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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.

Connor Harris
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Steven
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  • That explanation is not good because it would imply that all functions with a convergent Fourier series are analytic. – Giuseppe Negro Nov 27 '17 at 16:40
  • @GiuseppeNegro Can you provide an example where a convergent Fourier series wouldn't be analytic? I'm not quite understanding that conceptually. – Steven Nov 27 '17 at 16:41
  • 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. – Giuseppe Negro Nov 27 '17 at 16:44

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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.

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