Undergraduate Analysis

CP(O)

+ h) -

if h # 0,

f(x) - f'(x)h,

= o.

Conversely, if a number L and such a function f(x

+ h) h

f(x) _ L -

Ip

+

exist, we have

cp(h) h '

so that the limit as h -+ 0 exists and is equal to L. Thus L is uniquely determined and is equal to f'(x). The function cp(h) can be written conveniently in the form cp(h)

= hljl(h),

lim ljI(h) = 0,

where

'-0

namely we simply let ljI(h) = cp(h)jh if h # 0, and 1jI(0) = O. We can now prove the chain rule. Let k = k(h) = f(x and let y = f(x). Then g(f(x

+ h»

+ k) - g(y) g'(y)k + kljl(k)

+ h) -

f(x),

- g(f(x» = g(y =

(where lim ljI(k) = 0),

.-0

and consequently g(f(x

+ h»

- g(f(x»

h = g'(f(x» f(x

+ h~

- f(x)

+

f(x

+~-

f(x) ljI(k(h».

Taking the limit as h -+ 0, and using the fact that the functions 1jI and k are continuous at 0 and take on the value 0, we obtain the chain rule. We conclude with some standard derivatives. If f is a constant function, then f' (x) = 0 for all x. If f(x) = x, then f'(x) = 1. If n is a positive integer, and f(x) = x·, then f'(x) = nx·- '. This is proved by induction. It is true for n = 1. Assume it for n, and use the rule for the product of functions: The derivative of x"+ 1 is the derivative

[III, §l]

PROPERTIES OF THE DERIVATIVE

69

of x· . x and is equal to nx"-I· x

+ x' = (n +

l)x",

as desired. If f(x) = cg(x) where c is a constant and 9 is differentiable, then f'(x) = cg'(x). Immediate. The above remarks allow us to differentiate polynomials. If n is a positive integer, and f(x) = x-' = I/x', then we also have f'(x) = -nx-·- I . This follows at once from the rule for differentiating quotients. Both f and f' are of course defined only for x ". O. Finally, we shall also use the notation df/dx instead of f'(x). Furthermore, we allow the classical abuse of notation such that if y = f(u) and u = g(x), then dy dydu dx = dudx'

III, §1. EXERCISES I. Let

ex be an irrational number having the following property.

There exists a number c > 0 such Ihat for any rational number pjq (in lowesl form) with q > 0 we have

ex-£.I>.:.., l q q2 or equivalently,

c Iqex - pi >-. q (a) LeI f be the function defined for all numbers as follows. If x is not a rational number, then f(x) = O. If x is a rational number, which can be written as a fraction pjq, with integers q, p and if this fraction is in lowest form, q > 0, Ihen f(x) = ljq'. Show that f is differentiable at ex. (b) Let 9 be the function defined for all numbers as follows. If x is irrational, then g(x) = O. If x is rational, written as a fraction pjq in lowest form, q > 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)
[III, §2]

71

MEAN VALUE THEOREM

Theorem 2.3 (Mean Value Theorem). Let f be continuous on an interval [a, b] with a < b, and differentiable on the interval a < x < b. Then there exists c such that a < c < band f(b) - f(a) = j'(c)(b - a). Proof Let g(x)

= f(x)

-

=

f(~ ~(a) (x -

a).

Then g(b) = g(a) = f(a). We apply Lemma 2.2 to g, and obtain Theorem 2.3.

fib) f(o)

o

<

h

A function f on an interval is said to be (weakly) increasing if whenever x ;;; y we have f(x) ;;; f(y). It is said to be strictly increasing if whenever x < y we have f(x) < fly). We define (weakly) decreasing and strictly decreasing similarly. Corollary 2.4. Let f be continuous on [a, b] and differentiable on a < x < b. Assume j'(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)
such that g(y)

P] -+ [a, b]

= unique x e [a, b] such that fIx) = y. gof(x) = x

and

Thus

fog(y)=y.

We call 9 the inverse function off. Theorem 3.1. Let f be continuous, strictly increasing on [a, b]. Then the inverse function of f is continuous and strictly increasing.

[III, §3]

75

INVERSE FUNCTIONS

Proof Let g be the inverse function. That g is strictly increasing is obvious. We must prove continuity. Let y E [lX, Jl] (notation as above). Given €, and y = f(c), consider the closed interval of radius € centered at c. I

I

1

= C - € if a ~ c - €, and XI = a otherwise. + € ::::: b, and Xl = b otherwise. Then f(x l ) ~f(Xl).

Let XI c

Let Xl

=C +€

if

flx~)

flc) [(x,)

·-+-a+--.. ·~--r;---;x"'~--+6-We may assume a < b. We select (j equal to the minimum of and

f(c) - f(XI)'

except when this minimum is O. Suppose first that this minimum is not o. If Iy - yl < (j then the unique X such that y = f(x) lies in the interval XI < X < Xl' and hence Ig(y) - cl < €. If the minimum is 0, then either a = c or c = b, that is c is an end point. Say c = a. In that case, we disregard XI> 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 <

p. Let Yo = f(xo) and y = f(x). Then

g(y) - g(yo) X - Xo 1 y - Yo = f(x) - f(xo) = f(x) - f(xo>" 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
I

It is but an exercise to show that {i'+' = a'a' and aO = 1. Also (a"')' = ti". We leave the proofs to the reader. Ifa x = y, we sometimes write x

= log.y.

Note that we can now easily prove the fa{;t that every positive number has an n-th root. If a > 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 + - <-. +Inn

- - < log I n

(b) By the same technique, prove that

I

I

I

I

2

n

2

n-I

- + ... + - < log n < I + - + ... + --. 22. (a) For each integer n

~

I, let

Show that a nH < an' [Hint: consider an - a nH and use Exercise 21.] (b) Let bn = an - lin. Show that bn+ 1 > bn. (c) Prove that the sequences {an} and {bn} are Cauchy sequenoes. Their limit is called the Euler number y.

23. If 0 ~ x :;; 1/2, show that 10g(1 - x) ~ -x - x 2 • [Note: When you have Taylor's formula and series later, you can see that

x2 x 3 10g(1 - x) = -x - - - - - .... 2 3 The point is that -x is a good approximation to 10g(1 - x) when x is small.] 24. (a) Let s be a number. Define the binomial coefficients and

G) =

B(n, s)

1)(5 - 2)'''(5 - n

n~2.

for Prove the estimate IB(n, 5)1 :;; Isle~'(n

= 5(5 -

-

1)~l/n.

In particular,

lim sup IB(n, s)I"n :;; I.

+ 1)/n!

90

[IV, §3]

ELEMENTARY FUNCTIONS

Note that the above estimate applies as well if 5 is complex. (b) If 5 is not an integer ~ 0, show that lim IB(n, 5)1"· = I. 25. Let a be a real number > 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)

<

= F(d) - F(c). <

This holds whether c < d or d < c. From the rules for the derivative of the sum, we conclude that whenever f, g are continuous, we have f f(x) dx

+

fg(x)dx = f(f(x)

+ g(x»)dx,

and for any constant c we have

f cf(x) dx = c f f(x) dx.

The same formulas hold therefore when we insert the limits of integration, i.e. replace by in these relations, where we use the more usual notation instead of Thus

r:

r r: I:.

and

f(Cf) = f f· c

In particular, using c = -1, we conclude that

fU-g)=

f>- fg·

The above properties are known as the linearity of the integral.

104

THE ELEMENTARY REAL INTEGRAL

[V, §2]

V, §2. PROPERTIES OF THE INTEGRAL 1beorem 2.1. Let a, b be two numbers with a
Proof. Let q> = 9 - f. Then q> !?; 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
We let M;(f) be the maximum of f on [ai' ai+']' and miU) the minimum. We define the lower and upper sums of f with respect to the partition by n-I

n-l

L:U, P) =

L miU)(a;+1 -

i==O

ail

and

U:U, P)

= ;::0 L M;(f)(ai+l

-

aJ

106

[V, §2]

THE ELEMENTARY REAL INTEGRAL

Depending on a choice of c, e [a" aI+ 1 ], we define the Riemann sum ,,-1

R~(f, P)

= i=O L f(c,)(a'+1

- atJ.

Then trivially we have the inequalities

By the size of the partition P as above, we mean the maximum of the lengths of the subintervals, that is size(P)

= max , (0,+1

- atJ.

The proof of the next result will use uniform continuity rather than ordinary continuity. Theorem 2.7. Let f be COll/illUOUS all [a, b]. Givell E, there exists 0 such that if P is a partition of [a, b] with size(P) < 0, thell

U:lf, P) - L~(f, P) <

E.

The illlegral J,~(f) is equal 10 the least upper boulld of all lower sums alld also equal to the greatest 10IVer bOUlld of all upper sums. Finally, for any Riemalln sllln R~(f, P) with size(P) < 0 (as above), we have

r

f -

R~(f, p)

< E.

Proof By Theorem 4.6 of Chapter II. the function f is uniformly continuous, so given E there exists 0 such that if x, )' e [a, b] and Ix - yl < 0, then If(x) - f(y) I < Ej(b - a). Now let P be a partition of [a, b] of size < o. Then for all x E [a" a,+.J and c, E [a" 01+1] we have €

If(x) - f(c;lI < b _ a

o ~ Mi(f) -

and

€

m;(f) < -b- .

-a

Therefore

o ~ U:(f, P) -

L~(f, P) =

(*)

n-1

L (M;(f) E

11-1

< b'--

This proves the first statement.

m,(f))(a'+1 - a,)

i=O

L (0,+1

II i=O

- a,) =

E.

[V, §2)

PROPERTIES OF THE INTEGRAL

107

As to the second statement, concerning the integral, by Theorem 1.I (2) and Theorem 2.1, it follows that for every partition P, we have

In particular, from inequality (*) for every E, it follows that l~(f) is the unique number which is the least upper bound of all lower sums and the greatest lower bound of all upper sums. The final statement concerning Riemann sums then follows at once, since both the Riemann sum R:(f, P) and the integral f are squeezed between the upper and lower sums whose difference is < E, if size(P) < b. This concludes the proof.

J:'

V, §2. EXERCISES

J, 9 be continuous functions on [a, b] with a < b. Assume 9 positive. Show that there exists c E [a, b] such that

1. (a) Let

f• J(x)g(x) dx b

f• g(x) dx. b

= J(e)

(b) Bonnel mean value theorem (1849). Let J, 9 be continuous real valued functions on [a, b). Assume J positive monotone decreasing. Show that there exists a point C E [a, b] such that

f• J(x)g(x) dx = J(a) f• g(x) dx. b

,

First assume that J is C l , so J' ;:;; O. Let G(x) be the integral of 9 from a to x. Integrate by parts. Using the intermediate value theorem, show that there is some CI E [a, bl such that

f• J(x)g(x) dx b

= J(b)G(b)

+ G(cd(J(a) - J(b»).

Divide by J(a), use the hypothesis that J is decreasing to conclude that the right side is on the segment between G(b) and G(ed, so by the intermediate value theorem again, is equal to G(e) for some c, thus proving the result in this case. In general, possibly wait until Chapter X, §3, Exercise 7. Show that there exists a sequence {f,,} of C l functions with f,,(a) = J(a), f,,(b) = J(b), each f" is monotone decreasing, and {f,,} converges uniformly to f Use bump function~ to do this. The theorem is true for each f,,, with some Coo instead of c. B} Weierstrass Balzano, the sequence {coo} has a point of accumulation e E [a,bl which does what you want.

108

THE ELEMENTARY REAL INTEGRAL

d·

I

p.(x) = 2.

2. Let

Show that

I-,,

-d.(x' - 1)·). n. x I

if

p.(x)P.,(x) dx = 0

'I

and that

[V. §2)

In #; 11,

2 P.(x)' dx = - 21.

_I

11

+

3. Show Ihat

f-,'

if

xmP.(x) dx = 0

III

<

n.

Evaluate

f'

x·p.(x) dx.

-I

4. Let a < b. If I. 9 are conlinuous on [a. b). lei

(I. g)

= f.r(X)g(X) ,Ix.

Show that the symbol (I. g) satisfies the following properlies. (al If I" I,. 9 are conlinuous on [a. b). then


=

(I,. g) + (I,. g).

If r is a number. then (cf. g) = r(I. g). (b) We have
UID.

=

' ]'" [J.II(xW

+

I/q = 1. Prove lhal

Let q be a number such lhat I/p

dx

.

I(I. g) I ;;; UIU.M,· [Hill': If IIIlI p and IIgll, oF O. let 10 Chapter IV. §2.J

or

II

=

II IP/IIJ11: and v = Igl'/lIgll: and apply Exercise

[V. §3]

109

TAYLOR'S FORMULA

6. Notation being as in the preceding exercise, prove that

[Hilll: Let 1 denote the integral. Show that

and apply Exercise 5.] 7. Let f: J -> C be a complex valued function defined on an interval J. Write f = f. + if,. where fl. f, are real valued and continuous. Define the indefinite integral

ff(X)dX

= ff,(X)dX + i ff,(x)dx,

and similarly for the definite integral. Show that the integral is linear, and prove similar properties for it with change of variables and integrating by parts. 8. Show that for real a ,p 0 we have

Show that for every integer n ,p 0,

'f ·

o i"Xdx

= O.

V, §3. TAYLOR'S FORMULA Theorem 3.1. Let f be a function having n continuous derivatives on an interval J. Let a, b E J. Then

f(b) = f(a)

I'(a) I!

+ -(b -

a)

+ :.. +

jl"-t)(a) (b - ar- t (n - I)!

where R = n

t r (b - tr- jl")(t) dt. J. (n - I)! b

Proof. We start with f(b)

= f(a) +

f

I'(t) dt.

+ R..

llO

[V, §3]

THE ELEMENTARY REAL INTEGRAL

We integrate by parts, using induction. Assume the formula of the theorem proved for a certain n > 1. We let u(t)

= pn)(t)

and

dv(t)

= -(b -

tr-

I

dt.

The formula for n + I drops out.

Theorem 3.2. There exists a number c between a and b such that

Proof. Say a < b. Let M be the maximum of pn) on the interval [a, b] and let m be the minimum of pn) on this interval. Then we have the ine-

qualities

rb(b-tr-I m J. (n _ I)! dt;;; Rn

;;;

rb(b-tr- I M J. (n _ I)! dt.

The integrations are easily performed to give m

< R = (b _

n

a)n

<M = .

n! Since I(n) is assumed continuous, by the intermediate value theorem, we conclude that there is some c with a ;;; c ;;; b such that Rn

_ I(n)( )

(b _ a)n -

c,

n! thereby proving the theorem, if a < b. If b < a then the same argument can be applied but the above inequalities have to be reversed when n is odd, since then b - t is ;;; O. There is no change in the final conclusion. Examples. Computing the derivatives and evaluating at a = 0 yields the usual formulas for sin x, cos x and e% as foUows. . sm x

%

=x -

e =

I

x3 -3 + ... + (_I)m-' !

n-I

X 2m - 1

(2m-I)!

x + x + ... + -;-(n----:I"' ""')! + Rn(x).

+ R z +I(X) m'

[V, §3]

III

TA YLOR'S FORMULA

Since the sine and cosine in absolute value are always bounded by 1, we can use Theorem 3.2 to conclude that in these cases,

Ixl· IR.(x) I ;: : ; - •. n.

If 0 ;::::; x, then the remainder term for eX satisfies

If x ;::::; 0, then the remainder term for eX satisfies IR'(x)1

because eC

;::::;

Ixl· :s: -n., '

I for x ;::::; c ;::::; O.

Before doing the slightly harder case of the binomial expansion, we give a general corollary of Theorem 3.2. Theorem 3.3. Let f be of class en on a closed interval containing 0, say [-u, u] with u > O. Let K = max Ip·)(c)l. Then for all x in the interval,

fIx) = flO)

ce(-u,u)

x2

x n- 1

2!

(II - I)!

+ f'(O)x + f(2)(0) - + ... + r-1)(0)

alld IR.(x)l ;::::; Kx·/n!, so R.(x) = O(lxl·) for x

-->

+ R.(x),

o.

Proof Immediate from Theorem 3.2, taking a = 0 and b = x.

The polynomial , f(2)(0) 2 p._1(x)=f(0l+f(0)x+ 2! x

...

+

f(·-I)(O) .-1 +(n_I)!x

is called the Taylor poynomial of f, of degree;::::; II - I. In Exercise 3, you will prove a uniqueness statement about this polynomial. The binomial expansion We shall now consider the binomial expansion. Let fIx) = (I + x)',

112

[V, §3]

THE ELEMENTARY REAL INTEGRAL

where s is real ~ O. We may assume that s is not an integer; otherwise everything is trivial. Then j(·)(x) = s(s - 1)(s - 2)···(s -

II

+ 1)(1 + xy-·.

We consider the interval - 1 < x < I. We take a = 0 and b = x. Then

where

s) (k

=

sIs -

I) ... (s - k k!

+ 1)

is the generalized binomial coefficient. We estimate R.(x) and show that R.(x) --+ 0 as R.(x) =

r sIs -

Jo

I) ... (s (II - I)!

II

+

II --+ 00.

I) (x _ t r 1(1

Suppose first 0 ;£ x < 1. We estimate (1 then perform the integration, and find that

+ tr·

We have

+ tY-· dt.

by (1

+ tr· ;£ 2'.

We

(I)

From the estimate of the binomial coefficient in Exercise 24, Chapter IV, §2, it follows that R.lx) -+ 0 as II -+ 00 for 0 ;£ x < I. Suppose now that c is a number, 0 < c < 1, and consider the interval -1 < -c ;£ x ;£ O. We estimate

x - t g(t) = - - . 1+t When t = 0, we have g(O) = x. Also, g(x) = O. Taking the derivative of 9 shows that 9 is decreasing between x and O. Thus in any case, we find that x 1

t

+t

[V, §3]

113

TAYLOR'S FORMULA

whence X- t -1 r

n-l

< _cn-l -

+

Estimating the integral by Corollary 2.3 shows that

Again by the estimate of the binomial coefficient, it follows that Rn -+ 0 as n -+ 00. Note that our estimate is independent of x in the interval

-c < x <0. On the other hand, in both cases (1) and (2), when we keep n fixed, we have an estimate of Rn(x) in terms of Ixln , of the form

IRn(x)1 ;::;; Klxl n,

(3)

for some constant K. In case 0 5 x < 1, this is already stated in (1). In case x < 0, the constant K depends on the choice of number c such that -1 < c ;::;; x ;::;; O. We give a direct proof for inequality (3) in this case, and note that the proof is less delicate than for (2). We may estimate the term (1 + tr-" by some constant, because now n is fixed. The constant depends on c. The binomial coefficient is now fixed. We take the absolute value of the expression inside the integral sign, and pull out the appropriate constants. Then we perform the integration of what remains, namely IXI

f.o

I In (Ixl - tr' dt = ..:...n

to get estimate (3). At the beginning of the next section, we shall describe concepts and a notation to view estimates such as (3) in a broader context. The logarithm Rather than follow the general method with the remainder term, we shall indicate another way of gelling the Taylor formula for the logarithm. Theorem 3.4. For -1 < x 10g(1

+ x) =

5 1, one

x2 x - 2

has

x3

x"

3

n

+ - - ... + ( - l r ' - + R n+, (x),

114

[Y, §3]

THE ELEMENTARY REAL INTEGRAL

with the remainder term tn

x

Rn+l(x) = (_I)n

I

o

-1- dt.

