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Kyungpook Mathematical Journal 2018; 58(3): 519-531

Published online September 30, 2018

Copyright © Kyungpook Mathematical Journal.

Uniqueness of Entire Functions Sharing Polynomials with Their Derivatives

Pulak Sahoo*
Department of Mathematics, University of Kalyani, West Bengal-741235, India
e-mail : sahoopulak1@gmail.com

Gurudas Biswas
Department of Mathematics, Hooghly Women’s College, West Bengal-712103, India
e-mail : gdb.math@gmail.com

Received: May 25, 2017; Accepted: June 28, 2018

Abstract

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In this paper, we investigate the uniqueness problem of entire functions sharing two polynomials with their k-th derivatives. We look into the conjecture given by Lü, Li and Yang [Bull. Korean Math. Soc., 51(2014), 1281–1289] for the case F = fnP(f), where f is a transcendental entire function and P(z) = amzm+am−1zm−1+ …+a1z+a0(≢ 0), m is a nonnegative integer, am, am−1, …, a1, a0 are complex constants and obtain a result which improves and generalizes many previous results. We also provide some examples to show that the conditions taken in our result are best possible.

Keywords: entire function, derivative, uniqueness.

1. Introduction, Definitions and Results

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In this paper, by meromorphic (entire) function we shall always mean meromorphic (entire) function in the complex plane. We assume that the reader is familiar with the standard notations of Nevanlinna’s theory of meromorphic functions as explained in [7, 9, 17]. For a nonconstant meromorphic function f, we denote by T(r, f) the Nevanlinna Characteristic function of f and by S(r, f) any quantity satisfying S(r, f) = o{T(r, f)} for all r outside a possible exceptional set of finite logarithmic measure. The meromorphic function a is called a small function of f, if T(r, a) = S(r, f), where r → ∞ outside a possible exceptional set of finite measure.

Let k be a positive integer or infinity and a ∈ ℂ∪{∞}. We denote by Nk)(r, a; f) the counting function of all those a-points of f whose multiplicities are not greater than k and by N(k+1(r, a; f) the counting function of all those a-points of f whose multiplicities are greater than k.

Let f and g be two nonconstant meromorphic functions and Q1, Q2 be two polynomials or complex numbers. If f − Q1 and g − Q2 have the same zeros with the same multiplicities, then we say that f − Q1 and g − Q2 share the value 0 CM. Especially, if Q1 = Q2 = a, where a ∈ ℂ ∪ {∞}, then we say that f and g share the value a CM, when f − a and g − a have the same zeros with the same multiplicities. The uniqueness problem of entire and meromorphic functions sharing values, small functions, polynomials with their derivatives is an interesting topic of value distribution theory. Many mathematicians (see [5, 8, 16, 19, 20, 21]) worked on this topic and they gave many conjectures and results. In 1976, Rubel and Yang [14] first proved a result which is as follows.

Theorem A

If a nonconstant entire function f and its derivative f′ share two distinct finite values CM, then f = f′.

Theorem A suggests the following question.

Question 1

What can be said if a nonconstant entire function f shares one finite value CM with its derivative f′?

In 1996, Brück [2] presented the following conjecture relating to Question 1.

Conjecture 1

Let f be a nonconstant entire function. Suppose that ρ1(f), the first iterated order of f, is not a positive integer or infinite whereρ1(f)=lim supr→∞log log T(r,f)log rand if f and f′ share one finite value a CM, thenf′-af-a=c, for some nonzero constant c.

In 1996, Brück [2] proved that the conjecture is true if a = 0 or N(r, 0; f′) = S(r, f). In 1998, Gundersen and Yang [6] proved that the conjecture is true if f is of finite order and fails, in general, for meromorphic functions. In 2004, Chen and Shon [3] proved that the conjecture is true for entire function of order ρ1(f)<12. In 2005, Al-Khaladi [1] proved that the conjecture is true for meromorphic function f when N(r, 0; f′) = S(r, f).

Now it is natural to ask the following question.

Question 2

Whether Brück Conjecture holds if the function f is replaced by its n-th power fn?

