1 Introduction

A complex-valued harmonic function f that is harmonic in a simply connected domain \(\Omega \subset {\mathbb {C}}\) has the canonical representation

$$\begin{aligned} f = h + \overline{g}, \end{aligned}$$
(1.1)

where h and g are analytic in \(\Omega \) with \(g(z_0)=0\) for some prescribed point \(z_0 \in \Omega \). According to a theorem of Lewy [17], f is locally univalent, if and only if its Jacobian \(J_f(z) = |f_z(z)|^2-|f_{\bar{z}}(z)|^2 =|h^\prime (z)|^2-|g^\prime (z)|^2\) does not vanish, and is sense-preserving if the Jacobian is positive. Then \(h^\prime (z) \ne 0\) and the analytic function \(\omega =g^\prime /h^\prime \), called the second complex dilatation of f, has the property \(|\omega |<1\) in \(\Omega \). Throughout this paper, we will assume that f is locally univalent and sense-preserving, and we call f a harmonic mapping. Also, we assume \(\Omega = {\mathbb {D}}\subset {\mathbb {C}}\), and \(z_0 = 0\), where \({\mathbb {D}}\) is the open unit disk on the complex plane. The class of all sense-preserving univalent harmonic mappings of \({\mathbb {D}}\) with \(h(0)=g(0) = h^\prime (0) -1=0\) is denoted by \({\mathcal {S}}_{\mathcal {H}}\), and its subclass for \(g^\prime (0) = 0\) by \({\mathcal {S}}^0_{\mathcal {H}}\) (cf. [8]). Fundamental informations about harmonic mappings in the plane can be found in [11]. Note that each f satisfying (1.1) in \({\mathbb {D}}\) is uniquely determined by coefficients of the following power series expansions

$$\begin{aligned} h(z) = \sum _{n=0}^{\infty } a_n z^n, \quad g(z) = \sum _{n=1}^{\infty } b_n z^n \quad (z\in {\mathbb {D}}), \end{aligned}$$
(1.2)

with \(a_n \in {\mathbb {C}}, n=0,1,2,...\), and \(b_n \in {\mathbb {C}}\), \(n=1,2,3,...\). When \(f\in {\mathcal {S}}_{\mathcal {H}}\), then \(a_0 = 0, a_1 = 1\).

In [14], the authors studied the properties of a subclass \(\overline{{\mathcal {S}}}^\alpha _{\mathcal {H}}\) of \({\mathcal {S}}_{\mathcal {H}}\), consisting of all univalent anti-analytic perturbations of the identity in the unit disk with \(|b_1|=\alpha \), and in [15], the authors studied the class \(\widehat{{\mathcal {S}}}^{\alpha }\) of all \(f \in {\mathcal {S}}_{\mathcal {H}}\), such that \(|b_1|=\alpha \in (0,1)\) and \(h\in \mathcal {CV}\), where \(\mathcal {CV}\) denotes the well-known family of normalized, univalent functions which are convex.

The classical Schwarz–Pick estimate for an analytic function \(\omega \) which is bounded by one on the unit disk of the complex plane is the inequality

$$\begin{aligned} |\omega ^\prime (z)| \le \frac{1-|\omega (z)|^2}{1-|z|^2} \quad (|z| <1). \end{aligned}$$
(1.3)

Ruscheweyh [21] has obtained the best-possible estimates of higher order derivatives of bounded analytic functions on the disk. Similar estimates were derived by other methods and for different classes of analytic functions in one and several variables by Anderson and Rovnyak [1]

$$\begin{aligned} (1-|z|^2)^{n-1}\left| \frac{\omega ^{(n)}(z)}{n!}\right| \le \frac{1-|\omega (z)|^2}{1-|z|^2}\quad (n=1,2,...). \end{aligned}$$
(1.4)

The case \(z=0\) in (1.4) asserts that if

$$\begin{aligned} \omega (z) = c_0+c_1 z+c_2 z^2 +\cdots , \end{aligned}$$
(1.5)

then

$$\begin{aligned} |c_n| \le 1-|c_0|^2, \end{aligned}$$
(1.6)

for every \(n\ge 1\). This result is classical and due to Wiener; see [2, 16].

