Inequality of numerical radius

Question

I want to ask on how to prove this inequality $$|\langle Tx,x\rangle|\leq w(T)||x||^{2}.$$ Where $w(T)=(\sup|\langle Tx,x\rangle|;||x||=1)$

Also, about the inequality, I would go with it like this $$|\langle Tx,x\rangle |\leq ||Tx||||x||(CS)\leq ||T||||x||^{2}(Since~T~is~bounded)$$ I don't know how to obtain $w(T)$ from this. T is a bounded linear operator.

Answer 1

Let $x$ be in the domain of $T$ with $x\neq 0$. Then $y = \frac{x}{\Vert x \Vert}$ has norm 1. Hence \begin{align*} \vert\langle Ty, y \rangle\vert \leq w(T). \end{align*} Now multiply the inequality with $\Vert x \Vert^2 \neq 0$. This yields \begin{align*} \vert \langle Tx, x\rangle\vert \leq w(T) \Vert x \Vert^2. \end{align*}