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?
Answer
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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