SINGULAR INTEGRALS IN THE UNIT DISKGEORGE A.. GAL Received 23 January 2006; Revised 19 April 2006; Accepted 20 April 2006 The purpose of this paper is to prove several results in approxi
Trang 1SINGULAR INTEGRALS IN THE UNIT DISK
GEORGE A ANASTASSIOU AND SORIN G GAL
Received 23 January 2006; Revised 19 April 2006; Accepted 20 April 2006
The purpose of this paper is to prove several results in approximation by complex Pi-card, Poisson-Cauchy, and Gauss-Weierstrass singular integrals with Jackson-type rate, having the quality of preservation of some properties in geometric function theory, like the preservation of coefficients’ bounds, positive real part, bounded turn, starlikeness, and convexity Also, some sufficient conditions for starlikeness and univalence of analytic functions are preserved
Copyright © 2006 G A Anastassiou and S G Gal This is an open access article distrib-uted under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
1 Introduction
Let us consider the open unit diskD = {z ∈ C;|z| < 1}andA(D) = { f : D → C;f is
an-alytic onD, continuous on D, f (0) =0, f (0)=1} Therefore, if f ∈ A(D), we have
f (z) = z +∞
k =2a k z k, for allz ∈ D.
For f ∈ A(D) and ξ ∈ R,ξ > 0, let us consider the complex singular integrals
P ξ(f )(z) = 1
2ξ
+∞
−∞ f
ze iu
e −| u | /ξ du, z ∈ D,
Q ξ( f )(z) = π ξ
π
− π
f
ze iu
u2+ξ2du, z ∈ D, Q ξ ∗(f )(z) = π ξ
+∞
−∞
f
ze − iu
u2+ξ2 du, z ∈ D,
R ξ(f )(z) =2ξ3
π
+∞
−∞
f
ze iu
u2+ξ2 2du, z ∈ D,
W ξ(f )(z) =1
πξ
π
− π f
ze iu
e − u2/ξ du, z ∈ D,
W ξ ∗(f )(z) =1
πξ
+∞
−∞ f
ze − iu
e − u2/ξ du, z ∈ D.
(1.1)
Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2006, Article ID 17231, Pages 1 19
DOI 10.1155/JIA/2006/17231
Trang 2HereP ξ(f ) is said to be of Picard type, Q ξ(f ), Q ξ ∗(f ), and R ξ(f ) are said to be of
Poisson-Cauchy type, andW ξ(f ) and W ξ ∗(f ) are said to be of Gauss-Weierstrass type.
In the very recent papers [3–5], classes of convolution complex polynomials were in-troduced and their approximation properties regarding rates, global smoothness preser-vation properties, and some geometric properties like the preserpreser-vation of coefficients’ bounds, positivity of real part, bounded turn, starlikeness, convexity, and univalence were proved
The aim of this paper is to obtain similar properties for the above-defined complex singular integrals
2 Complex Picard integrals
In this section, we study the properties ofP ξ(f )(z).
