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A comprehensive study on fuzzy and crisp graph indices: generalized formulae, proximity and accuracy analysis

  • This article presents a set of generalized -dependent formulae for fuzzy and crisp versions of various graph indices, including the first and second Zagreb indices, harmonic index and Randic index. These formulae are applied to the identity graph of the commutative ring Z, and the resulting indices are calculated by using MATLAB software for 20 prime numbers. The generated data were applied for machine learning by using Python and Jupyter notebook to investigate the relationship between fuzzy and crisp indices. The article also includes the relationship between fuzzy and crisp indices in the form of six-degree polynomials and an error analysis.

    Citation: Muhammad Umar Mirza, Rukhshanda Anjum, Maged Z. Youssef, Turki Alsuraiheed. A comprehensive study on fuzzy and crisp graph indices: generalized formulae, proximity and accuracy analysis[J]. AIMS Mathematics, 2023, 8(12): 30922-30939. doi: 10.3934/math.20231582

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  • This article presents a set of generalized -dependent formulae for fuzzy and crisp versions of various graph indices, including the first and second Zagreb indices, harmonic index and Randic index. These formulae are applied to the identity graph of the commutative ring Z, and the resulting indices are calculated by using MATLAB software for 20 prime numbers. The generated data were applied for machine learning by using Python and Jupyter notebook to investigate the relationship between fuzzy and crisp indices. The article also includes the relationship between fuzzy and crisp indices in the form of six-degree polynomials and an error analysis.



    Consider D={z:|z|<1}, an open unit disc in the complex plane C, H as the collection of all analytic functions in D. Let A be the subclass of H consisting of all functions of the form

    f(z)=z+n=2anzn,zD

    and S be a subclass of A containing all univalent functions. A function fA is starlike if f(D) is starlike with respect to origin and the class of all starlike functions fA is denoted by ST. Similarly, a function fA is convex if f(D) is convex with respect to all points of f(D) and the class of all convex functions fA is denoted by CV. An analytic function p:DC of the form

    p(z)=1+n=1pnzn,|pn|2 (1.1)

    satisfying (p(z))>0 for all zD is known as a function with positive real part. The class of such functions, denoted by P, is known as the class of Carathéodory functions. Note that (p(z))>0 can be written as |arg(p(z))|<π2. Connections between ST and P and CV and P are as follows: a function fST if and only if zf(z)f(z)P,zD and a function fCV if and only if 1+zf(z)f(z)P,zD. Thus, the properties of ST and CV functions can be obtained from the properties of functions in the class P. Note that the Möbius function

    L0(z)=1+z1z=1+2z+2z2+...=1+2n=1zn,zD, (1.2)

    is analytic and univalent in the open unit disc D and it maps D onto the right half-plane and is in the class P.

    Even though many authors extensively explored the concept of starlikeness of a given order for a long time, Robertson [16] was the pioneer in introducing the concept of an analytic univalent functions mapping an open unit disc onto a starlike domain with respect to the boundary point. He constructed the subclass G of H of functions g,g(0)=1 mapping D onto a starlike domain with respect to g(1)=limr1g(r)=0 and (eiρg(z))>0 for some real ρ and all zD. Assume also that the constant function 1 belongs to the class G. He conjectured that the class G coincides with the class G,

    G={gH:g(z)0,g(z)=1+n=1dnzn,zD}, (1.3)

    such that

    (2zg(z)g(z)+1+z1z)>0,zD (1.4)

    and proved that GG. Later, this conjecture was confirmed by Lyzzaik [11] who proved GG. Furthermore, if gG,g1 then g is univalent close-to-convex in D, as proved by Robertson [16]. It is worth mentioning that the analytic characterization (1.4) was known earlier to Styer [19].

    In [3], a class G(M),M>1, consisting of all analytic and non-vanishing functions of the form (1.3), such that

    (2zg(z)g(z)+zP(z;M)P(z;M))>0,zD,

    which is a closely related function to the class G was introduced by Jakubowski [3]. Here,

    P(z;M)=4z((1z)2+4zM+1z)2,zD

    is the Pick function. The class G(1) was also considered in [3], where

    G(1)=(gH:g(z)0,(1+2zg(z)g(z))>0,zD).

    Todorov [20] linked the class G with a functional f(z)/(1z) for zD and obtained a structural formula and coefficient estimates. Silverman and Silvia [17] introduced an interesting class G(β)G,0<β1, consisting of all analytic function g of the form (1.3) and satisfying

    (zg(z)g(z)+(1β)1+z1z)>0,zD.

    Clearly, G(12)=G. For 1<A1 and A<B1, Jakubowski and Włodarczyk [4] defined the class G(A,B), of all g of the form (1.3), satisfying (2zg(z)g(z)+1+Az1Bz)>0,zD (see also the work of Sivasubramanian [18]). Related works on the class G were considered earlier by [1, 6–10, 14]. We remark at this point, that the function

    12ln1+z1z=z+z33++z2n+12n+1+

    is univalent in D. In this article, we are interested in introducing and investigating a new class as follows.

    Definition 1.1. Let Gc be the class consisting of all functions of the form (1.3) satisfying

    {2z g(z)g(z)+(1+z1z)c}>0, zD, (1.5)

    where 0<c2.

    If c=1, the class G1=G was introduced and investigated by Robertson [16]. For this new class of functions, we obtain representation theorem, interesting coefficient estimates and also certain differential subordination implications involving this new class.

    Example 1.1. The function

    g1(z)=11zexp{12z0((1+t1t)c1t)dt}

    is in the class Gc, where 0<c2.

