半純函數體中的函數方程 - 政大學術集成

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(1)國立政治大學應用數學系碩士學位論文. 政治大. 立 Equations in the Field of On Functional. ‧ 國. 學. Meromorphic Functions. ‧. 半純函數體中的函數方程 n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. 碩士班學生：葉長青撰指導教授：陳天進博士中華民國九十九年六月.

(2) Contents. 謝辭. i. Abstract. ii. 中文摘要. iii. 1 Introduction. 政治大 Some Special Functional Equations 立. 4. 學. ‧ 國. 2. 1. 3 Basic Theory of Value Distribution. 7. ‧. 4 Some Necessary Conditions for the Existence of Solution of Func15. er. io. sit. y. Nat. tional Equations. 5 The Distribution of Zeros and Poles of a Class of Meromorphic. al. n. Functions. References. Ch. engchi. i Un. v. 32. 38.

(3) 謝辭在政大待了六年，第一個要感謝的是我的指導老師，陳天進老師。還記得當初大四，正在準備研究所，老師撥出額外的時間，替我們（一群要考試的同學）複習線性代數，訓練我們的邏輯思考與寫證明要注意的地方。那時，我才知道我大學四年白混了。老師在教學上很嚴格，但是私底下對我卻非常照顧。能夠當您的學生，真是太棒了。第二個要感謝的是蔡炎龍老師。能夠遇到蔡老師真的是太幸運的。因為您，我才能夠找回高中三年對於程式的熱誠，我才能夠學習到這麼多資訊領域的知識。也感謝老師願意聆聽我不論是在課業上或是感情上的傾訴，並且給予我許多的建議、開導。. 立. 政治大. 感謝研究室的各位學長姊、同學們的照顧。特別是祐宇學長，在我剛進. ‧ 國. 學. 研究室的時候，每天都找我去吃飯，讓我感到相當溫馨。林祐宇我可能會成為的你競爭對手，你小心囉！. ‧. 最後，感謝爸、媽給予我最大的支持，讓我在政大的這六年可以安然的度. sit. y. Nat. 過。. io. n. al. er. 此篇論文謹獻給我親愛的家人、師長和朋友們。. Ch. engchi. i Un. v. 葉長青謹誌于. 國立政治大學應用數學系中華民國九十九年六月. i.

(4) Abstract In this thesis, we use the theory of value distribution to study the existence of solution of the following functional equation: p X. aj (z)fj (z)kj = 1,. j=1. where a1 (z), · · · , ap (z) are meromorphic functions. For some special case, new and old examples of the solutions are given. For the general case, a necessary condition for the existence of solution is considered.. 政治大. Moreover, we obtain a result on the distribution of zeros and poles of a class of meromorphic functions.. 立. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. ii. i Un. v.

(5) 中文摘要在這篇論文中，我們將利用值分佈的理論來探討下列函數方程解的存在性與其性質：. p X. aj (z)fj (z)kj = 1,. j=1. 其中 a1 (z), · · · , ap (z) 為半純函數。對某些特殊方程，除了文獻裡已知的結果外，我們亦提供其它的例子。一般而言，我們探討解存在的必要條件。另外，我們證明了某一類半純函數之零點與極點之分佈的結果。. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. iii. i Un. v.

(6) 1. Introduction In this thesis, we will use the theory of value distribution to study the existence. of solution of the following functional equation: p X. aj (z)fj (z)kj = 1,. (1.1). j=1. where a1 (z), · · · , ap (z) are meromorphic functions. First, we consider some special cases. In 1927, P. Montel [17] stated the following theorem.. Theorem 1.1. 政治大 [17] Let f, 立 g be two transcendental entire functions. Then if m and. ‧ 國. f n + gm = 1. ‧. cannot hold.. 學. n are integers ≥ 3, the functional equation. sit. y. Nat. n. al. er. io. Later in 1965, A.V. Jategaonkar [12] gave a complete proof of the theorem.. i Un. v. In 1966, Gross [4, 5] considered the functional equation. Ch. engchi. f n + gn ≡ 1. (1.2). and proved the following result.. Theorem 1.2 [5] For n > 2, there do not exist two non-constant entire solutions that satisfy (1.2).. Theorem 1.3 [4] For n > 3, there do not exist two non-constant meromorphic functions f and g that satisfy (1.2).. 1.

(7) Theorem 1.4 [5] For n = 2, all entire solutions of (1.2) for n = 2 are of the form f (z) = cos(η(z)) and g(z) = sin(η(z)), where η(z) is an entire function.. Theorem 1.5 [4] For n = 2, all meromorphic solutions of (1.2) for n = 2 are of the form f (z) =. 2β(z) 1 − β(z)2 , g(z) = , 1 + β(z)2 1 + β(z)2. where β is a meromorphic function.. 政治大. Moreover, Gross [5] considered an example of meromorphic solution of (1.2) for. 立. 學. ‧ 國. n = 3 as follow:. 1 (1 + 3−1/2 ℘0 ), 2℘ −1 (1 − 3−1/2 ℘0 ), g= 2℘. f=. ‧ sit. y. Nat. where ℘ is the Weierstrass ℘-function.. al. n. n = 3 as follows.. er. io. In 1966, Baker [1] gave a general form of meromorphic solution of (1.2) for. Ch. engchi. i Un. v. Theorem 1.6 [1] If f and g are non-constant meromorphic functions satisfying (1.2) for n = 3, then f and g are of the form f (z) = F (ω(z)), g(z) = cG(ω(z)) = cF (−ω(z)) = F (−c2 ω(z)), 1 + 3−1/2 ℘0 (z) 1 − 3−1/2 ℘0 (z) and , respec2℘(z) 2℘(z) tively. ω(z) is a entire function and c is a cube-root of unity. where F and G are elliptic functions. In 1970, Yang [22] considered a more general case of Theorem 1.1.. 2.

(8) Theorem 1.7 [22] The functional equation a(z)f n (z) + b(z)g m (z) = 1. (1.3). (a, b, f, g are meromorphic functions and m, n are integers ≥ 3) cannot hold, if T (r, a) = o(T (r, f )), T (r, b) = o(T (r, g)),. (1.4). unless m = n = 3. If f, g are entire and (1.4) holds, then (1.3) cannot hold even if m = n = 3. Another simple functional equation is f n + g n + hn ≡ 1. (1.5). 政治大 In 1985, Hayman [9] proved a general theorem and, in 1979, D. J. Newman 立. and M. Slater [19] gave a theorem which implies the following result.. ‧ 國. 學. Theorem 1.8 [9] For n ≥ 9, there do not exist three non-constant meromorphic. ‧. functions f, g, and h that satisfies (1.5).. y. Nat. n. al. Ch. er. io. h can not be three non-constant polynomials.. sit. Theorem 1.9 [19] Let f, g, and h be solutions of (1.5) with n = 6, then f, g, and. i Un. v. In 2002, Ishizaki [11] gave a new proof of the non-existence of transcendental. engchi. meromorphic solution of (1.5) for n ≥ 9.. For the functional equation (1.5), the solutions for 1 ≤ n ≤ 6 can be given explicitly which will be discussed in section 2. However, for n = 7, 8, the existence of solution of (1.5) is still open. For the general case of (1.1), Toda [21], K.-W. Yu and C.-C. Yang [25], and I. Lahiri and K.-W. Yu [13] gave a necessary condition for the existence of solution of (1.1) which will be discussed in section 4. Finally, we obtain a result on the distribution of zeros and poles of a class of meromorphic functions which will be proved in section 5.. 3.

(9) 2. Some Special Functional Equations In this section, we collect some new and old examples of solutions of the func-. tional equation (1.5): f n + g n + hn = 1 for 1 ≤ n ≤ 6. Example 2.1 (n=1) Let f = ez , g = e−z , h = −ez − e−z + 1, then f + g + h = 1.. 政治大. In fact, given meromorphic functions g and h and set f = 1 − g − h.Then f, g and. 立. h satisfy (1.5).. ‧ 國. 學. ‧. Example 2.2 (n=2) Let α(z) be a meromorphic function and defined √ α2 − 2 (α2 + 1)i √ f= √ , g= , h = 2α, 3 3 then. io. sit. y. α4 − 4α2 + 4 α4 + 2α2 + 1 6α2 − + 3 3 3. er. Nat. f 2 + g 2 + h2 =. = 1.. n. al. Ch. i Un. v. Therefore, f 2 + g 2 + h2 = 1 has infinitely many solutions.. engchi. Example 2.3 (n=3) Lehmer [14] showed that (1.5) has infinitely many solutions. In fact, if β(z) is a meromorphic function and let f (z) = 9β(z)4 , g(z) = −9β(z)4 + 3β(z), h(z) = −9β(z)3 + 1, Then f 3 + g 3 + h3 = (9β(z)4 )3 + (−9β(z)4 + 3β(z))3 + (−9β(z)3 + 1)3 = 729β(z)12 + (−729β(z)12 + 729β(z)9 − 243β(z)6 + 27β(z)3 ) + (−729β(z)9 + 243β(z)6 − 27β(z)3 + 1) = 1. 4.

