How piecewise smoothness of a function is related to the Dirichlet conditions?

Question

Do all piecewise smooth functions satisfy Dirichlet conditions for Fourier series representation? In the theorem of Fourier series can we write that being piecewise smooth is the sufficient condition for obtaining its Fourier series ?

Answer 1

Yes and no. The sufficient condition for a existence of a Fourier series of a function $f:\mathbb R\to \mathbb C$ is that $f$ has to be periodic and quadratic integrable. As example let $f$ be $2\pi$ periodic, then we have there exists the Fourier series if and only if $\int_0^{2\pi}|f(t)|^2dt<\infty$. In that case we have with $e_k(t):=e^{ikt}$ $$\lim_{n\to\infty} \underbrace{\sum_{k=-n}^n\frac 1{2\pi}\int_0^{2\pi} f(t)e^{-ikt}dt\:e_k}_{S_nf:=}=f$$ the convergence in the sence of the quadratic integrable functionspace $\mathcal L^2_{2\pi}$. That means that $$\int_0^{2\pi}|f(t)-lim_{n\to\infty}S_n(t)|^2dt=0$$ In other words $\lim_{n\to\infty}S_nf(x)=f(x)$ for allmost every $x\in \mathbb R$, but it don't have to hold for every $x$!

If $f$ is piecewise smooth in a way that the one-sided limits $f'(x+),f'(x-)$ and $f(x+),f(x-)$ exists for every $x\in \mathbb R$. Then is $f$ quadratic integrable, so the Fourier exists. But furthermore we have now for every $x\in\mathbb R$ $$\lim_{n\to\infty}S_nf(x)=\tfrac 1 2(f(x+)+f(x-)).\tag {*}$$ So yes, you can say that if a function is piecewise smooth then the Fourier series converges pointwise with $(*)$.