Let f be continuously differentiable on R. Let $f_n(x) = n\left(f\left(x + \frac1n\right) - f(x)\right)$. Let $f$ be continuously differentiable on $\mathbb R$. Let $f_n(x) = n\left(f\left(x + \frac1n\right) - f(x)\right)$. 
Then show that


*

*$f_n$ converges on $\mathbb R$, but not necessarily uniformly

*$f_n$ converges to the derivative of $f$ uniformly on $[0, 1]$.


I could show that $f_n$ converges to derivative of $f$. But could not show the rest.
Thanks
 A: For the first part take $f=x^3$ then $f_n(x)=3x^2+3x/n+1/n^2$
Clearly $f_n(x$) converges to $3x^2$
On the other hand, $f_n(x)\rightarrow f(x)$ uniformly on $\mathbb{R}$ means
$sup_{x\in \mathbb{R}}|f_n(x)−f(x)|\rightarrow 0$ as $n\rightarrow \infty$ when supremum is taken over all real numbers.
Now 
$sup_{x\in \mathbb{R}}|f_n(x)−f(x)|=sup_{x\in \mathbb{R}} ∣3x/n+1/n∣\geq |3n/n|=3 \nrightarrow 0$ as $n\rightarrow \infty$.
(The supremum is taken oven the  $\mathbb{R}$)
Hence it's not uniformly convergent.
For the 2nd part use mean value theorem and use the uniform continuity of $f$ in a closed bounded interval. 
A: Here's the answer to second part. 
By Lagrange's Mean Value Theorem (or Taylor expansion) we have 
$$f(x+\frac{1}{n})=f(x)+\frac{1}{n}f'(\theta_{x,n})$$ where $|\theta_{x,n}-x|<\frac{1}{n}$.  Then we have $f_n(x)=f'(\theta_{x,n})$ $\forall x\in$ $[0,1]$. Now $f'$ is uniformly continuous on $[0,1]$ since $[0,1]$ is compact. So given $\epsilon >0$ we have $\exists$ $\delta$$>0$ such that $|x-y|$$<\delta$ implies $|f'(x)-f'(y)|$$<$ $\epsilon$. Thus we have $\exists$ $N$$\in$$\mathbb{N}$ such that for all $n>N$ we have $|\theta_{x,n}-x|<\frac{1}{n}$$<$ $\delta$ This implies $\forall$ $x$ $\in$ $[0,1]$ and $n>N$ we have $|x-\theta_{x,n}|$$<$ $\delta$. This implies $|f'(x)-f'(\theta_{x,n})|$$<$ $\epsilon$ i.e. $|f_n(x)-f'(x)|$$<$ $\epsilon$ $\forall$ $x$ $\in$ $[0,1]$ $\forall $ $n>N$. Thus $f_n\rightarrow  f'$ uniformly.
