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Integral Transforms of Certain Subclasses of Analytic Functions
Journal of Mathematical Analysis and Applications 258, 466᎐489 Ž2001. doi:10.1006rjmaa.2000.7383, available online at http:rrwww.idealibrary.com on
Integral Transforms of Certain Subclasses of Analytic Functions Yong Chan Kim1 Department of Mathematics, College of Education, Yeungnam Uni¨ ersity, Gyongsan 712-749, Korea E-mail: [email protected]
and Frode Rønning School of Teacher Education, Sør-Trøndelag College, N-7004 Trondheim, Norway E-mail: [email protected] Submitted by Stephan Ruscheweyh Received September 21, 1999
Let A be the class of normalized analytic functions in the unit disk ⌬ and define the class P␥ Ž  . s f g A < ᭚ g ⺢ < Re e i Ž Ž 1 y ␥ . f Ž z . rz q ␥ f ⬘ Ž z . y  . 4 ) 0, z g ⌬ 4 . For a function f g A we define the integral transform
V Ž f .Ž z . s
H01Ž t .
f Ž tz . t
dt ,
where is a real-valued, nonnegative weight function with H01 Ž t . dt s 1. For f g P␥ Ž  . we will find conditions such that VŽ f . is starlike and also such that VŽ f . g P␥ Ž ␣ .. We also discuss how to write the integral transform as a convolution with a 2 F1 hypergeometric function. 䊚 2001 Academic Press Key Words: starlike functions; integral transforms. 1
The work of the first author was supported by KOSEF under Grant 981-0102-010-2. 466
0022-247Xr01 $35.00 Copyright 䊚 2001 by Academic Press All rights of reproduction in any form reserved.
467
INTEGRAL TRANSFORMS
1. INTRODUCTION Let A be the class of analytic functions in the unit disk ⌬ with the normalization f Ž0. s f ⬘Ž0. y 1 s 0. For f g A we define the integral transform
Ž 1.1.
V Ž f . Ž z . s
1
H0 Ž t .
f Ž tz . t
dt ,
where is a real-valued, nonnegative weight function normalized so that H01 Ž t . dt s 1. For  - 1 and 0 F ␥ F 1 we define the set P␥ Ž  . s f g A < ᭚ g ⺢ < Re e i Ž Ž 1 y ␥ . f Ž z . rz q ␥ f ⬘ Ž z . y  . 4 ) 0, z g ⌬4 . We shall mainly discuss the following problems. Ž1. For f g P␥ Ž  ., find conditions such that VŽ f . g S*, where S* denotes the usual class of starlike functions. Ž2. For f g P␥ Ž  ., find conditions such that VŽ f . g P␥ Ž ␣ .. To deal with these problems we shall use the duality theory for convolutions. Similar methods have earlier been used to discuss univalence and starlikeness Žalso of order ␣ ) 0. of VŽ f . in the case ␥ s 1, e.g., in the papers w4, 10, 12x. In w6x some aspects of the case ␥ - 1 are discussed, using other methods. Some special cases of Ž t . are particularly interesting such as
Ž t . s Ž 1 q c . t c , c ) y1, for which V is known as the Bernardi operator, and ␦
1 Ž c q 1. c Ž t . s t log ⌫Ž ␦ . t
ž /
␦ y1
, c ) y1, ␦ G 0,
which gives the Komatu operator w7x. We see that for ␦ s 1 the Komatu operator reduces to the Bernardi operator. We can also look at some of these integral operators as taking the convolution between the function f and a 2 F1 hypergeometric function. We use the notation 2 F1Ž a, b; c; z . for the usual Gaussian hypergeometric function, and define the convolution operator
Ž 1.2.
Ha, b , c Ž f . Ž z . [ z 2 F1 Ž a, b; c; z . ) f Ž z . ,
468
KIM AND RøNNING
where f Ž z . s Ý⬁ns 1 a n z n. We use the following basic facts about hypergeometric functions Žsee, e.g., w1x.. 2 F1 Ž a, b; c; z . s
⬁
⌫ Ž a q n.
Ž a. n s B Ž x, y . s
1
H0 Ž 1 y t .
Ž a. n Ž b . n
Ý ns0 Ž c . n n! ⌫ Ž a.
xy1 yy1
t
z n , < z < - 1,
,
⌫Ž x. ⌫Ž y.
dt s
⌫Ž x q y.
, x ) 0, y ) 0,
and ⬁
Ž a. n Ž b . n
Ý ns0 Ž c . n n!
s
⌫ Ž c. ⌫ Ž c y a y b.
, c ) b ) 0 and c ) a q b.
⌫ Ž c y a. ⌫ Ž c y b .
Then we can write Ha, b , c Ž f . Ž z . s s s
⬁
Ž a. n Ž b . n
Ý anq1 Ž c . n! n ns0
z nq1
⬁
⌫Ž c.
⌫ Ž a q n. ⌫ Ž b q n.
Ýa ⌫ Ž a . ⌫ Ž b . ns0 nq 1 ⌫ Ž c q n . ⌫ Ž n q 1 .
z nq 1
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1. =
⬁
Ý
⬁
a nq 1 z nq1
ns0
Ž c y a. m Ž 1 y a. m
Ý ms0 Ž c y a y b q 1 . m m! 1 bqny1
=
H0 t
Ž1 y t .
cy aybqm
dt .
This expression can be written as Ha, b , c Ž f . Ž z . s
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1. =
1 by 2
H0
t
Ž1 y t .
cy ayb
2 F1
ž
c y a, 1 y a ; 1 y t f Ž tz . dt , cyaybq1
/
469
INTEGRAL TRANSFORMS
where we assume a ) 0, b ) 0 and c ) a q b y 1. This means that the convolution Ž1.2. is an integral operator of the form Ž1.1. with
Ž t . s
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1. =Ž 1 y t .
cy ayb
2 F1
t by 1
c y a, 1 y a ; 1yt . cyaybq1
ž
/
Using the Euler integral representation Ž c ) b ) 0. 2 F1
Ž a, b; c; z . s
⌫Ž c.
1 by1
⌫ Ž b. ⌫ Ž c y b.
H0
t
Ž1 y t .
cyby1
Ž 1 y tz .
ya
dt
we see that with a s 1 we get H1, b , c Ž f . Ž z . s
⌫Ž c.
1 by1
H ⌫ Ž b. ⌫ Ž c y b. 0
t
Ž1 y t .
cy by1
f Ž tz . t
dt.
This integral operator is known as the Carlson᎐Shaffer operator Žsee w2x.. For the operator Ha, b, c , given a, b, c, and ␥ in a certain range, we get the sharp value of  for problem Ž1.. However, it is likely that the range of a, b, c, and ␥ , for which the result holds, is larger than the one we have obtained. Some related results for Ha, b, c are obtained in w11x. 2. MAIN RESULTS We start with functions from P␥ Ž  ., where ␥ may be any number in w0, 1x. For now we will assume ␥ ) 0. With Ž t . given we define ⌳␥ Ž t . s
1
Ht
Ž s . s 1r␥
ds.
Further we let g␥ Ž t . be the solution of the initial value problem
Ž 2.1.
2 t 1r ␥y1
d
Ž t 1r␥ Ž 1 q g Ž t . . . s ␥ dt
Ž1 q t .
2
, g Ž 0 . s 1.
Before we proceed to state our results we need some more information about the function g␥ . Solving Ž2.1. we see that g␥ Ž t . can be expressed as g␥ Ž t . s
2
␥
ty1 r␥
t
1r ␥y1
H0 Ž 1 q .
2
d y 1.
470
KIM AND RøNNING
Then we get 2
g␥X Ž t . s y
␥
1r ␥y1
t
ty1r␥y1 2
H0 Ž 1 q .
2
d q
2
1
␥ t Ž1 q t .2
.
For 0 - t - 1 we have
1r ␥y1
t
H0 Ž 1 q .
2
d )
1
t
Ž1 q t .
2
H0
1r ␥ y1
d s ␥
t 1r ␥
Ž1 q t .
2
,
which implies that g␥X Ž t . - 0, so g␥ Ž t . is strictly decreasing on Ž0, 1.. For any weight function Ž t . ) 0 with H01 Ž t . dt s 1 we therefore have H01 Ž t . g␥ Ž t . dt - 1. THEOREM 2.1.
Define  - 1 by
 1y
1
H0 Ž t . g
sy
␥
Ž t . dt ,
and let f g P␥ Ž  ., ␥ ) 0. Assume that lim t ª 0q t 1r ␥⌳␥ Ž t . s 0. Then VŽ f . g S* if and only if
Ž 2.2.
Re
1 1r ␥ y1
H0
t
⌳␥ Ž t .
ž
h Ž tz . tz
y
1
Ž1 q t .
2
/
dt G 0, z g ⌬ ,
where
Ž 2.3.
hŽ z . s
z Ž1 q
⑀y 1 2
Ž1 y z.
z.
2
, < ⑀ < s 1.
Theorem 2.1 is not very useful directly because in order to find out whether VŽ f . g S* we have to check the condition Ž2.2., and this is often difficult. Therefore, to get some applicable results we will identify some situations where Ž2.2. holds. The first result of this type is one where we can compare an unknown situation to a known one with a condition that is reasonably easy to check. We introduce the notation L⌳ ␥ Ž h . s inf
1 1r ␥ y1
H zg⌬ 0
t
⌳␥ Ž t . Re
and formulate the following result.
ž
h Ž tz . tz
y
1
Ž1 q t .
2
/
dt
471
INTEGRAL TRANSFORMS
Assume ⌳␥ is integrable on w0, 1x and positi¨ e on Ž0, 1.. If
THEOREM 2.2.
⌳␥ Ž t . log 1t is decreasing on Ž0, 1. then for 1r2 F ␥ F 1 we ha¨ e L⌳ ␥Ž h. G 0, where hŽ z . is as in Ž2.3.. The next results describe what we are able to prove about some of the specific operators mentioned in the Introduction. The first result deals with the Komatu operator. THEOREM 2.3. Let 1r2 F ␥ F 1 and g␥ Ž t . be defined by Ž2.1.. Define  s  Ž c, ␦ , ␥ . by

Ž c q 1. sy 1y ⌫Ž ␦ .
Ž 2.4.
␦
1 c
H0
1
␦ y1
ž /
t log
t
g␥ Ž t . dt.
Then for f g P␥ Ž  . and
Ž c q 1. Fc , ␦ Ž z . s ⌫Ž ␦ .
Ž 2.5.
␦
1 cy 1
H0
t
1
␦ y1
ž / log
t
f Ž tz . dt
we ha¨ e Fc, ␦ g S* for y1 - c F 1r␥ y 1 and ␦ G 1. The ¨ alue of  is sharp. We now turn to the operator Ha, b, c Ž f . for which we obtain the following result. THEOREM 2.4. Let 1r2 F ␥ F 1 and g␥ Ž t . be defined by Ž2.1.. Define  s  Ž a, b, c, ␥ . by
Ž 2.6.
 1y
⌫Ž c.
sy
⌫ Ž b . ⌫ Ž c y b . ⌫ Ž c y a y b q 1.
=
1 by 1
H0
t
Ž1 y t .
cy ayb
2 F1
ž
c y a, 1 y a ; 1 y t g␥ Ž t . dt. cyaybq1
/
Then for f g P␥ Ž  ., 0 - a F 1, 0 - b F 1r␥ , and c G a q b we ha¨ e Ha, b, c Ž f . g S*. The ¨ alue of  is sharp. In the case a s 1 we have the Carlson᎐Shaffer operator which we denote by Lb , c Ž f . Ž z . s
⌫Ž c.
