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Consider a Fourier Series

$$S(t) = \sum_{k=0}^{\infty}c_{k}e^{ikt}$$

where $c_{k}$ are complex coefficients such that the sum $\sum |c_{k}|$ is finite. I am also given that $\lim_{k\to\infty}k^{m}c_{k}=0$ for some fixed $m>0$.

Question: What can we say about the differentiability of $S$?

What I tried: If I can prove that $\sum k|c_{k}|<\infty$, then the sequence of derivatives of the partial sums $$f_{n}'(t)=\sum_{k=0}^{n}ikc_{k}e^{ikt}$$ must converge uniformly to a continuous limit, then $S$ would be differentiable. However, I am not sure how to apply the fact that $\lim_{k\to\infty}k^{m}c_{k}=0$ for some fixed $m>0$ - certainly $\sum k|c_{k}|<\infty$ implies the terms should go to 0 but how do I know the converse is true?

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  • $\begingroup$ If $\sum_k |c_k k^m| < \infty$ then integrate $m$-times $\sum_k c_k (ik)^m e^{ikt}$ you'll recover $S(t)$ $\endgroup$
    – reuns
    Aug 4 '19 at 3:58
  • $\begingroup$ How do you show $\sum_{k}|c_{k}k^{m}|<\infty$ given that $\lim_{m\to\infty}k^{m}c_{k}=0$ and $\sum_{k}|c_{k}|<0$? $\endgroup$ Aug 4 '19 at 4:52
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    $\begingroup$ I don't. $c_k = O(k^{-m})$ implies $\sum_k |c_k k^{m-2}| < \infty$ so $S(t)$ is $C^{m-2}$, and that $\sum_k c_k (ik)^{m-1} e^{ikt}$ converges in $L^2$ so that $S(t)$ is the $m-1$-th primitive of a $L^2$ function. The $m$-th derivative is only well-defined in the sense of distributions. $\endgroup$
    – reuns
    Aug 4 '19 at 17:27
  • $\begingroup$ Can we assume that $m$ is an integer? Otherwise, we can set $c_k=1/k$ if $k$ is a power of $2$, and $c=0$ otherwise. This does not satisfy $\sum_k k|c_k|\lt\infty$. $\endgroup$
    – Thijs
    Aug 7 '19 at 10:07
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    $\begingroup$ I think it might be time to ask the professor about whether $m$ is supposed to be an integer -- no use spending a ton of time on a problem that's not the one intended. $\endgroup$ Aug 7 '19 at 10:21
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Before I type anything else, I would like to point out that the assumption (8) below is insufficient to establish that

$\displaystyle \sum_0^\infty k \vert c_k \vert < \infty, \tag 0$

even when the condition (9) binds.  A counterexample is provided by the series

$R(t) = \displaystyle \sum_1^\infty \dfrac{1}{k^2} e^{ikt}; \tag{0.1}$

it is well-known that

$\displaystyle \sum_1^\infty \dfrac{1}{k^2} = \dfrac{\pi^2}{6}, \tag{0.2}$

see Showing $\sum _{k=1} 1/k^2 = \pi^2/6$ ; however, with

$c_k = \dfrac{1}{k^2}, \tag{0.3}$

we find

$\displaystyle \sum_1^\infty kc_k = \sum_1^\infty k \dfrac{1}{k^2} = \sum_1^\infty \dfrac{1}{k} = \infty; \tag{0.4}$

nevertheless, if adopt the stronger hypothesis that

$\displaystyle \sum_1^\infty k^m \vert c_k \vert < \infty, \tag{0.5}$

we will discover its suffiency, as is shown below.

I assume

$0 < m \in \Bbb Z, \tag 0$

that is, $m$ is a positive integer.

Consider the sequence of partial sums

$S_n(t) = \displaystyle \sum_{k = 0}^n c_ke^{ikt} \tag 1$

of the series

$S(t) = \displaystyle \sum_{k = 0}^\infty c_ke^{ikt}; \tag 2$

it is easy to see that $S_n(t)$ is a $C^\infty$ function for every $n \in \Bbb N$; indeed, the $S_n(t)$ are analytic, each being the sum of a finite number of analytic functions $c_ke^{ikt}$; furthermore for $n > p$ we have

$S_n(t) - S_p(t) = \displaystyle \sum_{p + 1}^n c_k e^{ikt}, \tag 3$

whence

$\vert S_n(t) - S_p(t) \vert = \vert \displaystyle \sum_{p + 1}^n c_k e^{ikt} \vert \le \sum_{p + 1}^n \vert c_k \vert, \tag 4$

since

$\vert e^{ikt} \vert = 1; \tag 5$

now taking $p$ and $n$ sufficiently large we may affirm that

$\displaystyle \sum_{p + 1}^n \vert c_k \vert < \epsilon \tag 6$

for any real

$\epsilon > 0; \tag 7$

this assertion of course follows easily from the hypothesis

$\displaystyle \sum_0^\infty \vert c_k \vert < \infty. \tag 8$

In light of these remarks, we conclude that the sequence of functions $S_n(t)$ converges uniformly in $t$; thus the limit function $S(t)$ is indeed continuous.

Note that we have not yet called upon the hypothesis that

$\displaystyle \lim_{k \to \infty} k^m c_k = 0.  \tag 9$

Now observe that the $S_n(t)$ (1), being finite sums, are each in fact differentiable functions of $t$; indeed,

$S_n'(t) = \displaystyle \sum_{k = 0}^n ikc_ke^{ikt}; \tag{10}$

also,

$\vert S_n'(t) - S_p'(t) \vert = \vert \displaystyle \sum_{p + 1}^n ikc_k e^{ikt} \vert \le \sum_{p + 1}^n k\vert c_k \vert < \epsilon \tag{11}$

for $n$, $p$ sufficiently large in light of our added assumption (0.5) with $m = 1$, and thus the sequence $S_n'(t)$ is Cauchy and hence it also is uniformly convergent. since $\epsilon$ is independent of $t$; these facts in concert are sufficient for the existence of a function $S'(t)$ such that

$ S'(t) = \displaystyle \lim_{n \to \infty} S_n'(t) = ( \lim_{n \to \infty} S_n(t))' = (S(t))', \tag{12}$

in accord with the standard theorem on convergence of sequences of derivatives.

The process described in the above may be continued for larger values of $m$, the result being similar to that attained so far, provided of course that (0.5) binds for the chosen value of $m$. Indeed, we may write

$S''(t) = \displaystyle \sum_1^\infty -k^2c_k e^{ikt}, \tag{13}$

and so forth. Higher derivatives of $S(t)$ may be expressed in an analogous manner, assuming (0.5) holds for appropriate values of $m$.

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    $\begingroup$ Appreciate the effort in this answer - One thing; why are we proving $|S'_{n}(t) - S_{p}'(t)|<\epsilon$ ? It seems this is only proving the sequence is Cauchy, not convergent. Are you fixing $p$ for now, then taking the limit as $p\to\infty$? $\endgroup$ Aug 12 '19 at 11:27
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For sufficiently large $ k $ ,$ |c_k |\le \frac {2}{|k^m|} $

Pick $ m=2$

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  • $\begingroup$ We are not given a choice of $m$ so I think we need to account for all possible values. Also how did you get the inequality ? $\endgroup$ Aug 7 '19 at 12:25

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