- #1

- 2,111

- 17

## Main Question or Discussion Point

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?

[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?