9 is differentiable at x and (f + g)'{x) = f'(x)
+ g'(x).
If c is a number, then (cf)'{x) = cf'(x).
[XVII, §3)
PROPERTIES OF THE DERIVATIVE
Proof. Let Al
469
= j'(x) and A2 = g'(x) so that f(x
+ h) -
f(x)
= Alh + Ihll/Jl(h),
g(x
+ h) -
g(x)
= A2h + Ihll/J2(h),
where lim l/J;(h) = O. Then .~O
(f + g)(x
+ h) -
(f + g)(x) = f(x =
=
Since lim (l/J,(h)
+ l/J2(h»
=
+ h) + g(x + h) - f(x) - g(x) A1h + A2 h + Ihl(l/J,(h) + l/Jih» (AI + A2)(h) + Ihl(l/J,(h) + l/J2(h».
0, it follows by definition that
.~O
Al
+ A2 = (f + g)'(x),
as was to be shown. The statement with the constant is equally clear. Product. Let FI x F2 .... G be a product, as defined in Chapter VII, §1. Let U be open in E and let f: U .... F I and g: U .... F 2 be maps differentiable at x E U. Then the product map fg is differentiable at x and (fg)'(x)
= j'(x)g(x) + f(x)g'(x).
Before giving the proof, we make some comments on the meaning of the product formula. The linear map represented by the right-hand side is supposed to mean the map VI->
U'(x)v)g(x) + f(x)(g'(x)v).
Note that j'(x): E .... F I is a linear map of E into F I> and when applied to vEE yields an element of Fl' Furthermore, g(x) lies in F 2, and so we can take the product U'(x)v)g(x) E G.
Similarly for f(x)(g'(x)v). In practice we omit the extra set of parentheses, and write simply j'(x)vg(x). .
470
DERIVATIVES IN VECTOR SPACES
[XVII, §3]
Proof We have f(x
+ h)g(x + h) - f(x)g(x) = f(x + h)g(x + h) - f(x + h)g(x) + f(x + h)g(x) - f(x)g(x) = f(x + h)(g(x + h) - g(x») + (f(x + h) - f(x»)g(x) = f(x + h)(g'(x)h + IhI1/t2(h») + (f'(x)h + Ihll/tl(h»)g(x) =f(x + h)g'(x)h + Ihlf(x + h)1/t2(h) + f'(x)hg(X) + Ihll/tl(h)g(x) = f(x)g'(x)h + f'(x)hg(x) + (f(x + h) - f(x»)g'(x)h + Ihlf(x + h)1/t2(h) + Ihll/tl(h)g(x).
The map h 1-+ f(x)g'(x)h
+ f'(x)hg(x)
is the linear map of E into G, which is supposed to be the desired derivative. It remains to be shown that each of the other three terms appearing on the right is of the desired type, namely o(h). This is immediate. For instance, l(f(x
+ h) -
f(x»)g'(x)hl
< If(x
+ h) -
f(x)llg'(x)llhl
and lim .-0
If(x
+ h)
- f(x) I Ig'(x) I = 0
because f is continuous, being differentiable. The others are equally obvious, and our property is proved. Example. Let J be an open interval in R and let
e1-+ A(e)
=
(a,tt))
and
t 1-+
X(t)
be two differentiable maps from J into the space of m x n matrices, and into R" respectively. Thus for each t, A(t) is an m x n matrix, and X(t) is a column vector of dimension n. We can form the product A(t)X(t), and thus the product map t 1-+
A(t)X(t),
which is differentiable. Our rule in this special case asserts that d
dt A(t)X(t) = A'(t)X(t)
+ A(t)X'(t)
[XVII, §3]
PROPERTIES OF THE DERIVATIVE
471
where differentiation with respect to t is taken componentwise both on the matrix A(t) and the vector X(t). Actually, this case is covered by the case treated in Chapter X, §5, since our maps go from an interval into vector spaces with a product between them. The product here is the product of a matrix times a vector. If m = I, then we deal with the even more special case where we take the dot product between two vectors. Chain rule. Let U be open in E and let V be open in F. Let f: U ..... V and g: V ..... G be maps. Let x E U. Assume that f is differentiable at x and 9 is differentiable at f(x). Then 9 0 f is differentiable at x and
(g 0 f)'(x) = g'(f(x») 0 j'(x). Before giving the proof, we make explicit the meaning of the usual formula. Note that j'(x): E ..... F is a linear map, and g'(f(x»): F ..... G is a linear map, and so these linear maps can be composed, and the composite is a linear map, which is continuous because both g'(f(x») and j'(x) are continuous. The composed linear map goes from E into G, as it should.
Proof Let k(h) g(f(x
= f(x + h) -
f(x). Then
+ h») - g(f(x») = g'(f(x»)k(h) + Ik(h)II/!,(k(h»)
with lim I/!,(k) = O. But
'-0
k(h) = f(x
+ h) -
f(x) = j'(x)h
+ Ihll/!2(h),
with lim I/!2(h) = O. Hence h-O
g(f(x
+ h») -
g(f(x») = g'(f(x»)j'(x)h
+ Ihlg'(f(X»)I/!2(h) + Ik(h)II/!,(k(h»).
The first term has the desired shape, and all we need to show is that each of the next two terms on the right is o(h). This is obvious. For instance, we have the estimate
Ik(h)1 < 1j'(x)llhl
+ Ihlll/!2(h)1
and
lim I/!.(k(h») = 0 h-O
from which we see that Ik(h)II/!,(k(h») = o(h). We argue similarly for the other term.
472
DERIVA TIVES IN VECTOR SPACES
[XVII, §3)
The chain rule of course can be expressed in terms of matrices when the vector spaces are taken to be R", Rm, and RS respectively. In that case, in terms of the Jacobian matrices we have
the multiplication being that of matrices. Maps with coordinates. Let U be open in E, let f: U -+ F, x ... x Fm' and let f = (f•• .. . ,f.J be its expression in terms of coordinate maps. Then f is differentiable at x if and only if each /; is differentiable at x, and if this is the case, then f'(x)
= (f~(x), ... ,f~(x».
Proof This follows as usual by considering the coordinate expression f(x
+ h) -
f(x) = (f,(x
+ h) -
f,(x), ... ,fm(x
+ h)
- fm(x».
Assume that Fl..x) exists, so that /; (x
+ h)
- /;(x) = fi (x)h
+ ... ,vJ + f(vl> ... ,vi, .. . ,vn ),
f(vl> ... ,cv" ... ,vJ = cf(v" ... ,vn ), for v" vi E E; and c E R. In this section, we study the case n = 2, in which case the map is said to be bilinear. Examples. The examples we gave previously for a product (as in Chapter VII, §1) are also examples of continuous bilinear maps. We leave
[XVII, §5]
THE SECOND DERIVATIVE
479
it to the reader to verify for bilinear maps (or multilinear maps) the condition analogous to that proved in §I for linear maps. Cf. Exercise I. Thus the dot product of vectors in Rn is continuous bilinear. The product of complex numbers is continuous bilinear, and so is the cross product in R3 . Other examples: The map L(E, F) x E
-+
F
given by
(A., v) ..-d(v) that we just considered. Also, if E, F, G are three spaces, then L(E, F) x L(F, G)
-+
L(E, G)
given by composition,
is continuous bilinear. The proof is easy and is left as Exercise 4. Finally, if A = (ai)
is a matrix of n 2 numbers au (i = I, ... ,n; j = I, ... ,n), then A gives rise to a continuous bilinear map
by the formula
where X, Yare column vectors, and IX = (XI> .•• ,xn) is the row vector called the transpose of X. We study these later in the section. Theorem 5.1. Let w: E 1 x E 2 -+ F be a continuous bilinear map. Then w is dijferentiable, and for each (Xl> X2) e E 1 x E 2 and every
we have
480
DERIVATIVES IN VECTOR SPACES
so that Dw: E, x E 2 and D3 w = O.
->
L(E,
X
[XVII, §5]
E 2 , F) is linear. Hence D 2w is constant,
Proof We have by definition
This proves the first assertion, and also the second, since each term on the right is linear in both (Xl' X2) = X and h = (h" h 2 )· We know that the derivative of a linear map is constant, and the derivative of a constant map is 0, so the rest is obvious. We consider especially a bilinear map kExE->F
and say that A is symmetric if we have A(v, w)
for all v,
WEE.
= A(w, v)
In general, a multilinear map
A: Ex··· x E
->
F
is said to be symmetric if
for any permutation u of the indices I, ... ,no In this section we look at the symmetric bilinear case in connection with the second derivative. We see that we may view a second derivative D2f(x) as a continuous bilinear map. Our next theorem will be that this map is symmetric. We need a lemma. Lemma 5.2. Let k E x E -> F be a bilinear map, and assume that there exists a map l/J defined for all sufficiently small pairs (v, w) E E x E with values in F such that
l/J(v, w)
lim (tI,
= 0,
w)-(O. 0)
and that
lA(v, w)1 Then A = O.
~
Il/J(v, w)llvllwl.
[XVII, §5]
481
THE SECOND DERIVATIVE
Proof This is like the argument which gave us the uniqueness of the derivative. Take v, WEE arbitrary, and let s be a positive real number sufficiently small so that o/J(sv, sw) is defined. Then IA(SV, sw)1 ;;;; Io/J(sv, sw)llsvllswl, whence
s21A(v, w)l;;;; s21t/J(sv,sw)llvllwl. Divide by S2 and let s
->
O. We conclude that A(v, w)
= 0, as desired.
Theorem 5.3. Let U be open in E and let f; U -> F be twice differentiable, and such that D2f is continuous. Then for each x E U, the bilinear map D2f(x) is symmetric, that is
D2f(x)(v, w) = D2f(x)(w, v) for all v, wEE. Proof Let x E U and suppose that the open ball of radius r in E centered at x is contained in U. Let v, WEE have lengths < rf2. We shall imitate the proof of Theorem 1.1, Chapter XV. Let g(x)
= f(x + v) -
f(x).
Then
f(x
+ v + w) = g(x
+
f(x
+ w) -
w) - g(x) =
= =
f(x
f: f: f
+ v) + f(x)
g'(x + tw)w dt [Df(x
+ v + tw) -
Df(x
+ tw)]w dt
fD 2f(X + SV + tw)vds· wdt.
Let
o/J(sv, tw) = D2f(x + SV + tw) - D2f(x).
482
[XVII, §5]
DERIVATIVES IN VECTOR SPACES
Then g(x + w) - g(x) =
s: s: + s: f:
D 2f(x)(v, w) ds de
t/t(sv, ew)v . w ds de
= D 2f(xXv, w)
+ Ip(v, w)
where I{J(v, w) is the second integral on the right, and satisfies the estimate IIp(v, w)1 ~ sup It/t(sv, ew)llvllwl· s. r
The sup is taken for 0
~
s
~
I and 0 < e :5; I. If we had started with
g,(x) = f(x + w) - f(x) and considered g,(x + v) - g,(x), we would have found another expression for the expression f(x
+ v + w) -
f(x
+ w) -
f(x
+ v) + f(x),
namely D 2f(xXw, v)
+ Ip,(v, w)
where 11p,(v, w)1 < sup It/t,(sv, ew)llvllwl. s••
But then D 2f(x)(w, v) - D 2f(x)(v, w)
= Ip(v, w) -
Ip,(v, w).
By the lemma, and the continuity of D 2fwhich shows that sup It/t(sv, ew)1 and sup It/t,(sv, tw)1 satisfy the limit condition of the lemma, we now conclude that D 2f(x)(w, v)
as was to be shown.
= D 2f(xXv, w),
[XVII, §5]
THE SECOND DERIVATIVE
483
We now give an interpretation of the second derivative in terms of matrices. LetA: Rn x Rn -+ R be a bilinear map, and let e" ... ,en be the unit vectors of Rn. If
and
are vectors, with coordinates v" w, E R (so these are numbers as coordinates) then
..1.(v, w) = ..1.(vle, =
+ ... + vnen, wlel + ... + wnen)
Lj v, wjA(e" ej) i,
the sum being taken for all values of i, j = 1, ... ,n. Let au = A(ei, ej). Then au is a number, and we let A = (aij) be the matrix formed with these numbers. Then we see that
..1.(v, w)
= L aUvi Wj. i. j
Let us view v, W as column vectors, and denote by 'v (the transpose of v) the row vector 'v = (VI> .•• ,vn ) arising from the column vector
v=c} Then we see from the definition of multiplication of matrices that
..1.(v, w)
= 'vAw,
which written out in full looks like
:
: =" a--v-w·.
a.n)(w,) o. •..
ann
Wn
I..JIJIJ
i,i
We say that the matrix A represents the bilinear mapA. It is obvious conversely that given an n x n matrix A, we can define a bilinear map by letting
.(v, w) 1-+ 'vAw.
484
DERIVATIVES IN VECTOR SPACES
[XVII, §5]
Let Ai/ Rn
X
Rn -> R
be the map such that
Then we see that the arbitrary bilinear map Acan be written uniquely in the form
A=
L a'jAij' i, j
In the terminology of linear algebra, this means that the bilinear maps {A'j} (i = 1, ... ,11 and j = 1, ... ,11) form a basis for L 2(Rn, R). We also sometimes write Aij = AI ® Aj where Ai is the coordinate function of Rn given by Ai(v"", ,vn) = Vi' Now let U be open in Rn and let g: U -> L 2(Rn, R) be a map which to each x E U associates a bilinear map g(x): Rn x Rn
->
R.
We can write g(x) uniquely as a linear combination of the A'j. That is, there are functions glj of x such that g(x) =
L gIJ{x)A'j'
i,i
Thus the matrix which represents g(x) is the matrix (gIJ{x»), whose ordinates depend on x.
c0-
Theorem 5.4. Let U be opell ill Rn alld let f: U -> R be a fUllctioll. Theil f is of class C 2 if alld ollly if all the partial derivatives of f of order ;:i; 2 exist and are contilluous. If chis is the case, chell D2f(x) is represellted by the macrix (D,DJ(x»). Proof. The first derivative Df(x) is represented by the vector (1 x matrix) grad f(x) = (Dt!(x), ... , D.J(x»), namely Df(x)v
= Dt!(X)VI + '" + Dnf(x)vn
11
[XVII, §5)
485
THE SECOND DERIVATIVE
if v = (v" ... ,Vn) is given in terms of its coordinates v, E R. Thus we can write
+ ... + Dnf(X)An
Df(x) = D.J(X)A,
where A, is the i-th coordinate function of Rn, that is
Thus we can view Df as a map of U into an n-dimensional vector space. In the case of such a map, we know that it is of class C' if and only if the partial derivatives of its coordinate functions exist and are continuous. In the present case, the coordinate functions of Df are D,f, ... ,Dn! This proves our first assertion. As to the statement concerning the representation of D 2 f(x) by the matrix of double partial derivatives, let wE Rn and write W in terms of its coordinates (wl>'" ,Wn), Wi E R. It is as easy as anything to go back to the definitions. We have Df(x
where qJ(h)
= o(h).
+ h) - Df(x)
= D 2f(x)h
+ 0 such that for all v, E E, we have IA(v" .. . ,Vn)I ;;;; Clvillv,I·· ·Ivnl.
2. Denote the space of continuous multilinear maps as above by L(E" ... ,En; F). If A is in this space, define IAI to be the greatest lower bound of all numbers C > 0 such that
for all v, E E,. Show that this defines a norm. 3. Consider the case of bilinear maps. We denote by L'(E, 1') the space of continuous bilinear maps of Ex E,...F. If AEL(E,L(E,F», denote by f, the bilinear map such that f,(v, w) = .l(vXw). Show thatlAI = If,l. 4. Let E, F, G be normed vector spaces. Show that the composition of mappings L(E, F) x L(F, G) -+ L(E, G)
given by (A, CJJ) 1-+ CJJ 0 A is continuous and bilinear. Show that the constant C of Exercise 1 is equal to 1. 5. Let f be a function of class C' on some open ball U in Rn centered at A. Show that f(X) = f(A)
+ Df(A) . (X -
A)
+ g(XXX -
A, X - A)
where g: U -+ L '(Rn, R) is a continuous map of U into the space of bilinear maps ofRn into R Show that one can select g(X) to be symmetric for each X E U.
XVII, §6. HIGHER DERIVATIVES AND TAYLOR'S FORMULA We may now consider higher derivatives. We define D"f(x) = D(DP-'fXx).
Thus D"f(x) is an element of L(E, L(E, ... ,L(E, F) .. .») which we denoted by LP(E, F). We say thatfis of class CP on U or is a CP map if D'f(x) exists for each x E U, and if D"f: U .... Lk(E, F)
is continuous for each k = 0, ... , p.
488
[XVII, §6)
DERIVATIVES IN VECTOR SPACES
We have trivially DOD'f(x) = D"J(x) if q + r = p and if DPf(x) exists. Also the p-th derivative DP is linear in the sense that DP(f + g) = D"J + DPg
and
DP(cf)
= cDPf
If AE U(E, F) we write
If q
+ r = p, we can evaluate A(VI> ... ,vp) in two steps, namely
We regard A(VI> ... , vo) as the element of U-O(E, F) given by
Lemma 6.1. Let V2"" ,vp be fixed elements of E. Assume that f is p times differentiable on U. Let g(x)
= DP- 1f(x)(V2' ... , vp).
Then 9 is differentiable on U and Dg(xXv)
= D"J(x)(v, V2,""
vp).
Proof The map g: U -+ F is a composite of the maps and where A is given by the evaluation at (V2"" ,vp). Thus A is continuous and linear. It is an old theorem that
namely the corollary of Theorem 3.1. Thus Dg(x)v
= (D"J(X)V)(V2"'"
vp),
which is precisely what we wanted to prove. Theorem 6.2. Let f be of class CP on U. Then for each x E U the map D"J(x) is multilinear symmetric.
[XVII, §6]
HIGHER DERIVATIVES AND TAYLOR'S FORMULA
489
Proof. By induction on p ~ 2. For p = 2 this is Theorem 5.3. In particular, if we let 9 = Dp-1f we know that for VI' V1 E E,
and since D"f = D1DP- 1f we conclude that
(0)
DPf(x)(vl>'" ,vp) = (D1DP-1f(x»)(vl> V1)' (V3,'" ,vp) =
(D1DP-1f(x»)(V1,
VI)'
(V3,'" ,vp)
= D"f(X)(V1' vI> V3, ... ,vp). Let q be a permutation of (2, ... ,p). By induction,
By the lemma, we conclude that
(00) From (0) and (00) we conclude that D"f(x) is symmetric because any permutation of (I, ... ,p) can be expressed as a composition of the permutations considered in (0) or (00). This proves the theorem. For the higher derivatives, we have similar statements to those obtained with the first derivative in relation to linear maps. Observe that if WE LP(E, F) is a multilinear map, and AE L(F, G) is linear, we may compose these Ex···xE~F~G
to get A0 w, which is a multilinear map of Ex··· x E -> G. Furthermore, wand A being continuous, it is clear that A0 w is also continuous. Finally, the map AO : LP(E, F)
->
LP(E, G)
given by "composition with A", namely
is immediately verified to be a continuous linear map, that is for WI> W 1 E LP(E, F) and C E R we have and
490
[XVII, §6]
DERIVATIVES IN VECTOR SPACES
and for the continuity,
SO
Theorem 6.3. Let f: U -+ F be p-times differentiable and let ).: F -+ G be a continuous linear map. Then for every x e U we have DP(}.o f)(x)
= }.
0
DPf(x).
Proof Consider the map x 1-+ DP-I(}.o f)(x). By induction,
By the corollary of Theorem 3.1 concerning the derivative D(}..o DP- 'f), namely the derivative of the composite map
we get the assertion of our theorem. If one wishes to omit the x from the notation in Theorem 6.3, then one must write
Occasionally, one omits the lower. and writes simply DP(}.o f)
= }. oD"f.
Taylor's formula. Let U be open in E and let f: U -+ F be of class CPo Let x e U and let ye E be such that the segment x + ty, 0 ;;;; t ;;;; I, is contained in U. Denote by ylk) the k-tuple (y, y, ... ,y). Then
f(x
Dr(x)y
+ y) = f(x) + ~
I!
+ ... +
DP-I'!(X)
(p- I)
y
(p - 1)1
where Rp =
1
5.o
(I - t)P-I (p - I)!
DPf(x
+ ty)ylp) dt.
+R P
[XVII, §6]
HIGHER DERIVATIVES AND TAYLOR'S FORMULA
491
Proof We can give a proof by integration by parts as usual, starting with the mean value theorem, f(x
+ y) =
f(x)
We consider the map t.-. Df(x product
+ L'Df(X + ty)y dt.
+ ty)y of the interval into F, and the usual RxF-+F
which consists in multiplying vectors of F by numbers. We let u = Df(x
+ ty)y
and
dv = dt, v = - (I - t)
This gives the next term, and then we proceed by induction, letting u
= D"f(x + ty)ylp)
and
dv
=
tr (p _ I)!
(I -
l
dt
at the p-th stage. Integration by parts yields the next term of Taylor's formula, plus the next remainder term. The remainder term R p can also be written in the form Rp =
f.
1
(1
- ty-' D"f(x + ty) dt . y(p).
o (p - I)!
The mapping
y.-.
f.
1
0
(I t)p-I (p-l)! D"f(x+ty)dt
is continuous. Iff is infinitely differentiable, then this mapping is infinitely differentiable since we shall see later that one can differentiate under the integral sign as in the case studied in Chapter X. Estimate of the remainder. Notation as in Taylor's formula, we can also write f(x
+ y) = f(x) + Df(x)y + ... + D"f(x)y(Pl + 8(y) l!
p!
