uniform convergence of $f(\theta)=\theta, \theta \in [-\pi, \pi]$

Question

I am mostly guessing using the theorem that because $f$ is Lipschitz continuous with $L= 2\pi$, $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sin{n \theta}}{n}$ converges uniformly to $f$. Is this start on the right track towards proving the uniform convergence on the interval $[-\pi,\pi]$?

Answer 1

When considering the uniform convergence of Fourier series, one must take into account the continuity of the periodic extension of $f$. This is because if a Fourier series converges uniformly on $[-\pi,\pi)$, then (by periodicity) it converges uniformly on $\mathbb{R}$; hence its sum is a continuous function on $\mathbb{R}$.

Given that $f(\theta)=\theta$ on $(-\pi,\pi)$, it is clear that the periodic extension is discontinuous at $\pi$: the limit from the left is $\pi$, the limit from the right is $-\pi$. (It may help to sketch the periodic extension.) Thus the Fourier series does not converge uniformly.