(10) Example 2.4 (n=4) Gross [4] gave the following example: Let f (z) = 21/4 (sin2 z − cos2 z + i sin z cos z), g(z) = (−1)1/4 (2i sin z cos z + sin2 z), h(z) = (−1)1/4 (2i sin z cos z − cos2 z). Then we have f 4 + g 4 + h4 = 1 Also Green [3] considered the following example: If w(z) is a non-constant. 政治大. entire function, then the three transcednental entire functions. 立f =8. g =(−8)−1/4 (e3w − e−w ), h =(−1)1/4 e2w ,. Nat. y. ‧. io. n. al. sit. f 4 + g 4 + h4 = 1. er. satisfy. (e3w + e−w ),. 學. ‧ 國. −1/4. i Un. v. Example 2.5 (n=5) Gundersen [8] gave the transcendental meromorphic solution as follows:. Ch. engchi. b3 b4 − b1 b2 1 , 1 ≤ k ≤ 4, c = ,d = Let ak = e and bk = a − 1 (b + b ) − (b + b ) k 3 4 1 2 p (c − b1 )(c − b2 ). Then b1 + b2 6= b3 + b4 , c = 6 0 and d 6= 0. Define 2kπi/5. u(z) = 1 +. 1 u(z)5 − 1 1 5 , v(z) = 1 + , w(z) = . c + dez c + de−z v(z)5 − 1. Let π. f = u, g = e 5 i vw, h = v. Then f 5 + g 5 + h5 = 1.. 5.

(11) Example 2.6 (n=6) Gundersen [7] constructed three meromorphic functions f, g and h satisfying the functional equation (1.5) as n = 6.. Consider the differential equation (F 0 )2 = K(F − A)(F − B)(F − C),. (2.1). where K 6= 0 is a constant, and F is a meromorphic solution of (2.1). Then by a result of F. Rellich [20], F is an elliptic function. Set w(z) = α. 政治大 w−b w+b u f立 = , g= , h=i . v v v. 學. ‧ 國. Finally, define. f 6 + g 6 + h6 = 1.. ‧. io. sit. y. Nat. n. al. er. Then. (4 + 2i)F (z) − (3 + 4i)A − (1 − 2i)B . (4 + 2i)F (z) − 5A + (1 − 2i)B. Ch. engchi. 6. i Un. v.

(12) 3. Basic Theory of Value Distribution In this section, we introduce and review some basic facts and notations in. complex analysis and value distribution which will be used throughout the rest of the thesis. For the sake of brevity, proofs are omitted because they are standard and can be found in [2, 6, 10, 23, 24]. In Nevanlinna’s value distribution theory, the following Poisson-Jensen’s formula plays a very important role.. 政治大 be the zeros and poles of f in |z| < R, 1 ≤ µ ≤ M ,. Theorem 3.1 (Poisson-Jensen’s formula) Let 0 < R < ∞ and f be meromorphic in |z| < R and aµ and bν. 立. 學. 2π. R2 − r 2 dϕ R2 − 2Rr cos(θ − ϕ) + r2 0

(13)

(14)

(15)

(16) N M X

(17) R(z − bν )

(18)

(19) R(z − aµ )

(20) X

(21) −

(22) . log

(23)

(24) 2 + log

(25)

(26) 2 R − aµ z

(27) R −b z

(28) Z. ν. ν=1. µ=1. io. n. al. sit. Nat. y. log |f (Reiϕ )|. er. 1 log |f (z)| = 2π. ‧. ‧ 國. 1 ≤ ν ≤ N , respectively. If z = reiθ , 0 ≤ r < R, and f (z) 6= 0, ∞, then we have. i Un. v. By taking z = 0 in Theorem 3.1, we get the Jensen’s formula.. Ch. engchi. Theorem 3.2 (Jensen’s formula) Under the assumption of Theorem 2.1, if f (0) 6= 0, ∞, then we have 1 log |f (0)| = 2π. Z 0. 2π. M X. N. X R R log + log . log |f (Re )|dϕ − |aµ | ν=1 |bν | µ=1 iθ. The assumption f (0) 6= 0, ∞ in Theorem 3.1 can be eliminated. In fact, for 0 ≤ r < ∞, let n(r, f ) denote the number of poles of f in |z| ≤ r counting multiplicities. Consider the Laurent expansion of f at the origin f (z) = cλ z λ + cλ+1 z λ+1 + · · · .. 7.

(29) 1 Note that λ = n(0, ) − n(0, f ). Consider the function f   f (z)( R )λ if z 6= 0 z g(z) =  c Rλ if z = 0, λ then we have the generalized Jensen’s formula.. Theorem 3.3 (generalized Jensen’s formula) Under the assumption of Theorem 3.1 without the condition f (0) 6= 0, ∞, then we have 1 log | cλ | = 2π. M X

(30)

(31) R 1 log

(32) f (Reiϕ )

(33) dϕ − log − n(0, ) log R | aµ | f µ=1. 2π. Z 0. N X. 政治大. R log + + n(0, f ) log R, | bν | ν=1. 立. ‧ 國. 學. where cλ is the first non-zero coefficient of the Laurent expansion of f at 0.. ‧. From now on, meromorphic function means meromorphic in the whole complex plane. First of all, we introduce the positive logarithmic function.. io. n. al.   log x log+ x = max{log x, 0} =  0. Ch. engchi. er. sit. y. Nat. Definition 3.4 For x ≥ 0,. i Un. v. if x ≥ 1 if o ≤ x < 1.. Obviously, log+ x is a continuous non-negative increasing function on [0, ∞) satis1 1 fying log x = log+ x − log+ and | log x| = log+ x + log+ . x x Let f be a meromorphic function, Nevanlinna [18] introduced the following notations.. Definition 3.5 For 0 < r < ∞, 1 m(r, f ) = 2π. Z. 2π. log+ |f (reiθ )|dθ.. 0. 8.

(34) Definition 3.6 For 0 < r < ∞, Z r n(t, f ) − n(0, f ) dt + n(0, f ) log r, N (r, f ) = t 0 where n(t, f ) denotes the number of poles of f in the disc | z | ≤ t counting multiplicities. N (r, f ) is called the counting function of f . For 0 ≤ r < ∞, n(r, f ) denotes the number of poles of f (z) in |z| ≤ r counting multiplicities; n(r, f ) denotes the number of poles of f (z) in |z| ≤ r ignoring multiplicities; nk) (r, 1/f ) (resp.n(k (r, 1/f )) denotes the number of zeros of f (z) in |z| ≤ r with order ≤ k(resp.≥ k) counting multiplicities; nk) (r, 1/f ) (resp.n(k (r, 1/f )) denotes the number of zeros of f (z) in |z| ≤ r with order ≤ k(resp.≥ k) ignoring multiplicities.. 立. 政治大. 學. ‧ 國. Definition 3.7 For 0 < r < ∞, the function T (r, f ) defined by T (r, f ) = m(r, f ) + N (r, f ). ‧. is called the (Nevanlinna) characteristic function of f .. Nat. sit. y. It is clear that T (r, f ) is a non-negative increasing function and a convex func-. io. er. tion of log r. Let f be given in Theorem 2.1. It follows from the integration by parts in Riemann-Stieltjes integral, we have Z R n(t, 1 ) − n(0, 1 ) M X R f f log = dt | aµ | t 0 µ=1. n. al. and. N X. Ch. engchi. R log = | bν | ν=1. Z 0. R. i Un. v. n(t, f ) − n(0, f ) dt. t. On the other hand, the generalized Jensen’s formula can be rewritten as. N X

(35)

(36) R log+

(37) f (Reiϕ )

(38) dϕ + log + n(0, f ) log R | b | ν 0 ν=1

(39)

(40) Z 2π M X

(41)

(42) 1 1 R 1 +

(43)

(44) = log

(45) dϕ + log + n(0, ) log R + log | cλ | .

