# Converse of Wiener's lemma

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 $f$. – Akhil Mathew Aug 16 '10 at 21:27
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. – Qiaochu Yuan Aug 17 '10 at 1:16
Akhil Mathew: You are right!! I forgot to say that the unit of $A$ belongs to $B$. – AD. Aug 17 '10 at 5:20
@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.\\ – AD. Aug 17 '10 at 5:45
@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 $C_2$. – AD. Aug 17 '10 at 5:46

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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Regarding (1) - Nice, but you do not really need to take the derivative to reach $x^2\in B$, in fact you do not need $B$ to be closed in $A$. (3) Works fine! – AD. Aug 17 '10 at 8:35
(Here are some proofs of the statement $x^2\in B$: Put $f(z)=(1−zx)^{−1}$, and look at --- 1) $\int_{-\pi}^\pi f(re^{it})e^{−2it}dt$. --- 2) $(f(z)+f(-z)-2)/2z^2$, and let $z\to0$. --- 3) $(f(z)-1-zx)/z^2$, and let $z\to0$.) – AD. Aug 17 '10 at 20:46
AD: I'm not sure you can define integrals unless you are working with a closed subspace. – Akhil Mathew Aug 17 '10 at 21:20
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). – AD. Aug 18 '10 at 4:48

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.

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