1 Introduction

Let \({\mathcal {A}}\) be the class of functions f analytic in the unit disk \({\mathbb {D}}=\{z\in {\mathbb {C}}:\, |z|<1\}\) with the normalization \(f(0)=0=f'(0)-1\). Let \({\mathcal {S}}\) denote the class of functions f from \({\mathcal {A}}\) that are univalent in \({\mathbb {D}}\). Then the logarithmic coefficients \(\gamma _n\) of \(f\in {{\mathcal {S}}}\) are defined by the formula

$$\begin{aligned} \frac{1}{2}\log \left( \frac{f(z)}{z}\right) =\sum _{n=1}^\infty \gamma _nz^n, \quad z\in {\mathbb {D}}. \end{aligned}$$
(1)

These coefficients play an important role for various estimates in the theory of univalent functions. When we require a distinction, we use the notation \(\gamma _n(f)\) instead of \(\gamma _n\). For example, the Koebe function \(k(z)=z(1-e^{i\theta }z)^{-2}\) for each \(\theta \) has logarithmic coefficients \(\gamma _n(k)=e^{in\theta }/n\), \(n\ge 1\). If \(f\in {{\mathcal {S}}}\) and \(f(z)=z+\sum _{n=2}^{\infty } a_{n}z^{n}\), then by (1) it follows that \(2\gamma _1=a_2\) and hence, by the Bieberbach inequality, \(|\gamma _1|\le 1\). Let \({{\mathcal {S}}}^{\star }\) denote the class of functions \(f\in {{\mathcal {S}}}\) such that \(f({\mathbb {D}})\) is starlike with respect to the origin. Functions \(f\in {{\mathcal {S}}}^{\star }\) are characterized by the condition \(\mathrm{Re}\, (zf'(z)/f(z))>0\) in \({\mathbb {D}}\). The inequality \(|\gamma _n|\le 1/n\) holds for starlike functions \(f\in {{\mathcal {S}}}\), but is false for the full class \({\mathcal {S}}\), even in order of magnitude. See [4, Theorem 8.4 on page 242]. In [6], Girela pointed out that this bound is actually false for the class of close-to-convex functions in \({\mathbb {D}}\) which is defined as follows: A function \(f\in \mathcal {A}\) is called close-to-convex, denoted by \(f\in \mathcal {K}\), if there exists a real \(\alpha \) and a \(g\in {{\mathcal {S}}}^{\star }\) such that

$$\begin{aligned} \mathrm{Re} \left( e^{i\alpha }\frac{zf'(z)}{g(z)} \right) > 0, \quad z\in {\mathbb {D}}. \end{aligned}$$

For \(0\le \beta <1\), a function \(f\in {{\mathcal {S}}}\) is said to belong to the class of starlike functions of order \(\beta \), denoted by \(f\in {{\mathcal {S}}}^{\star }(\beta )\), if \(\mathrm{Re} \left( zf'(z)/f(z)\right) >\beta \) for \(z\in {\mathbb {D}}\). Note that \({{\mathcal {S}}}(0)=:{{\mathcal {S}}}^{\star }\). The class of all convex functions of order \(\beta \), denoted by \({{\mathcal {C}}}(\beta )\), is then defined by \({{\mathcal {C}}}(\beta )=\{f\in \mathcal {S}:\, zf'\in {{\mathcal {S}}}^{\star }(\beta )\}\). The class \({{\mathcal {C}}}(0)=:{{\mathcal {C}}}\) is usually referred to as the class of convex functions in \({\mathbb {D}}\). With the class \({\mathcal {S}}\) being of the first priority, its subclasses such as \({{\mathcal {S}}}^{\star }\), \( {{\mathcal {K}}}\), and \({\mathcal {C}}\), respectively, have been extensively studied in the literature and they appear in different contexts. We refer to [4, 7, 10, 12] for a general reference related to the present study. In [5, Theorem 4], it was shown that the logarithmic coefficients \(\gamma _n\) of every function \(f\in {{\mathcal {S}}}\) satisfy

$$\begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}|^{2}\le \frac{\pi ^{2}}{6} \end{aligned}$$
(2)

and the equality is attained for the Koebe function. The proof uses ideas from the work of Baernstein [3] on integral means. However, this result is easy to prove (see Theorem 1) in the case of functions in the class \(\mathcal {U}:=\mathcal {U}(1)\) which is defined as follows:

