26
$\begingroup$

In a recent preprint On the invariant subspace problem in Hilbert spaces Per H. Enflo claims to have solved the invariant subspace problem, showing that every bounded linear operator on a separable complex Hilbert space has a closed non-trivial invariant subspace (CNTIS for short in what follows).

Right at the beginning, the author writes

without loss of generality, we can assume that $T$ is one to one, $\mathcal{R}(T)\neq H$, $\overline{\mathcal{R}(T)}=H$ and for every $y\neq0$ we have $y\in\overline{\operatorname{span}\{T^jy:j\geq1\}}$.

I understand why we can suppose three of the four assumptions, namely (eliminating the case $T=0$ for which every closed subspace is invariant):

  • if $T$ is not one to one, then $\ker T$ is a CNTIS;
  • $\overline{\mathcal{R}(T)}$ is always a closed invariant subspace, so if it is not the full space $H$ then it is a CNTIS;
  • $\overline{\operatorname{span}\{T^jy:j\geq1\}}$ is always a closed invariant subspace, so if it does not contain $y$ then it is not the full space $H$, hence is a CNTIS.

What I don't understand is why we can assume that $\mathcal{R}(T)\ne H$. I tried thinking why the case of a bijective $T:H\to H$ would be trivial, trying to exploit that $T^{-1}$ exists and is bounded too, but I couldn't come up with a way of showing that there exists a CNTIS in this case. In particular, I tried showing that $\overline{\operatorname{span}\{T^jy:j\geq1\}}$ is always a proper subspace, looking somehow for a vector which is orthogonal, but couldn't make any progress.

Question

Can someone explain how we can easily show that there exists a CNTIS for a bijective bounded linear operator $T:H\to H$?

$\endgroup$
7
  • 8
    $\begingroup$ Anyone know if the paper is correct? $\endgroup$ May 27, 2023 at 17:17
  • $\begingroup$ @mathworker21 I did not see any reaction yet. $\endgroup$ May 30, 2023 at 10:14
  • 2
    $\begingroup$ I am very much hoping for @TerryTao $\endgroup$ May 30, 2023 at 10:14
  • 6
    $\begingroup$ @JochenWengenroth I am personally hoping for ISP/hypercyclic specialists such as Sophie Grivaux. (The paper being referred to is not written in a style conducive to easy verification, in my opinion.) $\endgroup$
    – Yemon Choi
    May 31, 2023 at 3:26
  • 9
    $\begingroup$ @YemonChoi I have met the hypercyclic specialists these days and, to say the least, they are not convinced. However, Per Enflo promised a new version of the manuscript. $\endgroup$ Jun 14, 2023 at 7:17

2 Answers 2

28
$\begingroup$

(If a moderator wants to remove this/make it a comment, please do. For various reasons I do not want to have an account on this site) )

I believe one simply needs to know that the spectrum of $T$ is nonempty, so up to replacing $T$ by some $T - \lambda I$ one can assume wlog that $0$ is in the spectrum of $T$ (of course invariant subspaces are not changed by adding a multiple of the identity).

Since $T$ is injective, as you noticed, it cannot be surjective and this gives you the desired outcome?

$\endgroup$
7
$\begingroup$

Today (2023-11-16), Enflo gave a talk about the invariant subspace problem at Valencia which one can watch on youtube https://www.youtube.com/watch?v=LXOAOY-NEV8

I did not get more insight than from his preprint on the arxiv. However, he announced to put a new version there, soon (and apologized for the delay).

$\endgroup$
2
  • 1
    $\begingroup$ Looks like the relevant talk starts 1 hr 5 mins into the video. $\endgroup$
    – Nik Weaver
    Nov 16, 2023 at 17:37
  • 1
    $\begingroup$ Right, Nik, sorry that I forgot to mention this. $\endgroup$ Nov 16, 2023 at 19:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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