If $f\in C^0(\Bbb R)$ is $2\pi$-periodic surely $\int\limits_0^{2\pi}|f(x)|^pdx<\infty$ at least for $p=1$. Is it true for any $p\in[1,\infty]$? where $\|f\|_{L^{\infty}}=\text{supess}(|f|):=\inf\{\alpha\in\Bbb R\ :\ |f(x)|\le\alpha\ \ a.e.\}$
Since $f$ is continuous over $\Bbb R$ it is continuous over any compact set (in particular $[0,2\pi])$ so it is uniformly continuous and bounded and $\|f\|_{L^{\infty}}<\infty$
There is an example given by Kolmogorov of a function that is $L^1$ whose Fourier series diverges everywhere. But what if the function had to be $L^1$ and continuous? Would there still be a counter-example? (It would have to be not Holder continuous since Holder continuity+$L^1$+$2\pi$-periodicity $\implies$ Fourier series converges).
