1 Introduction

Let \({\mathcal {S}}\) denote the class of all functions f that are analytic and univalent in the open unit disk \(U=\{z \in {\mathbb {C}}: |z| < 1\}\) and normalized by \(f(0) = f'(0) - 1 = 0\).

A set \(D \subset {\mathbb {C}}\) is said to be starlike with respect to \(w_0\), an interior point of D, if the intersection of each half-line beginning at \(w_0\) with the interior of D is connected.

We denote by \({{\mathcal {S}}}{{\mathcal {T}}}\) the class of all starlike functions, i.e., the subclass of \({\mathcal {S}}\) consisting of functions that map U onto domains starlike with respect to \(w_0=0\) (briefly starlike domains). Recall that a function \(f \in {\mathcal {S}}\) is starlike if and only if

$$\begin{aligned} \textrm{Re}\ \frac{zf'(z)}{f(z)} > 0 \ \ \textrm{for} \ \ \textrm{all} \ z \in U. \end{aligned}$$

Let \({{\mathcal {C}}}{{\mathcal {V}}}\) denote the class of all functions \(f \in {\mathcal {S}}\) that are convex in U, i.e., such that f(U) is a convex domain.

Let \(\gamma : z=z(t), t \in [a,b],\) be a smooth, directed arc and suppose that a function f is analytic on \(\gamma \). Then the arc \(f(\gamma )\) is said to be

- starlike with respect to \(w_0 \notin f(\gamma )\) if \(\textrm{arg}(f(z(t))-w_0)\) is a nondecreasing function of t,

- convex if the argument of the tangent to \(f(\gamma )\) is a nondecreasing function of t.

In 1991, Goodman ([1, 2]) introduced geometrically defined classes \(\mathcal {UCV}\) and \(\mathcal {UST}\) of uniformly convex and uniformly starlike functions, respectively.

Recall that a function \(f \in {\mathcal {S}}\) is in the class \(\mathcal {UCV}\) (\(\mathcal {UST}\)) if for every circular arc \(\gamma \subset U\) with center \(\zeta \in U\), the arc \(f(\gamma )\) is convex (starlike with respect to \(f(\zeta )\)).

If we take \(\zeta \) such that \(|\zeta | \le k, 0 \le k \le 1\) we obtain the following natural extension of the concept of uniform starlikeness (see [11] and also [10, 12]). Namely, a function \(f \in {\mathcal {S}}\) is said to be k-uniformly starlike in U, if the image of every circular arc \(\gamma \) contained in U with center at \(\zeta \), where \(|\zeta | \le k\), is starlike with respect to \(f(\zeta )\).

We denote by \(k \text{- } \mathcal {UST}\) the class of all k-uniformly starlike functions. Notice that \({0 \text{- } \mathcal {UST}}={{\mathcal {S}}}{{\mathcal {T}}}\) and \({1 \text{- } \mathcal {UST}}=\mathcal {UST}.\) Moreover, it is clear that \(\mathcal {UST} \subset k \text{- } \mathcal {UST}\subset {{\mathcal {S}}}{{\mathcal {T}}}\) for every \(k \in [0,1]\).

Goodman obtained the analytic conditions for \(\mathcal {UCV}\) and \(\mathcal {UST}\) expressed by two complex variables. For an arbitrary k such that \(0 \le k \le 1\), the class \(k \text{- } \mathcal {UST}\) can be characterized (see [11]) as follows

$$\begin{aligned} f\in k \text{- } \mathcal {UST}\Leftrightarrow \textrm{Re}\ \frac{f(z)-f(\zeta )}{(z-\zeta )f'(z)} \ge 0\ \ \textrm{for} \ \ \textrm{all} \ z \in U, |\zeta | \le k. \end{aligned}$$

For \(k=1\), we get Goodman’s condition for uniform starlikeness.

It turned out that (see [3, 7])

$$\begin{aligned} f \in \mathcal {UCV} \Leftrightarrow \textrm{Re}\left\{ 1+\frac{zf''(z)}{f'(z)}\right\} > \left| \frac{zf''(z)}{f'(z)}\right| ,\ \ z \in U. \end{aligned}$$

Finding this analytic condition essentially simplified further investigations of uniformly convex functions ([3, 4, 7, 8]).

It is more difficult to investigate the class \(\mathcal {UST}\) (\(k \text{- } \mathcal {UST}\)) because of its characterization in terms of two complex variables. In particular checking whether a function belongs to the class \(\mathcal {UST}\) leads to very complicated computations. Hence only simple examples of uniformly starlike functions are known.

