R E S E A R C H
Open Access
Meromorphic functions that share a
polynomial with their difference operators
Bingmao Deng
1, Dan Liu
1, Yongyi Gu
2and Mingliang Fang
1**Correspondence:
mlfang@scau.edu.cn
1Institute of Applied Mathematics,
South China Agricultural University, Guangzhou, China
Full list of author information is available at the end of the article
Abstract
In this paper,we prove the following result: Letfbe a nonconstant meromorphic function of finite order,p be a nonconstant polynomial, and c bea nonzero constant. If f,
cf, and
ncf(n≥2) share∞andpCM, thenf≡
cf. Our result provides a
difference analogue oftheresult of Chang and Fang in 2004 (Complex Var. Theory Appl. 49(12):871–895, 2004).
MSC: Primary 30D35; secondary 39B32
Keywords: Uniqueness; Meromorphic functions; Difference operators
1 Introduction and main results
In this paper, we use the base notations of the Nevanlinna theory of meromorphic func-tions which are defined as follows [9, 18, 19].
Letf be a meromorphic function. Throughout this paper, a meromorphic function al-ways means meromorphic in the whole complex plane.
Definition 1
m(r,f) = 1 2π
2π
0
log+freiθdθ.
m(r,f) is the average of the positive logarithm of|f(z)|on the circle|z|=r.
Definition 2
N(r,f) = r
0
n(t,f) –n(0,f)
t dt+n(0,f)logr,
N(r,f) = r
0
n(t,f) –n(0,f)
t dt+n(0,f)logr,
wheren(t,f) (n(t,f)) denotes the number of poles off in the disc|z| ≤t, multiples poles are counted according to their multiplicities (ignore multiplicity).n(0,f) (n(0,f)) denotes the multiplicity of poles off at the origin (ignore multiplicity).
N(r,f) is called the counting function of poles of f, andN(r,f) is called the reduced counting function of poles off.
Definition 3
T(r,f) =m(r,f) +N(r,f).
T(r,f) is called the characteristic function off. It plays a cardinal role in the whole theory of meromorphic functions.
Definition 4 Letf be a meromorphic function. The order of growth off is defined as follows:
ρ(f) = lim r→∞
log+T(r,f) logr .
Ifρ(f) <∞, then we say thatf is a meromorphic function of finite order.
Definition 5 Leta,f be two meromorphic functions. IfT(r,a) =S(r,f), whereS(r,f) =
o(T(r,f)), asr→ ∞outside of a possible exceptional set of finite logarithmic measure. Then we say thatais a small function off. And we useS(f) to denote the family of all small functions with respect tof.
Definition 6 Letf andgbe two meromorphic functions, andpbe a polynomial. We say that f andg sharepCM, provided thatf(z) –p(z) and g(z) –p(z) have the same zeros counting multiplicity.And if f and g have the same poles countingmultiplicity,then we say that f and g share∞CM.
In this paper, we also use some known properties of the characteristic functionT(r,f) as follows [9, 18, 19].
Property 1 Letfj(j= 1, 2, . . . ,q) beqmeromorphic functions in|z|<Rand 0 <r<R. Then
T
r, q
j=1
T(r,fj)
≤ q
j=1
T(r,fj), T
r, q
j=1
fj
≤ q
j=1
T(r,fj) +logq
hold for 1≤r<R.
Property 2 Suppose thatf is meromorphic in|z|<R(R≤ ∞) andais any complex num-ber. Then, for 0 <r<R, we have
T
r, 1
f –a
=T(r,f) +O(1).
Property 2 is the first fundamental theorem.
Property 3 Suppose thatf is a nonconstant meromorphic function anda1,a2, . . . ,anare
n≥3 distinct values in the extended complex plane. Then
(n– 2)T(r,f) < n
j=1
N
r, 1
f–aj
Property 3 is the second fundamental theorem. For more properties aboutT(r,f), please see [9, 18, 19].
For a meromorphic functionf(z), we define its shift byfc(z) =f(z+c) and its difference operators by
cf(z) =f(z+c) –f(z), ncf(z) =nc–1
cf(z)
.cf(z)
.
In[10]the following result was proved.
Theorem 1 Let f be a nonconstant meromorphic function, andabea nonzero finite com-plex number.If f,f,and fshare a CM,then f ≡f.
In 2001, Li and Yang [12] considered the casewhen f,f, andf(n)share one value.
