# Generalized conditions for starlikeness and convexity of certain analytic functions

## Abstract

For analytic functions f(z) in the open unit disk $U$ with f (0) = 0 and f '(0) = 1, Nunokawa et al. (Turk J Math 34, 333-337, 2010)have shown some conditions for starlikeness and convexity of f(z). The object of the present paper is to derive some generalized conditions for starlikeness and convexity of functions f(z) with examples.

2010 Mathematics Subject Classification: Primary 30C45.

## 1 Introduction

Let $A$ denote the class of functions f(z) of the form

$f ( z ) = z + ∑ n = 2 ∞ a n z n$
(1.1)

which are analytic in the open unit disk $U= { z ∈ ℂ : ∣ z ∣ < 1 }$. Let $S$ be the subclass of $A$ consisting of functions f(z) which are univalent in $U$. A function $f ( z ) ∈S$ is said to be starlike with respect to the origin in $U$ if $f U$ is the starlike domain. We denote by $S *$ the class of all starlike functions f(z) with respect to the origin in $U$. Furthermore, if a function $f ( z ) ∈S$ satisfies $z f ′ ( z ) ∈ S *$, then f(z) is said to be convex in $U$. We also denote by $K$ the class of all convex functions in $U$. Note that $K⊂ S * ⊂S⊂A$.

To discuss the univalency of $f ( z ) ∈A$, Nunokawa [1] has given

Lemma 1.1 If $f ( z ) ∈A$ satisfies $f ″ ( z ) <1 z ∈ U$, then $f ( z ) ∈S$. Also, Mocanu [2] has shown that

Lemma 1.2 If $f ( z ) ∈A$ satisfies

$∣ f ′ ( z ) - 1 ∣ < 2 5 ( z ∈ U ) ,$

then $f ( z ) ∈ S *$.

In view of Lemmas 1.1 and 1.2, Nunokawa et al. [3] have proved the following results.

Lemma 1.3 If $f ( z ) ∈A$ satisfies

$∣ f ″ ( z ) ∣ ≦ 2 5 = 0 . 8 9 4 4 … ( z ∈ U ) ,$
(1.2)

Then $f ( z ) ∈ S *$.

Lemma 1.4 If $f ( z ) ∈A$ satisfies

$∣ f ″ ( z ) ∣ ≦ 1 5 = 0 . 4 4 7 2 … ∣ ( z ∈ U ) ,$
(1.3)

then $f ( z ) ∈K$.

The object of the present paper is to consider some generalized conditions for functions f(z) to be in the classes $S *$ or $K$.

## 2 Generalized conditions for starlikeness

We begin with the statement and the proof of generalized conditions for starlikeness.

Theorem 2.1 If $f ( z ) ∈A$ satisfies

$∣ f ( j ) ( z ) ∣ ≦ 2 5 - M ( z ∈ U ) ,$
(2.1)

for some j(j = 2, 3, 4, ...), then $f ( z ) ∈ S *$, where

$M = 0 ( j = 2 ) ∑ n = 2 j - 1 ∣ f ( n ) ( 0 ) ∣ ( j ≧ 3 ) .$
(2.2)

Proof For j = 2, the inequality (2.1) becomes (1.2) of Lemma 1.2. Thus, the theorem is hold true for j = 2. We need to prove the inequality for j 3. Note that

$f ″ ( z ) = ∫ 0 z f ‴ ( t ) d t + f ″ ( 0 ) .$
(2.3)

We suppose that $∣ f ‴ ( z ) ∣≦ N 3 ( z ∈ U ) .$ Then, (2.3) gives us that

$∣ f ″ ( z ) ∣ ≦ ∫ 0 ∣ z ∣ ∣ f ‴ ( ρ e i θ ) d ρ ∣ + ∣ f ″ ( 0 ) ∣ ≦ N 3 ∣ z ∣ + ∣ f ″ ( 0 ) ∣ < N 3 + ∣ f ″ ( 0 ) ∣ .$
(2.4)

Therefore, if f(z) satisfies

$∣ f ″ ( z ) ∣ < N 3 + ∣ f ″ ( 0 ) ∣ ≦ 2 5 z ∈ U ,$
(2.5)

then $f ( z ) ∈ S *$ by Lemma 1.3. This means that if f(z) satisfies

$∣ f ‴ ( z ) ∣ ≦ N 3 ≦ 2 5 - ∣ f ″ ( 0 ) ∣ ( z ∈ U ) ,$
(2.6)

then $f ( z ) ∈ S *$. Thus, the theorem is holds true for j = 3.