492
DERIVATIVES IN VECTOR SPACES
[XVII, §6)
where
le(y)1 < sup ID'1(x + ty~ - D'1(x)llyIP O~.~l
p.
and
· e(y) IIm-, IP y-o Y
0
= .
Proof We write D'1(x
+ ty) - D'1(x) = t/J(ty).
Since D'1 is continuous, it is bounded in some ball containing x, and
lim t/J(ty)
0
=
y-o
uniformly in t. On the other hand, the remainder R p given above can be written as
f
(I - t)p-I D'1(x)Y U(R", R) is a map, then for each x E U we can write g(x) =
L g(j)(X)A(}) (j)
where g(}) are the coordinate functions of g. This applies in particular when 9 = D"f for some p-times differentiable function f In that case, induction and the same procedure given in the bilinear case yield: Theorem 6.6. Let U be open in R" and let f: U -> R be a function. Then f is of class CP if and only if all the partial derivatives of f of order ~ p exist and are continuous. If this is the case, then D"f(x)
= L Dil ... DjJ(X)AiI ... jp lI)
and for any vectors v" ... ,v p E R" we have D"f(x)(v" ... ,vp )
= L Dj, ... Djpf(x)v'iI ... v,Jp' lJ)
Observe that there is no standard terminology generalizing the notion of matrix to an indexed set {ail'" Jp}
(which could be called a multimatrix) representing the multilinear map. The multimatrix {D J, ... DJpf(x}}
represents the p-th derivative D"f(x). In the notation of Chapter XV, §5, we can write also
where v·I . V = v.,D, I
+ ... + v·
In
Dn
is a partial differential operator with constant coefficients are the coordinates of the vector VI'
Vii' ... ,VI"
which
[XVll, §7J
495
PARTIAL DERIVATIVES
XVII, §6. EXERCISE I. Let U be open in E and V open in F. Let
J: U -+ V
g: V
and
-+
G
be of class CPo Let Xo E U. Assume that D'J(xo} = 0 for all k = 0, ... ,p. Show that D"(goJXxo} = 0 for 0 ~ k ~ p. [Hint: Induction.] Also prove that if D"g(J(xo)) = 0 for 0 ~ k ~ p, then (D'(g 0 J))(xo) = 0 for 0 ~ k ~ p.
XVII, §7. PARTIAL DERIVATIVES Consider a product E = E, x ... x En of complete normed vector spaces. Let VI be open in EI and let f:V,x··.xVn-F
be a map. We write an element XE VI x ... X V n in terms of its "coordinates," namely x = (x" ... ,xn ) with XI E VI' We can form partial derivatives just as in the simple case when E = Rn. Indeed, for x" ... ,XI_ .. XI+ .. •.. ,Xn fixed, we consider the partial map
of VI into F. If this map is differentiable, we call its derivative the partial derivative of f and denote it by DI fIx) at the point x. Thus, if it exists,
is the unique continuous linear map AE L(E;, F) such that fIx"~ ... ,XI
+ h, ... ,xn )
-
fIx"~
... ,Xn ) = A(h)
+ o(h),
for hEEl and small enough that the left-hand side is defined. Theorem 7.1. Let VI be open in E j (i = 1, ... ,n) and let f: V, x ...
X
Vn
-
F
be a map. This map is of class CP if and only if each partial derivative
DJ: V, x ... x V n - L(E;, F)
496
[XVII, §7]
DERIVATIVES IN VECTOR SPACES
is of class CP- I. If this is the case, and
then n
Df(x)v =
L Dd(x)v,. i;; 1
Proof We shall give the proof just for n = 2, to save space. We assume that the partial derivatives are continuous, and want to prove that the derivative of f exists and is given by the formula of the theorem. We let (x, y) be the point at which we compute the derivative, and let h = (hI> h2 ). We have
f(x
+ hI> Y + h2 )
-
f(x, y)
=
f(x + hI> Y + h 2 )
=
J: D,f(x + hI>
-
f(x + hi' y) + f(x + hI> y) - f(x, y)
Y + th 2 )h 2 dt + f DJ(x + thl> y)h l dt.
Since D2 f is continuous, the map I/J given by I/J(hl> th 2 ) = D,f(x + hI> Y + th 2 )
-
D,f(x, y)
satisfies lim I/J(hl> th 2 ) = O.
h-O
Thus we can write the first integral as
1 1
o D2 f(x
+ hI> Y + th 2 )h 2 dt
=
11
0 D,f(x, y)h 2 dt +
= D,j(x, y)h 2 +
f
In I/J(hl> th )h I
I/J(hl> th 2 )h 2 dt.
Estimating the error term given by this last integral, we find
fl/J(hl> th 2 )h 2 dt
th 2 )1
= o(h).
2
2
dt
[XVll, §7J
PARTIAL DERIVATIVES
497
Similarly, the second integral yields Dd(x, y)h,
+ o(h).
Adding these terms, we find that Df(x, y) exists and is given by the formula, which also shows that the map Df = f' is continuous, so f is of class C'. If each partial is of class C', then it is clear that f is C'. We leave the converse to the reader. Example. Let E, be an arbitrary space and let E 2 = Rm for some m so that elements of E2 can be viewed as having coordinates (y" ... oYm)' Let F = RS so that elements of F can also be viewed as having coordinates (z" ... ,zs)' Let U be open in E, x Rm and let
f: U
-+
RS
be a C' map. Then the partial derivative
may be represented by a Jacobian matrix. If (f" ... ,J.) are the coordinate functions of f, this Jacobian may be denoted by J7)(x, y), and we have
of,
of,
oy,
oYm
oj.
oj.
oy,
oYm
J'l'(x, y) =
For instance, let f(x, y, z) = (x 2y, sin z). We view R 3 as the product R x R 2 so that D 2 f is taken with respect to the space of the last two coordinates. Then D 2 f(x, y, z) is represented by the matrix
0)z . z) = (x0 cos 2
(2)
J I (x, y,
Of course, if we split R 3 as a product in another way, and compute D 2 f with respect to the second factor in another product representation, then the matrix will change. We could for instance split R 3 as R 2 x Rand thus take the second partial with respect to the second factor R. In that case, the matrix would be simply
498
DERIVATIVES IN VECTOR SPACES
[XVII, §7]
It will be useful to have a notation for linear maps of products into products. We treat the special case of two factors. We wish to describe linear maps kE, xE 2 -+F, xF 2 •
We contend that such a linear map can be represented by a matrix
All ( A2,
A'2) A22
where each Ai / Ej -+ F, is itself a linear map. We thus take matrices whose components are not numbers any more but are themselves linear maps. This is done as follows. Suppose we are given four linear maps Ai) as above. An element of E, x E 2 may be viewed as a pair of elements (v" V2) with v, E E, and V2 E E 2. We now write such a pair as a column vector
and define A.(v,. V2) to be
G::
~::)(~:) = G::~: : ~::~:)
so that we multiply just as we would with numbers. Then it is clear that A is a linear map of E, x E 2 into F, x F 2 • Conversely, let k E, x E 2 -+ F I X F 2 be a linear map. We write an element (v" V2) E E, x E 2 in the form (v" V2)
= (v" 0) + (0, V2)'
We also write A in terms of its coordinate maps A = (A" A2) where A,: E, x E 2 -+ F, and A2:E, x E 2 -+ F 2 are linear. Then A(v,. V2) = (AI (v,. V2). A2(V" V2») = (A,(v" 0)
+ A,(O. V2). A2(V" 0) + A2(0. V2»)'
The map
v, ...... AI(v" 0) is a linear map of E, into F, which we call A". Similarly, we let A,,(V,)
= A,(v" 0),
Adv2)
A2 ,(v,)
= A2(V" 0),
An(V2) = A2(0, V2)'
= A,(O, V2),
[XVII, §8]
DIFFERENTIATING UNDER THE INTEGRAL SIGN
499
Then we can represent A as the matrix
as explained in the preceding discussion, and we see that A(V"V2) is given by the multiplication of the above matrix with the vertical vector formed with VI and V2. Finally, we observe that if all Ai; are continuous, then the map A is also continuous, and conversely. We can apply this to the case of partial derivatives, and we formulate the result as a corollary. Corollary 7.2. Let U be open in £, x £2 and let I: U C'map. Let
-+
F1 X F2 be a
1= (f•.f2) be represented by its coordinate maps
I.:U-F.
and
12: U - F 2·
Then lor any x E U, the linear map DI(x) is represented by the matrix Dtf.(X) Dd.(x») ( Dtf,(x) D 2/2(x)·
I.
Proof. This follows by applying Theorem 7.1 to each one of the maps and 12, and using the definitions of the preceding discussion.
Observe that except for the fact that we deal with linear maps, all that precedes was treated in a completely analogous way for functions on open sets of n-space, where the derivative followed exactly the same formalism with respect to the partial derivatives.
XVII, §8. DIFFERENTIATING UNDER THE INTEGRAL SIGN The proof given previously for the analogous statement goes through in the same way. We need a uniform continuity property which is slightly stronger than the uniform continuity on compact sets, but which is proved in the same way. We thus repeat this property in a lemma.
500
[XVII, §8]
DERIVATIVES IN VECTOR SPACES
Lemma 8.1. Let A be a compact subset of a /lOrmed vector space, and let S be a subset of this normed vector space containing A. Let f be a contin1I0llS map defined on S. Given £ there exists [) such that if x E A and YES, and Ix - yl < 0, then If(x) - f(y)1 < £. Proof Given £, for each x E A we let r(x) be such that if yES and Iy - xl < r(x), then If(y) - f(x) I < £. Using the finite covering property of compact sets, we can cover A by a finite number of open balls Bi of radius [)i = r(Xi)f2, centered at Xi (i = I, ... ,n). We let
0= min Oi. If x E A, then for some i we have Iy - xd < r(x;), so that
Ix - xd
0 be such that Ig(x) - J(x)1 ~ e for all x. Assume that g has a fixed point x" and let x, be the fixed point ofJ. Show that lx, - x.l ~ ej(1 - h).
5. Let K be a continuous function of two variables, defined for (x, y) in the square a ~ x ~ h and a ~ y ~ h. Assume that IIKII ~ C for some constant C > O. Let J be a continuous function on [a, h) and let r be a real number satisfying the inequality Irl < C(h _ a) Show that there is one and only one function g continuous on [a, h) such that J(x)
= g(x) + r
f
K(t, x)g(t) dt.
6. (Newton's method) This method serves the same purpose as the shrinking lemma but sometimes is more efficient and converges more rapidly. It is used to find zeros of mappings. Let B, be a ball of radius r centered at a point x. E E. Let J: B, -+ E be a C' mapping, and assume that f" is bounded by some number C ~ 1 on B,. Assume that j'(x) is invertible for all x E B, and that Ij'(x) - 'I ~ C for all x E B,. Show that there exists a number 0 depending only on C and r such that if IJ(x.)1 ~ 0 then the sequence defined by
[XVIII, §IJ
505
THE SHRINKING LEMMA
lies in B, and converges to an element x such that fix) inductively that
= O.
[Hillt: Show
Ix.+, - x.1 ;;; Clf(x.)I, ,C
If(x.+I)I;;; Ix.+. -x.1 2' and hence putting rl = (C 30/2),
If(x,,)1 ;;; (C 3/2)'+2+ ...+2'-'0 2 '
Ix"+1 - xol ;;; qr,
+ r,2 + ... + rj,. ).
7. Apply Newton's method to prove the following statement. Assume that f: U - E is of class C 2 and that for some point Xo E U we have f(xo) = 0 and f'(x o) is invertible. Show that given y sufficiently close to 0, there exists x close to X o such that f(x) = )'. [Hillt: Consider Ihe map g(x) = f(x) - y.] [Note. The point of Ihe Newton method is that it often gives a procedure which converges much faster than the procedure of the shrinking lemma. Indeed, the shrinking lemma converges more or less like a geometric series. The Newton method converges with an expollem of 2".] 8. The following is a reformulation due to Tate of a theorem of Michael Shub. (a) Let II be a positive integer, and let f: R - R be a differentiable function such that f'(x) ;;; r > 0 for all x. Assume that f(x + I) = f(x) + II. Show that there exists a strictly increasing continuous map 0:: R -+ R satisfying
a(x + I) = a(x) + I such that
f(a(x)) = a(lIx). [Him: Follow Tate's proof. Show that f is continuous, strictly increasing, and let 9 be its inverse function. You want to solve a(x) = g(a(lIx». Let M be the set of all continuous functions which are increasing (not necessarily strictly) and satisfying a(x + I) = a(x) + I. On M, define the norm
lIall = sup la(x)l· O::i ... ~1
lei T: M - M be Ihe map such that
(Ta)(x) = g(a(lIx». Show that T maps Minto M and is a shrinking map. Show that M is complete, and that a fixcd point for T solves the problem.] Since one can write
IIX = a-'(fa(x»),
506
INVERSE MAPPING THEOREM
[XVIII, §2]
one says that the map x ..... nx is conjugate to f. Interpreting this on the circle, one gels the statement originally due to Shub that a differentiable function on the circle, with positive derivative, is conjugate to the n-th power for some n. (b) Show that the differentiability condition can be replaced by the weaker condition: There exist numbers "'" with 1 < " < " such that for all x ~ 0 we have 'IS;;;;' J(x
+ s) - J(x) ;;;;, "s.
Further problems involving similar ideas, and combined with another technique will be found at the end of the next section. It is also recommended that the first theorem on differential equations be considered simultaneously with these problems.
XVIII, §2. INVERSE MAPPINGS, LINEAR CASE Let A: E -> F be a continuous linear map. We continue to assume throughout that E, F are euclidean spaces, but what we say holds for complete normed spaces. We shall say that A is invertible if there exists a continuous linear map w: F -> E such that we A = idE and A 0 w = id F where idE and id F denote the identity mappings of E and F respectively. We usually omit the index E or F on id and write simply id or 1. No confusion can really arise, because for instance, w 0 A is a map of E into itself, and thus if it is equal to the identity mapping it must be that of E. Thus we have for every x E E and y E F:
W(A(X» = x
and
A(W(Y» = Y
by definition. We write A-I for the inverse of A. Consider invertible elements of L(E, E). If A., ware invertible in L(E, E), then it is clear that W 0 A is also invertible because (w 0 A)-1 = r i o w- I . For simplicity from now on, we shall write WA instead of W 0 A. Consider the special case).: R" -> R". The linear map A is represented by a matrix A = (ai)' One knows that A is invertible if and only if A is invertible (as a matrix), and the inverse of A, if it exists, is given by a formula, namely A
-I
1_
= Det(A) A ,
where A is a matrix whose components are polynomial functions of the components of A. In fact, the components of A are subdeterminants of A. The reader can find this in any text on linear algebra. Thus in this case, A is invertible if and only if its determinant is unequal to O.
[XVlll, §2J
507
INVERSE MAPPINGS, LINEAR CASE
Note that the determinant Det: Mat" X" -+ R is a continuous function, being a polynomial in the n2 coordinates of a matrix, and hence the set of invertible n x n matrices is open in Mat",". The next theorem gives a useful formula whose proof does not depend on coordinates.
Theorem 2.1. The sec of invertible elements of L(E, E) is open in L(E, E). If u E L(E, E) is such chac Iu I < I, chen I - u is invertible, and iCS inverse is given by che convergent series 00
(l - U)-I = I
+ u + u2 + ... = 2: un. n=O
Proof Since lui < I, and since for u, v E L(E, E) we have luvl ;;; lullvl, we conclude that lu"1 ;;; lui". Hence the series converges, being comparable to the geometric series. Now we have (I - uHI
+ u + u2 + ... + un) = I - U"+I = (I
+ u + ... + u"HI -
u).
Taking the limit as n -+ 00 and noting that u"+' -+ 0 as n -+ 00, we see that the inverse of I - u is the value of the convergent series as promised. We can reformulale whal we have jusl proved by stating that the open ball of radius I in L(E, E) centered at the identity I, consists of invertible elements. Indeed, if), E L(E, E) is such that I). - II < I, then we write ). = I - (l - ).) and apply our result. Let Uo be any invertible element of L(E, E). We wish to show that there exists an open ball of invertible elements centered at Uo' Let I
O'(u)h in view of the estimate
[XVIII, §2]
509
INVERSE MAPPINGS, LINEAR CASE
for some constant C. Thus the derivative is given by the formula as stated in the theorem. The fact that the map u ...... u- I is infinitely differentiable follows because the derivative is composed of inverses and continuous bilinear maps (composition), so that by induction ql is of class C' for every positive integer p. The theorem is proved.
Remark. In the case of matrices, the map A(x) ...... A(X)-I
where (x) = (Xi}) are the n2 components of the matrix A(x), can be seen to be Coo because the components of A(x) - 1 are given as polynomials in (x), divided by the determinant, which is not 0, and is also a polynomial. Thus one sees that this map is infinitely differentiable using the partial derivative criterion. However, even seeing this does not give the formula of the theorem describing the derivative of the inverse map, and this formula really would not be proved otherwise even in the case of matrices. Note that the formula contains the usual _u- 2 except that the noncommutativity of the product has separated this and placed u - I on each side of the variable v.
XVIII, §2. EXERCISES I. Let E be the space of 11 x
11
matrices with the usual norm IA I such that
IABI
~
IAIIBI.
Everything that follows would also apply to an arbitrary complete normed vector space with an associative product E x E -> E into itself, and an element I which acts like a multiplicative identity, such that II I = I. (a) Show that the series '" A" exp(A) = L n::O n! converges absolutely, and that Iexp(A) - I I < I if IA I is sufficiently small. (b) Show that the series 2
log(I + B)
= -B - -B + ... + (-I)n+ I -B" + ... I
2
11
converges absolutely if IBI < I and that in that case, Ilog(I
If II
+ B)I
~
IBI/(I -IBI).
- CI < I, show that the series 10gC=(C-I)-
converges absolutely.
(C - 1)2
2
+"'+(_1)"+1
(C - I)" 11
+."
510
INVERSE MAPPING THEOREM
[XVIII, §2)
(c) If IAI is suflicienily small show that log exp(A) = A and if IC - II < 1 show that exp log C = C. [Hinr: Approximate exp and log by the polynomials of the usual Taylor series, estimating the error terms.] (d) Show that if A, B commute,that is AB = BA, then exp(A
+ B) = exp A exp B.
State and prove the similar theorem for the log. (e) Let C be a matrix suflicienily close to I. Show that given an integer m > 0, there exists a matrix X such that xm = C, and that one can choose X so that XC = cx. 2 Let U be the open ball of radius I centered at I. Show that the map log: U - E is differenIiable. 3. Let V be the open ball of radius 1 centered at is differentiable.
o.
Show that the map expo V - E
4. Let K be a continuous function of two variables, defined for (x, y) in the square o ~ x ~ band 0 ~ y ~ b. Assume thatl/KI/ ~ C for some consIant C > o. LetJ be a continuous function on [0, b] and let r be a real number satisfying the inequality I
Irl < C(b _ a) . Show that there is one and only one function g continuous on [a, b] such that J(x) = g(x)
+r
r
K(t, x)g(t) dt.
•
(This exercise was also given in the preceding section. Solve it here by using Theorem 21.)
5. Exercises 5 and 6 develop a special case of a theorem of Anosov, by a proof due to Moser. First we make some definitions. Let A: R' _ R' be a linear map. We say that A is hyperbolic if there exist numbers b > I, c < I, and two linearly independent vectors v, w in R' such that Av = bv and Aw = cwo As an example, show that the matrix (linear map)
A=
G ~)
has this property. Next we introduce the C· norm. If J is a C' map, such that both J and bounded, we define the C' norm to be 1/ fII, = max(1/
J' are
n, 1/ !'II),
where 1/ 1/ is the usual sup norm. In this case, we also say that J is C'-bounded.
[XV III, SLJ
INVERSE MAPPINGS, LINEAR CASE
511
The theorem we are after runs as follows: Theorem. Let A: R' --> R' be a hyperbolic linear map. There exists 0 having the Jollowing property. IJ J: R' --> R' is a C 1 map such that
IIJ- All,
Joh = h 0 A. First prove a lemma. Lemma. Let M be the vector space oj continuous bounded maps oj R' into R'. Let T: M --> M be the map defined by Tp = P - A-' 0 P 0 A. Then T is a continuous linear map, and is invertible. To prove the lemma, write
where p. and p- are functions, and note that symbolically,
that is Tp· = (1 - S)p. where IISII < I. So lind an inverse for T on p•. Analogously, show that Tp- = (1 - So ')p- where IISoll < I. so that So T = So - 1 is invertible on p -. Hence T can be inverted componentwise. as it were. To prove the theorem. write J = A + 9 where 9 is C'-small. We want to solve for h = 1 + P with p E M, satisfying J 0 h = h 0 A. Show that this is equivalent to solving
Tp = -A-' ogoh, or equivalently,
This is then a fixed point condition for the map R: M
R(P) =
-->
M given by
-r'(r' ogo(1 + p».
Show that R is a shrinking map to conclude the proof. 6. One can formulate a variant of the preceding exercise (actually the very case dealt with by Anosov-Moser). Assume that the matrix A with respect to the standard basis of R' has integer coefficients. A vector Z E R' is called an integral vector if its coordinates are integers. A map p: R' --> R' is said to be periodic if
p(x
+ z) = p(x)
512
[XVIII, §3)
INVERSE MAPPING THEOREM
for all x E R2 and all integral vectors z. Prove: Theorem. Let A be hyperbolic, with integer coefficients. There exists Dhaving the Jollowing property. IJ g is a Co. periodic map. and IIgll, < D, and if
J = A + g. then there exists a periodic continuous map h satisJying the equation Joh
= hoA.
Note. With a bounded amount of extra work. one can show that the map h itself is CO-invertible, and so J = h 0 A 0 h' I.
