# Is this operator bounded?

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.

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 $$\langle Qx,x\rangle=\langle (T+S)^{-1/2}Ty,x\rangle=\langle Ty,(T+S)^{-1/2}x\rangle=\langle Ty,y\rangle\leqslant \langle (T+S)y,y\rangle\\= \langle (T+S)^{1/2}y,(T+S)^{1/2}y\rangle=\langle x,x\rangle.$$