Is the following continuously differentiable function uniformly continuous This particular question was asked in my real analysis quiz and my answers in it were not correct . So, I am asking them here .

Question: Let f be continuously differentiable on $\mathbb{R}$ . Let $f_n(x)=n (f(x+1/n)-f(x))$ . Then ,


*

*$f_n$ converges uniformly on $\mathbb{R}$ .


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


*$f_n$ converges to the derivative of f uniformly on [0,1] .
4.there is no guarantee that $f_n$ converges on any open interval .
Attempt: Continuity of $f_n$ implies that $f(x+1/n)-f(x) < \frac {\epsilon} {n}$  .  So, $f_n \to 0  $
as  $ n\to \infty$ . Also , convergence must be uniform by using defination of uniform convergence.So, (a) is correct ,(b) is wrong.
Derivative of 0 =0 . So, (c) is correct . (d) is wrong .
But answers given in answer key are different .

Answer:


 2,3

So , It is my humble request to tell me what mistake I am making and how to proceed to correct answer.
 A: Prove that $f_n (x) \to f'(x) $ as $n\to \infty$
For $(1)$ and $(2)$ take $f(x)=x^3$ and show the non-uniform convergence of $f_n$ on $\mathbb{R}$ .
For $(3)$
Claim:  $f_n$ converges uniformly to $f'$ on $[0,1]$
Proof : Note that $f$ is continously differentiable, i.e $f'$ is continous .
Since continous function on a compact set is uniformly continous , so $f'$ is uniformly continous on $[0,1]$.
So given $\epsilon \gt 0, \exists \delta \gt 0$   such that $\forall x,y\in [0,1]$ with $|x-y| \lt \delta$ , we have $|f'(x)-f'(y)| \lt \epsilon $
By Archimedan Property of Real Numbers , there exist $N \in \mathbb{N} $ such that $\forall n\gt N$ , have $\frac 1n \lt \delta$
Now ,let $x\in [0,1]$ be arbitary  and $n\gt N$
Then , $f(x+\frac 1n)-f(x)=\frac 1n f'(a)$   for some $x \lt a \lt x+\frac 1n$, by Mean Value Theorem
Thus $| f_n(x)-f'(x)|=|f'(a)-f'(x)|\lt \epsilon $ since $|x-a| \lt \frac 1n \lt \delta$
Thus the claim is proved.
$(4)$ is obviously false since the sequence always converges pointwise to the derivative, whatever be the domain.