$$\begin{aligned} \mathcal {U}(\lambda )=\left\{ f\in {{\mathcal {A}}}:\, \left| \left( \frac{z}{f(z)}\right) ^2f'(z)-1\right| < \lambda , \quad z\in {\mathbb {D}}\right\} , \end{aligned}$$

where \(\lambda \in (0,1]\). It is known that [1, 2, 11] every \(f\in \mathcal {U}\) is univalent in \({\mathbb {D}}\) and hence, \(\mathcal {U}(\lambda )\subset \mathcal {U} \subset \mathcal {S} \) for \(\lambda \in (0,1]\). The present authors have established many interesting properties of the family \(\mathcal {U}(\lambda )\). See [10] and the references therein. For example, if \(f\in {{\mathcal {U}}}(\lambda )\) for some \(0<\lambda \le 1\) and \(a_2 =f''(0)/2\), then we have the subordination relations

$$\begin{aligned} \frac{f(z)}{z}\prec \frac{1}{1+(1+\lambda )z+\lambda z^2}=\frac{1}{(1+z)(1+\lambda z)}, ~ z\in {\mathbb {D}}, \end{aligned}$$
(3)

and

$$\begin{aligned} \frac{z}{f(z)}+ a_2z \prec 1+2\lambda z + \lambda z^2, ~ z\in {\mathbb {D}}. \end{aligned}$$

Here \(\prec \) denotes the usual subordination [4, 7, 12]. In addition, the following conjecture was proposed in [10].

Conjecture 1 Suppose that \(f\in {{\mathcal {U}}}(\lambda )\) for some \(0<\lambda \le 1\). Then \(|a_n|\le \sum _{k=0}^{n-1}\lambda ^k\) for \(n\ge 2\).

In Theorem 1, we present a direct proof of an inequality analogous to (2) for functions in \({{\mathcal {U}}}(\lambda )\) and in Corollary 1, we obtain the inequality (2) as a special case for \({{\mathcal {U}}}\). At the end of Sect. 2, we also consider estimates of the type (2) for some interesting subclasses of univalent functions. However, Conjecture 1 remains open for \(n\ge 5\). On the other hand, the proof for the case \(n=2\) of this conjecture is due to [17] and an alternate proof was obtained recently by the present authors in [10, Theorem 1]. In this paper, we show that Conjecture 1 is true for \(n=3, 4\), and our proof includes an elegant proof of the case \(n=2\). The main results and their proofs are presented in Sects. 2 and 3.

2 Logarithmic coefficients of functions in \({{\mathcal {U}}}(\lambda )\)

Theorem 1

For \(0<\lambda \le 1\), the logarithmic coefficients of \(f\in \mathcal {U}(\lambda )\) satisfy the inequality

$$\begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}|^{2} \le \frac{1}{4}\left( \frac{\pi ^{2}}{6}+2\mathrm{Li\,}_{2}(\lambda )+\mathrm{Li\,}_{2}(\lambda ^{2})\right) , \end{aligned}$$
(4)

where \(\mathrm{Li}_2\) denotes the dilogarithm function given by

$$\begin{aligned} \mathrm{Li}_2(z)= \sum _{n=1}^\infty \frac{z^{n}}{n^2}= z\int _0^1\frac{\log (1/t)}{1-tz}\,dt. \end{aligned}$$

The inequality (4) is sharp. Further, there exists a function \(f \in \mathcal {U}\) such that \(|\gamma _{n}|>(1+\lambda ^n)/(2n)\) for some n.

Proof

Let \(f\in \mathcal {U}(\lambda )\). Then, by (3), we have

$$\begin{aligned} \frac{z}{f(z)}\prec (1-z)(1-\lambda z) \end{aligned}$$

which clearly gives

$$\begin{aligned} \sum _{n=1}^{\infty } \gamma _{n}z^{n}= \log \sqrt{\frac{f(z)}{z}}\prec \frac{ -\log (1-z)-\log (1-\lambda z)}{2}= \sum _{n=1}^{\infty } \frac{1}{2n}(1+\lambda ^{n})z^{n}. \end{aligned}$$
(5)

