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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?

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up vote 6 down vote accepted

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.

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hey, thanks! but the divergence of $f''$ is not so obvious to me, it is zero almost everywhere, except at countable many points, so why is its fourier-transformation divergent? – Stefan Jun 21 '12 at 0:07
@Stefan: the Fourier series of $f''$ computed either from $(4)$ or by differentiating $(1)$ twice is $$f''(t)=-4\pi^2\sum_{k=0}^\infty(-1)^k\sin((2k+1)2\pi t)$$ The terms to do not tend to $0$, so the series diverges. – robjohn Jun 21 '12 at 0:25

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 $L^2$.

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