Value-sharing and uniqueness of entire functions
We study the uniqueness of entire functions sharing a nonzero value and obtain some results improving the results obtained by Fang, J. F. Chen, X.Y. Zhang and W. C. Lin, et al.
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irk-123456789-1663212020-02-19T01:27:56Z Value-sharing and uniqueness of entire functions Wu, Chun Статті We study the uniqueness of entire functions sharing a nonzero value and obtain some results improving the results obtained by Fang, J. F. Chen, X.Y. Zhang and W. C. Lin, et al. Вивчається єдиність цілих функцій, що поділяють ненульове значення. Отримано дєякі результати, що поліпшують результати Фанга, Дж. Ф. Чена, С. Й. Жанга, В. Ц. Ліна та інших. 2014 Article Value-sharing and uniqueness of entire functions / Chun Wu // Український математичний журнал. — 2014. — Т. 66, № 12. — С. 1705–1717. — Бібліогр.: 19 назв. — англ. 1027-3190 http://dspace.nbuv.gov.ua/handle/123456789/166321 517.5 en Український математичний журнал Інститут математики НАН України |
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Статті Статті Wu, Chun Value-sharing and uniqueness of entire functions Український математичний журнал |
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We study the uniqueness of entire functions sharing a nonzero value and obtain some results improving the results obtained by Fang, J. F. Chen, X.Y. Zhang and W. C. Lin, et al. |
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Value-sharing and uniqueness of entire functions |
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Value-sharing and uniqueness of entire functions |
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Value-sharing and uniqueness of entire functions |
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Value-sharing and uniqueness of entire functions |
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Value-sharing and uniqueness of entire functions |
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value-sharing and uniqueness of entire functions |
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Інститут математики НАН України |
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Value-sharing and uniqueness of entire functions / Chun Wu // Український математичний журнал. — 2014. — Т. 66, № 12. — С. 1705–1717. — Бібліогр.: 19 назв. — англ. |
| series |
Український математичний журнал |
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AT wuchun valuesharinganduniquenessofentirefunctions |
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2025-07-14T21:08:54Z |
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2025-07-14T21:08:54Z |
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1837658103678500864 |
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UDC 517.5
Chun Wu (College Math. Sci., Chongqing Normal Univ., China)
VALUE-SHARING AND UNIQUENESS OF ENTIRE FUNCTIONS*
ПОДIЛ ДАНИХ ТА ЄДИНIСТЬ ЦIЛИХ ФУНКЦIЙ
We study the uniqueness of entire functions sharing a nonzero value and obtain some results improving the results due to
Fang, J. F. Chen, X. Y. Zhang and W. C. Lin et al.
Вивчається єдинiсть цiлих функцiй, що подiляють ненульове значення. Отримано деякi результати, що полiпшують
результати Фанга, Дж. Ф. Чена, С. Й. Жанга, В. Ц. Лiна та iнших.
1. Introduction and main results. In this paper, a meromorphic function will mean meromorphic
in the whole complex plane. We will use the standard notations of Nevanlinna’s value distribution
theory such as T (r, f), N(r, f), N(r, f), m(r, f) and so on, as explained in Hayman [1], Yang [2]
and Yi and Yang [3]. We denote by S(r, f) any function satisfying S(r, f) = o(T (r, f)), as r →∞
possibly outside a set r of finite linear measure.
Let a be a finite complex number, and k be a positive integer. We denote by Ek)(a, f) the set of
zeros of f − a with multiplicities at most k, where each zero is counted according to its multiplicity.
We denote by Ek)(a, f) the set of zeros of f − a with multiplicities are not greater than k, where
each zero is counted only once. In addition, we denote by N(k
(
r,
1
f − a
)(
or N (k
(
r,
1
f − a
))
the
counting function with respect to the set Ek)(a, f)
(
or Ek)(a, f)
)
.
Set
Nk
(
r,
1
f − a
)
= N
(
r,
1
f − a
)
+N (2
(
r,
1
f − a
)
+ . . .+N (k
(
r,
1
f − a
)
.
We define
Θ(a, f) = 1− lim
r→∞
N
(
r,
1
f − a
)
T (r, f)
and
δk(a, f) = 1− lim
r→∞
Nk
(
r,
1
f − a
)
T (r, f)
.
Let f and g be two nonconstant meromorphic functions, a be a finite complex number. We say
that f, g share the value a CM (counting multiplicities) if f, g have the same a-points with the same
multiplicities and we say that f, g share the value a IM (ignoring multiplicities) if we do not consider
the multiplicities. We denote by NL
(
r,
1
f − a
)
the counting function for a-points of both f and g
about which f has larger multiplicity than g , with multiplicity not be counted. Similarly, we have
the notation NL
(
r,
1
g − a
)
. Next we denote by N0(r, 1/F
′) the counting function of those zeros of
F ′ that are not the zeros of F (F − 1).
