Notes on analytic functions with a bounded positive real part

Abstract

For real α and β such that $0\le \alpha <1<\beta$, we denote by $\mathcal{S}(\alpha ,\beta )$ the class of normalized analytic functions f such that $\alpha <Re\{z{f}^{\prime }(z)/f(z)\}<\beta$ in . We find some properties, including inclusion properties, Fekete-Szegö problem and coefficient problems of inverse functions.

MSC:30C45, 30C55.

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction

Let ℋ denote the class of analytic functions in the unit disc $\mathbb{U}=\{z:|z|<1\}$ on the complex plane ℂ. Let denote the subclass of ℋ consisting of functions normalized by $f(0)=0$ and $f^{\prime }(0)=1$. Let denote the subclass of consisting of univalent functions. Denote by ${\mathcal{S}}^{\ast }$ and , the class of starlike functions and convex functions, respectively. It is well-known that $\mathcal{K}\subset {\mathcal{S}}^{\ast }\subset \mathcal{S}$.

We say that f is subordinate to F in , written as $f\prec F$ if and only if $f(z)=F(w(z))$ for some Schwarz function with $w(0)=0$ and $|w(z)|<1$, $z\in \mathbb{U}$. If $F(z)$ is univalent in , then the subordination $f\prec F$ is equivalent to $f(0)=F(0)$ and $f(\mathbb{U})\subset F(\mathbb{U})$.

We denote by ${\mathcal{S}}^{\ast }(A,B)$, the class of Janowski starlike functions, namely, the functions satisfying the subordination equation: $z{f}^{\prime }(z)/f(z)\prec (1+Az)/(1+Bz)$. Note that ${\mathcal{S}}^{\ast }(1,-1)={\mathcal{S}}^{\ast }$.

Now, we shall introduce the class of analytic functions used in the sequel.

Definition 1 Let α and β be real numbers such that $0\le \alpha <1<\beta$. The function $f\in \mathcal{A}$ belongs to the class $\mathcal{S}(\alpha ,\beta )$ if f satisfies the following inequality:

$$\alpha <Re\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}<\beta (z\in \mathbb{U}).$$

We remark that for given α, β ($0\le \alpha <1<\beta$), $f\in \mathcal{S}(\alpha ,\beta )$ if and only if f satisfies the following two subordination equations:

$$\frac{z{f}^{\prime }(z)}{f(z)}\prec \frac{1+(1-2\alpha )z}{1-z}\text{and}\frac{z{f}^{\prime }(z)}{f(z)}\prec \frac{1+(1-2\beta )z}{1-z},$$

(1)

since the functions $(1+(1-2\alpha )z)/(1-z)$ and $(1+(1-2\beta )z)/(1-z)$ map onto the right half plane, having real part greater than α, and the left half plane, having real part smaller than β, respectively. The above class $\mathcal{S}(\alpha ,\beta )$ is introduced by Kuroki and Owa [1]. They investigated coefficient estimates for $f\in \mathcal{S}(\alpha ,\beta )$ and found the necessary and sufficient condition for $f\in \mathcal{S}(\alpha ,\beta )$ using the following subordination.

Lemma 1 (Kuroki and Owa [1])

Let $f\in \mathcal{A}$ and $0\le \alpha <1<\beta$. Then $f\in \mathcal{S}(\alpha ,\beta )$ if and only if

$$\frac{z{f}^{\prime }(z)}{f(z)}\prec 1+\frac{\beta -\alpha }{\pi }ilog\left(\frac{1-e^{2\pi i\frac{1-\alpha }{\beta -\alpha }}z}{1-z}\right)(z\in \mathbb{U}).$$

Lemma 1 means that the function p defined by

$$p(z)=1+\frac{\beta -\alpha }{\pi }ilog\left(\frac{1-e^{2\pi i\frac{1-\alpha }{\beta -\alpha }}z}{1-z}\right)$$

(2)

maps the unit disk onto the strip domain w with $\alpha <Re(w)<\beta$. We note that the function $f\in \mathcal{A}$, given by

$$f(z)=zexp\{\frac{\beta -\alpha }{\pi }i{\int }_0^z\frac{1}{t}log\left(\frac{1-e^{2\pi i\frac{1-\alpha }{\beta -\alpha }}t}{1-t}\right)dt\},$$

is in the class $\mathcal{S}(\alpha ,\beta )$.

