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Publicacions Matem`atiques, Vol 43 (1999), 127–162. STRICTLY ANALYTIC FUNCTIONS ON p-ADIC ANALYTIC OPEN SETS Kamal Boussaf Abstract Let Kbe an algebraically closed complete ultrametric field. M. Krasner and P. Robba defined theories of analytic functions in K, but when Kis not spherically complete both theories have the disadvantage of containing functions that may not be expanded in Taylor series in some disks. On other hand, affinoid theories are only defined in a small class of sets (union of affinoid sets) [2], [13] and [17]. Here, we suppose the field Ktopologically separable (example Cp). Then, we give a new definition of strictly analytic functions over a large class of domains called analoid sets. Our theory uses the notion of T-sequence which caracterizes analytic sets in the sense of Robba. Thereby we obtain analytic functions satisfying the property of analytic continuation and which, however, will admit expansion in power series (resp. Laurent series) in any disk (resp. in any annulus). Moreover, the algebra of analytic functions will be stable by derivation. The process consists of defining a large class of analytic sets D, and a class of admissible sets making a covering of such a D, so that we obtain a sheaf on D. We finally give an example of differential equation whose solutions are strictly analytic functions in an analoid set. Such an example might not be involved in theories based on affinoid sets. I. Preliminaries Let Kbe an algebraically closed field complete for an ultrametric absolute value. We recall some standard notations and definitions. Given a∈Kand r>0, d(a, r) (resp. d(a, r−), resp. C(a, r)) denotes the circumferenced disk {x∈K||x−a|≤r}(resp. the noncircumferenced disk {x∈K||x−a|<r}, resp. the circle {x∈ K||x−a|=r}). We call a class of d(a, r) any non-circumferenced disk d(b, r−) with |a−b|≤r. Let r>0 and r >r ,Γ(a, r,r) (resp. ∆(a, r,r)) denotes the annulus {x∈K|r<|x−a|<r }(resp. {x∈K|r≤|x−a|≤r}).
128 K. Boussaf Let Dbe an infinite subset of K. Then R(D) denotes the set of rational functions h∈K(x) with no poles in D. This is a K-subalgebra of the algebra KDof all functions from Dinto K. Then R(D) is provided with the topology UDof uniform convergence on D, and is a topological group for this topology. H(D) denotes the completion of R(D) for this topology and its elements are named the analytic elements on D[4], [6], [11]. Definition. A set Dis said to be analytic if for every f∈H(D) for every a∈Dand r>0, the property f(x) = 0 whenever x∈d(a, r)∩D implies f(x) = 0 whenever x∈D. In Theorem 0 we recall the characterization of analytic sets by using T-sequences. Example and remarks. Let (an)n∈Nbe a sequence in Ksuch that |an|<|an+1|∀n∈N, lim n→∞ |an|= 1 and ∞ n=0 −log |an|<+∞. Let ρ∈]0,1[ and (rn)n∈Nbe a sequence in ]0,ρ[. Then by [6, Proposition 36.5], the open set D=d(0,2) \( n∈N d(an,r n−)) is analytic. Notice that Robba’s definition of analytic functions may extend to open sets like D. However, in a not spherically complete field such as Cp, this definition gives functions that may not be expanded in Taylor series in some disks. On the other hand, affinoid sets (resp. “connected” affinoid sets i.e.: infraconnected affinoid sets) are defined by Fresnel and Van Der Put in [7] and were used to construct a theory of analytic functions defined by a sheaf of analytic elements on affinoid sets. But, for example, if we consider the analytic set Ddefined above, we see that such a set can’t be covered by an increasing sequence of infraconnected affinoid sets. Therefore, by Fresnel and Van Der Put process, one obtains, for example, the characteristic function of the set d(0,2) \d(0,1−)asan analytic function on D([7, 1.7]). Hence, we see that the family of affinoid sets is too small to give a general theory of analytic functions, satisfying the principle of analytic continuation. The aim here is to construct a large family of analytic functions, rich in properties, and defined on a large class of open analytic sets named analoid sets, as are Robba’s sets, but avoiding the inconvenience of containing functions that may not be expanded in Taylor series in some disks in a non-spherically complete field. Particularly, we will see that
Strictly analytic functions on p-adic analytic open sets 129 strictly analytic functions satisfy the principle of analytic continuation on any analoid set (Theorem 28). Next, we recall some preliminaries definitions and particularly this of T-sequences. A sequence (an)n∈Nin Kis said to be an increasing distances sequence (resp. a decreasing distances sequence) if the sequence |an+1 −an|is strictly increasing (resp. decreasing) and has a limit l∈R∗+. A sequence (an)n∈Nis said to be a monotonous distances sequence if it is either an increasing distances sequence or a decreasing distances sequence. A sequence (an)n∈Nin Kis said to be an equal distances sequence if |an−am|=|am−aq|whenever n,m,q∈Nsuch that n=m=q. A set Din Kis said to be infraconnected if for every a∈D, the mapping Iafrom Dto R+defined by Ia(x)=|x−a|has an image whose closure in R+is an interval. (In other words, a set Dis not infraconnected if and only if there exist aand b∈Dand an annulus Γ(a, r1,r 2) with 0<r 1<r 2<|a−b|such that Γ(a, r1,r 2)∩D=∅.) As usual, given a set Ain Kand a point a∈K, we denote by δ(a, A) the distance from ato A. Let Dbe an infinite set in K, and let a∈D.IfDis bounded of diameter r, we denote by Dthe disk d(a, r), and if Dis not bounded, we put D=K. Then, D\Dis known to admit a partition of the form (d(ai,r− i)i∈J), with ri=δ(ai,D) for each i∈J. The disks d(ai,r− i)i∈J, are named the holes of D. 1. Monotonous filters. Let a∈ Dand S∈R∗ +be such that Γ(a, r, S)∩D=∅whenever r∈]0,S[ (resp. Γ(a, S, r)∩D=∅whenever r>S). We call an increasing (resp. a decreasing)filter of center a and diameter S,onDthe filter Fon Dthat admits for base the family of sets Γ(a, r, S)∩D(resp. Γ(a, S, r)∩ D). For every sequence (rn)n∈Nsuch that rn<r n+1 (resp. rn> rn+1) and lim n→∞ rn=S, it is seen that the sequence Γ(a, rn,S)∩D (resp. Γ(a, S, rn)∩D) is a base of Fand such a base is called a canonical base [6]. Given an increasing (resp. a decreasing) filter Fon Dof center aand diameter r, we will denote by PD(F) the set {x∈D||x−a|≥r} (resp. the set {x∈D||x−a|≤r}). Further PD(F) will be named the D-beach of F.
