Sharp Estimates and Symmetric Subclasses for Univalent Functions Defined by the nth Derivative of z/f(z)

1. Introduction

Let $\mathcal{A}$ be the family of functions $f\left(z\right)=z+{\sum }_{k\ge 2}{a}_k{z}^k$ analytic in the disk $\mathbb{U}=\{z:|z|<1\}$, and let $\mathcal{S}\subset \mathcal{A}$ consist of the univalent ones; [1] is the standard reference. The Koebe function $k\left(z\right)=z/{(1-z)}^2$ is the extremal function of many problems in $\mathcal{S}$. If $f\in \mathcal{A}$ has no zeros in $0<\left|z\right|<1$, the quotient

$$u_f\left(z\right):={\displaystyle \frac{z}{f\left(z\right)}}$$

extends analytically across the origin with $u_f\left(0\right)=1$.

A number of sufficient conditions for univalence are phrased through derivatives of $u_f$. Aksentev [2] proved in 1958 that

$$\left|{\displaystyle \frac{z^2{f}^{\prime }\left(z\right)}{f{\left(z\right)}^2}}-1\right|<1,z\in \mathbb{U},$$

(1)

guarantees $f\in \mathcal{S}$; the same sufficient condition was proved again, in 1972, by Ozaki and Nunokawa [3]. We call the set of $f\in \mathcal{A}$ that are zero-free in $0<\left|z\right|<1$ and satisfy (1) Aksentev’s class$\mathcal{V}$; it is denoted $\mathcal{U}$ in [4]. Nunokawa, Obradović and Owa [5] observed that $|u_f^{\prime \prime }|\le 1$ implies $f\in \mathcal{V}$, and Yang and Liu [6] found that the weaker hypothesis $|u_f^{\prime \prime }|\le 2$ still yields univalence. After a change of normalization, the Aksentev-type condition defines the class $\mathcal{U}\left(\lambda \right)$ of Fournier and Ponnusamy [7], for which coefficient characterizations and sections [8], logarithmic coefficients [9], growth and distortion [10,11] and further geometric properties [12] are available.

Al-Refai and Darus [13] replaced the second derivative by a derivative of arbitrary order. For an integer $n\ge 3$, a function $f\in \mathcal{A}$ with $f\left(z\right)\ne 0$ in $0<\left|z\right|<1$ is univalent whenever

$${\displaystyle \sum _{k=2}^{n-1}}{\displaystyle \frac{k-1}{k!}}\left|{\mu }_k\right|+{\displaystyle \frac{n-1}{n!}}\left|u_f^{\left(n\right)}\left(z\right)\right|\le 1(z\in \mathbb{U}),$$

(2)

where ${\mu }_k:=u_f^{\left(k\right)}\left(0\right)={\displaystyle \frac{d^k}{d{z}^k}}\left({\displaystyle \frac{z}{f\left(z\right)}}\right){|}_{z=0}$ ($k\ge 1$; the criterion uses $2\le k\le n-1$), and the constant 1 is best possible: it is attained by the Koebe function, by the functions $z/(1+az+z^2)$ with $\left|a\right|\le 2$, and by the functions $z/{\left(1\pm {\displaystyle \frac{z}{n-2}}\right)}^{n-1}$ (for $n=3$ these are the Koebe function and its rotation by $\pi$; in general $a_2=\mp {\displaystyle \frac{n-1}{n-2}}\ne 0$, so they are not in the class ${\mathcal{C}}_n$ introduced below). Putting $n=2$ in (2) gives back the Yang–Liu bound. Obradović and Ponnusamy [4] proved that (2) implies $f\in \mathcal{V}$ by a different argument. Writing

$${\sigma }_f:={\displaystyle \sum _{k=2}^{n-1}}{\displaystyle \frac{k-1}{k!}}\left|{\mu }_k\right|,{\mathrm{\Lambda }}_f:=\underset{z\in \mathbb{U}}{\sup }\left|u_f^{\left(n\right)}\left(z\right)\right|,$$

condition (2) is the same as ${\sigma }_f+{\displaystyle \frac{n-1}{n!}}{\mathrm{\Lambda }}_f\le 1$, and this is the form we shall use. Further criteria and geometric properties for bounded univalent functions are given in [14]. A meromorphic analogue of (2), in terms of the nth derivative of $1/\left(zf\right)$, together with growth and distortion estimates, is treated in [15].

The paper [13] establishes the criterion (2) and its sharpness. To the best of our knowledge, quantitative information about the class (2) itself, such as growth, distortion or coefficient estimates and statements on products, has not been published; the aim of this paper is to supply it.

Since (2) couples the numbers ${\mu }_k$ with the size of $u_f^{\left(n\right)}$, the sharpest statements are obtained when every ${\mu }_k$ vanishes and the whole allowance in (2) is given to ${\mathrm{\Lambda }}_f$. This is the class ${\mathcal{C}}_n$ of Definition 1 below: the zero-free normalized functions with $a_2=\dots =a_n=0$ for which the nth derivative of $z/f$ is bounded by $n!/(n-1)$ in $\mathbb{U}$. The vanishing of $a_2,\dots ,a_n$ is equivalent to that of ${\mu }_1,\dots ,{\mu }_{n-1}$ (Lemma 4), and members of ${\mathcal{C}}_n$ are univalent because they satisfy (2) (Remark 1). The restriction ${\mu }_1=\dots ={\mu }_{n-1}=0$ is a choice, not a consequence of the criterion; Section 3.4 treats the larger class in which only $a_2=0$ is imposed and shows exactly how much of the strength of the estimates for ${\mathcal{C}}_n$ is due to the extra conditions. Moreover, every $f\in {\mathcal{C}}_n$ begins like an n-fold symmetric function, $f=z+a_{n+1}{z}^{n+1}+\dots$, which is why symmetric subclasses and symmetric extremal functions occur naturally (Section 6).

The paper is organized as follows. Section 2 gives an exact integral representation of $z/f$ with estimates for its remainder, the formulas relating the Taylor coefficients of f and of ${(z/f)}^{\left(n\right)}$, and the parametrization of ${\mathcal{C}}_n$ by a ball of $H^{\infty }$. Section 3 contains the estimates: a pointwise majorant for Aksentev’s quantity on the whole class (2); growth, distortion and covering radius; coefficient bounds and the exact region of $(a_{n+1},a_{n+2})$; and sharp bounds on the larger class defined by $a_2=0$, together with the case $n=2$. Section 4 gives the product theorem with an exact radius. Section 5 compares the estimates with classical ones for $\mathcal{S}$ and for n-fold symmetric univalent functions. Section 6 describes the symmetric members of ${\mathcal{C}}_n$, and Section 7 summarizes the results.

2. Preliminaries

Everything in the later sections is derived from a handful of exact formulas for the function $u_f=z/f$. This section establishes them in the form needed, with complete proofs. Throughout, $n\ge 3$ is fixed, $f\in \mathcal{A}$ has no zero in $0<\left|z\right|<1$, we write $u=u_f$ and ${\mu }_k=u^{\left(k\right)}\left(0\right)$ for $k\ge 1$, and we assume ${\mathrm{\Lambda }}_f<\infty$ (which is automatic under (2)). For a function $\eta$ analytic and bounded in $\mathbb{U}$ we write ${\parallel \eta \parallel }_{\infty }={\sup }_{\mathbb{U}}\left|\eta \right|$, and $H^{\infty }=H^{\infty }\left(\mathbb{U}\right)$ denotes the Banach space of all such functions with this norm. The nth derivative of u is denoted

$$\eta \left(z\right):=u^{\left(n\right)}\left(z\right)=\sum _{m\ge 0}{\eta }_m{z}^m,\left|\eta \left(z\right)\right|\le {\mathrm{\Lambda }}_f(z\in \mathbb{U}).$$

Lemma 1

(Splitting of u). For every $z\in \mathbb{U}$,

$$u\left(z\right)=1+{\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \frac{{\mu }_k}{k!}}{z}^k+\mathsf{\Omega }\left(z\right),\mathsf{\Omega }\left(z\right):={\displaystyle \frac{z^n}{(n-1)!}}{\int }_0^1{(1-s)}^{n-1}\eta \left(sz\right)ds.$$

(3)

Moreover Ω is analytic in $\mathbb{U}$ and $\mathsf{\Omega }\left(z\right)=O\left(z^n\right)$ as $z\to 0$.

Proof. 

Step 0: u is analytic in $\mathbb{U}$. Since $f\in \mathcal{A}$, we have $f\left(z\right)/z=1+a_{2}z+a_3{z}^2+\dots$, which is analytic in $\mathbb{U}$ and equals $1\ne 0$ at $z=0$. By hypothesis f has no zero in $0<\left|z\right|<1$, so $f\left(z\right)/z$ has no zero in $\mathbb{U}$ at all, and therefore $u\left(z\right)=1/\left(f\left(z\right)/z\right)$ is analytic in $\mathbb{U}$, with $u\left(0\right)=1$. In particular all derivatives $u^{\left(k\right)}$ exist in $\mathbb{U}$, so ${\mu }_k=u^{\left(k\right)}\left(0\right)$ and $\eta =u^{\left(n\right)}$ are well defined, and $\eta$ is analytic in $\mathbb{U}$.

Step 1: Taylor’s formula with integral remainder. For $g:[0,1]\to \mathbb{C}$ with n continuous derivatives,

$$g\left(1\right)={\displaystyle \sum _{k=0}^{n-1}}{\displaystyle \frac{g^{\left(k\right)}\left(0\right)}{k!}}+{\displaystyle \frac{1}{(n-1)!}}{\int }_0^1{(1-s)}^{n-1}{g}^{\left(n\right)}\left(s\right)ds;$$

(4)

this is the standard Taylor formula, obtained by induction on n and integration by parts.

Step 2: the choice $g\left(s\right)=u\left(sz\right)$. Fix $z\in \mathbb{U}$. For $s\in [0,1]$ we have $\left|sz\right|<1$, so $g\left(s\right):=u\left(sz\right)$ is defined, and it is infinitely differentiable in the real variable s because u is analytic. Since u is complex differentiable, ${\displaystyle \frac{d}{ds}}u\left(sz\right)=z{u}^{\prime }\left(sz\right)$, and by induction

$$g^{\left(k\right)}\left(s\right)=z^k{u}^{\left(k\right)}\left(sz\right)(k\ge 0).$$

Hence $g\left(1\right)=u\left(z\right)$, $g\left(0\right)=u\left(0\right)=1$, $g^{\left(k\right)}\left(0\right)=z^k{\mu }_k$ for $k\ge 1$, and $g^{\left(n\right)}\left(s\right)=z^n\eta \left(sz\right)$. Substituting these into (4) gives

$$u\left(z\right)=1+{\displaystyle \sum _{k=1}^{n-1}}{\displaystyle \frac{{\mu }_k}{k!}}{z}^k+{\displaystyle \frac{z^n}{(n-1)!}}{\int }_0^1{(1-s)}^{n-1}\eta \left(sz\right)ds,$$

which is (3).

Step 3: Ω is analytic and $O\left(z^n\right)$. By (3), $\mathsf{\Omega }\left(z\right)=u\left(z\right)-1-{\sum }_{k=1}^{n-1}{\displaystyle \frac{{\mu }_k}{k!}}{z}^k$ is a difference of analytic functions, hence analytic in $\mathbb{U}$. On the compact disk $\left|w\right|\le {\displaystyle \frac{1}{2}}$ the continuous function $\eta$ is bounded, say by C; for $\left|z\right|\le {\displaystyle \frac{1}{2}}$ the points $sz$ lie in that disk, so $\left|\mathsf{\Omega }\left(z\right)\right|\le {\displaystyle \frac{{\left|z\right|}^n}{(n-1)!}}C{\int }_0^1{(1-s)}^{n-1}ds={\displaystyle \frac{C}{n!}}{\left|z\right|}^n$. Thus $\mathsf{\Omega }\left(z\right)=O\left(z^n\right)$.    □

Lemma 2

(Size of $\mathsf{\Omega }$ and of ${(\mathsf{\Omega }/z)}^{\prime }$). For all $z\in \mathbb{U}$,

$$\begin{array}{cc}\hfill \left|\mathsf{\Omega }\right(z\left)\right|& \le {\displaystyle \frac{{\mathrm{\Lambda }}_f}{n!}}{\left|z\right|}^n,\hfill \\ \hfill \left|{\left({\displaystyle \frac{\mathsf{\Omega }\left(z\right)}{z}}\right)}^{\prime }\right|& \le {\displaystyle \frac{n-1}{n!}}{\mathrm{\Lambda }}_f{\left|z\right|}^{n-2}.\hfill \end{array}$$

(5)

Proof. 

