I hope I don't make too many mistakes since this is my first post. I was trying to prove that $$M = \left[ f\in L^2[\Pi] : \sum_{n=-\infty}^{+\infty}\hat f(n) \quad converges \right]$$ is dense and first category set (Baire's category) in $L^2[\Pi]$.
With $ \ \Pi \ $ I mean the Torus and with $\hat f(n)$ I mean the Fourier coefficients of $f\in L^2[\Pi]$: $$\hat f(n)=\frac{1}{2\pi}\int_{\Pi}f(t)e^{-int}dt$$
When I say a set $E$ is first cathegory set I mean it's equal to a countable union of sets $E_n$ such that $int(\bar E_n)=\emptyset$ fo every n.
I know that a similar proof about the divergence set of the Fourier series of a $f \in L^1[\Pi]$ has been done using the linear and bounded operators $$L_n=\sum_{m=-n}^{n}\hat f(m)=\frac{1}{2\pi}\int_{\Pi}f(t)D_n(t)dt$$ and calculating their norm in the dual space of $L^2[\Pi] \;$ ($D_n(t)=\sum_{m=-n}^{n}e^{int}$ is the Dirichlet kernel).
I struggled with these for a couple of days and I couldn't manage to get an answer. hope someone can help me :).
Let me address only the non-trivial part of the question which is that $M$ is of first category.
Considering the map $$ T: (a_n)_{n\in \mathbb Z}\in \ell ^1(\mathbb Z) \ \mapsto \ \sum_{n\in \mathbb Z}a_ne^{int} \in L^2(\mathbb T), $$ it is clear that $M$ is precisely the range of $T$. It is also clear that $T$ is bounded and not surjective. So the result is a consequece of the following more general result:
Theorem. Let $X$ and $Y$ be Banach spaces and let $T:X\to Y$ be a non-surjective bounded map. Then the range of $T$ is of first category.
Proof. Let $B_X$ be the closed unit ball of $X$. Observing that $$ T(X) = \bigcup_{n\in \mathbb N} T(nB_X), $$ the proof will be concluded once we show that the closure of each $T(nB_X)$ has empty interior, and it is clearly enough to consider only the case $n=1$.
(If $X$ is reflexive this would be much easier because then $B_X$ is weakly compact, whence the same holds for $T(B_X)$, so $T(B_X)$ would be closed. However, since we are aiming at an application in which $X=\ell ^1$, we cannot assume that $X$ is reflexive.)
So let us assume by contradiction that $\overline{T(B_X)}$ has a nonempty interior.
It so happens that the standard proof of the Open Mapping Theorem (if $T:X\to Y$ is surjective then it is open), e.g. Theorem III.12.1 in Conway's "A Course in Functional Analysis", start out by using that $T$ is surjective to deduce that $\overline{T(B_X)}$ has a nonempty interior. After this crucial step, the surjectivity of $T$ is no longer needed and, with the sole information that $\overline{T(B_X)}$ has a nonempty interior, it is proved that $T$ is open.
In other words, we can borrow the argument of that proof to deduce that our $T$ is open, but this is an absurd since we are assuming that $T$ is not surjective. This concludes the proof. $\qquad \square$