General Mathematics Vol. 15, Nr. 2–3(2007), 164-176
Subordinations and integral means
inequalities
Tadayuki Sekine, Shigeyoshi Owa and Rikuo Yamakawa
Abstract
Applying the subordination theorem of J. E. Littlewood [1], and
Lemma of S. S. Miller and P. T. Monanu [2] to certain analytic functions, we show an integral means inequality. Further, we obtain an
integral means inequality for the first derivative.
2000 Mathematical Subject Classification: Primary 30C45
Key words and phrases: Integral means inequality, analytic function,
subordination, first derivative.
1
Introduction
Let A denote the class of functions f (z) of the form
(1)
f (z) = z +
∞
X
n=2
164
an z n
Subordinations and integral means inequalities
165
that are analytic in the open unit disk U = {z ∈ C : |z| < 1} .
Let g(z) denote the analytic function in U defined by
(2)
g(z) =
z
.
1−z
In this paper, we discuss the integral means inequalities of f (z) in A
and g(z) of the form (2), and f 0 (z)(f (x) ∈ A) and g 0 (z). Moreover we show
an estimate of f 0 (z).
We recall the concept of subordination between analytic functions. Given
two functions f (z) and g (z), which are analytic in U, the function f (z) is
said to be subordinate to g (z) in U if there exists a function w (z) analytic
in U with w (0) = 0 and |w (z)| < 1, such that f (z) = g (w (z)). We denote
this subordination by f (z) ≺ g (z). If g(z) is univalent in U, f (z) ≺ g (z)
if and only if f (0) = g(0) and f (U) ⊂ g(U).
We need the following subordination theorem of J. E. Littlewood.
Lemma A(Littlewood [1]) If f (z) and g (z) are analytic in U with f (z) ≺
g(z), then, for µ > 0 and z = reiθ (0 < r < 1)
Z 2π
Z 2π
µ
|f (z)| dθ 5
|g (z)|µ dθ.
0
0
Applying the lemma of Littlewood above, H. Silverman [5] showed the integral means inequalities for univalent functions with negative coefficients.
S. Owa and T. Sekine [3] proved integral means inequalities with coefficients inequalities for normalized analytic functions and polynomials(see
also Sekine et al. [4]).
In addition we need the following Lemma of S. S. Miller and P. T.
Mocanu.
166
Tadayuki Sekine, Shigeyoshi Owa and Rikuo Yamakawa
Lemma B(Miller and Mocanu [2]) Let g(z) = gn z n + gn+1 z n+1 + · · · be
analytic in U with g(z) 6= 0 and n = 1. If z0 = r0 eiθ0 (r0 < 1) and
|g(z0 )| = max |g(z)|
5
|z| |z0 |
then
(i)
z0 g 0 (z0 )
=k
g(z0 )
and
µ
(ii) Re
z0 g 00 (z0 )
g 0 (z0 )
¶
+ 1 = k,
where k = n = 1.
2
Integral means for f (z) and g(z)
Theorem 1. Let f (z) be in A and g(z) be the analytic f unction given by (2).
If the f unction f (z) satisf ies
(3)
¾
½
zf 00 (z)
2α+β +2γ −4δ
zf 0 (z)
0
−δ 0
>−
Re αf (z)+βzf (z)−γ
f (z)
f (z)
4
(z ∈ U)
f or α ∈ R, β = 0, β + 2γ = 0 and δ = 0, then, f or µ > 0 and z =
reiθ (0 < r < 1)
Z
2π
(4)
¯ ¡ iθ ¢¯µ
¯f re ¯ dθ 5
0
Z
2π
¯ ¡ iθ ¢¯µ
¯g re ¯ dθ.
0
Proof. By applying Lemma A, it suffices to show that
f (z) ≺
z
.
1−z
Subordinations and integral means inequalities
167
Let us define the function w(z) by
(5)
f (z) =
w(z)
1 − w(z)
(w(z) 6= 1) .
Hence we have an analytic function w(z) in U such that w(0) = 0. Further,
we prove that the analytic function w(z) satisfies |w(z)| < 1(z ∈ U) for
½
¾
zf 0 (z)
zf 00 (z)
0
Re αf (z) + βzf (z) − γ
−δ 0
f (z)
f (z)
½
¾
w(z)
zw0 (z)
zw0 (z)
zw00 (z)
zw0 (z)
= Re α
+β
−γ
−δ 0
−2δ
1−w(z)
(1−w(z))2
(1−w(z))w(z)
w (z)
1−w(z)
2α+β +2γ −4δ
> −
4
(z ∈ U) .
