Given a function $$f(x) = \begin{cases} \pi -x &\mbox{if } 0<x < 2\pi \\ 0 & \mbox {if } x= 0 \end{cases} $$ with $2\pi$-periodic continuation, I have obtained the Fourier Series:

$$f(x)=\sum_{k=-\infty}^{\infty}\frac{-ie^{ikx}}{k} = +\sum_{k=1}^{\infty} \frac{2}{k}\sin(kx)$$

I guess my zeroth question is whether I got this right, but my actual question is: how do I show that (or whether?) the Fourier Series converges towards my original function $f(x)$? I'm guessing I have to show uniform convergence of the series but I don't know how...

  • $\begingroup$ This will be a good reference: courses.mai.liu.se/GU/TATA57/Dokument/FourierSeries2.pdf $\endgroup$ – Sungjin Kim Jun 17 '17 at 12:25
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    $\begingroup$ You meant $\sum_{k=-\infty}^{\infty}\frac{i e^{ikx}}{k}$. A general proof is that $g(x) = f(x)- \sum_{k=-\infty}^{\infty}\frac{e^{ikx}}{k}$ is a piecewise $C^1$ function such that $\int_0^{2\pi} g(x) e^{i k x}dx = 0$ for every integer $k$. Thus $g$ is identically zero except maybe at a finite number of points. $\endgroup$ – reuns Jun 17 '17 at 13:16
  • $\begingroup$ In this special case, you can show that for $|r| < 1$, $-\log(1-r e^{ix}) = \sum_{k=1}^\infty \frac{r^k e^{ikx}}{k}$ follows from the geometric series. Taking the imaginary part and letting $r \to 1^-$, you'll get your result. $\endgroup$ – reuns Jun 17 '17 at 13:22
  • $\begingroup$ @i707107: Being a Swede it's nice to see a link to Swedish university. $\endgroup$ – md2perpe Jun 17 '17 at 21:49

Your function has left- and right-hand limits $f_l$, $f_r$ at every point, and corresponding left- and right-hand derivatives $$ f_{l}'(x) = \lim_{y\uparrow x}\frac{f_l(x)-f_l(y)}{x-y}, \\ f_{r}'(x) = \lim_{y\downarrow x}\frac{f_r(y)-f_r(x)}{y-x}. $$ So the Fourier series for $f$ converges everywhere to the mean of the left- and right-hand limits of $f$. Your function $f$ is everywhere defined so that it is the mean of the left- and right-hand limits. So the series converges to $f$ pointwise everywhere.

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