Firstly, we present the approximation properties
Theorem 2.1 Let f ∈ A(D) and ξ ∈ R, ξ > 0 Then
(i)P ξ(f )(z) is continuous on D, analytic on D, and P ξ(f )(0) = 0;
(ii)ω1(P ξ(f ); δ) D ≤ ω1(f ; δ) D , for all δ ≥ 0, where ω1(f ; δ) D =sup{| f (z1)− f (z2)|;
z1,z2∈ D, |z1− z2| ≤ δ};
(iii)|P ξ(f )(z) − f (z)| ≤ Cω2(f ; ξ) ∂D , for all z ∈ D, ξ > 0, where
ω2(f ; ξ) ∂D =supf
e i(x+u)
−2f
e iu
+ f
e i(x − u);x ∈ R,|u| ≤ ξ
Proof (i) Let z0,z n ∈ D be with lim n →∞ z n = z0 We get
P ξ(f )
z n
− P ξ( f )
z0 ≤ 1
2ξ
+∞
−∞
f
z n e iu
− f
z0e iue −| u | /ξ du
≤ 1
2ξ
+∞
−∞ ω1
f ;z n e iu − z0e iu
D e −| u | /ξ du
= 1
2ξ
+∞
−∞ ω1
f ;z n − z0
D e −| u | /ξ du
= ω1
f ;z n − z0
D
(2.2)
Passing to limit withn → ∞, it follows thatP ξ(f )(z) is continuous at z0∈ D, since f is
continuous onD It remains to prove that P ξ(f )(z) is analytic on D For f ∈ A(D), we
can write f (z) =∞ k =0a k z k,z ∈ D For fixed z ∈ D, we get f (ze iu)=∞ k =0a k e iku z k and since|a k e iku | = |a k |, for allu ∈ R, and the series ∞
k =0a k z k is absolutely convergent, it follows that the series∞
k =0a k e iku z kis uniformly convergent with respect tou ∈ R This immediately implies that the series can be integrated term by term, that is,
P ξ(f )(z) = 1
2ξ
∞
k =0
a k z k
∞
−∞ e iku e −| u | /ξ du
Also, sincea0=0, we getP ξ(f )(0) =0
Trang 3(ii) Letz1,z2∈ D, |z1− z2| ≤ δ We get
P ξ(f )
z1
− P ξ( f )
z2 ≤ 1
2ξ
+∞
−∞
f
z1e iu
− f
z2e iue −| u | /ξ du
≤ ω1
f ;z1− z2
D ≤ ω1(f ; δ) D
(2.4)
Passing to sup with|z1− z2| < δ, the desired inequality follows.
(iii) We have
P ξ(f )(z) − f (z) = 1
2ξ
+∞
−∞ f
ze iu
− f (z)
e −| u | /ξ du
= 1
2ξ
∞
0 f
ze iu
−2f (z) + f
ze − iu
e − u/ξ du,
(2.5)
which implies
P ξ(f )(z) − f (z) ≤ 1
2ξ
∞
0
f
ze iu
−2f (z) + f
ze − iue − u/ξ du, (2.6)
for allz ∈ D.
By the maximum modulus principle (see, e.g., [3, page 421]), we can take|z| =1, case when
f
ze iu
−2f (z) + f
ze − iu ≤ ω2(f ; u) ∂D, (2.7) which implies that for allz ∈ D we have
P ξ(f )(z) − f (z) ≤ 1
2ξ
+∞
0 ω2(f ; u) ∂D e − u/ξ du
= 1
2ξ
+∞
0 ω2
f ; u
ξ · ξ
∂D
e − u/ξ du
≤
1
2ξ
+∞
0
1 +u
ξ
2
e − u/ξ du
ω2(f ; ξ) ∂D ≤ Cω2(f ; ξ) ∂D
(2.8)
(for the last inequalities, see, e.g., [2, proof of Theorem 2.1(i), page 252])
Remark 2.2. Theorem 2.1(ii) and (iii) remain valid for f only continuous on D.
In what follows, we present some geometric properties ofP ξ(f )(z).
Theorem 2.3 If f (z) =∞ k =0a k z k , for all z ∈ D, then
P ξ(f )(z) =
∞
k =0
a k
Trang 4for all z ∈ D, that is, if f (0) = 0, then P ξ(f )(0) = 0 and if f (0)= 1, then P ξ(f )(0) =
1/(1 + ξ2) = 1, for all ξ > 0 Also,
a k
P ξ(f ) = a k( f )
1 +ξ2k2
≤a k( f ), ∀k =0, 1, . (2.10)
Proof In the proof ofTheorem 2.1(i), we can write
P ξ(f )(z) =
∞
k =0
a k z k
1
2ξ
+∞
−∞ e iku e −| u | /ξ du
But
1
2ξ
+∞
−∞ e iku e −| u | /ξ du
= 1
2ξ
+∞
−∞cos(ku) · e −| u | /ξ du =1
ξ
+∞
0 cos(ku)e − u/ξ du
=1
ξ · e − u/ξ −(1/ξ) cos(ku) + k sin(ku)
1/ξ2+k2
∞
1 +k2ξ2,
(2.12)
Now, recall that a function f ∈ A(D) is starlike if it is univalent and f (D) is a starlike
plane domain with respect to 0, and is convex if it is univalent onD and f (D) is a convex
plane domain
Also, let us introduce the following classes of analytic functions:
S1=
f ∈ A(D); f (z) = z +
∞
k =2
a k z k,
∞
k =2
ka k ≤1
,
S2=
f analytic in D, f (z) =
∞
k =1
a k z k,z ∈ D,a1 ≥ ∞
k =2
a k,
S3=f ∈ A(D);f (z) ≤1,∀z ∈ D
,
ᏼ=f : D −→ C;f is analytic on D, f (0) =1, Re f (z)
> 0, ∀z ∈ D
,
=f ∈ A(D); Re f (z)
> 0, ∀z ∈ D
,
S M =f ∈ A(D);f (z)< M, ∀z ∈ D
, M > 1.
