Journal of Inequalities in Pure and Applied Mathematics AN INEQUALITY ASSOCIATED WITH SOME ENTIRE FUNCTIONS SHAVKAT A. ALIMOV AND ONUR ALP ILHAN National University of Uzbekistan Department of Mathematical Physics Tashkent, Uzbekistan EMail: shavkat_alimov@hotmail.com volume 5, issue 3, article 67, 2004. Received 23 March, 2004; accepted 25 May, 2004. Communicated by: N.K. Govil EMail: onuralp@sarkor.uz Abstract Contents JJ J II I Home Page Go Back Close c 2000 Victoria University ISSN (electronic): 1443-5756 063-04 Quit The object of our paper is to determine the order of growth to infinity of some family of entire functions. For an arbitrary α > 0 we introduce the following function ∞ X zk Φ(z, α) = , α (k!) k=0 (1) α > 0, z ∈ C. Note that Φ(z, 1) = ez . It is easy to show that if α > 0 then the function Φ(z, α) is defined by series (1) for all z in the complex plane C. An Inequality Associated with Some Entire Functions Shavkat A. Alimov and Onur Alp Ilhan Proposition 1. The radius of convergence of the series (1) is equal to infinity. Proof. According to the Cauchy formula (see, e.g., [2, 2.6]) the radius of convergence of the series ∞ X cn z n n=0 is equal to R= lim 1 p n n→∞ −α In our case cn = (n!) following form (2) n! = √ 2πn |cn | . . We may use the Stirling formula (see [2, 12.33]) in the Title Page Contents JJ J II I Go Back Close Quit Page 2 of 15 n n e 1+ θn 11n , 0 < θn < 1, n = 1, 2, . . . J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au As a result we get 1 p = (n!)α/n n |cn | α/n n n √ θn = 2πn 1+ e 11n α/n n α θn α/2n α(ln n)/2n = (2π) e 1+ e 11n n α = (1 + εn ) → ∞, n → ∞, e where εn = o(1), n → ∞. An Inequality Associated with Some Entire Functions Shavkat A. Alimov and Onur Alp Ilhan Corollary 2. The function Φ(z, α), α > 0, is entire function of z. The function Φ(z, α) with α = 1q arises in estimates of the solutions of some Volterra type integral equations with kernel from Lp , where p1 + 1q = 1. We mention also the equation with convolution on the circle which these functions satisfy. For two arbitrary 2π-periodical functions f (θ) and g(θ) introduce their convolution Z 2π 1 (f ∗ g)(θ) = f (θ − ϕ)g(ϕ)dϕ. 2π 0 If we denote (3) fα (θ) = Φ(eiθ , α), then it is easy to check that this function satisfies the following equation (4) (fα ∗ fβ )(θ) = fα+β (θ), f1 (θ) = exp eiθ . Title Page Contents JJ J II I Go Back Close Quit Page 3 of 15 J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au It easy to show that every solution of equation (4) has the form (3). It is well known that for Φ(z, α) the following formula ln Φ(x, α) = αx1/α + o x1/α , x → +∞ is valid (see, e.g. [1, 4.1, Th. 68]). However, in some applications, an explicit estimate for the error of the above asymptotic approximation is desirable. We are going to prove the following inequality. Theorem 3. Let 0 < α ≤ 1. Then for all x ≥ 1 the inequality (5) ln Φ(x, α) ≤ αx1/α + 1−α ln x + ln(12α−2 ) α An Inequality Associated with Some Entire Functions Shavkat