Let $A, B ∈ \mathcal{B}(H)$ be selfadjoint operators on a Hilbert space and assume that $0 ≤ A ≤ B$. Prove that $\|A\| ≤ \|B\|$.
I tried to show the claim by using the fact that if $A$ is a bounded self-adjoint operator on a Hilbert space $H$, then $\|A\| = \sup_{\|x\|=1} \|Ax\|$ (and the same for $B$). And as we know that $0 ≤ A ≤ B$, it follows that $\|Ax\|≤ \|Bx\|$. So, $\|A\| = \sup_{\|x\|=1} \|Ax\| ≤ \sup_{\|x\|=1} \|Bx\| = \|B\|$. But I think that it is not okay.