1
$\begingroup$

Let $\overline{f}$ be a function on the whole real line, such that $\overline{f}$ is continuous and differentiable everywhere, and its derivative $\overline{f}'$ is also continuous everywhere. Now, restrict $\overline{f}$ to a function $f$ defined only on the interval $(0, \pi)$.

Does Fourier sine series of $f$ always converges to $f$ pointwise on $(0, \pi)$? I know that it does not converge uniformly on $(0, \pi)$.

What about Fourier cosine series? Does Fourier cosine series of $f$ always converge to $f$ uniformly on $(0, \pi)$?

My professor hasn't covered much in convergence, so I want to know more about the convergence of functions.

$\endgroup$
2
$\begingroup$

(Edit: extension method is free, first version unnecessarily assumed simple $π$-periodic extension from $(0,π)$ to $\mathbb R$)

For the full Fourier series there is uniform convergence on all closed intervals that contain no jumps and no kinks. See for instance the rather complex result cited in Wikipedia: Convergence of Fourier series.

The sine series obtained as Fourier sine series is an odd $2π$-periodic function. Thus it is obtained as the full Fourier series of the $2π$-periodic odd extension of $\bar f$, that is with $\bar f(x)=-\bar f(-x)=-\bar f(π-x)$. The no jumps condition demands $f(0)=0=f(π)$ from the original function. Differentiability will then be automatic. In general however the jump at the interval ends will produce an oscillation in the series known as "Gibbs phenomenon".

The cosine series obtained as Fourier cosine series is an even $2π$-periodic function. Thus it is obtained as the full Fourier series of the $2π$-periodic even extension of $\bar f$, that is with $\bar f(x)=\bar f(-x)=\bar f(π-x)$. Continuity is now automatic, one reason one uses DCT in JPEG, but differentiability is only given if the derivatives of $f$ at $0$ and $π$ were zero. I think that kinks do not form an obstacle for uniform convergence, as long as left and right limits of the derivative exist in every point.

$\endgroup$
  • $\begingroup$ Any function can be periodically extended to an odd or even function. You want to say for instance that the sine (resp. cosine) series converges uniformly if the odd (resp. even) periodic extension is absolutely continuous. $\endgroup$ – Paul Siegel Feb 26 '14 at 14:17
  • $\begingroup$ Not enough coffee. Yes, that is right, the extension method was not prescribed. The points on uniform convergence and jump points still stand. However, the mirrored continuation related to the cosine series is continuous, which is why DCT was chosen for (the old) JPEG. $\endgroup$ – LutzL Feb 26 '14 at 14:23
  • $\begingroup$ Changed and reordered the text to reflect that insight. $\endgroup$ – LutzL Feb 26 '14 at 14:37

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.