# When is fourier series non-differentiable?

• jostpuur
In summary, the conversation discusses the conditions for the Fourier coefficients that guarantee the non-differentiability of the Fourier series. The first condition mentioned is that if the sum of the absolute values of the coefficients is finite, then the Fourier series is continuously differentiable. However, it is noted that this does not necessarily imply differentiability, even if the sum of the absolute values is infinite. The original work by Hardy (1916) is referenced as a source for further information on this topic. Other examples are also mentioned as potential sources for studying the conditions for non-differentiability of Fourier series.

#### jostpuur

I know that if

$$\sum_{n=-\infty}^{\infty} |n| |\hat{f}(n)| < \infty$$

then

$$\sum_{n=-\infty}^{\infty} \hat{f}(n) e^{2\pi i nx}$$

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

$$\sum_{n=-\infty}^{\infty} |n| |\hat{f}(n)| = \infty$$

this will not obviously imply that

$$\lim_{\delta\to 0} \sum_{n=-\infty}^{\infty} \hat{f}(n) e^{2\pi inx} \frac{e^{2\pi in\delta} - 1}{\delta}$$

would diverge too.

What condition will suffice to prove that the Fourier series is not differentiable?