(46) iϕ 2π 0 f (Re ) | aµ | f µ=1. 1 2π. Z. 2π. 9.

(47) Therefore, we obtain 1 1 m(R, f ) + N (R, f ) = m(R, ) + N (R, ) + log | cλ | , f f that is, 1 T (R, f ) = T (R, ) + log | cλ | , f which is another form of the generalized Jensen’s formula and is also known as the Nevanlinna-Jensen’s formula.. Theorem 3.8 (Nevanlinna-Jensen’s formula) Let f be a meromorphic function , then, for r > 0,. 政治大. 1 T (r, f ) = T (r, ) + log | cλ | , f. 立. where cλ is the first non-zero coefficient of the Laurent expansion of f at 0.. ‧ 國. 學. By the Nevanlinna-Jensen’s formula, we can get the Nevanlinna’s first funda-. ‧. mental theorem.. y. Nat. io. sit. Theorem 3.9 (Nevanlinna’s First Fundamental Theorem) Let f be a mero-. er. morphic function and a be a finite complex number. Then, for r > 0, we have. a1l. iv C ) =hT (r, f ) + log |cλ |U +nε(a, r), f −a engchi. n T (r,. where cλ is the first non-zero coefficient of the Laurent expansion of. 1 at 0, and f −a. |ε(a, r)| ≤ log+ |a| + log 2.. Usually, Nevanlinna’s first fundamental theorem is written as T (r,. 1 ) = T (r, f ) + O(1). f −a. Now, we come to the most important theorem in the theory of value distribution, namely, Nevanlinna’s second fundamental theorem.. 10.

(48) Theorem 3.10 (Nevanlinna’s Second Fundamental Theorem) Let f be a nonconstant meromorphic function and aj ∈ C, 1 ≤ j ≤ q, be q distinct finite values (q ≥ 2). Then m(r, f ) +. q X. m(r,. j=1. 1 ) ≤ 2T (r, f ) − N1 (r) + S(r, f ), f − aj. where N1 (r) = 2N (r, f ) − N (r, f 0 ) + N (r,. 1 ) and f0. q X f0 f0 ) + O(1). S(r, f ) = m(r, ) + m(r, f f − aj j=1. 政治大 1. Given a ∈ C, by Nevanlinna’s first fundamental theorem,. 立. 1 ) = T (r, f ) − N (r, ) + O(1). f −a f −a. 學. ‧ 國. m(r,. Hence, Nevanlinna’s second fundamental theorem can be rewritten as follows.. ‧. Theorem 3.11 Let f be a non-constant meromorphic function and aj ∈ C∞ , 1 ≤. Nat. n. al. sit. N (r,. 1 ) − N1 (r) + S(r, f ), f − aj. er. io. (q − 2)T (r, f ) <. q X. y. j ≤ q, be q distinct values (q ≥ 3). Then. i n C U where N1 (r) and S(r, f ) are given h asein Theoremi 3.10. ngch j=1. Note that, in Theorem 3.11, if some aj = ∞, then N (r,. v. 1 ) should be read as f − aj. N (r, f ). Let n1 (t) = 2n(t, f ) − n(t, f 0 ) + n(t,. 1 ) and let n(t, f ) denote the number of f0. distinct poles of f in |z| ≤ t. Define Z r n(t, f ) − n(0, f ) N (r, f ) = dt + n(0, f ) log r, t 0 which is called the reduced counting function of f . Note that, if z0 is a pole of f of order k in |z| ≤ t, then z0 is counted k − 1 times by n1 (r). Similarly, for a finite. 11.

(49) value a, if z0 is a zero of f − a of order k in |z| ≤ t, then z0 is also counted k − 1 times by n1 (r). Hence, q. q X. X 1 1 ) − N1 (r) ≤ N (r, ). N (r, f − aj f − aj j=1 j=1. Therefore, we have the third form of Nevanlinna’s second fundamental theorem.. Theorem 3.12 Let f be a non-constant meromorphic function and aj ∈ C∞ , 1 ≤ j ≤ q, be q distinct values (q ≥ 3). Then (q − 2)T (r, f ) <. q X. N (r,. j=1. 1 ) + S(r, f ), f − aj. 政治大. where S(r, f ) is given as in Theorem 3.10.. 立. deficiency of a with respect to f is defined by δ(a, f ) = lim inf r→ ∞. 1 ) m(r, f −a. T (r, f ). = 1 − lim sup r→∞. 1 ) N (r, f −a. ‧. ‧ 國. 學. Definition 3.13 Let f be a non-constant meromorphic function and a ∈ C∞ . The. .. Nat. y. T (r, f ). sit. In Nevanlinna’s second fundamental theorem, the remainder term S(r, f ) is. er. io. a complicated object which can be estimated by using the method of logarithmic. al. n. iv n C hgrowth it clear, we need the concept of the meromorphic function. i U e n g ofc h. derivative. It turns out that S(r, f ) is small comparing to T (r, f ). In order to make. Classically, we use the maximum modulus to measure the growth of an entire function.. Definition 3.14 Let f be a meromorphic function. The order λ of f is defined to be log+ T (r, f ) λ = lim sup log r r→∞ and the lower order µ of f is defined to be µ = lim inf r→∞. log+ T (r, f ) . log r. 12.

(50) Definition 3.15 Let f (z) and a(z) be meromorphic functions. If T (r, a) = S(r, f ), then a(z) is called a small function of f (z). Let f be an entire function. Define, for r ≥ 0, M (r, f ) = max |f (z)|. |z≤r|. Then the relation between M (r, f ) and T (r, f ) is given as follows. Theorem 3.16 Let 0 ≤ r < R < ∞ and f be an entire function, we have T (r, f ) ≤ log+ M (r, f ) ≤. R+r T (R, f ). R−r. 政治大 T (r, f ) ≤ log M (r, f ) ≤ 3T (2r, f ). 立. In particular,. +. 學. ‧ 國. By Theorem 3.16, the order and lower order of an entire function are unambiguous. Now, we can state the properties of S(r, f ).. ‧. Lemma 3.17 Let f be a non-constant meromorphic function. If f is of finite. (r → ∞).. n. al. f0 ) = O(log r), f. er. io. m(r,. If f is of infinite order, then 0. m(r,. Ch. sit. y. Nat. order, then. i Un. v. e n g c h i(r → ∞, r 6∈ E),. f ) = O(log(rT (r, f ))), f. where E is a set of finite measure. Theorem 3.18 Let f be a non-constant meromorphic function and S(r, f ) be defined in Theorem 3.10. If f is of finite order, then S(r, f ) = O(log r),. (r → ∞).. If f is of infinite order, then S(r, f ) = O(log(rT (r, f ))), where E is a set of finite measure.. 13. (r → ∞, r 6∈ E),.

(51) In the thesis, we will denote by S(r, f ) any quantity satisfy S(r, f ) = o(T (r, f )) as r → ∞ if f is of finite order, and S(r, f ) = o(T (r, f )) as r → ∞, r 6∈ E if f is of infinite order, where E is a set of finite measure. f0 ) = S(r, f ). Moreover, Milloux [16] proved the followf. By Lemma 3.17, m(r, ing.. Theorem 3.19 Let f be a non-constant meromorphic function and k be a positive integer and let Ψ(z) =. k X. ai (z)f (i) (z),. i=1. 政治大 Ψ. where a1 (z), a2 (z), . . . , ak (z) are small functions of f . Then. 立. m(r,. f. ) = S(r, f ).. ‧ 國. 學. Now we record some well-known results on two meromorphic functions sharing. ‧. four or five small functions as follows.. y. sit. io. er. functions.. Nat. We have the generalization of second fundamental theorem for three small. n. al. i n C U h e nfunction. a2 (z) and a3 (z) are three distinct small g c h i Then. v. Theorem 3.20 [23] Let f be a non-constant meromorphic function and a1 (z),. T (r, f ) <. 3 X. N (r,. j=1. 1 ) + S(r, f ), f − aj. Finally, we state a Nevanlinna theorem about the linear combination of meromorphic functions which will be needed in section 4.. Theorem 3.21 [23] Suppose f1 , · · · , fn are linearly independent meromorphic functions satisfying the following identity n X. fj = 1. j=1. 14.

