Radii problems for certain analytic functions

Lemma 2.1

Let $h\in P(0\text{,}n)=P_n$ for $z\in E$ . Then

1. $\left|\frac{h^{\prime }(z)}{h(z)}\right|⩽\frac{2n|z{|}^{n-1}}{1-|z{|}^{2n}}$

2. $|{\mathit{zh}}^{\prime }(z)|⩽\frac{2n|z{|}^n\mathit{Reh}(z)}{1-|z{|}^{2n}}$

3. $\frac{1-|z{|}^n}{1+|z{|}^n}⩽\mathit{Reh}(z)⩽|h(z)|⩽\frac{1+|z{|}^n}{1-|z{|}^n}$ .

For (i) we refer to MacGraw (1963), (ii) will be found in Bernardi (1974) and for (iii) (see Shah, 1972).

Lemma 2.2

Let $p\in P(\gamma \text{,}n)$ for $z={\mathit{re}}^{i\theta }\in E$ . Then $$\frac{1+(2\gamma -1)r^n}{1+r^n}⩽\mathit{Rep}(z)⩽|p(z)|⩽\frac{1-(2\gamma -1)r^n}{1-r^n}$$ and $$|{\mathit{zp}}^{\prime }(z)|⩽\frac{2n(1-\gamma )r^n\mathit{Rep}(z)}{(1-r^n)[1+(1-2\gamma )r^n]}\text{.}$$

Theorem 3.1

Let $f\text{,}g\in A_n$ and let $\frac{f^{\prime }}{g^{\prime }}\in P_n$ , where $g\in V_k(\gamma \text{,}n)$ . Then $f\in V_k(\gamma \text{,}n)$ for $|z|<r_0$ , where

$$r_0^n=r_0^n(\gamma \text{,}n)=\frac{1-\gamma }{(n-\gamma +1)+\sqrt{n^2-2\gamma n+2n}}\text{.}$$ This result is sharp.

Proof

We can write $$f^{\prime }(z)=g^{\prime }(z)h(z)\text{,}\mathrm{where}g\in V_k(\gamma \text{,}n)\text{,}h\in P_n\text{,}z\in E\text{.}$$ Logarithmic differentiation yields, $$\frac{{({\mathit{zf}}^{\prime }(z))}^{\prime }}{f^{\prime }(z)}=\frac{{({\mathit{zg}}^{\prime }(z))}^{\prime }}{g^{\prime }(z)}+\frac{{\mathit{zh}}^{\prime }(z)}{h(z)}=H(z)+\frac{{\mathit{zh}}^{\prime }(z)}{h(z)}\text{,}h\in P_k(\gamma \text{,}n)=\left(\frac{k}{4}+\frac{1}{2}\right)\{(1-\gamma )H_1(z)+\gamma \}-\left(\frac{k}{4}-\frac{1}{2}\right)\{(1-\gamma )H_2(z)+\gamma \}+\frac{{\mathit{zh}}^{\prime }(z)}{h(z)}\text{,}$$ where $H_i\in P_n\text{,}i=1\text{,}2$ .