XVIII, §3. THE INVERSE MAPPING THEOREM Let U be open in E and let I: U -+ F be a C' map. We shall say that I is C'-invertible on U if the image of I is an open set V in F, and if there is a C' map g: V -+ U such that I and 9 are inverse to each other, that is for all x E U and y E V we have g(f(x»)
=x
.
and
I(g(y»)
=
y.
In considering mappings between sets, we used the same notion of invertibility without the requirements that the inverse map 9 be C'. All that was required when dealing with sets in general is that I, 9 are inverse to each other simply as maps. Of course, one can make other requirements besides the C' requirement. One can say that I is CO-invertible if the inverse map exists and is continuous. One can say that I is CP-invertible if I is itself CP and the inverse map 9 is also CPo In the linear case, we dealt with linear invertibility, which in some sense is the strongest requirement which we can make. It will turn out that if I is a C' map which is C'-invertible. and if I happens to be CP, then its inverse map is also CPo This is the reason why we emphasize C' at this point. However, it may happen that a C' map has a continuous inverse, without this inverse map being differentiable. For example: Let J: R - R be the map I(x) = x 3 • Then certainly I is infinitely differentiable. Furthermore, I is strictly increasing, and hence has an inverse mapping g: R _ R which is nothing else but the cube root: g(y) = y1/3. The inverse map 9 is not differentiable at 0, but is continuous at 0. Let
I:U-
V
and
g:V-W
be invertible CP maps. We assume that V is the image of I and W is the image of g. We denote the inverse of I by I-I and that of 9 by g'l. Then
[XV 111,
~jJ
THE INVERSE MAPPING THEOREM
513
it is clear that 9 0 I is CP-invertible, and that (g 0 f) - I = I - log - I, because we know that a composite of CP maps is also CPo Let I: U -+ F be a CP map, and let Xo E U. We shall say that I is locally CP-invertible at Xo if there exists an open subset U I of U containing Xo such that I is CP-invertible on U I' By our definition, this means that there is an open set V. of F and a CP map g: VI -+ U I such that log and 9 0 I are the respective identity mappings of VI and U I' It is clear that a composite of locally invertible maps is locally invertible. In other words. if
I: U -+
and
V
g: V -+ W
are CP maps, Xo E U, I(xo) = Yo, if I is locally CP-invertible at Xo, if 9 is locally CP-invertible at Yo, then 9 0 I is locally CP-invertible at Xo. It is useful to have a terminology which allows us to specify what is the precise Image of an invertible map. For this purpose, we shall use a word which is now standard in mathematics. Let U be open in E and let V be open in F. A map tp:
U
-+
V
will be called a CP-isomorphismif it is CP, and if there exists a CP map t/J:V-+U
such that cp, t/J are inverse to each other. Thus cp is Cp·invertible on U, and V is the image cp(U) on which the CP inverse of cp is defined. We write the inverse often as t/J = cp - I. If
U.l.v
and
V4W
are Cp·isomorphisms, then the composite go I is also a CP-isomorphism, whose inverse is given by I-log-I. The word isomorphism is also used in connection with continuous linear maps. In fact, a continuous linear map J.:E-+F
is said to be an isomorphism if it is invertible. Thus the word isomorphism always means invertible, and the kind of invertibility is then made explicit in the context. When it is used in relation to CP maps, invertibility means CPo invertibility. When it is used in connection with continuous linear maps, invertibility means continuous linear invertibility. These are the only two examples with which we deal in this chapter. There are other examples in mathematics. however. Let t/J: U -+ V be a continuous map which has a continuous inverse cp: V -+ U. In other words, t/J is a CO·invertible map. If U I is an open subset of U, then t/J(U.) = VI is an open subset of V because t/J = cp-I
514
INVERSE MAPPING THEOREM
[XVlIl, §3]
and 'P is continuous. Thus open subsets of U and open subsets of V correspond to each other under the associations and Let U be open in E. A CP map IjJ:U-+V
which is CP-invertible on U is also called a CP chart. If Q is a point of U, we call1jJ a chart at Q. If IjJ is not invertible on all of U but is CP-invertible on an open subset U 1 of U containing Q, then we say that IjJ is a local CPisomorphism at Q. If E = R" = F and the coordinates of R" are denoted by X" . .. ,xn' then we may view IjJ as also having coordinate functions,
In this case we say that 1jJ" ... ,ljJn are local coordinates (of the chart) at Q, and that they form a CP-coordinate system at Q. We interpret IjJ as a change of coordinate system from (XI' ... ,xn) to (1jJ I (x), ••• ,ljJn(x»), of class CP. This terminology is in accord with the change from polar to rectangular coordinates as given in examples following the inverse mapping theorem, and which the reader is probably already acquainted with. We give here another example of a chart which is actually defined on all of E. These are translations. We let
be the map such that tv(x) = given by
X
+ v. Then the derivative of tv is obviously Dt'(x)
=1
where 1 is the identity mapping. Observe that if U is an open set in E and vEE then tv(U) is an open set, which is called the translation of U by v. It is sometimes denoted by U v, and consists of all elements x + v with x E U. We have
if w, vEE, and
A map tv is called the translation by v. For instance, if U is the open ball centered at the origin, of radius Y, then tv(U) = Uv is the open ball centered at v, of radius Y.
[XVIII, §3]
515
THE INVERSE MAPPING THEOREM
When considering functions of one variable, real valued, we used Ihe derivative as a lest for invertibility. From the ordering properties of the real numbers, we deduced invertibility from the fact that the derivative was positive (say) over an interval. Furthermore, at a given point, if the derivative is not equal to 0, then the inverse function exists, and one has a formula for its derivative. We shall now extend this result to the general case, the derivative being a linear map. Theorem 3.1 (Inverse mapping theorem). Let U be open in E, let Xo E U, and let f: U .... F be a C ' map. Assume that the derivative f'(xo): E .... F is invertible. Then f is locally C ' -invertible at Xo' If cp is its local inverse, and y = f(x), then cp'(y) = f'(X)-1. Proof We shall first make some reductions of the problem. To begin with, let A = f'(xo), so that A is an invertible continuous linear map of E into F. If we form the composite
.1.- 1 of: U .... E, then the derivative of .1.-' 0 f at Xo is .1.-' of'(xo) = 1. If we can prove that .1.- 1 0 f is locally invertible, then it will follow that f is locally invertible, because f = A0 .1.- 1 0 f. This reduces the problem to the case where f maps U into E itself, and where f'(xo) = 1. Next, let f(xo) = yo. Let f.(x) = f(x + xo) - yo. Then f, is defined on an open set containing 0, and f, (0) = O. In fact, f, is the composite map "0 U f. E"Yo E U -xo~ -=+ - t o •
It will suffice to prove that f, is locally invertible, because f. and then
=L
yO 0
f
0
Txo
is the composite of locally invertible maps, and is therefore invertible. We have thus reduced the proof to the case when Xo = 0, f(O) = 0 and 1'(0) = 1, which we assume from now on. Let g(x) = x - f(x). Then g'(O) = 0, and by continuity there exists r > 0 such that if Ixl ~ r then
Ig'(x)1 ~ t· Furthermore, by the continuity of I' and Theorem 2.1, f'(x) is invertible for Ixl ~ r.
516
INVERSE MAPPING THEOREM
[XVIII, §3]
From the mean value theorem (applied between 0 and x) we see that Ig(x)1 ;;; !Ixl, and hence 9 maps the closed ball B,(O) into B'/2(0). We contend that given y e B"2(0) there exists a unique element x e B,(O) such that f(x) = y. We prove this by considering the map gy(x) = y
+x
- f(x).
If Iyl ;;; r/2 and Ixl ;;; r then Igy(x)1 ;;; r. and hence gy may be viewed as a mapping of the complete metric space B,(O) into itself. The bound of! on the derivative together with the mean value theorem shows that gy is a shrinking map, namely
x,.
for X2 e B,(O). By the shrinking lemma. it follows that gy has a unique fixed point. which is precisely the solution of the equation f(x) = y. This proves our contention. Let V, be the set of all elements x in the open ball B,(O) such that If(x)1 < rJ2. Then V, is open, and we let V, be its image. By what we have just seen. the map f: V, - V, is injective, and hence we have inverse maps f:V,-V"
f-' = cp:
V, - V,.
We must prove that V, is open and that cp is of class C'. Let x, e V, and let y, = f(x ,) so that ly,l < r/2. If y e E is such that Iyl < r/2 then we know that there exists a unique x e B,(O) such that f(x) = y. Writing x = x - f(x) + f(x) we see that Ix - x,l ;;; If(x) - f(x,)1 ;;; If(x) - f(x,)1
+ Ig(x) - g(x,)1 + !Ix - x,l.
Transposing on the other side, we find that (*)
Ix - xd ;;; 2If(x) - f(x,)I.
This shows that if y is sufficiently close to y" then x is close to x" and in particular. Ixl < r since Ix,l < r. This proves that x e V,. and hence that ye V,. so that V, is open. The inequality (*) now shows that cp = f- I is continuous. To prove differentiability. note that !'(Xl) is invertible because f(x) - f(x,)
=!,(x,)(x -
x,)
+ Ix -
xdt/J(x - x,)
[XVIII, §3)
THE INVERSE MAPPING THEOREM
517
where I/J is a map such that lim I/J(x - x,) = O. Substitute this in the expression (**)
r
'(y) -
r '(y,) -
/,(x,)-'(y - y,) = x - x, - /,(x,)-'(J(x) - f(x,».
Using the inequality (*), and a bound C for /,(x,)-', we obtain
1(**)1
=
I/'(x,)-'Ix - x,ll/J(x - x,)1
~
2C1y - y,lll/J(cp(y) - cp(y,»I.
Since cp = f-' is continuous, it follows from the definition of the derivative that cp'(y,) = /,(x,)-'. Thus cp' is composed of the maps cp, /" and
uinverse," namely cp'(y)
= /'(cp(YW',
and these maps are continuous. It follows that cp' is continuous, whence cp is of class C'. This proves the theorem. Corollary 3.2. If f is of class CP thell its local illverse is of class CPo Proof By induction, assume the statement proved for p - I. Then /' is of class CP-', the local inverse cp is of class CP-', and we know that the map u 1-+ u-, is C"'. Hence cp' is of class CP-', being composed of CP-' maps. This shows that cp is of class CP, as desired.
In some applications, one needs a refinement of the first part of the proof, given a lower bound for the size of the image of f when the derivative of f is close to the identity. We do this in a lemma, which will be used in the proof of the change of variable formula.
Lemma 3.3. Let V be opell ill E, alld let f: V -> E be of class C'. Assume that f(O) = 0, /,(0) = I. Let r > 0 alld assume that B,(O) c U. Let o < s < I, alld assume that I/'(z) - /,(x)1 ;;; s
for all x, z € B,(O). If y € E alld x € B,(O) such that f(x) = y.
Iyl
~ (I - s)r, thell there exists a ullique
518
[XVIII, §3]
INVERSE MAPPING THEOREM
Proof The map g, given by g,(x) = x - f(x) + Y is defined for Ixl ~ r and IYI ~ (I - s)r. Then g, maps B,(O) into itself, because from the esti· mate If(x) - xl
= If(x) -
f(O) - I'(O)xl ~ Ixl sup II'(z) - 1'(0)1 ~ sr,
we obtain Ig,(x) I ~ sr + (I - s)r = r. Furthermore, g, is a shrinking map, because from the mean value theorem we get
Ig,(x,) - g,(x2)1 = Ix! - X2 - (f(x!) - f(X2»)1
= lx,
- X2 - I'(OXx! - X2)
+ : U
-+
F = RJ
such that
q>(p, 0/, 02)
= (p cos 0. sin 02' p sin 0. sin 02' p cos O2),
The determinant of the Jacobian of q> is given by Det J",(p, 0" O2) = _p2 sin O2 and is not equal to 0 whenever O2 is not an integral multiple of 7[, For such points, the map q> is locally invertible. For instance, we write
x
= pcos 0. sin 02'
y
= psin 0/ sin 02'
z=pcos0 2·
Let V be the open set of all (x, y, z) such that x > 0, Y > 0, on V the inverse of q> is given by the map
Z
> O. Then
.p:V-+U such that
ljJ(x, y, z) = (Jx 2 + y2 + Z2, arcsin J : 2' arccos J 2 Z 2 2)' x+y x+y+z The open subset U I of U corresponding to V (that is ljJ(V») is the set of points (p, 0/,° 2 ) such that p
0 0,
Example 3. Let q>: R 2
-+
R 2 be given by
q>(x, y) = (x + x 2f(x, y), Y + y2g(X, y»)
520
INVERSE MAPPING THEOREM
[XVIII, §4]
where f, 9 are C' functions. Then the Jacobian of cp at (0,0) is simply the identity matrix:
Hence cp is locally C'-invertible at (0,0). One views a map rp as in this example as a perturbation of the identity map by means of the extra terms x'f(x, y) and y'g(x, y), which are very small when x, yare near O.
Example 4. The continuity of the derivative is needed in the inverse mapping theorem. For example, let
= x + 2x' sin(l/x) flO) = o.
f(x)
if x -F 0,
Then f is differentiable, but not even injective in any open interval con· taining O. Work it out as an exercise.
XVIII, §3. EXERCISES I. Let f: U - F be of class c' on an open set U of E. Suppose that the derivative of f at every point of U is invertible. Show that f(U) is open.
2. Let I(x, y) = (e" + e", e' - e') By computing Jacobians, show that f is locally invertible around every point of R 2 • Does I have a global inverse on R 2 itself? 3. Let I: R 2 - R' be given by I(x, y) = (e" cos y, e" sin y). Show that Df(x, }') is invertible for all (x, y) E R 2, that I is locally invertible at every point, but does not have an inverse defined on all of R2 • 4. Let I: R 2 _ R 2 be given by I(x, y) = (x' - y2, 2xy). Determine the points of R' at which f is locally invertible, and determine whether f has an inverse defined on all of R2 •
The results ofthe next section will be covered in a more general situation in §5. However, the case offunctions on n-space is suffiCiently important to warrant the repetition. Logically, however, the reader can omit the next section.
XVIII, §4. IMPLICIT FUNCTIONS AND CHARTS Throughout this section, we deal with maps which are assumed to be of class C', and thus we shall say invertible instead of saying C'·invertible, and similarly for locally invertible instead of saying locally C··invertible. We always take p ~ 1.
[XVIII, §4)
IMPLICIT FUNCTIONS AND CHARTS
521
We start with the most classical form of the implicit function theorem.
Theorem 4.1. Let f: J, x J 2
-
R be a function of two real variables,
defined on a product of open intervals J" J 2' Assume that f is of class cPo Let (a, b) E J, x J 2 and assume that f(a, b) = 0 but D,j(a, b) # O. Then the map
given by (x. Y) I-> (x, f(x, y» is locally CP invertible at (a, b).
Proof All we need to do is to compute the derivative of if! at (a, b). We write if! in terms of its coordinates, if! = (if!" if!2)' The Jacobian matrix of if! is given by oif!, oy J~(x,
y) = Oif!2 oy
0
-
of of ax oy
and this matrix is invertible at (a, b) since its determinant is equal to of/oy # 0 at (a, b). The inverse mapping theorem guarantees that if! is locally invertible at (a, b).
Corollary 4.2. Let S be the set of pairs (x. y) such thac f(x, y) = O. Then there exists an open set U, in R 2 containing (a, b) such that if!(S r. U ,) consists of all numbers (x, 0) for x in some open interval around a. Proof Since if!(a, b) = (a, 0), there exist open intervals V" V2 containing a and 0 respectively and an open set U, in R 2 containing (a, b) such that the map if!:U,-V,xV2
has an in verse
(both of which are C' according to our convention). The set of points (x, y) E U, such that f(x, y) = 0 then corresponds under if! to the set of points (x, 0) with x E ~, as desired.
522
tNVERSE MAPPtNG THEOREM
[XVIII, §4]
Theorem 4.1 gives us an example of a chart given by the two coordinate functions x and f near (a, b) which reflects better the nature of the set S. In elementary courses, one calls S the curve determined by the equation f = O. We now see that under a suitable choice of chart at (a, ·b), one can transform a small piece of this curve into a factor in a product space. As it were, the curve is straightened out into the straight line VI' Example I. In the example following the inverse mapping theorem, we deal with the polar coordinates (r,8) and the rectangular coordinates (x, y). In that case, the quarIer circle in the first quadrant was straightened out into a straight line as on the following picture: y
8
--I---I--x
r
In this case V I is the open first quadrant and VI is the open interval 0< 8 < n/2. We have I/J(S n VI) = {l} x ~. The function f is the function f(x, y) = x 2 + y2 - 1. The next theorem is known as the implicit function theorem. Theorem 4.3. Let f: J I X J 2 -+ R be a function of two variables, defined on a product of open intervals. Assume that f is ofclass CPo Let
and assume that f(a, b) = 0 but D,/(a, b) 'I O. Then there exists an open interval J in R containing a and a CP function
g:J
-+
R
such that g(a) = band f(x, g(x» = 0 for all x E J.
[XVIII, §4]
523
IMPLICIT FUNCTIONS AND CHARTS
Proof By Theorem 4.1 we know that the map
t/J:J 1 x J 2 -+R x R=R 2 given by (x,y)~(x,f(x,y»
is locally invertible at (a, b). We denote its local inverse by ({J, and note that ({J has two coordinates, ({J = «{J" ({J2) such that
({J(x, z) = (x, ({J2(X, z»)
for x E R,
ZE
R.
We let g(x) = ((J2(X,0). Since t/J(a, b) = (a, 0) it follows that ((J2(a,0) = b so that g(a) = b. Furthermore, since t/J, ({J are inverse mappings, we obtain
(x, 0) = t/J«{J(x, 0») = t/J(x, g(x») = (x, f(x, g(x»). This proves that f(x, g(x») = 0, as was to be shown. We see that Theorem 4.3 is essentially a corollary of Thcorem 4.1. We have expressed y as a function of x explicitly by means of g, starting with what is regarded as an impliciI relation f(x, y) = O. Example 2. Consider the function f(x, y) = x2 + y2 - 1. The equation f(x, y) = 0 is that of a circle, of course. If we take any point (a, b) on the circle such that b # 0, then D2 f(a, b) # 0 and the theorem states that we can solve for y in terms of x. The explicit function is given by
y=Jl-x 2
if b > 0,
y=-Jl-x 2
if b < O.
If on the other hand b = 0 and then a # 0, then DI f(a, b) # 0 and we can solve for x in terms of y by similar formulas. We shall now generalize Theorem 4.3 to the case of functions of several variables. Theorem 4.4. Let U be open in R· and let f: U -+ R be a CP function on U. Let (a, b) = (a" .. . ,a._" b) E U and assume that f(a, b) = 0 but D.f(a, b) # O. Then the map
t/J: U
-+
R·- 1
X
R = R·
524
[XVIII, §4]
INVERSE MAPPING THEOREM
given by (x, y) ...... (x,f(x, y») is locally CP invertible at (a, b).
[Note. We write (a, y) as an abbreviation for (ai' ... ,an-I' y).] Proof The proof is basically the same as the proof of Theorem 4.1. The map !/J has coordinate functions Xl' ... 'Xn - I and f lis Jacobian matrix is therefore
1
0
0
0
1
0
J",(x) =
0
0 of ox,
of oX n
of OX2
and is invertible since ils determinant is again Dnf(a, b) the theorem.
~
O. This proves
Corollary 4.5. Let S be the set of points P E U such that f(P) = () Then there exists an open set U t in U containing (a, b) such thac !/J(S () U,) consists of all points (x, 0) with X in some open set VI of Rn - '. Proof Clear, and the same as Ihe corollary of Theorem 4.1.
From Theorem 4.4 one can deduce the implicil function theorem for functions of several variables. Theorem 4.6. Let U be open in Rn and let f: U -+ R be a CP function on U. Let (a, b) = (a" . .. ,an-" b) E U and assume thac f(a, b) = 0 but Dnf(a, b) ~ O. Then there exists an open ball V in Rn-t centered at (a) and a CP function
g: V
-+
R
such that g(a) = band fix, g(x») for all x
E
V.
=0
[XVIII, §4]
IMPLICIT FUNCTIONS AND CHARTS
525
Proof The proof is exactly the same as that of Theorem 4.3, except that x = (x" . .. ,x n_ I) lies in Rn- '. There is no need to repeat it.
In Theorem 4.6, we see that the map G given by
x ...... (x, g(x»
= G(x)
or writing down the coordinates
gives a parametrization of the hypersurface defined by the equation f(x" .. . ,xn - I , y) = 0 near the given point. We may visualize this map as follows:
G
,
On the right we have the surface f(X) = 0, and we have also drawn the gradient at the point P = (a, b) as in the theorem. We are now in a position to prove a result which had been mentioned previously (Chapter XV, §2 and §4), concerning the existence of differentiable curves passing through a point on a surface. To get such curves, we use our parametrization, and since we have straight lines in any given direction passing through the point a in the open set V of Rn-I, all we need to do is map these straight lines into the surface by means of our parametrization G. More precisely: Corollary 4.7. Let U be open in Rn and let f: U -+ R be a CP function. Let P E U and assume that f(P) = 0 but grad f(P) # O. Let w be a vector ofRn which is perpendicular to grad f(P). Let S be the set of points X such thar f(X) = O. Then there exists a CP curve a:J
-+
S
defined on an open interval J containing the origin such thar a(O) = P and a'(O) = w. Proof Some partial derivative of fat P is not O. After renumbering the variables, we may assume that Dnf(P) # O. By the implicit function theorem, we obtain a parametrization G as described above. We write P
526
INVERSE MAPPING THEOREM
[XVIII, §5]
in terms of its coordinates, P = (a, b) = (a" ... ,an -" b) so that G(a) = P. Then G'(a) is a linear map
G'(a): Rn-' .... Rn. In fact, for any x the matrix
= (Xl' ... ,Xn -
l)
the derivative G'(x) is represented by
0 0 0
o o
0
og og ax, ax,
og OX n
which has rank n - I. From linear algebra we conclude that the image of G'(a) in Rn has dimension n - I. Given any vector v in Rn-l we can define a curve ex in S by lelling ex(t) = G(a
+ tv).