2 Bounds of the Fekete–Szegö and Other Functionals

Theorem 2.1

Let \(f\in \widehat{S}^{\alpha }, f=h+\bar{g}\) with the power series (1.2). Then

$$\begin{aligned} |b_n|\le \alpha + \frac{(1-\alpha ^2)(n-1)}{2}\quad (n=2,3,...). \end{aligned}$$
(2.1)

Proof

Making use of a relation \(g^\prime =\omega h^\prime \) and the power series expansions (1.2) and (1.5), we obtain

$$\begin{aligned} nb_n = \sum _{p=0}^{n-1}(p+1)a_{p+1}c_{n-p-1}\quad (n=2,3,...). \end{aligned}$$
(2.2)

Since \(h\in {\mathcal {CV}}\), \(|a_k|\le 1 \ (k=1,2,...)\). Applying this for (2.2), we have

$$\begin{aligned} |b_n| \ \le \ \dfrac{1}{n}\sum \limits _{p=0}^{n-1}(p+1)|c_{n-p-1}|. \end{aligned}$$

The fact \(g^\prime =\omega h^\prime \), for the case \(z=0\), implies that \(c_0=b_1\), so that by (1.6), we obtain \(|c_{n-p-1}| \le 1-|b_1|^2 = 1-\alpha ^2\). Therefore,

$$\begin{aligned} \begin{array}{rcl} |b_n| &{}\le &{}\alpha + \dfrac{1}{n}\sum \limits _{p=0}^{n-2}(p+1)(1-\alpha ^2)\\ &{}=&{}\alpha +\dfrac{(1-\alpha ^2)(n-1)}{2}.\end{array} \end{aligned}$$
(2.3)

Specially, we get

$$\begin{aligned} |b_2| \le \alpha +\frac{1-\alpha ^2}{2},\quad |b_3|\le 1+\alpha -\alpha ^2. \end{aligned}$$

For the case \(n=2\), the inequality is sharp, with the equality realized by the function

$$\begin{aligned} f(z) =\frac{z}{1-z} +\overline{\frac{z}{1-z}-\frac{1-\alpha }{1+\alpha }\log \frac{1+\alpha z}{1-z}}. \end{aligned}$$

We note that for \(\alpha \) close to 1, the above bounds are better than that obtained in [15]. \(\square \)

In conclusion, we obtain

Corollary 2.2

Let \(f\in \widehat{S}^{\alpha }, f=h+\bar{g}\) with the power series (1.2). Then

$$\begin{aligned} |b_n|\le \min \left\{ \alpha + \frac{(1-\alpha ^2)(n-1)}{2}, \frac{\alpha +\sqrt{(n-\alpha ^2)(n-1)}}{n}\right\} \quad (n=2,3,...). \end{aligned}$$
(2.4)

Theorem 2.3

Let \(f\in \widehat{S}^{\alpha }, f=h+\bar{g}\) with the power series (1.2). Then for \(\mu \in \mathbb {R}\)

$$\begin{aligned} |b_{3} - \mu b_2^2| \le \frac{1-\alpha ^2}{3}\left\{ 1+\frac{3}{4}|\mu |(1-\alpha ^2)+\left| 2-3\mu b_1\right| \right\} +\alpha \max \left\{ \frac{1}{3},|1-\mu b_1|\right\} , \end{aligned}$$
(2.5)

and

$$\begin{aligned} |b_{n+1} - b_n| \le 2\alpha +\big (1-\alpha ^2\big )\frac{2n-1}{2}. \end{aligned}$$
(2.6)

Proof

From the relation (2.2), we have

$$\begin{aligned} 2b_2 = c_1 + 2a_2 c_0, \quad 3b_3 = c_2+2a_2 c_1 + 3a_3 c_0. \end{aligned}$$

Then

$$\begin{aligned} |b_{3} - \mu b_2^2|= & {} \left| \frac{1}{3} c_2+\frac{2}{3} a_2 c_1 + a_3 c_0 - \mu \left( \frac{1}{2} c_1 + a_2 c_0\right) ^2\right| \\= & {} \left| \left( \frac{1}{3} c_2 -\frac{\mu }{4}c_1^2\right) + a_2 c_1\left( \frac{2}{3}-\mu c_0\right) + c_0\big (a_3-\mu c_0a_2^2\big )\right| \\\le & {} \frac{1}{3}\left| c_2 -\frac{3}{4} \mu c_1^2\right| +|a_2||c_1|\left| \frac{2}{3}-\mu c_0\right| +\alpha |a_3 - \mu c_0a_2^2|. \end{aligned}$$

