M. H. Al-Abbadi and M. Darus (2009) recently introduced a new generalized derivative
operator μλ1,λ2n,m, which generalized many well-known operators studied earlier
by many different authors. In this present paper, we shall investigate a new
subclass of analytic functions in the open unit disk U={z∈ℂ:|z|<1} which is defined by new generalized derivative operator. Some results on coefficient inequalities, growth and distortion theorems, closure theorems, and extreme points of analytic functions belonging to the subclass are obtained.

1. Introduction and Definitions

Let 𝒜(x) denote the class of functions of the form

f(z)=z+∑k=x+1∞akzk,akiscomplexnumber,
and x∈ℕ={1,2,3,...}, which are analytic in the open unit disc U={z∈ℂ:|z|<1} on the complex plane ℂ; note that 𝒜(1)=𝒜 and 𝒜(x)⊆𝒜(1). Suppose that 𝒮(x) denote the subclass of 𝒜(x) consisting of functions that are univalent in U. Further, let 𝒮α*(x) and 𝒞α(x) be the classes of 𝒮(x) consisting of functions, respectively, starlike of order α and convex of order α in U, for 0≤α<1. Let 𝒯(x) denote the subclass of 𝒮(x) consisting of functions of the form

f(z)=z-∑k=x+1∞|ak|zk,x∈ℕ={1,2,3,…},
defined on the open unit disk U={z∈ℂ:|z|<1}. A function f∈𝒯(x) is called a function with negative coefficient and the class 𝒯(1) was introduced and studied by Silverman [1]. In [1] Silverman investigated the subclasses of 𝒯(1) denoted by S𝒯*(α) and C𝒯(α) for 0≤α<1. That are, respectively, starlike of order α and convex of order α. Now (x)k denotes the Pochhammer symbol (or the shifted factorial) defined by

The authors in [2] have recently introduced a new generalized derivative operator μλ1,λ2n,m as follows.

Definition 1.1.

For f∈𝒜=𝒜(1), the generalized derivative operator μλ1,λ2n,m:𝒜→𝒜 is defined by
μλ1,λ2n,mf(z)=z+∑k=2∞(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)akzk,(z∈U),
where n,m∈ℕ0={0,1,2,…},λ2≥λ1≥0,andc(n,k)=(n+k-1n)=(n+1)k-1/(1)k-1.

Special cases of this operator include the Ruscheweyh derivative operator in the cases μλ1,0n,1≡μ0,0n,m≡μ0,λ2n,0≡Rn [3], the Salagean derivative operator μ1,00,m+1≡Sn [4], the generalized Ruscheweyh derivative operator μλ1,0n,2≡Rλn [5], the generalized Salagean derivative operator introduced by Al-Oboudi μλ1,00,m+1≡Sβn [6], and the generalized Al-Shaqsi and Darus derivative operator μβ,0λ,n+1≡Dλ,βn where (n=λ,m=n+1,λ1=β,andλ2=0) can be found in [7]. It is easily seen that μλ1,00,1f(z)=μ0,00,mf(z)=μ0,λ20,0f(z)=μλ1,11,1f(z)=f(z),μλ1,01,1f(z)=μ0,01,mf(z)=μ0,λ21,0f(z)=μ1,00,2f(z)=zf'(z), and also μλ1,0n-1,0f(z)=μ0,0n-1,mf(z) where n=1,2,3,….

By making use of the generalized derivative operator μλ1,λ2n,m, the authors introduce a new subclass as follows.

Definition 1.2.

For 0≤α<1,(n,m∈ℕ0={0,1,2,…})andλ2≥λ1≥0, let ℋλ1,λ2n,m(x,α) be the subclass of 𝒮(x) consisting of functions f satisfying
Re(z(μλ1,λ2n,mf(z))′μλ1,λ2n,mf(z))>α,(z∈U),
where
μλ1,λ2n,mf(z)=z+∑k=x+1∞(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)akzk,{x=1,2,3,…,(z∈U)}, and μλ1,λ2n,mf(z)≠0.

Further, we define the class 𝒯ℋλ1,λ2n,m(x,α) by
𝒯ℋλ1,λ2n,m(x,α)=ℋλ1,λ2n,m(x,α)∩𝒯(x),{x=1,2,3,…},
for 0≤α<1,(n,m∈ℕ0={0,1,2,…}), and λ2≥λ1≥0.