If there exists z0 ∈ U such that
max |w(z)| = |w(z0 )| = 1,
5
|z| |z0 |
then we have by Lemma B,
z0 w0 (z0 )
= k, Re
w(z0 ) = e ,
w(z0 )
iθ
µ
z0 w00 (z0 )
w0 (z0 )
¶
For½such a point z0 ∈ U, we obtain that
¾
z0 f 00 (z0 )
z0 f 0 (z0 )
0
−δ 0
Re αf (z0 ) + βz0 f (z0 ) − γ
f (z0 )
f (z0 )
= k − 1 (k = 1).
168
Tadayuki Sekine, Shigeyoshi Owa and Rikuo Yamakawa
½
w(z0 )
z0 w0 (z0 )
z0 w0 (z0 )
+β
−
γ
1 − w(z0 )
(1 − w(z0 ))2
(1 − w(z0 ))w(z0 )
¾
00
0
z0 w (z0 )
z0 w (z0 )
δ 0
− 2δ
w (z0 )
1 − w(z0 )
½
w(z0 )
kw(z0 )
kw(z0 )
Re α
−γ
+β
2
1 − w(z0 )
(1 − w(z0 ))
(1 − w(z0 ))w(z0 )
¾
00
z0 w (z0 )
kw(z0 )
δ 0
− 2δ
w (z0 )
1 − w(z0 )
½
¾
½
¾
w(z0 )
βkw(z0 )
Re (α − 2δk)
+ Re
1 − w(z0 )
(1 − w(z0 ))2
¾
¾
½
½
γk
δz0 w00 (z0 )
Re
− Re
1 − w(z0 )
w0 (z0 )
½
¾
δz0 w00 (z0 )
2δk − α
βk
γk
+
−
− Re
2
2(cos θ − 1)
2
w0 (z0 )
2δk − α βk γk
−
−
+ δ(1 − k)
2
4
2
α (β + 2γ)k
+δ
− −
2
4
α β + 2γ
− −
+δ
2
4
2α + β + 2γ − 4δ
−
(α ∈ R, β = 0, β + 2γ = 0, δ = 0) ,
4
= Re α
−
=
−
=
−
=
5
=
5
=
which contradicts the hypothesis (3) of the theorem. Therefore there is no
z0 ∈ U such that |w(z0 )| = 1. This implies that |w(z)| < 1 for all z ∈ U.
Thus we have that
f (z) ≺
z
,
1−z
which shows that
Z
2π
¯ ¡ iθ ¢¯µ
¯f re ¯ dθ 5
0
This completes the proof.
Z
2π
0
¯ ¡ iθ ¢¯µ
¯g re ¯ dθ.
Subordinations and integral means inequalities
169
Corollary 1. Let the function f (z) in A and the analytic function g(z)
given by (2) satisfy the conditions in Theorem 1. Then, for µ > 0 and
z = riθ (0 < r < 1)
Z
2π
¯ ¡ iθ ¢¯µ
¯f re ¯ dθ 5
2πrµ
µ
(1 + r2 ) 2
0
(
1+
∞
X
Q2n−1
j=0 (µ + 2j)
22n (n!)2
n=1
µ
r
1 + r2
¶2n )
.
Proof.
Z
2π
¯ ¡ iθ ¢¯µ
¯f re ¯ dθ 5
0
=
=
=
=
=
=
Z
2π
0
Z
rµ
2π
µ
(1 + r2 ) 2
0
Z
rµ
2π
µ
2r cos θ
1−
1 + r2
(
µ
(1 + r2 ) 2
0
Z
rµ
2π
µ
(1 + r2 ) 2
0
0
2π +
µ
(1 + r2 ) 2
µ
r2 ) 2
2π +
n=1
− µ2
n
∞ µ µ¶ µ
X
−
2
n=1
(
rµ
∞ µ
2π X
Z
(
rµ
dθ
)
dθ
¶n
∞ µ µ¶ µ
X
−2
2r cos θ
1+
−
n
1 + r2
n=1
2π +
µ
(1 + r2 ) 2
(1 +
(
¶ −µ
2
¶n
∞ µ µ¶ µ
X
−2
2r cos θ
−
1 + r2
n
n=0
(
rµ
¯
¯
¯ reiθ ¯µ
¯
¯
¯ 1 − reiθ ¯ dθ
n
∞ µ µ¶ µ
X
−
2
n=1
2n
¶µ
2r cos θ
−
1 + r2
2r
−
1 + r2
2r
−
1 + r2
¶n Z
)
dθ
)
¶n
dθ
)
2π
n
cos θ dθ
0
¶2n Z
2π
0
)
cos2n θ dθ
170
Tadayuki Sekine, Shigeyoshi Owa and Rikuo Yamakawa
=
=
=
3
(
rµ
(1 +
2π +
µ
r2 ) 2
1+
µ
(1 + r2 ) 2
2πr
(1 +
n=1
(
µ
2n
Q2n−1
µ
Q2n−1
µ
2
∞
X
1+
µ
r2 ) 2
∞
X
n=1
¶2n
r
1 + r2
n=1
(
2πrµ
∞ µ µ¶ µ
X
−
j=0 (µ + 2j)
22n (2n)!
j=0 (µ + 2j)
22n (n!)2
4(2n)! π
· 22n · n 2 ·
(2 n!) 2
r
1 + r2
r
1 + r2
¶2n
(2n)!