+t

To estimate the remainder, we have two cases: Case 1. 0

~

x

~

1. Then

In particular, the remainder approaches 0 in the stated interval, and the bound Ij(n + 1) can even be made independent of x, or as one says, uniformly in x. Case2.

-1<x~O.

Then

IRn+l(x)! ~ (n

Ixln +'

+ 1)(1 + x)

In this case, the remainder also approaches 0 as n -+ 00. If x stays away from -I, Le. -1 + 0 ~ x ~ 0 for some 0 > 0, then

Ixln +!

I

IRn +, (x) I ~ o(n + I) ~ o(n + 1) The arguments used to prove the above statements are easy, and are left as Exercise 1. Note that Case 1 gives us the formula for log 2: log 2 = 1 -

*

t + ! - + ....

Arctangent Theorem 3.5. On the interval - I x3 arctan x = x - -

xS

~

x

~

I, one has X 2m - 1

+ - - ... + (_I)m-1 352m-

where

1

+ R 2m+I' (x)

[V, §3]

TAYLOR'S FORMULA

115

The remainder has the estimate

Again we leave the proof as an exercise. Remark. Note the phenomenon that the Taylor formula for log and arctangent has a remainder tending to 0 at the end point x = I. Of course, the remainder tends to 0 much faster for Ixl < 1, because the powers of x then contribute to the smallness of the remainder as n -+ 00.

V, §3. EXERCISES 1. Prove Theorem 3.4 by integrating I I" - - = 1-1+12 -'" +(-lr'I"-' +(-1)"--

1+1

1+1

from 0 to x with -I < x < 1. Prove the estimates for the remainder to show that it tends to 0 as n --+ co. If 0 < c < I, show that this estimate can be made independent of x in the interval -c ~ x ~ c, and that there is a constant K such that the remainder is bounded by Klxl"+1. 2. Do the same type of things for the function 1/(1

+ (2 )

to prove Theorem 3.5.

3. Let I, g be polynomials of degrees ~ d. Let a > O. Assume that there exists C > 0 such that for all x with Ix I ~ a we have I/(x) - g(x) I ~ C1xld+ I.

Show that

Ih(x)1

~

I =

C1xl

d

g. (Show first that if h is a polynomial of degree + l , then h = 0.)

~

d such that

Exercise 3 shows that the polynomials obtained in Exercises I and 2 actually are the same as those obtained from the Taylor formula. 4. Let a > 1. Prove that a"

lim-=O.

"-CO n!

Conclude that the remainder terms in the Taylor expansions for the sine, cosine, and exponential function tend to 0 as n tends to infinity.

5. (a) Prove that log 2 = log(4/3) + log(3/2), or even better, 10 25 81 log 2 = 710g- - 210g- + 310g9 24 80'

116

THE ELEMENTARY REAL INTEGRAL

[V, §4]

(b) Find a rational number approximating log 2 to five decimals, and prove that it does so. The above trick is much more efficient than the slowly convergent expression of log 2 as the alternating series. 6. (a) Prove that u+v arctan u + arctan v = arctan - - . 1 - uv

(b) Prove that 1t/4 = arctan I = arctan(I/2) + arctan(If3). (c) Find a rational number approximating 1t/4 to 3 decimals. (d) You will do so even faster if you prove that 1t

4=

4 arctan(I/5) - arctan(I/239).

For all this, cf. my First Course in Calculus, fifth edition, Springer-Verlag, Chapter XIII, §5 and §6. 7. Let A > 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

<0

=

lub IJ(x)I,

also written

xeS

sup IJ(x)l.
We contend that this is a norm. If IIJ11
This is true for all

+ g(x) I ~ IJ(x) I + Ig(x)1

XES.

S;

M,

+ M 2•

Hence

IIJ + gll
+ g(x)1 ~ IIJ11
<0 ,

thus proving N 3. The norm in this example is called the sup norm.

[VI, §2]

NORM ED VECTOR SPACES

133

Observe that in Example 2, the argument can be used to deal with a more general situation. Again, let 8 be a non-empty set, let E be a normed vector space with norm I I, and let !?4(8, E) be the set of all bounded maps of 8 and E. Then !?4(8, E) is a vector space, and one can define a norm II II on it by the same formula that was used in Example 2. The proof that this is a norm is exactly the same as that given above. The space of bounded maps is perhaps the space most used throughout this book. A norm from a scalar product A norm on a vector space is often defined by a scalar product. By this we mean a product (v, w) 1-+ v·

W

=
from E x E into R satisfying the following conditions: SP 1. We have v· w

= w·

v for all v, wEE.

SP 2. We have for u, v, WEE, u . (v

+ w) = u . v + II' W.

SP 3. If X is a number, then (xv) . W = x(v . w) = v· (xw).

In addition, the scalar products we shall consider will be positive definite, that is they satisfy the additional property: SP 4. If v = 0 then v· v = 0, and

if v ¢ 0 then v· v > O.

Examples are given in the next section. As an abbreviation, we shall often write v 2 instead of v· v. However, we do not write v3 , or any other exponent. Using the properties of the scalar product, we find that (v

+ W)2 = v2 + 2v· W + w2 ,

(v -

W)2 =

v2

-

2v· W

+ w2 ,

as usual. The notation v· W will be useful when dealing with vectors of n-space, and
134

NORMED VECTOR SPACES

[VI, §2]

Theorem 2.1. Let E be a vector space with a positive definite scalar

product. Then I(v, w)1 2 ~ (v, v)(w, w). Proof Let x = (w, w) and y = - (v, w). Then by SP 4 we have

o ~ (xv + yw)·(xv + yw) = x 2v· V + 2xy(v· IV) + y2 IV . IV.

Substituting the values for x and y yields

o ~·(IV' W)2(V' v) -

2(w. w)(v . IV)'

+ (v' w)2(w. w).

If w = 0 then the inequality of the theorem is obvious, both sides being equal to O. If w ,p 0, then w· w ,p 0, and we can divide the last expression by w· w. We then obtain

o ~ (v' vXw· w) -

(v· w)2,

which proves the theorem. We define Ivl = ~. We can rewrite the inequality of Theorem 2.1 in the form

Iv,wl < Ivllwl by taking the square root of both sides. This inequality is known as the Schwarz inequality . Theorem 2.2. Thefunction

V 1-+

Ivl is a norm on E.

Proof. We clearly have N I. If a E R and vEE, then lavl = Jav·av

= Ja 2 v.v = lallvl, so that N 2 is satisfied. As to N 3, we have

Iv + wl 2 = (v + w)·(v + w) = v·v + 2v·w + w·w ::;; Ivl 2 + 21vllwl + Iwl 2 = (Ivl + Iw!)2,

[VI, §2]

135

NORMED VECTOR SPACES

using the Schwarz inequality in the second step. Taking the square root yields N 3. We do not go further here into the study of the scalar product. We merely wanted to show how it could be used to yield a norm on a vector space. We comment further in the next section on n-space. We shall use some geometric terminology with norms. Let E be a normed vector space, and let WEE. Let r > 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]

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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 ! <

lajl

+ !bjl ;:;; max la,l + max Ibil

;:;; IIAII

+ IIBII.

i

j

This is true for each j, and hence IIA

+ BII

= max laj

+ bj ! :S

IIAII

+

IIBII,

j

so the triangle inequality is satisfied. Finally, if c e R, IIcAIl

= max Ica,l = max lella,l = lei max lad = lelllAII· i

i

i

This proves that II II is a norm. We shall call it the sup or max norm. Still another norm useful in some applications is given in the exercises. Consider the sup norm on R 2 • The closed ball centered at the origin of radius r consists of the set of all points (x, y) with x, y e R such that Ixl ;:;; r and I y I ;:;; r. Thus with this norm, the closed ball is nothing but the square (inside and the boundary).

r r

140

[VI, §3]

NORMED VECTOR SPACES

The sphere of radius r with respect to the sup norm is then the perimeter of the square. Note that it has corners, i.e. it is not smooth. It is easy to verify directly that the euclidean norm and the sup norm are equivalent. In fact, if A = (a ..... ,an)' then

so that

IIAII

=

max lajl < IAI· j

On the other hand, let la.1 = max lail. Then ;

and consequently

thus showing that our two norms are equivalent. It is in fact true that any two norms on Rn are equivalent. See Theorem 4.3. Finite dimensional vector spaces. Instead of Rn we could also deal with finite dimensional vector spaces. Let E be a vector space over R, of dimension n. Let {e" ... ,en} be a basis of E. Each element veE can be written as a linear combination with

Xi

e R.

With respect to this basis, we can then define the sup norm with respect to the basis to be

IIvll

=

max Ix"" ;

just as we did for Rn. It is easily verified that this defines a norm. If we change the basis, the sup norm will also change. However, see Theorem 4.3. Example. Let E be the vector space of polynomial functions on the interval [0, I]. For each positive integer d, let Ed be the subspace of polynomial functions of degree ~ d. Then Ed has dimension d + 1. Let II II '" be the sup norm on Ed as in Example 2 of §2. On the other hand,

[VI, §3]

141

n-SPACE AND FUNCTION SPACES

let!. be the function !.(xl = x;. Then fo, ... Jd form a basis of Ed' and an arbitrary polynomial of degree d can be written as a linear combination

Define 1If11 = max la,l. Then Theorem 4.3 for finite dimensional vector ;

spaces implies that the two norm II II "" and II II are equivalent on Ed' The L l-norm. Let E be the vector space of continuous functions on [0, 1]. If fEE, define the L '-norm by IIf11.

= Ie' If(xll dx.

In Exercise 4 you will prove that this is indeed a norm. The L 2- norm. We shall now consider an example of a norm defined on a functions space by means of a scalar product. We let E be the space of continuous functions on the interval [0, 1]. Iff, gEE, we define
= Ie' f(x)g(x) dx.

The four properties of a positive definite scalar product are verified as immediate consequences of properties of the integral. Thus we obtain the corresponding norm, called the L 2- norm, 1If112

= (f,f)l/2 =

(I

I

0

f(x)2 dx

)1/2

.

Note that a continuous function is bounded on [0, 1] and hence that we can define the sup norm on the space E of continuous functions on [0, 1]. Let us denote the sup norm II II"". If fEE, and we let M= IIfll"", then

L I

f(X)2 dx ;;;

Ie'M

2

2 dx < M .

Hence IIfllz;;; IIf11",,· However, the two norms II liz and II II "" are not equivalent because there is no inequality going in the opposite direction. For instance, the function

142

NORMED VECTOR SPACES

[VI, §3]

whose graph is as follows:

has a sup norm equal to 1, but its U-norm is small. Taking such functions having a narrOwer and narrower peak, we can get functions having arbitrarily small U-norms, but with sup norms equal to 1. In Exercise 4, you will also prove that

IIflll;:i; IIfb but that the two norms L' and L 2 are not equivalent. The norms defined above, sup norms, CO-norms, L'-norm, and L 2 _ norm, and more generally norms coming from positive definite scalar products, are the basic norms in mathematics.

VI, §3. EXERCISES I. Let E, F be normed vector spaces, with norms denoted by I I. Let E x F be the set of all pairs (x, y) with x E E and y E F. Define addition eomponentwise: (x, y)

+ (x',

y')

= (x + x', Y + y'),

e(x, y) = (ex, ey),

for

CE

R. Show that E x F is a vector space. Define I(x, y)1 = max(lxl,lyl)·

Show that this is a norm on E x F. Generalize to 11 factors, i.e. if E" ... ,En are normed vector spaces, define a similar norm on E 1 X ••• x En (the set of IHuples (x I , ..• ,x.) with x, E E,). . 2. Let E = R", and for A

= (a"

... ,an) define n

IIAII = L: la,l· ;=1

Show that this defines a norm. Prove directly that it is equivalent to the sup norm.

[VI, §4]

COMPLETENESS

143

3. Using properties of the integral, prove in detail that the symbol
VI, §4. COMPLETENESS lei E be a normed vector space and let S be a subset. A sequence {x n } in S is said to converge in S if there exists v E S having the follow· ing property: Given E, there exists N such that for all n?; N we have IXn - vi < E. We then call v the limit of the sequence {xn }. This limit, if it exists, is uniquely determined, for if w is also a limit of the sequence, we select N, such that if n ;:>: N, then IX n - vi < E and N, such that if n ?; N, then IXn - wi < E. Take N = max(N., N,). If n?; N then

Iv - wi so Iv -

wi = 0 and v = w.

~

Iv - xnl + IXn - wi < 2£,

The limit is denoted by lim

Vn

= V.

A sequence {x n } in a normed vector space is called a Cauchy sequence if given E there exists N such that for all m, n ~ N we have

If a sequence converges then it is a Cauchy sequence. The reader need only copy the proof given in Chapter II, §1. In the case of real numbers, we proved, using the Archimedean axiom, that every Cauchy sequence of numbers has a limit. However, in an arbitrary normed vector space, this need not be the case. A normed vector space in which every Cauchy sequence has a limit is called complete, or also a Banach space. We shall see in a moment that R' is complete, and we shall meet later examples of complete function spaces.

144

[VI, §4]

NORMED VECTOR SPACES

Remark. If I b is a norm on E equivalent to I I, then convergent sequences, limits, and Cauchy sequences with respect to I b are the same as with respect to I I. This is verified at once. Example. Let E be a finite dimensional vector space over R, of dimension k. Let {e" ... ,ed be a basis, and take the sup norm on E with respect to this basis. Let {v" V2' ... } be a sequence of vectors in E, and write each Vn in terms of its coordinates: for n = 1,2, ... , Xnj

E

R.

Then we obtain k sequences of coordinates, namely corresponding to the columns:

Theorem 4.1. The sequence {vn } is a Cauchy sequence if and only the coefficient sequences {xn,}, ... ,{xn.J are Cauchy sequences in R.

if all

Proof. Essentially obvious. Suppose {vn } is a Cauchy sequence in E. Given (, there exists N such that if n, m ~ N then II Vn - Vm II < (. But then for each i = 1, ... ,k we have

and so the i-th coefficient sequence is Cauchy. Conversely, if every coefficient sequence is Cauchy, for each i we find N. such that whenever n, m ~ N, then Ixn , - xm.l < (. We let N be the maximum of N" ... ,N•. Then for m, n ~ N we get

so the sequence {vn} is Cauchy, thus proving Theorem 4.1. We continue with the space E, having the sup norm with respect to a basis. Theorem 4.2. The space E is complete. The sequence {vn } being as above,

if

lim

n-""

X n•

= Y.

[VI, §4]

COMPLETENESS

145

for i = I, ... ,k, then

lim

Vn

n-co

=

L• Yiej

i=1

and conversely.

Proof Given €, there exists N such that if n :1; N then Ix,,; - )'il < € for all i = I, ... ,k. Then IIv n - wll < E, if w = L: Yie,. The converse is equally obvious. Hence if (vnl is a Cauchy sequence in E its coefficient sequences converge to y" ... ,Y., respectively, and so Vn converges to W = LYje j •

So far, to be precise, Theorem 4.2 should really be stated as saying that E is complete with respect to the sup norm, or any norm equivalent to it. However, this restriction is unnecessary, as we now prove. Theorem 4.3. Let E be a finite dimensional vector space over R. Then any two norms on E are equivalent. In particular, every norm on R' is equivalent 10 the sup norm. Remark. Using Theorem 2.2 and Exercise 3 of Chapter VIII, one can give a formally much shorter proof of the present theorem. Proof. Let k = dim E be the dimension of E. We prove the theorem by induction on k. Suppose k = I and let (e d be a basis of E over R. Let II II be a norm on E. Then for x E R,

IIxe,1I

=

Ixllle,lI·

so any two norms are simply a constant multiple of each other, and hence are equivalent. We then prove the theorem by induction on k. Assume the theorem proved for k - I, k :1; 2. Fix a basis (e" ... ,e.) of E. It will suffice to prove that a given norm II II is equivalent to the sup norm II 110' One inequality is easy to prove. A vector VEE can be written as a linear combination v = "Ie, + ... + "kek, whence

IIvll

=

IIx,e, + ... + x.e.1I

~

Ix,llIe,1I + ... + Ix.llle.1I

~ C,lIvllo

146

[VI, §4]

NORMED VECTOR SPACES

where

Conversely, we must prove that there exists a number C > 0 such that for all vEE we have

IIvllo

~

CIIvlI.

Suppose no such constant exists. Given a positive integer m, there exists v # 0 in E such that

IIvllo> mllvll· If xl is the component of this vector v having maximum absolute value of all the components, we divide both sides of the preceding inequality by Ix}l. We let V m = xi IV. Then we still have

Furthermore, the j-th component of Vm is equal to 1, and all components of Vm have absolute value ~ I. Thus we have (*j

and

IIvm l < 11m.

For some fixed index j with 1 :::;; j :::;; k, there will be an infinite set J of integers m for which (*j is satisfied. We fix this integer j from now until the end of the proof. We let F be the subspace of consisting of all vectors whose j-th coordinate is equal to O. The norm on E induces a norm on F. By induction, the norm II II on F is equivalent to the sup norm on F, and in particular, there exists a number Cz > 0 such that for all IV E F we have

For each 111 E J we can write or with some element 10", E F. Given and m,ll E J, then

IIII'" -

10",11 ~

f,

take N such that 2/ N

112

IIv" - vmll < m - + II-

~

-N < f.

< f. If m.1l :?; N

[VI, §4]

147

COMPLETENESS

Hence {wm } is a Cauchy sequence with respect to II II, and by induction with respect to the sup norm on F. Since F is complete, it follows that {wm } converges to some element WE F with respect to the sup norm. (The limit is taken for III -> 00 and III E J as above.) Now we have for III E J:

By (*) and the convergence Will -> IV the limit of the right side is 0, so lIej + IVII = 0 and hence ej + IV = O. This is impossible because IV E F and ej is not in F. This contradiction proves the theorem. We now consider systematically the three norms on the space of continuous functions on a finite interval. The sup norm. The space Co([O, I]) is complete for the sup norm.

