Im stuck with this exercise
Show that there is a doble sided sequence $\ldots ,b_{-2},b_{-1},b_0,b_1,b_2,\ldots $ with $\lim_{n\to \pm \infty }b_n=0$ such that doesn't exists $f\in L^1((-\pi,\pi])$ with $\hat f(n)=b_n$ for all $n\in \Bbb Z $.
I dont have a clue about this exercise. My problem is that I doesn't have a clear way to know when a function is integrable just taking a look at it Fourier coefficients except in the case that the Fourier coefficients doesn't converge to zero.
This comes after a chapter in a book about convolution and the Poisson kernel, by example I know that if $f\in L^p$ then $\lim_{r\uparrow 1}\|f-f*P_r\|_p=0$ where $P_r(\zeta ):=\frac{1-r^2}{|1-r\zeta |^2}$ is the Poisson kernel for some chosen $r\in[0,1)$; also I know that $\|f*g\|_p\leqslant \|f\|_1\|g\|_p$ for $f\in L^1$ and $g\in L^p$; or that $\widehat{f*g}(k)=\hat f(k)\hat g(k)$ for $f,g\in L^1(\partial \Bbb D )$.
Then my idea was try to bound below $\|f*P_r\|_1$ and show that for suitable sequence of Fourier coefficients with the required conditions then $\lim_{r\uparrow 1}\|f*P_r\|_1=\infty $, thus I wrote $$ |f*P_r(t)|\geqslant |\operatorname{Re}(f*P_r)(t)|=\left|\sum_{k\in \Bbb Z }r^{|k|}\hat f(k)\cos(kt)\right| $$ but I dont find a way to bound below the quantity of the RHS. Maybe there is an easier path, honestly, I dont have a clue about this exercise, some help will be appreciated.
Say $c_0$ is the Banach space of all two-sided sequences $a$ with $\lim_{j\to\pm\infty}a_j=0$ and norm $||a||=\sup_j|a_j|$.
Suppose to the contrary that every $a\in c_0$ is the sequence of Fourier coefficients of some integrable function. Then we can define $T:c_o\to L^1(\Bbb T)$ by $$Ta\sim\sum a_ne^{int}.$$It follows from for example the Closed Graph Theorem that $T$ is bounded (if you don't know the CGT you can also use the Open Mapping Theorem to show that $T$ is bounded). But in fact very well known examples show that $T$ is not bounded (hint: Dirichlet kernel).