ANALYTIC FUNCTIONS OVER VALUED FIELDS

Let K be a non-archimedean, non-trivially (rank i) valued complete field. B,B denote the closed and open unit ball of K respectively. Necessary and sufficient conditions for analytic functions defined on B,B with values in K to be injective, necessary and sufficient conditions for fixed points, the problem of subordination are studied in this paper.

ANALYTIC FUNCTIONS OVER VALUED FIELDS R. BHASKARAN and V. KARUNAKARAN School of Mathematics Madurai Kamaraj University Madurai 625 021, INDIA (Received October 7, 1986 and in revised form June 5, 1987) ABSTRACT.Let K be a non-archimedean, non-trivially (rank i) valued complete field.B,B denote the closed and open unit ball of K respectively.Necessary and sufficient conditions for analytic functions defined on B,B with values in K to be injective, necessary and sufficient conditions for fixed points, the problem of subordination are studied in this paper.
KEY WORDS AND PHRASES.Non-archimedean valuation, Analytic, Univalent, Subordination,  Let K be a non-archimedean, non-trivially (rank 1) valued complete field.B,B denote respectively the closed and open unit ball of K.In the theory of functions of a single complex variable several deep results are proved using Cauchy's theorem in some form.The present study is aimed at analytic functions defined on B,B with values in K.In our case the stronger triangle inequality helps us to prove deeper results with less complications.For example, the local correspondence theorem, Schwarz' lemma, to cite a few.The local correspondence theorem is our starting point for the study of injective analytic functions on B and Lazard's theorem (Theorem 3.20,p.75,[i]),more precisely its consequence is the basis.First we list some results used in the later part.We then take up the problem of finding the necessary and sufficient condi- tions for analytic functions to be injective (one-to-one), the problem of subordination, conditions for the existence of fixed points etc.
We shall assume throughout this paper that K is algebraically closed so that the valuation is necessarily dense.

