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Let $f(x),f_n(x):[a,b]\to[c,d]$ s.t $f_n\to f$ uniformly on [a,b]. Let $g:[c,d]\to\mathbb R$ continious function. Prove that $g(f_n(x))\to g(f(x))$

Let $\epsilon>0$. If $f_n\to f$, $\exists n_0\in\mathbb N$ such that for all $n>n_0$ $|f_n(x)-f(x)|<\epsilon$. Since g is continious on [c,d], exists $\delta>0$ s.t if $|x-x_0|<\delta$ then $|g(x)-g(x_0)|<\epsilon$. In fact $\forall x\in [a,b]: f_n(x),f(x)\in[c,d]$ so if $n>n_0$ and $|f_n(x)-f(x)|<\delta$ then $|g(f_n(x))-g(f(x))|<\epsilon$. I'm sure something is really wrong with this proof (except it seems I proved it only for specific environment of $x_0$). What is it and how can my proof can be improved?

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What makes you sure that something is wrong? Would you mind splitting up the logic a little to make it more readable? I don't think the $|x-x_0| < \delta$ part is worded correctly; you should say that "for every $\epsilon > 0$ $\exists \delta > 0$ such that...". –  abiessu Aug 24 '13 at 17:55
    
Your title says that you want $g\circ f_n$ to converge uniformly, but you omitted this from the text. Note that your proof uses the fact that a continuous function on a closed interval is uniformly continuous. It is not specific to any $x_0$ when you realize that. –  Ted Shifrin Aug 24 '13 at 17:58
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up vote 3 down vote accepted

Since $f_n\to f$ uniformly, there exists, given $\eta >0$; $N=N_\eta$ such that $$|f_n(x)-f(x)|<\eta $$ whenever $n\geqslant N$, $x\in [a,b]$.

Since $g:[c,d]\to\Bbb R$ is continuous, it is uniformly continuous, whence given $\varepsilon >0$ there exists $\eta=\eta_\varepsilon >0$ such that for any $x,x'\in [c,d]$ with $|x-x'|<\eta$ we have $$|g(x)-g(x')|<\varepsilon$$

Combine the above to find, given $\varepsilon >0$; an $N=N_\varepsilon$ such that $$|g\circ f_n(x)-g\circ f(x)|<\varepsilon$$ whenever $n\geqslant N$.

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I'm assuming you meant $x, x' \in [c,d]$. –  abiessu Aug 24 '13 at 18:00
    
why the composition is less then epsilon? –  Coargu Aliquis Aug 24 '13 at 18:14
    
@CoarguAliquis That is for you to explain! Try to write out the proof. –  Pedro Tamaroff Aug 24 '13 at 18:16
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