Ramachandran et al. Advances in Difference Equations 2014, 2014:288

Lemma 2.1. [29] Let h be convex univalent in $\mathbb{U}$ with h(0) = 1 and $\Re(vh(z) + \mu) > 0$ $(v, \mu \in \mathbb{C})$ . If p is analytic in $\mathbb{U}$ with p(0) = 1 then <span id="page-6-3"></span> $$p(z) + \frac{zp'(z)}{vp(z) + \mu} \prec h(z) \quad (z \in \mathbb{U}) \quad \Rightarrow \quad p(z) \prec h(z) \quad (z \in \mathbb{U}). \tag{2.1}$$

Lemma 2.2. [30] Let h be convex in the open unit disk $\mathbb{U}$ and let $E \ge 0$ . Suppose B(z) $(z \in \mathbb{U})$ is analytic in $\mathbb{U}$ with $\Re(B(z)) > 0$ . If g(z) is analytic in $\mathbb{U}$ and h(0) = g(0). Then $$Ez^2g''(z) + B(z)g(z) < h(z) \quad \Rightarrow \quad g(z) < h(z). \tag{2.2}$$

Theorem 3.1. Let $c \ge 1$ , and h be convex univalent in $\mathbb{U}$ with h(0) = 1 and $\Re(h(z)) > 0$ . If a function $f \in \mathcal{A}$ satisfies the condition $$\frac{1}{1-\eta} \left[ \frac{z (B_{\kappa}^c f(z))'}{B_{\kappa}^c f(z)} - \eta \right] \prec h(z) \quad (0 \le \eta < 1; z \in \mathbb{U}), \tag{3.1}$$ then $$\frac{1}{1-\eta} \left[ \frac{z(B_{\kappa+1}^c f(z))'}{B_{\kappa+1}^c f(z)} - \eta \right] < h(z) \quad (0 \le \eta < 1; z \in \mathbb{U}).$$ (3.2) Proof Let $$p(z) = \frac{1}{1 - \eta} \left[ \frac{z(B_{\kappa + 1}^{c} f(z))'}{B_{\kappa + 1}^{c} f(z)} - \eta \right] \quad (z \in \mathbb{U}), \tag{3.3}$$ where p is an analytic function in $\mathbb{U}$ with p(0) = 1. By using (1.21), we get $$(1-\eta)p(z) + \eta = \kappa \frac{zB_{\kappa}^{c}f(z)}{B_{\kappa+1}^{c}f(z)} - (\kappa-1).$$ <span id="page-7-0"></span>Differentiating logarithmically with respect to z and multiplying by z, we obtain $$p(z) + \frac{zp'(z)}{(1-\eta)p(z) + \eta + \kappa - 1} = \frac{1}{1-\eta} \left[ \frac{zB_{\kappa}^{c}f(z)}{B_{\kappa+1}^{c}f(z)} - \eta \right].$$ The proof of the theorem follows now by an application of Lemma 2.1.

Theorem 3.2. Let $f \in A$ . If $B_{\kappa}^{c} f(z) \in k$ - $\mathcal{ST}(\eta)$ , then $B_{\kappa+1}^{c} f(z) \in k$ - $\mathcal{ST}(\eta)$ . Proof Let $$s(z) = \frac{z(B_{\kappa+1}^c f(z))'}{B_{\kappa+1}^c f(z)}.$$ From (1.21), we can write $$\kappa \frac{B_{\kappa}^{c} f(z)}{B_{\kappa+1}^{c} f(z)} = s(z) + \kappa - 1.$$ Taking logarithmic differentiation and multiplying by z, we get $$\frac{z(B_{\kappa}^c f(z))'}{B_{\kappa}^c f(z)} = s(z) + \frac{zs'(z)}{s(z) + \kappa - 1} \prec Q_{k,\eta}(z).$$ Since $Q_{k,n}(z)$ is convex univalent in $\mathbb{U}$ and $$\Re\left(Q_{k,\eta}(z)\right) > \frac{k+\eta}{k+1},$$ the proof of the theorem follows by Theorem 3.1 and condition (1.14).

