Open Math. 06; 4 789 800 Open Mathematics Open Access Research Article YingChun Li and ZhiHong Liu* Convolutions of harmonic right half-plane mappings OI 0.55/math-06-0069 Received March, 06; accepted July 7, 06. Abstract We first prove that the convolution of a normalied right half-plane mapping with another subclass of normalied right half-plane mappings with the dilatation.a C /=. C a/ is CH (convex in the horiontal direction) provided a or a 0. Secondly, we give a simply method to prove the convolution of two special subclasses of harmonic univalent mappings in the right half-plane is CH which was proved by Kumar et al. [, Theorem.]. In addition, we derive the convolution of harmonic univalent mappings involving the generalied harmonic right half-plane mappings is CH. Finally, we present two examples of harmonic mappings to illuminate our main results. Keywords Harmonic univalent mappings, Harmonic convolution, Generalied right half-plane mappings MSC 30C45, 58E0 Introduction and main results Assume that f uciv is a complex-valued harmonic function defined on the open unit disk U f C W jj < g, where u and v are real harmonic functions in U. Such function can be expressed as f h C g, where X X h./ C a n n and g./ b n n n are analytic in U, h and g are known as the analytic part and co-analytic part of f, respectively. A harmonic mapping f hcg is locally univalent and sense-preserving if and only if the dilatation of f defined by!./ g 0./=h 0./, satisfies j!./j < for all U. We denote by S H the class of all harmonic, sense-preserving and univalent mappings f h C g in U, which are normalied by the conditions h.0/ g.0/ 0 and h 0.0/. Let SH 0 be the subset of all f S H in which g 0.0/ 0. Further, let K H ; C H (resp. KH 0 ; C0 H ) be the subclass of S H (resp. SH 0 ) whose images are convex and close-to-convex domains. A domain is said to be convex in the horiontal direction (CH) if every line parallel to the real axis has a connected intersection with. For harmonic univalent functions X X f./ h./ C g./ C a n n C b n n and n n n n X X F./ H./ C G./ C A n n C B n n ; n YingChun Li College of Mathematics, Honghe University, Mengi 6699, Yunnan, China, E-mail liyingchunmath@63.com *Corresponding Author ZhiHong Liu College of Mathematics, Honghe University, Mengi 6699, Yunnan, China and School of Mathematics and Econometrics, Hunan University, Changsha 4008, Hunan, China, E-mail liuhihongmath@63.com 06 Li and Liu, published by e Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercial-Noerivs 3.0 License. ownload ate //8 84 AM
790 Y. Li, Z. Liu the convolution (or Hadamard product) of them is given by X X.f F /./.h H /./ C.g G/./ C a n A n n C b n B n n Many research papers in recent years have studied the convolution or Hadamard product of planar harmonic mappings, see [ ]. However, corresponding questions for the class of univalent harmonic mappings seem to be difficult to handle as can be seen from the recent investigations of the authors [3 7]. In [], Kumar et al. constructed the harmonic functions f a h a C g a K H in the right half-plane, which satisfy the conditions h a C g a =. / with! a./.a /=. a/. < a < ). By using the technique of shear construction (see [8]), we have h a./ n Ca. / and g a./ n a Ca (). / Obviously, for a 0, f 0./ h 0./ C g 0./ KH 0 is the standard right half-plane