DEMONSTRATIO MATHEMATICA
Vol. 49
No 2
2016
Sujoy Majumder
ON AN OPEN PROBLEM OF XIAO-BIN ZHANG
AND JUN-FENG XU
Communicated by N. Shanmugalingam
Abstract. The purpose of the paper is to study the uniqueness of meromorphic
functions sharing a nonzero polynomial. The results of the paper improve and generalize
the recent results due to X. B. Zhang and J. F. Xu [19]. We also solve an open problem as
posed in the last section of [19].
1. Introduction, definitions and results
In this paper, by meromorphic functions we shall always mean meromorphic functions in the complex plane.
Let f and g be two non-constant meromorphic functions and let a be
a finite complex number. We say that f and g share a CM, provided that
f ´ a and g ´ a have the same zeros with the same multiplicities. Similarly,
we say that f and g share a IM, provided that f ´ a and g ´ a have the same
zeros ignoring multiplicities. In addition, we say that f and g share 8 CM,
if 1{f and 1{g share 0 CM, and we say that f and g share 8 IM, if 1{f and
1{g share 0 IM.
We adopt the standard notations of value distribution theory (see [6]).
We denote by T prq the maximum of T pr, f q and T pr, gq. The notation
Sprq denotes any quantity satisfying Sprq “ opT prqq as r Ñ 8, outside of
a possible exceptional set of finite linear measure.
A finite value z0 is said to be a fixed point of f pzq if f pz0 q “ z0 . Throughout this paper, we need the following definition
N pr, a; f q
Θpa; f q “ 1 ´ lim sup
,
T pr, f q
rÑ8
where a is a value in the extended complex plane.
2010 Mathematics Subject Classification: Primary 30D35.
Key words and phrases: uniqueness, meromorphic function, small functions, nonlinear
differential polynomials.
DOI: 10.1515/dema-2016-0015
c Copyright by Faculty of Mathematics and Information Science, Warsaw University of Technology
162
S. Majumder
In 1959, W. K. Hayman (see [5], Corollary of Theorem 9) proved the
following theorem.
Theorem A. Let f be a transcendental meromorphic function and n p≥ 3q
is an integer. Then f n f 1 “ 1 has infinitely many solutions.
In 1997, C. C. Yang and X. H. Hua obtained the following uniqueness
result corresponding to Theorem A.
Theorem B. [14] Let f and g be two non-constant meromorphic functions,
n ≥ 11 be a positive integer. If f n f 1 and g n g 1 share 1 CM, then either
f pzq “ c1 ecz , gpzq “ c2 e´cz , where c1 , c2 and c are three constants satisfying
pc1 c2 qn`1 c2 “ ´1 or f ” tg for a constant t such that tn`1 “ 1.
In 2002, using the idea of sharing fixed points, M. L. Fang and H. L. Qiu
further generalized and improved Theorem B in the following manner.
Theorem C. [3] Let f and g be two non-constant meromorphic functions,
and let n ≥ 11 be a positive integer. If f n f 1 ´ z and g n g 1 ´ z share 0 CM, then
2
2
either f pzq “ c1 ecz , gpzq “ c2 e´cz , where c1 , c2 and c are three nonzero
complex numbers satisfying 4pc1 c2 qn`1 c2 “ ´1 or f “ tg for a complex
number t such that tn`1 “ 1.
For the last couple of years, a handful numbers of astonishing results
have been obtained regarding the value sharing of nonlinear differential
polynomials, which are mainly the k-th derivative of some linear expression
of f and g.
In 2010, J. F. Xu, F. Lu and H. X. Yi studied the analogous problem
corresponding to Theorem C, where in addition to the fixed point sharing
problem sharing of poles are also taken under supposition. Thus, the research
has somehow been shifted to wards the following direction.
Theorem D. [12] Let f and g be two non-constant meromorphic functions,
and let n, k be two positive integers with n ą 3k ` 10. If pf n qpkq and pg n qpkq
2
2
share z CM, f and g share 8 IM, then either f pzq “ c1 ecz , gpzq “ c2 e´cz ,
where c1 , c2 and c are three constants satisfying 4n2 pc1 c2 qn c2 “ ´1 or f ” tg
for a constant t such that tn “ 1.
Theorem E. [12] Let f and g be two non-constant meromorphic functions
satisfying Θp8, f q ą n2 , and let n, k be two positive integers with n ≥ 3k ` 12.
If pf n pf ´ 1qqpkq and pg n pg ´ 1qqpkq share z CM, f and g share 8 IM, then
f ” g.
Recently Xiao-Bin Zhang and Jun-Feng Xu [19] further generalized as
well as improved the results of [12] as follows.
On an open problem of Xiao-Bin Zhang and Jun-Feng Xu
163
Theorem F. [19] Let f and g be two transcendental meromorphic functions,
let ppzq be a nonzero polynomial with degppq “ l ≤ 5, n, k and m be three
positive integers with n ą 3k ` m ` 7. Let P pwq “ am wm ` am´1 wm´1 `
¨ ¨ ¨ ` a1 w ` a0 be a nonzero polynomial. If rf n P pf qspkq and rg n P pgqspkq share
p CM, f and g share 8 IM then one of the following three cases hold:
(1) f pzq ” tgpzq for a constant t such that td “ 1, where d “ GCDpn ` m,
. . . , n ` m ´ i, . . . , nq, am´i ­“ 0 for some i “ 1, 2, . . . , m,
(2) f and g satisfy the algebraic equation Rpf, gq ” 0, where Rpω1 , ω2 q “
ω1n pam ω1m ` am´1 ω1m´1 ` ¨ ¨ ¨ ` a0 q ´ ω2n pam ω2m ` am´1 ω2m´1 ` ¨ ¨ ¨ ` a0 q;
(3) P pzq reduces to a nonzero monomial, namely P pzq “ ai z i ı 0 for
some i P t0, 1, . . . , mu; if ppzq
is not a constant, then f “ c1 ecQpzq ,
ş
z
g “ c2 e´cQpzq , where Qpzq “ 0 ppzqdz, c1 , c2 and c are constants such
that a2i pc1 c2 qn`i rpn ` iqcs2 “ ´1, if ppzq is a nonzero constant b, then
f “ c3 ecz , g “ c4 e´cz , where c3 , c4 and c are constants such that
p´1qk a2i pc3 c4 qn`i rpn ` iqcs2k “ b2 .
Zhang and Xu made the following commend in Remark 1.2 [19]: “From
the proof of Theorem 1.3, we can see that the computation will be very
complicated when degppq becomes large, so we are not sure whether Theorem 1.3 holds for the general polynomial ppzq.” Also at the end of the paper,
following open problem was posed by the authors in [19].