(2.13)
According to, for example, [6, Exercise 4.9.1, page 97], iff ∈ S1, then|z f (z)/ f (z) −1<
1, for allz ∈ D, and therefore f is starlike (and univalent) on D.
According to [1, page 22 ], if f ∈ S2, then f is starlike (and univalent) on D.
By [7], if f ∈ S3, then f is starlike (and univalent) on D Also, it is well known that
is the class of functions with bounded turn (i.e.,|argf (z)| < π/2, for all z ∈ D) and that
f ∈ implies the univalency of f on D.
According to, for example, [6, Exercise 5.4.1, page 111], f ∈ S M implies thatf is
uni-valent in{z ∈ C;|z| < 1/M}
Trang 5We present the following.
Theorem 2.4 For all ξ > 0,
P ξ
S2
Proof ByTheorem 2.3, for f (z) =∞ k =1a k z k ∈ S2, we get
∞
k =2
a k
1 +ξ2k2
= ∞
k =2
a k
1 +ξ2· 1 +ξ2
1 +ξ2k2 ≤ 1
1 +ξ2
∞
k =2
a k ≤ a1
1 +ξ2 (2.15)
and sinceP ξ(f )(z) =∞ k =0(a k /(1 + ξ2k2))z k, it follows thatP ξ(f ) ∈ S2.
Let f (z) =∞ k =0a k z k ∈ ᏼ, that is, a0 =1 and if f (z)= U(x, y) + iV (x, y), z =x + iy ∈ D,
thenU(x, y) > 0, for all z = x + iy ∈ D.
We getP ξ(f )(0) = a0=1 and
P ξ( f )(z) = 1
2ξ
+∞
−∞ U
r cos(u + t), r sin(u + t)
e −| u | /ξ du
+i · 1
2ξ
+∞
−∞ V
r cos(u + t), r sin(u + t)
e −| u | /ξ du, ∀z = re it ∈ D,
(2.16)
which immediately implies
Re P ξ( f )(z)
= 1
2ξ
+∞
−∞ U
r cos(u + t), r sin(u + t)
e −| u | /ξ du > 0, (2.17)
Theorem 2.5 For all ξ > 0, (1 + ξ2)P ξ( S1)⊂ S1, (1 + ξ2)P ξ( S M)⊂ S M(1+ξ2 ), and (1 +
ξ2)P ξ(S3,ξ)⊂ S3, where
S3,ξ=
f ∈ S3; f (z) ≤ 1
1 +ξ2,∀z ∈ D
Proof Let f ∈ S1 ByTheorem 2.3, we obtain
1 +ξ2
P ξ(f )(z) =
∞
k =1
a k 1 +ξ2
if f (z) =∞ k =1a k z k ∈ S1 It follows that (1 +ξ2)P ξ(f )(0) = a1=1, that is,
1 +ξ2
P ξ(f )(z) = z +
∞
k =2
a k · 1 +ξ2
1 +ξ2k2z k,
∞
k =2
ka k 1 +ξ2
1 +ξ2k2 ≤
∞
k =2
ka k ≤1,
(2.20)
that is, (1 +ξ2)P ξ(f ) ∈ S1.
