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Kyungpook Mathematical Journal 2020; 60(2): 387-399

Published online June 30, 2020 https://doi.org/10.5666/KMJ.2020.60.2.387

Copyright © Kyungpook Mathematical Journal.

Set shared by an Entire Function with its k-th Derivative Us- ing Normal Families

Molla Basir Ahamed

Department of Mathematics, Kalipada Ghosh Tarai Mahavidyalaya, Bagdogra, West Bengal, 734014, India
e-mail : bsrhmd117@gmail.com

Received: April 7, 2018; Revised: June 27, 2019; Accepted: November 28, 2019

In this paper, we study a problem of a non-constant entire function f that shares a set with its k-th derivative f(k), where a, b and c are any three distinct complex numbers. We have found a gap in the statement of the main result of Chang-Fang-Zalcman [10], and with the help of their method, we have generalize their result in a more compact form. As an application, we generalize the famous Brück conjecture [9] with the idea of set sharing.

Keywords: normal families, entire functions, set sharing, derivative

It is well known that Nevanlinna theory plays an important role in considering the value distribution of meromorphic functions and non-trivial solutions of some complex differential equations. A function f is called meromorphic if it is analytic in the complex plane ℂ except at isolated poles. In what follows, we assume that the reader is familiar with the basic Nevanlinna Theory [18, 26]. It will be convenient to let E denote any set of positive real real numbers of finite linear measure, not necessarily the same at each occurrence. Let f and g be two meromorphic functions having the same set of a-points with the same multiplicities, we then say that f and g share the value a CM (counting multiplicities) and if we do not consider the multiplicities then f and g are said to share the value a IM.

When a = ∞, the zeros of fa means the poles of f.

Definition 1.1

For a non-constant meromorphic function f and any set , we define

E f ( S ) = a S { ( z , p ) × : f ( z ) = a , with multiplicity  p } , E ¯ f ( S ) = a S { ( z , 1 ) × { 1 } : f ( z ) = a } .

If then we simply say f and g share the set Counting Multiplicities (CM) (resp. Ignoring Multiplicities (IM)).

If contains one element only, then it coincides with the usual definition of CM(IM) sharing of values.

In 1926, Nevanlinna first showed that a non-constant meromorphic function on the complex plane ℂ is uniquely determined by the pre-images, ignoring multiplicities, of five distinct values (including infinity). A few years latter, he showed that when multiplicities are taken into consideration, four points are enough and in that case either the two functions coincide or one is the bilinear transformation of the other one.

Recall that the spherical derivative of a meromorphic function f on a plane domain is

f # ( z ) = f ( z ) 1 + f ( z ) 2 .

The sharing value problem between an entire function and its derivatives was first studied by Rubel-Yang [25]. They proved that if a non-constant entire function f and its derivative f′ share two distinct finite numbers a, b CM, then ff′.

In 1979, Mues-Steinmetz [23] improved the above theorem in the following manner.

Theorem A.([23])

Let f be a non-constant entire function. If f and f′ share two distinct values a, b IM then ff′.

Remark 1.1

Following example shows that the results of Rubel-Yang and Mues-Steinmetz are not in general true when value sharing is replaced by set sharing.

Example 1.1

Let S = { a 3 , 2 a 3 }, where a(≠ 0) be any complex number. Let f(z) = ez + a, then but ff′.

So for the uniqueness of an entire function and its derivative sharing a set, the cardinality of the range set should be at least three.

In this regard in 2003, using the properties of normal families, Fang-Zalcman [14] obtained the following result.

Theorem B.([14])

Let , where a, b are two non-zero distinct complex numbers satisfying a2b2, a ≠ 2b, a2ab + b2 ≠ 0. If for a non-constant entire function f, , then ff′.

In order to generalize the range set in the above theorem, in 2007 Chang-Fang-Zalcman [10] obtained the following result.

Theorem C.([10])

Let f be a non-constant entire function and let , where a, b and c are distinct complex numbers. If , then either

  • (1) ; or

  • (2) f ( z ) = C e - z + 2 3 ( a + b + c ) and (2abc)(2bca)(2cab) = 0; or

  • (3) f ( z ) = C e - 1 + i 3 z 2 + 3 ± i 3 6 ( a + b + c ) and a2 + b2 + c2abbcca = 0,

where is a non-zero constant.

