Understanding a simplifying assumption in Invariant Subspace Problem proof

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 $$T:H\to H$$?

• Anyone know if the paper is correct? May 27 at 17:17
• @mathworker21 I did not see any reaction yet. May 30 at 10:14
• I am very much hoping for @TerryTao May 30 at 10:14
• @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.) May 31 at 3:26
• @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. Jun 14 at 7:17

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?