Trang 6Let f ∈ S M We get
1 +ξ2
P ξ (f )(z) = 1 +ξ2
·
21ξ−∞+∞ f
ze iu
e iu e −| u | /ξ du
≤1 +ξ2 1
2ξ
+∞
−∞
f
ze iue −| u | /ξ du < M
1 +ξ2
, z ∈ D.
(2.21)
Also,P ξ(f )(0) =0 and (1 +ξ2)P ξ(f )(0) =1, which implies that (1 +ξ2)P ξ(f ) ∈ S M(1+ξ2 ). Now, let f ∈ S3,ξ We have
1 +ξ2
P ξ(f )(z) =1 +ξ2
· 1
2ξ
+∞
−∞ f
ze iu
e2iue −| u | /ξ du, (2.22) which implies
1 +ξ2
P ξ (f )(z) ≤ 1 +ξ2 1
2ξ ·
+∞
−∞
f
ze iue −| u | /ξ du ≤1, (2.23)
Remarks 2.6 (1) Since the constant (1 + ξ2) does not influence the geometric properties
ofP ξ(f ), it follows that for all ξ > 0 we have the following:
(i) if f ∈ S1, thenP ξ(f ) is starlike (and univalent) in D;
(ii) iff ∈ S M, then P ξ(f ) is univalent in {z ∈ C;|z| < 1/M(1 + ξ2)};
(iii) iff ∈ S3,ξ⊂ S3, thenP ξ(f ) is starlike and univalent in D.
(2) Since
P ξ (f )(z) = 1
2ξ
+∞
−∞ f
ze iu
e iu e −| u | /ξ du, (2.24)
it is obvious that the condition Re[f (z)] > 0, for all z ∈ D, does not imply Re[P ξ (f )(z)] >
0 onD.
In this case, we may follow the idea in, for example, [5, Theorem 3.4] to construct another singular integral as follows: for f ∈ A(D), we define S ξ(f )(z) =z
0Q n( u)du with
Q n( z) = 1
2ξ
+∞
−∞ f
ze it
Then, it is an easy task to show that (1 +ξ2)S ξ()⊂ , for all ξ > 0, and the following
estimate holds:
S ξ(f )(z) − f (z) ≤ Cω2(f ;ξ) ∂D, ∀z ∈ D, ξ > 0. (2.26) Since inf{1/(1 + ξ2);ξ ∈[0, 1]} =1/2, byTheorem 2.5, the following is immediate Corollary 2.7 P ξ(S3,1/2)⊂ S3and f ∈ S M implies that P ξ(f ) is univalent in {z ∈ C;|z| <
1/2M}, for all ξ ∈ [0, 1].
Trang 7Remark 2.8 Of course, if we consider, for example, ξ ∈[0, 1/2], then inf{1/(1 + ξ2);x ∈
[0, 1/2]} =4/5 and byTheorem 2.5we getP ξ( S3, 4/5) ⊂ S3andf ∈ S Mimplies thatP ξ(f )
is univalent in{z ∈ C;|z| < 4/5M}, for allξ ∈[0, 1/2].
ObviouslyS3,1/2⊂ S3,5/4and{z ∈ C;|z| < 1/2M} ⊂ {z ∈ C;|z| < 4/5M}
3 Complex Poisson-Cauchy integrals
In this section, we study the properties ofQ ξ(f ), Q ∗ ξ(f ), and R ξ( f ).
Firstly, we present the approximation properties
Theorem 3.1 (i) If f (z) =∞ k =0a k z k is analytic in D, then for all ξ > 0, Q ξ(f )(z),
Q ∗ ξ(f )(z), and R ξ(f )(z) are analytic in D and the following hold in D:
Q ξ(f )(z) =
∞
k =0
a k b k( ξ)z k, with b k( ξ) =2ξ
π
π
0
cosku
u2+ξ2du,
Q ∗ ξ(f )(z) = ∞
k =0
a k b ∗ k(ξ)z k, with b ∗ k(ξ) =2ξ
π
+∞
0
cosku
u2+ξ2du,
R ξ(f )(z) =
∞
k =0
a k c k( ξ)z k, with c k( ξ) =4ξ3
π
∞
0
cosku
u2+ξ2 2du.
