June 11, 2026
Let \(\Phi\) be a univalent function in \(\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}\), \(\Phi (\mathbb{D})\) is symmetric with respect to the real axis, starlike with respect to \(\Phi (0)=1\), and \(\Phi ^{\prime }(0)>0\). Let \(\mathcal{C}(\Phi )\) denote the class of Ma-Minda convex functions. In this article, we present the bounds on \(||a_{3}|-|a_{2}||\) for Taylor’s coefficients of the function \(f\) in the class \(\mathcal{C}(\Phi )\). We also establish the same bounds for the inverse coefficients. All the bounds we study here are sharp. We also present the conditions such that the bounds on \(|a_{3}|-|a_{2}||\) and \(|A_{3}|-|A_{2}||\) are invariant, where \(A_{2}\) and \(A_{3}\) are the first two coefficients of the Taylor series of the inverse functions of \(f\in \mathcal{C}(\Phi ).\) Thus provides examples of invariance and nonvariance among the subclasses of convex functions.
Let \(\mathcal{A}\) denote the class of analytic functions \(f\) defined in the open unit disk \(\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}\), normalized by the conditions \(f(0)=0\) and \(f^{\prime }(0)=1\), and having the Taylor expansion of the form \[f(z)=z+\sum_{n=2}^{\infty }a_{n}z^{n}. \label{1}\tag{1}\] Let \(\mathcal{S}\) represent a subclass of functions in \(\mathcal{A}\) that are univalent(one-to-one) in \(\mathbb{D}\). In 1985, de Branges [1] settled the celebrated Bieberbach conjecture by proving that for every function \(f\in \mathcal{S}\) given in the form \(\left( \ref{1}\right)\), the sharp estimate \(|a_{n}|\leq n,n\geq 2,\) holds, with equality attained only by the Koebe function \(k(z)=z/(1-z)^{2}\) or its rotations. This remarkable breakthrough naturally led to further investigations concerning the behavior of successive coefficients of univalent functions. In particular, it became interesting to ask whether the inequality \[\big||a_{n+1}|-|a_{n}|\big|\leq 1,\qquad n\geq 2,\]is valid for all \(f\in \mathcal{S}\). However, it was soon observed that this is not true even for \(n=2\). In fact, it was shown in [2] that the following sharp bounds hold: \[-1\leq |a_{3}|-|a_{2}|\leq \frac{3}{4}+e^{-\lambda _{0}}\left( 2e^{-\lambda _{0}}-1\right) =1.029\cdots ,\]where \(\lambda _{0}\) is the unique root in \(0<\lambda <1\) of the equation \(4\lambda =e^{\lambda }.\) Later, Hayman [3] established an important general result by proving that there exists an absolute constant \(C>0\) such that \[\big||a_{n+1}|-|a_{n}|\big|\leq C,\qquad n\geq 2,\]for all functions \(f\in \mathcal{S}\). Hayman’s original proof relied on his powerful method developed for the study of areally mean \(p\)-valent functions [4]. An alternative approach was later provided by Milin through the celebrated Lebedev–Milin inequalities, and a detailed exposition of these developments can be found in Duren’s classical monograph [2]. Despite the significance of Hayman’s estimate, little progress has been achieved in determining sharper bounds for \(C\). Ilina [5] proved in 1968 that \(C<4.26\cdots\), and subsequently Grispan [6], refining Milin’s method, improved this in 1976 by establishing that for \(n\geq 2\), \[-2.97\cdots <|a_{n+1}|-|a_{n}|<3.61\cdots .\]For many years, no further improvements were reported until the recent contribution of Obradović et al [7], who employed Grunsky inequalities to obtain a sharper estimate for the case \(n=3\), namely, \[|a_{4}|-|a_{3}|\leq 2.1033\cdots .\]
Thus, except for the sharp bounds known in the case \(n=2\), no sharp upper or lower estimates are currently available for the difference \(|a_{n+1}|-|a_{n}|\) when \(n\geq 3\) for functions belonging to the class \(\mathcal{S}\).
The coefficient problem for some subclasses of the class \(\mathcal{S}\) is settled by several authors. We first define some well-known subclasses of the class \(\mathcal{S}\).
An analytic function \(f\) is called starlike if the image domain \(f(\mathbb{D})\) is starlike with respect to the origin. The class of all univalent starlike functions is denoted by \(\mathcal{S}^{\ast }\). Analytically, a function \(f\) belongs to \(\mathcal{S}^{\ast }\) if and only if \[\Re \left( \frac{zf^{\prime }(z)}{f(z)}\right) >0,\qquad z\in \mathbb{D}.\]Similarly, an analytic function \(f\) is called convex if the image domain \(f(\mathbb{D})\) is convex. The class of all univalent convex functions is denoted by \(\mathcal{C}\). Analytically, a function \(f\) belongs to \(\mathcal{C}\) if and only if \[\Re \left( 1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}\right) >0,\qquad z\in \mathbb{D}.\]A function \(f\) is said to be close-to-convex if and only if there exists \(g\in \mathcal{S}^{\ast }\) such that \[\Re \left( \frac{zf^{\prime }(z)}{g(z)}\right) >0,\qquad z\in \mathbb{D}.\]The class of all close-to-convex functions in \(\mathbb{D}\) is denoted by \(\mathcal{K}\). In 1973, Pommerenke [4] conjectured that for \(f\in \mathcal{S}^{\ast }\), \(\bigl||a_{n+1}|-|a_{n}|\bigr|\leq 1,\;n\geq 2,\) and in 1978, Leung [8] proved this conjecture by showing that equality is attained for the function \[f(z)=\frac{z}{(1-\rho z)(1-\sigma z)},\qquad |\rho |=|\sigma |=1.\] In 1985, Koepf [9] showed that for \(f\in \mathcal{K}\), \(\bigl||a_{n+1}|-|a_{n}|\bigr|\leq 1\) for \(n=2\) and it is an open problem for \(n\geq 3.\) It seems to be a challenging problem to find the sharp upper and lower bounds for \(\bigl||a_{n+1}|-|a_{n}|\bigr|\), when \(f\in \mathcal{C}\), that is, for the convex functions. The only noteworthy results to date are attributed to Ming and Sugawa [10], where sharp upper bounds have been discovered when \(n\geq 2\) and has sharp lower bounds when \(n=2,3\). Determining the sharp lower bounds for \(n\geq 4\) is an open problem. By following this trend many other authors established the bounds on the difference of initial coefficients of different subclasses of univalent functions. Cho et al. [11] presented the bounds on \(||a_{3}|-|a_{2}||\) for the class of Bazilevic functions and for the class of non-Bazilevič functions the bounds on \(||a_{3}|-|a_{2}||\) were recently calculated in [12]. Moreover, for the class \(\mathcal{U}(\alpha ,\lambda )\) the bounds on the difference of initial coefficients were established in [13].