Theorem 2 Let f be an entire function,abea finite nonzero constant, andn≥2 bea positive integer.If f,f,and f(n)share a CM,then f assumes the form
f(z) =bewz–a(1 –w)
w , (1.1)
where b,w are two nonzero constants satisfying wn–1= 1.
Remark1 It is easy to see that the functions in(1.1)really share value a, since whenb= 0 andwn–1= 1, fromf(j)(z) =a,j= 0, 1,n, it follows thatbwewz=afor eachj= 0, 1,n. So, the functionsf(j)–a,j= 0, 1,n,have the same zeros counting multiplicity.
In 2004, Chang and Fang [1] considered the casewhen f,f, andf(n)share a small func-tion.
Theorem 3 Let f be an entire function,a be a nonzero small function of f,and n≥2be a positive integer.If f,f,and f(n)share a CM,then f ≡f.
Recently, value distribution in difference analogue of meromorphic functions has be-come a subject of some interest, see, e.g., [2–8, 11].
In 2012 and 2014, Chen et al. [2, 3] considered difference analogue of Theorem 1 and Theorem 2, andestablished the following result.
Theorem 4 Let f be a nonconstant entire function of finite order,and a(≡0)∈S(f)be a periodic entire function with period c.If f,cf,andncf (n≥2)share a CM,thencf ≡ ncf.
For other related results,the reader is referred to the references due toLatreuch, El Farissi, Belaïdi [11], El Farissi, Latreuch, Asiri [5], El Farissi, Latreuch, Belaïdi and Asiri [6].
By Theorems 3 and 4, it is natural to ask: Can we provide a difference analogue of The-orem 3? Or, can we delete the condition that ‘a(z) is a periodic entire function with period
c’ in Theorem 4?
In this paper,we study the problem and prove the following result.
Theorem 5 Let f be a nonconstant meromorphic function of finite order,and p be a non-constant polynomial.If f,cf,andncf (n≥2)share p and∞CM,then f ≡cf.
If f is an entire function, then f ,cf , andncf have no poles, obviouslyf,cf andncf share∞CM. By Theorem 5, weconsequentlyget the following result.
Corollary 1 Let f be a nonconstant entire function of finite order,and n≥2 bea positive integer.If f,cf,andncf share z CM,then f ≡cf.
Example 1 Let A, a, b, c be four finite nonzero complex numbers satisfying a= b,
n(≥2)∈Nsatisfying [eAc– 1]n–1= 1,eAc– 1 =aa–b, andg(z) be a periodic entire func-tion with periodc, and letf(z) =g(z)eAz+b. By simple calculation, we obtain
ncf(z) =cf(z) =
eAc– 1f(z) +a1 –eAc+ 1.
It is easy to see that f,cf, andncf (n≥2) shareaCM, andf =cf wheneAc= 2. This example shows that ‘p(z)cannotbe a constant’ in Theorem 5.
Example2 LetA,b,cbe three nonzero finite complex numbers satisfyingeAc= 1, and
f(z) =eAz+b,p(z) =b.It is easy to see that f,
cf, andncfsharep(z) CM. Butcf≡0≡f. This example also shows that ‘p(z)cannotbe a constant’ in Theorem 5.
Example3 LetA,cbe two nonzero finite complex numbers satisfyingeAc= 2 andf(z) =
eAzcot(πz
c ). By simple calculation, we obtain
f(z) =cf =ncf =eAzcot
πz c
.
Obviously, for any polynomial p,f,cf, andncf sharepand∞CM. This example satisfies Theorem 5.
In Examples1and2, we havecf ≡tf+a(1 –t)and f(z) =eAz+B+a, respectively, when
f,cf, andncf (n≥2) share a nonzero constantaCM. Hence we posed the following problem.
Problem 1 Assume that f is a nonconstant entire function of finite order,ais a nonzero constant, and thatf,cf, andcnf(n≥2) shareaCM. Whether or not,one of the following
two cases occurs:
(1) cf ≡tf+a(1 –t), wheretis a constant satisfyingtn–1= 1,
2 Some lemmas
Lemma 1([4, 7]) Let f be a meromorphic function of finite order,and cbea nonzero com-plex constant.Then
Tr,f(z+c)=T(r,f) +S(r,f).
Lemma 2([7, 8]) Let c∈C,kbea positive integer,and f bea meromorphic function of finite order.Then
m
r, k cf(z)
f(z)
=S(r,f).
