The Real And Complex Number Systems

By (*) and (**), we obtain that given ε > 0, there exists a positie integer N
such that as n ≥ N, we have
|f n (x n ) − f (x)| = |f n (x n ) − f (x n )| + |f (x n ) − f (x)|
< ε 2 + ε 2
= ε.
That is, we have proved that f n (x n ) → f (x) .
9.8 Let {f n } be a seuqnece of continuous functions defined on a compact
set S and assume that {f n } converges pointwise on S to a limit function f.
Prove that f n → f uniformly on S if, and only if, the following two conditions
hold.:
(i) The limit function f is continuous on S.
(ii) For every ε > 0, there exists an m > 0 and a δ > 0, such that n > m
and |f k (x) − f (x)| < δ implies |f k+n (x) − f (x)| < ε for all x in S and all
k = 1, 2, ...
Hint. To prove the sufficiency of (i) and (ii), show that for each x 0 in S
there is a neighborhood of B (x 0 ) and an integer k (depending on x 0 ) such
that
|f k (x) − f (x)| < δ if x ∈ B (x 0 ) .
By compactness, a finite set of integers, say A = {k 1 , ..., k r } , has the property
that, for each x in S, some k in A satisfies |f k (x) − f (x)| < δ. Uniform
convergence is an easy consequences of this fact.
Proof: (⇒) Suppose that f n → f uniformly on S, then by Theorem
9.2, the limit function f is continuous on S. In addition, given ε > 0, there
exists a positive integer N such that as n ≥ N, we have
|f n (x) − f (x)| < ε for all x ∈ S
Let m = N, and δ = ε, then (ii) holds.
(⇐) Suppose that (i) and (ii) holds. We prove f k → f uniformly on S as
follows. By (ii), given ε > 0, there exists an m > 0 and a δ > 0, such that
n > m and |f k (x) − f (x)| < δ implies |f k+n (x) − f (x)| < ε for all x in S
and all k = 1, 2, ...
Consider ∣ ∣f k(x0 ) (x 0 ) − f (x 0 ) ∣ ∣ < δ, then there exists a B (x 0 ) such that as
x ∈ B (x 0 ) ∩ S, we have
∣ fk(x0 ) (x) − f (x) ∣ ∣ < δ
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