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Vladikavkaz Mathematical Journal 2023, Volume 25, Issue 1, P. 81-92
YAK 517.53
DOI 10.46698/p5608-0614-8805-b
UNICITY ON ENTIRE FUNCTIONS CONCERNING THEIR DIFFERENCE OPERATORS AND DERIVATIVES
Rajeshwari, S.1 and Sheebakousar, B.2
1 Department of Mathematics, Bangalore Institute of Technology, Vishweshwarapura, Basavanagudi, Bangalore-560004, India; 2Presidency University, School of Engineering, Itagalpura, Rajanakunte, Yelahanka, Bangalore-560 064, India
E-mail: rajeshwari. s@presidencyuniversity. in, rajeshwaripreetham@gmail.com; sheeba.buzurg@gmail.com
Abstract. In this paper we study the uniqueness of entire functions concerning their difference operator and derivatives. The idea of entire and meromorphic functions relies heavily on this direction. Rubel and Yang considered the uniqueness of entire function and its derivative and proved that if f (z) and f '(z) share two values a,b counting multilicities then f (z) = f'(z). Later, Li Ping and Yang improved the result given by Rubel and Yang and proved that if f (z) is a non-constant entire function and a,b are two finite distinct complex values and if f (z) and f(k) (z) share a counting multiplicities and b ignoring multiplicities then f (z) = f(k)(z). In recent years, the value distribution of meromorphic functions of finite order with respect to difference analogue has become a subject of interest. By replacing finite distinct complex values by polynomials, we prove the following result: Let Af (z) be trancendental entire functions of finite order, k ^ 0 be integer and Pi and P2 be two polynomials. If Af (z) and f(k) share Pi CM and share P2 IM, then Af = f(k). A non-trivial proof of this result uses Nevanlinna's value distribution theory.
Key words: difference operator, shared values, finite order, uniqueness, entire function, polynomials. AMS Subject Classification: 30D35, 39A32.
For citation: Rajeshwari, S. and Sheebakousar, B. Unicity on Entire Functions Concerning Their Difference Operators and Derivatives, Vladikavkaz Math. J., 2023, vol. 25, no. 1, pp. 81-92. DOI: 10.46698/p5608-0614-8805-b.
The reader is presumed to be familiar with the fundamental notations and conclusions of Nevanlinna's value distribution theory of meromorphic functions [1, 2]. S(r, f) means that S(r, f) = o(T(r, f)) as r —y x outside of a possible exceptional set of finite logarithamic measure, and
stand for the exponents of convergence of zero sequence of f and the deficiency of f at the point a, respectively. For a nonconstant meromorphic function h, we denote by T(r, h) the Nevanlinna characteristic of h and by S(r, h) any quantity satisfying S(r, h) = o(T(r, h)), as r runs to infinity outside of a set E C (0, +x) of finite linear measure. We say that h is a small
1. Introduction and Main Results
1 — lim sup
© 2023 Rajeshwari, S. and Sheebakousar, B.
function of f, if T(r, h) = S(r, f). In the sequel, we denote by I a set of infinite linear measure not necessarily the same in all its occurrences.
We say that f and g share the value a IM (ignoring multiplicities), if f and g have the same a point. If f and g have the same a point with the same multiplicities, then we say f,g share the value a CM (counting multiplicities).
Definition 1 [3]. Let k be a nonnegative integer or infinity. For a € C U {to}, we denote by Ek(a; f) the set of all a points of f (z) where an a point of mulplicity m is counted m times if m ^ k and k + 1 times if m > k. If Ek(a; f) = Ek(a;g), then we say that f,g share the value a with weight k.
We write f and g share (a, k) to mean that f,g share the value a with weight k.
Rubel and Yang Chung-Chun [4] considered the uniqueness of an entire function and its derivative. They proved the following.
Theorem 1. Let f (z) be a non-constant entire function, let a,b be two finite distinct values. If f (z) and f '(z) share a,b CM, then f (z) = f'(z).
