3-D Shapes by Marina Cohen

By Marina Cohen

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Example text

It is clear that for any s, Chs = ζhs , C is invertible, C−1 hs = ζ−1 hs , C = |ζ| and C−1 = |ζ|−1 . Moreover sup s∈N fs −Chs fs − ζhs = sup hs hs s∈N < |ζ| = C . Corollary 4. Let (hs )s∈N be an orthogonal base for Eω . If (gs )s∈N is a sequence of vectors | hk , gs | gs = 0 for each k ∈ N, and that, of Eω satisfying: lim s→∞ |ωs | sup s∈N where Shk = ∑ s∈N hs − Shs |ωs |1/2 < 1, hk , gs gs for each k ∈ N, then (Shs )s∈N is an orthogonal base of Eω . ωs Remark 6. (1) Note that S defined above is a linear operator on Eω .

Set c0 (E∗ω ) := {(At )t∈N ∈ (E∗ω )N : lim At |ωt |1/2 = 0}. t→∞ Clearly c0 (E∗ω ) is a non-Archimedean Banach space over K when endowed the norm (At )t∈N = sup At |ωt |1/2 . t∈N Also, let |xt | < ∞}. 1/2 t∈N |ωt | Fω := {(xt )t∈N ∈ KN : sup In the same way Fω is a non-Archimedean Banach space over K with the norm (xt )t = sup t∈N |xt | |ωt |1/2 . In the context of the present situation we refer to c0 (E∗ω ) as the set of all sequences of elements in E∗ω which converge to 0, and to Fω as the set of bounded sequences of elements of K.

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. We study some sufficient conditions under which the family ( fs )s∈N is a base of Eω . Theorem 5. Let (hs )s∈N be an orthogonal base and let ( fs )s∈N be a sequence of vectors in Eω satisfying the following condition: There exists α ∈ (0, 1) such that for every x = ∑ xs hs ∈ Eω, s∈N sup |xs | fs − hs ≤ α .

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