[Infinite Order of Growth of Solutions of Second Order Linear Differential Equations
aa r X i v : . [ m a t h . C V ] J a n Infinite Order of Growth of Solutions of Second OrderLinear Differential Equations
Naveen Mehra and V. P. Pande
Abstract.
Considering differential equation f ′′ + A ( z ) f ′ + B ( z ) f = 0, where A ( z ) and B ( z )are entire complex functions, our results revolve around proving all non-trivial solutions areof infinite order taking various restrictions on coefficients A ( z ) and B ( z ).
1. Introduction and statement of main results
We consider the second order homogeneous linear differential equation, f ′′ + A ( z ) f ′ + B ( z ) f = 0 (1)where A ( z ) and B ( z ) are entire functions. It is well known result that all solutions of (1)are entire functions(see [ ]). All solutions are of finite order if and only if A ( z ) and B ( z )are polynomial(see [ ]). Thus, if at least one of the coefficients are not polynomial thensolutions can be of infinite order. It is wide area of research to find conditions on A ( z )and B ( z ) such that solutions are of infinite order. There are many results concerning thisproblem. Following theorem is the collection of such basic results which provide all non-trivial solutions of infinite order. Theorem . Suppose A ( z ) and B ( z ) are non-constant entire functions, then all non-trivial solutions of (1) are of infinite order, if one of the following holds:(i) [ ] ρ ( A ) < ρ ( B ) (ii) [ ] A ( z ) is a polynomial and B ( z ) is transcendental(iii) [ ] ρ ( B ) < ρ ( A ) ≤ . Consider the equation w ′′ + P ( z ) w = 0 (2)where P ( z ) is a polynomial of degree n . There are several papers in which the authors haveconsidered A ( z ) to be a solution of equation (2) and B ( z ) to satisfy different conditions sothat all non-trivial solutions are of infinite order. The next theorem is the collection of allthese results. Mathematics Subject Classification.
Key words and phrases. entire function, order of growth, complex differential equation, Fabry gap andFe´ j er gap.The research work of the first author is supported by research fellowship from Council of Scientific andIndustrial Research (CSIR), New Delhi. Theorem . Suppose A ( z ) is a solution of (2) and B ( z ) is a transcendental entirefunction satisfying one of the conditions mentioned below. Then all non-trivial solutions of (2) are of infinite order.(i) [ ] ρ ( B ) < (ii) [ ] µ ( B ) < and ρ ( A ) = ρ ( B ) (iii) [ ] µ ( B ) < + n +1) and ρ ( A ) = ρ ( B ) (iv) [ ] B ( z ) has Fabry gap and ρ ( A ) = ρ ( B ) . Motivated by above results we have replaced the conditions on B ( z ) to be transcendentalentire function satisfying T ( r, B ) ∼ log M ( r, B ) in a set E of positive upper logarithmicdensity, where the notation T ( r, B ) ∼ log M ( r, B ) meanslim r →∞ T ( r, B )log M ( r, B ) = 1 . Theorem . Let A ( z ) be a non-trivial solution of (2) , where P ( z ) = a n z n + · · · + a , a n = 0 and B ( z ) be transcendental entire function satisfying T ( r, B ) ∼ log M ( r, B ) in a set E of positive upper logarithmic density, then all non-trivial solutions of (1) are of infiniteorder. Kwon in his paper [ ] considered ρ ( A ) > < ρ ( B ) < /
