1. Introduction Let p∈N=1,2,… and denote Ap as the class of multivalent functions of the form(1)fz=zp+∑n=p+1∞anzn,which are analytic in the open unit disk: (2)U=z∈C:z<1.For two parameters α∈-p,p and β≥0, function f(z)∈Ap is said to be in class UST(p,α,β) of p-valent β-uniformly star-like functions of order α in U, if and only if(3)Rzf′zfz-α≥βzf′zfz-p,where R(·) denotes taking the real part of argument. On the other hand, function f(z)∈Ap is said to be in class UCV(p,α,β) of p-valent β-uniformly convex functions of order α in U, if and only if(4)R1+zf′′zf′z-α≥β1+zf′′zf′z-p.We note from (3) and (4) that (5)fz∈UCVp,α,β⟺zf′zp∈USTp,α,β.The classes UST(p,α,β) and UCV(p,α,β) were introduced recently by Khairnar and More [1]. Various subclasses of analytic and univalent or multivalent functions were studied in many papers (see, e.g., [2–4]). Recently, Nishiwaki and Owa in [5] introduced two classes MD(α,β) consisting of all functions f(z)∈A1, which satisfy (6)Rzf′zfz-α<βzf′zfz-1,and ND(α,β) consisting of all functions f(z)∈A1, which satisfy (7)R1+zf′′zf′z-α<βzf′′zf′z,where α≥1 and β≤0. We notice from definitions of these classes that (8)fz∈NDα,β⟺zf′z∈MDα,β.For each f(z)∈Ap, it is easily seen upon differentiating both sides of (1) q times with respect to z that(9)fqz=δp,qzp-q+∑n=p+1∞δn,qanzn-q,where p>q∈N0=N∪0, and δ(p,q) denotes q-permutations of p objects; that is, (10)δp,q=p!p-q!.
Let q,m∈N0 and p∈N such that p>q+m, and assume that (11)-δp-q,m≤α<δp-q,m, β≥0.Srivastava et al. in [6] introduced a subclass of the p-valent function class USm(p,q;α,β) consisting of functions f(z) of form (1), which satisfy the following analytic criterion: for all z∈U,(12)Rzmfq+mzfqz-α≥βzmfq+mzfqz-δp-q,m.Recently, many papers have discussed such aspects of analytic univalent or multivalent functions. For example, Aouf et al. in [7, 8] discussed a subclass of p-valent analytic function with negative coefficient by using higher-order derivatives and investigated many properties of distortion theorems, closure theorems, modified Hadamard products, and radii of close-to-convexity, starlikeness, and convexity. K. I. Noor and M. A. Noor in [9] defined some subclasses of analytic functions related to k-uniformly close-to-convex functions of higher order and studied the following: rate of growth of coefficients, inclusion relations, radius problems, and necessary conditions for univalency. Seoudy in [10] defined a class of p-valent functions defined by certain linear operator and obtained subordination and inclusion properties.
Following Srivastava et al. [6] and Nishiwaki and Owa [5], we define a new subclass MDm(p,q;α,β) of multivalent functions involving higher-order derivative.
Definition 1. Let q,m∈N0 and p∈N such that p>q+m, and α and β, two real numbers, satisfy that α≥δ(p-q,m) and β≤0. Function f(z)∈MDm(p,q;α,β), if and only if f(z)∈Ap and satisfies the following inequality:(13)Rzmfq+mzfqz-α<βzmfq+mzfqz-δp-q,m.
From the above definition, it is clear that MD1(1,0;α,β)=MD(α,β). In this paper, we obtain several properties including the coefficient inequalities, distortion theorems, extreme points, and integral means inequalities for this subclass MDm(p,q;α,β) of multivalent functions involving higher-order derivative.
2. Coefficient Inequalities We derive sufficient conditions for f(z) which are given by using coefficient inequalities.
Theorem 2. Let f(z) be a function of form (1). If the coefficients of f(z) satisfy(14)∑n=p+1∞Φn,p,q,m,α,βan≤δp,qα-2δp-q,m-α,where(15)Φn,p,q,m,α,β=δn,qδn-q,m+δp-q,m-α+δn-q,m-δp-q,m-α-2βδn-q,m-δp-q,m,then, f(z) is in class MDm(p,q;α,β).
