CC BY
31
2000 MSC 32A15, 32A20, 32A22, 32H30
NORMAL FAMILY AND THE SHARED POLYNOMIALS OF MEROMORPHIC FUNCTIONS Feng Lü1), Jun-Feng Xu2)
^Department of Mathematics,China University of Petroleum,
Dongying, Shandong 257061, China, e-mail: [email protected] 2) Department of Mathematics, Wuyi University,
Jiangmen, Guangdong 529020, China, e-mail: [email protected]
Abstract. In the paper, we study the uniqueness and the shared fixed-points of meromorphic functions and prove two main theorems which improve the results of Fang and Fang and Qiu.
Keywords: meromorphic functions, fixed-points, holomorphic coefficients, shared polynomials.
1 Introduction and main results
Schwick [8] was the first to draw a connection between values shared by functions in F (and their derivatives) and the normality of the family F. Specifically, he showed that if there exist three distinct complex numbers a1,a2, a3 such that f and f' share aj (j = 1, 2, 3) IM in D for each f Є F, then F is normal in D.
In 2006, Wang and Yi [9]proved a uniqueness theorem for entire functions that share a polynomial with their derivatives, as follows
Theorem A. Let f be a nonconstant entire function, let Q(z) be a polynomial of degree q > 1, and let k > q be an integer. If f and f' share Q(z) CM, and if f(k')(z) — Q(z) = 0 whenever f (z) — Q(z) = 0, then f = f.
According to Blochyis principle, numerous normality criteria have been obtained by starting from Picard type theorems. On the other hand, by Nevanlinna’s famous five point theorem and Montel’s theorem, it is interesting to establish normality criteria by using conditions known from a sharing values theorem.
In this note, we obtain the following normal family related to Theorem A.
Theorem 1.1 Let F be a family of holomorphic functions in a domain D; let Q(z) be a polynomial of degree q > 1, and let k > 2q +1 be an integer. If, for each f Є F , we have
f (z) = Q(z) ^ f(z) = Q(z) ^ f(k) = Q(z),
then F is normal in D.
In order to prove theorem 1.1, we need the following results, which are interesting in their own rights.
Proposition 1. Let F be a family of holomorphic functions in a domain D; let h(z) be a polynomial of degree q > 1; let k > q be an integer. If, for each f Є F, we have h(z) = 0 ^ f(z) = 0 and f(z) = 0^ f'(z) = h(z) =>- | f^{z)\ < M, where M is a positive number, then F is normal in D.
The author was supported by the NSF of China (10771121), the NSF of Guangdong Province (9452902001003278) and Excellent Young Fund of Department of Education of Guangdong (LYM08097).
Proposition 2. Let F be a family of holomorphic functions in a domain D; let Q(z) be a polynomial of degree q > 1; let k > 2q +1 be an integer. If, for each f G F, we have Q(z) — Q'(z) = 0 ^ f (z) = 0 and f (z) = 0 ^ f (z) = Q(z) — Q'(z) ^ f (k\z) = Q(z), then F is normal in D.
2 Some Lemmas
Lemma 2.1 [9] Let F be a family of functions meromorphic in a domain D, all of whose zeros have multiplicity at least k, and suppose that there exists A > 1 such that |f(k')(z)\ < A whenever f (z) =0, if F is not normal at z0 G D, then for each 0 < a < k there exist,
(a) points zn G D, zn ^ z0;
(b) functions fn G F, and
(c) positive number pn ^ 0 such that p-a fn(zn + pn() = gn(() ^ g(() locally uniformly, where g is a nonconstant meromorphic function in C, all of whose zeros have multiplicity at least k, such that g^(() < g^(0) = kA+1. In particular, if F is a family of holomorphic functions, then p(g) < 1.
Lemma 2.2 [2] Let g be a nonconstant entire function with p(g) < 1; let k > 2 be a positive integer; and let a be a nonzero finite value. If g(z) =0 ^ g'(z) = a, and g'(z) = a ^ g(k\z) = 0, then g(z) = a(z — z0), where z0 is a constant.