(52) Then for 1 ≤ j ≤ n, we have T (r, F1 ) ≤. n X. n. N (r,. j=1. X 1 1 )− N (r, fn ) + N (r, D) − N (r, ) + o(T (r)) fn D j=1. where D is the Wronskian of f1 , · · · , fn , and T (r) = max {T (r, fk )}, 1≤k≤n. E is a set with finite linear measure.. 4. Some Necessary Conditions for the Existence of Solution. 政治大. of Functional Equations. 立. ‧ 國. 學. In this section, we study the necessary condition for the existence of solution of the functional equation (1.1) considered by Toda [21], K.-W. Yu and C.-C. Yang [25],. ‧. and I. Lahiri and K.-W. Yu [13]. For completeness, we include their proofs.. y. Nat. Toda [21] first considered the necessary condition for the existence of entire. n. er. io. al. sit. solutions. Before stating the theorem, we need a lemma.. i Un. v. Lemma 4.1 [21] Let g0 , · · · , gp (≥ 1) be p+1 non-constant meromorphic functions in |z| < ∞ satisfying p X. Ch. engchi. αi gi = 1, α0 , · · · , αp 6= 0,. i=0. with constant coefficients and δ(∞, gi ) = 1 (i = 0, · · · , p). Then , we have p X. θp (0, gi ) ≤ p.. i=0. Here, θp (0, gi ) = 1 − lim sup r→∞. Np (r, 0, gi ) T (r, gi ). and Np (r, 0, gi ) =. X ak 6=0. log+. r + min(ρ0 , p) log r |ak |. 15. (4.1).

(53) where the summation is taken over the different zeros ak (6= 0) of gi counted min(ρk , p) times at ak , ρk (resp. ρ0 ) being the order of multiplicity of zero of gi at ak (resp. 0). P roof . Suppose g0 , · · · , gp are linearly independent, then the Wronskian ∆ = W (g0 , · · · , gp ) 6= 0. By differentiating both sides of (4.1), we have p X. (αi gi )(µ) = 0, µ = 1, . . . , p. (4.2). i=0. e = ∆/g0 , · · · , gp and ∆ fi = W (g0 , · · · , gi−1 , gi+1 , · · · , gp )gi /g0 , · · · , gp . Now let ∆ By (4.2). 政治大

(54)

(55)

(56)

(57)

(58)

(59)

(60) (p)

(61) (αp gp )

(62).

(63)

(64)

(65) α0 g0 ···

(66) 1 ..

(67) e = α0 , · · · , αp ∆

(68) .

(69) g0 , · · · , gp

(70)

(71) (α0 g0 )(p) · · ·. 立. αp gp .. .. αi−1 gi−1. ··· ···. (αi−1 gi−1 )(1) .. .. ···. (αi−1 gi−1 )(p) 0 (αi+1 gi+1 )(p) · · ·.

(72)

(73)

(74) g0

(75)

(76) . α0 , · · · , αi−1 , αi+1 , · · · , αp

(77)

(78) .. =

(79) .. g0 , · · · , gp

(80) .

(81)

(82) (p)

(83) g0. =. αi+1 gi+1. ···. y. 0 (αi+1 gi+1 )(1) · · · .. .. . . ···. sit. n. Ch. 1. er. io. al. ···. ‧. ‧ 國. 學. Nat.

(84)

(85)

(86) α0 g0

(87) ..

(88)

(89) . 1

(90) =

(91) .. g0 , · · · , gp

(92) .

(93)

(94)

(95) (α0 g0 )(p). e n g c h· ·i·. α0 , · · · , αi−1 , αi+1 , · · · , αp e ∆i gi. which implies gi =. ei ∆ . e αi ∆. 16. i Un. v. 1 gi+1 · · · .. 0 . ··· .. .. . . ···. ···. gi−1 .. . .. .. ···. gi−1 0 gi+1 · · ·. ···. (p). (p).

(96)

(97) gp

(98)

(99) ..

(100) .

(101)

(102) ..

(103) .

(104)

(105)

(106) gp(p)

(107).

(108)

(109) αp gp

(110)

(111) ..

(112)

(113) .

(114)

(115) ..

(116) .

(117)

(118) (αp gp )(p)

(119).

(120) Consider ei ∆ e i ) + m(r, 1 ) + m(r, αi ) ) ≤ m(r, ∆ e e αi ∆ ∆ e i ) + m(r, 1 ) + N (r, 1 ) + O(1) ≤ m(r, ∆ e e ∆ ∆ 1 e i ) + T (r, ) + O(1) ≤ m(r, ∆ e i ) + T (r, ∆) e + O(1) ≤ m(r, ∆ e ∆. m(r, gi ) = m(r,. e i ) + m(r, ∆) e + N (r, ∆) e + O(1) ≤ m(r, ∆ e + S(r) ≤ N (r, ∆) ≤. p X. Np (r, 0, gj ) + p. X. N (r, gj ) + S(r). j=0 p. j=0 p. ≤. p X. X (r, g ) + S(r), 政 N治大. Np (r, 0, gj ) + p. 立. j=0. where. ‧ 國. 學.    . j. j=0. O(1) , if gj are rational ! X S(r) = .  o T (r, gj ) ,r ∈ /E  . ‧. Therefore,. p. j=0. sit. y. Nat. T (r) ≡ max {m(r, gi ) + N (r, gi )}. 0≤i≤p. ≤. p X. p X. aNlp(r, 0, gj ) + p N (r, gj ) +i Nv (r, gi) + S(r)} n j=0 j=0 Ch U p hi e n gX c n. ≤ max {. p X. er. io. 0≤i≤p. Np (r, 0, gj ) + (p + 1). j=0. N (r, gj ) + S(r). (4.3). j=0. By definition of θp (0, gi ) and δ(∞, gi ), for arbitary ε > 0, there exists a positive number r0 such that for any given r ≥ r0 , we have Np (r, 0, gi ) < (1 − θp (0, gi ) + ε)T (r, gi ). (4.4). N (r, gi ) < (1 − δ(∞, gi ) + ε)T (r, gi ).. (4.5). and. 17.

(121) Since δ(∞, gi ) = 1, for all i = 0, · · · p, by (4.3), (4.4), (4.5), we have T (r) ≤. p X. (1 − θp (0, gj ) + ε)T (r, gj ) + (p + 1). j=0 p. ≤. X. p X. (1 − δ(∞, gj ) + ε)T (r, gj ). j=0. (1 − θp (0, gj ) + ε)T (r) + (p + 1). p X. εT (r). j=0. j=0. which imply 0≤. p X. (1 − θp (0, gj ) + ε)T (r) + (p + 1). p X. εT (r) − T (r). j=0. j=0 p. p. ! X X ε − 1 T (r) = (1 − θp (0, gj ) + ε) + (p + 1) ! 治 X 政 p+1− θ (0, g ) + (p + 1)ε + (p大 + 1) ε − 1 T (r) 立 ! j=0. j=0. p. =. p. 2. j. j=0. ‧ 國. p X. θp (0, gj ) + (p + 1)ε + (p + 1)2 ε T (r).. 學. = p−. j=0. θp (0, gj ) + (p + 1)ε + (p + 1)2 ε ≥ 0,. sit. y. Nat. p−. p X. ‧. Therefore, we obtain. j=0. io. n. al. er. which is true for all ε > 0. Hence,. i n CX θ (0, g ) ≤ p. U p j hengchi j=0 p. v. In the case when g0 , · · · , gp are linearly dependent. The original functional equation s X can be reduce to γi gi = 1, where g0 , · · · , gs are linearly independent with s < p i=0. as follows. By renumbering, we may assume that {g0 , · · · , gs } is the maximal linear independent subset of {g0 , · · · , gp }. Then gk =. s X. βk,j gj , s + 1 ≤ k ≤ p,. j=0. 18.