Open problem. What happens to Theorem 1.3 [19] if the condition “l ≤ 5”
is removed?
One of our objective to write this paper is to solve this open problem.
Now observing the above results, the following questions are inevitable.
Question 1. Is it possible to obtain the similar result corresponding to
the Theorem F if the sharing value is relaxed from CM to IM?
Question 2. Can the lower bound of n be further reduced in Theorem F?
In this paper, taking the possible answer of the above questions into
background we obtain the following results.
Theorem 1. Let f and g be two transcendental meromorphic functions,
let ppzq be a nonzero polynomial with degppq ≤ n ´ 1, n p≥ 1q, k p≥ 1q and
m p≥ 0q be three integers such that n ą 3k ` m ` 6 and P pwq be defined as
in Theorem F. If rf n P pf qspkq , rg n P pgqspkq share p CM and f , g share 8 IM
then the conclusion of Theorem F holds.
Theorem 2. Let f and g be two transcendental meromorphic functions,
let ppzq be a nonzero polynomial with degppq ≤ n ´ 1, n p≥ 1q, k p≥ 1q and
m p≥ 0q be three integers such that n ą 9k ` 4m ` 11 and P pwq be defined
164
S. Majumder
as in Theorem F. If rf n P pf qspkq , rg n P pgqspkq share p IM and f , g share 8
IM then the conclusion of Theorem F holds.
We now explain following definitions and notations which are used in the
paper.
Definition 1. [7] Let a P C Y t8u. For a positive integer p we denote by
N pr, a; f |≤ pq the counting function of those a-points of f (counted with
multiplicities) whose multiplicities are not greater than p. By N pr, a; f |≤ pq
we denote the corresponding reduced counting function.
In an analogous manner we can define N pr, a; f |≥ pq and N pr, a; f |≥ pq.
Definition 2. [9] Let k be a positive integer or infinity. We denote
by Nk pr, a; f q the counting function of a-points of f , where an a-point of
multiplicity m is counted m times if m ≤ k and k times if m ą k. Then
Nk pr, a; f q “ N pr, a; f q ` N pr, a; f |≥ 2q ` ¨ ¨ ¨ ` N pr, a; f |≥ kq.
Clearly N1 pr, a; f q “ N pr, a; f q.
Definition 3. [2] Let f and g be two non-constant meromorphic functions
such that f and g share the value a IM for a P C Y t8u. Let z0 be an
a-point of f with multiplicity p and also an a-point of g with multiplicity
q. We denote by N L pr, a; f q pN L pr, a; gqq, the reduced counting function of
those a-points of f and g, where p ą q ≥ 1 pq ą p ≥ 1q. Also we denote
p1
by N E pr, a; f q, the reduced counting function of those a-points of f and g,
where p “ q ≥ 1.
Definition 4. [8, 9] Let f and g be two non-constant meromorphic functions such that f and g share the value a IM. We denote by N ˚ pr, a; f, gq,
the reduced counting function of those a-points of f whose multiplicities
differ from the multiplicities of the corresponding a-points of g. Clearly
N ˚ pr, a; f, gq “ N ˚ pr, a; g, f q and N ˚ pr, a; f, gq “ N L pr, a; f q ` N L pr, a; gq.
Definition 5. [10] Let a, b1 , b2 , . . . , bq P C Yt8u. We denote by N pr, a; f |
g ‰ b1 , b2 , . . . , bq q the counting function of those a-points of f , counted
according to multiplicity, which are not the bi -points of g for i “ 1, 2, . . . , q.
2. Lemmas
Let F and G be two non-constant meromorphic functions defined in C.
We denote by H and V the functions as follows:
ˆ 2
˙ ˆ 2
˙
F
2F 1
G
2G1
(2.1)
H“
´
´
´
,
F1
F ´1
G1
G´1
ˆ
˙
ˆ
˙
F1
F1
G1
G1
(2.2)
V “
´
´
´
.
F ´1
F
G´1
G
On an open problem of Xiao-Bin Zhang and Jun-Feng Xu
165
Lemma 1. [13] Let f be a non-constant meromorphic function and
let an pzq pı 0q, an´1 pzq, . . . , a0 pzq be meromorphic functions such that
T pr, ai pzqq “ Spr, f q for i “ 0, 1, 2, . . . , n. Then
T pr, an f n ` an´1 f n´1 ` ¨ ¨ ¨ ` a1 f ` a0 q “ nT pr, f q ` Spr, f q.
Lemma 2. [18] Let f be a non-constant meromorphic function and let p, k
be positive integers. Then
(2.3)
Np pr, 0; f pkq q ≤ T pr, f pkq q ´ T pr, f q ` Np`k pr, 0; f q ` Spr, f q,
(2.4)
Np pr, 0; f pkq q ≤ kN pr, 8; f q ` Np`k pr, 0; f q ` Spr, f q.
Lemma 3. [11] If N pr, 0; f pkq | f ­“ 0q denotes the counting function of
those zeros of f pkq which are not the zeros of f , where a zero of f pkq is counted
according to its multiplicity, then
N pr, 0; f pkq | f ­“ 0q
≤ kN pr, 8; f q ` N pr, 0; f |ă kq ` kN pr, 0; f |≥ kq ` Spr, f q.
Lemma 4. ([6], Theorem 3.10) Suppose that f is a non-constant meromorphic function, k ≥ 2 is an integer. If
ˆ
1˙
f
,
N pr, 8, f q ` N pr, 0; f q ` N pr, 0; f pkq q “ S r,
f
then f “ eaz`b , where a ­“ 0, b are constants.
Lemma 5. [4] Let f pzq be a non-constant entire function and let k ≥ 2 be
a positive integer. If f pzqf pkq pzq ­“ 0, then f pzq “ eaz`b , where a ­“ 0, b are
constant.
Lemma 6. ([15], Theorem 1.24) Let f be a non-constant meromorphic
function and let k be a positive integer. Suppose that f pkq ı 0, then
N pr, 0; f pkq q ≤ N pr, 0; f q ` kN pr, 8; f q ` Spr, f q.
Lemma 7. [19] Let f and g be two non-constant meromorphic functions,
let P pwq be defined as in Theorem F and let k, m, n ą 2k ` m ` 1 be three
positive integers. If rf n P pf qspkq ” rg n P pgqspkq , then f n P pf q ” g n P pgq.
Lemma 8. ([17], Lemma 6) If H ” 0, then F , G share 1 CM. If further F ,
G share 8 IM then F , G share 8 CM.