    Proof. Taking logarithm on both sides and by a simple differentiation, one can easily get

    2g1(z)g1(z)+1z(1+z1z)c=1+zz(1z).

    Hence,

    {2zg1(z)g1(z)+(1+z1z)c}={1+z1z}>0

    implies g1Gc.

    Example 1.2. The function

    g2(z)=exp{12z0((1+t1t)c1t)dt}

    is in the class Gc, where 0<c2.

    Proof. Taking logarithm on both sides and by a simple differentiation, one can easily get

    2 g2(z)g2(z)+1z(1+z1z)c=1z.

    Therefore,

    {2z g2(z)g2(z)+(1+z1z)c}=1>0

    implies that g2Gc.

    Similarly, we can show that

    Example 1.3. The function

    g3(z)=11zexp{12z0((1+t1t)c1t)dt}

    is in the class Gc, where 0<c2.

    Example 1.4. The function

    g4(z)=11+zexp{12z0((1+t1t)c1t)dt}

    is in the class Gc, where 0<c2.

    The above examples show that there are many functions present in the class Gc proving that the class Gc is non empty.

    Theorem 2.1. Let 0<c2. Furthermore, let

    βc(z)=12cz0((1+t1t)c1t)dt. (2.1)

    A function g is in Gc if and only if there exists a starlike function sST such that

    g(z)=(s(z)z)12exp{c βc(z)}. (2.2)

    Proof. Suppose that sST and g is given by (2.2). Then, g is analytic and g(0)=1. Therefore, from (1.5), we have for some function g satisfying (2.2), there exists a starlike function s such that

    (g(z))2(zexp{2cβc(z)})=s(z).

    Hence,

    2logg(z)+logz2cβc(z)=logs(z).

    By a simple differentiation followed by simplification we get,

    2zg(z)g(z)+(1+z1z)c=zs(z)s(z),

    where

    (1+z1z)c=1+2cz+2c2z2+22c3+c3z3+2c4+2c23z4+.

    Therefore,

    {2z g(z)g(z)+(1+z1z)c}=(z s(z)s(z))>0.

    Hence, gGc.

    On the other hand, suppose gGc and

    s(z)=z g2(z)exp{2cβc(z)}.

    Then, s(0)=0,s(0)=1 and

    ( zs(z)s(z))=(2z g(z)g(z)+(1+z1z)c).

    The above expression is positive as gGc which implies sST.

    For the choices of c=1 and c=2 we get the following corollaries as listed below.

    Corollary 2.1. [16] A function g is in G1 if and only if there exists a function sST such that (g(z))2=(s(z)z)(1z)2.

    Corollary 2.2. A function g is in G2 if and only if there exists a function sST such that g(z)=(s(z)z)12 exp(21z).

    Theorem 2.2. (Herglotz representation theorem) Let 0<c2 and let g be an analytic function in D such that g(0)=1. Then, gGc if and only if

    g(z)=exp [ππlog(1zeit)dμ(t)12z0((1+t1t)c1t)dt], (2.3)

    where μ(t) is a probability measure on [π,π].

    Proof. Let 0<c2. If gGc, we can write

    2z g(z)g(z)+(1+z1z)c=ππ1+z eit1z eitdμ(t).

    By a simple integration and simplification we get,

    2 logg(z)=z0((1+t1t)c1t)dt2ππlog(1zeit)dμ(t).

    Upon simplification of the above equation, one can obtain (2.3). The converse part can be proved by similar lines as in the necessary part and hence the details are omitted.

    For c=1 and c=2 we get the following corollaries.

    Corollary 2.3. [16] Let g be an analytic function in D such that g(0)=1. Then, gG1 if and only if

    g(z)=(1z) exp(ππlog(1zeit)dμ(t)).

    Corollary 2.4. Let g be an analytic function in D such that g(0)=1. Then, gG2 if and only if

    g(z)=exp (21z+ππlog(1zeit)dμ(t)).

    Theorem 2.3. Let 0<c2. A function gGc if and only if there exists a function pP such that

    g(z)=1z exp (12(z0p(ζ)ζdζz0((1+t1t)c1t)dt)). (2.4)

    Proof. Let gGc. Then, by the definition of Gc, for some function pP,

    2g(z)g(z)+1z(1+z1z)c=p(z)z.

    Upon integration and simplification, we get

    logz(g(z))2=z0p(ζ)ζdζz0((1+t1t)c1t)dt

    which proves the necessary part of the theorem. Conversely, assume pP and p(0)=1 and let g be as in (2.4). Then, g is analytic in D and by applying simple calculations we can easily prove that gGc.

    For c=1 and c=2, we get the following corollaries.

    Corollary 2.5. Let gG1 if and only if there exists a function pP such that

    g(z)=1zz exp( 12z0p(ζ)ζdζ).

    Corollary 2.6. Let gG2 if and only if there exists a function pP such that

    g(z)=1z exp (12z0p(ζ)ζdζ2z1z).

    Theorem 2.4. Let 0<c2. A function gGc if and only if there exists a function p\in \mathcal{H} such that p\prec \frac{1+z}{1-z} and for z\in\mathbb{D} ,

    \begin{equation*} g(z) = \frac{1}{\sqrt z} \ \exp \ \left(\frac{1}{2}\left(\int\limits_0^z \frac{p(\zeta)}{\zeta}d\zeta- \int_0^z \left(\frac{\left(\frac{1+t}{1-t}\right)^c-1}{t}\right)dt \right)\right). \end{equation*}

    Proof. The proof follows from Theorem 2.3.