(3.1)
Also, if f is continuous on D, then Q ξ(f ), Q ξ ∗(f ), and R ξ(f ) are also continuous on D Here b1(ξ) > 0, for all ξ > 0, b1∗(ξ) = e − ξ , c1(ξ) =(1 +ξ)e − ξ , for all ξ > 0.
(ii)
Q ξ(f )(z) − f (z) ≤ C ω2(f ; ξ) ∂D
ξ , ∀x ∈ D, ξ ∈(0, 1],
Q ∗
ξ(f )(z) − f (z) ≤ C ω2(f ; ξ) ∂D
ξ , ∀z ∈ D, ξ ∈(0, 1],
R ξ(f )(z) − f (z) ≤ Cω1(f ; ξ) D, ∀z ∈ D, ξ ∈(0, 1].
(3.2)
(iii)
ω1
Q ∗ ξ(f ); δ
D ≤ ω1(f ; δ) D, ∀ξ ∈(0, 1],δ > 0,
ω1
Q ξ(f ); δ
D ≤ ω1(f ; δ) D, ∀ξ ∈(0, 1],∀δ > 0,
ω1
R ξ(f ); δ
D ≤ ω1(f ; δ) D, ∀ξ ∈(0, 1],δ > 0.
(3.3)
Proof (i) Let f (z) =∞ k =0a k z k,z ∈ D.
Reasoning as for the case of Picard-type integral inTheorem 2.1(i), we obtain
Q ξ(f )(z) =
∞
k =0
a k z k
ξ
π
π
− π e iku · 1
u2+ξ2du
Trang 8
ξ
π
π
− π e iku · 1
u2+ξ2du = ξ
π
π
− π
cosku
u2+ξ2du + i ξ
π
π
− π
sinku
u2+ξ2du
=2ξ π
π
0
cosku
u2+ξ2du = b k( ξ),
Q ∗ ξ(f )(z) = ∞
k =0
a k z k
ξ π
+∞
−∞ e iku · 1
u2+ξ2du
,
(3.5)
where
ξ π
+∞
−∞ e iku · 1
u2+ξ2du =2ξ
π
∞
0
cosku
u2+ξ2du = b ∗ k(ξ),
R ξ(f )(z) = ∞
k =0
a k z k
2ξ3
π
+∞
−∞
e iku
u2+ξ2 2du
,
(3.6)
where
2ξ3
π
+∞
−∞ e iku · 1
u2+ξ2 2du =4ξ3
π
∞
0
cosku
u2+ξ2 2du. (3.7)
The continuity of f on D implies the continuity of Q ξ(f ), Q ξ ∗(f ), and R ξ(f ) as in the
proof ofTheorem 2.1(i) forP ξ(f ).
It remains to show thatb1(ξ) > 0 and b ∗1(ξ) = e − ξ,c1(ξ) =(1 +ξ)e − ξ, for allξ > 0.
Indeed, firstly we have
b1(ξ) =2ξ
π
π
0
cosu
u2+ξ2du =2ξ
π
π/2
0
cosu
u2+ξ2du +
π
π/2
cosu
u2+ξ2du
=2ξ
π
π/2
0
cosu
u2+ξ2du −
π/2
0
sinu
(u + π/2)2+ξ2du
>2ξ
π
π/2
0
cosu −sinu
u2+ξ2 du
=2ξ
π
π/4
0
cosu −sinu
u2+ξ2 du +
π/2
π/4
cosu −sinu
u2+ξ2 du
:=2ξ
π I1+I2
.
(3.8)
Here
0< I1=
π/4
0
cosu −sinu
u2+ξ2 du >
π/4
0
cosu −sinu
π2/16
+ξ2du
= 16
π2+ 16ξ2[sinu + cos u] π/40 =16(
√
2−1)
π2+ 16ξ2 .
