For normal operator $T$, $\Vert T\Vert=\sup\left\{|\langle Tx,x\rangle|\colon \Vert x \Vert \leq 1\right\}$. In Rudin's Functional analysis, we have a proof which shows that for a normal operator (ref: Page 326 Theorem 12.25)
$\Vert T\Vert=\sup\left\{|\langle Tx,x\rangle|\colon \Vert x \Vert \leq 1\right\}$.
I have understood the proof but searching for an alternate one. Can someone provide reference for alternate proof.
Thank You.
 A: This elementary proof is mostly from Pedersen's Analysis Now, Page $98$, Proposition $3.2.25.$
Let $T$ be a normal operator on a complex Hilbert space $H$. Define the numerical radius of $T$ as $$w(T) = \sup_{\|x\| \le 1}{\left|\langle Tx,x\rangle\right|}$$
We wish to prove $w(T) = \|T\|$. Clearly $w(T) \le \|T\|$ by CSB. Notice that for all $x \in H$ holds $\left|\langle Tx,x\rangle\right| \le w(T)\|x\|^2$.
Verify that for all $x,y \in H$ we have
$$\langle T(x+y), x+y\rangle - \langle T(x-y), x-y\rangle = 2\langle Tx,y\rangle + 2\langle Ty,x\rangle$$
When $\|x\|= \|y\|= 1$ the parallelogram identity gives
\begin{align}\left|\langle T(x+y), x+y\rangle - \langle T(x-y), x-y\rangle\right| &\le w(T)(\|x+y\|^2 + \|x-y\|^2) \\
&= 2w(T)(\|x\|^2+\|y\|^2) \\
&\le4w(T)\end{align}
In particular, for $y = \frac{Tx}{\|Tx\|}$ we get
$$\left|\left\langle Tx, \frac{Tx}{\|Tx\|}\right\rangle + \frac{1}{\|Tx\|}\left\langle T^2x,x\right\rangle\right|  = \left|2\langle Tx,y\rangle + 2\langle Ty,x\rangle\right| \le 4w(T)$$
Considering the real part of the LHS gives
$$\|Tx\| + \frac{1}{\|Tx\|} \operatorname{Re}\langle T^2x,x\rangle \le 2w(T)$$
Fix $x \in H, \|x\| = 1$ and let $\langle T^2x, x\rangle = \left|\langle T^2x, x\rangle\right|e^{i\phi} $.
The previous inequality holds for any normal operator, so in particular for the operator $e^{-i\frac{\phi}2}T$ we get
$$\|Tx\| + \frac{1}{\|Tx\|}\left|\langle T^2x,x\rangle\right| \le 2w(T)$$
Therefore $\|Tx\| \le 2w(T)$ for all $x \in H, \|x\| = 1$. Taking the supremum over the unit sphere gives $\|T\| \le 2w(T)$.
Rearranging the above inequality also gives
\begin{align}
0 &\le 2w(T)\|Tx\| - \|Tx\|^2 -  \left|\langle T^2x, x\rangle\right|\\
&= - \big(w(T) - \|Tx\|\big)^2 + w(T)^2 - \left|\langle T^2x, x\rangle\right|\\
&\le  w(T)^2 - \left|\langle T^2x,x\rangle\right|
\end{align}
Taking the supremum over $\|x\| = 1$ gives $w(T^2) \le w(T)^2$.
Finally, recall that for normal operators we have $\|T^n\| = \|T\|^n, \forall n \in \mathbb{N}$.
Now $$\|T\|^{2^n} = \left\|T^{2^n}\right\| \le 2w\left(T^{2^n}\right) \le 2w(T)^{2^n}$$
so $\|T\| \le 2^{\frac1{2^n}} w(T), \forall n \in \mathbb{N}$. Letting $n\to\infty $ gives $\|T\| \le  w(T)$.

Note that this result is in general false for real Hilbert spaces. Namely, consider the normal operator $A : \mathbb{R}^2 \to \mathbb{R}^2$ given by the matrix $$A = \pmatrix{0 & 1 \\ -1 & 0}$$  
It satisfies $\langle Ax, x\rangle = 0, \forall x \in \mathbb{R}^2$ but $A \ne 0$.