mapping, where h 0./ and g. / 0./. / () Recently orff et al. studied the convolution of harmonic univalent mappings in the right half-plane (cf. [4, 5]). They proved that Theorem A ([4, Theorem 5]). Let f h C g ; f h C g SH 0 with h i C g i =. / for i ;. If f f is locally univalent and sense-preserving, then f f SH 0 and convex in the horiontal direction. Theorem B ([5, Theorem 3]). Let f n h C g SH 0 with h C g =. / and!./ g0./=h 0./ e i n. R; n N C /. If n ;, then f 0 f n SH 0 and is convex in the horiontal direction. Theorem C ([5, Theorem 4]). Let f h C g KH 0 with h./ C g./ =. with a. ; /. Then f 0 f SH 0 and is convex in the horiontal direction. / and!./. C a/=. C a/ We now begin to state the elementary result concerning the convolutions of f 0 with the other special subclass harmonic mappings. Theorem.. Let f h C g SH 0 with h./ C g./ =. / and!./. C a/=. C a/, then f 0 f SH 0 and is convex in the horiontal direction for a or a 0. The following generalied right half-plane harmonic univalent mappings were introduced by Muir [7] L c./ H c./ C G c./ C c C c. / C C c c. /. UI c > 0/ (3) Clearly, L./ f 0./, it was proved in [7] that L c.u/ fre.!/ > f h C g S H, then the above representation gives that =. C c/g for each c > 0. Moreover, if L c f h C ch0 C c The following Cohn s Rule is helpful in proving our main results. Cohn s Rule ([9, p.375]). Given a polynomial of degree n, let C g cg0 C c (4) p./ p 0./ a n;0 n C a n ;0 n C C a ;0 C a 0;0.a n;0 0/ (5) p./ p 0./ n p.=/ a n;0 C a n ;0 C C a ;0 n C a 0;0 n (6) ownload ate //8 84 AM
Convolutions of harmonic right half-plane mappings 79 enote by r and s the number of eros of p./ inside the unit circle and on it, respectively. If ja 0;0 j < ja n;0 j, then p./ a n;0p./ a 0;0 p./ is of degree n with r r and s s the number of eros of p./ inside the unit circle and on it, respectively. It should be remarked that Cohn s Rule and Schur-Cohn s algorithm [9, p. 378] are important tools to prove harmonic mappings are locally univalent and sense-preserving. Some related works have been done on these topics, one can refer to [, 3, 6, 3, 5, 6]. In [], the authors proved the following result. Theorem. Let f a h a C g a be given by (). If f n h C g is the right half-plane mapping given by h C g =. / with!./ e i n. R; n N C /, then f a f n S H is CH for a Πn nc ; /. In this paper, we will use a new method to prove the above theorem. The main difference of our work from [] is that we construct a sequence of functions for finding all eros of polynomials which are in U, and we will show that the dilatation of f a f n satisfies je!./j j.g a g/ 0 =.h a h/ 0 j < by using mathematical induction, which greatly simplifies the calculation compared with the proof of Theorem. in []. We also show that L c f a is univalent and convex in the horiontal direction for 0 < c. C a/=. a/, and derive the following theorem. Theorem.. Let L c H c C G c be harmonic mappings given by (3). If f a h a C g a is the right half-plane mappings given by (), then L c f a is univalent and convex in the horiontal direction for 0 < c.ca/=. a/. Recently, Liu and Li [6] defined a subclass of harmonic mappings defined by P c./ H c./ G c./ c C c. / C C c C c. / They proved the following result.. UI c > 0/ (7) Theorem E. ([6, Theorem 7]) Let P c./ be harmonic mappings defined by (7) and f n hcg with h