p-ADIC ANALYTIC INTERPOLATION

Let K be a complete ultrametric algebraically closed field. We study the Kernel of infinite van der Monde Matrices and show close connections with the zeroes of analytic functions. We study when such a matrix is invertible. Finally we use these results to obtain interpolation processes for analytic functions. They are more accurate if K is spherically complete. 1991 Mathematics subject classification: : 46S10 1. NOTATIONS, DEFINITIONS AND THEOREMS K denotes an algebraically closed field complete for an ultrametric absolute value. Given a E K, r > 0, we denote by d(a, r) (resp. d(a, r-)) the disk {x E K : r~ (resp. {x E K : a~ r~). Given r > 0 we denote by C(0,r) the circle d(0, r)~d(0, r’). Given ri, r2 E IR+ such that 0 ri T2, we denote by r(0, ri, r2) the set d(0, r-2)Bd(0, r1). 00 Given r > 0, we denote by A(d(0, r")) the algebra of the power series ~ bnxn n=o converging for |x| r. Given K-vector spaces E, F, ,C(E, F) will denote the space of the K-linear mappings from E into F. £ will denote the K-vector space of the sequences in K, and So will denote the subspace of the bounded sequences. The identically zero sequence will be denoted by (0). ~1 will denote the set of the sequences (an) such that limsup |03B1n| 1. So ~1 is seen n~~ to be a subspace of £ isomorphic to the space A(d(0, I ")), and obviously contains So. Let Moo be the set of the infinite matrices ( a; ~ j ) with coefficients in K. 03B4i,j will denote the Kronecker symbol. I~ will denote the infinite identical matrix defined as A,j = 03B4i,j. * Research partially supported by the Spanish Direccion General de Investigacion Cientifica y Technica (DGICYT, PS90-100)

Given K-vector spaces E, F, ,C(E, F) will denote the space of the K-linear mappings from E into F. £ will denote the K-vector space of the sequences in K, and So will denote the subspace of the bounded sequences. The identically zero sequence will be denoted by (0). 1 will denote the set of the sequences (an) such that limsup |03B1n| 1. So ~1 is seen n~t o be a subspace of £ isomorphic to the space A(d(0, I ")), and obviously contains So. Let Moo be the set of the infinite matrices ( a; ~ j ) with coefficients in K. 03B4i,j will denote the Kronecker symbol. I~ will denote the infinite identical matrix defined as A,j = 03B4i,j. * Research partially supported by the Spanish Direccion General de Investigacion Cientifica y Technica (DGICYT, PS90-100) In this paper, (an) will denote an injective sequence in d(o,1' ) such that an fl 0 for every n > o. and we will denote by M(an) the infinite matrix M = (at,~ ) defined as 03BBi,j = (ai)j, (i,j) ~ IN x N.
A matrix M = (a~~') E M~, will be said to be bounded if there exists A E R+ such that A whenever (i, j ) E IN x IN. M will be said to be line-vanishing if for each i E IN, we have lim a=,J = 0. A line-vanishing matrix M is seen to define a K-linear mapping 03C8M from So into £.
So the matrix M = M(an) clearly defines a K-linear mapping ~M from £i into £, m because given a sequence (bn) E ~1, the series ' is obviously convergent. n=0 Lemmas 1 and 2 are immediate : Let a = (an) and b = (bn) be two sequences in K. We will denote by a * b the n convolution product (cn) defined as cn = 03A3 ajbn-j. =o Theorem 3 : : Let (an) be a regular sequence of d(0,1') such that there exists g E A(d(0,1 ")) satis f ying (i) an is a zero of order 1 of g for all n E IN.
Remarks. 1. Mainly, the proof of Theorem 3 takes inspiration from that of Lemma 3 in [7]. However, in this lemma, the considered matrix, roughly, was P. Here the matrix we consider is a van der Monde matrix M and we look for P.
2. Given ~1~I the matrix P depends on g and therefore is not unique satisfying ( 1 ) -( 7). Indeed is not a ring because the multiplication of matrices is not always defined and even when it is defined, is not always associative. As a consequence, if P, P' satisfy MP = MP' = PM = P'M = I~, we cannot conclude P' = P.