This will be proved in Chapter VII, Theorem 3.2. The L 1 and L 2· norm s. The space Co([O, I]) is not complete for each one of these two norms, To see this, we have to exhibit for each of these norms a sequence which is Cauchy but does not have a limit in Co([O, I]). We do this for the L '-norm, and leave it as an exercise for the L 2- norm . Let

f.(x) =

1/.fi r: { '\In

if lin;;; x ;;; I, 'f 0 < < II

I

=X =

n.

The graph of f. is a truncation of the function fIx) = o;;; x ;;; I, as shown.

Olin

1/.fi,

defined for

148

[VI, §4]

NORMED VECTOR SPACES

From elementary calculus, wc assume elementary properties of the improper integral which allow us to evaluate integrals of some functions which are not continuous on [0, I], for instance:

f,

x-"2 dx = lim

o

f-O

f'

x-"2 dx = 2x"2

(

I' =

2.

0

In some sense the function! is the limit of the sequence Cr.}, but is not in CO([O, I]). In any case, we can apply the formalism of integration, and the usual inequalities hold, i.e. it is a routine maller to verify ad hoc that the following steps are valid. First we can define the V-norm ad hoc on differences! - I... namely (1)

III -

1.11, =

fa' II(x) - f.(x) I dx =

r" 2

(x-'/2 - y'n)dx

I

I

I

I

vlt

ynt

- )., - y'n - y'n . Hence for positive integers,

(2)

111.-1.,11,

Ill,

n, we have

~ IIf.-III,

+ III-I." II , ~ r.:+ e-

Given E, if jN> I/E, then the right side is ~ 2E, whence {f.} is L l _ Cauchy. Actually, equality (I) shows that {In} is L1-convergent to f. However, there is no continuous function 9 on [0. I] such that {f.} is L '-convergent to g. Because if there were such a function g, then as

n -Jo

00,

whence II! - gil, = 0, that is

fa' I!(x) -

g(x)l dx = O.

But .f. 9 are continuous on the half-open interval (0, I], and the usual argument shows that if there is some c E (0,11 such that f(c) ". g(c), then I!(x) - g(x)! > 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 <, there exists an element XES such that d(x,z) ;;; r - e] The center of the ball of minimal radius containing S is called the circumcenter of S. Let X be a metric space. By an isometry of X we mean a bijection 9: X->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<:ductifs sur un corps local I," Publications IHES, 41 (1972), pp. 5-25).] Thus the theorem was embedded and applied in fancy mathematics. Serre formulated and proved what we gave here as Exercise 6, the existence of the circumcenter. An exposition of Serre's remark was made by K. Brown, Buildings, Springer-Verlag, 1989, Chapter VI, §5, Theorem 2. Then completely elementary, beautiful, and powerful results could be extracted from the fancy context, as I have done here in Exercises 5, 6, and 7. Note how the exercises themselves are totally elementary, but even a discussion of their history is not. That's life, but one should not let the deep history hide or obstruct those results, accessible at the present level.

VI, §5. OPEN AND CLOSED SETS Let S be a subset of a normed vector space E. We shall say that S is open (in E) if given v E S there exists I' > 0 such that the open ball of radius I' centered at v is contained in S. Example 1. An open ball is an open sel. Indeed, let B be the open ball of radius I' > 0 centered at some point VEE. Given wEB, we have Iw-vl0 such that s+o
Iz - vi

~

Iz - wi + Iw - vi

~

0 + s < r.

Picture:

B

r

We emphasize that our notion of open set is relative to the given normed vector space in which the set lies. For instance, if we view R as a subspace of R k (consisting of all vectors whose i-th coordinate is 0 for i > I), then R is open in itself, but of course is not open in Rk •

152

NORMED VECTOR SPACES

[VI, §5]

Remark. IJ a set is open with respect to the given norm, it is also open with respect to any equivalent norm. This remark is immediate from the definitions. Prove it in detail as an exercise. For example, let E = Rk and let v E Rk• Consider the two norms II II and I I equal to the sup norm and the euclidean norm respectively. Then any open ball in one norm contains an open ball in the other norm, centered at the same point. The balls for the sup norm are squares for the euclidean norm.

We used open balls to define open sets. Note that a set S is open if and only ifJor each point x oj S there exists an open set V such that x E V and V is contained in S. Indeed, this condition is certainly satisfied if S is open, taking V to be the prescribed open ball. However, conversely, if this condition is satisfied, we can find an open ball B centered at x and contained in V, and then B c: V c: S, so that S is open. If x is a point of E, we define an open neighborhood of x to be any open set containing x.

Let V, V be open sets in E. Then V n V is open. Proof Given v E V n V, there exists an open ball B of radius r centered at v contained in V, and there exists an open ball B' of radius r' centered at v contained in V. Let 0 = min(r, r'). Then the open ball of radius 0 centered at v is contained in V n V, which is therefore open. By induction, it follows that if VI> . .. ,V n are open, then V, n ... n V n is open. Thus the intersection oj a finite number oj open sets is open. However, the intersection of an infinite number of open sets may not be open. For instance, let V n be the open interval -lin < x < lin in R. The intersection of all V n (n = I, 2, ...) is just the origin 0, and is not open in R. Note that our definition of open set is such that the empty set is open. Furthermore, the whole space E itself is open. Let I be some set, and suppose given Jor each i E I an open set Vi. Let V be the union oj the VI' that is the set oj all x such that x E Vi Jor some i.

[VI, §5]

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OPEN AND CLOSED SETS

Then U is open, because given x e U, we know that x e Ui for some i, so there exists an open ball B centered at x such that x e B CUi C U, whence U is open.

Example 2. Let S be an arbitrary subset of E, and for each xeS, let Bx be the open ball of radius 1. The union of all balls Bx for all xeS is open. Example 3. Let E = R, and let S be the set of integers n ;;; 1. For each n, let B. be the open interval centered at n of radius lin. The union of all B. for all n 2: I is an open set which looks like this:

o

) (

2

I

3

)

4

We define a closed set in a normed vector space E to be the complement of an open set. Thus a set S is closed if and only if given a point ye E, y ¢ S, there exists an open ball centered at y which does not intersect S. Let S be a subset of E. Let veE. We say that v is adherent to S if given £ there exists an element xeS such that Ix - vi < £. This means that the open ball of radius £ centered at v must contain some element of S for every £. In particular, if v e S, then v is adherent to S. We observe that every adherent point to S is the limit of a sequence in S. Indeed, if v is adherent to S, given n we can find x. e S such that Ix. - vi < lin, and the sequence {x.} converges to v. Given £, find N such that liN < £. If n ;;; N then Ix. - vi < lin ~ liN < £, so v is the limit of {x.}. Conversely, if veE is the limit of a sequence {x.} with x. e S for all n, then v is adherent to S, as follows at once from the definition. Theorem 5.1. Let S be a subset of a normed vector space E. Then S is closed if and only if S contains all its adherent points. Proof. Assume that S is closed. If v is an adherent point, then any open ball centered at v must contain some element of S by definition, and hence v cannot be in the complement of S. Hence v lies in S. Conversely, assume that S contains all its adherent points. Let y be in the complement of S. Then y is not adherent to S, and so there exists some open ball centered at y whose intersection with S is empty. Hence the complement of S is open, thereby proving the theorem.

Corollary 5.2. The set S is closed if and only if the following condition is satisfied. Every sequence {x.} of elements of S which converges in E has its limit in S.

154

NORMED VECTOR SPACES

[VI, §5]

Proof If a sequence of elements of S converges to an element vEE, then v is adherent to S. If S is closed, then v E S. Conversely, assume that every sequence in S which converges in E has its limit in S. Let v be an adherent point of S. Given n, there exists x. E S such that Ix. - vi < l/n. The sequence {x.} converges to v, and by hypothesis, v E S, hence S is closed.

Remark. In the Corollary, we are not asserting that every sequence in S has a limit in S. We are merely asserting that if a sequence in S has a limit in E, then that limit must be in S. We are not even asserting that every sequence in S has a convergent subsequence. For instance, let S be the set of positive integers n in R, that is S = Z +. Then S is closed in R, but no subsequence of S has a limit. Example 4. A closed interval is closed in R. The set of numbers consisting of l/n (all positive integers n) and 0 is closed in R. However, the set of all numbers l/n (for all positive integers n) is not closed in R. The number 0 lies in the complement, but any open ball centered at 0 contains some number l/n. Proposition 5.3. Let S, T be closed sets in E. Then S u T is closed. This can be verified directly, or better, follows formally from the analogous statement for open sets. Indeed, let us denote by 'CS = 'C£S the complement of S in E, that is the set of all x E E such that x ¢ S. Then 'C(S u T) = 'CS r. 'CT, and hence the complement of S u T is open, so that S u T is closed. By induction, a finite union of closed sets is closed.

In a similar way, one can prove that an infinite intersection of closed sets is closed, because an infinite union of open sets is open. If I is SOUle set, and for each i we have associated a closed set S" then the complement of the intersection

is the union of the sets

U 'CS"

and is open. Hence this intersection is

iel

closed. For example, let S 1 => S2 => ... => S. => . •. be a sequence of closed sets such that S. => Sn+ l' Then the intersection is closed. Note that this intersection may be empty. For instance, taking E = R, let Sn be the set of all numbers x such that x ~ n. Then Sn is closed, and the intersection of all Sn is empty.

[VI, §5]

OPEN AND CLOSED SETS

ISS

Theorem 5.4. Let E, F be normed vector spaces, and let E x F have the sup norm. Let U be open in E and V open in F. Then U x V is open in E x F. If S is closed in E and T is closed in F, chen S x T is closed in

Ex F. Proof. Let u e U and v e V. There exists an open ball B in E centered at u and contained in U, and there exists an open ball B' in F centered at v and contained in V. Let r be the minimum of the radii of Band B' and let B" B~ be the open balls of radius r in E and F respectively, centered at u and v respectively. By definition of the sup norm, B, x B;. is then the open ball of radius r centered at (u, v) in E x F, and is contained in U x V, thus showing that U x V is open. Now for the statement about closed sets, let (x, y) be in the complement of S x T. Thus x If S or y If T. Say x If S. There exists an open set W in E containing x whose intersection with S is empty. Let W' be an open set in F containing y. Then W x W' is an open set in E x F, whose intersection with S x T is empty. Hence S x T is closed. This proves the theorem. ( \

By i'nduction, the theorem extends to a finite number of factors. In particular: Corollary 5.5. Let S" ... ,S" be closed sets in R. Then

is closed in R".

For instance, if S; is the interval [0, I], then

is a closed n-cube in R". (For n = 2, it is the closed square, and for n = 3 it is what we would ordinarily call the closed cube.) There is a fairly large number of basic statements about closed sets whose proofs will be left as (simple) exercises. However, we state here some basic definitions, giving rise to these statements. Let S be a subset of a normed vector space E, and let S denote the set of all points in E which are adherent to S. We call S the closure of S. In Exercise I, you will prove that S is closed, and thus deserves its name. If S is a subset of T and SeT c S, then we say that S is dense in T. Example 4. You can easily verify the following assertions, in the simplest case when E = R or R".

156

NORMED VECTOR SPACES

[VI, §5]

(a) Let a < b be real numbers. The interval [a, b] is closed, and is the closure of the open interval (a, b). It is also the closure of the half open intervals [a, b) or (a, b]. (b) The infinite interval (a, (0) consisting of all x > 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
[VI, §5]

157

OPEN AND CLOSED SETS

Proof· Assume that f is continuous. Let V be open in F and let XES be such that f(x) E V. Put y = f(x). From the definition of being open, there exists € > 0 such that B,(y) c V. By the definition of continuity, there exists 0 > 0 such that

if v E S and Iv - xl < 0, then If(v) - yl <

€,

so f(v) E B,(y). Hence Bb) n S c r1(V), so r1(V) is open in S. Conversely, assume that the inverse image under f of an open set in F is open in S. Let XES and f(x) = y. Given €, the inverse image r1(B,(y» is open in S, so there exists 0> 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 <

if the limit exists, it

there exists /5 such that when-

f.

This applies a fortiori when xeS' so that w is also a limit for xeS'. If w' is also a limit, there exists /5 1 such that whenever xeS and Ix - vi < /5 1 then If(x) - w'l <

f.

If Ix - vi < min(/5, /51) and xeS, then Iw - wi ;;;; Iw - f(x)1

+ If(x) - w'l < 2f.

This holds for every f, and hence Iw - w'l W = w', as was to be shown. If f is a constant map, that is f(x) =

Wo

= 0,

W -

w'

= 0,

and

for all xeS, then

lim f(x) = wo' Indeed, given f, for any /5 we have If(x) If v e S, and if the limit

Wo

I=0 <

f.

limf(x)

exists, then it is equal to f(v). Indeed, for any /5, we have Iv - vi < /5, whence if w is the limit, we must have If(v) - wi < dor all f. This implies that f(v) = w. We define an element v of S to be 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. If v is isolated, then limf(x)

exists and is equal to f (v).

162

[VII, §1]

LIMITS

The rules for limits of sums, products and composite maps apply as before. We shall list them again with their proofs. Limit of a sum. Let S be a subset of a normed vector space and let v be adherent to S. Let f, 9 be maps of S into some normed vector space. Assume that Iimf(x) Then lim(f

=w

and

lim g(x)

= w'.

+ gXx) exists and is equal to w + w'.

Proof. Given have

£,

there exists

(j

such that if

If(x) -

wi <

£,

Ig(x) -

w'l <

£.

XES

and Ix - v I <

(j

we

Then If(x)

+ g(x) - w- w'l

This proves that w

~

If(x) -

wi + Ig(x)

-

w'l < 2£.

+ w' is the limit of f(x) + g(x) as x -+ v.

Limit of a product. We do not have a product as part of the given structure of a normed vector space. However, we may well have such products defined, for instance the scalar products. Thus we discuss possible products to which the limit theorem will apply. Let E, F, G be normed vector spaces. By a product of E x F -+ G we shall mean a map E x F -+ G denoted by (u, v) 1-+ uv, satisfying the following conditions: PR 1. If u, u' E E and v E F, then (u u(v + v') = uv + uv'.

+ u')v =

PR 2. If c E R, then (cu)v

= c(uv) = u(cv).

PR 3. For aU u, v we have

luvl

~

uv

+ u'v.

If v, v'

E

F, then

lullvl.

Example 1. The scalar product of vectors in n-space Condition PR 3 is nothing but the Schwarz inequality!

IS

a product.

Example 2. Let S be a non-empty set, and let E = t4(S, R) be the vector space of bounded functions on S, with the sup norm. Iff is bounded

[VII, §1]

163

BASIC PROPERTIES

and 9 is bounded, then one sees at once that the ordinary product f 9 is also bounded. In fact, let C I = IIfll and C 2 = IIgll, so that If(x)1 ~ C I and Ig(x)l :s; C 2 for all XES. Then If(x)g(x)1 < C I C 2

for all XES, whence IIfgll ~ 1If11 product are obviously satisfied.

= IIfllllgll

IIgli. The first two conditions of the

Example 3. If readers know about the cross product of vectors in R 3 , they can verify that it is a product of R 3 x R 3 -+ R 3 • Example 4. View the complex numbers C as a vector space over R. Then the product of complex numbers is a product satisfying our three conditions. The norm is simply the absolute value. Suppose given a product E x F -+ G. Let S be some set, and let f: S -+ E and g: S -+ F be mappings. Then we can form the product mapping by defining (fg)(x) = f(x)g(x). Note that f(x) E E and g(x) E F, so we can form the product f(x)g(x). We now formulate the rules for limits of products. Let S be a subset of some normed vector space, and let v be adherent to S. Let E x F -+ G be a product, as above. Let

f: S -+ E

and

g: S-+ F

be maps of S into E and F respectively. If limf(x)

=w

and

lim g(x) = z,

then lim f(x)g(x) exists, and is equal to wz. Proof. Given

£,

there exists () such that whenever Ix 1

£

1

£

If(x) -

wi < 21z1 + l'

Ig(x) -

zl < 21wl + l'

If(x) I < Iwl + 1.

vi < () we have

164

[VII, §l]

LIMITS

Indeed, each one of these inequalities holds for x sufficiently close to v, so they hold simultaneously for x sufficiently close to v. We have If(x)g(x) -

wzl

= If(x)g(x) - f(x)z ~

If(x)(g(x) -

~

If(x)llg(x)

+ f(x)z

z)1 + I(f(x) - zl + If(x) -

wzl w)zl wllzl

-

thus proving our assertion. The reader should have become convinced by now that it is no harder to work with normed vector spaces than with the real numbers. We have the corollaries as for limits of functions. If c is a number, then

lim cf(x) x-v

and

= c limf(x); X-I)

if f,J2 are maps of the same set S into a normed vector space, then Iim(J.{x) - fix)) = lim f,(x) - limf2(x).

We also have the result in case one limit is O. We keep the notation of the product E x F -> G, and again let f: S -> E and g: S -> F be maps. We assume thatfis bounded, and that lim g(x) = O. Then lim f(x)g(x) exists and is equal to O. x-v Proof· Let K > 0 be such that I f(x)1 ~ K for all XES. Given £, there exists /j such that whenever Ix - vi :::: /j we have Ig(x)1 < €/K. Then If(x)g(x)1 ~ If(x)llg(x)1 < K€/K = E, thus proving our assertion.

Limit of a composite map. Let S, T be subsets of normed vector spaces. Letf: S -> T and g: T -> F be maps. Let v be adherent to S. Assume that Iimf(x) exists alld is equal to w. Assume that w is adherent to T. Assume that

lim g(y)

[VII, §I]

165

BASIC PROPER TIES

exists and is equal to z. Then

lim g(f(x» exists and is equal to z. Proof. Given €, there exists 0 such that whenever yET and Iy - wi < 0 then Ig(y) - zl < €. With the above 0 being given, there exists 0 1 such that whenever XES and Ix - vi < 01 then If(x) - wi < 0. Hence for such x, Ig(f(x») -

zl <

€,

as was to be shown. Limit of a quotient. Let f: S -+ R or f: S -+ C be a function. Let v be adherent to S. Assume that lim fIx) exists, and is equal to z # O. Then lim Ilf(x) exists and is equal to liz. Proof The function WI-+ w- I is continuous on the set of real or complex numbers # O. For all x in S sufficiently close to v, the value fIx) is close to z, and hence # O. The function Xl-+ Ilf(x) for XES is the composite of f and the inverse WI-+ W-I. Therefore the desired limit follows from the limit for the composite map proved previously.

Limits of inequalities. Let S be a subset of a normed vector space, and let f: S -+ R, g: S -+ R be functions defined on S. Let v be adherent to S. Assume that the limits limf(x)

lim g(x)

and

exist. Assume that f(x) ;;; g(x) for all x sufficiently close to v in S. Then

lim f(x) ;;; lim g(x). x ..... v

X-ll

Proof. Let q:>(x) = g(x) - f(x). Then q:>(x) ~ 0 for all x sufficiently close to v, and by linearity it will suffice to prove that lim q:>(x) ~ O. Let x-v

y = lim
€,

we can find

XES

such that Iq:>(x) -

yl <

€.

