Motzkin factorization in algebras of analytic elements

In this work, we show that the Motzkin factorization requires the set of analytic elements to be a K-algebra. Hence, we propose a new presentation of this problem, using a very close relationship, between the Mittag-Leffler series and Motzkin product. 1991 Mathematics subject classification: : ,~6510 1.INTRODUCTION. Let K be an algebraically closed complete ultrametric field whose absolute value is denoted by ). I. Let D be a subset of K, whose closure is denoted by D. We denote by R(D) the K-algebra of the rational functions h(x) E K(x) with no pole in D, and we denote by H(D) the completed topological linear space of R(D) for the topology of uniform convergence on D. The elements of H(D) are called the Analytic Elements on D. We denote by Hb(D) the set of the bounded elements of H(D) and by the norm of uniform convergence defined on Hb(D), and then Hb(D) provided with this norm is a K-Banach algebra. Actually Hb(D) is equal to H(D) if and only if D is closed and bounded. More generally H(D) is provided with a structure of K-subalgebra of KD if and only if D satisfies the two following conditions : a) either D or K~D is bounded b) is included in the interior of D. We denote by A the set of D subset of K such that H(D) is a K-subalgebra of KD. It is seen that if D is open and satisfies b) then D is open too. Let us recall that a space H(D) does not contain characteristic functions different from 0 and 1 if and only if D is infraconnected (3~,~4~. A set D is said to be in f raconnected if for every a E D, the mapping from D to IR+ x --~ a~ has an image whose closure is an interval.


MOTZKIN FACTORIZATION IN ALGEBRAS OF ANALYTIC ELEMENTS Kamal Boussaf
Ann. Math. Blaise Pascal, Vol. 2, N ° 1, 1995, pp.73-91 Abstract. In this work, we show that the Motzkin factorization requires the set of analytic elements to be a K-algebra. Hence, we propose a new presentation of this problem, using a very close relationship, between the Mittag-Leffler series and Motzkin product. Let K be an algebraically closed complete ultrametric field whose absolute value is denoted by ). I. Let D be a subset of K, whose closure is denoted by D. We denote by R(D) the K-algebra of the rational functions h(x) E K(x) with no pole in D, and we denote by H(D) the completed topological linear space of R(D) for the topology of uniform convergence on D. The elements of H(D) are called the Analytic Elements on D. We denote by Hb(D) the set of the bounded elements of H(D) and by the norm of uniform convergence defined on Hb(D), and then Hb(D) provided with this norm is a K-Banach algebra. Actually Hb(D) is equal to H(D) if and only if D is closed and bounded.
More generally H(D) is provided with a structure of K-subalgebra of KD if and only if D satisfies the two following conditions : a) either D or K~D is bounded b) is included in the interior of D.
We denote by A the set of D subset of K such that H(D) is a K-subalgebra of KD. It is seen that if D is open and satisfies b) then D is open too. Let us recall that a space H(D) does not contain characteristic functions different from 0 and 1 if and only if D is infraconnected (3~,~4~. A set D is said to be in f raconnected if for every a E D, the mapping from D to IR+ x --~ a~ has an image whose closure is an interval. Notation. Let a E K, let D be a set in K. We denote by 6(a, D) the distance from a to D i.e ED}.
Let a and let a E 1R+. We will denote by d(a, r) the disk {x E at ~ r} by d(a, r') the disk {x E r} by C(a, r) the circle {x E a) = r} and by the set E K} For every a E C(a, r), d(a, r' ) is called a class of C(a, r). Let r' > r. We will denote by r(a, r, r') the set {x E Ixal r'}, and by ~(a, r, r') the set {x E h'1r r'}.
We now recall the definition of the holes.
Definitions. Let D be a closed subset of K. Let R be its diameter and let a E D. We put D = d(a, R) or D = K if R = oo and we now that D is in the form D B(~i~Id(03B1i, with ri ) n _ ~ whenever i # j and with rt = b(a=, D) whenever i E I. The disks d(a;, ri) (i E I) are named the holes of D.
