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Coefficient Inequalities for Strongly Close-to-Convex Functions
JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS ARTICLE NO.
205, 537]553 Ž1997.
AY975234
Coefficient Inequalities for Strongly Close-to-Convex Functions William Ma Integrated Studies Di¨ ision, Pennsyl¨ ania College of Technology, Williamsport, Pennsyl¨ ania 17701]5799
and David MindaU Department of Mathematical Sciences, Uni¨ ersity of Cincinnati, P.O. Box 210025, Cincinnati, Ohio 45221-0025 Submitted by H. M. Sri¨ asta¨ a Received December 20, 1995
Let f Ž z . s z q a2 z 2 q a3 z 3 q ??? be a normalized strongly close-to-convex function of order a ) 0 defined on the unit disk D. This means that there is a normalized convex univalent function w and b g R such that X
arg
f Ž z.
-
ib X
e w Ž z.
ap 2
for z g D. Then < a3 y a22 < q 13 < a x < 2 F 13 Ž 1 q 4a q 2 a 2 . and a3 y 23 a22 q 23 < a2 < 2 F
1 3
Ž 3 q 4a q 2 a 2 .
with equality if and only if f is a rotation of Fa Ž z . s
1 2 Ž1 q a .
ž
1qz 1yz
1q a
/
y1 .
Q 1997 Academic Press
* Research partially supported by a National Science Foundation grant. 537 0022-247Xr97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
538
MA AND MINDA
1. INTRODUCTION A holomorphic function f defined on the unit disk D s z : < z < - 14 is called strongly close-to-convex of order a ) 0 provided f is normalized Ž f Ž0. s 0, f X Ž0. s 1. and there exist b g R and a normalized convex univalent function w such that arg
fXŽ z. ib X
e w Ž z.
-
ap 2
for z g D. Let SCCŽ a ., 0 F a F 1, denote the class of all strongly closeto-convex functions of order a . The usual class of normalized close-to-convex functions is SCCŽ1.. For a s 0 we need to replace strict inequality by ‘‘less than or equal to.’’ Then SCCŽ0. s CV, the class of normalized convex univalent functions. Then SCCŽ0. ; SCCŽ a . for a ) 0. For a g w0, 1x all functions in SCCŽ a . are univalent. The function Fa Ž z . s
1 2Ž 1 q a .
ž
1qz 1yz
s z q Ž1 q a . z2 q
1q a
y1
/
1 3
Ž 3 q 4a q 2 a 2 . z 3 q ???
belongs to SCCŽ a . and is known to be extremal for a number of problems. For instance, if f Ž z . s z q a2 z 2 q a3 z 3 q ??? g SCCŽ a ., then < a2 < F 1 q a and < a3 < F 13 Ž 3 q 4a q 2 a 2 . . Equality holds in either quality if and only if f Ž z . s eyi u Fa Ž e i u z . for some u g R; that is, f is a rotation of Fa . See w2, Chap. 11x for the interesting history of the coefficient problem for SCCŽ a .. The goal of this paper is to establish two sharp coefficient inequalities for the class SCCŽ a .. The authors were led to such inequalities by their recent work relating to two-point distortion theorems for univalent function w5x. These inequalities yield sharp two-point distortion theorems for nonnormalized strongly close-to-convex functions of order a g w0, 1x. Our main result is the following theorem.
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
539
Suppose f Ž z . s z q a2 z 2 q a2 z 3 q ??? g SCCŽ a .. Then
THEOREM 1.
Ž i.
a3 y a22 q 13 < a2 < 2 F 13 Ž 1 q 4a q 2 a 2 .
and
Ž ii .
a3 y 23 a22 q 23 < a2 < 2 F 13 Ž 3 q 4a q 2 a 2 . .
Equality holds if and only if f is a rotation of Fa . Inequality Žii. implies the sharp bound on < a3 < over the class SCCŽ a .. For a s 0 the inequality Ži. was established for the class CV by Trimble w7x. Also, for a s 0 the coefficient bound < a2 < F 1 for the class CV together with inequality Ži. yields inequality Žii.. For a ) 0 the second inequality does not seem to be a consequence of the first. 2. REPRESENTATION FORMULA FOR SCCŽ a . A basic tool in the proof of our main theorem is an elementary representation for derivatives of functions in SCCŽ a . that was already noted in w1x. If f g SCCŽ a ., then w f X Ž z .rŽ e i bw X Ž z ..x1r a has positive real part in D if we select the branch at the origin with value eyi b r a . The fact that this function has positive real part in D implies that cosŽ bra . ) 0, so we may assume that u s bra g Žypr2, pr2.. Then pŽ z . s
1 cos Ž u .
ž
1r a
fXŽ z. e i bw X Ž z .
q i sin Ž u .
/
belongs to the class P of normalized Ž pŽ0. s 1. holomorphic functions defined on D with positive real part. Thus, if f g SCCŽ a ., then f X Ž z . s e i auw X Ž z . cos Ž u . p Ž z . y i sin Ž u . s w X Ž z . e i u cos Ž u . p Ž z . y i sin Ž u .
a
4
a
,
where a ) 0, < u < - pr2, w g CV, and p g P. Conversely, if f X has this form, then f g SCCŽ a .. For future reference we note the explicit representation of the function Fa and its rotations. For eyi u Fa Ž e i u . we have FaX Ž e i u z . s w X Ž e i u z . p Ž e i u z . s
1
Ž 1 y e iu z .
2
ž
a
1 q e iu z 1 y e iu z
a
/
,
where w Ž z . s zrŽ1 y z . g CV and pŽ z . s Ž1 q z .rŽ1 y z . g P.
540
MA AND MINDA
From the representation formula we derive formulas for the second and third coefficients in the power series of functions in SCCŽ a . in terms of coefficients of functions in P and CV. Suppose f Ž z . s z q a2 z 2 q a3 z 3 q ??? ,
w Ž z . s z q b 2 z 2 q b 3 z 3 q ??? , and p Ž z . s 1 q d1 z q d 2 z 2 q ??? . Then
e iu
cos Ž u . p Ž z . y i sin Ž u .
4
a
s 1 q e i u cos Ž u . p Ž z . y 1
4
a
s 1 q a cos Ž u . e i u d1 z q a cos Ž u . e i u d 2 q
a Ž a y 1. 2
cos 2 Ž u . e 2 i u d12 z 2 q ??? .
By using the representation formula we obtain 1 q 2 a2 z q 3a3 z 2 q ??? s fXŽ z. s Ž 1 q 2 b 2 z q 3b 3 z 3 q ??? .Ž 1 q a cos Ž u . e i u d1 z q ??? . s 1 q 2 b 2 q a cos Ž u . e i u d1 z q 3b 3 q a cos Ž u . e i u d 2 q
a Ž a y 1. 2
cos 2 Ž u . e 2 i u d12
q2 a cos Ž u . e i u b 2 d1 z 2 q ??? . Consequently, 2 a2 s 2 b 2 q a cos Ž u . e i u d1
Ž 1.
and 3a3 s 3b 3 q 2 a cos Ž u . e i u b 2 d1 q a cos Ž u . e i u d 2 q
a Ž a y 1. 2
cos 2 Ž u . e 2 i u d12 .
Ž 2.
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
541
Next, we make use of the fact that qŽ z. s 1 q
zwY Ž z .
wX Ž z .
s 1 q c1 z q c 2 z 2 q ???
belongs to P to express a2 and a3 in terms of the coefficients of two functions in P. From 1q
zwY Ž z .
wX Ž z .
s 1 q 2 b 2 z q Ž 6 b 3 y 4 b 22 . z 2 q ??? ,
we obtain c1 s 2 b 2 , c 2 s 6 b 3 y 4 b 22 , or 2 b 2 s c1 , 6 b 3 s c12 q c 2 . If we substitute these expressions into Ž1. and Ž2. we obtain 2 a2 s c1 q a cos Ž u . e i u d1 , 3a3 s
1 2
Ž 3.
Ž c2 q c12 . q a cos Ž u . e iu c1 d1
q a cos Ž u . e i u d 2 q
a Ž a y 1. 2
cos 2 Ž u . e 2 i u d12 .
Ž 4.
