Let $T$ be an invertible positive operator and $S$ be another positive operator on a complex Hilbert space.
We then study
$$ \Vert (T+S)^{-1/2}T(T+S)^{-1/2}\Vert$$


I would assume that this norm is bounded by one.


But I fail to see how one could actually show this? Cause the definition of the square root using the functional calculus is rather abstract.



Is this operator bounded?
Denote $Q=(T+S)^{-1/2}T(T+S)^{-1/2}$. The inequality $\|Q\|\leqslant 1$ is equivalent to $\langle Qx,x\rangle\leqslant \langle x,x\rangle$ for all vectors $x$. Denote $(T+S)^{-1/2}x=y$, we get $$\langle Qx,x\rangle=\langle (T+S)^{-1/2}Ty,x\rangle=\langle Ty,(T+S)^{-1/2}x\rangle=\langle Ty,y\rangle\leqslant \langle (T+S)y,y\rangle\\=
\langle (T+S)^{1/2}y,(T+S)^{1/2}y\rangle=\langle x,x\rangle.$$