PRELIMINARIES.
A function f:B+K is said to be analytic if there exists a sequence {a of elements We say that f(A) e An(B) is invertible if there exists g(A) e An(B) such A polynomial of the form A m + 8m_l m-I + + 80, m N, lil i, i=0,1 m-l, is called a Weierstrass polynomial.Clearly all the roots of the Weierstrass polynomial are in B. We state now Lazard's theorem and its consequence without proof A n THEOREM 2 1 (Lazard).Let f(A) a be any non-zero analytic function on B and o n m max(i fail fU).Then for every analytlc function on B there exist unique analytic functions q,r such that r is a polynomial of degree less than m and g=qf+r.THEOREM 2.2 Let f,m be as in Theorem 2.1.Then there exists an invertible element g in An(B) and elements 8o,81 8m_l of K such that 18iI I, i 0,I m-I and im-I In contrast to the case of An(B) we note that if f e An(B), then f need not be bounded (choose {an} in K such that n-I < lanl < n.Then f(A) An B " Z a defines an analytic function on By Lemma 6.41 p.235 [I] sup{If(A)l:AeB} o n sup lanl.By choice of {an}, sup lanl is not finite (i.e.f is not bounded).Infect An(B) we have f e if and only if lim suplan ll/n where f(A) e an For the proofs of the statements made in this section and for more details we refer to [i].
In this section we first establish the local correspondence theorem.This enables us to normalize the injective functions so that they have the form A + a2A2 + a3A3 + THEOREM 3.1 (Local Correspondence).If f(A) e An(B) and f(A)-w has a zero of order o m at A e B then there exists a neighbourhood N(w o) of w and a neighbourhood N(A o) of o o A in B such that each value in N(w o) is assumed exactly m times in the neighbourhood o N(A at m distinct points.o PROOF.Let f e An(B) and f(0) 0 so that 0 is a zero of order m.Then f(1) Am(am + am+iA + Let w e K be such that lwl < Ufi sup{fail: i > m}.Then we note that the Weierstrass polynomials for f and f-w are of the same degree (see Theorem 2.2 for the description of the degree of the Weierstrass polynomial).Let S (0) be a r suitable neighbourhood of 0 in which a + amll + is invertible.Choose e e K m such that 0 < lal < min(I/llfll,r,l).Define g(X) f(al)/am.g(1) is clearly analytic and 0 is a zero of order m for g.Let w be such that lw/aml Wgll, then the Weierstrass polynomial associated with g and g w/ TM have the same degree.Now g(%) xm(am + am+l % + (am + am+l + ...) is invertible and so g(X) has the required representation as a product of a Weierstrass polynomial and an invertible element of An(B).Hence %m is the Weierstrass polynomial of g.This means the Weierstrass poly- nomial of g w/a m is also of degree m.Since the zeros of an analytic function in B are the same as those of the Weierstrass polynomial it follows that there exist Ii,2, 'Xm in the unit ball B such that g(Xi w/a TM =0. (i.e.f(a% i) w, i=l, 2, m).Thus there are m points in the sphere Sial(0) at which f takes the value w.Thus for any w in Sllglllal m (0), we have m points in Sial(0) at which the value w is assumed by f.In order to make them points distinct choose a so that in Sial(0), f'(X) 0, 0. Suppose f(X w Define g(%) f(l+l w so that g is analytic and g(0) If is a zero of order m for f(%) w then 0 is a zero of order m for g.Hence by o o the above arguments there exist neighbourhoods V(O), U(0) in the range and domain respectively such that each point w in V(0) is taken exactly m times in U(0) by g.
In other words we have neighbourhoods U(o), V(w o) such that each w in V(w is taken exactly m times by f in U(A ).The proof is complete.Choose a e such that lXoJ < JaJ (possible by denseness of the valuation) and define g() f(aX).Clearly g An(B).Now g(Xo/a) f(X o) w o, and o/a e B. By Theorem 3.1 we have neighbour- hoods N(Wo of Wo and N(o/a) of o/a such that if w e N(w o), there are m points i'2 %m e N(%o/a) and g(Ai w, i=1,2 m.The proof is complete. Note: From the above result it follows that a one-to-one analytic function cannot have multiple zeros and in particular there is no loss in generality in assuming the normalization f(0) 0, f'(0) for one-to-one analytic mappings.This will be assumed henceforth.Now we are in a position to establish the necessary and sufficient conditions for an analytic function to be injective.
THEOREM 3.3.f e An(B) is one-to-one if and only if fail < i, i Z 2, where f(A) + a22 + a313 + PROOF.Let fail < i, i "2.Then If(l I) f(A2)l I(ll-12) + 2 ai(Ail -2 =i Ii 21.This means that f is one-to-one. Conversely let f be one-to-one.Since f(A) 1(l+a2k+...) by Corollary 2.2 it follows that p(1) I is the associated Weierstrass polynomial and l+a21+a312+... is invertible (has no zeros).Hence fail < i, i 2 (by the equivalent formulation mentioned in section 2).Note: Let f e An(B) be one-to-one.Then (f-f(0))/f'( 0) is an isometry.THEOREM 3.4.Let f(1) I + a212 + a313 + e An(B).Then f is one-to-one if and only if fail i, i 2. B PROOF.Let f e An(B).We define f e An(B) for e by f (I) f(1)/.
Clearly f e An(B).Also if f is one-to-one f is also one-to-one.Note that f.(1) I+a212 + a3213+... Hence if f is one-to-one, by Theorem 3.3, as f is also one-to-one, laiwi-ll < i, i 2 i.e. fail < I/II i-l, i > 2).Since the valuation is dense we can choose a sequence e K such that 0 <I < and n n lnl I. Arguing as above we have fail < i/IWn li-l, i > 2, n 1,2...This means B Again if [ai[ i, i a 2 and if 11,12 e it follows that If(1 I) f( 12) ai(1 -I) 111-121 and hence f is I(-) + one-to-one.
2 B Note: As before a one-to-one analytic function on in the normalized form is an isometry.
Let S (S denote the class of all normalized functions in An(B) (An(B)) that are one-to-one.We have the following properties of S and S S f, i.If f belongs to S or then (I)I 1 and so f'(1) 0.

S S
2. If f,g belong to S or then (fg) (I) f( 1)+g(1)-1 belongs to S or S Z a.b.1 also belongs to S or 3.If f,g belong to S or then (f *g) (I) I+ What is the corresponding result for f belonging to sO? B 7. If f belongs to S then for every o e there exists Io in B such that f(1o o" If f belongs to S then f(B) B.
8. If f and g are in S then (f g) (I) f(g(1)) also belongs to S Properties to 6 and 8 are easily verified.We shall verify 7. Let f e S and B o e Now f, f-o have Weierstrass polynomials of the same degree since they have the same norm.But I is the Weierstrass polynomial of f since f is an element of S.
Therefore the degree of the Weierstrass polynomial of f-7o is I, i.e. there exists a Define f(X) f(X) where 0 < II< I, X B. Then f e An(B).So by. the above argument there exists X 1 e B such that f(XI) Yo' i.e. f(X I) o" But B c f( ).

