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 ,

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

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

  3. $f_n$ 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 .



So , It is my humble request to tell me what mistake I am making and how to proceed to correct answer.

  • $\begingroup$ Why does continuity of $f_n$ implies what you wrote? A continuous function not necessarily satisfy the inequality you wrote. Also the derivative of a general continuously differentiable function doesn't have to be identically $0$, which seems to be implied by what you wrote. $\endgroup$ – Keen-ameteur Sep 7 '20 at 7:06
  • 1
    $\begingroup$ How about $f(x)=x^3$ . Then $f_n(x)-f'(x)=\frac 1{n^2}+\frac {3x}n $ and so for $x_n=n $ . we have $f_n(x_n)-f'(x_n) \gt 3$ showing non-uniform convergence of $f_n$ on $\mathbb{R}$ $\endgroup$ – user-492177 Sep 7 '20 at 7:54
  • $\begingroup$ @user710290 you are taking $f_n(x)−f′(x)$ but question talks about $f_n(x)−f(x)$. $\endgroup$ – Tim Sep 19 '20 at 7:18
  • $\begingroup$ The limit function of the sequence $\{f_n(x)\}$ is $f'(x)$ . That's why I am talking about $f_n(x)-f'(x)$ . I do not agree with what you said. $\endgroup$ – user-492177 Sep 19 '20 at 11:43
  • $\begingroup$ @user710290 got it !! So, 1,2 are sorted out .Can you please also help with (3) and (4)? $\endgroup$ – Tim Sep 19 '20 at 13:06

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.


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