Also note that various subclasses of ℋλ1,λ2n,m(x,α) and 𝒯ℋλ1,λ2n,m(x,α) have been studied by many authors by suitable choices of n,m,λ1,λ2, and x. For example,

class of convex function of order α with negative coefficients. Also

𝒯ℋ0,λ20,0(x,α)≡𝒯ℋλ1,00,1(x,α)≡𝒯ℋ0,00,m(x,α)≡𝒯ℋλ1,11,1(x,α)≡S𝒯*(x,α),𝒯ℋ0,λ21,0(x,α)≡𝒯ℋλ1,01,1(x,α)≡𝒯ℋ0,01,m(x,α)≡𝒯ℋ1,00,2(x,α)≡C𝒯(x,α),𝒯ℋλ1,00,2(x,α)≡P(x,λ1,α)(0≤λ1<1),𝒯ℋλ1,01,2(x,α)≡C(x,λ1,α)(0≤λ1<1).
The classes S𝒯*(x,α) and C𝒯(x,α) were studied by Chatterjea [8] (see also Srivastava et al. [9]), whereas the classes P(x,λ1,α) and C(x,λ1,α) were, respectively, studied by Altintaş [10] and Kamali and Akbulut [11]. When λ1=λ2=0 or m=1,λ2=0, or m=0,λ1=0 in the class ℋλ1,λ2n,m(x,α), we have the class Rn(α) introduced and studied by Ahuja [12]. Finally we note that when m=2,λ2=0 in the class ℋλ1,λ2n,m(x,α) we have the class 𝒦λn(x,α) introduced and studied by Al-Shaqsi and Darus [13].

2. Coefficient Inequalities

In this section, we provide a necessary and sufficient condition for a function f analytic in U to be in ℋλ1,λ2n,m(x,α) and in 𝒯ℋλ1,λ2n,m(x,α).

Theorem 2.1.

For 0≤α<1 and λ2≥λ1≥0, let f∈𝒮(x) be defined by (1.1). If
∑k=x+1∞(k-α)(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)|ak|≤1-α,x=1,2,…,
then f∈ℋλ1,λ2n,m(x,α), where n∈ℕ={1,2,…} and m∈ℕ0={0,1,2,…}.

Proof.

Assume that (2.1) holds true. Then we shall prove condition (1.5). It is sufficient to show that
|z(μλ1,λ2n,mf(z))′μλ1,λ2n,mf(z)-1|≤1-α,(z∈U).
So, we have that
|z(μλ1,λ2n,mf(z))′μλ1,λ2n,mf(z)-1|=|z(μλ1,λ2n,mf(z))′-μλ1,λ2n,mf(z)μλ1,λ2n,mf(z)|=|∑k=x+1∞((k-1)(1+λ1(k-1))m-1/(1+λ2(k-1))m)c(n,k)akzkz+∑k=x+1∞((1+λ1(k-1))m-1/(1+λ2(k-1))m)c(n,k)akzk||z|<1,≤∑k=x+1∞((k-1)(1+λ1(k-1))m-1/(1+λ2(k-1))m)c(n,k)|ak|1-∑k=x+1∞((1+λ1(k-1))m-1/(1+λ2(k-1))m)c(n,k)|ak|,
and expression (2.3) is bounded by (1-α).

Hence (2.2) holds if
∑k=x+1∞(k-1)(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)|ak|≤(1-α)[1-∑k=x+1∞(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)|ak|],
which is equivalent to
∑k=x+1∞(k-α)(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)|ak|≤(1-α),
by (2.1). Thus f∈ℋλ1,λ2n,m(x,α). Note that the denominator in (2.3) is positive provided that (2.1) holds.

Theorem 2.2.

Let f be defined by (1.2) and 1-∑k=x+1∞(k(1+λ1(k-1))m-1/(1+λ2(k-1))m)c(n,k)|ak|≥0(x=1,2,…). Then f∈𝒯ℋλ1,λ2n,m(x,α) if and only if (2.1) is satisfied.

Proof.