First we have to recall the Mittag-Leffler series for an analytic element on infraconnected closed set [9] [10] [13]. When D is not bounded, we denote by Ho(D) the subspace of the f E H(D) such that j lim ~ ~ |f(x)| = 0 | x | ~ ~x ~ D Let D be an infraconnected closed set and let f E H(D). We know that f admits a 00 unique series ~hn named the Mittag-Leffler series of f on D whose sum is equal to f, , n=0 where ho E H(ÎJ) and for each n > 1, there exists a hole Tn of D such that hn E Ho(K~Tn). ( The sequences (Tn), (hn) are injective [9], [10], [13]).
Moreover for each n E IN* we have ~hn~D = ~hn~KBTn and lim hn = 0.
The holes Tn are called the f -holes (assuming hn ~ 0) and hn is called the Mittag-Lefher term of f associated to the hole Tn.
If / E Hb(D) then ~h0~D = ~h0~ and we have sup ~hn~D. Besides, if D is not bounded and if f E H(D) then ho is a constant. Now given a f -hole T, the Mittag-Lefler series of f associated to T will.be denoted by fT. Thus in the Mittag-Leffler series of f above, we have hn = fTn whenever n E IN*.
In the same way we will always denote by 70 the element ho of H(D).
We say that f is semi-invertible in H(D) if it factorizes in the form Pg where P is a polynomial whose zeros belong to D and g is an element of H(D) invertible in H(D). Let us remember the valuation defined on K and the valuation function associated to an analytic element.
Let log be a real logarithm function of base 9 > 1. We define a valuation v in K* by v{x) = -log|x|.
Let a E K and let r E R+. We call the circular filter of center a of diameter r the filter ~' that admits as generating system the annuli r(a, r', r") with a E d(a,r), r' r r" [5][8].
Now let D be a set such that 7 is secant with D. The intersection of .~' with D is called circular filter of center a, of diameter r, on D [5], [8]. Then we know that for every f E H(D), has a limit D03C6a,r(f) along FD and the mapping D03C6a,r is a multiplicative semi-norm in H(D) [5], [6]. Thus we have Besides, given a polynomial or more generally an element f on H(D U d(a, r)} we have = = As a consequence, if r E |K| so does Now, let E IR be such that the circular filter F of center a, of diameter r = is secant with D. We put va(f, ) = -log(D03C6a,r(f)).
In particular when a = 0 we just put v( f , ~C) = When D n d(a, r-) ~ ~ or when D n (k' ~ d(a, r)) ~ ~ we see that v a ( f , -l o g ( r ) ) = l i m -v ( f ( x ) ) |x -a| ~ r |x -a| ~ r, x ~ D . ~ Now, let D be an infraconnected set and let T = d(a, r' ) be a hole of D. The circular filter .~' of center a, of diameter r is certainly secant with D. In particular, if D has diameter R > r then we have Henceforth, we suppose that D is infraconnected and closed. Lemma 2.1 : : Let T = d(a, r-}, with a E K , and r > 0, let E = K ~ T and let bET. Let g E H(E) be invertible in H(E). Then there exist a E K, q E TL, and h E H(E) invertible in H(E), satisfying -1~E 1, lim h(x) =1 and g{x) = a{x -b)q h(x). Besides a, q, h, are respectively unique, satis f ying those relations. Further, both a, q do not depend on b in T, belongs to Ho(E). 9 Proof : Without loss of generality we may obviously assume a = 0. It is easy to show that g is of the form g+g, with y and Hb(E). Let  We can now give the following definitions.
Definitions. Let E = j~ B with a and r > 0. Let f H(E) be invertible in H(E) and let A(.r 2014 be the factorization given in Lemma 2.1. The integer q will be named the index of f in the hole d(a,r") and will be denoted by The element f will be called a pure factor associated to d(a,r") if A = 1.
Let f belong to H(D). Let T be a hole of D and let h be a pure factor associated to T. Then f will be said to admit h as a Motzkin factor in the hole T if 2 0 1 4 belongs to H(D U T) and has no zero inside T.