For f Ž z . s z q a2 z 2 q a3 z 3 q ??? g SCCŽ a . the two functionals L Ž f . s Re a22 y a3 4 q 13 < a2 < 2 and M Ž f . s Re 3a3 y 2 a22 4 q 2 < a2 < 2 play important roles in the proof of Theorem 1. We use Ž1. and Ž2. to express LŽ f . in terms of b1 , b 2 , d1 , and d 2 . Now a a22 y a3 s b 22 y b 3 q cos Ž u . e i u b 2 d1 3 a y cos Ž u . e i u d 2 3 q
a Ž a q 2. 12
cos 2 Ž u . e 2 i u d12
542
MA AND MINDA
and 1 3
< a2 < 2 s
1 3
< b2 < 2 q
q
a 3
a2 12
cos 2 Ž u . < d1 < 2
cos Ž u . Re e i u b 2 d1 4 ,
so that L Ž f . s Re b 22 y b 3 4 q q q
a 3
cos Ž u . Re
a2 12
1 3
½
< b2 < 2 q
aq2 4
2a 3
cos Ž u . Re b 2 4 Re e i u d1 4 2
cos Ž u . Ž e i u d1 . y e i u d 2
5
cos 2 Ž u . < d1 < 2 .
Ž 5.
Similarly, we use Ž3. and Ž4. to obtain an expression for M Ž f . in terms of c1 , c 2 , d1 , and d 2 . From 3a3 y 2 a22 s
1 2
c 2 q a cos Ž u . e i u d 2 y
a 2
cos 2 Ž u . e 2 i u d12
Ž 6.
and 2 < a2 < 2 s
1 2
< c1 < 2 q a cos Ž u . Re e i u c1 d1 4 q
a2 2
cos 2 Ž u . < d1 < 2 ,
Ž 7.
we obtain MŽ f . s
1 2
Re c 2 4 q
1 2
< c1 < 2 q a cos Ž u . Re e i u c1 d1 4
½
q a cos Ž u . Re y q
a2 2
1 2
cos Ž u . e 2 i u d12 q e i u d 2
cos 2 Ž u . < d1 < 2 .
5 Ž 8.
3. TECHNICAL RESULTS In this section we gather together several preliminary estimates that we need for our proof of Theorem 1.
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
543
LEMMA 1. Suppose pŽ z . s 1 q d1 z q d 2 z 2 q ??? g P and u g Žypr2, pr2.. Then cos Ž u . 2 Re e i u d1 4 q Re
½
1 2
2
cos Ž u . Ž e i u d1 . y e i u d 2
5
F 4.
Equality holds if and only if u s 0 and pŽ z . s Ž1 q z .rŽ1 y z .. Proof. Since pŽ z . g P if and only if p Ž eyi u z . s 1 q eyi u d1 z q ey2 i u d 2 z 2 q ??? belongs to P, it suffices to show that N Ž p . s cos Ž u . 2 Re d1 4 q Re 12 cos Ž u . d12 y eyi u d 2 4 F 4 and equality implies u s 0 and pŽ z . s Ž1 q z .rŽ1 y z .. Next, we eliminate d 2 in N Ž p .. Since p g P w4x, d 2 y 12 d12 F 2 y 12 < d1 < 2 . This yields yRe eyi u d 2 4 F 2 y 12 < d1 < 2 y 21 Re eyi u d12 4 , so that Re 12 cos Ž u . d12 y eyi u d 2 4 F 2 y 12 < d1 < 2 q 21 Re Ž cos Ž u . y eyi u . d12 4 s 2 y 12 < d1 < 2 y 21 sin Ž u . Im d12 4 . Thus, N Ž p . F cos Ž u . 2 Re d1 4 q 2 y 12 < d1 < 2 y 12 sin Ž u . Im d12 4 . Recall that if p g P, then < d1 < F 2. Thus, we wish to show that cos Ž u . 2 Re d1 4 q 2 y 12 < d1 < 2 y 12 sin Ž u . Im d12 4 F 4 for < d1 < F 2. Set d1 s 2 re it , where 0 F r F 1 and t g Žyp , p x. Then we want to demonstrate that H Ž r . s cos Ž u . 4 r cos Ž t . q 2 y r 2 Ž 2 q 2 sin Ž u . sin Ž 2 t . . s 2 cos Ž u . 2 r cos Ž t . q 1 y r 2 Ž 1 q 2 sin Ž u . cos Ž t . sin Ž t . . F 4
544
MA AND MINDA
for 0 F r F 1. Note that H Ž 0 . s 2 cos Ž u . F 2 - 4. Next, we show that H Ž1. F 4 with equality only for u s t s 0. H Ž 1 . s 4 cos Ž u . cos Ž t . y 4 cos Ž u . sin Ž u . cos Ž t . sin Ž t . . We want to prove that ycos Ž u . sin Ž u . cos Ž t . sin Ž t . F 1 y cos Ž u . cos Ž t . and that equality implies u s t s 0. Set x s cosŽ u . g Ž0, 1x and y s cosŽ t . g wy1, 1x. Then " '1 y x 2 s sinŽ u . and " 1 y y 2 s sinŽ t .. We want to show
'
"xy'1 y x 2
'1 y y
2
F 1 y xy.
For y1 F y - 0 and 0 - x F 1 it is elementary to see that strict inequality holds, so we need only consider 0 F y F 1 and 0 F x F 1 and show xy'1 y x 2
'1 y y
2
F 1 y xy
with strict inequality unless x s y s 1. This is equivalent to G Ž x, y . s 1 y 2 xy q x 4 y 2 q x 2 y 4 y x 4 y 4 being nonnegative on the unit square 0 F x, y F 1 and GŽ x, y . s 0 only for x s y s 1. Now, G Ž 0, y . s 1 ) 0 2
G Ž 1, y . s 1 y 2 y q y 2 s Ž 1 y y . G 0 G Ž x, 0 . s 1 ) 0 2
G Ž x, 1 . s 1 y 2 x q x 2 s Ž 1 y x . G 0, so GŽ x, y . G 0 on the boundary of the square with equality only at Ž1, 1.. It is not difficult to show that if
G x
Ž x, y . s 0 s
G y
Ž x, y .
at an interior point of the square, then y s x. Since 2
G Ž x, x . s Ž 1 y x 2 . Ž 1 y x 4 . ) 0
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
545
for 0 - x - 1, we conclude that GŽ x, y . G 0 for 0 F x, y F 1 and equality implies x s y s 1. This proves that H Ž1. F 4 with equality only for u s t s 0. All that remains is to show H Ž r . - 4 for 0 - r - 1. If H X Ž r . / 0 for 0 - r - 1, we are done. Now, H X Ž r . s 4 cos Ž u . cos Ž t . y r Ž 1 q 2 sin Ž u . cos Ž t . sin Ž t . . . If H X Ž t 0 . s 0, then r0 s
cos Ž t . 1 q 2 sin Ž u . cos Ž t . sin Ž t .
.
Suppose r 0 g Ž0, 1.. Then H Ž r 0 . s 2 cos Ž u . 1 q r 0 cos Ž u . - 4 since r 0 - 1. This proves that H Ž r . F 4 and equality implies r s 1 and u s t s 0. It follows that N Ž p . F 4 and equality implies u s 0 and pŽ z . s Ž1 q z .rŽ1 y z .. Next, we prove Theorem 1Žii. for a special subclass of SCCŽ a .. For u g Žypr2, pr2. and s, t g Žyp , p x set
LEMMA 2.
ws Ž z . s qs Ž z . s 1 q
z 1 y eisz
z wsY Ž z .
wsX Ž z .
s
g CV, 1 q eisz 1 y eisz
s 1 q 2 e i s z q 2 e 2 i s z 2 q ??? g P and pt Ž z . s
1 q e iŽ ty u . z 1 y e iŽ ty u . z
s 1 q 2 e iŽ ty u . z q 2 e 2 iŽ ty u . z 2 q ??? g P .
If f s, t Ž z . s z q a2 z 2 q a3 z 3 q ??? g SCCŽ a . is determined from f s,X t Ž z . s wsX Ž z . e i u cos Ž u . pt Ž z . y i sin Ž u .
4
a
,
then 3a3 y 2 a22 q 2 < a2 < 2 F 3 q 4a q 2 a 2 and equality holds if and only if u s 0 and s s t; that is, if and only if f s, t is a rotation of Fa , f s, s Ž z . s eyi s Fa Ž e i s z .
546
MA AND MINDA
Proof. For f s, t we have c1 s 2 e i s
d1 s 2 e iŽ ty u .
c2 s 2 e 2 i s
d 2 s 2 e 2 iŽ ty u . .
