Proof that a particular subset of $L^2[\Pi]$ is dense and first category set (Baire's category) 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 :).
 A: 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.  $\qquad \square$
