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 :).

  • 2
    $\begingroup$ 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

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. $\qquad \square$

  • $\begingroup$ +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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.