# Showing Following Fourier series converges to sawtooth function

This question is originated from S/S Fourier Analysis Chapter 2 Exercise 8.

Problem says show sawtooth function$$f(x)= \begin{cases} -\frac{\pi}{2}-\frac{x}{2}, -\pi<x<0\\ \frac{\pi}{2}-\frac{x}{2}, 0<x<\pi\\ \end{cases}$$ has fourier series $$\frac{1}{2i}\sum_{n\neq 0} \frac{e^{inx}}{n}$$

It's just a simple calculation and we can show the series converges by Dirichlet's test.

But what about the converse, $$\frac{1}{2i}\sum_{n\neq 0} \frac{e^{inx}}{n}$$ (pointwise) converges to $$f(x)= \begin{cases} -\frac{\pi}{2}-\frac{x}{2}, -\pi<x<0\\ \frac{\pi}{2}-\frac{x}{2}, 0<x<\pi\\ \end{cases}$$

For me, with only tools learned until Chapter 2, I think the only way to showing is using uniqueness of the Fourier series with uniform convergence condition.

I think the series $$\frac{1}{2i}\sum_{n\neq 0} \frac{e^{inx}}{n}$$ will uniformly convergent on $[\pi+\delta,-\delta]$ and $[\delta,\pi-\delta]$

then we can show the series has same fourier series by interchanging sum and integral and it's done. But Weierstrass M-test fails on this series.

Is the conjecture true? Otherwise, What's the another way to showing convergence to sawtooth function?

• See Wikipedia for some results on pointwise convergence of Fourier series. – Jyrki Lahtonen Jul 10 '16 at 10:45
• @JyrkiLahtonen This question is asking about not convergence itself but convergence with specific value. Also Convergence can be checked by Dirichlet's test. Can you explain why that theorem can answer the question? – Maddy Jul 11 '16 at 7:20
• The most basic result (the only one I ever learned) says that when a periodic function has one-sided limits and derivatives at a given point, the Fourier series will, at that point, converge to the average of the one-sided limits of the function. This is mentioned in the Wikipage, but admittedly the authors of that article aim a bit higher :-). Anyway, sawtooth has one-sided limits and derivatives everywhere, so it applies. – Jyrki Lahtonen Jul 11 '16 at 7:24