This gives us $$\frac{1}{1-\gamma }\left\{\frac{{({\mathit{zf}}^{\prime }(z))}^{\prime }}{f^{\prime }(z)}-\gamma \right\}=\left(\frac{k}{4}+\frac{1}{2}\right)\left\{H_1(z)+\frac{1}{1-\gamma }\frac{{\mathit{zh}}^{\prime }(z)}{h(z)}\right\}-\left(\frac{k}{4}-\frac{1}{2}\right)\left\{H_2(z)+\frac{1}{1-\gamma }\frac{{\mathit{zh}}^{\prime }(z)}{h(z)}\right\}\text{.}$$ Now, for $i=1\text{,}2$ , we use Lemma 2.1, with $|z|=r$ , to have $$\mathit{Re}\left\{H_i(z)+\frac{1}{1-\gamma }\frac{{\mathit{zh}}^{\prime }(z)}{h(z)}\right\}⩾{\mathit{ReH}}_i(z)-\frac{1}{1-\gamma }\left|\frac{{\mathit{zh}}^{\prime }(z)}{h(z)}\right|⩾\frac{1-r^n}{1+r^n}-\frac{1}{(1-\gamma )}\frac{2{\mathit{nr}}^n}{(1-r^{2n})}=\frac{(1-\gamma ){(1-r^n)}^2-2\gamma r^n}{(1-\gamma )(1-r^{2n})}=\frac{(1-\gamma )-2(n-\gamma +1)r^n+(1-\gamma )r^{2n}}{(1-\gamma )(1-r^{2n})}\text{.}$$ Therefore, for $|z|<r_0$ where $r_0$ is as stated in (3.1), $$\mathit{Re}\left\{H_i(z)+\frac{1}{1-\gamma }\frac{{\mathit{zh}}^{\prime }(z)}{h(z)}\right\}>0\text{.}$$ This implies that $$\frac{1}{1-\gamma }\left\{\frac{{({\mathit{zf}}^{\prime }(z))}^{\prime }}{f^{\prime }(z)}-\gamma \right\}\in P_k(0\text{,}n)\mathrm{for}|z|<r_0\text{,}$$ which leads us to the required result that $f\in V_k(\gamma \text{,}n)$ for $|z|<r_0$ , where $r_0$ is as given in (3.1).

Sharpness can be seen by taking $$H_i(z)=h(z)=\frac{1-z^n}{1+z^n}\in P_n.\square$$

Theorem 3.2

Let $f\text{,}g\in A_n$ and $g\in P_2^{\prime }(\gamma \text{,}n)=P^{\prime }(\gamma \text{,}n)$ in E. If $\frac{f^{\prime }}{g^{\prime }}\in P(\gamma \text{,}n)$ , then $f\in V_2(0\text{,}n)$ for $|z|<r_1$ where $r_1$ is given by

$$r_1^n=\frac{1}{2n(1-\gamma )+\sqrt{4n^2{(1-\gamma )}^2+1}}\text{.}$$ This result is also sharp.

Proof

$$\mathrm{Let}\frac{f^{\prime }(z)}{g^{\prime }(z)}=h(z)\text{,}\mathrm{where}h\in P(\gamma \text{,}n)\text{.}$$ Then $$\frac{f^{″}(z)}{f^{\prime }(z)}=\frac{h^{\prime }(z)}{h(z)}+\frac{g^{″}(z)}{g^{\prime }(z)}\text{.}$$ That is $$\frac{{({\mathit{zf}}^{\prime }(z))}^{\prime }}{f^{\prime }(z)}=\frac{{\mathit{zh}}^{\prime }(z)}{h(z)}+\frac{{\mathit{zp}}^{\prime }(z)}{p(z)}+1\text{,}\mathrm{where}{g}^{\prime }=p\in P(\gamma \text{,}n)\mathrm{in}E\text{.}$$

Now, using Lemma 2.2, we have

$$\mathit{Re}\frac{{({\mathit{zf}}^{\prime }(z))}^{\prime }}{f^{\prime }(z)}⩾1-\frac{4n(1-\gamma )r^n}{(1-r^n)(1+r^n)}=\frac{1-4n(1-\gamma )r^n-r^{2n}}{1-r^{2n}}\text{.}$$ The right hand side of the above inequality (3.3) is positive for $|z|<r_1$ , where $r_1$ is given by (3.2).