Li Ping and Yang Chung-Chun [5] improved Theorem 1 and proved.
Theorem 2. Let f (z) be a non-constant entire function, and let a, b be two finite distinct complex values. If f (z) and f (k\z) share a CM, and share b IM. Then f (z) = f (k)(z).
The value distribution of meromorphic functions of finite order with respect to difference analogue has become a subject of some interests, see [6-16].
Theorem 3 [17]. Suppose that f(z) and g{z) are nonconstact meromorphic functions. If f,g share 0, 1, oo CM and N(r, j) +N(r, /) < (d + o(l))T(r, /) for r € I and re oo, where d is a positive number satisfying 0 < d < which I c (0, +oo) is a subset of infinite linear measure, then f = g or f.g = 1.
Theorem 4 [4]. Let f be a nonconstant entire function. If f shares two distinct finite values CM with f', then f = f'.
More results on uniqueness of f' with its n-th derivative f(n) were obtained by several authors (see [18-20]). In view of the progress on the difference analogues of classical Nevanlinna theory of meromorphic functions [21, 22], it is quite natural to investigate the uniqueness problems of meromorphic functions and their difference operators (see [23-26]).
Example 1. Let f (z) = e , where A = 0 is a constant. Then f(k) = AkeAz and Af = f (z + 1) - f (z) = (eA — 1)eAz. Clearly, A(f) and f'k) share 0 CM and to IM, and that p = 1. We can choose A such that eA — 1 = Ak, and so f ^ A(f).
Theorem 5 [27]. Let f (z) be a trancendental entite function of finite order, let n = 0 be a finite complex number, n ^ 1, k ^ 0 be two integers and let a, b be two distinct finite complex values. If f (z) and (A^jf (z)){k) share a CM and share b IM, then f (z) = (A^jf (z)){k).
Lemma 1 [9]. Let Af be a nonconstant meromorphic function of finite order, let n = 0 be a finite complex number. Then
for all r outside of a possible exceptional set E with finite logarithmic measure.
Lemma 2 [28, Lemma 4.3]. Let Af be a nonconstant meromorphic function. Suppose that the polynomials Pj, j = 0,1,...,q, q > p, and let P (Af) = a0(Af )p + ai(Af )p-1 + ... + ap (a0 = 0) is a polynomial of degree p with constant coefficient aj, j = 0,1,... ,p. Then
Lemma 3. Let Af and Ag be two non constant entire functions, and let Pi, P2 be two polynomials. If
H =_^___^_.0,
(A/-P1)(A/-P2) (Ag-Pi)(Ag-P2) '
and Af and Ag share Pi CM, and share P2 IM, then either
2T(r, A/) < ¥ (r, + JV (r, + S(r, A/)
or
Af = Ag.
< Integrating H which leads to
Ag-P2 = Af - P2 Ag-P1 Af - Pi'
where C is a nonzero constant.
If C = 1, then Af = Ag. If C = 1, then from above, we have
Pi - P2 _{C- 1) A/ - CP2 + Pi Ag - Pi " A/ - Pi
and
T (r, Af) = T (r, Ag) + S(r, Af) + S(r, Ag).
Obviously, c-gr1P2 / a and GPSl[2 / 6. It follows that N (r, ^ ^ = 0. Then by the
Second Fundamental Theorem,
1 \ / 1
2 T(r, Af) = N(r, Af) + N {r, J + N {r, J + N J
+ S(r, Af) < N (r, + IV (r, + 5(r, A/),
that is
2T(r, A/) < ¥ (r, + ¥ (r, + 5(r, A/). >
Lemma 4. Let Af be a transcendental entire function of finite order, k be positive integer, let Pi be a nonzero complex value or constant. If Af and f(k) share Pi CM, and N(r, j^j) = S(r, Af ), then one of the following cases must occur:
• f(k) = Hep, where p is a polynomial, and H ^ 0 is a small function of ep.