Lemma 2.2 : :
Let T = d(a,r"), let E = K B T with a K, and let f be a pure factor associated to r such that ~f -1~E 1. Then m(f, T) = 0.
Proof : Indeed, let q = m(f, T) and let f = (xa)qh. If q ~ 0, this contradicts the unicity of q and h shown in Lemma 2.1.
Denoting by gT the group of the invertible elements of H(K B T), by Lemma 2.1 we have Corollary 2.3. Corollary 2.3 : : Let T = d(a,r'). The set of the pure factors associated to T is a submultiplicative group of the group of . Further, every element of GT is of the form ah with h a pure factor associated to T and .1 E K*. First, we assume D n (K B d(a, r)) ~ 0. There does exist s diam(D)[ such that for every u ~]r, s[. Then by Proposition [5] there exist ui, ..., ut ~]r, s[ such that |f(x)| =D holds in all x E C(a, n) n D for every u ~]r, s[B{u1, ..., ut}. So we just take D' = (D n d(a, s)) Now we suppose that the condition D n (K B d(a, r)) ~ ~ is not satisfied. Since d(a, r-) ~ D and D03C6a,r don't depend on a in d(a, r), then there exists bED fl C(a, r) Hence there exists s ~]0, r[ such for every u ~]s, r[. Then by Proposition [5] there exists ui, ..., ut ~]s, r[ such that |f(x)| =D holds in all x E C(b, u) n D for every u ~]s, r[B{u1, ..., ut}. Thus we now take D' _ (D ~ d(b, r-)) ~ U d(a, r') . and then we check that D' is the set we are looking for. Theorem 2.5 : : Let T be a hole of D and let f have a Motzkin factor h in T . Then h is unique. Further, if T is not a f-hole, h is the polynomial of the zeros of f inside T. Besides, if E is another infraconnected set included in D admitting T as a hole, and if '"S ince T ~ D' it is seen that D' n (K B T) is an infraconnected closed bounded set included in D that admits T as a hole. If we seeas an element of H(D' n (K B T)), ' then by (1) and by uniqueness of Mittag-Leffler term we have (-) = 0. So 2 0 1 4 belongs to H(K) and therefore is a polynomial P. Since -; belongs to H(D') and has no zeros inside T, it is seen that m(h,T) = m(l,T), so we have lim l(x) h(x) = 1. Hence P = 1 and this proves that h is unique. Now we assume that T is not a f-hole. Hence f belongs to H(D U T). Let Q be the polynomial of the zeros of f inside T. Then f Q belongs to H(D U T) and has no zeros inside T [4j. Since its Motzkin factor h is unique, we have h = Q. The last statement about ~ is obvious, because -, clearly belongs to H(E U T) and has no zero inside T. This ends the proof of Theorem 2.5.
