MHB Is Continuously Differentiable Imply Fourier Coefficients in l^1?

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Here is this week's POTW:

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Let $f : \Bbb S^1\subset \Bbb C \to \Bbb C$ be a continuous map. Show that if $f$ is continuously differentiable on $\Bbb S^1$, then its Fourier coefficient sequence $\{\hat{f}_n\}_{n\in \Bbb Z}$ belongs to $\ell^1(\Bbb Z)$.-----

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This week's problem was correctly solved by Opalg. You can read his solution below.

Spoiler

Integrate by parts to get $$\widehat{f}_{\!n} = \frac1{2\pi}\int_0^{2\pi}f(x)e^{-inx}dx = \frac1{2\pi}\Bigl[\, f(x)\frac{e^{-inx}}{-in}\Bigr]_0^{2\pi} + \frac1{2\pi in}\int_0^{2\pi}f'(x)e^{-inx}dx = \frac1{in}\widehat{f'}_{\!n}. $$ Since $f'$ is continuous on the compact space $\Bbb{S}^1$ it is square-integrable. So by Parseval's theorem its Fourier coefficient sequence $\{\widehat{f'}_{\!n}\}$ is in $\ell^2(\Bbb {Z})$. It then follows from the Cauchy–Schwarz inequality that $$ \sum_{n\in\Bbb{Z}}|\widehat{f}_{\!n}| = \sum_{n\in\Bbb{Z}}\Bigl|\frac1{in}\widehat{f'}_n\Bigr| \leqslant \Bigl(\sum_{n\in\Bbb{Z}}\Bigl|\frac1{n^2}\Bigr|\Bigr)^{1/2} \Bigl(\sum_{n\in\Bbb{Z}}\bigl|\,\widehat{f'}_{\!n}\bigr|^2\Bigr)^{1/2} < \infty.$$