Apply now the estimate that holds for the coefficients of convex functions: \(|a_n| \le 1\ (n=2,3,...), |a_3 -\nu a_2^2| \le \max \{1/3,|1-\nu |\} \ (\nu \in \mathbb {R})\), and the relation (1.6). We obtain then

$$\begin{aligned} |b_{3} - \mu b_2^2|\le & {} \frac{1}{3}\left[ |c_2|+\frac{3}{4}\left| \mu c_1^2\right| \right] +|a_2||c_1|\left| \frac{2}{3}-\mu c_0\right| +\alpha |a_3 - \mu b_1 a_2^2|\\\le & {} \frac{1-\alpha ^2}{3}\left[ 1+\frac{3}{4}|\mu |(1-\alpha ^2)\right] +(1-\alpha ^2)\left| \frac{2}{3}-\mu b_1\right| \\&+\, \alpha \max \{1/3,|1-\mu b_1|\}. \end{aligned}$$

Next, by (2.2), we have

$$\begin{aligned} |b_{n+1} - b_n|= & {} \left| \frac{1}{n+1}\sum \limits _{p=1}^{n+1} p a_p c_{n+1-p}-\frac{1}{n}\sum \limits _{p=1}^n p a_p c_{n-p}\right| \\\le & {} |a_{n+1}c_0|+ |a_n c_0|+\frac{1}{n+1}\sum \limits _{p=1}^n p |a_p c_{n+1-p}| +\frac{1}{n}\sum \limits _{p=1}^{n-1} p |a_p c_{n-p}|\\\le & {} 2\alpha +\frac{1-\alpha ^2}{n+1}\sum \limits _{p=1}^n p+\frac{1-\alpha ^2}{n}\sum \limits _{p=1}^{n-1} p\\= & {} 2\alpha +(1-\alpha ^2)\frac{2n-1}{2}. \end{aligned}$$

The proof is now complete; however, the results are not sharp, for example, the function that realizes the accuracy of \(|b_2|\) in the previous theorem gives \(|b_2-b_1| = (1-\alpha ^2)/2 \le 2\alpha + (1-\alpha ^2)/2\), for any \(\alpha \in (0,1)\). The right-hand side is obtained from (2.6) for the case when \(n=1\). \(\square \)

Theorem 2.4

For \(f\in \widehat{S}^{\alpha }, f=h+\bar{g}\) and \(|z| = r <1\), it holds

$$\begin{aligned} \left| \frac{r}{1+r}-\frac{1-\alpha }{1+\alpha }\log \frac{1+r}{1-\alpha r}\right|\le & {} \ | g(z)| \ \le \ \frac{r}{1-r}+\frac{1-\alpha }{1+\alpha }\log \frac{1-r}{1+\alpha r}, \end{aligned}$$
(2.7)
$$\begin{aligned} \frac{2r}{1+r}-\frac{r(1-\alpha ^2)}{|r-\alpha |(1-\alpha r)}\le & {} \ \left| \frac{zg^{\prime \prime }(z)}{g^\prime (z)}\right| \ \le \ \frac{2r}{1-r}+\frac{r(1-\alpha ^2)}{|r-\alpha |(1-\alpha r)}, \end{aligned}$$
(2.8)
$$\begin{aligned} {{\mathrm{Re}}}\left( 1+\frac{zg^{\prime \prime }(z)}{g^\prime (z)}\right)> & {} \frac{r(\alpha ^2-1)}{|\alpha -r|(1-\alpha r)}+\frac{1-r}{1+r}. \end{aligned}$$
(2.9)

Proof

Applying the relation \(g^\prime =\omega h^\prime \), we estimate \(|g^\prime (z)|\) as follows [15]:

$$\begin{aligned} \frac{|\alpha -r|}{(1-\alpha r)(1+r)^2}\ \le \ |g^\prime (z)|\ \le \ \frac{\alpha +r}{(1+\alpha r)(1-r)^2}\quad (|z| =r <1). \end{aligned}$$

Then integrating along a radial line \(\zeta =te^{i\theta }\), the right-hand side of (2.7) is obtained immediately [15].