In 2008, Yang and Zhang [18] answered the above question by proving the following result.

Theorem B

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Let f be a nonconstant entire function, n ≥ 7 be an integer and let F = fn. If F and F′ share 1 CM, then F = F′ and f assumes the formf(z)=ce1nz, where c is a nonzero constant.

In 2010, Zhang and Yang [22] further improved Theorem B by considering k-th derivative of fn as follows.

Theorem C

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Let f be a nonconstant entire function and n, k be two positive integers such that n ≥ k + 1. If fn and (fn)(k)share the value 1 CM, then fn = (fn)(k)and f assumes the formf(z)=ceλnz, where c, λ are nonzero constants and λk = 1.

In 2011, Lü and Yi [11] considered polynomial sharing instead of value sharing and proved the following result.

Theorem D

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Let f be a transcendental entire function, n, k be two positive integers and Q ≢ 0 be a polynomial. If fn − Q and (fn)(k) − Q share the value 0 CM and n ≥ k + 1, then fn = (fn)(k)and f assumes the formf(z)=ceλnz, where c, λ are nonzero constants and λk = 1.

Regarding Theorem D one may ask the following question.

Question 3

What can be said if fn −Q1and (fn)(k) −Q2share the value 0 CM, where Q1, Q2are polynomials with Q1Q2 ≢ 0?

In 2014, Lü, Li and Yang [10] answered the above question for k = 1 and obtained the following result.

Theorem E

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Let f be a transcendental entire function and n ≥ 2 be an integer. If fn − Q1and (fn)′ − Q2share the value 0 CM, thenQ2Q1is a polynomial andf′=Q2nQ1f. Furthermore, if Q1 = Q2, thenf(z)=ce1nz, where Q1, Q2are polynomials with Q1Q2 ≢ 0, and c is a nonzero constant.

In the same paper the authors posed the following conjecture.

Conjecture 2

Let f be a transcendental entire function and n, k be two positive integers such that n ≥ k + 1. If fn − Q1and (fn)(k) − Q2share the value 0 CM, then(fn)(k)=Q2Q1fn. Furthermore, if Q1 = Q2, thenf(z)=ceλnz, where Q1, Q2are polynomials with Q1Q2 ≢ 0, and c, λ are nonzero constants such that λk = 1.

Recently Majumder [12] showed that the above conjecture is true for any positive integer k. The following two examples given in [12] respectively shows that the condition n ≥ k + 1 and f is transcendental in Conjecture 2 are essential.

Example 1

Let f(z) = e2z + z. Then f − Q1 and f′ − Q2 share 0 CM, but f′≢Q2Q1f, where Q1(z) = z + 1 and Q2(z) = 3.

Example 2

Let f(z) = z. Then f2 − Q1 and (f2)′ − Q2 share 0 CM, but (f2)′≢Q2Q1f2, where Q1(z) = 2z2 + z and Q2(z) = 2z2 + 4z.

In [10] the authors posed the following two questions.

Question 4

What can be said if in Conjecture 2 the condition “fn” be replaced by “P(f)” whereP(z)=∑i=0naizi?

Question 5

What can be said if in Conjecture 2 the condition “fn” be replaced by “f(z + c1)f(z + c2) …f(z + cn)” where cj (j = 1, 2, …, n) are constants?

Our aim to write this paper is to investigate the Conjecture due to Lü, Li and Yang by considering the function F = fnP(f) where f is a transcendental entire function and P(z)=∑i=0maizi, a0, a1, …, am(≠ 0) are complex constants. Though we are able to find out an affirmative solution of Question 4 as far as we know Question 5 remains open. The following is the main result of the paper.

Theorem 1

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Let f be transcendental entire function and n, m, k be positive integers such that n ≥ m+k+1. If fnP(f)−Q1and (fnP(f))(k)−Q2share 0 CM, then P(z) reduces to a nonzero monomial, namely P(z) = aizi for some i ∈ {0, 1, …, m} and(fn+i)(k)=Q2Q1fn+i. Furthermore, if Q1 = Q2, then f assumes the formf(z)=ceλn+iz, where Q1, Q2are polynomials with Q1Q2 ≢ 0 and c, λ are nonzero constants such that λk = 1.

