Let $$F_n(x)=\frac{a_0}{2}+\sum_{k=1}^\infty \left(a_n\cos(n\pi x)+b_n\sin(n\pi x)\right),$$ the ourier serie of a $2\pi-$periodic function $f:\mathbb R\longrightarrow \mathbb R$. I have a theorem in my course that says that if $f$ is $\alpha -$Holder continuous, then $F_n(x)\to F(x)$ for all $x\in [-\pi,\pi]$.
Remark : But if $f$ is continous, it' not enough.
Question) Is it the holder continuity on all $\mathbb R$ or on $[-\pi,\pi]$ only ? Because in exercise, they alway says that "since $f$ is Lipschitz on $(-\pi,\pi)($ the Fourier series converge. But even for $f(x)=x$ if $x\in [02\pi[$ and prolonge by $2\pi$ periodicity. But since a continuous function Holder continuous on compact, I'm confuse with this fact.
Actually, if the periodic extension of $f$ is Holder continuous at $x$, then the Fourier series converges to $f(x)$ at $x$ (and, of course, at all $2n\pi$ translates of $x$.) If $f$ has left- and right-hand limits at $x$, and $f$ is Holder continuous on the left and on the right of $x$ using these limiting values, then the Fourier series converges to mean of the left- and right-hand limits of $f$ at that $x$. You don't need to assume conditions at other points, except for integrability on the full interval.