Reamrk 1.2

We see from the next example that the conclusion of Theorem C need not hold if the CM shared set is replaced by an IM shared set.

Example 1.2.([10])

Let and f(z) = sinz or cos z. Clearly and f takes none of the forms (1) – (3) in Theorem C.

Remark 1.3

In Example 1.2, one may consider k-th derivative of f instead of first derivative, when k is any odd positive integer to get the same conclusion.

Remark 1.4

We have found a little gap in the statement of Theorem C as follows.

  • (i) In the statement of the Theorem C, the author should require that “f be a non-constant entire function having zeros of multiplicities ≥ 1.

  • It is affirmed that f has zeros, so it is natural to see that the possible form of the function should not be of the form in Theorem C as f has no zero in this particular form.

(ii) From the above discussions, one may note that a non-constant entire function f sharing an arbitrary set of three finite complex numbers a, b and c (CM) with its first derivative, must have some specific form.

It is natural to ask the following question:

Question 1.1

Is it possible to extend Theorem C to k-th derivative of f ?

If the answer of Question 1.1 is found to be affirmative, then it will be interesting to investigate on the following question:

Question 1.2

What can we say about the possible forms of the function f ?

Since f and f(k) share the set , one may observe that among all the possible relationship between f and f(k), clearly f(k)f is the most obvious one. So before going to state our main results, we want to discuss a natural question: What is the general solution of f(k)f ? The natural answer is f(z) = ℒθ(z) (see [1, 6]) where we define ℒθ(z) as follows

θ ( z ) = c 0 e z + c 1 e θ z + c 2 e θ 2 z + + c k - 1 e θ k - 1 z ,

where ci ∈ ℂ for i ∈ {0, 1, 2, . . . , k − 1} with ck−1 ≠ 0 and θ = cos ( 2 π k ) + i sin  ( 2 π k ).

To answer all the above questions affirmatively is the main motivation of writing this paper. We have tried to take care of the points mentioned in Remark 1.2. Following is the main result of this paper.

Theorem 1.1

Let f be a non-constant entire function, having zeros of multiplicity at least k and let , where a, b and c are distinct complex numbers. If , then f takes one of the following forms:

  • (1) f(z) = ℒθ(βz), where β is a root of the equation zk − 1 = 0,

  • (2) f ( z ) = θ ( η z ) + 2 3 ( a + b + c ), where η is a root of the equation zk + 1 = 0 and (2abc)(2bca)(2cab) = 0,

  • (3) f ( z ) = θ ( ζ z ) + 3 ± i 3 6 ( a + b + c ), where ζ(≠ 1) is a root of the equation z3k − 1 = 0 and a2 + b2 + c2abbcca = 0,

where ℒθ(z) is defined in (1.1).

Remark 1.5

The conclusion (2) of Theorem 1.1 can not be omitted, following example ensures this fact.

Example 1.3

Let k = 2, hence θ = −1. So we have η = ±i, and hence θ ( η z ) = c 0 e η z + c 1 e θ η z = 1 4 ( e i z + e - i z ), where c 0 = 1 4 = c 1. Let S = { a , 1 4 , 1 2 - a }, where a ∈ ℂ \ {1/4}. Let f ( z ) = 1 2 cos  z + cos 2 ( z 2 ), clearly it is of the form f ( z ) = θ ( η z ) + 2 3 ( a + b + c ). One can verify that , and the multiplicities of zeros of f(z) are at least 2.

We begin our investigation with the following lemmas, which are essential to prove our main results.

Lemma 2.1.([13])

The order of an entire function having bounded spherical derivative on is at most 1.

Lemma 2.2.([14])

Letbe a family of holomorphic functions in a domain D. Let k be a positive integer. Let a, b and c be three distinct finite complex numbers and M a positive number. If, for any f ∈ ℱ, the zeros of f are of multiplicityk and |f(k)(z)| ≤ M whenever f(z) ∈ {a, b, c}, thenis normal in D.

