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I'm taking an introductory course in Fourier analysis and I'm trying to solve the following problem

Prove that the Fourier series of a continuously differentiable function $f$ on the circle is absolutely convergent.
[Hint: Use the Cauchy-Schwarz inequality and Parseval's identity for $f'$.]

I solved it like this $$\sum_{n=-\infty}^{\infty} |\hat{f}(n)| \le \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2 = \dfrac{1}{2\pi}\int_{0}^{2\pi}|f(x)|^2 < \infty$$

I'm thinking that the equality holds because of Parseval's identity and that the integral is bounded because $f$ is bounded because it's continuously differentiable on the circle.

The problem is that I'm pretty sure that this solution is wrong since I haven't used the information in the hint or the fact that $f$ is differentiable. Indeed, if this solution is correct then this should hold for any continuous $f$? I'm just not sure why this solution is wrong.

The correct solution is available here (page 30)

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  • $\begingroup$ The first inequality in your proof is false. "$\geq$" would be right, but this does not help you. $\endgroup$ – PhoemueX Mar 9 '16 at 21:16
  • $\begingroup$ @PhoemueX Ah, of course! Thank you! $\endgroup$ – Dan Setterquist Mar 9 '16 at 21:25
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    $\begingroup$ If $f$ has derivative $g$, then how does $\hat{f}(n)$ relate to $\hat{g}(n)$? Once you get this, then you can use Parseval's theorem and Cauchy-Schwarz. $\endgroup$ – DisintegratingByParts Mar 10 '16 at 1:06

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