2. Kumar et. al. [ ] assumed that B ( z ) has Fabry gap.Recall that, for an entire function f ( z ) = ∞ X n =0 a λ n z λ n if ∞ X n =0 λ n diverges to ∞ we say that ithas Fabry gap. Theorem . Suppose that A ( z ) is an entire function of finite non-integral order with ρ ( A ) > , and that all the zeros of A ( z ) lies in the angular sector θ < argz < θ satisfying θ − θ < πp + 1 if p is odd, and θ − θ < (2 p − π p ( p + 1) if p is even, where p is the genus of A ( z ) . Let B ( z ) be an entire function satisfying theconditions mentioned below. Then all non-trivial solutions f of equation (1) are of infiniteorder.(i) [ ] 0 < ρ ( B ) < ,(ii) [ ] B ( z ) has Fabry gap. Motivated by above results we have replaced B ( z ) to be a transcendental entire functionhaving a multiply connected Fatou component. Theorem . Suppose that A ( z ) be an entire function of finite non-integral order with ρ ( A ) > , and that all the zeros of A ( z ) lies in the angular sector θ < argz < θ satisfying θ − θ < πp + 1 if p is odd, and θ − θ < (2 p − π p ( p + 1) NFINITE ORDER OF GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS3 if p is even, where p is the genus of A ( z ) and let B ( z ) be a transcendental entire functionwith a multiply connected Fatou component. Then all non-trivial solutions of equation (1) are of infinite order. In several papers like [ ], [ ], [ ], [ ] and [ ] authors have considered A ( z ) = h ( z ) e P ( z ) , where P ( z ) is a polynomial of degree n satisfying λ ( A ) < ρ ( A ) and B ( z ) withvarious conditions. Here we have exchanged the conditions of A ( z ) with B ( z ) and prove thefollowing result. Theorem . Let B ( z ) = h ( z ) e P ( z ) be a transcendental entire function satisfying λ ( B ) <ρ ( B ) = n , where P ( z ) is a polynomial of degree n and A ( z ) satisfies the properties mentionedbelow, then all non-trivial solutions of (1) are of infinite order.(a) A ( z ) has Fabry gap(b) A ( z ) satisfies T ( r, A ) ∼ log M ( r, A ) in a set E of positive upper logarithmic density(c) A ( z ) has multiply connected Fatou component. Following example shows that if we skip conditions of our theorem then we do get solutionof finite order.
Example . (a) f ′′ − e z f ′ + e z f = 0 has a solution e z − of finite order . T ( r, e z ) = rπ and log M ( r, e z ) = r ,thus T ( r, e z ) log M ( r, e z ) (i) A ( z ) = − e z is a solution of w ′′ − w = 0 and B ( z ) = e z does not satisfy the conditionof Theorem 3.(ii) B ( z ) = e z satisfy λ ( B ) < ρ ( B ) but A ( z ) does not satisfy the condition of Theorem6(b)(b) f ′′ + ( e z − f ′ + e z f = 0 has a solution e z of finite order . B ( z ) = e z satisfy λ ( B ) < ρ ( B ) but A ( z ) does not satisfies the condition of Theorem 6(a)i.e. it does not has Fabry gap.(c) f ′′ + Cz ∞ Y n =1 (cid:18) za n (cid:19) f ′ − Cz ∞ Y n =1 (cid:18) za n (cid:19) f = 0 has a solution z of order . A ( z ) = Cz ∞ Y n =1 (cid:18) za n (cid:19) where the a n satisfy < a < a < ... and grow so rapidly that a n +1 < A ( a n ) < a n +1 ,is constructed by Baker [ ] . It has multiply connected Fatou component but λ ( B ) = ρ ( B ) i.e. it does not satisfy the condition of Theorem 6(c).(d) f ′′ + Cz ∞ Y n =1 (cid:18) za n (cid:19) f ′ − Cz ∞ Y n =1 (cid:18) za n (cid:19) f = 0 has a solution z of order . B ( z ) has multiply connected Fatou component but ρ ( A ) ≤ i.e. it does not satisfy thecondition of Theorem 5.