Lemma 2.3.([15])

Let f be a non-constant meromorphic function of finite order ρ, and ε > 0 a constant. Then there exists a set E ⊂ [0, 2π) which has linear measure zero, such that if ψ0 ∈ [0, 2π) − E, then there is a constant R0 = R0(ψ0) > 0 such that for all z satisfying argz = ψ0 and |z| > R0, we have

| f ( k ) ( z ) f ( z ) | z k ( ρ + ɛ - 1 ) .

Lemma 2.4

Let f be an entire function, and suppose that |f(k)(z)| is unbounded on some ray arg z = θ. Then there exists an infinite sequence of points zn = rneθ where rn → ∞, such that f(k)(zn) → ∞ and

| f ( z n ) f ( k ) ( z n ) | ( 1 + o ( 1 ) ) z n k

as zn → ∞.

Proof

For k ≥ 1, one must have

f ( z n ) = i = 0 k - 1 z n i ( i ) ! f ( i ) ( 0 ) + { 0 z n 0 z 0 z ( k - 1 ) - times f ( k ) d z d z k - times } .

By applying the triangle inequality in (2.2), rest of the proof follows from the proof of [16, Lemma 4, page. 421].

Lemma 2.5.([22])

A class of functions f meromorphic in a domain D ⊂ ℂ is normal in D if and only if f# is uniformly bounded on any compact subset of D for .

Lemma 2.6.([20, 24])

Let f be an entire function of order at most 1 and k be a positive integer. Then

m ( r , f ( k ) f ) = o ( log r ) ,             a s r .

Lemma 2.7

Let α be a non-constant entire function and a, b and c are three distinct finite complex numbers. Then there does not exist an entire function f having zeros of multiplicityk, satisfying the differential equation

( f ( k ) - a ) ( f ( k ) - b ) ( f ( k ) - c ) ( f - a ) ( f - b ) ( f - c ) = e α .

Proof

Assume there exists an entire function f satisfying (2.3). Then we see that |f(i)(z)| ≤ max{a, b, c} whenever f(z) ∈ {a, b, c}, i ∈ {1, 2, ..., k}. As the zeros of f have multiplicity ≥ k, we have, by Lemma 2.2, that the family ℱw = {fw : w ∈ ℂ}, where fw(z) = f(w + z), is normal on the unit disc. Hence by Marty’s Theorem, we get that f#(w) = (fw)#(0) is uniformly bounded for all w ∈ ℂ. Therefore in view of Lemma 2.1, we get that f has order at most 1.

From (2.3), we obtain , where and ℬ are two constants. It is clear that , since α is non-constant.

Next we claim that abc ≠ 0. On contrary, let abc = 0. i.e., a = 0 or b = 0 or c = 0. Without any loss of generality, we may assume that a = 0. It follows from (2.3) that

f ( k ) ( f ( k ) - b ) ( f ( k ) - c ) f ( f - b ) ( f - c ) = e A z + .

Again we see that

f ( k ) ( f ( k ) - b ) ( f ( k ) - c ) f ( f - b ) ( f - c ) = ( f ( k ) ) 3 f ( f - b ) ( f - c ) - ( b + c ) ( f ( k ) ) 2 f ( f - b ) ( f - c ) + b c f ( k ) f ( f - b ) ( f - c ) = f ( k ) f f ( k ) f - b f ( k ) f - c - b + c b - c ( f ( k ) f - b - f ( k ) f - c ) + b c ( A 1 f ( k ) f + 1 f ( k ) f - b + 1 f ( k ) f - c ) ,

where ,ℬ1 and are constants. So there exists , ℬ2 and such that

m ( r , f ( k ) ( f ( k ) - b ) ( f ( k ) - c ) f ( f - b ) ( f - c ) ) A 2 m ( r , f ( k ) f ) + 2 m ( r , f ( k ) f - b ) + C 2 m ( r , f ( k ) f - c ) + O ( 1 ) .

Thus by Lemma 2.6, we deduce that

T ( r , e A z + ) = m ( r , e A z + ) = o ( log r ) ,

and this is not possible since .

Therefore, we must have abc ≠ 0. We set

g ( z ) = f ( z / A ) i . e . , g ( k ) ( z ) = 1 A k f ( k ) ( z / A ) .

