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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 :).

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    $\begingroup$ Welcome to MSE! I think this is a good first post, I upvoted your question so that it gets attention $\endgroup$ – projectilemotion Jan 13 at 21:10
  • $\begingroup$ For the dense part you can look at smooth functions. $\endgroup$ – Jose27 Jan 13 at 21:29
  • $\begingroup$ thank you @projectilemotion! Glad to be here. @Jose27 will do, but with smooth you mean $C^{\inf}$? Because i know that $C^{\inf}$ is dense in $L^2$ but I'm not sure about the correlation between my M and $C^{\inf}$. I didn't thought of that beacuse I was thinking to use the Ortonormal complete system ${e_n=e^{inx}}$ of $L^2$ which is also numerable and dense (if we use the rationals) and get two birds with one stone... was just dreaming probably. $\endgroup$ – Don Abbondio Jan 13 at 22:05
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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. $\qquad \square$

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  • $\begingroup$ +1. Quick question: Does the answer change if $M$ includes conditionally convergent series instead of only absolutely convergent ones? $\endgroup$ – Jose27 Jan 14 at 4:09
  • $\begingroup$ thank you @Ruy! Also thanks @Jose27, i got the dense part now :) $\endgroup$ – Don Abbondio Jan 14 at 9:55
  • $\begingroup$ @Jose, I think conditionally convergent doubly infinite series is a bit tricky do define. It would seem natural do adopt $$ \exists \lim_{N\to\infty}\sum_{n=-N}^N a_n, $$ but then $$ \cdots -1 -1 -1 + 0 +1 +1 +1 \cdots $$ would be summable!? $\endgroup$ – Ruy Jan 14 at 21:19
  • $\begingroup$ That definition reminds me of the Cauchy principal value for integrals. I'd probably go with both $\sum_{-\infty}^0$ and $\sum_0^{\infty}$ converge (the same way we do with integrals). $\endgroup$ – Jose27 Jan 14 at 21:34

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