2
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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, 2017 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, 2017 at 10:36
  • $\begingroup$ @Ranc What do you mean ? $\endgroup$
    – reuns
    Jul 31, 2017 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, 2017 at 11:15

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Is there an elementary proof of this(for continuous f) avoiding use of Carleson's theorem? $\endgroup$
    – Pmorphy
    Jul 31, 2017 at 11:55
  • $\begingroup$ No, as dar as I know. $\endgroup$ Aug 1, 2017 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.