Lemma 2.3 [2] Let F be a family of holomorphic functions in a domain D; let k > 2 be a positive integer; and let a be a function holomorphic in D, such that a(z) = 0 for z G D. If for every f G F, f (z) =0 ^ f'(z) = a(z) and f'(z) = a(z) ^ \f(k\z)\ < h, where h is a positive
number, then F is normal in D.
In order to prove theorem 1.1, we need some definitions.
Let A = {z : \z\ < r0}, let Q(z) be a polynomial of degree q > 1 and R(z) = Q(z) — Q'(z) =
zmP(z), P(z) = 0, when z G A. Define that Qa(z) = Q(z + a), where a is a constant, then
Ra(z) = Qa(z) — Q'a(z) = (z + a)mPa(z).
Define Xa = f jRa and Aa(0) ^ 0, where / is holomorphic function in A. Thus we get f = \af + Ra = \a\f + ^a1. By mathematic induction we get f(k) = \akf + ^ak (k > q + 2), where
^ak = Ra{Xka 1 + Pk-2[K]} + 2 + Pk-3[K]} + ■■■ + ^^ {^ + Pk-(q+2)[K]} (2.1)
and Pk-2[Xa],..., Pk-(q+2)[Aa] are differential polynomial in Xa with degree at most k — 2,..., k — (q + 2) respectively. Let ^ak(0) — Qa(0) = 0. Define Wa(0) = 0 where
(2.2)
Define <£a(0) = 0 where
— — [1 + (-j-)'Qa + —Q'a]Ra-------------QaR'a• (2-3)
Wa Wa Wa
Lemma 2.4 Let f (z) be analytic in the disc A = {z : \z\ < r0}; let a be a complex number
such that \a\ < r0; let k > q + 2 be a positive integer. If Qa, Ra, Xa, ^ak, Wa and pa are defined
as above; if f (0) = 0, \ f — Ra\z=o = 0, Ra = 0 ^ f (z) = 0 and
f (z) = 0 = f (z) = Ra ^ f(k)(z) = Qa,
then
f R
T(r, /) < LD[r, f] + Mlog |— -------i_U + iog|/(0)|, (2.4)
WaPa^ak — Qa)f
where
f f(k) f(k) f R f(k)
LD[r, f] = ,\/| uiir. —) + m(r, ——) + m(r, + m(r, —--------------------------------f-) + m(r, —-------—)]
f f f f — Ra f — Ra
a
(k-2)
f(k+1) W' \'
+ ,\/1///(/'• —----—) + ,\/2 mir. -p) + m(r, ) + ... + m(r, ——)]
f' — Ra Wa Xa Xa
+ M3[m(r, Ra) + m(r, R'a) + ■ ■ ■ + m(r, R(aq)) + m(r, Qa) + m(r, Q'a) + log 2], and M1, M2, M3 are positive numbers.
Lemma 2.5 [1] Let U(r) be a nonnegative, increasing function on an interval [R1,R2](0 < R1 < R2 < +ro); let a, b be two positive constants satisfying b > (a + 2)2; and let
U(r) < a{log + U(p) + log —-—} + b
p — r
whenever R1 < r < p < R2. Then, for R1 < r < R2,
R2
U(r) < 2alog —------------h 2b.
R2 — r
Lemma 2.6 [1] Let g(z) be a transcendental entire function. Then
limsup \z\g^(z) = +ro.
|z | —
3 Proof of Proposition 1
Let z0 G D. If h(z0) = 0, by Lemma 2.3, F is normal at z0. Now suppose that h(z0) = 0. Without loss of generality, we may assume that z0 = 0, A = {z : \z\ < $} G D and h(z) = zmb(z), where b(0) = 1 and b(z) = 0 (z G A). We shall prove that F is normal at z = 0.
Let = {F = : / G ?}. We know that if is normal at c = 0, then ? is normal at
z = 0. Thus we only need to prove F1 is normal at z = 0.
For each f G F, from h(z) = 0 ^ f (z) = 0, we get z = 0 is a zero of f. Thus we have
f (z) = a,nzn + an+1zn+1 + ... (an = 0) (n > 1),
and
f (z) — h(z) = nanzn 1 + (n + 1)an+1zn + ... — (zm + ...)