2 Inclusion properties

Theorem 1 For given $0\le \alpha <1<\beta$, let A and B be real numbers such that

$$\frac{2-\alpha -\beta }{\beta -\alpha }\le B<A\le \frac{\beta -2\alpha \beta +\alpha }{\beta -\alpha }.$$

(3)

Then ${\mathcal{S}}^{\ast }(A,B)\subset \mathcal{S}(\alpha ,\beta )$.

Proof At first, we note that

$$-1<\frac{2-\alpha -\beta }{\beta -\alpha }\text{and}\frac{\beta -2\alpha \beta +\alpha }{\beta -\alpha }<1.$$

For $f\in {\mathcal{S}}^{\ast }(A,B)$, we know that the following inequality holds:

$$\frac{1-A}{1-B}<Re\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}<\frac{1+A}{1+B}(z\in \mathbb{U}).$$

Therefore, it suffices to show that α and β satisfy the following inequalities:

$$\alpha \le \frac{1-A}{1-B}\text{and}\frac{1+A}{1+B}\le \beta .$$

(4)

Using inequality (4), we can derive that

$$1+B\ge \frac{2(1-\alpha )}{\beta -\alpha }\text{and}1+A\le \frac{2\beta (\beta -\alpha )}{\beta -\alpha }.$$

(5)

Also,

$$1-B\le \frac{2(\beta -1)}{\beta -\alpha }\text{and}1-A\ge \frac{2\alpha (\beta -a)}{\beta -\alpha }.$$

(6)

By the above inequalities (5) and (6), we can easily obtain the inequalities (3), so the proof of Theorem 1 is completed. □

Lemma 2 (Miller and Mocanu [2])

Let Ξ be a set in the complex plane ℂ and let b be a complex number such that $Re(b)>0$. Suppose that a function $\psi :{\mathbb{C}}^2\times \mathbb{U}\to \mathbb{C}$ satisfies the condition

$$\psi (i\rho ,\sigma ;z)\notin \Xi$$

for all real ρ, $\sigma \le -{|b-i\rho |}^2/(2Re(b))$ and all $z\in \mathbb{U}$. If the function $p(z)$ defined by $p(z)=b+b_{1}z+b_2{z}^2+\cdots$ is analytic in and if

$$\psi (p(z),z{p}^{\prime }(z);z)\in \Xi ,$$

then $Re(p(z))>0$ in .

Theorem 2 Let $f\in \mathcal{A}$, $1/2\le \alpha <1$ and $Re\{z{f}^{\prime }(z)/f(z)\}>\alpha$ in . Then

$$Re\left\{\frac{f(z)}{z}\right\}>\gamma (\alpha ):=\frac{1}{3-2\alpha }(z\in \mathbb{U}).$$

(7)

Proof Write $\gamma (\alpha ):=\gamma$ and note that $\frac{1}{2}\le \gamma <1$ for $\frac{1}{2}\le \alpha <1$. Let p be defined by

$$p(z)=\frac{1}{1-\gamma }(\frac{z{f}^{\prime }(z)}{f(z)}-\gamma ).$$

Then p is analytic in , $p(0)=1$ and

$$\frac{z{f}^{\prime }(z)}{f(z)}=1+\frac{(1-\gamma )z{p}^{\prime }(z)}{(1-\gamma )p(z)+\gamma }=\psi (p(z),z{p}^{\prime }(z)),$$

where

$$\psi (r,s)=1+\frac{(1-\gamma )s}{(1-\gamma )r+\gamma }.$$

(8)

Also,

$$\{\psi (p(z),z{p}^{\prime }(z)):z\in \mathbb{U}\}\subset \{w\in \mathbb{C}:Re(w)>\alpha \}:={\Omega }_{\alpha }.$$

Now for all real ρ, $\sigma \le -\frac{1}{2}(1+{\rho }^2)$,

$$\begin{array}{rcl}Re(\psi (i\rho ,\sigma ))& =& Re(1+\frac{(1-\gamma )\sigma }{(1-\gamma )i\rho +\gamma })=1+\frac{\gamma (1-\gamma )\sigma }{{\gamma }^2+{(1-\gamma )}^2{\rho }^2}\\ \le & 1-\frac{1}{2}\gamma (1-\gamma )\frac{1+{\rho }^2}{{\gamma }^2+{(1-\gamma )}^2{\rho }^2}.\end{array}$$

Now, we let

$$g(\rho )=\frac{1+{\rho }^2}{{\gamma }^2+{(1-\gamma )}^2{\rho }^2}.$$

(9)

