Differentiability of $\sum_{k=0}^{\infty}c_{k}e^{ikt}$ given $\lim_{k\to\infty}k^{m}c_{k}=0$

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?

• If $\sum_k |c_k k^m| < \infty$ then integrate $m$-times $\sum_k c_k (ik)^m e^{ikt}$ you'll recover $S(t)$ Aug 4 '19 at 3:58
• How do you show $\sum_{k}|c_{k}k^{m}|<\infty$ given that $\lim_{m\to\infty}k^{m}c_{k}=0$ and $\sum_{k}|c_{k}|<0$? Aug 4 '19 at 4:52
• 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. Aug 4 '19 at 17:27
• 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 $\sum_k k|c_k|\lt\infty$. Aug 7 '19 at 10:07
• 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. Aug 7 '19 at 10:21

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

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

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 $$m$$.

• 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 $p\to\infty$? Aug 12 '19 at 11:27

For sufficiently large $$k$$ ,$$|c_k |\le \frac {2}{|k^m|}$$

Pick $$m=2$$

• 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 ? Aug 7 '19 at 12:25