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.


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 $\forall x \in A.$


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)

  • $\begingroup$ Thanks for the instruction, I have made the relevant amendments. Is my proof valid now? $\endgroup$ – Alexy Vincenzo Nov 24 '13 at 8:52
  • $\begingroup$ You should probably justify why $g$ is uniformly continuous. Also I think you want to estimate $|g(f_{n}(x))-g(f(x))|$ not $|g(f_{m}(x))-g(f_{n}(x))|$. $\endgroup$ – user71352 Nov 30 '13 at 6:39
  • $\begingroup$ Cauchy Criterion for Uniform convergence: A sequence $(f_n)$ of functions converges uniformly on $E$ iff for each $\epsilon>0,$ there exists $K \in \mathbb{N}$ such that $\|f_n-f_m\|_E = sup\{|f_n(x)-f_m(x)|: x \in E\} < \epsilon, \forall n,m \geq K.$ Now, I have shown $|g(f_m(x))-g(f_n(x))| < \epsilon, $ whenever $m \geq n>N$ and $\forall x \in A.$ Why is it wrong to conclude that $\|gof_n-gof_m\|_A \leq \epsilon \ ?$ $\endgroup$ – Alexy Vincenzo Dec 4 '13 at 14:16

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