* This work is supported by Chongqing City Board of Education Science and technology project (KJ130632) and the
found of Chongqing Normal University (13XLB024).
c© CHUN WU, 2014
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 12 1705
1706 CHUN WU
Definition 1.1. Let k be a positive integer. Let f and g be two nonconstant meromorphic
functions such that f and g share the value 1 IM. Let z0 be a 1-point of f with multiplicity p and
a 1-point of g with multiplicity q. We denote by Nf>k
(
r,
1
g − 1
)
the reduced counting function of
those 1-points of f and g such that p > q = k. Ng>k
(
r,
1
f − 1
)
is defined analogously.
Definition 1.2. Let f and g satisfy Ep)(1, f) = Ep)(1, g), where p ≥ 2 is an integer. We denote
by N (p+1(r, 1; f |g 6= 1) the reduced counting function of those 1-points of f with multiplicities at
least p+ 1, which are not the 1-points of g. Also N(p+1(r, 1; g|f 6= 1) is defined analogously.
In [4], Fang got the following results.
Theorem A. Let f and g be two nonconstant entire functions and n, k be two positive integers
with n > 2k+ 4. If (fn)(k) and (gn)(k) share 1 CM, then either f(z) = c1e
cz, g(z) = c2e
−cz, where
c1, c2 and c are three constants satisfying (−1)k(c1c2)
n(nc)2k = 1 or f = tg for a constant t such
that tn = 1.
Theorem B. Let f and g be two nonconstant entire functions and n, k be two positive integers
with n > 2k + 8. If
(
fn(f − 1)
)(k)
and
(
gn(g − 1)
)(k)
share 1 CM, then f = g.
In 2008, J. F. Chen, X. Y. Zhang [5] improved the above result and obtained the following result.
Theorem C. Let f and g be two nonconstant entire functions and n, k be two positive integers
with n > 5k + 7. If (fn)(k) and (gn)(k) share 1 IM, then either f(z) = c1e
cz, g(z) = c2e
−cz, where
c1, c2 and c are three constants satisfying (−1)k(c1c2)
n(nc)2k = 1 or f = tg for a constant t such
that tn = 1.
Theorem D. Let f and g be two nonconstant entire functions and n, k be two positive integers
with n > 5k + 13. If
(
fn(f − 1)
)(k)
and
(
gn(g − 1)
)(k)
share 1 IM, then f = g.
In 2008, X. Y. Zhang, J. F. Chen and W. C. Lin [6] extended the above result by proving the
following result.
Theorem E. Let f(z) and g(z) be two entire functions, n, m and k be three positive integers
with n > 3m + 2k + 5, and P (z) = amz
m + am−1z
m−1 + . . . + a1z + a0 or P (z) = C, where
a0 6= 0, a1, . . . , am−1, am 6= 0, C 6= 0 are complex constants. If
[
fnP (f)
](k)
and
[
gnP (g)
](k)
share 1 CM, then either f(z) = c1e
λz, g(z) = e−λz, where λ1, λ2, λ are three constants satisfying
(−1)k(λ1λ2)
n(nλ)2kC2 = 1, or f(z) and g(z) satisfy the algebraic equation R(f, g) = 0, where
R(ω1, ω2) = ωn1 p(ω1)− ωn2 p(ω2).
In this paper we always use L(z) denoting a arbitrary polynomial of degree n, i.e.,
L(z) = anz
n + an−1z
n−1 + . . .+ a0 = an(z − c1)l1(z − c2)l2 . . . (z − cs)ls , (1.1)
where ai, i = 0, 1, . . . , n, an 6= 0 and cj , j = 1, 2, . . . , s, are finite complex number constants, and
c1, c2, . . . , cs are all the distinct zeros of L(z), l1, l2, . . . , ls, s, n are all positive integers satisfying
l1 + l2 + . . .+ ls = n and let l = max{l1, l2, . . . , ls}. (1.2)
Corresponding to the above results, some authors considered the uniqueness problems of entire
functions that have fixed points, see M. L. Fang and H. Qiu [7], W. C. Lin and H. X. Yi [8], J. Dou,
X. G. Qi and L. Z. Yang [9]. In this paper, we consider the existence of solutions of [L(f)](k) − P
and the corresponding uniqueness theorems, and we obtain the following results which generalize the
above theorems.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 12
VALUE-SHARING AND UNIQUENESS OF ENTIRE FUNCTIONS 1707
Theorem 1.1. Let f be a transcendental entire function. If n > k + s, then [L(f)](k) = P has
infinitely many solutions, where P 6≡ 0 is a polynomial.
Remark 1.1. It is easy to see that a polynomial Q(z)− P (z) has exactly max{m,n} solutions
(counting multiplicities), where degQ = m, degP = n, but a transcendental entire function may have
no solution. For example, let f(z) = eα(z) + P (z), then function f(z)− P (z) has no any solution,
where α(z) is an entire function.