Proof. Suppose that inequality (14) holds, and denote (16)Fz=zmfq+mzfqz-α-βzmfq+mzfqz-δp-q,m.From the definition, we can verify that if (17)Fz+δp-q,mFz-δp-q,m<1,then f(z)∈MDm(p,q;α,β). In fact, denoting(18)Ψ±z;f,m,p,q,α,β=zmfq+mz-αfqz±δp-q,mfqz-βeiθzmfq+mz-δp-q,mfqz,Σ±z;m,p,q,α=∑n=p+1∞δn,qδn-q,m±δp-q,m-αanzn-p,Ξ±m,p,q,α=∑n=p+1∞δn,qδn-q,m±δp-q,m-α·an,we have(19)Fz+δp-q,mFz-δp-q,m=Ψ+z;f,m,p,q,α,βΨ-z;f,m,p,q,α,β=δp,q2δp-q,m-α+Σ+z;m,p,q,α-βeiθΣ-z;m,p,q,0-αδp,q-Σ-z;m,p,q,α+βeiθΣ-z;m,p,q,0≤δp,q2δp-q,m-α+Ξ+m,p,q,α-βΞ-m,p,q,0αδp,q-Ξ-m,p,q,α+βΞ-m,p,q,0.Here, we use technology f(z)=eiθfz. If (14) satisfies, we drive that the last expression above is bounded by 1 which implies f(z)∈MDm(p,q;α,β). Thus, the proof of Theorem 2 is completed.
Example 3. Function f(z) given by (20)fz=zp+∑n=p+1∞δp,qα-2δp-q,m-αγ+p+1ξnn+γn+γ+1Φn,p,q,m,α,βznbelongs to class MDm(p,q;α,β) for γ>-(p+1) and ξn∈C with ξn=1.
As a special case of Theorem 2, as in [2], we can obtain the following corollary.
Corollary 4. Function f(z)∈A1 is in class MD(α,β), if (21)∑n=2∞n+1-α+n-1-α-2βn-1·an≤α-2-α.
In view of Theorem 2, we introduce subclass MDm∗(p,q;α,β) which consists of functions of the form(22)fz=zp+∑n=p+1∞anzn,whose Taylor-Maclaurin coefficients an are nonnegative and satisfy inequality (14). By the coefficient inequalities for classes MDm∗(p,q;α,β), we have the following theorem.
Theorem 5. If β1≤β2≤0, then (23)MDm∗p,q;α,β1⊂MDm∗p,q;α,β2.
Since MD1(1,0;α,β)=MD(α,β), we get the following corollary, which is a theorem in [2].
Corollary 6. If β1≤β2≤0, then (24)MDα,β1⊂MDα,β2.
3. Distortion Theorems Lemma 7. If f(z)∈MDm∗(p,q;α,β), then there exists p0∈N such that(25)∑n=p0+1∞an≤Ap0,where(26)Ap0=δp,qα-2δp-q,m-α-∑n=p+1p0Φn,p,q,m,α,βanΦp0+1,p,q,m,α,β,and Φ(n,p,q,m,α,β) is given in (15).
Proof. From the definition of Φ(n,p,q,m,α,β), there exists p0∈N such that function Φ(n,p,q,m,α,β) is increasing with respect to n when n>p0. According to Theorem 2, we have(27)∑n=p0+1∞Φn,p,q,m,α,βan≤δp,qα-2δp-q,m-α-∑n=p+1p0Φn,p,q,m,α,βan.From(28)Φp0+1,p,q,m,α,β∑n=p0+1∞an≤∑n=p0+1∞Φn,p,q,m,α,βan,we have(29)Φp0+1,p,q,m,α,β∑n=p0+1∞an≤δp,qα-2δp-q,m-α-∑n=p+1p0Φn,p,q,m,α,βan.This implies that inequality (25) holds.
Using the same argument, we obtain the following inequality.
Lemma 8. If f(z)∈MDm∗(p,q;α,β), then there exists p0∈N such that(30)∑n=p0+1∞nan≤Bp0,where(31)Bp0=p0+1δp,qα-2δp-q,m-α-∑n=p+1p0Φn,p,q,m,α,βanΦp0+1,p,q,m,α,β,and Φ(n,p,q,m,α,β) is defined by (15).
Theorem 9. Let f(z) be a function in class MDm∗(p,q;α,β). Then, for z=r<1, (32)fz≤rp+∑n=p+1p0anrn+Ap0rp0+1,fz≥rp-∑n=p+1p0anrn-Ap0rp0+1,where Ap0 and Bp0 are given in Lemmas 7 and 8, respectively.
Proof. Let f(z) be a function of form (22). For z=r<1, using Lemma 7, we have (33)fz≤zp+∑n=p+1p0an·zn+∑n=p0+1∞an·zn≤zp+∑n=p+1p0an·zn+zp0+1∑n=p0+1∞an≤rp+∑n=p+1p0anrn+Ap0rp0+1,fz≥zp-∑n=p+1p0an·zn-∑n=p0+1∞an·zn≥zp-∑n=p+1p0an·zn-zp0+1∑n=p0+1∞an≥rp-∑n=p+1p0anrn-Ap0rp0+1.
