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.


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

  • 7
    $\begingroup$ Anyone know if the paper is correct? $\endgroup$ May 27 at 17:17
  • $\begingroup$ @mathworker21 I did not see any reaction yet. $\endgroup$ May 30 at 10:14
  • 2
    $\begingroup$ I am very much hoping for @TerryTao $\endgroup$ May 30 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 at 3:26
  • 8
    $\begingroup$ @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 at 7:17

1 Answer 1


(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?


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