[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*}[/step]

[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: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.[/step]

[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: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]