1 by2
H ⌫ Ž b. ⌫ Ž c y b. 0
t
Ž1 y t .
cy by1
f Ž tz . dt ,
472
KIM AND RøNNING
where 0 - b - c. We write the result for the Carlson᎐Shaffer operator explicitly. Let 1r2 F ␥ F 1 and g␥ Ž t . be defined by Ž2.1.. Define
COROLLARY 2.5.  s  Ž b, c, ␥ . by
 1y
sy
⌫Ž c.
1 by1
H ⌫ Ž b. ⌫ Ž c y b. 0
t
Ž1 y t .
cy by1
g␥ Ž t . dt.
Then for f g P␥ Ž  ., 0 - b F 1r␥ , and c G 1 q b we ha¨ e L b, c Ž f . g S*. The ¨ alue of  is sharp. Another question that might be asked is the following. Let ␣ - 1 be given. Find the smallest  s  Ž ␣ . such that if f g P␥ Ž  . we have VŽ f . g P␥ Ž ␣ .. We prove the following result, which holds for all ␥ g w0, 1x. Let ␣ - 1 and 0 F ␥ F 1 be gi¨ en, and define  s  Ž ␣ .
THEOREM 2.6. by
Ž 2.7.
 1y
1
H0 Ž t .
sy
1 y Ž 1 q ␣ . rŽ 1 y ␣ . t 1qt
dt.
If f g P␥ Ž  . then VŽ f . g P␥ Ž ␣ .. The ¨ alue of  is sharp. The last result we will present says something about the location of f Ž z .rz, given that f g P␥ Ž  .. Hence, in this result there is no integral transform involved, except as a tool in the proof. THEOREM 2.7.  Ž ␣ , ␥ . by
Ž 2.8.
 1y
Let ␣ - 1 and 0 - ␥ F 1 be gi¨ en, and define  s
sy
1
1 Ž1y ␥ .r ␥
H ␥ 0
t
1 y Ž 1 q ␣ . rŽ 1 y ␣ . t 1qt
dt.
If f g P␥ Ž  . then for some g ⺢ we ha¨ e Re e i
f z
žŽ. / z
y ␣ ) 0, z g ⌬ .
The ¨ alue of  is sharp.
3. SOME EXAMPLES In this section we shall look at some examples where we see how our results improve earlier results. It should be said that in some of the
473
INTEGRAL TRANSFORMS
examples the results are not directly comparable, because the previous results deal mostly with the classes R␥ Ž  . s f g A < Re Ž 1 y ␥ . f Ž z . rz q ␥ f ⬘ Ž z . y  4 ) 0, z g ⌬ 4 instead of the larger classes P␥ Ž  .. 3.1. EXAMPLE. In w9x Ponnusamy investigated some problems that are similar to those discussed in this paper. The notation is somewhat different, but if we translate Theorem 5 in w9x to our notation we get the following. Let
␦ Ž␥ . s
1
H0
dt 1 q t␥
.
If g g R␥ Ž  ., with
s
y y Ž 1 y Ž 1r2␥ . . Ž 2 ␦ Ž ␥ . y 1 . 1 y Ž 1 y Ž 1r2␥ . . Ž 2 ␦ Ž ␥ . y 1 .
, ␥ G 1r2,
then G Ž z . s H01 g Ž tz . dtrt g S*. Here s 0.0573 . . . is a number that can be computed through a rather involved algorithm which we will not repeat. To compare, we use our Theorem 2.3 with c s 0 and ␦ s 1. Theorem 5 w9x
␥ 1r2 1
 s y s y0.0573 s
y y log 2 q 1r2 1 y log 2 q 1r2
Theorem 2.3
s s y0.3104 . . .
s
4 log 2 y 3 4 log 2 y 2 1 y 2 log 2 2 y 2 log 2
s y0.294 . . . s y0.629 . . .
In the proof of Theorem 2.3 we see that the extremal function belongs to R␥ Ž  ., so the result is sharp also in this class. 3.2. EXAMPLE. As far as we know, the class R␥ Ž  . was introduced by Chichra w3x, where it was proved that if f g R␥ Ž0. for some ␥ with Re ␥ G 0, then Re f Ž z .rz ) 0 for z g ⌬. Using Theorem 2.7 we see that this result may be extended in the case ␥ ) 0. Indeed, for f g R␥ Ž  . with  - 0, we can conclude that for some g ⺢ we have Re e i f Ž z .rz ) 0. Explicitly, we get for ␥ s 1 that the smallest possible value of  is Ž1 y 2 log 2.rŽ2 y 2 log 2. s y0.629 . . . and for ␥ s 1r2 we get  s Ž4 log 2 y 3.rŽ4 log 2 y 2. s y0.294 . . . . Here we see that the results are not directly comparable, because we cannot actually conclude that Re f Ž z .rz ) 0 by our methods.
474
KIM AND RøNNING
3.3. EXAMPLE. In w8x Owa and Nunokawa improved Chichra’s result. They showed that if Re Ž 1 y ␥ .
gŽ z.
q ␥ g ⬘Ž z . ) 
z
then Re
gŽ z. z
)␣,
where ␣ s  q Ž1 y  .Ž2 ␦ Ž␥ . y 1., and ␦ Ž␥ . is as in Example 3.2. Solving this equation for  we get ␥
s
Ž 3.1.
t ␣ y H01 11 y q t ␥ dt ␥
2H01 1 qt t ␥ dt
.
The result of Owa and Nunokawa corresponds to our Theorem 2.7, but also here with the difference that we work with P␥ Ž  . instead of R␥ Ž  .. Solving equation Ž2.8. with respect to  we get
Ž 3.2.
s
. r Ž1 y ␣ . t H01 t Ž1y ␥ .r ␥ 1 y Ž1 q1␣q dt t . Ž . Ž . 1 y 1 q ␣ r 1 y ␣ t H 1 t Ž1y ␥ .r ␥ dt y ␥
0
1q t
If we do the change of variables t s u␥ in Ž3.2. we see, after a little calculation, that the expressions Ž3.2. and Ž3.1. are equal. There is no comment in w8x that their value of  is sharp. In the proof of Theorem 2.6 and Theorem 2.7 we see that the extremal function belongs to R␥ Ž  ., and it has real coefficients. Therefore, this function will satisfy Re g Ž z .rz ) ␣ , so the result of Owa and Nunokawa is actually sharp. Their proof is based on a subordination theorem by Hallenbeck and Ruscheweyh w5x. 3.4. EXAMPLE. Let
Ž t . s ␦ Ž t . s ⌫ Ž ␦ .
y1
1
␦y1
, ␦ G 0,
ž / log
t
Žthe Komatu operator with c s 0. and F␦ Ž z . s V Ž f .Ž z .. In w6x it is shown ␦ that if Re f ⬘Ž z . )  then Re F␦X Ž z . G 1 y Ž 1 y  . Ž 1 y log 2 . 1 q
ž
1 2
␦ y1
/
[␣
475
INTEGRAL TRANSFORMS
for ␦ G 1. Comparing with the results we get from Theorem 2.6 we get the following table which for given ␦ and ␣ shows the values of  that we get using Corollary 3.1 in w6x and our Theorem 2.6 respectively. The case ␣ s 0. Corollary 3.1 w6x
␦ 1 2
1 y 3 log 2 3 y 3 log 2 1 y 5 log 2 5 y 5 log 2
Theorem 2.6 6y2
s y1.172 . . .
12 y 2 2 y 3 Ž 3 .
s y1.607 . . .
4 y 3 Ž 3 .
s y1.816 . . . s y4.078 . . .
The case ␣ s 1r2. ␦ 1 2
Corollary 3.1 w6x 4 y 6 log 2 6 y 6 log 2 6 y 10 log 2
Theorem 2.6 9y2
s y0.086 . . .
10 y 10 log 2
s y0.303 . . .
12 y 2 3 y 3 Ž 3 . 4 y 3 Ž 3 .
s y0.408 . . . s y1.539 . . .
4. PROOFS 4.1. Proof of Theorem 2.1. Let f g P␥ Ž  . and define GŽ z . s
Ž 1 y ␥ . f Žzz . q ␥ f ⬘ Ž z . y  1y
.
Then for some ␣ g ⺢ we have Re e i ␣ G Ž z . ) 0. With F Ž z . s VŽ f .Ž z ., we want to investigate starlikeness of F; i.e., we want to find conditions such that Re
zF⬘ Ž z . FŽ z.
) 0, z g ⌬ .
By the Duality Principle Žsee w10, 13x. we may restrict our attention to the function f g P␥ Ž  . for which G Ž z . s Ž1 q xz .rŽ1 q yz ., < x < s < y < s 1. Further, it is well known from convolution theory w13, p. 94x that F g S* m
1 z
Ž F ) h . Ž z . / 0, z g ⌬ ,
476
KIM AND RøNNING
where ⑀y 1 2
z Ž1 q
hŽ z . s
Ž1 y z.
z.
, < ⑀ < s 1.
2
Hence, F g S* if and only if Ž␥ / 0. 0/ s
1
Ž F ) h. Ž z .
z
1
1
H0 Ž t . 1 y tz dt )
s Ž1 y  .
1
H0 Ž t . 1
s Ž1 y  .
Ž t .
H0
s Ž1 y  .
H0 Ž t .
ž ž
1
Ž1 y  .
h Ž tz .
h Ž tz .
1
␥ z
1r ␥

q
tz
1
1y
z
/
H0
/
1
1
␥ z
1r ␥
z
H0
1 q yw
dt )
y g␥ Ž t . dt )
tz
w 1r␥y1 Ž 1 q xw .
1
1
␥ z 1r␥ 1
z
1
␥ z 1r␥
w 1r␥y1 h Ž tw . tw
z
H0
H0
dw q  )
w 1r␥y1 Ž 1 q xw . 1 q yw
w 1r␥y1 Ž 1 q xw . 1 q yw
dw y g␥ Ž t . dt )
1 q xz 1 q yz
hŽ z . z dw
dw
.
Using another well-known result from convolution theory w13, p. 23x we conclude that this holds if and only if Re Ž 1 y  .
1
H0
Ž t .
1
1
␥ z
1r ␥
z
H0
w 1r ␥y1 h Ž tw . tw
dw y g␥ Ž t . dt )
1 2
.
Inserting 1 s Ž1 y  .H01Ž t .Ž1 y g␥ Ž t .. dt on the right-hand side we can rewrite this as 1
Re
H0
Ž t .
1 t 1r ␥
t 1r ␥
␥ z 1r␥
w 1r␥y1 h Ž tw .
z
H0
tw
dw y
t 1r␥ Ž 1 q g␥ Ž t . . 2
dt ) 0.
With the change of variables u s tw we now get Re
1
H0
Ž t .
1
t 1r ␥
␥ z 1r␥
1
tz
H0
u1r ␥y1 h Ž u . u
du y
t 1r␥ Ž 1 q g␥ Ž t . . 2
Integrating by parts we get that this is equivalent to Re
1 1r ␥ y1
H0
t
⌳␥ Ž t .
ž
h Ž tz . tz
y
1
Ž1 q t .
2
/
dt G 0.
dt ) 0.
477
INTEGRAL TRANSFORMS
4.2. Proof of Theorem 2.2. We shall omit some details here because there are many similarities between this proof and the proof of Theorem 1 in w4x. With hŽ z . s z Ž1 q Ž ⑀ y 1. zr2.rŽ1 y z . 2 , < ⑀ < s 1, we want to prove that 1 1r ␥ y1
Ž 4.1.
⌳␥ Ž t . Re
t
H0
ž
h Ž tz . tz
y
1
Ž1 q t .
/
2
dt G 0
for z g ⌬. Using the same argument as in w4x we see that it is enough to prove Ž4.1. for < z < s 1. Let ⑀ in the expression of h be ⑀ s e i and minimize Re hŽ tz .rtz with respect to . Then we see that Ž4.1. follows if 2 y tz
1 1r ␥ y1
H0
⌳␥ Ž t . Re
t
Ž 1 y tz .
