In a recent preprint [On the invariant subspace problem in Hilbert spaces](https://arxiv.org/abs/2305.15442) *Per H. Enflo* claims to have solved the [invariant subspace problem](https://en.wikipedia.org/wiki/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 $T:H\to H$?



Understanding a simplifying assumption in Invariant Subspace Problem proof
(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?


