most recent 30 from http://mathoverflow.net 2013-05-23T09:19:30Z http://mathoverflow.net/feeds/question/80021 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80021/entire-calculus-and-clmc-algebras Stefan Waldmann 2011-11-04T08:53:55Z 2012-02-10T16:42:15Z

<p>If $\mathcal{A}$ is a complete locally convex (Hausdorff) associative unital algebra (over $\mathbb{C}$) one is interested in defining "transcendental" functions of a given algebra element $a \in \mathcal{A}$ like e.g. exponentials $\exp(a)$ by means of the power series expansion. This works fine for complete locally <em>multiplicatively</em> convex algebras. Recall that $\mathcal{A}$ is called lmc if there is a defining system of seminorms which are submultiplicative for the product. Equivalently, such an algebra is a (suitable) projective limit of Banach algebras. Then the (algebraic) polynomial calculus sending a polynomial $p \in \mathbb{C}[z]$ to the algebra element $p(a)$ extends by completion to an entire calculus $$ \mathcal{O}(\mathbb{C}) \ni f \mapsto f(a) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} a^n \in \mathcal{A}, $$ which is a continuous algebra homomorphism for a given $a$. Here $\mathcal{O}(\mathbb{C})$ is equipped with its usual Fréchet topology of locally uniform convergence. Equivalently and more convenient in this context, one can use the seminorms given by $p_R(f) = \sum_{n=0}^\infty \frac{|f^{(n)}(0)|}{n!} R^n$, from which one sees the continuity of the entire calculus on the nose.</p> <p>Now there are many lc algebras which are definitely not lmc like the Weyl algebra generated by the canonical commutation relations $[Q, P] = i\hbar \mathbb{1}$ (whatever lc topology you may put on it). </p> <p>My question is whether it is possible to have an entire calculus in the sense that there is a continuous algebra homomorphism extending the polynomial calculus to $\mathcal{O}(\mathbb{C})$ <em>without</em> having a lmc algebra but just a locally convex algebra. Can one give examples, reasonable conditions etc? </p>

http://mathoverflow.net/questions/80021/entire-calculus-and-clmc-algebras/88116#88116 Jochen Wengenroth 2012-02-10T16:42:15Z 2012-02-10T16:42:15Z

<p>At least for commutative and completely metrizable locally convex algebras there is a theorem of Mityagin, Rolewicz, and Zelasko (Studia Math. 21 (1962), 291-306) which says that the algebra has to be locally m-comvex if all entire functions operate on it (meaning that $f(a)=\sum\limits_{n=0}^\infty\frac{ f^{(n)}(0)}{n!} a^n$ converges for all entire functions $f$ and all elements of the algebra).</p>