2
y
t <1 y tz < 2
y
2
Ž1 q t .
2
G 0,
< z < s 1, z / 1. Further calculation gives the equivalent condition H
Ž␥ .
1 1r ␥
Ž y. s H t 0
⌳␥ Ž t .
3 y 4 Ž 1 q y . t q 2 Ž 4 y y 1. t 2 q 4 Ž y y 1. t 3 y t 4 2
Ž 1 q t 2 y 2 yt . Ž 1 q t . 2
G 0, where y s Re z. A series expansion of H Ž ␥ . Ž y . gives H Ž␥ . Ž y . s
⬁
Ý HkŽ␥ . Ž 1 q y . k ,
<1 q y < - 2,
ks0
where HkŽ ␥ . can be seen to be a positive multiple of
˜hŽk␥ . s H1 t 1r␥y1⌳␥ Ž t . sk Ž t . dt , 0
where sk Ž t . s
t kq 1
Ž1 q t .
2 kq4
ž
1 y 2t q
ky1 kq3
t2 .
/
We can see that sk Ž t . has one and only one zero in Ž0, 1.. We denote this zero by t k , and then we see that sk Ž t . ) 0 if 0 F t - t k and sk Ž t . - 0 if t k - t - 1. Let hŽk␥ . s
1 1r ␥ y1
H0
t
1 log sk Ž t . dt , t
478
KIM AND RøNNING
and ⌳␥ Ž t k .
˜ ␥ Ž t . s ⌳␥ Ž t . y ⌳
log Ž 1rt k .
log
1 t
.
˜ ␥ Ž t . sk Ž t . The assumption that ⌳␥ Ž t .rlogŽ1rt . is decreasing implies that ⌳ ) 0 and therefore 0F
⌳␥ Ž t k .
1 1r ␥ y1
H0
t
˜ ␥ Ž t . sk Ž t . dt s ˜hŽk␥ . y ⌳
log Ž 1rt k .
hŽk␥ . .
If we can prove that hŽk␥ . G 0 it follows that ˜ hŽk␥ . G 0 and further L⌳ ␥Ž h. G Ž1r2. 0, as desired. In Lemma 5.1 we show that h k G 0, and we will now show that for 1r2 F ␥ F 1 we have hŽk␥ . G hŽ1r2. . This implies that L⌳ ␥Ž h. G 0 k for 1r2 F ␥ F 1. We can write hŽk␥ . s
1
1
H0
t
2y1r ␥
1 t log sk Ž t . dt. t
With t k as above, set MkŽ ␥ . [ 1rt k2y 1r␥ G 1. Then 1
tk
I1Ž ␥ . [
H0
I2Ž ␥ . [
Ht
t
2y1r ␥
t
2y1r ␥
1
1
k
1 t log sk Ž t . dt G MkŽ ␥ . t
1 t log sk Ž t . dt [ MkŽ ␥ . I1 t
tk
H0
1 t log sk Ž t . dt F MkŽ ␥ . t
1
Ht
k
1 t log sk Ž t . dt [ MkŽ ␥ . I2 . t
Hence, hŽk␥ . s I1Ž ␥ . y I2Ž ␥ . G MkŽ ␥ . Ž I1 y I2 . s hŽ1r2. G 0. k 4.3. Proof of Theorem 2.3. Let 1r2 F ␥ F 1 and assume c F 1r␥ y 1, ␦ G 1. Then ␦
Ž t . s
1 Ž c q 1. c t log ⌫Ž ␦ . t
␦ y1
ž /
and ⌳␥ Ž t . s
Ž c q 1. ⌫Ž ␦ .
␦
1 cy 1r ␥
Ht
s
1
ž / log
s
␦ y1
ds.
479
INTEGRAL TRANSFORMS
It is easily verified that t 1r ␥⌳␥ Ž t . ª 0 when t ª 0q. Using Theorems 2.1 and 2.2 the result follows if we can prove that the function g Ž t . s ⌳␥ Ž t .rlog 1t is decreasing on Ž0, 1.. Differentiating we get
Ž c q 1. g ⬘Ž t . s ⌫Ž ␦ .
␦ 1 t
Ht1 s cy1r␥ Ž log 1s .
␦y1
ds y t cy1r␥ Ž log 1t .
Ž log 1t .
2
␦
.
If we can prove that 1 cy 1r ␥
Ž 4.2.
Ht
s
1
␦y1
ds y t
ž / log
cq1y1r ␥
s
␦
1
F0
ž / log
t
we have g ⬘Ž t . F 0, and we are done. We see that for t s 1 we have equality in Ž4.2., so the inequality holds if the left-hand side is increasing. Differentiating the left-hand side with respect to t we get yt cy 1r␥ log
1
␦y1
q Ž 1r␥ y 1 y c . t cy1r␥ log
ž / ž / t
s t cy 1r␥ log
1
1
␦
ž / t
q ␦ t cy1r␥ log
␦ y1
ž /
␦y1
␦ y 1 q Ž 1r␥ y 1 y c . log
t
1
1 t
t
,
which clearly is greater than or equal to zero when ␦ G 1 and c F 1r␥ y 1. To prove sharpness, let f g P␥ Ž  . be the function for which
Ž1 y ␥ .
f Ž z. z
q ␥ f ⬘Ž z . y  s Ž 1 y  .
1qz 1yz
.
Using a series expansion we see that we can write f Ž z . s z q 2Ž 1 y  .
⬁
zk
Ý
1 y ␥ q ␥k
ks2
.
Further, using
Ž 4.3.
1 cq ky1
H0
t
1
ž / log
t
␦y1
dt s
⌫Ž ␦ .
Ž c q k.
␦
we get from Ž2.5. that ␦ Ž 4.4. Fc , ␦ Ž z . s z q 2 Ž 1 y  . Ž 1 q c .
⬁
Ý ks2
zk ␦
Ž c q k. Ž1 y ␥ q ␥ k.
480
KIM AND RøNNING
and
Ž 4.5.
zFcX, ␦ Ž z . s z q 2 Ž 1 y  . Ž 1 q c .
␦
⬁
kz k
Ý ks2
␦
Ž c q k. Ž1 y ␥ q ␥ k.
.
Expanding g␥ Ž t . in Ž2.1. in a power series, we obtain g␥ Ž t . s 1 q 2
Ž 4.6.
k
⬁
Ž y1. Ž 1 q k .
Ý
1 q ␥k
ks1
tk
which, when inserted into Ž2.4. and using Ž4.3., gives
 1y
s y1 q 2 Ž 1 q c .
␦
⬁
Ž y1.
Ý
kq 1
Ž k q 1. ␦
Ž c q k q 1. Ž 1 q ␥ k .
ks1
from which we get 1
Ž 4.7.
1y
s 2Ž 1 q c .
⬁
␦
Ž y1.
Ý ks1
kq 1
Ž k q 1. ␦
Ž c q k q 1. Ž 1 q ␥ k .
.
Substituting z s y1 into Ž4.5. the right-hand side becomes, after a shift in the indices, y1 q 2 Ž 1 y  . Ž 1 q c .
␦
⬁
Ý ks1
Ž k q 1 . Ž y1.
kq 1
␦
Ž c q k q 1. Ž 1 q ␥ k .
.
Using Ž4.7. we see that this is zero, and since Fc, ␦ Žy1. / 0, this shows that the result is sharp. 4.4. Proof of Theorem 2.4. Let 1r2 F ␥ F 1. We introduce the notation FA , B , C Ž 1 y t . s2 F1 Ž A, B; C ; 1 y t . . With A s c y a, B s 1 y a, and C s c y a y b q 1 we can write
Ž t . s
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1.
t by1 Ž 1 y t .
cy ayb
FA , B , C Ž 1 y t . ,
and we get ⌳␥ Ž t . s
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1. =
1 by1y1r ␥
Ht
s
Ž1 y s.
cy ayb
FA , B , C Ž 1 y s . ds.
481
INTEGRAL TRANSFORMS
Again it is easily verified that t 1r ␥⌳␥ Ž t . ª 0 when t ª 0q. Define ⌳␥ Ž t .
gŽ t. s
log 1t
.
As in the proof of Theorem 2.3 we want to prove that g ⬘Ž t . - 0 on Ž0, 1.. Differentiating we get g ⬘Ž t . s
⌫Ž c.
h Ž t . rt
⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1.
Ž log 1t .
2
,
where hŽ t . s
1 by1y1r ␥
s
Ht
Ž1 y s.
cy ayb
y log Ž 1rt . t by1r␥ Ž 1 y t .
FA , B , C Ž 1 y s . ds cy ayb
FA , B , C Ž 1 y t . .
Then hŽ1. s 0 and it is enough to show that h⬘Ž t . G 0 on Ž0, 1.. Using d dt 2
F1
ž
c y a, 1 y a ; 1yt cyaybq1
Ž c y a. Ž 1 y a.
sy
cyaybq1
/
2 F1
ž
c y a q 1, 2 y a ; 1yt cyaybq2
/
we get h⬘ Ž t . s ylog Ž 1rt . t by1y1r␥ Ž 1 y t . =
žž
1
␥
qayc tqby
yt Ž 1 y t .
/
cy ayby1
1
␥
/
Ž c y a. Ž 1 y a. cyaybq1
FA , B , C Ž 1 y t . FAq 1, Bq1, Cq1 Ž 1 y t . .
If we assume that 0 - a F 1, 0 - b F 1r␥ and c G a q b we see that the inequality below holds for t g Ž0, 1. since the left-hand side will be positive and the right-hand side will be negative. c y a, 1 y a ; 1yt cyaybq1 c y a q 1, 2 y a ; 1yt 2 F1 cyaybq2 2 F1
ž
/
ž
Gy
Ž c y a. Ž 1 y a.
/ t Ž1 y t .
c y a y b q 1 Ž c y a y 1r␥ . t q 1r␥ y b
.
482
KIM AND RøNNING
This clearly implies h⬘Ž t . G 0 for t g Ž0, 1.. The proof of sharpness follows much the same track as in the proof of Theorem 2.3. As extremal function we again have f Ž z . s z q 2Ž 1 y  .
⬁
zk
Ý
1 y ␥ q ␥k
ks2
.
Let ⌸k [
1 bq ky2
H0
t
Ž1 y t .
cy ayb
2 F1
ž
c y a, 1 y a ; 1 y t dt. cyaybq1
/
Then we can write F Ž z . [ Ha, b , c Ž f . Ž z . ⬁
⌫Ž c.
s z q 2Ž 1 y  .
⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1.
Ý ks2
⌸k zk 1 y ␥ q ␥k
and
Ž 4.8.
zF⬘ Ž z . s z q 2 Ž 1 y  . =
⬁
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1.
k⌸ k z k
Ý
1 y ␥ q ␥k
ks2
.
With g␥ as in Ž4.6. we get from Ž2.6.
Ž 4.9.
1 1y
s2
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1.
=
⬁
Ý ks2
Ž y1.
kq 1
Ž k q 1 . ⌸ kq 1
1 q ␥k
.
Substituting z s y1 into Ž4.8. and using Ž4.9. we see, after a shift in the indices, that the right-hand side of Ž4.8. becomes zero. Clearly F Žy1. / 0, so this shows that the result is sharp. 4.5. Proof of Theorem 2.6. The idea of the proof is similar to the one used to prove Theorem 2 in w4x. Let F Ž z . s H01 Ž t . f Ž ttz . dt. Then we can write F⬘Ž z . s H01Ž t . f ⬘Ž tz . dt s f ⬘Ž z .)H01Ž t . dtrŽ1 y tz .. The assumption that f g P␥ Ž  . means that with
Ž 4.10.
gŽ z. s
Ž 1 y ␥ . f Ž z . rz q ␥ f ⬘ Ž z . y  1y
483
INTEGRAL TRANSFORMS
it is so that for some g ⺢ we have Re e i g Ž z . ) 0. Assume first that ␥ / 0. Then we get from Ž4.10. f ⬘Ž z . s
1
␥
Ž  q Ž1 y  . g Ž z. . y
1 y ␥ f Ž z.