Now, let \(\mathcal{B}\) represent the class of all analytic (holomorphic) functions \(\omega\) in \(\mathbb{D}\) with the property that \(\omega (0)=0\) and \(|\omega (z)|<1\) for \(z\in \mathbb{D}\). These functions are called Schwarz functions. A number of problems in geometric function theory can be answered in an easy and precised way by using the concept of subordination. An analytic function \(f\) is said to be subordinate to some other analytic function \(g\) if there exists \(\omega \in \mathcal{B}\) such that \(f(z)=g\left( \omega (z)\right)\) for \(z\in \mathbb{D}\). In the case, if \(g\) is univalent and \(f(0)=g(0)\), then \(f(\mathbb{D})\subset g(\mathbb{D})\).
In order to unify and generalize many well-known subclasses of convex and starlike functions, Ma and Minda [14] introduced a broad family of analytic function classes associated with a suitable analytic function \(\Phi\). Let \(\Phi\) be an analytic and univalent function in \(\mathbb{D}\) such that \(\Phi (\mathbb{D})\) is symmetric with respect to the real axis, starlike with respect to \(\Phi (0)=1\), and satisfies \(\Phi ^{\prime }(0)>0\). The class of Ma-Minda convex functions, denoted by \(\mathcal{C}(\Phi )\), consists of those functions \(f\in \mathcal{A}\) for which \[1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}\prec \Phi (z),\qquad z\in \mathbb{D},\]where the function \(\Phi (z)\) has series expansion of the form \[\Phi (z)=1+B_{1}z+B_{2}z^{2}+B_{3}z^{3}+\ldots .\]The Koebe’s \(1/4\) theorem states that there exists an inverse function \(f^{-1}\) for every univalent function \(f\) defined in \(\mathbb{D}\), at least on the disk with a radius 1/4, having the series expansion as follows\[f^{-1}(z)=z+\sum_{n=2}^{\infty }A_{n}z^{n}. \label{equ34}\tag{2}\] Since \(f(f^{-1}(z))=z\), so from (1 ) and (2 ), we have \[A_{2}=-a_{2},\;\;\;\text{and}\;\;\;A_{3}=2a_{2}^{2}-a_{3}. \label{A2}\tag{3}\] Libera and Zlotkiewicz [15] were the first to demonstrate that, for the class \(\mathcal{C}\) of convex functions, the inverse coefficients \(A_{n}\) retain the traditional inequality \(\left\vert a_{n}\right\vert \leq 1\) when \(2\leq n\leq 7\). Some invariance features among the family of strongly convex functions were demonstrated by Thomas and Verma [16] in 2016. In [17], the invariance property between the coefficient functionals of the subclass of convex functions associated with sigmoid functions is examined. Thomas recently provided a thorough explanation of this characteristic for a few coefficient functions for the class of convex functions and its subclasses. This characteristic is questioned for the subclass of convex functions associated with the cardioid domain, see [18].
Many authors have recently explored coefficient bounds for inverse functions (see [19], [20]). Specifically, Sim
and Thomas showed that \(-1\leq
|A_{3}|-|A_{2}|\leq 3\) for \(f\in \mathcal{S}\), [19]. They also studied the sharp bounds on coefficient differences for
some other subclasses of univalent functions.
In this article, we present the bounds on \(||a_{3}|-|a_{2}||\) for the class \(\mathcal{C}(\Phi )\). We also establish the bounds on \(||A_{3}|-|A_{2}||\).
The bounds being presented here are sharp.
The Caratheodory class, denoted by \(\mathcal{P}\), is the collection of holomorphic functions \(p\) in the unit disk \(\mathbb{D}=\left\{ z\in \mathbb{C}:|z|<1\right\}\) satisfying the condition \(Re\left( p(z)\right) >0\) for \(z\in \mathbb{D}\) and having series expansion of the form \[p(z)=1+\sum_{n=1}^{\infty }p_{n}z^{n}.\]We use the following lemma to prove our results.
Lemma 1. [21] Let \(p\in \mathcal{P}\). Then \[p_{1}=2t_{1},\]and \[p_{2}=2t_{1}^{2}+2(1-|t_{1}|^{2})t_{2},\]where \(t_{i}\in \overline{\mathbb{D}}\) for \(i\in \left\{ 1,2\right\}\). If \(|t_{1}|=1\), then there exists a unique function \(p\in \mathcal{P}\) given as \[p(z)=\frac{1+t_{1}z}{1-t_{1}z}.\]If \(t_{1}\in \mathbb{D}\) and \(|t_{2}|=1\), then there exists a unique function \(p\in \mathcal{P}\) defined as \[p(z)=\frac{1+(\overline{t_{1}}t_{2}+t_{1})z+t_{2}z^{2}}{1+(\overline{t_{1}}t_{2}-t_{1})z-t_{2}z^{2}}. \label{eq001}\tag{4}\]
In our first result, we establish the sharp bound on the difference of initial coefficients \(||a_{3}|-|a_{2}||\).