Fix $z\in \mathbb{U}$. For every $s\in [0,1]$ the point $sz$ lies in $\mathbb{U}$, so $\left|\eta \left(sz\right)\right|\le {\mathrm{\Lambda }}_f$.

Step 1: the first estimate. From (3),

$$\begin{array}{cc}\hfill \left|\mathsf{\Omega }\right(z\left)\right|& \le {\displaystyle \frac{{\left|z\right|}^n}{(n-1)!}}{\int }_0^1{(1-s)}^{n-1}\left|\eta \left(sz\right)\right|ds\le {\displaystyle \frac{{\left|z\right|}^n{\mathrm{\Lambda }}_f}{(n-1)!}}{\int }_0^1{(1-s)}^{n-1}ds\hfill \\ \hfill & ={\displaystyle \frac{{\left|z\right|}^n{\mathrm{\Lambda }}_f}{(n-1)!}}·{\displaystyle \frac{1}{n}}={\displaystyle \frac{{\mathrm{\Lambda }}_f}{n!}}{\left|z\right|}^n.\hfill \end{array}$$

Step 2: a formula for ${(\mathsf{\Omega }/z)}^{\prime }$. Because $\mathsf{\Omega }$ is analytic with $\mathsf{\Omega }=O\left(z^n\right)$ and $n\ge 2$ (Lemma 1), the quotient $\mathsf{\Omega }\left(z\right)/z$ is analytic in $\mathbb{U}$. By (3),

$${\displaystyle \frac{\mathsf{\Omega }\left(z\right)}{z}}={\displaystyle \frac{z^{n-1}}{(n-1)!}}J\left(z\right),J\left(z\right):={\int }_0^1{(1-s)}^{n-1}\eta \left(sz\right)ds.$$

Differentiation under the integral sign is permitted (the integrand and its z-derivative are continuous on $[0,1]\times \mathbb{U}$), so J is analytic in $\mathbb{U}$ and $J^{\prime }\left(z\right)={\int }_0^1{(1-s)}^{n-1}s{\eta }^{\prime }\left(sz\right)ds$. The product rule gives

$${\left({\displaystyle \frac{\mathsf{\Omega }\left(z\right)}{z}}\right)}^{\prime }={\displaystyle \frac{(n-1)z^{n-2}}{(n-1)!}}J\left(z\right)+{\displaystyle \frac{z^{n-1}}{(n-1)!}}{J}^{\prime }\left(z\right).$$

(6)

To rewrite the second term, put $h\left(s\right):=\eta \left(sz\right)$ and $p\left(s\right):={(1-s)}^{n-1}s$; then $h^{\prime }\left(s\right)=z{\eta }^{\prime }\left(sz\right)$, so

$$z{J}^{\prime }\left(z\right)={\int }_0^{1}p\left(s\right)h^{\prime }\left(s\right)ds.$$

Integrating by parts,

$${\int }_0^{1}p\left(s\right)h^{\prime }\left(s\right)ds={\left[p\left(s\right)h\left(s\right)\right]}_{s=0}^{s=1}-{\int }_0^1{p}^{\prime }\left(s\right)h\left(s\right)ds=-{\int }_0^1{p}^{\prime }\left(s\right)\eta \left(sz\right)ds,$$

because $p\left(0\right)=0$ (factor s) and $p\left(1\right)=0$ (factor ${(1-s)}^{n-1}$ with $n-1\ge 1$). Here $p^{\prime }\left(s\right)={(1-s)}^{n-1}-(n-1)s{(1-s)}^{n-2}$. Therefore

$$z^{n-1}{J}^{\prime }\left(z\right)=z^{n-2}·z{J}^{\prime }\left(z\right)=-z^{n-2}{\int }_0^1\left[{(1-s)}^{n-1}-(n-1)s{(1-s)}^{n-2}\right]\eta \left(sz\right)ds.$$

Inserting this in (6) and collecting the terms with $\eta \left(sz\right)$,

$${\left({\displaystyle \frac{\mathsf{\Omega }\left(z\right)}{z}}\right)}^{\prime }={\displaystyle \frac{z^{n-2}}{(n-1)!}}{\int }_0^1\left[(n-1){(1-s)}^{n-1}-{(1-s)}^{n-1}+(n-1)s{(1-s)}^{n-2}\right]\eta \left(sz\right)ds.$$

The bracket equals $(n-2){(1-s)}^{n-1}+(n-1)s{(1-s)}^{n-2}={(1-s)}^{n-2}\left[(n-2)(1-s)+(n-1)s\right]={(1-s)}^{n-2}(n-2+s)$, so

$${\left({\displaystyle \frac{\mathsf{\Omega }\left(z\right)}{z}}\right)}^{\prime }={\displaystyle \frac{z^{n-2}}{(n-1)!}}{\int }_0^1{(1-s)}^{n-2}(n-2+s)\eta \left(sz\right)ds.$$

(7)

Step 3: the second estimate. The weight ${(1-s)}^{n-2}(n-2+s)$ is non-negative on $[0,1]$, and $\left|\eta \left(sz\right)\right|\le {\mathrm{\Lambda }}_f$, so (7) yields

$$\left|{\left({\displaystyle \frac{\mathsf{\Omega }\left(z\right)}{z}}\right)}^{\prime }\right|\le {\displaystyle \frac{{\left|z\right|}^{n-2}{\mathrm{\Lambda }}_f}{(n-1)!}}{\int }_0^1{(1-s)}^{n-2}(n-2+s)ds.$$

With $t=1-s$ we have $n-2+s=n-1-t$, hence

$${\int }_0^1{(1-s)}^{n-2}(n-2+s)ds={\int }_0^1{t}^{n-2}(n-1-t)dt=(n-1)·{\displaystyle \frac{1}{n-1}}-{\displaystyle \frac{1}{n}}={\displaystyle \frac{n-1}{n}}.$$

Therefore $\left|{(\mathsf{\Omega }/z)}^{\prime }\right|\le {\displaystyle \frac{{\left|z\right|}^{n-2}{\mathrm{\Lambda }}_f}{(n-1)!}}·{\displaystyle \frac{n-1}{n}}={\displaystyle \frac{n-1}{n!}}{\mathrm{\Lambda }}_f{\left|z\right|}^{n-2}$, since $(n-1)!n=n!$.    □

The first estimate in (5) is used in the growth, distortion, covering and product theorems (Theorems 2, 3, 7) and in the proof of Proposition 1; the second is used only in Theorems 1, 3 and 7, where the quantity $z{\mathsf{\Omega }}^{\prime }-\mathsf{\Omega }=z^2{(\mathsf{\Omega }/z)}^{\prime }$ appears.

Lemma 3

(Coefficients of $\mathsf{\Omega }$). Writing $\mathsf{\Omega }\left(z\right)={\sum }_{j\ge 0}{\delta }_j{z}^{n+j}$, one has

$${\delta }_j={\eta }_j{\displaystyle \frac{j!}{(n+j)!}}(j\ge 0),$$

(8)

and $|{\eta }_j|\le {\mathrm{\Lambda }}_f$ for every $j\ge 0$. The identity (8) uses only that η is analytic in $\mathbb{U}$.

Proof. 

The formula for ${\delta }_j$. Fix $\rho \in (0,1)$. The Taylor series ${\sum }_m{\eta }_m{w}^m$ of the analytic function $\eta$ has radius of convergence at least 1, so it converges uniformly on the closed disk $\left|w\right|\le \rho$. For $\left|z\right|\le \rho$ and $s\in [0,1]$ we have $\left|sz\right|\le \rho$; hence ${\sum }_m{\eta }_m{s}^m{z}^m=\eta \left(sz\right)$ with uniform convergence in $(s,z)\in [0,1]\times \left\{\right|z|\le \rho \}$, and the series may be integrated termwise in (3):

$$\mathsf{\Omega }\left(z\right)={\displaystyle \frac{z^n}{(n-1)!}}\sum _{m\ge 0}{\eta }_m{z}^m{\int }_0^1{(1-s)}^{n-1}{s}^{m}ds.$$

The integral is a Beta integral, ${\int }_0^1{(1-s)}^{n-1}{s}^{m}ds=B(m+1,n)={\displaystyle \frac{\mathrm{\Gamma }(m+1)\mathrm{\Gamma }\left(n\right)}{\mathrm{\Gamma }(m+n+1)}}={\displaystyle \frac{m!(n-1)!}{(n+m)!}}$. Therefore

$$\mathsf{\Omega }\left(z\right)=\sum _{m\ge 0}{\eta }_m{\displaystyle \frac{m!}{(n+m)!}}{z}^{n+m}\left(\right|z|\le \rho ).$$

As $\rho <1$ was arbitrary this holds in $\mathbb{U}$, and by uniqueness of Taylor coefficients the coefficient of $z^{n+j}$ is ${\delta }_j={\eta }_{j}j!/(n+j)!$.

The bound $|{\eta }_j|\le {\mathrm{\Lambda }}_f$ is Cauchy’s estimate for the Taylor coefficients of a function bounded by ${\mathrm{\Lambda }}_f$ in $\mathbb{U}$ (apply the Cauchy integral formula on $\left|w\right|=\rho$ and let $\rho \to 1^{-}$).    □

For $j\ge 1$ the bound $|{\eta }_j|\le {\mathrm{\Lambda }}_f$ is not sharp in general (Wiener’s inequality [17,18] gives $|{\eta }_j|\le {\mathrm{\Lambda }}_f-{\left|{\eta }_0\right|}^2/{\mathrm{\Lambda }}_f$ for $j\ge 1$). The improvement is needed only for $j=n$, in the proof of Theorem 5(b), where it is proved.

We now single out the family studied in the rest of the paper.

Definition 1

(The class ${\mathcal{C}}_n$). For an integer $n\ge 3$,

$$\begin{array}{cc}\hfill {\mathcal{C}}_n:=\{f\in \mathcal{A}:& f\left(z\right)\ne 0(0<|z|<1),a_2=a_3=\dots =a_n=0,\hfill \\ & \underset{z\in \mathbb{U}}{\sup }\left|{\displaystyle \frac{d^n}{d{z}^n}}\left({\displaystyle \frac{z}{f\left(z\right)}}\right)\right|\le {\displaystyle \frac{n!}{n-1}}\}.\hfill \end{array}$$

Equivalently, ${\mathcal{C}}_n$ consists of the $f\in \mathcal{A}$ without zeros in $0<\left|z\right|<1$ for which $a_2=\dots =a_n=0$ and ${\mathrm{\Lambda }}_f\le n!/(n-1)$. We write $M_n:=n!/(n-1)$.

Lemma 4

(Identification of ${\mathcal{C}}_n$ and the tail identity). Let $f\in \mathcal{A}$ have no zero in $0<\left|z\right|<1$.

(a)${\mu }_1=\dots ={\mu }_{n-1}=0$ if and only if $a_2=\dots =a_n=0$.

(b)If these conditions hold, then

$$a_{n+1+j}=-{\eta }_j{\displaystyle \frac{j!}{(n+j)!}}=-{\delta }_j(0\le j\le n-1).$$

(9)

(c)Identity (9) fails for $j=n$: under the same conditions,

$$a_{2n+1}={\delta }_0^2-{\delta }_n={\displaystyle \frac{{\eta }_0^2}{{(n!)}^2}}-{\eta }_n{\displaystyle \frac{n!}{\left(2n\right)!}}.$$

(10)

Proof.(a), “if”. Suppose $a_2=\dots =a_n=0$. Then $f\left(z\right)=z\left(1+\vartheta \left(z\right)\right)$ with $\vartheta \left(z\right)={\sum }_{k\ge n}{a}_{k+1}{z}^k$ analytic in $\mathbb{U}$ and $\vartheta \left(z\right)=O\left(z^n\right)$. Since $1+\vartheta \left(0\right)=1\ne 0$, the function $1/(1+\vartheta )$ is analytic near 0, and for $\left|\vartheta \right(z\left)\right|<1$ it equals the geometric series $1-\vartheta +{\vartheta }^2-\dots$, each term beyond the first being $O\left(z^n\right)$. Hence $u\left(z\right)=1/(1+\vartheta \left(z\right))=1+O\left(z^n\right)$ near 0, i.e. the Taylor coefficients ${\mu }_k/k!$ of u vanish for $1\le k\le n-1$.

