[proofplan]
The proof has three parts. First we record that a closed two-sided ideal in a $C^*$-algebra is automatically self-adjoint, which makes the quotient involution well-defined. Second we check that the quotient norm and quotient multiplication make $A/I$ into a Banach algebra. Finally we prove the $C^*$-identity without assuming it in the quotient: the key estimate is obtained from a contractive approximate identity in $I$, which identifies the quotient norm of $a+I$ with the limit of $\|a-ae_\lambda\|_A$ and then passes the $C^*$-identity down from $A$.
[/proofplan]
[step:Show that the closed two-sided ideal is self-adjoint]
By [citetheorem:8571], the closed two-sided ideal $I\trianglelefteq A$ is a hereditary $C^*$-subalgebra of $A$. In particular, $I$ is closed under the involution inherited from $A$. Thus
\begin{align*}
I^*:=\{i^*:i\in I\}=I.
\end{align*}
[guided]
We need the involution on cosets to be independent of the representative. If $a+I=b+I$, then $a-b\in I$, and the desired equality $a^*+I=b^*+I$ is equivalent to $(a-b)^*\in I$. Thus the necessary structural fact is that $I$ is self-adjoint.
The theorem [citetheorem:8571] applies because the present hypotheses state that $A$ is a $C^*$-algebra and $I\trianglelefteq A$ is a closed two-sided ideal. Its conclusion says that $I$ is a hereditary $C^*$-subalgebra of $A$. Being a $C^*$-subalgebra means, in particular, that $I$ is closed under the involution inherited from $A$. Therefore
\begin{align*}
I^*:=\{i^*:i\in I\}=I.
\end{align*}
This is precisely the condition needed for the quotient involution to be well-defined.
[/guided]
[/step]
[step:Define the quotient involution and verify the algebraic operations]
Let $q:A\to A/I$ be the quotient map, defined by
\begin{align*}
q(a):=a+I.
\end{align*}
If $a+I=b+I$, then $a-b\in I$. Since $I=I^*$, we have
\begin{align*}
a^*-b^*=(a-b)^*\in I,
\end{align*}
so $a^*+I=b^*+I$. Therefore
\begin{align*}
(a+I)^*:=a^*+I
\end{align*}
is well-defined.
The quotient multiplication is well-defined because $I$ is two-sided. Indeed, if $a-a'\in I$ and $b-b'\in I$, then
\begin{align*}
ab-a'b'=(a-a')b+a'(b-b')\in I.
\end{align*}
The quotient [vector space](/page/Vector%20Space) operations are well-defined by the definition of the quotient vector space. Hence $A/I$ is a complex algebra.
The involution on $A/I$ is conjugate-linear, involutive, and anti-multiplicative because the involution on $A$ has those properties:
\begin{align*}
(\lambda(a+I)+\mu(b+I))^*=\overline{\lambda}(a+I)^*+\overline{\mu}(b+I)^*
\end{align*}
for all $\lambda,\mu\in\mathbb C$ and $a,b\in A$,
\begin{align*}
((a+I)^*)^*=a+I,
\end{align*}
and
\begin{align*}
((a+I)(b+I))^*=(b+I)^*(a+I)^*.
\end{align*}
[/step]
[step:Make the quotient into a Banach algebra]
The quotient norm is
\begin{align*}
\|a+I\|_{A/I}:=\inf_{i\in I}\|a+i\|_A.
\end{align*}
Because $I$ is a closed linear subspace of the [Banach space](/page/Banach%20Space) $A$, the quotient norm is a norm and $A/I$ is complete. Thus $A/I$ is a Banach space.
We now verify submultiplicativity. Let $a,b\in A$ and let $\varepsilon>0$. By the definition of the infimum, choose $i,j\in I$ such that
\begin{align*}
\|a+i\|_A\le \|a+I\|_{A/I}+\varepsilon
\end{align*}
and
\begin{align*}
\|b+j\|_A\le \|b+I\|_{A/I}+\varepsilon.
\end{align*}
Since $I$ is a two-sided ideal,
\begin{align*}
(a+i)(b+j)-ab=aj+ib+ij\in I.
\end{align*}
Therefore
\begin{align*}
\|(a+I)(b+I)\|_{A/I}\le \|(a+i)(b+j)\|_A.
\end{align*}
Using submultiplicativity in $A$, we obtain
\begin{align*}
\|(a+I)(b+I)\|_{A/I}\le (\|a+I\|_{A/I}+\varepsilon)(\|b+I\|_{A/I}+\varepsilon).