Actually we can consider 03C6M o E £(So, £) and then this is the identity in ~0. Next we can consider 03C8P' o03C8M E £( So , ~1) and this is the identity in ~0. But we cannot consider 03C8P' o (03C6M o 03C8p) because 03C8P' is not defined in ~1. In the same way, we cannot consider (~ p~ o o ~ p because 9 p~ o is only defined in So. We consider the matrix P and look for "inverses" M such that MP = PM = I~. 3. Let P, Q E M~ satisfy (1) - (7). Let £' = let £" = Then the restriction of 03C6M to £' (resp. £") is just the reciprocal of (resp. 03C8Q).
Conjecture. Under the hypothesis of Theorem 1, every matrix satisfying properties (1) - (7)  For each set D in K, we denote by H(D) the set of the analytic elements in D (i. e., the completion of the set of the rational functions with no pole in D).
For every r E~O,1(, f satisfies -log = v( f , -log r). Besides f is bounded in d(0,1") if and only if the sequence (bn) belongs to So. If f is bounded in d(0,1-), then = sup |bn| and -log = lim v( f , ). Lemma 4 : : Let J(t) E A(d(0,1-)) and let E (o,1) satisfy r1 r2. . If f admits q zeros in d(0, rl) (taking multiplicities into account) and t distinct zeros at,..., , at, of multiplicity order ~'j (1 j t) respectively in r(O, rl, , r2), then f satisfies First we suppose Ker03C6M ~ {(o)} and therefore we can assume b E Ker03C6M. Then, by Lemma 2, f satisfies f(aj) = 0 for every j E N. But for every r ~]0, 1[, we know that f belongs to H(d(U, r)) and has finitely many zeros in d(0, r). Hence we have lim |an| =1. n~R eciprocally, let the sequence (an) satisfy lim |an| = 1. By Proposition 5 in [4], we n~~0 0 know that there exists a not identically zero analytic function f (t) = ~ bntn E A(d(o,1~')) n=o 00 which admits each aj as a zero. Hence we have 03A3 bnanj = 0, and of course the sequence n=o (bn) belongs to ~1, hence to Ker03C6M. Now we suppose that Ker03C8M ~ (0) and we assume that the sequence (bn) belongs to Ker03C8M. In particular Ker03C6M ~ (0) and therefore lim |an| = 1. Without loss of generality n~w e may clearly assume |an+1| for all n E IN. Besides, by definition we have |a1| > 0. By Lemma 3 we know that inf v(bn) = lim lim n~IN |x|~1,x~D Now for each ~ > 0, let be the unique integer such that v(an) > ~ for every n q(p,) and v(an) ~c for every n > q(~c). By Lemma Since v(f, p) is bounded when p approaches 0, by (1) it is seen that ~ v( aj) must be j=1 00 bounded and therefore we have TI |an| > 0. n==i 00 Reciprocally we suppose 03A0 |an| > 0. We can easily check that lim |an| =  we see that lim 03BD(Pq( ), ) = 03A3 v(aj). But by (2)  3. PROVING THEOREM 3.
As an application of Corollary (of Theorem 5) in [8], we have this lemma. Lemma 6 : : Let f E A(d(0, 1-)) have a regular sequence of zeros (bn) and satisfy lim = +00. Then 1/f belongs to H(03A9(bn)). and by Lemma 4 the sequence of the zeros (an) satisfies N |03B1n| = 0, hence 4 M is injective. Tt=l Now we look for P. Since g admits each 03B1j as a simple zero, it factorizes in A(d(0, 1-1)) in the form 03C8j(x)(1x /03B1j) and we have 03C8j(03B1j) ~ 0. We put x = 0 3 C 8 j ( x ) 0 3 C 8 j ( 0 3 B 1 j ) . Then gj belongs to A(d(0,1")) and may be written as L 03BBn,jxn. We denote by P the matrix n=o and we will show this satisfies Properties (1) - (7). For convenience, we put D = Q( an). Since lim |g(x)| = +~, by Lemma  Now, we will check that P(b) E ~~ for all b E Eo. Let b := (bn) E Eo, let a := 00 j (an) = P(b) and let J(t) = L antn. For each j e IN we put = L bmgm(t). Then n=0 m=0 f j belongs to A(d(o,1')) for all j ~ IN. Let r 6JO, 1{. Like the family |03BBn,j| rn, the family tends to zero uniformly with respect to n when j tends to +00. That way, in we have ~f -= 0 and therefore f belongs to H(d(O,r). This is true for all r ~]0, 1[ and therefore f belongs to A(d(o,1')). Hence P(b) E ~1. This shows (5).
