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Am tyring to solve an exercise and i am stuck , i would appreciate if anyone has any idea on how to continue , thanks for your time !

The exercise: let $f:\Bbb R \to \Bbb R$ a $2\pi$-periodic function which:

a)is continuous at $0$ and integrable at $[-\pi,\pi]$.

b)$\hat{f}(n)\geq 0 $ for all $n \in \Bbb Z$ where $\hat{f}(n)$ is n-th Fourier coefficient of $f.$

Show that under these conditions the Fourier series of $f$ converges to $f(t)$ for Lebesgue almost everywhere in $[-\pi,\pi]$. Equivelently , $\sum_{n\in \Bbb Z} \hat{f}(n)e^{int}=f(t)$ for lebesgue almost everywhere.

What i have done so far:

1st)We know that if $t\in Leb(f)$ then $\sigma_n(f)(t)\to f(t)$ where $\sigma_n(f)(t)=(K_n*f)(t)$ with $K_n$ the n-th Fejer kernel.

2nd)Since f is continuous at $0$ we know that $\sigma_n(f)(0)\to f(0)$ from fejer's lemma. With this i get

that $\lim_{n\to \infty}\sum_{k=1}^{n} (1-\frac{k}{n+1})\hat{f}(k)=\frac{f(0)-\hat{f}(0)}{2}$. Because $\sigma_n (f)(0)=\sum_{k=-n}^{n} (1-\frac{|k|}{n+1}) \hat{f}(k)$.

3rd)I have expressed the $s_n(f)(t)=\sum_{k=-n}^{n} \hat{f}(k)e^{ikt}$ in terms of $\sigma_n (f)(t)$ like this

$s_{n+1} (f)(t)=\sigma_{n+1}(f)(t) +\sum_{k=-(n+1)}^{n+1}\frac{|k|}{n+2}\hat{f}(k)e^{ikt}$.

Now i want to take $n\to \infty$ but i need to show that the second term goes to $0$.But i dont know how to procceed

Any ideas? :)

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  • $\begingroup$ Do you know that the Cesàro sum of a convergent series converges to the same limit, even if that is $\pm \infty$? $\endgroup$ – Daniel Fischer Jan 29 '18 at 13:05
  • $\begingroup$ Yeap , i know that ... how is it going to help me ? $\endgroup$ – dem0nakos Jan 29 '18 at 13:10
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    $\begingroup$ Well, what is the connection between the Fourier series of $f$ at $0$ and $\sigma_n(f)(0)$? $\endgroup$ – Daniel Fischer Jan 29 '18 at 13:11
  • $\begingroup$ $\sigma_n(f)(0)$ is the cesaro sum of $s_n(f)(0)$ and i know that $\sigma_n(f)(0)\to f(0)$ i cant deduce that $s_n(f)(0) \to f(0)$. Am i missing anything ? $\endgroup$ – dem0nakos Jan 29 '18 at 13:15
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    $\begingroup$ Since $$s_n(f)(0) = \sum_{k = -n}^n \hat{f}(k)\,,$$ you know that $s_n(f)(0)$ converges. Either to a nonnegative real number or to $+\infty$. $\endgroup$ – Daniel Fischer Jan 29 '18 at 13:17
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Since the Fourier coefficients of $f$ are all nonnegative,

$$s_n(f)(0) = \sum_{k = -n}^n \hat{f}(k)$$

converges, either to a nonnegative real number or to $+\infty$. Hence the associated Cesàro sum converges to the same limit. But this Cesàro sum is $\sigma_n(f)(0)$, and we know that converges to $f(0)$. Thus the sequence of Fourier coefficients is summable, and the Fourier series of $f$ converges absolutely (and uniformly) everywhere. Since almost all points are Lebesgue points of $f$, $\sigma_n(f)$ converges almost everywhere to $f$, and once more using the fact that the Cesàro sum of a convergent series has the same value as the original series it follows that the Fourier series converges to $f$ almost everywhere.

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