Terminology: Banach spaces equipped with continuous associative product? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:12:40Z http://mathoverflow.net/feeds/question/99571 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99571/terminology-banach-spaces-equipped-with-continuous-associative-product Terminology: Banach spaces equipped with continuous associative product? Yemon Choi 2012-06-14T08:48:15Z 2013-01-11T17:56:05Z <p>This is admittedly a low-interest question mathematically, and is arguably a question I could resolve if I had time over the next few days to go and look through a large number of the Banach algebra/functional analysis books on my shelves and in the library. However, it strikes me that this is easily crowd-sourceable and that people may know of texts I am less familiar with. My reason for asking on MO rather than MSE is that I think it will get better answers here.</p> <p>So: the usual definition of a Banach algebra is that it is a (complex) algebra equipped with a complete vector-space norm, such that $\Vert ab\Vert\leq \Vert a\Vert \Vert b\Vert$ for all elements $a,b$.</p> <p>Now suppose we have a (complex) algebra $A$ equipped with a complete vector-space norm $\Vert\cdot\Vert$ and a constant $K>0$ such that $\Vert ab\Vert\leq K\Vert a\Vert \Vert b\Vert$ for all $a,b$. These are much rarer in the literature, most likely for the following reason: a standard exercise doled out to students is to show that there is an equivalent norm on $A$ for which multiplication is contractive, i.e. rendering $A$ (in this new norm) a Banach algebra in the usual sense. In this sense "one has nothing new".</p> <p>However, in some joint work I am writing up, I am toying with the idea of working in this greater generality, in order to let certain technical functorial constructions have more natural formulations. (In a bit more detail, it is to do with certain homologically flavoured constructions for Banach algebras and Banach bimodules more naturally living in a world where multiplication need not be contractive.)</p> <p>So my question is this:</p> <blockquote> <p>do these kinds of algebra have a standard name, and where are the established sources for such terminology?</p> </blockquote> <p>I have a dim recollection that they are given a name of their own in Zelazko's old book on Banach algebras, but I don't recall what the name was, and I can't find anything in Bonsall &amp; Duncan.</p> <p><em>Note:</em> I am not after arguments as to what terminology should or should not be, or observations about one definition being a "semigroup object in Ban$_1$" while the other is a "semigroup object in Ban". Rather, I need some idea of whether one choice of terminology is standard, and hence least likely to cause confusion/irritation to the intended audience, should I decide to pursue this course. </p> http://mathoverflow.net/questions/99571/terminology-banach-spaces-equipped-with-continuous-associative-product/99596#99596 Answer by Salvo Tringali for Terminology: Banach spaces equipped with continuous associative product? Salvo Tringali 2012-06-14T11:44:47Z 2012-06-14T15:46:55Z <p>To my knowledge, something in the same vein was first (?) considered in: C. Le Page (1967), <em>Sur quelques conditions impliquant la commutativité dans les algèbres de Banach,</em> C.R.A.S. Paris, Ser. A-B, 265, A235-A237 (<a href="http://ams.org/mathscinet-getitem?mr=226409" rel="nofollow">click</a>). In any event, if really necessary, I would refer to a norm (of ring-like structures) with that property as a quasi-norm wrt multiplication. Indeed, the kind of algebras that you're considering are somehow related to quasi-normed algebras (cf. <a href="http://en.wikipedia.org/wiki/Quasi-norm" rel="nofollow">Wiki</a>). But this is not the real answer that I would have liked to give.</p> http://mathoverflow.net/questions/99571/terminology-banach-spaces-equipped-with-continuous-associative-product/99604#99604 Answer by Nik Weaver for Terminology: Banach spaces equipped with continuous associative product? Nik Weaver 2012-06-14T12:42:25Z 2013-01-11T17:56:05Z <p>Yemon, I have used the term "weak Banach algebra" for such things. I don't think there is a standard term, though. I vaguely recall seeing people simply call them Banach algebras (probably in some older papers when the terminology in the subject hadn't really stabilized).</p> <p>(I ran into this issue when dealing with the Lipschitz algebra $Lip_0(X)$ for $X$ a complete finite diameter metric space. You really want to use Lipschitz number as the norm, even though this only makes it a weak Banach algebra. There's no real penalty for doing this, and the advantage is that it allows you to identify $X$ <em>isometrically</em> with the normal spectrum of $Lip_0(X)$.)</p> <p>Edit: I've just realized that <em>this</em> is what Gelfand meant by "normed ring". E.g., on the first page of his book <em>Commutative Normed Rings</em> (1960) he writes:</p> <p>"A <em>normed ring</em> is a complex Banach space in which an associative multiplication is defined that is permutable with the multiplication by complex numbers, distributive with respect to addition, and continuous in each factor."</p> <p>and there is a footnote which says "In another terminology, a <em>Banach algebra</em>."</p> <p>A few pages in he proves that you can always achieve $\|xy\| \leq \|x\|\|y\|$ by renorming.</p>