It's well known that the Fourier series converges uniformly when a function is $C^2$ and periodic on say $[0,2\pi]$. If the function is not periodic you can have lack of uniform convergence near the endpoints due to "Gibbs" phenomenon. However I would like to understand if and when you still have uniform convergence on compact subsets of $[0,2\pi]$? This seems to be geometrically obvious but I can't find a good source for it. Any suggestions would be appreciated. For instance $f(x) = x$ on $[0,1]$. It would appear that the Fourier series converges uniformly on compact subsets of $[0,2\pi]$ but this doesn't seem to follow from analysis of the Fourier coefficients. Isn't it the same as if I had done an odd extension of $f(x)$ so that I have $-|x|$ on $[-2\pi,2\pi]$ and find the fourier series for this new domain? In the latter case I clearly get uniform convergence and I don't think this should depend on the fact that I reflected and chose a different domain.

  • 3
    $\begingroup$ I think you might want to restrict to compact subsets of $(0, 2\pi)$. In this case you can surely find your answer on the wikipedia page on convergence of Fourier series. $\endgroup$ – Matt Young Sep 30 '10 at 4:41
  • $\begingroup$ When you say your function is non-periodic -- are you still assuming it is everywhere defined on the closed interval $[0,2\pi]$ and $C^2$ everywhere on that interval? $\endgroup$ – Yemon Choi Sep 30 '10 at 5:01
  • $\begingroup$ The fourier series of the sawtooth wave function is (up to constants) $\sum_{n\ge 1} \sin (xn)/n$, but the triangle wave has the series $\sum_{n\ge 1} \cos (xn)/n^2$, which has a quite different rate of convergence than the former series. $\endgroup$ – Florian Sep 30 '10 at 13:09
  • $\begingroup$ Yes $C^2$ up to the boundary on $[0,2]$, Yemon Choi. Florian: I think both of those series still converge uniformly though in the interior (the first one you show by using the Dirichlet series test). Matt Young: I did not see any reference to what you have mentioned actually. If I had found this I would have not asked the question here. $\endgroup$ – Dorian Sep 30 '10 at 14:16
  • $\begingroup$ Then I don't understand what you mean by "Isn't it the same as if I had done ...". $\endgroup$ – Florian Sep 30 '10 at 14:29

I assume you mean "...if and when you still have uniform convergence on compact subsets of (0,2π)? " This is in the nature of what is called a localization theorem. These go back to Riemann who proved that the convergence of the Fourier series of an $L^1$ function $f$ AT a point $x$ depends only on the behavior of $f$ in any small neighborhood of $x$. I'll stick my neck out a little and say that I wouldn't be too surprised if there is a generalization of Riemann's Localization Theorem that says that local uniform convergence near $x$ likewise only depends on the behavior of $f$ near $x$ (but that is pure conjecture on my part).

  • 2
    $\begingroup$ Right, and the claim should follow already from the explicit form of the localization theorem, see springerlink.com/content/f4q3v26220670389 (link superseded by new link below). $\endgroup$ – Piero D'Ancona Sep 30 '10 at 10:17
  • $\begingroup$ So this paper confirms the conjecture you've made? $\endgroup$ – Dorian Sep 30 '10 at 14:16
  • $\begingroup$ Thanks for that reference, Piero. It is not exactly easy reading, but it does seem to confirm what I conjectured. $\endgroup$ – Dick Palais Sep 30 '10 at 21:28
  • $\begingroup$ @PieroD'Ancona Would you happen to remember that reference? The Springer link is no longer working. $\endgroup$ – Todd Trimble Jul 31 '20 at 18:42
  • $\begingroup$ I suspect the paper was Riemann’s Localization Theorem. An Estimate for the Rate of Convergence, by S. A. Teliakovskii (Journal of Mathematical Sciences, Vol. 155, No. 1, 2008) available <a href="link.springer.com/content/pdf/10.1007/…> $\endgroup$ – Piero D'Ancona Aug 1 '20 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.