Theorem 3.3. Let $f \in \mathcal{A}$ . If $B_{\kappa}^{c}f(z) \in k$ - $\mathcal{UCV}(\eta)$ , then $B_{\kappa+1}^{c}f(z) \in k$ - $\mathcal{UCV}(\eta)$ . Proof By virtue of (1.12), (1.13), and Theorem 3.2, we obtain <span id="page-7-1"></span> $$\begin{split} B_{\kappa}^{c}f(z) \in k\text{-}\mathcal{UCV}(\eta) & \Leftrightarrow & z\big(B_{\kappa}^{c}f(z)\big)' \in k\text{-}\mathcal{ST}(\eta) \\ & \Leftrightarrow & B_{\kappa}^{c}zf'(z) \in k\text{-}\mathcal{ST}(\eta) \\ & \Rightarrow & B_{\kappa+1}^{c}zf'(z) \in k\text{-}\mathcal{ST}(\eta) \\ & \Leftrightarrow & B_{\kappa+1}^{c}f(z) \in k\text{-}\mathcal{UCV}(\eta) \end{split}$$ and hence the proof is complete.

Theorem 3.4. Let $f \in \mathcal{A}$ . If $B_{\kappa}^{c}f(z) \in \mathcal{UCC}(k, \eta, \beta)$ , then $B_{\kappa+1}^{c}f(z) \in \mathcal{UCC}(k, \eta, \beta)$ . Proof Since $$B_{\kappa}^{c}f(z)\in\mathcal{UCC}(k,\eta,\beta),$$ $$\frac{z(B_{k}^{c}f(z))'}{k(z)} \prec Q_{k,\eta}(z) \quad \text{for some } k(z) \in k\text{-}\mathcal{ST}(\beta).$$ For g(z) such that $B_{\kappa}^{c}g(z)=k(z)$ we have <span id="page-8-1"></span> $$\frac{z(B_{\kappa}^{c}f(z))'}{B^{c}\sigma(z)} \prec Q_{k,\eta}(z). \tag{3.4}$$ Letting $$h(z) = \frac{z(B_{\kappa+1}^c f(z))'}{B_{\kappa+1}^c g(z)}$$ and $H(z) = \frac{z(B_{\kappa+1}^c g(z))'}{B_{\kappa+1}^c g(z)}$ . We observe that h(z) and H(z) are analytic in $\mathbb{U}$ and h(0) = H(0) = 1. Now, by Theorem 3.2, <span id="page-8-0"></span> $$B_{k+1}^c g(z) \in k-\mathcal{ST}(\beta)$$ and $\Re(H(z)) > \frac{k+\beta}{k+1}$ . Also note that <span id="page-8-2"></span> $$z(B_{\kappa+1}^c f(z))' = (B_{\kappa+1}^c g(z))h(z). \tag{3.5}$$ Differentiating both sides of (3.5), we obtain $$\frac{z(z(B_{\kappa+1}^c f(z))')'}{B_{\kappa+1}^c g(z)} = z \frac{(B_{\kappa+1}^c g(z))'}{B_{\kappa+1}^c g(z)} h(z) + zh'(z) = H(z) \cdot h(z) + zh'(z). \tag{3.6}$$ Now using the identity (1.21), we obtain $$\frac{z(B_{\kappa}^{c}f(z))'}{B_{\kappa}^{c}g(z)} = \frac{B_{\kappa}^{c}(zf'(z))}{B_{\kappa}^{c}g(z)}$$ $$= \frac{z(B_{\kappa+1}^{c}zf'(z))' + (\kappa - 1)B_{\kappa+1}^{c}(zf'(z))}{z(B_{\kappa+1}^{c}g(z))' + (\kappa - 1)B_{\kappa+1}^{c}g(z)}$$ $$= \frac{\frac{z(B_{\kappa+1}^{c}zf'(z))'}{B_{\kappa+1}^{c}g(z)} + (\kappa - 1)\frac{B_{\kappa+1}^{c}(zf'(z))}{B_{\kappa+1}^{c}g(z)}$$ $$= \frac{\frac{z(B_{\kappa+1}^{c}zf'(z))'}{B_{\kappa+1}^{c}g(z)} + (\kappa - 1)\frac{B_{\kappa+1}^{c}(zf'(z))}{B_{\kappa+1}^{c}g(z)}$$ $$= h(z) + \frac{zh'(z)}{H(z) + \kappa - 1}.$$ (3.7) From (3.4), (3.6), and the above equation, we conclude that $$h(z) + \frac{zh'(z)}{H(z) + \kappa - 1} \prec Q_{k,\eta}(z).$$ On letting E = 0 and $B(z) = \frac{1}{H(z) + \kappa - 1}$ , we obtain $$\Re \big(B(z)\big) = \frac{\Re \big(H(z) + \kappa - 1\big)}{|H(z) + \kappa - 1|^2} > 0$$ and the above inequality satisfies the conditions required by Lemma 2.2. Hence $$h(z) \prec Q_{k,\eta}(z)$$ and so the proof is complete. <span id="page-9-2"></span>Using a similar argument to Theorem 3.4, we can prove the following theorem.