Theorem 1. Let \(f\in \mathcal{C}(\Phi )\) be of the form 1 with \(B_{1}>0\). Then \[|a_{3}|-|a_{2}|\leq \begin{cases} \frac{\left\vert B_{1}^{2}+B_{2}\right\vert -3B_{1}}{6}, & \text{if }\; |B_{1}^{2}+B_{2}|\geq 4B_{1}, \\ \frac{B_{1}}{6}, & \text{if }\;|B_{1}^{2}+B_{2}|<4B_{1},\end{cases}\]and \[|a_{3}|-|a_{2}|\geq \begin{cases} \frac{\left\vert B_{1}^{2}+B_{2}\right\vert -3B_{1}}{6}, & \text{if }\; B_{1}\geq 2|B_{1}^{2}+B_{2}|, \\[8pt] -\frac{B_{1}}{2}\sqrt{\frac{B_{1}}{B_{1}+|B_{1}^{2}+B_{2}|}}, & \text{if }\; 5B_{1}\leq 4|B_{1}^{2}+B_{2}|, \\[8pt] -\frac{B_{1}}{6}-\frac{3B_{1}^{2}}{8{|B_{1}^{2}+B_{2}|}+8{B_{1}}}, & \text{otherwise}.\end{cases}\]These inequalities are sharp.
Proof. Since \(f\in \mathcal{C}(\Phi )\), then there exists a function \(w\in \mathcal{B}\) such that \[1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}=\Phi \left( w(z)\right) .
\label{ma}\tag{5}\] Let \(p\in \mathcal{P}\). Then \[p(z)=\frac{1+w(z)}{1-w(z)}=1+p_{1}z+p_{2}z^{2}+p_{3}z^{3}+\cdots . \label{p}\tag{6}\]
Now, by comparing the coefficients of 5 and 6 , we have \[a_{2}=\frac{B_{1}}{4}p_{1},\quad \quad a_{3}:=\left( \frac{-B_{1}+B_{1}^{2}+B_{2}}{24}\right)
p_{1}^{2}+\frac{B_{1}}{12}p_{2}.
\label{a2}\tag{7}\] By using 7 , we get \[|a_{3}|-|a_{2}|=|bp_{1}^{2}+cp_{2}|-|ap_{1}|,\]with \[a:=\frac{B_{1}}{4},\quad \quad b:=\left(
\frac{-B_{1}+B_{1}^{2}+B_{2}}{24}\right) ,\quad \text{and}\quad c:=\frac{B_{1}}{12}.\]By using Lemma 1, we get \[\begin{align}
|a_{3}|-|a_{2}|& ={|4bt_{1}^{2}+c(2t_{1}^{2}+2(1-{|t_{1}|}^{2})t_{2})|}-2{|at_{1}|} \\
& \leq {m|t_{1}|}^{2}+2{c}\left( 1-{|t_{1}|}^{2}\right) |t_{2}|-2{|at_{1}|},
\end{align}\]where \(m:=|4b+2c|\). Since \(\mathcal{P}\) is rotationally invariant, so we can assume, \(t_{1}\in \lbrack 0,1]\) and by using the fact
that \(|t_{2}|\leq 1\), we have \[|a_{3}|-|a_{2}|\leq \left( {m}-2{c}\right) t_{1}^{2}-2{a}t_{1}+2{c}=\Upsilon
(x),\]where \[\Upsilon (x)=k_{2}x^{2}+k_{1}x+k_{0}, \label{equ2}\tag{8}\] with \[k_{2}={m}-2{c},\quad \quad k_{1}=-2{a},\quad \quad
k_{0}=2{c}.\]Here we have two cases:
\(\mathbf{A}_{1}:\) If \(k_{2}\leq 0\). Given that \(k_{1}<0\), thus by (8 ) we get, \(\Upsilon ^{\prime
}(x)=2k_{2}x+k_{1}<0\) and it yields that \(\Upsilon\) is a decreasing function. Therefore \[\Upsilon (x)\leq \Upsilon (0)=k_{0}=2c=\frac{B_{1}}{6}.\]\(\mathbf{A}_{2}:\) Now if \(k_{2}>0\), then \(\Upsilon\) is a quadratic function and it has positive leading coefficient so, \[\Upsilon (x)\leq \max \{\Upsilon (0),\Upsilon (1)\}.\]Let \(k_{2}+k_{1}<0\). Then \(\Upsilon (1)=k_{2}+k_{1}+k_{0}<k_{0}=\Upsilon (0)\) and thus \(\Upsilon (x)\leq \Upsilon (0)=2c=\frac{B_{1}}{6}\). Further, let \(k_{2}+k_{1}\geq 0\). Then \(\Upsilon (1)=k_{2}+k_{1}+k_{0}\geq k_{0}=\Upsilon
(0)\). Thus \[\Upsilon (x)\leq \Upsilon (1)=k_{2}+k_{1}+k_{0}=|4b+2c|-2a=\frac{\left\vert
B_{1}^{2}+B_{2}\right\vert -3B_{1}}{6}.\]From above discussion, we have \[|a_{3}|-|a_{2}|\leq
\begin{cases}
\frac{\left\vert B_{1}^{2}+B_{2}\right\vert -3B_{1}}{6}, & \text{if }\;
|B_{1}^{2}+B_{2}|\geq 4B_{1}, \\
\frac{B_{1}}{6}, & \text{if }\;|B_{1}^{2}+B_{2}|<4B_{1}.\end{cases}\]The bound is sharp for the function \(f\) defined by 5 with \(p(z)=(1+z)/(1-z)\), when
\(|B_{1}^{2}+B_{2}|\geq 4B_{1}\) and for \(|B_{1}^{2}+B_{2}|<4B_{1}\) the equality exists with \(p(z)=(1+z^{2})/(1-z^{2})\).
Now, we will find the lower bound on \(|a_{3}|-|a_{2}|\).