Again, by Rogosinski’s theorem (see [4, 6.2]), we obtain

$$\begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}|^{2} \le \sum _{n=1}^{\infty }\frac{1}{4n^{2}}(1+\lambda ^{n})^{2} =\frac{1}{4}\left( \sum _{n=1}^{\infty }\frac{1}{n^{2}} +2\sum _{n=1}^{\infty }\frac{\lambda ^{n}}{n^{2}}+ \sum _{n=1}^{\infty }\frac{\lambda ^{2n}}{n^{2}}\right) \end{aligned}$$

and the desired inequality (4) follows. For the function

$$\begin{aligned} g_{\lambda }(z)=\frac{z}{(1-z)(1-\lambda z)}, \end{aligned}$$

we find that \(\gamma _{n}(g_{\lambda }) =(1+\lambda ^n)/(2n)\) for \(n\ge 1\) and therefore, we have the equality in (4). Note that \(g_{1}(z)\) is the Koebe function \(z/(1-z)^2\).

From the relation (5), we cannot conclude that

$$\begin{aligned} |\gamma _{n}(f)|\le |\gamma _{n}(g_{\lambda })|=\frac{1+\lambda ^n}{2n} ~ \text{ for } f\in \mathcal {U}(\lambda ). \end{aligned}$$

Indeed for the function \(f_{\lambda }\) defined by

$$\begin{aligned} f_{\lambda }(z)= \frac{z}{(1-z)(1-\lambda z)(1+(\lambda /(1+\lambda )) z)} \end{aligned}$$
(6)

we find that

$$\begin{aligned} \frac{z}{f_{\lambda }(z)}= 1+ \frac{\lambda -(1+\lambda )^2}{1+\lambda } z +\frac{\lambda ^2}{1+\lambda }z^3 \end{aligned}$$

and

$$\begin{aligned} \left( \frac{z}{f_{\lambda }(z)}\right) ^2f_{\lambda }'(z)-1= -\frac{2\lambda ^2}{1+\lambda }z^3 =-\left( 1- \frac{(1+2\lambda )(1-\lambda )}{1+\lambda }\right) z^3 \end{aligned}$$

which clearly shows that \(f_{\lambda }\in \mathcal {U}(\lambda )\). The images of \({\mathbb {D}}\) under \(f_{\lambda }(z)\) for certain values of \(\lambda \) are shown in Fig. 1a–d. Moreover, for this function, we have

Fig. 1
figure 1

The image of \(\displaystyle f_{\lambda }(z)= \frac{z}{(1-z)(1-\lambda z)(1+(\lambda /(1+\lambda )) z)}\) under \({\mathbb {D}}\) for certain values of \(\lambda \)

Full size image
$$\begin{aligned} \log \left( \frac{f_{\lambda }(z)}{z}\right)= & {} -\log (1-z)-\log (1-\lambda z)-\log \left( 1+\frac{\lambda }{1+\lambda } z\right) \\= & {} 2\sum _{n=1}^{\infty } \gamma _{n}(f_{\lambda })z^n, \end{aligned}$$

where

$$\begin{aligned} \gamma _{n}(f_{\lambda }) =\frac{1}{2}\left( \frac{1+\lambda ^{n}}{n} + (-1)^{n}\frac{\lambda ^n}{(1+\lambda )^n}\right) . \end{aligned}$$

This contradicts the above inequality at least for even integer values of \(n\ge 2\). Moreover, with these \(\gamma _{n}(f_{\lambda })\) for \(n\ge 1\), we obtain

$$\begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}(f_{\lambda })|^{2}= & {} \frac{1}{4} \sum _{n=1}^{\infty } \left\{ \frac{(1+\lambda ^{n})^2}{n^2}+ 2\frac{(-1)^{n}}{n} \left( \left( \frac{\lambda ^{2} }{1+\lambda }\right) ^{n} \right. \right. \\&\left. \left. + \left( \frac{\lambda }{1+\lambda }\right) ^{n} \right) + \left( \frac{\lambda }{1+\lambda }\right) ^{2n} \right\} \end{aligned}$$

and by a computation, it follows easily that

$$\begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}(f_{\lambda })|^{2}= & {} \frac{1}{4}\left( \frac{\pi ^{2}}{6}+2\mathrm{Li\,}_{2}(\lambda )+\mathrm{Li\,}_{2}(\lambda ^{2})\right) \\&-\frac{1}{2} \log \left[ \left( 1+\frac{\lambda ^2}{1 +\lambda }\right) \left( 1+\frac{\lambda }{1+\lambda }\right) \right] +\frac{\lambda ^2}{4(1+2\lambda )}\\< & {} \frac{1}{4}\left( \frac{\pi ^{2}}{6}+2\mathrm{Li\,}_{2}(\lambda )+\mathrm{Li\,}_{2}(\lambda ^{2})\right) ~ \text{ for } 0<\lambda \le 1, \end{aligned}$$

and we complete the proof. \(\square \)

