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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.

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Observe that $x^\ast \in \{x' : \lVert x'\rVert \leq 1\}$ and use the definition of the supremum. – commenter Oct 7 '12 at 12:16
@commenter: Sincere thanks for taking time to edit my question and answer it. I am sorry I just begin to study C*-algebra and I am not quite familiar with the notations. Thanks for tha answer again. I realize this is not a good question and I don't know why I was stuck here... – ShinyaSakai Oct 7 '12 at 12:48
Dear ShinyaSakai. You are welcome. I think Dixmier is a very good book, but it might be a bit terse if you're just starting out with C*-algebras (many people consider it hard to read). Maybe you could read some less specialized book first? Many books on functional analysis contain introductions to C*-algebras, for example Conway or Pedersen. Concerning the notation with norms: It looks a bit better to use \|..\| or \lVert..\rVert in place of ||..||. On mathoverflow using * instead of \ast sometimes gives problems with the formatting, so I would suggest to use the latter. – commenter Oct 7 '12 at 13:00
Dear @commenter: Thanks again. I'll remember what you said on notations. As this book is suggested by my teacher, I think I will continue with it while reading the books you mentioned which are easier to follow. – ShinyaSakai Oct 7 '12 at 13:24

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

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