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Let $(f_n(x))_{n=1}^{\infty}$ a series of uniformly continuous functions, $\mathbb{R}\to\mathbb{R}$ which uniformly converges to the function $f$, and a continuous function $g$ :$\mathbb{R}\to\mathbb{R}$.

I need to give an example where $(g\circ f_n)_{n=1}^{\infty}$ is not uniformly converges and to prove that if $g$ would be uniformly continuous so $(g\circ f_n)_{n=1}^{\infty}$ would uniformly converges.

As an example I gave $g=\sin x$ and $f_n= \frac{1} {n+x^2}$, but I'm having a hard time proving the claim.

Any hints? Thanks!

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Is this a homework problem? – user38268 Dec 13 '11 at 14:54
No, as a part of studying to a quiz. – Jozef Dec 13 '11 at 15:38
Since $\sin x$ is uniformly continuous, it's not surprising you're having trouble with that example. – Robert Israel Dec 13 '11 at 18:22
up vote 3 down vote accepted

To prove the second claim, now assume that $g$ is uniformly continuous. By definition of uniformly continuous, for all $\epsilon>0$, there exists $\delta>0$ which depend only on $\epsilon$ such that $$|g(x)-g(y)|<\epsilon\mbox{ for all $x, y\in\mathbb{R}$ such that }|x-y|<\delta.$$ Since $f_n$ converges uniformly to $f$, for the above $\delta>0$, there exists an positive integer $N$ such that if $n\geq N$, then $$|f_n(x)-f(x)|<\delta\mbox{ for all }x\in\mathbb{R},$$ which implies that if $n\geq N$, then $$|g\circ f_n(x)-g\circ f(x)|=|g(f_n(x))-g(f(x))|<\epsilon\mbox{ for all }x\in\mathbb{R}.$$ This proves the claim.

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@Josef Note you do not need the uniform continuity (or even continuity) of the $f_n$. – David Mitra Dec 13 '11 at 14:42
I should have known this one! thanks a lot! – Jozef Dec 13 '11 at 20:06

For the counterexample:

Take $f_n(x)=x+{1\over n}$, $f(x)=x$, and $g(x)=x^2$. Then each $f_n$ is uniformly continuous, and $\{f_n\}$ converges uniformly to $f$.

Now, $g(f_n(x))=\left(x+{1\over n}\right)^2 $ converges pointwise to $h(x)=x^2$.

We have: $$ |(x+{\textstyle{1\over n}})^2 -x^2|= \left|{2 x \over n} +{1\over n^2}\right|, $$ which can be made arbitrarily large for any fixed $n$. This shows $g\circ f_n$ does not converge uniformly to $h$.

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