130 K. Boussaf We call a monotonous filter on Da filter which is either an increasing filter or a decreasing filter. Given a monotonous filter Fwe will denote by diam(F) its diameter. The field Kis said to be spherically complete if each nested sequence of disks has a nonempty intersection. The field Cp, for example, is not spherically complete. However, every algebraically closed complete ultrametric field admits a spherically complete algebraically closed extension [6]. Let Fbe an increasing (resp. a decreasing) filter of center aand diameter Son D. The filter Fis said to be pierced if for every r∈]0,S[, (resp. r>S), Γ(a, r, S)(resp. Γ(a, S, r)) contains some hole Tmof D. 2. Monotonous distances holes sequences. Let a∈ D. Let (Tm,i)1≤i≤s(m) m∈N be a sequence of holes of Dwhich satisfies δ(a, Tm,i)=dm(1 ≤i≤s(m),m∈N), dm<d m+1 (resp. dm> dm+1), and lim m→∞ dm=R>0. The sequence (Tm,i)1≤i≤s(m) m∈N is called an increasing (resp. decreasing) distances holes sequence that runs the increasing (resp. decreasing) filter Fof center a, of diameter R. The filter Fwill be named the increasing (resp. decreasing) filter associated to the sequence (Tm,i)1≤i≤s(m) m∈N . The D-beach of Fwill be also named the D-beach of (Tm,i)1≤i≤s(m) m∈N . Finally, an increasing (resp. decreasing) distances holes sequence will be called a monotonous distances holes sequence and the sequence (dm)m∈Nis called the monotony of the monotonous distances holes sequence. Let (Tm,i)1≤i≤s(m) m∈N be a monotonous distances holes sequence and for every (m, i)(i∈{1,... ,s(m)},m ∈N), let ρm,i = diam(Tm,i). The number lim inf m→∞ ( min 1≤i≤s(m)(ρm,i)) (resp. lim sup m→∞ ( max 1≤i≤s(m)(ρm,i))) will be called inferior limit-piercing (resp. superior limit-piercing) of the sequence (Tm,i)1≤i≤s(m) m∈N . If a monotonous holes sequence of diameter rhas an inferior limitpiercing ρ>0 and a superior limit-piercing ρ<r, it will be said correctly pierced. A set Dwill be said to be correctly pierced if every monotonous distances holes sequence of Dwith a not empty D-beach is correctly pierced.
Strictly analytic functions on p-adic analytic open sets 131 A set Dis said to be well pierced if δ(D,K \D)>0, i.e. the set of diameters of holes of Dhas a strictly positive lower bound. 3. Weighted sequences. We call a weighted sequence a sequence (Tm,i,q m,i)1≤i≤s(m) m∈N with (Tm,i)1≤i≤s(m) m∈N a monotonous distances holes sequence and (qm,i)1≤i≤s(m) m∈N a sequence of nonnegative integers. Given m∈Nand i∈{1,... ,s(m)}, qm,i is called the weight of Tm,i. The D-beach of (Tm,i)1≤i≤s(m) m∈N is also named the D-beach of (Tm,i,q m,i)1≤i≤s(m) m∈N . The monotonous filter associated to (Tm,i)1≤i≤s(m) m∈N , is also called filter associated to (Tm,i,q m,i)1≤i≤s(m) m∈N . A weighted sequence is said to be correctly pierced if its associated monotonous distances holes sequence is. For every m∈N,weput Ωm= max 1≤i≤s(m) qm,i log dm ρm,i + j=i 1≤j≤s(m) qm,j(log dm−log |am,j −am,i|) . The sequence (Ωm)m∈Nwill be called perturbations sequence of the weighted sequence. We will say that the weighted sequence has a perturbations sequence bounded by λ∈R+, if sup m∈N Ωm≤λ. More generally we say that the weighted sequence has a bounded perturbations sequence, if sup m∈N Ωm<+∞. A weighted sequence (Tm,i,q m,i)1≤i≤s(m) m∈N will be said to be idempotent if qm,i = 0 or 1 for all (m, i), (1 ≤i≤s(m),m∈N) and qm,i = 0 for infinitely many (m, i). Let S1and S2be two monotonous distances holes (resp. weighted) sequence. We will say that S1and S2are cofiltring if they are associated to the same monotonous filter of K. We see that two cofiltring sequences have the same centers and the same D-beach.
132 K. Boussaf Let S=(T m,i)1≤i≤k(m) m∈N and S =(T m,i)1≤i≤k (m) m∈N be two cofiltring monotonous distances holes sequences, of center a. The holes of the set D=K\(∪1≤i≤k(m) m∈N T m,i)∪(∪1≤i≤k (m) m∈N T m,i) form a monotonous distances sequence which is cofiltring to Sand S. We will denote it by S∪S. Thus, S∪S is in the form (Tm,i)1≤i≤k(m) m∈N , and we remark that a hole of S∪S is either a hole of Sor a hole of S. 4. T-sequences. Let S=(Tm,i,q m,i)1≤i≤s(m) m∈N be an increasing (resp. decreasing) weighted sequence and for all m∈N, let qm= s(m) i=1 qm,i. The weighted sequence Swill be said to be a T-sequence if it satisfies: lim m→∞ sup 1≤j≤s(m) dm ρm,j qm,j i=j 1≤j≤s(m) dm |am,i −am,j|qm,i m−1 n=1 dn dmqn =0 (resp. lim m→∞ sup 1≤j≤s(m) dm ρm,j qm,j i=j 1≤j≤s(m) dm |am,i −am,j|qm,i m−1 n=1 dm dnqn =0). Remark. A weighted sequence (Tm,i,q m,i)1≤i≤s(m) m∈N is a T-sequence if and only if lim m→∞ −sup 1≤j≤s(m) qm,j(log dm−log ρm,j) + i=j 1≤i≤s(m) qm,i(log dm−log |am,i −am,j|) + m−1 n=1 qn|log dm−log dn| =+∞.