Hence (2.4) and (2.3) mean that

( g ( k ) - a / A k ) ( g ( k ) - b / A k ) ( g ( k ) - c / A k ) ( g - a ) ( g - b ) ( g - c ) C e z ,

where C = e A 3 k 0. We see that (2.5) can be written as

( g ( k ) ) 3 + C 1 ( g ( k ) ) 2 + C 2 g ( k ) ( g - a ) ( g - b ) ( g - c ) - C e z = C 3 ( g - a ) ( g - b ) ( g - c ) ,

where are constants with . With ɛ = 1 3, Lemma 2.3 shows that there exists a set E ⊂ [0, 2π) of measure zero such that for each ψ0 ∈ [0, 2π) − E, there is a constant R0 = R0(ψ0) > 0 such that whenever arg z = ψ0 and |z| > R0,

| ( g ( k ) ) 3 + C 1 ( g ( k ) ) 2 + C 2 g ( k ) ( g - a ) ( g - b ) ( g - c ) | K z ,

for some positive constant K. We may now suppose that π/2 and 3π/2 are contained in the set E. Then [0, 2π) − E = E1E2, where E1 = {θ ∈ [0, 2π) : cosθ > 0} and E2 = {θ ∈ [0, 2π) : cosθ < 0}. Let θE1, then by (2.6) and (2.7), we get for sufficiently large r,

| C 3 ( g ( r e i θ ) - a ) ( g ( r e i θ ) - b ) ( g ( r e i θ ) - c ) | = | ( g ( k ) ( r e i θ ) ) 3 + C 1 ( g ( k ) ( r e i θ ) ) 2 + C 2 g ( k ) ( r e i θ ) ( g ( r e i θ ) - a ) ( g ( r e i θ ) - b ) ( g ( r e i θ ) - c ) - C e r e i θ | C e r cos θ - K r ,             as            r .

It follows that

g ( r e i θ ) a , b     or    c , as  r .

Let θE2. We claim that |g(k)(re)| is bounded as r → ∞. On contrary, we suppose that |g(k)(re)| is unbounded as r → ∞. Then by Lemma 2.4, there exists a sequence rn → ∞ such that |g(k)(re)| → ∞ and

| g ( r e i θ ) g ( k ) ( r e i θ ) | ( 1 + o ( 1 ) ) r n k .

With |g(k)(rne)| → ∞, we note that

| ( g ( r n e i θ ) - a ) ( g ( r n e i θ ) - b ) ( g ( r n e i θ ) - c ) ( g ( k ) ( r n e i θ ) ) 3 + C 1 ( g ( k ) ( r n e i θ ) ) 2 + C 2 g ( k ) ( r n e i θ ) | ( 1 + o ( 1 ) ) r n 3 k .

Since |g(k)(rne)| → ∞, so it follows from (2.6) that

| ( g ( r n e i θ ) - a ) ( g ( r n e i θ ) - b ) ( g ( r n e i θ ) - c ) r n 3 k C 3 | = | ( g ( k ) ( r n e i θ ) ) 3 + C 1 ( g ( k ) ( r n e i θ ) ) 2 + C 2 g ( k ) ( r n e i θ ) - C 3 r n 3 k C 3 C e r n e i θ | = r n - 3 k e - r n cos θ ( g ( k ) ( r n e i θ ) ) 3 + C 1 ( g ( k ) ( r n e i θ ) ) 2 + C 2 g ( k ) ( r n e i θ ) - C 3 C 3 C .

Thus from (2.6), (2.10) and (2.11), we get

1 - o ( 1 ) | r n 3 k ( g ( k ) ( r n e i θ ) ) 3 + C 1 ( g ( k ) ( r n e i θ ) ) 2 + C 2 g ( k ) ( r n e i θ ) - C 3 ( g ( r n e i θ ) - a ) ( g ( r n e i θ ) - b ) ( g ( r n e i θ ) - c ) | | r n 3 k C 3 ( g ( r n e i θ ) - a ) ( g ( r n e i θ ) - b ) ( g ( r n e i θ ) - c ) | + C r n 3 k e r n cos  θ 0 ,

which is absurd. Hence |g(k)(rne)| is bounded as r → ∞ for each θE2. For a positive integer k, we can always write

g ( r e i θ ) = i = 0 k - 1 z n i ( i ) ! f ( i ) ( 0 ) + ( e i θ ) k { 0 r 0 t 0 t ( k - 1 ) - times f ( k ) ( z ) d t d t k - times } .

