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I read the definition of $C^*$ algebra in Wikipedia where it says $\|x^* x \| = \|x\|\|x^*\|$ is equivalent to $\|xx^*\| = \|x\|^2$ but I do not know why. Can you show me how to derive $\|xx^*\| = \|x\|^2$ from $\|x^* x \| = \|x\|\|x^*\|$?

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You can take a look at this book by Doran and Belfi. The proof you are looking for takes most of chapter 3, some 15+ pages. Unfortunately, not all of it is shown in Google Books.

On a brief historical note, the question you are asking was asked by Gelfand and Naimark in 1943. Glimm and Kadison proved it for unital algebras in 1960, using deep results. The non-unital case is from the 1970s.

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For one direction, if we start by assuming $\|xx^*\| = \|x\|^2$ then as $C^*$-algebras are Banach algebras we get $ \|x\|^2 = \|xx^*\| \le \|x\|\|x^*\|$ so that $\|x\| \le \|x^*\|$. Then $\|x\| \le \|x^*\| \le \| (x^*)^*\| = \|x\|$ so that $\|x\| = \|x^*\|$. It then follows that $\|xx^*\| = \|x\|^2 = \|x\|\|x^*\|$.

The other direction is a fair bit tricker and I'm trying to remember how it goes.

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  • $\begingroup$ Thanks! "The other direction is nontrivial" as wikipedia says. I will appreciate your full answer! $\endgroup$ – Eden Harder Mar 25 '14 at 9:13

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