Lemma 9. [19] Let f , g be non-constant meromorphic functions, let n, k be
two positive integers with n ą k ` 2, and let P pwq be defined as in Theorem F.
Let αpzqpı 0, 8q be a small function with respect to f with finitely many
zeros and poles. If rf n P pf qspkq rg n P pgqspkq ” α2 , f and g share 8 IM, then
P pwq is reduced to a nonzero monomial, namely P pwq “ ai wi ı 0 for some
i P t0, 1, . . . , mu.
166
S. Majumder
Lemma 10. [15] Let fj pj “ 1, 2, 3q be a meromorphic and f1 be nonconstant. Suppose that
3
ÿ
fj ” 1
j“1
and
3
ÿ
j“1
N pr, 0; fj q ` 2
3
ÿ
N pr, 8; fj q ă pλ ` op1qqT prq,
j“1
as r Ñ `8, r P I, λ ă 1 and T prq “ max1≤j≤3 T pr, fj q. Then f2 ” 1 or
f3 ” 1.
Lemma 11. Let f , g be two transcendental meromorphic functions and let
n
n
pkq
qspkq
P pwq be defined as in Theorem F. Let F “ rf P pf
, G “ rg P pgqs
, where
p
p
ppzq is a non zero polynomial and np≥ 1q, kp≥ 1q and mp≥ 0q are integers
such that n ą 3k ` m ` 3. If f , g share 8 IM and H ” 0, then either
rf n P pf qspkq rg n P pf qspkq ” p2 or f n P pf q ” g n P pgq.
Proof. Since H ” 0, by Lemma 8 we get F and G share 1 CM. On integration
we get
1
bG ` a ´ b
(2.5)
”
,
F ´1
G´1
where a, b are constants and a ­“ 0. We now consider the following cases:
Case 1. Let b ­“ 0 and a ­“ b. If b “ ´1, then from (2.5) we have
´a
.
F ”
G´a´1
Therefore
N pr, a ` 1; Gq “ N pr, 8; F q “ N pr, 8; f q.
So in view of Lemmas 1 and 2 for p “ 1 and using the second fundamental
theorem we get
pn ` mq T pr, gq ≤ T pr, Gq ` Nk`1 pr, 0; g n P pgqq ´ N pr, 0; Gq
≤ N pr, 8; Gq ` N pr, 0; Gq ` N pr, a ` 1; Gq
` Nk`1 pr, 0; g n P pgqq ´ N pr, 0; Gq ` Spr, gq
≤ N pr, 8; gq ` Nk`1 pr, 0; g n P pgqq ` N pr, 8; f q ` Spr, gq
≤ N pr, 8; f q ` N pr, 8; gq ` Nk`1 pr, 0; g n q ` Nk`1 pr, 0; P pgqq ` Spr, gq
≤ 2 N pr, 8; gq ` pk ` 1qN pr, 0; gq ` T pr, P pgqq ` Spr, gq
≤ tk ` 3 ` mu T pr, gq ` Spr, gq,
which is a contradiction since n ą k ` 3.
On an open problem of Xiao-Bin Zhang and Jun-Feng Xu
167
If b ­“ ´1, from (2.5) we obtain that
˙
ˆ
´a
1
”
.
F ´ 1`
b
b2 rG ` a´b
b s
So
pb ´ aq
; Gq “ N pr, 8; F q “ N pr, 8; f q.
b
Using Lemmas 1, 2 and the same argument as used in the case when b “ ´1,
we can get a contradiction.
Case 2. Let b ­“ 0 and a “ b. If b “ ´1, then from (2.5) we have
N pr,
F G ” 1,
i.e.,
rf n P pf qspkq rg n P pgqspkq ” p2 ,
where rf n P pf qspkq and rg n P pgqspkq share p CM.
If b ­“ ´1, from (2.5) we have
1
bG
”
.
F
p1 ` bqG ´ 1
Therefore
1
; Gq “ N pr, 0; F q.
1`b
So in view of Lemmas 1 and 2 for p “ 1 and using the second fundamental
theorem, we get
N pr,
pn ` mq T pr, gq
ˆ
˙
1
≤ N pr, 8; Gq ` N pr, 0; Gq ` N r,
; G ` Nk`1 pr.0; g n P pgqq
1`b
´ N pr, 0; Gq ` Spr, gq
≤ N pr, 8; gq ` pk ` 1qN pr, 0; gq ` T pr, P pgqq ` N pr, 0; F q ` Spr, gq
≤ N pr, 8; gq ` pk ` 1qN pr, 0; gq ` T pr, P pgqq ` pk ` 1qN pr, 0; f q
` T pr, P pf qq ` kN pr, 8; f q ` Spr, f q ` Spr, gq
≤ tk ` 2 ` mu T pr, gq ` t2k ` 1 ` mu T pr, f q ` Spr, f q ` Spr, gq.
Without loss of generality, we suppose that there exists a set I with infinite
measure such that T pr, f q ≤ T pr, gq for r P I.
So for r P I we have
tn ´ 3k ´ 3 ´ mu T pr, gq ≤ Spr, gq,
which is a contradiction since n ą 3k ` 3 ` m.
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S. Majumder
Case 3. Let b “ 0. From (2.5) we obtain
G`a´1
.
a
If a ­“ 1 then from (2.6), we obtain
F ”
(2.6)
N pr, 1 ´ a; Gq “ N pr, 0; F q.
We can similarly deduce a contradiction as in Case 2. Therefore a “ 1 and
from (2.6) we obtain
F ” G,
i.e.,
rf n P pf qspkq ” rg n P pgqspkq .
Then by Lemma 7 we have
f n P pf q ” g n P pgq.
This completes the proof.
Lemma 12. Let f , g be two transcendental meromorphic functions, let ppzq
be a non-zero polynomial with degppq ≤ n ´ 1, where n and k be two positive
integers such that n ą k. Let rf n spkq , rg n spkq share p CM and f , g share
8 IM. Now when rf n spkq rg n spkq ” p2 ,
(i) if ppzq şis not a constant, then f “ c1 ecQpzq , g “ c2 e´cQpzq , where
z
Qpzq “ 0 ppzqdz, c1 , c2 and c are constants such that pncq2 pc1 c2 qn “ ´1,
(ii) if ppzq is a nonzero constant b, then f “ c3 edz , g “ c4 e´dz , where c3 , c4
and d are constants such that p´1qk pc3 c4 qn pndq2k “ b2 .
Proof. Suppose that
rf n spkq rg n spkq ” p2 .