Theorem 1.2. Let f(z) and g(z) be two nonconstant entire functions and n, k, l and p(≥ 2) be
four positive integers satisfying 5l > 4n + 5k + 7. If Ep)
(
1, (L(f))(k)
)
= Ep)
(
1, (L(g))(k)
)
, then
either f = b1e
bz + c, g = b2e
−bz + c or f and g satisfy the algebraic equation R(f, g) ≡ 0, where
b1, b2, b are three constants satisfying (−1)k(b1b2)
n(nb)2k = 1 and R(ω1, ω2) = L(ω1)− L(ω2).
Corollary 1.1. Let f(z) and g(z) be two nonconstant entire functions and n, k and l be three
positive integers satisfying 5l > 4n + 5k + 7. If
[
L(f)
](k)
and
[
L(g)
](k)
share 1 IM, then either
f = b1e
bz + c, g = b2e
−bz + c or f and g satisfy the algebraic equation R(f, g) ≡ 0, where b1, b2, b
are three constants satisfying (−1)k(b1b2)
n(nb)2k = 1 and R(ω1, ω2) = L(ω1)− L(ω2).
Remark 1.2. When l = n, l = n−1, c = 0, from Corollary 1.1 we can easily get Theorems C, D.
Corollary 1.2. Let f(z) and g(z) be two nonconstant entire functions and n, k and l be three
positive integers satisfying 2l > n + 2k + 4. If
[
L(f)
](k)
and
[
L(g)
](k)
share 1 CM, then either
f = b1e
bz + c, g = b2e
−bz + c or f and g satisfy the algebraic equation R(f, g) ≡ 0, where b1, b2,
b are three constants satisfying (−1)k(b1b2)
n(nb)2k = 1 and R(ω1, ω2) = L(ω1)− L(ω2).
Remark 1.3. When l = n, c = 0, from Corollary 1.2 we can easily get Theorem A. When
l = n− 1, l = n−m, c = 0, Corollary 1.2 promotes Theorems B, E.
Remark 1.4. If L(f) ≡ L(g), we obtain
anf
n + an−1f
n−1 + . . .+ a1f ≡ angn + an−1g
n−1 + . . .+ a1g.
Let h = f/g. If h is a constant, then substituting f = gh into above equation we deduce
ang
n(hn − 1) + an−1g
n−1(hn−1 − 1) + . . .+ a1g(h− 1) ≡ 0,
which implies hd = 1, d = (n, . . . , n− i, . . . , 1), an−i 6= 0 for some i = 0, 1, . . . , n−1. Thus f ≡ tg
for a constant t such that td = 1. If h is not a constant, then we know by above equation that f and
g satisfy the algebraic equation R(f, g) ≡ 0, where R(ω1, ω2) = L(ω1)− L(ω2).
Remark 1.5. Moreover, let L(z) is a generic polynomial of degree at least 5. Then from the
equation L(f) ≡ L(g), one can conclude that f ≡ g and no other nonconstant meromorphic solutions
f and g. In [15] Yang and Hua exhibits some classes of such polynomials. And some related definitions
and results, we refer the reader to [16, 17].
2. Some lemmas.
Lemma 2.1 (see [1]). Let f(z) be a nonconstant meromorphic function and a1(z), a2(z) be two
meromorphic functions such that T (r, ai) = S(r, f), i = 1, 2. Then
T (r, f) ≤ N(r, f) +N
(
1
f − a1
)
+N
(
r,
1
f − a2
)
+ S(r, f).
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 12
1708 CHUN WU
Lemma 2.2 (see [1]). Let f be a nonconstant meromorphic function, k be a positive integer,
and c be a nonzero finite complex number. Then
T (r, f) ≤ N(r, f) +N
(
r,
1
f
)
+N
(
r,
1
f (k) − c
)
−N
(
r,
1
f (k+1)
)
+ S(r, f) ≤
≤ N(r, f) +Nk+1
(
r,
1
f
)
+N
(
r,
1
f (k) − c
)
−N0
(
r,
1
f (k+1)
)
+ S(r, f),
where N0
(
r,
1
f (k+1)
)
denotes the counting function which only counts those points such that
f (k+1) = 0 but f(f (k) − c) 6= 0.
Lemma 2.3 (see [11]). Let an(6= 0), an−1, . . . , a0 be constants and f be a nonconstant mero-
morphic function, then
T
(
r, anf
n + an−1f
n−1 + . . .+ a0
)
= nT (r, f).
Lemma 2.4 (see [12]). Let f(z) be a nonconstant meromorphic function, s, k be two positive
integers. Then
Ns
(
r,
1
f (k)
)
≤ kN(r, f) +Ns+k
(
r,
1
f
)
+ S(r, f).
Clearly, N
(
r,
1
f (k)
)
= N1
(
r,
1
f (k)
)
.