Using the same argument, we can prove the following result.
Theorem 10. Let f(z) be a function in class MDm∗(p,q;α,β). Then, for z=r<1, (34)f′z≤prp-1+∑n=p+1p0nanrn-1+Bp0rp0,f′z≥prp-1-∑n=p+1p0nanrn-1-Bp0rp0,where An and Bn are given in Lemmas 7 and 8, respectively.
4. Extreme Points Theorem 11. Let fp(z)=zp and, for each n=p+1,p+2,…, define(35)fnz=zp+δp,qα-2δp-q,m-αΦn,p,q,m,α,βzn,where Φ(n,p,q,m,α,β) is defined by (15). Then f(z)∈MDm∗(p,q;α,β) if and only if it can be expressed in the form (36)fz=∑n=p∞λnfnz,where λn≥0 for all n=p,p+1,…, and ∑n=p∞λn=1.
Proof. Suppose that(37)fz=∑n=p∞λnfnz=λpfpz+∑n=p+1∞λnfnz=zp+∑n=p+1∞λnδp,qα-2δp-q,m-αΦn,p,q,m,α,βzn.Then,(38)∑n=p+1∞Φn,p,q,m,α,βλnδp,qα-2δp-q,m-αΦn,p,q,m,α,β=∑n=p+1∞λnδp,qα-2δp-q,m-α=δp,qα-2δp-q,m-α1-λp≤δp,qα-2δp-q,m-α.Thus, it follows from Theorem 2 that f(z)∈MDm∗(p,q;α,β).
Conversely, suppose that f(z)∈MDm∗(p,q;α,β). Since (39)an≤δp,qα-2δp-q,m-αΦn,p,q,m,α,β, n=p+1,p+2,…,we denote(40)λn=Φn,p,q,m,α,βδp,qα-2δp-q,m-αan, n=p+1,p+2,….And λp=1-∑n=p+1∞λn. By a simple calculation, we get f(z)∈MDm∗(p,q;α,β).
Corollary 12. The extreme points of MDm∗(p,q;α,β) are functions fp(z)=zp and (41)fnz=zp+δp,qα-2δp-q,m-αΦn,p,q,m,α,βzn,for each n=p+1,p+2,….
5. Integral Means Inequalities Assume that two functions f(z) and g(z) are analytic in U. We say that f(z) is subordinate to g(z), written as f(z)≺g(z), if there exists an analytic function w(z) in U with w(0)=1 and wz<1 such that f(z)=g(w(z)).
Lemma 13 (see [11]). 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),(42)∫02πfzμdθ≤∫02πgzμdθ.
Theorem 14. Let f(z)∈MDm∗(p,q;α,β) and fn(z) be given by (35). Suppose that(43)∑n=p+1∞an≤Φn,p,q,m,α,βδp,qα-2δp-q,m-α.If there exists function w(z), z∈U, that satisfied the condition (44)wz=Φn,p,q,m,α,βδp,qα-2δp-q,m-α∑n=p+1∞anzn-p1/n-p,then, for z=reiθ, 0<r<1, we have (45)∫02πfzμdθ≤∫02πfnzμdθ.
Proof. In order to obtain the result, it is necessary to prove the following inequality: (46)∫02π1+∑n=p+1∞anzn-pμdθ≤∫02π1+δp,qα-2δp-q,m-αΦn,p,q,m,α,βzn-pμdθ.From Lemma 13, it is sufficient to verify the subordination: (47)1+∑n=p+1∞anzn-p≺1+δp,qα-2δp-q,m-αΦn,p,q,m,α,βzn-p.Thus, there exists an analytic function w(z) in C such that (48)1+∑n=p+1∞anzn-p=1+δp,qα-2δp-q,m-αΦn,p,q,m,α,βwzn-p.We find that(49)wz=Φn,p,q,m,α,βδp,qα-2δp-q,m-α∑n=p+1∞anzn-p1/n-p,which readily yields w(0)=0 and (50)wz1/n-p=Φn,p,q,m,α,βδp,qα-2δp-q,m-α∑n=p+1∞anzn-p≤Φn,p,q,m,α,βδp,qα-2δp-q,m-α∑n=p+1∞anzn-p≤Φn,p,q,m,α,βδp,qα-2δp-q,m-αz∑n=p+1∞an≤z<1.This means that the hypotheses of w(z) are satisfied and the theorem is proved.