By the assumption f (z) = 0 ^ f (z) = h(z), we get
/ = —^m+1 + am+2zm+2 + ... (3.1)
m + 1
Hence we get is a family of holomorphic. functions in A. Next we prove VF = ^ 6 J i, F = 0 ^ \F'\ < M, where M = maxz&A \b(z)\ > 1.
Suppose that F(a0) = 0, then f (a0) = 0.
If a0 ^ 0, we get F'(a0) = - ”'m+i) = Kao)-
a0 a0
If a0 = 0, we get F'(a0) = b(a0) - ^ = 1 - ^ Thus we get F = 0 => |F'| < M.
Now we prove that F1 is normal at z = 0. Suppose on the contrary that F1 is not normal
at z = 0, then by Lemma 2.1, we can find zn ^ 0, pn ^ 0 and fn G F such that
-1 fn(zn + pnC) /o o\
*(0 = A* (*. + *.0»"9(<) (X2)
locally uniformly on C, where g is a nonconstant entire function such that g^(() < g^(0) = M +1. In particular p(g) < 1. Without loss of generality we assume that lim — = c E C. In the
n—O Pn
following we consider two cases.
Case 1: c = oo. Then zn ^ 0 and ► 0 as ??. —► oo. Set hn(() = p~1 _ Then by
zn zn
(3.2), we get
7 />\ -1 fn(zn + pn( pn (o o\
MC) = Pn , nm(1 + —0 ^g(0- (3-3)
(zn + pnS ) zn
We claim:
g(C) =0 ^ g (C) = 1 and g (C) = 1 ^ g(k\() =0.
Suppose that g((0) = 0, then by Hurwitz’s Theorem, there exist (n, (n ^ (0, such that (for n sufficiently large)
7 ,, \ -1 fn(zn + pnCn) n
hn(Cn) = pn ----------------------------——— = 0.
zn
Thus fn(zn + PnCn) = 0, by the assumption we have fn(zn + Pn(n) = (zn + PnCn)mb(z,n + pn(n), then we derive that
g\C0) = lim + Pn^ = lim b(zn + p„<„)(l + —Ci)m = &(0) = 1.
n—o zm n—o z^
nn
Thus g(() = 0 ^ g'(() = 1. Next we prove g'(() = 1 ^ g(k)(() = 0. By (3.3) we know
g'«)
fn(zn + pnZ ) __ fn(zn + pnC )
(~n + pnOmb{Zn + PnC) ^(1 + tO mb(Zn + PnC)
We suppose that g'(Co) = 1, obviously g' ^ 1, for otherwise g^(0) < g'(0) = 1 < M +1, which is a contradiction. Hence by Hurwitz’s Theorem, there exist (n, (n ^ (0, such that (for n sufficiently large)
______fn(-n ~l~ PnCn)______ _ ^
(zn PnCn)m^(~'n PnCn)
Thus f'2(zn + pn(n) = h(zn + PnÇn), by the assumption we get | fn\zn + pn(n)I < M. Then
\9{k)((o)\ = lim \^fik){zn + Pn(n)\ < lim = 0.
n^œ zm n^œ zm
^n ^n
Thus we prove the Claim. By Lemma 2.2, we get g = Z — b, where b is a constant. Thus we have g^(0) < 1 < M +1, which is a contradiction.
Case 2: c = to. We set
G««) = %T- (3-4)
pn
Then
fJZn + PniC — — ) )
G„(C) = + - *(C - *- - G(0
We know that z = 0 is zero of fn with multiplicity m +1, then we get 0 is a zero of G(Z) with
multiplicity m +1 and
G(m+1)(0) = lim G^ (0) = m! (3.5)
If G'(() = (m, we derive that G(() = . Hence we obtain g(() = ^qr¡;(( + c). It
follows that g^(0) < < M+ 1, a contradiction. Thus G'(() ^ (m. Using the same argument
as in the proof of Case 1, we get
, , (G{k)(()<M, k = m+l,
G(Z) = 0 & G'(Z) = Zm and G (Z) = Zm n
y > y > y > y G(k)(Z)=0, k > m +2.
