I am trying to prove what seems to be a standard result on the convergence of Fourier series,namely that if $ f:[-\pi,\pi] \longrightarrow \mathbb{C}$ is a continuous function then $$S_n(f)=\sum_{-n}^{n}\widehat{f}(k)e^{iks}$$ converges almost everywhere to $f$.

Here is my attempt: First we have to justify that $S_n(f)$ converges to something meaningfull . Since $ |\widehat{f}(k)|=|\frac{1}{2\pi}\int f(t)e^{-ikt}dt|\leq \frac{1}{2\pi}\int |f(t)|e^{-ikt}dt \leq \Vert f\Vert_{\infty}C\frac{1}{2\pi k}$ for some constant $C$,it is clear that the coefficients are square summable and we can use the weierstarss criterion. The value of the limit should coincide with $f$ because Fejer's theorem holds $\sigma_{n}(f)\longrightarrow f$ for continous functions and $ S_{n}(f) $ cannot tend somewhere else.

But why almost everywhere convergence?? What am i missing?

  • $\begingroup$ In general, the Fourier series converges to $f$ in $L^2([-\pi,\pi])$ but not in $L^\infty([-\pi,\pi])$ nor pointwise. Then I would look at a general sequence $f_n \in L^2$ converging to $0$. Does it converge a.e. to $0$, and does it have a subsequence converging a.e. to $0$ ? $\endgroup$ – reuns Jul 31 '17 at 10:09
  • $\begingroup$ How can you use Weirstrass M-test for this series (even if the coefficients are square summable)? $\endgroup$ – Ranc Jul 31 '17 at 10:36
  • $\begingroup$ @Ranc What do you mean ? $\endgroup$ – reuns Jul 31 '17 at 11:07
  • $\begingroup$ You know that $(\hat{f}(n))_{n \in \mathbb{Z}}$ is square summable. How do you show $\sum_{n\in \mathbb{Z}} \hat{f}(n) \mathrm{e}^{inx}$ converge? (How do you use Weirstrass' M-test?) $\endgroup$ – Ranc Jul 31 '17 at 11:15

Your bound on $\widehat f(k)$ is incorrect. The correct one is $$|\widehat f(k)|\le\frac{1}{2\,\pi}\int_{-\pi}^\pi|f(t)|\,dt\le\|f\|_\infty.$$

Since $f$ is continuous, $f\in L^2$. Carleson's theorem gives the result.

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  • $\begingroup$ Is there an elementary proof of this(for continuous f) avoiding use of Carleson's theorem? $\endgroup$ – Pmorphy Jul 31 '17 at 11:55
  • $\begingroup$ No, as dar as I know. $\endgroup$ – Julián Aguirre Aug 1 '17 at 12:59

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