(122) where βk,j are constants. Hence, 1=. p X. αi gi. i=0. =. s X. αj gj +. j=0. = =. k=s+1 p. s X j=0 s X. p X. αj +. X. αk. s X. βk,j gj. j=0. ! αk βk,j. gj. k=s+1. γj gj ,. j=0. where γj =. αj +. p X. ! αk βk,j .. 政治大. k=s+1. Clearly, we may assume that γj 6= 0, 0 ≤ j ≤ s. Thus, we have. 立X. γj gj = 1, γj 6= 0. 學. ‧ 國. s. j=0. By the result of the first part, we have. ‧. s X. θs (0, gj ) ≤ s. y. sit. Nat. j=0. io. j=0. al. n. p X. θp (0, gj ) =. s X. Ch ≤. j=0 s X. p X. θp (0, gj ) +. engchi X θs (0, gj ) +. v. θp (0, gj ). i Un. j=s+1 p. j=0. er. Obviously, for s < p, we have θp (0, gi ) ≤ θs (0, gi ) ≤ 1. Hence. θp (0, gj ). j=s+1. ≤ s + (p − s) = p q Next, we state the result of Toda [21]. Theorem 4.2 [21] Let f0 , · · · , fp (p ≥ 1) be p + 1 non-constant entire functions and let a0 , · · · , ap be p + 1 meromorphic functions (6= 0) in |z| < ∞ such that T (r, ai ) = o(T (r, fi )), (i = 0, · · · , p). 19.

(123) Then, if the following functonal equation p X. ai fini = 1. i=0. holds for some integers n0 , · · · , np (≥ 1), it must be p X 1 1 ≥ . ni p i=0. P roof . We know that T (r, f n ) ∼ nT (r, f ), (r → ∞). 政治大. Since. 立. n. al. er. 1 ), we have gi. io. By definition of Np (r,. C1 h. sit. T (r, ai fini ) ∼ ni T (r, fi ), (r → ∞). y. ‧. ≤ T (r, ai fini ) + S(r, fi ). Nat. Thus, we get. ≤ ni T (r, fi ) + T (r, ai ) + S(r, fi ). 學. ‧ 國. T (r, ai fini ) ≤ T (r, ai ) + ni T (r, fi ) = ni T (r, fi ) + S(r, fi ). engchi. iv n U 1. 1 1 1 ) + pN (r, ni ) = Np (r, ) + pN (r, ) ni ) ≤ Np (r, ai f i ai fi ai fi 1 1 Np (r, a f ni ) pN (r, fi ) Np (r, a1i ) i i ≤ ni ni + T (r, ai fi ) T (r, ai fi ) T (r, ai fini ). Np (r,. Therefore, Np (r, a f1ni ). N (r, f1i ) p p lim sup lim sup ≤ ni ≤ lim sup ni = T (r, ai fi ) ni r→∞ T (r, ai fi ) ni r→∞ r→∞ T (r, ai fi ) i i. pN (r, f1i ). By definition of θp , 1 − lim sup r→∞. Np (r, a f1ni ) i i. T (r, ai fini ). = θp (0, ai fini ) ≥ 1 −. 20. p . ni.

(124) Since each ai (z) is a small function, by Lemma 4.1, we have p X i=0. p 1− ni. ≤. p X. θp (0, ai fini ) ≤ p. i=0. which implies that p X 1 1 ≥ . ni p i=0. q 3 1 ≥ if n ≤ 6. It follows from Theorem 4.2, n 2 the functional equation admits no entire solution if n > 6. In general, we have the For the case p = 3 in (1.1), since. following consequence.. i=0. ‧. then. p X 1 1 < , ni p i=0. 學. p X. i. ‧ 國. Corollary 4.3 Let fi. 政治大 and 立 a be defined as in Theorem 4.5. If. ai (z)fini (z) = 1 cannot hold.. sit. y. Nat. io. lemma.. n. al. er. Before we prove the result of K.-W. Yu and C.-C. Yang, we need the following. Ch. engchi. i Un. v. Lemma 4.4 [25] Let F1 , · · · , Fp be p (≥ 2) linearly independent non-constant meromorphic functions. If p X. Fj = 1,. j=1. then, ". p X. p X. # 1 N (r, Fj ) + N (r, ) + S(r), T (r) ≤ 2(p − 1) Fj j=1 j=1 where T (r) = max T (r, Fj ) and S(r) = o 1≤j≤p. p X j=1. tional set of r of finite linear measure.. 21. ! T (r, Fj ) , with a possible excep-.

(125) P roof . Let D = W (F1 , F2 , · · · , Fp ). As in proof of Lemma 4.1, we have Fj =. ∆j , ∆. where ∆=. D Fj Dj and ∆j = F1 · · · Fp F1 · · · Fp. with Dj are the minor determinants of elements Fj , 1 ≤ j ≤ p. By the first fundamental theorem, 1 ) + O(1) ∆ 1 1 1 ≤ m(r, ∆j ) + m(r, ) + N (r, ) − N (r, ) + O(1) ∆ ∆ ∆ 1 1 ≤ m(r, ∆j ) + T (r, ) − N (r, ) + O(1) ∆ ∆ 1 ≤ m(r, ∆j ) + T (r, ∆) − N (r, ) + O(1) ∆ 1 ≤ m(r, ∆j ) + m(r, ∆) + N (r, ∆) − N (r, ) + O(1) ∆ 1 ≤ N (r, ∆) − N (r, ) + S(r) ∆. m(r, Fj ) ≤ m(r, ∆j ) + m(r,. 立. 政治大. ‧. ‧ 國. 學. (4.6). sit. y. Nat. We know that if z0 is not a pole and a zero of Fj , 1 ≤ j ≤ p, then z0 is not a pole. io. er. of ∆. Thus, the poles of ∆ come from the zeros and poles of Fj , 1 ≤ j ≤ p. But some of them may be canceled in the expression of ∆, that is, some zeros or poles. n. al. Ch. i Un. v. of F1 , · · · , Fj may not be a pole of ∆. Suppose that z0 is a pole or a zero of F1 and (n) F1 is not a pole and a zero of Fj , for all j = 2 · · · p. Then has a pole of order at F1 most p − 1 at z0 . By these simple argument we have " p # p X X 1 N (r, ∆) ≤ (p − 1) N (r, Fj ) + N (r, ) Fj j=1 j=1. engchi. 22.

(126) On the other hand, by the first fundamental theorem and (4.6), T (r) = max T (r, Fj ) 1≤j≤p. = max [m(r, Fj ) + N (r, Fj )] 1≤j≤p. 1 ≤ N (r, ∆) − N (r, ) + max N (r, Fj ) + S(r) 1≤j≤p ∆ # " p p X X 1 ≤ (p − 1) N (r, Fj ) + N (r, ) Fj j=1 j=1 1 + max N (r, Fj ) − N (r, ) + S(r) 1≤j≤p ∆ Since N (r, Fj ) = N (r,. 立. and. F∆ 1 治 )≤N (r, F ∆) + N (r, ) 政 ∆ 大 ∆ j. j. ‧. ‧ 國. 學.

(127)

(128) (1) (1)

(129) F1(1) /F1 · · · Fj−1 /Fj−1 Fj+1 /Fj+1 · · · Fp(1) /Fp

(130)

(131) (2) (2)

(132) F1(2) /F1 · · · Fj−1 /Fj−1 Fj+1 /Fj+1 · · · Fp(2) /Fp

(133) Fj ∆ =

(134) .. .. .. .. .. ... .

(135) . . . .

(136)

(137) (p−1) (p−1) (p−1)

(138) F1 /F1 · · · Fj−1 /Fj−1 Fj+1 /Fj+1 · · · Fp(p−1) /Fp.

(139)

(140)

(141)

(142)

(143)

(144)

(145) ,

(146)

(147)

(148)

(149)

(150). er. io. sit. y. Nat. we have. (4.7). " # X X 1 1 N (r, Fj ) − N (r, ) ≤ (p − 1) N (r, Fi ) + N (r, ) ∆ Fi i6=j i6=j " p # p X X 1 N (r, Fj ) + N (r, ) ≤ (p − 1) Fj j=1 j=1. n. al. Ch. engchi. i Un. v. (4.8). Combine (4.7) and (4.8), we have # " p p X X 1 T (r) ≤ 2(p − 1) N (r, Fj ) + N (r, ) + S(r) Fj j=1 j=1 q Remark. If δ(∞, Fj ) = 1, 1 ≤ j ≤ p in Lemma 4.4, then N (r, Fj ) = S(r) and. 23.

