Let $J$ be a closed ideal in a $C^{\ast}$-algebra $A$ and $a,b \in A$ such that $0 \leq a \leq b.$ If $b \in J$ then show that $a \in J.$
This exercise has been given as a corollary of the following propositions $:$
Proposition 1 $:$ 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 $|a| = (a^{\ast} a)^{\frac {1} {2}}.$
Proposition 2 $:$ Let $a$ and $b$ be elements in a $C^{\ast}$-algebra $A$ with $a^{\ast} a \leq b.$ Then for any $p \in \left (0, \frac {1} {2} \right )$ there exists $u \in A$ such that $a = u b^{p}.$
In view of the above propositions it turns out to me that if we can show that $a^{\ast} a \leq a^{\ast} b$ then we are through. But I don't think it's true. Is there any clever approach to this problem? Any help will be appreciated.
Thanks for your time.
Because $a\geq0$, you can write $a=c^*c$, where $c=a^{1/2}\in A$.
By Proposition 2, there exists $u\in A$ with $c=ub^{1/4}$. So $$ a=b^{1/4}u^*ub^{1/4}\in J, $$ since $b^{1/2}\in J$. Note that since $c=c^*$ we have that $u$ and $b$ commute, but we don't really need it here.