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How do I prove this lemma?

Lemma: Suppose that $X$ is a Hilbert $A$-module. For each $x\in X$ there exists a unique $y\in X$ such that $x=y \langle y,y \rangle $.

I don't know if this is needed but we already know Cohen's factorization theorem that states If $A$ is a Banach algebra with a bounded left or right approximate identity, then for all $a\in A$ there exist $b,c \in A$ such that $a=bc$.

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If $\langle x,x\rangle$ is invertible you can set $y=x\langle x,x\rangle^{1/3}$ –  userNaN Jun 26 '12 at 21:26
    
@Norbert: You mean $-1/3$, right? –  Jonas Meyer Jun 26 '12 at 21:49
    
@JonasMeyer Yes, you are right. Do you have any ideas for the general case? –  userNaN Jun 26 '12 at 21:50
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@Norbert: I cheated, i.e., I knew where to look it up. It is Proposition 2.31 in Raeburn and Williams's Morita equivalence and continuous trace $C^*$-algebras. They cite the article at the following link as the source of the proof they present (link to large pdf, where the result is found on the sixth page, a.k.a. page 145): archive.numdam.org/ARCHIVE/BSMF/BSMF_1996__124_1/… –  Jonas Meyer Jun 26 '12 at 21:54
    
@JonasMeyer Thanks! –  userNaN Jun 26 '12 at 22:06

1 Answer 1

up vote 3 down vote accepted

If you have access to the book Morita equivalence and continuous trace C*-algebras by I. Raeburn and D. Williams, you will find this proved there as Proposition 2.31 on page 21 (the proof is on the next page). In this 6 MB pdf file of the article "Déformations de C*-algèbres de Hopf" by E. Blanchard, the result is on the sixth page (labeled as page 145 in the publication).

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Thanks @t.b. for improving the book reference. –  Jonas Meyer Jul 3 '12 at 3:34

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