Then formulas Ž6. and Ž7. give 3a3 y 2 a22 s e 2 i s q 2 a cos Ž u . e iŽ2 ty u . y 2 a cos 2 Ž u . e 2 it s e 2 i s 1 y 2 i a cos Ž u . sin Ž u . e 2 iu and 2 < a2 < 2 s 2 q 4a cos Ž u . Re e iŽ tys. 4 q 2 a 2 cos 2 Ž u . s 2 1 q 2 a cos Ž u . cos Ž u . q a 2 cos 2 Ž u . , where u s t y s. Note that 3a3 y 2 a22
2
s 1 q 4a cos Ž u . sin Ž u . sin Ž 2 u . q 4a 2 cos 2 Ž u . sin 2 Ž u . . Ž 9 .
We wish to prove 3a3 y 2 a22 F 3 q 4a q 2 a 2 y 2 < a2 < 2 . Since f s, t g SCCŽ a ., < a2 < F 1 q a and so the right-hand side of the preceding inequality is positive. Therefore, it is enough to show 3a3 y 2 a22
2
F Ž 3 q 4a q 2 a 2 y 2 < a2 < 2 .
2
s 1 q 8 a 1 y cos Ž u . cos Ž u . q 4a 2 sin 2 Ž u . q 4 Ž 1 y cos Ž u . cos Ž u . .
2
q 16 a 3 sin 2 Ž u . 1 y cos Ž u . cos Ž u . q 4a 4 sin 4 Ž u . . From Ž9. we see that it is sufficient to show cos Ž u . sin Ž u . sin Ž 2 u . q a cos 2 Ž u . sin 2 Ž u . F 2 1 y cos Ž u . cos Ž u . q a sin 2 Ž u . q 4 Ž 1 y cos Ž u . cos Ž u . . q 4a 2 sin 2 Ž u . 1 y cos Ž u . cos Ž u . q a 3 sin 4 Ž u .
2
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
547
and that equality implies u s 0 and u s 0 Žthat is, s s t .. It is elementary that
a cos 2 Ž u . sin 2 Ž u . F a sin 2 Ž u . q 4 Ž 1 y cos Ž u . cos Ž u . .
2
q 4a 2 sin 2 Ž u . 1 y cos Ž u . cos Ž u . q a 3 sin 4 Ž u . and equality implies u s 0 and u s 0. Hence, all that remains is to verify that cos Ž u . sin Ž u . sin Ž 2 u . F 2 1 y cos Ž u . cos Ž u . or cos Ž u . sin Ž u . cos Ž u . sin Ž u . F 1 y cos Ž u . cos Ž u . . If we replace u by yu , then we see that this inequality was established in the proof of Lemma 1 and equality implies u s 0 s u. This completes the proof of Lemma 2. COROLLARY.
For u g Žypr2, pr2., s, t g Žyp , p x, and 0 - a F 1,
cos Ž 2 s . q 4a cos Ž u . cos Ž t y s . q 2 a cos Ž u . sin Ž u . sin Ž 2 t . q 2 a 2 cos 2 Ž u . F 1 q 4a q 2 a 2 . Equality holds if and only if u s 0 and s s t s 0, p . Proof. Since Theorem 1Žii. holds for f s, t , we conclude that M Ž f s, t . F 3a3 y 2 a22 q 2 < a2 < 2 F 3 q 4a q 2 a 2 . From formula Ž8. we find that M Ž f s, t . s 2 q cos Ž 2 s . q 4a cos Ž u . cos Ž t y s . q 2 a cos Ž u . cos Ž 2 t y u . y cos Ž u . cos Ž 2 t . q 2 a 2 cos Ž u . s 2 q cos Ž 2 s . q 4a cos Ž u . cos Ž t y s . q 2 a cos Ž u . sin Ž u . sin Ž 2 t . q 2 a 2 cos 2 Ž u . . This proves the desired inequality. If equality holds, then equality must hold in Lemma 2 so u s 0 and s s t. Also, Re 3a3 y 2 a22 4 s 3a3 y 2 a22 . For u s 0 and s s t we have 3a3 y 2 a22 s e 2 i s, so the preceding equality implies s s t s 0 or s s t s p .
548
MA AND MINDA
LEMMA 3. 2. Then
Suppose u g Žypr2, pr2., s g Žyp , p x, a ) 0, and < d1 < F
cos Ž 2 s . q 2 a Ž cos Ž u . Re e iŽ uys. d1 4 q 2 a cos Ž u . . y 12 a cos Ž u . < d1 < 2 q 12 a 2 cos 2 Ž u . < d1 < 2 q 12 a cos Ž u . sin Ž u . Im
½Že
iu
d1 .
2
5 F 1 q 4a q 2 a
2
.
Equality implies u s 0, and either s s 0 and d1 s 2 or s s p and d1 s y2. Proof. Since < e i u d1 < s < d1 <, it suffices to show cos Ž 2 s . q 2 a cos Ž u . Re eyi sd1 4 q 2 a cos Ž u . y 12 a cos Ž u . < d1 < 2 q 21 a 2 cos Ž u . < d1 < 2 q 12 a cos Ž u . sin Ž u . Im d12 4 F 1 q 4a q 2 a 2 . Set d1 s 2 re it , where 0 F r F 1 and t g Žyp , p x. Then we want to show cos Ž 2 s . q 4a r cos Ž u . cos Ž s y t . q 2 a cos Ž u . y 2 a r 2 cos Ž u . q 2 a 2 r 2 cos 2 Ž u . q 2 a r 2 cos Ž u . sin Ž u . sin Ž 2 t . F 1 q 4a q 2 a 2 . This inequality will hold if we can show H Ž r . s cos Ž 2 s . q 4a r cos Ž u . cos Ž s y t . q 2 a cos Ž u . y 2 a r 2 cos Ž u . q 2 a 2 cos 2 Ž u . q 2 a r 2 cos Ž u . sin Ž u . sin Ž 2 t . F 1 q 4a q 2 a 2 for 0 F r F 1, with strict inequality unless r s 1 and either s s t s 0 or s s t s p. Now H Ž 0 . s cos Ž 2 s . q 2 a cos Ž u . q 2 a 2 cos 2 Ž u . F 1 q 2 a q 2 a 2 - 1 q 4a q 2 a 2 and H Ž 1 . s cos Ž 2 s . q 4a cos Ž u . cos Ž s y t . q 2 a 2 cos 2 Ž u . q2 a cos Ž u . sin Ž u . sin Ž 2 t . F 1 q 4a q 2 a 2
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
549
by the Corollary to Lemma 2 with the proper conditions for equality. All that remains is to prove H Ž r . - 1 q 4a q 2 a 2 for 0 - r - 1. If H X Ž r . / 0 for 0 - r - 1, we are done. Suppose H X Ž r 0 . s 0 for some r 0 g Ž0, 1.. Then r0 s
cos Ž s y t . 1 y sin Ž u . sin Ž 2 t .
and as 0 - r 0 - 1 H Ž r 0 . s cos Ž 2 s . q 2 a cos Ž u . q 2 a 2 cos 2 Ž u . q 2 a r 0 cos Ž u . cos Ž s y t . F cos Ž 2 s . q 2 a cos Ž u . q 2 a 2 cos 2 Ž u . q 2 a r 0 cos Ž u . F 1 q 2 a Ž 1 q r0 . q 2 a 2 - 1 q 4a q 2 a 2 . This completes the proof.
4. PROOF OF THEOREM 1Ži. Because SCCŽ a . is rotationally invariant, it suffices to show that f g SCCŽ a . implies L Ž f . F 13 Ž 1 q 4a q 2 a 2 .
Ž 10 .
and equality implies f Ž z . s Fa Ž z . or f Ž z . s yFa Žyz .. We begin by showing that we may assume Re e i u d1 4 G 0 in the expression Ž5. for LŽ f .. Suppose Re e i u d14 - 0. Note that g Ž z . s yf Žyz . g SCCŽ a .. Also, g X Ž z . s f X Ž yz . s w X Ž yz . e i u Ž cos Ž u . p Ž yz . y i sin Ž u . .
a
where yw Ž yz . s z y b1 z q b 2 z 2 q ??? and p Ž yz . s 1 y d1 z q d 2 z 2 q ??? . Now, Re e i u Žyd1 .4 ) 0 and LŽ f . s LŽ g .; we replace f by g Ž z . s yf Žyz . when Re e i u d14 - 0. Thus, we want to prove Ž10. under the restriction Re e i u d14 G 0 and show that equality implies f s Fa .