Corollary 1

The logarithmic coefficients of \(f\in \mathcal {U}\) satisfy the inequality

$$\begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}|^{2}\le \sum _{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}. \end{aligned}$$
(7)

We have equality in the last inequality for the Koebe function \(k(z)=z(1-e^{i\theta }z)^{-2}\). Further there exists a function \(f \in \mathcal {U}\) such that \(|\gamma _{n}|>1/n\) for some n.

Remark 1

From the analytic characterization of starlike functions, it is easy to see that for \(f\in \mathcal {S}^{\star }\),

$$\begin{aligned} \frac{zf'(z)}{f(z)} -1=z\left( \log \left( \frac{f(z)}{z}\right) \right) ' = 2\sum _{n=1}^{\infty } n \gamma _{n}z^{n} \prec \frac{2z}{1-z} \end{aligned}$$

and thus, by Rogosinski’s result, we obtain that \(|\gamma _{n}|\le 1/n\) for \(n\ge 1\). In fact for starlike functions of order \(\alpha \), \(\alpha \in [0,1)\), the corresponding logarithmic coefficients satisfy the inequality \(|\gamma _{n}|\le (1-\alpha )/n\) for \(n\ge 1\). Moreover, one can quickly obtain that

$$\begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}|^{2}\le (1-\alpha )^2 \frac{\pi ^2}{6} \end{aligned}$$

if \(f\in \mathcal {S}^{\star }(\alpha )\), \(\alpha \in [0,1)\) (See also the proof of Theorem 2 and Remark 3). As remarked in the proof of Theorem 1, from the relation (7), we cannot conclude the same fact, namely, \(|\gamma _{n}|\le 1/n\) for \(n\ge 1\), for the class \(\mathcal {U}\) although the Koebe function \(k(z)=z/(1-z)^2\) belongs to \(\mathcal {U}\cap \mathcal {S}^{\star }\). For example, if we set \(\lambda =1\) in (6), then we have

$$\begin{aligned} \frac{z}{f_1(z)}=(1-z)^{2}\left( 1+\frac{z}{2}\right) =1-\frac{3}{2}z +\frac{z^{3}}{2}, \end{aligned}$$

where \(f_1\in {{\mathcal {U}}}\) and for this function, we obtain

$$\begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}(f_1)|^{2}= & {} \sum _{n=1}^{\infty } \left( \frac{1}{n}+(-1)^{n}\frac{1}{2^{n+1}}\right) ^{2} =\frac{\pi ^{2}}{6}+\frac{1}{12}\\&-\log \frac{3}{2}<\frac{\pi ^{2}}{6}. \end{aligned}$$

On the other hand, it is a simple exercise to verify that \(f_1 \notin \mathcal {S}^{\star }\). The graph of this function is shown in Fig. 1d.

Let \({{\mathcal {G}}}(\alpha )\) denote the class of locally univalent normalized analytic functions f in the unit disk \(|z| < 1\) satisfying the condition

$$\begin{aligned} \mathrm{Re} \left( 1+\frac{zf''(z)}{f'(z)}\right)<1+\frac{\alpha }{2} \quad \text{ for } |z|<1, \end{aligned}$$

and for some \(0<\alpha \le 1\). Set \({{\mathcal {G}}}(1)=:{{\mathcal {G}}}\). It is known (see [13, Equation (16)]) that \(\mathcal {G}\subset {{\mathcal {S}}}^{\star }\) and thus, functions in \({{\mathcal {G}}}(\alpha )\) are starlike. This class has been studied extensively in the recent past, see for instance [9] and the references therein. We now consider the estimate of the type (2) for the subclass \({{\mathcal {G}}}(\alpha )\).