A. Alimov and Onur Alp Ilhan is valid. Remark 1. The order in estimate (5) is precise, at least when α = 1/q, where q is natural, because in this case for all x ≥ 1 the inequality (6) ln Φ(x, α) ≥ αx1/α is true. As it easy to verify, for α = 1 the inequality (6) becomes equality. At first we prove the inequality (5) for α = 1q , where q is natural, and after that we use the interpolation technique to prove it for all α, 0 < α ≤ 1. Lemma 4. Let q be a natural number and Q(x) be the following polynomial (7) Q(x) = q−1 X k=0 Title Page Contents JJ J II I Go Back Close Quit Page 4 of 15 k (k + 1) x . [(k + q)!]1/q J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au Then there exists a constant c1 ≤ 2 so that Z ∞ 1 q (8) e− q t Q(t)dt ≤ c1 q 2 . 0 Proof. It follows from (7) that the inequality q q X X tk−1 ktk−1 Q(t) = k ≤ 1/q [(k + q − 1)!] k=1 k=1 (9) is valid for all t > 0. Then Z Z 1 − 1q tq (10) e Q(t)dt ≤ An Inequality Associated with Some Entire Functions 1 − 1q tq e 0 0 Shavkat A. Alimov and Onur Alp Ilhan ktk−1 dt k=1 q 1 X Z ≤ k k=1 q X k−1 t dt = 0 q X 1 = q. k=1 Further, for t ≥ 1 it follows from (9) that q Q(t) ≤ X k=1 q ktk−1 ≤ tq−1 X k=1 k = tq−1 q(q + 1) . 2 Using this estimate we get Z ∞ Z q(q + 1) ∞ − 1q tq q−1 − 1q tq (11) e Q(t)dt ≤ e t dt 2 1 1 q(q + 1) −1/q q(q + 1) = e < . 2 2 Title Page Contents JJ J II I Go Back Close Quit Page 5 of 15 J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au Taking into consideration (10) and (11) we may write Z ∞ Z 1 Z ∞ 1 q − 1q tq − 1q tq e Q(t)dt = e Q(t)dt + e− q t Q(t)dt 0 0 1 q(q + 1) ≤q+ ≤ 2q 2 , 2 and this inequality proves Lemma 4. We consider the auxiliary function Fq (x) = (12) ∞ X k=q An Inequality Associated with Some Entire Functions xk−q+1 , (k!)1/q x ≥ 0. Lemma 5. Let q ∈ N. Then with some constant c1 ≤ 2 the following inequality 1 q Fq (x) ≤ c1 q 2 e q x , (13) x ≥ 0, is valid. Proof. Consider the derivative of the function (12), which equals to (14) ∞ ∞ X X xk xk−q F (x) = (k − q + 1) = (k + 1) . 1/q 1/q (k!) [(k + q)!] k=q k=0 0 (15) Q(x) = k=0 Title Page Contents JJ J II I Go Back Close Quit By introducing the following polynomial q−1 X Shavkat A. Alimov and Onur Alp Ilhan Page 6 of 15 k (k + 1) x , [(k + q)!]1/q J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au and comparing (14) and (15) we get F 0 (x) − Q(x) = ∞ X (k + 1) k=q xk . [(k + q)!]1/q Further we use the following equality (16) ∞ X xk−q+1 xk q−1 (k + 1) = x (k + 1) [(k + q)!]1/q [(k + q)!]1/q k=q k=q ∞ X = xq−1 ∞ X Bk (q) k=q where Bk (q) = xk−q+1 , (k!)1/q k+1 . [(k + 1)(k + 2) · · · (k + q)]1/q Hence, according to definition (12) and equality (16), (17) F 0 (x) − Q(x) = xq−1 ∞ X k−q+1 Bk (q) k=q x . (k!)1/q It is clear, that Bk (q) ≤ 1. Then it follows from equality (17) that (18) F 0 (x) − Q(x) ≤ xq−1 F (x), x > 0. An Inequality Associated with Some Entire Functions Shavkat A. Alimov and Onur Alp Ilhan Title Page Contents JJ J II I Go Back Close Quit Page 7 of 15 In as much as 1 q eqx h i0 1 q e− q x F (x) = F 0 (x) − xq−1 F (x), J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au we get from the inequality (18) that h 1 q i0 1 q e− q x F (x) ≤ e− q x Q(x), x > 0. By integrating this inequality and taking into consideration that F (0) = 0 we get Z x 1 q − 1q xq e F (x) ≤ e− q t Q(t)dt, x > 0. 