Strictly analytic functions on p-adic analytic open sets 133 Theorem 0. Let Dbe a set in K. Then Dis analytic if and only if any T-sequence of Dhas an empty D-beach. Proof: By [6, Theorem 38.8], we know that Dis analytic if and only if Dis infraconnected and any T-sequence of holes of Dhas an empty D-beach. Thus we only have to show that if Dis not infraconnected, then Dadmits a T-sequence with a not empty D-beach. Let us suppose that Dis not infraconnected. Then there exist a,b∈Dand r1,r2∈R+ such that 0 <r 1<r 2<|a−b|and Γ(a, r1,r 2)∩D=∅. Hence, we see that every element of Γ(a, r1,r 2) belongs to a hole of D. So, it is easly seen that Γ(a, r1,r 2) admits a partition Pby a family of holes of D.By[12, Propositions 1.2 and 2.5] there exist t∈]0,r 2[, u∈d(a, r2) and an increasing idempotent T-sequence Swith holes in P, of center u and diameter t. Then since |b−u|≥t, we see that Sis a T-sequence of holes of Dwith a not empty D-beach. II. Strictly analytic functions 1. Polar and T-polar sequences. Let Dbe infraconnected, let a∈K, let r>0 and let ρ∈]0,r[. We call an increasing (resp. decreasing) polar sequence, of center a,of diameter rand separation ρevery sequence of the form (bm,i)1≤i≤k(m) m∈N satisfying: bm,i ∈ D\D,∀(m, i)(m∈N,i ∈{1,... ,k(m)}), with |bm,i −a|=|bm,j −a|=dmwhenever i,j∈{1,... ,k(m)},|bm,i −a|< |bm+1,j −a|(resp. |bm,i −a|>|bm+1,j −a|) whenever 1 ≤i≤k(m) and 1 ≤j≤k(m+ 1), lim m→∞ dm=rand inf (m,i)=(n,j)|bm,i −bn,j|=ρ. The sequence (dm)m∈Nis called the monotony of the polar sequence. We call the D-beach of (bm,i)1≤i≤k(m) m∈N the set D∩(K\d(a, r−)) (resp. D∩d(a, r)). A polar sequence (bm,i)1≤i≤k(m) m∈N is called T-polar sequence if for some σ∈]0,ρ] there exists a family (qm,i)1≤i≤k(m) m∈N of nonnegative integers such that (d(bm,i,σ−),q m,i)1≤i≤k(m) m∈N is a T-sequence. Remark. An element of a polar sequence is either an element of D\D or an element of a hole of D.
134 K. Boussaf Example. Suppose K=Cp. Let Γ = {x∈K;xpn= 1, for some n≥0}and let D=K\Γ. It is well known that elements of Γ form a sequence (bm,i)1≤i≤pm−1(p−1) m∈N where bm,i lies in the circle of center 1 and diameter p−1 pm(p−1) . Then by [5, Proposition II.4], (bm,i)1≤i≤pm−1(p−1) m∈N is a T-polar sequence. Notice that for all r<p −1 p(p−1) the set Dr={x∈K;|x−γ|≥r∀γ∈ Γ}is not analytic ([5]). 2. Definition of an analoid set. We build up our theory of strictly analytic functions on the so-called analoid sets. Definition. Dwill be said to be an analoid if Dsatisfies: 1) Every T-polar sequence admits an empty D-beach. 2) Every monotonous distances holes sequences with a not empty Dbeach has a superior limit-piercing strictly inferior to its diameter. 3) Dis open. Remark. An analoid set is analytic. Indeed, suppose that an analoid set Dadmits a T-sequence S=(Tm,i,q m,i)1≤i≤s(m) m∈N with a not empty D-beach. Let Rbe the diameter of S. For every (m, i)(m∈N, i∈{1,... ,s(m)}), let Tm,i =d(am,i,ρ − m,i). Since Dis analoid, Shas a superior limit-piercing ρ<R. Hence, without loss of generality we assume that diameter of holes of Sare upper bounded by σ∈]ρ, R[. Let (d(bm,j,σ−))1≤j≤l(m) m∈N be the sequence of disks of diameter σsuch that every hole Tm,i (m∈N,1≤i≤s(m)) is included in some d(bm,j,σ−) and that every disk d(bm,j ,σ−)(m∈N,1≤j≤l(m)) contains some Tm,i. For every (m, j)(m∈N,1≤j≤l(m)) we denote by Im,j the set of (m, i) such that Tm,i ⊂d(bm,j,σ−) and we put pm,j = (m,i)∈Im,j qm,i. Then, by [6, Proposition 35.4], the weighted sequence S=(d(bm,j,σ−),p m,j)1≤j≤l(m) m∈N is a T-sequence and therefore D admits a T-polar sequence with a not empty D-beach, a contradiction with the hypothesis “Dis analoid”.
Strictly analytic functions on p-adic analytic open sets 135 3. Examples. Definitions. Let Dbe a set that contains at least two points. Dis called quasi-connected ([11]) if for any two points x,y∈D, the set {|z−x|;z∈K\D, |z−x|≤|y−x|} is finite. A quasi-connected set Dis called regular ([10]) (resp. completely regular [13]) if for any two points x,y∈Dand any r∈|K∗|with r≤|y−x|, the set (K\D)∩d(a, r) can be covered by countably (resp. finitely) many open balls with radius r. We recall that the notion of regular quasi-connected set was first given by Krasner ([11]) when the residue class field of Kis not countable. Quasi-connected sets (and particulary regular and completely regular quasi-connected sets) are analoids. Indeed, on one hand, it is well known that quasi-connected sets are open. On the other hand, since for all x,y in a quasi-connected set D, the set {|z−x|;z∈K\D, |z−x|≤|y−x|} is finite, we see that Dhas neither monotonous distances holes sequences nor polar sequence with a not empty D-beach and consequently it is an analoid. But generally, analoid sets are not quasi-connected. Indeed, an analoid may have monotonous distances holes sequence and polars sequence with a not empty beach, which is not true for quasi-connected sets. Let (an)n∈Nbe an increasing distances sequence in the disk d(0,1) of limit 1 satisfying ∞ n=0 −log |an|<+∞and let (ρn)n∈Nbe a sequence in ]0,1[ such that lim sup n→∞ ρn<1. Let D=d(0,2) \( n∈N d(an,ρ − n)). Clearly the holes of Dare of the form d(an,ρ − n). According to [6, Proposition 36.5], such a set is analytic and then we see that Dis an analoid. Remarks. i) One can’t have an increasing covering of such a set D by admissible sets either in the sense of [Fresnel Van der Put and Morita] or in the sense of [Karlowski and Ullrich]. Consequently, one can’t define analytic functions on Din these different senses. ii) We notice that the class of analoid sets is not stable by intersection, but we define a subclass of “special” analoid sets which is so and which makes covering of every analoid set.
