# $a=cc^*c$ for some $c$. $a \in A$ a $C^*$ algebra.

Let $$A$$ be a $$C^*$$ algebra. Let $$a \in A$$, then there exists $$c \in A$$ such that $$a=cc^*c$$.

This fact is used from example (1) of Prop 4.25. How does one show this?

• The construction of $c$ is sketched in the proof of Prop. 4.25. Where do you have problems? – MaoWao Mar 16 at 18:17
• Ah, ok, I did not read the proof, and jumped to the example for some intuition. – Bryan Shih Mar 16 at 18:36

$$a^*a \ge 0$$ so the function $$t \mapsto \frac{t^{1/6}}{\sqrt{t+\frac1n}}$$ is continuous on $$\sigma(a^*a) \subseteq [0,\infty]$$ and satisfies $$0\mapsto 0$$.

Thus it makes sense to define a sequence $$(c_n)_n$$ as $$c_n = a\left(a^*a + \frac1n\right)^{-1/2}(a^*a)^{1/6} \in A$$

Denote $$d_{mn} = \left(a^*a + \frac1m\right)^{-1/2} - \left(a^*a + \frac1n\right)^{-1/2}$$ for $$m,n\in\mathbb{N}$$. We have \begin{align} \|c_m - c_n\|^2 &= \|a \,d_{mn}(a^*a)^{1/6}\|^2 \\ &= \|(a^*a)^{1/6}d_{mn}(a^*a)d_{mn}(a^*a)^{1/6}\|\\ &= \|{d_{mn}}^2(a^*a)^{4/3}\|\\ &= \|{d_{mn}}(a^*a)^{2/3}\|^2\\ &= \left\|\left(a^*a + \frac1m\right)^{-1/2}(a^*a)^{2/3} - \left(a^*a + \frac1n\right)^{-1/2}(a^*a)^{2/3}\right\|^2\\ &\xrightarrow{m,n\to\infty} \|(a^*a)^{1/6}-(a^*a)^{1/6}\|^2 \\ &= 0 \end{align}

because $$f_n(t) = \frac{t^{2/3}}{\sqrt{t+\frac1n}} \xrightarrow{n\to\infty} t^{1/6}$$ uniformly on $$\sigma(a^*a)$$.

Hence $$(c_n)_n$$ converges to an element $$c \in A$$. We claim that $$c$$ is the desired element.

\begin{align} cc^*c &= \lim_{n\to\infty} c_nc_n^*c_n \\ &= \lim_{n\to\infty} a\left(a^*a + \frac1n\right)^{-1/2}(a^*a)^{1/6}(a^*a)^{1/6}\left(a^*a + \frac1n\right)^{-1/2}(a^*a)\left(a^*a + \frac1n\right)^{-1/2}(a^*a)^{1/6}\\ &= \lim_{n\to\infty} a\cdot (a^*a)^{3/2}\left(a^*a + \frac1n\right)^{-3/2}\\ &= a \cdot 1\\ &= a \end{align} because $$g_n(t) = \frac{t^{3/2}}{\left(t+\frac1n\right)^{3/2}} \xrightarrow{n\to\infty} 1$$ uniformly on $$\sigma(a^*a)$$. The limit function $$1$$ does not satisfy $$0 \mapsto 0$$ but we can adjoin a unit to the algebra $$A$$ so that it makes sense.

Therefore $$a = cc^*c$$.

This proof was mostly from Pedersen's $$C^*$$-algebras and their Automorphism Groups, Lemma $$1.4.4.$$ and $$1.4.5.$$