Show that Quotient of a C*-algebra is a C*-algebra in its own right

Let $A$ be a commutative C*-algebra over $\mathbb{C}$ and let $J$ be an ideal of $A$ such that $J\in Closed(A)$ and that contains $x^*$ if it contains $x$, for all $x\in A$.

I know that $A/J$ is also a Banach algebra (Rudin FA 11.4) and also has an involution of the form $[x]^*:=[x^*]$ which is well defined (if $[x]=[y]$, $x-y\in J$, so that $x^*-y^* \in J$ and so $[x]^*=[y]^*$).

I would like to prove that $A/J$ is in fact a C*-algebra, that is, that $$||[x][x]^*||=||[x]||^2$$ (the B*-condition is equivalent to the C*-condition).

However I got stuck:

For one direction it is easy:

$$||[x][x]^*|| \leq ||[x]||||[x]^*|| = ||[x]||^2$$ using the fact that $||[x]^*|| \equiv \inf(\{||x^*-j|| | j\in J\}) = \inf(\{||x^*-j^*|| | j\in J\}) = \inf(\{||x-j|| | j\in J\}) \equiv ||[x]||$ because if $j\in J$ then $j^* \in J$ and $A$ is a C*-algebra, so that $||x-j||=||x^*-j^*||$.

For the other direction, $$||[x]||^2 = ||[x]|| ||[x]^*|| \leq ||x-j_1|| ||x^*-j_2|| \stackrel{A\,is\,a\,C*}{=} ||(x-j_1)(x^*-j_2^*)|| = ||x x^*-j_1 x^*-xj_2^*-j_1j_2^* ||$$ for all $(j_1,j_2)\in J^2$.

If $\varepsilon>0$, then $\exists j_\varepsilon\in J$ such that $$||x x^* -j_\varepsilon||<||x x^*+J||+\varepsilon$$ So now if I could find $(j_1,j_2)\in J^2$ such that $j_\varepsilon=-j_1 x^*-xj_2^*-j_1j_2^*$ then the proof would be complete.

However, I am not sure how to do that, if that is even possible.

• Do you need to restrict to commutative algebras? – C-Star-W-Star Jul 3 '15 at 13:21
• Maybe not, but Rudin FA 11.4 applies to commutative algebras, so otherwise one would have to rely on something else. – user118719 Jul 3 '15 at 13:26
• The checks you're doing don't rely on commutativity... – C-Star-W-Star Jul 3 '15 at 13:27
• I see, thanks. Still, I am stuck. – user118719 Jul 3 '15 at 13:28
• Do you have a little knowledge of approximate units? (It will be alot easier!) – C-Star-W-Star Jul 3 '15 at 13:55

Approximate Unit

Regard ball cone: $$\mathcal{B}_+:=\{A\in\mathcal{A}:\|A\|<1:A\geq0\}$$

Order elements: $$E,E'\in\mathcal{I}\cap\mathcal{B}_+:\quad E\leq E'$$

Then one has: $$I\in\mathcal{I}:\quad\|I-IE\|,\|I-EI\|\stackrel{E\to1}{\longrightarrow}0$$

(That is the hard part!)

Quotient Norm

Note that it holds: $$1-\sigma(E)\geq0\implies\|1-E\|\leq1$$

An estimate gives: $$\|A+\mathcal{I}\|\leq\|A-AE\|=\|A(1-E)\|\\ \leq\|A+I\|\cdot1+\|I-IE\|\to\|A+I\|$$

So the norm writes: $$\|A+\mathcal{I}\|=\lim_{E\to1}\|A-AE\|$$

That gives rise to: $$\|A+\mathcal{I}\|^2=\lim_{E\to1}\|A-AE\|^2=\lim_{E\to1}\|(1-E)A^*A(1-E)\|\\ \leq1\lim_{E\to1}\|A^*A-A^*AE\|=\|A^*A+\mathcal{I}\|=\|(A+\mathcal{I})^*(A+\mathcal{I})\|$$

Thus it satisfies: $$\|A+\mathcal{I}\|^2=\|(A+\mathcal{I})^*(A+\mathcal{I})\|$$

Concluding C*-algebra.

• Let me know if you're stuck somewhere. – C-Star-W-Star Jul 3 '15 at 15:06
• Thanks! By $E\leq E'$ you mean $E'-E$ is a positive operator (positive means self-adjoint and with non-negative real spectrum)? – user118719 Jul 3 '15 at 15:14
• Yes, exactly!! :) – C-Star-W-Star Jul 3 '15 at 15:15
• And be careful not to drop selfadjoint as there are nontrivial nilpotents with positive spectrum. – C-Star-W-Star Jul 3 '15 at 15:17
• I'm trying to understand why you order elements in $\mathcal{I}\cap\mathcal{B}_+$, and it seems (from googling "approximate units") that you are trying to say $\mathcal{I}\cap\mathcal{B}_+$ is a net, and then examine the limits of these nets. Is there a reason why you are looking at nets and not just good old sequences? I thought in metric spaces (of which Banach algebras are an example of) sequences are sufficient. All this machinery seems like a huge overkill to prove an inequality. – user118719 Jul 3 '15 at 20:22