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Results & Lemmas (16)
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Lemma 1.1
Lemma 1.1. (see [12, page 135, Corollary 3.4h.1]). Let Q be univalent in, and be analytic in a domain D containing. If is starlike, and P…
Lemma 1.1. (see [12, page 135, Corollary 3.4h.1]). Let Q be univalent in $\mathbb{U}$ , and $\varphi$ be analytic in a domain D containing $Q(\mathbb{U})$ . If $zQ'(z) \cdot \varphi[Q(z)]$ is starlike, and P is analytic in $\mathbb{U}$ with P(0) = Q(0) and $P(\mathbb{U}) \subset D$ , then
$$zP'(z) \cdot \varphi[P(z)] \prec zQ'(z) \cdot \varphi[Q(z)] \Longrightarrow P \prec Q,$$
(1.7)
and Q is the best dominant.
Lemma 1.2
Lemma 1.2. (see [12, page 135, Corollary 3.4h.2]). Let Q be convex univalent in, and let be analytic in a domain D containing. Assume that…
Lemma 1.2. (see [12, page 135, Corollary 3.4h.2]). Let Q be convex univalent in $\mathbb{U}$ , and let $\theta$ be analytic in a domain D containing $Q(\mathbb{U})$ . Assume that
$$\Re\left[\theta'[Q(z)] + 1 + \frac{zQ''(z)}{Q'(z)}\right] > 0.$$
(1.8)
If P is analytic in $\mathbb{U}$ with P(0) = Q(0) and $P(\mathbb{U}) \subset D$ , then
$$zP'(z) + \theta[P(z)] \prec zQ'(z) + \theta[Q(z)] \Longrightarrow P \prec Q, \tag{1.9}$$
and Q is the best dominant.
Theorem 2.1
Theorem 2.1. Let Q(z) be univalent and nonzero in, Q(0) = 1, and let zQ'(z)/Q(z) be starlike in. If a function satisfies the subordination…
Theorem 2.1. Let Q(z) be univalent and nonzero in $\mathbb{U}$ , Q(0) = 1, and let zQ'(z)/Q(z) be starlike in $\mathbb{U}$ . If a function $f \in \mathcal{A}_p$ satisfies the subordination
$$\frac{zf^{(q+1)}(z)}{f^{(q)}(z)} \prec \frac{zQ'(z)}{Q(z)} + p - q,$$
(2.2)
then
$$\frac{f^{(q)}(z)}{\lambda(p;q)z^{p-q}} < Q(z), \tag{2.3}$$
and Q is the best dominant.
Corollary 2.3
Corollary 2.3. Let. If satisfies (2.10) then (2.11)
Corollary 2.3. Let $-1 \le B < A \le 1$ . If $f \in \mathcal{A}_p$ satisfies
$$\frac{zf^{(q+1)}(z)}{f^{(q)}(z)} \prec \frac{z(A-B)}{(1+Az)(1+Bz)} + p - q,$$
(2.10)
then
$$\frac{f^{(q)}(z)}{\lambda(p;q)z^{p-q}} < \frac{1+Az}{1+Bz}.$$
(2.11)
Corollary 2.5
Corollary 2.5. (see [13]). If, then Similarly, choosing q=1 and, Theorem 2.1 yields the following corollary.
Corollary 2.5. (see [13]). If $f \in S_p^*(\phi)$ , then
$$\frac{f(z)}{z^p} < \frac{h_{\phi,p}}{z^p}.\tag{2.19}$$
Similarly, choosing q=1 and $Q(z)=k'_{\phi,p}/pz^{p-1}$ , Theorem 2.1 yields the following corollary.
Corollary 2.6
Corollary 2.6. (see [13]). If, then
Corollary 2.6. (see [13]). If $f \in C_p^*(\phi)$ , then
$$\frac{f'(z)}{z^{p-1}} < \frac{k'_{\phi,p}}{z^{p-1}}. (2.20)$$
Theorem 2.7
Theorem 2.7. Let Q(z) be convex univalent in and Q(0)=1. If satisfies then and Q is the best dominant.
