Exploring more from Equivalent operator norm as $|\langle Au,v\rangle|$.
$A$ is a linear bounded operator, and $H$ is a Hilbert space. Let $P := \sup \{|\langle Au,u\rangle| : u \in H,\ \|u\|=1\},$ and $Q:=\sup \{|\langle Au,v\rangle| : u,v \in H,\ \|u\|=\|v\|=1\}.$
1- Suppose $A$ is self-adjoint, to show: $$ P=Q . $$
I was able to show that $P \leq Q$, but couldn't proceed with the other direction!
2- Suppose $H$ is a complex Hilbert space, to show: $$Q \leq 2P .$$
I was able to show that $\langle A(x+\alpha y), x+\alpha y\rangle − \langle A(x-\alpha y), x-\alpha y\rangle = 2\overline\alpha\langle Ax,y\rangle+2\alpha\langle Ay,x\rangle,$ where $|\alpha|=1$. Couldn't proceed further with this equivalence. Thanks in advance for any help!
Notice that $$\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle = 2\langle Ax,y\rangle+2\langle Ay,x\rangle.$$ Since $A $ is self adjoint and $H $ is a real Hilbert space, $2\langle Ax,y\rangle+2\langle Ay,x\rangle=4\langle Ax,y\rangle$.
So $$\begin{align*} |4\langle Ax,y\rangle| &=|\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle |\\ &\leq P\|x+y\|^2+P\|x-y\|^2\\ &=2P(\|x\|^2+\|y\|^2). \end{align*}$$ If $\|x\|=\|y\|=1$, we have $$|\langle Ax,y\rangle|\leq P.$$
When $H$ is a complex Hilbert space, we have $$\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle = 2\langle Ax,y\rangle+2\langle Ay,x\rangle,$$ and $$\langle A(x+iy), x+iy\rangle − \langle A(x- iy), x- iy\rangle = -2i\langle Ax,y\rangle+2i\langle Ay,x\rangle.$$ Then by a similar calculation as above, one has $$|\langle Ax,y\rangle+\langle Ay,x\rangle|\,,\,|\langle Ay,x\rangle-\langle Ax,y\rangle|\leq 2P\:(\|x\|=\|y\|=1).$$ Since $$\begin{align*} |\langle Ax,y\rangle| &=\frac12|2\langle Ax,y\rangle|\\ &=\frac12|\langle Ax,y\rangle+\langle Ay,x\rangle-\langle Ay,x\rangle+\langle Ax,y\rangle|\\ &\leq\frac124P=2P,\end{align*}$$ the inequality follows.