# Approximate units from strictly positive elements in $C^{*}$-algebras.

The existence of a countable approximate unit in a $C^{*}$-algebra $B$ is equivalent to the existence of a strictly positive element $h\in B$. There are several ways to construct an approximate unit from $h$. My question is, does $h(h+\frac{1}{n})^{-1}$ constitute an approximate unit for $B$?

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Do you have something to do with Baaj-Julg picture of KK-theory? –  Kolya Ivankov Mar 17 '11 at 16:17
Sorry, of course you do ^_^ –  Kolya Ivankov Mar 17 '11 at 16:20
It's a small world... –  alterationx10 Mar 17 '11 at 16:36
Does anyone have a reference for this equivalence? Thanks... –  Sergio A. Yuhjtman Nov 5 '12 at 23:41

I think this works: Functional calculus shows that $h h (h+1/n)^{-1} \rightarrow h$. Then $h$ is strictly positive if and only if $Bh$ is dense in $B$ (See, for example, Jensen+Thomsen, "Elements of KK-Theory", Lemma 1.1.21). So for $b\in B$ and $\epsilon>0$, we can find $c\in B$ with $\|b-ch\|<\epsilon$, and for all $n$ sufficiently large, also $\|ch h(h+1/n)^{-1} - ch\| < \epsilon$. Thus \begin{align*} &\| b - bh(h+1/n)^{-1} \| \\&< \epsilon + \| ch - chh(h+1/n)^{-1}\| + \|chh(h+1/n)^{-1} - bh(h+1/n)^{-1}\| \\ &< 2\epsilon + \epsilon \|h(h+1/n)^{-1}\| < 3\epsilon. \end{align*} Thus we're done.