Read e-book online Analytic elements in p-adic analysis PDF

By Alain Escassut

ISBN-10: 9810222343

ISBN-13: 9789810222345

The behaviour of the analytic parts on an infraconnected set D in ok an algebraically closed entire ultrametric box is especially defined by means of the round filters and the monotonous filters on D, particularly the T-filters: zeros of the weather, Mittag-Leffler sequence, factorization, Motzkin factorization, greatest precept, injectivity, algebraic homes of the algebra of the analytic parts on D, difficulties of analytic extension. this can be utilized to the differential equation y'=hy (y,h analytic parts on D), analytic interpolation, p-adic workforce duality on meromorphic items and to the p-adic Fourier rework

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If R < diam(D) we put r" = min r'' and in the same way, T{a,R,r") fl D is not empty and is included in every set r(a ; -,r'-,r'') 0 D. Now if p = R = diam(D), let h £ D, and let r' = max r-. Then T(6, r', ,R) H D is not empty and is included in every set T(aj,r'j,r") C\ D. Definition: Let a £ D, let p = 6(a,D) and let R £]0,+oo[ be such that p < R < diam(D). e. ^" is the filter which admits for base the family of sets of the form Dn (f)T(al, r^r")) with a{ £ d(a,R),r'l < R < r'- (1 < i < i=l q , qe IN).

Gn + Wn+i, hn+1 = hn + Wn+\. We check that P - gn+1hn+1 = ( P n + 1 - /i„VTn+i - flfnV„+i) - Vn+iWn+i. By (10) and (6) it is seen that v{Vn+\Wn+i) > (n + 2)r and therefore by (11) we obtain i) at the order n + 1. By (10) and (6), we notice that ii) is clear at the order n + 1. Besides, ;iO)and(( iv) at the order n + 11 clearly cleai comes from iv) at the order n together with (6) and (10). By (8) it is seen that deg(gn+i) = deg(gn) = deg(j). Finally by (9) we have deg(hn+i) n ) , deg(Pn+i) - deg(gn)) and therefore by iii) at the »+l) < < m&x(deg(h^n), (P)-deg(go), dt order n we have deg(hn+i) > max(de#(P) - deg(g0), deg(Pn+i) - deg(gn)).

We now suppose that we cannot find disks Ai, A2 defined as above. ,. Thus Ti admits a center a G A and then, T\ is secant with d(a,r>i). Hence we have r\ < r2 because otherwise T\ would be equal to Ti. (a,r^~) of d(a,r2). 2(x — a) — r2. This finishes showing that the mapping T —► ipy is injective. Now we will show that this mapping is also surjective. Indeed let ip be an absolute value on F(x), and let r — inf ip(x — A). A£F We first suppose that there exists a G F such that ip(x — a) = r . Since tp is an absolute value, we check that r > 0 because if r — 0, we have 4>{h) = h(a) for every h G F(x) and then %j) is not an absolute value.

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Analytic elements in p-adic analysis by Alain Escassut


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