Theorem 2.7. Let Q(z) be convex univalent in $\mathbb{U}$ and Q(0)=1. If $f\in\mathcal{A}_p$ satisfies
$$\frac{f^{(q)}(z)}{\lambda(p;q)z^{p-q}} \cdot \left(\frac{zf^{(q+1)}(z)}{f^{(q)}(z)} - p + q\right) < zQ'(z),\tag{2.21}$$
then
$$\frac{f^{(q)}(z)}{\lambda(p;q)z^{p-q}} < Q(z), \tag{2.22}$$
and Q is the best dominant.
Theorem 2.9
Theorem 2.9. Let Q(z) be convex univalent in and Q(0) = 1. If satisfies then and Q is the best dominant.
Theorem 2.9. Let Q(z) be convex univalent in $\mathbb{U}$ and Q(0) = 1. If $f \in \mathcal{A}_p$ satisfies
$$\frac{zf^{(q+1)}(z)}{\lambda(p;q)z^{p-q}} < zQ'(z) + (p-q)Q(z), \tag{2.27}$$
then
$$\frac{f^{(q)}(z)}{\lambda(p;q)z^{p-q}} < Q(z), \tag{2.28}$$
and Q is the best dominant.
Corollary 2.10
Corollary 2.10. (see [17, Corollary 2.11]). Let Q(z) be convex univalent in, and Q(0) = 1. If satisfies then With p = 1, Corollary 2.10…
Corollary 2.10. (see [17, Corollary 2.11]). Let Q(z) be convex univalent in $\mathbb{U}$ , and Q(0) = 1. If $f \in \mathcal{A}_p$ satisfies
$$\frac{f'(z)}{z^{p-1}} < zQ'(z) + pQ(z), \tag{2.31}$$
then
$$\frac{f(z)}{z^p} \prec Q(z). \tag{2.32}$$
With p = 1, Corollary 2.10 yields the following corollary.
Corollary 2.11
Corollary 2.11. (see [17, Corollary 2.9]). Let Q(z) be convex univalent in, and Q(0) = 1. If satisfies then
Corollary 2.11. (see [17, Corollary 2.9]). Let Q(z) be convex univalent in $\mathbb{U}$ , and Q(0) = 1. If $f \in \mathcal{A}$ satisfies
$$f'(z) \prec zQ'(z) + Q(z), \tag{2.33}$$
then
$$\frac{f(z)}{z} < Q(z). \tag{2.34}$$
Theorem 2.12
Theorem 2.12. Let Q(z) be univalent and nonzero in, Q(0) = 1, and be starlike. If satisfies then and Q is the best dominant.
Theorem 2.12. Let Q(z) be univalent and nonzero in $\mathbb{U}$ , Q(0) = 1, and $zQ'(z)/Q^2(z)$ be starlike. If $f \in \mathcal{A}_p$ satisfies
$$\frac{\lambda(p;q)z^{p-q}}{f^{(q)}(z)} \cdot \left(\frac{zf^{(q+1)}(z)}{f^{(q)}(z)} - p + q\right) < \frac{zQ'(z)}{Q^2(z)},\tag{2.35}$$
then
$$\frac{f^{(q)}(z)}{\lambda(p;a)z^{p-q}} < Q(z), \tag{2.36}$$
and Q is the best dominant.