Suppose G(Z) is a polynomial. Let
G(Z )= bqZ q + bq-1Z q 1 + ■■■ + bm+1Z m+1 (bm+1 = 0) ■ (3.6)
From G(Z) = 0 & G'(Z) = Zm, we get
G(Z ) = Z (G' (Z ) — Z m)A. (3.7)
Thus, by (3.6) and (3.7) we have G(() = bq(q — g_(„\+1)(m+1 (q > m + 2) or G(() = A(m+1,
and from (3.5), we get G(() = ^qr[(m+1- Then G'(() = (m, a contradiction.
In the following we assume that G(Z) is a transcendental entire function.
Let us consider the family T = {tn : t.n(() = m+iÍi }; we see that tn is a entire function satisfying
f t.n(() <M, k = m + 1, tn(Z) = 0 & tn(Z) = Zm ^ ^ zx
n \tn(Z)=0, k > m +2.
By Lemma 2.3, we have T is normal on D1 = {Z : (1/2)m < |Z | < 2m}, thus there exists a M1 satisfying
' ,, = (2"‘)'”‘+2>"|G'((2”‘)"C)|
^9mj2(m+l)n _|_ j ((^(9m)ra(^) |2 — ^
Set r(z) = then r(z) is a transcendental entire function. We know that for each z £ C,
there exists a integer n such that z = (2m)nZ, where (1/2)m < |Z| < 2m. We can get
\z\r\z) < (2m)3m+44(C) + < (2m)3m+4M! + ^1. (3.8)
From Lemma 2.6, we get
limsup |z|H(z) = +to,
lzl^-ж
which contradicts with (3.8).
Thus, we prove that F\ is normal at z = 0. Hence F is normal at z = 0.
4 Proof of Proposition 2
Let z0 E D. If [Q(z) — Q'(z)]lz=z0 = 0, by Lemma 2.3, F is normal at z0. Now suppose that [Q(z) — Q'(z)]lz=z0 = 0. Without loss of generality, we may assume that z0 = 0, A = {z : Izl < 8} E D and R(z) = Q(z) — Q'(z) = zmP(z), where P(z) = 0 (z E A). We shall prove that F is normal at z = 0.
Suppose on the contrary that F is not normal at z = 0, then by Lemma 2.1, we can find zn ^ 0, pn ^ 0 and fn E F such that
9n(C) = fn(zn + PnC) ^ 9(Z) (4.1)
locally uniformly on C, where 9 is a nonconstant entire function. Without loss of generality, we assume that
lim — = с. E C.
n^^ Pn
First, we shall prove that g(Z) is a transcendental entire function. In fact, we only need to prove
that g(Z) = 0. The argument given in the proof of Proposition 1 shows that
9(Z) = 0 ^ 9'(Z) = 0,
thus 9 only has multiple zeros. Suppose (0 is a zero of 9(Z) with multiplicity s(> 2), then 9(s\(o) = 0. Thus there exists a positive number 8, such that
9(Z) = 0, 9'(Z) = 0, 9(s)(Z) = 0 (4.2)
on DO = {Z : 0 < IZ — ZoI < 8}. By (4.1) and Rouche theorem, there exist Zn,j(j = 1, 2,..., s) on
D5/2 = {Z : IZ — Zo I < 8/2} such that
9'n(Cn,j) fn(zn + pnZn,j) 0 (j 1, l2, ■■■, s)-
It follows from R(z) = 0 ^ f(z) = 0 and f(z) = 0 ^ f'(z) = R(z) that fn(zn + pnZn,j) =
R(zn + PnZn,j) = 0. Thus
9n(Zn,j) = pnfn(zn + pnZn,j) = pnR(zn + pnZn,j) = 0(j = 1, 2, ..., s),
so each Zn,j is a simple zero of 9n(Z), that is Zn,j = Zni(1 — i = j — s). On the other hand
lim 9n(Znj) = lim pnR(zn + PnZn,j) = 0
From (4.2), we get
lim Zn,j = Zo (j = 1, 2,...,s).