\end{align*}
Letting $\varepsilon\downarrow 0$ gives
\begin{align*}
\|(a+I)(b+I)\|_{A/I}\le \|a+I\|_{A/I}\|b+I\|_{A/I}.
\end{align*}
Thus $A/I$ is a Banach algebra.
[/step]
[step:Identify the quotient norm using an approximate identity for the ideal]
By [citetheorem:8570], the closed two-sided ideal $I$ has a contractive positive approximate identity $(e_\lambda)_{\lambda\in\Lambda}$ contained in $I$. Thus $e_\lambda\in I$, $0\le e_\lambda$, $\|e_\lambda\|_A\le 1$, and
\begin{align*}
\lim_{\lambda}\|ie_\lambda-i\|_A=0
\end{align*}
and
\begin{align*}
\lim_{\lambda}\|e_\lambda i-i\|_A=0
\end{align*}
for every $i\in I$. We claim that for each $a\in A$,
\begin{align*}
\|a+I\|_{A/I}=\lim_{\lambda}\|a-ae_\lambda\|_A.
\end{align*}
First, since $ae_\lambda\in I$, the element $a-ae_\lambda$ represents the coset $a+I$. Hence
\begin{align*}
\|a+I\|_{A/I}\le \|a-ae_\lambda\|_A
\end{align*}
for every $\lambda$, and so
\begin{align*}
\|a+I\|_{A/I}\le \liminf_{\lambda}\|a-ae_\lambda\|_A.
\end{align*}
Let $A^+$ denote the unitization of $A$, and let $1_{A^+}$ denote its identity element. Conversely, fix $i\in I$. Since $(a+i)e_\lambda\in A$ and $\|1_{A^+}-e_\lambda\|_{A^+}\le 1$ for positive contractions $e_\lambda$, we have
\begin{align*}
\|a-ae_\lambda\|_A\le \|(a+i)(1_{A^+}-e_\lambda)\|_A+\|i-ie_\lambda\|_A.
\end{align*}
The first term satisfies
\begin{align*}
\|(a+i)(1_{A^+}-e_\lambda)\|_A\le \|a+i\|_A.
\end{align*}
Taking the limit superior and using $\|i-ie_\lambda\|_A\to 0$, we get
\begin{align*}
\limsup_{\lambda}\|a-ae_\lambda\|_A\le \|a+i\|_A.
\end{align*}
Taking the infimum over $i\in I$ gives
\begin{align*}
\limsup_{\lambda}\|a-ae_\lambda\|_A\le \|a+I\|_{A/I}.
\end{align*}
The lower and upper inequalities prove the claimed formula.
[guided]
The quotient norm is an infimum over all representatives of the coset. Approximate identities let us choose representatives in a controlled way. The elements
\begin{align*}
a-ae_\lambda
\end{align*}
all represent $a+I$, because $ae_\lambda\in I$ by the right ideal property. Hence each one gives an upper representative norm:
\begin{align*}
\|a+I\|_{A/I}\le \|a-ae_\lambda\|_A.
\end{align*}
This proves
\begin{align*}
\|a+I\|_{A/I}\le \liminf_{\lambda}\|a-ae_\lambda\|_A.
\end{align*}
For the reverse inequality, we compare $a-ae_\lambda$ with an arbitrary representative $a+i$ of the same coset. Since $i\in I$, the approximate identity satisfies
\begin{align*}
\|i-ie_\lambda\|_A\longrightarrow 0.
\end{align*}
In the unitization $A^+$, we decompose
\begin{align*}
a-ae_\lambda=(a+i)(1_{A^+}-e_\lambda)-(i-ie_\lambda).
\end{align*}
Taking norms and using the triangle inequality gives
\begin{align*}
\|a-ae_\lambda\|_A\le \|(a+i)(1_{A^+}-e_\lambda)\|_A+\|i-ie_\lambda\|_A.
\end{align*}
Because $e_\lambda$ is a positive contraction, the continuous functional calculus gives
\begin{align*}
\|1_{A^+}-e_\lambda\|_{A^+}\le 1.
\end{align*}
Therefore
\begin{align*}
\|(a+i)(1_{A^+}-e_\lambda)\|_A\le \|a+i\|_A.
\end{align*}
Passing to the limit superior yields
\begin{align*}
\limsup_{\lambda}\|a-ae_\lambda\|_A\le \|a+i\|_A.