Next, we suppose that the theorem is true for j = 2, 3, 4, ..., (k - 1). Then, letting $∣ f ( k ) ( z ) ∣≦ N k ( z ∈ U ) ,$ we have that

$∣ f ( k - 1 ) ( z ) ∣ = ∫ 0 z f ( k ) ( t ) d t + f ( k - 1 ) ( 0 ) ≦ N k ∣ z ∣ + ∣ f ( k - 1 ) ( 0 ) ∣ < N k + ∣ f ( k - 1 ) ( 0 ) ∣ .$
(2.7)

Thus, if f(z) satisfies

$∣ f ( k - 1 ) ( z ) ∣ < N k + ∣ f ( k - 1 ) ( 0 ) ∣ ≦ 2 5 - ∑ n = 2 k - 2 ∣ f ( n ) ( 0 ) ∣ ,$
(2.8)

then $f ( z ) ∈ S *$. This is equivalent to

$∣ f ( k ) ( z ) ∣ ≦ N k ≦ 2 5 - ∑ n = 2 k - 1 ∣ f ( n ) ( 0 ) ∣ .$
(2.9)

Therefore, the theorem holds true for j = k. Thus, applying the mathematical induction, we complete the proof of the theorem.

Example 2.1 Let us consider a function

$f ( z ) = z + a 2 z 2 + a 3 z 3 + a 4 z 4 .$
(2.10)

Since

$∣ f ‴ ( z ) ∣ = 2 4 ∣ a 4 ∣ ,$

if f(z) satisfies

$2 4 ∣ a 4 ∣ ≦ 2 5 - 2 ∣ a 2 ∣ - 6 ∣ a 3 ∣ ,$

then $f ( z ) ∈ S *$. This is equivalent to

$5 ∣ a 2 ∣ + 3 5 ∣ a 3 ∣ + 1 2 5 ∣ a 4 ∣ ≦ 1 .$

Therefore, we put

$a 2 = e i θ 1 2 5 , a 3 = e i θ 2 9 5 , a 4 = e i θ 3 7 2 5 .$

Consequently, we see that the function

$f ( z ) = z + e i θ 1 2 5 z 2 + e i θ 2 9 5 z 3 + e i θ 3 7 2 5 z 4$

is in the class $S *$.

## 3 Generalized conditions for convexity

For the convexity of f(z), we derive

Theorem 3.1 If $f ( z ) ∈A$ satisfies

$∣ f ( j ) ( z ) ∣ ≦ 1 j ! 4 5 - P ( z ∈ U ) .$
(3.1)

for some j(j = 3, 4, 5, ...), then $f ( z ) ∈K$, where

$P = ∑ n = 2 j - 1 n ⋅ n ! ∣ f ( n ) ( 0 ) ∣ .$
(3.2)

Proof We have to prove for j 3. Note that

$( z f ′ ( z ) ) ″ = 2 f ″ ( z ) + z f ‴ ( z ) = 2 ∫ 0 z f ‴ ( t ) d t + f ″ ( 0 ) + z f ‴ ( z ) .$
(3.3)

If $∣ f ‴ ( z ) ∣≦ N 3 ( z ∈ U ) ,$ then we have that

$∣ ( z f ′ ( z ) ) ″ ∣ ≦ 2 ∫ 0 z f ‴ ( t ) d t + f ″ ( 0 ) + ∣ z f ‴ ( z ) ∣ ≦ 2 ∫ 0 z ∣ f ‴ ( ρ e i θ ) d ρ ∣ + 2 ∣ f ″ ( 0 ) ∣ + N 3 ∣ z ∣ ≦ 3 N 3 ∣ z ∣ + 2 ∣ f ″ ( 0 ) ∣ < 3 N 3 + 2 ∣ f ″ ( 0 ) ∣ .$
(3.4)

We know that $f ( z ) ∈K$ if and only if $z f ′ ( z ) ∈ S *$. Therefore, if

$3 N 3 + 2 f ″ ( 0 ) ≦ 2 5 ,$
(3.5)

then $z f ′ ( z ) ∈ S *$ by means of Lemma 1.3. Thus, if

$∣ f ‴ ( z ) ∣ ≦ N 3 ≦ 2 3 1 5 - ∣ f ″ ( 0 ) ∣ ( z ∈ U ) ,$
(3.6)

then $f ( z ) ∈K$. This shows that the theorem is true for j = 3.

