INEQUALITIES IN THE COMPLEX PLANE RÓBERT SZÁSZ Romania EMail: szasz_robert2001@yahoo.com Inequalities in the Complex Plane Róbert Szász vol. 8, iss. 1, art. 27, 2007 Received: 28 July, 2006 Title Page Accepted: 04 January, 2007 Contents Communicated by: H.M. Srivastava 2000 AMS Sub. Class.: 30C99. Key words: Differential subordination, Extreme point, Locally convex linear topological space, Convex functional. Abstract: A differential inequality is generalised and improved. Several other differential inequalities are considered. JJ II J I Page 1 of 12 Go Back Full Screen Close Contents 1 Introduction 3 2 Preliminaries 4 3 The Main Result 6 4 Particular Cases 9 Inequalities in the Complex Plane Róbert Szász vol. 8, iss. 1, art. 27, 2007 Title Page Contents JJ II J I Page 2 of 12 Go Back Full Screen Close 1. Introduction Let H(U ) be the set of holomorfic functions defined on the unit disc U = {z ∈ C : |z| < 1}. In [2, pp. 38 Example 2.4. d] and [3, pp. 192 Example 9.3.4] the authors have proved, as an application of the developed theory, the implication: If f ∈ H(U ), f (0) = 1 and Re(f (z) + zf 0 (z) + z 2 f 00 (z)) > 0, z ∈ U then Re f (z) > 0, z ∈ U. The aim of this paper is to generalise this inequality and to determine the biggest α ∈ R for which the implication f (0) = 1, Re(f (z)+zf 0 (z)+z 2 f 00 (z)) > 0, (∀) z ∈ U ⇒ Re f (z) > α, (∀) z ∈ U Inequalities in the Complex Plane Róbert Szász vol. 8, iss. 1, art. 27, 2007 Title Page Contents holds true. In this paper each many-valued function is taken with the principal value. JJ II J I Page 3 of 12 Go Back Full Screen Close 2. Preliminaries In our study we need the following definitions and lemmas: Let X be a locally convex linear topological space. For a subset U ⊂ X the closed convex hull of U is defined as the intersection of all closed convex sets containing U and will be denoted by co(U ). If U ⊂ V ⊂ X then U is called an extremal subset of V provided that whenever u = tx + (1 − t)y where u ∈ U, x, y ∈ V and t ∈ (0, 1) then x, y ∈ U. An extremal subset of U consisting of just one point is called an extreme point of U. The set of the extreme points of U will be denoted by EU. Lemma 2.1 ([1, pp. 45]). If J : H(U ) → R is a real-valued, continuous convex functional and F is a compact subset of H(U ), then max{J(f ) : f ∈ co(F)} = max{J(f ) : f ∈ F} = max{J(f ) : f ∈ E(co(F))}. In the particular case if J is a linear map then we also have: Inequalities in the Complex Plane Róbert Szász vol. 8, iss. 1, art. 27, 2007 Title Page Contents JJ II J I min{J(f ) : f ∈ co(F)} = min{J(f ) : f ∈ F} = min{J(f ) : f ∈ E(co(F))}. Page 4 of 12 Suppose that f, g ∈ H(U ). The function f is subordinate to g if there exists a function θ ∈ H(U ) such that θ(0) = 0, |θ(z)| < 1, z ∈ U and f (z) = g(θ(z)), z ∈ U. The subordination will be denoted by f ≺ g. Observation 1. Suppose that f, g ∈ H(U ) and g is univalent. If f (0) = g(0) and f (U ) ⊂ g(U ) then f ≺ g. When F ∈ H(U ) we use the notation Go Back s(F ) = {f ∈ H(U ) : f ≺ F }. Full Screen Close Lemma 2.2 ([1, pp. 51]). Suppose that Fα is defined by the equality α 1 + cz , |c| ≤ 1, c 6= −1. Fα (z) = 1−z If α ≥ 1 then co(s(Fα )) consists of all functions in H(U ) represented by α Z 2π 1 + cze−it f (z) = dµ(t) 1 − ze−it 0 where µ is a positive measure on [0, 2π] having the property µ([0, 2π]) = 1 and 1 + cze−it E(co(s(Fα ))) = t ∈ [0, 2π] . 