The Real And Complex Number Systems

Sequences of Functions
Uniform convergence
9.1 Assume that f n → f uniformly on S and that each f n is bounded on
S. Prove that {f n } is uniformly bounded on S.
Proof: Since f n → f uniformly on S, then given ε = 1, there exists a
positive integer n 0 such that as n ≥ n 0 , we have
|f n (x) − f (x)| ≤ 1 for all x ∈ S. (*)
Hence, f (x) is bounded on S by the following
|f (x)| ≤ |f n0 (x)| + 1 ≤ M (n 0 ) + 1 for all x ∈ S. (**)
where |f n0 (x)| ≤ M (n 0 ) for all x ∈ S.
Let |f 1 (x)| ≤ M (1) , ..., |f n0 −1 (x)| ≤ M (n 0 − 1) for all x ∈ S, then by
(*) and (**),
So,
|f n (x)| ≤ 1 + |f (x)| ≤ M (n 0 ) + 2 for all n ≥ n 0 .
|f n (x)| ≤ M for all x ∈ S and for all n
where M = max (M (1) , ..., M (n 0 − 1) , M (n 0 ) + 2) .
Remark: (1) In the proof, we also shows that the limit function f is
bounded on S.
(2) There is another proof. We give it as a reference.
Proof: Since Since f n → f uniformly on S, then given ε = 1, there exists
a positive integer n 0 such that as n ≥ n 0 , we have
|f n (x) − f n+k (x)| ≤ 1 for all x ∈ S and k = 1, 2, ...
So, for all x ∈ S, and k = 1, 2, ...
|f n0 +k (x)| ≤ 1 + |f n0 (x)| ≤ M (n 0 ) + 1 (*)
where |f n0 (x)| ≤ M (n 0 ) for all x ∈ S.
Let |f 1 (x)| ≤ M (1) , ..., |f n0 −1 (x)| ≤ M (n 0 − 1) for all x ∈ S, then by
(*),
|f n (x)| ≤ M for all x ∈ S and for all n
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