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if ${U_{n}}$ is an approximate unit for a $C^{*}$-algebra A. Is ${U_{n}^{2}}$ is an approximate unit for a $C^{*}$-algebra A? Thank in advance.

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Let $\{U_n\}$ be your approximate unit.

The $U_n$ are positive so that the $U_n^2=U_n^*U_n$ are also positive.

Also by the C*-equation we have $$\|U_n^2\|=\|U^*_nU_n\|=\|U_n\|^2\leq 1\Rightarrow \|U_n^2\|\leq 1.$$

You should be able to show/find that for any $a\in A$ $$\|U_naU_n-a\|\rightarrow 0\qquad(*)$$ and any $x\in A$ we have $$\|U_nx-xU_n\|\rightarrow 0.\qquad(**)$$

Note $$ \begin{align}\|U_n^2a-a\|&\leq \left\|U_n^2 a-U_naU_n\right\|+\|U_naU_n-a\| \\&=\left\|U_n(U_na)-(U_na)U_n\right\|+\|U_naU_n-a\|\rightarrow0. \end{align}.$$

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    $\begingroup$ thanks for your reply and your time. $\endgroup$ – reza Dec 10 '14 at 20:36
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Approximate units are bounded by $1$, so \begin{align} \|U_n^2X-X\|&\leq\|U_n^2X-U_nX\|+\|U_nX-X\|=\|U_n(U_nX-X)\|+\|U_nX-X\|\\ &\leq\|U_nX-X\|+\|U_nX-X\|=2\|U_nX-X\| \end{align}

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  • $\begingroup$ thanks. but for increasing the approximate unit U_{N}^{2} Does not hold, unless C*-algebra is commutative,true? $\endgroup$ – reza Dec 8 '14 at 19:46
  • $\begingroup$ Yes, you are right. I wasn't thinking about the monotonicity property. $\endgroup$ – Martin Argerami Dec 8 '14 at 21:28
  • $\begingroup$ thanks for your reply and your time. $\endgroup$ – reza Dec 9 '14 at 22:15

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