Definitions. We will call the f-supersequence o/ D the sequence of the holes (Tn)n~I such that either Tn is a f-hole or f belongs to H(D U T) and has at least one zero inside Tn. If f admits a Motzkin factor h in a hole T, it will be denoted by The index of h in T is called the Motz1cin index of f in T and denoted by m(/,T). For every hole, which does not belong to the f-supersequence, we put /~ = 1.  In both cases, f ° will be called the principal factor of f. For every x E K d 0 1), we have |x203A0 x -|x2|, and x -1| = |03BBn+1 x| hence j=2 = |x~03BBn+1|. Thus f n+1 -f n is not bounded in H(E}, and therefore the sequence (fn)n~IN* does not converge in H(E). According to Theorem 4 in [11] the 00 product " 03A0 fn" should converge to x2h in H(E}. Here we see that this is not true in the n=1 general case. Actually the proof given in [11] only shows the simple convergence of the sequence (In)' By Lemma 2.6, Lemma 3.4 is immediate. and a n E Tn. When n > N, f Tn is just in the form (1 + (In) with wn E Ho ( K ~ Tn) and I. Besides, as f has no zero in D, obviously f ° has no zero in D and therefore has no zero in D hence f ° is of the form A(l + 03C90(x)) with wo E H(D), ~03C90~D 1. Let us suppose q1 ~ 0. We may obviously assume ai = 0. Let Ti = d(0, r"). Thus, in Tl, h admits 0 as a zero of order ql if qi > 0 (resp. a pole of order -ql if qi 0) and has neither any zero nor any pole different from 0. Anyway, when x E Ti, we have N (2) Ih(x)1 = B|xq1| with B = |A|03A0 |03B1n|qn. n=2 First we suppose r ft Then there exists r' > r such that h has neither any zero nor any pole in 0394(0, r, r'), hence h has neither any zero nor any pole in d(0, r') 1 {0}. Therefore is of the form in all of d( 0, r') 1 {0}. In particular, this is true in 0394(0,r,r') n D and shows that D03C60,r(h) = Brq1 while = Br'Q1. . Hence we see that is not constant in ~(o, r, r') n D and therefore this contradicts relation (1). Now we suppose that r belongs to |K| and then by a classical linear change of variable, we may restrict ourselves to the case where r = 1. Hence we have Tl = d(o,1-). By (2) we have D03C60,1(h) = B. But by definition B belongs to |K| and therefore we may clearly assume B = 1 without loss of generality. Hence h is of the form P(x) Q(x). with (3) = 1 and P prime to Q. Let us suppose qi > 0. By definition of h, P has no zero different from 0 in Ti while Q has no zero in Ti. Hence Q satisfies (4) whenever x E Ti and then by (3) we have = 1, while obviously P(0) = 0. Hence by (3) it is seen that ~~P -= 1 and therefore by (4) we have = 1' But we know that for every g E R(D U d(0,1)) we have > > == ~g~D hence we see that > 1 and this contradicts (1).
We now suppose qi 0. By definition h is obviously invertible in R(D). Hence we put F = 1 h and we see that F satisfies ~F -1~D 1 and admits 0 as a unique zero in T1 while it has no pole in Ti. Hence the same process lets us get to the same contradiction and finishes showing that qn = 0 for every n > 1. Proof : Without loss of generality, we may obviously assume f(a) = 1. By Lemma 3.6, we already know that if f satisfies 1, then for every hole of the f -supersequence, we have m( f, T) = 0. Now we suppose that for every hole T of the f -supersequence we have m( f T) = 0 and will prove that 1. Indeed, by Lemma 1, for each hole of the f -supersequence, we have 1. Besides since f is invertible, f ° must also be invertible, hence it is of the form (1 + ~(x)), with 1. Then it is seen that

4.EXISTENCE OF MOTZKIN FACTOR AND MOTZKIN FACTORIZATION.
We will show all the semi-invertible elements to have Motzkin factorization, step after step, and first we consider rational functions. Proposition 4.1 : : Let f E R(D). . Then f admits Motzkin factorization. Proof : Since the number of zero and pole of f is finite, let Tl, ..., Tq be the f -supersequence of D. For each n =1, ..., q we denote by f Tn the rational function whose numerator ( resp. denominator ) is the polynomial of the zeros (resp. of the poles ) of f inside Tn. Finally we denote by f0 the rational function whose numerator ( resp. denominator ) is the polynomial whose zeros are the zeros (resp. the poles) of f in D~(KB), (resp. in KB). It is seen that f ° is the principal factor of f, and that for each n = 1, ...q, is the Motzkin factor of f associated to the hole Tn.  (1) and (2) we havẽ ( 0 3 C 9 n ( 1 + 0 3 C 8 n ) ) T n~D -~03C9n~D and finallỹ 03C6Tn~D -+ 03C8n))Tn~D = ~03C9n~D = 03 C 6 T n -1D . 00 In the same way we put 03A0(1 +wn) = 1 with (3) 1. It is seen that 03C8 belongs n; i 00 to Ho(K B ( U Tn)). Hence the Mittag-Leffler Theorem [1] applied to 03C8 shows that (4) Relation (6) does not depend on the hole T and shows that, for each fixed n E N*, the sequence is a Cauchy sequence which converges in H(K ~ Tn), to an element whose index is equal to 0, and this convergence is uniform with respect to n. For each n E N*, we put = lim Then it is seen that ' m ~ ~0 0 00 03A0 03C6n # lim 03A0(fm)Tn.
