- #1
jostpuur
- 2,112
- 19
I know that if
<br /> \sum_{n=-\infty}^{\infty} |n| |\hat{f}(n)| < \infty<br />
then
<br /> \sum_{n=-\infty}^{\infty} \hat{f}(n) e^{2\pi i nx}<br />
is continuously differentiable as a function of x.
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 \int d\mu(x)\psi(x) diverges, it doesn't mean that a limit of other integrals \lim_{n\to\infty} \int d\mu(x) \psi_n(x) would diverge too, even when \psi_n\to\psi point wisely. So this means that even if I know that
<br /> \sum_{n=-\infty}^{\infty} |n| |\hat{f}(n)| = \infty<br />
this will not obviously imply that
<br /> \lim_{\delta\to 0} \sum_{n=-\infty}^{\infty} \hat{f}(n) e^{2\pi inx} \frac{e^{2\pi in\delta} - 1}{\delta}<br />
would diverge too.
What condition will suffice to prove that the Fourier series is not differentiable?
<br /> \sum_{n=-\infty}^{\infty} |n| |\hat{f}(n)| < \infty<br />
then
<br /> \sum_{n=-\infty}^{\infty} \hat{f}(n) e^{2\pi i nx}<br />
is continuously differentiable as a function of x.
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 \int d\mu(x)\psi(x) diverges, it doesn't mean that a limit of other integrals \lim_{n\to\infty} \int d\mu(x) \psi_n(x) would diverge too, even when \psi_n\to\psi point wisely. So this means that even if I know that
<br /> \sum_{n=-\infty}^{\infty} |n| |\hat{f}(n)| = \infty<br />
this will not obviously imply that
<br /> \lim_{\delta\to 0} \sum_{n=-\infty}^{\infty} \hat{f}(n) e^{2\pi inx} \frac{e^{2\pi in\delta} - 1}{\delta}<br />
would diverge too.
What condition will suffice to prove that the Fourier series is not differentiable?