2. Preliminary Lemma
Next lemma is due to Gundersen[ ] and it gives an estimation of logarithmic derivatives. Lemma . [ ] Let f be a transcendental entire function of finite order ρ , let Γ = { ( k , j ) , ( k , j ) ... ( k m , j m ) } denote finite set of distinct pairs of integers that satisfies k i >j i ≥ , for i = 1 , , ...m , and let ǫ > be a given constant. Then the following threestatements holds: NAVEEN MEHRA AND V. P. PANDE (i) there exists a set E ⊂ [0 , π ) that has linear measure zero, such that if ψ ∈ [0 , π ) − E , then there is a constant R = R ( ψ ) > so that for all z satisfying argz = ψ and | z | ≥ R and for all ( k, j ) ∈ Γ , we have (cid:12)(cid:12)(cid:12)(cid:12) f ( k ) ( z ) f ( j ) ( z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ | z | ( k − j )( ρ − ǫ )) (3) (ii) there exists a set E ⊂ (1 , ∞ ) that has finite logarithmic measure, such that for allz satisfying | z | does not belongs to E ∪ [0 , and for all ( k, j ) ∈ Γ , the inequality(2) holds.(iii) there exists a set E ⊂ [0 , ∞ ) that has finite linear measure, such that for all zsatisfying | z | does not belongs to E and for all ( k, j ) ∈ Γ , we have (cid:12)(cid:12)(cid:12)(cid:12) f ( k ) ( z ) f ( j ) ( z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ | z | ( k − j )( ρ + ǫ ) . (4) Lemma . [ ] Let f ( z ) be a meromorphic function of finite order ρ . Given ξ > and δ , < δ < / , there is a constant K ( ρ, ξ ) such that for all r in a set F of lower logarithmicdensity greater than − ξ and for every interval J of length δr Z J (cid:12)(cid:12)(cid:12)(cid:12) f ′ ( re ιθ ) f ( re ιθ ) (cid:12)(cid:12)(cid:12)(cid:12) dθ < K ( ρ, ξ )( δ log 1 δ ) T ( r, f ) . Lemma . [ ] Let f ( z ) be an entire function satisfying T ( r, f ) ∼ log M ( r, f ) in a set E of positive upper logarithmic density. For given < c < and r ∈ E , the set I r = { θ ∈ [0 , π ) : log | f ( re ιθ ) | ≤ (1 − c ) log M ( r, f ) } has linear measure zero. Next lemma is extracted from proof of the theorem in [ ] Lemma . Let f ( z ) be an entire function of finite order satisfying T ( r, f ) ∼ log M ( r, f ) in a set log densE > and satisfies condition of Lemma 4 in a set F such that log dens ( F ) > − ξ , then | f ( z ) | > (1 − c ) log M ( r, f ) , where < c < / , r ∈ E ∩ F and θ ∈ [0 , π ] \ I r where I r = { θ ∈ [0 , π ) : log | f ( re ιθ ) | ≤ (1 − c ) log M ( r, f ) } . Let M ( r, f ) = max. {| f ( z ) | : | z | = r } and L ( r, f ) = min. {| f ( z ) | : | z | = r } , then following lemma gives relation between M ( r, f )and L ( r, f ) when f has at most finite number of poles. Lemma . [ ] Let f be a transcendental meromorphic function with at most finitely manypoles. If J ( f ) has only bounded components, then for any complex number a , there exist aconstant < d < and two sequences { r n } and { R n } of positive numbers with r n → ∞ and R n /r n → ∞ ( n → ∞ ) such that M ( r, f ) d ≤ L ( r, f ) , r ∈ G where G = ∪ ∞ n =1 { r : r n < r < R n } . NFINITE ORDER OF GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS5
3. Proof of the theorems
Before the proof of each of our result, we give some lemmas which will be useful in prov-ing our results.
Proof of Theorem 3.
