# 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 ,

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 .

2,3

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

• 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. – Keen-ameteur Sep 7 '20 at 7:06
• 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 $\mathbb{R}$ – user-492177 Sep 7 '20 at 7:54
• @user710290 you are taking $f_n(x)−f′(x)$ but question talks about $f_n(x)−f(x)$. – Tim Sep 19 '20 at 7:18
• 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. – user-492177 Sep 19 '20 at 11:43
• @user710290 got it !! So, 1,2 are sorted out .Can you please also help with (3) and (4)? – 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.