550
MA AND MINDA
Because w g CV, we have w7x Re b 22 y b 3 4 q 13 < b 2 < 2 F
1 3
Ž 11 .
and Re b 2 4 F < b 2 < F 1. Note that Re b 2 4 s 1 if and only if pŽ z . s zrŽ1 y z .. Since Re e i u d1 4 G 0, we obtain LŽ f . F
1 3 q q
2a
q
a 3
3
cos Ž u . Re e i u d1 4
cos Ž u . Re
a2
½
1 2
2
cos Ž u . Ž e i u d1 . y e i u d 2
cos 2 Ž u . < d1 < 2 q Re
12
½Že
iu
d1 .
2
5
5
.
By Lemma 1, cos Ž u . 2 Re e i u d1 4 q Re
½
1 2
2
cos Ž u . Ž e i u d1 . y e i u d 2
5
F4
and equality implies u s 0 and pŽ z . s Ž1 q z .rŽ1 y z .. Since p g P, < d1 < F 2. Thus, 1
4a
2
3
q
3
q
a 2 cos 2 Ž u . F
1
Ž 1 q 4a q 2 a 2 . . 3 3 Equality implies u s 0, pŽ z . s Ž1 q z .rŽ1 y z . and w Ž z . s zrŽ1 y z ., that is, f s Fa . Conversely, if f s Fa it is straightforward to verify that equality holds. LŽ f . F
5. PROOF OF THEOREM 1Žii. Since SCCŽ a . is rotationally invariant, it is enough to prove that f g SCCŽ a . implies M Ž f . F 3 q 4a q 2 a 2 with equality if and only if either f Ž z . s Fa Ž z . or f Ž z . s yFa Žyz .. We begin by bounding M Ž f . by a quantity depending only on c1 and d1. Since q g P w4x, d 2 y 12 d12 F 2 y 12 < d1 < 2 and so Re e i u d 2 4 F 2 y 12 < d1 < 2 q
1 2
Re e i u d12 4 .
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
551
Then 2
Re y 12 cos Ž u . Ž e i u d1 . q e i u d 2 F 2 y 12 < d1 < 2
½
5
q 12 sin Ž u . Im
½Že
iu
d1 .
2
5.
This yields MŽ f . F
1
Re c 2 4 q
2
1 2
< c1 < 2 q a cos Ž u . Re e i u c1 d1 4
q 2 a cos Ž u . y
a
q
2
a 2
cos Ž u . < d1 < 2
cos Ž u . sin Ž u . Im
½
Ž e i u d1 .
2
q
5
a2 2
cos 2 Ž u . < d1 < 2 .
Similarly, p g P so c 2 y 12 c12 F 2 y 12 < c1 < 2 , or Re c 2 4 F 2 y 12 < c 2 < 2 q 1 2
1 2
Re c12 4 ,
Re c 2 4 q 12 < c1 < 2 F 1 q 41 < c1 < 2 q s1q
1 2
1 4
Re c12 4
Re 2 c1 4 .
Thus, MŽ f . F 1 q
1 2
Re 2 c1 4 q a cos Ž u . Re e i u c1 d1 4
q 2 a cos Ž u . y q
a 2
a 2
cos Ž u . < d1 < 2
cos Ž u . sin Ž u . Im
½ Ž e i u d1 .
2
5
a2
q
2
˜Ž f . . cos 2 Ž u . < d1 < 2 s M
˜Ž f . F 3 q 4a q 2 a 2 when c1 s 2 e i s. For c1 s 2 e i s, Next, we show M ˜ Ž f . s 1 q 2 cos 2 Ž s . q 2 a cos Ž u . Re e iŽ uys. d1 4 M q 2 a cos Ž u . y q
a 2
a 2
cos Ž u . < d1 < 2
cos Ž u . sin Ž u . Im
½
Ž e i u d1 .
2
5
q
a2 2
cos 2 Ž u . < d1 < 2 .
552
MA AND MINDA
Since 1 q 2 cos 2 Ž s . s 2 q cos Ž 2 s . ,
˜Ž f . F 3 q 4a q 2 a 2 follows immediately from Lemma 3. the inequality M Equality implies u s 0 and either s s 0 Ž c1 s 2. and d1 s 2 or s s p Ž c1 s y2. and d1 s y2. This means either f Ž z . s Fa Ž z . or f Ž z . s yFa Žyz .. ˜Ž f . - 3 q 4a q 2 a 2 for < c1 < - 2. All that remains is to demonstrate M ˜Ž f . is a subharmonic function of c1 which is nonconstant and Since M ˜Ž f . F 3 q 4a q 2 a 2 for < c1 < s 2, we get strict inequality for < c1 < - 2. M 6. INVARIANT FORMULATION A not necessarily normalized analytic function f on D is called strongly close-to-convex of order a if there exists a not necessarily normalized convex univalent function w on D such that arg
fXŽ z. X
w Ž z.
-
ap 2
for z g D. This class of functions is linearly invariant in the sense that f Ž T Ž z . . y f Ž T Ž 0. . f X Ž T Ž 0. . T X Ž 0.
g SCC Ž a .
for any conformal automorphism T of D. Pommerenke w6x noted this linear invariance and gave a simple external geometric characterization when 0 - a F 1. There is an invariant formulation of Theorem 1 for nonnormalized strongly close-to-convex functions of order a . Recall the differential operators w3x D1 f Ž z . s Ž1 y < z < 2 . f X Ž z . 2
D2 f Ž z . s Ž1 y < z < 2 . f Y Ž z . y 2 z Ž1 y < z < 2 . f X Ž z . 3
2
D3 f Ž z . s Ž1 y < z < 2 . f Z Ž z . y 6 z Ž1 y < z < 2 . f Y Ž z . q 6 z 2 Ž1 y < z < 2 . f X Ž z . . Note that Dj f Ž0. s f Ž j. Ž0. Ž j s 1, 2, 3.. These differential operators are invariants in the sense that < Dj Ž S( f (T .< s < Dj Ž f .(T < for all Euclidean
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
553
motions S of C and conformal automorphisms T of D. The Schwarzian derivative Sf Ž z . s
fZ Ž z. fXŽ z.
3
y
2
ž
fYŽ z. fXŽ z.
2
/
is related by D3 f Ž z . D1 f Ž z . THEOREM 2.
y
3 2
D2 f Ž z .
ž
D1 f Ž z .
2
/
2
s Ž1 y < z < 2 . S f Ž z . .
Suppose f is strongly close-to-con¨ ex of order a on D. Then 1 D2 f Ž z .
Ž1 y < z < 2 . < S f Ž z . < q
2 D1 f Ž z .
2
F 2 Ž 1 q 4a q 2 a 2 .
and D3 f Ž z . D1 f Ž z .
y
ž
D2 f Ž z . D1 f Ž z .
2
/
q
D2 f Ž z . D1 f Ž z .
2
F 2 Ž 3 q 4a q 2 a 2 . .
These inequalities are both sharp; equality holds if and only if f s S( fa (T where S is a conformal automorphism of C and T is a conformal automorphism of D. Proof. For z s 0 these inequalities reduce to Theorem 1. The general case follows from the invariance of the differential operators together with the linear invariance of the class of functions.
REFERENCES 1. D. A. Brannan, J. G. Clunie, and W. E. Kirwan, On the coefficient problem for functions of bounded rotation, Ann. Acad. Sci. Fenn. Ser. A Vol. 1, No. 523 Ž1973.. 2. A. W. Goodman, ‘‘Univalent Functions,’’ Vol. II, Polygonal, Washington, NJ, 1983. 3. S. Kim and D. Minda, Two-point distortion theorems for univalent functions, Pacific J. Math. 163 Ž1994., 137]157. 4. W. Ma and D. Minda, Uniformly convex functions, II, Ann. Polon. Math. 58 Ž1993., 275]285. 5. W. Ma and D. Minda, Two-point distortion theorems for strongly close-to-convex functions, Complex Variables Theory Appl., to appear. 6. Ch. Pommerenke, On close-to-convex analytic functions, Trans. Amer. Math. Soc. 114 Ž1965., 176]186. 7. S. Y. Trimble, A coefficient inequality for convex univalent functions, Proc. Amer. Math. Soc. 48 Ž1975., 266]267.