Theorem 2

Let \(0<\alpha \le 1\) and \({{\mathcal {G}}}(\alpha )\) be defined as above. Then the logarithmic coefficients \(\gamma _{n}\) of \(f\in {{\mathcal {G}}}(\alpha )\) satisfy the inequalities

$$\begin{aligned} \sum _{n=1}^{\infty }n^{2}|\gamma _{n}|^{2}\le \frac{\alpha }{4(\alpha +2)} \end{aligned}$$
(8)

and

$$\begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}|^{2}\le \frac{\alpha ^2}{4}\,\mathrm{Li\,}_{2}\left( \frac{1}{(1+\alpha )^2} \right) . \end{aligned}$$
(9)

Also we have

$$\begin{aligned} |\gamma _{n}|\le \frac{\alpha }{2(\alpha +1)n} ~ \text{ for } n\ge 1. \end{aligned}$$
(10)

Proof

If \(f\in {{\mathcal {G}}}(\alpha )\), then we have (see eg. [8, Theorem 1] and [13])

$$\begin{aligned} \frac{zf'(z)}{f(z)} -1 \prec \frac{(1+\alpha )(1-z)}{1+\alpha -z}-1 = -\alpha \left( \frac{z/(1+\alpha )}{1-(z/(1+\alpha ))}\right) , \quad z\in {{\mathbb {D}}}, \end{aligned}$$
(11)

which, in terms of the logarithmic coefficients \(\gamma _{n}\) of f defined by (1), is equivalent to

$$\begin{aligned} \sum _{n=1}^{\infty }(-2n \gamma _{n})z^{n}\prec \alpha \sum _{n=1}^{\infty }\frac{z^n}{(1+\alpha )^{n}}. \end{aligned}$$
(12)

Again, by Rogosinski’s result, we obtain that

$$\begin{aligned} \sum _{n=1}^{\infty }4n^{2} |\gamma _{n}|^{2}\le \alpha ^2\sum _{n=1}^{\infty }\frac{1}{(1+\alpha )^{2n}}=\frac{\alpha }{\alpha +2} \end{aligned}$$

which is (8).

Now, since the sequence \(A_{n}=\frac{1}{(1+\alpha )^{n}}\) is convex decreasing, we obtain from (12) and [15, Theorem VII, p.64] that

$$\begin{aligned} |-2n \gamma _{n}|\le A_{1}=\frac{1}{1+\alpha }, \end{aligned}$$

which implies the desired inequality (10). As an alternate approach to prove this inequality, we may rewrite (11) as

$$\begin{aligned} \sum _{n=1}^{\infty } (2n \gamma _{n})z^{n} =z\left( \log \left( \frac{f(z)}{z}\right) \right) ' \prec \phi (z)=-\alpha \left( \frac{z/(1+\alpha )}{1-(z/(1+\alpha ))}\right) \end{aligned}$$

and, since \(\phi (z)\) is convex in \({\mathbb {D}}\) with \(\phi '(0)=-\alpha /(1+\alpha )\), it follows from Rogosinski’s result (see also [4, Theorem 6.4(i), p. 195]) that \(|2n \gamma _{n}|\le \alpha /(1+\alpha )\). Again, this proves the inequality (10).

Finally, we prove the inequality (9). From the formula (12) and the result of Rogosinski (see also [12, Theorem 2.2] and [4, Theorem 6.2]), it follows that for \(k\in {\mathbb {N}}\) the inequalities

$$\begin{aligned} \sum _{n=1}^kn^2\left| \gamma _n\right| ^2 \le \frac{\alpha ^2}{4}\sum _{n=1}^k \frac{1}{(1+\alpha )^{2n}} \end{aligned}$$

are valid. Clearly, this implies the inequality (8) as well. On the other hand, consider these inequalities for \(k=1, \ldots ,N\), and multiply the k-th inequality by the factor \(\frac{1}{k^2}-\frac{1}{(k+1)^2},\) if \(k=1, \ldots ,N-1\) and by \(\frac{1}{N^2}\) for \(k=N\). Then the summation of the multiplied inequalities yields

$$\begin{aligned} \sum _{k=1}^N\left| \gamma _k\right| ^2\le & {} \frac{\alpha ^2}{4} \sum _{k=1}^N\frac{1}{k^2 (1+\alpha )^{2k}}\\\le & {} \frac{\alpha ^2}{4}\sum _{k=1}^{\infty }\frac{1}{k^2 (1+\alpha )^{2k}}\\= & {} \frac{\alpha ^2}{4} \,\mathrm{Li\,}_{2}\left( \frac{1}{(1+\alpha )^2} \right) ~ \text{ for } N=1,2,\ldots , \end{aligned}$$

which proves the desired assertion (9) if we allow \(N\rightarrow \infty \). \(\square \)