Let α < β be such that β − α < π , r > S denote the closure of S . Denote S ( α, β ) = { z : α < argz < β } , S ( α, β, r ) = { z : α < argz < β } ∩ { z : | z | < r } . Let f be an entire function of order ρ ( f ) ∈ (0 , ∞ ). For simplicity, set ρ = ρ ( f ) and S = S ( α, β ). Then f is said to blow up exponentially in S if for any θ ∈ ( α, β )lim r →∞ log log | f ( re ιθ ) | log r = ρ and decays to zero exponentially in S if for any θ ∈ ( α, β )lim r →∞ log log | f ( re ιθ ) | − log r = ρ Lemma . [ ] Let f be a non-trivial solution of f ′′ + P ( z ) f = 0 , where P ( z ) = a n z n + · · · + a , a n = 0 . Set θ j = jπ − arg ( a n ) n +2 and S j = ( θ j , θ j +1 ) , where j = 0 , , , · · · , n + 1 and θ n +2 = θ + 2 π . Then f has the following properties:(i) In each sector S j , f either blows up or decays to zero exponentially.(ii) If, for some j , f decays to zero in S j , then it must blow up in S j − and S j +1 . However,it is possible for f to blow up in many adjacent sectors.(iii) If f decays to zero in S j , then f has at most finitely many zeros in any closed sub-sectorwithin S j − ∪ S j ∪ S j +1 .(iv) If f blows up in S j − and S j , then for each ǫ > , f has finitely many zeros in eachsector S ( θ j − ǫ, θ j + ǫ ) , and furthermore, as r → ∞ , n ( S ( θ j − ǫ, θ j + ǫ, r ) , , f ) = (1 + o (1)) 2 p | a n | π ( n + 2) r n +22 , where n ( S ( θ j − ǫ, θ j + ǫ, r ) , , f ) is the number of zeros of f in the region S ( θ j − ǫ, θ j + ǫ, r ) . Lemma . [ ] Let A ( z ) and B ( z ) be two entire functions such that for real constants α > , β > , θ , θ , where α > , β > and θ < θ , we have | A ( z ) | ≥ exp { (1 + o (1) α | z | β ) } , (5) | B ( z ) | ≤ exp { (1 + o (1) | z | β ) } (6) as z → ∞ in S ( θ , θ ) = { z : θ ≤ argz ≤ θ } . Let ǫ > be a given small constant, and let S ( θ + ǫ, θ − ǫ ) = { z : θ + ǫ ≤ argz ≤ θ − ǫ } . If f is a non-trivial solution of (1) with ρ ( f ) < ∞ , then the following conclusions hold:(i) There exists a constant b ( = 0) such that f ( z ) → b as z → ∞ in S ( θ + ǫ, θ − ǫ ) .Furthermore, | f ( z ) − b | ≤ exp {− (1 + o (1)) α | z | β } (7) as z → ∞ in S ( θ + ǫ, θ − ǫ ) . NAVEEN MEHRA AND V. P. PANDE (ii) For each integer k > , | f ( k ) ( z ) | ≤ exp {− (1 + o (1)) α | z | β } (8) as z → ∞ in S ( θ + ǫ, θ − ǫ ) . Proof. If ρ ( A ) < ρ ( B ), then it is already proved in [ ] that all non-trivial solutions of(1) are of infinite order. Now, suppose that ρ ( A ) > ρ ( B ).Assume ρ ( f ) < ∞ . Set θ j = jπ − arg ( a n ) n +2 and S j = ( θ j , θ j +1 ), where j = 0 , , , · · · , n + 1 and θ n +2 = θ + 2 π . Let θ ∈ S j . Since A ( re ιθ ) is a solution of equation (2), thus A ( z ) eitherblows up or decays to zero exponentially in each sector S j . Case 1:
Suppose A ( z ) blows up exponentially in each sector S j . Then, we havelim r →∞ log log | A ( re ιθ ) | log r = n + 22 . Then for any given constant ǫ ∈ (0 , πρ ( A ) ) and β ∈ (0 , ρ ( A ) − ρ ( B )3 ), we have | A ( z ) | ≥ exp {| z | n +22 − β }≥ exp { | z | n +22 − β + 12 | z | n +22 − β }≥ exp { | z | n +22 − β + 12 | z | n +22 − β }≥ exp { (1 + | z | − β ) 12 | z | n +22 − β }≥ exp { (1 + o (1)) α | z | n +22 − β } where α = . | B ( z ) | ≤ exp {| z | ρ ( B )+ β } ≤ exp {| z | ρ ( A ) − β } ≤ exp { o (1) | z | n +22 − β } as z → ∞ in S j ( ǫ ) = { z : θ j + ǫ < arg.z < θ j − ǫ } , j = 0 , , · · · , n + 1. Combining aboveinequalities for A ( z ), B ( z ) and Lemma 7, exist corresponding constants b j = 0 such that | f ( z ) − b j | ≤ exp {− (1 + o (1)) α | z | n +22 − β } as z → ∞ in S j (2 ǫ ), j = 0 , , · · · , n + 1. Therefore, f is bounded in the whole complex planeby the Phragmen-Lindelof principle. So f is a nonzero constant in the whole complex planeby Liouville’s theorem. But f cannot be non-constant, which gives rise to a contradiction. Case 2:
Suppose A ( z ) decays to zero in at least one sector S j . Then, we havelim r →∞ log log | A ( re ιθ ) | − log r = n + 22 . Then we get | A ( re ιθ ) | < exp ( − r n +22 − ξ )where r → ∞ and ξ is a positive constant. Since ρ ( f ) < ∞ by Lemma 1, we have (cid:12)(cid:12)(cid:12)(cid:12) f ( k ) ( z ) f ( z ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ | z | kρ ( f ) NFINITE ORDER OF GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS7 for | z | ≥ R = R ( θ ) > θ ∈ [0 , π ) /G , where G is a set with linear measure 0.Since we have T ( r, B ) ∼ log M ( r, B )in a set E having positive upper logarithmic density, by using Lemma 4, for 0 < c < / M ( r, B ) − c < | B ( z ) | (9)where r ∈ E ∩ F and θ ∈ [0 , π ] \ I r where I r and F as defined in Lemma 4.From equation 1 we get | B ( z ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12) f ′′ ( re ιθ ) f ( re ιθ ) (cid:12)(cid:12)(cid:12)(cid:12) + | A ( z ) | (cid:12)(cid:12)(cid:12)(cid:12) f ′ ( re ιθ ) f ( re ιθ ) (cid:12)(cid:12)(cid:12)(cid:12) M ( r, B ) − c < | B ( z ) | ≤ (1 + o (1)) r ρ ( f ) M ( r, B ) < (1 + o (1)) r ρ ( f ) for r > R ( θ ); r ∈ E ∩ F ∩ G and θ ∈ S j \ I r .But M ( r, B ) < (1 + o (1)) r ρ ( f ) is not possible for transcendental entire function B ( z ).Hence ρ ( f ) is infinite. (cid:3) Proof of Theorem 5.
Following lemma provides the information of the minimum modulus of entire function ofnon-integral order having zeros in definite sectors.
Lemma . [ ] Let f ( z ) be an entire function of finite non-integral order ρ and of genus p > . Suppose that for any given ǫ > , all the zeros of f ( z ) have their arguments in thefollowing subset of real numbers: S ( p, ǫ ) = { θ : | θ | ≤ π p + 1) − ǫ } if p is odd, and S ( p, ǫ ) = { θ : π p + ǫ ≤ | θ | ≤ π p + 1) − ǫ } if p is even. Then for any c > , there exists a real number R > such that | f ( − r ) | ≤ exp( − cr p ) for all r ≥ R . Proof.
Let us suppose that solution f of (1) is of finite order. Using Lemma 1(b), wehave (cid:12)(cid:12)(cid:12)(cid:12) f ( k ) ( re ιθ ) f ( re ιθ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ r kρ ( f ) (10)for r E ∪ [0 , E is a set with finite logarithmic measure and r > R ( θ ). Let usrotate the axes of the complex plane, assume that all the zeros of A ( z ) have their argumentsin the set S ( p, ǫ ) defined in Lemma 8 for some ǫ >
0. Hence by Lemma 8, there exists apositive real number R such that for all r > R , we havemin | z | = r | A ( z ) | ≤ | A ( − r ) | ≤ exp( − cr p ) < . (11)By Lemma 5, we have M ( r, B ) d ≤ | B ( re ιθ ) | (12) NAVEEN MEHRA AND V. P. PANDE for 0 < d < r ∈ G = ∪ ∞ n =1 { r : r n < r < R n } . From equation (1) we get, | B ( z ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12) f ′′ ( re ιθ ) f ( re ιθ ) (cid:12)(cid:12)(cid:12)(cid:12) + | A ( z ) | (cid:12)(cid:12)(cid:12)(cid:12) f ′ ( re ιθ ) f ( re ιθ ) (cid:12)(cid:12)(cid:12)(cid:12) Using (10), (11) and (12) for r > max { R ( θ ) , R } such that r ∈ G \ E ∪ [0 ,
1] and θ ∈ { θ :min | z | = r | A ( z ) | = | A ( z ) |} , we have M ( r, B ) d < | B ( z ) | ≤ (1 + o (1)) r ρ ( f ) M ( r, B ) < (1 + o (1)) r ρ ( f ) which is a contradiction for a transcendental entire function. (cid:3) Proof of Theorem 6.