g =. / and dilatation!./ e i n. R; n N C /. Then P c f n is univalent and convex in the horiontal direction for 0 < c =n. Similar to the approach used in the proof of Theorem E, we get the following result. Theorem.3. Let L c H c C G c be harmonic mappings given by (3) and f n h C g with h C g =. / and dilatation!./ e i n. R; n N C /. Then L c f n is univalent and convex in the horiontal direction for 0 < c =n. Remark.4. In Theorem. and Theorem.3, let c, clearly, L f 0, so Theorem. and Theorem.3 are generaliation of Theorem C and Theorem B, respectively. It also explains why Theorem B does not hold for n 3, since c, according to Theorem.3, it follows that 0 < c =n hold for n ;. Preliminary results In this section, we will give the following lemmas which play an important role in proving the main results. Lemma. ([5, Eq.(6)]). If f h C g SH 0 with h C g =. / the dilatation of f 0 f is given by and dilatation!./ g0./=h 0./, then e!./! C! C!!0 C!0!0 C!0 (8) ownload ate //8 84 AM
79 Y. Li, Z. Liu Lemma. ([, Lemma.]). Let f a h a C g a be defined by () and f h C g SH 0 be right half-plane mapping, where h C g =. / with dilatation!./ g 0./=h 0./.h 0./ 0; U/. Then e!, the dilatation of f a f, is given by e!./.a /!. C!/ C.a /!0. /. a/. C!/ C.a /! 0. / (9) Lemma.3. Let L c H c C G c be defined by (3) and f h C g SH 0 be right half-plane mapping, where h C g =. / with dilatation!./ g 0./=h 0./.h 0./ 0; U/. Then e!, the dilatation of L c f, is given by e!./ Œ. c/. C c/!. C!/ c!0. / Œ. C c/. c/. C!/ c! 0. / (0) Proof. By (4), we know that e!./.g cg0 / 0.h C ch 0 / 0 Similar calculation as in the proof of [6, Lemma 7] gives (0). Lemma.4 ([0, Theorem 5.3]). A harmonic function f h C g locally univalent in U is a univalent mapping of U onto a domain convex in the horiontal direction if and only if h g is a conformal univalent mapping of U onto a domain convex in the horiontal direction. Lemma.5 (See []). Let f be an analytic function in U with f.0/ 0 and f 0.0/ 0, and let where ; R. If Then f is convex in the horiontal direction. './. C e i /. C e i / ; () f 0./ Re > 0. U/ './ Lemma.6. Let L c H c C G c be a mapping given by (3) and f h C g be the right half-plane mapping with h C g =. /. If L c f is locally univalent, then L c f SH 0 and is convex in the horiontal direction. Proof. Recalling that L c H c C G c and Hence H c C G c. C c/. / ; h C g h g C c.h c C G c /.h g/ C c.h c h H c g C G c h G c g/; H c G c.h c G c /.h C g/.h c h C H c g G c h G c g/ Thus Next, we will show that S, we have H c h G c g Re ' C c.h0 g 0 / C.Hc 0 Gc 0 / C c.h g/ C.H c G c / () Cc.h g/c.h c G c / is convex in the horiontal direction. Letting './ =. / 8 < Re 8 < Re h Cc.h0 C g 0 / h 0 g 0 h 0 Cg 0 h Cc.h0 C g 0 /! C! C.H 0 c C G0 c / H 0 c G 0 c H 0 ccg 0 c ' C.Hc 0 C G0 c /!c ' C! c i i 9 = 9 = ; ; ownload ate //8 84 AM
Convolutions of harmonic right half-plane mappings 793 ( ) C c Re Œr./ C r. / c./. / C c Re fr./ C r c./g > 0; where r./!./ C!./ ; r c./! c./ C! c./. Therefore, by Lemma.5 and the equation (), we know that H c h G c g is convex in the horiontal direction. Finally, since we assumed that L c f is locally univalent and H c h G c g is convex in the horiontal direction, we apply Lemma.4 to obtain that L c f H c h C G c g is convex in the horiontal direction. 