(2.7)
Since f and g share 8 IM, (2.7) one can easily say that f and g are transcendental entire functions. We consider the following cases:
Case 1. Let degpppzqq “ lp≥ 1q. Since n ą k, it follows that N pr, 0; f q “
Oplog rq and N pr, 0; gq “ Oplog rq. Let
(2.8)
F1 “
rf n spkq
p
and G1 “
rg n spkq
.
p
From (2.7) we get
(2.9)
F1 G1 ” 1.
If F1 ” cG1 , where c is a nonzero constant, then by (2.9), F1 is a constant
and so f is a polynomial, which contradicts our assumption. Hence F1 ı G1 .
On an open problem of Xiao-Bin Zhang and Jun-Feng Xu
169
Let
Φ“
(2.10)
rf n spkq ´ p
.
rg n spkq ´ p
We deduce from (2.10) that
Φ ” eβ ,
(2.11)
where β is an entire function.
Let f1 “ F1 , f2 “ ´eβ G1 and f3 “ eβ . Here f1 is transcendental. Now
from (2.11), we have
f1 ` f2 ` f3 ” 1.
Hence by Lemma 6, we get
3
ÿ
N pr, 0; fj q ` 2
j“1
3
ÿ
N pr, 8; fj q ≤ N pr, 0; F1 q ` N pr, 0; eβ G1 q ` Oplog rq
j“1
≤ pλ ` op1qqT prq,
as r Ñ `8, r P I, λ ă 1 and T prq “ max1≤j≤3 T pr, fj q. So by Lemma 10,
we get either eβ G1 ” ´1 or eβ ” 1. But here the only possibility is that
eβ G1 ” ´1, i.e., rg n spkq ” ´e´β ppzq and so from (2.7), we obtain
F1 ” eγ1 G1 ,
i.e.,
rf n spkq ” eγ1 rg n spkq ,
where γ1 is a non-constant entire function. Now from (2.7) we get
(2.12)
1
pf n qpkq ” ce 2 γ1 ppzq,
1
pg n qpkq ” ce´ 2 γ1 ppzq,
where c “ ˘1.
Since N pr, 0; f q “ Oplog rq and N pr, 0; gq “ Oplog rq, so we can take
(2.13)
f pzq “ h1 pzqeαpzq ,
gpzq “ h2 pzqeβpzq ,
where h1 and h2 are nonzero polynomials and α, β are two non-constant
entire functions.
We deduce from (2.7) and (2.13) that either both α and β are transcendental entire functions or both are polynomials.
We consider the following cases:
Subcase 1.1. Let k ≥ 2. First we suppose both α and β are transcen1
1
h
1
dental entire functions. Let α1 “ α ` h11 and β1 “ β `
and β1 are transcendental entire functions.
1
h2
h2 .
Clearly both α1
170
S. Majumder
Note that
ˆ
1˙
rf n s
Spr, nα1 q “ S r, n ,
f
Moreover, we see that
ˆ
1˙
rg n s
Spr, nβ1 q “ S r, n .
g
N pr, 0; rf n spkq q ≤ N pr, 0; p2 q “ Oplog rq.
N pr, 0; rg n spkq q ≤ N pr, 0; p2 q “ Oplog rq.
From these and using (2.13), we have
(2.14)
N pr, 8; f n q ` N pr, 0; f n q ` N pr, 0; rf n spkq q “ Spr, nα1 q
ˆ
1˙
rf n s
“ S r, n
f
and
(2.15)
N pr, 8; g n q ` N pr, 0; g n q ` N pr, 0; rg n spkq q “ Spr, nβ1 q
ˆ
1˙
rg n s
“ S r, n .
g
Then from (2.14), (2.15) and Lemma 4, we must have
(2.16)
f “ eaz`b , g “ ecz`d ,
where a ­“ 0, b, c ­“ 0 and d are constants. But these types of f and g do not
agree with the relation (2.7).
Next we suppose α and β are both non-constant polynomials, since otherwise f , g reduces to a polynomials contradicting that they are transcendental.
1
1
Also from (2.7) we get α ` β ” C i.e., α ” ´β . Therefore degpαq “
degpβq.
Suppose that hi ’s i “ 1, 2 are non-constant polynomials. We deduce from
(2.13) that
(2.17)
1
1
1
1
1
1
rf n spkq ” Ahn´k
rhk1 pα qk ` Pk´1 pα , h1 qsenα ” ppzqenα ,
1
and
(2.18)
rg n spkq ” Bhn´k
rhk2 pβ qk ` Qk´1 pβ , h2 qsenβ ” ppzqenβ ,
2
1
1
1
1
where A, B are nonzero constants, Pk´1 pα , h1 q and Qk´1 pβ , h2 q are differ1
1
1
1
ential polynomials in α , h1 and β , h2 , respectively.
Since degppq ≤ n ´ 1, from (2.17) and (2.18), we conclude that both h1
and h2 are nonzero constants.
So we can rewrite f and g as follows:
(2.19)
f “ eγ ,
g “ eδ ,
1
1
where γ ` δ ” C and degpγq “ degpδq. Clearly γ ” ´δ .
On an open problem of Xiao-Bin Zhang and Jun-Feng Xu
171
If degpγq “ degpδq “ 1, then we again get a contradiction from (2.7).
Next we suppose that degpγq “ degpδq ≥ 2.
We deduce from (2.19) that
1
1
pf n q “ nγ enγ ,
2
1
3
1
2
pf n q “ rn2 pγ q2 ` nγ senγ ,
1
2
3
pf n q “ rn3 pγ q3 ` 3n2 γ γ ` nγ senγ ,
1
1
2
2
1
3
pf n qpivq “ rn4 pγ q4 ` 6n3 pγ q2 γ ` 3n2 pγ q2 ` 4n2 γ γ ` nγ pivq senγ ,
1
1
2
1
2
1
pf n qpvq “ rn5 pγ q5 ` 10n4 pγ q3 γ ` 15n3 γ pγ q2 ` 10n3 pγ q2 γ
2
3
3
1
` 10n2 γ γ ` 5n2 γ γ pivq ` nγ pvq senγ ,
.....................
1
1
2
1
2
1
rf n spkq “ rnk pγ qk ` Kpγ qk´2 γ ` Pk´2 pγ qsenγ .
Similarly we get
1
1
rg n spkq “ rnk pδ qk ` Kpδ qk´2 δ ` Pk´2 pδ qsenδ
1
1
2
1
“ rp´1qk nk pγ qk ´ Kp´1qk´2 pγ qk´2 γ ` Pk´2 p´γ qsenδ ,
1
where K is a suitably positive integer and Pk´2 pγ q is a differential polynomial
1
in γ .
