On the Weak Type of Meromorphic Functions
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1 International Mathematical Forum, 4, 2009, no 12, On the Weak Type of Meromorphic Functions Sanjib Kumar Datta Department of Mathematics University of North Bengal PIN: , West Bengal, India sanjib kr Arindam Jha Tarangapur NK High School Post: Tarangapur, Uttar Dinajpur PIN: , West Bengal, India d ajha@yahoocoin Abstract In this paper we introduce the definition of weak type of meromorphic functions and establish its integral representation We also investigate some growth properties related to the weak type of meromorphic functions Mathematics Subject Classification: 30D35, 30D30 Keywords: Order, lower order, type, weak type, integral representation, growth 1 Introduction, Definitions and Notations Let f be a meromorphic function of finite positive orde defined in the open complex plane C The type σ f of f is defined as follows : σ f = When f is entire one can easily verify that σ f = log M (r, f), 0 <ρ f < In the paper we introduce the following two definitions :
2 570 S K Datta and A Jha Definition 1 The weak type τ f of a meromorphic function f of finite positive lower orde is defined by For entire f, τ f = log M (r, f) τ f =, 0 <λ f < Definition 2 A meromorphic function f of finite non zero lower orde is said to be of weak type τ f if the integral )] k+1 0 > 0) converges for k>τ f and diverges for k<τ f In this paper we establish the equivalence of Definition 1 and Definition 2 We also study some growth properties related to the weak type of meromorphic functions We do not explain the standard definitions and notations of the theory of entire and meromorphic functions as those are available in [4] and [3] 2 Lemmas In this section we present some lemmas which will be needed in the sequel Lemma 1 ([1]) If f is meromorphic and g is entire then for all sufficiently large values of r, T (f g) {1+o (1)} Lemma 2 Let the integral Then Proof Since the integral T (r, g) T (M (r, g),f) log M (r, g) )] k+1 dr ( > 0) converges where 0 <λ f < exp {} =0 [exp (r λ k )] k+1 dr ( > 0) converges, then exp {} [exp ( )] k+1 dr < ε, if >R(ε)
3 Weak type of meromorphic functions 571 Therefore, exp{(r o) λ f }+r 0 Since exp {} increases with r, so exp {} dr < ε [exp (r λ k+1 exp{(r o) λ f }+r0 ie, for all large values of r, exp {} [exp (r dr exp {T (,f)} λ k+1 [ ( exp 0 [ )] k+1 ( exp o )] so that exp{(r o) λ f }+r 0 exp {} [exp (r dr exp {T (,f)} λ k+1 [ ( )] f k, )] exp 0 exp {T (,f)} [ ( )] k <εif >R(ε) exp 0 This proves the lemma exp {} ie, =0 [exp (r λ k 3 Theorems In this section we present the main results of the paper Theorem 1 Let f be a meromorphic function of lower orde and of weak type τ f Also let 0 < λ f < Then Definition 1 and Definition 2 are equivalent Proof Case 1 τ f = Definition 1 Definition 2
4 572 S K Datta and A Jha As τ f =, from Definition 1 we obtain for arbitrary G and for all sufficiently large values of r, > G ( ) ie, exp {} > [ exp ( )] G (1) If possible let the integral )] G+1 0 > 0) be converge Then by Lemma 2, exp {} [exp (r =0 λ G So for a sequence of values of r tending to infinity exp {} < [ exp ( )] G (2) Therefore from (1) and (2) we arrive at a contradiction Hence )] G+1 0 > 0) diverges whenever G is finite, which is the Definition 2 Definition 2 Definition 1 Let G be any positive number Since τ f =, from Definition 2, the divergence of the integral )] G+1 0 > 0) gives for arbitrary positive ε and for all sufficiently large values of r that exp {} > [ exp ( )] G ε which implies that ie, > (G ε), G ε Since G is arbitrary, it follows that Thus Definition 1 follows = Case 2 0 τ f <
