In Murphy's book there is the following remark:
Let $I$ be a closed ideal in a $C^*$-algebra $A$ and $J$ a closed ideal in $I$. Then $J$ is also a closed ideal in $A$.
In the book he uses that closed ideals are $C^*$-sub algebras (uses approximate units) and so any positive $b\in J$ has a root $b^{1/2}$ in $J$ and there is an approximate unit $u_\lambda\in I$. It follows for any $a\in A$ that $au_\lambda b\in J$, since $au_\lambda b^{1/2}\in I$ by virtue of $I$ being an ideal in $A$ and then $au_\lambda b^{1/2}b^{1/2}\in J$ from $J$ being an ideal in $I$.
Since $J$ is closed it follows that $\lim_\lambda au_\lambda b = ab$ lies in $J$. Since positive elements span the algebra it follows $J$ is a left ideal, since $J$ is also self adjoint its a right ideal too.
There are two parts of the proof that seem pointless to me:
- Why talk about roots? $a u_\lambda$ already lies in $I$ (no need to consider $au_\lambda b^{1/2}$!) so $au_\lambda b$ lies $J$. Since $u_\lambda$ is an approximate unit in $I$ and $J\subset I$ it follows $\lim_\lambda u_\lambda b = b$ and we could be done already.
- If you want to use roots then you don't need the approximate unit, if $b=b^{1/2}b^{1/2}$ then $ab=ab^{1/2}b^{1/2}$ where $ab^{1/2}\in I$ because $I$ is an ideal in $A$, it follows then $ab\in J$.
Am I missing something or is the proof actually two proofs mashed together?
I think you are right. The argument that if $b\in J^+$ then $ab^{1/2}b^{1/2}\in IJ=J$ works fine.