Then a(O) = G(a) = P. Furthermore, a'(t) = G'(a
+ tv)v, so that
ex'(O) = G'(a)v. Thus the velocity vector of a is the image of v under G'(a). The subspace of Rn consisting of all vectors perpendicular to grad f(P) has dimension n - I. We have already seen (easily) in Chapter XV, §2, that a'(O) is perpendicular to grad f(P). Hence the image of G'(a) is contained in the orthogonal complement of grad f(P). Since these two spaces have the same dimension, they are equal. This proves our corollary.
XVIII, §S. PRODUCT DECOMPOSITIONS We shall now generalize the results of the preceding section to the general case where dimension plays no role, only the product decompositions. The proofs are essentially the same, linear maps replacing the matrices of partial derivatives. Theorem 5.1. Let U be open in a product E x F, and let f: U .... G be a C'map. Let (a, b) be a point of U with a E E and b E F. Assume that
Dd(a, b): F .... G
[XVIII, §5]
527
PRODUCT DECOMPOSITIONS
is invertible (as continuous linear map). Then the map "': U
-+
E
given by
xG
(x, y) H (x, f(x, y»
is locally CP-invertible at (a, b).
Proof We must compute the derivative ""(a, b). Since Dzf(a, b) is invertible, let us call it A. If we consider the composite map A-Iof:U-+G~F
then its second partial derivative will actually be equal to the identity. If we can prove that the map (*)
(x, y) H (x, A- ,
0
f(x, y»
is locally invertible at (a, b), then it follows that'" is locally invertible because '" can be obtained by composing the map from (*) with an invertible linear map, namely (v, w) H (v, .low).
This reduces our problem to the case when G = F and D z f(a, b) is equal to the identity, which we assume from now on. In that case, the derivative ""(a, b) has a matrix representation in terms of partial derivatives, namely II D"'(a, b) = ( D, f(a, b)
° ) (I,
Dzf(a, b) =
D I f(a, b)
0)
Iz .
Let p. = D I f(a, b). Then the preceding matrix is easily seen to have as inverse the matrix II ( -p.
0)
12
representing a continuous linear map of E x F -+ E x F. Thus D"'(a, b) is invertible and we can apply the inverse mapping theorem to get what we want. Note the exact same pattern of proof as that of the simplest case of Theorem 4.1. The values of f are now vectors of course. Let c = f(a, b). Then c is an element of G. Let S be the set of all (x, y) E U such that f(x, y) = c. We
528
INVERSE MAPPING THEOREM
[XVIII. §5]
view S as a level sel of f. wilh level c. The map t/! is a charI al (a. b). and we see Ihal under this chart. we obtain the same kind of straightening out of S locally near (a.b) that we obtained in §4. We fonnulate it as a corollary. Corollary 5.2. Let the notation be as in Theorem 5.1. Let f(a, b) = c and leI S be the subsel of V consisling of all (x, y) such thaI f(x, y) = c. There exists an open set V I of V containing (a, b), and a C'·isomorphism t/!: VI -> VI x V2 with VI open in E, V2 open in G, such that t/!(S n V,) = VI x {c}.
In the chapter on partial derivatives. we saw that the partial D 2 f(a, b) could be represented by a matrix when we deal with euclidean spaces. Thus in Theorem 5.1, suppose E x F = Rn and write
We have the map
and the isomorphism
This isomorphism is represented by the matrix of.
of,
ox n _ m + I
ax"
ofm
ofm AX"
Jlj'(x" ... ,xn ) = OXn -
m+ 1
+" ...
evaluated at (a" ... ,on)' The last set of coordinates (x n- m ,xn) plays the role of the (y) in Theorem 5.1. The creepy nature of the coordinates arises first from an undue insistence on the particular ordering of the coordinates (x I' ... ,xn ) so that one has to keep track of symbols like n-m+l;
second, from the non·geometric nature of the symbols which hide the linear map and identify Rm occurring as a factor of Rn, and Rm occurring as the
[XVIII, §5]
529
PRODUCT DECOMPOSITIONS
space containing the image of f; third, from the fact that one has to evaluate this matrix at (a" ... ,an) and that the notation of,
oan - m+
1
of, oan
to denote ofm
iJan -
m+ 1
ofm oan
( Dn- m+; ,f,(a) Dn- m+,fm(a)
...
Dn~,(a») Dnfm(a)
is genuinely confusing. We were nevertheless duty bound to exhibit these matrices because that's the way they look in the literature. To be absolutely fair, we must confess to feeling at least a certain computational security when faced with matrices which is not entirely apparent in the abstract (geometric) formulation of Theorem 5.1. Putting in coordinates for Rn and Rm, we can then formulate Theorem 5.1 as follows. CoroUary5.3. Let a = (a" ... ,an) be a point of Rn. Let f" .. · Jm be CP functions defined on an open set of Rn containing a. Assume that the Jacobian matrix (DJ,(a») (i = 1, ... ,m and j = n - m + I, ... ,n) is invertible. Then the functions
form a CP coordinate system at a. Proof This is just another terminology for the result of Theorem 5.1 in the case of Rn = Rn- m x Rm.
We obtain an implicit mapping theorem generalizing the implicit function theorem. Theorem 5.4. Let U be open in a product E x F and let f: U --> G be a CP map. Let (a, b) be a point of U with a E E and b E F. Let f(a, b) = O. Assume that D2 f(a, b): F --> G is invertible (as continuous linear map). Then there exists an open ball V centered at a in E and a continuous map g: V --> F such that g(a) = band f(x, g(x») = 0 for all x E V. If V is a sufficiently small ball, then 9 is uniquely determined, and is of class CPo Proof The existence of 9 is essentially given by Theorem 5.1. If we denote the inverse map of o/J locally by rp, and note that rp has two components, rp = (rp" r(2) such that rp(x, z) = (x, rpix, z»),
530
INVERSE MAPPING THEOREM
[XVIII, §5]
then we let 9(x) =
lJ,(b)
such that a(a) = b. For each a E M define Ta by
Ta(x) = a(x) - I(x, a(x»). It is an exercise to show that for suitable choice of r < s < I the map T
maps M into itself, and is a shrinking map, whose fixed point is precisely 9. Thus starting with any map ex, the sequence ex, Tex, T'a, . ..
converges to 9 uniformly. If D,/(a, b) = A is not assumed to be I, then we let I, = A- I 0 I, and T is replaced by the map TI such that
T,a(x)
= a(x) - II (x, a(x») = a(x) -
A-1/(x, a(x»).
If the map I is given in terms of coordinates, then D,/(a, b) is represented by a partial Jacobian matrix, and its inverse can be computed explicitly in terms of the coordinates.
[XVIII, §5]
PRODUCT DECOMPOSITIONS
531
We now return to the aspect of the situation in Theorem 5.1 concerned with the straightening out of certain subsets. Such subsets have a special name, and we give the general definition concerning them. Let S be a subset of E. We shall say that S is a submanifo1d of E if the following condition is satisfied. For each point XES there exists a CPisomorphism
mapping an open neighborhood U of x in E onto a product of open sets VI in some space F I' V2 in some space F 2' such that
.p(S n U)
V, x {c}
=
for some point c in V2 • Thus the chart provides a CP-change of coordinates so that in the new space, .p(S n U) appears as a factor in the product. The chart .p at x gives rise to a map of S,
.pIS: S n
U ..... VI
simply by restriction; that is we view .p as defined only on S. The restriction of such a chart .p to S n U is usually called a chart for Sat x. It gives us a representation of a small piece of S near x as an open subset in some space Fl' Of course, there exist many charts for S at x. Theorem 5.1, and the theorems of the preceding section, give criteria for the level set of f to be a submanifold, namely that a certain derivative should be invertible. We shall now derive another criterion, starting from a parametrization of the set. Let E, be a closed subspace of E, and let E 2 be another closed subspace. We shall say that E is the direct sum of E I and E2 and we write
if the map given by is an invertible continuous linear map. If this is the case, then every element of E admits a unique decomposition as a sum
532
INVERSE MAPPING THEOREM
[XVIII, §5]
Example 1. We can wriIe R" as a direct sum of subspaces Rq x {OJ and {OJ x R' if q + s = n. Example 2. Let F be any subspace of R". Let FJ. be the subspace of all vectors WE R" which are perpendicular to all elements of F. Then from linear algebra (using the orthogonalization process) one knows that
R" = F® FJ. is a direct sum of F and its orthogonal complement. This type of decomposition is the most useful one when dealing with R" and a subspace. Example 3. Let VI' •.. ,vq be linearly independent elements of R". We can always find (in infinitely many ways if q #- n) elements vq + 1> ••• ,V" such that {VI> ... ,v"} is a basis of R". Let E, be the space generated by VI' •••• vq and E 2 the space generated by vq + I • . . . •V". Then
In Example 2, we select vq + h ... ,V" so that they are perpendicular to E,. We can also select them so that they are perpendicular to each other. When we have a direct sum decomposition E projections 7t 1 :
E -+ E I
and
7t 2 :
= E I ® E 2 then
we have
E -+ E 2
on the first and second factor respectively, namely 7t,(v I + V2) = v, and 7t,(VI + V2) = V2 if VI E E, and v2 E E 2 · When E 2 = Et is the orthogonal complement of E I then the projection is the orthogonal projection as we visualize in the following picture:
R
[XVIII, §5]
PRODUCT DECOMPOSITIONS
533
We have drawn the case dim E, = 2 and dim E 2 = I. Such decompositions are useful when considering tangent planes. For instance we may have a piece of a surface as shown on the next picture:
We may want to project it on the first two coordinates, that is on R2 x {O}, but usually we want to project it on the plane tangent to the surface at a point. We have drawn these projections side by side in the next picture: R
R' F
The tangent plane is not a subspace but the translation of a subspace. We have drawn both the subspace F and its translation F p consisting of all points w + P with WE F. We have a direct sum decomposition
Theorem 5.5. Let V be an open set in F and let
g: V -. E
534
INVERSE MAPPING THEOREM
[XVIII, §5]
be a CP map. Let a E V and assume that g'(a): F ..... E is an invertible continuous linear map between F and a closed subspace E I of E. Assume that E admits a direct sum decomposition E = E, Ei'> E2 · Then the map ~:
V x E2
.....
E
given by (x, y) t-+ g(x)
+Y
is a local CP-isomorphism at (a, 0).
Proof We need but to consider the derivative of ~, and obtain ~'(a, O)(v,
w) = g'(a)v
+w
for v E F and WE E 2 . Then ~'(a, 0) is invertible because its inverse is given by v, + V2 t-+ (it- 'v" V2) if VI E E" V2 E E 2 and it = g'(a). We can now apply the inverse mapping theorem to conclude the proof. From Theorem 5.5, we know that there exist open sets VI in F containing a, V2 in E2 containing 0, and U in E such that ~:
V, x V2
.....
U
is a CP-isomorphism, with inverse.p: U ..... VI X V2 . Then g(x)
= ~(x, 0).
Let S = g(VI). Then S is the image of VI under g, and is a subset of E parametrized by 9 in such a way that our chart .p straightens S out back into V, x (OJ, that is .p(S) = V, x {O}.
We note that Theorems 5.1 and 5.5 describe in a sense complementary aspects of the product situation. In one case we get a product through a map f which essentially causes a projection, and in the other case we obtain the product through a map 9 which causes an injection. At all times, the analytic language is adjusted so as to make the geometry always visible, without local coordinates. There is a Jacobian criterion for the fact that D 2 f(a, b) is invertible, as described in Chapter XVII, §7. We can also give a matrix criterion for the
[XVIII, §5]
535
PRODUCT DECOMPOSITIONS
hypothesis of Theorem 5.5. Let us consider the case when £ F = Rmwith m ;;; n. Then V is open in Rm and we have a map g: V -+
= R"
and
R".
The derivative g'(a) is represented by the actual Jacobian matrix
J. Rm be a mapping of class C' with m < n. Assume ['(xo) is surjective for some x o. Show that f is not injective. (Actually much more is true, but it's harder to prove.) 3. Let f: R --> R be a C I function such that ['(x) ", 0 for all x E R. Show that f is a C'-isomorphism of R with the image of f 4. Let U be open in R' and let f: U --> Rmbe Coo with ['(x): R' --> Rmsurjective for all x in U. Prove that flU) is open. 5. Let f: Rm --> R' be a C ' map. Suppose that x E Rm is a point at which Df(x) is injective. Show that there is an open set U containing x such that fCy) ", f(x) for all y E U. 6. Let [a, b] be a closed interval J and let f: J --> R' be a map of class CI. Sbow that the image f(J) has measure 0 in R'. By this we mean that given E, there exists a sequence of squares {SI. S" ... } in 'R' such that the area of the square S" is equal to some number K" and we have f(J)c
US,
Generalize this to a map f: J using cubes instead of squares.
-->
and
LK, <E.
R 3 , in which case measure zero is defined by
7. Let U be open in R' and let f: U --> R3 be a I]lap of class C" Let A be a compact subset of U. Show Ihal f(A) has measure 0 in R 3 • (Can you generalize this, 10 maps of Rm into R' when n > m?) 8. Let U be open in R' and let f: U --> Rm be a C ' map. Assume that m ~ nand lei a E U. Assume that f(a) = 0, and that the rank of the matrix (Djf,(a» is m, if (f" ... ,1m) are the coordinate functions of f Show that there exists an open subset U 1 of U containing a and a C'-isomorphism 'P: VI --> U, (where V, is open in R') such thai f(q>(x" ... ,x,» = (x,_m+ I ' ' ' ' ,x,).
9. Let I: R x R --> R be a C I function such that D,/(a, b) ", 0, and let g solve the implicit function theorem, so thai fIx. g(x» = 0 and g(a) = b. Show that g'(x)
= _ Dtf(x, g(x» D,f(x, g(x»)'
[XVIII, §5]
537
PRODUCT DECOMPOSITIONS
10. Generalize Exercise 9, and show that in Theorem 5.4, the derivative of 9 is given by
g'(x) = - (D 2 f(x, g(x»( I 0 D.J(x, g(x»). II. Let f: R .... R be of class C I and such that I j'(x) I ~ c < I for all x. Define g:
R2 .... R2
by
g(x, y) = (x
+ f(y), Y + f(x)).
Show that the image of 9 is all of R 2 • 12. Let f: R" .... R" be a C I map, and assume that I f'(x) I ~ c g(x) = x + f(x). Show that g: R" .... R" is surjective.
< I for all x E R". Let
13. Let).: E .... R be a continuous linear map. Let F be its kernel, that is the set of all WEE such that ).(w) = O. Assume F # E and let Vo E E, Vo f F. Let F, be the subspace of E generated by vo. Show that E is a direct sum F Ell F, (in particular, prove that the map (w, r) ..... W
+ rvo
is an invertible linear map from F x R onto E). 14. Let J(x, y) = (x cos y, x sin y). Show that the determinant of the Jacobian of J in the rectangle I < x < 2 and 0 < Y < 7 is positive. Describe the image of the rectangle under f. 15. Let S be a submanifold of E, and let PES. If and are two charts for S at P (where V" V 2 are open in R 3 ), show that there exists a local isomorphism between VI at !/J,(P) and V2 at!/J2(P), mapping !/JI(P) on !/J2(P), 16. Let!/J,: VI" S .... VI be a chart for S at P and let gl: VI .... V, "S be its inverse mapping. Suppose VI is open in F" and let x, E F, be the point such that
g,(x l ) = P. Show that the image of g'l(x,): F I
....
E is independent of the chart for S at P.
(It is called the subspace of E which is parallel to the tangent space of S at P.)
CHAPTER
XIX
Ordinary Differential Equations
XIX, §1. LOCAL EXISTENCE AND UNIQUENESS We link here directly with the shrinking lemma. and this section may be read immediately after the first section of the preceding chapter. We defined a vector field previously over an open set of R". We don't need coordinates here, so we repeat the definition. We continue to assume that E, F are euclidean spaces, and what we say holds more generally for complete normed vector spaces. Let U be open in E. By a vector field on U we mean a map f: U -+ E. We view this as associating a vector f(x) E E to each point x E U. We say the vector field is of class cP if f is of class C'. We assume p ;;; I throughout. and the reader who does not like p ;;; 2 can assume p = I. Let Xo E U and let f: U -+ E be a vector field (assumed to be of class C' throughout). By an integral Curve for the vector field, with initial condition xo, we mean a mapping ex: J
-+
U
defined on some open interval J containing 0 such that ex is differentiable. ex(O) = Xo and ex'(t) = f(ex(t» for all t E J. We view ex'(t) as an element of E (this is the case of maps from numbers to vectors). Thus an integral curve for f is a curve whose velocity vector at each point is the vector associated to the point by the vector 538
[XIX, §1]
LOCAL EXISTENCE AND UNIQUENESS
539
field. If one thinks of a vector field as associating an arrow to each point, then an integral curve looks like this:
--
•
/
-
- - ---,-/
Remark. Let ex: J
-+
U be a continuous map satisfying the condition
ex(t) = Xo
+
Lf(ex(u» duo
Then ex is differentiable, and its derivative is ex'(t) = f(ex(t». Hence ex is of class e' and is an integral curve for f Conversely, if ex is an integral curve for f with initial condition xo, then ex obviously satisfies our integral equation since indefinite integrals of a continuous map differ by a constant, and the initial condition determines this constant uniquely. Thus to find an integral curve, we shall have to solve the preceding integral equation. This wi1l be a direct consequence of the shrinking lemma. Theorem 1.1. Let U be open in E and let f: U -+ E be a e' vector field. Let Xo E U. Then there exists an integral curve ex: J -+ U with initial condition Xo. If J is sufficiently sma/l, this curve is uniquely determined. Proof Let a be a number > 0 and let Bo be the open ball of radius a centered at Xo' We select a sufficiently small so that f is bounded by a number e on Bo • We can do this because f is continuous. Furthermore, we select a so small that J' is bounded by a constant K > 1 on the closed ball Bo • Again we use the continuity of J'. Now select b > 0 such that be < a and also bK < 1. Let I b be the closed interval [-b, b]. Let M be the set of all continuous maps
such thaI ex(O) = Xo. Then M is closed in the space of all bounded maps with the sup norm. For each ex E M define a map Sex by (Sex)(t) = Xo
+
Lf(cx(U» duo
540
[XIX, §I]
ORDINARY DIFFERENTIAL EQUATIONS
We contend that Srx lies in M. First, it is clear that Srx(O) Srx is continuous. Next, for alit E lb, ISrx(t) - xol
~
= Xo
and that
bC
so Srx E M. Finally, for ex, PE M we have
Srx(t) - SP(t) =
f~ (f(rx(u)) -
f(P(u))) du,
whence by the mean value theorem, ISrx(t} - SP(t) I ~ bK suplrx(u) - P(u)1 UE/b
~
bKllrx -
PII.
This proves that S is a shrinking map, and by the shrinking lemma, S has a unique fixed point ex, that is Srx = rx. This means that rx satisfies the integral equation which makes it an integral curve of f, as was to be shown. We shall be interested in a slightly more general situation, and for future reference, we state explicitly the relationship between the constants which appeared in the proof of Theorem 1.1. These are designed to yield uniformity results later. Let V be an open set in some space, and let f: V x V
-+
E
be a map defined on a product of V with some set V. We say that f satisfies a Lipschitz condition on V uniformly with respect to V if there exists a number K > 0 such that
If(v, x) - f(v, y)1
~
Klx -
YI
for all v E V and x, Y E U. We call K a Lipschitz constant. If f is of class C 1 , then the mean value theorem shows that f is Lipschitz on some open neighborhood of a given point (vo, xo) in V x V, and continuity shows that f itself is bounded on such a neighborhood. It is clear that in the proof of Theorem 1.1, only a Lipschitz condition intervened. The mean value theorem was used only to deduce such a condition. Thus a Lipschitz condition is the natural one to take in the present situation. Furthermore, suppose that we find integral curves through each point x of U. Then these curves depend on two variables, namely the variable t
[XIX, §1]
LOCAL EXISTENCE AND UNIQUENESS
541
(interpreted as a time variable), and the variable x itself, the initial condition. Thus we should really write our integral curves as depending on these two variables. We define a local flow for f at X o to be a mapping
a:J x Uo -+ U where J is some open interval containing 0, and U 0 is an open subset of U containing xo, such that for each x in U 0 the map t 1-+ ait) = a(t, x)
is an integral curve for f with initial condition x, i.e. such that a(O, x) = x. As a mailer of notation, we have written ax to indicate that we view x as a certain parameter. In general, when dealing with maps with two arguments, say ",(t, x), we denote the separate mappings in each argument when the other is kept fixed by lfJit) or lfJ,(x). The choice of letters and the context will always be carefully specified to prevent ambiguity. The derivative of the integral curve will always be viewed as vector valued since the curve maps numbers into vectors. Furthermore, when dealing with flows, we sometimes use the notation a'(t, x)
to mean D,a(t, x) and do not use the symbol' for any other derivative except partial derivative with respect to t, leaving other variables fixed. Thus a'(t, x)
= a~(t) = D,a(t, x)
by definition. All other partials (if they exist) will be written in their correct notation, that is D 2 , ... and total derivatives will be denoted by D as usual. Example. Let U = E be the whole space, and let 9 be a constant vector field, say g(x) = v ,p 0 for all x E U. Then the flow a is given by a(t, x) = x
+ tv.