Then

$$g^{\prime }(\rho )=\frac{2(2\gamma -1)\rho }{{({\gamma }^2+{(1-\gamma )}^2{\rho }^2)}^2},$$

hence $g^{\prime }(\rho )=0$ occurs at only $\rho =0$ and g satisfies

$$g(0)=\frac{1}{{\gamma }^2}$$

and

$$\underset{\rho \to \mathrm{\infty }}{lim}g(\rho )=\frac{1}{{(1-\gamma )}^2}.$$

Since $1/2\le \gamma <1$, we have

$$\frac{1}{{\gamma }^2}\le g(\rho )<\frac{1}{{(1-\gamma )}^2},$$

hence we get

$$Re(\psi (i\rho ,\sigma ))\le 1-\frac{1}{2}\gamma (1-\gamma )\frac{1}{{\gamma }^2}=\frac{3\gamma -1}{2\gamma }=\alpha .$$

This shows that $Re\{\psi (i\rho ,\sigma )\}\notin {\Omega }_{\alpha }$. By Lemma 2, we get $Re(p(z))>0$ in , and this shows that inequality (7) holds and the proof of Theorem 2 is completed. □

Theorem 3 Let $f\in \mathcal{A}$, $1<\beta <3/2$ and $Re\{z{f}^{\prime }(z)/f(z)\}<\beta$ in . Then

$$Re\left\{\frac{f(z)}{z}\right\}<\delta (\beta ):=\frac{1}{3-2\beta }(z\in \mathbb{U}).$$

(10)

Proof Note that $\delta :=\delta (\beta )=\frac{1}{3-2\beta }>1$ for $\beta >1$. Let p be defined by

$$p(z)=\frac{1}{1-\delta }(\frac{z{f}^{\prime }(z)}{f(z)}-\delta ).$$

Then p is analytic in , $p(0)=1$ and

$$\frac{z{f}^{\prime }(z)}{f(z)}=\psi (p(z),z{p}^{\prime }(z)),$$

where ψ is given in (8). Also

$$\{\psi (p(z),z{p}^{\prime }(z)):z\in \mathbb{U}\}\subset \{w\in \mathbb{C}:Re(w)<\beta \}:={\Omega }_{\beta }.$$

Now, for all real ρ, $\sigma \le -\frac{1}{2}(1+{\rho }^2)$,

$$Re(\psi (i\rho ,\sigma ))\ge 1-\frac{1}{2}\delta (1-\delta )g(\rho ),$$

where $g(\rho )$ is given (9). Since

$$\frac{1}{{(1-\delta )}^2}<g(\rho )\le \frac{1}{{\delta }^2}$$

for all $\delta >1$, we have

$$Re(\psi (i\rho ,\sigma ))\ge \frac{3\delta -1}{2\delta }=\beta .$$

This shows that $Re\{\psi (i\rho ,\sigma )\}\notin {\Omega }_{\beta }$. By Lemma 2, we get $Re(p(z))>0$ in , and this is equivalent to

$$Re\left\{\frac{f(z)}{z}\right\}<\delta (z\in \mathbb{U}),$$

and the proof of Theorem 3 is completed. □

Combining the above Theorems 2 and 3, we can obtain the following result:

Theorem 4 Let $f\in \mathcal{A}$, $1/2\le \alpha <1<\beta <3/2$ and $\alpha <Re\{z{f}^{\prime }(z)/f(z)\}<\beta$ in . Then

$$\gamma (\alpha )<Re\left\{\frac{f(z)}{z}\right\}<\delta (\beta )(z\in \mathbb{U}),$$

where $\gamma (\alpha )$ and $\delta (\beta )$ is given (7) and (10).

3 Some coefficient problems

In this section, we investigate coefficient problems for functions in the class $\mathcal{S}(\alpha ,\beta )$. In [1], Kuroki and Owa investigated the coefficient of the function p given by (2); the function p can be written as

$$p(z)=1+\sum _{n=1}^{\mathrm{\infty }}{B}_n{z}^n,$$

where

$$B_n=\frac{2(\beta -\alpha )}{n\pi }sin\left(\frac{1-\alpha }{\beta -\alpha }\pi \right).$$

We denote by ${\mathcal{S}}_{\sigma }(\alpha ,\beta )$ the class of bi-univalent functions f such that $f\in \mathcal{S}(\alpha ,\beta )$ and the inverse function $f^{-1}\in \mathcal{S}(\alpha ,\beta )$. Srivastava et al. investigated the estimates on the initial coefficient for certain subclasses of analytic and bi-univalent functions in [3, 4]. Ali et al. have studied similar problems in [5].