3 Proofs of theorems Proof of Theorem.. By Theorem A, we know that f 0 f is convex in the horiontal direction. Now we need to establish that f 0 f is locally univalent. Substituting!./. C a/=. C a/ into (8) and simplifying, yields e!./ 3 C C3a C. C a/ C a q./ C C3a C. C a/ C a 3 q./. A/. B/. C /. A/. B/. C / ; (3) If a, then q./ 3 C 5 C C. C /. C / has all its three eros in U. By Cohn s Rule, so je!./j < for all U. If a 0, it is clearly that je!./j j j < for all U. If a < 0, we apply Cohn s Rule to q./ 3 C C3a C. C a/ C a. Note that j a j <, thus we get Since Q./ a 3q./ a 0 q./ 4 4 a ˇˇˇ 4 C a 3a ˇ 4 a ˇ ˇ ˇ we use Cohn s Rule on q./ again, we get q./ q./ C. C a a / 4 a C C a. a/ 4 a 4 C a 3a 4 a ˇ <.as a < 0/;! W 4 4Ca 3a q 4 a./! 4a.a /.4 C a a / C a a C.4 a / 4 C a a 4 a q./ Clearly, q./ has one ero at We show that j 0 j, or equivalently, 0 C a a 4 C a a j C a a j j4 C a a j. C a a /.4 C a a / 3.4 a /.a / 0 Therefore, by Cohn s Rule, q./ has all its three eros in U, that is A; B; C U and so je!./j < for all U. A new proof of Theorem. By Theorem A, it suffices to show that the dilatation of f a f n satisfies je!./j < for all U. Setting!./ e i n in (9), we have " nc e!./ n e i a n C. C an n/e # i C i.n a an/e a C. C an n/ei n C.n a n i p./ e an/ei nc p./ ; (4) where and p./ nc a n C. C an n/e i C.n a an/e i ; (5) p./ nc p.=/ ownload ate //8 84 AM
794 Y. Li, Z. Liu Firstly, we will show that je!./j < for a n n. In this case, substituting a into (4), we have nc nc ˇ nc n n nc e i C e i ˇ je!./j ˇ ˇ i nc n n e n nc n nc ei n C e i nc ˇ n e i ei nc n nc ei n n nc C n nc n nc ei n C e i nc ˇ j n e i j < Next, we will show that je!./j < for n nc < a <. Obviously, if 0 is a ero of p./, then = 0 is a ero of p./. Hence, if A ; A ; ; A nc are the eros of p./ (not necessarily distinct), then we can write Now for ja j j, A j A j e!./ n e i. A /. A /. A /. A /. A nc /. A nc /.j ; ; ; n C / maps U onto U. It suffices to show that all eros of (5) lie on U for n nc < a <. Since ja 0;0j j.n a an/e i j < ja nc;0 j, then by using Cohn s Rule on p./, we get p./ a nc;0p./ a 0;0 p./. a/. C n/œ. C a/. a/n 4 p./.n a an/e i p./ n n n C n C n C e i Since n. a/.cn/œ.ca/. a/n < a <, we have > 0. Let q nc 4./ n n ja 0; j j nc e i j < ja n; j, by using Cohn s Rule on q./ again, we obtain p./ a n;q./ a 0; q./ q./ n.n C 4/.n C / nc e i q./ n n C n C 4 n C n C 4 e i nc n C nc e i, since Let q./ n nc nc4 n C nc4 e i, then ja 0; j nc4 < ja n ;j, we get p 3./ a n ;q./ a 0; q./ q./.n C /.n C 6/.n C 4/ nc4 e i q./ n n C 4 n C 6 n 3 C n C 6 e i Let q 3./ n nc4 nc6 n 3 C nc6 e i, then ja 0;3 j nc6 < ja n ;3j, we obtain By using this manner, we claim that p k./ p 4./ a n ;3q 3./ a 0;3 q3./ q 3./.n C 4/.n C 8/.n C 6/ Œn C.k /.n C k/ Œn C.k / nc6 e i q3./ n 3 n C 6 n C 8 n 4 C n C 8 e i n kc n C.k / n C k where k ; 3; ; n. To prove the equation (6) is correct for all k N C.k /, it suffices to show p kc./ ŒnC.k / ŒnC.kC/ n k n C k.n C k/ nc.kc/ n n k C n C k e i (6).kC/ C nc.kc/ e i (7) ownload ate //8 84 AM