Lemma 2.5 (see [13]). Let f be a nonconstant meromorphic function, k(≥ 1) be a positive
integer and let ϕ( 6≡ 0,∞) be a small function of f. Then
T (r, f) ≤ N(r, f) +N
(
r,
1
f
)
+N
(
r,
1
f (k) − ϕ
)
−N
(
r,
1(
f (k)/ϕ
)′
)
+ S(r, f). (2.1)
Lemma 2.6 (see [14]). Let f(z) be a nonconstant meromorphic function and k be a positive
integer. Suppose that f (k) 6≡ 0, then
N
(
r,
1
f (k)
)
≤ N
(
r,
1
f
)
+ kN(r, f) + S(r, f).
Lemma 2.7 (see [18]). Let f, g share (1, 0). Then
(i) Nf>1
(
r,
1
g − 1
)
≤ N
(
r,
1
f
)
+N(r, f)−N0
(
r,
1
f ′
)
+ S(r, f),
(ii) Ng>1
(
r,
1
f − 1
)
≤ N
(
r,
1
g
)
+N(r, g)−N0
(
r,
1
g′
)
+ S(r, g).
Lemma 2.8. Let f(z) and g(z) be two nonconstant entire functions, k, p(≥ 2) be two positive
integers. If Ep)(1, f
(k)) = Ep)(1, g
(k)) and
∆ = 2δk+1(0, g) + δk+1(0, f) + δk+2(0, g) + δk+2(0, f) > 4,
then
1
f (k) − 1
=
bg(k) + a− b
g(k) − 1
, where a, b are two constants.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 12
VALUE-SHARING AND UNIQUENESS OF ENTIRE FUNCTIONS 1709
Proof. Let
Φ(z) =
f (k+2)
f (k+1)
− 2
f (k+1)
f (k) − 1
− g(k+2)
g(k+1)
+ 2
g(k+1)
g(k) − 1
. (2.2)
Clearly m(r,Φ) = S(r, f) + S(r, g). We consider the cases Φ(z) 6≡ 0 and Φ(z) ≡ 0.
Let Φ(z) 6≡ 0, then if z0 is a common simple 1-point of f (k) and g(k), substituting their Taylor
series at z0 into (2.2), we see that z0 is a zero of Φ(z). Thus, we have
N11
(
r,
1
f (k) − 1
)
=
= N11
(
r,
1
g(k) − 1
)
≤ N
(
r,
1
Φ
)
≤ T (r,Φ) +O(1) ≤ N(r,Φ) + S(r, f) + S(r, g). (2.3)
Noting that Ep)(1, f
(k)) = Ep)(1, g
(k)), we deduce from (2.2) that
N(r,Φ) ≤ N (p+1
(
r, 1; f (k)|g(k) 6= 1
)
+N (p+1
(
r, 1; g(k)|f (k) 6= 1
)
+
+N (2
(
r,
1
f (k)
)
+N (2
(
r,
1
g(k)
)
+N0
(
r,
1
f (k+1)
)
+N0
(
r,
1
g(k+1)
)
+
+NL
(
r,
1
f (k) − 1
)
+NL
(
r,
1
g(k) − 1
)
. (2.4)
Here N0
(
r,
1
f (k+1)
)
has the same meaning as in Lemma 2.2. From Lemma 2.2, we obtain
T (r, g) ≤ Nk+1
(
r,
1
g
)
+N
(
r,
1
g(k) − 1
)
−N0
(
r,
1
g(k+1)
)
+ S(r, g). (2.5)
Since
N
(
r,
1
g(k) − 1
)
= N11
(
r,
1
g(k) − 1
)
+N (2
(
r,
1
f (k) − 1
)
+
+Ng(k)>1
(
r,
1
f (k) − 1
)
+N (p+1(r, 1; g(k)|f (k) 6= 1). (2.6)
Thus we deduce from (2.3) – (2.6) that
T (r, g) ≤ N (p+1
(
r, 1; f (k)
∣∣ g(k) 6= 1
)
+ 2N (p+1
(
r, 1; g(k)
∣∣ f (k) 6= 1
)
+Nk+1
(
r,
1
g
)
+
+N (2
(
r,
1
f (k)
)
+N (2
(
r,
1
g(k)
)
+N0
(
r,
1
f (k+1)
)
+N (2
(
r,
1
f (k) − 1
)
+
+NL
(
r,
1
f (k) − 1
)
+NL
(
r,
1
g(k) − 1
)
+Ng(k)>1
(
r,
1
f (k) − 1
)
+ S(r, f) + S(r, g). (2.7)
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 12
1710 CHUN WU
From the definition of N0
(
r,
1
f (k+1)
)
, we see that
N0
(
r,
1
f (k+1)
)
+N (2
(
r,
1
f (k) − 1
)
+N(2
(
r,
1
f (k)
)
−N (2
(
r,
1
f (k)
)
≤ N
(
r,
1
f (k+1)
)
.