    Theorem 3.1. Let 0 < c \leq 2 and z\in \mathbb{D} . If g \in \mathcal{G}_c , we have the following sharp inequalities.

    \begin{equation} \left|c+d_1\right| \leq 1, \end{equation} (3.1)
    \begin{equation} \left| c^2+2d_2-d_1^2\right| \leq 1, \end{equation} (3.2)

    and

    \begin{equation} \left|2c^3+c+9d_3-9d_1d_2+3d_1^3\right| \leq 3. \end{equation} (3.3)

    Further, for \alpha \in \mathbb{R} , let

    \begin{equation} \mathcal{H}(\alpha, c) = 4d_{2}- 8 \alpha c d_{1}- 2d_{1}^{2}(1+ 2\, \alpha)+ 2 c^{2}( 1-2 \alpha). \end{equation} (3.4)

    Then,

    \begin{equation} |\mathcal{H}(\alpha, c)|\leq \begin{cases} 2\left(1-2\alpha |c+d_1|^2\right), \mathit{\text{ if }}\alpha \leq \frac{1}{2}, \\ 2\left(1-2(1-\alpha) |c+d_1|^2\right), \mathit{\text{ if }}\alpha \geq \frac{1}{2}. \end{cases} \end{equation} (3.5)

    Proof. Let

    p(z) = 2z\frac{g^{\prime}(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c, \ z\in\mathbb{D}.

    On expanding the right hand side of the above function p , we get

    \begin{equation} p(z) = 1+2\left(c+d_1\right)z +2\left(c^2+2d_2-d_1^2\right)z^2+\frac{2}{3}\left(2c^3+c+9d_3-9d_1d_2+3d_1^3\right)z^3+\cdots . \end{equation} (3.6)

    By making use of the known inequality |p_i| \leq 2 for all p \in \mathcal{P} , we can get the sharp inequalities given in (3.1)–(3.3). From (1.1) and (3.6) and from the known fact that

    |p_2-\alpha p_1^2|\leq \begin{cases} 2-\alpha\left|p_1\right|^2, \ \text{ if }\alpha \leq \frac{1}{2},\\ 2-\left(1-\alpha\right)\left|p_1\right|^2, \ \text{ if }\alpha \geq \frac{1}{2}, \end{cases}

    we can obtain (3.5)

    For c = 1 and c = 2, we have the following corollaries as stated below.

    Corollary 3.1. [2] Let z\in \mathbb{D} . If g \in \mathcal{G}_1 , we have the following inequalities.

    \begin{equation*} \left|1+d_1\right| \leq 1,\, \left| 1+2d_2-d_1^2\right| \leq 1 ,\, \left|1+3d_3-3d_1d_2+d_1^3\right| \leq 1. \end{equation*}

    Further,

    \begin{equation*} |\mathcal{H}(\alpha, 1)|\leq \begin{cases} 2\left(1-2\alpha |1+d_1|^2\right), \mathit{\text{ if }}\alpha \leq \frac{1}{2}, \\ 2\left(1-2(1-\alpha) |1+d_1|^2\right), \mathit{\text{ if }}\alpha \geq \frac{1}{2}. \end{cases} \end{equation*}

    All of these inequalities are sharp.

    Corollary 3.2. Let z\in \mathbb{D} . If g \in \mathcal{G}_2 , the following inequalities hold.

    \begin{equation*} \left|1+\frac{d_1}{2}\right| \leq \frac{1}{2}, \, \left| 4+2d_2-d_1^2\right| \leq 1,\, \left|6+3d_3-3d_1d_2+d_1^3\right| \leq 1. \end{equation*}

    Also,

    \begin{equation*} |\mathcal{H}(\alpha, 2)|\leq \begin{cases} 2\left(1-2\alpha |2+d_1|^2\right), \mathit{\text{ if }}\alpha \leq \frac{1}{2}, \\ 2\left(1-2(1-\alpha) |2+d_1|^2\right), \mathit{\text{ if }}\alpha \geq \frac{1}{2}. \end{cases} \end{equation*}

    All of these inequalities are sharp.

    Theorem 3.2. Let 0 < c \leq 2 and let the function g(z) be of the form (1.3) belong to the class \mathcal{G}_c . Then, for n = 2, 3, \cdots , the following estimates

    \begin{align*} &\left|n d_n-c(n-2) d_{n-1}+\cdots+\left[1+(-1)^{n-1}\right] \frac{c(c-1) \cdots(c-n+2)}{2(n-1)!}d_1\right.\\ &\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad +\left. \left[1-(-1)^n\right]\frac{c(c-1)(c-2) \cdots(c-n+1)}{2n!}\right|^2 & \end{align*}
    \begin{align*} \leq 1 + \sum\limits_{k = 1}^{n-1}\left|(k+1)d_k-c(k-1)d_{k-1}+\cdots +\left[{1+3(-1)^{k-1}}\right] \, \frac{c(c-1)(c-2)\cdots(c-k+2)}{2(k-1)!}d_1 \right.\\\quad \quad +\left.\left[1+\left(-1\right)^{k}\right]\frac{c(c-1)\cdots(c-k+1)}{2 k!}\right|^2 \end{align*}

    hold.

    Proof. Let the function g of the form (1.3) belong to the class \mathcal{G}_c . Then, there exists a function p\in \mathcal{P} such that,

    \begin{equation} p\left(z\right) = 2z \ \frac{g^{\prime}(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c, \ z\in\mathbb{D}. \end{equation} (3.7)

    Since p\in \mathcal{P} , there exists a function \omega of the form

    \omega\left(z\right) = \frac{p\left(z\right)-1}{p\left(z\right)+1}, \ z\in\mathbb{D},

    where \omega is analytic in \mathbb{D}, \omega\left(0\right) = 0, \left|\omega \left(z\right)\right|\leq 1 for z\in\mathbb{D} . Furthermore,

    \begin{equation} p(z) = \frac{1+\omega(z)}{1-\omega(z)}. \end{equation} (3.8)

    From (3.7) and (3.8), we have

    \left((1-z)^cg(z)+2z\left(1-z\right)^c g'(z)+\left(1+z\right)^c g(z)\right)\omega(z) = \left(1-z\right)^c \left(2zg'(z)-g(z)\right)+\left(1+z\right)^c g(z).