By using Lemma 1, we get \[|a_{3}|-|a_{2}|=\left\vert me^{i\theta }t_{1}^{2}+2ce^{i\phi
_{1}}(1-|t_{1}|^{2})t_{2}\right\vert -2|at_{1}|,\]where \[m=\left\vert 4b+2c\right\vert ,\quad \quad \theta =arg(4b+2c),\quad \quad
\phi _{1}=arg(2c).\]Since \(\mathcal{P}\) is rotationally invariant so we can assume, \(t_{1}\in
\lbrack 0,1]\) and \(t_{2}=re^{i\phi _{2}}\) with \(r\in \lbrack 0,1]\). Thus \[|a_{3}|-|a_{2}|=\left\vert me^{i\theta }t_{1}^{2}+2rce^{i\phi
}(1-t_{1}^{2})\right\vert -2at_{1},\]with \(\phi =\phi _{1}+\phi _{2}\). As \(\left\vert e^{i\phi }\right\vert =1\), so \[|a_{3}|-|a_{2}|=\left\vert
me^{i(\theta -\phi
)}t_{1}^{2}+2rc(1-t_{1}^{2})\right\vert -2at_{1}.\]Now \[\begin{align}
|a_{3}|-|a_{2}|=& \sqrt{m^{2}t_{1}^{4}+4mcrt_{1}^{2}(1-t_{1}^{2})\cos
(\theta -\phi )+4c^{2}r^{2}(1-t_{1}^{2})^{2}}-2at_{1}, \notag \\
\geq & \left\vert mt_{1}^{2}-2rc(1-t_{1}^{2})\right\vert -2at_{1},
\label{equ3}
\end{align}\tag{9}\] as \(\cos (\theta -\phi )\geq -1\). If \(b=c=0\), then \[|a_{3}|-|a_{2}|\geq -2at_{1}\geq -2a=-\frac{B_{1}}{2}.\]Now, let
\(b\neq 0,\;c\neq 0\). Then from 9 , we have two cases:
B\(_{1}\): If \(mt_{1}^{2}-2rc(1-t_{1}^{2})\leq 0\), then \[t_{1}\leq \sqrt{\frac{2rc}{m+2rc}}:=\zeta _{1}.\]So, from (9 ) and by using the fact, \(r\in \lbrack 0,1]\), we have \[\begin{align}
|a_{3}|-|a_{2}|& \geq -(m+2rc)t_{1}^{2}-2at_{1}+2rc \\
& \geq -(m+2rc)\zeta _{1}^{2}-2a\zeta _{1}+2rc \\
& =-2a\sqrt{\frac{2rc}{m+2rc}}=-2a\sqrt{\frac{r}{\frac{m}{2c}+r}} \\
& \geq -2a\sqrt{\frac{1}{\frac{m}{2c}+1}} \\
& ={-2a}\sqrt{\frac{2c}{m+2c}} \\
& =-\frac{B_{1}}{2}\sqrt{\frac{B_{1}}{B_{1}+|B_{1}^{2}+B_{2}|}}.
\end{align}\]B\(_{2}\): If \(mt_{1}^{2}-2rc(1-t_{1}^{2})>0\), then \(t_{1}>\zeta _{1}\), and we define \[\varphi (t_{1},r)=(m+2rc)t_{1}^{2}-2at_{1}-2rc.\]Since, for \(a^{2}\leq 2c(m+2c)\) and by using the fact that \(r\in \lbrack
0,1]\), we have \[\frac{\partial \varphi }{\partial t_{1}}=2(m+2r{c})t_{1}-2{a}\geq 2\sqrt{2rc(m+2rc)}-2a\geq 0.\]It implies, \(\varphi\) is an increasing function and from (9 ), we have \[|a_{3}|-|a_{2}|\geq -2{a}\sqrt{\frac{2r{c}}{m+2r{c}}}\geq -2{a}\sqrt{\frac{2{c}}{m+2{c}}}=-\frac{B_{1}}{2}\sqrt{\frac{B_{1}}{B_{1}+|B_{1}^{2}+B_{2}|}},\]as \(r\in \lbrack 0,1]\). Now, for \(a\geq m+2c\), we deduce \[\frac{\partial \varphi }{\partial t_{1}}=2(m+2r{c})t_{1}-2{a}\leq 2m+4r{c}-2{a}\leq 2m+4{c}-2{a}\leq
0.\]It implies, \(\varphi\) is a decreasing function and from (9 ), we get \[|a_{3}|-|a_{2}|\geq m-2{a}=\frac{\left\vert B_{1}^{2}+B_{2}\right\vert
-3B_{1}}{6}.\]At the end, for \(a^{2}>2c(m+2c)\) and \(a<m+2c\), we have \[\zeta _{1}<\frac{{a}}{m+2{c}}:=\zeta _{2}<1,\]and \(\frac{\partial \varphi }{\partial r}=-2c(1-t_{1}^{2})\leq 0\) for \(r\in
\lbrack 0,1]\) and \(t_{1}\in \lbrack \zeta _{1},1]\). Thus \[\varphi (t_{1},r)\geq \varphi (t_{1},1)=(m+2{c})t_{1}^{2}-2{a}t_{1}-2{c}=h(t_{1}).\]Since \(h^{\prime }(t_{1})=0\) yields \(t_{1}=\zeta _{2}\). Thus from (9 ) \[|a_{3}|-|a_{2}|\geq
-2{c-\frac{{a}^{2}}{m+2{c}}}=-\frac{B_{1}}{6}-\frac{3B_{1}^{2}}{8{|B_{1}^{2}+B_{2}|}+8{B_{1}}}.\]From above discussion, we have \[|a_{3}|-|a_{2}|\geq
\begin{cases}
\frac{\left\vert B_{1}^{2}+B_{2}\right\vert -3B_{1}}{6}, & \text{if }\;
B_{1}\geq 2|B_{1}^{2}+B_{2}|, \\[8pt]
-\frac{B_{1}}{2}\sqrt{\frac{B_{1}}{B_{1}+|B_{1}^{2}+B_{2}|}}, & \text{if }\;
5B_{1}\leq 4|B_{1}^{2}+B_{2}|, \\[8pt]
-\frac{B_{1}}{6}-\frac{3B_{1}^{2}}{8{|B_{1}^{2}+B_{2}|}+8{B_{1}}}, & \text{otherwise}.\end{cases}\]When \(B_{1}\geq 2|B_{1}^{2}+B_{2}|\), the inequality is sharp for the function \(f\) defined by 5 with \(p(z)=(1+z)/(1-z)\).
