Let $B(H)$ the bounded linear functions on a Hilbert space $H$, endowed with the operator norm $\|T\|_{op}=\sup\limits_{\|h\|\le 1}\|T(h)\|_H$. We can write $\|T(h)\|_H^2=\langle T(h),T(h)_H\rangle$ for $h\in H$.
I have seen the equality $\|T\|_{op}=\sup\limits_{\|h\|\le 1}|\langle T(h),h\rangle |$ somewhere in a proof (for $T^*=T$), but I have no idea why this equality holds. How to prove $\|T\|_{op}=\sup\limits_{\|h\|\le 1}|\langle T(h),h\rangle |$? Thank you.
For non-normal $T $ the equality fails; for example, for $T=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ the left-hand-side is 1, while the right-hand-side is 1/2.
For $T $ normal, $\|T\|$ agrees with the spectral radius. That means that $\lambda \|T\|\in\sigma(T)$ for some $\lambda\in\mathbb T$. So $T-\lambda \|T\|\,I$ cannot be bounded below (as $T$ is normal, it doesn't have residual spectrum). Thus there exists a sequence of approximate unit eigenvectors $\{h_n\}$ such that $$Th_n-\lambda\|T\|\,h_n\to0.$$it follows that $$\langle Th_n,h_n\rangle\to\lambda\|T\|,$$and then $$|\langle Th_n,h_n\rangle|\to\|T\|.$$