10 MATI ABEL AND REYNA MAR´ IA PEREZ-TISCARE˜´ NO Then x λ is quasi-invertible in for some α Ix. Since E, then x λ is quasi-invertible also in E. Hence, λ spE(x). Consequently, spE(x) α∈Ix spEα (x). Now, we suppose that λ C\{0} is such that λ spE(x). Then x λ is quasi-invertible in E. If x λ and y =( x λ )q −1 Eβ, then there exists γ I such that and Eγ. Then, x λ , y and hence x λ is quasi-invertible in Eγ. This means that λ spEγ (x). Moreover, γ Ix because x λ Eγ. Hence, λ α∈Ix spEα (x). This means that α∈Ix spEα (x) spE(x). Then statement (1) holds. By (1) is true that spE(x) spEα(x) for every α Ix. So ρE(x) ρEα (x) for every α Ix. Hence ρE(x) inf α∈Ix ρEα (x). Suppose that ρE(x) inf α∈Ix ρEα (x). Then there exists a λ such that ρE(x) λ inf α∈Ix ρEα (x). Hence, λ / spE(x) and λ (0,ρEα (x)) spEα (x) for every α Ix, which is not possible. Consequently, ρE(x) = inf α∈Ix ρEα (x) in this case. b) Let x QinvE. Then there is an element y E such that y is the quasi- inverse of x. Similarly as above we can assume that x, y for some γ I. Therefore x is quasi-invertible in Eγ. Thus x QinvEγ α∈I QinvEα. Let now x α∈I QinvEα. Then there exists an index β I such that x QinvEβ. Therefore, x is quasi- invertible in E. Hence, x QinvE. The proof for invertible elements in E is similar. 2. Let (Eα,τα) be a Q-algebra for every α I. Since holds (3), then (E, τ) is a Q-algebra by the statement b). Let now the spectral radius ρEα of be a seminorm on for each α I. Since is a Q-algebra, then ρEα is upper semicontinuous at zero by Lemma 2.3, that is, for each ε 0 = ρEα (θ) and α I there is a neighborhood of zero in (Eα,τα) such that ε ρEα (x) for every x Uα. Then the absolutely k-convex hull Vk = Γk(U) of U = α∈I for some k (0, 1] is a neighborhood of zero11 in (E, τ). If x Vk, then x = n ν=1 μν 11Because U Vk for every β I.
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