Theorem - Let $\mathcal{A}$ be a Banach algebra and $\mathcal{I} \subseteq \mathcal{A}$ a closed ideal. Then $\mathcal{A}/\mathcal{I}$ is Banach algebra under the quotient norm.

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Proof: Denote the quotient norm by $\|\cdot\|_q$ .

By the parent entry we know that $\mathcal{A}/\mathcal{I}$ is a Banach space under the quotient norm. Thus, we only need to show the normed algebra inequality:

Using the fact that $\mathcal{A}$ is a Banach algebra and the definition of quotient norm we have that: \begin{eqnarray*} \|ab+\mathcal{I}\|_q & = & \inf_{z \,\in\, ab + \mathcal{I}} \|z\| = \inf_{u \,\in\, a + \mathcal{I}\atop v \,\in\, b + \mathcal{I}} \|uv\| \\ & \leq & \inf_{u \,\in\, a + \mathcal{I}\atop v \,\in\, b + \mathcal{I}} \|u\|\|v\| \leq \inf_{u \,\in\, a + \mathcal{I}}\|u\| \inf_{v \,\in\, b + \mathcal{I}}\|v\| \\ & = & \|a+ \mathcal{I}\|_q \|b+ \mathcal{I}\|_q \end{eqnarray*} $\square$