As a consequence, the sequence (fm)O is also convergent in H(D), and actually in Hb(D ).
Let 03C60 be its limit. Then we have this factorization : § = II 03C6n. It is seen that this is Let (Tm)m~I be the I-sequence of D. We notice that when i) is satisfied, § is obviously invertible.
First we suppose i) satisfied and will show that so is ii). By the Mittag-Leffler Theorem we have ______ (2) 11 -~ ~03C6 -03C6(a)~D.
But it is seen that (~ 2014 = Hence by (2) we see that In the same way we have (03C6 -03C6(a))0 = 03C60-03C6(a) and then by the Mittag-Leffler Theorem [1] we have _ (4) ~~~a " 1.  (5), statement ii) is clearly proven. Now we will show that each one of the statements ii) and iii) separately implies i). We suppose ii) satisfied. Hence we have (6) mEl If D is bounded, by statement ii) and by (6) we obtain i). Now let D be not bounded. We now suppose iii) satisfied. Hence we have (8) 1 for all m E I.
If D is bounded we have ~03C60 -03C60(a)~D 1 hence by (8) (8) and (1) we m~I see that |03BD -1) 1 hence by (8) we obtain i) again. Finally, i) is implied as well by ii) as by iii). Obviously by (1), i) implies 1 and therefore we may apply Proposition 4.2. Now we suppose that either ii) or iii) is satisfied. Hence so is i) and so are u) and v) by Proposition 4.2.
Finally, we will show w) and at the same time we will finish proving the equivalence between ii) and iii). Let 03C8 = (03C60(a))-103C6. We may apply Proposition 4.2 to 03C8 and we have (9) ~03C80 -1~D = ~03C80 -1~D. Proof: We suppose that a = 0. Since D E A and D is unbounded, we have K B D is bounded [4]. So, since f is semi-invertible, there exists r > 0 such that the set B = {x E > r} is included in D, and such that f(x) ~ 0 whenever Now, we suppose f to be semi-invertible and will show it to have Motzkin factorization. By Lemma 3.4 we may clearly suppose that f is invertible without loss of generality.
First, we suppose that there exists NI in Rj such that (1) M whenever x E D.
Let h E R(D) satisfy (2) h~D M 2, and let h 03A0 hTn be the Motzkin factorization By (3) and (4) [4] there exists a quasi-minorated element / Hb(B) satisfying (1) lim /(J?) = 0 and such that :r/ does not belong to H(B). Since / Hb(B) |x|~w e can take it such that 1. Without loss of generality, we may assume that 0 belongs to a hole of B. Let T = d(a, r") be another hole of B, and let F = x(1+f) (x-a) . Then it is seen that JF belongs to Hb(B) and is invertible in Hb(B) because both x (x-a) , 1 + f are invertible in Hb(B). Hence F admits Motzkin factorization. In particular, we see that = 201420142014-. . However we check that (:r 2014 a)F does not belong to -H~~B) because (x-a) (a-2014 a)F = x(1 + f) and by hypothesis, xf does not belong to H(B).
In the same way, let (?==2014. Since F is invertible in Hb(B), so is G. But then we see that 201420142014G does belong to Hb(B) and has no zero in B, but obviously its inverse does not belong to H(B). Therefore 201420142014G is not semi-invertible in H(B). Thus, there exist invertible elements h, g in H(B) such that hg is not semi-invertible, although it does belong to H(B). This contradicts Theorem 1 in [11] which apparently states that f fT extends to an element of H(D U T).   Proof : The product is direct because for each element, Motzkin factorization is unique.
Thus x is the set of the invertible elements whose Motzkin factorization is finite. Since every element of g has Motzkin factorization, it obviously belongs to the closure of ~l.