In this paper, we give many examples of members of the class of uniformly starlike functions \(\mathcal {UST}\) and we establish its connections with some subclasses of convex functions. For instance, we get that all functions convex of order 3/4 are uniformly starlike.

2 Some Members of the Class of Uniformly Starlike Functions

Goodman ([2]) gave some examples of uniformly starlike functions. He proved that

$$\begin{aligned} \frac{z}{1-Az} \in \mathcal {UST} \Leftrightarrow |A| \le \frac{1}{\sqrt{2}} \end{aligned}$$

and

$$\begin{aligned} \left( |A|\le \frac{\sqrt{2}}{2n}, n > 1\right) \Rightarrow z + Az^n \in \mathcal {UST}. \end{aligned}$$

It was proven in [11] that

$$\begin{aligned} f(z) = z + Az^2 \in k \text{- } \mathcal {UST}\Leftrightarrow |A| \le \sqrt{3}/\sqrt{4(k^2+3)}. \end{aligned}$$

For \(k=1\) we get

$$\begin{aligned} f(z) = z + Az^2 \in \mathcal {UST} \Leftrightarrow |A| \le \sqrt{3}/{4}. \end{aligned}$$

This result was mentioned by Goodman, but without a proof.

We have also more general result (see [11]):

If \(0 < k \le 1\) and for some integer \(n \ge 2\)

$$\begin{aligned} |A| \le \frac{1}{n}\,\sqrt{\frac{n+1}{n+1+(n-1)k^2}}, \end{aligned}$$

then \(f(z) = z + Az^n \in k \text{- } \mathcal {UST}.\)

For \(k=1\) we get the result of Merkes and Salmassi ([5]), which improves the bound \(|A| \le 1/(\sqrt{2}n)\) obtained by Goodman.

Merkes and Salmassi ([5]) proved the following result.

Theorem 2.1

Let \(f \in {\mathcal {S}}\). If for all \(z \in U\), \(w \in U\)

$$\begin{aligned} \textrm{Re}\,\frac{f'(w)}{f'(z)} \ge 0, \end{aligned}$$

then \(f \in \mathcal {UST}.\)

Using this, we get the following sufficient condition for uniform starlikeness

Theorem 2.2

Let \(f(z) = z + \sum _{n=2}^{\infty }\,a_nz^n\), \(z \in U\). If

$$\begin{aligned} |\textrm{Arg}{f'(z)}| \le \frac{\pi }{4}\ \ \ for\ \ z \in U, \end{aligned}$$

then \(f \in \mathcal {UST}.\)

Proof

Let f be analytic in U and normalized by \(f(0)=f'(0)-1=0\). Assume that \(|\textrm{Arg}{f'(z)}| \le \pi /4, z \in U.\) This condition implies that \(\textrm{Re}f'(z) > 0\) for \(z \in U\), hence \(f \in {\mathcal {S}}\). Moreover if \(z \in U\), \(w \in U\) then

$$\begin{aligned} \left| \textrm{Arg}\frac{f'(w)}{f'(z)}\right| = |\textrm{Arg}f'(w) - \textrm{Arg}f'(z)| \le |\textrm{Arg}f'(w)| + |\textrm{Arg}f'(z)| \le \frac{\pi }{2}. \end{aligned}$$

This is equivalent to

$$\begin{aligned} \textrm{Re}\,\frac{f'(w)}{f'(z)} \ge 0\ \ \textrm{for} \ \ \textrm{all} \ z,w \in U. \end{aligned}$$

Thus by Theorem 2.1, \(f \in \mathcal {UST}\).

This result generates many examples of uniformly starlike functions. \(\square \)

Example 2.1

Let

$$\begin{aligned} P_1(z) = \sqrt{\frac{1+z}{1-z}}, \ \ z \in U. \end{aligned}$$

Then \(P_1(U) = \{w \in {\mathbb {C}}: |\textrm{Arg}\, w| < \pi /4\}.\)

If \(f_1'(z) = P_1(z)\), then in view of Theorem 2.2, \(f_1 \in \mathcal {UST}\). Thus the function

$$\begin{aligned} f_1(z) = 2\arctan \sqrt{\frac{1+z}{1-z}} - \sqrt{1-z^2} +1 -\frac{\pi }{2}, \ \ z \in U, \end{aligned}$$

is uniformly starlike in U.