In order to prove the left-hand side of (2.7), we note first that g is univalent. Let \(\Gamma = g(\{z: |z|=r\})\) and let \(\xi _1\in \Gamma \) be the nearest point to the origin. By a rotation we may assume that \(\xi _1>0\). Let \(\gamma \) be the line segment \(0\le \xi \le \xi _1\) and suppose that \(z_1=g^{-1}(\xi _1)\) and \(L=g^{-1}(\gamma )\). With \(\zeta \) as the variable of integration on L, we have that \(\mathrm{d}\xi =g^\prime (\zeta )\mathrm{d}\zeta >0\) on L. Hence

$$\begin{aligned} \xi _1= & {} \int \limits _0^{\xi _1}\mathrm{d}\xi = \int \limits _0^{z_1}g^\prime (\zeta )\mathrm{d}\zeta = \int \limits _0^{z_1}|g^\prime (\zeta )||\mathrm{d}\zeta | \ge \int \limits _0^r|g^\prime (te^{i\theta })|\mathrm{d}t\\\ge & {} \int \limits _0^r\frac{|\alpha -r|}{(1-\alpha r)(1+r)^2}\mathrm{d}r = \left| \frac{r}{1+r}-\frac{1-\alpha }{1+\alpha }\log \frac{1+r}{1-\alpha r}\right| . \end{aligned}$$

From the relation \(g^\prime = \omega h^\prime \), we obtain

$$\begin{aligned} \frac{zg^{\prime \prime }(z)}{g^\prime (z)}= \frac{z\omega ^\prime (z)}{\omega (z)}+ \frac{zh^{\prime \prime }(z)}{h^\prime (z)}. \end{aligned}$$
(2.10)

Since h is convex, so is univalent, then it holds [13, p. 118]

$$\begin{aligned} \frac{2r}{1+r}\le \left| \frac{zh^{\prime \prime }(z)}{h^\prime (z)}\right| \le \frac{2r}{1-r}\quad (|z|=r). \end{aligned}$$
(2.11)

Moreover, \(\omega \) satisfies [12, p. 320]

$$\begin{aligned} \left| \frac{\omega (z)-\omega (0)}{1-\overline{\omega (0)}\omega (z)}\right| \le |z|\quad (|z|=r), \end{aligned}$$
(2.12)

from which it follows

$$\begin{aligned} \left| \omega (z) - \frac{\omega (0)(1-r^2)}{1-|\omega (0)|^2 r^2}\right| \le \frac{r(1-|\omega (0)|^2)}{1-|\omega (0)|^2r^2}. \end{aligned}$$
(2.13)

We note that \(|\omega (0)|=|c_0|=|b_1|=\alpha \), so that by (2.13) we have

$$\begin{aligned} \frac{|r-\alpha |}{1-\alpha r}\le |\omega (z)| \le \frac{r+\alpha }{1+\alpha r}. \end{aligned}$$
(2.14)

Taking into account (2.10), (2.11), and (2.14) and the Schwarz–Pick inequality (1.3), we obtain for \(|z|=r<1\),

$$\begin{aligned} \left| \dfrac{zg^{\prime \prime }(z)}{g^\prime (z)}\right|\le & {} \left| \dfrac{z\omega ^\prime (z)}{\omega (z)}\right| + \left| \dfrac{zh^{\prime \prime }(z)}{h^\prime (z)}\right| \\\le & {} \dfrac{r(1-|\omega (z)|^2)}{|\omega (z)|(1-r^2)}+\dfrac{2r}{1-r}\\\le & {} \dfrac{r(1-r^2)(1-\alpha ^2)}{(1-r^2)|r-\alpha |(1-\alpha r)}+\dfrac{2r}{1-r}\\= & {} \dfrac{r(1-\alpha ^2)}{|r-\alpha |(1-\alpha r)}+\dfrac{2r}{1-r}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} \left| \dfrac{zg^{\prime \prime }(z)}{g^\prime (z)}\right|\ge & {} \left| \dfrac{zh^{\prime \prime }(z)}{h^\prime (z)}\right| -\left| \dfrac{z\omega ^\prime (z)}{\omega (z)}\right| \\\ge & {} \dfrac{2r}{1+r}-\dfrac{r(1-|\omega (z)|^2)}{|\omega (z)|\big (1-r^2\big )}\\\ge & {} \dfrac{2r}{1+r}-\dfrac{r(1-r^2)(1-\alpha ^2)}{\big (1-r^2\big )|r-\alpha |(1-\alpha r)}\\= & {} \dfrac{2r}{1+r}-\dfrac{r\big (1-\alpha ^2\big )}{|r-\alpha |(1-\alpha r)}. \end{aligned}$$

