INTEGRAL OPERATORS ON THE UCD(
α
)-CLASS
by Daniel Breaz
Abstract. We consider the class of functions f =z+a2z2 +a3z3 +..., that are analytically and univalent in the unit disk.
We consider the class of functions UCD(α), α ≥0 with the properties
( )
z zf( ) ( )
z z Uf′ ≥ ′′ ∀ ∈
Re α .
In this paper we present some results from integrals operators at this class.
1.Introduction
Theorem A. [4] A sufficient condition for a function f of the form
( )
z =z+a2z2 +a3z3 +...f (1)
so that it belongs to the UCD
( )
α ,α ≥0 class is( )
[
1 1]
12
≤ ⋅ − +
∑
k∞= k α k ak . (2)2.Main results
Theorem 1. Let be f ∈S ,f
( )
z =z+a2z2+a3z3+...,z∈U . We suppose that the coefficients of function f verify the condition( )
[
1 1]
12
≤ ⋅ − +
∑
k∞= k α k ak .We consider the integral operator
( )
=∫
( )
z
dt t
t f z F
0
, which is the Alexander operator.
Proof
Let be f of the form (1), verifying the condition (2). We have:
( )
(
)
=
+ + + = + + + = + + +
=
∫
∫
z z
z
t a t a t dt t
a t a dt
t t a t a t z F
0 3
3 2 2
0
2 3 2 0
3 3 2 2
3 2
1 K K
K
... 3 2
3 3 2 2
z a z a z+ +
= .
Denoting
2 , ≥
= k
k a b k
k ,
we can write,
( )
3 3 ...2
2 + +
+
=z b z b z z
F
.
In that following, we evaluate the relation (2), for the function
F
, with the coefficientsb
k.( )
[
+ −]
⋅ =∑
[
+( )
−]
⋅ =∑
[
+( )
−]
⋅ ⋅ ≤∑
∞= ∞= k ∞= k k a k ak k b
k k
k
k k
k k
k
1 1
1 1
1 1
1
2 2
2
α α
α
( )
[
1 1]
[
1( )
1]
12 2
< ⋅ − + ≤
⋅ −
+
∑
∑
∞= k∞= k kk k k a
a
k α
α .
According to the relation (2), from the Theorem A, we obtain:
( )
z UUCD
F∈ α , α >0 , ∈ .
Theorem 2. Let be
f
∈
A
,( )
3 3 ...2
2 + +
+
=z a z a z z
f ,α ≥0, z∈U.We suppose
that the condition (2) is satisfied and we consider the Libera operator,
( )( )
=∫
( )
z
dt t f z z f L
0
2
.
Then L∈UCD
( )
α , 0α ≥ , z∈U . ProofLet be f of the form (1) verifying the condition (2).
( )
(
)
= + + +
= + + + =
∫
z z
t a t a t z dt t
a t a t z z F
0 4
3 3
2 2
0
3 3 2 2
4 3
2 2 2
K K
=
+ + +
= 2 3 3 4 K
2
4 3
2 2
z a z a z z
∑
∞= + + = + ++ =
2 3
3 2 2
1 2 4
2 3
2
k
k k
z k
a z
z a z a
z K .
Denoting
1 2
+ =
k a bk k ,
we can write
( )( )
3 3 ...2
2 + +
+
=z b z b z z
f
L .
We evaluate the relation (2), for the function obtained after integration with the coefficients bk..
Since 2k ≥ , we have
( )
[
]
[
( )
]
[
( )
]
≤+ ⋅ ⋅ − + =
+ ⋅ − + =
⋅ −
+
∑
∑
∑
∞=2 ∞=2 ∞=2 1 2 11 1
2 1 1
1 1
k
k k
k k
k
k a k
k k
a k
k b
k
k α α α
( )
[
]
13 2 3 2 1
1
2
< ≤ ⋅ ⋅ − + ≤
∑
∞=
k
k
a k
k α .
that implies
( )
f UCD( )
αL ∈ , α ≥0, so, the Libera operator preserves this class.
Theorem 3. Let be f ∈A, f
( )
z =z+a2z2 +a3z3+...,z∈U . We suppose that the condition (2) is satisfied and we consider the Bernardi operator,( )
= +∫
z f( )
t ⋅t −dtz z F
0
1
1 γ
γγ , 1γ ≥− .
Then, we have F∈UCD
( )
α , 0α ≥ , z∈U . ProofLet be f of the form (1), verifying the condition (2). We have
( )
= +∫
z(
+ + +)
⋅ − = +∫
z(
t +a t + +a t + +)
dt=z dt t t
a t a t z z F
0
2 3 1 2 0
1 3
3 2 2
1 1
K
K γ γ γ γ γ
= + + + + + + = + + + + +
= + + K + + + K
3 2 1 1 2 1
1 3 3
2 2 1 0 2 2 1 γ γ γ γ γ γ γ γ γ γ γ γ γ γ z a z a z z t a t z z
∑
∞= ⋅ ++ ⋅ + = + ++ + ++ + = 2 3 3 2 2 1 3 1 2 1 k k k z k a z z a z a z γγ γγγγ K .