Example 2.2

Let

$$\begin{aligned} P_2(z) = 1 + \frac{2}{\pi ^2} \left( \log \frac{1+\sqrt{z}}{1-\sqrt{z}}\right) ^2,\ \ z \in U. \end{aligned}$$

This function plays an important role in the class \(\mathcal {UCV}\) of uniformly convex functions. It is known (see [3, 7]) that \(f \in \mathcal {UCV}\) if and only if

$$\begin{aligned} 1 + \frac{zf''(z)}{f'(z)} \prec P_2(z) \ \ \textrm{in} \ \ U. \end{aligned}$$

The image of the unit disk U under \(P_2\) is bounded by the parabola

$$\begin{aligned} \{w=u+iv: v^2=2u-1\}. \end{aligned}$$

Moreover \(|\textrm{Arg}P_2(z)| < \pi /4\) for z in U. Thus \(f'_2(z) = P_2(z)\) for \( z \in U,\) implies \(f _2 \in \mathcal {UST}\).

Using standard methods of integration after some calculations we get that

$$\begin{aligned} f_2(z)= & {} z + \frac{2}{\pi ^2}\left[ (z-1)\left( \log \frac{1+\sqrt{z}}{1-\sqrt{z}}\right) ^2 +4\sqrt{z}\left( \log \frac{1+\sqrt{z}}{1-\sqrt{z}}\right) +4\log (1-z)\right] , \\ {}{} & {} z \in U, \end{aligned}$$

belongs to the class \(\mathcal {UST}\).

Example 2.3

Let

$$\begin{aligned} P_3(z) = \frac{1+Az}{1-Az},\ \ z \in U, \end{aligned}$$

where \(0 < A \le \sqrt{2}-1.\)

Then \(P_3\) maps U onto the disk

$$\begin{aligned} \left\{ w \in {\mathbb {C}}: \left| w - \frac{1+A^2}{1-A^2}\right| < \frac{2A}{1-A^2}\right\} . \end{aligned}$$

For \(0 < A \le \sqrt{2}-1\) the disk \(P_3(U)\) is contained in the region \(\{w \in {\mathbb {C}}: |\textrm{Arg}\, w| < \pi /4\}.\) Hence if \(f_3'(z)=P_3(z)\) then \(f_3 \in \mathcal {UST}\), since \(|\textrm{Arg}f_3'(z)| < \pi /4\) for \(z \in U.\)

Thus the function

$$\begin{aligned} f_3(z)=-z-\frac{2}{A}\log (1-Az),\ z \in U\ (0 < A \le \sqrt{2}-1), \end{aligned}$$

is in the class \(\mathcal {UST}\).

Example 2.4

Let

$$\begin{aligned} P_4(z) = \frac{1}{\sqrt{1-z}},\ \ z \in U. \end{aligned}$$

This function maps the unit disk onto the domain bounded by the hyperbola

$$\begin{aligned} \{w=u+iv: u^2-v^2=1/2, u>0\} \end{aligned}$$

and clearly \(P_4(U) \subset \{w \in {\mathbb {C}}: |\textrm{Arg}\, w| < \pi /4\}.\) Thus

$$\begin{aligned} f_4(z)= 2-2\sqrt{1-z}= z + \frac{1\cdot 1}{1\cdot 4}z^2 +\frac{1\cdot 1\cdot 3}{1\cdot 4\cdot 6}z^3+ \frac{1\cdot 1\cdot 3\cdot 5}{1\cdot 4\cdot 6\cdot 8}z^4+\cdots \end{aligned}$$

is uniformly starlike in U, because of \(f_4'(z)=P_4(z)\) for \( z \in U.\)

Example 2.5

Let

$$\begin{aligned} P_5(z)=\sqrt{1+z},\ \ z \in U. \end{aligned}$$

The image of the unit disk U under \(P_5\) is bounded by the lemniscate

$$\begin{aligned} \{w=u+iv: (u^2+v^2)^2-2(u^2-v^2)=0, u>0\} \end{aligned}$$

and is contained in the region \(\{w \in {\mathbb {C}}: |\textrm{Arg}\, w| < \pi /4\}.\) Hence the function

$$\begin{aligned} f_5(z)=\frac{2}{3}\left[ (1+z)^{3/2}-1\right] =z + \frac{1}{4}z^2-\frac{1\cdot 1}{4\cdot 6}z^3+ \frac{1\cdot 1\cdot 3}{4\cdot 6\cdot 8}z^4-\dots , \ \ z \in U, \end{aligned}$$

belongs to the class \(\mathcal {UST}\), since \(f_5'(z)=P_5(z)\) for \(z \in U.\)

3 Uniformly Starlike Functions and Subclasses of Convex Functions

Let \(\varphi \) be an analytic univalent function which satisfies the following conditions: \(\textrm{Re}\,\varphi (z) > 0\) for \(z \in U\), \(\varphi (0)=1, \varphi '(0) > 0\) and \(\varphi (U)\) is convex and symmetric with respect to the real axis.