205, 537]553 Ž1997.
AY975234
Coefficient Inequalities for Strongly Close-to-Convex Functions William Ma Integrated Studies Di¨ ision, Pennsyl¨ ania College of Technology, Williamsport, Pennsyl¨ ania 17701]5799
and David MindaU Department of Mathematical Sciences, Uni¨ ersity of Cincinnati, P.O. Box 210025, Cincinnati, Ohio 45221-0025 Submitted by H. M. Sri¨ asta¨ a Received December 20, 1995
Let f Ž z . s z q a2 z 2 q a3 z 3 q ??? be a normalized strongly close-to-convex function of order a ) 0 defined on the unit disk D. This means that there is a normalized convex univalent function w and b g R such that X
arg
f Ž z.
-
ib X
e w Ž z.
ap 2
for z g D. Then < a3 y a22 < q 13 < a x < 2 F 13 Ž 1 q 4a q 2 a 2 . and a3 y 23 a22 q 23 < a2 < 2 F
1 3
Ž 3 q 4a q 2 a 2 .
with equality if and only if f is a rotation of Fa Ž z . s
1 2 Ž1 q a .
ž
1qz 1yz
1q a
/
y1 .
Q 1997 Academic Press
* Research partially supported by a National Science Foundation grant. 537 0022-247Xr97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
538
MA AND MINDA
1. INTRODUCTION A holomorphic function f defined on the unit disk D s z : < z < - 14 is called strongly close-to-convex of order a ) 0 provided f is normalized Ž f Ž0. s 0, f X Ž0. s 1. and there exist b g R and a normalized convex univalent function w such that arg
fXŽ z. ib X
e w Ž z.
-
ap 2
for z g D. Let SCCŽ a ., 0 F a F 1, denote the class of all strongly closeto-convex functions of order a . The usual class of normalized close-to-convex functions is SCCŽ1.. For a s 0 we need to replace strict inequality by ‘‘less than or equal to.’’ Then SCCŽ0. s CV, the class of normalized convex univalent functions. Then SCCŽ0. ; SCCŽ a . for a ) 0. For a g w0, 1x all functions in SCCŽ a . are univalent. The function Fa Ž z . s
1 2Ž 1 q a .
ž
1qz 1yz
s z q Ž1 q a . z2 q
1q a
y1
/
1 3
Ž 3 q 4a q 2 a 2 . z 3 q ???
belongs to SCCŽ a . and is known to be extremal for a number of problems. For instance, if f Ž z . s z q a2 z 2 q a3 z 3 q ??? g SCCŽ a ., then < a2 < F 1 q a and < a3 < F 13 Ž 3 q 4a q 2 a 2 . . Equality holds in either quality if and only if f Ž z . s eyi u Fa Ž e i u z . for some u g R; that is, f is a rotation of Fa . See w2, Chap. 11x for the interesting history of the coefficient problem for SCCŽ a .. The goal of this paper is to establish two sharp coefficient inequalities for the class SCCŽ a .. The authors were led to such inequalities by their recent work relating to two-point distortion theorems for univalent function w5x. These inequalities yield sharp two-point distortion theorems for nonnormalized strongly close-to-convex functions of order a g w0, 1x. Our main result is the following theorem.
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
539
Suppose f Ž z . s z q a2 z 2 q a2 z 3 q ??? g SCCŽ a .. Then
THEOREM 1.
Ž i.
a3 y a22 q 13 < a2 < 2 F 13 Ž 1 q 4a q 2 a 2 .
and
Ž ii .
a3 y 23 a22 q 23 < a2 < 2 F 13 Ž 3 q 4a q 2 a 2 . .
Equality holds if and only if f is a rotation of Fa . Inequality Žii. implies the sharp bound on < a3 < over the class SCCŽ a .. For a s 0 the inequality Ži. was established for the class CV by Trimble w7x. Also, for a s 0 the coefficient bound < a2 < F 1 for the class CV together with inequality Ži. yields inequality Žii.. For a ) 0 the second inequality does not seem to be a consequence of the first. 2. REPRESENTATION FORMULA FOR SCCŽ a . A basic tool in the proof of our main theorem is an elementary representation for derivatives of functions in SCCŽ a . that was already noted in w1x. If f g SCCŽ a ., then w f X Ž z .rŽ e i bw X Ž z ..x1r a has positive real part in D if we select the branch at the origin with value eyi b r a . The fact that this function has positive real part in D implies that cosŽ bra . ) 0, so we may assume that u s bra g Žypr2, pr2.. Then pŽ z . s
1 cos Ž u .
ž
1r a
fXŽ z. e i bw X Ž z .
q i sin Ž u .
/
belongs to the class P of normalized Ž pŽ0. s 1. holomorphic functions defined on D with positive real part. Thus, if f g SCCŽ a ., then f X Ž z . s e i auw X Ž z . cos Ž u . p Ž z . y i sin Ž u . s w X Ž z . e i u cos Ž u . p Ž z . y i sin Ž u .
a
4
a
,
where a ) 0, < u < - pr2, w g CV, and p g P. Conversely, if f X has this form, then f g SCCŽ a .. For future reference we note the explicit representation of the function Fa and its rotations. For eyi u Fa Ž e i u . we have FaX Ž e i u z . s w X Ž e i u z . p Ž e i u z . s
1
Ž 1 y e iu z .
2
ž
a
1 q e iu z 1 y e iu z
a
/
,
where w Ž z . s zrŽ1 y z . g CV and pŽ z . s Ž1 q z .rŽ1 y z . g P.
540
MA AND MINDA
From the representation formula we derive formulas for the second and third coefficients in the power series of functions in SCCŽ a . in terms of coefficients of functions in P and CV. Suppose f Ž z . s z q a2 z 2 q a3 z 3 q ??? ,
w Ž z . s z q b 2 z 2 q b 3 z 3 q ??? , and p Ž z . s 1 q d1 z q d 2 z 2 q ??? . Then
e iu
cos Ž u . p Ž z . y i sin Ž u .
4
a
s 1 q e i u cos Ž u . p Ž z . y 1
4
a
s 1 q a cos Ž u . e i u d1 z q a cos Ž u . e i u d 2 q
a Ž a y 1. 2
cos 2 Ž u . e 2 i u d12 z 2 q ??? .
By using the representation formula we obtain 1 q 2 a2 z q 3a3 z 2 q ??? s fXŽ z. s Ž 1 q 2 b 2 z q 3b 3 z 3 q ??? .Ž 1 q a cos Ž u . e i u d1 z q ??? . s 1 q 2 b 2 q a cos Ž u . e i u d1 z q 3b 3 q a cos Ž u . e i u d 2 q
a Ž a y 1. 2
cos 2 Ž u . e 2 i u d12
q2 a cos Ž u . e i u b 2 d1 z 2 q ??? . Consequently, 2 a2 s 2 b 2 q a cos Ž u . e i u d1
Ž 1.
and 3a3 s 3b 3 q 2 a cos Ž u . e i u b 2 d1 q a cos Ž u . e i u d 2 q
a Ž a y 1. 2
cos 2 Ž u . e 2 i u d12 .
Ž 2.
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
541
Next, we make use of the fact that qŽ z. s 1 q
zwY Ž z .
wX Ž z .
s 1 q c1 z q c 2 z 2 q ???
belongs to P to express a2 and a3 in terms of the coefficients of two functions in P. From 1q
zwY Ž z .
wX Ž z .
s 1 q 2 b 2 z q Ž 6 b 3 y 4 b 22 . z 2 q ??? ,
we obtain c1 s 2 b 2 , c 2 s 6 b 3 y 4 b 22 , or 2 b 2 s c1 , 6 b 3 s c12 q c 2 . If we substitute these expressions into Ž1. and Ž2. we obtain 2 a2 s c1 q a cos Ž u . e i u d1 , 3a3 s
1 2
Ž 3.
Ž c2 q c12 . q a cos Ž u . e iu c1 d1
q a cos Ž u . e i u d 2 q
a Ž a y 1. 2
cos 2 Ž u . e 2 i u d12 .
Ž 4.