Next, we assume that theorem is true for j = 3, 4, 5, ..., (k - 1). Then, letting $∣ f ( k ) ( z ) ∣ ≦ N k ( z ∈ U ) ,$ we obtain that

$( z f ′ ( z ) ) ( k - 1 ) = ∣ ( k - 1 ) f ( k - 1 ) ( z ) + z f ( k ) ( z ) ∣ = ( k - 1 ) ∫ 0 z f ( k ) ( t ) d t + f ( k - 1 ) ( 0 ) + z f ( k ) ( z ) ≦ ( k - 1 ) ∫ 0 ∣ z ∣ ∣ f ( k ) ( ρ e i θ ) d ρ ∣ + ∣ f ( k - 1 ) ( 0 ) ∣ + ∣ z ∣ f ( k ) ( z ) .$
(3.7)

Now, we consider $∣ f ( k ) ( z ) ∣ ≦ N k ( z ∈ U ) ,$. Then, (3.7) implies that

$( z f ′ ( z ) ) ( k - 1 ) ≦ k N k ∣ z ∣ + ( k - 1 ) f ( k - 1 ) ( 0 ) < k N k + ( k - 1 ) f ( k - 1 ) ( 0 ) .$
(3.8)

Since, if

$z f ′ ( z ) ( k - 1 ) ≦ 1 ( k - 1 ) ! 4 5 - ∑ n = 2 k - 2 n ⋅ n ! f ( n ) ( 0 ) ,$

then $f ( z ) ∈K$ (or $z f ′ ( z ) ∈ S *$), if f(z) satisfies that

$k N k + ( k - 1 ) f ( k - 1 ) ( 0 ) ≦ 1 ( k - 1 ) ! 4 5 - ∑ n = 2 k - 2 n ⋅ n ! f ( n ) ( 0 ) ,$
(3.9)

that is, that

$N k ≦ 1 k ! 4 5 - ∑ n = 2 k - 1 n ⋅ n ! f ( n ) ( 0 ) ,$
(3.10)

then $f ( z ) ∈K$. Thus, the result is true for j = k. Using the mathematical induction, we complete the proof the theorem.

Example 3.1 We consider the function

$f ( z ) = z + a 2 z 2 + a 3 z 3 + a 4 z 4 .$

Then, if f(z) satisfies

$2 4 ∣ a 4 ∣ ≦ 1 2 4 4 5 - 8 ∣ a 2 ∣ - 1 0 8 ∣ a 3 ∣ ,$

then $f ( z ) ∈K$. Since

$2 5 ∣ a 2 ∣ + 2 7 5 ∣ a 3 ∣ + 1 4 4 5 ∣ a 4 ∣ ≦ 1 ,$

we consider

$a 2 = e i θ 1 4 5 , a 3 = e i θ 2 8 1 5 , a 4 = e i θ 3 8 6 4 5 .$

With this conditions, the function

$f ( z ) = z + e i θ 1 4 5 z 2 + e i θ 2 8 1 5 z 3 + e i θ 3 8 6 4 5 z 4$

belongs to the class $K$.

If we use the same technique as in the proof of Theorem 2.1 applying Lemma 1.4, then we have

Theorem 3.2 If $f ( z ) ∈A$ satisfies

$∣ f ( j ) ( z ) ∣ ≦ 1 5 - M ( z ∈ U )$
(3.11)

for some j (j = 2, 3, 4, ...), then $f ( z ) ∈K$, where M is given by (2.2).

## References

1. 1.

Nunokawa M: On the order of strongly starlikeness of strongly convex functions. Proc Jpn Acad 1993, 68: 234–237.

2. 2.

Mocanu PT: Some starlikeness conditions for analytic function. Rev Roum Math Pures Appl 1988, 33: 117–124.

3. 3.

Nunokawa M, Owa S, Polatoglu Y, Caglar M, Duman EY: Some sufficient conditions for starlikeness and convexity. Turk J Math 2010, 34: 333–337.

## Acknowledgements

This paper was completed when the first author was visiting Department of Mathematics, Kinki University, Higashi-Osaka, Osaka 577-8502, Japan, from Atat ürk University, Turkey, between February 17 and 26, 2011.

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Correspondence to Shigeyoshi Owa.

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Uyanik, N., Owa, S. Generalized conditions for starlikeness and convexity of certain analytic functions. J Inequal Appl 2011, 87 (2011). https://doi.org/10.1186/1029-242X-2011-87