2
$\begingroup$

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 $\forall x \in A.$

$\endgroup$
1
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.