# Entire calculus and clmc algebras

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 multiplicatively 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.

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).

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 $\mathcal{O}(\mathbb{C})$ without having a lmc algebra but just a locally convex algebra. Can one give examples, reasonable conditions etc?

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Sorry if I misunderstand, but it seems like you are not imposing any continuity conditions on the multiplication operation. At the very least, you'd need the multiplication function $(x,y) \mapsto xy$ to be jointly continuous in the two variables $(x,y)$. –  Zen Harper Nov 8 '11 at 2:21
Oh, that is misunderstanding. Of course, I want $\mathcal{A}$ to be a locally convex algebra... I'll fix it. Thanks... –  Stefan Waldmann Nov 8 '11 at 8:19

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).