Theorem 3.5. Let $f \in \mathcal{A}$ . If $B_{\kappa}^{c}f(z) \in \mathcal{UQC}(k, \eta, \beta)$ , then $B_{\kappa+1}^{c}f(z) \in \mathcal{UQC}(k, \eta, \beta)$ . <span id="page-9-0"></span>Now we examine the closure properties of the integral operator $L_{\nu}$ .

Theorem 3.6. Let $\gamma > -\frac{k+\eta}{k+1}$ . If $B_k^c \in k$ - $\mathcal{ST}(\eta)$ so is $L_{\gamma}(B_k^c)$ . Proof From the definition of $L_{\gamma}(f)$ and the linearity of the operator $B_{\kappa}^{c}$ we have <span id="page-9-1"></span> $$z(B_{\kappa}^{c}L_{\gamma}(f))' = (\gamma + 1)B_{\kappa}^{c}f(z) - \gamma B_{\kappa}^{c}L_{\gamma}(f). \tag{3.8}$$ Substituting $\frac{z(B_\kappa^2 L_\gamma(f(z)))'}{B_\kappa^2 \nu L_\gamma(f(z))} = p(z)$ in (3.8) we may write $$p(z) = (\gamma + 1) \frac{B_{\kappa}^{c} f(z)}{B_{\kappa}^{c} L_{\gamma}(f(z))} - \gamma.$$ (3.9) On differentiating (3.9) we get $$\frac{z(B_{\kappa}^{c}(f(z)))'}{B_{\kappa}^{c}(f(z))} = \frac{z(B_{\kappa}^{c}L_{\gamma}f(z))'}{B_{\kappa}^{c}L_{\gamma}(f(z))} + \frac{zp'(z)}{p(z) + \gamma} = p(z) + \frac{zp'(z)}{p(z) + \gamma}.$$ <span id="page-9-3"></span>By Lemma 2.1, we have $p(z) \prec Q(k, \eta)$ , since $\Re(Q(k, \eta) + \gamma) > 0$ . This completes the proof of Theorem 3.6. By a similar argument we can prove Theorem 3.7 as below.

Theorem 3.7. Let $\gamma > -\frac{k+\eta}{k+1}$ . If $B_{\kappa}^c \in k$ - $\mathcal{UCV}(\eta)$ so is $L_{\gamma}(B_{\kappa}^c)$ .

