Example of continuous periodic function with divergent Fourier series

Question

Let $f:[-\pi,\pi]\to\Bbb C$ be continuous such that $f(-\pi)=f(\pi)$. Let $a_n = \displaystyle \int_{-\pi}^\pi f(\theta) \overline{\exp(in\theta)} \frac{\mathrm d\theta}{2\pi}$ for $n \in \Bbb Z$ be its Fourier coefficients. It is clear that the Fourier series $\displaystyle \sum_{n \in \Bbb Z} a_n \exp(in\theta)$ converges to $f(\theta)$ almost everywhere, and that for every null set $E$ there is such an $f$ such that the Fourier series diverges exactly on $E$.

So, in particular, there is such an $f$ such that its Fourier series diverges at $\theta=0$.

However, I searched examples of such functions to no avail, since they are behind paywalls. The closest I got to such an example is in the article Beispiele Stetiger Funktionen mit Divergenter Fourierreihe ("Examples of Continuous Functions with Divergent Fourier Series"), where it is said:

Bekanntlich hat P. du Bois-Reymond zuerst die Existenz einer überall stetigen Funktion erwiesen, deren Fouriersche Reihe an einer Stelle divergiert.

Herr H. A. Schwarz gab dann ein einfacheres Beispiel.

translated:

"It is well known that P[aul] du Bois-Reymond was the first to demonstrate the existence of an everywhere continuous function with a Fourier series that diverges at a point.

Then, H[ermann] A[mandus] Schwarz gave an easier example."

So my question is, could one give an explicit example of such an $f$ whose Fourier series diverges (only) at $\theta=0$?

The function $f:[0,\pi]\to\Bbb R$ defined by $f(x)=\sum\limits_{k=1}^\infty\frac{\sin\left[(2^{k^3}+1)\frac x2\right]}{k^2}$ has a Fourier series divergent at zero. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Apr 27 '19 at 07:15
Perhaps related: https://en.wikipedia.org/wiki/Uniform_boundedness_principle#Example:_pointwise_convergence_of_Fourier_series — I was suspended for talking, Apr 27 '19 at 07:16
@Mattos Pedro Tamaroff refers to Körner's Fourier Analysis which is behind paywall — Kenny Lau, Apr 27 '19 at 07:23
@TheSimpliFire If you substitute $x=0$ then don't you get $\sum 0 = 0$? — Kenny Lau, Apr 27 '19 at 07:25
@KennyLau $f(0)=0$ but its Fourier series at zero tends to $+\infty$. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Apr 27 '19 at 07:27
@TheSimpliFire would you consider making it an answer? Thanks! — Kenny Lau, Apr 27 '19 at 07:28
Why the function doesn't satisfy Fourier's Theorem? I know it is continuous but I suppose I can't derive it term by term to check if the derivate piecewise continuous. Can you help me @KennyLau ? — Cleto Pereira, Jul 14 '20 at 14:58

ə̷̶̸͇̘̜́̍͗̂̄︣͟ · Answer 1 · 2021-07-06T13:15:23.850

One such example (Fejér) from $[0,\pi]$ to $\Bbb R$ is given^[1]: $$f(x)=\sum\limits_{k=1}^\infty\frac{\sin\left[(2^{k^3}+1)\frac x2\right]}{k^2}$$ which is even and periodic with period $2\pi$. Since $$\left\vert\frac{\sin\left[(2^{k^3}+1)\frac x2\right]}{k^2}\right\vert\le\frac1{k^2}$$ by the Weierstrass $M$-test, $f$ is convergent and continuous (sum is bounded above by $\pi^2/6$). In the proof, let $$S_n=\frac\pi2\sum_{k=0}^n a_k$$ where $a_k$ are the coefficients of the Fourier series. Also let $$\lambda_{n,m}=\int_0^{\pi} \sin \left[ (2m + 1) \frac{t}{2} \right] \cos nt \ dt \text{ and } \sigma_{n,m} = \sum_{k=0}^n \lambda_{k,m},$$ and it is proven that $$S_{2^{p^3-1}} \ge \frac{1}{p^2} \sigma_{2^{p^3-1},2^{p^3-1}} \ge \frac{1}{2p^2} \ln(2^{p^3-1}) = \frac{p^3-1}{2p^2} \ln 2$$ which clearly tends to $+\infty$ as $p\to\infty$. This means that the sequence $(S_n)$ diverges and the Fourier series of $f(x)$ at zero is also divergent.

_Reference

_{[1] Merx. J-P. (2016). A Continuous Function With Divergent Fourier Series. Available from: http://www.mathcounterexamples.net/continuous-function-with-divergent-fourier-series/.}

@Ilya $f(x)$ is defined only in $[0,\pi]$ and, therefore, isn't odd. In order to get the example, we need to stablish an even extension of it. — Cleto Pereira, Jul 08 '20 at 17:30

Example of continuous periodic function with divergent Fourier series

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