(a), “only if”, and (b). Suppose ${\mu }_1=\dots ={\mu }_{n-1}=0$. Then (3) reads $u=1+\mathsf{\Omega }$, and since $u=z/f$,

$$f\left(z\right)\left(1+\mathsf{\Omega }\left(z\right)\right)=z(z\in \mathbb{U}).$$

(11)

Write $f\left(z\right)={\sum }_{p\ge 1}{a}_p{z}^p$ with $a_1=1$, and $\mathsf{\Omega }\left(z\right)={\sum }_{j\ge 0}{\delta }_j{z}^{n+j}$ (Lemma 3). The coefficient of $z^m$ in the product $f\left(z\right)\mathsf{\Omega }\left(z\right)$ is the sum of ${\delta }_j{a}_p$ over all pairs $(j,p)$ with $j\ge 0$, $p\ge 1$ and $(n+j)+p=m$, that is, over $j\ge 0$ with $n+j\le m-1$, taking $p=m-n-j$. Comparing the coefficients of $z^m$ on both sides of (11), for every $m\ge 2$ (where the right side contributes 0),

$$a_m+\sum _{\begin{array}{c}j\ge 0\\ n+j\le m-1\end{array}}{\delta }_j{a}_{m-n-j}=0.$$

(12)

Range $2\le m\le n$. The sum in (12) is empty, because it would require $n+j\le m-1\le n-1$ with $j\ge 0$. Hence $a_m=0$ for $2\le m\le n$. This proves “only if” in (a).

Range $n+1\le m\le 2n$. In (12) the index $p=m-n-j$ satisfies $1\le p\le m-n\le n$. By the previous paragraph $a_p=0$ for $2\le p\le n$, so only the term with $p=1$ survives. It occurs for $j=m-n-1$, which satisfies $0\le j\le n-1$ because $n+1\le m\le 2n$, and it equals ${\delta }_{m-n-1}{a}_1={\delta }_{m-n-1}$. Hence $a_m=-{\delta }_{m-n-1}$. Writing $m=n+1+j$ with $0\le j\le n-1$ and using (8) gives (9).

(c). Take $m=2n+1$ in (12). Now j ranges over $0\le j\le n$ (from $n+j\le 2n$), and $p=m-n-j=n+1-j$ runs through $n+1,n,\dots ,1$ as $j=0,1,\dots ,n$. For $1\le j\le n-1$ we have $2\le p\le n$, so $a_p=0$. The remaining two terms are: $j=0$, $p=n+1$, equal to ${\delta }_0{a}_{n+1}=-{\delta }_0^2$ by (9); and $j=n$, $p=1$, equal to ${\delta }_n{a}_1={\delta }_n$. So (12) reads $a_{2n+1}-{\delta }_0^2+{\delta }_n=0$, i.e. $a_{2n+1}={\delta }_0^2-{\delta }_n$. Finally ${\delta }_0={\eta }_0/n!$ and ${\delta }_n={\eta }_{n}n!/\left(2n\right)!$ by (8).    □

Remark 1.

By Lemma 4(a), every $f\in {\mathcal{C}}_n$ has ${\mu }_1=\dots ={\mu }_{n-1}=0$, hence ${\sigma }_f=0$, and (2) reduces to ${\mathrm{\Lambda }}_f\le n!/(n-1)$, which is precisely the last requirement in Definition 1. So ${\mathcal{C}}_n$ lies inside the class defined by (2) and its members are univalent [13]. It is the part of that class where the whole allowance in (2) is spent on ${\mathrm{\Lambda }}_f$ and none on the ${\mu }_k$; this is what makes (9) a one-to-one dictionary between the first n non-trivial coefficients of f and the first n Taylor coefficients of η, with no cross terms. Note that (2) says nothing about ${\mu }_1$, so ${\mu }_1=0$ is a genuine extra requirement in ${\mathcal{C}}_n$. Univalence of the members of ${\mathcal{C}}_n$ is used in this paper only in the sharpness part of Corollary 1; the univalence of $F_s$ in Theorem 7 comes from Aksentev’s theorem, and none of the estimates of Section 3 needs univalence.

Lemma 5

(Monomial test functions). Let $p\ge 2$ be an integer, $0<c\le 1$, and $f\left(z\right)=z/(1+c{z}^p)$. Then $f\in \mathcal{A}$, f has no zero in $0<\left|z\right|<1$, and $u_f\left(z\right)=1+c{z}^p$. Moreover:

(a)if $p<n$, then ${\mathrm{\Lambda }}_f=0$, ${\mu }_p=cp!$, and ${\mu }_k=0$ for all other $k\in \{1,\dots ,n-1\}$;

(b)if $p\ge n$, then ${\mu }_k=0$ for $1\le k\le n-1$, $a_2=\dots =a_p=0$, $a_{p+1}=-c$, and ${\mathrm{\Lambda }}_f=cp!/(p-n)!$.

Proof. 

Since $|c{z}^p|<1$ in $\mathbb{U}$, the denominator $1+c{z}^p$ does not vanish there; the expansion $f\left(z\right)=z(1-c{z}^p+c^2{z}^{2p}-\dots )$ shows $f\in \mathcal{A}$, $a_k=0$ for $2\le k\le p$ and $a_{p+1}=-c$. Obviously $u_f=1+c{z}^p$, so $u_f^{\left(k\right)}\left(0\right)=cp!$ for $k=p$ and 0 for the other $k\ge 1$. Finally $u_f^{\left(n\right)}\left(z\right)\equiv 0$ if $p<n$, while for $p\ge n$ it equals $c{\displaystyle \frac{p!}{(p-n)!}}{z}^{p-n}$, whose supremum over $\mathbb{U}$ is $cp!/(p-n)!$.    □

The next proposition says that ${\mathcal{C}}_n$ is as large as a ball of $H^{\infty }$, and it turns every sharpness question about ${\mathcal{C}}_n$ into a question about bounded analytic functions.

Proposition 1

(Parametrization of ${\mathcal{C}}_n$). Let $B_n$ be the closed ball of radius $M_n=n!/(n-1)$ in $H^{\infty }\left(\mathbb{U}\right)$. The map

$$\mathsf{\Phi }:{\mathcal{C}}_n\to B_n,\mathsf{\Phi }\left(f\right)={\left({\displaystyle \frac{z}{f}}\right)}^{\left(n\right)},$$

is a bijection. Its inverse is $\eta \mapsto z/(1+{\mathsf{\Omega }}_{\eta })$, where ${\mathsf{\Omega }}_{\eta }\left(z\right)={\displaystyle \frac{z^n}{(n-1)!}}{\int }_0^1{(1-s)}^{n-1}\eta \left(sz\right)ds$.

Proof. 

If $f\in {\mathcal{C}}_n$ and $\eta =\mathsf{\Phi }\left(f\right)$, then $\eta \in B_n$ by definition, and Lemma 1 with ${\mu }_1=\dots ={\mu }_{n-1}=0$ gives $z/f=1+{\mathsf{\Omega }}_{\eta }$. Hence $\mathsf{\Phi }$ is injective and its inverse has the stated form.

Conversely let $\eta \in B_n$ and define ${\mathsf{\Omega }}_{\eta }$ as above; it is analytic in $\mathbb{U}$. The first estimate in (5) (whose proof only uses $\left|\eta \right|\le M_n$) gives $|{\mathsf{\Omega }}_{\eta }\left(z\right)|\le {\displaystyle \frac{M_n}{n!}}{\left|z\right|}^n={\displaystyle \frac{{\left|z\right|}^n}{n-1}}\le {\displaystyle \frac{1}{2}}$, because $n\ge 3$. Hence $u:=1+{\mathsf{\Omega }}_{\eta }$ has no zeros, and $f:=z/u\in \mathcal{A}$ has no zeros in $0<\left|z\right|<1$. Since ${\mathsf{\Omega }}_{\eta }=O\left(z^n\right)$, we get $u=1+O\left(z^n\right)$ and so $a_2=\dots =a_n=0$ by Lemma 4(a). By termwise integration (Lemma 3), ${\mathsf{\Omega }}_{\eta }\left(z\right)={\sum }_j{\eta }_j{\displaystyle \frac{j!}{(n+j)!}}{z}^{n+j}$, and differentiating n times gives $u^{\left(n\right)}=\eta$. Thus $\sup |u^{\left(n\right)}|\le M_n$, $f\in {\mathcal{C}}_n$ and $\mathsf{\Phi }\left(f\right)=\eta$.    □

3. Estimates for the Class

Several results of this section are short consequences of the representation of Section 2: Theorems 2 and 4 and Corollary 2 follow from the first estimate in (5) and Cauchy’s estimate for the coefficients of $\eta$, and are recorded because their extremal functions are explicit. The substantial results are the distortion theorem, which needs the derivative estimate of Lemma 2, Theorem 5, which needs the Schwarz–Pick and Wiener inequalities, and the product theorem (Theorem 7).

3.1. A Pointwise Majorant for Aksentev’s Quantity

Every function satisfying (2) also lies in Aksentev’s class $\mathcal{V}$; this inclusion is due to Obradović and Ponnusamy [4]. For $T_f:=z^2{f}^{\prime }/f^2-1$ one has the general bound $|T_f{|\le |z|}^2$ on all of $\mathcal{V}$ (Remark 2). What the criterion adds is an intermediate majorant in terms of the two quantities ${\sigma }_f$ and ${\mathrm{\Lambda }}_f$ that enter (2); it is attained, and it is much smaller than ${\left|z\right|}^2$ on ${\mathcal{C}}_n$.

Theorem 1.

Let $n\ge 3$ and let $f\in \mathcal{A}$, with $f\left(z\right)\ne 0$ for $0<\left|z\right|<1$, satisfy (2). Then for all $z\in \mathbb{U}$

$$\left|{\displaystyle \frac{z^2{f}^{\prime }\left(z\right)}{f{\left(z\right)}^2}}-1\right|\le {\sigma }_f{\left|z\right|}^2+{\displaystyle \frac{n-1}{n!}}{\mathrm{\Lambda }}_f{\left|z\right|}^n\le {\left|z\right|}^2.$$

(13)

In particular $f\in \mathcal{V}$. Both inequalities are sharp: for $E_0\left(z\right)=z/(1+z^n/(n-1))$ (here ${\sigma }_f=0$) equality holds in the first one at every $z\in \mathbb{U}$, and for $Q_2\left(z\right)=z/(1+z^2)$ equality holds in the second one at every $z\in \mathbb{U}$.

Proof. 