Convolutions of harmonic right half-plane mappings 795 n kc nc.k / Let q k./ n k C nck nck e i, then q k./ n kc nc.k / q k.=/ C nck nck e i n kc. Since ja 0;k j j nck e i j nck < ja n kc;kj, by using Cohn s Rule on q k./, we deduce that p kc./ a n kc;kq k./ a 0;k q k./ q k./ ŒnC.k / ŒnC.kC/.n C k/ nck e i q k./ n k n C k nc.kc/ n.kc/ C nc.kc/ e i Setting n k.k / in (6), we have p n./ 3n.3n 4/.3n /! 3n e i 3n Then 0 3n e i 3n is a ero of p n./, and 3n e i j 0 j ˇ 3n ˇ j3n j C je i j 3n 3n C 3n So 0 lies inside or on the unit circle jj, by Lemma, we know that all eros of (5) lie on U. This completes the proof. Proof of Theorem.. In view of Lemma.6, it suffices to show that L c f a is locally univalent and sensepreserving. Substituting!./! a./.a /=. a/ into (0), we have e!./ Œ. c/. C c/ a a. C a a / c.a /. a/. / Œ. C c/. c/. C a a / c.a /. a/. / Œ. c/. C c/ Œ.a /. a/ C.a /.a /c. / Œ. C c/. c/ Œ. a/ C.a /. a/.a /c. / 3 Ca ccac Ca Cc C ccac Cc C Ca ccac a. c/ Cc Cc Ca ccac Cc a. c/ Cc 3 (8) Next we just need to show that je!./j < for 0 < c. C a/=. a/, where < a <. We shall consider the following two cases. Case. Suppose that a 0. Then substituting a 0 into (8) yields e!./ c Cc C c. / Cc c Cc C c Cc. / c Cc c Cc Then two eros and. c/=.cc/ of the above numerator lie in or on the unit circle for all 0 < c, so we have je!./j <. Case. Suppose that a 0. From (8), we can write e!./ 3 Ca ccac Ca Cc C ccac Cc C Ca ccac a. c/ Cc Cc Ca ccac Cc a. c/ Cc p./ 3 p./. A/. B/. C /. A/. B/. C / We will show that A; B; C U for 0 < c. C a/=. a/. Applying Lemma to p./ 3 C a c C ac C c C C a c C ac C c a. c/ C c ownload ate //8 84 AM
796 Y. Li, Z. Liu Note that j a. c/ j < for c > 0 and < a <, we get Cc p./ a 3p./ a 0 p a. c/./ p./ C Cc p./. C c C a ac/. C c a C ac/ C c 6ac C c Ca a c a c. C c/. C c/ C c C 6ac c a a c C a c. C c/. C c C a ac/. C c a C ac/ Cc a ac cca Cac C C. C c/ C c C a ac C c C a ac. C c C a ac/. C c a C ac/ C a c. a/. /. C c/ C a C c. a/ So p./ has two eros and Ca c. a/ which are in or on the unit circle for 0 < c. C CaCc. a/ a/=. a/. Thus, by Lemma, all eros of p./ lie on U, that is A; B; C U and so je!./j < for all U. 4 Examples In this section, we give interesting examples resulting from Theorem. and Theorem.. Example 4.. In Theorem., note that f h C g SH 0.a C /=. C a/. By shearing with h C g =. / and the dilatation!./ Solving these equations, we get h 0./ C g 0./. / and g 0./!./h 0./ h 0./ C a. / 3. C / and g 0./.a C /. / 3. C / Integration gives and then, By the convolutions, we have So that and h./ C C a 4 g./ C a 4. / C a log 8 a log. / 8 f 0 f h 0 h C g 0 g h./ C h0./ h 0 h 4 C C a 8 g 0 g 4 C a 8. / C a 6 log. / a 6 log g./ C C C C ; (9) C (0) g0./ C. C a/. / 3. C / C.a C /. / 3. C / Images of U under f 0 f for certain values of a are drawn in Figure and Figure (a)-(d) by using Mathematica. By Theorem., it follows that f 0 f is locally univalent and convex in the horiontal direction for a ; 05; 0;, but it is not locally univalent for a 0. ownload ate //8 84 AM
Convolutions of harmonic right half-plane mappings 797 Fig.. Images of U under f 0 f for a 0. 0.75 0.70 0.65 0.60 0.55 0.50 0.45 0.70 0.65 0.60 0.55 0.50 0.45 0.40 (a) a 0 (b) Partial enlarged view for a 0 Fig.. Images of U under f 0 f for various values a. (a) a (b) a 05 (c) a 0 (d) a ownload ate //8 84 AM
