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find an upper and lower limit for the number of the times the following functions are differentiable and that their derivative is continious

$$f(x) = \sum_1^\infty \frac{e^{inx}}{n^4}$$ $$g(x) = \sum_1^\infty \frac{e^{inx}}{n^{4+\epsilon}} \quad 0 \lt\epsilon\lt1$$ $$h(x) = \sum_{-\infty}^\infty \frac{e^{inx}}{2^{\lvert n\rvert}}$$

I know that if f is k times differentiable and that the k derivative is continious then the Fourier coefficients $\hat{f(n)} = o(\lvert n^{-k}\rvert)$

so i can conclude that the an upper limit for $f(x)$ is 3, and for $g(x)$ is 4

but I dont get an upper limit for $h(x)$

and how to find lower limits? thx

Daniel Katzan
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    You should also know that if $\sum|n\hat f(n)|<\infty$ then $f$ is continuously differentiable (and the Fourier series for $f'$ is obtained in the obvious way from the series for $f$). – David C. Ullrich Aug 08 '15 at 15:32

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Hint.

$h$ is indefinitely differentiable. You can prove it by induction using the fact that for $k \in \mathbb N$ $$h_k(x) = \sum_{-\infty}^\infty (i n)^k\frac{e^{inx}}{2^{\lvert n\rvert}}$$ is normally convergent and the theorem related to differentiability of sequences of functions.

mathcounterexamples.net
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  • does $h_k$ stand for the k deritative of h? can you please explain how to show that for k=1? – Daniel Katzan Aug 08 '15 at 16:44
  • $h_k$ is indeed the $k$-derivative of $h$. Do you know the theorem mentioned at following link https://en.m.wikipedia.org/wiki/Uniform_convergence#To_differentiability? – mathcounterexamples.net Aug 08 '15 at 16:49
  • yes I know i...I understand how to go from there and complete the proof, but why does $h_1(x) = \sum_{-\infty}^\infty (i n)\frac{e^{inx}}{2^{\lvert n\rvert}}$ I also know that $\hat{f'(n)} = in\hat {f(n)}$.. but still not sure how to show the equality – Daniel Katzan Aug 08 '15 at 16:55
  • You can prove using the ratio test that $h_1$ is normally convergent. With the theorem that I mentioned in the link, you can prove that $h$ is differentiable and $h_1$ is its derivative. You can then proceed by induction. – mathcounterexamples.net Aug 08 '15 at 17:07
  • yeh. I got a bit confused and mixed up some stuff.. but I got it now, thx – Daniel Katzan Aug 08 '15 at 17:43