142 K. Boussaf Theorem 8. The presheaf His a sheaf for the G-topology on D. Proof: This is a consequence of the following lemma. Lemma 9. Let U1and U2be correctly pierced closed analytic sets such that U1∩U2=∅.Letf1∈H(U1),f2∈H(U2)such that f1/U1∩U2=f2/U1∩U2. Then, there exists f∈H(U1∪U2)such that f/Uj=fj(j=1,2). Proof: By Lemma 5 a hole of U1∩U2is either a hole of U1or a hole of U2. Hence Lemma 9 holds by [14, Theorem 8.3]. Remark. In a next paper we will study properties of this sheaf. Let’s now prove that an analoid admits a covering by an increasing sequence of D-admissible sets. Lemma 10. Let S=(d(am,i,ρ − m,i),q m,i)1≤i≤s(m) m∈N , and S=(d(bm,i,ρ − m,i),q m,i)1≤i≤s(m) m∈N be two weighted idempotent sequences of diameter Rand let δ>0and r<Rsuch that ∀m,n∈N,∀i∈ {1,... ,s(m)},∀j∈{1,... ,s(n)}i=j, we have δ≤ρm,i ≤r,δ≤ ρ m,i ≤rand |am,i −an,j|=|bm,i −bn,j|. Then Sis a T-sequence if and only if Sis a T-sequence. Proof: Let m∈Nand let j∈{1,... ,s(m)}. We put Am,j = i=j 1≤i≤s(m) dm |am,i −am,j|qm,i = i=j 1≤i≤s(m) dm |bm,i −bm,j|qm,i and if (dn)n∈Nis increasing (resp. decreasing) we put Bm= m−1 n=1 dn dmqn (resp. Bm= m−1 n=1 dm dnqn ).
Strictly analytic functions on p-adic analytic open sets 143 By definition, S(resp. S)isaT-sequence if and only if lim m→∞ sup 1≤j≤s(m)dm ρm,j qm,j Am,jBm=0(1) (resp. lim m→∞ sup 1≤j≤s(m)dm ρ m,j qm,j Am,jBm=0.(2) According to the hypothesis of the lemma, obviously there exist α,β∈R+and N∈Nsuch that αqm,i ≤dm ρm,i qm,i ≤βqm,i ,and(3) αqm,i ≤dm ρ m,i qm,i ≤βqm,i ,∀i∈{1,... ,s(m)}.(4) We now suppose that Sis a T-sequence; consequently Ssatisfies (1). But since each qm,i lies in {0,1}(m∈N,1≤i≤s(m)), by (1) and (3), for every sequence (xm)m∈Nsuch that xm∈{1,... ,s(m)},wehave (5) lim m→∞ Am,xmBm=0. In particular, if (xm)m∈Nis such that dm ρ m,xmqm,xm Am,xm= sup 1≤j≤s(m)dm ρ m,j qm,j Am,j, we see that using (4) and (5) we have lim m→∞ sup 1≤j≤s(m)dm ρ m,j qm,j Am,jBm=0. We have a symmetric proof when Sis a T-sequence. Definition. Two cofiltring weighted sequences S=(d(am,i,ρ − m,i),q m,i)1≤i≤s(m) m∈N and S=(d(am,i,ρ − m,i),q m,i)1≤i≤s(m) m∈N will be said to be similar if they have the same inferior and superior limit-piercing.
144 K. Boussaf Corollary 11. Let Sand Sbe similar weighted sequences. Then S is correctly pierced if and only if Sis correctly pierced. Besides, if Sis correctly pierced, then Sis an idempotent T-sequence if and only if Sis an idempotent T-sequence. Lemma 12. Let U1,U2be infraconnected sets such that δ(U1,U 2)≤min(diam(U1),diam(U2)). Then U1∪U2is infraconnected. Proof: Let a,b∈U1∪U2and let r1,r2∈Rsuch that 0 <r 1<r 2< |a−b|. We just have to check that (U1∪U2)∩Γ(a, r1,r 2)=∅. If a,b∈U1or a,b∈U2, then since both U1and U2are infraconnected, we have U1∩Γ(a, r1,r 2)=∅or U2∩Γ(a, r1,r 2)=∅and therefore (U1∪U2)∩Γ(a, r1,r 2)=∅. Now we assume that a∈U1and b∈U2. First suppose, r1<diam(U1). Then there exists a∈U1such that r1<|a−a|≤diam(U1). But since U1is infraconnected and a,a∈U1, we have U1∩Γ(a, r1,min(r2,|a− a|)) =∅. Consequently (U1∪U2)∩Γ(a, r1,r 2)=∅. Now suppose r1≥diam(U1). We first assume U2∩ U1=∅; then for every b∈U2∩ U1, we have Γ(a, r1,r 2)=Γ(b,r 1,r 2). So, since U2is infraconnected and 0<r 1<r 2<|b−b|, we have U2∩Γ(b,r 1,r 2)=∅and therefore (U1∪U2)∩Γ(a, r1,r 2)=∅. Finally, we assume U2∩ U1=∅. Then it is clear that δ(U1,U 2)= inf x∈U 2 (|x−a|). But as r1≥diam(U1) and δ(U1,U 2)≤min(diam(U1), diam(U2)), there exists b∈U2such that |b−a|<r 2. Since U2is infraconnected, we have Γ(b,|b−a|,r 2)∩U2=∅; but as Γ(a, |b−a|,r 2)= Γ(b,|b−a|,r 2)⊂Γ(a, r1,r 2), therefore (U1∪U2)∩Γ(a, r1,r 2)=∅. We have a symmetric proof when a∈U2and b∈U1. This ends the proof of Lemma 12. Lemma 13. Let U1and U2be infraconnected sets such that U1∪U2 is infraconnected. Then U1⊂ U2or U2⊂ U1. Proof: Suppose that U1⊂ U2and U2⊂ U1. By ultrametricity it is seen that δ( U1, U2)>max(diam( U1),diam( U2)) and δ( U1, U2)=|a−b| for all a∈U1and b∈U2. Let r1,r2∈R+be such that diam( U1)< r1<r 2<|a−b|; then we see that Γ(a, r1,r 2)∩( U1∪ U2)=∅. But this contradicts the hypothesis that U1∪U2is infraconnected.