But

166

yl. Hence

",(x) - y ~ I",(x) -


for every

[VII, §l]

LIMITS

£.

y > ",(x) - £ > -£

and

£

This implies that y > 0, as desired.

As before, we have a second property concerning limits of inequalities which guarantees the existence of a limit. Let S be a subset ofa normed vector space, and let f, g be as in the preceding assertion. Assume in addition that w

= limf(x) = lim g(x). x-u

x-v

Let h: S -+ R be another function such that f(x)

~

hex)

~

g(x)

for all x sufficiently close to v. Then lim hex) exists, and is equal to the limit of f (or g) as x -+ v.

x-.

Proof. Given

£,

there exists /j such that whenever

Ig(x) -

wi <

and

£

If(x) -

Ix - vi < /j we have wi <

£,

and consequently

o < g(x) -

f(x) ~ If(x) -

wi + Ig(x)

-

wi < 2£.

But

Iw - h(x)1

~

Iw - g(x) I + Ig(x) - h(x)1

<

£

+ g(x)

- f(x)

<£+2£=3£,

as was to be shown. We have limits with infinity only when dealing with real numbers, as in the following context. Let S be a set of numbers, containing arbitrarily large numbers. Let f: S -+ F be a map of S into a normed vector space. We say that Jim f(x)

[VII, §I]

167

BASIC PROPERTIES

exists if there exists w E F such that given E, there exists B > 0 such that for all XES, X ~ B we have If(x) - wi < £. This generalizes the notion of limit of a sequence, the set S being then taken as the set of positive integers. The theorems concerning limits of sums and products apply to the limits as x ... 00, replacing the condition "there exists fJ such that for Ix - v I < fJ" by "there exists B such that for x

~

B"

everywhere. The proofs are otherwise the same, and need not be repeated. We summarize systematically the limits which occur: lim,

lim,

lim.

Any statement of the usual type for one of these can be formulated for the others and proved in a similar way. For this purpose, one occasionally needs the condition for completeness formulated in terms of mappings rather than sequences as follows.

Theorem 1.2. Let S be a subset of a normed vector space E. Let f: S ... F be a map of S into a normed vector space F, and assume that F is complete. Let v be adherent to S. The following conditions are equivalent: (a)

The limit lim f(x) exists.

(b)

Cauchy criterion. Given E, there exists fJ such that whenever x, yES and

Ix - vi < fJ, (c)

Iy -

vi < fJ,

then If(x) - f(y)1 < E. For every sequence {xn } in S converging to v, the limit lim f(x n ) exists.

Proof. Assume (a). We deduce (b) by a 2E-argument. If w is the limit in (a), then If(x) - f(y)J ~ If(x) -

wi + Iw -

f(y)l,

so we get (b) at once. Next assume (b). Let {xn } be a sequence in S which converges to v. Then {f(xn )} is a Cauchy sequence in F. Indeed, given E, let fJ be as in

168

[VII, §l]

LIMITS

(b). Then for all m, n sufficiently large, we have

IXn - vi < b

and

Hence If(x",) - f(xnll < €, so the sequence {f(xn)} is Cauchy. Since F is complete, the sequence {f(xn )} has a limit, thereby proving (c). Assume (c). There is a sequence {x n } in S converging to v, and by assumption, {f(xn )} converges to an element w in F. We need to prove that w is the limit of fix) as x --+ v. If not, there exists € > 0 such that for all positive integers n there is an element Yn E S satisfying

vi < lin

IYn -

but

If(Yn) - wi ;;;;

€.

Let w' = lim f(Yn) which exists by assumption. Then we conclude that Iw' - wi;;;; €. However, the sequence ( ,xn, Yn .... ) converges to v, so by assumption the sequence ( ... ,f(xn),f(Yn) ) converges, both to wand to w', so w = w'. a contradiction which proves (a). This concludes the proof of Theorem 1.2. See Theorem 2.1 for the equivalence between (a) and (c) in the context of continuity. We conclude this section with comments on the dependence of limits on the given norm. Let E be a vector space, and let I I.. I b be norms on E. Assume that these norms are equivalent. Let S be a subset of E, and f: S --+ F a map of S into some normed vector space F. Let v be adherent to S. We can then define the limit with respect to each norm. We contend that if the limit exists with respect to one norm, then it exists with respect to the other and the limits are equal.

Suppose the limit off(x) as x --+ v, XES, exists with respect to let this limit be w. Let C, > 0 be such thaI

for all then

uE E.

Given

€,

there exists b, such that if XES and

If(x) - wi <

Let b

= b,/C,.

Suppose that XES and

€.

Ix -

Ix - vi, < b"

vl2

< b. Then

I I" and

Ix - vi, < b,

[VII, §1]

169

BASiC PROPERTIES

and consequently we also have I f(x) - wi < £. This proves that w is also the limit with respect to I ". Similarly, if we have two equivalent norms on F, we see that the limit is also the same whether taken with respect to one of these norms or the other. Finally, we note that there exist some associations satisfying PR 1 and PR 2, and only a modified version of PR 3, namely: PR 3C. There exists a number C > 0 such that for all u, v we have

luvl

~

C1ullvl.

The study of such a generalized product can be reduced to the other one by defining a new norm on E, namely

Ixl.

=

C1xl·

Condition PR 3 is then satisfied for this new norm. Thus all the statements involving limits of products apply. Actually, in practice, PR 3 is satisfied by most of the natural norms one puts on vector spaces, and the natural products one takes of them.

VII, §1. EXERCISES 1. A subset S of a normed vector space E is said to be convex if given x, yES the

points (1 - I)x

+ Iy,

o~ I ~

I,

are contained in S. Show that the closure of a convex set is convex. 2. Let S be a set of numbers containing arbitrarily large numbers (that is, given an integer N > 0, there exists XES such that x ~ N). Let f; S .... R be a function. Prove that the following conditions are equivalent: (a) Given £, there exists N such that whenever x, yES and x, y ~ N then If(x) - f(y) I <

£.

(b) The limit lim !(x)

exists.

(Your argument should be such that it applies as well to a map f; S .... F of S into a complele normed vector space.)

170

[VII, §2]

LIMITS

Exercise 2 is applied most often in dealing with improper integrals, letting

S'"9 = o

lim B-co

S89 . 0

3. Let F be a normed vector space. Let E be a vector space (not normed yet) and let L: E -+ F be a linear map, that is satisfying L(x + y) = L(x) + L(y) and L(cx) = cL(x) for all c E R, x, Y E E. Assume that L is injective. For each x E E, define Ixl = IL(x)l. Show that the function X 1-+ Ixl is a norm on E. 4. Let P s be the vector space of polynomial functions of degree

~5

on the interval [0, I]. Show that Psis closed in the space of all bounded functions on [0, I] with the sup norm. [Hillt: Iff(x) = asx s + ... + ao is a polynomial, associate to it the point (as •... ,ao) in R6 , and compare the sup norm on functions with the norm on R6 .]

5. Let E be a complete normed vector space and let F be a subspace. Show that the closure of F in E is a subspace. Show that this closure is complete. 6. Let E be a normed vector space and F a subspace. Assume that F is dense in E and that every Cauchy sequence in F has a limit in E. Prove that E is complete.

VII, §2. CONTINUOUS MAPS Let S be a subset of a normed vector space, and let f: S -+ F be a map of S into a normed vector space. Let v E S. We shall say that f is continuous at v if

limf(x) exists, and consequently is equal to f(v). Put another way,j is continuous at v if and only if given E, there exists b such that whenever Ix - v I < b we have

If(x) - f(v)1 <

E.

Here, as often in practice, we omit the XES when the context makes it clear. Let So be a subset of S. We say that f is continuous on So (or relatively continuous on So) if f is continuous at every element of So. For convenience, we now take S = So' We emphasize that in the definition of continuity, we definitely take v E S. In §1, when investigating properties of limits, we took v adherent to S, but not necessarily in S. From the properties of limits, we get analogous properties for con. tinuous maps.

[VII, §2]

171

CONTINUOUS MAPS

Sum. If f, g: S - F are continuous at v, then f

+ g is continuous at v.

Product. Let E, F, G be normed vector spaces, and E x F _ G a product. Let f: S - E and g: S - F be continuous at v. Then the product map f g is continuous at v. Composite maps. Let S, T be subsets of normed vector spaces, and let F be a normed vector space. Let f:S-T

and

g: T-F

be maps. Let v E Sand w = f(v). Assume that f is continuous at v and g is continuous at w. Then g 0 f is continuous at v.

Quotient. Let f: S - R or f: S - C be continuous at v, and f(v) l' O. Then Ilf is continuous at v. All these follow from the corresponding property of limits. For the composite, one can also repeat the proof of Theorem 4.2, Chapter II. Iffis a continuous map and c is a number, then cf is a continuous map. Thus the set of continuous maps of S into a normed vector space F is itself a vector space, which will be denoted by CO(S, F). One can characterize continuity entirely by means of limits of certain sequences, as in Theorem 1.2(c). Theorem 2.1. Let S be a subset ofa normed vector space, and let f: S - F be a map of S into a normed vector space F. Let v E S. The map f is continuous at v if and only if,for every sequence {xn } of elements of S which converges to v, we have lim f(x n) = f(v). Proof. Assume that f is continuous at v. Given E, there exists {) such that if Ix - vi < {) then If(x) - f(v)1 < E. With this {) given, there exists N such that for n > N we have IX n - v I < {), and hence If(x n ) - f(v)1 < E, thus proving the desired limit. Conversely, assume that the limit stated in the theorem holds for every sequence {xn } in S converging to v. It will suffice to prove: Given E, there exists N such that whenever Ix - vi < liN then If(x) - f(v)1 < E. Suppose this is false. Then for some E and for every positive integer n there exists Xn E S such that IXn - vi < lin but If(x n) - f(v)1 > E. The sequence {xn } converges to v, and we have a contradiction which proves ihat f must be continuous at v.

Again let S be a subset of a normed vector space, and let v be adherent to S. Suppose that v is not in S. Let f: S _ F be a map of S as before, and

172

[VII, §2]

LIMITS

assume that limf(x)

= w.

If we define fat v to be f(v) = w, then it follows from the definition of continuity that we have extended our map f to a continuous map on the set Su {v}. We say that f has been extended by continuity. In a moment, we shall define the notion of uniform continuity. In Exercises 7 and 8 you will see how a uniformly continuous map on S can always be extended by continuity to the closure of S. Exercises 10 and 12 show how this extension is possible for continuous linear maps. The situation with continuous linear maps is especially important. See Theorem 1.2 of Chapter X, where the matter will be treated from scratch. Equivalent norms. We make the same remark with respect to continuity that we made previously with respect to limits corresponding to equivalent norms. Let I II and I b be equivalent norms on a vector space E. Let S be a subset of E and f: S --+ F a map of S into a normed vector space. Let vEE, v adherent to S. If f is continuous at v with respect to I II, then f is also continuous at v with respect to lb. This comes from the fact that equivalent norms give rise to the same limits whenever the limits exist. The next theorem deals with maps into a product space. If Fl>' .. ,F. are normed vector spaces, we can form F = FIx ... X F. with the sup norm. A map f: S --+ F is given by coordinate mappings fl' ... ,f. such that f(x) = (f1(X), ... ,J.(x», and Ii maps S into F i • We shall deal especially with the case when F = R' and Ii are called the coordinate functions off. Theorem 2.2. Let S be a subset ofa normed vector space. Let f: S --+ F = FIx ...

X

F.

be a map of S into a product of normed vector spaces, and let

be its representation in terms of coordinate mappings. Let v be adherent to S. Then limf(x)

[VII, §2] exists

173

CONTINUOUS MAPS

if and only if lim};(x)

exists for each i = 1, ... ,k. If that is the case and = (WI>'" ,w.) with W; e F h then W; = lim!.{x).

W

is the limit off(x),

W

Proof· This theorem is essentially obvious. We nevertheless give the proof in detail. Suppose

= W = (wl>'"

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 <

E.

... ,y.). By definition, !y; -

Ix -

wd < E whenever

vi < 0,

so that Wi =

Conversely, if

Wi =

lim };(x).

limf,(x), for all i = 1, ... ,k, then given

0; such that whenever Ix -

vi <

E

there exists

0; we have

If.{x) -

wd <

E.

Let 0 = min 0;. By definition, when Ix - vi < 0 each 1J.(x) - wd < E for i = 1, ... ,k and hence If(x) - wi < f, so w is the limit off(x) as x -> v. Our theorem is proved. Coronary 2.3. The map f is continuous if and only J. is continuous, i = 1, ... ,k.

if each coordinate map

Proof. Clear.

Uniform continuity Suppose f is continuous on S. The 0 occurring in the definition of continuity depends on v. That is, for each ve S, there exists o(v) such that

174

[VII, §2]

LIMITS

if Ix - vi < b(V) then If(x) - f(v)1 < E. When one can select this b independently of v, then the map f is called uniformly continuous. Thus f is defined to be uniformly continuous if given E, there exists b such that whenever x, yES and Ix - yl < b then If(x) - f(y)1 < E. The principal criterion for uniform continuity is given in the chapter on compactness. Here we give some examples.

Example 1. Let f(x) = I/x. Then f is continuous on R+ (the set of real numbers > 0), but is not uniformly continuous. You can see this by looking at values f(x) and flY) for x, y near O. Given E, numbers x and y > 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 < <. Then Ix - vol < 0, and so

Proof Fix

Vo E

I!(v) - !(vo) 1 ~ I!(v) - /(x)l

+ I/(x) -

/(vo) 1 < 2<, qed.

In Exercise 7, the set S' is contained in S essentially by definition, but is not necessarily equal to S. The next exercise gives a condition under which S' = S. 8. Prove Theorem 2.6. IProo! By the preceding exercise, it suffices to prove that for every v E S, the limit lim [(x) (x E S, X --> v) exists. By basic properties of limits, it suffices to prove that given a sequence {x,,) in S converging to v, then lim [(x,,) exists. By unifonn continuity of [, given e there exists 0 such that for :'::".l' E S, Ix - .1'1 < 0 we have I[(x) - [(.I'll < e. There exists N such that for II ;;; N we have Ix" - vi < 0/2. Hence for Ill, II ;;; N, we have Ix" - x",1 < 0, so I[(x,,) - [(x",)1 < e. Hence the sequence {[(x,,)} is Cauchy. Sincc F is complete, lim [(x,,) exists, thus concluding the proof. 9. Let S, T be closed subsets of a nonned vector space, and let A = S v T. Let [: A --> 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 <

E,

I I nx - Nc

=-:S;-<E.

thus proving that {fn} converges uniformly to 0 on T.

Example 2. Let {In} be the sequence of functions whose graph is shown below:

Each function I. has a peak forming a triangle, and its values are equal to 0 for x ;;; Cn, where {cn} is a sequence of numbers decreasing to O. Then for each x > 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 <

£.

Then If(x) -

I.(x)1

~ If(x) -

<

£

fm(xll +

+ IIfm -

Ifm(x) - I.(x) 1

fnll

< 2£. This shows that IIf - I.II < 2£, whence {I.} converges to f unifonnly. If in addition the functions I. are all bounded, then

If(x)l ~ 2< + IfN(X)i ~ 2< + IIfNIIThis shows that f is bounded, and proves that @(S, F) is complete, also completing the proof of the theorem. Theorem 3.2. Let S be a non-empty subset of a normed vector space. Let (I.} be a sequence of continuous maps of S into a normed vector space F, and assume that {fn} converges uniformly to a map f: S -+ F. Then f is continuous. Proof. Let v E S. Select n so large that IIf - I.II < £. For this choice of n, using the continuity of I. at v, select f> such that whenever Ix - vi < f> 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 <

£.

The

If(x) - f(v)1 < 3£,

and our theorem is proved. From Theorem 3.2, we can conclude at once that the convergence of the functions in Example 3 above cannot be uniform, because the limit function is not continuous. Corollary 3.3. Let S be a non-empty subset of a normed vector space, and let BCo(S, F) be the vector space of bounded continuous maps of S into a complete normed vector space F. Then BCo(S, F) is complete (for the sup norm).

[VII, §3]

LIMITS IN FUNCTION SPACES

183

Proof. Any Cauchy sequence in BCo(S, F) has a limit in £J(S, F), by Theorem 3.1, and this limit is continuous by Theorem 3.2. This proves the Corollary.

Corollary 3.4. The space of bounded continuous maps BCo(S, F) is closed in the space of bounded maps (for the sup norm). Proof. Any adherent point to the space BCO(S, F) is a limit of a sequence of bounded continuous maps, and Theorem 3.2 shows that this limit is in the space. As usual, any notion involving n -+ 00 has analogues for other types of limits. Before we formulate the analogues, we make some general remarks about the way notation concerning sequences indexed by positive integers has a parallel notation in the case of a variable which may range over a set which is not the positive integers. So we formalize the notion of a family of objects, and a family of mappings. Let X be a set, and let T be a set. By a family of elements of X indexed by T we simply mean a mapping f: T -+ X. Thus to each element t e T we have associated an element f(t) e X. We may think of (j(t») as parametrized by the set T, and write j, instead of f(t). Let S be a set and let X = Maps(S, F) where F is a normed vector space. We shall be concerned with families of maps (j,) indexed by T, with each j,: S -+ F being a map of S into F. From such a family, we obtain a mapping f:TxS-+F

defined by

f(t, x) = j,(x).

Conversely, given a mapping depending on two variables f: T x S -+ F, for each t e T we can define a map J,: S -+ F by letting j,(x) = f(t, x). Now let S be a subset of a normed vector space E and let T be any set. Let f: T x S -+ F be a map into a normed vector space F. We view f as depending on two variables t e T and xeS. Let v be adherent to S. Assume that for each t e T the limit lim f(t, x) exists. If v e S then this limit necessarily is equal to f(t, v). If v ¢ S, then we define f(t, v) = j,(v) = lim j,(x). x-v

Then the map j, such that x 1-+ j,(x) = f(t, x) may be viewed as defined on the union of S and its boundary point v. When the above limit exists, we are in a situation already considered when each j, has been extended by continuity to the boundary point v.

184

[VII, §3]

LIMITS

On the other hand, suppose that T is a subset of a normed vector space, and that w is adherent to T. We may consider the family of mappings {!.},e T' Suppose that the limit lim f(t, x)

= lim !.(x)

exists for each xeS. Then we may also define the limit value f(w, x) = fw(x) = lim !.(x) = lim f(t, x).

We shall say that the limit lim f(t, x) exists uniformly for xeS if given"

,-w

there exists li such that whenever It - wi < li, then If(t, x) - f(w, x)1 < "

for all

xeS.