(151) N (r, Fj ) = S(r). In this case, (4.6) and (4.7) can be reduced as follows. 1 ) + O(1) ∆ 1 ≤ m(r, ∆j ) + T (r, ) + O(1) ∆. m(r, Fj ) ≤ m(r, ∆j ) + m(r,. ≤ m(r, ∆j ) + T (r, ∆) + O(1) ≤ m(r, ∆j ) + m(r, ∆) + N (r, ∆) + O(1) ≤ N (r, ∆) + S(r) and T (r) = max T (r, Fj ) 1≤j≤p. 政治大. = max [m(r, Fj ) + N (r, Fj )] 1≤j≤p. p X. N (r,. j=1 p. X 1 1 N (r, kj ) + S(r) k fj j=1 j. n. al. Ch. Then it reduced to Theorem 4.2.. engchi. sit. io. p X 1 T (r) + S(r) ≤ (p − 1) k j=1 j. er. Nat. p X 1 1 ≤ (p − 1) T (r, kj ) + S(r) k fj j=1 j. ‧. ≤ (p − 1). 1 ) + S(r) Fj. y. ≤ (p − 1). 學. ‧ 國. ≤ N (r, ∆) + S(r) 立. i Un. v. Now, we come to a result of K.-W. Yu and C.-C. Yang. Theorem 4.5 [25] Let f1 , . . . , fp be p (≥ 2) non-constant meromorphic functions and let a1 , . . . , ap be p meromorphic functions (6= 0) such that T (r, aj ) = o(T (r, fj )), 1 ≤ j ≤ p. Then, if the fj and aj (1 ≤ j ≤ p) satisfy equation p X. k. aj (z)fj j (z) = 1,. j=1. 24.

(152) for some integer k1 , . . . , kp , then p X 1 1 ≥ . k 4(p − 1) j j=1. P roof . By Lemma 4.4, we need to consider N (r,. 1 k aj f j j. k. ) and N (r, aj fj j ) as follows.. With a simple calculation, we have N (r,. 1 k aj fj j. ) ≤ N (r,. 1 1 1 ) + N (r, kj ) aj kj fj. ≤ T (r,. 1 1 1 ) + T (r, kj ) aj kj fj. 1 T (r, f ) + O(1) 政治 k 大 1 kj j. ≤ T (r, aj ) +. j. 立 ≤ k T (r) + S(r). (4.9). j. ‧ 國. 學. and. 1 N (r, fjkj ) kj 1 k ≤ T (r, aj ) + T (r, fj j ) + O(1) kj 1 ≤ T (r) + S(r) kj. k. Nat. n. al. (4.10). er. io. sit. y. ‧. N (r, aj fj j ) ≤ N (r, aj ) +. By (4.9), (4.10) and Lemma 4.4, we obtain. Ch. p e nX c1 h i g. !. T (r) ≤ 4(p − 1). j=1. kj. i Un. v. T (r) + S(r). p X 1 1 Thus, we have ≥ and the proof is finished. k 4(p − 1) j=1 j. q. Finally, we consider the following result of I. Lahiri and K.-W. Yu. Before we start, we need some lemmas. Lemma 4.6 Let f be a meromorphic function and p be a positive integer. Then we have Np (r, 0, f ) ≤ pN (r, 0, f ).. 25.

(153) q. P roof . Clearly, It follows from the definition of Np (r, 0, f ).. Lemma 4.7 [15] Let p ≥ 2 and F1 , · · · , Fp be linearly independent non-constant meromorphic functions. If p X. Fj (z) = 1,. j=1. then we have T (r, F1 ) <. p X. p. Np−1 (r,. j=1. X 1 ) + Ap N (r, Fj ) + o(T (r)), Fj j=1.  1     2    2p − 3 where T (r) = max T (r, Fj ) with r ∈ / E and Ap = 3 1≤j≤p   √   2p + 1 − 2 2p    2. 政治大. 立. if p = 2 if p = 3, 4, 5 . if p ≥ 6. ‧ 國. 學. P roof . By Theorem 3.21, we have. ‧. p X. p. sit. y. Nat. X 1 1 T (r, F1 ) < N (r, ) − N (r, Fj ) + N (r, D) − N (r, ) + o(T (r))) Fj D j=1 j=2 where D is the Wronskian of F1 , · · · , Fp . Let p. er. io. p X. X 1 1 N (r) = N (r, Fj ) + N (r, D) − N (r, ) N (r, ) − Fj D j=1 j=1. n. and. al. Ch. p. engchi. i Un. v. p. X 1 N (r) = Np−1 (r, ) + Ap N (r, Fj ). F j j=1 j=1 ∗. X. It suffice to show that N (r) ≤ N ∗ (r). (4.11). We say that z is a b-point of f if f (z) − b = 0. Then we define the following functions:   m, z is a b-point of f with multiplicity m ≥ 1, b µf (z) =  0, z is not a b-point of f ,. 26.

(154)   1, z is a b-point of f with multiplicity m ≥ 1, µbf (z) =  0, z is not a b-point of f , and    m, z is a b-point of f with multiplicity m ≤ n − 1,   νfb (z) = n − 1, z is a b-point of f with multiplicity m > n − 1,     0, z is not a b-point of f . Let µ=. p X. µ0Fi. −. p X. µ∞ Fi. +. µ∞ D. µ0D. −. ∗. and µ =. νF0 i. + Ap. p X. µ∞ Fi .. j=1. j=1. j=2. j=1. p X. In order to prove (4.11), it is reduced to show that µ(z) ≤ µ∗ (z). Now, consider the following two cases.. 立. 政治大. In this case, z is also not a pole of D. This implies that = 0,. p X. µ∞ Fj. = 0, and. p X. µ∞ Fj = 0.. ‧. µ∞ D (z). 學. ‧ 國. Case 1. z is not a pole of Fj , 1 ≤ j ≤ p.. j=1. j=2. Nat. (p−1). (4.12). y. with multiplicity at least µ0Fj (z) − νF0 j (z), the number of p X 0 0 zeros of D with multiplicity is at least µFj (z) − νFj (z) . Hence we have. al. er. io. sit. Since z is a zero of Fi. iv p n C X µ0D (z) ≥ h e n µ0Fg(z) (z) . c h− iνF0 U. n. j=1. j. j. (4.13). j=1. By (4.12), (4.13) and the definitions of µ and µ∗ , we obtain that µ(z) ≤ µ∗ (z).. Case 2. z is a pole of some Fj . We consider two subcases: Subcase 1. z is not a pole of F1 . Without loss of generality, we may assume that z is a zero of Fi with multicity mi , i = 1, · · · , k, 1 ≤ k ≤ n − 2 and a pole of Fi. 27.

(155) with multicity mi , i = k+1, · · · , n. Since F1 +· · ·+Fn = 1, we have D = (−1)n+1 D1 , where D1 =. F10 .. .. ··· .. .. (n−1). ···. F1 Let q=. k X. (mi −. νF0 i (z)). n−1 X. −. i=1. 0 Fn−1 .. . (n−1). Fn−1. mi −. i=k+1. (n + k)(n − k − 1) 2. If q ≥ 0, then z is a zero of D with multiplicity at least q. Otherwise, z is a pole of D with multiplicity as least −q. Thus, we have. 政治大. µ0D (z) − µ∞ D (z) ≥ q. 立. This implies that. ‧ 國. νF0 i +. (n + k)(n − k − 1) ≥ µ(z) + µ∞ Fn 2. io. i=1. al. n. Since µ∞ Fn ≥ 1, we have. µ(z) ≤. Ch k X. i Un. 2. i=1. and µ (z) =. v. e(nn+gk)(n c h−i k − 1) − 1,. νF0 i +. ∗. y. Nat. ⇒. k X. sit. i=1. k n−1 X X (n + k)(n − k − 1) 0 0 ≥ µ∞ (z) − µ (z) + µ − µ∞ D D Fi Fi 2 i=1 i=k+1. er. −. µ∞ D (z). ‧. k X. 學. ⇒. k n−1 X X (n + k)(n − k − 1) 0 ≥ (mi − νFi (z)) − mi − 2 i=1 i=k+1. νF0 i +. µ0D (z). k X. νF0 i + An (n − k).. i=1. It is clearly that. (n + k)(n − k − 1) − 1 ≤ An (n − k). Thus, we have 2 µ(z) ≤ µ∗ (z).. Subcase 2. z is a pole of F1 . Without loss of generality, we may assume that z is a pole of F1 , Fk+1 , · · · , Fn with multicity mi respectively, and a pole of Fi with. 28.

