Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of functions that converges uniformly to $f$ on $A$ and $|f_n(x)| \leq M, \forall x\in A, \forall n \in \mathbb{N}.$ If $g$ is continuous on $[-M,M],$ does $(g o f)_{n \in \mathbb{N}}$ converge uniformly on $A$ ?
May I verify if my proof is correct? Thank you.
Proof:
Claim: $(f_n)_{n \in \mathbb{N}}$ is a sequence that converges in $[-M,M] .$
Proof :$\forall \epsilon > 0\exists K \in \mathbb{N}: |f_n(x) – f(x)| < \epsilon, \forall x \in A.$ $ \implies |f(x)| = |f(x) – f_n(x) +f_n(x)| \leq \epsilon + M, \forall x \in A, \forall n > K.$
Now, $g$ is uniformly continuous on $[-M,M]$ and $(f_n)_{n \in \mathbb{N}}$ is a convergent sequence of functions in $[-M,M]$
Given $\epsilon > 0, \exists \delta> 0$ such that $|g(f_m(x))-g(f_n(x))| < \epsilon$ whenever $|f_m(x) -f_n(x)| < \delta. $
Since $(f_n)_{n \in \mathbb{N}}$ is uniformly convergent, given $\delta > 0, \exists N \in \mathbb{N}$ such that $|f_m(x) -f_n(x)| < \delta,$ whenever $m\geq n>N$ and $\forall x \in A. $ It follows that $|g(f_m(x))-g(f_n(x))| < \epsilon, $ whenever $m \geq n>N$ and $\forall x \in A.$
Best Answer
I think you need to use uniform continuity of $g$ to get $\|g\circ f_{n}-g\circ f\|_{L^{\infty}(A)}\to0$ not just continuity (not that you don't have this in the question I just didn't see you use this). Other than that the proof seems good.
Since [-M,M] is closed bounded interval, by Uniform Continuity Theorem, since g is continuous on a closed bounded interval, then g is uniformly continuous on [-M, M] (The proof is using contradiction, learnt long before this chapter)