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Let $A$ be a commutative Banach algebra with unit. It is well known that if the Gelfand transform $\hat{x}$ of $x\in A$ is non-zero, then $x$ is invertible in $A$ (the so called Wiener Lemma in the case when $A$ is the Banach algebra of absolutely convergent Fourier series).

As a converse of the above, let $B$ be a Banach space contained in $A$ and suppose $B$ is closed under inversion - i.e.: If $x\in B$ and $x^{-1}\in A$ then $x^{-1}\in B$.

(1) Prove that $B$ is a Banach algebra.

(2) Must $A$ and $B$ have the same norm? If not are the norms similar?

(3) Do $A$ and $B$ have the same maximal ideal space?

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    I don't think (1) is true. Pick any Banach algebra $A$ with a noninvertible element $f$ whose square $f^2$ is not a multiple of $f$ and consider the subspace spanned by $f$.2010-08-16
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    I do not understand 1) how these statements constitute a converse of what you have written, and 2) what the first two questions mean. Do you mean that B is a Banach subalgebra of A? And I don't know what "same" norm means.2010-08-17
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    Akhil Mathew: You are right!! I forgot to say that the unit of $A$ belongs to $B$.2010-08-17
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    @Qiaochu Yuan: 1) I see why you ask - for a given Banach algebra the converse of Wiener's Lemma is obvious if an element is invertible then $\hat{x}$ is non-zero. To see what I mean I suggest the to phrase it like this: Wieners Lemma states that IF you work in a commutative unitary Banach algebra THEN the necessary condition of inversion implies inversion. Here we say that inversion in $B$ implies that $B$ is a Banach algebra - provided there is a larger Banach algebra $A$ where inversion really takes place.\\2010-08-17
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    @Qiaochu Yuan: 2) Same norm would be: $\|x\|_B =\|x\|_A$ for all $x\in B$.\\ Similar norm would be: $C_1\|x\|_B \leq \|x\|_A\leq C_2 =\|x\|_B$ for all $x\in B$ for some uniform constants $C_1$ and $C_2$.2010-08-17
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    Replace "Banach" with "quasi-Banach". Is it possible to deduce (1)?2010-08-17
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    When you say $B$ is a Banach space contained in $A$, do you mean that it is a linear subspace of $A$, equipped with a complete norm that is stronger than (i.e. majorizes some constant multiple of) the given norm of $A$?2012-01-08
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    @YemonChoi I do not see if that is necessary. BTW, looking at it now again I see that there are obvious counterexamples to (3) e.g. $B=\mathbb{C}$ and $A=$ "what ever"...2012-01-08
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    @YemonChoi I think any contribution is interesting, at least it might widen my knowledge :)2012-01-08
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    @AD: (3) is obviously false, *as was pointed out in Akhil Matthew's answer*2012-01-08
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    @AD: when you say "I do not see that it is necessary" - my point was that you should make it clear in your question what actual question you are asking. When you say "a Banach space $B$ contained in $A$" what do you *mean*, **precisely**?2012-01-08
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    @YemonChoi Yes, I will think over it all, perhaps I remove it all.2012-01-08
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    @YemonChoi I just re-read Akhil Mattew's answer (I read it some time ago) - and sure (3) was disproved there!2012-01-08

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