(156) multicity mi , i = 2, · · · , k, 2 ≤ k ≤ n − 1. Since F1 + · · · + Fn = 1, we have D = D2 , where D2 =. F20 .. .. ··· .. .. Fn0 .. .. (n−1). ···. Fn(n−1). F2 Let q=. k X. (mi −. νF0 i (z)). −. i=2. n X. mi −. i=k+1. (n − k)(n + k − 1) 2. If q ≥ 0, then z is a zero of D with multiplicity at least q. Otherwise, z is a pole of D with multiplicity as least −q. Thus, we have µ0D (z) − µ∞ D (z) ≥ q This implies that. 政治大. 立. k X. νF0 i +. io. i=1. al. n. and. ∗. (n − k)(n + k − 1) ≥ µ(z). 2 k X. Ch. µ (z) =. y. Nat. Then we have. sit. i=2. n k X X (n − k)(n + k − 1) 0 0 − µ∞ ≥ µ∞ (z) − µ (z) + µ Fi D D Fi 2 i=1 i=k+1. er. νF0 i +. ‧ 國. k X. ‧. ⇒. 學. k n X X (n − k)(n + k − 1) 0 ∞ 0 µD (z) − µD (z) ≥ (mi − νFi (z)) − mi − 2 i=2 i=k+1. i Un. v. νF0 i + An (n − k + 1).. engchi. i=2. Since (n + k − 1)(n − k)/2 ≤ An (n − k + 1) for 1 ≤ k ≤ n − 1. Hence, µ(z) ≤ µ∗ (z). q. which completes the proof. Combine Lemma 4.6 and Lemma 4.7, we have. Lemma 4.8 [13] Let p be a positive integer and p ≥ 2. Let F1 , · · · , Fp be p linearly independent non-constant meromorphic functions. If p X. Fj (z) = 1,. j=1. 29.

(157) then we have p. p X. X 1 T (r) ≤ (p − 1) N (r, ) + Ap N (r, Fj ) + o(T (r)), F j j=1 j=1 where T (r) and Ap is defined as in Lemma 4.7.. P roof . Without loss of generality, we may assume that T (r) = max T (r, Fi ) = T (r, F1 ). Therefore, by Lemma 4.7, p. p X. X 1 Np−1 (r, ) + Ap N (r, Fj ) + o(T (r)). T (r) ≤ (p − 1) Fj j=1 j=1 Furthermore, by Lemma 4.6,. 立. T (r) ≤ (p − 1). 政治X 大 1 p. p. X. N (r,. j=1. Fj. N (r, Fj ) + o(T (r)).. ) + Ap. j=1. ‧ 國. 學. q. ‧. Now, we can use Lemma 4.6, Lemma 4.7 and Lemma 4.8 to prove the result of I. Lahiri and K.-W. Yu.. sit. y. Nat. n. al. er. io. Theorem 4.9 [13] Let p be a positive integer and p ≥ 2. Let f1 , · · · , fp be p. v. non-constant meromorphic functions and a1 , · · · , ap be p non-constant meromorphic functions such that. Ch. engchi. i Un. T (r, aj ) = o(T (r, fj )), 1 ≤ j ≤ p, as r → +∞ and r ∈ / E. If fj and aj , 1 ≤ j ≤ p, satisfy p X. k. aj (z)fj j (z) = 1. j=1. for some positive integers k1 , · · · , kp , then p X 1 1 ≥ k p − 1 + Ap j=1 j. where Ap is defined as in Lemma 4.7.. 30.

(158) k. P roof . Suppose that Fj = aj fj j , 1 ≤ j ≤ p are linearly independent. Let ! p X T (r) = max T (r, Fj ) and S(r) = o T (r, fj ) . As in Theorem 4.5, we have the 1≤j≤p. j=1. following inequality. N (r,. 1 k aj fj j. ) ≤ N (r,. 1 1 1 ) + N (r, kj ) aj kj fj. ≤ T (r,. 1 1 1 ) + T (r, kj ) aj kj fj. ≤ T (r, aj ) + ≤ and. 立. k. 1 k T (r, fj j ) + O(1) kj. 1 T (r) + S(r) kj. (4.14). 政治大 1. N (r, aj fj j ) ≤ N (r, aj ) +. k. ‧. ‧ 國. 學. N (r, fj j ) kj 1 k ≤ T (r, aj ) + T (r, fj j ) kj 1 ≤ T (r) + S(r) kj. (4.15). n. al. er. io. sit. y. Nat. By (4.14), (4.15) and Lemma 4.8, we obtain that ! p X 1 T (r) ≤ (p − 1 + Ap ) T (r) + S(r). k j=1 j p X 1 1 Thus, we have ≥ . k p − 1 + Ap j=1 j. Ch. engchi. i Un. v. k. In the case Fj = aj fj j , 1 ≤ j ≤ p are linearly dependent, we may find a maximal linearly independent subset {Fj1 , · · · , Fjq } of {Fj } such that b1 (z)Fj1 (z) + · · · + bq (z)Fjq (z) = 1 where bl (z) are constant multiply of aj (z) such that bl (z) = o(T (r, Fjl )). By the case above and Ap is increasing with p, we have q p X X 1 1 1 1 ≥ ≥ ≥ . k k q − 1 + A p − 1 + A j,l q p j=1 j l=1. q. Thus, we finish the proof.. 31.

(159) 5. The Distribution of Zeros and Poles of a Class of Meromorphic Functions In this section, we will prove our main result of this thesis. Namely, theorem 5.3. below. Now, we consider the following problem: Given meromorphic functions f1 , · · · , fp and positive integers k1 , · · · , kp . Set F (z) =. p X. k. Rj (z)fj j (z),. 政治大 where R (z), · · · , R (z) are rational functions. What is the value distribution of 立 j=1. 1. p. F (z)?. ‧ 國. 學. In fact, we will prove that, under some conditions, F (z) has infinitely many. identically.. ‧. zeros or poles or some of partial sums (one of which may be F ) are equal to zero. sit. y. Nat. io. al. n. following result.. er. The first result on this problem was proved by Toda. In fact, Toda proved the. Ch. engchi. i Un. v. Theorem 5.1 [21] Let f1 , · · · , fp be p (≥ 2) non-constant entire functions and let k. k1 , · · · , kp be p integers not less than one such that at least one of fiki /fj j (i 6= j) is transcendental and. p X 1 1 < . k p−1 j=1 j. Then, for rational functions Rj (z)(6= 0)(j = 1, · · · , p) F (z) ≡. p X. k. Rj (z)fj j (z). j=1. has infinitely many zeros or some of partial sums (one of which may be F ) are equal to zero identically.. 32.

(160) Later, K.-W. Yu and C.-C. Yang [25] considered the above problem for meromorphic functions and state the following result without proof. For completeness, we provide a proof.. Theorem 5.2 [25] Let f1 , · · · , fp be p (≥ 2) non-constant transcendental meromorphic functions and let k1 , · · · , kp be p integers not less than one and at least one f ki of ikj is transcendental such that fj p X 1 1 < . k 4(p − 1) j j=1. Then we have. 政X 治大 F (z) = R (z)f , p. 立. j. kj j. j=1. ‧ 國. 學. where Rj (z) (6= 0), 1 ≤ j ≤ p are rational functions, has infinitely many zeros or poles, or some of the partial sums of F are equal to zero identically.. ‧. P roof . Suppose that the assertion is false. This means that F (z) has finitely many. y. Nat. sit. zeros and poles and has no partial sum is equal to zero. In particular, F (z) 6= 0.. n. al. er. io. We may assume that F (z) has zeros {a1 , · · · , an } and poles {b1 , · · · , bm }, counting n m Y Y (z − bs )F (z)/ (z − at ) has no zeros and poles. Then we multicity. Therefore, s=1. can write F (z) as. Ch. t=1. engchi. i Un. v. F (z) = R(z)ef (z) , where R(z) =. m n Y Y (z − at )/ (z − bs ) is rational and f (z) is entire. Let t=1. s=1. gj (z) = −. f (z) , 1 ≤ j ≤ p. kj. It is obvious that gj (z) is entire, 1 ≤ j ≤ p. From F (z) =. p X j=1. 1=. p X. R−1 Rj (fj egj )kj. j=1. 33. k. Rj (z)fj j , we get.

