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Let's say I have a piecewise continuous function which has the fourier series $\sum_n\ c_{n}e^{inx}$ and I assume that $ \sum_n\ n|c_{n}|$ converges, then I know the following holds:

The fourier series of the derivatives $\sum_n\ inc_{n}e^{inx}$ is absolutely convergent and thus from the Weierstrass M-test I know $\sum_n\ inc_{n}e^{inx}$ is uniformly convergent.

Thus, $ f' (x)=\sum_n\ inc_{n}e^{inx}$ is a continuous $2 \pi $-periodic function.

We can say then that $ f(x)=\sum_n\ c_{n}e^{inx}$ is a continuously differentiable $2 \pi $-periodic function.

My question is this:

Does the converse also holds?

To be more precise:

If $ f(x)=\sum_n\ c_{n}e^{inx}$ is a continuously differentiable $2 \pi $-periodic function, does it follow that $\sum_n\ n|c_{n}|$ converges?

My thought on the matter are such:

I think the converse does not hold, since we've used the Weierstrass M-test which is kind of a one-way ticket.

I attempted at constructing a counter-example. Since I want $\sum_n\ n|c_{n}|$ to diverge, it suffices to find $c_{n}$ such that $c_{n} \sim \frac{1}{ n^{2} } $.

Thus, if I look at $ f' (x)=\sum_n\ inc_{n}e^{inx}$, it is clear that I need a continuous $2 \pi $-periodic function such that its fourier coefficients decay rate is $\frac{1}{ n } $.

Unfortunately, every continuous function I tried had a fourier coefficients decay rate of $\frac{1}{ {n^{2} } } $. The ones that I know of that have a decay rate of $ \frac{1}{ n } $ are discontinuous functions.

Your thoughts on the matter are greatly appreciated!

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  • $\begingroup$ In the second paragraph of your thoughts you want a sequence that decays like $1/n^2$ yet later on you want $1/n$ ? $\endgroup$ – Justpassingby Dec 1 '15 at 7:54
  • $\begingroup$ For the derivative: $1/n$ $\endgroup$ – zokomoko Dec 1 '15 at 7:55
  • $\begingroup$ Equivalently, you want a continuous function whose Fourier coefficients are not in $\ell^1$. As you suspected, such functions exist (lots of them). For example, there are continuous functions whose Fourier series diverges at a given point; this is discussed in pretty much any book on the subject. (In fact, the Fourier coefficients of a continuous function need not decay faster than any given $\ell^2$ sequence, but this is much harder to prove.) $\endgroup$ – user138530 Dec 1 '15 at 8:43
  • $\begingroup$ Thank you for the comment Christian, while it might be a novice question, doesn't that contradict Dirichlet's theorem? That the fourier series converges to f(x) at point of continuouity $\endgroup$ – zokomoko Dec 1 '15 at 8:45
  • $\begingroup$ I don't know what exactly you mean by Dirichlet's theorem, but it's certainly not the case that the Fourier series has to converge at points of continuity: en.wikipedia.org/wiki/… $\endgroup$ – user138530 Dec 1 '15 at 9:18
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Here is a counterexample. This web page https://en.wikipedia.org/wiki/Wiener_process#Wiener_representation gives the randomly created function $$ W(t) = \zeta_0 t + \frac{\sqrt 2}{\pi} \sum_{n=1}^\infty \frac {\zeta_n \sin(n \pi t)}n $$ where $(\zeta_n)$ is a sequence of independent random variables distributed Gaussian mean 0, variance 1. This page says this is one way to create Brownian motion on $[0,1]$. Thus $$f(t) = W(t) - W(1) t = \frac{\sqrt 2}{\pi} \sum_{n=1}^\infty \frac {\zeta_n \sin(n \pi t)}n $$ is continuous with probability 1. But one can show that $$ \sum_{n=1}^\infty \frac{|\zeta_n|}{n} $$ diverges with probability 1 (hint: first show that $ \sum_{n=1}^\infty \frac{|\zeta_n|-\mathbb E|\zeta_n|}{n} $ converges using Kolmogorov's three series theorem https://en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem).

Further notes: Another place to look for counterexamples is with random functions of the form $$ f(x) = \sum_{n=-\infty}^\infty r_n a_n e^{i\pi n x} $$ where $r_n$ are identically distributed random variables taking values plus or minus one with probability $1/2$. If you look in the book Random Fourier Series with Applications to Harmonic Analysis by Michael B. Marcus & Gilles Pisier, or in the book Some Random Series of Functions, 2nd Edition by Jean-Pierre Kahane, you will find conditions on the sequence $a_n$ such that the random function is continuous almost surely. Then out of those sequences, it should be relatively straightforward to find an example that is not absolutely summing.

Also, look at https://en.wikipedia.org/wiki/Wiener_algebra, but that only tells you the name of the space of functions whose coefficients are absolutely summing - it doesn't provide examples of continuous functions not in the Wiener algebra.

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  • $\begingroup$ Thank you for your insightful respone! As for an explicit example, I thought of looking at continuous functions with vertical tangents ( such as abs(x) ) . But could not find any that had a computeable fourier series. I mean, the coefficients were too hard to calculate $\endgroup$ – zokomoko Dec 2 '15 at 1:49
  • $\begingroup$ I'm fairly sure that abs(x) has absolutely summing Fourier coefficients. $\endgroup$ – Stephen Montgomery-Smith Dec 2 '15 at 2:19
  • $\begingroup$ Sorry, I meant sqrt(abs(x)).. a function with verrical tangent $\endgroup$ – zokomoko Dec 2 '15 at 6:03
  • $\begingroup$ Try writing it as a cosine series. Then split the resulting integral as a sum of $\int_0^{1/n}$ and $\int_{1/n}^\pi$. Estimate the latter integral by integrating by parts a couple of times, and then estimate the integral that is left over. I think you will find that the Fourier coefficients are $O(n^{-3/2})$. $\endgroup$ – Stephen Montgomery-Smith Dec 2 '15 at 10:17
  • $\begingroup$ I just made an edit, giving Brownian motion as a counterexample. $\endgroup$ – Stephen Montgomery-Smith Dec 2 '15 at 13:44
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Suppose $h$ is a periodic continuous function $[0,2\pi]$ for which $\sum_n |c_n| =\infty$. Define $$ f(x) = \int_0^x \left[h(t)-\frac{1}{2\pi}\int_0^{2\pi}h(u)du\right]dt. $$ Then $f$ is continuously differentiable and periodic on $[0,2\pi]$ with Fourier coefficients $c_n(f)$ for which \begin{align} inc_n(f) & = -\frac{1}{2\pi}\int_{0}^{2\pi}f(y)(-ine^{-iny})dy \\ & = -\frac{1}{2\pi}\int_{0}^{2\pi}f(y)\frac{d}{dy}e^{-iny}dy \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}h(y)e^{-iny}dy = c_n(h),\;\;\; n\ne 0. \end{align} So your problem reduces to considering the Fourier series of a periodic continuous function $h$. It is known that there is a periodic continuous function $h$ such that the Fourier series for $h$ diverges at $0$. Such an $h$ provides the counter example that you want because $\sum_n|c_n(h)| =\infty$ must occur.

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