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I found this question:

Let $A$ be a commutative Banach algebra over $\mathbb{C}$, and $a\in A$ such that for some $\epsilon >0$ we have $|ax|>\epsilon|x|$ for all $x\ne 0$. Prove that there is an isometric embedding $A\to B$ of Banach algebras such that the image of $a$ is invertible.

Since $a$ is not a zero-divisor, $A\to A_a$ is really an inclusion, and my idea is to define a norm in $A_a$ which extends the norm of $A$ and then let $B$ be the completion of $A_a$.

I think my idea is equivalent to the question, actually if $B\supset A$ is as the question required, then $B$ contains a copy of $A_a$(and$\bar{A_a}$), and the norm of $B$ restricted to $A_a$(and$\bar{A_a}$) is what I am asking for.

I also notice the following: suppose $b\in A$, $x\in A_a$ and $bx\in A$, then I have $$|bx|\leq |b||x|\implies|x|\geq \frac{|bx|}{|b|}$$ so I have $$|x|\geq \sup_{b\in A,bx\in A}\frac{|bx|}{|b|}$$

What I want to ask is

1.How do I even know that the supremum on the right exists?

2.How to define a norm in $A_a$ which extends that in $A$?

Any hints or help are appreciated, thank you.

For the first question, I have the following proof:

First, I can replace $a$ by $\frac{a}{\epsilon}$ and assume that $\epsilon=1$. Suppose $b\in A$ and $bx\in A$, and since $x\in A_a$, $a^nx\in A$ for some $n\in \mathbb{N}$. Then I have $a^nx\cdot b=bx\cdot a^n$, from which I can obtain: $$|bx| \leq|bx\cdot a^n|\leq |a^nx||b|\implies \frac{|bx|}{|b|} \leq |a^nx|$$ and hence $\frac{|bx|}{|b|}$is bounded.

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    Are you assuming $A$ is commutative? – 2017-02-10
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    @Eric Wofsey What is the difference if I assume $A$ is commutative? – 2017-02-10
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    @Jonas Meyer , But why$|bx|/|b|$ is bounded above? $b$ may not be of the form $a^n$, for example we may have $b^2=a$ – 2017-02-10
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    Well the usual algebraic theory of localization is only for commutative rings...there are localizations of noncommutative rings but their theory is a lot more complicated and my (admittedly poor) understanding is that a lot of things can go wrong. For instance, as far as I know it is not true in general that if $A$ is a noncommutative ring and $a\in A$ is a non-zero divisor then the map from $A$ to the localization that inverts $a$ is injective. – 2017-02-10
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    @Eric Wofsey, I see, then I think I only need the result when $A$ is commuatative, thank you – 2017-02-10

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