Corollary 2

The logarithmic coefficients \(\gamma _{n}\) of \(f\in \mathcal {G}:=\mathcal {G}(1)\) satisfy the inequalities

$$\begin{aligned} \sum _{n=1}^{\infty }n^{2}|\gamma _{n}|^{2}\le \frac{1}{12} ~ \text{ and } ~ \sum _{n=1}^{\infty }|\gamma _{n}|^{2}\le \frac{1}{4}\,\mathrm{Li\,}_{2}\left( \frac{1}{4} \right) . \end{aligned}$$

The results are the best possible as the function \(f_0(z)=z-\frac{1}{2}z^{2}\) shows. Also we have \(|\gamma _{n}|\le 1/(4n)\) for \(n\ge 1\).

Remark 2

For the function \(f_0(z)=z-\frac{1}{2}z^{2}\), we have that \(\gamma _{n}(f_0)= -\frac{1}{n2^{n+1}}\) for \(n=1,2,\ldots \) and thus, it is reasonable to expect that the inequality \(|\gamma _{n}|\le \frac{1}{n2^{n+1}}\) is valid for the logarithmic coefficients \(\gamma _{n}\) of each \(f\in \mathcal {G}.\) But that is not the case as the function \(f_{n}\) defined by \(f_{n}'(z)=(1-z^{n})^{\frac{1}{n}}\) shows. Indeed for this function we have

$$\begin{aligned} 1+\frac{zf_n''(z)}{f_n'(z)} =\frac{1-2z^n}{1-z^n} \end{aligned}$$

showing that \(f_n\in \mathcal {G}\). Moreover,

$$\begin{aligned} \log \frac{f_{n}(z)}{z}=-\frac{1}{n(n+1)}z^{n}+ \cdots , \end{aligned}$$

which implies that \(|\gamma _{n}(f_n)|= \frac{1}{2n(n+1)}\) for \(n=1,2,\ldots \), and observe that \(\frac{1}{2n(n+1)}>\frac{1}{n2^{n+1}}\) for \(n=2,3, \ldots \). Thus, we conjecture that the logarithmic coefficients \(\gamma _{n}\) of each \(f\in \mathcal {G}\) satisfy the inequality \(|\gamma _{n}|\le \frac{1}{2n(n+1)}\) for \( n=1,2,\ldots \). Clearly, Corollary 2 shows that the conjecture is true for \(n=1\).

Remark 3

Let \(f\in \mathcal {C} (\alpha )\), where \(0\le \alpha <1\). Then we have [18]

$$\begin{aligned} \frac{zf'(z)}{f(z)}-1 \prec G_\alpha (z)-1 =\sum _{n=1}^{\infty }\delta _n z^n, \end{aligned}$$
(13)

where \(\delta _n\) is real for each n,

$$\begin{aligned} G_\alpha (z)= \left\{ \begin{array}{ll} \displaystyle \frac{(2\alpha -1)z}{(1-z)[(1-z)^{1-2\alpha } -1]} &{} \text{ if } \alpha \ne 1/2,\\ \displaystyle \frac{-z}{(1-z)\log (1-z)} &{} \text{ if } \alpha = 1/2, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \beta (\alpha )=G_{\alpha }(-1)=\inf _{|z|<1} G_{\alpha }(z)= \left\{ \begin{array}{ll} \displaystyle \frac{1-2\alpha }{2[2^{1-2\alpha } -1]} &{} \text{ if } 0\le \alpha \ne 1/2<1,\\ \displaystyle \frac{1}{2\log 2} &{} \text{ if } \alpha = 1/2 \end{array} \right. \end{aligned}$$

so that \(f\in {{\mathcal {S}}}^{\star }(\beta (\alpha ))\). Also, we have [16]

$$\begin{aligned} \frac{f(z)}{z}\prec \frac{K_\alpha (z)}{z}= \left\{ \begin{array}{ll} \displaystyle \frac{(1-z)^{2\alpha -1} -1}{1-2\alpha } &{} \text{ if } 0\le \alpha \ne 1/2<1,\\ \displaystyle -\frac{\log (1-z)}{z} &{} \text{ if } \alpha = 1/2, \end{array} \right. \end{aligned}$$

and \(K_\alpha (z)/z\) is univalent and convex (not normalized in the usual sense) in \({\mathbb {D}}\).