Moreover

$$\begin{aligned} 1+ \frac{zg^{\prime \prime }(z)}{g^\prime (z)} \ =\ \frac{z\omega ^\prime (z)}{\omega (z)}+1+ \frac{zh^{\prime \prime }(z)}{h^\prime (z)} \end{aligned}$$
(2.15)

and h is convex, therefore

$$\begin{aligned} {{\mathrm{Re}}}\left( 1+\dfrac{zg^{\prime \prime }(z)}{g^\prime (z)}\right)= & {} {{\mathrm{Re}}}\dfrac{z\omega ^\prime (z)}{\omega (z)}+{{\mathrm{Re}}}\left( 1+ \dfrac{zh^{\prime \prime }(z)}{h^\prime (z)}\right) \nonumber \\> & {} {{\mathrm{Re}}}\dfrac{z\omega ^\prime (z)}{\omega (z)}+\dfrac{1-r}{1+r}. \end{aligned}$$
(2.16)

By the above, (1.3) and (2.14), we have

$$\begin{aligned} {{\mathrm{Re}}}\left( 1+\frac{zg^{\prime \prime }(z)}{g^\prime (z)}\right) > \frac{r(\alpha ^2-1)}{|\alpha -r|(1-\alpha r)}+\frac{1-r}{1+r}, \end{aligned}$$
(2.17)

as asserted. \(\square \)

3 Estimates of the Bloch’s Constant

A harmonic function f is called the Bloch function if

$$\begin{aligned} \mathfrak {B}_f = \sup \limits _{z,w \in {\mathbb {D}},z\ne w}\frac{|f(z)-f(w)|}{\varrho (z,w)} <\infty , \end{aligned}$$
(3.1)

where

$$\begin{aligned} \varrho (z,w) = \frac{1}{2}\log \left( \frac{1+\left| \frac{z-w}{1-\bar{z} w}\right| }{1-\left| \frac{z-w}{1-\bar{z} w}\right| }\right) =\mathrm{artanh}\left| \frac{z-w}{1-\bar{z} w}\right| \end{aligned}$$

denotes the hyperbolic distance in \({\mathbb {D}}\), and \({\mathfrak {B}}_f\) is called the Bloch’s constant of f. The harmonic Bloch’s constant was studied by Colonna [9]. Colonna established that the Bloch’s constant \({\mathfrak {B}}_f\) of a harmonic mapping \(f= h+\bar{g}\) can be expressed in terms of moduli of the derivatives of h and g

$$\begin{aligned} {\mathfrak {B}}_f= & {} \sup _{z\in {\mathbb {D}}}(1-|z|^2)\left( |h^\prime (z)|+|g^\prime (z)|\right) \nonumber \\= & {} \sup _{z\in {\mathbb {D}}}(1-|z|^2)|h^\prime (z)|(1+|\omega (z)|), \end{aligned}$$
(3.2)

which agrees with the well-known notion of the Bloch’s constant for analytic functions. Moreover, the function f is Bloch if and only if both h and g are, and

$$\begin{aligned} \max ({\mathfrak {B}}_h, {\mathfrak {B}}_g) \le {\mathfrak {B}}_f \le {\mathfrak {B}}_h+{\mathfrak {B}}_g. \end{aligned}$$

Colonna also obtained the best-possible estimate of the Bloch’s constant for the family of harmonic mappings of \({\mathbb {D}}\) into itself. Recently, the Bloch’s constant was studied by many authors, see, for example [3, 4, 19]. Very interesting results in this direction were obtained in [5–7, 18, 20, 22]. Our aim is to determine the bounds for the Bloch’s constant in the classes \(\overline{{S}}^{\alpha }\) and \(\widehat{S}^{\alpha }\).

Theorem 3.1

Let \(f=h+\bar{g}\) with \(h(z) = z/(1-Bz), -1 < B < 1\), and let \(|B| = A,\ 0 \le A < 1\). Then the Bloch’s constant \({\mathfrak {B}}_f\) is bounded by

$$\begin{aligned} {\mathfrak {B}}_f \le (1+\alpha )\frac{(1+r_0)^3(1-r_0)^2}{(1-Ar_0)^2(1+\alpha r_0)}, \end{aligned}$$
(3.3)

where \(r_0\) is given by

$$\begin{aligned} r_0=\frac{\alpha (1+3A)-3-A+\sqrt{(1 + \alpha )(1 + A)(9 - 7 A + \alpha (-7 + 9 A))}}{4\alpha +2A(\alpha -1)}. \end{aligned}$$
(3.4)