Let \(5B_{1}\leq 4|B_{1}^{2}+B_{2}|\). Then by Lemma 1, the equality exists for the function \(f\) defined by 5 , where \(p(z)\) is given by 4 with \[t_{1}=\sqrt{\frac{{B_{1}}}{{|B_{1}^{2}+B_{2}|}+{B_{1}}}},\quad \text{and}\quad t_{2}=\begin{cases}
-\frac{B_{1}^{2}+B_{2}}{{|B_{1}^{2}+B_{2}|}}, & \text{if }\;
B_{1}^{2}+B_{2}\neq 0, \\[1.2em]
1, & \text{if }\;B_{1}^{2}+B_{2}=0.\end{cases}\]For \(B_{1}^{2}+B_{2}=0\), the calculations are obvious and \(|a_{3}|-|a_{2}|=-\frac{B_{1}}{2}\).
Now for \(B_{1}^{2}+B_{2}\neq 0\), \[\begin{align}
|a_{3}|-|a_{2}|& =\left\vert bp_{1}^{2}+cp_{{2}}|-|ap_{1}\right\vert \\
& =\frac{1}{6}\left\vert \left( B_{1}^{2}+B_{2}\right)
t_{1}^{2}+B_{1}(1-t_{1}^{2})t_{2}\right\vert -\frac{1}{2}\left\vert
B_{1}t_{1}\right\vert ,
\end{align}\]and after some simplifications, we get \[|a_{3}|-|a_{2}|=-\frac{B_{1}}{2}\sqrt{\frac{B_{1}}{B_{1}+|B_{1}^{2}+B_{2}|}}.\]Lastly, for \(5B_{1}>4|B_{1}^{2}+B_{2}|\) and
\(B_{1}<2|B_{1}^{2}+B_{2}|\), the inequality is sharp for the function \(f\) defined by 5 , where \(p(z)\) is given by 4 with \[t_{1}=\frac{3{B_{1}}}{2{|B_{1}^{2}+B_{2}|}+2{B_{1}}},\quad \text{and}\quad
t_{2}=-\frac{B_{1}^{2}+B_{2}}{{|B_{1}^{2}+B_{2}|}}.\]Now \[\begin{align}
36|bp_{1}^{2}+cp_{2}|^{2}& =\left\vert \left( B_{1}^{2}+B_{2}\right)
t_{1}^{2}+B_{1}(1-t_{1}^{2})t_{2}\right\vert ^{2} \\
& =\left\vert B_{1}^{2}+B_{2}\right\vert ^{2}t_{1}^{4}+2Re\left( B_{1}(\overline{B_{1}^{2}+B_{2}})t_{1}^{2}(1-t_{1}^{2})t_{2}\right)
+B_{1}^{2}(1-t_{1}^{2})^{2} \\
& =\left\vert B_{1}^{2}+B_{2}\right\vert ^{2}t_{1}^{4}-2\left\vert
B_{1}^{3}+B_{1}B_{2}\right\vert
t_{1}^{2}(1-t_{1}^{2})+B_{1}^{2}(1-t_{1}^{2})^{2} \\
& =\left( \left\vert B_{1}^{2}+B_{2}\right\vert
t_{1}^{2}-B_{1}(1-t_{1}^{2})\right) ^{2} \\
& =\left( \frac{9B_{1}^{2}}{4{|B_{1}^{2}+B_{2}|}+4{B_{1}}}-B_{1}\right) ^{2}.
\end{align}\]Since \(5B_{1}>4|B_{1}^{2}+B_{2}|\), it implies \(\frac{9B_{1}^{2}}{4{|B_{1}^{2}+B_{2}|}+4{B_{1}}}-B_{1}>0\) and thus \[|bp_{1}^{2}+cp_{2}|=\frac{3B_{1}^{2}}{8{|B_{1}^{2}+B_{2}|}+8{B_{1}}}-\frac{B_{1}}{6}.\]Therefore \[|a_{3}|-|a_{2}|=|bp_{1}^{2}+cp_{2}|-|ap_{1}|=-\frac{B_{1}}{6}-\frac{3B_{1}^{2}}{8{|B_{1}^{2}+B_{2}|}+8{B_{1}}}.\]It completes the proof. ◻
In this theorem, we present the sharp bounds on \(||A_{3}|-|A_{2}||\).
Theorem 2. Let \(f\in \mathcal{C}(\Phi )\) be of the form 1 with \(B_{1}>0\). Then \[|A_{3}|-|A_{2}|\leq \begin{cases} \frac{{|2B_{1}^{2}-B_{2}|}-3{B_{1}}}{6}, & \text{if }\; |2B_{1}^{2}-B_{2}|\geq 4B_{1}, \\ \frac{B_{1}}{6}, & \text{if }\;|2B_{1}^{2}-B_{2}|<4B_{1}.\end{cases}\]and \[|A_{3}|-|A_{2}|\geq \begin{cases} \frac{\left\vert 2B_{1}^{2}-B_{2}\right\vert -3B_{1}}{6}, & \text{if }\; B_{1}\geq |4B_{1}^{2}-2B_{2}|, \\[8pt] -\frac{B_{1}}{2}\sqrt{\frac{B_{1}}{B_{1}+|2B_{1}^{2}-B_{2}|}}, & \text{if }\; 5B_{1}\leq |8B_{1}^{2}-4B_{2}|, \\[8pt] -\frac{B_{1}}{6}-\frac{3B_{1}^{2}}{8{|2B_{1}^{2}-B_{2}|}+8{B_{1}}}, & \text{otherwise}.\end{cases}\]All of these bounds are sharp.