So we must have

g ( r e i θ ) | i = 0 k - 1 z n i ( i ) ! f ( i ) ( 0 ) | + | { 0 r 0 t 0 t ( k - 1 ) - times f ( k ) ( z ) d t d t k - times } | | i = 0 k - 1 z n i ( i ) ! f ( i ) ( 0 ) | + r k ,

where ℳ = ℳ(θ) is a positive constant depending on θ.

Hence by (2.8) and (2.12), for every θ ∈ [0, 2π) − E, there exists a positive constant ℒ = ℒ(θ) such that for z = re with r > r0,

| g ( z ) z k | .

Since the order of g is at most 1, it follows from (2.8), (2.13), the Phragěn-Lindelöf Theorem, and Liouville’s Theorem, that g is a polynomial of degree at most k, which is impossible by (2.5). This completes the proof.

Lemma 2.8

Let where a, b and c be any three distinct finite complex numbers and a non-zero constant. If Ef (S) = Ef(k) (S), where f is an entire function having zeros of multiplicitiesk and satisfying f(k) ≢ 0 and

( f ( k ) - a ) ( f ( k ) - b ) ( f ( k ) - c ) ( f - a ) ( f - b ) ( f - c ) A ,

then f must take one of the following forms:

  • (1) f(z) = ℒθ (βz), where β is a root of the equation zk − 1 = 0,

  • (2) f ( z ) = θ ( η z ) + 2 3 ( a + b + c ), where η is a root of the equation zk + 1 = 0 and (2abc)(2bca)(2cab) = 0,

  • (3) f ( z ) = θ ( ζ z ) + 3 ± i 3 6 ( a + b + c ), where ζ(≠ 1) is a root of the equation z3k − 1 = 0 and a2 + b2 + c2abbcca = 0,

where is a non-zero constant.

Proof

From the proof of Lemma 2.7, we note that f has order at most 1. Since f and f(k) have the same order and f having zeros of multiplicities ≥ k satisfying f(k) ≢ 0 and also , so one can deduce that the form is

f ( k ) ( z ) = c 0 α k e α z + c 1 α k e α θ z + + c k - 1 α k e α θ k - 1 z = α k θ ( α z ) , ( say )

where ci ∈ ℂ, for i ∈ {0, 1, 2, . . . , k − 1} with ck−1 ≠ 0, α ∈ ℂ − {0},

θ = cos ( 2 π k ) + i sin ( 2 π k )

and

θ ( α z ) = c 0 e α z + c 1 e α θ z + + c k - 1 e α θ k - 1 z .

Integrating (2.15) k-times, we get

f ( z ) = θ ( α z ) + Q k - 1 ( z ) ,

where is a polynomial of degree ≤ k − 1.

Using (2.15) and (2.16), we get from (2.14)

( α 3 k - A ) ( θ ( α z ) ) 3 + ( 1 α 2 k - 3 A Q k - 1 - A 1 ) ( θ ( α z ) ) 2 + ( 2 α 3 k - 3 A Q k - 1 2 - 2 A 1 Q k - 1 - A 2 ) θ ( α z ) + ( 3 - A Q k - 1 3 - A 1 Q k - 1 2 - A 2 Q k - 1 - A 3 ) 0 ,

where ℒ1 = −(a + b + c), ℒ2 = ab + bc + ca and ℒ3 = −abc. It follows that

α 3 k - A , 1 α 2 k = A ( 3 Q k - 1 + 1 ) , 2 α k = A ( 3 Q k - 1 2 + 2 1 Q k - 1 + 2 ) , 3 = A ( Q k - 1 3 + 1 Q k - 1 2 + 2 Q k - 1 + 3 ) .

We now discuss the following three possible cases.

Case 1

Let α ∈ {z : zk − 1 = 0}. Then from (2.18) and (2.19), we get and . Thus we see that

f ( z ) = θ ( β z ) ,

where β is a root of the equation zk − 1 = 0.