    Let

    \omega(z) = \sum\limits_{n = 1}^{\infty}\omega_n z^n.

    Considering the expansion of the function of g as power series, we get

    (2+\sum_{n = 1}^{\infty}\left(d_n-cd_{n-1}+\frac{c(c-1)}{2!}d_{n-2}+\cdots+(-1)^{n-1} \, \frac{c(c-1)\cdots(c-n+2)}{(n-1)!}d_1+\\ (-1)^{n} \, \frac{c(c-1)\cdots(c-n+1)}{n!}\right)z^n\\+2\sum_{n = 1}^{\infty}\left(nd_n-c(n-1)d_{n-1}+\frac{c(c-1)}{2!}(n-2)d_{n-2}+\cdots+(-1)^{n-1} \, \frac{c(c-1)\cdots(c-n+2)}{(n-1)!}d_1\right)z^n\\+ \sum_{n = 1}^{\infty}\left(d_n+cd_{n-1}+\frac{c(c-1)}{2!}d_{n-2}+\cdots+\frac{c(c-1)\cdots(c-n+2)}{(n-1)!}d_1+\frac{c(c-1)\cdots(c-n+1)}{n!}\right)z^n)\left(\sum_{n = 1}^{\infty}\omega_n z^n\right)\\ = 2\sum_{n = 1}^{\infty}\left(nd_n-c(n-1)d_{n-1}+\frac{c(c-1)}{2!}(n-2)d_{n-2}+\cdots+(-1)^{n-1} \, \frac{c(c-1)\cdots(c-n+2)}{(n-1)!}d_1 \right)z^n\\+ \sum_{n = 1}^{\infty}\left(d_n+cd_{n-1}+\frac{c(c-1)}{2!}d_{n-2}+\cdots+\frac{c(c-1)\cdots(c-n+2)}{(n-1)!}d_1 +\frac{c(c-1)\cdots(c-n+1)}{n!}\right)z^n\\- \sum_{n = 1}^{\infty}\left(d_n-cd_{n-1}+\frac{c(c-1)}{2!}d_{n-2}+\cdots+(-1)^{n-1} \, \frac{c(c-1)\cdots(c-n+2)}{(n-1)!}d_1+(-1)^{n} \, \frac{c(c-1)\cdots(c-n+1)}{n!}\right)z^n .

    For z \in \mathbb{D} , this can be again simplified to bring into the form,

    \begin{eqnarray} \left(1+ \left(\sum\limits_{n = 1}^{\infty}\left((n+1)d_n-c(n-1)d_{n-1}+\cdots+\left[{1+3(-1)^{n-1}}\right]\frac{c(c-1)(c-2)\cdots(c-n+2)}{2(n-1)!}d_1\right.\right.\right.\\ + \left.\left.\left. \left[1+\left(-1\right)^{n}\right]\frac{c(c-1)\cdots(c-n+1)}{2 n!}\right)z^n\right) \left(\sum\limits_{n = 1}^{\infty}\omega_n z^n\right)\right)\\ \, = \sum\limits_{n = 2}^{\infty}\left(nd_n-c(n-2)d_{n-1}+\cdots+\left[1+(-1)^{n-1}\right]\frac{c(c-1)\cdots(c-n+2)}{2(n-1)!}d_1 \right. \\+ \left. \left[1-\left(-1\right)^{n}\right] \frac{c(c-1)(c-2)\cdots(c-n+1)}{2(n)!}\right)z^n. \end{eqnarray} (3.9)

    For \ n = 2, 3, \cdots , let

    \begin{equation}p_n(c) = (n+1)d_n-c(n-1)d_{n-1}+\cdots+\left[{1+3(-1)^{n-1}}\right] \, \frac{c(c-1)(c-2)\cdots(c-n+2)}{2(n-1)!}d_1 \\ + \left[1+\left(-1\right)^{n}\right] \frac{c(c-1)\cdots(c-n+1)}{2 n!} \end{equation} (3.10)

    and

    \begin{equation}s_n(c) = nd_n-c(n-2)d_{n-1}+\cdots+\left[1+(-1)^{n-1}\right] \, \frac{c(c-1)\cdots(c-n+2)}{2(n-1)!}d_1 \\ +\left[1-\left(-1\right)^{n}\right] \frac{c(c-1)(c-2)\cdots(c-n+1)}{2(n)!}. \end{equation} (3.11)

    From (3.9)–(3.11), we have

    \begin{equation} \sum\limits_{n = 1}^{\infty}\omega_n z^n+\sum\limits_{n = 2}^{\infty}\left(p_1(c)\omega_{n-1}+\cdots p_{n-1}(c)\omega_1\right)z^n = \sum\limits_{n = 1}^{\infty}s_n(c)z^n, \,\,\,z\in\mathbb{D}. \end{equation} (3.12)

    Equating the coefficient of z , we have

    2 \omega_1 = d_1+2 c.

    Since |\omega_1|\leq1 , we obtain

    \left|\frac{d_1}{2}+c\right|\leq1

    and for n = 2, 3, \cdots ,

    \omega_n+p_1(c)\omega_{n-1}+\cdots+p_{n-1}\omega_1 = s_n(c).