Proof. From 3 , we have \[A_{2}=-\frac{B_{1}}{4}p_{1},\quad \quad A_{3}:=\left( \frac{2B_{1}^{2}+B_{1}-B_{2}}{24}\right) p_{1}^{2}-\frac{B_{1}}{12}p_{2}.\]Now, by using 7 , we get \[|A_{3}|-|A_{2}|=|bp_{1}^{2}+cp_{2}|-|ap_{1}|,\]where \[a:=\frac{B_{1}}{4},\quad \quad b:=\left( \frac{2B_{1}^{2}+B_{1}-B_{2}}{24}\right) ,\quad \text{and}\quad
c:=-\frac{B_{1}}{12}.\]By utilizing Lemma 1, we have \[\begin{align}
|A_{3}|-|A_{2}|& ={|4bt_{1}^{2}+c(2t_{1}^{2}+2(1-{|t_{1}|}^{2})t_{2})|}-2{|at_{1}|} \\
& \leq {m|t_{1}|}^{2}+2{c}\left( 1-{|t_{1}|}^{2}\right) |t_{2}|-2{|at_{1}|},
\end{align}\]with \(m:=|4b+2c|\). As we know, \(\mathcal{P}\) is rotationally invariant, it implies, \(t_{1}\in \lbrack 0,1]\) and by using the
inequality, \(|t_{2}|\leq 1\), we get \[|A_{3}|-|A_{2}|\leq \left( {m}-2{c}\right) t_{1}^{2}-2{a}t_{1}+2{c}=\Upsilon
(x),\]with \[\Upsilon (x)=k_{2}x^{2}+k_{1}x+k_{0}, \label{equ12}\tag{10}\] where \[k_{2}={m}-2{c},\quad \quad k_{1}=-2{a},\quad \quad
k_{0}=2{c}.\]Now, there are two cases:
\(\mathbf{A}_{1}:\) Let \(k_{2}\leq 0\). For \(k_{1}<0\), so (10 ) implies, \(\Upsilon ^{\prime
}(x)=2k_{2}x+k_{1}<0\) and thus we have that \(\Upsilon\) is decreasing. Therefore \[\Upsilon (x)\leq \Upsilon (0)=k_{0}=2c=\frac{B_{1}}{6}.\]\(\mathbf{A}_{2}:\) If \(k_{2}>0\), then \(\Upsilon\) is a quadratic function with positive leading coefficient, thus \[\Upsilon
(x)\leq \max \{\Upsilon (0),\Upsilon (1)\}.\]Here, if \(k_{2}+k_{1}<0\), it implies \(\Upsilon
(1)=k_{2}+k_{1}+k_{0}<k_{0}=\Upsilon (0)\) and it yields \(\Upsilon (x)\leq
\Upsilon (0)=2c\). On the other hand, if \(k_{2}+k_{1}\geq 0\), then \(\Upsilon
(1)=k_{2}+k_{1}+k_{0}\geq k_{0}=\Upsilon (0)\). Therefore \[\Upsilon (x)\leq \Upsilon (1)=k_{2}+k_{1}+k_{0}=|4b+2c|-2a=\frac{{|2B_{1}^{2}-B_{2}|}-3{B_{1}}}{6}.\]Thus, from the above discussion, we have \[|A_{3}|-|A_{2}|\leq
\begin{cases}
\frac{{|2B_{1}^{2}-B_{2}|}-3{B_{1}}}{6}, & \text{if }\;
|2B_{1}^{2}-B_{2}|\geq 4B_{1}, \\
\frac{B_{1}}{6}, & \text{if }\;|2B_{1}^{2}-B_{2}|<4B_{1}.\end{cases}\]The inequality is sharp for the function \(f\) defined by 5 with \(p(z)=(1+z)/(1-z)\), when \(|2B_{1}^{2}-B_{2}|\geq 4B_{1}\) and for \(|2B_{1}^{2}-B_{2}|<4B_{1},\) the extremal function \(f\)
is given by 5 with \(p(z)=(1+z^{2})/(1-z^{2})\).
Now, for the lower bound on \(|A_{3}|-|A_{2}|\) by using Lemma 1, we have \[|A_{3}|-|A_{2}|=\left\vert me^{i\theta
}t_{1}^{2}+2ce^{i\phi
_{1}}(1-|t_{1}|^{2})t_{2}\right\vert -2|at_{1}|,\]with \[m=\left\vert 4b+2c\right\vert ,\quad \quad \theta =arg(4b+2c),\quad \quad
\phi _{1}=arg(2c).\]Since \(\mathcal{P}\) possesses the property of rotational invariance, thus we can have, \(t_{1}\in \lbrack 0,1]\) and \(t_{2}=re^{i\phi
_{2}}\) where \(r\in
\lbrack 0,1]\). Therefore \[|A_{3}|-|A_{2}|=\left\vert me^{i\theta }t_{1}^{2}+2rce^{i\phi
}(1-t_{1}^{2})\right\vert -2at_{1},\]with \(\phi =\phi _{1}+\phi _{2}\). As \(\left\vert e^{i\phi }\right\vert =1\), so \[|A_{3}|-|A_{2}|=\left\vert
me^{i(\theta -\phi
)}t_{1}^{2}+2rc(1-t_{1}^{2})\right\vert -2at_{1}.\]It implies \[\begin{align}
|A_{3}|-|A_{2}|=& \sqrt{m^{2}t_{1}^{4}+4mcrt_{1}^{2}(1-t_{1}^{2})\cos
(\theta -\phi )+4c^{2}r^{2}(1-t_{1}^{2})^{2}}-2at_{1}, \notag \\
\geq & \left\vert mt_{1}^{2}-2rc(1-t_{1}^{2})\right\vert -2at_{1},
\label{equ13}
\end{align}\tag{11}\] since \(\cos (\theta -\phi )\geq -1\). Let \(b=c=0\). Then \[|A_{3}|-|A_{2}|\geq -2at_{1}\geq -2a=-\frac{B_{1}}{2}.\]On
the other hand, let \(b\neq 0,\;c\neq 0\). Then from 11 , there are two cases:
B\(_{1}\): Let \(mt_{1}^{2}-2rc(1-t_{1}^{2})\leq 0\). Then \[t_{1}\leq \sqrt{\frac{2rc}{m+2rc}}:=\zeta _{1}.\]By using (11 ) and the fact that, \(r\in \lbrack 0,1]\), we deduce \[\begin{align}
|A_{3}|-|A_{2}|& \geq -(m+2rc)t_{1}^{2}-2at_{1}+2rc \\
& \geq -(m+2rc)\zeta _{1}^{2}-2a\zeta _{1}+2rc \\
& =-2a\sqrt{\frac{2rc}{m+2rc}}=-2a\sqrt{\frac{r}{\frac{m}{2c}+r}} \\
& \geq -2a\sqrt{\frac{1}{\frac{m}{2c}+1}} \\
& ={-2a}\sqrt{\frac{2c}{m+2c}} \\
& =-\frac{B_{1}}{2}\sqrt{\frac{B_{1}}{B_{1}+|2B_{1}^{2}-B_{2}|}}.
