If $T\geq 0$ then $|\langle x,STx\rangle|\leq\|S\|\langle x,Tx\rangle$

Question

I am trying to prove that on a complex Hilbert space, if $T\geq 0$ and $S$ is any bounded linear operator, then $T\geq 0$ then $|\langle x,STx\rangle|\leq\|S\||\langle x,Tx\rangle|$.

Answer 1

This is not true. Let $\theta>0$ be sufficiently small, $$ T=\pmatrix{\cos\theta&\sin\theta\\ \sin\theta&\cos\theta} \quad\text{and}\quad S=\pmatrix{\cos\theta&\sin\theta\\ 0&0}. $$ Then $T>0$ and $\|S\|=1$, but $|\langle e_1,STe_1\rangle|=1>\cos\theta=\|S\|\langle e_1,Te_1\rangle$.