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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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    @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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For now, I shall assume that $B$ is a closed subspace of $A$ (and is treated simply as a topological vector space for the moment).

(1) Let $x \in B$; we need to prove that $x^2 \in B$. From this elementary algebra will imply that $B$ is closed under multiplication. Now, when $t$ is close to zero, we have $1-tx$ invertible in $A$, hence in $B$; so $1+tx + t^2x^2 + \dots$ lies in $B$ for $t$ close to zero. Taking the second derivative at $0$ (recall that $B$ is a closed subspace) shows that $x^2 \in B$.

I don't fully understand what you are asking in (2) yet, so:

(3) No. Take $A$ to be any Banach algebra not equal to the complex numbers and $B$ the subalgebra spanned by the identity.

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    In a Banach space you can, look up the Bochner integral. In complete linear spaces there are sort of Riemann integrals discovered by S. Rolewicz (but these are strange and must be handled gently).2010-08-18
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I think the OP needs to give a precise formulation of what he/she means by "a Banach space inside a Banach algebra", because this affects the answers to the questions in non-trivial ways.

Contrary to what the OP seems to claim in comments to Akhil Matthew's answer, if $B$ is not closed in $A$, then it can be an inverse-closed subspace without the given Banach norm on $B$ being submultiplicative. However, by Akhil's argument, $B$ is a subalgebra of $A$.

It is still not clear to me what precisely is meant by (2). So I shall just point out for the record that if we let $A({\mathbb T})$ be the space of all continuous functions ${\mathbb T}\to {\mathbb C}$ with absolutely convergent Fourier series, equipped with pointwise product, this is a Banach algebra and it is a unital subalgebra of $C({\mathbb T})$. Clearly the norm on $A({\mathbb T})$ is not equivalent to the sup-norm inherited from $C({\mathbb T})$; but $A({\mathbb T})$ is inverse-closed in $C({\mathbb T})$, by Gelfand's version of Wiener's $1/f$ lemma.

Note also in (3) that one can have commutative, semisimple, unital Banach algebras $A$ and $B$, and an injective, contractive, unital algebra homomorphism $B\to A$ with dense range, such that $A$ and $B$ have different maximal ideal spaces. The example I have in mind is due, I think, to Honary, but I am out of the office right now and can't look this up.