Quotients of C*-algebras

Preliminaries. Let $\epsilon$ be any strictly positive number. We can consider $$f_{\epsilon }(x)=\frac{x}{x+\epsilon },g_{\epsilon }(x)=\frac{x}{(x+\epsilon {)}^2}$$ on $[0,\mathrm{\infty }[$. $f_{\epsilon }$ is an increasing function. $g_{\epsilon }(x)$ reaches its maximum at $x=\epsilon$.

Proposition. Any closed ideal $I$ in a C*-algebra $A$ is closed with respect to the adjoint operation.

Proposition. When $X$ is a compact space and $I$ is a closed ideal of $C(X)$, there exists a closed subset $K$ of $X$ such that $I=\{f\in C(X)\mid x\in K\Rightarrow f(x)=0\}$.

Proposition. Let $A$ be a C*-algebra, $I$ a closed (bilateral) ideal of $A$ and $x$ an element of $A$. We have $\parallel [x]{\parallel }^2=\parallel [x^{*}x]\parallel$ in the quotient Banach algebra $A/I$. Proof. Step 1. First assume that $x$ is normal and hence the C*-algebra $C_x$ generated by $x$ is commutative. The composition of the inclusion $C_x\to A$ with the projection $A\to A/I$ induces an injection $\iota$ of $C_x/(C_x\cap I)$ into $A/I$. Let $D$ denote the closure of the image of $\iota$. $C_x$ is a C*-algebra by the proposition above. To show that $D$ is a C*-algebra (hence the statement of this proposition for $x$), it is enough to show that $\iota$ is an isometry.

We have the (contractive) Gelfand transform $\Gamma :D\to C(M_D)$ where $M_D$ is the maximal ideal space of $D$. The composition $\Gamma \iota :C_x\to C(M_D)$ induces a continuous mapping $(\Gamma \iota {)}^{\#}:M_D\to M_x$ where $M_x$ is the maximal ideal space of $C_x$. If this mapping is surjective, $\Gamma \iota$ becomes an isometry and so is $\iota$.

As $D$ is commutative, ``$f\in D$ is invertible'' is equivalent to ``$\Gamma (f)$ is invertible.'' If $(\Gamma \iota {)}^{\#}$ was not surjective, there would exist a non-invertible element $y$ in $C_x$ such that $\Gamma \iota y=1$. Then we should have $\iota y$ invertible in $D$, which is a contradiction. Hence $(\Gamma \iota {)}^{\#}$ is surjective.

Step 2. Let $x$ be any element of $A$. Put $\delta =\parallel [x^{*}x]\parallel$. We are going to establsh $\parallel [x]{\parallel }^2\le \delta$.

Step 2-1. Suppose $x^{*}x\neg \in I$ so that $\delta >0$ and we have $$x=x\frac{\delta }{x^{*}x+\delta }+x\frac{x^{*}x}{x^{*}x+\delta }.$$ For the first term in the right hand side, its norm squared is equal to the norm of $$(x\frac{\delta }{x^{*}x+\delta }{)}^{*}x\frac{\delta }{x^{*}x+\delta }={\delta }^2{g}_{\delta }(x^{*}x),$$ which is equal to $\delta /4$. Hence the norm of the first term of the RHS in $$[x]=[x]\frac{\delta }{[x^{*}x]+\delta }+[x]\frac{[x^{*}x]}{[x^{*}x]+\delta }.$$ is majored by $\sqrt{\delta }/2$. For the second term, by the step 1 the Banach subalgebra of $A/I$ generated by $[x^{*}x]$ is a C*-algebra and we can do functional calculus there. We have $\parallel f_{\delta }([x^{*}x])\parallel =1/2$. Hence we obtain $$\parallel [x]\parallel \le \frac{\sqrt{\delta }}{2}+\parallel [x]\parallel .\frac{1}{2},$$ that leads to $\parallel [x]\parallel \le \delta$.

Step 2-2. When $x^{*}x\in I$, for any strictly positive real number $\epsilon$ we have $$[x]=[x]\frac{\epsilon }{[x^{*}x]+\epsilon }+[x]\frac{[x^{*}x]}{[x^{*}x]+\epsilon }=[x]\frac{\epsilon }{[x^{*}x]+\epsilon }$$ in $A/I$. When $\epsilon <\parallel x^{*}x\parallel$, we have $$\parallel [x]\frac{\epsilon }{[x^{*}x]+\epsilon }\parallel \le \parallel x\frac{\epsilon }{x^{*}x+\epsilon }\parallel =\frac{\sqrt{\epsilon }}{2}.$$ By taking the limit $\epsilon \to 0$, we get $\parallel [x]\parallel =0$.