Ritt''''''''s Second Theorem described polynomial solutions of the functional equation P (f ) = Q(g), where P, Q are polynomials. In this paper, using techniques of value distribution theory into account the special properties of L - functions, we describe solutions of the above equation for L - functions and a class of polynomials of Fermat-Waring type. Namely, use Lemma 2.1, Lemma 2.2, and Lemma 2.5, we study conditions to equations in the Theorem 1.1 have solutions on sets of L - functions in the extended Selberg class. Then we apply the obtained results from the Theorem 1.1, and use Lemma 2.3, Lemma 2.4, and Lemma 2.6 to study the uniqueness problem for L - functions sharing finite set in the Theorem 1.2.
TNU Journal of Science and Technology 225(13): 31 - 37 UNIQUENESS OF L - FUNCTIONS IN THE EXTENDED SELBERG CLASS Nguyen Duy Phuong TNU - Defense and Security Training Centre ABSTRACT Ritt's Second Theorem described polynomial solutions of the functional equation P (f ) = Q(g), where P, Q are polynomials In this paper, using techniques of value distribution theory into account the special properties of L - functions, we describe solutions of the above equation for L - functions and a class of polynomials of Fermat-Waring type Namely, use Lemma 2.1, Lemma 2.2, and Lemma 2.5, we study conditions to equations in the Theorem 1.1 have solutions on sets of L - functions in the extended Selberg class Then we apply the obtained results from the Theorem 1.1, and use Lemma 2.3, Lemma 2.4, and Lemma 2.6 to study the uniqueness problem for L - functions sharing finite set in the Theorem 1.2 Keyword: Function equations; polynomials of Fermat-Waring type; shared sets; sets of zeros; L functions Received: 24/3/2020; Revised: 21/8/2020; Published: 27/8/2020 TÍNH DUY NHẤT CỦA L – HÀM TRONG LỚP SELBERG MỞ RỘNG Nguyễn Duy Phương Trung tâm Giáo dục Quốc phòng An ninh – ĐH Thái Ngun TĨM TẮT Định lí Ritt thứ hai cho ta nghiệm đa thức phương trình hàm P (f) = Q (g), P, Q đa thức Trong báo này, sử dụng kỹ thuật lý thuyết phân phối giá trị có tính đến thuộc tính đặc biệt L - hàm, chúng tơi nghiên cứu phương trình hàm đa thức cho L - hàm lớp đa thức loại Fermat-Waring Cụ thể, sử dụng Bổ đề 2.1, Bổ đề 2.2 Bổ đề 2.5, nghiên cứu điều kiện để phương trình Định lý 1.1 có nghiệm tập L - hàm lớp Selberg mỏ rộng Sau đó, chúng tơi áp dụng kết thu từ Định lý 1.1 sử dụng Bổ đề 2.3, Bổ đề 2.4 Bổ đề 2.6 để nghiên cứu vấn đề cho L - hàm nhận chung tập hữu hạn Định lý 1.2 Từ khóa: Phương trình hàm; đa thức loại Fermat – Waring; tập chia sẻ; tập không điểm; L - hàm Ngày nhận bài: 24/3/2020; Ngày hoàn thiện: 21/8/2020; Ngày đăng: 27/8/2020 Email: phuongnd@tnu.edu.vn https://doi.org/10.34238/tnu-jst.2891 http://jst.tnu.edu.vn; Email: jst@tnu.edu.vn 31 Nguyen Duy Phuong TNU Journal of Science and Technology Introduction In 1922, Ritt ([1]) studied functional equation P (f ) = Q(g), where P, Q are polynomials, and described it’s polynomial solutions Since the paper of Ritt ([1]), the functional equation P (f ) = Q(g), where P, Q are polynomials, has been investigated by many authors (see [2]- [6]) Pakovich [5] studied the functional equation P (f ) = Q(g), where P, Q are polynomials, and f, g are entire functions Khoai-An-Hoa [6] considered the functional equation of the form P (f ) = Q(g), where