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? :)
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.