Suppose $f$ is continuously differentiable on the unit circle. Show that the Fourier series of $f$ converges absolutely (thus uniformly) to $f$.

Let the Fourier series of $f$ be given by $$ f(x) \sim \sum_{n=-\infty}^{\infty} c_{n}e^{inx}$$

We want to show that $$ \sum_{n=-\infty}^{\infty} |c_{n}e^{inx}| \to f(x) $$

Now from Plancherel's identity, $$f'(x) = \int \tilde{f}'(s) e^{2 \pi i s x} \ dx$$

What would be the next steps from here? Expand $f'(x)$ into a Fourier series?


Yes, you can expand $f'$ into Fourier series, and apply Parseval's identity to it, arriving at $\sum n^2|\widehat f(n)|^2<\infty$. The Cauchy-Schwarz inequality yields the convergence of $\sum |\widehat{f}(n)|$.

The fact that the Fourier series converges to $f$ (rather than another function) can be shown by repeating the above for the function $f_N(x)=f(x)-\sum_{n=1}^N \widehat{f}(n)e^{inx}$. The relevant sums tend to zero as $N\to\infty$, which implies $f_N\to 0$ uniformly.

  • 1
    $\begingroup$ Could you elaborate on your second paragraph? I fail to see how $f_N \rightarrow 0$ uniformly follows without assuming pointwise convergence of the Fourier series to $f$ in the first place. $\endgroup$ – balu Nov 24 '17 at 10:35

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