For f Ž z . s z q a2 z 2 q a3 z 3 q ??? g SCCŽ a . the two functionals L Ž f . s Re a22 y a3 4 q 13 < a2 < 2 and M Ž f . s Re 3a3 y 2 a22 4 q 2 < a2 < 2 play important roles in the proof of Theorem 1. We use Ž1. and Ž2. to express LŽ f . in terms of b1 , b 2 , d1 , and d 2 . Now a a22 y a3 s b 22 y b 3 q cos Ž u . e i u b 2 d1 3 a y cos Ž u . e i u d 2 3 q
a Ž a q 2. 12
cos 2 Ž u . e 2 i u d12
542
MA AND MINDA
and 1 3
< a2 < 2 s
1 3
< b2 < 2 q
q
a 3
a2 12
cos 2 Ž u . < d1 < 2
cos Ž u . Re e i u b 2 d1 4 ,
so that L Ž f . s Re b 22 y b 3 4 q q q
a 3
cos Ž u . Re
a2 12
1 3
½
< b2 < 2 q
aq2 4
2a 3
cos Ž u . Re b 2 4 Re e i u d1 4 2
cos Ž u . Ž e i u d1 . y e i u d 2
5
cos 2 Ž u . < d1 < 2 .
Ž 5.
Similarly, we use Ž3. and Ž4. to obtain an expression for M Ž f . in terms of c1 , c 2 , d1 , and d 2 . From 3a3 y 2 a22 s
1 2
c 2 q a cos Ž u . e i u d 2 y
a 2
cos 2 Ž u . e 2 i u d12
Ž 6.
and 2 < a2 < 2 s
1 2
< c1 < 2 q a cos Ž u . Re e i u c1 d1 4 q
a2 2
cos 2 Ž u . < d1 < 2 ,
Ž 7.
we obtain MŽ f . s
1 2
Re c 2 4 q
1 2
< c1 < 2 q a cos Ž u . Re e i u c1 d1 4
½
q a cos Ž u . Re y q
a2 2
1 2
cos Ž u . e 2 i u d12 q e i u d 2
cos 2 Ž u . < d1 < 2 .
5 Ž 8.
3. TECHNICAL RESULTS In this section we gather together several preliminary estimates that we need for our proof of Theorem 1.
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
543
LEMMA 1. Suppose pŽ z . s 1 q d1 z q d 2 z 2 q ??? g P and u g Žypr2, pr2.. Then cos Ž u . 2 Re e i u d1 4 q Re
½
1 2
2
cos Ž u . Ž e i u d1 . y e i u d 2
5
F 4.
Equality holds if and only if u s 0 and pŽ z . s Ž1 q z .rŽ1 y z .. Proof. Since pŽ z . g P if and only if p Ž eyi u z . s 1 q eyi u d1 z q ey2 i u d 2 z 2 q ??? belongs to P, it suffices to show that N Ž p . s cos Ž u . 2 Re d1 4 q Re 12 cos Ž u . d12 y eyi u d 2 4 F 4 and equality implies u s 0 and pŽ z . s Ž1 q z .rŽ1 y z .. Next, we eliminate d 2 in N Ž p .. Since p g P w4x, d 2 y 12 d12 F 2 y 12 < d1 < 2 . This yields yRe eyi u d 2 4 F 2 y 12 < d1 < 2 y 21 Re eyi u d12 4 , so that Re 12 cos Ž u . d12 y eyi u d 2 4 F 2 y 12 < d1 < 2 q 21 Re Ž cos Ž u . y eyi u . d12 4 s 2 y 12 < d1 < 2 y 21 sin Ž u . Im d12 4 . Thus, N Ž p . F cos Ž u . 2 Re d1 4 q 2 y 12 < d1 < 2 y 12 sin Ž u . Im d12 4 . Recall that if p g P, then < d1 < F 2. Thus, we wish to show that cos Ž u . 2 Re d1 4 q 2 y 12 < d1 < 2 y 12 sin Ž u . Im d12 4 F 4 for < d1 < F 2. Set d1 s 2 re it , where 0 F r F 1 and t g Žyp , p x. Then we want to demonstrate that H Ž r . s cos Ž u . 4 r cos Ž t . q 2 y r 2 Ž 2 q 2 sin Ž u . sin Ž 2 t . . s 2 cos Ž u . 2 r cos Ž t . q 1 y r 2 Ž 1 q 2 sin Ž u . cos Ž t . sin Ž t . . F 4
544
MA AND MINDA
for 0 F r F 1. Note that H Ž 0 . s 2 cos Ž u . F 2 - 4. Next, we show that H Ž1. F 4 with equality only for u s t s 0. H Ž 1 . s 4 cos Ž u . cos Ž t . y 4 cos Ž u . sin Ž u . cos Ž t . sin Ž t . . We want to prove that ycos Ž u . sin Ž u . cos Ž t . sin Ž t . F 1 y cos Ž u . cos Ž t . and that equality implies u s t s 0. Set x s cosŽ u . g Ž0, 1x and y s cosŽ t . g wy1, 1x. Then " '1 y x 2 s sinŽ u . and " 1 y y 2 s sinŽ t .. We want to show
'
"xy'1 y x 2
'1 y y
2
F 1 y xy.
For y1 F y - 0 and 0 - x F 1 it is elementary to see that strict inequality holds, so we need only consider 0 F y F 1 and 0 F x F 1 and show xy'1 y x 2
'1 y y
2
F 1 y xy
with strict inequality unless x s y s 1. This is equivalent to G Ž x, y . s 1 y 2 xy q x 4 y 2 q x 2 y 4 y x 4 y 4 being nonnegative on the unit square 0 F x, y F 1 and GŽ x, y . s 0 only for x s y s 1. Now, G Ž 0, y . s 1 ) 0 2
G Ž 1, y . s 1 y 2 y q y 2 s Ž 1 y y . G 0 G Ž x, 0 . s 1 ) 0 2
G Ž x, 1 . s 1 y 2 x q x 2 s Ž 1 y x . G 0, so GŽ x, y . G 0 on the boundary of the square with equality only at Ž1, 1.. It is not difficult to show that if
G x
Ž x, y . s 0 s
G y
Ž x, y .
at an interior point of the square, then y s x. Since 2
G Ž x, x . s Ž 1 y x 2 . Ž 1 y x 4 . ) 0
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
545
for 0 - x - 1, we conclude that GŽ x, y . G 0 for 0 F x, y F 1 and equality implies x s y s 1. This proves that H Ž1. F 4 with equality only for u s t s 0. All that remains is to show H Ž r . - 4 for 0 - r - 1. If H X Ž r . / 0 for 0 - r - 1, we are done. Now, H X Ž r . s 4 cos Ž u . cos Ž t . y r Ž 1 q 2 sin Ž u . cos Ž t . sin Ž t . . . If H X Ž t 0 . s 0, then r0 s
cos Ž t . 1 q 2 sin Ž u . cos Ž t . sin Ž t .
.
Suppose r 0 g Ž0, 1.. Then H Ž r 0 . s 2 cos Ž u . 1 q r 0 cos Ž u . - 4 since r 0 - 1. This proves that H Ž r . F 4 and equality implies r s 1 and u s t s 0. It follows that N Ž p . F 4 and equality implies u s 0 and pŽ z . s Ž1 q z .rŽ1 y z .. Next, we prove Theorem 1Žii. for a special subclass of SCCŽ a .. For u g Žypr2, pr2. and s, t g Žyp , p x set
LEMMA 2.
ws Ž z . s qs Ž z . s 1 q
z 1 y eisz
z wsY Ž z .
wsX Ž z .
s
g CV, 1 q eisz 1 y eisz
s 1 q 2 e i s z q 2 e 2 i s z 2 q ??? g P and pt Ž z . s
1 q e iŽ ty u . z 1 y e iŽ ty u . z
s 1 q 2 e iŽ ty u . z q 2 e 2 iŽ ty u . z 2 q ??? g P .
If f s, t Ž z . s z q a2 z 2 q a3 z 3 q ??? g SCCŽ a . is determined from f s,X t Ž z . s wsX Ž z . e i u cos Ž u . pt Ž z . y i sin Ž u .
4
a
,
then 3a3 y 2 a22 q 2 < a2 < 2 F 3 q 4a q 2 a 2 and equality holds if and only if u s 0 and s s t; that is, if and only if f s, t is a rotation of Fa , f s, s Ž z . s eyi s Fa Ž e i s z .
546
MA AND MINDA
Proof. For f s, t we have c1 s 2 e i s
d1 s 2 e iŽ ty u .
c2 s 2 e 2 i s
d 2 s 2 e 2 iŽ ty u . .