In theorem, we shall solve the Fekete-Szegö problem for $f\in S(\alpha ,\beta )$. We need the following lemma:

Lemma 3 (Keogh and Merkers [6])

Let $p(z)=1+c_{1}z+c_2{z}^2+\cdots$ be a function with positive real part in . Then, for any complex number ν,

$$|c_2-\nu c_1^2|\le 2max\{1,|1-2\nu |\}.$$

The following result holds for the coefficient of $f\in S(\alpha ,\beta )$.

Theorem 5 Let $0\le \alpha <1<\beta$ and let the function f given by $f(z)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_n{z}^n$ be in the class $\mathcal{S}(\alpha ,\beta )$. Then, for a complex number μ,

$$\begin{array}{rcl}|a_3-\mu a_2^2|& \le & \frac{\beta -\alpha }{2\pi }\sqrt{2-2cos(\frac{1-\alpha }{\beta -\alpha }\cdot 2\pi )}\\ \cdot max\{1;|\frac{1}{2}+(1-2\mu )\frac{\beta -\alpha }{\pi }i+(\frac{1}{2}-(1-2\mu )\frac{\beta -\alpha }{\pi }i)e^{2\pi i\frac{1-\alpha }{\beta -\alpha }}\left|\right\}.\end{array}$$

(11)

Proof Let us consider a function q given by $q(z)=z{f}^{\prime }(z)/f(z)$. Then, since $f\in \mathcal{S}(\alpha ,\beta )$, we have $q(z)\prec p(z)$, where

$$p(z)=1+\frac{\beta -\alpha }{\pi }ilog\left(\frac{1-e^{2\pi i\frac{1-\alpha }{\beta -\alpha }}z}{1-z}\right).$$

Let

$$h(z)=\frac{1+p^{-1}(q(z))}{1-p^{-1}(q(z))}=1+h_{1}z+h_2{z}^2+\cdots (z\in \mathbb{U}).$$

Then h is analytic and has positive real part in the open unit disk . We also have

$$q(z)=p\left(\frac{h(z)-1}{h(z)+1}\right)(z\in \mathbb{U}).$$

(12)

We find from equation (12) that

$$a_2=\frac{1}{2}{B}_1{h}_1$$

and

$$a_3=\frac{1}{4}{B}_1{h}_2-\frac{1}{8}{B}_1{h}_1^2+\frac{1}{8}{B}_2{h}_1^2+\frac{1}{8}{B}_1^2{h}_1^2,$$

which imply that

$$a_3-\mu a_2^2=\frac{1}{4}{B}_1(h_2-\nu h_1^2),$$

where

$$\nu =\frac{1}{2}(1-\frac{B_2}{B_1}-B_1+2\mu B_1).$$

Applying Lemma 3, we can obtain

$$\begin{array}{rcl}|a_3-\mu a_2^2|& =& \frac{1}{4}|B_1||h_2-\nu h_1^2|\\ \le & \frac{1}{2}\cdot max\{1;|1-2\nu |\}.\end{array}$$

(13)

And substituting

$$B_1=\frac{\beta -\alpha }{\pi }i(1-e^{2\pi i\frac{1-\alpha }{\beta -\alpha }})$$

(14)

and

$$B_2=\frac{\beta -\alpha }{2\pi }i(1-e^{4\pi i\frac{1-\alpha }{\beta -\alpha }})$$

(15)

in (13), we can obtain the result as asserted. □

Using the above Theorem 5, we can get the following result.

Corollary 1 Let the function f, given by $f(z)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_n{z}^n$, be in the class $\mathcal{S}(\alpha ,\beta )$. Also let the function $f^{-1}$, defined by

$$f^{-1}(f(z))=z=f(f^{-1}(z)),$$

(16)

be the inverse of f. If

$$f^{-1}(w)=w+\sum _{n=2}^{\mathrm{\infty }}{b}_n{w}^n(|w|<r_0;r_0>\frac{1}{4}),$$