The condition n ≥ m+k+1 in Theorem 1 is essential as shown by the following example.

Example 3

Let f(z) = ez − 1 and P(f) = f2 + 3f + 3. Then fP(f) − Q1 and (fP(f))′ − Q2 share 0 CM, but (fP(f))′≢Q2Q1fP(f), where Q1(z) = z + 1 and Q2(z) = 3z + 6.

The following example shows that the hypothesis of transcendental of f in Theorem 1 is necessary.

Example 4

Let f(z) = z − 1 and P(f) = f + 1. Then f3P(f) − Q1 and (f3P(f))′ − Q2 share 0 CM, but (f3P(f))′≢Q2Q1f3P(f), where Q1(z) = z4 + 3z2 and Q2(z) = −14z3 − 9z2 − 1.

2. Lemmas

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In this section we present some lemmas which will be needed in the sequel.

Lemma 1

([15]) Let f be a nonconstant meromorphic function and an(z)(≢ 0), an−1(z), …, a1(z), a0(z) be meromorphic functions such that T(r, ai(z)) = S(r, f) for i = 0, 1, 2, …, n. Then

T(r,anfn+an-1fn-1+…+a1f+a0)=nT(r,f)+S(r,f).

Lemma 2

([4]) Suppose that f is a transcendental meromorphic function and that

fnP*(f)=Q*(f),

where P*(f) and Q*(f) are differential polynomials in f with functions of small proximity related to f as the coefficients and the degree of Q*(f) is at most n. Then

m(r,∞;P*(f))=S(r,f).

Lemma 3

([7]) Let f be a nonconstant meromorphic function and let 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¯(r,a1;f)+N¯(r,a2;f)+S(r,f).

Lemma 4

([13]) Let f be a nonconstant meromorphic function and n, k, m be positive integers such that n ≥ k + 1. If fnP(f) = {fnP(f)}(k), then P(z) reduces to a nonzero monomial, namely P(z) = aizi for some i ∈ {0, 1, …, m}; and fn+i ≡ (fn+i)(k), where f assumes the formf(z)=ceλn+iz, where c is a nonzero constant and λk = 1.

3. Proof of the Theorem

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Proof of the Theorem 1

Let F*=FQ1, G*=GQ2, where F = fnP(f) and G = (fnP(f))(k). Clearly F* and G* share 1 CM except for the zeros of Qi(z), where i = 1, 2 and so N̄(r, 1; F*) = N̄(r, 1;G*) + S(r, f). Let

W=F*′(F*-G*)F*(F*-1).

We now consider the following two cases.

Case 1

Let W ≢ 0. It is obvious that m(r, ∞; W) = S(r, f).

Let z0 be zero of f with multiplicity p0(≥ 1) which is a zero of P(f) with multiplicity q0(≥ 1) such that Qi(z0) ≠ 0, where i = 1, 2. Then from (3.1), we obtain

W(z)=O ((z-z0)np0+q0-k-1).

Since n ≥ m+k+1 and f is transcendental entire, we see that N(r, ∞; W) = S(r, f). Consequently T(r, W) = S(r, f).

Now from (3.1) we see that

1F*=1WF*′F*(F*-1)(1-G*F*).

Therefore, it follows from above that m(r, 0; F*) = S(r, f), and hence m(r, 0; f) = S(r, f). Also

N(r,0;f)≤N(r,0;W)≤T(r,1W)≤T(r,W)+O(1)=S(r,f).

Hence T(r, f) = S(r, f), a contradiction.

Let z1 be zero of f with multiplicity p1(≥ 1) which is not a zero of P(f) and Qi(z1) ≠ 0 for i = 1, 2. Then as before we obtain

W(z)=O ((z-z1)np1-k-1),T(r,W)=S(r,f) and m(r,0;f)=S(r,f).

We now discuss the following two subcases.

