If $I$ is a prime ideal in a $C^*$-algebra $A$ and $S_1AS_2 \subseteq I$, then either $S_1 \subseteq I$ or $S_2 \subseteq I$

Question

Consider the following fragment from Murphy's "$C^*$-algebras and operator theory":

Can someone explain why we have $S_j \subseteq A S_j A$?

I can prove this if $S_j$ is a sub $C^*$-algebra of $A$ or if $A$ is unital.

Attempt:

Let $(u_\lambda)$ be an approximate unit for $A$. If $x\in S_j$, then somehow we should be able to write $x$ as a norm-limit of some net in $AS_jA$. Maybe we can prove something like $$x=\lim_\lambda u_\lambda^{1/2} x u_\lambda^{1/2}$$

But I don't see why that should hold.

Answer 1

Since $\|u_\lambda\|\leq1$, as long as we approximate on the left and write simultaneously, we obtain $x=\lim_\lambda u_\lambda xu_\lambda$ for all $x$. Indeed, fix $x\in A$ and $\varepsilon>0$. There exists $\lambda_0$ such that $\|xu_\lambda-x\|<\varepsilon$ and $\|u_\lambda x-x\|<\varepsilon$ for $\lambda\geq \lambda_0$. For such $\lambda$, we have $$\|u_\lambda xu_\lambda-x\|\leq\|u_\lambda\|\|xu_\lambda-x\|+\|u_\lambda x-x\|<2\varepsilon.$$