LOCALLY PSEUDOCONVEX INDUCTIVE LIMIT 5 the sequence (fk(x)) is a Cauchy sequence in A. Since A is complete, then (fk(x)) converges for every x ∈ R to an element yx ∈ A. Therefore pn(fk(x) − yx) = lim m→∞ pn(fk(x) − fm(x)) ε for every x ∈ R and n ∈ N whenever k N. Let f : R → A be a function such that f(x) = yx for each x ∈ R. Hence, (fk) converges to f in the topology τi. We shall prove now that f is continuous on R. For this, let x ∈ R and (xl) be a sequence in R such that lim l→∞ xl = x. Then lim l→∞ fk(xl) = fk(x) for each k because fk is continuous on R. Hence, for any ε 0 and k, n ∈ N there are M 0 such that pn(fk(xl) − fk(x)) ε 3 whenever l M and N 0 such that pn(fk(x) − f(x)) ε 3 for each x ∈ R whenever k N. Let now k N. Then pn(f(xl) − f(x)) = pn(f(xl) + fk(x) − fk(x) + fk(xl) − fk(xl) − f(x)) pn(fk(xl) − f(xl)) + pn(fk(x) − f(x)) + pn(fk(xl) − fk(x)) ε 3 + ε 3 + ε 3 = ε for each n ∈ N whenever l M. This means, that f is continuous on R and its support is compact (because f(x) = θA for every x ∈ R \ [−i, i]). Consequently, K(R,A [−i, i]) is a locally pseudoconvex F -algebra. Moreover, K(R,A [−i, i]) ⊂ K(R,A [−(i + 1),i + 1]) for each i ∈ N and K(R,A) = i∈N K(R,A [−i, i]). Therefore K(R,A) is a LFp-algebra4 without the unit element if we endow K(R,A) with the locally pseudoconvex inductive limit topology, defined by the canonical maps ϕi from K(R,A [−i, i]) into K(R,A). Now, from K(R,A [−i, i]) × A ⊂ K(R,A [−(i + 1),i + 1]) × A for each i ∈ N and K(R,A) × A = i∈N K(R,A [−i, i] × A = i∈N (K(R,A [−i, i]) × A) follows that K(R,A) × A = −→ (K(R,A [−i, i]) × A). We consider for each i the topology τi on K(R,A [−i, i]) × A which is defined by kn-homogeneous seminorms Qi,n, where Qi,n(f, a) = qi,n(f) + pn(a) for each f ∈ K(R,A [−i, i]), a ∈ A and n ∈ N. 4The topology on K(R, A [−i, i]), induced by τi+1 and by τ, coincides with the topology τi because qi,n(f) = q(i+1),n(f) for each f ∈ K(R, A [−i, i]) and n ∈ N and every neighborhood O of zero in (K(R, A [−i, i]),τi) defines a neighbourhood U of (K(R, A),τ) such that U ∩ K(R, A) = O.

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