␥
z
,
which leads to F⬘ Ž z . s
1
␥ y
s
1
␥
Ž t .
1
Ž  q Ž 1 y  . g Ž z . . )H
0
1 y ␥ f Ž z.
␥
)
1
H0
z
Ž t . 1 y tz
g Ž z.)  q Ž1 y  .
1
H0
dt
1 y tz
dt
Ž t . 1 y tz
dt y
1 y ␥ FŽ z.
␥
z
and further
Ž 4.11.
Ž1 y ␥ .
FŽ z. z
Ž t .
1
q ␥ F⬘ Ž z . s g Ž z . )  q Ž 1 y  .
H0
1 y tz
dt .
In the case ␥ s 0 we just write f Ž z. z
s Ž1 y  . g Ž z. q 
from which we get FŽ z. z
s g Ž z.)  q Ž1 y  .
1
H0
Ž t . 1 y tz
dt ,
which is Ž4.11. with ␥ s 0. Clearly F g P␥ Ž ␣ . m G Ž z . [ Ž F Ž z . y ␣ z .r Ž1 y ␣ . g P␥ Ž0.. With this function G the equality Ž4.11. can be written
Ž1 y ␥ .
GŽ z. z
q ␥ G⬘ Ž z . s g Ž z . )
y␣ 1y␣
q
1y
1
H 1y␣ 0
Ž t . 1 y tz
Since Re e i g Ž z . ) 0 for some g ⺢ it follows by duality that
Ž1 y ␥ .
GŽ z. z
q ␥ G⬘ Ž z . / 0
dt .
484
KIM AND RøNNING
if and only if
Ž 4.12.
Re
y␣ 1y␣
1y
q
Ž t .
1
H 1y␣ 0
1
dt )
1 y tz
2
, z g ⌬.
Using Re 1rŽ1 y tz . ) 1rŽ1 q t . we get that the left-hand side of Ž4.12. is greater than 1y y␣
q
1y␣ 1y
Ž t .
1
H0
dt .
1qt
Using the definition of rŽ1 y  . in Ž2.7. we get after some computation
 y Ž 1 q ␣ . r2 1y
1
sy
H0
Ž t . 1qt
dt
from which we get
y␣ 1y
q
1
H0
Ž t . 1qt
dt s
1 1y␣ 2 1y
,
which shows that Ž4.12. holds. Further, if Ž4.12. holds we deduce, using duality, that Ž1 y ␥ .G Ž z .rz q ␥ G⬘Ž z . is contained in a half plane not containing the origin, which means that G g P␥ Ž0., and hence F g P␥ Ž ␣ .. To prove sharpness we take, as in the proof of Theorem 2.3, f Ž z . s z q 2Ž 1 y  .
⬁
zk
Ý
1 y ␥ q ␥k
ks2
.
Then we can write
k z k
⬁
F Ž z . s V Ž f . Ž z . s z q 2 Ž 1 y  .
Ý ks2
1 y ␥ q ␥k
,
where k s H01 Ž t . t ky 1 dt. With this notation we get from Ž2.7. that 1 1y
y1s
 1y
s y1 q
1
H0 Ž t .
sy 1
H0 Ž t .
ž
1q
ž
1y
1q␣ 1y␣
1q␣ 1y␣
/
t dtr Ž 1 q t .
/
t dtr Ž 1 q t .
485
INTEGRAL TRANSFORMS
and 1 1y
s s
2
1
H 1y␣ 0
Ž t . t dt 1qt
⬁
2 1y␣
Ý Ž y1. k k . ks2
We see that
Ž1 y ␥ .
FŽ z. z
q ␥ F⬘ Ž z . s 1 q 2 Ž 1 y  .
⬁
Ý k z ky 1 , ks2
which for z s y1 takes the value 1 y 2Ž 1 y  .
1y␣
⬁
Ý Ž y1. k k s 1 y 2 Ž 1 y  . 2 Ž 1 y  .
s ␣.
ks2
This shows that the result is sharp. 4.6. Proof of Theorem 2.7. With ␦ s 1 and c s Ž1 y ␥ .r␥ we see, using Ž4.3., that the Komatu operator applied to the function f Ž z . s z q Ý⬁ks 2 a k z k can be written FŽ z. s z q
⬁
Ý ks2
ak z k 1 y ␥ q k␥
,
which gives
Ž1 y ␥ .
FŽ z. z
q ␥ F⬘ Ž z . s 1 q
⬁
Ý ak z ky 1 s ks2
The result now follows directly from Theorem 2.6.
5. A TECHNICAL LEMMA LEMMA 5.1.
Let hk s
1
H0
1 t log sk Ž t . dt , t
f Ž z. z
.
486
KIM AND RøNNING
where sk Ž t . s
t kq 1
Ž1 q t .
ž
2 kq4
1 y 2t q
ky1 kq3
t2 .
/
Then h k G 0 for k s 1, 2, . . . . Proof. With JnŽ k .
s
1
H0
log
t kq n
Ž1 q t .
2 kq4
dt
we can write ky1
h k s J 2Ž k . y 2 J 3Ž k . q
kq3
J4Ž k . .
From w12x we have the recursion formula
Ž 5.1.
JnŽ k . s
kqn kq3yn
Žk. Jny1 y
1 kq3yn
Žk. Iny1 , k q 3 ) n,
where InŽ k . s
t kq n
1
H0
Ž1 q t .
2 kq3
dt.
In w12x we also find a recursion formula for InŽ k .,
Ž 5.2.
InŽ k . s
y1
Ž k q 2 y n. 2
2 kq2
kqn
q
kq2yn
Žk. Iny1 , k q 2 ) n.
Using Ž5.1. and Ž5.2. we get, after some calculation, that for k G 2, hk s
k 4 q 4 k 3 q 11k 2 q 22 k q 6 2
Ž k y 1 . k 2 Ž k q 1 . Ž k q 3 . 2 2 kq2 q
2 k 3 y 22 k y 4
Ž k y 1. k Ž k q 1. Ž k q 3. 2
I0Ž k . y
2 k
J 0Ž k . .
487
INTEGRAL TRANSFORMS
To complete the proof we apply the same technique as in w12x. With the change of variable tu s 1 we can write hk s
k 4 q 4 k 3 q 11k 2 q 22 k q 6 2
Ž k y 1 . k 2 Ž k q 1 . Ž k q 3 . 2 2 kq2 q y
2 k 3 y 22 k y 4
H1
Ž k y 1. k 2 Ž k q 1. Ž k q 3. 2 k
⬁
H1
Ž log u . u kq 2 Ž 1 q u.
2 kq4
u kq 1
⬁
Ž 1 q u.
2 kq3
du
du.
The second integral above can be written Žwe omit some details since the calculations are very similar to those in w12x.: ⬁
H1
Ž log u . u kq2 2 kq4
Ž 1 q u.
du s
⬁
H1
s
Ž log u . u 1
H1
kq2
d y
ž
2k q 3
u kq 1
⬁
2k q 3 q
Ž 1 q u . y 2 kq3 Ž
kq 2
Ž 1 q u. ⬁
H 2k q 3 1
2 kq3
/
du
Ž log u . u kq 1 Ž 1 q u.
.
2 kq3
du
s L3 q L4 . Writing h k s L1 q L2 y 2k Ž L3 q L4 . we get L2 y
2 k
L3 s
ž
2 k 3 y 22 k y 4
Ž k y 1. k Ž k q 1. Ž k q 3. 2
=
u kq 1
⬁
H1
s
Ž 1 q u.
2 kq3
y
k Ž 2 k q 3.
/
du
2 k 4 y 42 k 2 y 68 k y 12
Ž k y 1. k 2
2
u kq 1
⬁
H Ž k q 1. Ž k q 3. Ž 2 k q 3. 1
Ž 1 q u.
2 kq3
du.
The polynomial 2 k 4 y 42 k 2 y 68 k y 12 is seen to be positive for k G 6. In w12x it is shown that L4 s
kq2
⬁
H 2k q 3 1
Ž log u . u kq 1 Ž 1 q u.
2 kq3
du -
2 Ž k q 2. 2
Ž 2 k q 3. 2
1 2 kq3
,
488
KIM AND RøNNING
which means that L1 y
2 k
L4 )
k 2 q 4 k 3 q 11k 2 q 22 k q 6 2
Ž k y 1. k Ž k q 1. Ž k q 3. 2 2
2 kq2
y
2 Ž k q 2.
1 2
k Ž 2 k q 3. 2
2 kq2
.
From this we find, after some computation, that L1 y 2k L4 ) 0 if and only if 2 k 6 q 16 k 5 q 81k 4 q 256k 3 q 409k 2 q 282 k q 54 ) 0. Hence, L1 y 2k L4 ) 0 for all k. We have now proved that h k ) 0 for k G 6, and by direct computation we find h1 s h2 s h3 s h4 s h5 s
233 960
y
667 100800
7 log 2 20 y
2669 2903040
log 2
y
12457 82790400
s 0.00010682 . . .
105
s 0.0000156618 . . .
log 2 756
y
597941 22140518400
s 0.000002519 . . .
log 2 4620 y
s 0.0000004324 . . .
log 2 25740
s 0.00000007784 . . . .
REFERENCES 1. G. E. Andrews, R. Askey, and R. Roy, ‘‘Special Functions,’’ Cambridge Univ. Press, Cambridge, UK, 1999. 2. B. C. Carlson and D. B. Shaffer, Starlike and prestarlike hypergeometric functions, SIAM J. Math. Anal. 15 Ž1984., 737᎐745. 3. P. N. Chichra, New subclasses of the class of close-to-convex functions, Proc. Amer. Math. Soc. 62, No. 1 Ž1977., 37᎐43. 4. R. Fourier and St. Ruscheweyh, On two extremal problems related to univalent functions, Rocky Mountain J. Math. 24, No. 2 Ž1994., 529᎐538. 5. D. J. Hallenbeck and St. Ruscheweyh, Subordination by convex functions, Proc. Amer. Math. Soc. 52, No. 1 Ž1975., 191᎐195. 6. Y. C. Kim and A. Lecko, Univalence of certain integral operators and differential subordination, Mathematica Ž Cluj . 1 Ž2001.. 7. Y. Komatu, On analytic prolongation of a family of operators, Mathematica Ž Cluj . 32, No. 55 Ž1990., 141᎐145. 8. S. Owa and M. Nunokawa, Application of a subordination theorem, J. Math. Anal. Appl. 188 Ž1994., 219᎐226.
INTEGRAL TRANSFORMS
489
9. S. Ponnusamy, Differential subordination and starlike functions, Complex Variables 19 Ž1992., 185᎐194. 10. S. Ponnusamy and F. Rønning, Duality for Hadamard products applied to certain integral transforms, Complex Variables 32 Ž1997., 263᎐287. 11. S. Ponnusamy and F. Rønning, Geometric properties for convolutions of hypergeometric functions and functions with the derivative in a halfplane, Integral Transforms Special Funct. 8, Nos. 1᎐2 Ž1999., 121᎐138. 12. S. Ponnusamy and F. Rønning, Integral transforms of functions with the derivative in a halfplane, Israel J. Math. 114 Ž1999., 177᎐188. 13. St. Ruscheweyh, ‘‘Convolutions in Geometric Function Theory,’’ Les Presses de l’Universite ´ de Montreal, ´ Montreal, ´ 1982.