(17)

then

$$|b_2|\le \frac{2(\beta -\alpha )}{\pi }sin\left(\frac{1-\alpha }{\beta -\alpha }\pi \right)$$

and

$$\begin{array}{rcl}|b_3|& \le & \frac{\beta -\alpha }{2\pi }\sqrt{2-2cos(\frac{1-\alpha }{\beta -\alpha }\cdot 2\pi )}\\ \cdot max\{1;|\frac{1}{2}-3\frac{\beta -\alpha }{\pi }i+(\frac{1}{2}+3\frac{\beta -\alpha }{\pi }i)e^{2\pi i\frac{1-\alpha }{\beta -\alpha }}\left|\right\}.\end{array}$$

Proof Relations (16) and (17) give

$$b_2=-a_2\text{and}{b}_3=2a_2^2-a_3.$$

Thus, we can get the estimate for $|b_2|$ by

$$|b_2|=|a_2|\le \frac{2(\beta -\alpha )}{\pi }sin\left(\frac{1-\alpha }{\beta -\alpha }\pi \right),$$

immediately. An application of Theorem 5 (with $\mu =2$) gives the estimates for $|b_3|$, hence the proof of Corollary 1 is completed. □

Next, we shall estimate on some initial coefficient for the bi-univalent functions $f\in {\mathcal{S}}_{\sigma }(\alpha ,\beta )$.

Theorem 6 Let f be given by $f(z)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_n{z}^n$ be in the class ${\mathcal{S}}_{\sigma }(\alpha ,\beta )$. Then

$$|a_2|\le \frac{|B_1|\sqrt{|B_1|}}{|B_1^2+B_1-B_2|}$$

(18)

and

$$|a_3|\le |B_1|+|B_2-B_1|,$$

(19)

where $B_1$ and $B_2$ are given by (14) and (15).

Proof If $f\in {\mathcal{S}}_{\sigma }(\alpha ,\beta )$, then $f\in \mathcal{S}(\alpha ,\beta )$ and $g\in \mathcal{S}(\alpha ,\beta )$, where $g=f^{-1}$. Hence

$$Q(z):=\frac{z{f}^{\prime }(z)}{f(z)}\prec p(z)\text{and}L(z):=\frac{z{g}^{\prime }(z)}{g(z)}\prec p(z),$$

where $p(z)$ is given by (2). Let

$$h(z)=\frac{1+p^{-1}(Q(z))}{1-p^{-1}(Q(z))}=1+h_{1}z+h_2{z}^2+\cdots$$

and

$$k(z)=\frac{1+p^{-1}(L(z))}{1-p^{-1}(L(z))}=1+k_{1}z+k_2{z}^2+\cdots .$$

Then h and k are analytic and have positive real part in . Also, we have

$$Q(z)=p\left(\frac{h(z)-1}{h(z)+1}\right)\text{and}L(z)=p\left(\frac{k(z)-1}{k(z)+1}\right).$$

By suitably comparing coefficients, we get

$$a_2=\frac{1}{2}{B}_1{h}_1,$$

(20)

$$2a_3-a_2^2=\frac{1}{2}{B}_1{h}_2-\frac{1}{4}{B}_1{h}_1^2+\frac{1}{4}{B}_2{h}_1^2,$$

(21)

$$-a_2=\frac{1}{2}{B}_1{k}_1$$

(22)

and

$$3a_2^2-2a_3=\frac{1}{2}{B}_1{k}_2-\frac{1}{4}{B}_1{k}_1^2+\frac{1}{4}{B}_2{k}_1^2,$$

(23)

where $B_1$ and $B_2$ are given by (14) and (15), respectively. Now, considering (20) and (22), we get

$$h_1=-k_1.$$

(24)

Also, from (21), (22), (23) and (24), we find that

$$a_2^2=\frac{B_1^3(h_2+k_2)}{4(B_1^2+B_1-B_2)}.$$

(25)

Therefore, we have

$$\left|a_2^2\right|\le \frac{{|B_1|}^3}{4|B_1^2+B_1-B_2|}(|h_2|+|k_2|)\le \frac{{|B_1|}^3}{|B_1^2+B_1-B_2|}.$$

This gives the bound on $|a_2|$ as asserted in (18). Now, further computations from (21), (23), (24) and (25) lead to

$$a_3=\frac{1}{8}(B_1(h_2+3k_2)+2h_1^2(B_2-B_1)).$$

This equation, together with the well-known estimates:

$$|h_1|\le 2,|h_2|\le 2\text{and}|k_2|\le 2$$

lead us to inequality (19). Therefore, the proof of Theorem 6 is completed. □