By \({{\mathcal {C}}}{{\mathcal {V}}}(\varphi )\), we denote the subclass of convex functions defined by

$$\begin{aligned} 1+\frac{zf''(z)}{f'(z)} \prec \varphi (z) \ \ \textrm{in}\ \ U. \end{aligned}$$

In particular

$$\begin{aligned} {{\mathcal {C}}}{{\mathcal {V}}}={{\mathcal {C}}}{{\mathcal {V}}}(\varphi _0), \ \ \textrm{where}\ \ \varphi _0(z)=\frac{1+z}{1-z}, \\ \mathcal {UCV}={{\mathcal {C}}}{{\mathcal {V}}}(\varphi _1), \ \ \textrm{where}\ \ \varphi _1(z)=1 + \frac{2}{\pi ^2} \left( \log \frac{1+\sqrt{z}}{1-\sqrt{z}}\right) ^2, \end{aligned}$$

and for

$$\begin{aligned} \varphi _2(z)=\frac{1+(1-2\alpha )z}{1-z},\ \ \ 0 \le \alpha < 1, \end{aligned}$$

the class \({{\mathcal {C}}}{{\mathcal {V}}}(\varphi _2)\) coincides with the class \({{\mathcal {C}}}{{\mathcal {V}}}(\alpha )\) consisting of all functions that are convex of order \(\alpha \), thus

$$\begin{aligned} {{\mathcal {C}}}{{\mathcal {V}}}(\varphi _2)={{\mathcal {C}}} {{\mathcal {V}}}(\alpha )=\left\{ f \in {\mathcal {S}}: \textrm{Re}\left( 1+\frac{zf''(z)}{f'(z)}\right) > \alpha ,\ z \in U\right\} . \end{aligned}$$

We need the following result of Suffridge [9]

Theorem 3.1

Let h be starlike in U with \(h(0)=0\) and let \(p(z)=a+a_nz^n + a_{n+1}z^{n+1}+ \dots \) be analytic in U. If

$$\begin{aligned} zp'(z) \prec h(z), \end{aligned}$$

then

$$\begin{aligned} p(z) \prec q(z)= a+ n^{-1}\int \limits _{0}^{z} h(t)t^{-1}\,{\text {d}}t. \end{aligned}$$

The function q is convex and is the best (a, n)-dominant.

Corollary 3.1

(Corollary 3.1d.1 in [6]) Let h be starlike in U, with \(h(0)=0\) and let \(p(z)=a+a_nz^n + a_{n+1}z^{n+1}+ \dots \) be analytic in U with \(a\ne 0\). If

$$\begin{aligned} \frac{zp'(z)}{p(z)} \prec h(z), \end{aligned}$$

then

$$\begin{aligned} p(z) \prec q(z)= a\,\exp \left[ n^{-1}\int \limits _{0}^{z} h(t)t^{-1}\,{\text {d}}t\right] \end{aligned}$$

and q is the best (a, n)-dominant.

Theorem 3.2

Let \(f \in {{\mathcal {C}}}{{\mathcal {V}}}(\varphi )\). If

$$\begin{aligned} \left| \textrm{Arg}\left\{ \exp \int \limits _{0}^{z}\frac{\varphi (t)-1}{t}{\text {d}}t\right\} \right| \le \frac{\pi }{4}\ \ \ for\ \ z \in U, \end{aligned}$$

then \(f \in \mathcal {UST}.\)

Proof

If \(f \in {{\mathcal {C}}}{{\mathcal {V}}}(\varphi )\) then

$$\begin{aligned} 1+\frac{zf''(z)}{f'(z)} \prec \varphi (z) \ \ \textrm{in}\ \ U \end{aligned}$$

or

$$\begin{aligned} z(\log f'(z))' \prec \varphi (z)-1 \ \ \textrm{in}\ \ U. \end{aligned}$$

Since \(\varphi (z)-1\) is convex, therefore is starlike, we get by Theorem 3.1

$$\begin{aligned} \log f'(z) \prec \int \limits _{0}^{z}\frac{\varphi (t)-1}{t}{\text {d}}t\ \ \textrm{in}\ \ U. \end{aligned}$$

From Corollary 3.1, we may also conclude that

$$\begin{aligned} f'(z) \prec \exp \int \limits _{0}^{z}\frac{\varphi (t)-1}{t}{\text {d}}t\ \ \textrm{in}\ \ U. \end{aligned}$$