Differentiate $u=z/f$: $u^{\prime }=1/f-z{f}^{\prime }/f^2$, so $z{u}^{\prime }=u-z^2{f}^{\prime }/f^2$, that is

$${\displaystyle \frac{z^2{f}^{\prime }\left(z\right)}{f{\left(z\right)}^2}}-1=u\left(z\right)-1-z{u}^{\prime }\left(z\right)=:T\left(z\right).$$

With $u\left(z\right)=1+{\sum }_{k\ge 1}{\displaystyle \frac{{\mu }_k}{k!}}{z}^k$ one has $T\left(z\right)=-{\sum }_{k\ge 2}(k-1){\displaystyle \frac{{\mu }_k}{k!}}{z}^k$. Split $T=T_1+T_2$ at $k=n$. Since every term of $T_1$ has $k\ge 2$ and $\left|z\right|<1$,

$$|T_1\left(z\right)|=|{\displaystyle \sum _{k=2}^{n-1}}(k-1){\displaystyle \frac{{\mu }_k}{k!}}{z}^k|\le {\displaystyle \sum _{k=2}^{n-1}}{\displaystyle \frac{k-1}{k!}}|{\mu }_k{\left|\right|z|}^k\le {\sigma }_f{\left|z\right|}^2.$$

By Lemma 1, ${\sum }_{k\ge n}{\displaystyle \frac{{\mu }_k}{k!}}{z}^k=\mathsf{\Omega }\left(z\right)$, so $T_2\left(z\right)=-{\sum }_{k\ge n}(k-1){\displaystyle \frac{{\mu }_k}{k!}}{z}^k=\mathsf{\Omega }\left(z\right)-z{\mathsf{\Omega }}^{\prime }\left(z\right)=-z^2{\left(\mathsf{\Omega }\left(z\right)/z\right)}^{\prime }$. The second estimate in (5) gives $|T_2\left(z\right)|\le {\displaystyle \frac{n-1}{n!}}{\mathrm{\Lambda }}_f{\left|z\right|}^n$. Adding the two bounds proves the first inequality in (13). For the second, ${\left|z\right|}^n\le {\left|z\right|}^2$ and (2) give ${\sigma }_f{\left|z\right|}^2+{\displaystyle \frac{n-1}{n!}}{\mathrm{\Lambda }}_f{\left|z\right|}^n\le {\left|z\right|}^2\left({\sigma }_f+{\displaystyle \frac{n-1}{n!}}{\mathrm{\Lambda }}_f\right)\le {\left|z\right|}^2$. Since ${\left|z\right|}^2<1$, (1) holds.

For $E_0$, $u=1+z^n/(n-1)$ and $T\left(z\right)={\displaystyle \frac{z^n}{n-1}}-{\displaystyle \frac{n{z}^n}{n-1}}=-z^n$, while ${\sigma }_{E_0}=0$ and ${\mathrm{\Lambda }}_{E_0}=n!/(n-1)$ (Lemma 5(b)); so the right-hand side of the first inequality is ${\left|z\right|}^n=\left|T\left(z\right)\right|$. For $Q_2$ one has $u=1+z^2$, $T\left(z\right)=-z^2$, ${\mu }_2=2$, $\sigma =1$, $\mathrm{\Lambda }=0$ (Lemma 5(a)), and the right-hand side of the second inequality is ${\left|z\right|}^2$.    □

Remark 2.

Let $f\in \mathcal{V}$, that is, $|T_f|<1$ in $\mathbb{U}$. As in the proof, $T_f=u-1-z{u}^{\prime }=-{\sum }_{k\ge 2}(k-1){\displaystyle \frac{{\mu }_k}{k!}}{z}^k$ is analytic in $\mathbb{U}$ with $T_f\left(0\right)=T_f^{\prime }\left(0\right)=0$. Schwarz’s lemma applied to $T_f$ and then to $T_f/z$ gives $|T_f{\left(z\right)|\le |z|}^2$ forevery$f\in \mathcal{V}$, with equality at one point only for $T_f=e^{i\theta }{z}^2$ (e.g. $Q_2$). Hence the second inequality in (13) and its sharpness are the Schwarz bound for $\mathcal{V}$; together with the inclusion (2)$\Rightarrow f\in \mathcal{V}$ of [4] it is known in substance. The new content of Theorem 1 is the first inequality, in terms of ${\sigma }_f$ and ${\mathrm{\Lambda }}_f$, and its sharpness through $E_0$. For $n=2$ the same computation gives $|T_f|\le {\displaystyle \frac{1}{2}}{\mathrm{\Lambda }}_f{\left|z\right|}^2$ under $|u^{\prime \prime }|\le {\mathrm{\Lambda }}_f$.

The classes cut out by (2) for different values of n do not form a chain, as two explicit functions show; neither lies in ${\mathcal{C}}_n$ (Example 1 has ${\mu }_2\ne 0$, and Example 2 violates (2) for $n=3$). Example 1 satisfies (2) for $n=3$ but not the Yang–Liu condition, and Example 2 does the reverse.

Example 1.

Let $f\left(z\right)=z/(1+0.2z^2+0.4z^3)$. Then $u_f=1+0.2z^2+0.4z^3$, ${\mu }_2=0.4$ and $u_f^{\prime \prime \prime }\equiv 2.4$, so for $n=3$ the left side of (2) equals $\frac{1}{2}\left(0.4\right)+\frac{1}{3}\left(2.4\right)=1$, and f satisfies (2). On the other hand ${\sup }_{\mathbb{U}}|u_f^{\prime \prime }|=\sup |0.4+2.4z|=2.8>2$, so the Yang–Liu condition ($n=2$) fails.

Example 2.

Let $f\left(z\right)=z/(1+0.15z^4)$. Then $u_f^{\prime \prime }\left(z\right)=1.8z^2$, so ${\sup }_{\mathbb{U}}\left|u_f^{\prime \prime }\right|=1.8\le 2$ and the Yang–Liu condition holds. For $n=3$, however, ${\mu }_2=0$ and $u_f^{\prime \prime \prime }\left(z\right)=3.6z$, giving $\frac{1}{3}·3.6=1.2>1$, so (2) fails.

Consequently the classes for different n share the ceiling $|T_f{|\le |z|}^2$ of $\mathcal{V}$ rather than forming an increasing chain.

3.2. Growth, Distortion and Covering Radius

Theorem 2

(Growth). Let $f\in {\mathcal{C}}_n$ and $r=\left|z\right|<1$. Then

$$|{\displaystyle \frac{z}{f\left(z\right)}}-1|\le {\displaystyle \frac{r^n}{n-1}}\mathrm{and}{\displaystyle \frac{r}{1+{\displaystyle \frac{r^n}{n-1}}}}\le \left|f\left(z\right)\right|\le {\displaystyle \frac{r}{1-{\displaystyle \frac{r^n}{n-1}}}}.$$

(14)

Both bounds on $\left|f\right(z\left)\right|$ are sharp for every $n\ge 3$: equality holds at every fixed r for

$$E_0\left(z\right)={\displaystyle \frac{z}{1+{\displaystyle \frac{z^n}{n-1}}}},$$

at $\arg z=0$ for the lower bound and at $\arg z=\pi /n$ for the upper bound.

Proof. 

For $f\in {\mathcal{C}}_n$ we have ${\mu }_1=\dots ={\mu }_{n-1}=0$ (Lemma 4), so $z/f-1=\mathsf{\Omega }$, and the first estimate in (5) with ${\mathrm{\Lambda }}_f\le n!/(n-1)$ gives $|z/f-1|\le r^n/(n-1)$. As $n\ge 3$, $r^n/(n-1)<1$, hence

$$1-{\displaystyle \frac{r^n}{n-1}}\le \left|{\displaystyle \frac{z}{f\left(z\right)}}\right|\le 1+{\displaystyle \frac{r^n}{n-1}},$$

and inverting yields the bounds for $\left|f\right(z\left)\right|$.

For sharpness apply Lemma 5(b) with $p=n$ and $c=1/(n-1)\le \frac{1}{2}$: the function $E_0$ lies in $\mathcal{A}$, has no zero in $0<\left|z\right|<1$, satisfies $a_2=\dots =a_n=0$, and ${\mathrm{\Lambda }}_{E_0}={\displaystyle \frac{n!}{n-1}}$, so $E_0\in {\mathcal{C}}_n$ (on the boundary of the family). For $z=r$ we get $|E_0\left(r\right)|=r/(1+r^n/(n-1))$, the lower bound; for $z=r{e}^{i\pi /n}$ we have $z^n=-r^n$ and $|E_0\left(z\right)|=r/(1-r^n/(n-1))$, the upper bound. This holds for each $r\in [0,1)$ and each $n\ge 3$.    □

The first inequality in (14) is also sharp (equality for $E_0$ at every z). Letting $r\to 1$ gives ${\sup }_{\mathbb{U}}\left|f\right|\le {\displaystyle \frac{n-1}{n-2}}$, sharp by $E_0$ at $z=r{e}^{i\pi /n}$, $r\to 1$; so ${\mathcal{C}}_n$ consists of bounded univalent functions (compare [14]).

Figure 1 shows, for several n, the image of a circle close to $\partial \mathbb{U}$ under $E_0$; the curves are simple, as expected for a univalent map.

Figure 2 confirms numerically that the two bounds in (14) are attained at the predicted angles.

Figure 3 plots both bounds against r, showing how the admissible region for $\left|f\right(z\left)\right|$ narrows as n increases.

Example 3.

Take $n=3$ and $E_0\left(z\right)=z/(1+z^3/2)=z-\frac{1}{2}{z}^4+\frac{1}{4}{z}^7-\dots$. Then $u=1+z^3/2$, ${\mu }_1={\mu }_2=0$, $u^{\prime \prime \prime }\equiv 3=3!/(3-1)$, so $E_0\in {\mathcal{C}}_3$. For $r=0.8$ the bounds in (14) are $0.8/(1+0.8^3/2)=0.6369\dots$ and $0.8/(1-0.8^3/2)=1.0753\dots$; direct evaluation gives $|E_0\left(0.8\right)|=0.6369\dots$ and $|E_0\left(0.8e^{i\pi /3}\right)|=1.0753\dots$.

The growth theorem is a direct consequence of (5). The distortion theorem below needs the derivative estimate in an essential way.

Theorem 3

(Distortion). Let $f\in {\mathcal{C}}_n$ and $r=\left|z\right|<1$. Then

$${\displaystyle \frac{1-r^n}{{\left(1+{\displaystyle \frac{r^n}{n-1}}\right)}^2}}\le \left|f^{\prime }\left(z\right)\right|\le {\displaystyle \frac{1+r^n}{{\left(1-{\displaystyle \frac{r^n}{n-1}}\right)}^2}}.$$

(15)

Both bounds are sharp: for $E_0$ equality holds at every r, at $\arg z=0$ in the lower bound and at $\arg z=\pi /n$ in the upper one.

Proof. 

Let $u=z/f=1+\mathsf{\Omega }$. From $f=z/u$ we get $f^{\prime }=(u-z{u}^{\prime })/u^2$, and $u-z{u}^{\prime }=1+\mathsf{\Omega }-z{\mathsf{\Omega }}^{\prime }=1-z^2{(\mathsf{\Omega }/z)}^{\prime }$. By the second estimate in (5) and ${\mathrm{\Lambda }}_f\le n!/(n-1)$, $|z^2{(\mathsf{\Omega }/z)}^{\prime }|\le {\displaystyle \frac{n-1}{n!}}{\mathrm{\Lambda }}_f{r}^n\le r^n$, hence $1-r^n\le |u-z{u}^{\prime }|\le 1+r^n$. By the first estimate, $1-{\displaystyle \frac{r^n}{n-1}}\le \left|u\right|\le 1+{\displaystyle \frac{r^n}{n-1}}$, which gives (15).

For $E_0$ we have $u=1+z^n/(n-1)$ and $u-z{u}^{\prime }=1-z^n$, so $E_0^{\prime }\left(z\right)=(1-z^n)/{(1+z^n/(n-1))}^2$. At $z=r$ this is the lower bound and at $z=r{e}^{i\pi /n}$ (where $z^n=-r^n$) the upper bound.    □

Corollary 1

(Covering radius). Every $f\in {\mathcal{C}}_n$ satisfies $f\left(\mathbb{U}\right)\supset \{w:|w|<\frac{n-1}{n}\}$, and the radius ${\displaystyle \frac{n-1}{n}}$ cannot be increased: $E_0\left(\mathbb{U}\right)$ does not contain $E_0\left(1\right)={\displaystyle \frac{n-1}{n}}$.

Proof. 

Fix $0<r<1$ and let $m\left(r\right)={\min }_{\left|z\right|=r}\left|f\left(z\right)\right|\ge r/(1+r^n/(n-1))$ by Theorem 2. The set $f\left({\mathbb{U}}_r\right)$, ${\mathbb{U}}_r=\left\{\right|z|<r\}$, is open (open mapping theorem) and contains 0, and its boundary lies in $f\left(\right\{\left|z\right|=r\left\}\right)$, because $\partial f\left({\mathbb{U}}_r\right)\subset \overline{f\left({\mathbb{U}}_r\right)}\setminus f\left({\mathbb{U}}_r\right)\subset f\left(\overline{{\mathbb{U}}_r}\right)\setminus f\left({\mathbb{U}}_r\right)$. If some w with $\left|w\right|<m\left(r\right)$ were not in $f\left({\mathbb{U}}_r\right)$, the segment from 0 to w would meet $\partial f\left({\mathbb{U}}_r\right)$ at a point of modulus $<m\left(r\right)$, which is impossible. Hence $f\left(\mathbb{U}\right)\supset \left\{\right|w|<m(r\left)\right\}$, and letting $r\to 1^{-}$ gives the radius $\lim r/(1+r^n/(n-1))={\displaystyle \frac{n-1}{n}}$.

