On Coefficient Estimates for a Certain Class of Analytic Functions

Abstract

In this paper, we consider a subclass $\mathcal{SQ}$ of normalized analytic functions f satisfying $\Re \sqrt{f^{\prime }\left(z\right)}>1/2$. For the functions in the class $\mathcal{SQ}$, we determine upper bounds for a number of coefficient estimates, among which are initial coefficients, the second Hankel determinant, and the Zalcman functional. Upper estimates for higher-order Schwarzian derivatives are also obtained.

1. Introduction

Let $\mathcal{A}$ be the family of all analytic and normalized functions

$$f\left(z\right)=z+\sum _{n=2}^{\infty }{a}_n{z}^n$$

(1)

defined on the open unit disk $\mathcal{U}=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$.

The class of Schwarz functions $\omega$, which are analytic in $\mathcal{U}$ and satisfy $\left|\omega \right(z\left)\right|<1$, $\omega \left(0\right)=0$, is denoted by $\mathcal{B}$. If $\omega \in \mathcal{B}$, then its power series expansion is given by

$$\omega \left(z\right)=\sum _{n=1}^{\infty }{c}_n{z}^n.$$

(2)

For two functions f and g analytic in $\mathcal{U}$, we say that f is subordinate to g, written $f\prec g$, if $\omega \in \mathcal{B}$ exists such that

$$f\left(z\right)=g\left(\omega \right(z\left)\right),z\in \mathcal{U}.$$

If, in particular, g is univalent in $\mathcal{U}$, then $f\prec g$ if and only if $f\left(0\right)=g\left(0\right)$ and $f\left(\mathcal{U}\right)\subset g\left(\mathcal{U}\right)$.

Let $f\in \mathcal{A}$ given by (1). The Hankel determinant $H_2\left(1\right)=a_3-a_2^2$ is the well known Fekete-Szegö functional, which is also a particular case of the Zalcman functional $a_{n+m-1}$$-a_n{a}_m$ [1]. The second Hankel determinant is given by $H_2\left(2\right)=a_2{a}_4-a_3^2$. For related results to upper bounds of the Hankel determinant and the Zalcman functional, see for example [2,3,4,5,6,7,8,9].

The Schwarzian derivative for $f\in \mathcal{A}$ is defined by

$$S\left(f\right)\left(z\right)={\left(\frac{f^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)}^{\prime }-\frac{1}{2}{\left(\frac{f^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)}^2,z\in \mathcal{U}.$$

The higher-order Schwarzian derivatives are defined inductively (see [10,11]) as follows:

$${\sigma }_3\left(f\right)=S\left(f\right)$$

(3)

$${\sigma }_{n+1}\left(f\right)={\left({\sigma }_n\left(f\right)\right)}^{\prime }-(n-1){\sigma }_n\left(f\right)\frac{f^{″}}{f^{\prime }},n\ge 4.$$

(4)

Let $S_n={\sigma }_n\left(f\right)\left(0\right)$. If $f\in \mathcal{A}$ is of the form (1), then

$$\begin{array}{ccc}\hfill S_3& =& 6(a_3-a_2^2)\hfill \\ \hfill S_4& =& 24(a_4-3a_2{a}_3+2a_2^3)\hfill \\ \hfill S_5& =& 24(5a_5-20a_2{a}_4-9a_3^2+48a_3{a}_2^2-24a_2^4).\hfill \end{array}$$

(5)

The related results for higher-order Schwarzian derivatives may be found in [12,13].

Denote by $\mathcal{SQ}$ the class of analytic functions f satisfying

$$\Re \sqrt{f^{\prime }\left(z\right)}>\frac{1}{2},z\in \mathcal{U}$$

(6)

or in terms of subordination

$$f^{\prime }\left(z\right)\prec \frac{1}{{(1-z)}^2},z\in \mathcal{U}.$$

(7)

Recently, several authors have investigated various coefficient estimates for functions belonging to different subclasses of univalent functions (see, for example [14,15,16,17,18,19], to mention only a few).

