Why does $\lVert x\rVert$ equal $\sup_{\lVert x'\rVert \leq 1} \lVert xx'\rVert$ in a C*-algebra

Question

This is on page 9 of Dixmier's C*-algebra

Let $A$ be a C*-algebra. For each $x \in A$, we have $$\lVert x\rVert = \sup_{\lVert x'\rVert \leq 1}\lVert xx'\rVert.$$

To prove this, the author says

It is clear that $\lVert x'\rVert \leq 1$ implies $\lVert xx'\rVert \leq \lVert x\rVert$. To show that $\lVert x\rVert \leq \sup_{\lVert x'\rVert \leq 1} \lVert xx'\rVert$, we can assume that $\lVert x \rVert =1$; then $\lVert x^\ast\rVert =1$ and $\sup_{\lVert x'\rVert \leq 1} \lVert xx'\rVert \geq \lVert xx^\ast \rVert = \lVert x\rVert^2 =1$.

I am confusioned by the last sentenced, which states $\sup_{\lVert x'\rVert \leq 1} \lVert xx'\rVert \geq \lVert xx^\ast\rVert$. But why does it hold?

Would anyone please give some explanation?

Thanks a lot.

Answer 1

We have $\sup ||xx'||\geq ||x\frac{x^\ast}{||x||}||=\frac{||xx^\ast||}{||x||}=||x||$