• T (r, ep ) = S(r, Af).
< Since Af is a transcendental entire function of finite order, Af and f(k) share Pi CM, then there is a polynomial p such that
Af - Pi = ep(f(k)) - Piep. (1)
Set g = f(k). It follows by (1) that
g = (gep)(k) - (Piep)(k). (2)
Then we rewrite (2) as
i+M! = Dt,, ,3,
g
where
D = (4) geP
Note that N ^r, jjyj^j = ^ (r> a) = /)> by Lemma 1 we have
(gep )(k)\
T(r, D) =T r, ■
AgeeP
Next we discuss two cases.
Case 1: e-p — D ^ 0. Rewrite (3) as
gep(e-p — D) = (Pi ep)(k). (5)
When D = 0, (5) implies
g = Hep. (6)
Here H ^ 0 is a small function of ep.
When D ^ 0, it follows from (5) that N^r,-^—^ = S(r,f). Then use the Second Fundamental Theorem to ep we can obtain
/ 1 \ 1
T(r, ep) = T(r, e~p)+0(l) < N(r, ep)+N ^r, — +N ^r, p.p _ D j +0( 1) = S(r, Af).
Case 2: e-p—D = 0. It implies that T(r, ep) = T(r, e-p)+O(1) = S(r, Af), a contradiction. From above discussions, we get T(r,ep) = S(r, Af). >
Theorem 6. Let Af (z) be a transcendental entire functions of finite order, k be integer such that k ^ 0 and let P1 and P2 be two polynomials. If Af (z) and f(k) share P1 CM and share P2 IM, then Af = f(k).
< If Af = f(k), there is nothing to prove. Solve Af # f(k). Since Af is a transcenedental entire function of finite order, Af and f(k) share Pi CM, then we get
Af — P1
e , (7)
where Q is a polynomial.
Since Af and f(k) share Pi CM and share P2 IM, then by second fundamental theorem and Lemma 1 we have
T{r, Af) < N (r, + N (r, + S(r, Af)
* " (r' + ¥ (r' 7^) * " (r' atV) + A/)
< T(r, Af — f(k)) + S(r, Af) < m(r, Af — f(k)) + S(r, Af) ^m(r,Af)+m(r,l-^pj + S(r,Af) < T(r, /) + S(r, /),
which implies
N Äfh^) < N (r' = T^+A/)-
A/ - P2J V ' eQ - 1 Then by (8), (12) and (14)
T(r, Af) = N (r, Af\Pi) +N(Af- P2) + S(r, Af). (8)
According to (7) and (8) we have
T(r, Af) = T{r, Af - /«) + S(r, Af) < N (r, A/^/(fc)) + S(r>A/) (9)
and
T(r, eQ) = m(r, eQ) = m ^r, ^Zp^j ^ m (r' A/-Pi) + A/)' ^ Then it follows from (7) and (9) that
m (r' Af^p) = m (r' Af-'/w) < m (r' A/-/w) + m(r'e° " 1}
< T(r,eQ) + S(r, Af).