\end{align*}
Since $i\in I$ was arbitrary, taking the infimum over all $i\in I$ gives
\begin{align*}
\limsup_{\lambda}\|a-ae_\lambda\|_A\le \|a+I\|_{A/I}.
\end{align*}
Combining the lower and upper estimates proves
\begin{align*}
\|a+I\|_{A/I}=\lim_{\lambda}\|a-ae_\lambda\|_A.
\end{align*}
[/guided]
[/step]
[step:Prove the C star identity in the quotient]
Let $a\in A$. Applying the norm formula from the previous step to $a$ gives
\begin{align*}
\|a+I\|_{A/I}^2=\lim_{\lambda}\|a-ae_\lambda\|_A^2.
\end{align*}
Since $A$ is a $C^*$-algebra, the $C^*$-identity in $A$ gives
\begin{align*}
\|a-ae_\lambda\|_A^2=\|(a-ae_\lambda)^*(a-ae_\lambda)\|_A.
\end{align*}
Expanding the product in $A^+$ and using $e_\lambda=e_\lambda^*$, we obtain
\begin{align*}
(a-ae_\lambda)^*(a-ae_\lambda)=(1_{A^+}-e_\lambda)a^*a(1_{A^+}-e_\lambda).
\end{align*}
This element represents the same coset as $a^*a+I$, because
\begin{align*}
(1_{A^+}-e_\lambda)a^*a(1_{A^+}-e_\lambda)-a^*a
\end{align*}
is a finite sum of terms each containing a factor $e_\lambda\in I$, and hence belongs to $I$. Therefore
\begin{align*}
\|a^*a+I\|_{A/I}\le \|(a-ae_\lambda)^*(a-ae_\lambda)\|_A.
\end{align*}
Taking the limit inferior gives
\begin{align*}
\|a^*a+I\|_{A/I}\le \|a+I\|_{A/I}^2.
\end{align*}
For the reverse inequality, put $c:=a^*a\in A$. Since
\begin{align*}
\|a-ae_\lambda\|_A^2=\|(1_{A^+}-e_\lambda)c(1_{A^+}-e_\lambda)\|_A,
\end{align*}
it is enough to bound the limit superior of the right-hand side by $\|c+I\|_{A/I}$. Fix $i\in I$. We decompose
\begin{align*}
(1_{A^+}-e_\lambda)c(1_{A^+}-e_\lambda)=(1_{A^+}-e_\lambda)(c+i)(1_{A^+}-e_\lambda)-(1_{A^+}-e_\lambda)i(1_{A^+}-e_\lambda).
\end{align*}
Because $e_\lambda$ is a positive contraction, $\|1_{A^+}-e_\lambda\|_{A^+}\le 1$. Hence
\begin{align*}
\|(1_{A^+}-e_\lambda)(c+i)(1_{A^+}-e_\lambda)\|_A\le \|c+i\|_A.
\end{align*}
Also
\begin{align*}
(1_{A^+}-e_\lambda)i(1_{A^+}-e_\lambda)=(i-e_\lambda i)(1_{A^+}-e_\lambda),
\end{align*}
so
\begin{align*}
\|(1_{A^+}-e_\lambda)i(1_{A^+}-e_\lambda)\|_A\le \|i-e_\lambda i\|_A\longrightarrow 0.
\end{align*}
Taking the limit superior gives
\begin{align*}
\limsup_\lambda \|(1_{A^+}-e_\lambda)c(1_{A^+}-e_\lambda)\|_A\le \|c+i\|_A.
\end{align*}
Since $i\in I$ was arbitrary, taking the infimum over $I$ yields
\begin{align*}
\|a+I\|_{A/I}^2\le \|a^*a+I\|_{A/I}.
\end{align*}
Together with the previous inequality, this proves
\begin{align*}
\|(a+I)^*(a+I)\|_{A/I}=\|a+I\|_{A/I}^2.
\end{align*}
[/step]
[step:Conclude that the quotient is a C star algebra]
We have shown that $A/I$ is a complex Banach algebra, that the formula
\begin{align*}
(a+I)^*=a^*+I
\end{align*}
defines a conjugate-linear involutive anti-automorphism, and that the quotient norm satisfies
\begin{align*}
\|(a+I)^*(a+I)\|_{A/I}=\|a+I\|_{A/I}^2
\end{align*}
for every $a\in A$. Therefore $A/I$, equipped with the quotient norm, quotient multiplication, and quotient involution, is a complex $C^*$-algebra.
[/step]