The sharpness can seen by considering $$\begin{array}{cc}\hfill & g(z)={\int }_0^z\left(\frac{1-(2\gamma -1)t^n}{1-t^n}\right)\mathit{dt}\text{,}\hfill \\ \hfill & h(z)=\frac{1-(2\gamma -1)z^n}{1-z^n}\hfill \end{array}$$ and $$f(z)={\int }_0^z{\left(\frac{1-(2\gamma -1)t^n}{1-t^n}\right)}^2\mathit{dt}\text{.}$$ The inclusion results for the classes $S_{\alpha }^{\ast }(\gamma \text{,}n)\text{,}{C}_{\alpha }(\gamma \text{,}n)\text{,}{K}_{\alpha }(\beta \text{,}\gamma )$ and $C_{\alpha }^{\ast }(\beta \text{,}\gamma )$ , with $n=1$ , have been studied by Noor (1991). We here deal with the converse case in general. □

Theorem 3.3

Let $f\in S_{\alpha }^{\ast }(\gamma \text{,}n)\text{,}\alpha ⩾0$ . Then $f\in S_{\alpha +1}^{\ast }(\gamma \text{,}n)$ for $|z|<r_0(\alpha \text{,}\gamma )$ is given as

$$r_0^n=r_0^n(\alpha \text{,}\gamma )=\frac{1+\alpha }{(1-\gamma +n)+\sqrt{{(1-\gamma +n)}^2-(1+\alpha )(1-2\gamma -\alpha )}}\text{.}$$

Proof

Since $f\in S_{\alpha }^{\ast }(\gamma \text{,}n)$ , so we can write it as $$\frac{{({\mathit{zD}}^{\alpha }f(z))}^{\prime }}{D^{\alpha }f(z)}=H(z)=(1-\gamma )h(z)+\gamma \text{,}$$ where $H\in P(\gamma \text{,}n)$ and so $h\in P(0\text{,}n)=P_n$ . Now using (1.7), we have $$\frac{D^{\alpha +1}f(z)}{D^{\alpha }f(z)}=\frac{1}{1+\alpha }\left\{\frac{{({\mathit{zD}}^{\alpha }f(z))}^{\prime }}{D^{\alpha }f(z)}+\alpha \right\}=\frac{1}{1+\alpha }\{(1-\gamma )h(z)+\gamma +\alpha \}\text{.}$$ Differentiating both sides logarithmically, we have $$\frac{z{(D^{\alpha +1}f(z))}^{\prime }}{D^{\alpha +1}f(z)}=\frac{{({\mathit{zD}}^{\alpha }f(z))}^{\prime }}{D^{\alpha }f(z)}+\frac{(1-\gamma ){\mathit{zh}}^{\prime }(z)}{(1-\gamma )h(z)+\gamma +\alpha }=(1-\gamma )h(z)+\gamma +\frac{(1-\gamma ){\mathit{zh}}^{\prime }(z)}{(1-\gamma )h(z)+\gamma +\alpha }\text{,}$$ or $$\frac{1}{1-\gamma }\left\{\frac{z{(D^{\alpha +1}f(z))}^{\prime }}{D^{\alpha +1}f(z)}-\gamma \right\}=h(z)+\frac{{\mathit{zh}}^{\prime }(z)}{(1-\gamma )h(z)+\gamma +\alpha }\text{.}$$ Therefore, $$\mathit{Re}\left\{\frac{1}{1-\gamma }\left(\frac{z{(D^{\alpha +1}f(z))}^{\prime }}{D^{\alpha +1}f(z)}-\gamma \right)\right\}=\mathit{Re}\left\{h(z)+\frac{{\mathit{zh}}^{\prime }(z)}{(1-\gamma )h(z)+\gamma +\alpha }\right\}\text{.}$$

By using Lemma 2.1 for $|z|=r<1$ , we have $$\mathit{Re}\left\{\frac{1}{1-\gamma }\left(\frac{z{(D^{\alpha +1}f(z))}^{\prime }}{D^{\alpha +1}f(z)}-\gamma \right)\right\}⩾\mathit{Reh}(z)\left\{1-\frac{2{\mathit{nr}}^n}{\frac{1-r^{2n}}{(1-\gamma )\left(\frac{1-r^n}{1+r^n}\right)+\gamma +\alpha }}\right\}=\mathit{Reh}(z)\left\{\frac{(1+\alpha )-2(1-\gamma +n)r^n+(1-2\gamma -\alpha )r^{2n}}{(1+\alpha )-2(1-\gamma )r^n+(1-2\gamma -\alpha )r^{2n}}\right\}\text{.}$$ The right hand side of above inequality is positive if $r^n<r_0^n$ and so is given by (3.4).