0 According to Lemma 4 Z An Inequality Associated with Some Entire Functions x 1 q e− q t Q(t)dt ≤ c1 q 2 , x > 0, 0 and consequently 1 q F (x) ≤ c1 q 2 e q x , x > 0. Shavkat A. Alimov and Onur Alp Ilhan Title Page Contents Lemma 6. Let q be a natural number and P (x) be the following polynomial (19) Pq (x) = q−1 X k=0 xk . (k!)1/q Then the estimate (20) is valid. JJ J II I Go Back Close Quit 1 q Pq (x)e− q x ≤ q, x > 0, Page 8 of 15 J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au Proof. It is clear that for any p > 0 the maximum of the function fp (x) = xp e−x , x ≥ 0, equals to max fp (x) = fp (p) = pp e−p . Then k − 1q xq maxx e x≥0 =q k/q maxy k/q −y e y≥0 =q k/q k/q k e−k/q = k k/q e−k/q . q Hence, xk − 1q xq k k/q e−k/q e ≤ . (k!)1/q (k!)1/q (21) Taking into account the Stirling formula (2) 1/q θk 1/q 1/2q k/q −k/q (k!) = (2πk) k e 1+ ≥ (2πk)1/2q k k/q e−k/q , 11k and using estimate (21) we get xk − 1q xq k k/q e−k/q e = (2πk)−1/2q ≤ 1. ≤ 1/q 1/2q k/q −k/q (k!) (2πk) k e Pq (x)e = q−1 X k=0 Shavkat A. Alimov and Onur Alp Ilhan Title Page Contents JJ J II I Go Back Close Then according to definition (19) − 1q xq An Inequality Associated with Some Entire Functions q−1 xk − 1q xq X e ≤ 1 = q. (k!)1/q k=0 Quit Page 9 of 15 J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au Lemma 7. Let α = inequality 1 q and q ∈ N. Then with some constant c2 < 3 the following 1 Φ x, q (22) 1 q ≤ c2 q 2 xq−1 e q x , x ≥ 1, is valid. Proof. Obviously, Φ x, 1 q = ∞ X k=0 k x = (k!)1/q q−1 X k=0 k ∞ X xk−q+1 x + xq−1 , 1/q (k!)1/q (k!) k=q An Inequality Associated with Some Entire Functions x ≥ 1. Hence, taking into account definitions (12) and (19), we may write 1 (23) Φ x, = P (x) + xq−1 Fq (x). q We may estimate the function in the right hand side of (23) by inequalities (20) and (13): 1 q 1 q 1 q 1 Φ x, ≤ qe q x + xq−1 c1 q 2 e q x ≤ (1 + c1 )q 2 xq−1 e q x , x ≥ 1, q where c1 ≤ 2, according to Lemma 5. Shavkat A. Alimov and Onur Alp Ilhan Title Page Contents JJ J II I Go Back Close Quit Page 10 of 15 We proved estimate (22) for integers q ≥ 1 only. Using this estimate we may prove it for an arbitrary q ≥ 1 by complex interpolation. For this purpose we J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au introduce the following function f (ζ) = f (ζ, b) = b (24) ζ−1 −bζ e ∞ X bkζ , (k!)ζ k=0 where ζ = ξ + iη, ξ > 0, −∞ < η < ∞, b ≥ 1. Lemma 8. Let 0 < ξ ≤ 1. Then with some constant c0 ≤ 12 the inequality (25) |f (ξ + iη)| ≤ c0 , ξ2 0 < ξ ≤ 1, −∞ < η < ∞, b > 0, An Inequality Associated with Some Entire Functions Shavkat A. Alimov and Onur Alp Ilhan is valid. Proof. According to definition (24), f (ξ + iη) = b ξ+iη−1 −b(ξ+iη) e ∞ X bk(ξ+iη) , (k!)