·
(n!)2
)
)
¶2n )
.
Integral means for the first derivative
The proof for the first derivative is similar.
Theorem 2. Let f (z) be in A and g(z) be the analytic f unction given by (2).
If the f unction f (z) satisf ies
¾
½
zf 000 (z)
α − 4β + 6γ
zf 00 (z)
0
− γ 00
>
(6)
Re αf (z) + β 0
f (z)
f (z)
4
for α = 0, 2β − γ 5 0 and γ = 0, then, for µ > 0 and z = reiθ (0 < r < 1)
Z
2π
(7)
¯ 0 ¡ iθ ¢¯µ
¯f re ¯ dθ 5
Z
0
2π
¯ 0 ¡ iθ ¢¯µ
¯g re ¯ dθ.
0
Proof. By Lemma A, it suffices to show that
f 0 (z) ≺
1
.
(1 − z)2
Let us define the function w(z) by
(8)
f 0 (z) =
1
(1 − w(z))2
(w(z) 6= 1) .
Subordinations and integral means inequalities
171
Hence we have an analytic function w(z) in U such that w(0) = 0. Further,
we prove that the analytic function w(z) satisfies |w(z)| < 1(z ∈ U) for
½
zf 00 (z)
zf 000 (z)
Re αf (z) + β 0
− γ 00
f (z)
f (z)
¾
0
½
= Re α
>
1
zw0 (z)
+
2β
−γ
(1 − w(z))2
1 − w(z)
α − 4β + 6γ
4
µ
zw00 (z)
3zw0 (z)
+
w0 (z)
1 − w(z)
¶¾
(z ∈ U).
If there exists z0 ∈ U such that
max |w(z)| = |w(z0 )| = 1,
5
|z| |z0 |
then we have by Lemma B,
z0 w0 (z0 )
w(z0 ) = e ,
= k, Re
w(z0 )
iθ
µ
z0 w00 (z0 )
w0 (z0 )
¶
= k − 1 (k = 1).
For such a point z0 ∈ U, we obtain that
½
¾
z0 f 00 (z0 )
z0 f 000 (z0 )
0
Re αf (z0 ) + β 0
− γ 00
f (z0 )
f (z0 )
½
= Re α
µ
1
z0 w0 (z0 )
+
2β
−γ
(1 − w(z0 ))2
1 − w(z0 )
µ
z0 w00 (z0 ) 3z0 w0 (z0 )
+
w0 (z0 )
1 − w(z0 )
¶¾
¶
µ
¶
µ
¶
µ
¶
α
βkw(z0 )
γz0 w00 (z0 )
γkw(z0 )
= Re
+2Re
−Re
−3Re
(1−w(z0 ))2
1−w(z0 )
w0 (z0 )
1−w(z0 )
172
Tadayuki Sekine, Shigeyoshi Owa and Rikuo Yamakawa
µ
¶
z0 w00 (z0 )
α cos θ
3
− βk + γ −Re 0
=
+ γk
2(cos θ − 1)
w (z0 )
2
5
α
3
− βk + γ(1 − k) + γk
4
2
´
α ³γ
=
+
−β k+γ
4
2
5
α
1
−β+ γ+γ
4
2
=
α − 4β + 6γ
4
(α = 0, 2β − γ 5 0, γ = 0),
which contradicts the hypothesis (6) of the Theorem 2. Therefore there is
no z0 ∈ U such that |w(z0 )| = 1. This implies that |w(z)| < 1 for all z ∈ U.
Thus we have that
f 0 (z) ≺
1
,
(1 − z)2
which shows that
Z
2π
¯ 0 ¡ iθ ¢¯µ
¯f re ¯ dθ 5
Z
0
2π
¯ 0 ¡ iθ ¢¯µ
¯g re ¯ dθ.
0
This completes the proof.
Corollary 2. Let the function f (z) in A and the analytic function g(z)
given by (2) satisfy the conditions in Theorem 2. Then, for µ > 0 and
z = riθ (0 < r < 1)
Z
2π
0
¯ 0 ¡ iθ ¢¯µ
¯f re ¯ dθ 5
2π
(1 + r2 )µ
(
1+
∞
X
n=1
Q2n−1
j=0 (µ + j)
(n!)2
µ
r
1 + r2
¶2n )
.