Based on the results obtained in previous research, in this paper, we investigate the initial coefficient bounds, the Zalcman functional, and the second Hankel determinant for functions in the class $\mathcal{SQ}$. Bounds for the higher-order Schwarzian derivatives for the class $\mathcal{SQ}$ are also obtained.

In order to prove our results, the next lemmas for Schwarz functions will be used.

Lemma 1 ([20]).

Let $\omega \left(z\right)=c_{1}z+c_2{z}^2+\dots$ be a Schwarz function. Then, for any real numbers $\alpha ,\beta$ such that

$$(\alpha ,\beta )\in \left\{\left|\alpha \right|\le \frac{1}{2},-1\le \beta \le 1\right\}\cup \left\{\frac{1}{2}\le \left|\alpha \right|\le 2,\frac{4}{27}{\left(\right|\alpha |+1)}^3-\left(\right|\alpha |+1)\le \beta \le 1\right\}$$

the following estimate holds:

$$|c_3+\alpha c_1{c}_2+\beta c_1^3| \le 1.$$

Lemma 2 ([21]).

Let $\omega \left(z\right)=c_{1}z+c_2{z}^2+\dots$ be a function in the class $\mathcal{B}$. Then, the next estimates hold

$$\begin{array}{c}\hfill |c_1| \le 1,|c_2| \le 1 - |c_1{|}^2\\ \hfill {\displaystyle |c_3| \le 1 - |c_1{|}^2-\frac{|c_2{|}^2}{1 + |c_1|}}\\ \hfill |c_4| \le 1 - |c_1{|}^2-{\left|c_2\right|}^2.\end{array}$$

The next result obtained by Efraimidis will be also needed.

Lemma 3 ([22]).

Let $\omega \left(z\right)=c_{1}z+c_2{z}^2+\dots$ be a Schwarz function. Then, for any complex number λ, the following estimates hold:

$$|c_2+\lambda c_1^2| \le max\left\{1,\left|\lambda \right|\right\}$$

(8)

$$|c_4+(1+\lambda )c_1{c}_3+c_2^2+(1+2\lambda )c_1^2{c}_2+\lambda c_1^4| \le max\left\{1,\left|\lambda \right|\right\}.$$

(9)

2. Coefficient Estimates

In this section, we obtain sharp bounds for the first five Taylor coefficients for functions in the class $\mathcal{SQ}$.

Theorem 1.

Let $f\in \mathcal{SQ}$ be of the form (1). Then, the first five initial coefficients of f are bounded by one. The estimates are sharp.

Proof. 

Assume that f is in $\mathcal{SQ}$. Then, from (7), we obtain that there exists a Schwarz function $\omega$ of the form (2) such that

$$f^{\prime }\left(z\right)=\frac{1}{{(1-\omega \left(z\right))}^2},z\in \mathcal{U}.$$

(10)

Making use of the series (1) and (2) into (10) and equating the coefficients, we obtain

$$\begin{array}{ccc}\hfill a_2& =& c_1\hfill \\ \hfill a_3& =& \frac{1}{3}(2c_2+3c_1^2)\hfill \\ \hfill a_4& =& \frac{1}{2}(c_3+3c_1{c}_2+2c_1^3)\hfill \\ \hfill a_5& =& \frac{1}{5}(2c_4+6c_1{c}_3+3c_2^2+12c_1^2{c}_2+5c_1^4).\hfill \end{array}$$

(11)

It is obvious that $|a_2| \le 1$. Since

$$a_3=\frac{2}{3}(c_2+\frac{3}{2}{c}_1).$$

the bound $|a_3| \le 1$ follows easily from (8) with $\lambda =3/2$. For the fourth coefficient, we have

$$|a_4|=\frac{1}{2}|(c_3+2c_1{c}_2+c_1^3)+(c_1{c}_2+c_1^3)|\le \frac{1}{2}(|c_3+2c_1{c}_2+c_1^3| + |c_1{c}_2+c_1^3|).$$