(ξ+iη) k=0 and hence |f (ξ + iη)| ≤ b ξ−1 −bξ e ∞ X bkξ = bξ−1 e−bξ Φ(bξ , ξ), ξ (k!) k=0 where the function Φ is defined by equality (1). Putting ξ = 1/q we get 1 1 (1−q)/q −b/q 1/q f + iη ≤ b e Φ b , . (26) q q Title Page Contents JJ J II I Go Back Close Quit Page 11 of 15 J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au According to Lemma 7 for all integers q ≥ 1 the following inequality 1/q 1 (27) Φ b , ≤ c2 q 2 b(q−1)/q eb/q , b ≥ 1, q is fulfilled. Hence, if q ∈ N then it follows from (26) and (27) that 1 f ≤ c2 q 2 , −∞ < η < ∞, (28) + iη q where c2 ≤ 3. Let us suppose now that 1/(q + 1) < ξ < 1/q. We may use the PhragmenLindelöf theorem (see [3, XII.1.1]) and applying it to (28) we get for some t, 0 < t < 1, the following estimate (29) 2(1−t) 2t |f (ξ + iη)| ≤ c2 (1 + q) q , 1−t t ξ= + , q+1 q In as much as 1 + q ≤ 2q and q ≤ 1/ξ we have (1 + q)2(1−t) q 2t ≤ 22(1−t) q 2 ≤ 4/ξ 2 . In that case it follows from the inequality (29) that |f (ξ + iη)| ≤ 4c2 . ξ2 An Inequality Associated with Some Entire Functions Shavkat A. Alimov and Onur Alp Ilhan Title Page −∞ < η < ∞. Contents JJ J II I Go Back Close Quit Page 12 of 15 This inequality coincides with required inequality (25). J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au Proof of Theorem 3. Follows immediately from Lemma 8 and from definitions (1) and (24): ∞ X xk 1/α 1/α Φ(x, α) = = x(1−α)/α eαx f (α, x1/α ) ≤ 4c0 α−2 x(1−α)/α eαx , α (k!) k=0 where c0 < 3. Obviously, this inequality is equivalent to (5). In closing we prove the inequality (6) (see Remark 1). Proposition 9. Let q ∈ N. Then 1 q 1 Φ x, ≥ eqx , q An Inequality Associated with Some Entire Functions x ≥ 0. Proof. Denote 1 g(x) = Φ x, q Title Page . Contents Obviously, ∞ X ∞ ∞ X xk−1 X xk−1 xk−1 ≥ k ≥ g (x) = k (k!)1/q (k!)1/q [(k − q)!]1/q k=q k=q k=1 0 =x q−1 ∞ X k=0 JJ J II I Go Back Close xk = xq−1 g(x). 1/q (k!) Quit Page 13 of 15 Hence, (30) Shavkat A. Alimov and Onur Alp Ilhan g 0 (x) − xq−1 g(x) ≥ 0, x > 0. J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au In as much as 1 q 1 q e q x [e− q x g(x)]0 = g 0 (x) − xq−1 g(x), we get from the inequality (30) that h 1 q i0 −qx e g(x) ≥ 0, x > 0. Then since g(0) = 1 we have 1 q e− q x g(x) ≥ 1. Hence, g(x) ≥ e 1 q x q An Inequality Associated with Some Entire Functions Shavkat A. Alimov and Onur Alp Ilhan , x > 0. Title Page Contents JJ J II I Go Back Close Quit Page 14 of 15 J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au References [1] G. POLYA AND G. SZEGO, Aufgaben und Lehrsatze aus der Analysis, 2 Band, Springer-Verlag, 1964. [2] E.T. WHITTAKER AND G.N. WATSON, A Course of Modern Analysis, Fourth Edition, Cambridge University Press, 1927. [3] A. ZYGMUND, Trigonometric Series, vol.2, Cambridge University Press, 1959. An Inequality Associated with Some Entire Functions Shavkat A. Alimov and Onur Alp Ilhan Title Page Contents JJ J II I Go Back Close Quit Page 15 of 15 J. Ineq. Pure and Appl. Math. 5(3) Art. 67, 2004 http://jipam.vu.edu.au