Integral Transforms of Certain Subclasses of Analytic Functions Yong Chan Kim1 Department of Mathematics, College of Education, Yeungnam Uni¨ ersity, Gyongsan 712-749, Korea E-mail: [email protected]
and Frode Rønning School of Teacher Education, Sør-Trøndelag College, N-7004 Trondheim, Norway E-mail: [email protected] Submitted by Stephan Ruscheweyh Received September 21, 1999
Let A be the class of normalized analytic functions in the unit disk ⌬ and define the class P␥ Ž  . s f g A < ᭚ g ⺢ < Re e i Ž Ž 1 y ␥ . f Ž z . rz q ␥ f ⬘ Ž z . y  . 4 ) 0, z g ⌬ 4 . For a function f g A we define the integral transform
V Ž f .Ž z . s
H01Ž t .
f Ž tz . t
dt ,
where is a real-valued, nonnegative weight function with H01 Ž t . dt s 1. For f g P␥ Ž  . we will find conditions such that VŽ f . is starlike and also such that VŽ f . g P␥ Ž ␣ .. We also discuss how to write the integral transform as a convolution with a 2 F1 hypergeometric function. 䊚 2001 Academic Press Key Words: starlike functions; integral transforms. 1
The work of the first author was supported by KOSEF under Grant 981-0102-010-2. 466
0022-247Xr01 $35.00 Copyright 䊚 2001 by Academic Press All rights of reproduction in any form reserved.
467
INTEGRAL TRANSFORMS
1. INTRODUCTION Let A be the class of analytic functions in the unit disk ⌬ with the normalization f Ž0. s f ⬘Ž0. y 1 s 0. For f g A we define the integral transform
Ž 1.1.
V Ž f . Ž z . s
1
H0 Ž t .
f Ž tz . t
dt ,
where is a real-valued, nonnegative weight function normalized so that H01 Ž t . dt s 1. For  - 1 and 0 F ␥ F 1 we define the set P␥ Ž  . s f g A < ᭚ g ⺢ < Re e i Ž Ž 1 y ␥ . f Ž z . rz q ␥ f ⬘ Ž z . y  . 4 ) 0, z g ⌬4 . We shall mainly discuss the following problems. Ž1. For f g P␥ Ž  ., find conditions such that VŽ f . g S*, where S* denotes the usual class of starlike functions. Ž2. For f g P␥ Ž  ., find conditions such that VŽ f . g P␥ Ž ␣ .. To deal with these problems we shall use the duality theory for convolutions. Similar methods have earlier been used to discuss univalence and starlikeness Žalso of order ␣ ) 0. of VŽ f . in the case ␥ s 1, e.g., in the papers w4, 10, 12x. In w6x some aspects of the case ␥ - 1 are discussed, using other methods. Some special cases of Ž t . are particularly interesting such as
Ž t . s Ž 1 q c . t c , c ) y1, for which V is known as the Bernardi operator, and ␦
1 Ž c q 1. c Ž t . s t log ⌫Ž ␦ . t
ž /
␦ y1
, c ) y1, ␦ G 0,
which gives the Komatu operator w7x. We see that for ␦ s 1 the Komatu operator reduces to the Bernardi operator. We can also look at some of these integral operators as taking the convolution between the function f and a 2 F1 hypergeometric function. We use the notation 2 F1Ž a, b; c; z . for the usual Gaussian hypergeometric function, and define the convolution operator
Ž 1.2.
Ha, b , c Ž f . Ž z . [ z 2 F1 Ž a, b; c; z . ) f Ž z . ,
468
KIM AND RøNNING
where f Ž z . s Ý⬁ns 1 a n z n. We use the following basic facts about hypergeometric functions Žsee, e.g., w1x.. 2 F1 Ž a, b; c; z . s
⬁
⌫ Ž a q n.
Ž a. n s B Ž x, y . s
1
H0 Ž 1 y t .
Ž a. n Ž b . n
Ý ns0 Ž c . n n! ⌫ Ž a.
xy1 yy1
t
z n , < z < - 1,
,
⌫Ž x. ⌫Ž y.
dt s
⌫Ž x q y.
, x ) 0, y ) 0,
and ⬁
Ž a. n Ž b . n
Ý ns0 Ž c . n n!
s
⌫ Ž c. ⌫ Ž c y a y b.
, c ) b ) 0 and c ) a q b.
⌫ Ž c y a. ⌫ Ž c y b .
Then we can write Ha, b , c Ž f . Ž z . s s s
⬁
Ž a. n Ž b . n
Ý anq1 Ž c . n! n ns0
z nq1
⬁
⌫Ž c.
⌫ Ž a q n. ⌫ Ž b q n.
Ýa ⌫ Ž a . ⌫ Ž b . ns0 nq 1 ⌫ Ž c q n . ⌫ Ž n q 1 .
z nq 1
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1. =
⬁
Ý
⬁
a nq 1 z nq1
ns0
Ž c y a. m Ž 1 y a. m
Ý ms0 Ž c y a y b q 1 . m m! 1 bqny1
=
H0 t
Ž1 y t .
cy aybqm
dt .
This expression can be written as Ha, b , c Ž f . Ž z . s
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1. =
1 by 2
H0
t
Ž1 y t .
cy ayb
2 F1
ž
c y a, 1 y a ; 1 y t f Ž tz . dt , cyaybq1
/
469
INTEGRAL TRANSFORMS
where we assume a ) 0, b ) 0 and c ) a q b y 1. This means that the convolution Ž1.2. is an integral operator of the form Ž1.1. with
Ž t . s
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1. =Ž 1 y t .
cy ayb
2 F1
t by 1
c y a, 1 y a ; 1yt . cyaybq1
ž
/
Using the Euler integral representation Ž c ) b ) 0. 2 F1
Ž a, b; c; z . s
⌫Ž c.
1 by1
⌫ Ž b. ⌫ Ž c y b.
H0
t
Ž1 y t .
cyby1
Ž 1 y tz .
ya
dt
we see that with a s 1 we get H1, b , c Ž f . Ž z . s
⌫Ž c.
1 by1
H ⌫ Ž b. ⌫ Ž c y b. 0
t
Ž1 y t .
cy by1
f Ž tz . t
dt.
This integral operator is known as the Carlson᎐Shaffer operator Žsee w2x.. For the operator Ha, b, c , given a, b, c, and ␥ in a certain range, we get the sharp value of  for problem Ž1.. However, it is likely that the range of a, b, c, and ␥ , for which the result holds, is larger than the one we have obtained. Some related results for Ha, b, c are obtained in w11x. 2. MAIN RESULTS We start with functions from P␥ Ž  ., where ␥ may be any number in w0, 1x. For now we will assume ␥ ) 0. With Ž t . given we define ⌳␥ Ž t . s
1
Ht
Ž s . s 1r␥
ds.
Further we let g␥ Ž t . be the solution of the initial value problem
Ž 2.1.
2 t 1r ␥y1
d
Ž t 1r␥ Ž 1 q g Ž t . . . s ␥ dt
Ž1 q t .
2
, g Ž 0 . s 1.
Before we proceed to state our results we need some more information about the function g␥ . Solving Ž2.1. we see that g␥ Ž t . can be expressed as g␥ Ž t . s
2
␥
ty1 r␥
t
1r ␥y1
H0 Ž 1 q .
2
d y 1.
470
KIM AND RøNNING
Then we get 2
g␥X Ž t . s y
␥
1r ␥y1
t
ty1r␥y1 2
H0 Ž 1 q .
2
d q
2
1
␥ t Ž1 q t .2
.
For 0 - t - 1 we have
1r ␥y1
t
H0 Ž 1 q .
2
d )
1
t
Ž1 q t .
2
H0
1r ␥ y1
d s ␥
t 1r ␥
Ž1 q t .
2
,
which implies that g␥X Ž t . - 0, so g␥ Ž t . is strictly decreasing on Ž0, 1.. For any weight function Ž t . ) 0 with H01 Ž t . dt s 1 we therefore have H01 Ž t . g␥ Ž t . dt - 1. THEOREM 2.1.
Define  - 1 by
 1y
1
H0 Ž t . g
sy
␥
Ž t . dt ,
and let f g P␥ Ž  ., ␥ ) 0. Assume that lim t ª 0q t 1r ␥⌳␥ Ž t . s 0. Then VŽ f . g S* if and only if
Ž 2.2.
Re
1 1r ␥ y1
H0
t
⌳␥ Ž t .
ž
h Ž tz . tz
y
1
Ž1 q t .
2
/
dt G 0, z g ⌬ ,
where
Ž 2.3.
hŽ z . s
z Ž1 q
⑀y 1 2
Ž1 y z.
z.
2
, < ⑀ < s 1.
Theorem 2.1 is not very useful directly because in order to find out whether VŽ f . g S* we have to check the condition Ž2.2., and this is often difficult. Therefore, to get some applicable results we will identify some situations where Ž2.2. holds. The first result of this type is one where we can compare an unknown situation to a known one with a condition that is reasonably easy to check. We introduce the notation L⌳ ␥ Ž h . s inf
1 1r ␥ y1
H zg⌬ 0
t
⌳␥ Ž t . Re
and formulate the following result.
ž
h Ž tz . tz
y
1
Ž1 q t .
2
/
dt
471
INTEGRAL TRANSFORMS
Assume ⌳␥ is integrable on w0, 1x and positi¨ e on Ž0, 1.. If
THEOREM 2.2.
⌳␥ Ž t . log 1t is decreasing on Ž0, 1. then for 1r2 F ␥ F 1 we ha¨ e L⌳ ␥Ž h. G 0, where hŽ z . is as in Ž2.3.. The next results describe what we are able to prove about some of the specific operators mentioned in the Introduction. The first result deals with the Komatu operator. THEOREM 2.3. Let 1r2 F ␥ F 1 and g␥ Ž t . be defined by Ž2.1.. Define  s  Ž c, ␦ , ␥ . by

Ž c q 1. sy 1y ⌫Ž ␦ .
Ž 2.4.
␦
1 c
H0
1
␦ y1
ž /
t log
t
g␥ Ž t . dt.
Then for f g P␥ Ž  . and
Ž c q 1. Fc , ␦ Ž z . s ⌫Ž ␦ .
Ž 2.5.
␦
1 cy 1
H0
t
1
␦ y1
ž / log
t
f Ž tz . dt
we ha¨ e Fc, ␦ g S* for y1 - c F 1r␥ y 1 and ␦ G 1. The ¨ alue of  is sharp. We now turn to the operator Ha, b, c Ž f . for which we obtain the following result. THEOREM 2.4. Let 1r2 F ␥ F 1 and g␥ Ž t . be defined by Ž2.1.. Define  s  Ž a, b, c, ␥ . by
Ž 2.6.
 1y
⌫Ž c.
sy
⌫ Ž b . ⌫ Ž c y b . ⌫ Ž c y a y b q 1.
=
1 by 1
H0
t
Ž1 y t .
cy ayb
2 F1
ž
c y a, 1 y a ; 1 y t g␥ Ž t . dt. cyaybq1
/
Then for f g P␥ Ž  ., 0 - a F 1, 0 - b F 1r␥ , and c G a q b we ha¨ e Ha, b, c Ž f . g S*. The ¨ alue of  is sharp. In the case a s 1 we have the Carlson᎐Shaffer operator which we denote by Lb , c Ž f . Ž z . s
⌫Ž c.
1 by2
H ⌫ Ž b. ⌫ Ž c y b. 0
t
Ž1 y t .
cy by1
f Ž tz . dt ,
472
KIM AND RøNNING
where 0 - b - c. We write the result for the Carlson᎐Shaffer operator explicitly. Let 1r2 F ␥ F 1 and g␥ Ž t . be defined by Ž2.1.. Define
COROLLARY 2.5.  s  Ž b, c, ␥ . by
 1y
sy
⌫Ž c.
