proof of quotients in $C^{*}$-algebras

Proof: We have that $ℐ$ is self-adjoint (http://planetmath.org/InvolutaryRing), since it is a closed ideal of a $C^{*}$-algebra (http://planetmath.org/CAlgebra) (see this entry (http://planetmath.org/ClosedIdealsInCAlgebrasAreSelfAdjoint)). Hence, the involution in $𝒜$ induces a well-defined involution in $𝒜/ℐ$ by ${(x+ℐ)}^{*}:=x^{*}+ℐ$.

Recall that, since $ℐ$ is closed, the quotient norm is indeed a norm in $𝒜/ℐ$ that makes $𝒜/ℐ$ a Banach algebra (see this entry (http://planetmath.org/QuotientsOfBanachAlgebras)). Thus we only have to prove the $C^{*}$ to prove that $𝒜/ℐ$ is a $C^{*}$-algebra.

Recall that $C^{*}$-algebras have approximate identities (http://planetmath.org/CAlgebrasHaveApproximateIdentities). Notice that $ℐ$ itself is a $C^{*}$-algebra and pick an approximate identity $(e_{\lambda })$ in $ℐ$ such that

• each $e_{\lambda }$ is positive.

• $\parallel e_{\lambda }\parallel \le 1$

We will only prove the case when $𝒜$ has an identity element $e$. For the non-unital case, one can consider $𝒜$ as a $C^{*}$-subalgebra of its minimal unitization and the same proof will still work.

Let $\parallel \cdot {\parallel }_q$ denote the quotient norm in $𝒜/ℐ$. We claim that for every $x\in 𝒜$:

${\parallel x+ℐ\parallel }_q=\underset{\lambda }{lim}\parallel x(e-e_{\lambda })\parallel$(1)

We will prove the above equality as a lemma at the end of the entry. Assuming this result, it follows that for every $a\in 𝒜$

${\parallel x+ℐ\parallel }_q^2=lim{\parallel x(e-e_{\lambda })\parallel }^2=lim\parallel (e-e_{\lambda })x^{*}x(e-e_{\lambda })\parallel \le lim\parallel (e-e_{\lambda })\parallel \parallel x^{*}x(e-e_{\lambda })\parallel$

Since each $e_{\lambda }$ is positive and $\parallel e_{\lambda }\parallel \le 1$ we know that its spectrum lies on the interval $[0,1]$. Hence $e-e_{\lambda }$ is also positive and its spectrum also lies on the interval $[0,1]$. Thus, $\parallel e-e_{\lambda }\parallel \le 1$. Therefore:

${\parallel x+ℐ\parallel }_q^2\le lim\parallel (e-e_{\lambda })\parallel \parallel x^{*}x(e-e_{\lambda })\parallel \le lim\parallel x^{*}x(e-e_{\lambda })\parallel ={\parallel x^{*}x+ℐ\parallel }_q$

Since $𝒜/ℐ$ is a Banach algebra, we also have ${\parallel x^{*}x+ℐ\parallel }_q\le {\parallel x+ℐ\parallel }_q^2$ and so

$${\parallel x+ℐ\parallel }_q^2={\parallel x^{*}x+ℐ\parallel }_q$$

which proves that $𝒜/ℐ$ is a $C^{*}$-algebra. $\mathrm{\square }$

$$

We now prove equality (1) as a lemma.

Lemma - Suppose $𝒜$ is a $C^{*}$-algebra with identity element $e$. Let $ℐ\subset 𝒜$ be a closed ideal and $(e_{\lambda })$ be an approximate identity in $ℐ$ such that each $e_{\lambda }$ is positive and $\parallel e_{\lambda }\parallel \le 1$. Then

$${\parallel x+ℐ\parallel }_q=\underset{\lambda }{lim}\parallel x(e-e_{\lambda })\parallel$$

for every $x$ in $𝒜$.

Proof: Since $y(e-e_{\lambda })⟶0$ for every $y\in ℐ$ it follows that

$lim\; sup\parallel x(e-e_{\lambda })\parallel$$=$$lim\; sup\parallel x-x{e}_{\lambda }-y+y{e}_{\lambda }\parallel$

$=$$lim\; sup\parallel (x-y)(e-e_{\lambda })\parallel$

$\le$$\parallel x-y\parallel$

Therefore, taking the infimum over all $y\in ℐ$ we obtain:

$$lim\; sup\parallel x(e-e_{\lambda })\parallel \le \underset{y\in ℐ}{inf}\parallel x-y\parallel ={\parallel x+ℐ\parallel }_q$$

Also, since $x{e}_{\lambda }\in ℐ$,

$$lim\; inf\parallel x(e-e_{\lambda })\parallel \ge \underset{y\in ℐ}{inf}\parallel x-y\parallel ={\parallel x+ℐ\parallel }_q$$

and this proves the lemma. $\mathrm{\square }$