Then by (10) and (11)
T{r,eQ) =m(r, +S(r,Af). (12)
On the other hand, (1) can be rewritten as
A/-Pi (13)
(11)
(14)
m 1 r' + N = ¥ Ä/bO + ¥ + A/)
< * (r, + m (r, + S(r, A/),
" (r' = ¥ (r' + A/)' (15)
and then
~N I r
A f-P2
N (r' A/-P2) = T(r'e9) + A/)-
(16)
Set
(A/y(A/-/W) ^ (A/-PI)(A/-P2) 1 J
and
V (/<"-«)(/<"-ft)' 1 '
Easy to know that 0 is an entire function by Lemma 1 and Lemma 2 we have T(r,«) = m.(r,<P) = m. i r, JMi^t^M + s(,, A/)
< ™ (Af-MV-*)) m 1 " Sj) + S(r'A/) = A/)'
that is
T (r,0) = S (r, A/). (19)
Obiously Let d = P, — k(P, — P2), k = 0, by Lemmas 1 and 2, we obtain
mlriV-fr. 1 f (A,Y
A/) V (P'2 — Pi)0 \Af — P, A/ — A/
*m ('• j)+ m ('• {irk - Ä))+m (n W)+s(r'A/) =s(r-A)
and
(20)
1 N ( (Af )'{Af — / (k))
171 ri in—i = 171 \ri
A/ -d) (A/ — P,)(A/ — P2)(A/ — d) J
( i-fW\ | ( (A/y(A/-/W) ^ |
^ m r' rmV> (A/-Pl)(A/-P2)(A/-d) + A/) = A/)-
(21)
Set
6 = (/)(fc+1)___m)
9 (/(fc)-Pi)(/(fc)-P2) (Af - P^Af - P2y [ J
We discuss two cases
Case 1: 0 = 0. Integrating both side of (22) which leads to
A/-Pi -Cf^)-P^ where c is a non zero constant. Then by Lemma 3 we see that
2 T{r, Af) < TV (r> A/l_Pi) + ^ (r, ^pj + ^(r, A/), (24)
which contradicts with (8).
Case 2: 0 ^ 0. By (9), (19) and (22) we can obtain
m(r, Af) = m(r, Af - /<*>) + S(r, Af) = m ^r, ti^l^llA^j + S(r, Af)
= m (V, + S(r, Af) < T (V, + A/) (25)
< T(r, 4-<j>)+ T(r, 0) + S(r, Af) < T(r, tP)+N (r, + ^(r, A/).
On the other hand
/„,,, / (, /(fc+1)(A/-/^)
(/(k) - Pi) (/(k) - P (2))y
= m [r> (fW-Pl)(fW-PW)) + A/)
/ /(fc+1) \ / A/ - /(fc) \ ^ m r, 7TT- + m r, —7TT-
V /(fc)-p2y v 'fW-Pij
<m(r, + S(r, A/) = ¥ (r, a/1_P2) + A/).
Hence combining (25) and (26) we obtain
T(r, A/) < m ( r, + N (r, ) + S(r, A/)
< ™ ( — ) + N (r, y^y) + S{r, A/) < (T(r, /«)) + S(r, A/).
(26)
T(r, A/) < 2AT (r> + S(r, A/). (27)
Next, case 2 is divided into two subcases.
Subcase 2.1. Pi = 0. Then by (7) and Lemma 1 we get
m(r, eQ) = m ^r, ^ j = S(r, A/). (28)
Then by (16), (27) and (28) we can have T(r, Af) = S(r, Af) a contradiction. Subcase 2.2. P2 = 0. Then by (16), (27) and (28) and Lemma 1 we get
(29)
From the fact that
T(r,f(fc)) < T(r, Af) + S(r, Af), (30)
which follows from (29) that
T (r, Af )= T (r,f(k)) + S (r, Af), (31)
By second Nevanlinna Fundamental theorem, Lemma 1, (8) and (31) we have
2T(r, Af) < 2T(r, f(k)) + S(r, Af)
< ¥ T^bi)+37 ym)+w (r> juhi)+ A/) * ¥ (r' aT^) +N (r' ¿/)+ T (r' T^b)" m (r' 7®h)+ A/)
< 2T(r, Af) - m (r> y^T^) + S(r, A/)-
Thus
m(r'JW^d)=S(r'Af)- (32)
From the First Fundamental Theorem, Lemma 1, (20) to (21), (31), (32) and Af is a transcendental entire function of finite order, we obtain
m (r' W^) ^ m (r' + m (r' + A/)
= - + N - N JWI~d) + s^ A
^ {r> ¿7) -N {r> T^tt) + ^= T ¿7) -T jw^)+ A/)
= T(r, A) — T(r, f(k)) + S(r, Af) = S(r, Af).