\end{align}\]B\(_{2}\): Now, let \(mt_{1}^{2}-2rc(1-t_{1}^{2})>0\). Then \(t_{1}>\zeta _{1}\), and we introduce a function
\(\varphi\) such that \[\varphi (t_{1},r)=(m+2rc)t_{1}^{2}-2at_{1}-2rc.\]Now, by using \(a^{2}\leq 2c(m+2c)\) and the fact that \(r\in \lbrack 0,1]\), we get \[\frac{\partial \varphi }{\partial t_{1}}=2(m+2r{c})t_{1}-2{a}\geq 2\sqrt{2rc(m+2rc)}-2a\geq 0.\]It implies, \(\varphi\) is an
increasing function and by using (11 ), we get \[|A_{3}|-|A_{2}|\geq -2{a}\sqrt{\frac{2r{c}}{m+2r{c}}}\geq
-2{a}\sqrt{\frac{2{\;c}}{m+2{c}}}=-\frac{B_{1}}{2}\sqrt{\frac{B_{1}}{B_{1}+|2B_{1}^{2}-B_{2}|}},\]since \(r\in \lbrack 0,1]\). on the other hand, if \(a\geq m+2c\), then we get \[\frac{\partial \varphi }{\partial t_{1}}=2(m+2r{c})t_{1}-2{a}\leq 2m+4r{\;c}-2{a}\leq 2m+4{c}-2{a}\leq 0.\]It implies, \(\varphi\) is a decreasing function and from (11
), we get \[|A_{3}|-|A_{2}|\geq m-2{a}=\frac{\left\vert 2B_{1}^{2}-B_{2}\right\vert
-3B_{1}}{6}.\]Lastly, when \(a^{2}>2c(m+2c)\) and \(a<m+2c\), we get \[\zeta _{1}<\frac{{a}}{m+2{c}}:=\zeta _{2}<1,\]and \(\frac{\partial \varphi }{\partial r}=-2c(1-t_{1}^{2})\leq 0\) for \(r\in
\lbrack 0,1]\) and \(t_{1}\in \lbrack \zeta _{1},1]\). Therefore \[\varphi (t_{1},r)\geq \varphi (t_{1},1)=(m+2{c})t_{1}^{2}-2{a}t_{1}-2{c}=h(t_{1}).\]As \(h^{\prime }(t_{1})=0\) implies \(t_{1}=\zeta _{2}\). Therefore by using (11 ), we deduce \[|A_{3}|-|A_{2}|\geq
-2{c-\frac{{a}^{2}}{m+2{c}}}=-\frac{B_{1}}{6}-\frac{3B_{1}^{2}}{8{|2B_{1}^{2}-B_{2}|}+8{B_{1}}}.\]Now, all the above discussion implies \[|A_{3}|-|A_{2}|\geq
\begin{cases}
\frac{\left\vert 2B_{1}^{2}-B_{2}\right\vert -3B_{1}}{6}, & \text{if }\;
B_{1}\geq |4B_{1}^{2}-2B_{2}|, \\[8pt]
-\frac{B_{1}}{2}\sqrt{\frac{B_{1}}{B_{1}+|2B_{1}^{2}-B_{2}|}}, & \text{if }\;
5B_{1}\leq |8B_{1}^{2}-4B_{2}|, \\[8pt]
-\frac{B_{1}}{6}-\frac{3B_{1}^{2}}{8{|2B_{1}^{2}-B_{2}|}+8{B_{1}}}, & \text{otherwise}.\end{cases}\]For \(B_{1}\geq |4B_{1}^{2}-2B_{2}|\), the equality exits for the function \(f\) given by 5 with \(p(z)=(1+z)/(1-z)\).
Now, Let \(5B_{1}\leq |8B_{1}^{2}+4B_{2}|\). Then Lemma 1 implies that the bound is sharp for the function \(f\) given
by 5 , with \(p(z)\) is defined by 4 where \[t_{1}=\sqrt{\frac{{B_{1}}}{{|2B_{1}^{2}-B_{2}|}+{B_{1}}}},\quad \text{and}\quad t_{2}=\begin{cases}
\frac{2B_{1}^{2}-B_{2}}{{|2B_{1}^{2}-B_{2}|}}, & \text{if }\;2B_{1}^{2}\neq
B_{2}, \\[1.2em]
1, & \text{if }\;2B_{1}^{2}=B_{2}.\end{cases}\]Let \(2B_{1}^{2}=B_{2}\). Then it is obvious \(|A_{3}|-|A_{2}|=-\frac{B_{1}}{2}\).
