Let $J$ be a closed ideal in a $C^{\ast}$-algebra $A$ and $a \in A.$ If $a^{\ast} a \in J$ then $a \in J.$
This result has been left as a corollary of the following proposition.
Proposition $:$ Let $A$ be a $C^{\ast}$-algebra and $a \in A.$ Then for any $p \in (0,1)$ there exists $u \in A$ such that $a = u |a|^{p},$ where $|a| = (a^{\ast} a)^{\frac {1} {2}}.$
So according to this proposition if we can show that $|a|^{p} \in J$ for some $p \in (0,1)$ then we are through. What we only know is that $|a|^2 \in J.$ How do I conclude it? Any help in this regard would be highly appreciated.
Thanks for your time.
We apply Weierstrass approximation theorem to get hold a sequence of polynomials $\{p_n\}_{n \geq 1}$ with constant terms equal to $0$ which converges uniformly to the function $g : t \mapsto t^{\frac {1} {4}}$ on $[0,\|a\|^2].$ Let $\phi$ be the continuous functional calculus for the element $|a|^2.$ Then $\{\phi (p_n)\}_{n \geq 1}$ will be a sequence of polynomials in $|a|^2$ converging to $|a|^{\frac {1} {2}}$ in norm. Since $|a|^2 \in J$ and $J$ is closed it follows that $|a|^{\frac {1} {2}} \in J.$ But then $a \in J$ by the given proposition since $J$ is an ideal (two sided and hence a left ideal).