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.



uniformity for Banach algebras - a three space property?
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. 