Theorem 2.13
Theorem 2.13. Let Q(z) be univalent and nonzero in, Q(0) = 1,, and zQ'(z)/[Q(z)(Q(z)+p-q-1)] be starlike in. If satisfies (2.40) then and Q…
Theorem 2.13. Let Q(z) be univalent and nonzero in $\mathbb{U}$ , Q(0) = 1, $Q(z) \neq q - p + 1$ , and zQ'(z)/[Q(z)(Q(z)+p-q-1)] be starlike in $\mathbb{U}$ . If $f \in \mathcal{A}_p$ satisfies
$$\frac{1 + (zf^{(q+2)}(z)/f^{(q+1)}(z)) - p + q + 1}{(zf^{(q+1)}(z)/f^{(q)}(z)) - p + q + 1} < 1 + \frac{zQ'(z)}{Q(z)(Q(z) + p - q - 1)},$$
(2.40)
then
$$\frac{zf^{(q+1)}(z)}{f^{(q)}(z)} - p + q + 1 < Q(z), \tag{2.41}$$
and Q is the best dominant.
Corollary 2.14
Corollary 2.14. (see [18, page 123, Corollary 6]). Let. If satisfies (2.48) then. For A = 0, B = b and A = 1, B = -1, Corollary 2.14 gives…
Corollary 2.14. (see [18, page 123, Corollary 6]). Let $-1 \le B < A \le 1$ . If $f \in \mathcal{A}$ satisfies
$$\frac{1 + (zf''(z)/f'(z))}{zf'(z)/f(z)} < 1 + \frac{(A-B)z}{(1+Az)^2},$$
(2.48)
then $f \in S^*[A, B]$ .
For A = 0, B = b and A = 1, B = -1, Corollary 2.14 gives the results of Obradovič and Tuneski [19].
Theorem 2.15
Theorem 2.15. Let Q(z) be univalent and nonzero in, Q(0) = 1,, and let zQ'(z)/[Q(z)+p-q-1] be starlike in. If satisfies (2.49) then and Q…
Theorem 2.15. Let Q(z) be univalent and nonzero in $\mathbb{U}$ , Q(0) = 1, $Q(z) \neq q - p + 1$ , and let zQ'(z)/[Q(z)+p-q-1] be starlike in $\mathbb{U}$ . If $f \in \mathcal{A}_p$ satisfies
$$1 + \frac{zf^{(q+2)}(z)}{f^{(q+1)}(z)} - \frac{zf^{(q+1)}(z)}{f^{(q)}(z)} < \frac{zQ'(z)}{Q(z) + p - q - 1},$$
(2.49)
then
$$\frac{zf^{(q+1)}(z)}{f^{(q)}(z)} - p + q + 1 < Q(z), \tag{2.50}$$
and Q is the best dominant.
Theorem 2.16
Theorem 2.16. Let Q(z) be a convex function in, and Q(0) = 1. If satisfies then and Q is the best dominant.
Theorem 2.16. Let Q(z) be a convex function in $\mathbb{U}$ , and Q(0) = 1. If $f \in \mathcal{A}_p$ satisfies
$$\frac{zf^{(q+1)}(z)}{f^{(q)}(z)} \left[ 2 + \frac{zf^{(q+2)}(z)}{f^{(q+1)}(z)} - \frac{zf^{(q+1)}(z)}{f^{(q)}(z)} \right] < zQ'(z) + Q(z) + p - q - 1, \tag{2.53}$$
then
$$\frac{zf^{(q+1)}(z)}{f^{(q)}(z)} - p + q + 1 < Q(z), \tag{2.54}$$
and Q is the best dominant.
Theorem 2.17
Theorem 2.17. Let Q(z) be a convex function in, with Q(0) = 1. If satisfies then and Q is the best dominant.
Theorem 2.17. Let Q(z) be a convex function in $\mathbb{U}$ , with Q(0) = 1. If $f \in \mathcal{A}_p$ satisfies
$$\frac{zf^{(q+1)}(z)}{f^{(q)}(z)} \left[ 1 + \frac{zf^{(q+2)}(z)}{f^{(q+1)}(z)} - \frac{zf^{(q+1)}(z)}{f^{(q)}(z)} \right] < zQ'(z), \tag{2.59}$$
then
$$\frac{zf^{(q+1)}(z)}{f^{(q)}(z)} - p + q + 1 < Q(z), \tag{2.60}$$
and Q is the best dominant.
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