5 Weak type of meromorphic functions 573 Subcase (a) 0 <τ f < Let f be of type τ f, where 0 < τ f < Then according to the Definition 1, for arbitrary positive ε and for a sequence of values of r tending to infinity we obtain that < (τ f + ε) )] τf +ε ie, exp {} < [ exp ( [ ( )] exp {} exp r λ τf +ε f ie, < [exp (r λ k [exp (r λ k exp {} 1 ie, < [exp (r λ k [exp (r λ k (τ f +ε) (3) exp{t (r,f )} Therefore )] k+1 0 > 0) converges for k>τ f Again by Definition 1, we obtain for all large values of r that So for k<τ f, we get from (4) that > (τ f ε) ie, exp {} > [ exp ( )] τf ε (4) exp {} [exp ( )] k > 1 [exp ( )] k (τ f ε) exp{t (r,f )} Therefore )] k+1 0 > 0) diverges for k<τ f Hence )] k+1 0 > 0) converges for k>τ f and diverges for k<τ f Subcase (b) τ f =0 When f is of weak type τ f = 0, Definition 1 gives for a sequence of values of r tending to infinity that <ε exp{t (r,f )} Then as before we obtain that )] k+1 0 > 0) converges for k>0 and diverges for k<0 Thus combining Subcase (a) and Subcase (b), Definition 2 follows
6 574 S K Datta and A Jha Definition 2 Definition 1 Since f is of weak type τ f, by Definition 2, for arbitrary positive ε the integral )] τ f +ε+1 0 > 0) converges Then by Lemma 2, exp {} [exp (r λ τ f +ε =0 So we obtain for a sequence of values of r tending to infinity that exp {} [exp (r λ τ f +ε < ε ie, exp {} < ε [ exp ( )] r λ τf +ε f ie, < log ε +(τ f + ε) ie, τ f + ε Since ε (> 0) is arbitrary, it follows from above that τ f (5) On the otherhand the divergence of the integral implies that exp {} [exp (r λ τ f ε =, ie, for all sufficiently large values of r, exp {} > [ exp ( ie, > (τ f ε) )] τf ε )] τ f ε+1 dr ( > 0) As ε (> 0) is arbitrary, it follows that τ f (6) So from (5) and (6) we obtain that = τ f This proves the theorem In the following theorem we obtain a relationship between τ f and σ f
7 Weak type of meromorphic functions 575 Theorem 2 Let f be a meromorphic function such that λ f and ρ f are both finite Also let f be of regular growth ie, λ f = ρ f Then τ f = σ f Proof Since f is of regular growth, we get that σ f = = = τ f (7) On the otherhand Definition 7 (cf[2]) implies that the integral exp{t (r,f )} [exp( )] k+1 0 > 0) converges for k>σ f and diverges for k<σ f From Definition 2 it follows that the integral )] k+1 0 > 0) is convergent for k>τ f and diverges for k<τ f Also all the quantities in the expression [ exp {} [exp ( )] k+1 exp {} [exp (r ρ k+1 are of non negative type So [ ] exp {} exp {} [exp (r λ k+1 [exp (r ρ k+1 0 > 0) 0 ie, exp {} [exp ( )] k+1 dr Hence from (7) and (8) we obtain that Thus the theorem is established ] exp {} [exp (r ρ k+1 dr for > 0 ie, τ f σ f (8) τ f = σ f Theorem 3 If f be a meromorphic function of regular growth ie, λ f = ρ f, then the following quantities are all equivalent (i) σ f = (iii), (ii) τ f =, and (iv)
