Exploring more from Equivalent operator norm as $|⟨,⟩|$ 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!
 A: 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.