Indeed, D,a(t, x) = v and since an integral curve is uniquely determined, with initial condition a(O, x) = x, it follows that the flow is precisely the one we have written down. The integral curves look like straight lines. In Exercise 4, we shall indicate how to prove that this is essentially the most general situation locally, up to a change of charts.
542
ORDINARY DIFFERENTIAL EQUATIONS
[XIX, §1]
We shall raise the question later whether the second partial D z rJ.(t, x) exists. It will be proved as the major result of the rest of this chapter that whenever f is CP then a itself, as a flow depending on both t and x, is also of class CPo Finally, to fix one more notation, we let I b be the closed interval [ - b, b] and we let J b be the open interval - b < t < b. If a > 0 we let B.(x) be the open ball of radius a centered at x, and we let B.(x) be the closed ball of radius a centered at X. The next theorem is practically the same as Theorem 1.1, but we have carefully stated the hypotheses in terms of a Lipschitz condition, and of the related constants. We also consider a time-dependent vector field. By this we mean a map
f: J x U
->
E
where J is some open interval containing O. We think of f(t, x) as a vector associated with x also depending on time t. An integral curve for such a time-dependent vector field is a differentiable map
a:Jo-U defined on an open interval J o containing 0 and contained in J, such that
a'(t) = f(t, aCt»). As before, a(O) is called the initial condition of the curve. We shall need time-dependent vector fields for applications in §4. We also observe that if f is continuous then a is of class C 1 since a' is the composite of continuous maps. By induction, one concludes that if f is of class CP then a is of class Cp+ '. We shall consider a flow for this time-dependent case also, so that we view a flow as a map
a:J o x U o - U where U 0 is an open subset of U containing X o and J 0 is as above, so that for each x the curve
t ..... aCt, x) is an integral curve with initial condition x (i.e. a(O, x) = x). Theorem 1.2. Let J be an open interval containing O. Let U be open in E. Let X o E U. Let 0 < a < I be such that the closed ball Bz.(x o) is COIItained in U. Let
f: J x U -> E
[XIX, §1]
LOCAL EXISTENCE AND UNIQUENESS
543
be a continuous map, bounded by a constant C > 0 and satisfying a Lipschitz condition on U with Lipschitz constant K > 0 uniformly with respect to J. If b < alC and b < 11K then there exists a uniquefiow ex: J b
X
B.(xo) -+ U.
If f is of class CP, then so is each integral curve ex x ' Proof Let x E B.(xo). Let M be the set of continuous maps
such that ex(O) = x. Then M is closed in the space of bounded maps under the sup norm. For each ex E M we define Sex by
Sex(t) = x
+ Lf(U, ex(u»
duo
Then Sex is certainly continuous and we have Scx(O) = x. Furthermore,
ISex(t) - xl ;;;; bC < a so that Sex(t) E
B2 .(xo) and Sex lies in M.
Finally for ex, P E M we have
IISex - SPII ;;;; b sup If(u, ex(u» - f(u, P(u»1 ue1b
;;;; bKllex - PII. This proves that S is a shrinking map, and hence S has a unique fixed point which is the desired integral curve. This integral curve satisfies ex(O) = x, and so depends on X. We denote it by exx ' and we can define ex(t, x) = exit), as a function of the two variables t and x. Then ex is a flow. This proves our theorem. Remark 1. There is no particular reason why we should require the integral curve to be defined on an interval containing 0 such that ex(O) = Xo. One could define integral curves over an arbitrary interval (open) and prescribe ex(t o) = Xo for some point to in such an interval. The existence and uniqueness of such curves locally follows either directly by the same method, writing
ex(t) = ex(t o) +
r' flu, ex(u» J,.
du,
or as a corollary of the other theorem, noting that an interval containing to can always be translated from an interval containing O.
544
[XIX, §l]
ORDINARY DIFFERENTIAL EQUATIONS
Combining Ihe local uniqueness wilh a simple leasl upper bound argumenl, we shall obtain the global uniqueness of integral curves. Theorem 1.3. Let f: J x U the open set U of E. Let
-+
E be a time-dependent vector field over
and be IWO integral curves wilh Ihe same initial condition are equal on J, n J 2 .
Xo'
Then a, and a2
= ait) for < b. Then T contains some b > 0 by the local uniqueness theorem. If T is not bounded from above, the equality of a,(1) and a2(1) for all t > 0 follows at once. If T is bounded from above, let b be ils least upper bound. We must show that b is the right end point of J, n J 2' Suppose this is nOI the case. Define curves P" P2 near 0 by Proof Let T be the set of numbers b such Ihat a, (t)
o~ t
and Then PI' P2 are integral curves of f with the initial conditions a,(b) and aib) respectively. The values P,(I) and P2(1) are equal for small negative t because b is a least upper bound of T. By continuity it follows Ihat a,(b) = a2(b), and finally we see from the local uniqueness theorem that P,(t) = P2(1) for all t in some neighborhood of 0, whence a, and a2 are equal in a neighborhood of b, contradicting the fact that b is a least upper bound of T. We can argue in the same way toward the left end points, and thus prove the theorem. It follows from Theorem 1.3 that the union of the domains of all integral curves of f with a given initial condilion Xo is an open interval which we denote by J(xo). Its end points are denoted by 1+(Xo) and C(xo) respectively. We allow by convention + 00 and - 00 as end points. Let !iP(f) be the subset of R x U consisting of all points (t, x) such that C(x) < t < t+(x).
A global flow for f is a mapping a: !iP(f) -+ U
such that for each x E U the partial map a.: J(x) -+ U, given by a.(t) = aCt, x)
[XIX, §1]
LOCAL EXISTENCE AND UNIQUENESS
545
defined on Ihe open inlerval J(x), is an inlegral curve for f wilh inilial condilion x. We define §)(f) to be the domain of the flow. We shall see in §4 that §)(f) is open and that if f is CP then the flow a is also CP on its domain. Remark 2. A time-dependent vector field may be viewed as a timeindependent vector field on some other space. Indeed, let f be as in Theorem 1.2. Define J:JxU-+RxE
by
!(I, x)
= (1,f(I, x»)
and view! as a time-independent vector field on J x U. Let so that D, a(1, S, x) = !(a(l, s, x»),
a(o, s, x) =
a be its flow,
(s, x).
We note that a has its values in J x U and thus can be expressed in terms of two components. In fact, it follows at once that we can write a in the form a(l, S, x) = (I
+ s, ail, s, x»).
Then a2 satisfies the differential equati9n
as we see from the definition of J Let fJ(l, x) = ail, 0, x). Then fJ is a flow for f, i.e. satisfies the differential equation D, fJ(l, x)
= f( I, fJ(l, x»),
fJ(O, x) = x.
Given x E U, any value of I such that a is defined at (I, x) is also such that a is defined at (I, 0, x) because ax and fJx are integral curves of the same vector field, with the same initial condition, hence are equal. Thus the study of time-dependent vector fields is reduced to the study of timeindependent ones.
546
ORDINARY DIFFERENTIAL EQUATIONS
[XIX, §l]
Remark 3. One also encounters vector fields depending on parameters, as follows. Let V be open in some space F and let
g: J x V x U
-+
E
be a map which we view as a time-dependent vector field on U, also depending on parameters in V. We define G: J x V x U
-+
F x E
by G(t, z, y) = (0, g(t, z, y»
for
V, and y E U. This is now a time-dependent vector field on V x U. A local flow for G depends on three variables, say P(t, z, y), with initial condition P(O, z, y) = (z, y). The map P has two components, and it t E J, Z E
is immediately clear that we can write P(t, z, y) = (z, a(/, z, y»
for some map a depending on three variables. Consequently a satisfies the differential equation D , :(t,
Z,
y)
= g(t, z, a(/, z, y»,
:(0, z,y) = y,
which gives the flow of our original vector field 9 depending on the parameters Z E V. This procedure reduces the study of differential equations depending on parameters to those which are independent of parameters.
XIX, §1. EXERCISES I. Let f be a C' vector field on an open set U in E. If f(xo) = 0 for some Xo E U, if a: J ..... U is an integral curve for f, and there exists some 10 E J such that a(t o) = Xo, show that a(l) = Xo for aliI E J. (A point Xo such that f(x o) = 0 is called a critical
point of the vector field.) 2. Let f be a C' vector field on an open set U of E. Let a: J -+ U be an integral curve for f Assume that all numbers I > 0 are contained in J, and that there is a point P in U such that lim :(t) = P. Prove that f(P) = O. (Exercises 1 and 2 have many applications, notably when g. In this case we see that P is a critical point of the vector field.)
f = grad 9 for some function
3. Let U be open in R" and let g: U ..... R be a function of class C'. Let Xo E U and assume that Xo is a critical point of 9 (that is g'(xo) = 0). Assume also that D'g(xo)
[XIX, §l]
LOCAL EXISTENCE AND UNIQUENESS
547
is negative definite. By definition, take this to mean that there exists a number c > 0 such that for all vectors v we have D'g(xo)(v, v);;;;
-clvl'.
Prove that if XI is a point in the ball B,(xo) of radius 1', centered at x o , and if I' is sufficiently small. then the integral curve" of grad 9 having X, as initial condition is defined for all t !1; 0 and lim ,,(t)
= X o'
[Hint: Let ljJ(t) = (,,(t) - xo)· ("(I) - xo) be the square of the distance from <x(t) to Xo' Show thatljJ is strictly decreasing, and in fact satisfies 1jJ'(t) ~ -2ClljJ(t),
where
C1
> 0 is near c, and is chosen so that D'g(x)(v, v) ~
-c,lvl'
for all X in a sufficiently small neighborhood of X O' Divide by ljJ(t) and integrate to see that log 1jJ(1) - log 1jJ(0)
~ -CI.
Alternatively. use the mean value theorem on ljJ(t,) -1jJ(t ,) to show that this difference has to approach 0 when I, < t, and t,.t, are large.] 4. Let U be open in E and let J: U -> E be a C' vector field on U. Let X o € U and assume that J(x o) = v oF O. Let" be a local flow for J at Xo. Let F be a subspace of E which is complementary to the one-dimensional space generated by v, that is the map
given by (t, y) ..... tv + y is an invertible continuous linear map. (a) If E = R" show that such a subspace exists. (b) Show that the map p: (t, y) ..... <x(I, Xo + y) is a local C' isomorphism at (0.0). You may assume that D,,, exists and is continuous. and that D,,,(O. x) = id. This will be proved in §4. Compute DP in terms of D," and D,,,. (c) The map q: (t, y) ..... X o + Y + tv is obviously a C ' isomorphism, because it is composed of a translation and an invertible linear map. Define locally at Xo the map 'I' by 'I' = p. q- '. so that by definition, q>(xo
+ Y + tv)
Using the chain rule, show that for all
X
=
,,(t, Xo + y).
near X o we have
Dq>rx)v = J(q>(x»
548
ORDINARY DIFFERENTIAL EQUATIONS
[XIX, §2]
If we view 'P as a change of chart near xo, then this result shows that the vector field f when transported by this change of chart becomes a constant vector field with value v. Thus near a point where a vector field does not vanish, we can always change the chart so that the vector field is straightened out. This is illustrated in the following picture:
• In this picture, we have drawn the flow, which is horizontal on the left, the vector field being constant. In general, suppose 'P: V o -+ V o is a C' isomorphism. We say that a vector field 9 on Vo and a vector field f on V 0 correspond to each other under 'P, or that f is transported to Vo by 'P if we have the relation f('P(x» = Dcp(x)g(x).
which can be regarded as coming from the following diagram:
Vo -c:--+l V 0
•
In the special case of our Exercise, 9 is the constant map such that g(x)
= v for all
xeJlo·
XIX, §2. APPROXIMATE SOLUTIONS As before, we let f: J xU ..... E be a time-dependent vector field on U. We now investigate the behavior of the flow with respect to its second argument, i.e. with respect to the points of U. Let J 0 be an open subinterval of J containing 0 and let 0 such that I f(t, a(t»)1 ~ C for all t in (b o - £, bo)· Then bo = b.
552
ORDINARY DIFFERENTIAL EQUATIONS
[XIX, §3]
Pl'oq( Suppose bo < b. From the integral expression for a, namely
a(c)
= a(c o) +
we see that for c" c2 in (b o -
£,
J' flu, a(u» du I.
bo) we have
This is the Cauchy criterion, and hence the limit
lim a(c) ,- bo
exists and is equal to an element Xo of U by hypothesis (i). By the local existence theorem, there exists an integral curve {3 defined on an open interval containing bo such that {3(b o) = Xo and {3'(c) = ftc, {3(c». Then {3' = a' on an open interval to the left of b o and hence a, {3 differ by a constant on this interval. Since their limits as I ---> bo are equal, this constant is O. Thus we have extended the domain of definition of a to a larger interval, as was to be shown.
Remark. Theorem 2.4 has an analogue giving a criterion for the integral curve being defined all the way to the left end point of J, and we shall use Theorem 2.4 in both contexts as a criterion for the integral curve to be defined on all of J.
XIX, §3. LINEAR DIFFERENTIAL EQUATIONS We shall consider a special case of differential equations, both for its own sake and for applications to the general case afterwards. We let L be a vector space as usual which in applications will be a space of continuous linear maps. We let E be some space, and assume given a product LxE--+E,
written
(A, w) ...... AW,
that is a bilinear map satisfying the condition IAwl Let J be an open interval, and let
~
IAllwl.
A: J -> L
be a continuous map. We consider the differential equation A'(c) = A(c}A(c)
[XIX, §3]
553
LINEAR DIFFERENTIAL EQUATIONS
corresponding to the time-dependent vector field on E given by (t, w) ...... A(t)w. In the applications. we have two cases: I. The product given by composition of mappings. namely
L(E. E) x L(E, E)
--+
L(E. E)
for some space E, so that .low = A 0 w for .l., W E L(E, E). 2. The product given by applying linear maps to vectors. namely L(E, E) x E
--+
E.
In the first case. suppose E = Rn• Then we can think of A(t) as an n x n matrix, and of the solution as an n x n matrix also. say B(t). so that our differential equation can be written B'(t) = A(t)B(t).
the product being multiplication of matrices. In the second case. we think of .lo(t) as a curve in Rn• which we write X(t), and the differential equation looks like X'(t) = A(t)X(t).
or in terms of coordinates,
x~(t)
= an,(t}x,(t) + ... + a..(t)xn(t).
It is clear that the solutions of our differential equation .l.'(t) = A(t).l.(t) form a vector space. One of the main facts which is always true in this linear case is that the integral curves are defined on the full interval J. This will be proved below, and we consider the slightly more general case when the equation depends on parameters.
Theorem 3.1. Let J be an open interval of R containing O. and let V be open in some space. Let A: J x V
--+
L
be a continuous map. and let L x E --+ E be a product. Let element of E. Then there exists a unique map
A: J x V --+ E.
Wo
be a fixed
554
[XIX, §3]
ORDINARY DIFFERENTIAL EQUATIONS
which, for each x
E
V, is a solution of the differential equation
}.'(t, x) = A(t, x)}.(t, x),
}'(O, x) =
Woo
This map }. is continuous. Proof Let us first fix x E V. Consider the differential equation }.'(t, x) = A(t, x)J.(t, x)
with initial condition }'(O, x) = Woo This is a differential equation on E, with time-dependent vector field f given by f(t, v) = A(t, x)v
for VEE. We want to prove that the integral curve is defined on all of l, and for this we shall use Theorem 2.4. Suppose that t ...... }.(t, x) is not defined on all of l. We look to the right, and let bo be the right end point of a maximal subinterval of 1 on which it is defined. If 1 has a right end point b then bo < b. (Of course, if 1 goes to infinity on the right, there is no b.) Now the map t ...... A(t, x) is bounded on every compact subinterval of J. In particular, we see that our vector field satisfies the Lipschitz condition of Theorem 2.4. Condition (i) is also satisfied, trivially, because our vector field is defined on the entire space E. This leaves condition (ii) to verify. We omit the index x for simplicity of notation, and on the interval ~ t < bo we have
°
}.(t) =
Wo
+ f~ A(u)}.(u) du
so that [}.(I)I
~ jWol + K
D}.(U)' du,
where K is a bound for the map t ...... A(t) on the compact interval [0, bol By Theorem 2.1, it follows that}. is bounded on the interval ~ t < bo, whence
°
f(t, }.(t))
= A(t)}.(t)
is bounded on this interval. Thus condition (ii) is satisfied, and our assumption that bo is not the right end point of J is contradicted. This proves that}. is defined on all of J.
[XIX, §3]
555
LINEAR DIFFERENTIAL EQUATIONS
We now consider A as a map with two variables, t and x, and shall prove its continuity, say at a point
Let c > 0 be so small that the interval I = [to - C. to + c] is contained in J. Let VI be an open ball centered at Xo and contained in V such that A is uniformly continuous and bounded on I x VI' (The existence of this ball is an immediate consequence of the compactness of I. cr. Lemma 8.1 of Chapter XVII, §8, where this is proved in detail.) For (t, x) E 1 X V, we have JA(t. x) - A{rO' xo)1 ~ IA(t, x) - A(t. xo)1
+ IA(t, xo)
- A(t o, xo)l.
The second term on the right is small when t is close to to because A is continuous, being differentiable. We investigate the first term on the right, and shall estimate it by viewing A(t, x) and A(t, xo) as approximate integral curves of the differential equation satisfied by A(t. xl. We find: IX(t. xo) - A(t, x)A(r, xo)1
~
IX{r, xo) - A(r, xo)A(t, xo)1
+ ~
IA(r, XO)A(t, x o) - A(r, x)A(r, xo)1
IA(r, xo) - A(t, x)IIA(t, xo)l·
By the uniform continuity of A and the fact that AU, xo) is bounded for t in the compact interval I, we conclude: Given i, there exists 0 such that if Ix-xol" • ,In • An open rectangle is a product as above, where the intervals J i are open. We shall usually deal with closed rectangles in what follows, and so do not use the adjective ..closed" unless we start dealing explicitly with other types of rectangles. If Pi is a partition of the closed interval J i , then we call (P 1>' •• ,Pn) = P a partition of the rectangle. In 2-space, a rectangle together with a partition looks like this: d
c b
565
566
[XX, §I]
MULTIPLE INTEGRALS
We view P as dividing the rectangle into subrectangles. Namely, if Si is a subinterval of the partition Pi' for each i = I, ... •n, then we call
S, x ...
X
Sn
a subrectangle of the partition P. Let R = J, x··· x I n be a rectangle, expressed as a product of intervals 1;. We define the volume of R to be v(R)
= I(J ,) .. ·IUn)
where lUi) is the length of 1;. If J i = [ai' b.J, then I(J i )
= bi
ai
-
so that v(R)
= (b ,
- a,)' .. (b n
-
an)'
We define the volume of an open rectangle similarly. The volume is equal to 0 if some interval J i consists of only one point. Let f be a bounded real valued function on a rectangle R. Let P be a partition of R. We can define the lower and upper Riemann sums by
LR(P,f) = UR( P,f)
I
=I
inf(f)v(S). s su p(f)v( S), s
where infif) is the greatest lower bound of all values f(x), for xeS. sUPif) is the least upper bound of all values f(x) for xeS, and the sum is taken over all subrectangles S of the partition P. If R is fixed throughout a discussion, we omit the subscript R and write simply L(P, f) and U(P, f). Let P' = (P', •... ,P~) be another partition of R. We shall say that P' is a refinement of P if each P; is a refinement of Pi (i = I, ... ,n). We recall that Pi being a refinement of Pi means that every number occurring in the sequence Pi also occurs in the sequence P;. If P, P' are two partitions of R, then it is clear that there exists a partition P" which is a refinement of both P and P'. This is achieved for intervals simply by inserting all points of a partition into the other. and then doing it for each interval occurring as a factor of the rectangle, in n-space. We have the usual lemma. Lemma 1.1. If P' is a refinement of P chen
L(P,f)
~
L(P',f)
~
U(P', f)
~
U(P, f).
[XX, §I]
567
ELEMENTARY MULTIPLE INTEGRATION
Proof The middle inequality is obvious. Consider the inequality relating L(P', f) and L(P,f). We can obtain P' from P by inserting a finite number of points in the partitions of the intervals occurring in P. By induction, we are thus reduced to the case when P' is obtained from P by inserting one point in some partition Pi for some i = I, ... ,n. For simplicity of notation, assume i = 1. The rectangles of P are of type
where Si is a subinterval of Pi' One of the intervals of Ph say T, is then split into two intervals T, T" by the insertion of a point in P. All the subrectangles of P' are the same as those of P, except when T occurs as a first factor. Then the rectangle S= T
X
S2
X •••
x S.
is replaced by two rectangles, namely S'
=T
X
S2
X •••
x S.
S" = T"
and
X
S2
X •••
x S•.
The term inf(f)v(S) S
in the lower sum L(P, f) is then replaced by the two terms inf(f)v(S') S'
+ inf(f)v(S"). S"
We have /(T) = /(T) + /(T"), and hence inf(f)v(S) = inf(f)/(T')/(S2)" ·/(S.) S
+ inf(f)/(T")/(S2)"
S
;:;; inf(f)v(S') S'
·/(S.)
