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Let $A$ be a commutative, unital Banach algebra and $I \subset A$ an ideal such that $I$ with the relative norm is a uniform Banach algebra and $A / I$ with the quotient norm is uniform as well. Does it follow that $A$ is uniform?

I expect there to be a counter example involving the Banach algebra $C(X)$ with $X$ a compact Hausdorff space but couldn't quite manage to construct one yet.

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Seeing as you have recently edited one of your answers to this question: was there something wrong with my original construction? –  Yemon Choi Jul 29 '11 at 3:30
    
I merely deleted an incorrect statement of mine (which had nothing to do with the question anyway..). –  santker heboln Aug 22 '11 at 23:57
    
I saw this old question in the sidebar while answering a new one - would you like to accept my answer, or are you still waiting for a better one? –  Yemon Choi Jul 3 at 0:34
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3 Answers 3

up vote 3 down vote accepted

I think A has to be isomorphic to a uniform algebra, by the following argument.

Let q be the quotient HM from A onto A/I. Let rA be the spectral radius in A, note that if x \in I then || x ||= rI(x)=rA(x).

Let a\in A have norm 1. I claim that rA(a) \geq 1/3.

For, let r > rA(a). Since the spectral radius can't be increased by a homomorphism, we have

|| q(a) || = rA/I(q(a)) < r;

then, since q is a quotient homomorphism, there exists b \in A such that q(b)=q(a) and || b || < r.

Since a-b \in I we have

rA(a-b) = || a- b || > 1 -r.

But since A is commutative the spectral radius rA is subaddititive, hence rA(a-b) \leq rA(a) + rA(b) < r + r = 2r.

Therefore 1-r < 2r, i.e. r > 1/3. It follows that rA(a)\geq 1/3. By rescaling, we deduce that ||a|| \geq rA(a) \geq || a||/3 , and thus the Gelfand transform of A is injective with closed range, as claimed.

I hope that does the trick. It's a nice problem, I haven't seen it before, but I'd be very surprised if the argument above - if correct - is either new or best possible.

EDIT: I've been asked to expand on some of the steps in the argument above.

Firstly: if q is a quotient map from a B space X to Y, then by defn, for every y\in Y and every \epsilon>0 there exists x\in X with q(x)=y and || x || \leq (1+\epsilon)|| y ||.

In this case X=A, Y=A/I and y=q(a). We know that || q(a) || = r(q(a)) < r, so choosing \epsilon appropriately, we can find b\in A such that || q(b) || < r.

Secondly: we end up showing that r > 1/3. But by definition, r was anything strictly greater than r_A(a). It follows that r_A(a) must be at least 1/3; for if it weren't, there would be room in between 1/3 and r_A(a) to insert some r which satisfies 1/3 > r > r_A(a), and we've just seen that's not possible.

It might help to look at the argument in a vaguer but more intuitive way (the 1/3 is a slight distraction). Suppose you could find an element a in A which had large norm but very small spectral radius. Then its image in A/I would also have very small spectral radius, hence by your assumption it would have small norm in A/I. By definition of the quotient norm, that means a is very close to I (in the sense of the distance from a point to a closed subspace) and so there exists a' \in I which is very close to a. In particular, a' should have large norm (since a does) and hence have large spectral radius by the assumption on I. But now a and a' are elements of A which are very close together, yet one has very small spectral radius and the other has large spectral radius. That shouldn't be possible, since the spectral radius is dominated by the norm.

Making everything precise above, one gets essentially the original argument I gave. It just so happens that large=1 and very small = 1/3 does the job.

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Yes you can use that as the definition. There is a book "Uniform Frechet Algebras" by H. Goldmann if you are interested. Many properties known from uniform Banach algebras translate more or less in this case e.g. the Gelfand transform is isometric isomorphic etc. A uniform Banach algebra is also characterized as being isomorphic to a closed, pointseparating subalgebra of C(X) for some compact Hausdorff space X (the spectrum..). In the uniform Frechet algebra case something similiar is true but not for a compact Hausdorff space but for a hemicompact space X etc. These algebras have of course fascinating applications in complex analysis (after all the algebra of holomorphic functions on a complex space / manifold is a (nuclear) frechet algebra).

Thanks for the explanation of the proof. I overlooked the point of the argument that r was chosen arbitrarily.

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Thanks for your answer Yemon Choi.

I have a few things I don't understand in your argument. I don't understand this step:

"[...] then, since q is a quotient homomorphism, there exists b \in A such that q(b)=q(a) and || b || < r. [...]"

sorry but why is ||b|| < r?

"Therefore 1-r < 2r, i.e. r > 1/3." agreed.

But:

"It follows that rA(a)\geq 1/3"

could you elaborate how this follows now, I don't see it.

About the origin of the problem: I don't know if this is a known problem, but I did a quite extensive search with no results. It came up in my research on Frechet algebras since I wanted to know if uniformity for Frechet algebras is a three space property (...). Since every (uniform) Frechet algebra is the projective limit of a sequence of (uniform) Banach algebras I looked at Banach algebras first.

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I'm now mildly interested on `uniformity' of Frechet algebras - does that mean being an inverse limit (in Arens-Michael sense) of uniform algebras? –  Yemon Choi Oct 31 '09 at 20:59
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