Equivalently, viewing {!.},eT as a family of maps from S into F, i.e. a map t ...... !. from T into Maps(S, F), then lim!. =fw, the limit being taken with respect to the sup norm. In other words, given " there exists li such that whenever It - wi < li, then II!. - fwll < ". A theorem like Theorem 3.2 can be formulated formally in terms of limits as follows. Letting

f =

lim f.,

we want to prove that f(v)

.

= ~i~ f(x) = ~i~ Cli_~ fn(X»).

If we could interchange the limits, then

and since each fn is continuous, the right-hand side is equal to lim f.(v).

[VII, §3]

185

LIMITS IN FUNCTION SPACES

On the other hand, the sequence {f,,} is assumed to converge uniformly, and in particular pointwise. Thus lim f,,(v) = f(v). The whole problem is therefore to interchange the limits. The argument given in Theorem 3.2, namely the splitting argument with 3£, is standard for this purpose, and uses the uniform convergence. An interchange of limits is in general not valid. Example: Let

be defined on the set S of all (x, y) such that x 'f- 0, y ¢ O. Then lim fIx, y) = - I

.-0

and hence

lim lim fIx, y) = - 1. y... O x-o

On the other hand, we see similarly that lim lim f(x, y) = 1.

x ..... o

},-o

A similar example with infinity can be cooked up, if we let

m-n f(m,n) = - m+n be defined for positive integers m, n. Then lim lim f(m, n) = -I,

while

lim lim f(m, n) = 1.

Theorem 3.5. Let Sand T be subsets of normed vector spaces. Let f be a map defined on T x S, having values in some complete normed vector space. Let v be adherellt to Sand w adherent to T. Assume that: (i)

.-v

lim fIt, x) exists for each

t E

T.

(ii) lim fIt, x) exists uniformly for

,-w

XES.

Then the limits lim lim f(t,x), (_I"

x.....v

all exist and are equal.

lim lim fIt, x),

,\"-V 1-11'

lim

«(.xl-(.... v)

fIt, x),

186

[VII, §3)

LIMITS

Prol!f. First by (i) and (ii), we may define for e E T and XES: 1.(1)) = /(1, v) = lim f(l, x)

f.,.(x) = lim f(I, x).

and

We shall prove lhat lim I.(v) exists by using Cauchy's criterion. For all t

XES

and

I, I' E

'II'

T we have

If,(v) - f,,(v)J ~ If,(v) - f,(x)J

By (ii), given

E,

there exists 0, such that if

Ie - 1\'/ < 15, then for all (I)

+ If,(x) -

XES

I,

f,,(x)J t'

E

+ If,,(x) -

f,,(v)l·

T and

and

we have

1.f.(X) - 1;,.(x)! <

1.f..(X) - .f.,.(x)1 < E,

and

E

and consequently I./;(.x) - .f..(x)1 < 2E.

(2)

By Ii) there exists ,5,(1, (3 )

I')

such that if

I.f.(x) -(,(l')1 < E

XES

and Ix - '" < O2 , then

II.·(x) - I..(I')1 <

and

E.

It follows from (2) and (3) that II.(v) - I..(v)1 < 4E, whence {I.(vl} converges as I-\\" say lim I.(v) = L. This proves our first limit.

,-", To see that f.,.Lx) -> L as x -> v. we note that the inequality 1J;,.(x) -

LI

~

11;,.(x) -

I.(x)l

+ II.(x) -

I.(v)1

+ II.(v) - LI

We first select 03 such that if II - \\'1 < 15" then 1.1;(/') - LI < E. We know from (1) that if Ie - 11"1 < 15" then for ali XES, If,(x) - /.,.(x)1 < e. Let I be such that II - 11'1 < min(ol,o3). Having chosen this I. there exists some 0 such that, by (i), if Ix - vi < 15, then is valid for all

lET.

If,(x) - f,(v)J

< 3e.

This shows that for such x, we have I.f..(x) - LI < 4E. whence

lim 1;,.(x) :-: ..... v

= L.

[VII, §3]

187

LIMITS IN FUNCTION SPACES

Finally, to see that !(t,x) approaches L as (I,X) space, we write

1/(1, x) - LI

~ If,(x) - Iw(x)l

->

(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
p.

p.

log(t 2

+ lJ)'J2 = ! log(t 2 + lJ)

by a polynomial P. This works because the Taylor formula for the log converges uniformly for lJ ~ u ~ 2A -lJ. Then take a sufficiently large number of terms from the Taylor formula for the exponential function, say a polynomial Q, and use Q 0 P to solve your problems. cr. Exercise 7 of Chapter V, §3. 10. Give another proof for the preceding fact, by using the sequence of polynomials {p.}, starting with Po(t) = 0 and Jetting p.+1(t) = p.(t)

Show that {p.} tends to

0

p.W).

uniformly on [0, 1], showing by induction that

o~ 0 whence 0 ~

+ 1(t -

- p.(t) " 20 , 2+ n0

0 - p.(t) ~ 2/n.

VII. §4. COMPLETION OF A NORMED VECTOR SPACE In this book, we deal with concrete normed vector spaces, and especiaUy with the sup norm. As we have seen in §3, the space of bounded maps is complete, and so is the space of continuous maps, under this sup norm.

[VII, §4]

COMPLETION OF A NORMED VECTOR SPACE

189

This suffices for the applications we have in mind. However, it may be useful to the reader to see a general technique to construct a completion, especially since this technique emphasizes the notion of Cauchy sequence, and is also used to construct the real numbers as a "completion" of the rational numbers. Thus we give this construction in the present section. Theorem 4.1. Let F be a subspace of a normed vector space F,. Assume that F is dense in F, , and that every Cauchy sequence in F has a limit in F,. Then F, is complete. Proof Let {x.} be a Cauchy sequence in F,. By hypothesis, for each positive integer n there exists Y. E F such that Ix. - y.1 < lin. Then {Y.} is Cauchy. To prove this, given E there exists N such that if m, n ~ N, then Ix. - xml < E. Then for m, n ~ N, we have

Iy. - Yml

~

Iy. - x.1 + Ix. - xml + IXm- Yml I

I

~-+E+-.

-n Let N, be an integer

~

m

N and

~

liE. Then for m, n ~ N, we find

Iy. - Yml

~

3E,

which proves that {Y.} is Cauchy. By assumption on F, the sequence {Y.} converges to an element v E F,. Pick N2 ~ N, such that for n ~ N2 we have Iy. - vi < E. Then for n ~ N2 we also have

Ix. - vi ::: Ix. - y.1 + Iy. - vi ::: 2E, which proves that {x.} converges to v, and concludes the proof of the theorem. In Theorem 4.1, we are given the space F, with a norm, and the subspace F. However, in some applications we want to construct F" before having a norm on F" and we want to extend the norm by continuity, as in the Exercises of Chapter VII, §2. We can do this in the following general context. Let F be a normed vector space. Let CS(F) be the set of all Cauchy sequences in F. It is immediately verified that CS(F) is itself a vector space, i.e. the sum of two Cauchy sequences is Cauchy, and if {x.} is Cauchy in F, c is a number, then {ex.} is Cauchy. The norms {Ix.1} form a sequence of real numbers ~ 0, and we claim that {lx.1} is a Cauchy sequence in R. This is immediate, because given E, there exists N such

190

LIMITS

that if m, n ;;; N then Ix. - xml <

E,

[VII, §4]

and then by the triangle inequality,

Hence {lx.1} converges to a real number;;; o. Proposition 4.2. Let

~ =

{x.} be a Cauchy sequence in F. Define I~I =

lim Ix.l.

Then this definition defines a seminorm on CS(F). Proof This is immediate from the triangle inequality and the property that the limit of a sum is the sum of the limits, and the limit of a constant times a sequence is the constant times the limit of the sequence. We leave the details to the reader.

Let F' = CS(F) be the space of Cauchy sequences in F with the seminorm of Proposition 4.2. We define a null sequence in F to be a sequence ( = {z.} such that lim z. = 0, or equivalently lim Iz.1 = O. Note that for such a null sequence, we have 1(1 = 0 by definition, and conversely. Again using the fact that the limit of a sum is the sum of the limits, we see that the set of all null sequences NS(F) is a subspace of F', i.e. of CS(F). We have almost solved our basic problem with the space F', except that we have defined a seminorm on F' rather than a norm. However, we do have a natural embedding of F in F', i.e. an injective linear map which is such that the seminorm on F' is equal to the given norm on F. Namely, to each x E F we associate the Cauchy sequence S(x) = (x, x, x, ... ) such that x. = x for all n. The map X 1-+ S(x)

is a linear map. If S(x) = 0, then x = O. Thus x 1-+ S(x) actually gives an injective linear map of F into F'. From the way we defined the seminorm on CS(F) and on F' we see that

Ixl = IS(x)l. Theorem 43. Given a normed vector space F, the space CS(F) = F' is a vector space with a seminorm, containing F in a natural way (actually S(F»), such that the seminorm on F' is equal to the norm on elements of F, and such that F is dense in F' and every Cauchy sequence in F (that is, S(F)) has a limit in F'. The space F' is complete.

[VII, §4]

COMPLETION Of A NORMED VECTOR SPACE

191

Since we defined the notion of being dense, limit, and completeness only for norms, we must add here a word of explanation in the context of semi norms. The notion of limit can be defined with a seminorm. We say that v E F' is a limit of a sequence {x n } in F if given E there exists N such that for II ~ N we have Ix" - vi < E. The only thing to notice here is that the limit of a sequence is not necessarily unique. Then we can define the notion of being dense, namely every element of F' is the limit of a sequence in F. We define completcness similarly for seminorms. The proof of Thcorem 4.3 is then immediate. Namely, let ¢ = {x,,} be a Cauchy sequence in F. Directly from the definition, we see that IS(x,,) - ¢I

=

lim Ix" - xml so If--O) lim IS(x,,) - ¢I

It!-("I)

= O.

Hence S(F) is dense in F' and lim S(x,,) = ¢. 11---%

The proof of Theorem 4.1 applies to the semi norm on F', and therefore

F' is complete with respect to the seminorm. Of course, having only a semi norm has some inconvenience. To obtain a space with a norm, we have to use a standard device of considering only equivalence classes as follows. We define two sequences {x,,}, {y,,} in F to be equivalent if {x" - y,,} is a null sequence. We could also define this equivalence by the property that there exists a null sequence {zn} such that Yn = x n + Zn for all II. Since NS(F) is a subspace, it is immediately verified that the above relation is an equivalence relation, denoted by ¢ ;; " for two sequences ¢, If. Furthermore, the following properties are satisfied.

'I,

If ~J ;; '11 and ~2 ;; '/2, then ~I + ¢2;; + '12' If c is a number and ¢ ;; 'I, then c~ ;; c". (b) Let F, be the set of equivalence classes of Cauchy sequences in F. Then F, is a vector space. (c) Let ~ = {x,,} and 11 = {Yn} be equivalent Cauchy sequences in F. Then I¢I = 1'11. Thus the seminorm is actually defined on the space of equivalence classes, and it is in fact a norm on F,. (a)

The statement that ~;; '1 implies lsi = I'll is again a consequence of the property that the limit of a sum is the sum of the limits. Thus we see that the semi norm on CS(F) = F' depends only on the equivalence class of a Cauchy sequence, which proves the equality I¢I = I'll in (c). The seminorm on F' is actually a norm on F" because if ¢ = {x n } is a sequence in F such that I¢I = 0, then ~ is a null sequence, and so is in the zero equivalence class. The embedding of F in F' actually yields an embedding of F in FI , because if x E F and SIx) is a null sequence, then Ixl < E for all E > 0, so x = O. Define S, (x) to be the equivalence class of S(x). Then the map x ....... S, (x) is an injective linear map of F into F,. Furthermore, from

192

[VII, §4]

LIMITS

property (cl above, we see that Ixl = 181 (xll. Thus norm-preserving injective linear map of F into Fl' usually not make a distinction between an element S,(xl in Fl' The space F1 is called the complelion of F with norm.

we obtain a natural In practice, we shall x E F and its image respect to the given

Remark. The real numbers could be defined from the rational numbers in a manner similar to the above, as equivalence classes of Cauchy sequences of rational numbers. The elementary school definition of a real number as an infinite decimal merely represents a real number as the limit of a Cauchy sequence from the partial decimals n

a+

L a,/Hi, k=1

where a is an integer. However, there are other representations of a real number as a limit of a sequence of rational numbers, for instance by trecimal or binary expansions. The real number is the equivalence class of all such Cauchy sequences which have the same limit. For example, one can express the rational number 2/3 as the decimal .66666···. Both ways are merely different representations of the same number. The number I has the decimal expansion .99999", and also a trecimal expansion, that is

" 9/1(f 1 = lim L " ..... CO .Ie=1

and

I = lim

L"

2/3'.

"-CO .Ie=1

We view I as the equivalence class of all its representations by convergent Cauchy sequences. Of course, the number I is known from earlier considerations about counting. However, for other real numbers, even a number like .,J2, or 32/5, let alone for e and n, the definition of such numbers as equivalence classes of Cauchy sequences of rational numbers is the most natural one. The next theorem puts together the above discussion. Theorem 4.4. Let F be a normed vector space. Let Fl be the space of eqUivalence classes of Cauchy sequences ill F, with the norm 011 F1 defined as above, that is, if e = {xn}, thell lei = lim Ix"l. Then SI(Fl is dellse in F l , alld every Cauchy sequence in F has a limit ill Fl' Helice F1 is complete. Proof. Let e = {xn} be a Cauchy sequence in F. Directly from the definition, we see that e = lim 8 1 (xnl, so 8. (F) is dense in Fl and the Cauchy sequence in F has a limit in F l namely its own class. That F l is complete now follows from Theorem 4.1, so Theorem 4.4 is proved.

CHAPTER

VIII

Compactness

VIII, §1. BASIC PROPERTIES OF COMPACT SETS Let S be a subset of a normed vector space E. Let {x n } be a sequence in S. By a point of accumulation of {x n } (in E) we mean an element VEE such that given E there exist infinitely many integers n such that IXn - v I < E. We may say also that given an open set U containing v, there exist infinitely many n such that Xn E U. Similarly, we define the notion of a point of accumulation of an infinite set S. It is an element vEE such that given an open set U containing v, there exist infinitely many elements of S lying in U. In particular, a point of accumulation of S is adherent to S. We define the notion of a compact set by the property of the Weierstrass-Bolzano theorem. A set S in E is said to be compact if every sequence of elements of S has a point of accumulation in S. This property is equivalent to the following properties, which could be taken as alternate definitions: (a) Every infinite subset of S has a point of accumulation in S. (b) Every sequence of elements of S has a convergent subsequence whose limit is in S. The equivalence between the definition and these properties is more a matter of language than anything else. We prove it in detail. Note by the way that if a set is compact with respect to the given norm, it is compact with respect to any other equivalent norm. Suppose S is compact, and let T be an infinite subset of S. Then T contains a denumerable set, which we enumerate, and which is then nothing but a sequence {x n }, such that X n if' X m whenever n if' m. This sequence 193

194

[VIII, § I]

COMPACTNESS

has a point of accumulation v E S. Given c, there exist infinitely many II such that Ix" - vi < E, and these infinitely many II give rise to infinitely many x" having this property, so that v is a point of accumulation for S. This proves (a). Assume (a). Let {x"l be a sequence of elements of S. If the set consisting of all x" is finite, then there exists an infinite set of integers, say I, such that for all II E I, the elements x" are all equal to the same element x. We can order the elements of I as 111 < 112 < ... so that the corresponding elements of the sequence {x"" x"" . ..} form a subsequence, which obviously converges to x. If on the other hand the set consisting of all x" is infinite, it has a point of accumulation v in S. We select II, such that Ix", - vi < 1/1. We then select "2 > II) such that Ix", - vi < 1/2. Inductively, suppose we have found II, < 112 < ... < Ilk such that 1

Ix n - "1 < j J

We select

Ilk + 1

>

Ilk

for j = I.... ,k.

such that I

Ix"•• , - vi < k +

1

Then the subsequence {x"" x"" .. . J converges to v, thus proving (b). Finally, if we assume (b), then given any sequence in S. it has a convergent subsequence whose limit is an element of S, and this limit is then a point of accumulation of the given sequence, thus proving that the set is compact. Theorem 1.1. A compact set is closed alld boullded. Proof Let S be compact and let v be in its closure, that is, v is adherent to S. Given 11, there exists x" E S such that I x" - vi < 1/11. The sequence {x"l converges to v. It has a convergent subsequence whose limit is in S. By the uniqueness of the limit, it follows that v E S, and hence S is closed. If S is not bounded, for each 11 there exists x" E S such that Ix,,1 > 11. Then the sequence {xn } does not have a point of accumulation in S. Indeed, if v were such a point of accumulation, consider m > 21 vi. Then

Ix", - vi ~ Ix",1 - Ivl ~ 111 - Ivl > ; . These inequalities contradict the fact that for infinitely many m we must have x", close to v. Hence S is bounded. Theorem 1.2. A closed subset of a compact set is compact.

[VIII, §I]

BASIC PROPERTIES OF COMPACT SETS

195

Proof Let S be a closed subset of a compact set K. Let T be an infinite subset of S. Then T has a point of accumulation in K. But a point of accumulation of T is adherent to T, hence to S, and since S is closed, it must lie in S. Hence S is compact.

Theorem 1.3. Let S be a compact set, and let S, ::> S2 ::> ... ::> Sn ::> ... be a sequence of non-empty closed subset such that Sn::> Sn+l' Then the intersection of all Sn for all 11 = 1, 2, ... is not empty. Proof Let X n E Sn' The sequence {xn} has a point of accumulation in S. Call it v. Then v is also a point of accumulation for each subsequence {x,} with k E; 11, and hence lies in the closure of Sn for each 11. But Sn is assumed closed, and hence v E Sn for all 11. This proves the theorem.

Theorem 1.4. Let S be a compact set in the normed vector space E, and let T be a compact set in the 1I0rmed vector space F. Theil S x T is compact in E x F (with the sup norm). Proof Let Zn = (xn, Yn) be the terms of a sequence in E x F with Sand Yn E T. The sequence {xn} has a subsequence {xn.l convergent to a limit v E S. The corresponding sequence {Yn.} (k = 1,2, ...) has a subsequence {YII' } (j = 1,2, ...) convergent to a limit IV E T. Then let X nE

I

It is clear that Zno -+ }

as j

-+ 00,

(v, w)

thus proving our theorem.

Notationally, the triple indices are somewhat disagreeable. We shall now reproduce the preceding proof using a terminology which avoids these repeated indices, so that readers may use this better notation if they wish. There exists an infinite subset J, of Z+ and there exists v E S such that lim X n = V.