As a special case, when $\alpha =0\text{,}n=1$ and $\gamma =0$ , we obtain the radius of convexity $2-\sqrt{3}$  for starlike functions. □

Theorem 3.4

Let for $\alpha ⩾0\text{,}f\in C_{\alpha }(\gamma \text{,}n)$ in E. Then $f\in C_{\alpha +1}(\gamma \text{,}n)$ for $|z|<r_0=r_0(\alpha \text{,}\gamma )$ , where $r_0$ is given by (3.4).

Proof

By using the definition of $C_{\alpha }(\gamma \text{,}n)$ , we have $$\begin{array}{cc}\hfill & f\in C_{\alpha }(\gamma \text{,}n)\iff D^{\alpha }f\in C(\gamma \text{,}n)\mathrm{in}E\text{.}\hfill \\ \hfill & \iff z{(D^{\alpha }f)}^{\prime }\in S^{\ast }(\gamma \text{,}n)\mathrm{in}E\text{.}\hfill \\ \hfill & \iff D^{\alpha }({\mathit{zf}}^{\prime })\in S^{\ast }(\gamma \text{,}n)\mathrm{in}E\text{.}\hfill \\ \hfill & \iff {\mathit{zf}}^{\prime }\in S_{\alpha +1}^{\ast }(\gamma \text{,}n)\mathrm{in}|z|<r_0\text{.}\hfill \\ \hfill & \iff D^{\alpha +1}({\mathit{zf}}^{\prime })\in S^{\ast }(\gamma \text{,}n)\mathrm{in}|z|<r_0\text{.}\hfill \\ \hfill & \iff z{(D^{\alpha +1}f)}^{\prime }\in S^{\ast }(\gamma \text{,}n)\mathrm{in}|z|<r_0\text{.}\hfill \\ \hfill & \iff D^{\alpha +1}f\in C(\gamma \text{,}n)\mathrm{in}|z|<r_0\text{.}\hfill \\ \hfill & \iff f\in C_{\alpha +1}(\gamma \text{,}n)\mathrm{in}|z|<r_0\text{,}\hfill \end{array}$$ which is the required result. □

Theorem 3.5

Let for $\alpha ⩾0\text{,}f\in K_{\alpha }(\beta \text{,}\gamma )$ in E. Then $f\in K_{\alpha +1}(\beta \text{,}\gamma )$ for $|z|<r_0=r_0(\alpha \text{,}\gamma )$ , where $r_0$ is given by (3.4).

Proof

Since $f\in K_{\alpha }(\beta \text{,}\gamma )$ , there exists $g\in S^{\ast }(\gamma \text{,}n)$ such that

$$\frac{z{(D^{\alpha }f(z))}^{\prime }}{D^{\alpha }g(z)}=(1-\beta )h(z)+\beta \text{,}h\in P(0\text{,}n)\text{.}$$ Also, since $g\in S^{\ast }(\gamma \text{,}n)$ , we can write

(3.6)

$$\frac{z{(D^{\alpha }g(z))}^{\prime }}{D^{\alpha }g(z)}=(1-\gamma )H(z)+\gamma \text{,}H\in P(0\text{,}n)\text{.}$$