798 Y. Li, Z. Liu Example 4.. In Theorem., by () and (4), we have L c f a ha./ C ch 0 a C c./ C C c Re " Ca. / C. C c/. / C c ga./ c. a/. C a/. / 3 # c. a/. C / C. C c/. C a/. / 3 cg 0 a./ " C a Ca C c 8 #. /. C a/. / 3 c.a / < a Ca C c 9 = C i Im. C c/. / ; If we take a 05; c 6, in view of Theorem., we know that L 6 f 05 is univalent and convex in the horiontal direction. The image of U under f 05 ; L 6 and L 6 f 05 are shown in Figure 3 (a)-(c), respectively. Fig. 3. Images of U under f 05 ; L 6 and L 6 f 05..0.0 0.5 0.5 -.0-0.5 0.5.0 -.0-0.5 0.5.0-0.5-0.5 -.0 (a) f 05 -.0 (b) L 6.0 0.5 -.0-0.5 0.5.0-0.5 -.0 (c) L 6 f 05 If we take a 05; c 6, then L 6 f 05 Re 7 C 3 08 (. C / 3 C i Im C 6. / 3 7. / The image of U under L 6 f 05 is shown in Figure 4 (a). Figure 4 (b) is a partial enlarged view of Figure 4 (a) showing that the images of two outer most concentric circles in U are intersecting and so L 6 f 05 is not univalent. ) ownload ate //8 84 AM
Convolutions of harmonic right half-plane mappings 799 Fig. 4. Images of the unit disk U under L 6 f 05..0 0.004 0.5 0.00 -.0-0.5 0.5.0-0.07-0.070-0.068-0.066-0.5-0.00-0.004 -.0 (a) L 6 f 05 (b) Partial enlarged view of L 6 f 05 Remark 4.3. This example shows that the condition 0 < c. C a/=. a/ in Theorem. is sharp. Acknowledgement The authors would like to thank the referees for their helpful comments. According to the hints of the referees the authors were able to improve the paper considerably. The research was supported by the Project Education Fund of Yunnan Province under Grant No. 05Y456, the First Bath of Young and Middle-aged Academic Training Object Backbone of Honghe University under Grant No. 04GG00. References [] Kumar R., orff M., Gupta S., Singh S., Convolution properties of some harmonic mappings in the right-half plane, Bull. Malays. Math. Sci. Soc., 06, 39(), 439 455 [] Kumar R., Gupta S., Singh S., orff M., On harmonic convolutions involving a vertical strip mapping, Bull. Korean Math. Soc., 05, 5(), 05 3 [3] Li L., Ponnusamy S., Convolutions of slanted half-plane harmonic mappings, Analysis (Munich), 03, 33(), 59 76 [4] Li L., Ponnusamy S., Convolutions of harmonic mappings convex in one direction, Complex Anal. Oper. Theory, 05, 9(), 83 99 [5] Boyd Z., orff M., Nowak M., Romney M., Wołoskiewic M., Univalency of convolutions of harmonic mappings, Appl. Math. Comput., 04, 34, 36 33 [6] Liu Z., Li Y., The properties of a new subclass of harmonic univalent mappings, Abstr. Appl. Anal., 03, Article I 79408, 7 pages [7] Muir S., Weak subordination for convex univalent harmonic functions, J. Math. Anal. Appl., 008, 348, 689 69 [8] Muir S., Convex combinations of planar harmonic mappings realied through convolutions with half-strip mappings, Bull. Malays. Math. Sci. Soc., 06, OI0.007/s48-840-06-0336-0 [9] Nagpal S., Ravichandran V., Univalence and convexity in one direction of the convolution of harmonic mappings, Complex Var. Elliptic Equ., 04, 59(9), 38 34 [0] Rosihan M. Ali., Stephen B. Adolf., Subramanian, K. G., Subclasses of harmonic mappings defined by convolution, Appl. Math. Lett., 00, 3, 43 47 [] Wang Z., Liu Z., Li Y., On convolutions of harmonic univalent mappings convex in the direction of the real axis, J. Appl. Anal. Comput., 06, 6, 45 55 [] Sokół Janus, Ibrahim Rabha W., Ahmad M. Z., Al-Janaby, Hiba F., Inequalities of harmonic univalent functions with connections of hypergeometric functions, Open Math., 05, 3, 69 705 [3] Kumar R., Gupta S., Singh S., orff M., An application of Cohn s rule to convolutions of univalent harmonic mappings, arxiv306.5375v [4] orff M., Convolutions of planar harmonic convex mappins, Complex Var. Theorey Appl., 00, 45, 63 7 ownload ate //8 84 AM
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