# Question about product of ideals in a $C^*$-algebra

Consider the following fragments from Murphy's book '$$C^*$$-algebras and operator theory' I'm trying to understand why $$B \cap I = BIB$$.

Attempt:

The inclusion $$BIB \subseteq B\cap I$$ is trivial since $$B$$ is hereditary and $$I$$ is an ideal. To show the other inclusion, it suffices to show that $$(B\cap I)^+ \subseteq BIB$$ since the positive elements of the $$C^*$$-algebra $$B \cap I$$ (this is a $$C^*$$-subalgebra because $$B \cap I$$ is a closed ideal of $$B$$) linearly span $$B\cap I$$.

Fix $$a \in B \cap I$$. Then $$a^{1/2} \in B \cap I$$.

Let $$(u_\lambda)$$ be an approximate unit for $$B$$. Then $$a = \lim_\lambda u_\lambda a = \lim_\lambda {u_\lambda} a^{1/2}a^{1/2} \in BIB$$

Is this correct?

Your argument is fine, but I don't think that the inclusion $$BIB\subset B\cap I$$ is "trivial". It is even a bit less "trivial" as the other one.
It is obvious that $$BIB\subset I$$. But you need to show that $$ayb\in B$$ when $$a,b\in B$$ and $$y\in I$$. This requires an approximate unit (as Murphy says): given an approximate unit $$\{u_\lambda\}$$ in $$B$$, we can write $$y=\sum_{j=1}^4 c_j y_j$$ with $$y_j\geq0$$ for all $$j$$. Since $$0\leq u_\lambda y_j u_\lambda\leq \|y_j\|\,u_\lambda^2\in B$$, we get $$u_\lambda y_j u_\lambda\in B$$ by means of $$B$$ being hereditary. Thus $$u_\lambda y u_\lambda\in B$$ for all $$\lambda$$. Then $$ayb=\lim_\lambda a(u_\lambda y u_\lambda)b\in B$$ as for each $$\lambda$$ the three elements in the product are in $$B$$. As $$BIB$$ is the closed linear span of elements $$ayb$$ as above, we get $$BIB\subset B$$.
• Yes, you are right but Murphy proved that a $C^*$-subalgebra is hereditary iff $bab'\in B$ for all $b,b'\in B, a\in A$ already. But I should have said I used that result. If I assume that result is it ok? – user745578 Jul 29 at 6:45
• Yes, with that and that $I\cap B$ is a closed algebra it is ok. – Martin Argerami Jul 29 at 12:53