Does the Fourier series for the $2\pi$ periodic function $$f(x) = \begin{cases} -1 \ \text{for} \ x \in (-\pi, 0) \\\\ 0 \ \text{for} \ x = 0 \\\\ 1 \ \text{for} \ x \in (0,\pi) \end{cases} $$ converge uniformly or pointwise?
$f(x)$ is discontinuous in $x=0$ so the Fourier series cannot converge uniformly, but why not?
The Fourier series will converge towards $f(x)$ in all continuous points and in the discontinuity ($x=0$) it will converge towards $\frac{f(0^+) + f(0^-)}{2} = \frac{1-1}{2} = 0 $. So indeed, the Fourier series does converge towards $f(x)$ in all points, but why does it not do it uniformly? How can it be shown with the definition of uniform convergence for series?
The series $\sum_{n=1}^\infty f_n(x), x \in I$ is said to converge uniformly towards $f(x)$ on $I$ if there for any $\varepsilon > 0$ exists an $N_0 \in \mathbb{N}$ such that $$ \bigg| f(x) - \sum_{n=1}^N f_n(x) \bigg| < \varepsilon $$ for all $x\in I$ and all $N\geq N_0$.
When is this definition violated in the case above?
This does not answer your question exactly, but it may clarify some things for you. The first step is to calculate the Fourier coefficients . $$\boxed{f(x)=\frac{1}{2}a_0+\sum_{n\ge 1} a_n\cos(nx)+b_n\sin(nx)}$$ convegence Uniformily if $\sum a_n $ and $\sum b_n $ absolument convege
we have function f odd then $$a_0=a_n=0$$ $$b_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx) dx = \frac{2}{\pi}\sin(nx) dx = \frac{2}{\pi}\left({\frac{1-(-1)^n}{n}}\right)$$
therfore : $$\boxed{f(x)=\sum_{n\ge 1} 2\left({\frac{1-(-1)^n}{n\pi}}\right)\sin(nx)}$$
we have : $ \sum |b_n| $ Not absolument converge
finnaly : f(x) does not convergence uniform