Lemma 3([18, 19]) Let n≥2be a positive integer.Suppose that fi(z) (i= 1, 2, . . . ,n)are meromorphic functions and gi(z) (i= 1, 2, . . . ,n)are entire functions satisfying
(i) ni=1fi(z)egi(z)≡0,
(ii) the orders offiare less than those ofegk–gl for1≤i≤n,1≤k<l≤n.
Then fi(z)≡0 (i= 1, 2, . . . ,n).
The following lemma is well known.
Lemma 4 Let the function f satisfy the following difference equation:
f(w+ 1) =α(w)f(w) +β(w)
in the complex plane.
Then the following formula holds:
f(w+k) =f(w) k–1
j=0
α(w+j) + k–1
l=0
β(w+l) k–1
j=l+1
α(w+j) (2.1)
for every w∈Cand k∈N+.
Formula (2.1) has many applications. For example, many solvable difference equations are essentially solved by using it (see [15–17]), and by using such obtained formulas the behavior of their solution can be studied (see, for example, recent papers [13, 14]; see also many related references therein). As another simple application, by using a linear change of variables, the following corollary is obtained:
Corollary 2 Let the function f satisfy the following difference equation:
f(w+c) =α(w)f(w) +β(w)
in the complex plane.
Then the following formula holds:
f(w+kc) =f(w) k–1
j=0
α(w+jc) + k–1
l=0
β(w+lc) k–1
j=l+1
α(w+jc)
From the ideas of Chang and Fang [1] and Chen and Li [3], we prove the following lemma.
Lemma 5 Let f be a nonconstant meromorphic function of finite order,p(≡0) be a poly-nomial,and n≥2be an integer.Suppose that
n
cf(z) –p(z)
f(z) –p(z) =e
α(z), cf(z) –p(z)
f(z) –p(z) =e
β(z), (2.2)
whereαandβare two polynomials, and that
Tr,eα+Tr,eβ=S(r,f). (2.3)
Thencf ≡tf+b(1 –t),where t,b are constants satisfying tn–1= 1and b= 0.Moreover,if
t= 1,then p(z)≡b.
Proof Firstly, we prove thatfcannot be a rational function. Otherwise, suppose thatf(z) =
P(z)/Q(z), whereP(z) andQ(z) are two co-prime polynomials. It follows from (2.2) thatf(z) andcf(z) share∞CM. We claim thatQ(z) is a constant. Otherwise, suppose that there existsz0such thatQ(z0+c) = 0. Sincef(z) andcf(z) share∞CM, and
cf(z) =
P(z+c)
Q(z+c)–
P(z)
Q(z)=
P(z+c)Q(z) –P(z)Q(z+c)
Q(z)Q(z+c) . (2.4)
We deduce that all zeros ofQ(z+c) must be the zeros ofQ(z). Otherwise, suppose that there existsz1such thatQ(z1+c) = 0 butQ(z1)= 0, then it follows from (2.4) thatz1+cis
a pole ofcf but not the pole off, which contradicts withf(z) andcf(z) share∞CM. Then we get
Q(z0+c) = 0 ⇒ Q(z0) = 0 ⇒ Q(z0–c) = 0 ⇒ · · · ⇒ Q(z0–lc) = 0.
This implies thatQ(z) has infinitely many zeros,which isa contradiction. Thus, the claim is proved.
Thenf is a nonconstant polynomial, suppose that
f(z) =akzk+ak–1zk–1+· · ·+a1z+a0, p(z) =bmzm+· · ·+b1z+b0.
Then we getcf(z) =f(z+c) –f(z),and obviously,degncf(z)≤degcf(z) <degf(z). Then it follows from (2.2) thatα(z),β(z) are constants,and we let eα(z)=a,eβ(z)=b. So we have
ncf(z) –p(z) =af(z) –p(z), cf(z) –p(z) =b
f(z) –p(z).
Then we getdegf =k≤degp=msincedegn
cf(z)≤degcf(z) <degf(z). Ifk<m, then we get a=b= 1. This implies f ≡cf(z)≡ncf, which contradicts with degcf(z) < degf(z). Ifk=mthen we geta=b=bk/(ak–bk),cf(z)≡ncf,and hence f(z) is a con-stant,which isa contradiction.