Thus from Theorem 2.2, if

$$\begin{aligned} \left| \textrm{Arg}\left\{ \exp \int \limits _{0}^{z}\frac{\varphi (t)-1}{t}{\text {d}}t\right\} \right| \le \frac{\pi }{4}\ \ \ \textrm{for}\ \ z \in U, \end{aligned}$$

then \(f \in \mathcal {UST}.\) \(\square \)

Corollary 3.2

$$\begin{aligned} {{\mathcal {C}}}{{\mathcal {V}}}(\varphi ) \subset \mathcal {UST}, \end{aligned}$$

provided that

$$\begin{aligned} \left| \textrm{Im}\int \limits _{0}^{z}\frac{\varphi (t)-1}{t}{\text {d}}t\right| \le \frac{\pi }{4},\ \ z \in U. \end{aligned}$$

Theorem 3.3

$$\begin{aligned} {{\mathcal {C}}}{{\mathcal {V}}}(3/4) \subset \mathcal {UST}. \end{aligned}$$

Proof

Let \(f \in {{\mathcal {C}}}{{\mathcal {V}}}(\alpha ), \, 0 \le \alpha < 1\). Then from the subordination

$$\begin{aligned} 1+\frac{zf''(z)}{f'(z)} \prec \frac{1+(1-2\alpha )z}{1-z} \ \ \textrm{in}\ \ U \end{aligned}$$

we get

$$\begin{aligned} f'(z) \prec \frac{1}{(1-z)^{2(1-\alpha )}} \ \ \textrm{in}\ \ U. \end{aligned}$$

We have

$$\begin{aligned} \left| \textrm{Arg}\frac{1}{(1-z)^{2(1-\alpha )}}\right| \ = 2(1-\alpha )|\textrm{Arg}(1-z)| \le 2(1-\alpha )\frac{\pi }{2}. \end{aligned}$$

Hence for \(3/4 \le \alpha < 1\) the function \(f'\) satisfies the condition

$$\begin{aligned} |\textrm{Arg}\, f'(z)| \le \sup _{|z|=1}\left| \textrm{Arg}\frac{1}{(1-z)^{2(1-\alpha )}}\right| \le \pi /4, \ \ z \in U. \end{aligned}$$

By Theorem 3.2 we get

$$\begin{aligned} {{\mathcal {C}}}{{\mathcal {V}}}(\alpha ) \subset {{\mathcal {C}}}{{\mathcal {V}}}(3/4) \subset \mathcal {UST}\ \ \textrm{for} \ \ 3/4 \le \alpha < 1. \end{aligned}$$

\(\square \)

Theorem 3.4

$$\begin{aligned} {{\mathcal {C}}}{{\mathcal {V}}}(\varphi _3) \subset \mathcal {UST}, \end{aligned}$$

where

$$\begin{aligned} \varphi _3(z)=1+Az\ \ for\ \ z \in U \ \ and\ \ 0< A \le \pi /4. \end{aligned}$$

Proof

If \(f \in {{\mathcal {C}}}{{\mathcal {V}}}(\varphi _3)\) then

$$\begin{aligned} 1+\frac{zf''(z)}{f'(z)} \prec 1 + Az \ \ \textrm{in}\ \ U. \end{aligned}$$

Hence

$$\begin{aligned} f'(z) \prec \exp (Az) \ \ \textrm{in}\ \ U \end{aligned}$$

and since

$$\begin{aligned} \left| \textrm{Arg}\left[ \exp \left( Ae^{i\theta }\right) \right] \right| = |A\sin \theta | \le A \le \frac{\pi }{4} \end{aligned}$$

we get

$$\begin{aligned} |\textrm{Arg}f'(z)| \le \frac{\pi }{4}. \end{aligned}$$

Thus by Theorem 3.2, f is uniformly starlike in U.

Theorem 3.4 can be written in an equivalent form as follows: \(\square \)

Corollary 3.3

Let \(f \in {\mathcal {S}}\). If

$$\begin{aligned} \left| \frac{zf''(z)}{f'(z)}\right| \le \frac{\pi }{4},\ \ z \in U, \end{aligned}$$

then \(f \in \mathcal {UST}.\)

Proof

Let \(f \in {\mathcal {S}}\). Then for \(z \in U\)

$$\begin{aligned} \left| \frac{zf''(z)}{f'(z)}\right| \le \frac{\pi }{4} \Rightarrow \frac{zf''(z)}{f'(z)} \prec \frac{\pi }{4}z \ \ \textrm{in}\ \ U. \end{aligned}$$

Hence \(f \in {{\mathcal {C}}}{{\mathcal {V}}}(\varphi _3)\) with \(A=\pi /4\), so by Theorem 3.4 the result follows. \(\square \)