1 by1
H ⌫ Ž b. ⌫ Ž c y b. 0
t
Ž1 y t .
cy by1
g␥ Ž t . dt.
Then for f g P␥ Ž  ., 0 - b F 1r␥ , and c G 1 q b we ha¨ e L b, c Ž f . g S*. The ¨ alue of  is sharp. Another question that might be asked is the following. Let ␣ - 1 be given. Find the smallest  s  Ž ␣ . such that if f g P␥ Ž  . we have VŽ f . g P␥ Ž ␣ .. We prove the following result, which holds for all ␥ g w0, 1x. Let ␣ - 1 and 0 F ␥ F 1 be gi¨ en, and define  s  Ž ␣ .
THEOREM 2.6. by
Ž 2.7.
 1y
1
H0 Ž t .
sy
1 y Ž 1 q ␣ . rŽ 1 y ␣ . t 1qt
dt.
If f g P␥ Ž  . then VŽ f . g P␥ Ž ␣ .. The ¨ alue of  is sharp. The last result we will present says something about the location of f Ž z .rz, given that f g P␥ Ž  .. Hence, in this result there is no integral transform involved, except as a tool in the proof. THEOREM 2.7.  Ž ␣ , ␥ . by
Ž 2.8.
 1y
Let ␣ - 1 and 0 - ␥ F 1 be gi¨ en, and define  s
sy
1
1 Ž1y ␥ .r ␥
H ␥ 0
t
1 y Ž 1 q ␣ . rŽ 1 y ␣ . t 1qt
dt.
If f g P␥ Ž  . then for some g ⺢ we ha¨ e Re e i
f z
žŽ. / z
y ␣ ) 0, z g ⌬ .
The ¨ alue of  is sharp.
3. SOME EXAMPLES In this section we shall look at some examples where we see how our results improve earlier results. It should be said that in some of the
473
INTEGRAL TRANSFORMS
examples the results are not directly comparable, because the previous results deal mostly with the classes R␥ Ž  . s f g A < Re Ž 1 y ␥ . f Ž z . rz q ␥ f ⬘ Ž z . y  4 ) 0, z g ⌬ 4 instead of the larger classes P␥ Ž  .. 3.1. EXAMPLE. In w9x Ponnusamy investigated some problems that are similar to those discussed in this paper. The notation is somewhat different, but if we translate Theorem 5 in w9x to our notation we get the following. Let
␦ Ž␥ . s
1
H0
dt 1 q t␥
.
If g g R␥ Ž  ., with
s
y y Ž 1 y Ž 1r2␥ . . Ž 2 ␦ Ž ␥ . y 1 . 1 y Ž 1 y Ž 1r2␥ . . Ž 2 ␦ Ž ␥ . y 1 .
, ␥ G 1r2,
then G Ž z . s H01 g Ž tz . dtrt g S*. Here s 0.0573 . . . is a number that can be computed through a rather involved algorithm which we will not repeat. To compare, we use our Theorem 2.3 with c s 0 and ␦ s 1. Theorem 5 w9x
␥ 1r2 1
 s y s y0.0573 s
y y log 2 q 1r2 1 y log 2 q 1r2
Theorem 2.3
s s y0.3104 . . .
s
4 log 2 y 3 4 log 2 y 2 1 y 2 log 2 2 y 2 log 2
s y0.294 . . . s y0.629 . . .
In the proof of Theorem 2.3 we see that the extremal function belongs to R␥ Ž  ., so the result is sharp also in this class. 3.2. EXAMPLE. As far as we know, the class R␥ Ž  . was introduced by Chichra w3x, where it was proved that if f g R␥ Ž0. for some ␥ with Re ␥ G 0, then Re f Ž z .rz ) 0 for z g ⌬. Using Theorem 2.7 we see that this result may be extended in the case ␥ ) 0. Indeed, for f g R␥ Ž  . with  - 0, we can conclude that for some g ⺢ we have Re e i f Ž z .rz ) 0. Explicitly, we get for ␥ s 1 that the smallest possible value of  is Ž1 y 2 log 2.rŽ2 y 2 log 2. s y0.629 . . . and for ␥ s 1r2 we get  s Ž4 log 2 y 3.rŽ4 log 2 y 2. s y0.294 . . . . Here we see that the results are not directly comparable, because we cannot actually conclude that Re f Ž z .rz ) 0 by our methods.
474
KIM AND RøNNING
3.3. EXAMPLE. In w8x Owa and Nunokawa improved Chichra’s result. They showed that if Re Ž 1 y ␥ .
gŽ z.
q ␥ g ⬘Ž z . ) 
z
then Re
gŽ z. z
)␣,
where ␣ s  q Ž1 y  .Ž2 ␦ Ž␥ . y 1., and ␦ Ž␥ . is as in Example 3.2. Solving this equation for  we get ␥
s
Ž 3.1.
t ␣ y H01 11 y q t ␥ dt ␥
2H01 1 qt t ␥ dt
.
The result of Owa and Nunokawa corresponds to our Theorem 2.7, but also here with the difference that we work with P␥ Ž  . instead of R␥ Ž  .. Solving equation Ž2.8. with respect to  we get
Ž 3.2.
s
. r Ž1 y ␣ . t H01 t Ž1y ␥ .r ␥ 1 y Ž1 q1␣q dt t . Ž . Ž . 1 y 1 q ␣ r 1 y ␣ t H 1 t Ž1y ␥ .r ␥ dt y ␥
0
1q t
If we do the change of variables t s u␥ in Ž3.2. we see, after a little calculation, that the expressions Ž3.2. and Ž3.1. are equal. There is no comment in w8x that their value of  is sharp. In the proof of Theorem 2.6 and Theorem 2.7 we see that the extremal function belongs to R␥ Ž  ., and it has real coefficients. Therefore, this function will satisfy Re g Ž z .rz ) ␣ , so the result of Owa and Nunokawa is actually sharp. Their proof is based on a subordination theorem by Hallenbeck and Ruscheweyh w5x. 3.4. EXAMPLE. Let
Ž t . s ␦ Ž t . s ⌫ Ž ␦ .
y1
1
␦y1
, ␦ G 0,
ž / log
t
Žthe Komatu operator with c s 0. and F␦ Ž z . s V Ž f .Ž z .. In w6x it is shown ␦ that if Re f ⬘Ž z . )  then Re F␦X Ž z . G 1 y Ž 1 y  . Ž 1 y log 2 . 1 q
ž
1 2
␦ y1
/
[␣
475
INTEGRAL TRANSFORMS
for ␦ G 1. Comparing with the results we get from Theorem 2.6 we get the following table which for given ␦ and ␣ shows the values of  that we get using Corollary 3.1 in w6x and our Theorem 2.6 respectively. The case ␣ s 0. Corollary 3.1 w6x
␦ 1 2
1 y 3 log 2 3 y 3 log 2 1 y 5 log 2 5 y 5 log 2
Theorem 2.6 6y2
s y1.172 . . .
12 y 2 2 y 3 Ž 3 .
s y1.607 . . .
4 y 3 Ž 3 .
s y1.816 . . . s y4.078 . . .
The case ␣ s 1r2. ␦ 1 2
Corollary 3.1 w6x 4 y 6 log 2 6 y 6 log 2 6 y 10 log 2
Theorem 2.6 9y2
s y0.086 . . .
10 y 10 log 2
s y0.303 . . .
12 y 2 3 y 3 Ž 3 . 4 y 3 Ž 3 .
s y0.408 . . . s y1.539 . . .
4. PROOFS 4.1. Proof of Theorem 2.1. Let f g P␥ Ž  . and define GŽ z . s
Ž 1 y ␥ . f Žzz . q ␥ f ⬘ Ž z . y  1y
.
Then for some ␣ g ⺢ we have Re e i ␣ G Ž z . ) 0. With F Ž z . s VŽ f .Ž z ., we want to investigate starlikeness of F; i.e., we want to find conditions such that Re
zF⬘ Ž z . FŽ z.
) 0, z g ⌬ .
By the Duality Principle Žsee w10, 13x. we may restrict our attention to the function f g P␥ Ž  . for which G Ž z . s Ž1 q xz .rŽ1 q yz ., < x < s < y < s 1. Further, it is well known from convolution theory w13, p. 94x that F g S* m
1 z
Ž F ) h . Ž z . / 0, z g ⌬ ,
476
KIM AND RøNNING
where ⑀y 1 2
z Ž1 q
hŽ z . s
Ž1 y z.
z.
, < ⑀ < s 1.
2
Hence, F g S* if and only if Ž␥ / 0. 0/ s
1
Ž F ) h. Ž z .
z
1
1
H0 Ž t . 1 y tz dt )
s Ž1 y  .
1
H0 Ž t . 1
s Ž1 y  .
Ž t .
H0
s Ž1 y  .
H0 Ž t .
ž ž
1
Ž1 y  .
h Ž tz .
h Ž tz .
1
␥ z
1r ␥

q
tz
1
1y
z
/
H0
/
1
1
␥ z
1r ␥
z
H0
1 q yw
dt )
y g␥ Ž t . dt )
tz
w 1r␥y1 Ž 1 q xw .
1
1
␥ z 1r␥ 1
z
1
␥ z 1r␥
w 1r␥y1 h Ž tw . tw
z
H0
H0
dw q  )
w 1r␥y1 Ž 1 q xw . 1 q yw
w 1r␥y1 Ž 1 q xw . 1 q yw
dw y g␥ Ž t . dt )
1 q xz 1 q yz
hŽ z . z dw
dw
.
Using another well-known result from convolution theory w13, p. 23x we conclude that this holds if and only if Re Ž 1 y  .
1
H0
Ž t .
1
1
␥ z
1r ␥
z
H0
w 1r ␥y1 h Ž tw . tw
dw y g␥ Ž t . dt )
1 2
.
Inserting 1 s Ž1 y  .H01Ž t .Ž1 y g␥ Ž t .. dt on the right-hand side we can rewrite this as 1
Re
H0
Ž t .
1 t 1r ␥
t 1r ␥
␥ z 1r␥
w 1r␥y1 h Ž tw .
z
H0
tw
dw y
t 1r␥ Ž 1 q g␥ Ž t . . 2
dt ) 0.
With the change of variables u s tw we now get Re
1
H0
Ž t .
1
t 1r ␥
␥ z 1r␥
1
tz
H0
u1r ␥y1 h Ž u . u
du y
t 1r␥ Ž 1 q g␥ Ž t . . 2
Integrating by parts we get that this is equivalent to Re
1 1r ␥ y1
H0
t
⌳␥ Ž t .
ž
h Ž tz . tz
y
1
Ž1 q t .
2
/
dt G 0.
dt ) 0.
477
INTEGRAL TRANSFORMS
4.2. Proof of Theorem 2.2. We shall omit some details here because there are many similarities between this proof and the proof of Theorem 1 in w4x. With hŽ z . s z Ž1 q Ž ⑀ y 1. zr2.rŽ1 y z . 2 , < ⑀ < s 1, we want to prove that 1 1r ␥ y1
Ž 4.1.
⌳␥ Ž t . Re
t
H0
ž
h Ž tz . tz
y
1
Ž1 q t .
/
2
dt G 0
for z g ⌬. Using the same argument as in w4x we see that it is enough to prove Ž4.1. for < z < s 1. Let ⑀ in the expression of h be ⑀ s e i and minimize Re hŽ tz .rtz with respect to . Then we see that Ž4.1. follows if 2 y tz
1 1r ␥ y1
H0
⌳␥ Ž t . Re
t
Ž 1 y tz .