Thus, we get
It's easy to see that N(r,4) = S(r, Af) and ((12)) can be rewritten as
4 =
Pi — d (/)(fc+1) d/(fc+1)
f~d _1
f{k) - d
(34)
f(k) - p1 f(k)
Then by (33) and (34) we can get
T(r, ^) = m(r, ^) + N(r, ^) = S(r, Af ). (35)
By (7), (25) and (35) we get
N(r, a/_Pi) =S(r,Af). (36)
Moreover, by (7), (31) and (36), we have
m(r'7^) =5(r'A/)' (37)
which implies
N (r' ¿7) = 171 (r' ÂT^Pï) ^ m (r' /M ) =A/)- (38)
Then by (7) we obtain T(r, Af ) = S(r, Af ), a contradiction. So, by (12), (16), (27) and the Second Fundamental Theorem of Nevanlinna, we can get
T(r, Af) < 2m (r> A/X_pJ + S(r, Af) < 2m (V, ^ + S(r, Af) < 2T(r, /<*>) - 2N (r, -^y) + S(r, Af) < IV (r, + N (r,
+ (r, ^ - 2N (r, + S(r, A f)T(r, Af) - N (r, -^y) + S(r, /),
which deduces that
N{r,J^)=s(r,Af)- (39)
It follows from the second theorem of Nevanlinna that
T(r, /(*>) < N (r, + N (r, j^j^) + S(r, Af)
^ ¥ (r' /w-pJ + A/) ^ T(r'/(fc)) + A/)'
which implies that
f(k)\ — 1Vlr _
/(*) _
Similarly
1
r(r, /(fc)) = iV (r, + S(r, A/).
r(r, /(fc)) = N (r, f{k)l_p^j + S{r, A/).
(40)
(41)
f(k) - P2 Then, by (27), we get
T (r, Af ) = 2T(r,f(fc)) + S (r, Af). (42)
By (25) and (26) we have
T(r, 0) = T(r, f(fc)) + S(r, Af). (43)
When case 1 occurs, we apply Lemma 4 and obtain
f(k) = He*. (44)
Here H ^ 0 is a small function of e*. Rewrite (16) as
/(fc+D(A/ - Pl)(Af - P2) - A/(/W - Pl) (/W - P2)
Combining (27) with (44) we get
5
X = f (fc+i)(Af - Pi)(Af - P2) - Af'(f(k) - Pi) (f(k) - P2) = £ Siei4, (46)
i=0
and
5
Y = (Af - Pi)(Af - P2) - Af'(f(k) - Pi) (f(k) - P2) = £ Yjejt, (47)
j=0
where Si and Yj are small functions of e*, S5 ^ 0 and Y6 ^ 0.
If X and Y are two mutually prime polynomials in e*, then we can get T(r, 0) = 6T(r, e*) + S(r, Af). It follows from (16), (41)-(43) that T(r, Af), a contradiction.
If X and Y are not two mutually prime polynomials in e*, it's easy to see that the degree of Y is large than X. Then submitting (38) into (12) implies
H = P2 (48)
and
t = Pi z + P2, (49)
where Pi =0 and P2 are polynomials.
According to (45), (48), (49) and by simple calculation, we must have
C
= -, (50)
Y f (k) - p2 ' v !
where C is a non-zero constant. Put (44) into (16) we have
c ((f )(k) - f (k+i) - PX) -Af'
(51)
(/(k) - Pi) (/(k) - P2) (A/ - Pi)(A/ - P2) "
We claim that /(k) = /(k+i).
