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Thanks a lot for your help for the previous question: Let $f(x)=\sum\limits_{n=1}^\infty \frac{1}{n} \sin(\frac{x}{n})$. Where is $f$ defined? Is it continuous? Differentiable? Twice-Differentiable?

$f(x)=\sum\limits_{n=1}^\infty \frac{1}{n} \sin\left(\frac{x}{n}\right)$

So I've fully understood how the function is defined for all $x\in\mathbb{R}$ and continuous (through the concept of the uniform convergence using the Mean Value Theorem and the Weierstrass M-Test), but I'm now stuck at its differentiabiltiy and twice-differentibility.

I first naively assumed something similar to Differentiability of a Uniformly Convergent Series; my basic idea is that if a series is uniformly convergent then it's also differentiable, using *Term-by-Term Differentiablity Theorem, but the answer above the posting says that it cannot be generalized.

How should I prove $f(x)$ in my case is differentiable (and/or twice-differentiable as well)?


* Term-by-Term Differentiability Theorem: Let $f_n$ be differentiable funcitons defined on an interval A, and assume $\sum\limits_{n=1}^\infty f'_n(x)$ converges unifomly to a limit $g(x)$ on A. If there exists a point $x_0 \in [a,b]$ where $\sum\limits_{n=1}^\infty f_n(x_0)$ converges, then the series $\sum\limits_{n=1}^\infty f_n(x)$ converges uniformly to a differentiable function $f(x)$ satisfying $f'(x)=g(x)$ on A. In other words, $f(x) = \sum\limits_{n=1}^\infty f_n(x)$ and $f'(x)=\sum\limits_{n=1}^\infty f'_n(x)$

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  • $\begingroup$ Well ... what is the series of derivatives? $\endgroup$ Commented Jun 5, 2018 at 16:53

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Sketch: The cool thing here is that you don't have to prove the series defining $f$ is uniformly convergent, or even pointwise convergent, everywhere. Since the series obviously converges at $x_0=0,$ we are off to a good start in using the differentiation theorem.

Now the series of derivatives is

$$\sum_{n=1}^{\infty}\frac{\cos (x/n)}{n^2}.$$

That series converges uniformly on $\mathbb R$ by an easy application of Weierstrass M. Thus, invoking the differentiation theorem, we have the series defining $f$ uniformly convergent on every bounded subinterval of $\mathbb R$ as well as the result

$$f'(x)= \sum_{n=1}^{\infty}\frac{\cos (x/n)}{n^2},\,\,x\in \mathbb R.$$

The series of second derivatives now awaits your consideration.

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  • $\begingroup$ Hi, thank you for the answer. Please understand that I haven't had a chance to have a formal math education so I'm self-teaching math for some reason, mainly for health concerns (I guess I need to write this comment ahead every time I post a question). I don't fully understand how I can use the differentiation theorem given the fact that the series converges. Can you please explain further by editing the answer and refer to the formal name of the differentiation theorem (if it's different than ordinary differentiation rules such as continuity, function value = limit value, and so on)? $\endgroup$ Commented Jun 5, 2018 at 17:15
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    $\begingroup$ Also, the link I referred says that uniform convergence doesn't necessarily imply its differentiability, but it seems as you're using the fact that the series converges uniformly by Weierstrass-M Test to invoke the differentiation theorem (which I don't exactly know what it is); is it all right to do so? $\endgroup$ Commented Jun 5, 2018 at 17:18

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