Case1.β(z) is a nonconstant polynomial. It follows from the second equation in (2.2)
It follows from (2.5) and (2.6) that
ncf(z) =
Sinceβ(z)is a nonconstant polynomial, we have that
where li (0≤i≤m) are constants satisfyinglm= 0 and m≥1. Obviously, for anyj∈ {0, 1, . . . ,n– 1}, we have
β(z+jc) =lmzm+ (lm–1+mlmjc)zm–1+· · ·+ m
t=0
lt(jc)t. (2.11)
From(2.8) and (2.11), we get
μn(z) =enlmzm+Pn,0(z)+λ
n–1,0e(n–1)lmz
m+P
n–1,0(z)+· · ·+λ
n–1,n–1e(n–1)lmz
m+P n–1,n–1(z)
+· · ·+λ1,0elmz
m+P
1,0(z)+· · ·+λ
1,n–1elmz
m+P
1,n–1(z), (2.12)
where Pi,j(z) are polynomials with degree less than m fori∈ {1, 2, . . . ,n}, j∈ {0, 1, . . . ,
Cni– 1}.
It follows from the first equation in (2.2) that
ncf(z) –eα(z)f(z) =p(z)1 –eα(z). (2.13)
From(2.7) and (2.13), we have
μn(z) –eα(z)
f(z) =p(z)1 –eα(z)–ν
n(z). (2.14)
From(2.3),(2.10)and since T(r,p) =S(r,f), we have that
T(r,p) +Tr,eα+Tr,eβ+T(r,μn) +T(r,νn) =S(r,f). (2.15)
Ifμn(z) –eα(z)≡0,by(2.14),(2.15), Property1, and Property2, we obtain
T(r,f) =T
r,p(z)(1 –e
α(z)) –νn(z)
μn(z) –eα(z)
=S(r,f),
which isa contradiction.
Henceμn(z) –eα(z)≡0. Combining this with (2.12), we get
ePn,0(z)enlmzm+λ
n–1,0ePn–1,0(z)+· · ·+λn–1,n–1ePn–1,n–1(z)
e(n–1)lmzm
+· · ·+λ1,0eP1,0(z)+· · ·+λ1,n–1eP1,n–1(z)
elmzm–eα(z)≡0. (2.16)
Next, we consider three subcases.
Case1.1.degα(z) >m. Then, for any 1≤i≤n, 1≤k<j≤n, we have
ρeα(z)–ilmzm=ρeα(z)=degα(z) >m, ρejlmzm–klmzm=m.
SincePi,j(z) are polynomials with degree less thanmfori∈ {1, 2, . . . ,n},j∈ {0, 1, . . . ,Cni– 1}, then fori= 1, 2, . . . ,n– 1,
ρ Ci
n–1
j=0
λi,jePi,j(z)
≤m– 1, ρePn,0(z)≤m– 1.
Case1.2.degα(z) <m. Then, for any 1≤i≤n, 1≤k<j≤n, we have
ρeα(z)–ilmzm=ρe–ilmzm=m, ρejlmzm–klmzm=m.
SincePi,j(z) are polynomials with degree less thanmfori∈ {1, 2, . . . ,n},j∈ {0, 1, . . . ,Cni– 1}, then fori= 1, 2, . . . ,n– 1,
ρ Ci
n–1
j=0
λi,jePi,j(z)
≤m– 1, ρePn,0(z)≤m– 1.
By(2.16)and using Lemma3, we obtainePn,0≡0, which is a contradiction.
Case1.3.degα(z) =m. Setα(z) =dzm+α∗(z), whered= 0 anddegα∗(z) <m. Rewrite (2.16) as
ePn,0(z)enlmzm+λ
n–1,0ePn–1,0(z)+· · ·+λn–1,n–1ePn–1,n–1(z)
e(n–1)lmzm
+· · ·+λ1,0eP1,0(z)+· · ·+λ1,n–1eP1,n–1(z)
elmzm–eα∗(z)edzm
≡0. (2.17)
Ifd=jlm, for anyj= 1, 2, . . . ,n, we have
ρedzm–jlmzm=ρe(d–jlm)zm=m, ρeα∗(z)<m.
Combining this with (2.17),by usingLemma 3, we get a contradiction.
Ifd=jlm, for somej= 1, 2, . . . ,n– 1, without loss of generality, we assume thatj= 1, then (2.17) can be rewritten as
ePn,0(z)enlmzm+λ
n–1,0ePn–1,0(z)+· · ·+λn–1,n–1ePn–1,n–1(z)
e(n–1)lmzm
+· · ·+λ1,0eP1,0(z)+· · ·+λ1,n–1eP1,n–1(z)–eα
∗(z)
elmzm≡0.
