By P. Ciarlet

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In contrast with the classical case, strict inequality may occur in Proposition 19. Indeed, let K = (Q p , | · |), ωt = p−t , ats = pt , hence |ωt | = pt , |ats | = p−t . (1) ∀s, lim |ats | |ωt |1/2 = lim t t 1 = 0. pt/2 (2) ∑ t,s |ats |2 |ωt | pt = ∑ 2t+s |ωs | t,s p =∑ t,s 2 1 pt+s = C > 1. On the other hand, ∑ s But since Aes 2 1 =∑ sup |ats |2 |ωt | |ωs | p |ω | t s s 1 1 =∑ sup t s |ωs | t p 1 =∑ s |ωs | 1 = 1 1− p p . = p−1 p p2 p > 1 it follows that < . p−1 p − 1 (p − 1)2 Proposition 20.

7 Bibliographical Notes This chapter is entirely devoted to non-Archimedean bounded linear operators. Most of the results of this chapter can be found in Diagana et al. [4, 12, 17], and Diarra [24, 25]. 2). This chapter is devoted to unbounded linear operators on free Banach spaces (respectively, on non-Archimedean Hilbert spaces Eω ). As for non-Archimedean bounded linear operators, some of the results go along with the classical line and others deviate from it. For the most part, the statements of the results are inspired by their classical counterparts.

Example 17 is a particular case of the next example. Example 18. Let (K, |·|) be a non-Archimedean valued field. Suppose that |ωt | = 1 for each t ∈ N. Define the linear operator A : Eω → Eω by: A = ∑ ats (es ⊗ et ), where ats = ϑt for t,s all t, s ∈ N with |ϑt | = 0 for each t ∈ N and lim |ϑt | = 0. t→∞ Proposition 18. The operator A defined above is bounded and does not have an adjoint. Proof. The proof is similar to that of Proposition 17, and therefore left to the reader. 4 Perturbation of Bases Let an orthogonal base (hs )s∈N be given and consider a sequence of vectors ( fs )s∈N , not necessarily an orthogonal base, in Eω such that the difference fs − hs is small in a certain sense.

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