Combining Theorem 2 from [ ] and Lemma 2.2 from [ ], we get Lemma . Let g ( z ) = ∞ P n =0 a λ n z λ n be an entire function of finite order with Fabry gap,and let u ( z ) be an entire function with ρ ( u ) ∈ (0 , ∞ ) . Then for any given ǫ ∈ (0 , ς ) , where ς = min (1 , ρ ( u )) , there exists a set K ⊂ (1 , ∞ ) satisfying log denseK ≥ ξ , where ξ ∈ (0 , is a constant such that for all | z | = r ∈ K , log M ( r, u ) > r ρ ( u ) − ǫ , log m ( r, g ) > (1 − ξ ) log M ( r, g ) , where M ( r, u ) = max {| u ( z ) | : | z | = r } , m ( r, g ) = min {| g ( z ) | : | z | = r } and M ( r, g ) = max {| g ( z ) | : | z | = r } . Lemma . [ ] Let f ( z ) be a non-constant entire function. Then there exists a realnumber R such that for all r ≥ R there exists z r with | z r | = r satisfying (cid:12)(cid:12)(cid:12)(cid:12) f ( z r ) f ′ ( z r ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ r. (13) Lemma . [ ] Let A ( z ) = h ( z ) e P ( z ) be an entire function with λ ( A ) < ρ ( A ) = n , where P ( z ) is a polynomial of degree n . Then for every ǫ > there exists E ⊂ [0 , π ) of linearmeasure zero such that(i) for θ ⊂ E + \ E there exists R > such that | A ( re ιθ ) | ≥ exp((1 − ǫ ) δ ( P, θ ) r n ) (14) for r > R .(ii) for θ ⊂ E − \ E there exists R > such that | A ( re ιθ ) | ≤ exp((1 − ǫ ) δ ( P, θ ) r n ) (15) for r > R . Proof.
Suppose f is a solution of finite order of equation (1). Then by Lemma 1 we get (cid:12)(cid:12)(cid:12)(cid:12) f ′′ ( re ιθ ) f ′ ( re ιθ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ r ρ ( f ) (16)where θ ∈ [0 , π ) \ E , where E is a set of linear measure 0 for r ≥ r ( θ ) > (cid:12)(cid:12)(cid:12)(cid:12) f ( z r ) f ′ ( z r ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ r (17) NFINITE ORDER OF GROWTH OF SOLUTIONS OF SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS9 for r ≥ r such that | z r | = r .Since λ ( B ) < ρ ( B ), choosing θ ∈ [0 , π ) \ E , where E is a set of linear measure 0 such that δ ( P, θ ) < | B ( re ιθ ) | ≤ exp((1 − ǫ ) δ ( P, θ ) r n ) (18)for r > r . From equation (1), we get | A ( re ιθ ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12) f ′′ ( re ιθ ) f ′ ( re ιθ ) (cid:12)(cid:12)(cid:12)(cid:12) + | B ( re ιθ ) | (cid:12)(cid:12)(cid:12)(cid:12) f ( re ιθ ) f ′ ( re ιθ ) (cid:12)(cid:12)(cid:12)(cid:12) (19)(a) Since A ( z ) is a transcendental entire function with Fabry gap, using Lemma 9 there exista set H ⊂ (1 , ∞ ) satisfying log densH ≥ ξ , where ξ ∈ (0 ,
1) for r ∈ H such that M ( r, A ) (1 − ξ ) < | A ( z ) | . (20)Using (16), (17), (18), (19) and (20) we will get M ( r, A ) (1 − ξ ) ≤ r ρ ( f ) + r exp((1 − ǫ ) δ ( P, θ ) r n ) r ∈ H , r > max. { r , r , r } and θ ∈ [0 , π ) \ ( E ∪ E ). M ( r, A ) (1 − ξ ) < r ρ ( f ) (1 + o (1)) , which is a contradiction for very large r .(b) Using (9), (16), (17), (18) and (19) we will get M ( r, A ) (1 − c ) < r ρ ( f ) (1 + o (1)) , for r ∈ E ∩ F , r > max. { r , r , r } and θ ∈ [0 , π ) \ ( E ∪ E ∪ I r ).which is a contradiction for very large r .(c) Using (12), (16), (17), (18) and (19) for , we get for 0 < d < r ∈ G = ∪ ∞ n =1 { r : r n
Email address : [email protected] V. P. Pande, Department of Mathematics, Kumaun University, S.S.J Campus, Almora-263601, Uttarakhand, India
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