Norm of bounded operator on a complex Hilbert space. It is fairly easy to show that for a bounded linear operator $T$ on a Hilbert space $H$ $$||T||=\sup_{||x||=1,||y||=1}|\langle y, Tx \rangle |.$$
If $H$ is a complex Hilbert space, can you show that
$$||T||=\sup_{||x||=1}|\langle x, Tx \rangle |\;?$$
 A: Yes, if the operator is self-adjoint. Here is a proof from Conway's book, A Course in Functional Analysis:
Let $M = \{\sup |\langle Ax, x\rangle| : \|x\| = 1\}$. If $\|x\| = 1$, then $|\langle Ax,x\rangle | \leq \|A\|$ (since this holds for all pairs of vectors, so certainly works if we use the same in each slot), so $M \leq \|A\|$.
Now let $h,g$ be unit vectors. Self-adjointness gives $$\langle A(h \pm g) , h \pm g\rangle = \langle Ah,h\rangle \pm 2\mbox{Re}\langle Ah,g\rangle + \langle Ag,g \rangle.$$ Subtracting these equations gives $$\langle A(h + g), h + g \rangle - \langle A(h-g), h-g \rangle = 4 \mbox{Re} \langle Ah,g \rangle.$$ Since $|\langle Af, f \rangle| \leq M\|f\|^2$ for any $f$, by Cauchy-Schwarz, we have using the parallelogram law that \begin{align*}&4\mbox{Re}\langle Ah,g\rangle \\\leq &M(\|h + g\|^2 + \|h-g\|^2) \\ =&2M(\|h\|^2 + \|g\|^2) \\ = &4M. \end{align*}
To finish up, choose $e^{i\theta}$ so that $\mbox{Re}(e^{i\theta}\langle Ah, g\rangle) = |\langle Ah,g \rangle|$. Then $$|\langle Ah, g \rangle| = \mbox{Re}(\langle Ae^{i\theta}h, g\rangle ) \leq M.$$ Take the supreumum over all $h,g$ to finish.
A: Not when it isn't true, e.g. for the operator $T(x,y)=(y,0)$ on $\mathbb C^2$ with the standard inner product.  In that case the $\sup$ is $\frac{1}{2}$ but the operator has norm $1$.
In general, that $\sup$ gives you the numerical radius, which is equivalent but often unequal to the operator norm.
A: Yes, of course. Use the polarization identity:
$\newcommand{\z}{\langle}\newcommand{\x}{\rangle}$
$$\z x+y,T(x+y)\x + i \z x+iy,T(x+iy)\x - \\
- \z x-y,T(x-y)\x -i \z x-iy,T(x-iy)\x = \,\dots
$$