Theorem 3.8. Let $\gamma > -\frac{k+\eta}{k+1}$ . If $B_k^c \in \mathcal{UCC}(k, \eta, \beta)$ so is $L_{\gamma}(B_k^c)$ . Proof By definition, there exists a function <span id="page-9-5"></span> $$K(z) = B_{\nu}^{c} g(z) \in k - \mathcal{ST}(\eta),$$ so that <span id="page-9-4"></span> $$\frac{z(B_{\kappa}^{c}(f(z)))'}{B_{\kappa}^{c}(g(z))} \prec Q_{k,\eta}(z) \quad (z \in \mathbb{U}). \tag{3.10}$$ Now from (3.8) we have $$\frac{z(B_{\kappa}^{c}f)'}{B_{\kappa}^{c}(g(z))} = \frac{z(B_{\kappa}^{c}L_{\gamma}(zf'))' + \gamma B_{\kappa}^{c}L_{\gamma}(zf'(z))}{z(B_{\kappa}^{c}L_{\gamma}(g(z)))' + \gamma B_{\kappa}^{c}L_{\gamma}(g(z))} \\ = \frac{\frac{z(B_{\kappa}^{c}(zf'(z)))'}{B_{\kappa}^{c}L_{\gamma}(g(z))} + \frac{\gamma B_{\kappa}^{c}(zf'(z))}{B_{\kappa}^{c}L_{\gamma}(g(z))}}{\frac{z(B_{\kappa}^{c}L_{\gamma}(g(z)))'}{B_{\kappa}^{c}L_{\gamma}(g(z))} + \gamma}.$$ (3.11) Since $B_{\kappa}^c g \in k\text{-}\mathcal{ST}(\eta)$ , by Theorem 3.6, we have $L_{\gamma}(B_{\kappa}^c g) \in k\text{-}\mathcal{ST}(\eta)$ . Taking $\frac{z(B_{\kappa}^c L_{\gamma}(g(z)))'}{B_{\kappa}^c L_{\gamma}(g)} = H(z)$ , we note that $\Re(H(z)) > \frac{k+\eta}{k+1}$ . Now for $h(z) = \frac{z(B_{\kappa}^c L_{\gamma}(f(z)))'}{B_{\kappa}^c L_{\gamma}(g(z))}$ we obtain <span id="page-10-4"></span><span id="page-10-3"></span> $$z(B_{\nu}^{c}L_{\nu}(f(z)))' = h(z)B_{\nu}^{c}L_{\nu}(g(z)). \tag{3.12}$$ Differentiating both sides of (3.12) yields $$\frac{z(B_{\kappa}^{c}(zL_{\gamma}(f))')'}{B_{\kappa}^{c}L_{\gamma}(g)} = zh'(z) + h(z)\frac{z(B_{\kappa}^{c}L_{\gamma}(g))'}{B_{\kappa}^{c}L_{\gamma}(g)} = zh'(z) + H(z)h(z).$$ (3.13) Therefore from (3.11) and (3.13) we obtain <span id="page-10-5"></span> $$\frac{z(B_{\kappa}^{c}f(z))'}{B_{\kappa}^{c}g} = \frac{zh'(z) + H(z)h(z) + \gamma h(z)}{H(z) + \gamma}.$$ (3.14) This in conjunction with (3.10) leads to $$h(z) + \frac{zh'(z)}{H(z) + \gamma} < Q(k, \eta)(z).$$ (3.15) Let us take $B(z) = \frac{1}{H(z) + \gamma}$ in (3.15) and observe that $\Re(B(z)) > 0$ as $\gamma > -\frac{k + \eta}{k + 1}$ . Now for A = 0 and B as described we conclude the proof since the required conditions of Lemma 2.2 are satisfied. A similar argument yields the following.

Theorem 3.9. Let $\gamma > -\frac{k+\eta}{k+1}$ . If $B_{\kappa}^c \in \mathcal{UQC}(k, \eta, \beta)$ so is $L_{\gamma}(B_{\kappa}^c)$ .