1
1
1
2
Since degpγq ≥ 2, we observe that degppγ qk q ≥ k degpγ q and so pγ qk´2 γ
1 k´2 2
1
is either a nonzero constant or degppγ q γ q ≥ pk ´ 1q degpγ q ´ 1. Also we
see that
1
1
2
1
1
degppγ qk q ą degppγ qk´2 γ q ą degpPk´2 pγ qq por degpPk´2 p´γ qqq.
Now from (2.12), we see that rf n spkq and rg n spkq share 0 CM and so the
polynomials
1
1
2
1
nk pγ qk ` Kpγ qk´2 γ ` Pk´2 pγ q
and
1
1
2
1
p´1qk nk pγ qk ´ Kp´1qk´2 pγ qk´2 γ ` Pk´2 p´γ q
must be identical but this is impossible for k ≥ 2.
Actually the terms
1
1
nk pγ qk ` Kpγ qk´2 γ
2
1
can not be identical for k ≥ 2.
Subcase 1.2. Let k “ 1.
Now from (2.7) we get
1
where p21 “
1
f n´1 f g n´1 g ” p21 ,
(2.20)
1 2
p .
n2
1
and p´1qk nk pγ qk ´ Kp´1qk´2 pγ qk´2 γ
2
172
S. Majumder
First we suppose that both α and β are transcendental entire functions.
Let h “ f g. Clearly h is a transcendental entire function. Then from (2.20)
we get
ˆ 1˙
ˆ 1
1˙
g
1h 2 1 h 2
”
´ h´n p21 .
(2.21)
´
g
2h
4 h
Let
1
α2 “
1
g
1h
´
.
g
2h
From (2.21) we get
α22
(2.22)
ˆ 1˙
1 h 2
”
´ h´n p21 .
4 h
First we suppose that α2 ” 0. Then we get
ˆ 1˙
1 h 2
´n 2
h p1 ”
4 h
and so T pr, hq “ Spr, hq, which is impossible. Next we suppose that α2 ı 0.
Differentiating (2.22) we get
1 ˆ 1 ˙1
1h h
1
1
1
2α2 α2 ”
` n h h´n´1 p21 ´ 2h´n p1 p1 .
2h h
Applying (2.22) we obtain
ˆ
˙
1
1
1 ˆˆ 1 ˙1
1
1 ˙
h 2
α2 2
1h
h
h α2
1
´n
(2.23)
h
´n p1 ` 2p1 p1 ´ 2 p1 ”
´
.
h
α2
2h
h
h α2
First we suppose that
1
´n
1
h 2
α
1
p1 ` 2p1 p1 ´ 2 2 p21 ” 0.
h
α2
Then there exists a non-zero constant c such that α22 ” ch´n p21 and so from
(2.22) we get
ˆ 1˙
1 h 2
´n 2
pc ` 1qh p1 ”
.
4 h
If c “ ´1, then h will be a constant. If c ­“ ´1, then we have T pr, hq “ Spr, hq,
which is impossible. Next we suppose that
1
1
h
α
1
´n p21 ` 2p1 p1 ´ 2 2 p21 ı 0.
h
α2
On an open problem of Xiao-Bin Zhang and Jun-Feng Xu
173
Then by (2.23) we have
n T pr, hq “ n mpr, hq
ˆ
ˆ
1 ˆˆ 1 ˙1
1
1 ˙˙
h
h α2
n1 h
´
` m r,
≤ m r, h
2h
h
h α2
(2.24)
1
1 `` 1 ˘1
h
1h
2 h
h
˙
1
´
˘
h1 α2
h α2
` Op1q
ˆ
ˆ
˙
1
1
1 ˆˆ 1 ˙1
1 ˙˙
1
1h
h
h α2
h 2
α2 2
1
´
` m r, n p1 ` 2p1 p1 ´ 2 p1
≤ T r,
2h
h
h α2
h
α2
≤ N pr, 0; α2 q ` Spr, hq ` Spr, α2 q.
From (2.22) we get
1
T pr, α2 q ≤ n T pr, hq ` Spr, hq.
2
Now from (2.24) we get
1
n T pr, hq ≤ Spr, hq,
2
which is impossible.
Thus α and β are both polynomials. Also from (2.7) we can conclude
1
1
that αpzq ` βpzq ” C for a constant C and so α pzq ` β pzq ” 0. We deduce
from (2.7) that
1
1
1
1
1
1
rf n s ” nrhn1 α ` hn´1
h1 senα ” ppzqenα ,
1
(2.25)
and
h2 senβ ” ppzqenβ .
rg n s “ nrhn2 β ` hn´1
2
(2.26)
Since degppq ≤ n ´ 1, from (2.25) and (2.26) we conclude that both h1 and
h2 are nonzero constant. So we can rewrite f and g as follows:
f “ eγ2 ,
(2.27)
g “ e δ2 .
Now from (2.7) we get
1
1
n2 γ2 δ2 enpγ2 `δ2 q ” p2 .
(2.28)
Also from (2.28) we can conclude that γ2 pzq ` δ2 pzq ” C for a constant C
1
1
1 1
and so γ2 pzq ` δ2 pzq ” 0. Thus from (2.28) we get n2 enC γ2 δ2 ” p2 pzq. By
computation we get
1
γ2 “ cppzq,
(2.29)
1
δ2 “ ´cppzq.
Hence
γ2 “ cQpzq ` b1 ,
(2.30)
where Qpzq “
şz
0
δ2 “ ´cQpzq ` b2 ,
ppzqdz and b1 , b2 are constants. Finally we take f and g as
f pzq “ c1 ecQpzq ,
gpzq “ c2 e´cQpzq ,
where c1 , c2 and c are constants such that pncq2 pc1 c2 qn “ ´1.
174
S. Majumder
Case 2. Let ppzq be a nonzero constant b.
In this case we see that f and g have no zeros and so we can take f and
g as follows:
(2.31)
f “ eα ,
g “ eβ ,
where αpzq, βpzq are two non-constant entire functions.
We now consider the following two subcases:
Subcase 2.1. Let k ≥ 2.
We see that
N pr, 0; rf n spkq q “ 0.
From this and using (2.31) we have
(2.32)
f n pzqrf n pzqspkq ­“ 0.
Similarly we have
(2.33)
g n pzqrg n pzqspkq ­“ 0.
Then from (2.32), (2.33) and Lemma 5 we must have
(2.34)
f “ eaz`b ,
g “ ecz`d ,
where a ­“ 0, b, c ­“ 0 and d are constants.
Subcase 2.2. Let k “ 1.
