2
$\begingroup$

Problem: Give a simple explanation of why, if the Fourier coefficients $a_{k}=b_{k}=0$ for all sufficiently large $k \gg 1$, then the Fourier series converges to an analytic function.

Attempted Explanation: Since the Fourier series is an infinite sum, if the series did not converge, then the function would blow up. Since $a_{k}=b_{k}=0$, and $k>>1$, the high valued k's finely tune the oscillating function to an analytic function.

This explanation was apparently not good because I only got 0.2/1 credit for the problem, thus I am wondering what a better explanation would be for it.

$\endgroup$
  • $\begingroup$ That explanation is not good because it would imply that all functions with a convergent Fourier series are analytic. $\endgroup$ – Giuseppe Negro Nov 27 '17 at 16:40
  • $\begingroup$ @GiuseppeNegro Can you provide an example where a convergent Fourier series wouldn't be analytic? I'm not quite understanding that conceptually. $\endgroup$ – Steven Nov 27 '17 at 16:41
  • $\begingroup$ Try computing the Fourier series of $f(x)=|x|,\quad x\in (-\pi, \pi )$. Show that it is convergent. Recall that $f$ is not analytic. $\endgroup$ – Giuseppe Negro Nov 27 '17 at 16:44
3
$\begingroup$

So there is some $N>0$ such that for all $k>N$ the coefficients are zero, you mean? Then it is easy since you have a finite sum of analytic functions. Finite sums of analytic functions are analytic.

$\endgroup$
  • $\begingroup$ That makes sense, thank you. Never thought about a finite sum of analytic functions being an analytic function. $\endgroup$ – Steven Nov 27 '17 at 16:38

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.