P and Q are Yi’s polynomials and then apply the obtained results to study the uniqueness problem for meromorphic functions sharing two subsets Now let us recall some basic notions Let C denote the complex plane By a meromorphic function we mean a meromorphic function in the complex plane C We assume that the reader is familiar with the notations of Nevanlinna theory and of L-functions in the Selberg class(see [7-16]) In this paper, we discuss the Ritt’s Second Theorem for L functions and meromorphic functions As a application, we present a uniqueness theorem for L - functions when the functions share values in a finite set 225(13): 31 - 37 The functional equation c = q + c1 (q ≥ 3) xq + a y +b has a pair (L1 , L2 ) of non-constant L - function solutions if and only if L1 = L2 , a = b, c = 1, c1 = The functional equation xq y q = a has no non-constant L - function solutions (L1 , L2 ) Now let a, b, c ∈ C, a 6= 0, b 6= 0, c 6= 0, q be a positive integer, and consider polynomials without multiple zero given by P (z) = z q + a, Q(z) = z q + b, (1.1) we obtain the following result Theorem 1.2 Let L1 and L2 be two nonconstant L - functions Let P , Q be polynomials of the form (1.1), and S, T are respective sets of zeros of P (z), Q(z) Then L1 = L2 and P = Q if one of the following conditions is satisfied: q ≥ and E L1 (S) = E L2 (T ); q ≥ and EL1 (S) = EL2 (T ), a 6= −1, b 6= −1; Now let us describe the main results of the q ≥ and EL1 (S) = EL2 (T ), a = b 6= −1 paper The Ritt’s Second Theorem for L - functions is as following Some lemmas Theorem 1.1 Let a, b, c, c1 ∈ C, a 6= 0, b 6= 0, c 6= 0, q be a positive integer We need some lemmas Then Lemma 2.1 [10] Let f be a non-constant meromorphic function on C and let a1 , a2 , , The functional equation aq be distinct points of C ∪ {∞} Then xq + a = c(y q + b)(q ≥ 3) has a pair (L1 , L2 ) of non-constant L - function solutions if and only if L1 = L2 , a = b, c = 32 (q − 2)T (r, f ) ≤ q X i=1 N (r, ) + S(r, f ), f − http://jst.tnu.edu.vn; Email: jst@tnu.edu.vn Nguyen Duy Phuong TNU Journal of Science and Technology where S(r, f ) = o(T (r, f )) for all r, except for a set of finite Lebesgue measure 225(13): 31 - 37 Proof of Theorems Now we use the above Lemmas to prove the Lemma 2.2 [6] For any nonconstant main result of the paper meromorphic function f, Proof of Theorem 1.1 The sufficient con1 dition of the theorem is easily seen Now we N (r, ′ ) ≤ N (r, ) + N (r, f ) + S(r, f ) f f show the necessary condition Assume that xq + a = c(y q + b)(q ≥ 3) (3.1) Lemma 2.3 [6] Let f and g be two nonconstant meromorphic functions If E f (1) = has a pair (L1 , L2 ) of non-constant L - funcE g (1), then one of the following three relation solutions Then tions holds: Lq1 + a − cb = cLq2 Suppose a − cb 6= Then, by Lemma 2.1, and note that N (r, L1 ) = S(r, L1 ), T (r, f ) ≤ N2 (r, f ) + N2 (r, ) + N2 (r, g) f N (r, L2 ) = S(r, L2 ), we obtain 1 + N2 (r, ) + 2(N (r, f ) + N (r, )) qT (r, L1 ) + S(r, L1 ) = T (r, Lq1 ) g f + N (r, g) + N (r, ) + S(r, f ) + S(r, g), g ) ≤ N (r, L1 ) + N (r, and the same inequality holds for T (r, g); L1 f g ≡ 1; +N (r, q ) + S(r, L1 ) L1 + a − bc f ≡ g 1 ) + N (r, ) ≤ N (r, L1 L2 Lemma 2.4 [13] Let L be an non-constant +S(r, L1 ) ≤ T (r, L1 ) + T (r, L2 ) + S(r, L1 ) L - function Then T (r, L) = dπL