There exist an infinite subset J 2 of J I and

WET

limYn = w.

such that

196

[VIII, §1]

COMPACTNESS

The sequence {z.} (n E J 2) then converges to (v, w), which is in S x T, thus proving the theorem. By induction, we conclude that a finite product of compact sets is compact. We use this fact immediately to get a converse of Theorem 1.1 in an important special case. Theorem 1.5. A subset of RO is compact bounded.

if and

only

if

it is closed and

Proof We already know by Theorem 1.1 that a compact subset of RO is closed and bounded, so we must prove the converse. Let S be closed and bounded in RO. There exists C > 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 <' and a subsequence {g(y"l}"EJ with an infinite subset J of Z+, such that

IgCI',,) - <'I

~ <'

for all

II E

J.

Since S is compact, there exists an infinite subset J' of J such that (g(.1''')}''EJ' converges to some clement v' E S, and lv' - vi ~ <'. so in particular, v' '! v. However, by the continuity of I we get . I(I") = lim f(g(y,,») = lim .1'" = IIeJ'

\I'

=

f(v).

fleJ'

Since I is injective, we must have v' = v, a contradiction which proves the theorem. Remark, If S is not compact, the conclusion that 9 is continuous is not true in general. Counterexamples are routinely constructed. For instance, let S consist of {O} u (1, 2J, and T = [0, 1]. Let g(O) = 0 and g(y) = y + I if .r > 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<

I - Am

+ CAm+' < I,

then for such A we get 1/,(Az.)1 < ICol, thereby contradicting the hypothesis that II(zo) I ~ 1/(/)1 for all complex numbers I. The left-hand inequality is of course obvious since 0 < A < 1. The right-hand inequality amounts to CAm + 1 < Am, or equivalently CA < 1, which is certainly satisfield for sufficiently small A. This concludes the proof. Remark. The idea of the proof is quite simple. We have our polynomial

and Cm '" O. If 9 = 0, we simply adjust cmzm so as to subtract a term in the same direction as Co, to shrink Co toward the origin. This is done by extracting the suitable moth root as above. Since 9 '" 0 in general, we have to do a slight amount of juggling to show that the third term is very small compared to cmzm, and that it does not disturb the general idea of the proof in an essential way.

[VIII, §4]

RELATION WITH OPEN COVERINGS

203

VIII, §4. RELATION WITH OPEN COVERINGS Theorem 4.1. Let S be a compact set in normed vector space E. Let r be a number > 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 <-

and

lx, - xml < r and this contradicts the property of our sequence

{xn } because

Xk

lies in the ball B(x m). This proves the theorem.

Let S be a subset of a normed vector space, and let I be some set. Suppose that for each i E I we are given an open set V;. We denote this association by {V;};El and call it a family of open sets. The union of the family is the set V consisting of all x E E such that x E V; for some i E I. We say that the family covers S if S is contained in this union, that is every XES is contained in some V;. We then say that the family (V;LEI is an open covering of S. If J is a subset of I, we call the family (Vj}}E) a subfamily, and if it covers S also, we call it a subcovering of S. In particular, if

is a finite number of the open sets V;, we say that it is a finite subcovering of S if S is contained in the finite union UII

U .. · U

V"'n

Theorem 4.2. Let S be a compact subset of a normed vector space, and let (ViLEI be an open covering ofS. Then there exists afinite subcovering, that is a finite number of open sets V;" ... ,Vi. whose IInion covers s. Proof We prove first that there exists a positive integer n such that for each XES, the ball BI/n(x) is contained in some Vi' Suppose this assertion is false. Then for each n there exists xn E S such that BI/n(xn) is not contained in Vi for all i. By compactness, there is a subsequence {xnJ which converges to an element v of S. For this v, there exists some V} containing v, and a ball BI/N(V) contained in ~ for some integer N. For

204

[VIII, §4]

COMPACTNESS

all n sufficiently large, and in particular n > 2N, we have IXn and hence

-

vi < Ij2N,

contradicting the supposition that our assertion is false. Thus the assertion is true with some positive integer n. By Theorem 4.1 there exists a finite number of balls BI,n(YI),'" ,BI'n(Ym) covering S, and each one of these balls is contained in some V" so a finite number Vi" ... ,u'm covers S, thus proving the theorem.

From Theorem 4.2, we get another proof that a continuous map on a compact set S is uniformly continuous, as follows. Given E, for each XES, there exists o(x) such that if yES and Iy - xl < o(x), then If(y) - f(x)! < E. For each XES let B(x) be the open ball centered at x, of radius o(x), and B'(x) the open ball of radius o(x)j2. Then the union of the balls B'(x) for all XES is an open covering of S, and thus there is a finite number of points XI' ... 'X n E S such that B'(x.), ... ,B'(xn) contains S. Let <=

v

.

mm

»)

(O(X I ) O(Xn 2 , ... , 2 .

Let x, y be any pair of points of S such that Ix - YI < 0. Then x is in some B'(Xk), that is

Since IY - xl <

°;;

O(Xk)J2, it follows that Y E B(Xk)' Hence

If(Y) - f(x)1 < If(Y) - f(xk)1

+ If(Xk) -

f(x) I < 2£.

This proves the uniform continuity. The property concerning the finite coverings is equivalent to the property of compactness. Theorem 4.3. Let S be a subset of a normed vector space, and assume that any open covering of S has a finite subcovering. Then S is compact.

[VIII, §4]

RELATION WITH OPEN COVERINGS

205

Proof. We must prove that any infinite subset T of S has a point of accumulation in S. Suppose this is not the case. Given XES, there exists an open set U x containing x but containing only a finite number of the elements of T. The family {U xlxeso is an open covering of S. Let {U x ,"" ,U x .} be a finite subcovering. We conclude that there is only a finite number of elements of T lying in the finite union

Ux,V"'vUXn ' This is a contradiction, which proves our theorem.

VIII, §4. EXERCISES 1. Let {V J' ... , V.,} be an open covering of a compact subsct 5 of a normed vector space. Prove that there exists a number r > 0 such that if x, Y € 5 and Ix - YI < r then x and yare contained in U i for some i.

2. Let {5, II e( be a family of compact subsets of a normed vector space E. Suppose the intersection 5, is empty. Prove that there is a finite number of

n

Ie (

indices i 1 • ... ,in such that 5·'I n'" n S·I" is empty. This is the "dual" property of the finite covering property.

3. Let 5 be a compact set and let R be the set of continuous real valued functions on S. Let I be a subset of R containing 0, and having the following properties: (i) If[, 9 € I, then [ + 9 € I. Iii) If [€ I and h € R, then lif € J. Such a subset is called an ideal of R. Let Z be the set of points x € 5 such that [(x) = 0 for all [€ I. We call Z the set of zeros of I. (a) Prove that Z is closed, expressing Z as an intersection of closed sets. (b) Let [€ R be a function which vanishes on Z, i.e. [(x) = 0 for all x € Z. Show that [ can be uniformly approximated by elements of I. [Him: Given " let C be the closed set of elements x € S such that I[(x) I <:; ,. For each x € C, there exists 9 € I such that 9 .;. 0 in a neighborhood of x. Cover C with a finite number of them, corresponding to functions g" ... ,g,. Let 9 = g; + ... + g;. Then 9 € I. Furthermore, 9 has a minimum on C, and for II large, the function [

IIg

I

+ IIg

is close to [on C, and its absolute value is < Justify all the details of this proof.]

E

on the complement of C in 5.

CHAPTER IX

Series

IX, §1. BASIC DEFINITIONS Let E be a normed vector space. Let {v.} be a sequence in E. The expressIon

'"

~>. 12=1 is called the series associated with the sequence, or simply a series. We call

•

SI! =

L VA: =

Vi

+ ... +

Vn

11:=1

its n-th partial sum. If lim s. exists, we say that the series converges, and

.-'"

we define the infinite sum to be this limit, that is

'" Vk = L 11:=1

lim

"-IX)

SI!_

In this case, the limit is called the sum of the series. Thus the sum of the series, if it exists, is defined as a limit of a certain sequence, and consequently, the theorems concerning limits of sequences apply to series. Notably, we have:

If

'" LV. 12=1 206

and

[IX, §1]

BASIC DEFINITIONS

207

are two series in E, and if both converge, then ~

~

~

L (vn + wn) = "=1 L vn + n=1 L wn·

n=1

~

If c is a number, and if

L Vn converges, then "=1 ~

~

L Vn = L CVn •

c

n=1

11=1

If E, F, G are normed vector spaces, and E x F ~

~

n= 1

n= 1

-+

G is a product, and

if

L vn and L w are convergent series in E and F respectively, then n

where sn

=

VI

+ ... + Vn

and t n = WI

+ ... + W n •

Note that sntn = L v/wj the sum being taken for i,j = 1, ... ,no The whole point of this chapter is to determine criteria for the convergence of series. As a matter of notation, one sometimes writes

omitting the n = 1 and OCJ if the context makes it clear. Of course, if a sequence is given for integers n > 0, we can write the sum of a series as ~

L Vn'

"=0

Similarly, we let ~

L Vn =

n=k

whenever it exists.

lim (Vk n- co

+ Vk+ I + ... + Vn)

208

[IX, §2]

SERIES

The convergence of a series I

Vn

depends only on

for k large. Indeed, if for some k ;;:; I the preceding series converges, then

converges also, as one sees at once by the theorem concerning limits of sums. Thus we may say that the convergence of the given series depends only on the behavior of Vn for n sufficiently large. For the same reason, if we have two series I Vn and I V~ such that Vn = v~ for all but a finite number of n, then one series converges if and only if the other converges. Indeed, if V n = v~ for all n > 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 +<

c;;: = (n + 1)'+' =

(n )1+' n

+I

and this ratio approaches I as n --+ 00. Thus the ratio test does not yield anything. We can compare other series with I I/n' for s > 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)

<

l.

45;40

Around each fraction Ojq, I jq, ... , qjq consider the interval of radius f(q)jq. For q ~ qOt the set S is contained in the union of such intervals....] 18. Let a be a real number. Assume that there is a number C > 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 <

Cn ,

it follows that Sn+ 1

;;;; Sn

and thus

and similarly,

i.e. the Sn form a decreasing sequence, and the Cn form an increasing sequence. Since Cn = Sn - Cn and Cn ::: 0, it follows that cn ~ Sn so that we have the following inequalities:

Given

E,

there exists N such that if n > 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<

------

------

--------"-~-------2Jt

31<

41<

51<

61<

The function f represents what is called a dampened oscillation. Let an be the area (with a plus sign) between the n-th arch of the curve and the x-axis. Then

sin x --dx = a l - a2 o x

i

nn

+ a3 -

...

+ (-I)n+l an ·

This is the other standard example of an alternating series, besides the alternating harmonic series. The limiting value is more difficult to Obtain, and we show how to get it in Exercise 2 of Chapter XIII, §3. Remark. The final statement in Theorem 3.1 allows us to estimate the tail end of an alternating series. Indeed, if we take the sum starting with

220

[IX, §3]

SERIES

the moth term, then all the hypotheses are still satisfied. Thus

L'" (-I)'a,

/c:=m

;;;a m·

The proof of Theorem 3.1 also shows that for positive integers m ;;; II, we have n

L

(-I)'a, ;;; am'

k==m

This gives a useful estimate in certain applications, when it is necessary to estimate certain tail ends uniformly and accurately. The same estimate will be obtained by a more powerful method in Theorem 3.3. Let (I(k) j, {g(k)) be sequences of numbers, say, but see Remark 2 below. The main method for dealing with series of the form Lf(k)g(k)

is based on Abel's theorem, which is analogous to integration by parts. Theorem 3.2 (Summation by parts). Lec che parcial sums of che g(k) be G(k) = g(1)

+ ... + g(k),

so chac

G(k

+ I) -

G(k)

= g(k + I).

Theil "-1

n

L f(k)g(k) = f(II)G(II) - L G(k)(f(k

+

I) - f(k).

Ie=!

Ie=!

Proof If II = I we interpret the sum on the right as being O. Suppose 2. Then

II ~

0-1

f(II)G(II) - L G(k)(f(k

+

I) - f(k)

Ie=!

= f(/I)G(II) -

"-1

L G(j)f(j j=1

o

= L f(k)G(k) kcl

+ I) +

"-1

L G(k)f(k) k=!

0

L G(k - I)f(k) k=2

[IX, §3]

221

NON-ABSOLUTE CONVERGENCE

=

I•

f(k)(G(k) - G(k - 1»

if we define G(O) = 0

1<=1

=

•

I

f(k)g(k),

1<=1

which proves the theorem.

Remark 1. Note how the formalism of the partial sum follows the formalism of integration by parts, that is

fudv=uv- fVdU. This is the reason why we wrote the second term on the right with a minus sign. For some purposes, it is useful to change signs and write the formula in the form

n

I

f(k)g(k)

=

f(Il)G(Il)

+

k=l

"-1

I

G(k)(f(k) - f(k

+ I).

k=1

This is the way we shall use it in Theorem 3.3, where the f(k) will be assumed positive decreasing, so that each difference f(k) - f(k + 1) is ~ O.

Remark 2. In the statement of Theorem 3.2, we purposely left out where f and g take on values. Although we said they were functions before the theorem, in fact, the proof is purely formal, and requires only that they take values in a vector space E, with a product of E x E into another vector space F so we can take sums. For instance F could be a normed vector space, and E could be R itself. Or E could be the space of real valued functions on some set, and the product is the ordinary product of functions. Or both f and g could take values in a vector space with a scalar product, in which case by f(k)g(k) we mean the scalar product. All these cases arise in practice. The same remark applies to Theorem 3.3 below. Additionally, although it's a minor point, we observe that we don't even need the commutativity of the product, and f, g could be maps into different vector spaces. All we need is the distributive law. In this case, we have to be careful to put objects on the appropriate side, and the formula has to be written "-I

n

I

k=1

f(k)g(k) = f(Il)G(Il) -

I

1<=1

(f(k

+ 1) -

f(k»G(k),

222

[IX, §3]

SERIES

with all the g(k) and G(k) on the same side (the right side). Still, we preferred to write the formula in Theorem 3.2 with an order which fits the formula for integration by parts. As an application of Theorem 3.2, we prove a result which will reprove Theorem 3.1 more structurally. Theorem 3.3. Let {a.} be a decreasing sequence of numbers ~ O. Let (g(k)} be a sequence in a normed vector space, with partial sums G(n). Assume that these partial sums are bounded, i.e. there exists C ~ 0 such that for all n. IG(n)1 ~ C Then

•

I a,g(k) k=1

~ Cal·

In fact, for n ~ m ~ I, we have

I•

a,g(k) ~ 2Cam .

k=m

Proof By Theorem 3.2 and the triangle inequality, we get n

I a,g(k) k=1

n-l

~ Ca.

+

I qa, - a

H ,)

~ Ca,.

k=1

which proves the first statement. The second is then immediate, since it amounts only to a renumbering of the sequence. The second estimate in the theorem is useful in that form, because it gives us an estimate for the norms (or absolute values in case of numbers) of the tail ends of the sequence, how fast they tend to 0, namely essentially as fast as the m-th term. Example. We reprove Theorem 3.1 as follows. With the notation of this theorem, we let g(k) = (- I)Hl, so the partial sums g(l)

+ ... + g(n)

are bounded, and in fact they are equal to 1 or O. Then the statement of Theorem 3.1 is a special case of Theorem 3.3. Warning. Just because the partial sums are bounded does not imply the convergence of a series, as you see from I (-If. In the application

[IX, §3]

223

NON-ABSOLUTE CONVERGENCE

to Theorem 3.1, what makes the series converge is that the numbers a. are not only decreasing, but are decreasing to O. . Note that Theorem 3.3 applies when the values g(k) are in the space of complex numbers, which is a complete normed vector space. Exercise 2 will give you an example when it is useful to have the theorem in the stated generality. Special Remark. In Theorem 3.3, we did not assume that (a.} is decreasing to O. In several examples and exercises, this will be the case, but for one important case when it isn't, see Abel's theorem, Exercise 7 of §6. There is still another theorem which is sometimes useful, with a mixture of summation and integration by parts, as follows. Theorem 3.4. Let f be a C' function on tile positive reals. Let (a.}, (r.} be sequences of numbers, say, willl rk ;;; 0, and {rd increasing 10 00. For positive B ;;; r" define tile counting function S.(B)

= L:

a•.

rk~B

Tilen

L:

aJ(r.)

= S.(B)f(B) - fB S.(r)!'(r) dr.

rk~8

rl

Proof Subtracting the left side from Su(B)f(B) we have

=

'~B a.(f(B) -

f(r.» =

- L: fB g.(r) dr

r rk~B

=

'I

'.~B g.(r) dr =

1

'~B

r

where

rI

aJ'(r) dr

g.(r)

= {~

f'(r) k

if r < if r;;;

r., r.,

a.J'(r) dr

8

=

Sa(r)f'(r) dr,

which proves the theorem.

'I

L

One last word on series of the type a.b.. Exercise 7 of §2 gives one method of proving convergence. Summation by parts gives another method which covers all cases in this section. These are the two standard methods which one tries in any given situation.

[IX, §3]

SERIES

224

IX, §3. EXERCISES I. leI

a, ;,; 0 for all n.

Assume Ihal L a, converges. Show Ihal L ~Jn converges.

2. Show Ihat for x rea~ 0 < x < 2", L e'~/n converges. Conclude Ihat and converge in the same interval. 3. A series of numbers L a, is said to converge absolulely if L la,l converges. Determine which of the following series converge absolutely, and which just converge.

sin n

(-I)"

(a)

L n 1+11'

(b) L ( - I ) " n

[Hint for (b): Show that among three consecutive positive integers, for at least one of them, say n, one has Isin nl ;,; 172.] (c)

L(-I)"j;+! -..fi n

"

n

(d)

L 2' n2

-

4n

sin n (e) L. 2n2 _ n

(f)

L (-I)' 2n' + n _

2' + I (g)L 3'_4

(h)

L nS _ n' + I

. I (1) L ( - I ) " -

. (J)

L (-I)" n(log n)'

5

ncosn

log n

I

4. For which values of x does the following series converge?

5. Let la,} be a sequence of real numbers such that L a, converges. Let lb,} be a sequence of real numbers which converges monotonically to infinity. (This means that lb,) is an unbounded sequence such that b,+ 1 ;,; b, for all n.) Show that

Does this conclusion still hold if we only assume thaI the partial sums of L a, are bounded? 6. The Cantor set. Let K be the subset of [0, I] consisting of all numbers having a trecimal expansion

f

an

"=1 3"

where a,

= 0 or a, = 2.

This set is called the Cantor sel.

[IX, §4]

ABSOLUTE CONVERGENCE IN VECTOR SPACES

225

(a) Show that the numbers a" in the trecimal expansion of a given number in K are uniquely determined. (b) Show that K is compact. 7. Peano curve. Let K be the Cantor set. Let S = [0, 1] x [0. I] be the unit square. Let f: K - S be the map which to each element L: aj3" of the Cantor set assigns the pair of numbers

b,.)

