1
$\begingroup$

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 .

Answer:

2,3

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

$\endgroup$
6
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.