The above inequality and Lemma 2.6 give
N0
(
r,
1
f (k+1)
)
+N (2
(
r,
1
f (k) − 1
)
≤ N
(
r,
1
f (k+1)
)
−N(2
(
r,
1
f (k)
)
+N (2
(
r,
1
f (k)
)
≤
≤ N
(
r,
1
f (k)
)
−N(2
(
r,
1
f (k)
)
+N (2
(
r,
1
f (k)
)
+ S(r, f) ≤ N
(
r,
1
f (k)
)
+ S(r, f). (2.8)
Substituting (2.8) in (2.7), we get
T (r, g) ≤ N (p+1
(
r, 1; f (k)
∣∣ g(k) 6= 1
)
+ 2N (p+1
(
r, 1; g(k)
∣∣ f (k) 6= 1
)
+Nk+1
(
r,
1
g
)
+
+N (2
(
r,
1
f (k)
)
+N (2
(
r,
1
g(k)
)
+N
(
r,
1
f (k)
)
+NL
(
r,
1
f (k) − 1
)
+
+NL
(
r,
1
g(k) − 1
)
+Ng(k)>1
(
r,
1
f (k) − 1
)
+ S(r, f) + S(r, g). (2.9)
Since Ep)(1, f
(k)) = Ep)(1, g
(k)), we have
pN (p+1
(
r, 1; g(k)
∣∣ f (k) 6= 1
)
+NL
(
r,
1
g(k) − 1
)
≤ N
(
r,
1
g(k) − 1
)
−N
(
r,
1
g(k) − 1
)
.
From Lemma 2.6, we obtain
N
(
r,
1
g(k) − 1
)
−N
(
r,
1
g(k) − 1
)
+N
(
r,
1
g(k)
)
−N
(
r,
1
g(k)
)
+
+N0
(
1
g(k+1)
)
≤ N
(
r,
1
g(k+1)
)
≤ N
(
r,
1
g(k)
)
+ S(r, g).
This shows
N
(
r,
1
g(k) − 1
)
−N
(
r,
1
g(k) − 1
)
≤ N
(
r,
1
g(k)
)
−N0
(
1
g(k+1)
)
.
Therefore,
pN (p+1
(
r, 1; g(k)|f (k) 6= 1
)
+NL
(
r,
1
g(k) − 1
)
≤ N
(
r,
1
g(k)
)
−N0
(
1
g(k+1)
)
.
From this, we have
2N (p+1
(
r, 1; g(k)|f (k) 6= 1
)
+NL
(
r,
1
g(k) − 1
)
≤
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VALUE-SHARING AND UNIQUENESS OF ENTIRE FUNCTIONS 1711
≤ N
(
r,
1
g(k)
)
≤ Nk+1
(
r,
1
g
)
+ S(r, g). (2.10)
Because
N (p+1
(
r, 1; f (k)|g(k) 6= 1
)
+NL
(
r,
1
f (k) − 1
)
≤ N (2
(
r,
1
f (k) − 1
)
≤
≤ N
(
r,
f (k)
f (k+1)
)
+ S(r, f) ≤ T
(
r,
f (k+1)
f (k)
)
+ S(r, f) ≤
≤ N
(
r,
f (k+1)
f (k)
)
+ S(r, f) ≤ N
(
r,
1
f (k)
)
+ S(r, f). (2.11)
Combining (2.9) – (2.11), Lemmas 2.4 and 2.7, we get
T (r, g) ≤ 2Nk+1
(
r,
1
g
)
+Nk+1
(
r,
1
f
)
+Nk+2
(
r,
1
g
)
+Nk+2
(
r,
1
f
)
+ S(r, f) + S(r, g).
Without loss of generality, we suppose that there exists a set I with infinite measure such that
T (r, f) ≤ T (r, g) for r ∈ I. Hence
T (r, g) ≤
{
2
[
1− δk+1(0, g)
]
+
[
1− δk+1(0, f)
]
+
+
[
1− δk+2(0, g)
]
+
[
1− δk+2(0, f)
]
+ ε
}
T (r, g) + S(r, g)
for r ∈ I and 0 < ε < ∆− 4, that is {∆− 4− ε}T (r, g) ≤ S(r, g), i.e., ∆− 4 ≤ 0 or ∆ ≤ 4, which
is a contradiction to our hypotheses ∆ > 4.
Hence, we get Φ(z) ≡ 0. Then by (2.2), we have
f (k+2)
f (k+1)
− 2f (k+1)
f (k) − 1
≡ g(k+2)
g(k+1)
− 2g(k+1)
g(k) − 1
.
By integrating two sides of the above equality, we obtain
1
f (k) − 1
=
bg(k) + a− b
g(k) − 1
, (2.12)
where a(6= 0) and b are constants.
Lemma 2.8 is proved.
