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Let $f: \Bbb R \to \Bbb R$ be a derivable function. $f'$ is uniformly continuous in $\Bbb R$

Prove that $[n(f(x+1/n)-f(x))]$ converges uniformly to $f'(x)$

I'm having a hard time seeing why does $f'$ being uniformly continuous help me and I tried to assume negatively that it doesn't converge uniformly but that didn't really get me anywhere.

Thanks

edit: Can anyone think of a derivable function that its derivative isn't uniformly continuous and doesn't maintain:$[n(f(x+1/n)−f(x))]$ converges uniformly to $f'(x)$ ?

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    $\begingroup$ Use the mean value theorem and then the uniform continuity of the derivative. $\endgroup$ – Svetoslav Jul 31 '15 at 14:47
  • $\begingroup$ Thank you Svetoslav! Can you (or anyone else) think of a derivable function that its derivative isn't uniformly continuous and doesn't maintain :$[n(f(x+1/n)−f(x))]$ converges uniformly to $f′(x)$ $\endgroup$ – lfc Jul 31 '15 at 14:58
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    $\begingroup$ @lfc: the one I have seen is $f(x)=\frac{\sin(e^x)}{1+x^2}$. It is uniformly continuous with unbounded derivative. If we relax the requirement that $f$ be defined on $\mathbb R$ then $f(x)=\sqrt x$ defined on $R^+$ will do. $\endgroup$ – Matematleta Jul 31 '15 at 15:55
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The difference quotient is equal to $f'(\xi)$ for some $\xi\in(x,x+1/n)$. Now because of the uniform continuity of the derivative you have that for each $\epsilon >0\exists \delta >0$ that $|f'(x)-f'(y)|\leq \epsilon \forall x,y: |x-y|\leq \delta$. Chose $1/n\leq \delta$ and you have it.

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For each $n\in N$ there is a $c_{n}\in (x,x+\frac{1}{n})$ such that $$\tag1n(f(x+1/n)-f(x))=f'(c_{n})$$

Let $\epsilon>0$.

Since $f'$ is uniformly continuous on $\mathbb R$ we may choose $\delta >0$ such that

$\vert x-y\vert <\delta \Rightarrow \vert f'(x)-f'(y)\vert <\epsilon$.

Now, choose $N\in \mathbb N$ so large that $n>N\Rightarrow \frac{1}{n}<\delta $. Then as soon as $n>N$ we have $$\tag2\vert f'(c_{n})-f'(x)\vert <\epsilon $$

Comparing $(1)$ and $(2)$ now gives the result.

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