r log r + O(r), where dL = Similarly P K i=1 λi be the degree of L - function and qT (r, L2 ) + S(r, L2 ) ≤ T (r, L2 ) + T (r, L1 ) K, λi are respectively the positive integer and +S(r, L2 ) positive real number in the functional equation of the definition of L - functions; Therefore dL N (r, L ) = π r log r + O(r), N (r, L) = q(T (r, L1 )+q(T (r, L2 )) ≤ 2(T (r, L1 )+T (r, L2 )) S(r, L) +S(r, L1 ) + S(r, L2 ), Lemma 2.5 [16] Let L1 , , LN be distinct (q−2)(T (r, L1 )+T (r, L2 ) ≤ S(r, L1 )+S(r, L2 ) non-constant L - functions Then L1 , , LN This is a contradiction to the assumption are linearly independent over C that q ≥ So a − cb = Then Lq1 = cLq2 Lemma 2.6 [13] Let L be an non-constant From this L2 = tL1 , tq c = Applying L - functions and a ∈ C Then equation Lemma 2.5 we have L1 = L2 and therefore L = a has infinitely many solutions t = 1, a = b, c = http://jst.tnu.edu.vn; Email: jst@tnu.edu.vn 33 TNU Journal of Science and Technology Nguyen Duy Phuong 225(13): 31 - 37 The sufficient condition of the theorem is Note that L2 − ei , i = 1, , k, always has easily seen Now we show the necessary con- zeros Then, if s0 be a zero of L2 − ei , then dition Assume that Q(L2 (s0 )) c = 0, Q(L2 (s0 )) + = 0, c c1 P (L1 (s0 )) c1 = q + c1 (3.2) xq + a y +b (s0 − 1)q(m1 −m2 ) (Lq20 (s0 ) + b(s0 − 1)qm2 ) c1 (Lq10 (s0 ) + a(s0 − 1)qm1 ) has a pair (L1 , L2 ) of non-constant L - function solutions Then = (3.6) c = q + c1 Lq1 + a L2 + b Since (3.6) and c, c1 6= we get Q(L2 (s0 )) = − cc1 6= Therefore Lq20 (s0 ) + b(s0 − 1)qm2 6= So (s0 − 1)q(m1 −m2 ) = It follows that We shall prove that c1 = Suppose, to the m1 > m2 Consider (3.5) Write contrary, c1 6= Note that when considering L-functions, these functions have only one P (z) = (z − a1 ) (z − aq ), possible pole at s = Write P (L1 ) = (L1 − a1 ) (L1 − aq ) L10 (s) L20 (s) From (3.5), (3.6) and note that L1 − , L1 (s) = , L2 (s) = , m (s − 1) (s − 1)m2 i = 1, , q, always has zeros we have L2 has pole at s = Thus m2 > By m1 ≥ 0, m2 ≥ 0, m1 > m2 > and (3.4), (3.5) we have a m i contradiction where Li (s), and (s − 1) , i = 1, 2, has no common zero From this and (3.2) we get Thus c1 = Therefore P (L1 ) = CQ(L2 ) qm qm Applying Part i) we get L1 = L2 , a = b, c(s − 1) (s − 1) = q q L (s) + a(s − 1)qm1 L (s) + b(s − 1)qm2 c = 1, c1 = 10 20 +c1 Suppose, to the contrary, functional equa(3.3) tion xq y q = a m1 m2 = has a non-constant L - function solution (3.4) (L , L ) Then By c1 6= we have We recall that P (x) = xq + a, Q(y) = y q + b Lq1 Lq2 = a Put R(y) = Q(y) + cc1 Suppose that R(z) has distinct zeros e1 , e2 , , ek with respec- Note that L1 , L2 have only one possible pole tive multiplicities l1 , l2 , , lk , ≤ k ≤ q, so at s = Onthe other hand L1 , L2 have infinitely zeros Therefore, there is a s0 6= that l1 + · · · + lk = q Then we get such that L1 (s0 ) = So a = A contradiction to assumption that a = c Q(L2 ) = Q(L2 ) + = R(L2 ) c1 P (L1 ) c1 Proof of Theorem 1.2 = (L2 − e1 )l1 (L2 − ek )lk , Lq Lq Proof of Part Set F = −a1 , G = −b2 , T (r) = T (r, f ) + T (r, g), S(r) = S(r, f ) + (s − 1)q(m1 −m2 ) (Lq20 (s) + b(s − 1)qm2 ) S(r, g) By E L1 (S) = E L2 (S) it follow that q c1 (L10 (s) + a(s − 1)qm1 ) E F (1) = E G (1) Then, applying Lemma 2.3 = (L2 − e1 )l1 (L2 − ek )lk (3.5) to the F , G we consider the following