The Real And Complex Number Systems

That is, f n + g n → f + g uniformly on S.
Remark: There is a similar result. We write it as follows. If f n → f
uniformly on S, then cf n → cf uniformly on S for any real c. Since the proof
is easy, we omit the proof.
(b) Let h n (x) = f n (x) g n (x) , h (x) = f (x) g (x) , if x ∈ S. Exercise 9.2
shows that the assertion h n → h uniformly on S is, in general, incorrect.
Prove that it is correct if each f n and each g n is bounded on S.
Proof: Since f n → f uniformly on S and each f n is bounded on S, then
f is bounded on S by Remark (1) in the Exercise 9.1. In addition, since
g n → g uniformly on S and each g n is bounded on S, then g n is uniformly
bounded on S by Exercise 9.1.
Say |f (x)| ≤ M 1 for all x ∈ S, and |g n (x)| ≤ M 2 for all x and all n. Then
given ε > 0, there exists a positive integer N such that as n ≥ N, we have
and
|f n (x) − f (x)| <
|g n (x) − g (x)| <
which implies that as n ≥ N, we have
ε
2 (M 2 + 1)
ε
2 (M 1 + 1)
|h n (x) − h (x)| = |f n (x) g n (x) − f (x) g (x)|
for all x ∈ S
for all x ∈ S
= |[f n (x) − f (x)] [g n (x)] + [f (x)] [g n (x) − g (x)]|
≤ |f n (x) − f (x)| |g n (x)| + |f (x)| |g n (x) − g (x)|
ε
<
2 (M 2 + 1) M ε
2 + M 1
2 (M 1 + 1)
< ε 2 + ε 2
= ε
for all x ∈ S. So, h n → h uniformly on S.
9.4 Assume that f n → f uniformly on S and suppose there is a constant
M > 0 such that |f n (x)| ≤ M for all x in S and all n. Let g be continuous
on the closure of the disk B (0; M) and define h n (x) = g [f n (x)] , h (x) =
g [f (x)] , if x ∈ S. Prove that h n → h uniformly on S.
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