Strictly analytic functions on p-adic analytic open sets 145 Lemma 14. Let U1,U2be infraconnected sets such that U1∪U2is infraconnected. Then a hole of U1∪U2is either a hole of U1or a hole of U2. Besides, if a hole of U1∪U2is included in U1and U2then it is included in a hole of U1and in a hole of U2. Proof: Let T=d(a, r−) be a hole of U1∪U2. By Lemma 13, we can assume without loss of generality that U1⊂ U2. First we suppose that T⊂ U1. Then it is obvious that Tis simultaneously included in a hole of U1and in a hole of U2. Since r= inf x∈U1∪U2 (|x−a|) = min( inf x∈U1 (|x−a|),inf x∈U2 (|x−a|)), it is seen that Tis either a hole of U1or a hole of U2. Now if we suppose that T⊂ U1, then it is clear that δ(a, U1)>r. But since δ(a, U1)=δ(a, U1) = inf x∈U1 (|x−a|), then obviously we have r= inf x∈U2 (|x−a|) and as T⊂ U2,Tis a hole of U2. Lemma 15. Let Dbe an analoid and let U1,U2be D-admissible sets without increasing T-sequence. Further, we assume that U1∪U2is correctly pierced. a) If U1∩U2=∅, then U1∪U2is D-admissible. b) If δ(U1,U 2) = diam(U1) = diam(U2), then U1∪U2is D-admissible. Proof: By Lemma 12, if U1∩U2=∅or δ(U1,U 2) = diam(U1)= diam(U2) then U1∪U2is infraconnected. Besides, since U1∪U2is bounded, closed and correctly pierced, we only have to show that it is well pierced, analytic, that every circled hole of Dincluded in U1∪U2 is strictly included in a hole of U1∪U2and that if Dis peripherally circled, then U1∪U2 D. The two last statements are obvious, because on one hand a circled hole of Dincluded in U1∪U2is included in a hole of U1∪U2. But by Lemma 14, a hole of U1∪U2is a hole of U1or a hole of U2. Then, since both U1and U2are D-admissible sets, we have a strict inclusion. On the other hand, if Dis peripherally circled, then both U1and U2are strictly included in D. Hence, U1∪U2is strictly included in D. Since U1and U2are well pierced, then by Lemma 14 so is U1∪U2. We will show that U1∪U2is analytic. Indeed let us suppose that this is not true. So, by Theorem 0, U1∪U2admits a T-sequence Swith a not empty U1∪U2-beach. But since U1∪U2is correctly pierced, by
146 K. Boussaf Lemma 3, we may assume that Sis an idempotent T-sequence with a bounded perturbations sequence. Lemmas 14 and 4 show that there exists a T-sequence S1or S2whose holes are holes of U1or holes of U2respectively. For example let us suppose S1to be this T-sequence. Obviously S1is decreasing because U1 does not admit increasing T-sequences. Since U1is analytic, by Theorem 0 we see that S1has an empty U1-beach. Hence the U1∪U2-beach of Sis included in U2. Let abe an element of the U1∪U2-beach of S, then a∈U2. a) We assume U1∩U2=∅and take b∈U1∩U2. In particular we have |a−b|>diam(S). By Lemma 14, we see that from certain rank, the holes of S1are included in holes of U2. To such a hole of U2we associate the sum of the weights of holes of S1that it contains. Hence, we obtain a weighted sequence of holes of U2which is, by [6, Proposition 35.4], a T-sequence. Thus U2admits a T-sequence with a not empty U2-beach, which contradicts the hypothesis that U2is analytic. b) Now we assume that (1) δ(U1,U 2) = diam(U1) = diam(U2). Since S1is decreasing, it is clear that (2) diam(S1)<diam(U1). As a∈U2, by (2), we have δ(U1,U 2)≤diam(S1) which is a contradiction with (1). Thus U1∪U2is analytic and therefore it is D-admissible. Notation. Henceforth, Kis supposed topologically separable. It is well known that such a field is not spherically complete ([14]) and we see that Cpsatisfies such a conditions. Definition. We will call prepierced filter on Devery monotonous filter on Dwith center, less thin than a polar sequence. Remark. Let Dbe an infraconnected set of Kand let λ<diam(D). Then, since Kis separable, the family of disks d(a, λ−) included in D which contain elements of D\Dis countable. Moreover, the family of disks included in Dwhose centers are centers of prepierced filters of diameter λis countable too.