8 576 S K Datta and A Jha Proof (i) (ii) In view of Theorem 2, as f is of regular growth σ f = = τ f = (ii) (iii) Since f is of regular growth ie, λ f = ρ f we get that τ f = = (iii) (iv) In view of Theorem 2 and the condition λ f = ρ f it follows that = = = τ f = σ f = (iv) (i) As f is of regular growth ie, λ f = ρ f, we obtain that Thus the theorem follows = = σ f Theorem 4 Let f be a meromorphic function and g be an entire function satisfying (i)0<λ g ρ g <, (ii) τ g > 0, (iii) σ g <, (iv)0<λ f g < and (v)0<τ f g < Then T (r,f g) (a) T (r,g) τ f g σ g if λ f g = ρ g and T (r,f g) (b) T (r,g) τ f g τ g if λ f g = λ g Proof From the definitions of τ f g,τ g and σ g we get for arbitrary positive ε and for all sufficiently large values of r, (τ f g ε) g (9) T (r, g) (τ g ε) r λg (10)
9 Weak type of meromorphic functions 577 T (r, g) (σ g + ε) r ρg (11) Also for a sequence of values of r tending to infinity, (τ f g + ε) g (12) Hence from (9) and (11) we obtain for all sufficiently large values of r that T (r, g) (τ f g ε) g (σ g + ε) r ρg Since λ f g = ρ g and ε (> 0) is arbitrary it follows from above that τ f g T (r, g) σ g This proves Theorem 4 (a) Also for a sequence of values of r tending to infinity we get from (10) and (12) that T (r, g) (τ f g + ε) g (τ g ε) r λg Since ε (> 0) is arbitrary and λ f g = λ g, we obtain from above that This proves Theorem 4 (b) τ f g T (r, g) τ g Corollary 1 If in addition g be of regular growth ie, λ g = ρ g then = τ f g T (r, g) σ g = τ f g τ g Proof Since g be of regular growth ie, λ g = ρ g, Corollary 1 follows from Theorem 2 and Theorem 4 Theorem 5 Let f be a meromorphic function and g be an entire function with (i)0<λ g = ρ f g <, (ii) σ fog < and (iii) τ g > 0 Then σ fog T (r, g) τ g
10 578 S K Datta and A Jha Proof From the definition of σ fog we get for all sufficiently large values of r that (σ fog + ε) g (13) Now from (10) and (13) it follows for all sufficiently large values of r that T (r, g) (σ fog + ε) g (τ g ε) r λg Since ε (> 0) is arbitrary and λ g = ρ f g, we obtain from above that σ fog T (r, g) τ g Thus the theorem is established Theorem 6 If f be meromorphic and g be entire such that (i)0<ρ g <, (ii) σ g <, (iii) ρ g = λ f, (iv)0<λ f ρ f < and (iv) τ f > 0, then log ρ f σ g τ f Proof In view of Lemma 1 and T (r, g) log + M (r, g), we get for all sufficiently large values of r that log o (1) + log T (M (r, g),f) ie, log o (1) + (ρ f + ε) log M (r, g) (14) Now from the definition of σ g we obtain for all sufficiently large values of r that log M (r, g) (σ g + ε) r ρg (15) Also from the definition of τ f we have for all sufficiently large values of r, (τ f ε) (16) So from (14), (15) and (16) it follows for all sufficiently large values of r that log o (1) + (ρ f + ε)(σ g + ε) r ρg (τ f ε) Since ε (> 0) is arbitrary and ρ g = λ f we get from above that log ρ f σ g τ f This proves the theorem
11 Weak type of meromorphic functions 579 Theorem 7 Let f be meromorphic and g be entire satisfying (i)0<λ g <, (ii) ρ f <, (iii) σ g < and (iv) τ g > 0 Also let g be of regular growth ie,λ g = ρ g Then log ρ f T (r, g) Proof In view of (10), (14) and (15) we obtain for all sufficiently large values of r that log T (r, g) o (1) + (ρ f + ε)(σ g + ε) r ρg (τ g ε) r λg Since ε (> 0) is arbitrary and λ g = ρ g,theorem 7 follows from Theorem 2 References [1] Bergweiler, W : On the Nevanlinna Characteristic of a composite function, Complex Variables 10 (1988), pp [2] Datta, S K : A note on the order and type of a meromorphic function, J Indian Acad Math, Vol 27, No 2 (2005), pp [3] Hayman, WK : Meromorphic functions, The Clarendon Press, Oxford, 1964 [4] Valiron, G : Lectures on the general theory of integral functions, Chelsea Publishing Company (1949) Received: July, 2008
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