The Real And Complex Number Systems

Hence, {g (x) x n } converges uniformly on (1 − δ, 1) .
So, from above sayings, we have proved that the sequence of functions
{g (x) x n } converges uniformly on [0, 1] .
Remark: It is easy to show the followings by definition. So, we omit the
proof.
(1) Suppose that for all x ∈ S, the limit function f exists. If f n → f
uniformly on S 1 (⊆ S) , then f n → f uniformly on S, where # (S − S 1 ) <
+∞.
(2) Suppose that f n → f uniformly on S and on T. Then f n → f uniformly
on S ∪ T.
9.7 Assume that f n → f uniformly on S and each f n is continuous on S.
If x ∈ S, let {x n } be a sequence of points in S such that x n → x. Prove that
f n (x n ) → f (x) .
Proof: Since f n → f uniformly on S and each f n is continuous on S, by
Theorem 9.2, the limit function f is also continuous on S. So, given ε > 0,
there is a δ > 0 such that as |y − x| < δ, where y ∈ S, we have
|f (y) − f (x)| < ε 2 .
For this δ > 0, there exists a positive integer N 1 such that as n ≥ N 1 , we
have
|x n − x| < δ.
Hence, as n ≥ N 1 , we have
|f (x n ) − f (x)| < ε 2 . (*)
In addition, since f n → f uniformly on S, given ε > 0, there exists a
positive integer N ≥ N 1 such that as n ≥ N, we have
|f n (x) − f (x)| < ε 2
for all x ∈ S
which implies that
|f n (x n ) − f (x n )| < ε 2 . (**)
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