XI e
In other words B B 4. SUBORDINATION.

S
Let f e An(B), g e If f(b) c g(B) and f(0) g(0), we say that f is subordinate to g.We denote this by f < g (the same as the classical notation).
In this section we study the following question: Let fg and B {: II r}, 0 < r < I.Is it true that f(B c g(Br) 0 <r r r The answer to this question is in the affirmative and we establish this through a sequence of auxiliary results.Among these we have an important result viz., the Schwarz lemma which is of independent interest in itself.
LEMMA 4.1.Let f e An(B) be such that If(X)l I, X e B and f(0) 0. Then PROOF.We know that f(X) p(X)g(X) by Corollary 2.2 where p(X) X m + 8re_lAin-l+ + 81X and g(X) ao + alX + are Weierstrass polynomial and the invertible element associated to f. g(X) being invertible we have Ig(X) laol, X e B. Consider the polynomial pl(X) xm+Sm_l Xm-I + + 81X-I.This is again a Weierstrass poly- nomial and so has all its roots in B (K being algebraically closed), i.e. there exists Xo e K such that p(Xo I. Now i Z If(Xo) Ip(Xo) Ig(Xo) laol.Thus If ]p(X) ]g(X)] IX] la IX].Again for X e B, If'(X)] Ip'(X)g(X)+p(X)g'(X)] o (]p'(X)()l, [p()s'()'()l) 1.
The proof is now complete.Ill I e

Let X B
Then there exists such that f(1)= g().But f < g r implies that f(0) g(0) 0 and If(1)l < i. Therefore by Lemma 4.2 II Ig(Y)l= If(X) If(1)l .<Ill <. r.In other words f(B g(Br).By a fixed point (as usual) we mean % e B or such that f(X) X for f belonging to either An(B) or An(B).We give necessary and sufficient conditions for f in An(B) or An(B) to have fixed points.
THEOREM 5.1.Let f(%) + a2k + belong to An(B).f has no fixed points in B other than 0 if and only if fail la21, i 3.
PROOF.If fail < la21, i > 3 and o s0 is a fixed point for f then it follows that la21 0. This means that fail < la21 cannot happen.This is a contradiction.
Conversely let f have no fixed points other than 0 i.e. f(A)-k has no zeros in B other than 0. Therefore a2+a3X+.., is invertible and so fail < la21, i 3. The proof is complete.
As before we can get the following theorem.
THEOREM 5.2.Let f(X) + a22 + belong to An(B) and f be not the identity map.f has no fixed points other than 0 if and only if fail la21, i 3. PROOF.If fail la21 i 3, arguing as in Theorem 5.1 we have that f has no fixed points in BO.For the converse choose g K such that I < and l,mll as m- m m Define fm () f(mk)/m' e B. Now f m(k) belongs to An(B) and has no fixed points in B. Since f () R + Z a.
by Theorem 5.1 we have lai i.e. fail < la21/[m Ii-2 i > 3 m 2 In other words lai[ la21 i > 3 The proof is complete.
function on B. Let An(B) stand for the space o n of all analytic functions on B. If f An(B) then f is bounded and we let ,f,suplanl.An(B) with this norm is a normed linear space over K I Further lgl[ max(lqll l[fll,UrH).
Further 8o81 m-I and g(A) are unique.Note:If f e An(B) and f has no zeros in B then f is invertible.B A function f: K is said to be analytic if there exists a sequence {an} belongs to S if and only if I is the associated Weierstrass polynomial.
LEMMA 4.2.If f e An(B), f(0) 0 and If(X)l 1, then If(X) IX < II < i. Define fl(a) f(a), e B. Then fl e An(B), fl(0) 0 and Ifl(e) i. Hence by Lemma 4.1 it follows that Ifl(=)l & I=I, e B. In particular If(X)l Ifl(I/)l & lll/ll.Choose a sequence e K such that IXI < I < 1 and lnl I. Then we have If((X) & Ill x being arbitrary it follows that If(1) IXI X e B Again a similar argument as in Lemma 4.1 shows that If'(X)l & I, X e NOTE.Lemmas 4.1 and 4.2 are the analogues of the classical Schwarz Lense.