Now, let \(2B_{1}^{2}\neq B_{2}\). Then \[\begin{align}
|A_{3}|-|A_{2}|& =\left\vert bp_{1}^{2}+cp_{{2}}|-|ap_{1}\right\vert \\
& =\frac{1}{6}\left\vert \left( 2B_{1}^{2}-B_{2}\right)
t_{1}^{2}-B_{1}(1-t_{1}^{2})t_{2}\right\vert -\frac{1}{2}\left\vert
B_{1}t_{1}\right\vert
\end{align}\]and after some calculations, we deduce \[|A_{3}|-|A_{2}|=-\frac{B_{1}}{2}\sqrt{\frac{B_{1}}{B_{1}+|2B_{1}^{2}-B_{2}|}}.\]At the end, for \(5B_{1}>|8B_{1}^{2}-4B_{2}|\) and \(B_{1}<|4B_{1}^{2}-2B_{2}|\), the extremal function \(f\) is given by 5 , where \(p(z)\) is defined by 4 with \[t_{1}=\frac{3{B_{1}}}{{|4B_{1}^{2}-2B_{2}|}+2{B_{1}}},\quad \text{and}\quad
t_{2}=\frac{2B_{1}^{2}-B_{2}}{{|2B_{1}^{2}-B_{2}|}}.\]Now \[\begin{align}
36|bp_{1}^{2}+cp_{2}|^{2}& =\left\vert \left( 2B_{1}^{2}-B_{2}\right)
t_{1}^{2}-B_{1}(1-t_{1}^{2})t_{2}\right\vert ^{2} \\
& =\left\vert 2B_{1}^{2}-B_{2}\right\vert ^{2}t_{1}^{4}-2Re\left( B_{1}(\overline{2B_{1}^{2}-B_{2}})t_{1}^{2}(1-t_{1}^{2})t_{2}\right)
+B_{1}^{2}(1-t_{1}^{2})^{2} \\
& =\left\vert 2B_{1}^{2}-B_{2}\right\vert ^{2}t_{1}^{4}-2\left\vert
2B_{1}^{3}-B_{1}B_{2}\right\vert
t_{1}^{2}(1-t_{1}^{2})+B_{1}^{2}(1-t_{1}^{2})^{2} \\
& =\left( \left\vert 2B_{1}^{2}-B_{2}\right\vert
t_{1}^{2}-B_{1}(1-t_{1}^{2})\right) ^{2} \\
& =\left( \frac{9B_{1}^{2}}{{|8B_{1}^{2}-4B_{2}|}+4{B_{1}}}-B_{1}\right)
^{2}.
\end{align}\]As, \(5B_{1}>|8B_{1}^{2}-4B_{2}|\), it gives \(\frac{9B_{1}^{2}}{{|8B_{1}^{2}-4B_{2}|}+4{B_{1}}}-B_{1}>0\), therefore \[|bp_{1}^{2}+cp_{2}|=\frac{3B_{1}^{2}}{8{|2B_{1}^{2}-B_{2}|}+8{B_{1}}}-\frac{B_{1}}{6}.\]It implies \[|A_{3}|-|A_{2}|=|bp_{1}^{2}+cp_{2}|-|ap_{1}|=-\frac{B_{1}}{6}-\frac{3B_{1}^{2}}{8{|2B_{1}^{2}-B_{2}|}+8{B_{1}}}.\]It completes the proof. ◻
Remark 3. From Theorem 1 and Theorem 2, it is seen that the upper bounds on \(|a_{3}|-|a_{2}|\) and \(|A_{3}|-|A_{2}|\) are the same when \(4B_{1}>\max \left\lbrace |B_{1}^{2}+B_{2}|, \;|2B_{1}^{2}-B_{2}|\right\rbrace\).
Now we present example of functions in which this property hold.
Example 1. Let
\(\Phi _{e}(z)=e^{z}=1+z+\frac{1}{2}z^{2}+\cdots ,\)
\(\Phi _{\sin }(z)=1+\sin (z)=1+z-\frac{1}{6}z^{3}+\cdots ,\)
\(\Phi _{L}(z)=\sqrt{1+z}=1+\frac{1}{2}z-\frac{1}{8}z^{2}+\cdots ,\)
\(\Phi _{l}(z)=z+\sqrt{1+z^{2}}=1+z+\frac{1}{2}z^{2}+\cdots ,\)
\(\Phi _{RL}(z)=\sqrt{2}-(\sqrt{2}-1)\sqrt{\frac{1-z}{1+2(\sqrt{2}-1)z}}=1+\frac{5-3\sqrt{2}}{2}z+\frac{71-51\sqrt{2}}{8}z^{2}+\cdots ,\)
\(\Phi _{C}(z)=1+\frac{4z}{3}+\frac{2z^{2}}{3},\)
\(\Phi _{\mathcal{BS}}(z)=1+\frac{z}{1-\alpha z^{2}}=1+z+\alpha z^{3}+\cdots\)
Then these functions are the subordinating functions for the classes \(\mathcal{C}_{e},\) \(\mathcal{C}_{s},\) \(\mathcal{C}_{L},\) \(\mathcal{C}_{l},\; \mathcal{C}_{RL},\mathcal{C}_{C},\) therefore the upper bounds on \(|a_{3}|-|a_{2}|\) and \(|A_{3}|-|A_{2}|\) are the same for functions in these classes.
Remark 4. From Theorem 1 and Theorem 2, it can also be observed that the lower bounds on \(|a_{3}|-|a_{2}|\) and \(|A_{3}|-|A_{2}|\) are the same when \(B_{2}=\frac{B_{1}^2}{2}.\)
In the following example we present a few functions in which this property hold.
Example 2. Let
\(\Phi _{e}(z)=e^{z}=1+z+\frac{1}{2}z^{2}+\cdots ,\)
\(\Phi _{l}(z)=z+\sqrt{1+z^{2}}=1+z+\frac{1}{2}z^{2}+\cdots.\)
Then these functions are the subordinating functions for the classes \(\mathcal{C}_{e}\) and \(\mathcal{C}_{l}\), thus the lower bounds on \(|a_{3}|-|a_{2}|\) and \(|A_{3}|-|A_{2}|\) are same for functions in these classes.