The first part used only the open mapping theorem; univalence enters only in the sharpness part. For sharpness note that $E_0\in {\mathcal{C}}_n$ is univalent in $\mathbb{U}$ (Remark 1), extends continuously to $\overline{\mathbb{U}}$ (the denominator does not vanish on $\left|z\right|=1$ since $n\ge 3$), and $E_0\left(1\right)={\displaystyle \frac{n-1}{n}}$. Suppose $E_0\left(z_0\right)=E_0\left(1\right)$ with $|z_0|<1$. As $E_0^{\prime }\left(z_0\right)=(1-z_0^n)/{(\dots )}^2\ne 0$ (note that $E_0^{\prime }\left(z\right)=0$ exactly when $z^n=1$, so $E_0$ is not locally univalent at the boundary point $z=1$, which is why a limiting argument is used instead of univalence on a larger disk), $E_0$ maps a neighborhood $V⋐\mathbb{U}$ of $z_0$ onto a neighborhood of $E_0\left(1\right)$. For r close to 1 the point $E_0\left(r\right)$ lies in that neighborhood, so $E_0\left(r\right)=E_0\left(v\right)$ with $v\in V$, and univalence gives $r=v\in V$, which is false for r close to 1. Therefore $E_0\left(1\right)\notin E_0\left(\mathbb{U}\right)$.    □

3.3. Coefficient Bounds

For $f\in {\mathcal{C}}_n$ identity (9) makes the first n non-trivial coefficients $a_{n+1},\dots ,a_{2n}$ proportional to ${\eta }_0,\dots ,{\eta }_{n-1}$, and Lemma 3 bounds each ${\eta }_j$. The next two theorems are direct consequences.

Theorem 4.

Let $f\in {\mathcal{C}}_n$. For every $j=0,1,\dots ,n-1$,

$$|a_{n+1+j}|\le {\displaystyle \frac{1}{(n-1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+j}{j}}\right)}},$$

(16)

and equality holds for

$$E_j\left(z\right)={\displaystyle \frac{z}{1+{\displaystyle \frac{z^{n+j}}{(n-1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+j}{j}}\right)}}}}.$$

(17)

For $E_j$, $a_k=0$ for $2\le k\le n+j$ and for $n+j+2\le k\le 2(n+j)$, and $a_{n+j+1}=-1/\left[(n-1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+j}{j}}\right)\right]$.

Proof. 

By (9), Lemma 3 and the definition of ${\mathcal{C}}_n$,

$$|a_{n+1+j}|=|{\eta }_j|{\displaystyle \frac{j!}{(n+j)!}}\le {\mathrm{\Lambda }}_f{\displaystyle \frac{j!}{(n+j)!}}\le {\displaystyle \frac{n!}{n-1}}·{\displaystyle \frac{j!}{(n+j)!}}={\displaystyle \frac{1}{(n-1)\left(\genfrac{}{}{0pt}{}{n+j}{j}\right)}},$$

because $\left({\displaystyle \genfrac{}{}{0pt}{}{n+j}{j}}\right)={\displaystyle \frac{(n+j)!}{n!j!}}$.

For sharpness let $p=n+j$ and $c=1/\left[(n-1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+j}{j}}\right)\right]\le \frac{1}{2}$ in Lemma 5(b). Then ${\mathrm{\Lambda }}_{E_j}=c{\displaystyle \frac{(n+j)!}{j!}}={\displaystyle \frac{n!}{n-1}}$, so $E_j\in {\mathcal{C}}_n$, and $a_{n+j+1}=-c$. Since $0<c\le {\displaystyle \frac{1}{2}}$ and $\left|z\right|<1$, we have $|c{z}^p|<1$, so the geometric series ${\displaystyle \frac{1}{1+c{z}^p}}={\sum }_{m\ge 0}{(-c{z}^p)}^m$ converges, and

$$E_j\left(z\right)=z\sum _{m\ge 0}{(-c)}^m{z}^{pm}=z-c{z}^{p+1}+c^2{z}^{2p+1}-\dots$$

with $p=n+j$. Thus the only non-zero coefficients of $E_j$ are $a_{pm+1}={(-c)}^m$ ($m\ge 0$); in particular $a_k=0$ for $2\le k\le p$ and for $p+2\le k\le 2p$, and $a_{p+1}=-c$.    □

For the logarithmic coefficients write $\log \left(f\left(z\right)/z\right)=2{\sum }_{k\ge 1}{\ell }_k{z}^k$; the branch is fixed by $\log \left(f\left(z\right)/z\right)=0$ at $z=0$, and it exists because $f\left(z\right)/z=1/u_f\left(z\right)$ is analytic and zero-free in $\mathbb{U}$.

Corollary 2.

Let $f\in {\mathcal{C}}_n$. For every $j=0,1,\dots ,n-1$,

$$|{\ell }_{n+j}|\le {\displaystyle \frac{1}{2(n-1)\left({\displaystyle \genfrac{}{}{0pt}{}{n+j}{j}}\right)}},$$

(18)

with equality for $E_j$ of (17).

Proof. 

On ${\mathcal{C}}_n$, $u=1+\mathsf{\Omega }$ and so $\log \left(f\left(z\right)/z\right)=-\log (1+\mathsf{\Omega }\left(z\right))=-\mathsf{\Omega }\left(z\right)+\frac{1}{2}\mathsf{\Omega }{\left(z\right)}^2-\dots$. Since ${\mathsf{\Omega }}^2=O\left(z^{2n}\right)$, for $j\le n-1$ the coefficient of $z^{n+j}$ on the right is $-{\delta }_j$, which equals $a_{n+1+j}$ by (9). Thus $2{\ell }_{n+j}=a_{n+1+j}$ and (18) follows from (16). Equality passes to $E_j$ because it holds in (16).    □

Example 4.

Let $n=3$ and $j=1$. Then (16) gives $|a_5|\le 1/\left[2\left({\displaystyle \genfrac{}{}{0pt}{}{4}{1}}\right)\right]=\frac{1}{8}$, attained by $E_1\left(z\right)=z/(1+z^4/8)=z-\frac{1}{8}{z}^5+O\left(z^9\right)$. For the same function $\log (E_1\left(z\right)/z)=-\frac{1}{8}{z}^4+O\left(z^8\right)$, so $2{\ell }_4=-\frac{1}{8}$ and $|{\ell }_4|=\frac{1}{16}$, the value given by (18).

Table 1 lists the numerical values of the bounds in Theorem 4 and Corollary 2 for $3\le n\le 6$.

The following theorem goes beyond the diagonal bounds of Theorem 4: it uses the parametrization of Proposition 1, and treats $a_{2n+1}$, the first coefficient at which the identity (9) breaks down.

Theorem 5.

Let $n\ge 3$.

(a)For $f\in {\mathcal{C}}_n$, $|a_{n+1}|\le {\displaystyle \frac{1}{n-1}}$ and

$$|a_{n+2}|\le {\displaystyle \frac{1-{(n-1)}^2{\left|a_{n+1}\right|}^2}{n^2-1}}.$$

(19)

Conversely, every pair $(\alpha ,\beta )$ with $\left|\alpha \right|\le {\displaystyle \frac{1}{n-1}}$ and $\left|\beta \right|\le {\displaystyle \frac{1-{(n-1)}^2{\left|\alpha \right|}^2}{n^2-1}}$ equals $(a_{n+1},a_{n+2})$ for some $f\in {\mathcal{C}}_n$.

(b)For $f\in {\mathcal{C}}_n$, $|a_{2n+1}|\le {\displaystyle \frac{1}{{(n-1)}^2}}$, with equality for $E_0$.

Proof. 

Preliminaries. Let $f\in {\mathcal{C}}_n$ and $\eta =\mathsf{\Phi }\left(f\right)$. By Proposition 1, $\eta \in B_n$, so $\phi :=\eta /M_n={\sum }_{m\ge 0}{\phi }_m{z}^m$ is analytic in $\mathbb{U}$ with $\left|\phi \right|\le 1$, and ${\eta }_m=M_n{\phi }_m$. Since ${\mu }_1=\dots ={\mu }_{n-1}=0$ (Lemma 4(a)), the identity (9) applies, and with $M_n=n!/(n-1)$,

$$a_{n+1}=-{\displaystyle \frac{{\eta }_0}{n!}}=-{\displaystyle \frac{{\phi }_0}{n-1}},a_{n+2}=-{\displaystyle \frac{{\eta }_1}{(n+1)!}}=-{\displaystyle \frac{n!{\phi }_1}{(n-1)(n+1)!}}=-{\displaystyle \frac{{\phi }_1}{(n-1)(n+1)}}.$$

(20)

We shall use the following form of the Schwarz–Pick inequality: if $\psi$ is analytic in $\mathbb{U}$ with $\left|\psi \right|\le 1$, then

$$|{\psi }^{\prime }{\left(0\right)|\le 1-|\psi \left(0\right)|}^2.$$

(21)

Indeed, if $\left|\psi \right(0\left)\right|=1$ then $\psi$ is constant by the maximum modulus principle and ${\psi }^{\prime }\left(0\right)=0$. Otherwise put $b=\psi \left(0\right)$, $\left|b\right|<1$, and $\tau \left(w\right)={\displaystyle \frac{w-b}{1-\overline{b}w}}$, a Möbius automorphism of the unit disk with $\tau \left(b\right)=0$. Then $\tau \circ \psi$ maps $\mathbb{U}$ into the closed unit disk and vanishes at 0, so Schwarz’s lemma gives ${|(\tau \circ \psi )}^{\prime }\left(0\right)|\le 1$. Since ${\tau }^{\prime }\left(w\right)={\displaystyle \frac{1-{\left|b\right|}^2}{{(1-\overline{b}w)}^2}}$, we get ${\tau }^{\prime }\left(b\right)={\displaystyle \frac{1}{1-{\left|b\right|}^2}}$ and ${(\tau \circ \psi )}^{\prime }\left(0\right)={\displaystyle \frac{{\psi }^{\prime }\left(0\right)}{1-{\left|b\right|}^2}}$, which is (21).

(a), necessity. Since $|{\phi }_0|\le 1$, (20) gives $|a_{n+1}|=|{\phi }_0|/(n-1)\le {\displaystyle \frac{1}{n-1}}$. Applying (21) to $\psi =\phi$ gives $|{\phi }_1|\le 1-|{\phi }_0{|}^2$. By (20), $|{\phi }_0|=(n-1)|a_{n+1}|$ and $|{\phi }_1|=(n^2-1)\left|a_{n+2}\right|$, so

$$(n^2-1)|a_{n+2}|\le 1-{(n-1)}^2{\left|a_{n+1}\right|}^2,$$

which is (19).