Now, the subordination relation (13), in terms of the logarithmic coefficients \(\gamma _{n}\) of f defined by (1), is equivalent to

$$\begin{aligned} 2\sum _{n=1}^{\infty }n \gamma _{n}z^{n}\prec G_\alpha (z)-1 = \sum _{n=1}^{\infty }\delta _n z^n, \quad z\in {\mathbb {D}}, \end{aligned}$$

and thus,

$$\begin{aligned} \sum _{n=1}^kn^2\left| \gamma _n\right| ^2 \le \frac{1}{4}\sum _{n=1}^k \delta _n^2 \quad \text{ for } \text{ each } k\in {\mathbb {N}}. \end{aligned}$$
(14)

Since f is starlike of order \(\beta (\alpha )\), it follows that

$$\begin{aligned} \frac{zK_\alpha '(z)}{K_\alpha (z)}-1= G_\alpha (z)-1 \prec 2(1-\beta (\alpha ))\frac{z}{1-z} \end{aligned}$$

and therefore, \(|\delta _n|\le 2(1-\beta (\alpha ))\) for each \(n\ge 1\). Again, the relation (14) by the previous approach gives

$$\begin{aligned} \sum _{k=1}^N\left| \gamma _k\right| ^2 \le \frac{1}{4} \sum _{k=1}^N\frac{\delta _k^2}{k^2} \le (1-\beta (\alpha ))^2 \sum _{k=1}^N\frac{1}{k^2} \end{aligned}$$

for \(N=1,2,\ldots ,\) and hence, we have

$$\begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}|^{2}\le \frac{1}{4}\sum _{n=1}^{\infty } \frac{\delta _n^2}{n^2}\le (1-\beta (\alpha ))^2 \sum _{n=1}^{\infty } \frac{1}{n^2} = (1-\beta (\alpha ))^2 \frac{\pi ^2}{6} \end{aligned}$$

and equality holds in the first inequality for \(K_\alpha (z)\). In particular, if f is convex then \(\beta (0)=1/2\) and hence, the last inequality reduces to

$$\begin{aligned} \sum _{n=1}^{\infty }|\gamma _{n}|^{2}\le \frac{\pi ^2}{24} \end{aligned}$$

which is sharp as the convex function \(z/(1-z)\) shows.

3 Proof of Conjecture 1 for \(n=2,3, 4\)

Theorem 3

Let \(f\in \mathcal {U}(\lambda )\) for \(0<\lambda \le 1\) and let \(f(z)=z+a_{2}z^{2}+a_{3}z^{3}+\cdots \). Then

$$\begin{aligned} |a_{n}|\le \frac{1-\lambda ^{n}}{1-\lambda } ~ \text{ for } 0<\lambda <1 \text{ and } n=2,3, 4, \end{aligned}$$
(15)

and \(|a_{n}|\le n \) for \(\lambda =1 \) and \(n\ge 2\). The results are the best possible.

Proof The case \(\lambda =1 \) is well-known because \(\mathcal {U}=\mathcal {U}(1)\subset \mathcal {S}\) and hence, by the de Branges theorem, we have \(|a_{n}|\le n\) for \(f\in \mathcal {U}\) and \(n\ge 2\). Here is an alternate proof without using the de Branges theorem. From the subordination result (3) with \(\lambda =1 \), one has

$$\begin{aligned} \frac{f(z)}{z}\prec \frac{1}{(1-z)^{2}}=\sum _{n=1}^{\infty } nz^{n-1} \end{aligned}$$

and thus, by Rogosinski’s theorem [4, Theorem 6.4(ii), p. 195], it follows that \(|a_{n}|\le n \) for \(n\ge 2\).

