Math 1270 HonorsODE I

Fall, 2008Remark on uniform convergence of series

1 Comparison test for uniform convergence

In the introductory notes we discussed uniform convergence and norm. There is oneimportant, and easy to understand result about uniform convergence which we need,but did not discuss. This is in B&S, but not until Chapter 9 (pg. 268), so perhapsnone of you have seen it before.First, recall the comparison test for convergence of an innite series of numbers

(B&S, 3.7.7, pg. 93):

Lemma 1 Suppose that fung and fMng are sequences of real numbers, with 0 un Mn for each positive integer n: If

1Xn=0

Mn converges, then1Xn=0

un converges.

There is also a comparison test for uniform convergence of a series of functions:In B&S it is given on page 268, and called Weierstrass M -test.

Theorem 2 Let ffng be a sequence of functions, and fMng a sequence of positivenumbers, such that in some interval a t b, jun (t)j Mn, for each n . If theseries of numbers

1Xn=0

Mn converges, then the series of functions1Xn=0

fn (t) converges

uniformly for t 2 [a; b] ;to a function (t).

A theorem in our earlier set of notes, and in B&S, pg. 234, implies that if eachof the functions fn is continuous on [a; b], then the limit (t) of the series is alsocontinuous.

As a simple example, let fn (t) = tn for 0 t 5: Let Mn = 5nn! . The series1Xn=1

5n

n!

1

converges (to e5 ). Hence1Xn=1

tn

n!

converges for any t 2 [0; 5] (which we already knew), and this convergence is uniformin [0; 5]. It is also uniform in [0; T ] for any T: But it is not uniform in [0;1). Thisis because, for any given n, limt!1 t

n

n!=1.

As a second example, let fn (t) =(cos t)n

n!, and let Mn = 1n! . Then jfn (t)j Mn

for each n; and for any t. Hence,1Xn=0

fn (t) =

1Xn=0

(cos t)n

n!converges uniformly in any

interval [a; b] . In fact, it converges uniformly on (1;1). .

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