2
y
t <1 y tz < 2
y
2
Ž1 q t .
2
G 0,
< z < s 1, z / 1. Further calculation gives the equivalent condition H
Ž␥ .
1 1r ␥
Ž y. s H t 0
⌳␥ Ž t .
3 y 4 Ž 1 q y . t q 2 Ž 4 y y 1. t 2 q 4 Ž y y 1. t 3 y t 4 2
Ž 1 q t 2 y 2 yt . Ž 1 q t . 2
G 0, where y s Re z. A series expansion of H Ž ␥ . Ž y . gives H Ž␥ . Ž y . s
⬁
Ý HkŽ␥ . Ž 1 q y . k ,
<1 q y < - 2,
ks0
where HkŽ ␥ . can be seen to be a positive multiple of
˜hŽk␥ . s H1 t 1r␥y1⌳␥ Ž t . sk Ž t . dt , 0
where sk Ž t . s
t kq 1
Ž1 q t .
2 kq4
ž
1 y 2t q
ky1 kq3
t2 .
/
We can see that sk Ž t . has one and only one zero in Ž0, 1.. We denote this zero by t k , and then we see that sk Ž t . ) 0 if 0 F t - t k and sk Ž t . - 0 if t k - t - 1. Let hŽk␥ . s
1 1r ␥ y1
H0
t
1 log sk Ž t . dt , t
478
KIM AND RøNNING
and ⌳␥ Ž t k .
˜ ␥ Ž t . s ⌳␥ Ž t . y ⌳
log Ž 1rt k .
log
1 t
.
˜ ␥ Ž t . sk Ž t . The assumption that ⌳␥ Ž t .rlogŽ1rt . is decreasing implies that ⌳ ) 0 and therefore 0F
⌳␥ Ž t k .
1 1r ␥ y1
H0
t
˜ ␥ Ž t . sk Ž t . dt s ˜hŽk␥ . y ⌳
log Ž 1rt k .
hŽk␥ . .
If we can prove that hŽk␥ . G 0 it follows that ˜ hŽk␥ . G 0 and further L⌳ ␥Ž h. G Ž1r2. 0, as desired. In Lemma 5.1 we show that h k G 0, and we will now show that for 1r2 F ␥ F 1 we have hŽk␥ . G hŽ1r2. . This implies that L⌳ ␥Ž h. G 0 k for 1r2 F ␥ F 1. We can write hŽk␥ . s
1
1
H0
t
2y1r ␥
1 t log sk Ž t . dt. t
With t k as above, set MkŽ ␥ . [ 1rt k2y 1r␥ G 1. Then 1
tk
I1Ž ␥ . [
H0
I2Ž ␥ . [
Ht
t
2y1r ␥
t
2y1r ␥
1
1
k
1 t log sk Ž t . dt G MkŽ ␥ . t
1 t log sk Ž t . dt [ MkŽ ␥ . I1 t
tk
H0
1 t log sk Ž t . dt F MkŽ ␥ . t
1
Ht
k
1 t log sk Ž t . dt [ MkŽ ␥ . I2 . t
Hence, hŽk␥ . s I1Ž ␥ . y I2Ž ␥ . G MkŽ ␥ . Ž I1 y I2 . s hŽ1r2. G 0. k 4.3. Proof of Theorem 2.3. Let 1r2 F ␥ F 1 and assume c F 1r␥ y 1, ␦ G 1. Then ␦
Ž t . s
1 Ž c q 1. c t log ⌫Ž ␦ . t
␦ y1
ž /
and ⌳␥ Ž t . s
Ž c q 1. ⌫Ž ␦ .
␦
1 cy 1r ␥
Ht
s
1
ž / log
s
␦ y1
ds.
479
INTEGRAL TRANSFORMS
It is easily verified that t 1r ␥⌳␥ Ž t . ª 0 when t ª 0q. Using Theorems 2.1 and 2.2 the result follows if we can prove that the function g Ž t . s ⌳␥ Ž t .rlog 1t is decreasing on Ž0, 1.. Differentiating we get
Ž c q 1. g ⬘Ž t . s ⌫Ž ␦ .
␦ 1 t
Ht1 s cy1r␥ Ž log 1s .
␦y1
ds y t cy1r␥ Ž log 1t .
Ž log 1t .
2
␦
.
If we can prove that 1 cy 1r ␥
Ž 4.2.
Ht
s
1
␦y1
ds y t
ž / log
cq1y1r ␥
s
␦
1
F0
ž / log
t
we have g ⬘Ž t . F 0, and we are done. We see that for t s 1 we have equality in Ž4.2., so the inequality holds if the left-hand side is increasing. Differentiating the left-hand side with respect to t we get yt cy 1r␥ log
1
␦y1
q Ž 1r␥ y 1 y c . t cy1r␥ log
ž / ž / t
s t cy 1r␥ log
1
1
␦
ž / t
q ␦ t cy1r␥ log
␦ y1
ž /
␦y1
␦ y 1 q Ž 1r␥ y 1 y c . log
t
1
1 t
t
,
which clearly is greater than or equal to zero when ␦ G 1 and c F 1r␥ y 1. To prove sharpness, let f g P␥ Ž  . be the function for which
Ž1 y ␥ .
f Ž z. z
q ␥ f ⬘Ž z . y  s Ž 1 y  .
1qz 1yz
.
Using a series expansion we see that we can write f Ž z . s z q 2Ž 1 y  .
⬁
zk
Ý
1 y ␥ q ␥k
ks2
.
Further, using
Ž 4.3.
1 cq ky1
H0
t
1
ž / log
t
␦y1
dt s
⌫Ž ␦ .
Ž c q k.
␦
we get from Ž2.5. that ␦ Ž 4.4. Fc , ␦ Ž z . s z q 2 Ž 1 y  . Ž 1 q c .
⬁
Ý ks2
zk ␦
Ž c q k. Ž1 y ␥ q ␥ k.
480
KIM AND RøNNING
and
Ž 4.5.
zFcX, ␦ Ž z . s z q 2 Ž 1 y  . Ž 1 q c .
␦
⬁
kz k
Ý ks2
␦
Ž c q k. Ž1 y ␥ q ␥ k.
.
Expanding g␥ Ž t . in Ž2.1. in a power series, we obtain g␥ Ž t . s 1 q 2
Ž 4.6.
k
⬁
Ž y1. Ž 1 q k .
Ý
1 q ␥k
ks1
tk
which, when inserted into Ž2.4. and using Ž4.3., gives
 1y
s y1 q 2 Ž 1 q c .
␦
⬁
Ž y1.
Ý
kq 1
Ž k q 1. ␦
Ž c q k q 1. Ž 1 q ␥ k .
ks1
from which we get 1
Ž 4.7.
1y
s 2Ž 1 q c .
⬁
␦
Ž y1.
Ý ks1
kq 1
Ž k q 1. ␦
Ž c q k q 1. Ž 1 q ␥ k .
.
Substituting z s y1 into Ž4.5. the right-hand side becomes, after a shift in the indices, y1 q 2 Ž 1 y  . Ž 1 q c .
␦
⬁
Ý ks1
Ž k q 1 . Ž y1.
kq 1
␦
Ž c q k q 1. Ž 1 q ␥ k .
.
Using Ž4.7. we see that this is zero, and since Fc, ␦ Žy1. / 0, this shows that the result is sharp. 4.4. Proof of Theorem 2.4. Let 1r2 F ␥ F 1. We introduce the notation FA , B , C Ž 1 y t . s2 F1 Ž A, B; C ; 1 y t . . With A s c y a, B s 1 y a, and C s c y a y b q 1 we can write
Ž t . s
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1.
t by1 Ž 1 y t .
cy ayb
FA , B , C Ž 1 y t . ,
and we get ⌳␥ Ž t . s
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1. =
1 by1y1r ␥
Ht
s
Ž1 y s.
cy ayb
FA , B , C Ž 1 y s . ds.
481
INTEGRAL TRANSFORMS
Again it is easily verified that t 1r ␥⌳␥ Ž t . ª 0 when t ª 0q. Define ⌳␥ Ž t .
gŽ t. s
log 1t
.
As in the proof of Theorem 2.3 we want to prove that g ⬘Ž t . - 0 on Ž0, 1.. Differentiating we get g ⬘Ž t . s
⌫Ž c.
h Ž t . rt
⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1.
Ž log 1t .
2
,
where hŽ t . s
1 by1y1r ␥
s
Ht
Ž1 y s.
cy ayb
y log Ž 1rt . t by1r␥ Ž 1 y t .
FA , B , C Ž 1 y s . ds cy ayb
FA , B , C Ž 1 y t . .
Then hŽ1. s 0 and it is enough to show that h⬘Ž t . G 0 on Ž0, 1.. Using d dt 2
F1
ž
c y a, 1 y a ; 1yt cyaybq1
Ž c y a. Ž 1 y a.
sy
cyaybq1
/
2 F1
ž
c y a q 1, 2 y a ; 1yt cyaybq2
/
we get h⬘ Ž t . s ylog Ž 1rt . t by1y1r␥ Ž 1 y t . =
žž
1
␥
qayc tqby
yt Ž 1 y t .
/
cy ayby1
1
␥
/
Ž c y a. Ž 1 y a. cyaybq1
FA , B , C Ž 1 y t . FAq 1, Bq1, Cq1 Ž 1 y t . .
If we assume that 0 - a F 1, 0 - b F 1r␥ and c G a q b we see that the inequality below holds for t g Ž0, 1. since the left-hand side will be positive and the right-hand side will be negative. c y a, 1 y a ; 1yt cyaybq1 c y a q 1, 2 y a ; 1yt 2 F1 cyaybq2 2 F1
ž
/
ž
Gy
Ž c y a. Ž 1 y a.
/ t Ž1 y t .
c y a y b q 1 Ž c y a y 1r␥ . t q 1r␥ y b
.
482
KIM AND RøNNING
This clearly implies h⬘Ž t . G 0 for t g Ž0, 1.. The proof of sharpness follows much the same track as in the proof of Theorem 2.3. As extremal function we again have f Ž z . s z q 2Ž 1 y  .
⬁
zk
Ý
1 y ␥ q ␥k
ks2
.
Let ⌸k [
1 bq ky2
H0
t
Ž1 y t .
cy ayb
2 F1
ž
c y a, 1 y a ; 1 y t dt. cyaybq1
/
Then we can write F Ž z . [ Ha, b , c Ž f . Ž z . ⬁
⌫Ž c.
s z q 2Ž 1 y  .
⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1.
Ý ks2
⌸k zk 1 y ␥ q ␥k
and
Ž 4.8.
zF⬘ Ž z . s z q 2 Ž 1 y  . =
⬁
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1.
k⌸ k z k
Ý
1 y ␥ q ␥k
ks2
.
With g␥ as in Ž4.6. we get from Ž2.6.
Ž 4.9.
1 1y
s2
⌫Ž c. ⌫ Ž a. ⌫ Ž b . ⌫ Ž c y a y b q 1.
=
⬁
Ý ks2
Ž y1.
kq 1
Ž k q 1 . ⌸ kq 1
1 q ␥k
.
Substituting z s y1 into Ž4.8. and using Ž4.9. we see, after a shift in the indices, that the right-hand side of Ž4.8. becomes zero. Clearly F Žy1. / 0, so this shows that the result is sharp. 4.5. Proof of Theorem 2.6. The idea of the proof is similar to the one used to prove Theorem 2 in w4x. Let F Ž z . s H01 Ž t . f Ž ttz . dt. Then we can write F⬘Ž z . s H01Ž t . f ⬘Ž tz . dt s f ⬘Ž z .)H01Ž t . dtrŽ1 y tz .. The assumption that f g P␥ Ž  . means that with
Ž 4.10.
gŽ z. s
Ž 1 y ␥ . f Ž z . rz q ␥ f ⬘ Ž z . y  1y
483
INTEGRAL TRANSFORMS
it is so that for some g ⺢ we have Re e i g Ž z . ) 0. Assume first that ␥ / 0. Then we get from Ž4.10. f ⬘Ž z . s
1
␥
Ž  q Ž1 y  . g Ž z. . y
1 y ␥ f Ž z.