Considering Subcase 1.2 one can easily get
(2.35)
f “ eaz`b ,
g “ ecz`d ,
where a ­“ 0, b, c ­“ 0 and d are constants.
Finally we can take f and g as
f “ c3 edz ,
g “ c4 e´dz ,
where c3 , c4 and d are nonzero constants such that p´1qk pc3 c4 qn pndq2k “ b2 .
This completes the proof.
Lemma 13. Let f and g be two transcendental meromorphic functions,
let ppzq be a nonzero polynomial with degppq ≤ n ´ 1, let n and k be two
positive integers with n ą k ` 2. Let P pwq be defined as in Theorem F and
rf n P pf qspkq , rg n P pgqspkq share p CM and also f , g share 8 IM. Suppose
that rf n P pf qspkq rg n P pgqspkq ” p2 , then P pzq reduces to a nonzero monomial,
namely P pzq “ ai z i ı 0 for some i P t0, 1, . . . , mu; ifş ppzq is not a constant,
z
then f “ c1 ecQpzq , g “ c2 e´cQpzq , where Qpzq “ 0 ppzqdz, c1 , c2 and c
2
n`i
2
are constants such that ai pc1 c2 q rpn ` iqcs “ ´1, if ppzq is a nonzero
constant b, then f “ c3 ecz , g “ c4 e´cz , where c3 , c4 and c are constants such
that p´1qk a2i pc3 c4 qn`i rpn ` iqcs2k “ b2 .
Proof. The proof of lemma follows from Lemmas 9 and 12.
On an open problem of Xiao-Bin Zhang and Jun-Feng Xu
175
Lemma 14. [1] Let f and g be two non-constant meromorphic functions
sharing 1 IM. Then
p2
N L pr, 1; f q ` 2N L pr, 1; gq ` N E pr, 1; f q ´ N f ą1 pr, 1; gq ´ N gą1 pr, 1; f q
≤ N pr, 1; gq ´ N pr, 1; gq.
Lemma 15. [1] Let f , g share 1 IM. Then
N L pr, 1; f q ≤ N pr, 0; f q ` N pr, 8; f q ` Spr, f q.
Lemma 16. [1] Let f , g share 1 IM. Then
1
(i) N f ą1 pr, 1; gq ≤ N pr, 0; f q ` N pr, 8; f q ´ N0 pr, 0; f q ` Spr, f q,
1
(ii) N gą1 pr, 1; f q ≤ N pr, 0; gq ` N pr, 8; gq ´ N0 pr, 0; g q ` Spr, gq.
Lemma 17. Suppose that f and g be two non-constant meromorphic functions and let P pωq be defined as in Theorem F. Let F “ rf n P pf qspkq ,
G “ rg n P pgqspkq , where np≥ 1q, mp≥ 0q and kp≥ 1q, are integers. If f ,
g share 8 IM and V ” 0, then F ” G.
Proof. Suppose that V ” 0. Then by integration we obtain
1
1
1´
” Ap1 ´ q.
F
G
If z0 is a pole of f then it is a pole of g. Hence from the definition of F and
G we have F pz1 0 q “ 0 and Gpz1 0 q “ 0. So A “ 1 and hence F ” G.
Lemma 18. Suppose that f and g be two non-constant meromorphic functions. Let F , G be defined as in Lemma 17 and H ı 0, where np≥ 1q, mp≥ 0q
and kp≥ 1q are three integers such that n ` m ą 3k ` 3. If f , g share 8 IM
and F , G share 1 IM, then
`
˘
n ` m ´ 3k ´ 3 N pr, 8; f q
≤ 2pk ` m ` 1qtT pr, f q ` T pr, gqu ` Spr, f q ` Spr, gq.
Similar result holds for g also.
Proof. Suppose that 8 is an e.v.P of f and g then the lemma follows
immediately.
Next suppose that 8 is not an e.v.P of f and g. Since H ı 0 from
Lemma 17 we have V ı 0. We suppose that z0 is a pole of f with multiplicity
q and a pole of g with multiplicity r. Clearly z0 is a pole of F with multiplicity
pn ` mqq ` k and a pole of G with multiplicity pn ` mqr ` k. Noting that f ,
g share 8 IM from the definition of V it is clear that z0 is a zero of V with
multiplicity at least n ` m ` k ´ 1. Now using the Milloux theorem (see [6],
p. 55), and Lemma 1, we obtain from the definition of V that
mpr, V q “ Spr, f q ` Spr, gq.
176
S. Majumder
Also by Lemma 15 we get
N ˚ pr, 1; F, Gq “ N L pr, 1; F q ` N L pr, 1; Gq
≤ N pr, 0; F q ` N pr, 8; F q ` N pr, 0; Gq ` N pr, 8; Gq
` Spr, f q ` Spr, gq
≤ 2pk ` 1qN pr, 8, f q ` pk ` m ` 1qT pr, f q
` pk ` m ` 1qT pr, gq ` Spr, f q ` Spr, gq.
Thus using Lemmas 1 and 2 we get
`
˘
n ` m ` k ´ 1 N pr, 8; f q
≤ N pr, 0; V q ≤ T pr, V q ` Op1q
≤ N pr, 8; V q ` mpr, V q ` Op1q
≤ N pr, 0; F q ` N pr, 0; Gq ` N ˚ pr, 1; F, Gq
` Spr, f q ` Spr, gq
≤ Nk`1 pr, 0; f n P pf qq ` Nk`1 pr, 0; g n P pgqq ` kN pr, 8; f q
` kN pr, 8; gq ` N ˚ pr, 1; F, Gq ` Spr, f q ` Spr, gq
≤ Nk`1 pr, 0; f n q ` Nk`1 pr, 0; P pf qq ` NK`1 pr, 0; g n q
` Nk`1 pr, 0; P pgqq ` 2kN pr, 8; f q ` N ˚ pr, 1; F, Gq
` Spr, f q ` Spr, gq
≤ pk ` 1qN pr, 0; f q ` N pr, 0; P pf qq ` pk ` 1qN pr, 0; gq
` N pr, 0; P pgqq ` 2kN pr, 8; f q ` N ˚ pr, 1; F, Gq
` Spr, f q ` Spr, gq.
This gives
`
˘
n ` m ´ 3k ´ 3 N pr, 8; f q
≤ 2pk ` m ` 1qtT pr, f q ` T pr, gqu ` Spr, f q ` Spr, gq.
This completes the proof of lemma.