S
+ inf(f)v(S"). S"
This proves that L(P, f) ;:;; L(P', f). The inequality concerning the upper sum is proved the same way. We define the lower integral LR(f) to be the least upper bound of all numbers LR(P, f), and the upper integral UR(f) to be the greatest lower bound of all numbers UR(P,f). We say that f is Riemann integrable (or simply integrable) if
568
MULTIPLE INTEGRALS
[XX, §1]
in which case we define its integral JR(f) to be equal to the lower or upper integral; it does not matter which. Example. Let f be the constant funclion 1. Let R
= [a" b,]
x ... x [an' bnl
Let P = (P" . .. ,Pn ) be a partition of R. Each Pi can be written in the form
where
The subrectangles of the partition are of the type
The lower sum is equal to the upper sum, and is equal to the repeated sum
"n
kl
i.. =O
it =0
L ... L (c'.h+'
- c'.h)",(cn.i.+, - cn.iJ
We evaluate the last sum first, and note that
,
L (cn.i.+, j,,=O
- cn.i ) = bn - an'
By induction, we find that JR(I) = (b , - a,) .. · (b N - aN)
= v(R).
From the definitions of the least upper bound and greatest lower bound, we obtain at once an (E, P)-characterization of the integrability of f, namely: f is integrable such that
011
R
if alld ollly if,
givell
E,
there exists a partitioll P of R
IU(P,f) - L(P,f)1 < (.
Furthermore, we also note that if the preceding inequality holds for P, then it holds for every partition P' which is a refinement of P.
[XX, §1]
ELEMENTARY MULTIPLE INTEGRATION
569
Theorem 1.2. The integrable functions on R form a vector space. The
integral satisfies the following properties: (NT I. The map f (NT 2. If f
~
1-+
.
IRf is linear. ~
0, then IRf
0.
Proof The first assertion follows from the fact that for each subrectangle S of a partition of R we have
+ inf(g)
inf(f) s
~
inf(f s
+ g)
~
suP(f s
+ g)
s
sup(f) s
+ sup(g),
+ g)
U(P, f) + U(P, g).
~
s
and hence for the partition P,
L(P, f) + L(P, g)
~
L(P, f + g)
~
U(P, f
~
Also for any number c ~ 0, inf(cf) = c inf(f). s s The linearity follows at once. As for (NT 2, if f inf(f) ~ s
so that L(P, f) ~
~
0 then
°
°
for all partitions P. Property (NT 2 follows at once.
From (NT ( and (NT 2 we have a strengthening of (NT 2, namely:
If f, 9 are integrable and f Indeed, we have 9 -
f >
~
g, then IR(f)
~
I R(g)·
0, so IR(g - f) ~ 0, and by linearity,
whenoe our assertion. We now want to integrate over more general sets than rectangles. A subset K of Rn will be said to be negligible if given £, there exists a finite number of rectangles R I> •.. ,R m which cover K (that is whose union contains K) and such that
570
[XX, §I]
MULTIPLE INTEGRALS
It is clear that in this definition, we may take the rectangles to be either open or closed. Furthermore, a negligible subset is clearly bounded. Its closure is also negligible, and is compact. A function f on a rectangle R will be said to be admissible if it is bounded and continuous except possibly on a negligible subset of R. It is trivial that a finite union of negligible sets is negligible. Hence a finite sum of admissible functions on R is admissible, and in fact, the set of admissible functions forms a vector space. It is also clear that the product of two admissible functions is admissible, and if f, 9 are admissible, Ihen so are max(j; g), min(f, g), and If I· We define the size of a partition P of R to be < b if the sides of all subrectangles of P have a length < b.
Theorem 1.3. Every admissible function on R is integrable. Given an admissible function, and £, there exists b such that if P is a partition of R of size < b, then U(P, f) - L(P, f)
""Sm are the subrectangles of P which intersect R I , ... ,R k then
Let 0 < min(o" O2 ), To compare the upper and lower sum of f with respect to this partition, we distinguish the subrectangles S according as S is one of SI' ... ,Sm or is not. We obtain:
U(P, f) - L(P, f) = =
L [sup(f) s
f
s
inf(f)]v(S)
s
[SUP(f) - inf(f)]v(Sj)
j=15j
5J
+ L [sup(f) S'*SJ
s
inf(f)]v(S) S
;;; 211 f112£ + £ L v(S) 5,*5,
;;; 211f112£
+ w(R).
This proves that f is integrable. . Furthermore, suppose we change the values of f on D to those of another function g. The lower sums L(P, f) and L(P, g) then differ only in those terms m
m
L inf(f)v(Sj) j= 1 5J
and
L inf(g)v(S) j= 1 5J
572
[XX, §l]
MULTIPLE INTEGRALS
which are estimated by IIf1l2£ and IIg1l2£ respectively. Thus for £ smal~ the lower sums are close together. Since these lower sums are also close to the respective integrals, it follows that I R(f) = I R(g). This proves the theorem. A subset A of Rn will be said to be admissible if it is bounded, and if its boundary is a negligible set. We denote the boundary of a set A by oA. The verification of the following properties is left to the reader as an exercise: o(A u B) c (oA
U
oB),
o(A n B) c (oA
o(A - B) c (oA
where we denote by A - B the set of all x
oB),
oB)
U
E
U
A such that x
rt B.
Hence:
Lemma 1.5. A finite union of admissible sets is admissible, a finite intersection of admissible sets is admissible, and
if A,
B are admissible. so is
A-B.
Let A be a subset of RP and B a subset of Rq. Then A x B is a subset of and
Rp+q,
o(A x B) = (oA x B)
U
(A x oB).
This is immediately verified. By induction, we find that o(A I
X •••
x An)
= union of AI
x ... x oA, x ... x An
the union taken for all i = 1, ... ,n if AI' ... ,A n are subsets of euclidean spaces. We can apply this to the case of a rectangle R = [at>
btl
x ... x [an'
bJ
and find that its boundary is the union of sets [at> b l ] x ... x {a,} x ... x [an, bn]
and
The boundary of a rectangle obviously is negligible. For instance, we can cover a set [at> b l ] x ... x {c} x ... x [an, bn]
[XX, §1]
ELEMENTARY MULTIPLE INTEGRATION
573
by one rectangle
where J is an interval of length ( containing c, so that the volume of this rectangle is arbitrarily small. It is also nothing but an exercise to show that if A, B are admissible, then A x B is admissible. A function f on R· is said to be admissible if it is admissible on every rectangle. Let f be admissible, and equal to 0 outside the rectangle S. Let R be a rectangle containing S. We contend that IR(f) = Is(f). To prove this, write R = [a" b l ] x ... x [a., b.],
We view (ai> Ci> d h b i) as forming a partition Pi' and let P = (P" ... ,p.) be the corresponding partition of R. Then S appears as one of the subrectangles of the partition P of R. Let 9 be equal to f except on the boundary of S, where we define 9 to be equal to O. If P' is any partition of R which is a refinement of P, and S' is a subrectangle of P', then either S' is a subrectangle of S, or S' does not intersect S, or S' has only boundary points in common with S. Hence for each P' we find that
where P's is the partition of S induced by P' in the natural way. From this it follows at once that IR(g) = ISO.
1Af is linear.
574
MULTIPLE INTEGRALS
[XX, §I]
INT 3. We have IAf = IAfAINT 4. For every rectangle S we have IS s" then welelt z =Sz - s, and we see that y lies in B n B'. Finally, consider the set B' - B consisting of all x E B' such that x f/ B. It is the set of all x written as above, with t, + t z > I. Let Sz = t, + t z - I. An element x E B' - B can then be written
with 0 ~ t, ~ 1 and 0 < Sz ~ (1' (The condition Sz ~ (, comes from the fact that Sz + (1 - t z ) = t,.) Conversely, any element x written with I, and Sz satisfying 0 < Sz ~ (, lies in B' - B, as one sees immediately. Hence, except for boundary points, we conclude that 8' - B = (B - B')
+ vz.
Consequently, 8' - Band B - B' have the same volume. Then Vol B
= Vol(B -
8')
+ Vol(B n
B')
= Vol(B' -
B)
+ Vol(B n
B')
= Vol B'.
This proves the lemma. From the lemma, we conclude that for any number c, VoIO(v,
+ cVz, Vz, ... ,Vn) =
Volo(v" Vz, .. · ,Vn)'
Indeed, if c = 0 this is obvious, and if c "# 0 then c VolO(v,
+ CVz, VZ, ... ,Vn) = =
VOIO(VI
+ CV z , CVz,'"
Vo lOt VI ~
CV 1 , ... 'Vn) -
,Vn) C
V0 IO( VI'
\ Vl, ... ,Vn)'
We can then cancel c to get our conclusion. To prove the linearity of VOiD with respect to its first variable, we may assume that vz,'" ,Vn are linearly independent, otherwise both sides of
[XX. §4)
CHANGE OF VARIA3LES
589
(00) are equal to O. Let v, be so chosen that {VI>'" .vn } is a basis of Rn. Then by induction. and what has been proved above.
VolO(c,v,
+ ... + CnVn • V2.···. Vn)
= Volo(c,v,
+ ... + Cn-'Vn-I> V2.···. Vn)
= VoIO(c,VI> V2 •...• vn ) = c, Volo(v, •... •vn ).
From this the linearity follows at once. and the theorem is proved. Corollary 4.3. Let S be the ullit cube spalllled by the ullit vectors ill Rn. Let A: Rn -+ Rn be a lillear map. Theil VolA(S) = I Det(A)I·
Proof If VI .... •Vn are the images of el>'" .en under A. then A(S) is the block spanned by VI>' •• •vn • If we represent A by the matrix A = (a;j). then Vi
and hence Det(vi •... •vtl )
=
ail el
+ ... + aj,.e"
= Det(A) = Del(A).
This proves the corollary.
Corollary 4.4. If R is allY rectallgle ill Rn alld A: Rn -+ Rn is a lillear map.
thell VolA(R)
= I Det(A) IVol(R).
Proof. After a translation. we can assume that the rectangle is a block. If R = A,(S) where S is the unit cube. then
whence by Corollary 4.3. VolA(R)
= IDet(A
0
A,)I
= IDet(A) Det(A,)1 = IDet(A)1 Vol(R).
The next theorem extends Corollary 4.4 to the more general case where the linear map A is replaced by an arbitrary C'-invertible map. The proof then consists of replacing the linear map by its derivative and estimating the error thus introduced. For this purpose. we define the Jacobian determinant f:,.Jlx)
= DetJf(x) = Det j'(x)
where J fix) is the Jacobian matrix. and j'(x) is the derivative of the map f: U -+ RO.
590
[XX, §4]
MUL TJPLE INTEGRALS
Theorem 4.5. Lee R be a recrallgle ill R", canrailled ill some opell sec U. Lee (: U -> R" be a C' map, which is C'-illverrible Oil U. Then Vol/(R) =
Dllr
l.
Proof. When I is linear, this is nothing but Corollary 4.4 of the preceding theorem. We shall prove the general case by approximating/ by its derivative. Let us first assume that R is a cube for simplicity. Let P be a parI ilion of R, obtained by dividing each side of R into N equal segments for large N. Then R is partitioned into N" subcubes which we denote by SJ (j = I, ... ,N"). We let OJ be the center of Sj' We have Vol feR) =
L Vol f(S) j
because lhe images J(Sj) have only negligible sets in common. We investigate f(S) for eachj. LeI C be a bound for IF(x)' 'I, x E R. Such a bound exists because x ...... IF(x)' 'I is cominuous on R which is compact. Given £, we take N so large thaI for x, Z E Sj we have If'(z) - F(x)l
LeI Aj = F(o) wherc
OJ
< £Ie.
is the ccmer of thc cube Sj' Then
IA~ I . 0 f'(z) A~' 0 r(x)1 J J.
R" be an invertible linear map. and let B be a ball centered at the origin in R". Show that
r
JB
e-(Ty.Ty)
dy
=
f
e-(x.x)
dxldet T-11.
T(B)
(The symbol (,) denotes the ordinary dot product in R".) Taking the limit as the ball's radius goes to infinity, one gets
598
[XX, §4]
MULTIPLE INTEGRALS
3. Let B.(r) be the closed ball of radius r in R·. centered at the origin. with respeel to the euclidean norm. Find its volume v,,(r). [Hint: First note that v,,(r)
We may assume n
~
= r"v,,(I).
2. The ball B.(I) consists of all (x" . .. ,x.) such that
Put (x" X2) = (x. y) and let g be the characteristic function of B.(I). Then
v,,(l)
=
f,f,U•.
_,g(x.Y.x3 •... ,x.)dX3 ... dX.] dxdy
where R._ 2 is a rectangle of radius I centered at the origin in (n - 2)-space. If x 2 + y2 > I then g(x. Y. X3 •. .. ,x.) = O. Let D be the disc of radius I in R2 • If x 2 + T ;;; 1. then g(x. Y. X3.·.· ,x.) viewed as function of (X3 •... ,x.) is the characteristic function of the ball
Hence the inner integral is equal to
J.
g(x.y.x3.···,x.) dx3··· dx•
= (1
- x2
-
y2)(.-2)/2v,,_2(1)
R"-l.
so that
Using polar coordinates, the last integral is easily evaluated. and we find: 11·
V2 (I) = n
n!
and
1·3·5···(2n -I)
Suppose that r is a function such that rex r(/2) = Show that
-fi.
v,,(I)
ro + n12)"
+ I) =
xr(x). r(1) = I. and
[XX, §4]
CHANGE OF VARIABLES
4. Determine the volume of the region in
a· defined by the inequality
Ix.! + ... + Ix.1 5. Determine the volume of the region in
~ r.
a'· = a' x
Iz.! + ... + Iz.1 where
599
~
...
X
a' defined by
r,
z, = (x" y,) and Iz,1 = Jxf + yf is the euclidean norm in
6. (Spherical coordinates) (a) Define f:
a'.
a 3 -+ a 3 by
Xl
= r COS 8•.
X2
= r sin 8. cos 82 ,
X3
= r sin 8 1 sin 8 2 .
Show that
Show that f is invertible on the open set
0
I, and 11 + I, ;::> 1. (This is just a triangle.) Find the area of A by integration. (b) Let v" v, be linearly independent vectors in R'. Find the area of the set of points 11V1 + I,V, with 0;::> I, and I, + I, ~ I, in terms of Oot(v" v,). 9. Let VI> • •• ,v" be linearly independent vectors in R". Find the volume of the solid consisting of all points
with 0 ;::> I, and I,
+ ... + I" ;::>
1.
10. Let B. be the closed ball of radius a > 0, centered at the origin. In n-space, let X = (x I ' ... • x.) and let r = IXI. where I I is the euclidean norm. Take
0< a < I, and let A. be the annulus consisting of all points X with a ;::> I X I ;::> 1. Both in the case n = 2 and n = 3 (i.e. in the plane and in 3-space), compute the integral
I
•
=
fA.IXI
_I_dx, ... dx.
"
Show that this integral has a limit as a .... o. Thus, contrary to what happens in I-space, the function f(X) = III X I can be integrated in a neighborhood of O. [Hint: Use polar or spherical coordinates. Actually, using n-dimensional spherical coordinates, the result also holds in II-space.] Show further that in 3-space, the function III XI' can be similarly integrated near O. II. Let B be the region in the first quadrant of R' bounded by the curves xy = I, xy = 3, x' - y' = I. and x' - y' = 4. Find the value of the integral
ff(X' + y') dx dy B
by making the substitution u = x' ing the change of variables formula.
r
and v = xy. Explain how you are apply-
602
[XX, §5]
MULTIPLE INTEGRALS
12. Prove that
where A denotes the half plane x
~
a > O. [Hint: Use the transformation
and
y
= vx.]
13. Find the integral
fff
xyz dx dy dz
taken over the ellipsoid
14. Let [ be in the Schwartz space on R·. Define a normalization of the Fourier
transform by r
(y)
=
f
[(x),-""" dx.
R"
Prove that the function h(x) = ,-••> is self dual, that is h" = h. 15. Let B he an /I x n non-singular real matrix. Define ([ 0 B)(x) = [(Bx). Prove that the dual of [ 0 B is given by (fo B)"{y) = _1_r('B-1y),
IIBII where IIBII is the absolute value of the determinant of B. 16. For bE R' define jj,(x) = [(x - b). Prove that (jj,)" (y)
= ,-,...."!" (y).
XX, §5. VECTOR FIELDS ON SPHERES Let S be the ordinary sphere of radius I, centered at the origin. By a tangent vector field on the sphere, we mean an association
F:S->R' which to each point X of the sphere associates a vector F(X) which is tan---+ gent to the sphere (and hence perpendicular to OX). The picture may be drawn as follows:
[XX, §5]
VECTOR FIELDS ON SPHERES
603
,
,
___ ... _ -
,, , __ ---,I-
,o
,
_
x + F(X)
For simplicity of expression, we omit the word tangent, and from now on speak only of vector fields on the sphere. We may think of the sphere as the earth, and we think of each vector as representing the wind at the given point. The wind points in the direction of the vector, and the speed of the wind is the length of the arrow at the point. We suppose as usual that the vector field is smooth. For instance, the vector field being continuous would mean that if P, Q are two points close by on the sphere, then F(P) and F( Q) are arrows whose lengths are close, and whose directions are also close. As F is represented by coordinates, this means that each coordinate function is continuous, and in fact we assume that the coordinate functions are of class ct. Theorem 5.1. Given any vector field on the sphere, there exists a point P on the sphere such that F(P) = O. In terms of the interpretation with the wind, this means that there is some point on earth where the wind is not blowing at all. To prove Theorem 5.1, suppose to the contrary that there is a vector field such that F(X) '" 0 for all X on the sphere. Define
F(X) E(X) = IF(X) I' that is let E(X) be F(X) divided by its (euclidean) norm. Then E(X) is a unit vector for each X. Thus from the vector field F we have obtained a vector field E such that all the vectors have norm 1. Such a vector field is called a unit vector field. Hence to prove Theorem 5.1, it suffices to prove:
Theorem 5.2. There is no unit vector field on the sphere. The proof which follows is due to Milnor (Math. Monthly, October,
1978).
604
[XX, §5]
MULTIPLE INTEGRALS
Suppose there exists a vector field E on the sphere such that
IE(X)I
= I
for all X. For each small real number t, define G,(X) = X
+ tE(X).
Geometrically, this means that G,(X) is the point obtained by starting at X, going in the direction of E(X) with magnitude t. The distance of X + tE(X) from the origin 0 is then obviously
JI + t
2
•
Indeed, E(X) is parallel (tangent) to the sphere, and so perpendicular to X itself. Thus
IX
+ tE(X)!2 = (X + tE(X»2 = X 2 +
t 2E(X)2
= I + t 2,
since both X and E(X) are unit vectors.
Lemma 5.3. For all t sufficiently small, the image G,(S) of the sphere IInder G, is equal to the whole sphere of radius + t2•
Jl
Proof This amounts to proving a variation of the inverse mapping theorem, and the proof will be left as an exercise to the reader.
We now extend the vector field E to a bigger region of 3-space, namely the region A between two concentric spheres, defined by the inequalities
a ::: IXI
;;a b.
This extended vector field is defined by the formula E(rU) = rE(U)
for any unit vector U and any number r such that a
;;a r ;;a b.
[XX, §5]
VECTOR FIELDS ON SPHERES
605
II follows that the formula G,(X) = X
+ eE(X)
also given in terms of unit vectors U by G,(rU)
=
rU
+ eE(rU) = rG,(U)
defines a mapping which sends the sphere of radius r onto the sphere of radius r 1 + eZ by the lemma, provided that e is sufficiently small. Hence it maps A onto the region between the spheres of radius
J
aJI
+ eZ
and
bJI
+ eZ•
By the change of volumes under dilations, it is then clear that Volume G,(A) = (JI
+ eZ )3
Volume(A).
JI
+ eZ still involves a square root, and is Observe that taking the cube of not a polynomial in e. On the other hand, the Jacobian matrix of G, is
as you can verify easily by writing down the coordinates of E(X), say E(X) = (gl(X, y, z), gz(x, y, z), g3(X, y, z»).
Hence the Jacobian determinant has the form
where fo, ... /3 are functions. Given the region A, this determinant is then positive for all sufficiently small values of e, by continuity, and the fact that the determinant is 1 when e = o. For any regions A in 3-space, the change of variables formula shows that the volume of G,(A) is given by the integral Vol G,(A) =
Iff
6 G,(x, y, z) dy dx dz.
A
If we perform the integration, we see that
606
MULTIPLE INTEGRALS
[XX, §5]
where Ci
=
Iff
fi(x, y, z) dy dx dz.
A
Hence Vol G,(A) is a polynomial in t of degree 3. Taking for A the region between the spheres yields a contradiction which concludes the proof.
XX, §S. EXERCISE Prove the statements depending on inverse mapping theorems which have been left as exercises in the above proof.
CHAPTER
XXI
Differential Forms
XXI, §1. DEFINITIONS We recall first two simple results from linear (or rather multilinear) algebra. We use the notation EI') = E x Ex··· x E, r times.
Theorem A. Let E be a finite dimensional vector space over the reals. For each positive integer r there exists a vector space f\ E and a multilinear alternating map
denoted by (u" ...•u,) ...... U, 1\ •.. 1\ u" having the following property: If {v" ... ,vn } is a basis of E, then the elements { V i 1 1\ • •• 1\
form a basis of
v·I.. }'
i1 < i2 < ... < i,.,
I\' E.
We recall that alternating means that some i #- j.
11, 1\ .•• 1\ 11,
=0
if
1/,
= lIj
for
Theorem B. For each pair of positive integers (r. s), there exists a lInique product (bilinear map)
607
608
[XXI, §I]
DIFFERENTIAL FORMS
such that
if u., ... ,Un W h
(UI A .•. A
u,)
.. . , Ws E
E then
X (WI A ... A W s )-- UI A .•. A U, A WI A ••• A W S '
This product is associative.
