Does $S\leq T$ imply $\|S\|_{op}\leq \|T\|_{op}$?

Question

Let $T:V\to V$ and $S:V\to V$ be bounded self adjoint operators on a complex Hilbert space $V$. Is it true that if $0\leq T-S$ then $\|S\|_{op}\leq \|T\|_{op}$ with respect to the operator norm? Thanks!

Answer 1

In general is false: consider $V= \mathbb{C}$ and $T $ as the identity and $S(z) = -\lambda z $ for some fixed $\lambda>1$. Then $T-S = I+\lambda I$ that has positive eigenvalues $1+\lambda>0$ hence is definite positive but $||T|| = 1$ while $||S||=\lambda>1$.

It is true if $S\geq0$ though. In this case $T -S\geq 0$ implies that $q_T(x)=\langle Tx,x\rangle\geq q_S(x)=\langle Sx,x\rangle$ hence $\sup_{||x||= 1} q_T(x)\geq \sup_{||x||= 1} q_S(x) $ and $\inf_{||x||= 1} q_T(x)\geq \inf_{||x||= 1} q_S(x) $.

From spectral theory we know that $\sup_{||x||= 1} q_T(x) = \max \sigma(T)$ and $\inf_{||x||= 1} q_T(x) = \min \sigma(T)$ It follows that since $S\geq 0$ also $T\geq 0$. Also we know that the spectral radius $r_T = ||T||= \max |\sigma(T)| = \max \sigma(T)$ and $r_S = ||S|| = \max \sigma(S)$, it follows that $||S||\leq ||T||$.

Answer 2

It is true when $0\leq T\leq S$.

By functional calculus on $C^*(1,S)$, we have $$S\leq \lVert S\rVert 1,$$ so $$0\leq T\leq \lVert S\rVert 1.$$ Again by functional calculus on $C^*(1,T)$, $$\lVert T\rVert \leq \lVert S\rVert.$$