Lemma 19. Suppose that f and g be two non-constant meromorphic functions. Let F , G be defined as in Lemma 17 and H ı 0, where np≥ 1q, mp≥ 0q
and kp≥ 1q are three integers such that n ` m ą k ` 1. If f , g share 8 IM
and F , G share 1 CM, then
`
˘
n ` m ´ k ´ 1 N pr, 8; f q
≤ pk ` m ` 1qtT pr, f q ` T pr, gqu ` Spr, f q ` Spr, gq.
Similar result holds for g also.
Proof. The proof follows from Lemma 18.
On an open problem of Xiao-Bin Zhang and Jun-Feng Xu
177
3. Proofs of the Theorems
Proof of Theorem 1. Let
F “
rf n P pf qspkq
ppzq
and G “
rg n P pgqspkq
.
ppzq
Note that since f and g are transcendental meromorphic functions, ppzq is
a small function with respect to both rf n P pf qspkq and rg n P pgqspkq . Also F ,
G share 1 CM except the zeros of ppzq and f , g share 8 IM.
Case 1. Let H ı 0.
From (2.1) it can be easily calculated that the possible poles of H occur
at (i) multiple zeros of F and G, (ii) those 1 points of F and G whose
multiplicities are different, (iii) those poles of F and G whose multiplicities
1
1
are different, (iv) zeros of F pG q which are not the zeros of F pF ´1qpGpG´1qq.
Since H has only simple poles we get
(3.1)
N pr, 8; Hq
≤ N ˚ pr, 8; F, Gq ` N ˚ pr, 1; F, Gq ` N pr, 0; F | ≥ 2q ` N pr, 0; G| ≥ 2q
1
1
` N 0 pr, 0; F q ` N 0 pr, 0; G q ` Spr, f q ` Spr, gq,
1
1
where N 0 pr, 0; F q is the reduced counting function of those zeros of F which
1
are not the zeros of F pF ´ 1q and N 0 pr, 0; G q is similarly defined.
Let z0 be a simple zero of F ´ 1 but ppz0 q ­“ 0. Then z0 is a simple zero
of G ´ 1 and a zero of H. So
(3.2)
N pr, 1; F | “ 1q ≤ N pr, 0; Hq ≤ N pr, 8; Hq ` Spr, f q ` Spr, gq.
Note that N ˚ pr, 1; F, Gq “ 0 and N ˚ pr, 8; F, Gq ≤ N pr, 8; f q. Now using
(3.1) and (3.2) we get
(3.3)
N pr, 1; F q
≤ N pr, 1; F | “ 1q ` N pr, 1; F | ≥ 2q
≤ N pr, 8; f q ` N pr, 0; F | ≥ 2q ` N pr, 0; G| ≥ 2q
1
1
` N pr, 1; F | ≥ 2q ` N 0 pr, 0; F q ` N 0 pr, 0; G q ` Spr, f q ` Spr, gq.
Now in view of Lemma 3 we get
(3.4)
1
N 0 pr, 0; G q ` N pr, 1; F |≥ 2q
1
≤ N pr, 0; G | G ­“ 0q ≤ N pr, 0; Gq ` N pr, 8; gq ` Spr, gq.
Hence using (3.3), (3.4), Lemmas 1, 2 and 19 we get from the second
178
S. Majumder
fundamental theorem that
(3.5)
pn ` mqT pr, f q
≤ T pr, F q ` Nk`2 pr, 0; f n P pf qq ´ N2 pr, 0; F q ` Spr, f q
≤ N pr, 0; F q ` N pr, 8; F q ` N pr, 1; F q ` Nk`2 pr, 0; f n P pf qq
1
´ N2 pr, 0; F q ´ N0 pr, 0; F q
≤ N pr, 8, f q ` N pr, 8; gq ` N pr, 0; F q ` Nk`2 pr, 0; f n P pf qq
1
` N pr, 0; F | ≥ 2q ` N pr, 0; G| ≥ 2q ` N pr, 1; F | ≥ 2q ` N 0 pr, 0; G q
´ N2 pr, 0; F q ` Spr, f q ` Spr, gq
≤ 3 N pr, 8; f q ` Nk`2 pr, 0; f n P pf qq ` N2 pr, 0; Gq ` Spr, f q ` Spr, gq
≤ 3 N pr, 8; f q ` Nk`2 pr, 0; f n P pf qq ` k N pr, 8; gq ` Nk`2 pr, 0; g n P pgqq
` Spr, f q ` Spr, gq
≤ p3 ` kq N pr, 8; f q ` pk ` 2q N pr, 0; f q ` T pr, P pf qq ` pk ` 2q N pr, 0; gq
` T pr, P pgqq ` Spr, f q ` Spr, gq
≤ pk ` m ` 2q tT pr, f q ` T pr, gqu ` p3 ` kqN pr, 8; f q ` Spr, f q ` Spr, gq
p3 ` kqpk ` m ` 1q
≤ pk ` m ` 2q tT pr, f q ` T pr, gqu `
tT pr, f q ` T pr, gqu
n`m´k´1
` Spr, f q ` Spr, gq

„
p3 ` kqpk ` m ` 1q
tT pr, f q ` T pr, gqu ` Spr, f q ` Spr, gq.
≤ k`m`2`
n`m´k´1
In a similar way we can obtain
(3.6) pn ` mqT pr, gq
„

p3 ` kqpk ` m ` 1q
≤ k`m`2`
tT pr, f q ` T pr, gqu ` Spr, f q ` Spr, gq.
n`m´k´1
Adding (3.5) and (3.6) we get

„
p6 ` 2kqpk ` m ` 1q
n ´ m ´ 2k ´ 4 ´
tT pr, f q ` T pr, gqu
n`m´k´1
≤ Spr, f q ` Spr, gq,
i.e.
rpn ` m ` 2qpn ´ m ´ 3k ´ 7q ` 4k ` 12stT pr, f q ` T pr, gqu
≤ Spr, f q ` Spr, gq,
which is a contradiction since n ą 3k ` m ` 6.
On an open problem of Xiao-Bin Zhang and Jun-Feng Xu
179
Case 2. Let H ” 0. Then by Lemma 11 we have either
rf n P pf qspkq rg n P pgqspkq ” p2 ,
(3.7)
or
f n P pf q ” g n P pgq.
(3.8)
From (3.8) we get
(3.9)
f n pam f m ` am´1 f m´1 ` . . . ` a0 q
” g n pam g m ` am´1 g m´1 ` . . . ` a0 q.
