Existence of a approximate unit $U_{n}^{2}$ for a $ C^{*}$-algebra $

Question

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.

score 1 · Answer 1 · answered Dec 10 '14 at 17:15

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

thanks for your reply and your time. – reza Dec 10 '14 at 20:36 — reza, Dec 10 '14 at 20:36

score 0 · Answer 2 · answered Dec 04 '14 at 04:46

0

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}

answered Dec 04 '14 at 04:46

Martin Argerami

205,756

thanks. but for increasing the approximate unit U_{N}^{2} Does not hold, unless C*-algebra is commutative,true? – reza Dec 08 '14 at 19:46
Yes, you are right. I wasn't thinking about the monotonicity property. – Martin Argerami Dec 08 '14 at 21:28
thanks for your reply and your time. – reza Dec 09 '14 at 22:15

Existence of a approximate unit $U_{n}^{2}$ for a $ C^{*}$-algebra $

2 Answers2