Let be
k a
bk k
++ ⋅
= γγ 1 .
We evaluate the relation (2) for the function F
( )
z =z+b2z2 +b3z3+... We have:( )
[
]
[
( )
]
[
( )
]
≤ ++ ⋅ ⋅ − + = ++ ⋅ − + = ⋅ − +∑
∑
∑
∞=2 ∞=2 ∞=2 11 1 1 1 1 1 1 k k k k k k k a k k k a k k b k k γγ α γγ α α
( )
[
]
∑
[
( )
]
∑
∞= ≤ ∞= + − ⋅ ≤ + + ⋅ ⋅ − + ≤ 2 2 1 1 1 1 1 1 1 k k kk k k a
a k
k α
γ γ
α , so:
( )
αUCD F∈ . Theorem 4. Let
( )
= +∫
z f( )
t ⋅t −dtz z F 0 1 1 γ γγ .
If f is of the form (1), f∈S, α >0 and Ref′
( )
z ≥α zf ′′( ) ( )
z ∀ z∈U, then( ) ( )
1+γ F′ z +zF′′( )
z ≥α z[
(
2+γ)
F′′+zF ′′′]
.Proof
( )
∫
( )
−( )
=∫
( )
⋅ − + ⇔ ⋅ + = z z dt t t f z F z dt t t f z z F 0 1 0 1 11 γ γ γ
γγ γ .
After successive derivations, we obtain:
( )
z z F( ) ( )
z f z z F( )
z zF( ) ( )
z f zF z = + + ⋅ + ⇔ ⋅ = ⋅ + + ⋅ ⋅ +γγ − γ γ− γγ γ γ γ 1 ' 1 ' 1 1 1
1 ⇒
( )
+ +′( )
+ +′′( )
=( )
⇔ ′( )
+ + ⋅ ⋅ ′′( )
=( )
⇔⋅
+ f z F z z F z f z
z F z z F z F ' 1 1 ' 1 1 '
1 γ γ γ γ
γ
( )
z F( )
z zF( )
z f( )
zF = ′′
+′′′ + ′′ + + ′′ ⇔ γ γ 1 1 1 . So, we have f
( )
z F( )
z zF ′′′( )
zIf Re f′
( )
z ≥α zf ′′( ) ( )
z ∀z∈U,α >0, then( )
f( )
z zf( )
zUCD
f ∈ α ⇒ ′ ≥α ′′ .
In this inequality, we put the expressions of
f
′
andf
′′
, and we obtain:( )
( )
( )
′′′( )
⇔+ + ′′
++ ≥
′′ + +
′z zF z z F z zF z
F
γ γγ
α
γ 1
1 1
2 1
1
( ) ( )
( )
(
+) ( )
′′ + ′′′( )
⇔+ ≥ ′′ + ′ + +
⇔ F z zF z z 2 F z z2F z
1 1
1
1 γ
γ α γ
γ
( ) ( )
+ F z +zF( )
z ≥ z[
(
+) ( )
F z +z F ′′′( )
z]
⇔ 2
" 2 "
'
1 γ α γ .
Remark 5. If f is of the form (1), f ∈S,α >0 and Re f′
( )
z ≥α zf ′′( ) ( )
z ∀z∈U, then( )
z zf
α
1 '
log ≤ .
Proof
According to the Theorem 4 we have:
( )
( )
( )
′′′( )
⇔+ + ′′
++ ≥
′′ + +
′z zF z z F z z F z
F 2
1 1 1
2 1
1
γ γγ
α γ
( )
( )
( )
( )
( )
( )
⇔′
′′
+ + ′ ⋅
⋅ = ′′
+ + ′
′′′ + + ′′ ++ ⋅ ⋅ ≥
⇔ z F z zF z
z F z z
F
z F z z
F z
γ α
γ γ γγ
α
1 1 log
1 1
1 1 1
2 1
( )
[
f z]
[
f( )
z]
zz
⋅ ≤ ′ ′ ⇔ ′ ′ ⋅ ⋅ ≥
⇔1 α log log α1 .
References
1.Miller S.S., Mocanu P.T.- Integral operators on certain classes of analytic functions, Univalent Functions, Fractional Calculus, and their Applications, Halstead Pres, John Wiley (1989), 153-166.
3. Miller S.S., Mocanu P.T.- Differential subordinations and univalent functions, Michigan, Math. J. 28(1981), 157-171.
4. Rosy T., Stephen A.B., Subramanian K.G., Silverman H.- Classes of convex functions, Internat. J. Math.& Math. Sci., vol 23, No. 12(2000), 819-825.
5. Silverman H.- Univalent functions with negative coefficients, Proc. Amer. Math. Soc. 51(1975), 109-116.