␥
z
,
which leads to F⬘ Ž z . s
1
␥ y
s
1
␥
Ž t .
1
Ž  q Ž 1 y  . g Ž z . . )H
0
1 y ␥ f Ž z.
␥
)
1
H0
z
Ž t . 1 y tz
g Ž z.)  q Ž1 y  .
1
H0
dt
1 y tz
dt
Ž t . 1 y tz
dt y
1 y ␥ FŽ z.
␥
z
and further
Ž 4.11.
Ž1 y ␥ .
FŽ z. z
Ž t .
1
q ␥ F⬘ Ž z . s g Ž z . )  q Ž 1 y  .
H0
1 y tz
dt .
In the case ␥ s 0 we just write f Ž z. z
s Ž1 y  . g Ž z. q 
from which we get FŽ z. z
s g Ž z.)  q Ž1 y  .
1
H0
Ž t . 1 y tz
dt ,
which is Ž4.11. with ␥ s 0. Clearly F g P␥ Ž ␣ . m G Ž z . [ Ž F Ž z . y ␣ z .r Ž1 y ␣ . g P␥ Ž0.. With this function G the equality Ž4.11. can be written
Ž1 y ␥ .
GŽ z. z
q ␥ G⬘ Ž z . s g Ž z . )
y␣ 1y␣
q
1y
1
H 1y␣ 0
Ž t . 1 y tz
Since Re e i g Ž z . ) 0 for some g ⺢ it follows by duality that
Ž1 y ␥ .
GŽ z. z
q ␥ G⬘ Ž z . / 0
dt .
484
KIM AND RøNNING
if and only if
Ž 4.12.
Re
y␣ 1y␣
1y
q
Ž t .
1
H 1y␣ 0
1
dt )
1 y tz
2
, z g ⌬.
Using Re 1rŽ1 y tz . ) 1rŽ1 q t . we get that the left-hand side of Ž4.12. is greater than 1y y␣
q
1y␣ 1y
Ž t .
1
H0
dt .
1qt
Using the definition of rŽ1 y  . in Ž2.7. we get after some computation
 y Ž 1 q ␣ . r2 1y
1
sy
H0
Ž t . 1qt
dt
from which we get
y␣ 1y
q
1
H0
Ž t . 1qt
dt s
1 1y␣ 2 1y
,
which shows that Ž4.12. holds. Further, if Ž4.12. holds we deduce, using duality, that Ž1 y ␥ .G Ž z .rz q ␥ G⬘Ž z . is contained in a half plane not containing the origin, which means that G g P␥ Ž0., and hence F g P␥ Ž ␣ .. To prove sharpness we take, as in the proof of Theorem 2.3, f Ž z . s z q 2Ž 1 y  .
⬁
zk
Ý
1 y ␥ q ␥k
ks2
.
Then we can write
k z k
⬁
F Ž z . s V Ž f . Ž z . s z q 2 Ž 1 y  .
Ý ks2
1 y ␥ q ␥k
,
where k s H01 Ž t . t ky 1 dt. With this notation we get from Ž2.7. that 1 1y
y1s
 1y
s y1 q
1
H0 Ž t .
sy 1
H0 Ž t .
ž
1q
ž
1y
1q␣ 1y␣
1q␣ 1y␣
/
t dtr Ž 1 q t .
/
t dtr Ž 1 q t .
485
INTEGRAL TRANSFORMS
and 1 1y
s s
2
1
H 1y␣ 0
Ž t . t dt 1qt
⬁
2 1y␣
Ý Ž y1. k k . ks2
We see that
Ž1 y ␥ .
FŽ z. z
q ␥ F⬘ Ž z . s 1 q 2 Ž 1 y  .
⬁
Ý k z ky 1 , ks2
which for z s y1 takes the value 1 y 2Ž 1 y  .
1y␣
⬁
Ý Ž y1. k k s 1 y 2 Ž 1 y  . 2 Ž 1 y  .
s ␣.
ks2
This shows that the result is sharp. 4.6. Proof of Theorem 2.7. With ␦ s 1 and c s Ž1 y ␥ .r␥ we see, using Ž4.3., that the Komatu operator applied to the function f Ž z . s z q Ý⬁ks 2 a k z k can be written FŽ z. s z q
⬁
Ý ks2
ak z k 1 y ␥ q k␥
,
which gives
Ž1 y ␥ .
FŽ z. z
q ␥ F⬘ Ž z . s 1 q
⬁
Ý ak z ky 1 s ks2
The result now follows directly from Theorem 2.6.
5. A TECHNICAL LEMMA LEMMA 5.1.
Let hk s
1
H0
1 t log sk Ž t . dt , t
f Ž z. z
.
486
KIM AND RøNNING
where sk Ž t . s
t kq 1
Ž1 q t .
ž
2 kq4
1 y 2t q
ky1 kq3
t2 .
/
Then h k G 0 for k s 1, 2, . . . . Proof. With JnŽ k .
s
1
H0
log
t kq n
Ž1 q t .
2 kq4
dt
we can write ky1
h k s J 2Ž k . y 2 J 3Ž k . q
kq3
J4Ž k . .
From w12x we have the recursion formula
Ž 5.1.
JnŽ k . s
kqn kq3yn
Žk. Jny1 y
1 kq3yn
Žk. Iny1 , k q 3 ) n,
where InŽ k . s
t kq n
1
H0
Ž1 q t .
2 kq3
dt.
In w12x we also find a recursion formula for InŽ k .,
Ž 5.2.
InŽ k . s
y1
Ž k q 2 y n. 2
2 kq2
kqn
q
kq2yn
Žk. Iny1 , k q 2 ) n.
Using Ž5.1. and Ž5.2. we get, after some calculation, that for k G 2, hk s
k 4 q 4 k 3 q 11k 2 q 22 k q 6 2
Ž k y 1 . k 2 Ž k q 1 . Ž k q 3 . 2 2 kq2 q
2 k 3 y 22 k y 4
Ž k y 1. k Ž k q 1. Ž k q 3. 2
I0Ž k . y
2 k
J 0Ž k . .
487
INTEGRAL TRANSFORMS
To complete the proof we apply the same technique as in w12x. With the change of variable tu s 1 we can write hk s
k 4 q 4 k 3 q 11k 2 q 22 k q 6 2
Ž k y 1 . k 2 Ž k q 1 . Ž k q 3 . 2 2 kq2 q y
2 k 3 y 22 k y 4
H1
Ž k y 1. k 2 Ž k q 1. Ž k q 3. 2 k
⬁
H1
Ž log u . u kq 2 Ž 1 q u.
2 kq4
u kq 1
⬁
Ž 1 q u.
2 kq3
du
du.
The second integral above can be written Žwe omit some details since the calculations are very similar to those in w12x.: ⬁
H1
Ž log u . u kq2 2 kq4
Ž 1 q u.
du s
⬁
H1
s
Ž log u . u 1
H1
kq2
d y
ž
2k q 3
u kq 1
⬁
2k q 3 q
Ž 1 q u . y 2 kq3 Ž
kq 2
Ž 1 q u. ⬁
H 2k q 3 1
2 kq3
/
du
Ž log u . u kq 1 Ž 1 q u.
.
2 kq3
du
s L3 q L4 . Writing h k s L1 q L2 y 2k Ž L3 q L4 . we get L2 y
2 k
L3 s
ž
2 k 3 y 22 k y 4
Ž k y 1. k Ž k q 1. Ž k q 3. 2
=
u kq 1
⬁
H1
s
Ž 1 q u.
2 kq3
y
k Ž 2 k q 3.
/
du
2 k 4 y 42 k 2 y 68 k y 12
Ž k y 1. k 2
2
u kq 1
⬁
H Ž k q 1. Ž k q 3. Ž 2 k q 3. 1
Ž 1 q u.
2 kq3
du.
The polynomial 2 k 4 y 42 k 2 y 68 k y 12 is seen to be positive for k G 6. In w12x it is shown that L4 s
kq2
⬁
H 2k q 3 1
Ž log u . u kq 1 Ž 1 q u.
2 kq3
du -
2 Ž k q 2. 2
Ž 2 k q 3. 2
1 2 kq3
,
488
KIM AND RøNNING
which means that L1 y
2 k
L4 )
k 2 q 4 k 3 q 11k 2 q 22 k q 6 2
Ž k y 1. k Ž k q 1. Ž k q 3. 2 2
2 kq2
y
2 Ž k q 2.
1 2
k Ž 2 k q 3. 2
2 kq2
.
From this we find, after some computation, that L1 y 2k L4 ) 0 if and only if 2 k 6 q 16 k 5 q 81k 4 q 256k 3 q 409k 2 q 282 k q 54 ) 0. Hence, L1 y 2k L4 ) 0 for all k. We have now proved that h k ) 0 for k G 6, and by direct computation we find h1 s h2 s h3 s h4 s h5 s
233 960
y
667 100800
7 log 2 20 y
2669 2903040
log 2
y
12457 82790400
s 0.00010682 . . .
105
s 0.0000156618 . . .
log 2 756
y
597941 22140518400
s 0.000002519 . . .
log 2 4620 y
s 0.0000004324 . . .
log 2 25740
s 0.00000007784 . . . .
REFERENCES 1. G. E. Andrews, R. Askey, and R. Roy, ‘‘Special Functions,’’ Cambridge Univ. Press, Cambridge, UK, 1999. 2. B. C. Carlson and D. B. Shaffer, Starlike and prestarlike hypergeometric functions, SIAM J. Math. Anal. 15 Ž1984., 737᎐745. 3. P. N. Chichra, New subclasses of the class of close-to-convex functions, Proc. Amer. Math. Soc. 62, No. 1 Ž1977., 37᎐43. 4. R. Fourier and St. Ruscheweyh, On two extremal problems related to univalent functions, Rocky Mountain J. Math. 24, No. 2 Ž1994., 529᎐538. 5. D. J. Hallenbeck and St. Ruscheweyh, Subordination by convex functions, Proc. Amer. Math. Soc. 52, No. 1 Ž1975., 191᎐195. 6. Y. C. Kim and A. Lecko, Univalence of certain integral operators and differential subordination, Mathematica Ž Cluj . 1 Ž2001.. 7. Y. Komatu, On analytic prolongation of a family of operators, Mathematica Ž Cluj . 32, No. 55 Ž1990., 141᎐145. 8. S. Owa and M. Nunokawa, Application of a subordination theorem, J. Math. Anal. Appl. 188 Ž1994., 219᎐226.
INTEGRAL TRANSFORMS
489
9. S. Ponnusamy, Differential subordination and starlike functions, Complex Variables 19 Ž1992., 185᎐194. 10. S. Ponnusamy and F. Rønning, Duality for Hadamard products applied to certain integral transforms, Complex Variables 32 Ž1997., 263᎐287. 11. S. Ponnusamy and F. Rønning, Geometric properties for convolutions of hypergeometric functions and functions with the derivative in a halfplane, Integral Transforms Special Funct. 8, Nos. 1᎐2 Ž1999., 121᎐138. 12. S. Ponnusamy and F. Rønning, Integral transforms of functions with the derivative in a halfplane, Israel J. Math. 114 Ž1999., 177᎐188. 13. St. Ruscheweyh, ‘‘Convolutions in Geometric Function Theory,’’ Les Presses de l’Universite ´ de Montreal, ´ Montreal, ´ 1982.