Find a linear operator $V$ such that $S=VT$

Question

Let $H$ be a separable Hilbert space. Let $S,T \in L(H)$ be such that $\|Sx\| \leq \|Tx\|$ for all $x \in H$. How do we show that there is $V\in L(H)$ such that $S=VT$?

Answer 1

Hint: We can define a linear map $V_0:{\rm range}(T)\to H$ satisfying the required property and also $\|V_0\|\le 1$.
Then extend it by Hahn-Banach.

Answer 2

Hints: Let $M$ be the range of $T$. Define $V$ on $M$ by $V(Tx)=Sx$. Show that this is well defined and that $V$ is continuous. Extend $V$ to the closure of $M$ by continuity. Recall that $H=\overline {M} +M^{\perp}$. Define $V$ to be $0$ on $M^{\perp}$; this defined $V$ on $H$ by additivity. Check that $V$ is continuous on $H$. [The extended map is nothing but $V\circ P$ where $P$ is the projection on $M$]. By definition we have $S=V\circ T$.