'" b'.+1 '" ( L2"''-'2n

'

where bm = a..,/2. Show that f is continuous, and is surjective. [It is then possible to extend f to a continuous map of the interval [0, I] onto the square. This is called a Peano curve. Note that the interval has dimension 1 whereas its image under the continuous map f has dimension 2. This caused quite a sensation at the end of the nineteenth century when it was discovered by Peano.]

IX, §4. ABSOLUTE CONVERGENCE IN VECTOR SPACES Let E be a complete normed vector space. Let we can form the series

L v. be a series in E.

Then

L'" Iv.1

"=1

of the norms of each term. Letting a. = Iv.', we see that L a. is a series of numbers;;; 0, to which we can apply the criteria developed in §2. Theorem 4.1. If L I v.1 converges, then

L v. converges also.

This is easily seen, for if

is the partial sum, then for

In ;;;

Is. - sml

=

n we have

•

L

k=m+1

v. ;;;

•

L

k=m+1

Iv.l.

Given E, there exists N such that if In, n ;;; N and say In ;;; n then the expression on the right is < E. Thus {s.} is a Cauchy sequence, and converges since E is assumed complete.

L

LV•.

converges absoWhenever the series Iv. I converges, we say that lutely. We have just seen that absolute convergence Implies convergence, whence the terminology is justified.

226

[IX, §4)

SERIES

I

In defining the value of a series sums

we take the limit of the partial

Vn'

It is thus important to consider the order in which the terms V n occur. For instance, if we consider the series (-I)" and try to sum it by

I

putting parentheses this way:

(-I + 1)+(-1 + 1)+'" we obtain the value O. On the other hand, putting the parentheses this way

-I +

(I - I)

+

I) + ...

(I -

we obtain the value -I. The partial sums actually oscillate between -I and 0, and the series does not converge. We know that the series L:(-I)n(I/II) converges. However, by reordering the terms, we can obtain a series which does not converge. For instance, consider the following ordering: IIIII

-I + - + - - - + - + - + ... + 2

4

3

6

8

I

I

211t

+.

--

5

I

2111

+2

I

I

2112

7

+ ... + - - - + ...

We select the sequence nl < n2 < n3 < ... as follows. Having chosen some II., we pick nk+ I such that the sum I

-::--"7

2n.

+2

1

+ ... + - 2nk+ I

is greater than 2, say. When we subtract the odd term immediately afterwards, what remains is still > I. Thus the partial sums become arbitrarily large. The preceding phenomena are due to the presence of negative terms in the series, as shown by the next theorem.

Theorem 4.2. Let E be a complete normed vector space, and let I Vn be an absolutely convergent series in E. Then the series obtained by any rearrangement of the terms also converges absolutely, to the same limit. Proof. The rearrangement of the series is determined by a permutation of the positive integers Z +. That is, there exists a bijective mapping

[IX, §4]

ABSOLUTE CONVERGENCE IN VECTOR SPACES

227

U: Z + -+ Z + such that the rearranged series can be written in the form

.,

L 11=

Given

£,

va(n)"

J

there exists N such that if In,

11

> N and say In

IVml + ... + Iv,,1 <

(1)

~ 11

then

£.

Select N I > 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<=1

Un

n=1

This proves that the limit of the rearranged series is the same as the limit of the original series, as desired. There is a variation to the above theorem.

228

[IX, §4)

SERIES

Theorem 4.3. Let E be a complete normed vector space. Let {vn .. } be a doubly indexed family of elements of E, with n, m E Z+. Assume: (i)

For each n, the series

L'"

Ivn .. I converges.

m=l

(ii)

The repeated series

,,~ C~1 IV

nm

l) converges,

Then the series taken in reverse direction, that is

cOIwerges, so lio the series without tlte norm signs, Qlld

L(L v",,,) = L(L V"m) = n

m

In

for some wEE.

W

II

Furthermol'e, given c, there exists No such that for all M, N have N

I\' -

~

No we

At

L L Vnm ,,=1 m::!

<

E,

Proof This theorem is actually analogous to Theorem 3.5 of Chapter VII. Here instead of points adherent to sets, we deal with" infinity." We may view the family {v"m} as a mapping Z+ x Z+ -+ E, and so is the parlial sums mapping f: Z+

X

Z+

N

-+

E

given by

f(N, M) =

At

L m=} LV"",.

11=1

Thus the variables (I,x) of Chapter VII, Theorem 3.5, are replaced with the variables (N. M). To apply that reference. we have only to verify that the two conditions of that theorem are satisfied. The first one is a consequence of the hypothesis that for each N, lim f(N, M) /11 .... 00

[IX, §5]

ABSOLUTE AND UNIFORM CONVERGENCE

229

exists, even with absolute convergence of the series. So we have only to verify the second hypothesis of uniform convergence, that is that the family UN} is uniformly convergent. Say N < N'. For all positive integers M, we have ~

By the second hypothesis, given then N'

€

N'

M

I Ilv.ml n=N+l m=l

there exists No such that if N, N' ~ No

'"

I I IV.ml < € n=N+l m=l

This yields precisely the uniformly convergence of the sequence UN}' Thus we may apply Theorem 3.5 of Chapter VII in the present context to conclude the proof.

IX, §S. ABSOLUTE AND UNIFORM CONVERGENCE We can apply the preceding results to sequences of functions, Let S be a set, and let U.} be a sequence of functions on S. We form the partial sums

s.

=

f, + ,.. + I.

so that s. is a function, s.(x) =

We shall say that the series I if the series

f, (x) + ' ., + I.(x),

In (also written I

In(x)} converges absolutely

L11.(x)1 converges for each XES. We shall say that the series If. converges uniformly if the sequence of functions {s.} converges uniformly. In most instances, the functions I. are bounded, In this case, we can use the sup norm, and uniform, absolute convergence on S is the same as the convergence of the series I 111.11. This is but a special case of that discussed in the preceding section,

230

[IX, §5]

SERIES

We see that the convergence of a series of functions is determined by the convergence of series of numbers, except that estimates now depend on x, and to prove uniform convergence, we must show that these estimates can be made in such a way that they do not depend 'on x. Example I. The series L (-I)·x·ln converges uniformly for 0 5: x ;::; I, absolutely for x < I, but not absolutely for x = 1. Proof For 0;::; x < I, the series L x·ln is dominated by L x·, which is a geometric series and converges. For x = I, the series L lin diverges. Finally, to verify the uniform convergence on the whole interval [0, I], we sum by parts. For the tail end, Theorem 3.3 gives the estimate

.

L

I

I (-Ij'x'ln ;::; -x" ;::; -.

k=rn

m

m

Hence given e, pick N > 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<-<-+ -5:2 2 - n n n- n

Let

s. (x )

=

~ (-I)'(x k2

L., k=1

+ k) •

Using Theorem 3.3, we conclude that for m, n large, we have lis. - s..11 < e, whence the convergence is uniform. However, the convergence is not absolute, because we can compare the series with L lin from below to see that L (x + n}/n 2 diverges. In the absolutely convergent case, we have a standard test called the Weierstrass test. Theorem 5.1. Let {!.) be a sequence of bounded functions such that IIf.11 ;::; M. for suitable numbers M., and assume that L M. converges. Then L!. converges uniformly and absolutely. If each f. is continuous on some set S, then L!. is continuous. Proof Immediate from the definitions, the comparison test, and Theorem 3.2 of Chapter VII, for the continuity statement.

[IX, §5]

ABSOLUTE AND UNIFORM CONVERGENCE

231

Example 3. The series

is uniformly and absolutely convergent for all x because

and we know that function J(x}.

I

l/n 2 converges. Thus the series defines a continuous

The next theorem will not be used in this book, but provides an example of uniformly convergent series, so we include it. Theorem 5.2 (Tietze extension theorem). Let A be a closed subset oj a normed vector space E and let J be a continuous (real valued) Junction on A. Then there exists a continuous Junction J* on E whose restriction to A is equal to J. If J has values in [0, I], then we can choose J* to have values in [0, I] also. Proof Assume first that J has values in [0, I]. If C, D are disjoint closed subsets of E, we denote by gC,D a function with values in [0, I] such that g(C) = and g(D) = I. Such a function exists by Exercise I(c) of Chapter VII, §2.

°

We shall now define functions In on A and gn on E. We let fo = J and define sets A o , Bo by the conditions: A o = (x E A such that J(x) ~

tl,

Bo = (x

n

E

A such that J(x) ?;

We let go = tgAo.Bo and define we have defined I.; we have

J1 = fo -

go. Inductively, suppose that

mWn}, that J.(x) ?; WHrJ.

An = (x E A such that J.(x) ;:;; Bn

= (x E A

such

We then define

and let

1.+1 = I. -

gn' (Here of course, we understand by gn its restric-

232

[IX, §5J

SERIES

lion A.) Then in particular:

1.+1

=

f - (go + ... + gn)·

We have

0 =< gn < = l(l)n 33

(*)

and

o ~fn ~ H)n.

The first inequality is clear. The second is proved by induction. It is clear for n = O. Let n > 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<~
Conclude the same for 7. Let

(1

L (sin nx)/n and L (cos nx)/n.

be the set of sequences"

= {a.}. a. E R, such

that

L" la.1

11=1

converges. This is the space of §2, Exercise 15, with the (I-norm

Lla.l·

"""1 =

(a) Show that the closed ball of radius I in (1 is not compact. (b) Let" = {a.} be an element of (l, and let A be the set of all sequences P= {b.} in (, such that Ib.1 ~ la.1 for all n. Show that every sequence of elements of A has a point of accumulation in A, and hence that A is compact.

234

[IX, §6]

SERIES

8. Let F be the complete normed vector space or continuous runctions on [0,211] with the sup norm. For cx = {a,,} in fl, let w

L(cx)

=L

a. cos nx.

n=1

Show that L is a continuous linear map or f' into F, and that IIL(cx)1I ~ IIcxll l ror all cx E fl. 9. For z E C (complex numbers) and Izi # I, show thaI the rollowing limit exists and give the values:

I)

z. J(z) = lim ( -.-1 . 'I-CO Z + Is it possible to define J(z) when Izi = I in such a way to make J continuous?

10. For z complex, let J(z)

z·

= lim -I-,,' +Z II-eo

(a) What is the domain or definition of J, that is ror which complex numbers z does the limit exist? (b) Give explicitly the values of J(z) ror the various z in the domain of f II. (a) For z complex, show that the series co

J, (I -

ZIl-1

z·)(1 - z·+1)

converges to 1/(1 - z)' ror Izi < I and 10 l/z(1 - z)' ror Izi > I. [Hint: This is mostly a question of algebra. Formally, factor out I/z, then at first add I and subtract I in the numerator, and use a partial fraction decomposition, pushing the thing through algebraically, before you worry about convergence. Use partial sums.] (b) Prove that the convergence is uniform ror Izi ~ c < I and Izl ;;; b > I.

IX, §6. POWER SERIES Perhaps the most important series are power series, namely Or

where a. E R and x E Rand Z E C. The number a. are called the coefficients of the series. We are interested in criteria for the absolute convergence of the series.

[IX, §6]

235

POWER SERIES

Lemma 6.1. LeI {a,,1 be numbers che series

~

0 and lec r be a number> 0 such chac

'\' a r"

L.- "

converges. Then the series converges also for all numbers x such chac o;;; x ;;; r. Proof Obvious, from the comparison test.

I

Corollary 6.2. If {anI is a sequence of ",,,nbers, and lhen a"z" converges absolutely and uniformly for Izi

I

la"lr" converges,

;;; r.

Proof By definition. Example 1. For any r test:

> 0 the series L. r" jn! converges, by the ratio r"+1

n!

r

----=--, (n + I)! r" n + 1 which goes to 0 as n --t 00, so the comparison test works. This implies that the series L. z" jll! converges absolutely for all z, and uniformly for Izi ;;; r. Similarly, the series

and Z2

_4

2!

4!

cc

~2J1

,,=0

(2n)!

1--+=---··· = 2:(-1)"--converge absolutely for all :, and uniformly for

Izi ;;; r.

I

Theorem 6.3. LeI a"z" be a power series. If it does IlOI COil verge absolulely for all z, lhell lhere exists a number R such lhal lhe series COIlconverges absolulely for Izi < R and does nOI cOllverge absolutely for

Izi > R. Proof Suppose the series does not converge absolutely for all z. Let R be the least upper bound of those numbers r ~ 0 such that

Ila"lr"

236

[IX, §6]

SERIES

converges. Then L la.llzl· diverges if Izi > R and converges if Izi < R by Theorem 6.2, so our assertion is obvious. The number R in Theorem 6.3 is called the radius of convergence of the power series. If the power series converges absolutely for all z, then we say that its radius of convergence is infinity. When the radius of convergence is 0, then the series converges absolutely only for z = 0. We now assume that you know about the lim sup of a sequence of real numbers. By convention, if this sequence is not bounded from above, we say that the lim sup is 00. If the sequence is bounded from above, the lim sup was defined in the exercises of Chapter II, §l. We recall the definition. Let {x.} be a sequence of real numbers, bounded from above. Then t = lim sup x. is the number satisfying either one of the following two properties: The number t is the largest point of accumulation of the sequence. Given £, there exist infinitely many n such that only a finite number of n such that x. ~ I + £.

I -

£ ~

x., and there is

The main theorem about the radius of convergence is the following. Theorem 6.4. Let convergence. Then

L a.z·

be a power series, and let r be its radius of

~r =

lim sup 10.1'/·.

If r = 0, this relation is to be interpreted as meaning that the sequence {la.I'/·} is not bounded. If r = 00, it is to be interpreted as meaning that lim sup 10.1'/· = 0. Proof Let t = lim sup 10.1'/·. Suppose first that t # 0, 00. Given £ > 0, there exist only a finite number of n such that I0.1'/· ~ t + £. Thus for all but a finite number of n, we have

10.1 ~ (t + £)., whence the series L a.z· converges absolutely if Izi < lilt + £), by comparison with the geometric series. Therefore the radius of convergence r satisfies r ~ 1/(1 + £) for every £ > 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 <

E/(b - a)

to conclude the proof.

Theorem 7.2. Let {f,,} be a sequence of differentiable functions on an interval [a, b] with a < b. Assume that each f~ is continuous, and that

240

[IX, §7)

SERIES

the sequence (f~} converges uniformly to a function g. Assume also that there exists one point Xo E [a, b) such that the sequence of (fn(XO)} converges. Then the sequence (fn} converges uniformly to a function f, which is differentiable, and f' = g. Proof. For each n there exists a number fn(x)

=

r

f:

+ cn ,

Cn

all x

such that E

[a, b).

Let x = xo. Taking the limit as n -+ CD shows that the sequence of numbers (cn } converges, say to a number c. For an arbitrary x, we take the limit as n -+ CD and apply Theorem 7.1. We see that the sequence (I.} converges pointwise to a function f such that f(x) = !:g

+ c.

On the other hand, this convergence is uniform, because

r r r f: -

so III. -

fll

~ (b - a)

g =

(f: - g)

~ (b -

a) IIf: - gil,

1If: - gil + ICn - cl. This proves our theorem.

Remark 1. The essential assumption in Theorem 7.2 is the uniform convergence of the derived sequence (f~}. As an incidental assumption, one needs a pointwise convergence for the sequence (fn}, and it turns out that pointwise convergence at one point is enough to make the argument go through, although in practice, the pointwise convergence is usually obvious for all x E [a, b). Remark 2. Before going on with the applications of Theorem 7.2, we make some remarks on the proof. Assume for instance that {I.} converges uniformly to f. We want to prove that f is differentiable and f' = g. It is in a way natural to try to do this directly without integration, that is look at the sequence of difference quotients at a point x (fixed), u(k, h) = J.(x

+~ -

J.(x)

What we want amounts formally to an interchange of limits lim lim u(k, h) h-O k-co

= lim

lim u(k, h).

k-cn h-O

[IX, §7]

DIFFERENTIATION AND INTEGRATION OF SERIES

241

But we don't know that the conditions for an interchange of limits are satisfied. Hence trying to carry out the proof according to this pattern doesn't work. We go around the difficulty by using the fundamental theorem of calculus, expressing /" as an integral of f: plus the constant of integration. Use of the integral gives us a version of uniformity, independently of the formalism concerning interchanges of limits. We obtain the theorem for differentiation of series as corollary.

Corollary 7.3. Let L fn be a series of differentiable functions with continuous derivatives on the interval [a, b], a < b. Assume that the derived series Lf~ converges uniformly on [a, b], and that L/" converges pointwise for one poin!. Let f = L/'" Then f is differentiable, and j' = Lf~. Proof Apply Theorem 7.2 to the sequence of partial sums of the series. The corollary states that under the given hypotheses, we can differentiate a series term by term. Again, we emphasize that the uniform convergence of the derived series is essential.

Example 1. The function •

f(x)

= L Sin ~

2

X

n

is a continuous function. If we try to differentiate term by term, we obtain the series L cos n2 x. It is at first not obvious if this series converges. It can be shown that it does not.

Example 2. The series f(x) =

'" sin nx L 3 ,,-1 n

converges absolutely and uniformly for x E R. Differentiating term by term, we obtain the series g(x)

=L

cos nx n

2'

which converges absolutely and uniformly for x E R. Hence by Corollary 7.3, we get j'(x) = g(x). We can replace 3 by any integer ~ 3, and a similar argument applies.

242

[IX, §7]

SERIES

Example 3. Let co

g(x) =

I "=1

cos 1tX 2'

It

Differentiating term by term yields the series If(X)

=I

-sin nx n

.

This series does not converge absolutely. By Exercise 6 of §5, you should know that it converges uniformly on every interval [0, 21t - 0] with 0> 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'·


0

Two functions f, 9 are called orthogonal if

f(x)g(x) dx.


"'.(x) = sin nx.

and

CP.(x) = cos nx

O. Let

(Take n ~ I except for CPo = I.) Show that the functions CPo, CP., orthogonal. [Hin!: Use the formula sin nx cos mx = j[sin(n

+ m)x + sin(n

and similar ones.] Find the norms of CP., CPo,

- m)x]

"'m'

4. Let (a.l be a sequence of numbers such that that the series f(x) =

"'m are pairwise

L a.

converges absolutely. Prove

L a. cos nx

converges uniformly. Show that (f, CPo) = 0,

(f, I/Jm)

=0

for all m,

5. (Borel, 1890's) Let {a.} be a sequence of numbers. Show that there exists an infinitely differentiable function 9 defined on some open interval containing 0 such that g'·)(O) ~ a•.