The inequality $|c_3+2c_1{c}_2+c_1^3| \le 1$ follows from Lemma 1 with $\alpha =2$ and $\beta =1$. Applying (8) with $\lambda =1$, we obtain

$$|c_1{c}_2+c_1^3| = |c_1\left|\right|c_2+c_1^2| \le 1.$$

Finally, we have $|a_4| \le 1$. Observe that

$$|a_5| =\frac{2}{5}\left|(c_4+3c_1{c}_3+c_2^2+5c_1^2{c}_2+2c_1^4)+\frac{1}{2}(c_2^2+2c_1^2{c}_2+c_1^4)\right|$$

$$\le \frac{2}{5}|c_4+3c_1{c}_3+c_2^2+5c_1^2{c}_2+2c_1^4| + \frac{1}{5}|c_2^2+2c_1^2{c}_2+c_1^4|.$$

From (9) with $\lambda =2$, we immediately obtain $|c_4+3c_1{c}_3+c_2^2+5c_1^2{c}_2+2c_1^4| \le 2$. For the bound of the second term, the triangle inequality and the inequality of $|c_2|$ in Lemma 2 give

$$|c_2^2+2c_1^2{c}_2+c_1^4| \le |c_2{|}^2+2|c_1{|}^2|c_2| + |c_1{|}^4\le (1 - |c_1{|}^2{)}^2+2|c_1{|}^2(1 - |c_1{|}^2) + |c_1{|}^4=1$$

and therefore $|a_5| \le 1$.

The estimates for all five coefficients are sharp for the function $f\left(z\right)=\frac{z}{1-z}$. □

3. Bounds for Hankel Determinant and Zalcman Functional

In this section, the bounds for Hankel determinants $H_2\left(1\right),H_2\left(2\right)$, and the Zalcman functional $a_4-a_2{a}_3$ and $a_5-a_3^2$ are obtained.

Theorem 2.

Let $f\in \mathcal{SQ}$ be of the form (1). Then,

$$|H_2\left(1\right)|\le \frac{2}{3}\mathit{and}|H_2\left(2\right)|\le \frac{4}{9}.$$

The bounds are sharp.

Proof. 

Suppose that $f\in \mathcal{SQ}$ has the form (1). The first inequality follows easily:

$$|H_2\left(1\right)| = |a_3-a_2^2| =\frac{2}{3}\left|c_2\right|\le \frac{2}{3}.$$

Making use of (11), we have

$$|H_2\left(2\right)| = |a_2{a}_4-a_3^2| =\frac{1}{18}|9c_1{c}_3+3c_1^2{c}_2-8c_2^2|.$$

By triangle inequality, we obtain

$$|H_2\left(2\right)|\le \frac{1}{18}\left(9|c_1||c_3|+3|c_1{|}^2|c_2|+8|c_2{|}^2\right).$$

Applying the inequalities for $|c_2|$ and $|c_3|$ in Lemma 2, we receive

$$|H_2\left(2\right)|\le \frac{1}{18}\left[9|c_1|\left(1 - |c_1{|}^2-\frac{|c_2{|}^2}{1 + |c_1|}\right)+3|c_1{|}^2|c_2| + 8|c_2{|}^2\right]$$

$$=\frac{1}{18}\left[9|c_1|(1 - |c_1{|}^2) + |c_2{|}^2\frac{8 - |c_1|}{1 + |c_1|}+3|c_1{|}^2|c_2|\right]$$

$$\le \frac{1}{18}\left[9|c_1|(1 - |c_1{|}^2) + (1 - |c_1{|}^2{)}^2\frac{8 - |c_1|}{1 + |c_1|}+3|c_1{|}^2(1 - |c_1{|}^2)\right]$$

$$=\frac{2}{9}(-|c_1{|}^4 - |c_1{|}^2 + 2)\le \frac{4}{9}.$$

If $c_2=1$ and $c_k=0,k\ne 2$, then $a_2=0,a_3=2/3$, and $a_4=0$. This shows that the equality in the assertion of our theorem holds for the function given by (10) with $\omega \left(z\right)=z^2$. □

Theorem 3.