1 − ze−it Observation 2. If L : H(U ) → H(U ) is an invertible linear map and F ⊂ H(U ) is a compact subset, then L(co(F)) = co(L(F)) and the set E(co(F)) is in one-to-one correspondence with EL(co(F)). Inequalities in the Complex Plane Róbert Szász vol. 8, iss. 1, art. 27, 2007 Title Page Contents JJ II J I Page 5 of 12 Go Back Full Screen Close 3. The Main Result Theorem 3.1. If f ∈ H(U ), f (0) = 1; m, p ∈ N∗ ; ak ∈ R, k = 1, p and q (3.1) Re m f (z) + a1 zf 0 (z) + · · · + ap z p f (p) (z) > 0, z ∈ U then ∞ X Inequalities in the Complex Plane ! Pm m−1 k Cm Cm+n−k−1 (3.2) 1 + inf Re zn z∈U P (n) n=1 ! ∞ Pm m−1 k X C C k=0 m m+n−k−1 n < Re f (z) < 1 + sup Re z , P (n) z∈U n=1 k=0 Róbert Szász vol. 8, iss. 1, art. 27, 2007 z∈U Title Page Contents where P (x) = 1 + a1 x + a2 x(x − 1) + · · · + ap x(x − 1) · · · (x − p + 1). Proof. The condition of the theorem can be rewritten in the form q 1+z m f (z) + a1 zf 0 (z) + · · · + ap z p f (p) (z) ≺ , 1−z J I Go Back f (z) + a1 zf 0 (z) + · · · + ap z p f (p) (z) ≺ 1+z 1−z m . p (p) f (z) + a1 zf (z) + · · · + ap z f Z (z) = 0 2π 1 + ze−it 1 − ze−it Full Screen Close According to the results of Lemma 2.2, where µ([0, 2π]) = 1. II Page 6 of 12 which is equivalent to 0 JJ m dµ(t) = h(z), If f (z) = 1 + ∞ X bn z n , z∈U n=1 then 0 p (p) f (z) + a1 zf (z) + · · · + ap z f (z) = 1 + ∞ X bn P (n)z n . n=1 On the other hand ! Z m Z 2π ∞ n 2π X X 1 + ze−it m−1 n k C C z e−int dµ(t), dµ(t) = 1 + m m+n−k−1 −it 1 − ze 0 0 n=1 k=0 with Cpq = 0 if q > p. The equalities Cpq = 0 if q > p imply also that: n X m−1 k Cm Cm+n−k−1 = m X The above two developments in power series imply that: ! Z ∞ ∞ m X X X m−1 k zn 1+ bn P (n)z n = 1 + Cm Cm+n−k−1 n=0 n=1 and Róbert Szász vol. 8, iss. 1, art. 27, 2007 Title Page Contents m−1 k Cm+n−k−1 . Cm k=0 k=0 Inequalities in the Complex Plane 2π e−int dµ(t) 0 k=0 JJ II J I Page 7 of 12 Go Back Full Screen bn = 1 P (n) m X m−1 k Cm Cm+n−k−1 !Z 2π e−int dµ(t). 0 k=0 Consequently, f (z) = 1 + ∞ X n=1 1 P (n) m X k=0 ! m−1 k Cm Cm+n−k−1 z n Z 0 2π e−int dµ(t). Close If m Z 2π 1 + ze−it dµ(t), z ∈ U, µ([0, 2π]) = 1 , B = h ∈ H(U ) h(z) = 1 − ze−it 0 q m 0 p (p) f (z) + a1 zf (z) + · · · + ap z f (z) > 0, z ∈ U C = f ∈ H(U ) Re then the correspondence L : B → C, L(h) = f defines an invertible linear map and according to Observation 2 the extreme points of the class C are ! m ∞ X X 1 m−1 k z n e−int , z ∈ U, t ∈ [0, 2π). ft (z) = 1 + Cm Cm+n−k−1 P (n) n=1 k=0 Inequalities in the Complex Plane Róbert Szász vol. 8, iss. 1, art. 27, 2007 Title Page This result and Lemma 2.1 implies the assertion of Theorem 3.1. Contents JJ II J I Page 8 of 12 Go Back Full Screen Close 4. Particular Cases If we put p = 2, a1 = a2 = m = 1 in Theorem 3.1 then we get: Corollary 4.1. If f ∈ H(U ), f (0) = 1 and (4.1) Re(f (z) + zf 0 (z) + z 2 f 00 (z)) > 0, z ∈ U, then π(e2π + 1) 2πeπ > Re f (z) > , e2π − 1 e2π − 1 and these results are sharp in the sense that (4.2) sup Re f (z) = z∈U f ∈C π(e2π + 1) e2π − 1 Inequalities in the Complex Plane z∈U Róbert Szász vol. 8, iss. 1, art. 27, 2007 and Title Page Contents π inf Re f (z) = z∈U f ∈C 2πe . −1 e2π Proof. Theorem 3.1 implies the following inequalities: ! ! ∞ ∞ X X 2 2 z n < Re f (z) < 1 + sup Re zn . 1 + inf Re 2 2 z∈U n +1 n +1 z∈U n=1 n=1 The minimum and maximum principle for harmonic functions imply that ! ! ∞ ∞ X X 2 2 n int sup Re z = sup Re e n2 + 1 n2 + 1 z∈U t∈[0,2π] n=1 n=1 ! ! ∞ ∞ X X 2 2 n int z = inf Re e . inf Re z∈U t∈[0,2π] n2 + 1 n2 + 1 n=1 n=1 JJ II J I Page 9 of 12 Go Back Full Screen Close By considering the integral Z In = |z|=n+ 12 eizt dz, (z 2 + 1)(e2πiz − 1) t ∈ [0, 2π), using the equality limn→∞ In = 0 and residue theory we deduce that ! ∞ X 2 π(et + e2π−t ) ikt e = , t ∈ [0, 2π) 1 + Re k2 + 1 e2π − 1 k=1 Inequalities in the Complex Plane Róbert Szász vol. 8, iss. 1, art. 27, 2007 and so we get 2π π π(e + 1) 2πe > Re(f (z)) > 2π , 2π e −1 e −1 z ∈ U. Title Page Contents If we put m = 2, a1 = 0, a2 = 4, Theorem 3.1 implies Corollary 4.2. If f ∈ H(U ), f (0) = 1 and p (4.3) Re f (z) + 4z 2 f 00 (z) > 0, z ∈ U, JJ II J I Page 10 of 12 then (4.4) ∞ X (−1)k · k Re f (z) > 1 + 4 , 2 (2k − 1) k=1 Go Back z∈U Full Screen Close and this result is sharp. Proof. Theorem 3.1 and the minimum principle imply that ∞ X eikt · k Re f (z) > 1 + 4 inf Re t∈[0,2π] (2k − 1)2 k=1 ! . It is easy to observe that ∞ ∞ ∞ X X X keikt eikt eikt 2 = + (2k − 1)2 2k − 1 n=1 (2k − 1)2 n=1 k=1 Z 1X Z 1Z ∞ 2 k−1 ikt = (x ) e dx + 0 Z = (4.5) 0 Since Re 0 k=1 1 it e dx + 1 − x2 eit Z eit −1 ≥ , 2 it 1−x e 1 + x2 0 1Z 0 1 0 ∞ 1X (x2 y 2 )k−1 eikt dxdy k=1 it e dxdy, 1 − x2 y 2 eit Inequalities in the Complex Plane t ∈ [0, 2π). x ∈ [0, 1], t ∈ [0, 2π) Róbert Szász vol. 8, iss. 1, art. 27, 2007 Title Page and eit −1 ≥ , x, y ∈ [0, 1], t ∈ [0, 2π), 2 2 it 1−x y e 1 + x2 · y 2 by integrating we get that Z 1 Z 1 1 eit dx ≥ − dx Re 2 it 2 0 1+x 0 1−x e and Z 1 Z 1 1 eit Re dxdy ≥ − dxdy. 2 2 it 2 2 0 1−x y e 0 1+x y In the derived inequalities, equality occurs if t = π, this means that ∞ ∞ X X (−1)k · k keikt inf Re = t∈[0,2π) (2k − 1)2 (2k − 1)2 k=1 k=1 Re and the inequality (4.4) holds true. Contents JJ II J I Page 11 of 12 Go Back Full Screen Close References [1] D.J. HALLENBECK AND T.H. MAC GREGOR, Linear Problems and Convexity Techniques in Geometric Function Theory, Pitman Advanced Publishing Program, 1984. [2] S.S. MILLER AND P.T. MOCANU, Differential Subordinations, Marcel Decker Inc. New York. Basel(2000). [3] P.T. MOCANU, TEODOR BULBOACĂ AND G. Şt. SĂLĂGEAN, Toeria Geoemtrică a Funcţiilor Univalente, Casa Cărţii de Ştiinţă1, Cluj-Napoca, 1991. Inequalities in the Complex Plane Róbert Szász vol. 8, iss. 1, art. 27, 2007 Title Page Contents JJ II J I Page 12 of 12 Go Back Full Screen Close