uniformity for Banach algebras - a three space property? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T10:25:21Z http://mathoverflow.net/feeds/question/3482 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3482/uniformity-for-banach-algebras-a-three-space-property uniformity for Banach algebras - a three space property? santker heboln 2009-10-30T23:12:35Z 2011-07-29T03:12:08Z <p>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?</p> <p>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.</p> http://mathoverflow.net/questions/3482/uniformity-for-banach-algebras-a-three-space-property/3495#3495 Answer by Yemon Choi for uniformity for Banach algebras - a three space property? Yemon Choi 2009-10-31T00:14:40Z 2009-10-31T20:58:07Z <p>I think A has to be <em>isomorphic</em> to a uniform algebra, by the following argument.</p> <p>Let q be the quotient HM from A onto A/I. Let r<sub>A</sub> be the spectral radius in A, note that if x \in I then || x ||= r<sub>I</sub>(x)=r<sub>A</sub>(x).</p> <p>Let a\in A have norm 1. I claim that r<sub>A</sub>(a) \geq 1/3.</p> <p>For, let r > r<sub>A</sub>(a). Since the spectral radius can't be increased by a homomorphism, we have</p> <p>|| q(a) || = r<sub>A/I</sub>(q(a)) &lt; r;</p> <p>then, since q is a quotient homomorphism, there exists b \in A such that q(b)=q(a) and || b || &lt; r.</p> <p>Since a-b \in I we have</p> <p>r<sub>A</sub>(a-b) = || a- b || > 1 -r.</p> <p>But since A is commutative the spectral radius r<sub>A</sub> is subaddititive, hence r<sub>A</sub>(a-b) \leq r<sub>A</sub>(a) + r<sub>A</sub>(b) &lt; r + r = 2r.</p> <p>Therefore 1-r &lt; 2r, i.e. r > 1/3. It follows that r<sub>A</sub>(a)\geq 1/3. By rescaling, we deduce that ||a|| \geq r<sub>A(a)</sub> \geq || a||/3 , and thus the Gelfand transform of A is injective with closed range, as claimed.</p> <p>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.</p> <p><b>EDIT:</b> I've been asked to expand on some of the steps in the argument above.</p> <p>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 ||.</p> <p>In this case X=A, Y=A/I and y=q(a). We know that || q(a) || = r(q(a)) &lt; r, so choosing \epsilon appropriately, we can find b\in A such that || q(b) || &lt; r. </p> <p>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.</p> <p>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.</p> <p>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. </p> http://mathoverflow.net/questions/3482/uniformity-for-banach-algebras-a-three-space-property/3538#3538 Answer by santker heboln for uniformity for Banach algebras - a three space property? santker heboln 2009-10-31T13:15:28Z 2011-07-29T03:12:08Z <p>Thanks for your answer Yemon Choi.</p> <p>I have a few things I don't understand in your argument. I don't understand this step:</p> <p>"[...] then, since q is a quotient homomorphism, there exists b \in A such that q(b)=q(a) and || b || &lt; r. [...]"</p> <p>sorry but why is ||b|| &lt; r?</p> <p>"Therefore 1-r &lt; 2r, i.e. r > 1/3." agreed.</p> <p>But: </p> <p>"It follows that rA(a)\geq 1/3" </p> <p>could you elaborate how this follows now, I don't see it.</p> <p>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 inductive limit of a sequence of (uniform) Banach algebras I looked at Banach algebras first.</p> http://mathoverflow.net/questions/3482/uniformity-for-banach-algebras-a-three-space-property/3610#3610 Answer by santker heboln for uniformity for Banach algebras - a three space property? santker heboln 2009-10-31T22:16:48Z 2009-10-31T22:16:48Z <p>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).</p> <p>Thanks for the explanation of the proof. I overlooked the point of the argument that r was chosen arbitrarily.</p>