Using (1.7), we have $$\frac{z{(D^{\alpha +1}f(z))}^{\prime }}{D^{\alpha +1}g(z)}=\frac{D^{\alpha +1}({\mathit{zf}}^{\prime }(z))}{D^{\alpha +1}g(z)}=\frac{\frac{1}{1+\alpha }z{(D^{\alpha }({\mathit{zf}}^{\prime }(z)))}^{\prime }+\frac{\alpha }{1+\alpha }{D}^{\alpha }({\mathit{zf}}^{\prime }(z))}{\frac{1}{1+\alpha }z{(D^{\alpha }g(z))}^{\prime }+\frac{\alpha }{1+\alpha }{D}^{\alpha }g(z)}=\frac{\frac{z{(D^{\alpha }({\mathit{zf}}^{\prime }(z)))}^{\prime }}{D^{\alpha }g(z)}+\alpha \frac{D^{\alpha }({\mathit{zf}}^{\prime }(z))}{D^{\alpha }g(z)}}{\frac{z{(D^{\alpha }g(z))}^{\prime }}{D^{\alpha }g(z)}+\alpha }\text{.}$$

By using (3.5) and (3.6), we have

$$\frac{z{(D^{\alpha +1}f(z))}^{\prime }}{D^{\alpha +1}g(z)}=\frac{\frac{z{(D^{\alpha }({\mathit{zf}}^{\prime }(z)))}^{\prime }}{D^{\alpha }g(z)}+\alpha \{(1-\beta )h(z)+\beta \}}{(1-\gamma )H(z)+\gamma +\alpha }\text{.}$$

From (3.5), we have $$z{(D^{\alpha }f(z))}^{\prime }=D^{\alpha }g(z)\{(1-\beta )h(z)+\beta \}\text{.}$$ Differentiating both sides, we have $$z{(z{(D^{\alpha }f(z))}^{\prime })}^{\prime }=(1-\beta ){\mathit{zh}}^{\prime }(z)(D^{\alpha }g(z))+{(D^{\alpha }g(z))}^{\prime }\{(1-\beta )h(z)+\beta \}\text{.}$$ That is,

$$\frac{z{(D^{\alpha }{(\mathit{zf}(z))}^{\prime })}^{\prime }}{D^{\alpha }g(z)}=(1-\beta ){\mathit{zh}}^{\prime }(z)+\{(1-\gamma )H(z)+\gamma \}\{(1-\beta )h(z)+\beta \}\text{.}$$ Using (3.8) in (3.7), we obtain $$\left\{\frac{1}{1-\beta }\left(\frac{z{(D^{\alpha +1}f(z))}^{\prime }}{D^{\alpha +1}g(z)}-\beta \right)\right\}=\left\{h(z)+\frac{{\mathit{zh}}^{\prime }(z)}{(1-\gamma )H(z)+\gamma +\alpha }\right\}\text{.}$$ Now $$\mathit{Re}\left\{\frac{1}{1-\beta }\left(\frac{z{(D^{\alpha +1}f(z))}^{\prime }}{D^{\alpha +1}f(z)}-\beta \right)\right\}=\mathit{Reh}(z)\left\{1-\frac{\frac{2{\mathit{nr}}^n}{1-r^n}}{(1-\gamma )\frac{1-r^n}{1+r^n}+\gamma +\alpha }\right\}\text{.}$$ We note that right hand side is positive for $|z|<r_0=r_0(\alpha \text{,}\gamma )$ given by (3.4) and also $g\in S_{\alpha }^{\ast }(\gamma \text{,}n)$ for $|z|<r_0$ . Hence $f\in K_{\alpha +1}(\beta \text{,}\gamma )$ for $|z|<r_0$ . □

Theorem 3.6

Let $f\in C_{\alpha }^{\ast }(\beta \text{,}\gamma )$ for $\alpha ⩾0$ , $z\in E$ . Then $f\in C_{\alpha +1}^{\ast }(\beta \text{,}\gamma )$ for $|z|<r_0=r_0(\alpha \text{,}\gamma )$ , where $r_0$ is given by (3.4).