Subcase 1

Let n > m+ k + 1. Then from (3.3), we obtain

N(r,0;f)≤N(r,0;W)≤T(r,1W)≤T(r,W)+O(1)=S(r,f).

Hence from (3.4) and (3.5), we get T(r, f) = S(r, f), a contradiction.

Subcase 2

Let n = m + k + 1. Then from (3.3) we see that

N(2(r,0;f)≤N(r,0;W)≤T(r,1W)≤T(r,W)+O(1)=S(r,f).

Then (3.4) gives

T(r,f)=N1)(r,0;f)+S(r,f).

Let

F=fnP(f) =amfn+m+am-1fn+m-1+…+a1fn+1+a0fn =Fm+Fm-1+…+F1+F0,         say.

Since F − Q1 and F(k) − Q2 share 0 CM, there exists an entire function α, such that

F(k)-Q2=eα(F-Q1).

First we assume that eα is not constant. Differentiating (3.8), we get

F(k+1)-Q2′=α′eα(F-Q1)+eα(F′-Q1′).

From (3.8) and (3.9) we obtain

F(k+1)F-α′F(k)F-F(k)F′=Q1F(k+1)-(α′Q1+Q1′)F(k)-Q2F′+(Q2′-α′Q2)F+α′Q1Q2+Q2Q1′-Q1Q2′.

From (3.8) we see that

T(r,eα)≤(k+2)T(r,F)+O(log r)+S(r,F)=(n+m)(k+2)T(r,f)+S(r,f).

Since T(r, α′) = S(r, eα), it follows that T(r, α′) = S(r, f). Now from (3.7), we deduce for i ∈ {0, 1, …, m} that

Fi′=ai(n+i)fn+i-1f′ =fm{ai(n+i)fn+i-m-1f′}Fi′′=ai(n+i)(n+i-1)fn+i-2(f′)2+ai(n+i)fn+i-1f″ =fm{ai(n+i)(n+i-1)fn+i-m-2(f′)2+ai(n+i)fn+i-m-1f″}Fi′′′=ai(n+i)(n+i-1)(n+i-2)fn+i-3(f′)3  +3ai(n+i)(n+i-1)fn+i-2f′ f″+ai(n+i)fn+i-1f‴ =fm{ai(n+i)(n+i-1)(n+i-2)fn+i-m-3(f′)3  +3ai(n+i)(n+i-1)fn+i-m-2f′ f″+ai(n+i)fn+i-m-1f‴}

and so on.

Thus in general we have

Fi(k)=fm∑λiaλifl0λi(f′)l1λi…(f(k))lkλi,

where l0λi, l1λi,…, lkλi are nonnegative integers satisfying ∑j=0kljλi=n+i-m,n+i-m-k≤l0λi≤n+i-m-1 and aλi are constants for i ∈ {0, 1, …, m}. Also we have

Fi(k+1)=fm∑λibλifp0λi(f′)p1λi…(f(k+1))pk+1λi,

where p0λi, p1λi,…, pk+1λi are nonnegative integers satisfying ∑j=0k+1pjλi=n+i-m,n+i-m-k-1≤p0λi≤n+i-m-1 and bλi are constants for i ∈ {0, 1, …, m}.

Thus we have from (3.7)

F(k)=fm{∑λmaλmfl0λm(f′)l1λm…(f(k))lkλm+∑λm-1aλm-1fl0λm-1(f′)l1λm-1…(f(k))lkλm-1+…+∑λ1aλ1fl0λ1(f′)l1λ1…(f(k))lkλ1+∑λ0aλ0fl0λ0(f′)l1λ0…(f(k))lkλ0}

and

F(k+1)=fm{∑λmbλmfp0λm(f′)p1λm…(f(k+1))pk+1λm+∑λm-1bλm-1fp0λm-1(f′)p1λm-1…(f(k+1))pk+1λm-1+…+∑λ1bλ1fp0λ1(f′)p1λ1…(f(k+1))pk+1λ1+∑λ0bλ0fp0λ0(f′)p1λ0…(f(k+1))pk+1λ0}.