Then formulas Ž6. and Ž7. give 3a3 y 2 a22 s e 2 i s q 2 a cos Ž u . e iŽ2 ty u . y 2 a cos 2 Ž u . e 2 it s e 2 i s 1 y 2 i a cos Ž u . sin Ž u . e 2 iu and 2 < a2 < 2 s 2 q 4a cos Ž u . Re e iŽ tys. 4 q 2 a 2 cos 2 Ž u . s 2 1 q 2 a cos Ž u . cos Ž u . q a 2 cos 2 Ž u . , where u s t y s. Note that 3a3 y 2 a22
2
s 1 q 4a cos Ž u . sin Ž u . sin Ž 2 u . q 4a 2 cos 2 Ž u . sin 2 Ž u . . Ž 9 .
We wish to prove 3a3 y 2 a22 F 3 q 4a q 2 a 2 y 2 < a2 < 2 . Since f s, t g SCCŽ a ., < a2 < F 1 q a and so the right-hand side of the preceding inequality is positive. Therefore, it is enough to show 3a3 y 2 a22
2
F Ž 3 q 4a q 2 a 2 y 2 < a2 < 2 .
2
s 1 q 8 a 1 y cos Ž u . cos Ž u . q 4a 2 sin 2 Ž u . q 4 Ž 1 y cos Ž u . cos Ž u . .
2
q 16 a 3 sin 2 Ž u . 1 y cos Ž u . cos Ž u . q 4a 4 sin 4 Ž u . . From Ž9. we see that it is sufficient to show cos Ž u . sin Ž u . sin Ž 2 u . q a cos 2 Ž u . sin 2 Ž u . F 2 1 y cos Ž u . cos Ž u . q a sin 2 Ž u . q 4 Ž 1 y cos Ž u . cos Ž u . . q 4a 2 sin 2 Ž u . 1 y cos Ž u . cos Ž u . q a 3 sin 4 Ž u .
2
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
547
and that equality implies u s 0 and u s 0 Žthat is, s s t .. It is elementary that
a cos 2 Ž u . sin 2 Ž u . F a sin 2 Ž u . q 4 Ž 1 y cos Ž u . cos Ž u . .
2
q 4a 2 sin 2 Ž u . 1 y cos Ž u . cos Ž u . q a 3 sin 4 Ž u . and equality implies u s 0 and u s 0. Hence, all that remains is to verify that cos Ž u . sin Ž u . sin Ž 2 u . F 2 1 y cos Ž u . cos Ž u . or cos Ž u . sin Ž u . cos Ž u . sin Ž u . F 1 y cos Ž u . cos Ž u . . If we replace u by yu , then we see that this inequality was established in the proof of Lemma 1 and equality implies u s 0 s u. This completes the proof of Lemma 2. COROLLARY.
For u g Žypr2, pr2., s, t g Žyp , p x, and 0 - a F 1,
cos Ž 2 s . q 4a cos Ž u . cos Ž t y s . q 2 a cos Ž u . sin Ž u . sin Ž 2 t . q 2 a 2 cos 2 Ž u . F 1 q 4a q 2 a 2 . Equality holds if and only if u s 0 and s s t s 0, p . Proof. Since Theorem 1Žii. holds for f s, t , we conclude that M Ž f s, t . F 3a3 y 2 a22 q 2 < a2 < 2 F 3 q 4a q 2 a 2 . From formula Ž8. we find that M Ž f s, t . s 2 q cos Ž 2 s . q 4a cos Ž u . cos Ž t y s . q 2 a cos Ž u . cos Ž 2 t y u . y cos Ž u . cos Ž 2 t . q 2 a 2 cos Ž u . s 2 q cos Ž 2 s . q 4a cos Ž u . cos Ž t y s . q 2 a cos Ž u . sin Ž u . sin Ž 2 t . q 2 a 2 cos 2 Ž u . . This proves the desired inequality. If equality holds, then equality must hold in Lemma 2 so u s 0 and s s t. Also, Re 3a3 y 2 a22 4 s 3a3 y 2 a22 . For u s 0 and s s t we have 3a3 y 2 a22 s e 2 i s, so the preceding equality implies s s t s 0 or s s t s p .
548
MA AND MINDA
LEMMA 3. 2. Then
Suppose u g Žypr2, pr2., s g Žyp , p x, a ) 0, and < d1 < F
cos Ž 2 s . q 2 a Ž cos Ž u . Re e iŽ uys. d1 4 q 2 a cos Ž u . . y 12 a cos Ž u . < d1 < 2 q 12 a 2 cos 2 Ž u . < d1 < 2 q 12 a cos Ž u . sin Ž u . Im
½Že
iu
d1 .
2
5 F 1 q 4a q 2 a
2
.
Equality implies u s 0, and either s s 0 and d1 s 2 or s s p and d1 s y2. Proof. Since < e i u d1 < s < d1 <, it suffices to show cos Ž 2 s . q 2 a cos Ž u . Re eyi sd1 4 q 2 a cos Ž u . y 12 a cos Ž u . < d1 < 2 q 21 a 2 cos Ž u . < d1 < 2 q 12 a cos Ž u . sin Ž u . Im d12 4 F 1 q 4a q 2 a 2 . Set d1 s 2 re it , where 0 F r F 1 and t g Žyp , p x. Then we want to show cos Ž 2 s . q 4a r cos Ž u . cos Ž s y t . q 2 a cos Ž u . y 2 a r 2 cos Ž u . q 2 a 2 r 2 cos 2 Ž u . q 2 a r 2 cos Ž u . sin Ž u . sin Ž 2 t . F 1 q 4a q 2 a 2 . This inequality will hold if we can show H Ž r . s cos Ž 2 s . q 4a r cos Ž u . cos Ž s y t . q 2 a cos Ž u . y 2 a r 2 cos Ž u . q 2 a 2 cos 2 Ž u . q 2 a r 2 cos Ž u . sin Ž u . sin Ž 2 t . F 1 q 4a q 2 a 2 for 0 F r F 1, with strict inequality unless r s 1 and either s s t s 0 or s s t s p. Now H Ž 0 . s cos Ž 2 s . q 2 a cos Ž u . q 2 a 2 cos 2 Ž u . F 1 q 2 a q 2 a 2 - 1 q 4a q 2 a 2 and H Ž 1 . s cos Ž 2 s . q 4a cos Ž u . cos Ž s y t . q 2 a 2 cos 2 Ž u . q2 a cos Ž u . sin Ž u . sin Ž 2 t . F 1 q 4a q 2 a 2
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
549
by the Corollary to Lemma 2 with the proper conditions for equality. All that remains is to prove H Ž r . - 1 q 4a q 2 a 2 for 0 - r - 1. If H X Ž r . / 0 for 0 - r - 1, we are done. Suppose H X Ž r 0 . s 0 for some r 0 g Ž0, 1.. Then r0 s
cos Ž s y t . 1 y sin Ž u . sin Ž 2 t .
and as 0 - r 0 - 1 H Ž r 0 . s cos Ž 2 s . q 2 a cos Ž u . q 2 a 2 cos 2 Ž u . q 2 a r 0 cos Ž u . cos Ž s y t . F cos Ž 2 s . q 2 a cos Ž u . q 2 a 2 cos 2 Ž u . q 2 a r 0 cos Ž u . F 1 q 2 a Ž 1 q r0 . q 2 a 2 - 1 q 4a q 2 a 2 . This completes the proof.
4. PROOF OF THEOREM 1Ži. Because SCCŽ a . is rotationally invariant, it suffices to show that f g SCCŽ a . implies L Ž f . F 13 Ž 1 q 4a q 2 a 2 .
Ž 10 .
and equality implies f Ž z . s Fa Ž z . or f Ž z . s yFa Žyz .. We begin by showing that we may assume Re e i u d1 4 G 0 in the expression Ž5. for LŽ f .. Suppose Re e i u d14 - 0. Note that g Ž z . s yf Žyz . g SCCŽ a .. Also, g X Ž z . s f X Ž yz . s w X Ž yz . e i u Ž cos Ž u . p Ž yz . y i sin Ž u . .
a
where yw Ž yz . s z y b1 z q b 2 z 2 q ??? and p Ž yz . s 1 y d1 z q d 2 z 2 q ??? . Now, Re e i u Žyd1 .4 ) 0 and LŽ f . s LŽ g .; we replace f by g Ž z . s yf Žyz . when Re e i u d14 - 0. Thus, we want to prove Ž10. under the restriction Re e i u d14 G 0 and show that equality implies f s Fa .