If $f\in \mathcal{SQ}$ is of the form (1), then the next inequalities hold

$$|a_4-a_2{a}_3| \le \frac{8\sqrt{3}}{27}\mathit{and}|a_5-a_3^2|\le 0.7789\dots$$

Proof. 

Assume that $f\in \mathcal{SQ}$. From (11), we obtain

$$|a_4-a_2{a}_3| =\frac{1}{6}|3c_3+5c_1{c}_2|.$$

Then, by triangle inequality, we have

$$|a_4-a_2{a}_3| \le \frac{1}{6}\left(3\right|c_3| + 5|c_1\left|\right|c_2\left|\right).$$

In view of Lemma 2, we obtain

$$\begin{gathered} |a_4-a_2{a}_3| \le \frac{1}{6}\left[3\left(1 - |c_1{|}^2-\frac{|c_2{|}^2}{1 + |c_1|}\right)+5|c_1||c_2|\right] \\ =\frac{1}{6}\left(3-3|c_1{|}^2-\frac{3|c_2{|}^2}{1 + |c_1|}+5|c_1||c_2|\right). \end{gathered}$$

(12)

Writing $|c_1| =x$ and $|c_2| =y$ in (12), we obtain

$$|a_4-a_2{a}_3| \le g(x,y)$$

where

$$g(x,y)=\frac{1}{6}\left(3-3x^2-\frac{3y^2}{1+x}+5xy\right).$$

Since $|c_2| \le 1-|c_1{|}^2$, the region of variability of $(x,y)$ coincides with

$$D=\left\{(x,y):0\le x\le 1,0\le y\le 1-x^2\right\}.$$

Therefore, we need to find the maximum value of $g(x,y)$ over the region D. The critical points of $g(x,y)$, given by the system

$$\left\{\begin{array}{c}{\displaystyle -6x+\frac{3y^2}{{(1+x)}^2}+5y=0}\hfill \\ {\displaystyle -\frac{6y}{1+x}+5x=0}\hfill \end{array}\right.$$

are $(0,0)$ and ${\displaystyle (\frac{22}{75},\frac{1067}{3375})}$. Elementary calculations show that $(0,0)$ is a maximum point and $g(0,0)=1/2$. On the boundary of D, we have

$$\begin{array}{ccc}\hfill g(x,0)& =& \frac{1}{6}(3-3x^2)=\frac{1}{2}(1-x^2)\le \frac{1}{2}\hfill \\ \hfill g(0,y)& =& \frac{1}{6}(3-3y^2)=\frac{1}{2}(1-y^2)\le \frac{1}{2}\hfill \\ \hfill g(x,1-x^2)& =& \frac{4}{3}(x-x^3)\le \frac{8\sqrt{3}}{27}=0.5132\dots ,\mathrm{for}x=\frac{1}{\sqrt{3}}.\hfill \end{array}$$

From all these inequalities, we obtain

$$g(x,y)\le \frac{8\sqrt{3}}{27},\mathrm{for}\mathrm{all}(x,y)\in D$$

which is the desired bound for $|a_4-a_2{a}_3|$. The Schwarz function

$$\omega \left(z\right)=z\frac{z+c_1}{1+c_{1}z}=c_{1}z+(1-c_1^2)z^2-c_1(1-c_1^2)z^3+\dots ,$$

where ${\displaystyle c_1=\frac{1}{\sqrt{3}},c_2=\frac{2}{3}}$, and ${\displaystyle c_3=-\frac{2}{3\sqrt{3}}}$, which shows that this inequality is sharp.