Strictly analytic functions on p-adic analytic open sets 147 Lemma 16. Let a∈Kand r>0. Then a partition of d(a, r−)by non-circumferenced disks is a singleton or infinite. Proof: Suppose that the partition is not reduced to a singleton. Let d(a, ρ−) be the element of the partition containing a. Then, we see that ρ<r. Since Kis algebraically closed, its valuation group is dense in R. Then, let b,c∈Ksuch that ρ<|b−a|<|c−a|<r. Let d(b, ρ− b) and d(c, ρ− c) are the elements of the partition containing band crespectively. Obviously we have d(b, ρ− b)∩d(c, ρ− c)=∅. Hence, the partition of d(a, r−) is infinite. Lemma 17. Let Dbe an analoid. Let d(a, r−)be such that d(a, r) D.Letλ∈]0,r[and λ/∈|K|.Let(d(an,ρ n−))n∈Nbe the family of holes of Dincluded in d(a, r−), of diameter superior or equal to λ.Let(d(bn,µ n−))n∈Nbe the family of holes of Dincluded in d(a, r−), of diameter strictly inferior to λ.Let(cn)n∈Nbe a sequence of elements of Dsuch that, for all n∈N,cnis center of a prepierced filter of Dof diameter λ.Let(d(dn,λ −))n∈Nbe the family of disks which contain an element of D\D. If (sn)n∈Nand (tn)n∈Nare sequences in R+of limits zero, such that d(a, r−)= n∈N (d(an,ρ n(1 + sn)−) ∪d(bn,λ(1 + tn)−)∪d(cn,λ −)∪d(dn,λ −)), then there exists m∈Nsuch that d(a, r−)=d(am,ρ m(1 + sm)−)or d(a, r−)=d(bm,λ(1 + tm)−). Proof: Let T1={d(an,ρ n(1 + sn)−); n∈N}, let T2={d(bn,λ(1 + tn)−); n∈N}, let T3={d(cn,λ −); n∈N}and let T4={d(dn,λ −); n∈ N}. We put T=T1∪T 2∪T 3∪T 4. We will denote by Rthe relation defined on Tby URVif there exists W∈T such that U⊂Wand V⊂W. This relation is obviously seen to be an equivalence relation on T. For every U∈T, we put U= V∈U Vwhere Uis the equivalence class of U. We will show that for each U∈T, there exists V∈Usuch that V= Uand that there is only one equivalence class with respect to the relation R.
148 K. Boussaf Suppose that for certain U∈T we have V Ufor all V∈U. Therefore there exists a sequence (Vα(n))n∈Nin U, strictly increasing with respect to the inclusion, whose limit diameters is equal to diam( U). Without loss of generality we may assume that the sequence (Vα(n))n∈N either is in T1or is in T2or is in T3or is in T4. First, we assume that Vα(n)∈T 1,∀n∈Nand we write Vα(n)=d(aα(n),ρ α(n)(1 + sα(n))−),∀n∈N. Then, since lim n→+∞sn= 0, we see that (1) lim n→∞ ρα(n)= diam( U). By [6, Theorem 3.1], we may extract from (aα(n) )n∈Na sequence (aβ(n) )n∈N which either is convergent or is an equal distances sequence or is a monotonous distances sequence. Since ρn≥λfor all n∈Nand d(an,ρ − n)∩d(am,ρ m)=∅, for all n=m, clearly the sequence (aβ(n))n∈Ncan’t be convergent. If (aβ(n))n∈Nis an equal distances sequence of value A, then since (Vα(n))n∈Nis a strictly increasing sequence, we obviously have (2) A<diam( U). Moreover, as the holes of Dare disjointed, we have (3) ρβ(n)≤A, ∀n∈N. Then we see that (2) and (3) contradict (1). If (aβ(n))n∈Nis a monotonous distances sequence, then by (1) necessarily the sequence (d(aβ(n),ρ − β(n))n∈Nis an increasing distances holes sequences, of diameter diam( U) and of superior limit-piercing diam( U). Then, since the sequence (d(aβ(n),ρ − β(n))n∈Nhas a not empty D-beach (because Vβ(n)⊂ U⊂d(a, r−) D,∀n∈N), we see that this contradicts the hypothesis “Dis an analoid”. Second, suppose that Vα(n)∈T 2,∀n∈N. On one hand, since (Vα(n))n∈Nis strictly increasing, we see that diam( ˆ U)>λ. On the other hand, since lim n→+∞tn= 0, we have diam( ˆ U)=λ, which is impossible.
Strictly analytic functions on p-adic analytic open sets 149 Finally, (Vα(n))n∈Ncan’t be a strictly increasing sequence for the inclusion in T3(resp. T4) because their elements are of diameter λ.Thus this finishes proving that for each U∈T, there exists V∈Usuch that V= U. Hence we see that d(a, r−) admits a partition by a family of elements of T. Let us suppose that this partition is not reduced to a singleton. Then, by Lemma 16 this partition is infinite. Let t∈]λ, r[. By [12, Proposition 2.5] we deduce the existence of an idempotent increasing T-sequence S=(d(um,r− m),1)1≤i≤k(m) m∈N of diameter tand whose elements belong to T. We will show that Sis correctly pierced. Indeed, we notice that every element of Thas a diameter superior to λ, and therefore the inferior limit-piercing of Sis not zero. Moreover, by definition, every hole of T3 and T4is of diameter λ. We also remark that, since lim n→+∞tn= 0, then every monotonous distances sequence in T2has a superior limit-piercing equal to λ. On other hand, since lim n→+∞sn= 0 and D∩(K\d(a, r−)) =∅ every increasing distances sequence of holes in T1has a not empty Dbeach, and therefore it is correctly pierced (because Dis). Consequently, as t>λand as T=T1∪T 2∪T 3∪T 4,Sis correctly pierced too. Let bbe a center of Sand let m∈N. •If d(um,r− m)∈T 1, then there exists γ(m)∈Nsuch that d(um,r− m)=d(aγ(m),ρ γ(m)(1 + sγ(m))−) and therefore we put d(vm,r− m)=d(aγ(m),ρ − γ(m)). •If d(um,r− m)∈T 2, then there exists γ(m)∈Nsuch that d(um,r− m)=d(bγ(m),λ(1 + tγ(m))−) and therefore we put d(vm,r− m)=d(bγ(m),λ −). •If d(um,r− m)∈T 3, then there exists γ(m)∈Nsuch that d(um,r− m)=d(cγ(m),λ −). In this case, by hypothesis, d(cγ(m),λ −) is the disk of centers of a prepierced filter of diameter λ. Then we see that there exists vm∈ D\Dsuch that (4) |um−vm|<|um−b|. So, if vmbelongs to a hole of Dof diameter ρ, we put d(vm,r m −)= d(vm,ρ −); else (i.e. vm∈D\D), we put d(vm,r m −)=d(vm,λ −). •If d(um,r− m)∈T 4then there does exist γ(m)∈Nsuch that d(um,r− m)=d(dγ(m),λ−) and then we put d(vm,r− m)=d(dγ(m),λ−).