Finally we will prove the last statement of the theorem. Let 03C6(x) = 03A3 vnxn. The n=0 function § belongs to A(d(o,1' )) and is invertible in A(d(o,1 ~')) thanks to the inequality |03BD0| > |03BDn| whenever n > 0. Hence the function G(x) = is easily seen to satisfy i), ii), iii), iv) like g. Then G factorizes in A(d(o, 1-)) and can be written as x/03B1j) with = Hence we put Gj(x) = 0 3 C 6 j ( x ) 0 3 C 6 j ( 0 3 B 1 j ) = gj(x)03C6(x) 03C6(03B1j) . Now it is clearly seen that the power series of Gj is E By definition, the matrix Q satisfies the n==o same properties as P. But when § is not a constant function, for each fixed j E N, we do not have = 03BBn,j for all n E N. Hence Q is different from P. As a consequence we see that 03C8M is not surjective, it would be an automorphism of £o and therefore 03C8P would also be an automorphism of So and it would be unique. This ends the proof of Theorem 3.

PROVING THEOREMS 4 AND 5
Notation. For each integer q E IN*, we will denote by the group of the q-roots of 1.

Proof.
Let r = |as|. Since limn~~ |an| ---.1, the circle r) contains finitely many terms of the sequence (an). Without loss of generality we may assume r whenever n d, lanl ~ > r whenever n > t and janj = r, whenever n = t, ... , t (with obviously I s t). Whatever q E 1N, ( E Q(q) are, it is seen that we have a; | = lasl for all j t and |03B6hasa; | = for all j > t. In the residue class field k of K, for every every q-root v of 1 in k, we have 03BDh ~ 03B3j whenever j = I, ... , t, whenever h =1, ... , q -1. Now let ( be a q-th root of 1 in K. Then by classical properties of the polynomials, we have |03B6h -aj as| = 1, hence |03B6has -aj| = |as| = r whenever h = 1, ... , q -1, whenever j = l,..., t. This completes the proof of Lemma 7. Lemma 8 : Let (an) be a regular sequence and let p = inf |an -am|.
(3) (an) is a subsequence of (bn),  for all (n, m) E Et x Et such that n # m. We will choose qt such that both (at+i), (,8t+1) are satisfied. Indeed, by Lemma 7 we can take a prime integer u such that, given 03B6t E G(u), we have aj| ( = max(|at|, |aj|) for all j E N, for all h =1, ... , u -1, bnl ] = max(|at|, |b'n|) for all n st, for all h = 1,... , u -1. Thus we can take qt = u and we see that both (~t+1 ) are satisfied. Hence we can construct the sequence (qt) by induction and, therefore, the sequence (b~) satisfying (6) is now constructed. Then it is easily checked that the sequence (bn) so obtained satisfies (1) Now let (bn) be the sequence of the zeros of g. Clearly satisfies (1) and (4) and also satifies |b"n -b'm [ = max(|b"n|, |b'm|) whenever n, m E N and b"m| = max(|b"n|,|bnm|) whenever n ~ m. Now we put b2n = bn and b2n+1 = b"n. The sequence clearly satisfies (1), (2), (3), (4) and also satisfies (5) because the zeros of h are the bn while those of g are the bn. Thus the zeros of f are just the bn, and then, by (7), we have lim |f(x)| = +00. Proof of Theorem 4. Without loss of generality we may obviously assume |03B1n+1| whenever n E IN. Let p = |03B10|. Hence by hypothesis each disk d(03B1q,03C1-) contains no point an for each n ~ q. Let D = For each n E IN, let Tn be the hole d(an, p!) of D. Since |03B1n| = 0, it is shortly checked that the sequence (Tn, 1) is a T-sequence of D ( [8]). Then, since K is spherically complete, by [4], Theorem 4, there exists g E admitting each an as a simple 00 zero and having no zero else in d( o,1 "). Therefore, as 03A0 |03B1n| = 0, is is seen that g n=0 satisfies lim |g(x)| = +~. Now we can apply Theorem 3, which shows that the matrix 00 M = M(an) admits inverses P. Then the sequence (an) satisfies 03A3 an03B1nj = bj for every n=0 j E IN and this clearly ends the proof of Theorem 4.
Proof of Theorem 5. By Lemma 8, there exists a regular sequence (In) of d(0,l") such that (an) is a subsequence of (in ) together with an analytic function g E admitting each 1m as a simple zero and having no other zero in d(0,l"), satisfying lim |g(x)| = + oo with p = inf |03B3n -03B3m|. Then, by Theorem