I have been reading one article about inequalities, and this equality is stated, $$||T\pm S||^{2}=||(T\pm S)^{*}(T\pm S)||$$ where $T,S\in B(H)$. How should I go about proving this? I know that $$||T\pm S||=\sup_{x\in H,||x||=1}||(T+S)x||=\sup_{x\in H,||x||=1}\sqrt{\langle (T+S)x,x\rangle}$$ Squaring both sides, doesnt look like it leads me to the solution...
Thank you so much for helping me with this!
In a Hilbert space $H$, the equality : $$\|A\|^2 = \|A^*A\|$$ holds for any $A \in B(H)$. Just apply this to $A = T \pm S$.
This is because for $x \in H$, you have : \begin{align*} \|Ax\|^2 &= \langle Ax, Ax\rangle\\ &= \langle A^*Ax, x\rangle\\ &\leq \|A^*A\|\|x\|^2 \leq \|A^*A\| \leq \|A^*\|\|A\|. \end{align*} Now take the supremum on $x$ to get $\|A\|^2 \leq \|A^*A\| \leq \|A^*\|\|A\| = \|A\|^2$. This chain of inequalities is then an equality.