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

Question

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?

Answer 1

$a^*a \ge 0$ 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 $1.4.5.$