[Hine: The following procedure was shown to me by Tate. Given n <; 0 and " there exists a function f = f",< which is Coo on -I < x < I such that:

(I) f(O) = f'(0) = ... = !,"-I)(O) = 0 and f'·'(O) (2) If""(x) I ~ dor k = 0, ... ,n - I and Ixl ~ I.

= 1.

Indeed, let cP be a C
=I

if Ixl ~ £/2, I if ,{2 ~ Ixl ~ " cp(x) = 0 if '~Ix I ~ 1.

cp(x)

o ~ cp(x) ~

Integrate cp from 0 to x, n times to get f(x). Then let '. be chosen so that converges. Put <0

g(x)

= L aJ•. ,Jx). """0

L la.I'.

[IX, §7]

DIFFERENTIATION AND INTEGRATION OF SERIES

245

For k ;;; 0 the series

converges unifonnly on Ixl ;;; 1, as one sees by decomposing the sum from 0 to k and from k + 1 to 00, because for n > k we have

6. Given a Coo function g: [a. b] - R from a closed interval. show that 9 can be extended to a Coo function defined on an open interval containing [a. b]. 7. Let n ;;; 0 be an integer. Show that the series

converges for all x. Prove that y

= J,,(x)

is a solution to Bessel's equation

n')

1 ( 1 - - y=O. y"+-y'+ x x'

CHAPTER

X

The Integral in One Variable

Let Fa be a subspace of a normed vector space F. It occurs in many contexts in mathematics that one is given a linear map L: Fa ..... E which one wants to extend to the closure of Fa. It occurs especially in the context of integration theory, which we study in this chapter. We begin by systematizing the general framework.

X, §1. EXTENSION THEOREM FOR LINEAR MAPS Let F, E be normed vector spaces, and L: F ..... E a linear map. We contend that the following two conditions are equivalent: L is continuous.

There exists a number C> 0 such that IL(x)1 ~ Cjxl for all x

E

F.

Assume the first, and even assume that L is continuous only at O. Given I, there exists 0 such that whenever x E F and Ixl < 0 we have IL(x)1 < 1. Now given an arbitrary x E F, x ¥- 0, we have lox/2lxll < 0, whence IL(<5x/2Ixlll < I. Taking the numbers out of L yields IL(x)1

2

< ;514

We take C = 2/0. Conversely, assume the second condition. Given let <5 = l/e. If Ix - YI < 0, then

lL(x -

y)1

= IL(x)

- L(Y)I ~

C1x - yl <

l,

whence L is not only continuous but uniformly continuous.

246

l,

[X, §l]

EXTENSION THEOREM FOR LINEAR MAPS

247

A number C as above is called a bound for L. If B is the unit ball centered at the origin in F, then we see that L(B) is bounded by C, whence the name for C. In view of the linearity it is clear that there cannot be a number C, such that IL(x)1 ~ C, for all x E F unless L = O. Hence in the case of linear maps, we say that a linear map L is bounded if it is continuous. We mean by this that it takes bounded values on bounded sets. There is of course some impropriety in this usage in view of the general definition of bounded mappings, but it is standard usage and the reader will find no genuine trouble arising from it. Proposition 1.1. Let F be a normed vector space, and let Fo be a subspace. Then the closure of Fo in F is a subspace of F. This is nothing but an exercise: If 1"0 denotes the closure of Fo' and v E 1"0' then v = lim X n for some sequence of elements X n E Fo . If w = lim Yn with Yn E Fo , then v + w = lim(xn + Yn) also lies in 1"0. Furthermore, cv = c lim

Xn

= lim(cxn )

lies in 1"0' so 1"0 is a subspace. Suppose that we are given a linear map L: Fo --+ E instead of being given a linear map on F, and assume that L is continuous. We wish to extend L by continuity to the closure of Fo . This can be done in the following case. Theorem 1.2. Let F be a normed vector space, Let L: Fo --+ E be a continuous linear map of space E, and assume that E is complete. Then to a continuous linear map

and let Fo be a subspace. Fo into a normed vector L has a unique extension

L: 1"0 --+ E of the closure of Fo into E. If C is a bound for L, then C is also a bound for L. Proof Let v E Fo and let v = lim X n with X n E Fo . We contend that the sequence {L(x n )} in E is a Cauchy sequence. Given E, there exists N such that if m, n > N then E

IXn - vi < c' Then Ix. - xml ~ Ix. IL(xn)

£

Ixm-vl
vi + Iv - xml < 2£/C. Consequently -

L(xm)1 = IL(x n

thus proving our contention.

-

xm)1

~

2£,

248

THE INTEGRAL IN ONE VARIABLE

[X, §2]

Since E is assumed complete, the Cauchy sequence {L(x n )} converges to an element w in E, Suppose {x~} is another sequence of elements of Fa converging to v, and let {L(x~)} converge to w' in E. Then Iw - w'l ;:;; Iw - L(xn)1 + IL(x n) - L(x~)1 + IL(x~ - w'l· Furthermore, lL(xJ - L(x~)1 ;:;; Clxn - x~l. From the definition of convergence, it is then clear that for al1 11 sufficiently large, Iw - w'l < 3£. This is true for every £, and hence Iw - w'l = 0, w = w', This means that the limit of {L(x n )} is independent of the choice of sequence {x n } in Fa approaching v. _ We define L(v) = lim L(xn ). If v happens to be in Fa, then L(v) = L(v) because, for instance, we can take X n = v for all n. If v = lim X n and v' = lim x~ with X n , x~ E Fa, then

v

+ v'

=

lim(xn + x~.

Hence L(v

+ v') = lim L(xn + x~) = lim(L(xn) + L(x~» = Jim L(xJ + lim L(x~ = L(v) + L(v'),

Also, I(cv) = lim cL(xn ) = c lim L(xn ) = cI(v).

Hence L is linear. Final1y, C is also a bound for

r, because

lL(v)1 = IlimL(xn)1 = lim I L(x n ) I· Since !L(xn)l;:;; C1xnl, it follows from the theorem on inequalities of limits that lL(v)1 ;:;; C1vl, as desired.

X, §2. INTEGRAL OF STEP MAPS We now wish to develop systematically the theory of the elementary Riemann integral. In practice, one needs first the integral both of real valued and complex valued functions. Of course, a complex valued function can be written I, + if2 where II> 12 are real valued, so the integr.ll can be reduced to coordinate functions which are real valued. Next, one needs also to consider the integral of maps of an interval into some euclidean space Rk , Such maps are curves. It is however bothersome to write such a map always in terms- of coordinate functions, and we may think of R k

[X, §2]

249

INTEGRAL OF STEP MAPS

merely as a vector space, with a norm, and complete. It turns out that the theory actually is valid in this generality. In first reading, the reader may think of functions only. We denote the space where our maps take values by a neutral letter E, to cover both the real and complex case, and the case of vector spaces if the reader wants it so. This latter case is in fact useful later when we consider the calculus of maps from one space into another. Of course, some statements concerning the integral are related to the ordering of the real numbers: if a function is positive, then its integral is positive. In such cases, we always specify that the function is real valued. Let a, b be numbers, a ::; b. By a partition P of the interval [a, b] we shall mean a finite sequence of numbers (ao, ... ,an) such that

Let I: [a,b]-+E be a map (function). We shall say that 1 is a step map with respect to the partition P if there exist elements WI' ... 'Wn E E such that I(t)

= W,

if

ai-l

< t < aj ,

i = 1, ... ,n.

Thus 1 has constant value on each open interval determined by the partition. We don't care what value 1 has at the end points of each interval [a/_" aJ. If a, = a,_" we let

We say that 1 is a step map on [a, b] if it is a step map with respect to some partition. Let 1 be a step map with respect to the partition P as above. We define

n

I

=

(a, - a,_I)w/

i= J

and call this value the integral of 1 with respect to the partition P. If 1 is real valued, then the graph of 1 has the usual shape shown on the next figure, and the integral is the naive sum of the areas of the rectangles,

) I

,..---,

, ,

a

= 00

0.

,I

,,:02 , ,I

,~, ,

I

I

I

I

,

r""1 I

,

250

THE INTEGRAL IN ONE VARIABLE

[X, §2]

with a plus or minus sign according as the constant value of f over an interval is positive or negative. Suppose that f is a step map with respect to another portition Q of [a, b]. We contend that I P(f) = I Q(f).

To prove this, consider first the partition obtained from P by inserting one more point c between the points of P:

with

We observe that if ak < t < aUI and f(t) = constant value on each of the intervals and

WUh

c < t <

then f has this same

ak+ It

if ak < c or c < au I' Consequently, the integral of f with respect to the partition P, is equal to (0)

(a l - ao)wl

+ ... + (c -

ak)wk+ I

+ (ak+l

-

C)WUI

+ ... + (an

- an-I)Wn,

This sum differs from the sum for I P(f) only in that the one term - ak)wU I is replaced by the two terms as shown. However

(au I

and this shows that I p.(f) = I P(f). A partition R is said to be a refinement of P if every point of the partition P is also a point of the partition R. Inserting a finite number of points and using induction, we conclude that if R is a refinement of P, then I R(f) = I p(f). H Q is another partition, then P and Q have a common refinement Indeed, if Q = (b o, ... Pm), then we can insert inductively b o, ... ,bm to obtain this refinement, which we denote by R. Then I P(f) = I R(f) = I Q(f).

This shows that our integral does not depend on the partition. We shall therefore denote the integral of f by I(f).

[X, §2]

INTEGRAL OF STEP MAPS

251

It is clear that a step map f is bounded, because f takes on only a finite number of values, and the maximum of the norms of these values is a bound for f. We have also an obvious bound for l(f). Preserving the preceding notations, and letting II fII be the sup norm, we find n

1/(1)1 ~

L lai -

i=l

n

ai-d

Iwd ~

L (ai -

ai-l)lIfll

i=l

< (b - a)lIf11. Except for showing that step maps form a vector space, we have proved: Lemma 2.1. The set of step maps of [a, b] into E is a subspace of the space of all bounded maps of [a, b] into E. Denote it by St([a, b], E). The map I: St([a, b], E) .... E is a linear map with bound b - a, that is

1/(f)1 < (b - a)lIfli. Proof· Let f, 9 be step maps. Suppose that f is a step map with respect to the partition P and 9 is a step map with respect to the partition Q. Let R be a common refinement of P and Q. Then both f and 9 are step maps with respect to R. Let R = (co, ... ,c,) and suppose that f has a constant value WJ+l on Cj < t < Cj+l and 9 has a constant value Vj+l on

Then f + 9 has a constant value Vj+ 1 + W j + 1 on this open interval. If d is a number, then df has a constant value dw j + 1 on this interval. Hence the set of all step maps is a vector space. Combined with the preceding results, the linearity of 1 is obvious. This proves the lemma. Let F be the space of all bounded maps from [a, b] into E. Let Fo be the subspace of step maps. We can apply the linear extension theorem to 1 and thus we know that there is a unique linear map of Po with values in E which extends I. We shall denote this linear map again by I, and call it the integral. We see that the integral is defined on all bounded maps which are uniform limits of step maps. To emphasize the dependence of 1 on the interval [a, b] we also write l(f) = 1:(1), or also

252

THE INTEGRAL IN ONE VARIABLE

[X, §3]

Lemma 2.2. Let f be a step map on [a, b]. Let a ~ c ~ b. Then f is a step map on [a, c] and on [c, b], and I~(f) = I~(f)

+ I~(f).

Proof. Let P be a partition of [a, b] with respect to which f is a step map. Let P, be the refinement of P obtained by inserting c in P. The state· ment of Lemma 2.2 is then clear from the sum (*).

Lemma 2.3. Let E = R be the real numbers. If f is a step function on [a, b] such that f;;; 0 (that is f(t) ;;; 0 for all t) then I~(f) ~ O. If f, g are step functions on [a, b] such that f ~ g, then

Proof. If f(t) ;;; 0 for all t, and P is a partition with respect to which f is a step function, then n

l~(f) =

L (a, -

ai_l)wi

i= 1

and WI ;;; 0 for all i. Thus the integral is a sum of terms each of which is ;;; 0, and is consequently ;;; O. If f ~ g, we apply what we have just proved to g - f ;;; 0 and use the linearity of the integral,

X, §3. APPROXIMATION BY STEP MAPS To get the integral defined on continuous maps, it suffices to show that these are contained in the closure of the space of step maps. Theorem 3.1. Every continuous map of [a, b] into E can be uniformly approximated by step maps. The closure of 8t([a, b], E) contains CO([a, b], E).

Proof By Theorem 2.5 of Chapter VIII, we know that a continuous map f on [a, b] is uniformly continuous. Given £, choose 0 such that if

x, ye [a, b]

[X, §3]

APPROXIMATION BY STEP MAPS

253

and Ix - yl < {) then 1f(x) - f(Y)1 < €. Let P = (ao, ... ,an) be the partition of [a, b] such that each interval [a;_1> aJ has length (b - a)/Il, and choose Il SO large that (b - a)/Il < {). If a;_ 1 ~ t < a;, define g(t)

= f(a;-.).

Then for all t in [a, b] we have Ig(l) - f(t)1 <

€,

and 9 is a step map, thus proving Theorem 3.1. Note. The proof of Theorem 2.5, Chapter VII, is self-contained and very simple, based on nothing else than the Weierstrass-Bolzano theorem. See also Theorem 2.7 of Chapter V. Thus the present theorem can be proved immediately after the Weierstrass-Bolzano theorem, and the theory of integration can thus be developed very early in the game. The closure of the space of step maps contains a slightly wider class of functions which are useful in practice, for instance in the study of Fourier series. It is the class of piecewise continuous maps. A map f: [a, b] --+ E is said to be piecewise continuous if there exists a partition

P = (ao, ... ,an) of [a, b] and for each i

= I, ... ,Il a continuous map /;: [a;_I' a;]

--+

E

such that we have f(t) = /;(1)

if a;_, < t < a;.

The graph of a piecewise continuous function looks like this:

Essentially the same argument which was used to prove Theorem 3.1 can be used to prove that a piecewise continuous map can be uniformly approximated by step maps.

254

[X, §3]

THE INTEGRAL IN ONE VARIABLE

Note that instead of saying that there exists a continuous map fi having the property stated above, we can say equivalently that f should be continuous on any open subinterval ai-I < t < ai' and that the limits lim f(t)

and

lim f(t)

1-°'-1 '>QI-I

should exist. These limits are usually denoted by lim f(t) '-0/-1

and

lim f(t).

+

We leave it as an exercise for the reader to prove that the piecewise continuous maps form a subspace of the space of bounded maps. It will be convenient to have a name for the closure of the space of step maps in the space of bounded maps. We shall call it the space of regulated maps. Thus a map is regulated if and only if it can be uniformly approximated by step maps. We denote the space of regulaled maps by Reg([a, b], E),

or

St([a, b], E).

X, §3. EXERCISES I. If I is a continuous real valued function on [a, b), show that one can approximate

I

uniformly by step functions whose values are less than or equal to those of I. and also by step functions whose values are greater than or equal to those of f. The integrals of these step functions are then the standard lower and upper Riemann sums.

2. Show that the product of two regulated maps is regulated. The product of two piecewise continuous maps is piecewise continuous.

3. On the space of regulated maps define

I: [a. b) -> C,

show that

III is regulated, and

Show that this is a seminorm (all properties of a norm except that 11/11. may be 0 without I itself being 0).

lilli, ;;; 0 but

4. Let F be the vector space of real valued regulated functions on an interval [a, b]. We have the sup norm on F. We have the seminorm of Exercise 3. It is called the L l·sem inorm. Prove that the continuous functions are dense in F, for the L I-seminorm. In other words, prove that given IE F, there exists a continuous function g on [a, b) such that III - gil. < E. [Hint: First approximate I by a step function. Then approximate a stcp function by a continuous function obtained by changing a step function only near its discontinuities.]

[X, §4]

PROPERTIES OF THE INTEGRAL

255

5. On the space of regulated functions as in Exercise 4, define the scalar product
=

f

I(x)g(x) dx.

The seminorm associated with this scalar product is called the L 2-seminorm. (Cf. Exercise 11 of Chapter VI, §2.) Show that the continuous functions are dense in F for the L 2 -seminorm. 6. The space F still being as in Exercise 4 or 5, show that the step functions are dense in F for the L'·seminorm and the U-seminorm. 7. Let F be the space of regulated functions on [a, b] once more. Let Coo = eW([a, b]) be the space of infinitely differentiable real valued functions on [a, b]. Prove that Coo is (a) L'-dense and (b) L 2 -<1ense in F. [Hint: First approximate by step functions, then smooth out the corners by using bump functions which are 0 in a l5-interval around a corner, and 1 outside a 215interval around each corner. Pick 15 sufficiently small.] Note. Exercises 4 through 7 are quite useful because there is a certain type of result which is proved by the following technique. One proves the resuit first for very smooth functions, say Coo, and then one extends the result to a wider class of very kinky functions. Examples of this procedure will be met in the theory of Fourier series. In a subsequent course, the same technique applies to prove approximation results about a fancier integral, the Lebesgue integral. See the appendix of §4 and my Real and Functional Analysis, Chapter VI, §6. Actually, the very definition of the Lebesgue integral comes from this approximation technique, by describing more precisely the behavior of the limit of an L I or L 2 Cauchy sequence of step maps or of continuous maps. The net result is that such a sequence converges pointwise .. almost everywhere" in a precise sense, and absolutely uniformly outside sets of arbitrarily small measure. See the appendix of §4.

X, §4. PROPERTIES OF THE INTEGRAL The integral being defined as a limit, we can immediately formulate the properties stated in §2 for the integral applied to limits of step maps. We consider regulated maps from [a, b] into the complete normed vector space E. From §1, we conclude: Theorem 4.1. If Un} is a uniformly convergent sequence of regulated maps converging to f, then

Let f be regulated on an interval J. If a < b are numbers of this interval,

256

THE INTEGRAL IN ONE VARIABLE

[X, §4]

we define

Theorem 4.2. For any three numbers.a, b, c in J, we then have

Proof. Suppose first that a ;;;; c ;;;; b. If {fn} is a sequence of step maps on [a, b] converging to f uniformly on [a, b], then it also converges to f uniformly on [a, c] and on [c, b]. From the basic properties of limits, and Lemma 2.2, we conclude that our relation is valid. Say now that a < b < c. Then

and hence

f=J:-J:.

Our formula follows from the definitions.

Theorem 4.3. The integral is linear, that is if f, g are regulated on [a, b] then and for any number K.

This follows by the fact that the extension theorem yields a linear map. (It is an exercise to verify the last property when E = C and K is complex.)

Theorem 4.4. Let f, g be regulated real valued functions on [a, b]. If

f

~

0, then

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 <

Ihillfll,

and this goes to 0 as h -+ O.

Theorem 4.6. Let f be a regulated real valued function on [a, b] and assume a < b. Let a :>; 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 =