And then, by usingthe same argument as above, we get a contradiction. Hence,d=nlm. Rewrite (2.17) as
ePn,0(z)–eα∗(z)enlmzm+λ
n–1,0ePn–1,0(z)+· · ·+λn–1,n–1ePn–1,n–1(z)
e(n–1)lmzm +· · ·+λ1,0eP1,0(z)+· · ·+λ1,n–1eP1,n–1(z)
elmzm≡0.
Using the same argument as inCase1.1 andusing Lemma3, we obtain
λ1,0eP1,0(z)+· · ·+λ1,n–1eP1,n–1(z)≡0. (2.18)
Then, by (2.9) and (2.18), we get
n–1
t=0
λ1,teβ(z+tc)=nc–1e β(z)=
n–1
t=0
(–1)n–1–tCnt–1eβ(z+tc)≡0. (2.19)
Ifm≥2, then for anyt= 0, 1, . . . ,n– 1, we have
From(2.19) with (2.20), we get
eqn–1(z)elmzm+(lm–1+mlm(n–1)c)zm–1– (n– 1)eqn–2(z)elmzm+(lm–1+mlm(n–2)c)zm–1
+· · ·+ (–1)n–1eq0(z)elmzm+(lm–1+mlmc)zm–1≡0.
Using the same argument asCase1.1, we obtain a contradiction.
Hencem= 1. Thusβ(z) =l1z+l0, wherel1= 0. Then, for anyn≥1, we deduce that
nc–1eβ(z)=el1c– 1n–1eβ(z).
Hence, it follows from (2.19) that (el1c– 1)n–1≡0, which yieldsel1c= 1. Then, for any t∈N+, we have
eβ(z+tc)=el1z+tl1c+l0=el1z+l0el1ct=eβ(z). (2.21)
By the second equation in (2.2) and (2.21), we get
cf(z) =eβ(z)f(z) +p(z)
1 –eβ(z)=eβ(z)f(z) +1 –eβ(z)b1(z),
2cf(z) =eβ(z)cf(z) +cp(z)
1 –eβ(z)
=eβ(z)eβ(z)f(z) +p(z)1 –eβ(z)+cp(z)
1 –eβ(z)
=e2β(z)f(z) +1 –eβ(z)p(z)eβ(z)+cp(z)
=e2β(z)f(z) +1 –eβ(z)b 2(z).
By mathematical induction, it is easy to get, for any integert≥2,
tcf(z) =etβ(z)f(z) +1 –eβ(z)bt(z),
whereb1(z) =p(z),bt(z) =p(z)e(t–1)β(z)+cbt–1=
t–1
i=0e(t–1–i)
β(z)i cp(z). Hence,
ncf(z) =enβ(z)f(z) +1 –eβ(z)b
n(z), (2.22)
wherebn(z) =ni=0–1e(n–1–i)β(z)i cp(z).
Fromthe first equation in (2.2) and (2.22), we have
eα(z)–enβ(z)f(z) =1 –eβ(z)bn(z) –p(z)1 –eα(z). (2.23)
Ifeα(z)–enβ(z)≡0,then by(2.15),(2.23), Property1, and Property2, we have
T(r,f) =T
r,(1 –e
β(z))bn(z) –p(z)(1 –eα(z))
eα(z)–enβ(z)
=S(r,f),
Henceeα(z)–enβ(z)≡0. It follows from (2.22) and (2.23) that
By mathematical induction, it is easy to get, for any integert≥2,
Hence,
ncf(z) =e(n–1)βcf(z) +
1 –eβbn(z), (2.25)
wherebn(z) = n–1
i=1 icp(z)e(n–1–i)β.
Using the same argument as the above, it is easy to geteα=enβ. Then it follows from (2.2)and eα=enβthat
ncf(z) =e(n–1)βcf(z) +
1 –e(n–1)βp(z). (2.26)
Ifcf(z)≡cnf(z), it follows from (2.26) thate(n–1)β= 1. Combining (2.25) and (2.26), we have
1 –eβ
n–1
i=1
icp(z)e(n–1–i)β=1 –eβbn(z) =1 –e(n–1)βp(z). (2.27)
Ifp(z) is a constant, then the left-hand side of equation (2.27) is equal to 0, and hence
p(z)≡0,which isa contradiction.