(a), sufficiency. Let $(\alpha ,\beta )$ satisfy $\left|\alpha \right|\le {\displaystyle \frac{1}{n-1}}$ and $\left|\beta \right|\le {\displaystyle \frac{1-{(n-1)}^2{\left|\alpha \right|}^2}{n^2-1}}$, and put $c:=-(n-1)\alpha$, so $\left|c\right|\le 1$. If $\left|c\right|=1$, the hypothesis on $\beta$ forces $\beta =0$; take $\phi \equiv c$, so ${\phi }_0=c$ and ${\phi }_1=0$. If $\left|c\right|<1$, put

$$\kappa :=-{\displaystyle \frac{(n^2-1)\beta }{1-{\left|c\right|}^2}},\mathrm{so}\mathrm{that}\left|\kappa \right|\le {\displaystyle \frac{(n^2-1)}{1-{\left|c\right|}^2}}·{\displaystyle \frac{1-{\left|c\right|}^2}{n^2-1}}=1,$$

and define $\phi \left(z\right)={\displaystyle \frac{c+\kappa z}{1+\overline{c}\kappa z}}$. This is $T_c\left(\kappa z\right)$ with $T_c\left(w\right)={\displaystyle \frac{c+w}{1+\overline{c}w}}$, a Möbius automorphism of the unit disk (because $\left|c\right|<1$); since $\left|\kappa z\right|<1$, we have $\left|\phi \right|<1$ in $\mathbb{U}$. Moreover ${\phi }_0=\phi \left(0\right)=c$, and since $T_c^{\prime }\left(0\right)=1-{\left|c\right|}^2$, ${\phi }_1={\phi }^{\prime }\left(0\right)={\kappa (1-|c|}^2)=-(n^2-1)\beta$. In both cases $\phi$ is analytic with $\left|\phi \right|\le 1$, so $M_n\phi \in B_n$, and $f:={\mathsf{\Phi }}^{-1}\left(M_n\phi \right)\in {\mathcal{C}}_n$ by Proposition 1. By (20), $a_{n+1}=-{\phi }_0/(n-1)=\alpha$ and $a_{n+2}=-{\phi }_1/\left((n-1)(n+1)\right)=\beta$.

(b). By (10) and (8), $a_{2n+1}={\delta }_0^2-{\delta }_n$ with

$${\delta }_0={\displaystyle \frac{{\eta }_0}{n!}}={\displaystyle \frac{{\phi }_0}{n-1}},{\delta }_n={\eta }_n{\displaystyle \frac{n!}{\left(2n\right)!}}={\phi }_n{\displaystyle \frac{n!}{n-1}}·{\displaystyle \frac{n!}{\left(2n\right)!}}={\displaystyle \frac{{\phi }_n}{(n-1)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}},$$

because ${\displaystyle \frac{\left(2n\right)!}{{(n!)}^2}}=\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)$. We need $|{\phi }_n|\le 1-|{\phi }_0{|}^2$. Let $\epsilon =e^{2\pi i/n}$ and $\psi \left(z\right):={\displaystyle \frac{1}{n}}{\sum }_{k=0}^{n-1}\phi \left({\epsilon }^{k}z\right)$. As an average of values of $\phi$, $\left|\psi \right|\le 1$. Since ${\sum }_{k=0}^{n-1}{\epsilon }^{km}$ equals n if $n\mid m$ and 0 otherwise, $\psi \left(z\right)={\sum }_{q\ge 0}{\phi }_{nq}{z}^{nq}$. Put $\chi \left(\zeta \right)={\sum }_{q\ge 0}{\phi }_{nq}{\zeta }^q$; this series converges for $\left|\zeta \right|<1$ (take z with $z^n=\zeta$ in the series of $\psi$), so $\chi$ is analytic in $\mathbb{U}$, and $\chi \left(\zeta \right)=\psi \left(z\right)$ for any z with $z^n=\zeta$, whence $\left|\chi \right|\le 1$. Now $\chi \left(0\right)={\phi }_0$ and ${\chi }^{\prime }\left(0\right)={\phi }_n$, so (21) for $\chi$ gives $|{\phi }_n|\le 1-|{\phi }_0{|}^2$. With $x:=|{\phi }_0|\in [0,1]$ we obtain

$$|a_{2n+1}|\le |{\delta }_0{|}^2+\left|{\delta }_n\right|\le {\displaystyle \frac{x^2}{{(n-1)}^2}}+{\displaystyle \frac{1-x^2}{(n-1)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}.$$

This is a convex combination, with weights $x^2$ and $1-x^2$, of $A={\displaystyle \frac{1}{{(n-1)}^2}}$ and $B={\displaystyle \frac{1}{(n-1)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}$. Since $\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)\ge 2n>n-1$ we have $B<A$, so the combination is at most A, i.e. $|a_{2n+1}|\le {\displaystyle \frac{1}{{(n-1)}^2}}$.

For equality, the function $E_0\left(z\right)={\displaystyle \frac{z}{1+z^n/(n-1)}}=z{\sum }_{m\ge 0}{\left(-{\displaystyle \frac{z^n}{n-1}}\right)}^m$ (geometric series, $|z^n/(n-1)|<1$) has, as the coefficient of $z^{2n+1}$ (the term $m=2$) equal to ${\displaystyle \frac{1}{{(n-1)}^2}}$.    □

(a)).Remark 3 (Closed form of the extremal functions of The region in (a) depends only on $\left|\alpha \right|$ and $\left|\beta \right|$. By Schwarz–Pick equality, a pair on the boundary $|{\phi }_1|=1-|{\phi }_0{|}^2$ with $|{\phi }_0|<1$ arises exactly from a Möbius parameter $\phi \left(z\right)={\displaystyle \frac{c+\kappa z}{1+\overline{c}\kappa z}}$, $\left|c\right|<1$, $\left|\kappa \right|=1$, that is, $\eta =M_n\phi$. For $c\ne 0$ write $\phi ={\displaystyle \frac{1}{\overline{c}}}\left(1-{\displaystyle \frac{1-{\left|c\right|}^2}{1+pz}}\right)$ with $p=\overline{c}\kappa$. Then ${\mathsf{\Omega }}_{\eta }$ can be integrated in elementary terms:

$${\mathsf{\Omega }}_{\eta }\left(z\right)={\displaystyle \frac{n}{n-1}}{\displaystyle \frac{z^n}{\overline{c}}}\left({\displaystyle \frac{1}{n}}{-(1-|c|}^2)I_n\left(pz\right)\right),$$

(22)

$$I_n\left(y\right)={\displaystyle \frac{1}{y^n}}\left[{(1+y)}^{n-1}\log (1+y)+{\displaystyle \sum _{k=1}^{n-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k}}\right){\displaystyle \frac{{(-1)}^k{(1+y)}^{n-1-k}\left({(1+y)}^k-1\right)}{k}}\right],$$

(23)

where $I_n\left(y\right)={\int }_0^1{\displaystyle \frac{{(1-s)}^{n-1}}{1+sy}}ds$ (substitute $t=1+sy$) and the principal logarithm is used ($\left|y\right|<1$). The extremal function is $f=z/(1+{\mathsf{\Omega }}_{\eta })$. We checked (22) against numerical integration for $n=2,3,4$ (05_verify_revision.py). So the extremal functions of(a)are explicit, involving a logarithm and a polynomial in z; those of all other results are the monomial functions $E_j$ and $Q_2$.

Remark 4

(Bounds for ${\mu }_k$ on the whole class). On the whole class (2) one has, for $2\le k\le n-1$, $|{\mu }_k|\le k!/(k-1)$, because every summand of ${\sigma }_f$ is non-negative and ${\sigma }_f\le 1$. This is an immediate restatement of the criterion, and it is sharp: $Q_k\left(z\right)=z/(1+z^k/(k-1))$ satisfies (2) with equality and ${\mu }_k=k!/(k-1)$ by Lemma 5(a). For $k=2$, since ${\mu }_2=2(a_2^2-a_3)$, it reads $|a_2^2-a_3|\le 1$. This inequality is not specific to the class (2): it holds for every univalent function, because $1/f(1/\zeta )=\zeta -a_2+(a_2^2-a_3){\zeta }^{-1}+\dots$ belongs to the class Σ and the area theorem bounds its coefficient of ${\zeta }^{-1}$ by 1 [1]. The case $k=2$ is therefore classical (the Koebe function gives equality), and we do not present it as a Fekete–Szego-type result.

3.4. A Larger Class: $a_2=0$ and the Criterion Itself

Condition (2) does not force ${\mu }_2,\dots ,{\mu }_{n-1}$ to vanish, and the restriction ${\mu }_2=\dots ={\mu }_{n-1}=0$ defining ${\mathcal{C}}_n$ is a choice, not a consequence of the criterion. To measure what it buys, let ${\mathcal{D}}_n$ be the set of zero-free $f\in \mathcal{A}$ with $a_2=0$ (equivalently ${\mu }_1=0$, because $u^{\prime }\left(0\right)=-a_2$) that satisfy (2). Then ${\mathcal{C}}_n\subset {\mathcal{D}}_n$.

Theorem 6.

Let $n\ge 3$, $f\in {\mathcal{D}}_n$ and $r=\left|z\right|<1$. Then $|z/f-1|\le r^2$,

$${\displaystyle \frac{r}{1+r^2}}\le \left|f\left(z\right)\right|\le {\displaystyle \frac{r}{1-r^2}},{\displaystyle \frac{1-r^2}{{(1+r^2)}^2}}\le \left|f^{\prime }\left(z\right)\right|\le {\displaystyle \frac{1+r^2}{{(1-r^2)}^2}},$$

and $f\left(\mathbb{U}\right)\supset \left\{\right|w|<{\displaystyle \frac{1}{2}}\}$. All these estimates are sharp, with extremal function $Q_2\left(z\right)=z/(1+z^2)\in {\mathcal{D}}_n$, and ${\displaystyle \frac{1}{2}}$ cannot be replaced by a larger number.

Proof. 

By Lemma 1 with ${\mu }_1=0$, $u-1={\sum }_{k=2}^{n-1}{\displaystyle \frac{{\mu }_k}{k!}}{z}^k+\mathsf{\Omega }$. Put $w_k={\displaystyle \frac{k-1}{k!}}\left|{\mu }_k\right|$ and $\omega ={\displaystyle \frac{n-1}{n!}}{\mathrm{\Lambda }}_f$, so ${\sum }_{k=2}^{n-1}{w}_k+\omega \le 1$ by (2). Since ${\displaystyle \frac{r^k}{k-1}}\le r^2$ for $k\ge 2$ (the function $k\mapsto r^k/(k-1)$ decreases) and ${\displaystyle \frac{r^n}{n-1}}\le r^2$, and $\left|\mathsf{\Omega }\right|\le \omega {\displaystyle \frac{r^n}{n-1}}$ by (5),

$$|u-1|\le {\displaystyle \sum _{k=2}^{n-1}}{w}_k{\displaystyle \frac{r^k}{k-1}}+\omega {\displaystyle \frac{r^n}{n-1}}\le r^2\left({\displaystyle \sum _{k=2}^{n-1}}{w}_k+\omega \right)\le r^2.$$

This gives the bounds for $\left|f\right|=r/\left|u\right|$. By Theorem 1, $|u-z{u}^{\prime }-1|\le r^2$, and $f^{\prime }=(u-z{u}^{\prime })/u^2$ gives those for $|f^{\prime }|$. For $Q_2$ we have $u=1+z^2$, ${\mu }_2=2$, $\sigma =1$, $\mathrm{\Lambda }=0$, so (2) holds with equality and $Q_2\in {\mathcal{D}}_n$; $|Q_2\left(r\right)|=r/(1+r^2)$, $|Q_2\left(ir\right)|=r/(1-r^2)$, and $Q_2^{\prime }\left(z\right)=(1-z^2)/{(1+z^2)}^2$ equals the lower bound at $z=r$ and the upper bound at $z=ir$. The covering statement follows as in Corollary 1 from ${\min }_{\left|z\right|=r}\left|f\right|\ge r/(1+r^2)\to {\displaystyle \frac{1}{2}}$. Finally $Q_2\left(z\right)={\displaystyle \frac{1}{2}}$ iff ${(z-1)}^2=0$, so ${\displaystyle \frac{1}{2}}\notin Q_2\left(\mathbb{U}\right)$.    □

The covering radius ${\displaystyle \frac{1}{2}}$ is smaller than ${\displaystyle \frac{n-1}{n}}$ for every $n\ge 3$, and $r^2$ replaces $r^n/(n-1)$: the strength of the estimates of ${\mathcal{C}}_n$ comes precisely from the vanishing of ${\mu }_2,\dots ,{\mu }_{n-1}$, and Theorem 6 shows that nothing better than the $n=2$ values survives without it.

