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4 typo and clarification

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 $\mathcal{O}$ \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})$without having a lmc algebra but just a locally convex algebra. Can one give examples, reasonable conditions etc? 3 clarification If$\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}$is called lmc if there is a defining system of seminorms which are submultiplicative for the product. Then the (algebraic) polynomial calculus sending a polynomial$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$\mathcal{O}$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? 2 TeXbug, stupid keyboard... 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. 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$\mathcal{O}$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. Can one give examples, reasonable conditions etc?

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