Lemma 2.9. Let f(z) and g(z) be two nonconstant entire functions, k be a positive integer. If
f (k) and g(k) share a nonzero polynomial 1 IM and
∆ = δk+2(0, g) + δk+2(0, f) + δk+1(0, f) + 2δk+1(0, g) > 4,
then
1
f (k) − 1
=
bg(k) + a− b
g(k) − 1
, where a, b are two constants.
Proof. Since f (k) and g(k) share the value 1 IM, we have N (p+1(r, 1; g(k) | f (k) 6= 1) = 0.
Proceeding as in the proof of Lemma 2.8, we obtain conclusion of Lemma 2.9.
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1712 CHUN WU
Lemma 2.10. Let f(z) and g(z) be two nonconstant entire functions, k be a positive integer. If
f (k) and g(k) share a nonzero polynomial 1 CM and
∆ = δk+2(0, g) + δk+2(0, f) > 1,
then
1
f (k) − 1
=
bg(k) + a− b
g(k) − 1
, where a, b are two constants.
Proof. Since f (k) and g(k) share the value 1 CM, we have NL
(
r,
1
F − 1
)
= NL
(
r,
1
G− 1
)
=
= 0 and N (p+1
(
r, 1; g(k)|f (k) 6= 1
)
= 0. Proceeding as in the proof of Lemma 2.8, we obtain
conclusion of Lemma 2.10.
Lemma 2.11 (see [19]). Let f(z) be a nonconstant entire function and k(≥ 2) be a positive
integer. If ff (k) 6= 0, then f = eaz+b, where a 6= 0, b are constants.
3. Proof of theorems. 3.1. Proof of Theorem 1.1. Because f is a transcendental entire, we
get T (r, P ) = o(T (r, f)). Suppose that z0 6∈ {z : P (z) = 0} is a zero of L(f) with its multiplicity
l ≥ k + 2, then z0 is a zero of (L(f)(k)/P )′ with its multiplicity l − k − 1 ≥ 1. From Lemmas 2.3
and 2.5, we have
nT (r, f) = T (r, L(f)) + S(r, f) ≤
≤ N
(
r,
1
L(f)
)
+N
(
r,
1
L(f)(k) − P
)
−N
(
r,
1
(L(f)(k)/P )′
)
+ S(r, f) ≤
≤ Nk+1
(
r,
1
L(f)
)
+N
(
r,
1
L(f)(k) − P
)
−N0
(
r,
1
(L(f)(k)/P )′
)
+ S(r, f) ≤
≤ Nk+1
(
r,
1
(f − c1)l1
)
+ . . .+Nk+1
(
r,
1
(f − cs)ls
)
+N
(
r,
1
L(f)(k) − P
)
+ S(r, f) ≤
≤ (k + s)T (r, f) +N
(
r,
1
L(f)(k) − P
)
+ S(r, f).
Thus, we get
(n− k − s)T (r, f) ≤ N
(
r,
1
L(f)(k) − P
)
+ S(r, f).
Noting that n > k + s, we get
[
L(f)](k)
]
= P has infinitely many solutions.
3.2. Proof of Theorem 1.2. Let L(z) and l be given by (1.1), (1.2), respectively. Without loss of
generality, we suppose that an = 1, l = l1, and c = c1, we get
Θ(0, L(f)) = 1− lim
r→∞
N
(
r,
1
L(f)
)
T
(
r, L(f)
) ≥ 1− lim
r→∞
∑s
j=1
N
(
r,
1
f − cj
)
nT (r, f)
≥ 1− s
n
≥ l − 1
n
.
Similarly, we have Θ
(
0, L(g)
)
≥ l − 1
n
.
Moreover, we have
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VALUE-SHARING AND UNIQUENESS OF ENTIRE FUNCTIONS 1713
δk+1
(
0, L(f)
)
= 1− lim
r→∞
Nk+1
(
r,
1
L(f)
)
T (r, L(f))
≥
≥ 1− lim
r→∞
∑s
j=2
Nk+1
(
r,
1
(f − cj)lj
)
+Nk+1
(
r,
1
(f − c)l
)
nT (r, f)
≥
≥ 1− lim
r→∞
(s− 1)T (r, f) + (k + 1)T (r, f) + S(r, f)
nT (r, f)
≥
≥ 1− s+ k
n
≥ l − k − 1
n
.
Similarly, we obtain
δk+1
(
0, L(g)
)
≥ 1− s+ k
n
≥ l − k − 1
n
,
δk+2
(
0, L(g)
)
≥ 1− s+ k + 1
n
≥ l − k − 2
n
,
δk+2
(
0, L(f)
)
≥ 1− s+ k + 1
n
≥ l − k − 2
n
.
Because 5l > 4n+ 5k + 7, we get
∆ = 2δk+1
(
0, L(g)
)
+ δk+1
(
0, L(f)
)
+ δk+2
(
0, L(g)
)
+ δk+2
(
0, L(f)
)
> 4.
