Inequalities for numerical radius of complex Hilbert space operator

Question

Let $A$ be a (complex) Hilbert space operator. Show that $\omega(A) \leq ||A|| \leq 2 \omega(A)$, where $\omega(A)$ is the numerical radius of $A$ given by $\omega(A) := \displaystyle{\sup_{\|x\|=1}|\langle Ax,x\rangle|}$.

I already showed $||A|| \leq 2 \omega(A)$ by using the polarization identity. I am stuck on showing $\omega(A) \leq ||A||$. I think I am missing something rather obvious.

Answer 1

It is obvious when you use the Cauchy-Schwarz inequality: $$ \omega(A) = \sup_{\lVert x \rVert = 1} \lvert \langle Ax, x \rangle \rvert \leq \sup_{\lVert x \rVert = 1} \lVert Ax \rVert \lVert x \rVert \leq \sup_{\lVert x \rVert = 1} \lVert Ax \rVert \sup_{\lVert y \rVert = 1} \lVert y \rVert = \sup_{\lVert x \rVert = 1} \lVert Ax \rVert = \lVert A \rVert $$