Does $S\leq T$ imply $\|S\|_{op}\leq \|T\|_{op}$? 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!
 A: 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 $C^*(1,T)$,
$$\lVert T\rVert \leq \lVert S\rVert.$$
A: 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
$||S||\leq ||T||$.