550
MA AND MINDA
Because w g CV, we have w7x Re b 22 y b 3 4 q 13 < b 2 < 2 F
1 3
Ž 11 .
and Re b 2 4 F < b 2 < F 1. Note that Re b 2 4 s 1 if and only if pŽ z . s zrŽ1 y z .. Since Re e i u d1 4 G 0, we obtain LŽ f . F
1 3 q q
2a
q
a 3
3
cos Ž u . Re e i u d1 4
cos Ž u . Re
a2
½
1 2
2
cos Ž u . Ž e i u d1 . y e i u d 2
cos 2 Ž u . < d1 < 2 q Re
12
½Že
iu
d1 .
2
5
5
.
By Lemma 1, cos Ž u . 2 Re e i u d1 4 q Re
½
1 2
2
cos Ž u . Ž e i u d1 . y e i u d 2
5
F4
and equality implies u s 0 and pŽ z . s Ž1 q z .rŽ1 y z .. Since p g P, < d1 < F 2. Thus, 1
4a
2
3
q
3
q
a 2 cos 2 Ž u . F
1
Ž 1 q 4a q 2 a 2 . . 3 3 Equality implies u s 0, pŽ z . s Ž1 q z .rŽ1 y z . and w Ž z . s zrŽ1 y z ., that is, f s Fa . Conversely, if f s Fa it is straightforward to verify that equality holds. LŽ f . F
5. PROOF OF THEOREM 1Žii. Since SCCŽ a . is rotationally invariant, it is enough to prove that f g SCCŽ a . implies M Ž f . F 3 q 4a q 2 a 2 with equality if and only if either f Ž z . s Fa Ž z . or f Ž z . s yFa Žyz .. We begin by bounding M Ž f . by a quantity depending only on c1 and d1. Since q g P w4x, d 2 y 12 d12 F 2 y 12 < d1 < 2 and so Re e i u d 2 4 F 2 y 12 < d1 < 2 q
1 2
Re e i u d12 4 .
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
551
Then 2
Re y 12 cos Ž u . Ž e i u d1 . q e i u d 2 F 2 y 12 < d1 < 2
½
5
q 12 sin Ž u . Im
½Že
iu
d1 .
2
5.
This yields MŽ f . F
1
Re c 2 4 q
2
1 2
< c1 < 2 q a cos Ž u . Re e i u c1 d1 4
q 2 a cos Ž u . y
a
q
2
a 2
cos Ž u . < d1 < 2
cos Ž u . sin Ž u . Im
½
Ž e i u d1 .
2
q
5
a2 2
cos 2 Ž u . < d1 < 2 .
Similarly, p g P so c 2 y 12 c12 F 2 y 12 < c1 < 2 , or Re c 2 4 F 2 y 12 < c 2 < 2 q 1 2
1 2
Re c12 4 ,
Re c 2 4 q 12 < c1 < 2 F 1 q 41 < c1 < 2 q s1q
1 2
1 4
Re c12 4
Re 2 c1 4 .
Thus, MŽ f . F 1 q
1 2
Re 2 c1 4 q a cos Ž u . Re e i u c1 d1 4
q 2 a cos Ž u . y q
a 2
a 2
cos Ž u . < d1 < 2
cos Ž u . sin Ž u . Im
½ Ž e i u d1 .
2
5
a2
q
2
˜Ž f . . cos 2 Ž u . < d1 < 2 s M
˜Ž f . F 3 q 4a q 2 a 2 when c1 s 2 e i s. For c1 s 2 e i s, Next, we show M ˜ Ž f . s 1 q 2 cos 2 Ž s . q 2 a cos Ž u . Re e iŽ uys. d1 4 M q 2 a cos Ž u . y q
a 2
a 2
cos Ž u . < d1 < 2
cos Ž u . sin Ž u . Im
½
Ž e i u d1 .
2
5
q
a2 2
cos 2 Ž u . < d1 < 2 .
552
MA AND MINDA
Since 1 q 2 cos 2 Ž s . s 2 q cos Ž 2 s . ,
˜Ž f . F 3 q 4a q 2 a 2 follows immediately from Lemma 3. the inequality M Equality implies u s 0 and either s s 0 Ž c1 s 2. and d1 s 2 or s s p Ž c1 s y2. and d1 s y2. This means either f Ž z . s Fa Ž z . or f Ž z . s yFa Žyz .. ˜Ž f . - 3 q 4a q 2 a 2 for < c1 < - 2. All that remains is to demonstrate M ˜Ž f . is a subharmonic function of c1 which is nonconstant and Since M ˜Ž f . F 3 q 4a q 2 a 2 for < c1 < s 2, we get strict inequality for < c1 < - 2. M 6. INVARIANT FORMULATION A not necessarily normalized analytic function f on D is called strongly close-to-convex of order a if there exists a not necessarily normalized convex univalent function w on D such that arg
fXŽ z. X
w Ž z.
-
ap 2
for z g D. This class of functions is linearly invariant in the sense that f Ž T Ž z . . y f Ž T Ž 0. . f X Ž T Ž 0. . T X Ž 0.
g SCC Ž a .
for any conformal automorphism T of D. Pommerenke w6x noted this linear invariance and gave a simple external geometric characterization when 0 - a F 1. There is an invariant formulation of Theorem 1 for nonnormalized strongly close-to-convex functions of order a . Recall the differential operators w3x D1 f Ž z . s Ž1 y < z < 2 . f X Ž z . 2
D2 f Ž z . s Ž1 y < z < 2 . f Y Ž z . y 2 z Ž1 y < z < 2 . f X Ž z . 3
2
D3 f Ž z . s Ž1 y < z < 2 . f Z Ž z . y 6 z Ž1 y < z < 2 . f Y Ž z . q 6 z 2 Ž1 y < z < 2 . f X Ž z . . Note that Dj f Ž0. s f Ž j. Ž0. Ž j s 1, 2, 3.. These differential operators are invariants in the sense that < Dj Ž S( f (T .< s < Dj Ž f .(T < for all Euclidean
STRONGLY CLOSE-TO-CONVEX FUNCTIONS
553
motions S of C and conformal automorphisms T of D. The Schwarzian derivative Sf Ž z . s
fZ Ž z. fXŽ z.
3
y
2
ž
fYŽ z. fXŽ z.
2
/
is related by D3 f Ž z . D1 f Ž z . THEOREM 2.
y
3 2
D2 f Ž z .
ž
D1 f Ž z .
2
/
2
s Ž1 y < z < 2 . S f Ž z . .
Suppose f is strongly close-to-con¨ ex of order a on D. Then 1 D2 f Ž z .
Ž1 y < z < 2 . < S f Ž z . < q
2 D1 f Ž z .
2
F 2 Ž 1 q 4a q 2 a 2 .
and D3 f Ž z . D1 f Ž z .
y
ž
D2 f Ž z . D1 f Ž z .
2
/
q
D2 f Ž z . D1 f Ž z .
2
F 2 Ž 3 q 4a q 2 a 2 . .
These inequalities are both sharp; equality holds if and only if f s S( fa (T where S is a conformal automorphism of C and T is a conformal automorphism of D. Proof. For z s 0 these inequalities reduce to Theorem 1. The general case follows from the invariance of the differential operators together with the linear invariance of the class of functions.
REFERENCES 1. D. A. Brannan, J. G. Clunie, and W. E. Kirwan, On the coefficient problem for functions of bounded rotation, Ann. Acad. Sci. Fenn. Ser. A Vol. 1, No. 523 Ž1973.. 2. A. W. Goodman, ‘‘Univalent Functions,’’ Vol. II, Polygonal, Washington, NJ, 1983. 3. S. Kim and D. Minda, Two-point distortion theorems for univalent functions, Pacific J. Math. 163 Ž1994., 137]157. 4. W. Ma and D. Minda, Uniformly convex functions, II, Ann. Polon. Math. 58 Ž1993., 275]285. 5. W. Ma and D. Minda, Two-point distortion theorems for strongly close-to-convex functions, Complex Variables Theory Appl., to appear. 6. Ch. Pommerenke, On close-to-convex analytic functions, Trans. Amer. Math. Soc. 114 Ž1965., 176]186. 7. S. Y. Trimble, A coefficient inequality for convex univalent functions, Proc. Amer. Math. Soc. 48 Ž1975., 266]267.