Proof of operator norm with the equality of involution

Question

Given the equality $$\|A^*A\| = \sup_{\| x \| = \| y \| = 1} | (Ax, (A^*)^*y) | $$

How do show that it is equal to $\|A\|^2$

Is it by using Cauchy-Schwarz inequality such that $(x, y) \leq \|x\|\|y\|$.

It gives that $$\sup_{\| x \| = \| y \| = 1} | (Ax, (A^*)^*y) | \leq \sup_{\| x \| = \| y \| = 1} | \|A\|\|A\| = \|A\|^2.$$ Since it is the supremum, therefore we can discard the $<$ sign

Is this proof reasonble ?

Answer 1

Your proof shows that $||A^*A|| \leq ||A||^2$. You still have to show $||A||^2 \leq ||A^*A||$.

Now $sup_{||x||=1}||(Ax,Ax)|| \leq ||A^*A||$ by the first equality in your post. The result thus follows.