10 MATI ABEL AND REYNA MAR´ IA PEREZ-TISCARE˜´ NO Then x λ is quasi-invertible in Eα for some α ∈ Ix. Since Eα ⊂ 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 ∈ Eα and y =( x λ )q −1 ∈ Eβ, then there exists γ ∈ I such that Eα ⊆ Eγ and Eβ ⊆ Eγ. Then, x λ , y ∈ Eγ 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 ∈ Eγ 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 Eβ ⊂ 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 Eα be a seminorm on Eα for each α ∈ I. Since Eα 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 Uα 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 Uα for some k ∈ (0, 1] is a neighborhood of zero11 in (E, τ). If x ∈ Vk, then x = n ν=1 μν uν 11Because Uβ ⊂ U ∩ Eβ ⊂ Vk ∩ Eβ for every β ∈ I.

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