cases: 34 http://jst.tnu.edu.vn; Email: jst@tnu.edu.vn Nguyen Duy Phuong TNU Journal of Science and Technology 225(13): 31 - 37 Lq Lq Case T (r, F ) ≤ N2 (r, F ) + N2 (r, ) F Case F.G = This mean −a1 −b2 = or Lq1 Lq2 = A, A 6= Applying Theorem 1.1 we obtain a contradiction Lq Lq Case F = G This mean −a1 = −b2 or 1 +N2 (r, G)+N2 (r, )+2(N (r, F )+N (r, )) Lq = CLq Applying Theorem 1.1 we get G F L1 = L2 and therefore a = b +N (r, G) + N (r, ) + S(r), G Proof of Part We first prove that L1 +a = c(L2 + b) We consider the following cases: T (r, G) ≤ N2 (r, F ) + N2 (r, ) F Case L1 (s), L2 (s) are both entire func1 sets S, T CM, +N2 (r, G)+N2 (r, )+2(N (r, G)+N (r, )) tions and share the respective G G where S = a1 , , aq with P (z) = (z − a1 ) (z − aq ), and T = b1 , , bq with +N (r, F ) + N (r, ) + S(r) (3.7) F Q(z) = (z − b1 ) (z − bq ) Then, we obtain an entire function Noting that l(s) = N (r, L1 ) = S(r, L1 ), N (r, L2 ) = S(r, L2 ), N (r, F ) = N (r, L1 ) = S(r, L1 ), N (r, F ) = N (r, L1 ) = S(r, L1 ), N2 (r, F ) = 2N (r, L1 ) = S(r, L1 ), N2 (r, G) (L1 − a1 ) (L1 − aq ) , (L2 − b1 ) (L2 − bq ) with l(s) 6= 0, ∞ for all s ∈ C By the First Fundamental Theorem, T (r, ) = T (r, L2 ) + O(1), i = 1, , q L − bi 1 ) = 2N (r, ) F L1 Denote the order of a meromorphic function f by ρ(f ), then it follows that 1 ) (3.8) N2 (r, ) = 2N (r, G L2 ρ( ) = ρ(L2 ) = From (3.7), (3.8) we obtain L − bi = 2N (r, L2 ) = S(r, L2 ), N2 (r, T (r, F ) = qT (r, L1 ) ≤ 4N (r, 1 ) + 3N (r, ) + S(r) L1 L2 ≤ 4T (r, L1 ) + 3T (r, L1 ) + S(r), T (r, G) = qT (r, L2 ) ≤ 3N (r, 1 ) + 4N (r, ) + S(r) L1 L2 Moreover, ρ(L1 − ) = ρ(L1 ) = 1, i = 1, , q Since the order of a finite product of functions of finite order is less then or equal to the maximum of the order of these factors (see [17]), we have ρ(l) ≤ This implies that l(s) is of the form l(s) = eAs+B where A, B are constants Since ≤ 3T (r, L1 ) + 4T (r, L1 ) + S(r) Therefore qT (r) ≤ 7T (r) + S(r), (q − 7)T (r) ≤ S(r) lim Li (s) = 1, s→+∞ we get (L1 − a1 ) (L1 − aq ) P (1) = s→+∞ (L2 − b1 ) (L2 − bq ) Q(1) lim l(s) = lim s→+∞ This is a contradiction to the assumption This implies that A = 0, that is, l(s) = c that q ≥ http://jst.tnu.edu.vn; Email: jst@tnu.edu.vn 35 Nguyen Duy Phuong TNU Journal of Science and Technology 225(13): 31 - 37 Case L1 (s) or L2 (s) has a pole at s = q ≥ and EL1 (S) = EL2 (T ), a 6= −1, with multiplicity m1 (≥ 0) or m2 (≥ 0), re- b 6= −1; spectively Set q ≥ and EL1 (S) = EL2 (T ), a = b 6= −1 m (s − 1) (L1 − a1 ) (L1 − aq ) , l(s) = (L2 − b1 ) (L2 − bq ) References where m = q(m2 − m1 ) is an integer We use the arguments similar to the Case So we [ 1] J Ritt, "Prime and composite polynoconclude that mials," Trans Amer Math Soc., vol 23, no 1, pp 51-66, 1922 l(s) = eAs+B , where A, B are constants [ 2] H K Ha, H A Vu, and N H Pham, "On functional equations for mero−m As+B −m morphic functions and applications," lim (s − 1) e = lim (s − 1) l(s) s→+∞ s→+∞ Archiv der Mathematik, vol 109, no 6, pp 539–554, 2017 (L1 − a1 ) (L1 − aq ) P (1) = lim = s→+∞ (L2 − b1 ) (L2 − bq ) Q(1) [ 3] H K Ha, and C C Yang, "On the So A = 0, m = 0, that is, l(s) = c Thus functional equation P (f ) = Q(g), L1 + a = c(L2 + b) Applying Theorem 1.1 Value Distribution Theory and Related we get L1 = L2 , a = b Topics," Advanced Complex