Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

1 Answer 1

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.