Show that the Fourier series for $f$ doesn't converge uniformly on $[-\pi,\pi]$.

Question

Let $\quad f(x)=\begin{cases}-1,&-\pi<x<0\\{}\\1,&0<x<\pi\end{cases}$

Show that the Fourier series for $f$ doesn't converge uniformly on $[-\pi,\pi]$.

My approach:

The Fourier series for $f$ is: $S_n(f)(x)=\frac{1}{\pi}\left[\int_{0}^{\pi}D_n(t-x)dt-\int_{-\pi}^{0}D_n(t-x)dt\right]$. Where $D_n$ is the Dirichlet kernel defined like $\quad D_n(t)=\frac{sin(n+1/2)t}{2sin(t/2)}, t\in[-\pi,0)\cup(0,\pi]$

I tried to suppose that the series converges and get to a contradiction without success. Any suggestions would be great!

Answer 1

Every uniformly convergent Fourier series has a continuous sum, but $f$ is not continuous.