Remark 5

(The case $n=2$). For $n=2$ the class ${\mathcal{C}}_2$ is the Yang–Liu class $\left\{\right|u^{\prime \prime }|\le 2\}$ with $a_2=0$, and ${\mathcal{C}}_2={\mathcal{D}}_2$. The proofs of Sections 2–4 go through for $n=2$ with three changes: in Proposition 1 the bound $|{\mathsf{\Omega }}_{\eta }|\le {\displaystyle \frac{{\left|z\right|}^n}{n-1}}\le {\displaystyle \frac{1}{2}}$ is replaced by $|{\mathsf{\Omega }}_{\eta }{|\le |z|}^2<1$; the restriction $c\le {\displaystyle \frac{1}{2}}$ in Lemma 5 and Theorem 4 becomes $c\le 1$; and in Corollary 1 the function $E_0=Q_2$ has poles at $\pm i$ on the boundary of the disk of convergence of the series, so the sharpness of ${\displaystyle \frac{1}{2}}$ is read off from $Q_2\left(z\right)={\displaystyle \frac{1}{2}}\iff z=1$. Univalence of ${\mathcal{C}}_2$ is the theorem of Yang and Liu [6]. Thus the formulas of Theorems 2 and 3 and Corollary 1 for $n=2$ are those of Theorem 6, with $s_2=3^{-1/2}$ in Theorem 7. The comparisons of Section 5 with n-fold symmetric functions degenerate for $n=2$ and are stated for $n\ge 3$.

4. A Product Theorem

Obradović and Ponnusamy [4] proved that if $g,h\in \mathcal{S}$ and $F\left(z\right)=g\left(z\right)h\left(z\right)/z$, then $F_r\left(z\right)=r^{-1}F\left(rz\right)$ belongs to Aksentev’s class (their $\mathcal{U}$, defined by (1)) for every $r\in (0,r_0]$, where $r_0\approx 0.3263$ is the radius reported there. We prove the corresponding statement for ${\mathcal{C}}_n$, with a radius that is sharp.

Theorem 7.

Let $n\ge 3$, let $f_1,f_2\in {\mathcal{C}}_n$, and put $F\left(z\right)=f_1\left(z\right)f_2\left(z\right)/z$ and $F_s\left(z\right)=s^{-1}F\left(sz\right)$ for $0<s<1$. Then $F_s\in \mathcal{A}$, $F_s$ has no zero in $0<\left|z\right|<1$, and for all $z\in \mathbb{U}$

$$\left|{\displaystyle \frac{z^2{F}_s^{\prime }\left(z\right)}{F_s{\left(z\right)}^2}}-1\right|\le 2s^n{\left|z\right|}^n+{\displaystyle \frac{2n-1}{{(n-1)}^2}}{s}^{2n}{\left|z\right|}^{2n}.$$

(24)

Consequently, if

$$s\le s_n:={\left({\displaystyle \frac{n-1}{2n-1}}\right)}^{1/n},$$

then $F_s\in \mathcal{V}$, and in particular $F_s$ is univalent in $\mathbb{U}$. The radius $s_n$ cannot be enlarged, not even for univalence: for every $s\in (s_n,1)$ the pair $f_1=f_2=E_0$ gives ${\sup }_{\mathbb{U}}|z^2{F}_s^{\prime }/F_s^2-1|>1$, and $F_s$ is not univalent in $\mathbb{U}$.

Proof. 

Write $z/f_i\left(z\right)=1+{\mathsf{\Omega }}_i\left(z\right)$ ($i=1,2$). Then $z/F\left(z\right)=(1+{\mathsf{\Omega }}_1)(1+{\mathsf{\Omega }}_2)=1+\Xi$ with $\Xi ={\mathsf{\Omega }}_1+{\mathsf{\Omega }}_2+{\mathsf{\Omega }}_1{\mathsf{\Omega }}_2$, and $z/F_s\left(z\right)=\left(sz\right)/F\left(sz\right)=1+\Xi \left(sz\right)$. Let $W_s\left(z\right)=1+\Xi \left(sz\right)$. As in the proof of Theorem 1, $z^2{F}_s^{\prime }/F_s^2-1=W_s-1-z{W}_s^{\prime }$, and with $w=sz$ this equals $(\Xi -w{\Xi }^{\prime })\left(w\right)$. Now

$$\Xi -w{\Xi }^{\prime }=({\mathsf{\Omega }}_1-w{\mathsf{\Omega }}_1^{\prime })+({\mathsf{\Omega }}_2-w{\mathsf{\Omega }}_2^{\prime })+{\mathsf{\Omega }}_1({\mathsf{\Omega }}_2-w{\mathsf{\Omega }}_2^{\prime })-w{\mathsf{\Omega }}_1^{\prime }{\mathsf{\Omega }}_2.$$

For $f_i\in {\mathcal{C}}_n$ we have $\mathrm{\Lambda }\le M_n$, so by (5)

$$|{\mathsf{\Omega }}_i|\le {\displaystyle \frac{{\left|w\right|}^n}{n-1}},|{\mathsf{\Omega }}_i-w{\mathsf{\Omega }}_i^{\prime }|=|w^2{({\mathsf{\Omega }}_i/w)}^{\prime }{|\le |w|}^n,$$

and therefore $|w{\mathsf{\Omega }}_i^{\prime }|\le |{\mathsf{\Omega }}_i|+|{\mathsf{\Omega }}_i-w{\mathsf{\Omega }}_i^{\prime }|\le {\displaystyle \frac{n}{n-1}}{\left|w\right|}^n$. Hence

$$|\Xi -w{\Xi }^{\prime }{|\le 2|w|}^n+{\displaystyle \frac{{\left|w\right|}^{2n}}{n-1}}+{\displaystyle \frac{{n\left|w\right|}^{2n}}{{(n-1)}^2}}={2\left|w\right|}^n+{\displaystyle \frac{2n-1}{{(n-1)}^2}}{\left|w\right|}^{2n},$$

and (24) follows from $\left|w\right|=s\left|z\right|$.

Put $t=s^n$. For $\left|z\right|<1$ the right-hand side of (24) is $<2t+{\displaystyle \frac{2n-1}{{(n-1)}^2}}{t}^2$, and

$$2t+{\displaystyle \frac{2n-1}{{(n-1)}^2}}{t}^2\le 1\iff t\le t_0:={\displaystyle \frac{n-1}{2n-1}}.$$

(The positive root of ${\displaystyle \frac{2n-1}{{(n-1)}^2}}{t}^2+2t-1=0$ is ${\displaystyle \frac{{(n-1)}^2}{2n-1}}\left({\displaystyle \frac{n}{n-1}}-1\right)={\displaystyle \frac{n-1}{2n-1}}$, because the discriminant equals $4n^2/{(n-1)}^2$.) Thus $s\le s_n$ gives (1) for $F_s$. It is univalent by Aksentev’s theorem, and $F_s\in \mathcal{A}$ without zeros in $0<\left|z\right|<1$ is clear from the definition.

For sharpness take $f_1=f_2=E_0$. Then $z/F_s\left(z\right)={(1+w^n/(n-1))}^2$ with $w=sz$, and a direct computation gives $z^2{F}_s^{\prime }/F_s^2-1=-2w^n-{\displaystyle \frac{2n-1}{{(n-1)}^2}}{w}^{2n}$. For $z=\rho \in (0,1)$ this is $-2t{\rho }^n-{\displaystyle \frac{2n-1}{{(n-1)}^2}}{t}^2{\rho }^{2n}$, whose modulus tends to $2t+{\displaystyle \frac{2n-1}{{(n-1)}^2}}{t}^2$ as $\rho \to 1^{-}$, and this exceeds 1 when $t>t_0$.

Finally we show that $F_s$ is not univalent for $s>s_n$. For $f_1=f_2=E_0$ and $c={\displaystyle \frac{1}{n-1}}$ we have $F\left(z\right)=E_0{\left(z\right)}^2/z=z{(1+c{z}^n)}^{-2}$, and differentiation gives

$$F^{\prime }\left(z\right)={(1+c{z}^n)}^{-2}-2nc{z}^n{(1+c{z}^n)}^{-3}={\displaystyle \frac{1-(2n-1)c{z}^n}{{(1+c{z}^n)}^3}}.$$

Since $(2n-1)c={\displaystyle \frac{2n-1}{n-1}}$, $F^{\prime }$ vanishes at every z with $z^n={\displaystyle \frac{n-1}{2n-1}}$, in particular at the positive number $s_n$ (because $s_n^n=t_0$). As $F_s^{\prime }\left(z\right)=F^{\prime }\left(sz\right)$, for $s\in (s_n,1)$ the point $z_0=s_n/s$ lies in $\mathbb{U}$ and $F_s^{\prime }\left(z_0\right)=0$. A non-constant analytic function whose derivative vanishes at $z_0$ is not injective in any neighborhood of $z_0$, so $F_s$ is not univalent in $\mathbb{U}$.    □

Remark 6.

(a)Limit.Since ${\displaystyle \frac{n-1}{2n-1}}={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2(2n-1)}}\to {\displaystyle \frac{1}{2}}$, we have

$$n\log s_n=\log {\displaystyle \frac{n-1}{2n-1}}⟶-\log 2,\mathrm{hence}n(1-s_n)\to \log 2(n\to \infty ),$$

so $s_n\to 1$ and $1-s_n\sim \left(\log 2\right)/n$.

(b)Interpretation.Over the class ${\mathcal{C}}_n$ the radius $s_n$ is exact both for Aksentev’s inequality and for univalence of the dilated product, and this exact value gives the relative size of the room that remains, $n(1-s_n)\to \log 2$. (How this limit should be read is discussed at the start of Section 5.)

Table 2 lists the radius $s_n$ for small n.

5. Comparison with Classical Results

The class ${\mathcal{C}}_n$ is contained in $\mathcal{S}$, so every classical bound for $\mathcal{S}$ [1] holds on ${\mathcal{C}}_n$, and the bounds proved here are smaller or equal for the trivial reason that the class is smaller. Table 3 therefore carries no information beyond the size of the gap; none of its rows is a sharpening of a known theorem, and the rows about coefficients and products compare statements on classes of very different size. The same reading applies to the radius $s_n$: that $s_n\to 1$ only reflects that $|z/f-1|\le r^n/(n-1)$ becomes a trivial restriction as n grows, and the informative statement is the exactness of $s_n$ in Theorem 7 for each fixed n. The theorems of this paper assume $n\ge 3$; the case $n=2$ is covered by Remark 5. What is known for $n=2$ concerns the criterion itself (Yang–Liu, Nunokawa–Obradović–Owa) and, for the related class $\mathcal{U}\left(\lambda \right)$, other quantities [7,8,9,10,12].

Since every member of ${\mathcal{C}}_n$ is $z+O\left(z^{n+1}\right)$, the class ${\mathcal{S}}_n$ of n-fold symmetric univalent functions, with the same kind of expansion, is the closer reference.

Definition 2

(The class ${\mathcal{S}}_n$). For an integer $n\ge 1$ let ${\mathcal{S}}_n:=\{f\in \mathcal{S}:f\left(e^{2\pi i/n}z\right)=e^{2\pi i/n}f\left(z\right)\mathrm{for}\mathrm{all}z\in \mathbb{U}\}$ (${\mathcal{S}}_1=\mathcal{S}$, and ${\mathcal{S}}_2$ is the class of odd univalent functions). Every $f\in {\mathcal{S}}_n$ has the form $f\left(z\right)=g{\left(z^n\right)}^{1/n}$ with $g\in \mathcal{S}$ [1], hence only the powers $z^{kn+1}$ occur in f.

Applying the classical bounds for $\mathcal{S}$ [1] to g gives the left column of Table 4, attained by the n-fold Koebe function $z/{(1-z^n)}^{2/n}$. Neither ${\mathcal{S}}_n$ nor ${\mathcal{C}}_n$ contains the other: $|a_{n+1}|\le {\displaystyle \frac{1}{n-1}}<{\displaystyle \frac{2}{n}}$ on ${\mathcal{C}}_n$, so the n-fold Koebe function is not in ${\mathcal{C}}_n$, while members of ${\mathcal{C}}_n$ with $a_{n+2}\ne 0$ are not n-fold symmetric. Consequently the table compares extremal values, not classes.

Proposition 2.