We only prove the right-hand side, since the other side can be justified using similar arguments in proof of Theorem 2.1. Since f∈𝒯ℋλ1,λ2n,m(x,α) by condition (1.5), we have that
Re(z(μλ1,λ2n,mf(z))′μλ1,λ2n,mf(z))=Re{z-∑k=x+1∞(k(1+λ1(k-1))m-1/(1+λ2(k-1))m)c(n,k)|ak|zkz-∑k=x+1∞((1+λ1(k-1))m-1/(1+λ2(k-1))m)c(n,k)|ak|zk}>α.
Choose values of z on real axis so that z(μλ1,λ2n,mf(z))′/μλ1,λ2n,mf(z) is real. Letting z→1- through real values, we have that
1-∑k=x+1∞(k(1+λ1(k-1))m-1/(1+λ2(k-1))m)c(n,k)|ak|zk1-∑k=x+1∞((1+λ1(k-1))m-1/(1+λ2(k-1))m)c(n,k)|ak|zk>α.
Thus we obtain
∑k=x+1∞(k-α)(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)|ak|≤1-α,
which is (2.1). Hence the proof is complete.

The result is sharp with the extremal function f given by

Let the function f given by (1.2) be in the class 𝒯ℋλ1,λ2n,m(x,α). Then
|ak|≤(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k),
where 0≤α<1,λ2≥λ1≥0,k≥x+1, and x=1,2,3,…. Equality holds for the function given by (2.9).

Proof.

Since f∈𝒯ℋλ1,λ2n,m(x,α), then condition (2.1) gives
|ak|≤(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k),
for each k=x+1 where x=1,2,3,….

Clearly the function given by (2.9) satisfies (2.10), and therefore, f given by (2.9) is in 𝒯ℋλ1,λ2n,m(x,α) for this function; the result is clearly sharp.

3. Growth and Distortion Theorems

In this section, growth and distortion theorems will be considered and covering property for function in the class will also be given.

Theorem 3.1.

Let the function f given by (1.2) be in the class 𝒯ℋλ1,λ2n,m(x,α). Then for 0<|z|=r<1,
r-(1-α)(1+λ2x)m(x+1-α)(1+λ1x)m-1c(n,x+1)rx+1≤|f(z)|≤r+(1-α)(1+λ2x)m(x+1-α)(1+λ1x)m-1c(n,x+1)rx+1,x=1,2,…,
where 0≤α<1,m∈ℕ0={0,1,2…},andn∈ℕ={1,2,…}.

Proof.

We only prove the right-hand side inequality in (3.1), since the other inequality can be justified using similar arguments. Since f∈𝒯ℋλ1,λ2n,m(x,α) by Theorem 2.2, we have that
∑k=x+1∞(k-α)(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)|ak|≤(1-α).
Now
(x+1-α)(1+λ1x)m-1(1+λ2x)mc(n,x+1)(∑k=x+1∞|ak|)=∑k=x+1∞(x+1-α)(1+λ1x)m-1(1+λ2x)mc(n,x+1)|ak|,≤∑k=x+1∞(k-α)(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)|ak|,≤1-α,x=1,2,3,….
And therefore,
∑k=x+1∞|ak|≤(1-α)(1+λ2x)m(x+1-α)(1+λ1x)m-1c(n,x+1),x=1,2,….
Since
f(z)=z-∑k=x+1∞|ak|zk,x=1,2,…,
then we have that
|f(z)|=|z-∑k=x+1∞|ak|zk|.
After that,
|f(z)|≤|z|+|z|x+1∑k=x+1∞|ak||z|k-(x+1)≤r+rx+1∑k=x+1∞|ak|.
By aid of inequality (3.4), it yields the right-hand side inequality of (3.1). Thus, this completes the proof.

Theorem 3.2.

Let the function f given by (1.2) be in the class 𝒯ℋλ1,λ2n,m(x,α). Then for 0<|z|=r<1,
1-(x+1)(1-α)(1+λ2x)m(x+1-α)(1+λ1x)m-1c(n,x+1)rx≤|f′(z)|≤1+(x+1)(1-α)(1+λ2x)m(x+1-α)(1+λ1x)m-1c(n,x+1)rx,x=1,2,…,
where 0≤α<1,λ2≥λ1≥0,m∈ℕ0={0,1,2,…},andn∈ℕ={1,2,…}.

Proof.