The proofs for these two statements will be briefly summarized in an E the r-thalternating product (or exterior product) appendix. We call E = R. Elements of E which can be written of E. If r = 0, we define in the form U I A .•. A U, are called decomposable. Such elements generate
!'I
!'I
If
!'I E.
Now let E* denote the space of linear maps from Rn into R. We call E* the dual space of Rn. It is the space which we denoted by L(Rn, R). If AI> ... ,An are coordinate functions on Rn, that is
then each Ai is an element of the dual space, and in fact {AI>'" ,An} is a basis of this dual space. Let U be an open set in Rn. By a differential form of degree r on U (or an r-form) we mean a map (I);
U
--+
!'I E*
from U into the roth alternating product of E*. We say that the form is of class CP if the map is of class CPo For convenience in the rest of the book, we assume that we deal only with forms of class CO:>, although we shall sometimes make comments about the possibility of generalizing to forms of lower order of differentiability. Since {AI> ... ,An} is a basis of E*, we can express each differential form in terms of its coordinate functions with respect to the basis
{A.II A namely for each x
E
••• A
A·I,}
(il < ... < i,),
U we have
W(X) =
L};, "'i.(x)Ai.
A ... A
Ai.
(i)
where fii) = /;.... i. is a function on U. Each such function has the same order of differentiability as w. We caIl the preceding expression the standard form of w. We say that a form is decomposable if it can be written as just one term
[XXI, §1]
609
DEFINITIONS
Every differenlial form is a sum of decomposable ones. We agree to the convention that functions are differential forms of degree O. It is clear that the differential forms of given degree form a vector space, denoted by O'(U). As with the forms of degree ~ 1, we assume from now on that all maps and all functions mentioned in the rest of this book are C"', unless otherwise specified. Let f be a function on U. For each x E U the derivative
f'(X): Rn --+ R is a linear map, and thus an element of the dual space. Thus
1': U --+ E* is a differential form of degree 1, which is usually denoted by df [Note. If f was of class C', then df would be only of class C'- I. Having assumed f to be C"', we see that df is also of class C"'.] Let Ai be the i-th coordinate function. Then we know that
dA;(X) = A,(X) = Ai for each x E U because A'(X) = A for any continuous linear map A. Whenever {XI"" ,xn} are used systematically for the coordinates of a point in Rn, it is customary in the literature to use the notation
dA;(X) = dx;. This is slightly incorrect, but is useful in formal computations. We shall also use it in this book on occasions. Similarly, we also write (incorrectly)
w=
L fm dXi,
1\ •.• 1\
dXi,
(i)
instead of the correct
w(X) =
L j(,)(X)A"
1\ •.. 1\
A",
(ij
In terms of coordinates, the map df (or 1') is given by
df(x)
= f'(x) = Dd(x)A I + ... + Dnf(X)An
where D;j(x) = Of/OXi is the i-th partial derivative. This is simply a restatement of the fact that if H = (h" ... ,hn ) is a vector, then
j'(x)H
= -of
ox
I
hi
of + ... + 0hn Xn
610
[XXI, §I]
DIFFERENTIAL FORMS
which was discussed long ago. Thus in older notation, we have of
of
= :;dx, + ... +:;- dxn · uXl UX
df(x)
n
Let ro and 1/1 be forms of degrees rand s respectively, on the open set U. For each x E U we can then take the alternating product ro(x) 1\ I/I(x) and we define the alternating product ro 1\ 1/1 by (ro
I/I)(x) = ro(x)
1\
1\
I/I(x).
If f is a differential form of degree 0, that is a function, then we define
f
1\
ro = fro
where (fro)(x) = f(x)ro(x). By definition, we then have co
1\
N
= fro
1/1.
1\
We shall now define the exterior derivative dro for any differential form roo We have already done it for functions. We shall do it in general first in terms of coordinates, and then show that there is a characterization independent of these coordinates. If ro = '" '(') d)'·II ~ Jh
1\'"
1\
d)'·I"
(i)
we define dro
= '" L. d'(,) !Ill
1\
d)"IJ
1\'"
1\
d)'·I,.'
(i)
Example. Suppose n = 2 and ro is a I-form, given in terms of the two coordinates (x, y) by ro(x, y) = f(x, y) dx + g(x, y) dy.
Then dro(x, y) = df(x, y) of
= ( ox dx
1\
= (Of
oy
+ dg(x, y)
f) +o oy dy
of
= oy dy
dx
1\
dx
1\
og
+ ox dx
_ Og) dy ox
dx
1\
dx
1\
1\
dy
Og) + (Og ox dx + oy dy dy
1\
dy
[XXI, §l]
611
DEFINITIONS
because the terms involving dx numerical example, take
dx and dy
1\
w(x, y) = y dx
1\
dy are equal to O. As a
+ (x 2 y) dy.
Then dw(x, y) = dy
dx
1\
+ (2xy) dx
= (I - 2xy) dy
1\
1\
dy
dx.
Theorem 1.1. The map d is linear, and satisfies dew
t/J) =
1\
dw
1\
t/J + (-l)'w
1\
dt/J
if r = deg w. The map d is uniquely determined by these properties, and by the fact that for a function J, we have df = f'. Proof The linearity of d is obvious. Hence it suffices to prove the formula for decomposable forms. We note that for any function f we have d(fw) = df
1\ W
+ f dw.
Indeed, if w is a function g, then from the derivative of a product we get d(fg) = f dg + g df If w
= g dA.-
'I
1\'"
1\
dA'I~
where g is a function, then d(fw)
= d(fg dA;,
1\ ... 1\
+ g df) 1\ =fdw + df 1\ W, = (f dg
dAiJ
dAi'
= d(fg)
1\ ... 1\
dAi'
1\ ..• 1\
dA;,
=
g dkJI
1\ ... 1\
dkJ~
=
gl{J
1\
dAi,
as desired. Now suppose that w =fdA'II
1\'"
1\
dA'l ..
and
,I,
~
=fw
with i, < ... < i, and j, < ... < j. as usual. If some i, = j., then from the definitions we see that the expressions on both sides of the equality in the theorem are equal to O. Hence we may assume that the sets of indices i" ... ,i, and j, , ... J. have no element in common. Then
dew
1\
I{J) = 0
612
[XXI, §1]
DIFFERENTIAL FORMS
by definition, and
dew
1\
if!) = d(fgw
1\
l{J) = d(fg)
W 1\ l{J
1\
= (g df + f dg)
1\
W 1\ l{J
=dw
1\
if! +fdg
W
1\
l{J
= dw
1\
if! + (-I)'iw
1\
dg
= dw
1\
if!
1\
+ (-IYw
1\
1\
lj,
dif!,
thus proving the desired formula, in the present case. (We used the fact that dg 1\ w = (-IYw 1\ dg whose proof is left to the reader.) The formula in the general case follows because any differential form can be expressed as a sum of forms of the type just considered, and one can then use the bilinearity of the product. Finally, d is uniquely determined by the formula, and its effect on functions, because any differential form is a sum of forms of type f dA" 1\ ••• 1\ dA" and the formula gives an expression of d in terms of its effect on forms of lower degree. By induction, if the value of d on functions is known, its value can then be determined on forms of degree ~ I. This proves the theorem.
XXI, §1. EXERCISES I. Show that ddf = 0 for any function f, and also for a I-form.
2. Show that ddw
= 0 for any differential form w.
3. In 3-space, express dw in standard form for each one of the following w: (a) w
= x dx + y dz
(b) w
(c) w = (sin x) dy + dz
(d) w
= xy dy + x dz = e'dx + y dy + e" dz
4. Find the standard expression for dw in the following cases: (a) w = x 2y dy - xy2 dx (b) w = ex'dx (c) w = f(x, y) dx where f is a function.
1\
dz
5. (a) Express dw in standard fonn if w
= x dy 1\ dz + Y dz
1\
dx
+ z dx
1\
dy.
1\
dx
+ h dx
1\
dy.
(b) Let f, g, h be functions, and let w
=f dy 1\ dz + g dz
Find the standard fonn for dw. 6. In n-space, find an (n - I)-form w such that dw = dx,
1\ ••• 1\
dx•.
[XXI, §2]
STOKES' THEOREM FOR A RECTANGLE
613
7. Let w be a fonn of odd degree on ·U, and let f be a function such that f(x) "" 0 for all x E U, and such that d(fw) = O. Show that w A dw = O. 8. A form w on U is said to be exact if there exists a form I/J such that w = dI/J. If WI' (1)2 are exact, show that WI A (t)2 is exact. 9. Show that the form 1 r
w(x, y, z) = -. (x dy
A
dz
+ y dz
is closed but not exact. As usual, r' = x' the complement of the origin in R3 •
A
dx
+ y' + z'
+ z dx
A
dy)
and the fonn is defined on
XXI, §2. STOKES' THEOREM FOR A RECTANGLE Let (J) be an n-forrn on an open set U in n-space. Let R be a rectangle in U. We can write (J) in the form w(x)
= f(x) dX 1
A ••• A
dx n ,
wherefis a function on U. We then define
where the integral of the function is the ordinary integral, defined in the previous chapter. Stokes' theorem will relate the integral of an (n - I)-form l/J over the boundary of a rectangle, with the integral of dl/J over the rectangle itself. However, we need the notion of oriented boundary, as the following example will make clear. Example. We consider the case n = 2. Then R is a genuine rectangle, R
= [a, b]
x [c, d].
I~
d
II
R c
a
b
614
[XXI, §2]
DIFFERENTIAL FORMS
The boundary consists of the four line segments
[a, b] x {c},
{b} x [c, d],
{a} x [c, d],
[a, b] x {d}
which we denote by I~,
respectively. These line segments are to be viewed as having the beginning point and end point determined by the position of the arrows in the preceding diagram. Let
w =fdx
+ gdy
be a I-form on an open set containing the above rectangle. Then
dw
= df 1\ dx + dg
1\
dy
and by definition,
dw
= of oy dy
1\
dx
og + ox dx
1\
dy
= (Og ox -
of) oy dx
1\
dy.
Then by definition and repeated integration,
f dw = [ JR
biJ x ... x [an, bn]. We parametrize the i-th pieces of the boundary of R by the maps
a}: R , -+ V defined by
Observe that omitting the variable Xi on the left leaves II - I variables, but that we number them in a way designed to preserve the relationship with the original n variables. We define the boundary of u to be the chain n
au = I
j~
(-I)i(u? - (1). ,
Example. We consider the case II = 2. Then R is a genuine rectangle, R = [a, b] x [c, d].
.~
d
.? c
.:
R
.g a
b
622
[XXI, §4]
DIFFERENTIAL FORMS
We then find: u?(y) = (a, y),
u:(y)
= (b, y),
= (x, c),
uHx)
= (x, d).
ug(x) Then U
o+
=
-U 1
+ U 11 -
0 U2
1 U2
is the oriented boundary, and corresponds to going around the square counterclockwise. In general, consider the identity mapping I: R
->
R
on the rectangle. Let 1/1 be an n-form. We may view 0°R as 01, so that
r
JiJOR
1/1
=
r 1/1 = ±(-1);[1 1/1 -
Jal
;= I
If
11/1]. II
If u: R -> V is a simplex, and w is an (n - I)-form on V, then
r w = JfJor u*(w)
Jao
R
as one sees at once by considering the composite map up
= u IP. 0
Stokes' theorem for simplices. LeI V be open in Rm and let w be an (n - I){orm on V. LeI u: R -> V be an n-simplex in V. Then
Proof Since du*w = u* dw, it will suffioe to prove that for any (n - I)-form 1/1 on R we have
1
dl/l =
R
r
JiJOR
1/1.
[XXI, §4]
STOKES' FORMULA FOR SIMPLICES
623
This is nothing else but Stokes' theorem for rectangles, so Stokes' theorem for simplices is simply a combination of Stokes' theorem for rectangles together with the formalism of inverse images of forms. In practice one parametrizes certain subsets of euclidean space by simplices, and one can then integrate differential forms over such subsets. This leads into the study of manifolds, which is treated in my Real and Functional Analysis. In the exercises, we indicate some simple situations where a much more elementary approach can be taken.
XXI, §4. EXERCISES I. Instead of using reclangles, one can use triangles in Slokes' theorem. Develop this parallel theory as follows. Let Vo, ... ,v, be elements of R" such that Vi - Vo (i = I, ... ,k) are linearly independent. We define the triangle spanned by vo, ... ,i', to consist of all poinls
with realI, such that 0 ;;! Ii and 10 + ... + I, = I. We denote Ihis triangle by T, or T(vo ....•v,). (a) Let w, Vi - Vo for i I•... ,k. Let S be the set of points
=
=
with 5i ~ 0 and 5, + ... + 5, ;;! I. Show thai T(vo,··· ,v,) is the translation of S by vo. Define the oriented boundary of the triangle T to be the chain
,
fJ"T
= L (-I)JT(vo, ... $], ... •v,). j=O
(b) Assume that k = n, and thai T is contained in an open set U of R". Let w be an (n - I)-form on U. In analogy to Stokes' theorem for rectangles, show thai
fdw=i W. JT ~T The subsequent exercises do not depend on anything fancy, and occur in R 2 • Essentially you don't need to know anything from this chapter. 2. Let A be the region of R 2 bounded by the inequalities
and
g,(x) ;;! y ;;! g2(X)
624
DIFFERENTIAL FORMS
[XXI, §4]
where 9" 92 are continuous functions on [a, b]. Let C be the path consisting of the boundary of this region, oriented counterclockwise, as on the following picture:
b
a
Show that if P is a continuous function of two variables on A, then
Prove a similar statement for regions defined by similar inequalities but with reo spect to y. This yields Green's theorem in special cases. The general case of Green's theorem is that if A is the interior of a closed piecewise C' path C oriented counterclockwise and m is a (·form then
f.m = II dm A
In the subsequent exercises, you may assume Green's theorem. 3. Assume that the function J satisfies Laplace's equation,
on a region A which is the interior of a curve C, oriented counterclockwise. Show
that
fcoy
oj dx - oj dy = O.
ox
4. If F = (Q, P) is a vector field, we recall that its divergence is defined to be div F = oQlox + oPloY. If C is a curve, we say that C is parametrized by arc length if 1IC'(5)1I = I (we then use 5 as the parameter). Let C(5)
= 19,(5),92(5»)
be parametrized by arc length. Define the unit normal vector at 5 to be the vector N(5) = 19,(5), -9',(5)).
[XXI, §4]
STOKES' FORMULA FOR SIMPLICES
625
Verify that this is a unit vector. Show that if F is a vector field on a region A, which is the interior of the closed curve C, oriented counterclockwise, and parametrized by arc length, then
If
(div F)dydx
= LF.N ds.
•
If C is not parametrized by arc length, we define the unit normal vector by N(r) n(r) = IN(r)l' where I N(r)1 is the euclidean norm. For any function J we define the nonnal derivalive (the directional derivative in the normal direction) to be
D.J = (grad J). n. so for any value of the parameter r, we have
(D.J)(r) = grad J(C(r»· n(r). 5. Prove Green's formulas for a region A bounded by a simple closed curve C. always assuming Green's theorem.
+ g6J] dx dy = fcgD.J ds. H.(g6J - J6g) dx dy = fc 0, let x
= r cos 8 and y = r sin 8.
'fJ(r)
=
I -2 1Cr
(dX)2 (d)2 dr + It dr. Let 'fJ be the function of r defined by
f J = -2I f2' J(r cos 8, r sin 8)r d8 c,.
tty
0
where C. is the circle of radius r, parametrized as above. Assume that J satisfies Laplace's equation
626
[XXI, §4]
DIFFERENTIAL FORMS
Show that 'P(r) does not depend on r and in fact f(O, 0)
= _I
if
21tr c,
[Hint: First take 'P'(r) and differentiate under the integral, with respect to r. Let D, be the disc of radius r which is the interior of C,. Using Exercise 4, you will find that 'P'(r)
I = 2xr
If
div grad f(x, y) dy dx
Dr
Taking the limit as r
I = 2xr
If
f (02f 02 ) dy dx ox 2 + oy2
Dr
-+
0, prove the desired assertion.]
= O.
Appendix
We shall give brief reviews of the proofs of the algebraic theorems which have been quoted in this chapter. We first discuss" formal linear combinations." Let S be a set. We wish to define what we mean by expressions
where {Ci} are numbers, and is;} are distinct elements of S. What do we wish such a "sum" to be like? Well, we wish it to be entirely determined by the "coefficients" Ci' and each "coefficient" Ci should be associated with the element Si of the set S. But an association is nothing but a function. This suggests to us how to define" sums" as above. For each S E S and each number C we define the symbol CS
to be the function which associates C to sand 0 to z for any element Z # s. If b, C are numbers, then clearly b(cs) = (bc)s
and
(b
ZE
S,
+ c)s = bs + cs.
We let T be the set of all functions defined on S which can be written in the form
where Ci are numbers, and Si are distinct elements of S. Note that we have no problem now about addition, since we know how to add functions.
629
630
APPENDIX
We contend that if 5" ... ,sn are distinct elements of S, then
are linearly independent. To prove this. suppose such that
Ct, .. "Cn
are numbers
(the zero function). Then by definition, the left-hand side takes on the value C; at Si and hence C; = O. This proves the desired linear independence. In practice, it is convenient to abbreviate the notation, and to write simply 5; instead of Is;. The elements of T, which are called formal linear combinations of elements of S, can be expressed in the form
and any given element has a unique such expression, because of the linear independence of 5" .•••sn' This justifies our terminology. We now come to the statements concerning multilinear alternating products. Let E, F be vector spaces over R. As before, let
E(')=Ex···xE• taken r times. Let
f: E(') -> F be an r-multilinear alternating map. Let V" •.. ,Vn be linearly independent elements of E. Let A = (a;i) be an r x n matrix and let
Then
f(u" ... ,u,) = f(a"vt
+ ... + a'nVn' ... •a"v, + ... + a,nvn)
=
L !(a1.d(l)va
=
L al.aO) ... o,.tI{r)!(V
• •
(l)t . . .
,a r •
va(r})
{I, ... ,n}.
In. this sum, all terms will be 0 whenever u is not an injective mapping, that IS whenever there is some pair i. j with i '# j such that u(i) = u0),
631
APPENDIX
because of the alternating property of f From now on, we consider only injective maps cr. Then (cr(l),.,. ,cr(r)} is simply a permutation of some r-tuple (il' .. ' ,i,) with i, < .. , < i,. We wish to rewrite this sum in terms of a determinant. For each subset S of (I, ... ,n} consisting of precisely r elements, we can take the r x r submatrix of A consisting of those elements a;j such that j E S. We denote by DetiA)
the determinant of this submatrix. We also call it the subdeterminant of A corresponding to the set S. We denote by P(S) the set of maps cr: (I, ... ,r}
->
{I, ... ,n}
whose image is precisely the set S. Then
L
Dets(A) =
DE
£s(cr)a.,DIIl·'· a,.Dl'I'
P{S)
and in terms of this notation, we can write our expression for f(u t> in the form (1)
f(ut> ... ,u,)
.••
,u,)
= L Dets(A)f(vs) s
where Vs denotes (Vi" ... ,Vi) if i, < ... < i, are the elements of the set S. The first sum over S is taken over all subsets of of I, ... ,n having precisely r elements. Theorem A. Let E be a vector space over R, of dimension n. Let r be an integer 1 ::: r ::: n. There exists a finite dimensional space E and an r-multilinear alternating map E(') -> E denoted by
!'I
!'I
(u" ... ,U,)I-+UI
1\ ... 1\ U,
satisfying the following properties:
AP 1. If F is a vector space over Rand g: El')
-> F is an r-multilinear alternating map, then there exists a unique linear map
such that for all u t>
•••
,u, E E we have
g(ut> ... ,u,) = g.(u.
1\ ... 1\
u,).
632
APPENDIX
AP 2. If {v" ... ,vn } is a basis of E, then the set of elements I
Vii A ... 1\ Vi"
is a basis of
~
h < ... < i,
~ II,
1\ E.
Proof For each subset S of {I,... ,n} consisting of precisely r elements, we select a letter ts. As explained at the beginning of the section, these letters ts form a basis of a vector space whose dimension is equal to the
binomial coefficient
G)'
It is the space of formal linear combinations
of these letters. Instead of ts, we could also write t(i) = ti, "", with i, < ... < i,. Let {v" ... ,vn } be a basis of E and let u" ... ,u, be elements of E. Let A = (au) be the matrix of numbers such that
Define U, 1\ ••• 1\ U,
=
L Dets{A)ts. s
We contend that this product has the required properties. The fact that it is multilinear and alternating simply follows from the corresponding property of the determinant. We note that if S = {i l , ... ,i,} with i, < ... < i" then
A standard theorem on linear maps asserts that there always exists a unique linear map having prescribed values on basis elements. In particular, if g: E(') -+ F is a multilinear alternating map, then there exists a unique linear map
such that for each set S, we have g.(t s ) = g(vs)
= g(Vi
I
, ... ,v,, )
if i ... ,i, are as above. By formula (I), it follows that "
g(u" ... ,u,) = g.{u, " ...
1\
for all elements u" ... ,U, of E. This proves AP I.
u,)
633
APPENDIX
As for AP 2, let {w" ... ,wn } be a basis of E. From the expansion of (1), it follows that the elements {w s }, i.e. the elements {WI, 1\ ••• 1\ w;J with all possible choices of r-tuples (i" ... ,i,) satisfying i I < ... < i, are generators of 1\ E. The number of such elements is precisely (;). Hence they must be linearly independent, and form a basis of 1\ E, as was to be shown.
Theorem B. For each pair of positive integers (r, s) there exists a unique bilinear map
such that
if u" ... , u" W" ... ,W, E E then
(UI 1\ ••• 1\
u,)
X (WI 1\ ••• 1\
w,) ......