    From the Eqs (3.9)–(3.12), we have

    \left(1+\sum\limits_{k = 1}^{n-1}p_k (c)\right)\left(\sum\limits_{k = 1}^{\infty}\omega_k z^k\right) = \sum\limits_{k = 1}^{n}s_k(c)z^k+\sum\limits_{k = n+1}^{\infty}E_k z^k,

    where E_k are the appropriate coefficients. Since |\omega(z)| < 1 for z\in\mathbb{D} ,

    \left|\sum\limits_{k = 1}^{n}s_k (c)z^k+\sum\limits_{k = n+1}^{\infty}E_kz^k\right|^2 < \left|1+\sum\limits_{k = 1}^{n-1}p_k (c)z^k\right|^2.

    By simplifying this, we have

    \sum\limits_{k = 1}^{n}|s_k(c)|^2\leq 1+\sum\limits_{k = 1}^{n-1}|p_k(c)|^2.

    Since |s_k(c)|^2\geq0 for k = 1, 2, \cdots, n-1, we obtain

    |s_n(c)|^2\leq1+\sum\limits_{k = 1}^{n-1}|p_k(c)|^2, n = 2,3,\cdots.

    This essentially completes the proof of Theorem 3.2.

    Let us consider the class \mathcal{B} defined by \mathcal{B} = \left\{\omega \in\mathcal{H}: \left|\omega (z)\right|\leq 1, z\in \mathbb{D}\right\} and \mathcal{B}_0 be the subclass of \mathcal{B} consisting of functions \omega such that \omega(0) = 0. The elements of \mathcal{B}_0 are also known as Schwarz functions.

    Lemma 3.1. [5] If \omega \in \mathcal{B}_0 is of the form \omega(z) = \sum\limits_{n = 1}^{\infty}\omega_n z^n, \ z\in\mathbb{D}, then for \nu\in\mathbb{C} ,

    \left|\omega_2-\nu\omega_1^2\right|\leq \max\left\{1, |\nu|\right\}.

    Lemma 3.2. If \omega\in\mathcal{B}_{0} is of the form \omega(z) = \sum\limits_{n = 1}^{\infty}\omega_n z^n, \ z\in\mathbb{D}, then for any real numbers q_1 and q_2 , the following sharp estimates holds:

    \begin{equation} \left|\omega_3+q_1 \ \omega_1 \ \omega_2+q_2 \ \omega_1^3\right|\leq H(q_1, q_2), \end{equation} (3.13)

    where

    H(q_1, q_2): = \begin{cases} 1 \mathit{\text{ if }}(q_1, q_2)\in D_1\cup D_2\\ |q_2| \mathit{\text{ if }}(q_1, q_2)\in \cup_{k = 3}^7 D_k\\ \frac{2}{3}\left(|q_1|+1\right)\left(\frac{|q_1|+1}{3\left(|q_1|+1+q_2\right)}\right)^{\frac{1}{2}} \mathit{\text{ if }}(q_1, q_2)\in D_8\cup D_9\\ \frac{q_2}{3}\left(\frac{q_1^2-4}{q_1^2-4q_2}\right)\left(\frac{q_1^2-4}{3\left(q_2-1\right)}\right)^{\frac{1}{2}} \mathit{\text{ if }}(q_1, q_2)\in D_{10}\cup D_{11}/\left\{\pm 2, 1\right\}\\ \frac{2}{3}\left(|q_1|-1\right)\left(\frac{|q_1|-1}{3\left(|q_1|-1-q_2\right)}\right)^{\frac{1}{2}} \mathit{\text{ if }}(q_1, q_2)\in D_{12} \end{cases}

    and the sets D_k, k = 1, 2, \cdots are defined in [15].

    Now we obtain a few upper bounds for early coefficients and for the Fekete-Szegö functional in the class \mathcal{G}_c .

    Theorem 3.3. Let g\in\mathcal{G}_c, 0 < c\leq2 . Then,

    \begin{equation} |c+d_1|\leq1, \end{equation} (3.14)
    \begin{equation} |d_1|\leq 1 + c, \end{equation} (3.15)
    \begin{equation} |c^2+2d_2-d_1^2|\leq1, \end{equation} (3.16)
    \begin{equation} |d_2|\leq 1 + c, \end{equation} (3.17)
    \begin{equation} |3d_3-3d_1 d_2+d_1^3|\leq \frac{4c^3+2c+3}{6} \end{equation} (3.18)

    and

    \begin{equation} |d_3|\leq \frac{3+2c+18c^2+10c^3}{18}. \end{equation} (3.19)

    Furthermore, for \delta\in\mathbb{R}

    \begin{equation} |d_2-\delta d_1^2|\leq \frac{1}{2} \ max\left\{1, 2|1-\delta|\right\}+c|2\delta-1|+c^2|-\delta|. \end{equation} (3.20)

    Proof. From the class \mathcal{G}_c , there exists \omega\in\mathcal{B}_0 of the form \omega(z) = \sum\limits_{n = 1}^\infty\omega_n z^n, \ z\in\mathbb{D} , such that

    \begin{equation} 2z\frac{g'(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c = \frac{1+\omega (z)}{1-\omega (z)}. \end{equation} (3.21)

    Since

    \begin{equation} 2z\frac{g'(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c \\ = 1+2\left(c+d_1)z\right)z+2\left(c^2+2d_2-d_1^2\right)z^2+\frac{2}{3}\left(2c^3+c+9d_3-9d_1 d_2+3d_1^3\right)+\cdots \end{equation} (3.22)

    and

    \begin{equation} \frac{1+\omega (z)}{1-\omega (z)} = 1+2\omega_1 z + \left(2\omega_{1}^2+2\omega_2\right)z^2+\left(2\omega_1^3+4\omega_1\omega_2+\omega_3\right)z^3+\cdots. \end{equation} (3.23)