(161) Since Rj R−1 is rational which implies that T (r, Ri R−1 ) = O(log r) as r → ∞, j = 1, · · · , p. p X 1 1 By assumption, < , some of fj egj must be rational. Otherwise, all k 4(p − 1) j=1 j gj fj e are transcendental meromorphic function, we have. T (r, Rj R−1 ) = o(T (r, fj egj )), for all 1 ≤ j ≤ p. p X 1 1 By Theorem 4.5, ≥ which is impossible. Actually, the number s of k 4(p − 1) j=1 j j such that fj egj is rational is less than p − 2. In fact, if s ≥ p − 1, it forces that all. of fi egi are rational. In this case. 立. 治政大 (f e ) f = i. ki i k fj j. gi ki. 學. ‧ 國. (fj egj )kj. is rational, which is a contradiction. Thus,. Suppose that f1 e , · · · , fs e are rational, so that. j=1. n. al. p X. Ch. p X. k. Rj fj j is not equal to zero identically.. j=s+1. e n g c hs i. Rj R−1 (fj egj )kj = 1 −. X. j=s+1. i Un. v. Rj R−1 (fj egj )kj. j=1. ≡ B(z), where B(z) is a nonzero rational function. We can write (5.1) as follows p X. Pj (fj egj )kj = 1. j=s+1. where Pj =. R−1 Rj (fj egj )kj is rational. er. io. and is not equal to 1 identically since Therefore,. s X. y. gs. sit. Nat g1. ‧. 1 ≤ s ≤ p − 2.. Rj R−1 is rational, s + 1 ≤ j ≤ p. B 34. (5.1).

(162) Since fj egj are transcendental and Pj (6= 0) are rational for all j = s + 1, · · · , p. It follows that T (r, Pj ) = o(T (r, fj egj )), j = s + 1, · · · , p. By Theorem 4.2 we have, p p X X 1 1 1 1 > ≥ > k k 4(p − s − 1) 4(p − 1) j=s+1 j j=1 j p X 1 1 < . Thus, we obtain the theorem. which contradicts to k 4(p − 1) j=1 j. q. Finally, we state and prove our main theorem as follows.. 政治大. Main Theorem Let f1 , · · · , fp be p (≥ 2) non-constant transcendental mero-. 立. p X 1 1 < . k p − 1 + Ap j=1 j. k. Rj (z)fj j ,. er. io. F (z) =. p X. sit. y. Nat. Then we have. ‧. ‧ 國. 學. morphic functions and let k1 , · · · , kp be p integers not less than one and at least one f ki of ikj is transcendental such that fj. n. a i v infinitely many zeros or where Rj (z) (6= 0), 1 ≤ j ≤ lp are rational functions, n has Ch poles, or some of the partial sums ofeF nare c h i toUzero identically. g equal j=1. P roof . Assume that the statement is false. This means that F (z) has only finitely many zeros and poles and has all partial sums of F are not identically zero. This implies that F (z) 6= 0. We may assume that F (z) has zeros {a1 , · · · , an } and poles m n Y Y {b1 , · · · , bm }, counting multiplicity. Therefore, (z − bs )F (z)/ (z − at ) has no s=1. t=1. zeros and poles. Then we can write F (z) as F (z) = R(z)ef (z) , where R(z) =. n m Y Y (z − at )/ (z − bs ) is rational and f (z) is entire. t=1. s=1. 35.

(163) Let gj (z) = −. f (z) , 1 ≤ j ≤ p. kj. It is obvious that gj (z) is entire, 1 ≤ j ≤ p. From F (z) =. p X. k. Rj (z)fj j , we get. j=1. 1=. p X. R−1 Rj (fj egj )kj. j=1. By assumption, Rj R−1 is rational which implies that T (r, Rj R−1 ) = O(log r) as r → ∞, j = 1, · · · , p.. 政治大. p X 1 1 < and Theorem 4.9, some of fj egj must be rational. k p − 1 + Ap j=1 j gj Otherwise, all fj e are transcendental, we have. By. 立. ‧ 國. 學. T (r, Rj R−1 ) = S(r, fj egj ), 1 ≤ j ≤ p.. ‧. p X 1 1 By Theorem 4.9, ≥ which is impossible. Also, the number s of k p − 1 + A j p j=1 j such that fj egj is rational is less than p − 2. If s ≥ p − 1, it force that all of fi egi. sit. y. Nat. are rational. In this case. n. Ch. is rational, which is a contradiction. Thus,. er. io. al. fini (fj egj )kj = nj (fj egj )nj fj. engchi. i Un. v. 1 ≤ s ≤ p − 2.. g1. gs. Suppose that f1 e , · · · , fs e are rational, so that. s X. R−1 Rj (fj egj )kj is rational. j=1 p. and is not equal to 1 identically since. X. k. Rj fj j is not equal to zero identically.. j=s+1. Therefore, p X. −1. gj kj. Rj R (fj e ). = 1−. j=s+1. s X j=1. ≡ B(z),. 36. Rj R−1 (fj egj )kj.

(164) where B(z) is a nonzero rational function. We have. p X. Pj (fj egj )kj = 1. j=s+1. with Pj =. Rj R−1 is rational, s + 1 ≤ j ≤ p. B. Since fj egj are transcendental and Pj are rational, for j = s + 1, · · · , p. It shows that T (r, Pj ) = S(r, fj egj ), j = s + 1, · · · , p. By Theorem 4.9, we have. 政治大. p X 1 1 ≥ . k (p − s) − 1 + A j p−s j=s+1. 學. n. al. engchi. 37. sit. i Un. where Ap is increasing in p which is a contradiction.. Ch. y. >. Nat. io. 1 p − s − 1 + Ap−s 1 > , p − 1 + Ap. ‧. p p X X 1 1 > k k j=1 j j=s+1 j. er. So. ‧ 國. 立. v. q.

(165) References [1] I. N. Baker, On a class of meromorphic functions, Proc. Amer. Math. Soc., 17 (1966), pp. 819–822. [2] C.-T. Chuang and C.-C. Yang, Fix-points and factorization of meromorphic functions, World Scientific Publishing Co. Inc., Teaneck, NJ, 1990. Translated from the Chinese. [3] M. L. Green, Some Picard theorems for holomorphic maps to algebraic varieties, Amer. J. Math., 97 (1975), pp. 43–75.. 政治大. [4] F. Gross, On the equation f n + g n = 1, Bull. Amer. Math. Soc., 72 (1966),. 立. pp. 86–88.. ‧ 國. 學. [5] F. Gross, On the functional equation f n + g n = hn , Amer. Math. Monthly, 73 (1966), pp. 1093–1096.. ‧. [6] F. Gross, Factorization of meromorphic functions, Mathematics Research. Nat. sit. y. Center, Naval Research Laboratory, Washington, D. C., 1972.. n. al. er. io. [7] G. G. Gundersen, Meromorphic solutions of f 6 + g 6 + h6 ≡ 1, Analysis (Munich), 18 (1998), pp. 285–290.. Ch. engchi. i Un. v. [8] G. G. Gundersen, Meromorphic solutions of f 5 + g 5 + h5 ≡ 1, Complex Variables Theory Appl., 43 (2001), pp. 293–298. The Chuang special issue. [9] W. Hayman, Warings Problem f¨ ur analytische Funktionen, Bayer. Akad. Wiss. Math.-Natur. Kl. Sitzungsber., (1984), pp. 1–13 (1985). [10] W. K. Hayman, Meromorphic functions, Oxford Mathematical Monographs, Clarendon Press, Oxford, 1964. [11] K. Ishizaki, A note on the functional equation f n +g n +hn = 1 and some complex differential equations, Comput. Methods Funct. Theory, 2 (2002), pp. 67– 85.. 38.

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(167) [23] C.-C. Yang and H.-X. Yi, Uniqueness theory of meromorphic functions, vol. 557 of Mathematics and its Applications, Kluwer Academic Publishers Group, Dordrecht, 2003. [24] L. Yang, Value distribution theory, Springer-Verlag, Berlin, 1993. Translated and revised from the 1982 Chinese original. [25] K.-W. Yu and C.-C. Yang, A note for waring’s type of equations for a ring of meromorphic functions, Indian Journal of Pure and Applied Mathematics, 33 (2002), pp. 1495–1502.. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. 40. i Un. v.

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