Equivalent operator norm as $|\langle Au,v\rangle|$

Question

Suppose $A$ is a bounded linear operator on Hilbert space $H$. We know that $\|A\|_{op} := \sup \{\|Au\| : u \in H,\ \|u\| = 1\}$. Curious to know if we can write this: $$\|A\|_{op} = \sup \{|\langle Au,v\rangle| : u,v \in H,\ \|u\|=\|v\|=1\}?$$

Got motivated from Equivalent Definitions of the Operator Norm

Answer 1

Observe that $|\langle Au, v \rangle| \le \|A\|$ for $u,v$ with $\|u\|=\|v\|=1$. For the other direction, one observes that for $w=\frac{Av}{\|Av\|},u=v, \|Av\|\ne 0$,$\|v\|=1$, $$\left|\left\langle Av, w\right\rangle\right|=\|Av\|$$

Answer 2

One direction follows from the Cauchy Schwartz inequality: $$|\langle Au,v\rangle| \leq \|Au\|\cdot\|v\| \\ \implies \sup\{|\langle Au,v\rangle| :\|u\|=\|v\|=1 \} \leq \sup \{\|Au\|: \|u\| = 1\} = \|A\|_{op}$$ The other direction is requires a little more work. $$ \text{Let } S = \left\{\left|\left\langle Au,\frac{Au}{\|Au\|}\right\rangle\right| : \|u\|= 1\right\} \subset \{|\langle Au,v\rangle| :\|u\|=\|v\|=1\}$$ Then we note that the supremum of S must be less than the supremum of $\{|\langle Au,v\rangle| :\|u\|=\|v\|=1\}$. However we can note further that: $$ \left|\left\langle Au,\frac{Au}{\|Au\|}\right\rangle\right| = \frac{1}{\|Au\|} \langle Au,Au\rangle = \|Au\|\\ \implies \sup S = \sup \{\|Au\| : \|u\| = 1\} = \|A\|_{op}$$ And we get that $$ \|A\|_{op} \leq \sup\{|\langle Au,v\rangle| :\|u\|=\|v\|=1\} $$