150 K. Boussaf Hence we have obtained a weighted sequence S=(d(vm,r− m),1)1≤i≤k(m) m∈N which is cofiltring to Sand satisfying (5) |um−un|=|vm−vn|,∀m, n ∈N. We first see that Sis increasing and has a not empty D-beach because d(a, r−) D. But since Dis correctly pierced and 0 <λ<t,Sis also correctly pierced. Now we see that both Sand Sare cofiltring, correctly pierced, satisfy (5) and that Sis an idempotent T-sequence. Then by Lemma 10, S is an idempotent T-sequence. Therefore, by Lemma 3, we can assume that Shas a bounded perturbations sequence. As each hole of Seither is a hole of Dor contains an element of D\D, then Lemma 4 shows that Dadmits either a T-sequence with a not empty D-beach or a Tpolar sequence with a not empty D-beach. But this contradicts the hypothesis that Dhas no T-polar sequences with a not empty D-beach. So, the partition of d(a, r−) is reduced to a singleton, and therefore, there exists T∈T such that T=d(a, r−). But since r>λ, we see that there exists m∈Nsuch that either d(a, r−)=d(am,ρ m(1 + sm)−)or d(a, r−)=d(bm,λ(1 + tn)−). This ends the proof of Lemma 17. Lemma 18. Let Dbe an analoid, let a,b∈Dand let r>0such that r≤|a−b|. Then there exists a D-admissible Ua,r containing a, of diameter r, without increasing T-sequences and such that every increasing distances sequence of holes of Ua,r is correctly pierced. Proof: Let λ∈]0,min(r, δ(a, K \D))[ be such that λ/∈|K|. Let (Fn)n∈Nbe the sequence of prepierced filters of diameter λ, secant with d(a, r−). Let (d(an,ρ n−))n∈Nbe the family of holes of Dincluded in d(a, r−), of diameter superior or equal to λ, let (d(bn,µ n−))n∈Nbe the family of holes of D, included in d(a, r−), of diameter strictly inferior to λ, let (cn)n∈N be such that for all n∈N,cnis center of Fnand let (d(dn,λ −))n∈Nbe the family of disks of diameter λwhich contain elements of (D\D)∩d(a, r−). For all n∈N, we put un=|a−an|.Ifd(an,ρ n−) is circled, we have un>ρ n. So, we may choose εn∈]0,1 n+1 [ satisfying ρn(1 + εn)<u n.If d(an,ρ n−) is not circled, we put εn= 0.
Strictly analytic functions on p-adic analytic open sets 151 For all n∈N, we put T1n=d(an,ρ n(1 + εn)−). Let T1={T1n;n∈ N}, let T2={d(bn,λ −); n∈N}, let T3={d(cn,λ −); n∈N}, let T4= {d(dn,λ −); n∈N}and let T=T1∪T 2∪T 3∪T 4. We define Ua,r as follows: Ua,r =d(a, r−)\ n∈N (T1n∪d(bn,λ −)∪d(cn,λ −)∪d(dn,λ −)). By construction we have a∈Ua,r and diam(Ua,r)≤r. Let us suppose that diam(Ua,r)<r. Then for r1,r2∈] max(λ, diam(Ua,r)),r[ and for u∈Ksuch that r1<|u−a|<r 2<|a−b|, the disk d(u, |a−u|−) is a union of holes of T. By Lemma 17, there exists T∈T such that d(u, |a−u|−)=T. More precisely, as λ<|u−a|, we have T∈T 1. Thus, the annulus Γ(a, r1,r 2) admits a partition Pby elements of T1. Let t∈]r1,r 2[. By [12, Proposition 2.5] there exists an idempotent increasing T-sequence S, of diameter twhose elements lie in T1. By hypothesis we have lim n→∞ εn= 0, so this sequence is similar to a sequence SDof holes of D. The D-beach of SDcontains bbecause |a−b|>r. Hence, since Dis correctly pierced, so is SD. Consequently, as Sis a T-sequence, using Corollary 11, we see that SDis a T-sequence, which is absurd because Ddoesn’t admit T-sequences with a not empty D-beach. Thus diam(Ua,r)=r. We will check that a hole of Ua,r is an element of T. Indeed, let Tbe a hole of Ua,r. It is clearly seen that Tis a union of holes of T. Then by Lemma 17, there exists T∈T such that T=T. We deduce that the diameters of holes of Ua,r are superior to λand consequently, Ua,r is well pierced. Besides, Ua,r is closed by construction. Next, given a monotonous distances (resp. weighted) sequence Sof holes of Ua,r, we may denote by S1(resp. S2, resp. S3, resp. S4) the subsequence which consists of the holes of Slying in T1(resp. T2, resp. T3, resp. T4). We will show that Ua,r is correctly pierced. Then, since Ua,r is well pierced, we only have to prove that every monotonous distances sequence Sof holes of Ua,r, of diameter ρ, either has an empty Ua,r-beach or has a superior limit-piercing strictly inferior to ρ. Without loss of generality we may suppose that Sis of center 0. If S1is infinite, then it is similar to a sequence SDof holes of D.AsUa,r ⊂Dand as Dis an analoid, we see that if Shas a not empty Ua,r-beach, then SDand S1 are correctly pierced.
158 K. Boussaf Definition. A set D⊂Kis said to be circled if at least one of the following statements is satisfied: i) Dadmits a circled hole. ii) Dis peripherally circled. Otherwise it is said uncircled. Proposition 27. Let Dbe an analoid. The following statements are equivalent: a) A(D)=H(D). b) Dis D-admissible. Proof: Clearly b) implies a). If Dis not D-admissible then, since it is an analoid, either Dis not closed or is not bounded or is not well pierced or is circled. First suppose that Dis not closed and, without loss of generality, that 0 ∈D\D(resp. we suppose that Dis unbounded). Let (an)n∈N be a sequence in K∗such that lim n→∞ n !|an|= 0. Then, such a sequence satisfies lim n→∞ |an| sn= 0 (resp. lim n→∞ |an|sn= 0) for all s>0. Hence we see that the series f= ∞ n=0 an xn(resp. f= ∞ n=0 anxn) belongs to A(K\{0}) (resp. A(K)). In particular the restriction f/D of fto Dbelongs to A(D), but it is well known that f/D is not in H(D). If Dis not well pierced, by [6, Theorem 19.7] H(D) is not stable by derivation. Hence, according to Proposition 22, H(D) is strictly included in A(D). Finally, suppose that Dadmits a circled hole that we suppose (without loss of generality) equal to d(0,r−) (resp. we suppose that Dis peripherally circled of diameter S). Then let (an)n∈Nbe a sequence in K∗ satisfying (1) (|an| rn)n∈N(resp. (|an|Sn)n∈N) is unbounded. (2) lim n→∞ n !|an|=r(resp. lim n→∞ n !|an|=1 S).
