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 :).

• Welcome to MSE! I think this is a good first post, I upvoted your question so that it gets attention – projectilemotion Jan 13 at 21:10
• For the dense part you can look at smooth functions. – Jose27 Jan 13 at 21:29
• 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. – Don Abbondio Jan 13 at 22:05

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. $$\qquad \square$$

• +1. Quick question: Does the answer change if $M$ includes conditionally convergent series instead of only absolutely convergent ones? – Jose27 Jan 14 at 4:09
• thank you @Ruy! Also thanks @Jose27, i got the dense part now :) – Don Abbondio Jan 14 at 9:55
• @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!? – Ruy Jan 14 at 21:19
• 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). – Jose27 Jan 14 at 21:34