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Mathematics
Calculus
When is fourier series non-differentiable?
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[QUOTE="jostpuur, post: 2644997, member: 68508"] I know that if [tex] \sum_{n=-\infty}^{\infty} |n| |\hat{f}(n)| < \infty [/tex] then [tex] \sum_{n=-\infty}^{\infty} \hat{f}(n) e^{2\pi i nx} [/tex] is continuously differentiable as a function of [itex]x[/itex]. Now I'm interested to know what kind of conditions exist for Fourier coefficients such that they guarantee the non-differentiability. It is a fact that just because some abstract integral [itex]\int d\mu(x)\psi(x)[/itex] diverges, it doesn't mean that a limit of other integrals [itex]\lim_{n\to\infty} \int d\mu(x) \psi_n(x)[/itex] would diverge too, even when [itex]\psi_n\to\psi[/itex] point wisely. So this means that even if I know that [tex] \sum_{n=-\infty}^{\infty} |n| |\hat{f}(n)| = \infty [/tex] this will not obviously imply that [tex] \lim_{\delta\to 0} \sum_{n=-\infty}^{\infty} \hat{f}(n) e^{2\pi inx} \frac{e^{2\pi in\delta} - 1}{\delta} [/tex] would diverge too. What condition will suffice to prove that the Fourier series is not differentiable? [/QUOTE]
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Mathematics
Calculus
When is fourier series non-differentiable?
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