Properties of Uniformly Convergent Series

Properties of Uniformly Convergent Series

Uniform convergence derives its main importance from two facts:

1. If a series of continuous terms is uniformly convergent, its sum is also continuous (Theorem 2, below).

2. Under the same assumptions, termwise integration is permissible (Theorem 3). This raises two questions:

1. How can a converging series of continuous terms manage to have a discontinuous sum? (Example 2)

2. How can something go wrong in termwise integration? (Example 3) Another natural question is:

3. What is the relation between absolute convergence and uniform convergence? The surprising answer: none. (Example 5)

These are the ideas we shall discuss. If we add finitely many continuous functions, we get a continuous function as their sum.

Example 2 will show that this is no longer true for an infinite series, even if it converges absolutely. However, if it converges uniformly, this cannot happen, as follows.

THEOREM 2 Continuity of the Sum

Let the series

a f m (z) ⫽ f 0 (z) ⫹ f 1 (z) ⫹ Á

m⫽0

be uniformly convergent in a region G. Let F (z) be its sum. Then if each term f m (z) is continuous at a point z 1 in G, the function F (z) is continuous at z 1 .

PROOF Let be s n (z) the n th partial sum of the series and R n (z) the corresponding remainder: s n ⫽ f 0 ⫹ f 1 ⫹Á⫹ f n , R n ⫽ f n⫹1 ⫹ f n⫹2 ⫹Á . Since the series converges uniformly, for a given P⬎0 we can find an N⫽N (P) such that

ƒR N (z) ƒ ⬍ for all z in G.

Since s N (z) is a sum of finitely many functions that are continuous at z 1 , this sum is

continuous at z 1 . Therefore, we can find a d⬎0 such that P

ƒs N (z) ⫺ s N (z 1 )ƒ⬍

for all z in G for which ƒz⫺z 1 ƒ ⬍ d.

Using F⫽s N ⫹ R N and the triangle inequality (Sec. 13.2), for these z we thus obtain

ƒ F(z) ⫺ F (z 1 )ƒ⫽ƒs N (z) ⫹ R N (z) ⫺ [s N (z 1 )⫹R N (z 1 )] ƒ

P P P ⬉ƒs N (z) ⫺ s N (z 1 )ƒ⫹ƒR N (z) ƒ ⫹ ƒ R N (z 1 )ƒ⬍ ⫹ ⫹ ⫽ P.

SEC. 15.5 Uniform Convergence. Optional

701 EXAMPLE 2 Series of Continuous Terms with a Discontinuous Sum

Consider the series

(x real).

This is a geometric series with q⫽ 1 2 ) times a factor x >(1 ⫹ x 2 . Its nth partial sum is

2 ) n (1 ⫹ x d . We now use the trick by which one finds the sum of a geometric series, namely, we multiply s n (x) by

(1 ⫹ x 2 ) n⫹1 d . Adding this to the previous formula, simplifying on the left, and canceling most terms on the right, we obtain

The exciting Fig. 368 “explains” what is going on. We see that if x⫽ 0 , the sum is

s (x) ⫽ lim n:⬁ s n (x) ⫽ 1 ⫹ x 2 , but for x⫽ 0 we have s n (0) ⫽ 1 ⫺ 1 ⫽ 0 for all n, hence s (0) ⫽ 0. So we have the surprising fact that the sum

is discontinuous (at x⫽ 0), although all the terms are continuous and the series converges even absolutely (its terms are nonnegative, thus equal to their absolute value!).

Theorem 2 now tells us that the convergence cannot be uniform in an interval containing x⫽ 0. We can also

verify this directly. Indeed, for x⫽ 0 the remainder has the absolute value ƒR 1

n (x) ƒ ⫽ ƒ s (x) ⫺ s n (x) ƒ ⫽

2 (1 ⫹ x ) n

and we see that for a given P (⬍1) we cannot find an N depending only on such that P ƒR n ƒ⬍P for all n⬎N (P) and all x, say, in the interval

Fig. 368. Partial sums in Example 2

Termwise Integration

This is our second topic in connection with uniform convergence, and we begin with an

CHAP. 15 Power Series, Taylor Series

EXAMPLE 3 Series for Which Termwise Integration Is Not Permissible

Let 2 u

m (x) ⫽ mxe ⴚmx and consider the series

in the interval

0 ⬉ x ⬉ 1. The nth partial sum is

s n ⫽ u 1 ⫺ u 0 ⫹ u 2 ⫺ u 1 ⫹Á⫹ u n ⫺ u nⴚ1 ⫽ u n ⫺ u 0 ⫽ u n . Hence the series has the sum F (x) ⫽ lim n:⬁ s n (x) ⫽ lim n:⬁ u n (x) ⫽ 0 (0 ⬉ x ⬉ 1) . From this we obtain

F (x) dx ⫽ 0.

On the other hand, by integrating term by term and using f 1 ⫹ f 2 ⫹Á⫹ f n ⫽ s n , we have

a f m (x) dx ⫽ lim n:⬁ m⫽1 冮 0 m⫽1 a 冮 f m (x) dx ⫽ lim 0 n:⬁ 冮 s n (x) dx. 0

Now s n ⫽ u n and the expression on the right becomes

ⴚnx 2 1 lim 1 ⴚn

n:⬁ u 冮 (x) dx ⫽ lim n 0 n:⬁ nxe dx ⫽ lim n:⬁ (1 ⫺ e )⫽ , 冮 0 2 2

but not 0. This shows that the series under consideration cannot be integrated term by term from x⫽ 0 to x⫽ 1.

䊏 The series in Example 3 is not uniformly convergent in the interval of integration, and

we shall now prove that in the case of a uniformly convergent series of continuous functions we may integrate term by term.

THEOREM 3 Termwise Integration

be a uniformly convergent series of continuous functions in a region G. Let C be any path in G. Then the series

m⫽ 0 冮 C 冮 C 冮 C

is convergent and has the sum

F (z) dz.

PROOF From Theorem 2 it follows that F (z) is continuous. Let s n (z) be the nth partial sum of the given series and R n (z) the corresponding remainder. Then F⫽s n ⫹ R n and by integration,

F (z) dz ⫽ s n (z) dz ⫹

R n (z) dz.

SEC. 15.5 Uniform Convergence. Optional

Let L be the length of C. Since the given series converges uniformly, for every given P⬎0 we can find a number N such that ƒR n (z) ƒ ⬍ P >L for all n⬎N and all z in G. By

applying the ML-inequality (Sec. 14.1) we thus obtain

Since R n ⫽ F⫺s n , this means that

` F (z) dz ⫺ s n 冮 (z) dz 冮 `⬍P

for all n⬎N .

Hence, the series (4) converges and has the sum indicated in the theorem. 䊏 Theorems 2 and 3 characterize the two most important properties of uniformly convergent

series. Also, since differentiation and integration are inverse processes, Theorem 3 implies

THEOREM 4 Termwise Differentiation

Let the series f 0 (z) ⫹ f 1 (z) ⫹ f 2 (z) ⫹ Á

be convergent in a region G and let F (z)

be its sum. Suppose that the series f 0 r (z) ⫹ f 1 r (z) ⫹ f 2 r (z) ⫹ Á converges uniformly

in G and its terms are continuous in G. Then

F r (z) ⫽ f 0 r (z) ⫹ f 1 r (z) ⫹ f 2 r (z) ⫹ Á

for all z in G.