By Lemma 2.8, we can have
1
L(f)(k) − 1
=
bL(g)(k) + a− b
L(g)(k) − 1
. (3.1)
Next, we consider the following three cases:
Case 1. b 6= 0 and a = b. Then from (3.1) we obtain
1
L(f)(k) − 1
=
bL(g)(k)
L(g)(k) − 1
. (3.2)
1.1. If b = −1, then it follows from (3.2) that
[
L(f)
](k)[
L(g)
](k) ≡ 1.
That is [
(f − c)l(f − c2)l2 . . . (f − cs)ls
](k)[
(g − c)l(g − c2)l2 . . . (g − cs)ls
](k)
= 1. (3.3)
1.1.1. When s = 1, we can rewrite (3.3)[
(f − c)n
](k)[
(g − c)n
](k)
= 1.
Since 5l > 4n+ 5k + 7, l = n, then n > 5k + 7. So f − c 6= 0, g − c 6= 0, according Lemma 2.11,
we have
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1714 CHUN WU
f = b1e
bz + c, g = b2e
−bz + c,
where b1, b2, b are three constants satisfying (−1)k(b1b2)
n(nb)2k = 1.
1.1.2. When s ≥ 2, we notice that 5l > 4n+5k+7. Hence l > 5k+7. Suppose that z0 is a l-fold
zero of f − c, we know that z0 must be a (l − k)-fold zero of
[
(f − c)l(f − c2)l2 . . . (f − cs)ls
](k)
.
Noting that g is an entire function, it follows from (3.3), which is a contradiction. Hence f − c 6= 0,
g − c 6= 0. So we get f = eα(z) + c, where α(z) is a nonconstant entire function. Thus we have
[f i](k) =
[
(eα + c)i
](k)
= pi
(
α′, α′′, . . . , α(k)
)
eiα, i = 1, 2, . . . , n, (3.4)
where pi, i = 1, 2, . . . , n, is differential polynomials about α′, α′′, . . . , α(k). Obviously, pi 6≡ 0,
T (r, pi) = S(r, f), i = 1, 2, . . . , n, we get from (3.3) and (3.4) that
N
(
r,
1
pne(n−1)α + . . .+ p1
)
= S(r, f).
According to Lemmas 2.1, 2.3 and f = eα + c, we get
(n− 1)T (r, f − c) = T (r, pne
(n−1)α + . . .+ p1) + S(r, f) ≤
≤ N
(
r,
1
pne(n−1)α + . . .+ p1
)
+N
(
r,
1
pne(n−1)α + . . .+ p2eα
)
+ S(r, f) ≤
≤ N
(
r,
1
pne(n−2)α + . . .+ p2
)
+ S(r, f) ≤
≤ (n− 2)T (r, f − c) + S(r, f),
which is a contradiction.
1.2. If a = b 6= −1, then (3.2) that can be written as
L(g)(k) =
−1
b
· 1
L(f)(k) − (1 + b)/b
. (3.5)
From (3.5) we get
N
(
r,
1
L(f)(k) − (1 + b)/b
)
= N(r, g) = S(r, f). (3.6)
By (3.6) and Lemma 2.2, we have
nT (r, f) = T (r, L(f)) +O(1) ≤
≤ Nk+1
(
r,
1
L(f)
)
+N
(
r,
1
L(f)(k) − (1 + b)/b
)
+ S(r, f) ≤
≤ Nk+1
(
r,
1
(f − c)l
)
+Nk+1
(
r,
1
(f − c2)l2 . . . (f − cs)ls
)
+ S(r, f) ≤
≤ (k + s)T (r, f) ≤ (k + n− l + 1)T (r, f) + S(r, f),
which is a contradiction, because 5l > 4n+ 5k + 7.
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VALUE-SHARING AND UNIQUENESS OF ENTIRE FUNCTIONS 1715
Case 2. b 6= 0 and a 6= b. We discuss the following we subcases:
2.1. Suppose that b = −1, then a 6= 0 and (3.1) can be rewritten as
L(f)(k) =
a
a+ 1− L(g)(k)
. (3.7)
From (3.7) we obtain
N
(
r,
1
a+ 1− L(g)(k)
)
= N(r, f) = S(r, g). (3.8)
From (3.8), Lemmas 2.2 and 2.4, we get
nT (r, g) = T (r, L(g)) +O(1) ≤ Nk+1
(
r,
1
L(g)
)
+ S(r, g).
Next, by using the argument as in Case 1.2, we have a contradiction.
2.2. Suppose that b 6= −1, then (3.1) be rewritten as
L(f)(k) − b+ 1
b
=
−a
b2
1
L(g)(k) + (a− b)/b
. (3.9)
From (3.9), we have
N
(
r,
1
L(f)(k) − (b+ 1)/b
)
= N(r, g). (3.10)
From (3.10), Lemmas 2.2, 2.4 and in the same manner as in Case 1.2, we can get a contradiction.