    By comparing the corresponding coefficients from (3.21)–(3.23), we have the following:

    \begin{equation} 2\left(c+d_1\right) = 2\omega_1, \end{equation} (3.24)
    \begin{equation} 2\left(c^2+2d_2-d_1^2\right) = 2\left(\omega_1^2+\omega_2\right) \end{equation} (3.25)

    and

    \begin{equation} \frac{2}{3}\left(2c^3+c+9d_3-9d_1d_2+3d_1^3\right) = 2\omega_1^2+4\omega_1\omega_2+\omega_3. \end{equation} (3.26)

    Since |\omega_1|\leq1 , from (3.24),

    \begin{equation*} |c+d_1|\leq1 \end{equation*}

    and hence

    \begin{equation*} |d_1|\leq 1 + c. \end{equation*}

    From (3.25) and by Lemma 3.1, we have

    \begin{equation*} \left|c^2+2d_2-d_1^2\right| = \left|\omega_2+\omega_1^2\right|\leq1 \end{equation*}

    and

    \begin{equation*} 2d_2 = \omega_2+2\omega_1^2-2c\omega_1, \end{equation*}

    which in turn gives

    \begin{equation*} 2\left|d_2\right|\leq \max\left\{1, 2\right\}+2c\left|\omega_1\right| \end{equation*}

    and hence

    \begin{equation*} \left|d_2\right|\leq 1+c. \end{equation*}

    From (3.26) and by Lemma 3.2, we get

    \begin{equation*} \left|3d_3-3d_1d_2+d_1^3\right|\leq \frac{1}{2}+\frac{2}{3} c^3+\frac{1}{3}c. \end{equation*}

    Hence, we obtain

    \begin{equation*} \left|3d_3-3d_1d_2+d_1^3\right|\leq \frac{4c^3+2c+3}{6}. \end{equation*}

    Substituting the formulas for d_1 and d_2 , we obtain

    \begin{eqnarray*} |3d_3 | & = &\left|\frac{1}{2}\left(\omega_3+7\omega_1\omega_2+6\omega_1^3\right)-\frac{3}{2}c\left(\omega_2+2\omega_1^2\right)+3c^2\omega_1-3\alpha\omega_1+\frac{1}{3}c^3+\frac{1}{3}c\right|\\ &\leq & \frac{1}{2}\left|\omega_3+7\omega_1\omega_2+6\omega_1^3\right|+\frac{3}{2}c\left|\omega_2+2\omega_1^2\right|+3c^2\left|\omega_1\right|+3c\left|\omega_1\right|+\frac{1}{3}c^3+\frac{1}{3}c \\ &\leq& \frac{1}{2} H\left(7, 6\right)+\frac{3}{2} c \max\left\{1, |-2|\right\}+3c^2+3c+\frac{1}{3}c^3+\frac{1}{3}c. \end{eqnarray*}

    Therefore,

    \begin{equation*} \left|d_3\right|\leq \frac{3+38c+18c^2+2c^3}{18}. \end{equation*}

    Furthermore, we get

    \begin{equation*} \left|d_2 - \delta d_1^2\right|\leq \frac{1}{2}\left|\omega_2+2\left(1-\delta\right)\omega_1^2\right|+c\left(2\delta-1\right)\left|\omega_1\right|+c^2\left|-\delta\right|. \end{equation*}

    The proof of the theorem is completed by virtue of Lemma 3.1.

    It can be remarked here that if \nu is a real number, Lemma 3.1 can be improved in the following way and can be found in [2].

    Lemma 3.3. If \omega \in \mathcal{B}_0 is of the form \omega(z) = \sum\limits_{n = 1}^{\infty}\omega_n z^n, \ z\in\mathbb{D}, then

    \begin{equation} \left|\omega_2-\nu\omega_1^2\right|\leq \begin{cases}-\nu, \ \nu\leq-1,\\1, \ -1\leq\nu\leq1,\\\nu, \ \nu\geq1. \end{cases} \end{equation} (3.27)

    For \nu < -1 or \nu > 1 , equality holds if and only if \omega\left(z\right) = z or one of its rotations. For -1 < \nu < 1 , equality holds if and only if \omega(z) = z^2, \ z\in\mathbb{D} or one of its rotations. For \nu = -1 equality holds if and only if \omega(z) = \frac{z(\lambda+z)}{(1+\lambda z)}, z\in\mathbb{D} or one of its rotations, while for \nu = 1 equality holds if and only if w(z) = \frac{-z(\lambda+z)}{(1+\lambda z, )}, 0\leq\lambda\leq1, z\in\mathbb{D} or one of its rotations.