Noting that (4.2) and g'n(Z) — pnR(zn + pn() has s zeros Zn,j(j = 1, 2,s) in D$/2, then Zo is a zero of g'(Z) of multiplicity s, and thus g(s\Zo) = 0. This is a contradiction. Hence g(Z) = 0 and g(Z) is a transcendental entire function.
Now we consider five cases.
Case 1: There exist infinitely many {nj} such that
(znj + Pnj C) = R(znj + Pnj C).
It follows that g'n.(Z) = pnjR(znj + pnjZ). Let j ^ to, we deduce that g'(Z) = 0, which contradicts that g is transcendental.
RfQ f/
Case 2: There exist infinitely many {/?,.,•} such that 4)nj{znj+ pnjZ) = 0) where t^n = / •
Thus we have
,(k) nj
i I i gnj (Z ) _________ yo ( i gnj (Z )
( Znj + Pnj Z) P\ znj + Pnj Z ) I = Q ( ZUj + Pnj Z )
pnj pnj
and
gS(Z ) Q(zn, + Pn, Z )pnk-,m+'n
<4(0 + p,,,0(^- + <)”>'
Noting that k > 2q +1 > 2m + 1, let j ^ to, we deduce that g(k\Z) = 0, which contradicts that g is transcendental.
Case 3: There exist infinitely many {nj} such that <£nj (znj + pnjZ) = 0, where
ipn = — [i + (+ ~j~Q']R ——QR'.
Wn Wn Wn
and Wn is defined as above. Let
'£i V,
zz
r(0 = Æ(”‘+1’[(— + Cr^PiU,, + p„:()gii’(C) + (— + crp + A„C)ft(‘:+ll(C)
Then
— pn,j mQ' (znj + Pnj Z )g'nj (Z ) — pn,j {m+l)Q(znj + Pnj Z )gj (Z )]•
r(C)Q(-.., + pn,0 +Q'(zni+pni()pi7-
+ OmP(znj + PnjOgnjiO - Qi^ij + PnjOpnj {m+1)9hAC)
-|_ (IÜL -|- . -f p. Q(znj + PnjQPnj ( Pl(znj + PnjO
Pnj J J 9nj(0 (+ () P (znj + pnj Z)
where R'(z) = zm-1Pi(z).
Thus, let j ^ to, we get g(k\Z) = 0, which contradicts that g is transcendental.
Case 4: There exist infinitely many {nj} such that pknj (znj + pnjZ) = Q(znj + pnjZ) where
Pkn = R{Xkn 1 + Pk-2[Xn]} + R {x!n 2 + Pk-3[Xn]} + ■■■ + R(q^ {x!n (q+1) + Pk-(q+2)[Xn]},
and Xn = fn~R. Thus, let j ^ to, we get
,9_}k-{m+l)r,c + Qmpp + = Q.
gg
n
j
Hence д' = 0 or (с + ()тР(0)(^)т + і?^(0) = 0, which contradicts that д is a transcendental entire function.
Case 5: There exist finitely many {nj } such that fj (znj + pnjZ) = R(znj + pnj Z), Wnj (znj + Pnj Z) = 0, <£nj (Znj + Pnj Z) = 0 and pknj (Znj + Pnj Z) = Q(Znj + Pnj Z).
For all n we may suppose that fnj (znj + Pnj Z) Ф R(znj + Pnj Z), Ф^ (znj + Pnj Z) Ф 0,
Pnj (znj + Pnj Z) Ф 0 and Pknj (znj + Pnj Z) Ф Q(znj + Pnj Z).
Take Zo Є C such that g(j')(Zo) = 0 (j = 0,1, ...,k). In case c = to, choose Zo to satisfy the
additional conditions that Z0 = —c and
(с + СоГР(0)(4ттГ + я(т)(0) ^0.
g(Zo)
Noting that k > 2q +1 > 2m + 1, this facts imply that Kn =^ 0 as n ^ to, so that logKn ^ —to as n ^ to.
