Convergence when the derivative is uniformly continuous

Question

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 $f'(x)$ ?

Answer 1

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.

Answer 2

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.