1549901369-Elements_of_Real_Analysis__Denlinger_

560 Chapter 9 • Sequences and Series of Functions

Since this is true Vm, n ;::: n 0 , {Ln} is a Cauchy sequence, and so it must
converge. Let

L = n->oo lim Ln.

The proof will be complete when we prove that lim f ( x) = lim Ln ( = L).

x-+xo n-+oo

Let c > 0. Vx ES - {xo}, and Vn EN,

lf(x) - LI ::=; lf(x) - fn(x)I + lfn(x) - Lnl + ILn - LI.

Since f n ---+ f uniformly on S - { Xo},

£

3 n1 EN 3 n 2: n1 ::::} llf n - fll < 3

£

::::} lfn(x) - f(x)I < 3 Vx ES - {xo}.

Since Ln---+ L, 3n2 EN 3 n 2: n2::::} ILn - LI< ~·
Choose any n 3 ;::: max{ n 1 , n2} and hold this n 3 fixed throughout the re-
mainder of the proof. Since lim f n 3 ( x) = Ln 3 , we can choose o > 0 such

that

X-+XQ

0 < Ix - Xo I < O ::::} If n 3 ( X) - Ln 3 I < ~ ·

(as well as lfn 3 (x) - f(x)I <~and ILn 3 - LI<~)

::::} If ( X) - LI ::=; If n 3 ( X) - Ln 3 I + If n 3 ( X) - f ( X) I + I Ln 3 - LI

< ~ + ~ + ~ = £.

Therefore, lim f(x) = L. •
X-+Xo

Corollary 9.3.6 (Uniform Convergence Preserves Continuity) If f n ---+
f uniformly on a set S ~JR, and if each f n is continuous on S, then f is con-
tinuous on S.

Proof. Suppose fn ---+ f uniformly on S and each fn is continuous on S.

Then Vxo ES, lim f(x) = lim [ lim fn(x)] since fn---+ f (uniformly) on S;
x-+xo x-+xo n-+oo

= n-+oo lim [ lim x-+xo fn(x)] by Theorem 9.3.5;

= lim [fn(xo)] since fn is continuous at xo;

n->oo

= f (xo) since f n ---+ f on S.

Therefore, f is continuous at x 0. •

00
Corollary 9.3. 7 If I: fk = f uniformly on a set S of real numbers, and each
k=O
function fk is continuous on S, then the limit function f is continuous on S.