Function $f$ such that Fourier-series converges uniformly, but the series of the derivatives are divergent

Question

I am studying Fourier-transformation right now, and I am asking if there exists a function $f$ such that is Fourier-series converges uniformly, the Fourier-series of $f'$ only in $L_2$ and that $f''$ is divergent?

Answer 1

The function $$ f(t)=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\sin((2k+1)2\pi t)\tag{1} $$ is the zig-zag function $$ f(t)=\left\{\begin{array}{}\frac{\pi^2}{2}t&\text{for }-\frac14\le t\lt\frac14\\\frac{\pi^2}{2}\left(\frac12-t\right)&\text{for }\hphantom{\,-}\frac14\le t\lt\frac34\end{array}\right.\tag{2} $$ Since $f$ is continuous, its Fourier series converges uniformly.

$f'$ is not continuous $$ f'(t)=\left\{\begin{array}{}\hphantom{\,-}\frac{\pi^2}{2}&\text{for }-\frac14\lt t\lt\frac14\\-\frac{\pi^2}{2}&\text{for }\hphantom{\,-}\frac14\lt t\lt\frac34\end{array}\right.\tag{3} $$ Since $f'$ is not continuous, its Fourier series cannot converge uniformly. However, since $f'\in L^2$, its Fourier series converges in $L^2$. $$ f''(t)=\pi^2\delta\left(t+\frac14\right)-\pi^2\delta\left(t-\frac14\right)\tag{4} $$ where $\delta$ is the dirac delta. The fourier series for $f''$ does not even tend to $0$, so it diverges.

Answer 2

The series $$f(t)=\sum_{n\in\mathbb Z}\frac{1}{n^2}e^{int}$$ converges uniformly. The distributional derivative $f'$ belongs to $L^2$, but the corresponding series does not converge pointwise. For $f''$ the series diverges in $L^2$.