UI 1\ ••. 1\ U, 1\ WI 1\ ••• 1\ W,.
This product is associative. Proof For each r-tuple (u" ... ,u,) consider the map of E(') into
1\+' E
given by
This map is obviously s-multilinear and alternating. Consequently, by AP 1 of Theorem A, there exists a unique linear map 9(u)
- 914 ...... 14,.·. IA, E \
-
such thaI for any elements WI'
...
--+
A,+, E
I \
,w, E E we have
Now the association (u) ...... g(u) is clearly an r-multilinear alternating map of E(') into L(IV. E, E>, and again by AP I of Theorem A, there exists a unique linear map
!'I+'
such that for all elements u" ... ,U, E E we have 9 141 •••• ,14 .. -- 9 *(u 1
To obtain the desired product the association
1\ ••• 1\ U r ) .
1\ E x IV. E --+ 1\+' E,
we simply take
634
APPENDIX
It is bilinear, and is uniquely determined since elements of the form u, 1\ ••• 1\ u, generate !'I E, and elements of the form w, 1\ •.• 1\ W s generate jI{ E. This product is associative, as one sees at once on decom-
posable elements, and then on aU elements by linearity. This proves Theorem B. Let E, F be vector spaces, finite dimensional over R, and let J.: E -+ F be a linear map. If J1: F --+ R is an element of the dual space F*, i.e. a linear map of F into R, then we may form the composite linear map
which we visualize as E
--+A F --+• R.
We denote this composite J1 0 A by A*(J1). It is an element of E*.
Theorem C. Let J.: E --+ F be a linear map. For each r there exists a unique linear map A*: /'\ F*
--+!'I E*
having the following properties: (i) A*(W 1\ "') = A*(W) 1\ A*("')forwE/'I P, "'EjI{ F*. (ii) If J1E F* then A*(J1) = J1 0 A, and A* is the identity on P = R.
If
Proof The composition of mappings
given by
is obviously multilinear and alternating. Hence there exists a unique linear map /'I F* --+ /'I E* such that
Property (i) now follows by linearity and the fact that decomposable elements J11 1\ ..• 1\ J1, generate /'I F*. Property (ii) comes from the definition. This proves Theorem C.
Index
Beginning point 389 Bernoulli polynomial 325 Abel's theorem 220, 223, 239 Bessel function 347, 364 Absolute convergence 213, 229 Big oh 117 of integrals 329 Bijective 6 of series 213, 229 Bilinear 478 Absolute value, real 23 Binomial coefficients and complex 97 expansion II, 89, 111 Absolutely integrable 329 Block 584 Acceleration 269 Bonnet mean value theorem 107 Accumulation point 35, 36, 132, 157, 193 Bound for linear map 247, 456 Adherent 41, 153 Boundary 159 Admissible 570, 572, 573 Boundary of chain 621 Algebraic closure 20 I Boundary point 158 Alternating product 607 Bounded 29, 131, 135 Anosov's theorem 511 Bounded from above or below 29 Approach 42, 51, 160 Bounded linear map 178, 247 Approximation by step maps 252, Bounded metric 136 307,319 Bounded set 29, 135 Approximation of identity 286 Bounded variation 260 Arbitrarily large 51 Bruhat-Tits 149, 150 Arcsine 95 Bump function 81 Arctan 95, 114 Area of sphere 600 c Associative 17 Asymptotic 119 C' 73, 468 Asymptotic expansion 118 C'-invertible 512 C' 73, 468, 482, 487 C' chart 514 B C' coordinate system 514 C'-isomorphism 513 Ball 135 C' norm 137 Banach space 143 A
637
638
INDEX
en functions 82, 244, 268 Cantor set 224 Cauchy criterion 167 Cauchy sequence 39, 143, 149 Cesaro kernel 312 Cesaro summation 314 Chain 428, 620 Chain rule 67, 282, 396, 471 Change of variables 490, 593 Characteristic function 574 Chart 514, 531 Circumcenter 150 Class C" 73, 468, 482, 487 Close together 441,442 Closed 153 Closed ban 135 Closed chain 428 Closed in a subset 156 Closed interval 24 Closed rectangle 565 Closure 155, 157 Coefficients 56, 235 Commutative 18 Commute with limits 61 Compact 193 Comparison test 209 Complement 4 Complete 143 Completeness axiom 29 Completion 189, 192 Complex numbers 95 Composite 6, 164 Conjugate 96 Connected 389 Connected along a partition 419 Conservation law 391 Conservative 390 Constant term 56 Contained 4 Continuous 59, 156, 170 Continuous linear map 178, 246, 455 Continuous path 388 Converge 34, 39, 143, 149, 206, 327 Converge absolutely 213,224,225, 229 Converge uniformly 179 Convergence of integrals 327 Convergence of products 214 Convex 64, 65, 72 Convolution 283, 311,355 Coordinate function 172, 472 Coordinate system 514 Cover 203 Critical point 546
Cube 579 Curve 268 Curve integral
395,418,438
D
Dampened oscillation 219 Decimal expansion 214 Decomposable form 608 Decreasing 36, 71 Degree 56, 20 I Degree of differential operator 374 Dense 155 Denumerable II Dependence on initial conditions 557 Derivation 410 Derivative 66, 99, 268 Differentiable 66, 73, 380, 382, 463, 465 Differential equations 538 Differential form 608 Differential operator 374 Differentiating series 241 Differentiating under integral sign 276, 336, 500 Dirac family 347 Dirac sequence 284, 313 Dircctional derivative 383 Dirichlet kernel 311 Disc 138 Distance 25, 135, 176 Distance function 135 Distributive 19 Diverge 209 Domain of flow 545 Dual space 608 E
Eigenfunction 306 Eigenvector 413 Element 3 Elkies estimate 325 Empty 3 End point 389 Enumerate 11 Equidistribution 310 Equivalence relation 15, 32 Equivalent norms 135 Euclidean norm 137 Euler number 89 Euler's relation 387 Even 27 Exact form 613
INDEX
Exponential 78 Exponential of matrices 509 Extend by continuity 172, 176 Extension theorem for linear maps 247 Exterior derivative 610 F Family 183 Family of open sets 203 Fejer kernel 312 Finite 12 Finite sequence 12 Finite subcovering 203 Fixed point theorem 150, 502 Flow 541 Formal linear combination 630 Fourier coefficients 294, 295 Fourier series 296, 303 Fourier transform 342, 354 Free homotopy 452 Function 41 Functional equations 358 Fundamental lemma of integration 265 Fundamental solution of heat equation 349 G Gamma function 346 Generate vector space 130 Global flow 544 Gradient 380 Graph 41 Greatest lower bound (glb) 30 Green's function 378 Green's theorem 624 H
Half-closed 24 Half-open 24 Harmonic 316, 375 Heat equation 350 Heat kernel 348, 372 Heat operator 349 Hermitian product 291 Hessian 486 Homologous 425, 429 Homotopic 444 Homotopy 444 Homotopy relative to end points 448
639
Hyperbolic 510 Hyperplane 385 Hypersurface 411
Ideal of continuous functions 205 Identity map 6 Image 4 Imaginary part 96, 99 Implicit functions 522, 527 Improper integral 326 Increasing 35, 71 Indefinite integral 103 Induction 8 Inf 30 Infinite 12 Infinite interval 24 Infinity 50 Initial condition 538, 557 Injective 5 Integers 27 Integrability theorem 430 Integrable 471 Integral 101, 249, 395, 568 Integral along a path 418, 438 Integral curve 538 Integral equation 504,510 Integral mean value theorem 107 Integral operator 285, 290, 306 Integral test 210 Integration by parts 105 Integration of series 239 Interchange of derivative and integral 276, 337, 340 Interchange of integrals 277, 339, 342 Interchange of limits J85 Interior point 158 Intermediate value theorem 62 Intersection 4 Interval 24 Inverse 74, 96 Inverse image 156 Inverse image of a form 616 Inverse mapping 7 Inverse mapping theorem 502, 515 Inverse path 398 Inversion function 74 Invertible 76, 512 Invertible linear map 506 Irrational 28 Isolated 158, 16 J Isometry 150 Isomorphism 76
640
INDEX
J Jacobian 467 Jacobian determinant
Minimum 61 Monotone 76 Multilinear 478 Multiple integral 568, 581 Multiplication 18
589
K Kernel function 286 Khintchine Theorem 216 Kinetic energy 391
N
L
{,
216, 233 215 L'-norm 141,254,329 L2 -norm 141, 255, 299 Lagrange multiplier 412 Landau kernel 288 Laplace operator 316 Lattice point problem 366 Leading coefficient 56 Least upper bound (Iub) 29· Lebesgue integral 262 Length 293 Level surface 384 Li(x) 123 Limit 41,42,50, 143, 160 Limit inferior 41 Limit superior 40 Linear combination 130 Linear differential equation 552 Lipschitz condition 74, 174, 320, 540, 578 Lipschitz constant 74, 540 Little oh 117 Local flow 541 Local isomorphism 513 Locally integrable 418,438 Locally invertible 513 Logarithm 83, 133 Lower integral 567 Lower sum 105 (2
M
Mapping 4 Matrix representing bilinear map Maximum 61 Meim value theorem 71, 475 Measure 262 Measure less than epsilon 263 Measure zero 215, 577 Metric space 136 Midpoint 149
483
Natural numbers 3, 8 Negative 22 Negligible 569 Newton quotient 66 Newton's law 390 Newton's method 504 Non-degenerate block 584 Norm 131, 293 Normal derivative 625 Normed vector space 132 Null sequence 190 Null space 292 Numbers 17
o o(g) 117 0(11) 67, 463 Odd 27 Onto 5 Open ball 135 Open covering 203 Open disc 138 Open in a subset 156 Open interval 24 Open neighborhood 152 Open set 51 Opposite path 398 Order of magnitude 116 Ordering 21 Oriented boundary 613 Oriented volume 584 Orthogonal 138, 292 Orthogonal family 301 Orthonormal 301 Oscillation 576 Oscillatory integral 318, 336
P p's and q's 29,32,69,216-217 Parallelogram law 137, 305 Parametrized 269 Partial derivative 275, 371, 495 Partial differential operator 324, 406 Partial sum 206
641
INDEX
Partition of interval 105, 249 Partition of rectangle 565 Parts (summation by) 220 Path 388 Path integral 398,418,438 Pathwise connected 448 Peano curve 225 Periodic 93 Permutation
5
Perpendicular 138, 292 Piecewise C I 389 Piecewise continuous 253, 388 Point of accumulation 35, 36, 132, 157, 193 Pointwise convergence
179, 317
Poisson inversion 352 Poisson kernel 315 Poisson summation formula 358 Polar coordinates 518,595 Polar form 98 Polynomials 54, 287 Positive definite 133, 292 Positive integer 8 Positive number 21 Potential energy 390 Potential function 392 Power series 234 Prime numbers 124 Product 162 Producl decomposition 526 Product of matrices 458 Products (infinite) 214 Projection 294 Proper 4 Pylhagoras theorem 293
Q Quadratic form
413
R Radius of convergence 235 Rapidly decreasing at infinity 353 Ratio test 209 Rational number 27 Real number 17 Real part 96, 99 Rectangle 565 Rectangular path 433 Refinement 250, 566 Regulated map 253 Relatively continuous 170 Relatively uniformly continuous 199 Remainder in Taylor formula 110
Repeated integral 581 Residue 431 Riemann integrable 567 Riemann-Lebesgue lemma Riemann-Stieltjes 259 Riemann sum 106 Root 56, 201
S Sawtooth function 345 Scalar product 133 Schanuel's tbeorem 217 Schwartz space 353 Schwarz inequality 135, 294 Second derivative 477 Self conjugate 173 Seminorm 136 Semiparallelogram law 149 Sequence 12, 32 Series 206 Shrinking lemma 502 Shrinking map 503 Shub's theorem 505 Simplex 620 Sine and cosine 90 Size of partition 106, 570 Sphere 135, 600 Spherical coordinates 599 Square roo I 23 Squeezing process 48 Standard 608 Star shaped 449 Step map 249, 252 Stieltjes integral 259 Stirling's formula 120 Stokes' theorem 614, 622 Strictly convex 72 Strictly decreasing 35, 71 Striclly increasing 35, 71 Strictly monotone 75 Subcovering 203 Subdivision 434 Submanifold 531 Subrectangle 566 Subseq uence 13 Subset 4 Subspace 130 Sufficiently close 43 Sufficiently large 35 Sum of series 206 Summation by parts 220 Sup 30 Sup norm 132, 139
318
642 Support 362 Surjective 5 Symmetric second derivative 481
tNDEX
Upper integral 567 Upper sum 105 V
T Tangent plane 386 Tangent vector 602 Tate's lemma 503 Taylor formula 109, 408, 490 Taylor polynomial Ill, 409 Taylor series, arbitrary 244 Theta series 358 Tietze extension 231 Total family 300 Translation 514 Trecimal expansion 192, 224 Triangle inequality 131 Trigonometric degree 311 Trigonometric polynomials 306, 311 U
Uniform convergence 179, 229 Uniform distribution 310 Uniform limit 180 Uniformly Cauchy 179 Uniformly continuous 63, 174, 198 Union 4 Unit element 9 Unit vector 295, 625
Value 4 Variation 260 Vector 130 Vector field 390, 538 Vector field on sphere 602 Vector space 129 Velocity 269 Volume 575, 584 Volume of n-ball 598 W
Wallis product 121 Weakly increasing 71 Weierstrass approximation theorem 287 Weierstrass-Bolzano theorem 38, 193 Weierstrass test 230 Well-ordering 8 Winding number 422
z Zero 18 Zeros of ideal 205 Zeta function 359
Undergraduate Texts in Mathematics Abbott: Understanding Analysis. Anglin: Mathematics: A Concise History and Philosophy. Readings in Mathematics. Anglin/Lambek: The Heritage of Thales. Readings in Mathematics. ApOSIOI: Introduction to Analytic Number Theory. Second edition. Armstrong: Basic Topology. Armstrong: Groups and Symmetry. Axler: Linear Algebra Done Right. Second edition. Beardon: Limits: A New Approach to Real Analysis. BaklNewman: Complex Analysis. Second edition. BanchofflWermer: Linear Algebra Through Geometry. Second edition. Berberian: A First Course in Real Analysis. Bix: Conics and Cubics: A Concrete Introduction to Algebraic Curves. Br~maud: An Introduction to Probabilistic Modeling. Bressoud: Factorization and Primalily Testing. Bressoud: Second Year Calculus. Readings ill Mathema/ics. Brickman: Mathematical Introduction to Linear Programming and Game Theory. Browder: Mathematical Analysis: An Introduction. Buchmann: Introduction to Cryptography. Buskeslvan Rooij: Topological Spaces: From Distance to Neighborhood. Callahan: The Geometry of Spacetime: An Introduction to Special and General Relavitity. Carter/van BrunI: The LebesgueStieltjes Integral: A Practical Introduction. Cederberg: A Course in Modem Geometries. Second edition.
Cbilds: A Concrete Introduction to Higher Algebra Second edition. Chung/AiISahlia: Elementary Probability Theory: With Stochastic Processes and an Introduction to Mathematical Finance. Fourth edition. Cox/Little/O'Shea: Ideals, Varieties, and Algorithms. Second edition. Croom: Basic Concepts of Algebraic Topology. Curtis: Linear Algebra: An Introductory Approach. Fourth edition. Daepp/Gorkin: Reading, Writing, and Proving: A Closer Look at Mathematics. Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory. Second edition. Dixmier: General Topology. Driver: Why Math? EbbinghauslFlumITbomas: Mathematical Logic. Second edition. Edgar: Measure, Topology, and Fractal Geometry. Elaydi: An Introduction to Difference Equations. Second edition. ErdiSslSuranyi: Topics in the Theory of Numbers. Estep: Practical Analysis in One Variable. Exner: An Accompaniment to Higher Mathematics. Exner: Inside Calculus. FineIRosenberger: The Fundamental Theory of Algebra Fischer: Intermediate Real Analysis. FlaniganJKazdan: Calculus Two: Linear and Nonlinear Functions. Second edition. Fleming: Functions of Several Variables. Second edition. Foulds: Combinatorial Optimization for Undergraduates. Foulds: Optimization Techniques: An Introduction. (continued after index)
Undergraduate Texts in Mathematics (continued from page ii)
Franklin: Methods of Mathematical Economics. Frazier: An Introduction to Wavelets Through Linear Algebra Gamelin: Complex Analysis. Gordon: Discrete Probability. HairerlWanner: Analysis by Its History. Readings in Mathematics.
Halmos: Finite-Dimensional Vector Spaces. Second edition. Halmos: Naive Set Theory. HammerlinIHoffmann: Numerical Mathematics. Readings in Mathematics.
HarriS/HirsUMossinghoff: Combinatorics and Graph Theory. Hartshorne: Geometry: Euclid and Beyond. Hijab: Introduction to Calculus and Classical Analysis. Hilton/Holton/Pedersen: Mathematical Reflections: In a Room with Many Mirrors. Hilton/Hollon/Pedersen: Mathematical Vistas: From a Room with Many Windows. looss/Josepb: Elementary Stability and Bifurcation Theory. Second edition. Isaac: The Pleasures of Probability. Readings in Mathematics.
James: Topological and Uniform Spaces. Janich: Linear Algebra. Janich: Topology. Janich: VeClor Analysis. Kemeny/Snell: Finite Markov Chains. Kinsey: Topology of Surfaces. Klambauer: Aspects of Calculus. Lang: A First Course in Calculus. Fifth edition. Lang: Calculus of Several Variables. Third edition. Lang: Introduction to Linear Algebra. Second edition. Lang: Linear Algebra. Third edition.
Lang: Short Calculus: The Original Edition of "A First Course in Calculus." Lang: Undergraduate Algebra. Second edition. Lang: Undergraduate Analysis. Laubenbacher/Pengelley: Mathematical Expeditions. Lax/BursteinlLax: Calculus with Applications and Computing. Volume J. LeCuyer: College Mathematics with
APL. LidllPilz: Applied Abstract Algebra. Second edilion. Logan: Applied Partial Differential Equations. LovaszlPelilcinIVesztergombi: Discrete Mathematics. Macld·Strauss: Introduction to Optimal Control Theory. Malitz: Introduction to Mathematical Logic. MarsdenlWeinslein: Calculus I, II, III. Second edition. Martin: Counting: The Art of Enumerative Combinatorics. Martin: The Foundations of Geometry and the Non-Euclidean Plane. Martin: Geometric Constructions. Martin: Transformation Geometry: An Introduction to Symmetry. MillmanlParker: Geometry: A Metric Approach with Models. Second edition. Moschovalds: Notes on Set Theory. Owen: A First Course in the Mathematical Foundations of Thermodynamics. Palka: An Introduction to Complex Function Theory. Pedrick: A First Course in Analysis. PeressinilSullivanlUbJ: The Mathematics of Nonlinear Programming. PrenowitzlJanlosciak: Join Geometries.
Undergraduate Texts in Mathematics Priestley: Calculus: A Liberal Art. Second edition. Protter/Morrey: A First Course in Real Analysis. Second edition. Protter/Morrey: Intennediate Calculus. Second edition. Pugh: Real Mathematical Analysis. Roman: An Introduction to Coding and Information Theory. Ross: Elementary Analysis: The Theory of Calculus. Samuel: Projective GeomelTy. Readings in Mathematics. Saxe: Beginning Functional Analysis Scharlau/Opolka: From Fcnnatto Minkowski. Schiff: The Laplace Transfonn: Theory and Applications. Sethuraman: Rings, Fields, and Vector Spaces: An Approach to Geometric Constructability. Sigler: Algebra. Silvermanffate: Rational Points on Elliptic Curves. Simmonds: A Brief on Tensor Analysis. Second edition. Singer: Geometry: Plane and Fancy.
Singerrrhorpe: Lecture Notes on Elementary Topology and GeomelTy. Smith: Linear Algebra Third edition. Smith: Primer of Modem Analysis. Second edition. SlanlonlWhite: Constructive Combinatorics. Stillwell: Elements of Algebra: GeomelTy, Numbers, Equations. Stillwell: Elements of Number Theory. Stillwell: Mathematics and Its History. Second edition. Stillwell: Numbers and GeomelTy. Readings in Mathematics. Strayer: Linear Programming and Its Applications. Toth: Glimpses of Algebra and GeomelTy. Second Edition. Readings in Mathematics. Troutman: Variational Calculus and Optimal Control. Second edition. Valenza: Linear Algebra: An Introduction to Abstract Mathematics. WhyburnlDuda: Dynamic Topology. Wilson: Much Ado About Calculus.
UNDERGRADUATE TEXTS IN MATHEMATICS
This is a logilolly sell-lontoined introdultion to analysis, suitable fa< students who have hod !wo years of 1011Ulus_ The book lenters around those properties that have to do with unifa<m lonvergenle and uniform limits in the lOntext of differentiation and integration. TopilS dislussed indude the dossilOI test for lonvergenle of series, fourier series, polynomial approximation, the Poisson kernel, the construction of harmonil fun[tions on the disk, ordinary differential equations, lurve integrals, derivatives in vellor spoles, multiple integrals, and others. One oflhe author's main lOnlerns is to olhieve a bolonle between lonuete examples and generoltheorems, augmented by a variety of interesting exerlises. Some new mOleriol has been added in this selond edition, for example: a new lhopter on the global version of inlegrotion of lomlly integrable veltor fields; a brief dislUssion of LI-Comby sequences, introduling students 10 the lebesgue inlegrol; ma<e material on Dirol sequenles and families, induding a section on the heat kernel; a ma<e systemotil dis,us~on of orders of magnitude; and a number of new exe"ises.
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