Case 3. b = 0 and a 6= 0. Then (3.1) can be rewritten as
L(g)(k) = aL(f)(k) + (1− a), (3.11)
L(g) = aL(f) + (1− a)p1(z), (3.12)
where p1 is a polynomial with its degp1 ≤ k. If a 6= 1, then (1− a)p1 6≡ 0. This together with (3.12)
and Lemma 2.1, we get
nT (r, g) = T (r, L(g)) +O(1) ≤ N
(
r,
1
L(g)
)
+N
(
r,
1
L(f)
)
+ S(r, g) ≤
≤
s∑
i=1
N
(
r,
1
g − ci
)
+
s∑
j=1
N
(
r,
1
f − cj
)
+ S(r, g) ≤
≤ s
[
T (r, g) + T (r, f)
]
+ S(r, g). (3.13)
Because n = l+ l2 + . . .+ ls, we get n− l = l2 + . . .+ ls ≥ s− 1, i.e., n− l ≥ s− 1, n− s ≥ l− 1.
From 5l > 4n+ 5k+ 7, we have l− 1 > 4(n− l) + 5k+ 6 > 4(s− 1) + 5k+ 6, so n− s ≥ l− 1 >
> 4(s− 1) + 5k + 6, i.e., n− s > 4(s− 1) + 5k + 6, thus, s <
n− 5k − 2
5
. Thus
nT (r, g) <
n− 5k − 2
5
[
T (r, g) + T (r, f)
]
+ S(r, g). (3.14)
On the other hand, from (3.12) and Lemma 2.3, we see that T (r, g) = T (r, f) + S(r, g).
Substituting this into (3.14), we deduce that
3n+ 10k + 4
5
T (r, g) < S(r, g), which is a contradiction.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 12
1716 CHUN WU
Thus a = 1, and so it follows from (3.12) that L(f) = L(g).
Next we consider the case when f and g are polynomials. Suppose that f − c and g − c have u
and v pairwise distinct zeros, respectively. Then f − c and g − c are of the forms
f − c = k1(z − a1)n1(z − a2)n2 . . . (z − au)nu ,
g − c = k2(z − b1)m1(z − b2)m2 . . . (z − bv)mv ,
so that
[f − c]l = kl1(z − a1)ln1(z − a2)ln2 . . . (z − au)lnu , (3.15)
[g − c]l = kl2(z − b1)lm1(z − b2)lm2 . . . (z − bv)lmv , (3.16)
where k1 and k2 are nonzero constants, nil > 5k + 7, mjl > 5k + 7, ni, mj , i = 1, 2, . . . u,
j = 1, 2, . . . v, are positive integers. Differentiating (3.12), we get
L(g)(k+1) = aL(f)(k+1). (3.17)
From (3.15), (3.16) and (3.17), we have
(z − a1)ln1−k−1(z − a2)ln2−k−1 . . . (z − au)lnu−k−1ξ1(z) =
= (z − b1)lm1−k−1(z − b2)lm2−k−1 . . . (z − bv)lmv−k−1ξ2(z), (3.18)
where ξ1(z) and ξ2(z) are polynomials deg ξ1 = (n − l)
∑u
i=1
ni + (u − 1)(k + 1) and deg ξ2 =
= (n − l)
∑v
j=1
mj + (v − 1)(k + 1). It follows that 5l > 4n + 5k + 7, we have 2l − n >
> 3(n− l) + 5k + 7 > 5k + 7.
Then
(2l − n)ni > 5k + 7, (2l − n)mj > 5k + 7, i = 1, 2, . . . , u, j = 1, 2, . . . , v.
So that
u∑
i=1
[
nil − (k + 1)
]
−
u∑
i=1
ni(n− l) =
u∑
i=1
[
ni(2l − n)− (k + 1)
]
> u(4k + 6) > (u− 1)(k + 1),
i.e.,
u∑
i=1
[
nil − (k + 1)
]
> (n− l)
u∑
i=1
ni + (u− 1)(k + 1).
Similarly,
v∑
j=1
[
mjl − (k + 1)
]
> (n− l)
v∑
j=1
mj + (v − 1)(k + 1).
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VALUE-SHARING AND UNIQUENESS OF ENTIRE FUNCTIONS 1717
Thus from (3.18) we deduce that there exists z0 such that L(f(z0)) = L(g(z0)) = 0, where z0
has multiplicity greater than 5k + 7. This together with (3.12), we deduce p1(z) ≡ 0, which also
prove the claim.
Therefore from (3.11) and (3.12) we get a = 1 and so L(f) ≡ L(g). Hence, this completes the
proof of Theorem1.2.
3.3 Proof of Corollary 1.1 (1.2). By using Lemma 2.9 (2.10) and the condition 5l > 4n+ 5k +
+7(2l > n+2k+4), proceeding as in the proof of Theorem 1.2, we can similarly prove Corollary 1.1
(1.2). We omit the details here.
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Received 07.11.12
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