Question

Prove that if a series of continuous functions converges uniformly, then the sum function is also continuous. 

Answer 1

We say $\sum^{∞}_{n=i}f_n$ converges uniformly to f if $S_N = \sum^N_{n=i} f_n$ converges uniformly to f.

Since each $f_n$ is continuous, as sum of continuous function is continuous => $S_N$ is continuous if $N \in \mathbb{N}$ . Since $S_N$ converges uniformly to f, given any ε>0 if $N_0 \in \mathbb{N}$ such that if $N ≥N_0$

$|S_N^* -f(x)| < \frac{ε}{3} \;if \;x \in X$

Take any $x_0 \in X$ since $S_{N_0}$ is continuous $S_0$ if $δ>0$ such that if $|x_0-x|< δ = 1$ =>

$|S_{N_0}(x_0) -S_{N_0}(x)| < \frac{ε}{3}$

Now

$|f(x_0) -f(x)| ≤ |f(x_0) -S_{N_0}(x_0)| + |S_{N_0}(x_0) -S_{N_0}(x)| + |S_{N_0}(x) -f(x)| \\ ≤ \frac{ε}{3} + \frac{ε}{3} + \frac{ε}{3} \\ ≤ ε$

$x \in (x_0- δ, x_0+ δ)$

By continuity of $S_{N_0}$

$|S_{N_0}(x) -S_{N_0}(x_0)| < \frac{ε}{3}$

if $|x_0-x| < δ$

If $x_0 \in x$ given any ε>0

If S>0 such that

$|x_0-x| < δ$ => $|f(x) -f(x_0)|< ε$

Hence f is continuous.