    We can improve the results obtained in (3.20), in view of Lemma 3.3 as follows: For \delta\in\mathbb{R} , we get,

    \begin{equation} \left|d_2-\delta d_1^2\right|\leq \begin{cases} (1+c)-2(1+c)^2\delta, \ \delta\leq0,\\ \frac{1-2c+2(c^2-2c)\delta}{2}, \ 0\leq\delta\leq\frac{1}{2},\\ \frac{1-2c+2(c^2+2c)\delta}{2}, \ \frac{1}{2}\leq\delta\leq\frac{3}{2},\\\left(1+c\right)^2\delta-2(1+c), \ \delta\geq\frac{3}{2}. \end{cases} \end{equation} (3.28)

    Theorem 3.4. Let 0 < r < 1 . If g\in\mathcal{G}_c then for |z| = r < 1 ,

    \begin{equation} \sqrt{\frac{-f_\alpha(-r)}{r}}\leq\left|g(z)\right| e^{-2 c \beta_c'(r)} \leq \sqrt{\frac{f_\alpha(-r)}{r}}. \end{equation} (3.29)

    Proof. Let us define

    \begin{equation} f(z) = z \ (g(z))^2 \exp\left\{-2 z \beta_c^{\prime}(z)\right\}. \end{equation} (3.30)

    Note that the function g is non-vanishing in \mathbb{D} . Therefore, f is analytic in \mathbb{D} and a simple computation shows that

    \begin{equation} z\frac{f'(z)}{f(z)} = 2z\frac{g^{\prime}(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c, \, \, z\in\mathbb{D}. \end{equation} (3.31)

    From (3.31), we can see that g\in \mathcal{G}_c if and only if f\prec\frac{1+z}{1-z}, z\in\mathbb{D} . By applying the result of Ma and Minda [12], we infer that

    \begin{equation} -f_\alpha(-r)\leq\left|f(z)\right|\leq f_\alpha(-r),\ |z| = r. \end{equation} (3.32)

    Hence,

    \begin{equation} -f_\alpha(-r)\leq\left|z \ (g(z))^2 \ \exp\left\{-2 z \beta_c^{\prime}(z)\right\} \right|\leq f_\alpha (-r),\ |z| = r, \end{equation} (3.33)

    which gives (3.29).

    For c = 1 and c = 2 we have the following corollaries.

    Corollary 3.3. For 0 < r < 1 , if g\in\mathcal{G}_1 then we have for |z| = r < 1

    \begin{equation} \sqrt{\frac{-f_\alpha(-r)}{r}}(1-r)\leq\left|g(z)\right|\leq \sqrt{\frac{f_\alpha(-r)}{r}}(1+r). \end{equation} (3.34)

    Corollary 3.4. For 0 < r < 1 , if g\in\mathcal{G}_2 then we have for |z| = r < 1 ,

    \begin{equation} \sqrt{\frac{-f_\alpha(-r)}{r}}\exp\frac{2}{1-r}\leq\left|g(z)\right|\leq \sqrt{\frac{f_\alpha(-r)}{r}}\exp\frac{2}{1+r}. \end{equation} (3.35)

    Theorem 3.5. Let g\in \mathcal{G}_c . Then,

    \begin{equation} |d_1| < 1,\,|d_2| < 1,\,|d_3| < 1,\,|d_4| < 1 \end{equation} (3.36)

    and

    \begin{equation} |d_2^2-d_3|\leq1. \end{equation} (3.37)

    Proof. Since g\in \mathcal{G}_c , there is a Schwarz function \omega satisfying that

    \begin{equation} 2z\frac{g^{\prime}(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c = \frac{1+\omega(z)}{1-\omega(z)} = 1+c_1z+c_2z^2+\cdots. \end{equation} (3.38)

    If \omega(z) = z , we have

    \begin{equation} 2z\frac{g^{\prime}(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c = 1+2z+2z^2+2z^3+\cdots. \end{equation} (3.39)

    Equating the coefficients using (3.22),

    d_1 = 1-c,\, \, d_2 = 1-c,\, \, d_3 = \frac{6-11c+6c^2-2c^3}{9}, \,d_4 = 1-\frac{32}{36}c-\frac{1}{9}c^2+\frac{5}{9}c^3-\frac{23}{36}c^4.

    Since 0 < c\leq2 , (3.36) holds.

    Also, since 0 < c\leq2 ,

    \left|d_2^2-d_3\right| = \left|(1-c)^2-\left(1-\frac{8}{9}c+\frac{5}{9}c^3\right)\right|
    = \left|c^2-\frac{10}{9}c-\frac{5}{9}c^3\right|

    implies that (3.37) holds.

    In this section we are connecting the class \mathcal{G}_c , 0 < c\leq2 by taking the equivalent condition of \mathcal{G}_c , 0 < c\leq2 associated with L_0(z) = \frac{1+z}{1-z} . Note that the definition \mathcal{G}_c, 0 < c\leq2 can be rewritten in the equivalent form as g \in \mathcal{G}_c if

    \begin{equation} 2z\frac{g'(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c\prec L_0(z), \ z\in\mathbb{D}. \end{equation} (4.1)

    To prove the differential subordination results, we need the following lemma which is stated below.

    Lemma 4.1. [13] Let \tau be univalent in \mathbb{D} , \psi and \phi be analytic in a domain D containing \tau(\mathbb{D}) with \phi(\omega)\neq0 when \omega\in \tau(\mathbb{D}) . Let T(z) = z \tau' \phi(\tau(z)) and \kappa(z) = \psi(\tau(z))+T(z) for z\in \mathbb{D} and satisfy either T is starlike univalent in \mathbb{D} or \kappa is convex univalent in \mathbb{D} . Also, assume that \Re \left\{\frac{z\kappa'(z)}{T(z)}\right\} > 0, z\in \mathbb{D}. \\ If p\in \mathcal{H} with p(0) = \tau(0), p(\mathbb{D})\subset D , and

    \psi(p(z))+zp'(z)\phi(p(z))\prec\psi(\tau(z))+z\tau'(z)\phi(\tau(z)), \ z\in \mathbb{D}

    then p\prec\tau and \tau is the best dominant.