(3.9)
Trang 9Also,I2< 0 and
I2 = − I2=
π/2
π/4
sinu −cosu
u2+ξ2 du ≤ 1
π2/16
+ξ2·
π/2
π/4[sinu −cosu]du
= 16
π2+ 16ξ2[−cosu −sinu] π/2 π/4 =16(
√
2−1)
π2+ 16ξ2 ,
(3.10)
which implies I1+I2≥0 Therefore, it follows that b1(ξ) > (2ξ/π)[I1+I2]≥0, for all
ξ > 0 Now let
b ∗1(ξ) =2ξ
π
∞
0
cosu
u2+ξ2du =
byv = u ξ
=2
π ·
∞
0
cos(uξ)
u2+ 1 du. (3.11) Applying now the classical residue theorem to f (z) = e iz /(z2+ 1), it is immediate that
∞
0 (cos(uξ)/(u2+ 1))du =(π/2)e − ξ, which impliesb ∗1(ξ) =(2/π) ·(π/2)e − ξ = e − ξ, for all
ξ > 0 For c1(ξ) =(4ξ3/π) ·0∞(cosu/(u2+ξ2)2)du, applying the residue theorem to f (z)=
e iz /(z2+ξ2)2, we immediately get
∞
0
cosu
u2+ξ2 2du = π
that is,c1(ξ) =(1 +ξ)e − ξ, for allξ > 0.
(ii) We can write
Q ξ(f )(z) − f (z) = ξ
π
π
0
f
ze iu
−2f (z) + f
ze − iu
u2+ξ2 du − f (z)E(ξ), (3.13) where
E(ξ) = E(ξ) =1−2ξ
π
π
0
du
u2+ξ2 =1−2
πarctg
π
ξ ≤ 2
(for the last estimate|E(ξ)| ≤(2/π2)ξ, see, e.g., [2, page 257])
Passing to modulus, it follows that
Q ξ(f )(z) − f (z) ≤ ξ
π
π
0
f
ze iu
−2f (z) + f
ze − iu
u2+ξ2 du + f DE(ξ)
≤ ξ π
π
0
ω2(f ; u) ∂D
u2+ξ2 du + f D ·E(ξ)
≤ C ξ
π · ω2(f ; ξ) ∂D ·
π
0
1 +u
ξ
2 1
u2+ξ2du.
(3.15)
Reasoning as in the proof of Theorem 3.1 [2, pages 257-258], we arrive at the desired estimate
ForQ ξ ∗(f )(z), we have
Q ∗ ξ(f )(z) − f (z) = ξ
π
∞ f
ze iu
−2f (z) + f
ze − iu
Trang 10which implies
Q ∗
ξ(f )(z) − f (z) ≤ ξ
π
∞
0
f
ze iu
−2f (z) + f
ze − iu
≤ C ξ π
∞
0
ω2(f ; u) ∂D
u2+ξ2 du = C ξ
π
∞
0
ω2
f ; (u/ξ) · ξ
∂D
u2+ξ2 du
≤ Cω2(f ; ξ) ∂D · ξ
π
∞
0
1 +u
ξ
2
· 1
u2+ξ2du ≤ C ω2(f ; ξ) ∂D
(3.17)
ForR ξ(f )(z), we obtain
R ξ(f )(z) − f (z) ≤2ξ3
π
+∞
−∞
f
ze iu
− f (z)
u2+ξ2 2 du
≤2π ξ3
+∞
−∞
ω1
f ; |z| ·e iu −1
D
u2+ξ2 2 du
≤ C2ξ
3
π
+∞
−∞
ω1
f ; |u|D
u2+ξ2 2 du
≤ C2ξ
3
π
∞
0 ω1
f ; u
ξ · ξ
D · 1
u2+ξ2 2du
≤ Cω1(f ; ξ) D2ξ
3
π
∞
0
1 +u
ξ
· 1
u2+ξ2 2du
= Cω1(f ; ξ) D
1 +2ξ 2
π
∞
0
u
u2+ξ2 2du
,
(3.18)
where
2ξ2
π
∞
0
u du
u2+ξ2 2 =2ξ2
π ·1
2
∞
ξ2
dv
v2 = ξ2
π ·−1 v
∞
ξ2= 1
which proves the estimate forR ξ(f )(z) too.