Case 2

Let α ∈ {z : zk+1 = 0}. Then from (2.18) and (2.19), we see that and Q k - 1 = - 2 3 1. It follows from (2.21) that

2 1 3 - 9 1 2 + 27 3 = 0

which in turn implies that

( 2 a - b - c ) ( 2 b - c - a ) ( 2 c - a - b ) = 0.

In this case, we get

f ( z ) = θ ( η z ) 2 3 ( a + b + c ) ,

where η is a root of the equation zk + 1 = 0.

Case 3

Let α ∉ {z : zk −1 = 0}∪{z : zk +1 = 0}. Then by (2.18) and (2.19), we get

Q k - 1 = 1 - α k 3 α k 1 .

Then by (2.18), (2.20) and (2.22), we obtained

2 = ( 1 ) 2 3 .

Next by (2.18), (2.21), (2.22) and (2.23), we also get

( 1 - α 3 k ) 3 = 1 27 ( 1 - α 3 k ) 1 3 .

Subcase 3.1

If α3k ≠ 1, then ℒ3 = (ℒ1)3/27. This with (2.23) shows that a = b = c, which is not possible.

Subcase 3.2

Hence α3k − 1 = 0. i.e., α k = - 1 ± i 3 2. Thus we have Q k - 1 = - 3 ± i 3 6 1. Simplifying (2.23), we get a2 + b2 + c2abbcca = 0.

Thus we have

f ( z ) = θ ( ζ z ) + 3 ± i 3 6 ( a + b + c ) ,

where ζ(≠ 1) is a root of the equation z3k − 1 = 0.

Since , therefore it is clear that

( f ( k ) - a ) ( f ( k ) - b ) ( f ( k ) - c ) ( f - a ) ( f - b ) ( f - c ) e α ( z ) ,

where α is an entire function. We note that by Lemma 2.7, α is a constant. Then we set . Thus (3.1) changes to

( f ( k ) - a ) ( f ( k ) - b ) ( f ( k ) - c ) ( f - a ) ( f - b ) ( f - c ) A .

Next we discuss the following two cases.

Case 1

If f(k) ≠ 0, then by Lemma 2.8, we see that f takes one of the three forms (1)–(3). So we are done.

Case 2

If f(k) vanishes at some point z0 ∈ ℂ. i.e., f(k)(z0) = 0.

Differentiating both sides of (3.2), we get

{ 3 ( f ( k ) ) 2 - 2 ( a + b + c ) f ( k ) + ( a b + b c + c a ) } f ( k + 1 ) A { 3 f 2 - 2 ( a + b + c ) f + ( a b + b c + c a ) } f .

Let f(k)(z0) = 0 and kn. So we may assume

f ( z ) = f ( z 0 ) + A n ( z - z 0 ) n +

Clearly we have f(k)(z) = Bn(zz0)nk +. . . and f′ (z) = nA(zz0)n−1 +. . .. We see that L.H.S of (3.3) vanishes at z0 with order nk while R.H.S of (3.3) vanishes with the order at least n − 1, which is not possible.

In 1996, the following conjecture was proposed by Brück [9].

Conjecture 4.1.([9])

Let f be a non-constant entire function. Suppose that ρ1(f) is not a positive integer or infinite. If f and f′ share one finite value a CM, then

f - a f - a = c ,

for some non-zero constant c, where ρ1(f) is the first iterated order of f which is defined by

ρ 1 ( f ) = lim sup r log log  T ( r , f ) log  r .

Many authors (for the case of differences see [19, 21] and for the cases of derivatives or differential polynomials see [2, 3, 4, 5, 6, 7, 8] and [11, 12, 17]) have studied the conjecture under some additional conditions. But the main conjecture is still open. In this direction, it is interesting to ask the following two questions.

Question 4.1

Does the conjecture hold if one considers a set having three arbitrary finite complex numbers instead of a value ?

Question 4.2

Is it possible to replace first derivative f′ by a more general derivative f(k) ?

Remark 4.1

Note that Lemma 2.8 answers the above questions in some sense.

Acknowledgements

The author would like to thank the referees for their constructive comments that led to better presentation of the paper.

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