Proof

Applying the distortion theorem

$$\begin{aligned} |h^\prime (z)| \le \frac{1}{(1-Ar)^2}\quad (|z| = r), \end{aligned}$$

and (3.2), we find

$$\begin{aligned} {\mathfrak {B}}_f= \sup \limits _{z\in {\mathbb {D}}}(1-|z|^2)|h^\prime (z)|(1+|\omega (z)|) \le (1+\alpha )\sup _{0\le r<1}\frac{(1+r)(1-r^2)}{(1-Ar)^2(1+\alpha r)}. \end{aligned}$$

Setting

$$\begin{aligned} q(r) =\frac{(1+r)(1-r^2)}{(1-Ar)^2(1+\alpha r)}, \end{aligned}$$

we observe that \(q^\prime (r) = 0\), if and only if

$$\begin{aligned} (1 + r)[(2\alpha +\alpha A -A)r^2+(3+A-\alpha -3\alpha A)r+\alpha -1-2A] =0. \end{aligned}$$

The last equation has solution in the interval (0, 1) at the point \(r_0\) given by (3.4), and the function q attains its maximum at \(r_0\). \(\square \)

Setting \(B=0\) in the above theorem, we obtain the estimate of \({\mathfrak {B}}_f\) in the class \(\bar{S}^{\alpha }\), below.

Corollary 3.2

For \(f\in \bar{S}^{\alpha }, f=h+\bar{g}\), the Bloch’s constant \({\mathfrak {B}}_f\) is bounded by

$$\begin{aligned} {\mathfrak {B}}_f \le (1+\alpha )\frac{(1+r_0)^3(1-r_0)^2}{(1+\alpha r_0)}, \end{aligned}$$
(3.5)

where \(r_0\) is given by

$$\begin{aligned} r_0 = \frac{\alpha -3 +\sqrt{9+2\alpha -7\alpha ^2}}{4\alpha }. \end{aligned}$$
(3.6)

Remark

By the fact that the Bloch’s constant is finite, we already have that the family of harmonic mappings with \(h(z) \equiv z\), and \(|b_1|=\alpha \), is a normal family. A function f is normal, if the constant \(\sigma _f\) is finite, where

$$\begin{aligned} \sigma _f = \sup \limits _{z\in {\mathbb {D}}}\frac{(1-|z|^2)|f^\prime (z)|}{1+|f(z)|}, \end{aligned}$$

see [10]. Indeed, since the quantity |f(z)| in \(\overline{{\mathcal {S}}}^\alpha \) is bounded [15]

$$\begin{aligned} |f(z)| \ge \left\{ \begin{array}{ll} \displaystyle \left( 1-\frac{1}{\alpha } \right) r -\left( 1-\frac{1}{\alpha ^2} \right) \log (1+\alpha r)&{} \quad \mathrm{for} \quad \alpha \ne 0, \\ \displaystyle r-\frac{r^2}{2}&{} \quad \mathrm{for} \quad \alpha =0, \end{array}\right. \end{aligned}$$
(3.7)

therefore, by (3.5) and (3.7), we obtain

$$\begin{aligned} \sigma _f \le \left\{ \begin{array}{ll} (1+\alpha )\frac{(1+r)(1-r^2)}{1+\alpha r}\frac{1}{1+\left( 1-\frac{1}{\alpha } \right) r -\left( 1-\frac{1}{\alpha ^2} \right) \log (1+\alpha r)}&{} \quad \mathrm{for} \quad \alpha \ne 0, \\ (1+\alpha )\frac{(1+r)(1-r^2)}{1+\alpha r}\frac{1}{1+r-\frac{r^2}{2}}&{} \quad \mathrm{for} \quad \alpha =0, \end{array}\right. \end{aligned}$$
(3.8)

where \(r=r_0\) is given by (3.6), and we see that in both cases \(\sigma _f\) is finite.

Remark

The univalent Bloch functions can be described in terms of geometry of their images; they are precisely those functions whose images do not contain disks of arbitrarily large radius [10]. Therefore, we suppose that the functions from the class \(\widehat{S}^{\alpha }\) may not be the Bloch functions. Indeed, reasoning similarly as in the Theorem 3.1 we note that in the class \(\widehat{S}^{\alpha }\) we have \(h(z) = z/(1-z)\), then \(|h^\prime (z)| \le 1/(1-r)^2\). Thus

$$\begin{aligned} {\mathfrak {B}}_f = (1+\alpha )\sup _{0\le r<1} \frac{(1+r)^2}{(1-r)(1+\alpha r)}, \end{aligned}$$

and the function \(p(r) =(1+r)^2/[(1-r)(1+\alpha r)]\) increases in the whole interval (0, 1), with infinity as the supremum.