(iii) Letz1,z2∈ D be with |z1− z2| ≤ δ We get
Q ∗
ξ(f )
z1
− Q ∗ ξ(f )
z2 ≤ ξ π
+∞
−∞
f
z1e iu
− f
z2e iu
u2+ξ2 du
≤ ω1
f ;z1− z2
D
ξ π
+∞
−∞
du
u2+ξ2 ≤ ω1(f ; δ) D,
(3.20)
where from passing to supremum afterz1,z2it follows thatω1(Q ∗ ξ(f ); δ) D ≤ ω1(f ; δ) D
Trang 11Q ξ( f )
z1
− Q ξ(f )
z2 ≤ ξ π
π
− π
f
z1e iu
− f
z2e iu
u2+ξ2 du
≤ ω1
f ;z1− z2
D · ξ π
π
− π
du
u2+ξ2
≤ ω1(f ; δ) D · ξ
π
+∞
−∞
du
u2+ξ2 = ω1(f ; δ) D
(3.21)
The reasonings forR ξ(f ) are similar, which proves the theorem.
In what follows, we present some geometric properties of complex Poisson-Cauchy integrals
Theorem 3.2 (i) If f (z) =∞ k =0a k z k , z ∈ D, and T ξ(f )(z)=∞ k =0A k z k is any from Q ξ( f ),
Q ∗ ξ(f ), and R ξ(f ), then
A k ≤ a k, ∀k =0, 1, . (3.22)
(ii) If f (z) =∞ k =1a k z k , z ∈ D, is univalent in D and f (D) is convex, then for any ξ > 0,
Q ξ(f )(z) is close-to-convex on D.
(iii) For all ξ > 0, with the notation in Section 2 , Q ∗ ξ(ᏼ)⊂ ᏼ, Rξ(ᏼ)⊂ ᏼ;
1
b1(ξ) · Q ξ
S3,b1 (ξ)
⊂ S3, 1
b1∗(ξ) · Q ∗ ξ
S3,b∗
1 (ξ)
⊂ S3, 1
c1(ξ) · R ξ
S3,c 1 (ξ)
⊂ S3, 1
b1(ξ) Q ξ
S M
⊂ S M/ | b1 (ξ)|, 1
b1∗(ξ) Q
∗
ξ
S M
⊂ S M/ | b ∗1 (ξ)|, 1
c1(ξ) R ξ(S M)⊂ S M/ | c1 (ξ)|,
(3.23)
where S3,a= { f ∈ S3;| f (z)| ≤ |a|} and S B = { f ∈ A(D); | f (z)| < B, z ∈ D}.
Proof (i) With the notations in the statement ofTheorem 3.1(i), for allk =0, 1, 2, ., we
obtain
b k( ξ) ≤2ξ
π
π
0
|cosku|
u2+ξ2 du ≤2ξ
π
π
0
du
u2+ξ2
=2ξ
π ·1
ξarctg
u ξ
π
0= 2
πarctg
π
ξ ≤1,
b ∗
k(ξ) ≤2ξ
π ·1
ξarctg
u ξ
∞
0 =1,
c k(ξ) ≤4ξ3
π
∞
0
du
u2+ξ2 2 =1,
(3.24)
which immediately implies (i)
...The continuity of f on D implies the continuity of Q ξ(f ), Q ξ ∗(f ), and R ξ(f ) as in the< /i>
proof ofTheorem...
(for the last inequalities, see, e.g., [2, proof of Theorem 2.1(i), page 252])
Remark 2.2. Theorem 2.1(ii) and (iii) remain valid for f only continuous on D.
In what...
The reasonings forR ξ(f ) are similar, which proves the theorem.
In what follows, we present some geometric properties of complex Poisson-Cauchy integrals