    Theorem 4.1. Let g be an analytic function with g(0) = 1 and let 0 < c\leq2 . If g satisfies

    \begin{equation} 2z\frac{g'(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c\prec 1+\frac{2z}{1-z^2}, \ z\in \mathbb{D} \end{equation} (4.2)

    then

    \begin{equation} p(z): = \left(g(z)\right)^2 \exp\left\{-2 z \beta_c^{\prime}(z)\right\}\prec L_0(z), \ z\in \mathbb{D}. \end{equation} (4.3)

    Proof. Let \psi(\omega) = 1, \omega\in\mathbb{C} and \phi(\omega): = \frac{1}{\omega}, \omega\in \mathbb{C}\setminus \left\{0\right\} . Note that L_0(\mathbb{D}): = \mathbb{C}\setminus\left\{0\right\} . Thus,

    \begin{equation} T(z) = zL_0'(z)\phi(L_0(z)) = \frac{zL_0'(z)}{L_0(z)} = \frac{2z}{1-z^2} \end{equation} (4.4)

    is analytic and also well defined. Also, we have

    \begin{equation} \Re\left\{z\frac{T'(z)}{T(z)}\right\} = \Re\left\{\frac{1+z^2}{1-z^2}\right\} > 0, \ z\in\mathbb{D}. \end{equation} (4.5)

    This implies that T is a starlike univalent function. From this, for a function \kappa(z): = \psi(L_0(z))+T(z) = 1+T(z), z\in \mathbb{D} , we get

    \Re\left\{z\frac{\kappa'(z)}{T(z)}\right\} = \Re \left\{z\frac{T'(z)}{T(z)}\right\} > 0, \ z\in\mathbb{D}.

    From Lemma 4.1,

    1+z\frac{p'(z)}{p(z)}\prec1+z\frac{L_0'(z)}{L_0(z)} = 1+\frac{2z}{1-z^2}, \ z\in\mathbb{D},

    which implies that p\prec L_0 . Let us take the analytic function g with g(0) = 1 and g(z)\neq0 for z\in\mathbb{D} satisfying (4.2). For a function as in (4.3), we can notice that p(0) = L_0(0) = 1, p(z)\neq0 for z\in\mathbb{D} and p is analytic. The proof of Theorem 4.1 is completed by observing that

    1+z\frac{p'(z)}{p(z)} = 2z\frac{g'(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c, \, \ z\in\mathbb{D}.

    Theorem 4.2. Let g(z) be an analytic function with g(0) = 1 and let 0 < c\leq2 . If

    \begin{equation} 2z\frac{g'(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c\prec\frac{1+z}{1-z}+\frac{2z}{1-z^2}, \ z\in \mathbb{D} \end{equation} (4.6)

    then

    \begin{equation} p(z): = z\left(g(z)\right)^2 \exp\left\{-2 z \beta_c^{\prime}(z)\right\} \left(\int_{0}^{z}\left(g(\zeta)\right)^2 \exp\left\{-2 \zeta \beta_c^{\prime}(\zeta)\right\} d\zeta\right)^{-1}\prec L_0 (z), \ z\in \mathbb{D}. \end{equation} (4.7)

    Proof. Let \psi(\omega) = \omega, \omega\in\mathbb{C} and \phi(\omega): = \frac{1}{\omega}, \omega\in \mathbb{C}\setminus \left\{0\right\} . Note that L_0(\mathbb{D}): = \mathbb{C}\setminus \left\{0\right\} and \psi and \phi are analytic in D . Thus, T defined by (4.4) is analytic and univalent and satisfies (4.5). Hence, \kappa(z) = \psi(L_0(z))+T(z) = L_0(z)+T(z), \ z\in \mathbb{D} . By using (4.5) we get,

    \Re \left\{z\frac{\kappa'(z)}{T(z)}\right\} = \Re\left\{z\frac{L_0'(z)}{T(z)}\right\}+\Re \left\{z\frac{T'(z)}{T(z)}\right\} = \Re \left\{L_0(z)\right\}+\Re \left\{z\frac{T'(z)}{T(z)}\right\} > 0, z\in \mathbb{D}.

    Note that, p is also analytic with p(0) = L_0(0) = 1 and p(z)\neq0 for z\in\mathbb{D} . From Lemma 4.1, we have

    p(z)+z\frac{p'(z)}{p(z)}\prec L_0(z)+z\frac{L_0'(z)}{L_0(z)} = \frac{1+z}{1-z}+\frac{2z}{1-z^2}, \ z\in \mathbb{D}.

    This essentially gives us that p\prec L_0 . Let us take the analytic function g with g(0) = 1 and g(z)\neq0 for z\in\mathbb{D} satisfying (4.6). For a function defined as in (4.7), we can observe that

    \begin{eqnarray*} p(0)& = &\lim\limits_{z\to 0}z\left(g(z)\right)^2 \exp\left\{-2 z \beta_c^{\prime}(z)\right\} \left(\int_{0}^{z}\left(g(\zeta)\right)^2 \exp\left\{-2 \zeta \beta_c^{\prime}(\zeta)\right\}d\zeta\right)^{-1} \\ & = & (g(0))^2 \lim\limits_{z\to 0}\left(\int_{0}^{z}\left(g(\zeta)\right)^2 \exp\left\{-2 \zeta \beta_c^{\prime}(\zeta)\right\}d\zeta\right)^{-1} = 1 = L_0(0). \end{eqnarray*}

    Therefore, p(z)\neq0 and analytic for all z\in \mathbb{D} . The proof of the theorem is then completed by noting that

    p(z)+z\frac{p'(z)}{p(z)} = 2z\frac{g'(z)}{g(z)}+\left(\frac{1+z}{1-z}\right)^c, \ z\in \mathbb{D}.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors are thankful to the referee(s) for many insightful suggestions and comments in the original manuscript. The second author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2019R1I1A3A01050861).

    The authors declare that they have no competing interests.



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