Using (3.7), (3.11) and (3.12) in (3.10), we obtain

fn+mP*(f)=Q*(f),

where Q*(f) is a differential polynomial in f of degree n + m and

P*(f)={∑λmbλmfp0λm(f′)p1λm…(f(k+1))pk+1λm+∑λm-1bλm-1fp0λm-1(f′)p1λm-1…(f(k+1))pk+1λm-1+…+∑λ1bλ1fp0λ1(f′)p1λ1…(f(k+1))pk+1λ1+∑λ0bλ0fp0λ0(f′)p1λ0…(f(k+1))pk+1λ0}P(f)-α′{∑λmaλmfl0λm(f′)l1λm…(f(k))lkλm+∑λm-1aλm-1fl0λm-1(f′)l1λm-1…(f(k))lkλm-1+…+∑λ1aλ1fl0λ1(f′)l1λ1…(f(k))lkλ1+∑λ0aλ0fl0λ0(f′)l1λ0…(f(k))lkλ0}P(f)-f′{∑λmaλmfl0λm-1(f′)l1λm…(f(k))lkλm+∑λm-1aλm-1fl0λm-1-1(f′)l1λm-1…(f(k))lkλm-1+⋯+∑λ1aλ1fl0λ1-1(f′)l1λ1…(f(k))lkλ1+∑λ0aλ0fl0λ0-1(f′)l1λ0…(f(k))lkλ0}P1(f)=A(f′)k+1+R*(f)

is a differential polynomial in f of degree n + m, where A is a suitable constant, P1(f) = am(n+m)fm+am−1(n+m−1)fm−1+ …+a1(n+1)f +a0n and R*(f) is a differential polynomial in f. Actually every monomial of R*(f) has the form

Ri(α′)fq0λi(f′)q1λi…(f(k+1))qk+1λi,

where q0λi, q1λi,…, qk+1λi are nonnegative integers satisfying ∑j=0k+1qjλi=n+2i-m, n+2i-m-k≤q0λi≤n+2i-m-1 and Ri(α′) are polynomials in α′ with constant coefficients for i ∈ {0, 1, …, m}.

First we suppose that P*(f) ≢ 0. Then by Lemma 2, we get m(r, ∞; P*) = S(r, f) and so T(r, P*) = S(r, f). Consequently, T(r,P*′)=S(r,f).

Note that from (3.14)

P*′(f)=A(k+1)(f′)kf″+Bα′(f′)k+1+S*(f)

is a differential polynomial in f, where B is a suitable constant and S*(f) is a differential polynomial in f. Actually every monomial of S*(f) has the form

Si(α′)fr0λi(f′)r1λi…(f(k+1))rk+1λi,

where r0λi, r1λi,…, rk+1λi are nonnegative integers satisfying ∑j=0k+1rjλi=n+2i-m, n+2i-m-k≤r0λi≤n+2i-m-1 and Si(α′) are polynomials in α′ with constant coefficients for i ∈ {0, 1, …, m}.

Let z2 be simple zero of f. Then from (3.14) and (3.15), we obtain

P*(f(z2))=A(f′(z2))k+1,

and

P*′(f(z2))=A(k+1)(f′(z2))kf″(z2)+Bα′(f′(z2))k+1.

This shows that z2 is a zero of P*f″-(c1P*′-c2α′P*)f′, where c1 and c2 are suitable constants. Let

Φ=P*f″-(c1P*′-c2α′P*)f′f.

Clearly Φ ≢ 0 and T(r, Φ) = S(r, f).

From (3.16) we have

f″=α1f+β1f′,

where

α1=ΦP*,   β1=c1P*′P*-c2α′

and

T(r,α1)=S(r,f),   T(r,β1)=S(r,f).

(3.14) and (3.18) together gives

P*′=(β1c1+c2c1α′) P* =A (β1c1+c2c1α′) (f′)k+1+(β1c1+c2c1α′) R*(f).

Using (3.15) and (3.17), we get

P*′=A(k+1)α1f(f′)k+{A(k+1)β1+Bα′}(f′)k+1+S*(f).