Otherwise, combining (22), (44) and (51) we can get T(r, e*) = S(r, A/). It follows from (16) and (27) that T(r, A/) = S(r, A/), a contradiction. Hence, it is a easy work to verify that
Pi = 1 (52)
and
/(k) = P2ez-P2 = Aez, (53)
where A is a nonzero constant and furthermore
A/ = Ae2z - PiAez + Pi. (54)
Then rewrite (27) as
A / — / (fc)
= ê - 1. (55)
/(k) - Pi
Put (49), (52)-(54) into (55) and a direct calculation deduces
A = P2 = ePl = 1. (56)
It follows from (1), (28), (52) and (56) that
H = -Pi(ero - 1)n = 1. (57)
Since A/ and /(k) share P2 IM and (41), (42) and (56) we get
e2z - Piez + (Pi - 1) = (ez - 1)2, (58)
i. e.,
Pi = 2. (59)
It follows from (57) that
ero = (-2)-i/n + 1. (60)
But we cannot get (2) from (60), a contradiction. When case 2 occurs we know that m(r,e*) = m(r,eQ) + O(1) = S(r, A/). Then by (16) and (27) we deduce T(r, A/) = S(r, A/) a contradiction. >
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Received November 13, 2021 Rajeshwari Srinivas
Department of Mathematics, Bangalore Institute of Technology, Vishweshwarapura, Basavanagudi, Bangalore-560004, India, Assistant Professor
E-mail: rajeshwari. s@presidencyuniversity. in, rajeshwaripreetham@gmail. com
https://orcid.org/0000-0002-7681-0830
Sheebakousar Buzurg
Presidency University, School of Engineering,
Itagalpura, Rajanakunte, Yelahanka, Bangalore-560 064, India,
Research Scholar
E-mail: sheeba.buzurg@gmail.com
https://orcid.org/0000-0001-9210-6707
Владикавказский математический журнал 2023, Том 25, Выпуск 1, С. 81-92
ЕДИНСТВЕННОСТЬ ЦЕЛЫХ ФУНКЦИЯХ ОТНОСИТЕЛЬНО ИХ РАЗНОСТНЫХ ОПЕРАТОРОВ И ПРОИЗВОДНЫХ
Раджешвари С.1, Шибакусар Б.2
1 Бангалорский технологический институт, Индия, Бангалор-560004, Вишвешварапура, Басаванагуди;
2 Президентский университет, Инженерная школа, Индия, Бангалор-560 064, Итагалпура, Раджанакунте, Елаханка E-mail: rajeshwari. s@presidencyuniversity. in, rajeshwaripreetham@gmail. com;
sheeba.buzurg@gmail.com
Аннотация. В этой статье мы изучаем единственность целых функций относительно их разностного оператора и производных. Представление о целых и мероморфных функциях сильно зависит от этого направления. Рубель и Янг рассмотрели единственность целой функции и ее производных; они доказали, что если f (z) и f'(z) разделяют два значения a, b с учетом кратностей, то f (z) = f '(z). Позже Ли Пинг и Янг улучшили результат Рубеля и Янга: если f (z) — непостоянная целая функция, а a и b — два конечных различных комплексных значения, и если f (z) и f(z) разделяют a с учетом кратностей и b — без учета кратностей, то f(z) = f(z). В последние годы проявляется значительный интерес к распределению значений мероморфных функций конечного порядка относительно разностного аналога. Заменив различные конечные комплексные значения многочленами, устанавливается следующий результат: пусть Af (z) — трансцендентная целая функция конечного порядка, k ^ 0 — целое число, а P1 и P2 — два многочлена; если Af (z) и fразделяют P1 с учетом кратностей и P2 игнорируя кратности, то Af = f(k). Нетривиальное доказательства этого результата использует теорию распределения значений Неванлинны.
Ключевые слова: разностный оператор, разделяемые значения, конечный порядок, единственность, целая функция, многочлены.
AMS Subject Classification: 30D35, 39A32.
Образец цитирования: Rajeshwari S. and Sheebakousar B. Unicity on Entire Functions Concerning Their Difference Operators and Derivatives // Владикавк. мат. журн.—2023.—Т. 25, № 1.—C. 81-92 (in English). DOI: 10.46698/p5608-0614-8805-b.
