Let $A$ be a bounded operator on a Hilbert space, $$r(A) \:=\:\sup_{\|v\|\,=\,1}{v^*Av}$$ be the numerical radius and $\|A\|_2$ the standard operator norm. Wikipedia (see #13) claims: $$r(A)\leq\|A\|_2\leq2r(A)$$ Technically, it claims this just for finite-dimensional Hilbert spaces, but there's no way the argument doesn't generalize with that uniform bound.
The first inequality is trivial. The second one ought to be straightforward: take $\{x_n\}_n,\{y_n\}_n$ sequences of unit vectors with $y_n^*Ax_n\to\|A\|_2$ and then do something (polarization?) to bound it by $y_n^*Ay_n+x_n^*Ax_n$. But I'm not seeing the trick. What am I missing?
If $A$ is normal, then $r(A)=\|A\|_2$, since the suprema are attained at the maximum eigenvalue.
For other matrices, $A=\frac{A-A^*}{2}+\frac{A+A^*}{2}$, where each summand is normal.