Now, we continue with the estimate of $|a_5-a_3^2|$. From (11), we obtain

$$|a_5-a_3^2| =\frac{1}{5}\left|2c_4+6c_1{c}_3+\frac{16}{3}{c}_1^2{c}_2+\frac{7}{9}{c}_2^2\right|.$$

Using the triangle inequality and the inequalities for $|c_4|,|c_3|$ in Lemma 2, we obtain

$$\begin{gathered} |a_5-a_3^2| \le \frac{1}{5}\left[2(1 - |c_1{|}^2 - |c_2{|}^2) + 6|c_1|\left(1 - |c_1{|}^2-\frac{|c_2{|}^2}{1 + |c_1|}\right)+\frac{16}{3}|c_1{|}^2|c_2| + \frac{7}{9}{|c_2|}^2\right] \\ =\frac{1}{5}\left[2+6|c_1| - 2|c_1{|}^2-6|c_1{|}^3-\frac{6|c_1||c_2{|}^2}{1+|c_1|}-\frac{11}{9}|c_2{|}^2+\frac{16}{3}|c_1{|}^2|c_2|\right]. \end{gathered}$$

(13)

Writing $|c_1|=x$ and $|c_2|=y$ in (13), we have $|a_5-a_3^2| \le h(x,y)$, where

$$h(x,y)=\frac{1}{5}\left(2+6x-2x^2-6x^3-\frac{6x{y}^2}{1+x}-\frac{11}{9}{y}^2+\frac{16}{3}{x}^{2}y\right).$$

Taking into account the inequality $|c_2|\le 1-|c_1{|}^2$, the region of variability of $(x,y)$ coincides with

$$D=\left\{(x,y):0\le x\le 1,0\le y\le 1-x^2\right\}.$$

Thus, we we need to find the maximum value of $h(x,y)$ over the region D. The solutions of the system

$$\left\{\begin{array}{c}{\displaystyle 6-4x-18x^2-\frac{6y^2}{{(1+x)}^2}+\frac{32}{3}xy=0}\hfill \\ {\displaystyle -\frac{12xy}{1+x}-\frac{22}{9}y+\frac{16}{3}{x}^2=0}\hfill \end{array}\right.$$

are the critical points of $h(x,y)$. The maximum of $h(x,y)$ is attained in $(0.5285\dots ,0.2259\dots )$, and its value is $h(0.5285\dots ,0.2259\dots )=0.7789\dots$ On the boundary of the region D, we have

$$\begin{array}{ccc}\hfill h(x,0)& =& \frac{1}{5}(2-2x^2-6x^3+6x)\le \frac{32(10+7\sqrt{7})}{1215}=0.7511\dots \hfill \\ \hfill h(0,y)& =& \frac{1}{5}(2-\frac{11}{9}{y}^2)\le \frac{2}{5}=0.4\hfill \\ \hfill h(x,1-x^2)& =& \frac{1}{45}(7+106x^2-113x^4)\le \frac{80}{113}=0.7079\dots \hfill \end{array}$$

It follows that $h(x,y)\le 0.7789\dots$ for $(x,y)\in D$, which is the desired bound for $|a_5-a_3^2|$. □

4. Bounds for Higher-Order Schwarzian Derivatives

In this section, we investigate the upper bounds of $|S_3|,|S_4|$, and $|S_5|$, where $S_3,S_4,S_5$ are given by (5).

Theorem 4.

Let $f\in \mathcal{SQ}$ be given by (1). Then, the following estimates hold:

$$|S_3| \le 4,|S_4| \le 12,|S_5| \le 73.176\dots$$

Proof. 

Let $f\in \mathcal{SQ}$. From (11), we have

$$|S_3| =6|a_3-a_2^2| =4|c_2| \le 4$$

For $c_2=1$ and $c_k=0,c_k\ne 2$, we obtain $a_2=0$ and $a_3=2/3$. This shows that the equality holds for the function given by (10) with $\omega \left(z\right)=z^2$. Further

$$|S_4| =24|a_4-3a_2{a}_3+2a_2^3| =12|c_3-c_1{c}_2|.$$

The inequality $|c_3-c_1{c}_2| \le 1$ follows from Lemma 1 with $\alpha =-1$ and $\beta =0$. Hence, $|S_4| \le 12$. If $c_3=1$ and $c_k=0,k\ne 3$, then $a_2=0,a_3=0$ and $a_4=1/2$. This means that equality holds for the function given by (10) with $\omega \left(z\right)=z^3$.

