A question on Hilbert spaces concerning kernel and partial isometry

Question

Let $\mathcal{H}$ be an infinite-dimensional Hilbert space, and let $T:\mathcal H\to \mathcal H$ be a continuous linear operator with $\operatorname{Ker} T\neq\{0\}$ and closed range $\mathcal{R}(T)$. Suppose that $T$ is a partial isometry, by which we mean that $T$ is an isometry on the orthogonal complement of its kernel, namely on $(\operatorname{Ker} T)^\perp$. Further, assume that $\operatorname{Ker}T\subset \mathcal{R}(T)\neq \mathcal{H}$. Let $I$ denote the identity operator on $\mathcal H$.

Does there exist $x\in(\operatorname{Ker}(T-I))^\perp$ such that $((T-I)y, x)=0$ for all $y\in(\operatorname{Ker}T)^\perp$, with $(\cdot,\cdot)$ denoting the inner product on $\mathcal H\,$?

Any answer or hint will be appreciated.

Answer 1

Let's focus on the required orthogonality condition $$0\:=\:((T-I)y,x)\:=\:(y,(T^*-I)x)\quad\forall\,y\in(\operatorname{Ker}T)^\perp,$$ it says that $\,(T^*-I)x\in\operatorname{Ker}T,$ hence the elements $x$ we are seeking have to satisfy $$TT^*x\:=\:Tx\tag{1}\,.$$ Remark:
This equation is equivalent to the symmetric version $\,T^*Tx=T^*x\,$ since partial isometries are also characterised by $\,T^*TT^*=T^*.$

$(1)$ is satisfied by eigenvectors to the eigenvalue $1$, noting that $\,\operatorname{Ker}(T-I)=\operatorname{Ker}(T^*-I)\,$ generally holds for partial isometries, but they do not have to exist.
There may exist non-zero solutions within the orthogonal complement $(\operatorname{Ker}(T-I))^\perp\,$: Looking at the assumptions $\,\{0\}\neq\operatorname{Ker}T\subset\mathcal{R}(T)\neq\mathcal{H}\,$ and using $\,\mathcal R(T)=(\operatorname{Ker}T^*)^\perp$ one has $$\,\{0\}\neq\operatorname{Ker}T\ \perp\ \tag{2} \operatorname{Ker}T^*\neq\{0\}\,.$$ This yields the orthogonal decomposition $$\mathcal{H}\:=\:\operatorname{Ker}T\oplus\operatorname{Ker}T^*\oplus\mathcal S\quad\text{with }\ \mathcal S=\mathcal R(T)\cap\mathcal R(T^*)\,,$$ and $\,\mathcal S=\{0\}\,$ may occur (as in the example below). Insert $\,x=k+k_*+s\,$ as ansatz into $(1)$, with $\,k\in\operatorname{Ker}T,\ k_*\in\operatorname{Ker}T^*, \ s\in\mathcal S$, to obtain $$k+s = TT^*(k+k_*+s) = T(k+k_*+s) = Tk_*+Ts\tag{3}\,.$$ Assuming $\,Ts=s\,$ (thus leaving the most general setting by now) gives the condition $\,Tk_*=k$, and $x=k+ k_*$ will do the job.
The set of all these $x$ forms a closed subspace of $\operatorname{Ker}T\oplus\operatorname{Ker}T^*$ (which may be trivial ... )

An example of a partial isometry admitting such elements:
Let $e_j$ denote the elements of an ONB in a countably-infinite $\mathcal H$ and define $$Te_j = \begin{cases} e_{j+1} & \text{$j\,$ is odd} \\ 0 & \text{$j\,$ is even}\end{cases}\quad\implies\; T^*e_j = \begin{cases} e_{j-1} & \text{$j\,$ is even} \\ 0 & \text{$j\,$ is odd}\end{cases}$$ Then

• $\operatorname{Ker}T\:=\:\operatorname{span}\{e_2,e_4,\ldots\} \:=\:\mathcal{R}(T)$
• $\operatorname{Ker}T^*\,=\:\operatorname{span}\{e_1,e_3, e_5,\ldots\} \:=\:\mathcal{R}(T^*)$
• $\mathcal S=\{0\}\,$ and $\,\operatorname{Ker}(T-I) = \operatorname{Ker}(T^*-I) = \{0\}$

and all $\,x=(T^*+I)k\,$ do the job with $k$ running through $\operatorname{Ker}T$.

Addendum:
Due to $(2)$ the corresponding orthogonal projectors satisfy $$(I-TT^*)(I-T^*T) = 0 = (I-T^*T)(I-TT^*)$$ whence $T(T^*)^2T = T^*T^2T^*$ which is the orthogonal projector onto $\mathcal S$. This might prove helpful in the search for further solutions of $(3)$ with $\,Ts\neq s$.