By (3.19) and (3.20), we have

(Ac1β1-A(k+1)β1+Ac2c1α′-Bα′) (f′)k+1-A(k+1)α1f(f′)k+(β1c1+c2c1α′) R*(f)-S*(f)≡0.

Since α1 ≢ 0, from (3.21) we get

N1)(r,0;f)=S(r,f).

Therefore from (3.6) and (3.22) we have

T(r,f)=S(r,f),

a contradiction.

Next we suppose that P*(f) ≡ 0. Then Q*(f) ≡ 0 by (3.13), where

Q*(f)=Q1F(k+1)-(α′Q1+Q1′)F(k)-Q2F′+(Q2′-α′Q2)F+α′Q1Q2+Q2Q1′-Q1Q2′.

So from (3.10) it follows that

F(k+1)F-α′F(k)F-F(k)F′≡0,

i.e.,

F(k+1)F(k)≡α′+F′F.

Integrating we obtain F(k) = c3Feα, where c3 is a nonzero constant. Substituting the values of F and F(k) into (3.8) we obtain

(c3-1)fnP(f)=Q2-Q1eαeα.

Clearly c3 ≠ 1 and all zeros of Q2 − Q1eα have the multiplicities at least n. Since n = m + k + 1, by Lemma 3 we get

T(r,eα)≤N¯(r,0;eα)+N¯(r,∞;eα)+N¯ (r,Q2Q1;eα)+S(r,eα) ≤1nN (r,Q2Q1;eα)+S(r,eα) ≤1nT(r,eα)+S(r,eα),

which contradicts to the assumption that eα is a nonconstant entire function.

Next we assume that eα is a constant, say c4. Then from (3.8), we have

F(k)-c4F≡Q2-c4Q1.

Since n = m + k + 1, it follows that

N(r,0;f)=S(r,f)

and hence by (3.4) we have

T(r,f)=S(r,f),

a contradiction.

Case 2

Let W ≡ 0. Then from (3.1) we get F* = G*,

i.e.,   (fnP(f))(k)=Q2Q1fnP(f).

If Q1 and Q2 are same polynomials then using Lemma 4 we can get the conclusion of the theorem. Next we assume that Q1 and Q2 are distinct. We show that P(z) reduces to a nonzero monomial of the form P(z) = aizi for some i ∈ {0, 1, …, m}. If not, we may assume that P(z) = amzm + am−1zm−1 + …+ a1z + a0, where at least two of am, am−1, …, a1, a0, namely ap, aq, p ≠ q are nonzero. As f is entire and n ≥ m + k + 1, from (3.25) we see that 0 is a Picard exceptional value of f. So we have f(z) = eα, where α is a nonconstant entire function. It is easy to see that for i ∈ {0, 1, …, m},

ai{(fn+i)(k)-Q2Q1fn+i}=[ti(α′, α″,…,α(k))-aiQ2Q1] e(n+i)α =si(α′, α″,…,α(k))e(n+i)α,

where si(α′, α″, …, α(k)) are differential polynomials in α′, α″, …, α(k) with rational coefficients. Using (3.25) and (3.26), we obtain

sm(α′, α″,…, α(k))emα+sm-1(α′, α″,…, α(k))e(m-1)α+…+s1(α′, α″,…, α(k))eα+s0(α′, α″,…, α(k))≡0.

Since T(r, si) = S(r, f) (i = 0, 1, …, m), by Borel theorem on the combination of entire functions (see Theorem 1.52 of [17]), (3.27) gives si = 0 for i ∈ {0, 1, …, m}. As ap, aq ≠ 0, from (3.26), we have

(fn+p)(k)=Q2Q1fn+p and (fn+q)(k)=Q2Q1fn+q.

Thus we get two different forms of f simultaneously, a contradiction. Hence P(z) = aizi for some i ∈ {0, 1, …, m}. Therefore, (fn+i)(k)=Q2Q1fn+i for some i ∈ {0, 1, …, m}. This completes the proof of Theorem 1.

Acknowledgements

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The authors are grateful to the referee for his/her valuable suggestions and comments towards the improvement of the paper.

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