So, we may consider \(f\in \mathcal {U}(\lambda )\) with \(0<\lambda <1\). The result for \(n=2\), namely, \(|a_{2}|\le 1+\lambda \) is proved in [10, 17] and thus, it suffices to prove (15) for \(n=3, 4\) although our proof below is elegant and simple for the case \(n=2\) as well. To do this, we begin to recall from (3) that

$$\begin{aligned} \frac{f(z)}{z}\prec \frac{1}{(1-z)(1-\lambda z)}=1+\sum _{n=1}^{\infty } \frac{1-\lambda ^{n+1}}{1-\lambda }z^{n} \end{aligned}$$

and thus

$$\begin{aligned} \frac{f(z)}{z} = \frac{1}{(1-z\omega (z))(1-\lambda z \omega (z))}, \end{aligned}$$

where \(\omega \) is analytic in \({\mathbb {D}}\) and \(|\omega (z)|\le 1\) for \(z\in {\mathbb {D}}\). In terms of series formulation, we have

$$\begin{aligned} \sum _{n=1}^{\infty }a_{n+1}z^{n}=\sum _{n=1}^{\infty } \frac{1-\lambda ^{n+1}}{1-\lambda }\omega ^n(z)z^n. \end{aligned}$$

We now set \(\omega (z)=c_{1}+c_{2}z +\cdots \) and rewrite the last relation as

$$\begin{aligned} \sum _{n=1}^{\infty }(1-\lambda )a_{n+1}z^{n}=\sum _{n=1}^{\infty } (1-\lambda ^{n+1})(c_{1}+c_{2}z +\cdots )^{n}z^{n}. \end{aligned}$$
(16)

By comparing the coefficients of \(z^{n}\) for \(n=1,2,3\) on both sides of (16), we obtain

$$\begin{aligned} \left\{ \begin{array}{ll} (1-\lambda )a_{2} &{} = (1-\lambda ^{2})c_{1}\\ (1-\lambda )a_{3}&{} = (1-\lambda ^{2})c_{2}+(1-\lambda ^{3})c_{1}^{2}\\ (1-\lambda )a_{4} &{} = (1-\lambda ^{2})\left( c_{3}+\mu c_{1}c_{2}+\nu c_{1}^{3}\right) , \end{array} \right. \end{aligned}$$
(17)

where

$$\begin{aligned} \mu = 2\frac{1-\lambda ^3}{1-\lambda ^2} ~ \text{ and } ~ \nu = \frac{1-\lambda ^4}{1-\lambda ^2}. \end{aligned}$$

It is well-known that \(|c_1|\le 1\) and \(|c_{2}|\le 1-|c_{1}|^{2}\). From the first relation in (17) and the fact that \(|c_1|\le 1\), we obtain

$$\begin{aligned} (1-\lambda )|a_{2}|=(1-\lambda ^{2})|c_{1}|\le 1-\lambda ^{2}, \end{aligned}$$

which gives a new proof for the inequality \(|a_{2}|\le 1+\lambda \).

Next we present a proof of (15) for \(n=3\). Using the second relation in (17), \(|c_1|\le 1\) and the inequality \(|c_{2}|\le 1-|c_{1}|^{2}\), we get

$$\begin{aligned} (1-\lambda )|a_{3}|\le & {} (1-\lambda ^{2})|c_{2}|+(1-\lambda ^{3})|c_{1}|^{2}\\\le & {} (1-\lambda ^{2})(1-|c_{1}|^{2})+(1-\lambda ^{3})|c_{1}|^{2}\\= & {} 1-\lambda ^{2}+(\lambda ^{2}-\lambda ^{3})|c_{1}|^{2}\\\le & {} 1-\lambda ^{3}, \end{aligned}$$

which implies \(|a_3|\le 1+\lambda +\lambda ^2\).

Finally, we present a proof of (15) for \(n=4\). To do this, we recall the sharp upper bounds for the functionals \( \left| c_3+\mu c_1c_2+\nu c_1^3\right| \) when \(\mu \) and \(\nu \) are real. In [14], Prokhorov and Szynal proved among other results that

$$\begin{aligned} \left| c_3+\mu c_1c_2+\nu c_1^3\right| \le |\nu | \end{aligned}$$

if \( 2\le |\mu |\le 4\) and \(\nu \ge (1/12)(\mu ^2+8)\). From the third relation in (17), this condition is fulfilled and thus, we find that

$$\begin{aligned} (1-\lambda )|a_{4}| =(1-\lambda ^{2})\left| c_{3}+\mu c_{1}c_{2}+\nu c_{1}^{3}\right| \le (1-\lambda ^{2})\left( \frac{1-\lambda ^4}{1-\lambda ^2}\right) =1-\lambda ^4 \end{aligned}$$

which proves the desired inequality \(|a_4|\le 1+\lambda +\lambda ^2+\lambda ^3\). \(\square \)