For every integer $n\ge 3$ and $0<r<1$:

$$\begin{array}{cc}& \left(\mathrm{i}\right){\displaystyle \frac{r}{1+\frac{r^n}{n-1}}}>{\displaystyle \frac{r}{{(1+r^n)}^{2/n}}},\left(\mathrm{ii}\right){\displaystyle \frac{r}{1-\frac{r^n}{n-1}}}<{\displaystyle \frac{r}{{(1-r^n)}^{2/n}}},\hfill \\ & \left(\mathrm{iii}\right){\displaystyle \frac{n-1}{n}}>4^{-1/n},\left(\mathrm{iv}\right){\displaystyle \frac{1}{n-1}}<{\displaystyle \frac{2}{n}}.\hfill \end{array}$$

Proof. 

Put $t=r^n\in (0,1)$ and $p=2/n\in (0,1)$.

(ii) The function $t\mapsto {(1-t)}^p$ is concave, hence lies below its tangent at 0: ${(1-t)}^{2/n}\le 1-{\displaystyle \frac{2t}{n}}<1-{\displaystyle \frac{t}{n-1}}$, the last step because ${\displaystyle \frac{2}{n}}>{\displaystyle \frac{1}{n-1}}\iff n>2$. Inverting gives (ii), since $r^n/(n-1)<1$.

(i) The function $t\mapsto {(1+t)}^p$ is concave, hence lies above its chord on $[0,1]$: ${(1+t)}^{2/n}>1+(2^{2/n}-1)t$ for $0<t<1$. It suffices to have $2^{2/n}-1\ge {\displaystyle \frac{1}{n-1}}$. For $n\ge 4$, $2^{2/n}=e^{2\ln 2/n}>1+{\displaystyle \frac{2\ln 2}{n}}$ and ${\displaystyle \frac{2\ln 2}{n}}\ge {\displaystyle \frac{1}{n-1}}\iff n(2\ln 2-1)\ge 2\ln 2$, which holds for $n\ge 4$ since $4(2\ln 2-1)=1.545>1.386=2\ln 2$. For $n=3$, $2^{2/3}-1=0.587>{\displaystyle \frac{1}{2}}$. Inverting ${(1+t)}^{2/n}>1+{\displaystyle \frac{t}{n-1}}$ gives (i).

(iii) The claim is ${\left(1+{\displaystyle \frac{1}{n-1}}\right)}^n<4$. For $n\ge 4$, ${\left(1+{\displaystyle \frac{1}{n-1}}\right)}^{n-1}<e$ and $1+{\displaystyle \frac{1}{n-1}}\le {\displaystyle \frac{4}{3}}$, so the product is $<{\displaystyle \frac{4}{3}}e<3.63<4$. For $n=3$, ${(3/2)}^3=3.375<4$.

(iv) is equivalent to $n<2n-2$, i.e. $n>2$.    □

6. Symmetric Members of ${\mathcal{C}}_n$

Each $f\in {\mathcal{C}}_n$ begins like an n-fold symmetric function, $f=z+a_{n+1}{z}^{n+1}+\dots$, and differs from one first at the coefficients $a_{n+2},\dots ,a_{2n}$ (Theorem 5). Which members are actually symmetric is answered by the parametrization of Proposition 1. Here ${\epsilon }_m=e^{2\pi i/m}$.

Proposition 3

(Symmetric members and $\eta$). Let $f\in {\mathcal{C}}_n$, $\eta =\mathsf{\Phi }\left(f\right)={\sum }_{j\ge 0}{\eta }_j{z}^j$ and $m\ge 1$. Then $f\left({\epsilon }_{m}z\right)={\epsilon }_{m}f\left(z\right)$ for all $z\in \mathbb{U}$ if and only if ${\eta }_j=0$ unless $m\mid n+j$. In particular, f is n-fold symmetric if and only if $\eta \left(z\right)=g\left(z^n\right)$ with g analytic in $\mathbb{U}$, ${\parallel g\parallel }_{\infty }={\parallel \eta \parallel }_{\infty }$; and $g\mapsto {\mathsf{\Phi }}^{-1}\left(g\left(z^n\right)\right)$ is a bijection of the closed ball $B_n$ of radius $M_n$ in $H^{\infty }$ onto ${\mathcal{C}}_n\cap {\mathcal{S}}_n$.

Proof. 

Since $u_f=z/f$, $f\left(\epsilon z\right)=\epsilon f\left(z\right)\iff u_f\left(\epsilon z\right)=u_f\left(z\right)$. By Lemma 1 (with ${\mu }_1=\dots ={\mu }_{n-1}=0$) and Lemma 3, $u_f=1+{\sum }_{j\ge 0}{\delta }_j{z}^{n+j}$ with ${\delta }_j={\eta }_{j}j!/(n+j)!$, so ${\delta }_j=0\iff {\eta }_j=0$. Comparing coefficients, $u_f\left(\epsilon z\right)=u_f\left(z\right)$ iff ${\delta }_j({\epsilon }^{n+j}-1)=0$ for all j, which is the stated condition. For $m=n$ it reads ${\eta }_j=0$ unless $n\mid j$, i.e. $\eta \left(z\right)=g\left(z^n\right)$ with $g\left(\zeta \right)={\sum }_q{\eta }_{nq}{\zeta }^q$, and $g\left(\zeta \right)=\eta \left(z\right)$ whenever $z^n=\zeta$, so ${\parallel g\parallel }_{\infty }={\parallel \eta \parallel }_{\infty }$. Conversely, for $g\in B_n$ the function $\eta =g\left(z^n\right)$ lies in $B_n$, so $f={\mathsf{\Phi }}^{-1}\left(\eta \right)\in {\mathcal{C}}_n$ by Proposition 1; it is n-fold symmetric by the first part and univalent by Remark 1, hence $f\in {\mathcal{S}}_n$. Injectivity comes from that of $\mathsf{\Phi }$.    □

The extremal functions are symmetric: $\mathsf{\Phi }\left(E_j\right)=M_n{z}^j$, so $E_j=z/(1+c{z}^{n+j})$ is $(n+j)$-fold symmetric, and among them only $E_0$ is n-fold symmetric. Since $E_0$ attains the bounds of Theorems 2, 3 and 5(b) and the covering radius, these remain sharp on ${\mathcal{C}}_n\cap {\mathcal{S}}_n$.

Theorem 8

(The pair $(a_{n+1},a_{2n+1})$ on ${\mathcal{C}}_n\cap {\mathcal{S}}_n$). Let $n\ge 3$. For $f\in {\mathcal{C}}_n\cap {\mathcal{S}}_n$,

$$|a_{n+1}|\le {\displaystyle \frac{1}{n-1}},|a_{2n+1}-a_{n+1}^2|\le {\displaystyle \frac{1-{(n-1)}^2{\left|a_{n+1}\right|}^2}{(n-1)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}},$$

(25)

and every pair $(\alpha ,\beta )$ with $\left|\alpha \right|\le {\displaystyle \frac{1}{n-1}}$ and $|\beta -{\alpha }^2|\le {\displaystyle \frac{1-{(n-1)}^2{\left|\alpha \right|}^2}{(n-1)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}}$ is attained by some $f\in {\mathcal{C}}_n\cap {\mathcal{S}}_n$. For $|a_{n+1}|<{\displaystyle \frac{1}{n-1}}$, equality in the second relation holds exactly for $\eta \left(z\right)=M_n{\displaystyle \frac{c+\kappa z^n}{1+\overline{c}\kappa z^n}}$, $\left|c\right|<1$, $\left|\kappa \right|=1$.

Proof. 

By Proposition 3, $\eta \left(z\right)=M_n\phi \left(z^n\right)$ with $\phi ={\sum }_q{\phi }_q{\zeta }^q$ analytic and $\left|\phi \right|\le 1$ in $\mathbb{U}$; so ${\eta }_0=M_n{\phi }_0$, ${\eta }_n=M_n{\phi }_1$. By (9) and (10),

$$a_{n+1}=-{\displaystyle \frac{{\eta }_0}{n!}}=-{\displaystyle \frac{{\phi }_0}{n-1}},a_{2n+1}-a_{n+1}^2=-{\eta }_n{\displaystyle \frac{n!}{\left(2n\right)!}}=-{\displaystyle \frac{{\phi }_1}{(n-1)\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}},$$

since $M_{n}n!/\left(2n\right)!=1/\left((n-1)\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)\right)$. Now (25) follows from $|{\phi }_0|\le 1$ and (21), $|{\phi }_1|\le 1-|{\phi }_0{|}^2$. Conversely, given $(\alpha ,\beta )$ put $c=-(n-1)\alpha$ and choose $\phi$ with ${\phi }_0=c$ and ${\phi }_1=-(n-1)\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)(\beta -{\alpha }^2)$ as in the sufficiency part of Theorem 5(a); then $f={\mathsf{\Phi }}^{-1}\left(M_n\phi \left(z^n\right)\right)$ is in ${\mathcal{C}}_n\cap {\mathcal{S}}_n$ by Proposition 3 and has the given pair. Equality in (21) with $|{\phi }_0|<1$ holds iff $\phi$ is a Möbius automorphism of $\mathbb{U}$.    □

For fixed $\alpha$ the admissible $\beta$ fill a disk centered at ${\alpha }^2$; its largest modulus over the region is ${\displaystyle \frac{1}{{(n-1)}^2}}$, reached at $\alpha =\pm {\displaystyle \frac{1}{n-1}}$, $\beta ={\alpha }^2$ by $E_0$ and its rotations, so Theorem 5(b) is sharp already on ${\mathcal{C}}_n\cap {\mathcal{S}}_n$.

Remark 7.

If $f_1,f_2\in {\mathcal{C}}_n$ are n-fold symmetric, so are $F=f_1{f}_2/z$ and its dilations $F_s$. The extremal pair $f_1=f_2=E_0$ of Theorem 7 is of this kind, hence the radius $s_n$ is exact already for products of n-fold symmetric members of ${\mathcal{C}}_n$, and for $s>s_n$ the derivative of the symmetric function $F_s$ vanishes at the n points $s_n{\epsilon }_n^k/s\in \mathbb{U}$.

7. Conclusions

We studied the subclass ${\mathcal{C}}_n$ of the Al-Refai–Darus class (2), in which the whole allowance of the criterion is spent on the nth derivative of $z/f$. The central fact is that $f\mapsto {(z/f)}^{\left(n\right)}$ is a bijection of ${\mathcal{C}}_n$ onto the closed ball of radius $n!/(n-1)$ in $H^{\infty }$ (Proposition 1), which rests on the integral representation of $z/f$ and the estimates for its remainder (Lemmas 1–4). It gives sharp growth, distortion and covering-radius estimates, with the covering radius $(n-1)/n$ (Theorems 2, 3, Corollary 1); sharp bounds for $a_{n+1},\dots ,a_{2n}$ and ${\ell }_n,\dots ,{\ell }_{2n-1}$ (Theorem 4, Corollary 2); the exact region of $(a_{n+1},a_{n+2})$, with extremal functions written in closed form (Theorem 5, Remark 3); and a product theorem whose radius $s_n$ is exact both for Aksentev’s inequality and for univalence (Theorem 7). The majorant of Theorem 1 for Aksentev’s quantity holds on the whole class (2); its upper bound ${\left|z\right|}^2$ is the Schwarz-lemma bound valid for all of Aksentev’s class, and the new information is the intermediate bound with ${\sigma }_f$ and ${\mathrm{\Lambda }}_f$. Theorem 6 shows what is lost when ${\mu }_2,\dots ,{\mu }_{n-1}$ are allowed to be non-zero: the estimates of ${\mathcal{C}}_n$ degrade to those with $r^2$ in place of $r^n/(n-1)$.

As for symmetry, the extremal functions $z/(1+c{z}^{n+j})$ are $(n+j)$-fold symmetric, the n-fold symmetric members of ${\mathcal{C}}_n$ are in bijection with the same ball of $H^{\infty }$ through $\eta \left(z\right)=g\left(z^n\right)$, and the exact region of $(a_{n+1},a_{2n+1})$ on them is (25) (Proposition 3, Theorem 8). The comparison with n-fold symmetric univalent functions (Proposition 2) shows that every extremal value for ${\mathcal{C}}_n$ in Table 4 lies strictly inside the corresponding value for ${\mathcal{S}}_n$; since the classes are not nested, this records a gap and is not an improvement of a known result.