Subordinations and integral means inequalities
173
Proof.
Z
2π
¯ 0 ¡ iθ ¢¯µ
¯f re ¯ dθ 5
Z
0
1
=
(1 + r2 )µ
1
=
(1 + r2 )µ
1
=
(1 + r2 )µ
1
=
(1 + r2 )µ
1
=
(1 + r2 )µ
1
=
(1 + r2 )µ
1
=
(1 + r2 )µ
2π
=
(1 + r2 )µ
2π
=
(1 + r2 )µ
2π
0
Z
2π
µ
0
Z
2π
2r cos θ
1−
1 + r2
(
0
Z
2π
(
0
∞ µ
2π X
2π +
0
(
2π +
n=1
(
2π +
2π +
2n
¶µ
∞ µ
X
−µ
∞
X
n=1
∞
X
n=1
¶µ
)
dθ
2r cos θ
−
1 + r2
2r
−
1 + r2
2r
−
1 + r2
2n
r
1 + r2
Q2n−1
µ
n=1
1+
n
¶µ
∞ µ
X
−µ
n=1
(
−µ
n
¶µ
∞ µ
X
−µ
n=1
(
dθ
¶µ
¶n
∞ µ
X
2r cos θ
−µ
−
1+
n
1 + r2
n=1
Z
1+
¶−µ
¶µ
¶n
∞ µ
X
−µ
2r cos θ
−
n
1 + r2
n=0
(
(
¯
¯µ
¯
¯
1
¯
¯
¯ (1 − reiθ )2 ¯ dθ
(µ + j)
(2n)!
j=0
Q2n−1
(µ + j)
(n!)2
j=0
µ
¶n Z
)
dθ
)
¶n
dθ
)
2π
n
cos θ dθ
0
¶2n Z
¶2n
2π
)
cos2n θ dθ
0
4(2n)! π
· 22n · n 2 ·
(2 n!) 2
r
1 + r2
r
1 + r2
¶2n
(2n)!
·
(n!)2
¶2n )
Putting µ = 1 in Corollary 2, we have the following.
.
)
)
174
Tadayuki Sekine, Shigeyoshi Owa and Rikuo Yamakawa
Corollary 3. Let the function f (z) in A and the analytic function g(z)
given by (2) satisfy the conditions in Theorem 2. Then, for 0 < r < 1
|f (z)| 5
2π
.
1 − r2
Proof.
2π
|f (z)| 5
1 + r2
2π
=
1 + r2
2π
=
1 + r2
2π
=
1 + r2
(
1+
2π
=
1 + r2
=
Q2n−1
(
1+
∞
X
(2n)!
n=1
(
1+
∞
X
22n
1+
∞
X
µ
1−
µ
2π
.
1 − r2
2 n
(1)n
n
(1)n
µ
2r
1 + r2
¶−1
¶2n )
¶2 !n )
r
1 + r2
2r
1 + r2
¶2 !
¶2 )− 12
r
1 + r2
õ
õ
((1)n )2
2r
1 + r2
1 − r2
1 + r2
2
¡1¢
1
, 1; 1;
2
(
¡1¢
r
1 + r2
((1)n )2
n=1
Ã
õ
(n!)2
n=1
(
µ
(1 + j)
(n!)2
j=0
n=1
2π
F
=
1 + r2
2π
=
1 + r2
∞
X
¶2 !n )
¶2 ! n )
Subordinations and integral means inequalities
175
References
[1] J. E. Littlewood, On inequalities in the theory of functions, Proc. London Math. Soc., (2) 23 (1925), 481-519.
[2] S. S. Miller, P. T. Mocanu, Second order differential inequalities in the
complex plane, J. Math. Anal. Appl., 65(1978), 289-305.
[3] S. Owa, T. Sekine, Integral means for analytic functions, J. Math. Anal.
Appl., 304(2005), 772-782.
[4] T. Sekine, S. Owa, R. Yamakawa, Integral means of certain analytic
functions, General Math., 13(2005), 99-108.
[5] H. Silverman, Integral means for univalent functions with negative coefficients, Houston J. Math., 23(1997), 169-174.
Research Unit of Mathematics
College of Pharmacy, Nihon University
7-1 Narashinodai 7-chome, Funabashi-shi, Chiba 274-8555, Japan
E-mail : tsekine@pha.nihon-u.ac.jp
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502, Japan
E-mail : owa@math.kindai.ac.jp
176
Tadayuki Sekine, Shigeyoshi Owa and Rikuo Yamakawa
Shibaura Institute of Technology
Minuma, Saitama-shi, Saitama 337-8570, Japan
E-mail : yamakawa@sci.shibaura-it.ac.jp