Analysis and Application, vol 3, pp 201-208, Proof of Part We use the arguments simi2004 lar to the Proof of Part and then applying Moreover Theorem 1.1 we get L1 = L2 Conclusion We study conditions to equations: c = q + c1 , xq + a = c(y q + b), q x +a y +b xq y q = a [ 4] F Pakovich, "On the functional equations F (A(z)) = G(B(z)), where A, B are polynomials and F, G are continuous functions," Math Proc Camb Phil Soc., vol 143, pp 469-472, 2007 [ 5] F Pakovich, "On the equation P (f ) = Q(g), where P, Q are polynomials and f, g are entire functions," Amer J Math., vol 132, no 6, pp 591-1607, 2010 have solutions on sets of L - functions in the extended Selberg class Since we give [ 6] some sufficient conditions for a finite set S to be a uniqueness range set of L - functions in the extended Selberg class Namely, let P (z) = z q +a, Q(z) = z q +b and let S, T are respective sets of zeros of P (z), Q(z) Then L1 = L2 and P = Q if one of the following conditions is satisfied: [ 7] q ≥ and E L1 (S) = E L2 (T ); 36 H K Ha, H A Vu, and X L Nguyen, "Strong uniqueness polynomials of degree and unique range sets for powers of meromorphic function," Int J Math., vol 29, no 5, pp 1-19, 2018, doi: https://doi.org/10.1142/S0129167X18500374 F Gross, and C.C Yang, "On preimage and range sets of meromorphic http://jst.tnu.edu.vn; Email: jst@tnu.edu.vn Nguyen Duy Phuong TNU Journal of Science and Technology functions," Proc Japan Acard Ser A Math Sci., vol 58, no 1, pp 17-20, 1982 [ 8] [ 9] [10] [11] [12] 225(13): 31 - 37 unique range sets," Kodai Math J., vol 18, pp 515-522, 1995 [13] J Steuding, Value-Distribution of LFunctions, Lecture Notes in MatheH Fujimoto, "On uniqueness of meromatics, vol 1877, Springer, 2007 morphic functions sharing finite sets," Amer J Math., vol 122, pp 1175[14] A D Wu, and P C Hu, "Uniqueness 1203, 2000 theorems for Dirichlet series," Bull Aust Math Soc., vol 91, pp 389–399, G Frank, and M Reinders, "A unique 2015 range set for meromorphic functions with 11 elements," Complex Variables Theory Appl., vol 37, no 1-4, pp 185- [15] F Pakovich, "Prime and composite Laurent polynomials," Bull Sci 193, 1998 Math., vol 133, pp 693-732, 2009 W K Hayman, Meromorphic Func[16] J Kaczorowski, G Molteni, and A tions, Clarendon, Oxford, 1964 Perelli, "Linear independence of Lfunctions," Forum Math., vol 18, pp B Q Li, "A result on value distri1–7, 2006 bution of L-functions," Proc Amer Math Soc., vol 138, no 6, pp [17] C C Yang, and H X Yi, Unique2071–2077, 2010 ness Theory of Meromorphic Functions, Mathematics and Its ApplicaE Mues, and M Reinders, "Meromortions, vol 557, Springer, 2003 phic functions sharing one value and http://jst.tnu.edu.vn; Email: jst@tnu.edu.vn 37 ... obtain the following result Theorem 1.2 Let L1 and L2 be two nonconstant L - functions Let P , Q be polynomials of the form (1.1), and S, T are respective sets of zeros of P (z), Q(z) Then L1 = L2 ... Then Lq1 = cLq2 Lemma 2.6 [13] Let L be an non-constant From this L2 = tL1 , tq c = Applying L - functions and a ∈ C Then equation Lemma 2.5 we have L1 = L2 and therefore L = a has infinitely... function in the complex plane C We assume that the reader is familiar with the notations of Nevanlinna theory and of L- functions in the Selberg class(see [7-16]) In this paper, we discuss the Ritt’s