TITLE: Norm of bounded operator on a complex Hilbert space.
QUESTION [5 upvotes]: 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
$$||T||=\sup_{||x||=1}|\langle x, Tx \rangle |\;?$$

REPLY [3 votes]: 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.