Let $f(x)=\chi_{[a,b]}(x)$ be the characteristic function of the interval $[a,b]\subset [-\pi,\pi]$.
Show that if $a\neq -\pi$, or $b\neq \pi$ and $a\neq b$, then the Fourier series does not converge absolutely for any $x$. [Hint: It suffices to prove that for many values of $n$ one has $|\sin n\theta_0|\ge c \gt 0$ where $\theta_0=(b-a)/2.$]
However, prove that the Fourier series converges at every point $x$.
I've computed the Fourier series and got $\frac{b-a}{2\pi}+\sum_{n\neq 0}\frac{e^{-ina}-e^{-inb}}{2\pi in}e^{inx}.$
Also, $|e^{-ina}-e^{-inb}|=2|\sin n\theta_0|$, and $\theta_0\in (0,\pi)$, so I can see that for infinitely many values of $n$, we have $|\sin n\theta_0|\ge c \gt 0$. But this does not guarantee $\sum_{n\neq 0}|\frac{e^{-ina}-e^{-inb}}{2\pi in}e^{inx}|\ge \sum \frac{c}{n}$, and in fact we might have this inequality only for the squares of integers, in which case the right hand side converges. So how does the hint solve the problem?
Moreover, for the second problem, to show that the Fourier series converges at every point, I think I need to use Dirichlet's test, using $1/n$ as the decreasing sequence to $0$, but how can I show that $\frac{e^{-ina}-e^{-inb}}{2\pi in}e^{inx}$ has bounded partial sums?
I would greatly appreciate any help.
Best Answer
Suppose $a \ne -\pi$ or $b \ne \pi$ and $a\ne b$. Then the function you are talking about must be discontinuous.
Suppose the series did converge absolutely for some $x$. That would mean $$ \sum_n \left|\frac{e^{-ina}-e^{-inb}}{2\pi in}\right| < \infty. $$ But that would force the uniform convergence of the Fourier series everywhere by the Weierstrass M-test. But uniform convergence would imply that the periodic extension of the limit function $\chi_{[a,b]}$ must be continuous everywhere, which only happens in the case that $a=-\pi$ and $b=\pi$, or $a=b$.
I'm not familiar with you text, but you should have some pointwise convergence theorem that shows the Fourier series converges to the mean of the left and right hand limits for your function.