We continue with the estimate for $|S_5|$. Taking into account (11), we obtain

$$|S_5| =24|2c_4-4c_1{c}_3-c_2^2+2c_1^2{c}_2|.$$

By the triangle inequality, we have

$$|S_5| \le 24(2|c_4|+4|c_1\left|\right|c_3|+|c_2{|}^2+2|c_1{|}^2|c_2\left|\right).$$

Applying the inequalities for $|c_3|$ and $|c_4|$ in Lemma 2, we obtain

$$\begin{gathered} |S_5| \le 24\left[2(1 - |c_1{|}^2 - |c_2{|}^2) + 4|c_1|\left(1 - |c_1{|}^2 - \frac{|c_2{|}^2}{1 + |c_1|}\right) + |c_2{|}^2+2|c_1{|}^2|c_2|\right] \\ =24\left(2-2|c_1{|}^2+4|c_1|-4|c_1{|}^3-|c_2{|}^2 - \frac{4|c_1||c_2{|}^2}{1 + |c_1|}+2|c_1{|}^2|c_2|\right). \end{gathered}$$

(14)

If we replace $|c_1|=x$ and $|c_2|=y$ in (14), then $|S_5| \le k(x,y)$ where

$$k(x,y)=24\left(2-4x^3-2x^2+4x-y^2-\frac{4x{y}^2}{1+x}+2x^{2}y\right).$$

Since $|c_2| \le 1-|c_1{|}^2$, the region of variability of $(x,y)$ coincides with

$$D=\left\{(x,y):0\le x\le 1,0\le y\le 1-x^2\right\}.$$

In order to obtain the upper bound of $|S_5|$, we need to find the maximum value of $k(x,y)$ over the region D. The critical points of $k(x,y)$ are the solution of the system

$$\left\{\begin{array}{c}{\displaystyle 1-x-3x^2-\frac{y^2}{{(1+x)}^2}+xy=0}\hfill \\ {\displaystyle -\frac{4xy}{1+x}-y+x^2=0}\hfill \end{array}\right.$$

The maximum value of $k(x,y)$ is attained in $(0.44402\dots ,0.088414\dots )$. In this case, $k(0.44402\dots ,0.088414\dots )$ = 73.176…Next, we verify the behaviour of the function $k(x,y)$ on the boundary of D:

$$\begin{array}{ccc}\hfill k(x,0)& =& 24(2-2x^2-4x^3+4x)\le \frac{8(35+13\sqrt{13})}{9}=72.7752\dots \hfill \\ \hfill k(0,y)& =& 24(2-y^2)\le 48\hfill \\ \hfill k(x,1-x^2)& =& 24(1+6x^2-7x^4)\le \frac{384}{7}=54.8571\dots \hfill \end{array}$$

In view of the above inequalities, we obtain $k(x,y)\le 73.176\dots$ Finally, the proof of the theorem is completed. □

5. Conclusions

In this paper, we investigate a number of coefficient problems for the class $\mathcal{SQ}$. The upper bounds for the initial coefficients, the second Hankel determinant, the Zalcman functional, and the higher-order Schwarzian derivatives have been derived. In our research, we have used the relationship between the coefficients of functions in the considered class $\mathcal{SQ}$ and the coefficients for the corresponding Schwarz functions. The results obtained in this note could be the subject of further investigation related with the Fekete–Szegö type functional such as $a_3-\mu a_2^2$, $a_2{a}_4-\mu a_3^2$ or $a_4-\mu a_2{a}_3$.