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 $T:H\to H$?