Ifp(z) is a nonconstant polynomial, letd=degp(z)≥1, then the left-hand side of equa-tion (2.27) is a polynomial with degree less thand, but the right-hand side of the equation is a polynomial with degreed,which isa contradiction.
Hencecf(z)≡cnf(z), ande(n–1)β= 1.
Ifeβ= 1 andp(z) is a nonconstant polynomial, then it follows from (2.25)–(2.26) that
bn(z)≡0.Thus
n–1
i=1
icp(z)e(n–1–i)β≡0. (2.28)
Letp(z) =amzm+am–1zm–1+· · ·+a0. It follows thatdegicp(z) =m–iifm≥i. Ifm≥2, then the left-hand side of (2.28) is a polynomial with degreem– 1≥1,which isa contra-diction.
Hencem= 1, that is,p(z) =a1z+a0. Thuscp(z) =a1c= 0. It follows from (2.28) that
a1ce(n–2)β= 0,which isa contradiction.
From the above discussion, we obtain that ifeβ= 1, thenp(z) (≡b) is a nonzero constant, hence
cf(z) =eβf(z) +p(z)
1 –eβ=eβf(z) +b1 –eβ=tf(z) +b(1 –t),
wheret=eβsatisfyingtn–1= 1.
Thus, Lemma 5 is proved.
Lemma 6(Hadamard’s factorization theorem [18]) Let f be an entire function of finite orderρ(f)with zeros{z1,z2, . . .} ⊂C\{0}and a k-fold zero at the origin.Then
f(z) =zkα(z)eβ(z),
3 Proof of Theorem 5
Proof Since the order of f is finite, and f, cf, cnf share∞ and p(z) CM, obviously (n
cf(z) –p(z))/(f(z) –p(z)) and (cf(z) –p(z))/(f(z) –p(z)) have no zeros and poles.By
Lemmas1and6, we have
ncf(z) –p(z)
f(z) –p(z) =e
α(z), cf(z) –p(z)
f(z) –p(z) =e
β(z), (3.1)
whereα(z)andβ(z)are two polynomials with degree≤ρ(f).
Using the same discussion as in Lemma 5, we deduce thatfcannot be a rational function. Hence,f is a transcendental meromorphic function, andT(r,p) =S(r,f).
SetF(z) :=f(z) –p(z), thenT(r,f) =T(r,F) +S(r,f) andT(r,p) =S(r,F). Obviously, we have
f(z) =F(z) +p(z), cf(z) =cF(z) +cp(z), ncf(z) =ncF(z) +ncp(z).
Rewrite (3.1) as
n
cF(z) +ncp(z) –p(z)
F(z) =e
α(z), cF(z) +cp(z) –p(z)
F(z) =e
β(z). (3.2)
Since p(z)is a nonconstantpolynomial, it follows thatncp(z) –p(z)≡0 andcp(z) –
p(z)≡0. Set
φ(z) :=(p(z) – n
cp(z))cF(z) – (p(z) –cp(z))ncF(z)
F(z) . (3.3)
Next, we consider two cases.
Case1.φ(z)≡0. Then, byT(r,p) =S(r,F), Lemma 1, and Lemma 2, we get
m(r,φ) =S(r,F). (3.4)
By (3.2)–(3.3), we can rewriteφ(z) as
φ(z) =p(z) –ncp(z)eβ(z)–p(z) –cp(z)
eα(z). (3.5)
Sincep(z) is a polynomial, we deduce thatN(r,φ) =S(r,F). Hence, we get
T(r,φ) =m(r,φ) +N(r,φ) =S(r,F). (3.6)
Sinceφ(z)≡0, by (3.5) we have
p(z) –ncp(z)e β(z)
φ(z)= 1 +
p(z) –cp(z) eα(z)
Then by (3.6), (3.7),T(r,p) =S(r,F), Property 2, and Property 3, we have
Hence by(3.6), Property1, and the previous inequality, we get
Tr,eβ=S(r,F). (3.8)
By simple calculation, we can rewrite (3.10) as follows:
a constant, which is a contradiction.
This completes the proof of Theorem5.
Acknowledgements
Research supported by the NNSF of China (Grant No. 11371149; 11701188) and the Graduate Student Overseas Study Program from South China Agricultural University (Grant No. 2017LHPY003).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Author details
1Institute of Applied Mathematics, South China Agricultural University, Guangzhou, China.2School of Mathematics and
Information Science, Guangzhou University, Guangzhou, China.
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Received: 21 August 2017 Accepted: 9 May 2018
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