This operator has to be self-adjoint Let $T$ be a self-adjoint strict contraction ($||Tx||<||x||$ for all $x\in H\setminus\{0\}$ where $H$ is a Hilbert space). One observe that the expression $T\sqrt{I-T^2}^{-1}$ is well-defined on the range of $\sqrt{(I-T^2)}^{-1}$. Can someone help me to prove that $T\sqrt{I-T^2}^{-1}$ is self-adjoint on this domain?
Thanks a lot
 A: The operator $I-T^{2}$ is positive because
$$
         ((I-T^{2})x,x) = \|x\|^{2}-\|Tx\|^{2} \ge 0
$$
with equality iff $x=0$. So there is a unique $\sqrt{I-T^{2}} \ge 0$ that is non-negative and commutes with every bounded operator that commutes with $T$. This square root is injective because
$$
         \|\sqrt{I-T^{2}}x\|^{2}=((I-T^{2})x,x)=\|x\|^{2}-\|Tx\|^{2}.
$$
The range of $\sqrt{I-T^{2}}$ is dense in $H$ because
$$
          \mathcal{R}(\sqrt{I-T^{2}})^{\perp}=\mathcal{N}(\sqrt{I-T^{2}})=\{0\}.
$$
Let $S=\sqrt{I-T^{2}}^{-1}$ on $\mathcal{D}(S)=\mathcal{R}(\sqrt{I-T^{2}})$. Then $S$ is a densely-defined selfadjoint linear operator, which follows from general results relating inverses and adjoints:
$$
           S^{\star}=(\sqrt{I-T^{2}}^{-1})^{\star}=(\sqrt{I-T^{2}}^{\star})^{-1}=S.
$$
Because $T$ commutes with $\sqrt{I-T^{2}}$, then
$$
              T\sqrt{I-T^{2}}=\sqrt{I-T^{2}}T \\
                  Tx = \sqrt{I-T^{2}}TSx, \;\;\; x\in\mathcal{D}(S)
$$
It follows that $T : \mathcal{D}(S)\rightarrow \mathcal{D}(S)$ and
$$
                     STx = TSx, \;\;\; x\in\mathcal{D}(S).
$$
It is conceivable that $STx$ is defined for more $x$ than is $TSx$, but it turns out this is not the case. Indeed, suppose $Tx \in \mathcal{D}(S)$; equivalently, there exists $y$ such that
$$
                    Tx = \sqrt{I-T^{2}}y,
$$
which can be rewritten as
$$
                   T^{2}x = T\sqrt{I-T^{2}}y=\sqrt{I-T^{2}}Ty,\\
                (I-T^{2})x = x-\sqrt{I-T^{2}}Ty, \\
                \sqrt{I-T^{2}}\left\{\sqrt{I-T^{2}}x+Ty\right\}=x \\
                        \implies x \in \mathcal{D}(S).
$$
So $x \in \mathcal{D}(S)$ iff $Tx \in \mathcal{D}(S)$ and, for any such $x$,
$$
                  TSx = STx,\;\;\; x\in\mathcal{D}(S).
$$
The operator $TS$ is symmetric on $\mathcal{D}(S)$ because $x,y\in\mathcal{D}(S)$ implies
$$
            (TSx,y) = (STx,y) = (Tx,Sy)=(x,TSy).
$$
Therefore, $TS \preceq (TS)^{\star}$, meaning that the graph of $TS$ is a subspace of the graph of $(TS)^{\star}$. Conversely, suppose $y \in \mathcal{D}((TS)^{\star})$. Then
$$
               ((TS)x,y) = (x,(TS)^{\star}y),\;\;\; x\in\mathcal{D}(S),\\
                (Sx,Ty) = (x,(TS)^{\star}y),\;\;\; x\in\mathcal{D}(S), \\
       \implies Ty \in \mathcal{D}(S^{\star})=D(S) \mbox{ and } STy = (TS)^{\star}y.
$$
By what has been shown, $Ty \in \mathcal{D}(S)$ iff $y \in \mathcal{D}(S)$ and, in that case, $TSy = STy = (TS)^{\star}y$. Therefore
$$
                     (TS)^{\star} \preceq TS.
$$
So $(TS)^{\star}=(TS)$ follows from $TS \preceq (TS)^{\star} \preceq TS$.