Strictly analytic functions on p-adic analytic open sets 159 Then such a sequence satisfies lim n→∞ |an| sn=0∀s>r (resp. lim n→∞|an|sn=0 ∀s<S). Hence the serie f= ∞ n=0 an xn(resp. f= ∞ n=0 anxn) belongs to A(K\d(0,r)) (resp. A(d(0,S−))) and consequently its restriction to D belongs to A(D). But according to (1), we see that f/D is not in H(D) and this ends the proof of Proposition 27. 6. Analytic continuation. In Theorem 28 we will show that strictly analytic functions on an analoid of Ksatisfy the property of analytic continuation. Theorem 28. Let Dbe an analoid in K,f∈A(D),a∈Dand r>0.Iff(x)=0for all x∈d(a, r)∩Dthen f(x)=0for all x∈D. Proof: Indeed, let b∈D. We will show that f(b) = 0. By Theorem 19, there exists a D-admissible Ua,b which contains aand b. It is obvious that f(x) = 0 for all x∈d(a, r)∩Ua,b. So, since f/Ua,b ∈H(Ua,b) and since Ua,b is analytic, we see that f(x) = 0 for all x∈Ua,b. In particular, we have f(b)=0. 7. The differential equation y=fy in algebras A(D). Let Dbe an analoid and let f∈A(D). We denote by E(f) the differential equation y=fy with y∈A(D) and by S(f) the K-vector space of solutions h∈A(D). Theorem 29. S(f)has dimension 0or 1. Proof: Assume that S(f) has a not identically zero solution g. Let a∈Dsuch that g(a)= 0. Let hbe another not identically zero solution. Let b∈D, then by Theorem 19, there exists a D-admissible set Ua,b containing aand b. For every l∈A(D), let la,b be the restriction of lto Ua,b. Then ga,b,ha,b are solutions of the equation y=fa,byin H(Ua,b). But since g(a)= 0, by [6, Theorem 55.4], we have h(b)=h(a) g(a)g(b). Therefore, if we put λ=h(a) g(a), we see that h(x)=λg(x) for all x∈D. Example. Let (an)n∈Nand (αn)n∈Nbe two sequences in Ksatisfying |an|<|an+1|,|αn|<|αn+1|∀n∈N, lim n→∞|an|= 1, lim n→∞|αn|= 2, ∞ n=0 −log |an|<+∞and ∞ n=0 (log 2 −log |αn|)<+∞. Let (ρn)n∈Nbe a
160 K. Boussaf sequence in ]0,1[ such that lim n→∞ρn= 0 and choose (λn)n∈Nand (µn)n∈N two sequences in Ksuch that 0 <lim n→∞ |λn| ρn , lim n→∞ µn= 0 and |λn|<ρ n ∀n∈N. For every n∈N, we put bn=an+λnand βn=αn+µn. Let D=d(0,3) \" n∈N d(an,ρ − n)) {αn;n∈N}{βn;n∈N}#. According to [6, Lemma 4 and Proposition 36.5], we check that Ddoesn’t admit T-polar sequences and therefore is analoid. Let f= ∞ n=0 λn (x−an)(x−bn)+ ∞ n=0 µn (x−αn)(x−βn). Since any Dadmissible set Uis well pierced, and since lim n→∞ λn= lim n→∞ µn= 0, it is easy to show that the restriction of fto Uis an element of H(U) and consequently f∈A(D). Let E(f) be the differential equation y=fy in A(D). The function g= ∞ n=0"x−bn x−an#∞ n=0"x−αn x−βn#lies in A(D) and is a solution of E(f). Indeed, first we show that given a D-admissible set U, the product hm= m n=0"x−bn x−an#converges uniformly on U.AsUis well pierced we may choose σ>0 such that diameter of holes of Uare lower bounded by σ. Since lim n→∞ |an−bn|= 0, there exists N∈Nsuch that |an−bn|<σ∀n≥N. Then, we see that we have |x−bn x−an |=1∀x∈U, ∀n≥N. Therefore, for m≥Nwe have hm+1 −hmU=hNU$$$$ x−bm+1 x−am+1 −1$$$$U . So, as hNis trivially bounded on Uand as $$$$ x−bm+1 x−am+1 −1$$$$U =$$$$ am+1 −bm+1 x−am+1 $$$$U ≤|am+1 −bm+1| σ we see that lim m→∞ hm+1 −hmU= 0. Therefore h= ∞ n=0"x−bn x−an#is an analytic element on any D-admissible set Uand consequently h∈A(D).
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162 K. Boussaf 13. Y. Morita, Analytic functions on an open subset of P1(k), J. Reine Angew. Math. 311/312 (1979), 361–383. 14. Ph. Robba, Fonctions analytiques sur les corps valu´es ultram´etriques complets. Prolongement analytique et alg`ebres de Banach ultram´etriques, Ast´erisque 10 (1973), 109–220. 15. M. C. Sarmant and A. Escassut,T-suites idempotentes, Bull. Sci. Math. (2) 106(3) (1982), 289–303. 16. M. C. Sarmant and A. Escassut, The equation y=ωy and meromorphic products, in “p-adic functional analysis,” Lecture Notes in Pure and Applied Math. 137, Dekker, New York, 1992, pp. 157–175. 17. J. Tate, Rigid analytic spaces, Invent. Math. 12 (1971), 257–289. Laboratoire de Math´ematiques Pures Universit´e Blaise Pascal (Clermont-Ferrand) Complexe Scientifique des C´ezeaux 63177 Aubiere Cedex FRANCE e-mail: b[email protected]clermont.fr Primera versi´o rebuda el 29 de gener de 1998, darrera versi´o rebuda el 19 de juny de 1998