Let h “ fg . If h is a constant, then substituting f “ gh into (3.9) we deduce
that
am g n`m phn`m ´ 1q ` am´1 g n`m´1 phn`m´1 ´ 1q ` ¨ ¨ ¨ ` a0 g n phn ´ 1q ” 0,
which implies hd “ 1, where d “ GCDpn`m, . . . , n`m´i, . . . , nq, am´i ­“ 0
for some i “ 0, 1, . . . , m. Thus f ” tg for a constant t such that td “ 1, where
d “ GCDpn ` m, . . . , n ` m ´ i, . . . , nq, am´i ­“ 0 for some i “ 0, 1, . . . , m.
If h is not a constant, then we know by (3.9) that f and g satisfying the
algebraic equation Rpf, gq “ 0, where Rpω1 , ω2 q “ ω1n pam ω1m ` am´1 ω1m´1 `
. . . ` a0 q ´ ω2n pam ω2m ` am´1 ω2m´1 ` . . . ` a0 q.
Remaining part of the theorem follows from (3.7) and Lemma 13. This
completes the the proof of the theorem.
Proof of Theorem 2. In this case F and G share 1 IM. Also f and g
share 8 IM.
Case 1. Let H ı 0. Here we see that
(3.10)
1q
NE pr, 1; F |“ 1q ≤ N pr, 0; Hq ≤ N pr, 8; Hq ` Spr, F q ` Spr, Gq.
Now using Lemmas 3, 14, 15, 16, (3.1) and (3.10), we get
(3.11)
N pr, 1; F q
1q
p2
≤ NE pr, 1; F q ` N L pr, 1; F q ` N L pr, 1; Gq ` N E pr, 1; F q
≤ N pr, 8; f q ` N pr, 0; F | ≥ 2q ` N pr, 0; G| ≥ 2q ` N ˚ pr, 1; F, Gq
p2
1
` N L pr, 1; F q ` N L pr, 1; Gq ` N E pr, 1; F q ` N 0 pr, 0; F q
1
` N 0 pr, 0; G q ` Spr, f q ` Spr, gq
≤ N pr, 8; f q ` N pr, 0; F | ≥ 2q ` N pr, 0; G| ≥ 2q ` 2N L pr, 1; F q
p2
1
1
` 2N L pr, 1; Gq ` N E pr, 1; F q ` N 0 pr, 0; F q ` N 0 pr, 0; G q
` Spr, f q ` Spr, gq
180
S. Majumder
≤ N pr, 8; f q ` N pr, 0; F | ≥ 2q ` N pr, 0; G| ≥ 2q
` N F ą1 pr, 1; Gq ` N Gą1 pr, 1; F q ` N L pr, 1; F q ` N pr, 1; Gq
1
1
´ N pr, 1; Gq ` N 0 pr, 0; F q ` N 0 pr, 0; G q ` Spr, f q ` Spr, gq
≤ 4 N pr, 8; f q ` N2 pr, 0; F q ` N pr, 0; F q ` N2 pr, 0; Gq ` N pr, 1; Gq
1
1
´ N pr, 1; Gq ` N 0 pr, 0; G q ` N 0 pr, 0; F q ` Spr, f q ` Spr, gq
1
≤ 4 N pr, 8; f q ` N2 pr, 0; F q ` N pr, 0; F q ` N2 pr, 0; Gq ` N pr, 0; G |G ­“ 0q
1
` N 0 pr, 0; F q ` Spr, f q ` Spr, gq
≤ 5N pr, 8; f q ` N2 pr, 0; F q ` N pr, 0; F q ` N2 pr, 0; Gq ` N pr, 0; Gq
1
` N 0 pr, 0; F q ` Spr, f q ` Spr, gq.
Hence, using (3.11), Lemmas 1, 2 and 18 we get from second fundamental
theorem that
(3.12)
pn ` mqT pr, f q
≤ N pr, 0; F q ` N pr, 8; F q ` N pr, 1; F q ` Nk`2 pr, 0; f n P pf qq
1
´ N2 pr, 0; F q ´ N0 pr, 0; F q
≤ 6N pr, 8, f q ` N2 pr, 0; F q ` 2 N pr, 0; F q ` Nk`2 pr, 0; f n P pf qq
` N2 pr, 0; Gq ` N pr, 0; Gq ´ N2 pr, 0; F q ` Spr, f q ` Spr, gq
≤ 6N pr, 8; f q ` Nk`2 pr, 0; f n P pf qq ` 2 N pr, 0; F q ` N2 pr, 0; Gq
` N pr, 0; Gq ` Spr, f q ` Spr, gq
≤ 6N pr, 8; f q ` Nk`2 pr, 0; f n P pf qq ` 2 kN pr, 8; f q ` 2 Nk`1 pr, 0; f n P pf qq
` k N pr, 8; gq ` Nk`2 pr, 0; g n P pgqq ` kN pr, 8; gq ` Nk`1 pr, 0; g n P pgqq
` Spr, f q ` Spr, gq
≤ p4k ` 6q N pr, 8; f q ` p3k ` 4qN pr, 0; f q ` 3T pr, P pf qq
` p2k ` 3q N pr, 0; gq ` 2T pr, P pgqq ` Spr, f q ` Spr, gq
12pk ` m ` 1q
≤
tT pr, f q ` T pr, gqu ` 4kT pr, f q ` p3k ` 4qN pr, 0; f q
n ` m ´ 3k ´ 3
` 3T pr, P pf qq ` p2k ` 3q N pr, 0; gq ` 2T pr, P pgqq ` Spr, f q ` Spr, gq
"
*
24pk ` m ` 1q
≤ 9k ` 5m ` 7 `
T prq ` Sprq.
n ` m ´ 3k ´ 3
In a similar way we can obtain
"
*
24pk ` m ` 1q
(3.13) pn ` mq T pr, gq ≤ 9k ` 5m ` 7 `
T prq ` Sprq.
n ` m ´ 3k ´ 3
On an open problem of Xiao-Bin Zhang and Jun-Feng Xu
181
Combining (3.12) and (3.13), we see that

„
pn ´ 9k ´ 4m ´ 7qpn ` m ´ 3k ´ 3q ´ 24pk ` m ` 1q
T prq ≤ Sprq.
(3.14)
n ` m ´ 3k ´ 3
When n ą 9k ` 4m ` 11, (3.14) leads to a contradiction.
Case 2. Let H ” 0. Then by Lemma 11 we have either
(3.15)
f n P pf q ” g n P pgq,
or
(3.16)
rf n P pf qspkq rg n P pgqspkq ” p2 .
Remaining part follows from Theorem 1. This completes the proof of the
theorem.
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S. Majumder
DEPARTMENT OF MATHEMATICS
KATWA COLLEGE
KATWA, WEST BENGAL-713130, INDIA
E-mail: [email protected]
Received April 16, 2014; revised version June 17, 2014.