For n = 1, 2, 3,..., put
hn(z) = fn(zn + PnZo + z)
Since zn + PnZ0 ^ 0 as n ^ to, it follows that (for sufficiently large n) hn is defined and holomorphic on |c| < Denote
an zn + P'nZ0-
Then, for sufficiently large n, hn(0) = 0, h'n(0) — Ran (0) = 0. By the assumption we get
hn(z) = 0 ^ h'n(z) = Ran ^ hin)(z) = Qan.
Let a = an and f (z) = hn(z) in Lemma 2.4, then we get
hn( an) — fn(0) = 0,гфап (0) — Wn(an) = °> ^a„ (0) — ^n(an) = °>
[pa„k — Qa„ ] lz=0 [pkn — Q]|z=an = 0,
thus hn(z) satisfies the assumption of Lemma 2.4.
Now applying Lemma 2.4 with r0 = |, and noting that the last three terms in (2.4) are bounded for 0 < r < 1/3, we obtain that, for sufficiently large n and 0 < r < 1/3,
h' hnk) hnk) h" — R'
T(r, hn) < А/, ///ir. -?) + m(r, ^-) + m(r, ^-) + m(r, J-----------------------------------^) + m(r, ' „ )]
hn hn h'n h'n — Ran h'n — Ran
h(k+l) ф' \' \(k-2)
+ '/,/,/ir. —————) + M2[m(r, ^) + m(r, -^) + ... + m(r, ^—)].
hn — Ran ^Pan Xan Xan
We can obtain, for 0 < r <t < 1/3,
T(r, hn) < Ck{ 1 + log+- + log ----------------b log +T(r, hn)
r t — r (4.3)
+ log +T (T,h'n)+ log +T(T,^n ) + log +T(T,Xan ) }.
Observe that T(r,h!n) = т{т,Нп) < т(т,Ьп) + ?n(r, ^2L), hence for 1/4 < r < p < 1/3 with t = (r + p)/2. From the above we obtain
T(r, hn) < Ck( 1 + log------h log+T(p, hn)).
P — r
By Lemma 2.5 it then follows that T(1/4,hn) < A, where A is a constant independent of n. Thus fn(z) is uniformly bounded for sufficiently large n and Izl < 1/8. However, from Pnfn(zn + PnZo) = g'n(Zo) ^ g"(Zo) = 0 we see that f (z) cannot bounded in Izl < 1/8. This is a contradiction, so the proof is complete.
5 Proof of Theorem 1.1
Let S = {g = f — Q : f G F} and R(z) = Q(z) — Q'(z). Obviously, S is normal in D if and only if F is normal in D. It follows from our assumption that, for any g G S, we have
g = 0 ^ g' = R ^ g(k) = Q. (5.1)
Let z0 G D. Now we prove that S is normal at z0. Let {gn} C S be a sequence.
If R(z0) = 0, then there exists a positive number b such that As = {z G D : lz—z0I < b} C D
and R(z) = 0 in As. Then by Lemma 2.3, {gn} is normal at z0.
If R(z0) = 0, then there exists a positive number b such that As = {z G D : lz—z0I < b} C D
and R(z) = 0 in As\{z0}. Suppose {gn} has a subsequence say, without loss of generality, itself, such that gn(z0) = 0, then {gn} is normal at z0 by Proposition 1. Suppose gn(z0) = 0 for all
but finite many of {gn}, then {gn} is normal at z0 by Proposition 2.
Thus F is normal in D and hence Theorem 1.1 is proved.
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НОРМАЛЬНОЕ СЕМЕЙСТВО И РАСПРЕДЕЛЕННЫЕ МНОГОЧЛЕНЫ МЕРОМОРФНЫХ ФУНКЦИЙ
Фенг Лю1', Жан Фенг Ксю2)
1 Китайский Нефтяной Университет,
Донгиин, Шаньдун 257061, Китай, e-mail: [email protected] 2 Вууй Университет,
Цзянмэнь, Гуангдонг 529020, Китай, e-mail: [email protected]
Аннотация. В работе изучается единственность и разделение неподвижной точки мероморф-ных функций. Доказаны две основные теоремы, улучшающие результаты Ванга и Кью.
Ключевые слова: мероморфная функция, неподвижная точка, распределенные мно-
гочлены.