Since f∈𝒯ℋλ1,λ2n,m(x,α), by Theorem 2.2, we have that
∑k=x+1∞(k-α)(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)|ak|≤1-α,x=1,2,3,….
Now
(x+1-α)(1+λ1x)m-1(1+λ2x)mc(n,x+1)(∑k=x+1∞k|ak|)=∑k=x+1∞(x+1-α)(1+λ1x)m-1(1+λ2x)mc(n,x+1)k|ak|≤(x+1)∑k=x+1∞(k-α)(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)|ak|≤(x+1)(1-α),(k≥x+1,x=1,2,3,…).
Hence
∑k=x+1∞k|ak|≤(x+1)(1-α)(1+λ2x)m(x+1-α)(1+λ1x)m-1c(n,x+1),x=1,2,….
Since
f′(z)=1-∑k=x+1∞k|ak|zk-1,x=1,2,…,
then we have that
1-|z|x∑k=x+1∞k|ak||z|k-(x+1)≤|f′(z)|≤1+|z|x∑k=x+1∞k|ak||z|k-(x+1),
and therefore,
1-rx∑k=x+1∞k|ak|≤|f′(z)|≤1+rx∑k=x+1∞k|ak|,x=1,2,….
By using the inequality (3.11) in (3.14), we get Theorem 3.2. This completes the proof.

4. Extreme Points

The extreme points of the class 𝒯ℋλ1,λ2n,m(x,α) are given by the following theorem.

Theorem 4.1.

Let fx(z)=z and
fk(z)=z-(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k)zk,
where 0≤α<1,λ2≥λ1≥0,(n,m∈ℕ0),k=x+1,x+2,…, and x=1,2,3,….

Then f∈𝒯ℋλ1,λ2n,m(x,α) if and only if it can be expressed in the form
f(z)=∑k=x∞δkfk(z),
where δk≥0 and ∑k=x∞δk=1.

Proof.

Suppose that f can be expressed as in (4.2). Our goal is to show that f∈𝒯ℋλ1,λ2n,m(x,α).

By (4.2), we have that
f(z)=∑k=x∞δkfk(z)=δxfx(z)+∑k=x+1∞δkfk(z)=δxfx(z)+∑k=x+1∞δk[z-(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k)zk]=∑k=x∞δkz-∑k=x+1∞δk(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k)zk=z-∑k=x+1∞δk(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k)zk.
Now
f(z)=z-∑k=x+1∞|ak|zk=z-∑k=x+1∞δk(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k)zk,
so that
|ak|=δk(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k).
Now, we have that
∑k=x+1∞δk=1-δx≤1,x=1,2,3,….
Setting
∑k=x+1∞δk=∑k=x+1∞δk(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k)(k-α)(1+λ1(k-1))m-1c(n,k)(1-α)(1+λ2(k-1))m≤1,
we arrive to
∑k=x+1∞(k-α)(1+λ1(k-1))m-1(1-α)(1+λ2(k-1))mc(n,k)|ak|≤1.
And therefore,
∑k=x+1∞(k-α)(1+λ1(k-1))m-1(1+λ2(k-1))mc(n,k)|ak|≤1-α,x=1,2,3,….
It follows from Theorem 2.2 that f∈𝒯ℋλ1,λ2n,m(x,α).

Conversely, let us suppose that f∈𝒯ℋλ1,λ2n,m(x,α); our goal is, to get (4.2). From (4.2) and using similar last arguments, it is easily seen that
f(z)=z-∑k=x+1∞|ak|zk=z-∑k=x+1∞δk(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k)zk,
which suffices to show that
|ak|=δk(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k).
Now, we have that f∈𝒯ℋλ1,λ2n,m(x,α), then by previous Theorem 2.3,
|ak|≤(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k).
That is
(k-α)(1+λ1(k-1))m-1c(n,k)|ak|(1-α)(1+λ2(k-1))m≤1.
Since ∑k=x∞δk=1, we see δk≤1, for each k=x,x+1,x+2,…, and x=1,2,3,….

We can set that
δk=(k-α)(1+λ1(k-1))m-1c(n,k)|ak|(1-α)(1+λ2(k-1))m.
Thus, the desired result is that
|ak|=δk(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k).
This completes the proof of the theorem.

Corollary 4.2.

The extreme points of 𝒯ℋλ1,λ2n,m(x,α) are the functions
fx(z)=z,fk(z)=z-(1-α)(1+λ2(k-1))m(k-α)(1+λ1(k-1))m-1c(n,k)zk,
where 0≤α<1,(n,m∈ℕ0={0,1,2,…}),λ2≥λ1≥0, and k=x+1,x+2,…,(x=1,2,3,…).

Acknowledgments

This work is fully supported by UKM-GUP-TMK-07-02-107, Malaysia. The authors are also grateful to the referee for his/her suggestions which helped us to improve the contents of this article.

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