"well-known" inequality for numerical radius of an operator Let $H$ be a Hilbert space. For $T\in\mathcal{B}(H)$, define the numerical radius of $T$ as $\omega(T):=\displaystyle{\sup_{\|x\|=1}|\langle Tx,x\rangle|}$. I am trying to prove that $\|T\|\leq2\omega(T)$ but I just don't see it. My text simply says that this is "easy" and all the literature online simply refer to this estimate as "well-known". 
I know that, for self-adjoint operators, it is $\|T\|=\omega(T)$ and it can be proven elementary. I thought that maybe writing $T=X+iY$ with $X,Y$ self-adjoint could help, but then I couldn't estimate the numerical radii of $X,Y$ related to $\omega(T)$.
 A: First, by Cauchy-Schwarz, we have for all $x \in H$ such that $\lVert x \rVert = 1$
\begin{equation*}
\lvert \langle Tx, x \rangle \rvert \leq \lVert Tx \rVert \underbrace{\lVert x \rVert}_{=1} \leq \lVert T \rVert
\end{equation*}
By the polarization identity, we have for all $(x, y) \in H^2$
$$
4\langle Tx, y \rangle
= \langle T(x + y), x + y \rangle - \langle T(x - y), x - y \rangle
+ i\langle T(x + iy), x + iy \rangle - \langle T(x - iy), x - iy\rangle.
$$
Applying Cauchy-Schwarz we get
$$
4\lvert \langle Tx, y \rangle \rvert
\le w(T) \big[\lVert x + y\rVert^2 + \lVert x - y \rVert^2 + \lVert x + iy \rVert^2 + \lVert x - iy \rVert^2\big]
$$
Finally, by triangular inequality:
\begin{equation*}
4\lvert \langle Tx, y \rangle \rvert \leq 4w(T)[\lVert x \rVert^2 + \lVert y \rVert^2]
\end{equation*}
Thus, if we take $\lVert x \rVert = \lVert y \rVert = 1$:
\begin{equation*}
\lvert \langle Tx, y \rangle \rvert \leq 2w(T)
\end{equation*}
As this is true for all $x, y \in H$ such that $\lVert x \rVert = \lVert y \rVert = 1$, you have: $\lVert T \rVert \leq 2w(T)$.
