Topological Indices and f-Polynomials on Some Graph Products

Abstract

We obtain inequalities involving many topological indices in classical graph products by using the f-polynomial. In particular, we work with lexicographic product, Cartesian sum and Cartesian product, and with first Zagreb, forgotten, inverse degree and sum lordeg indices.

1. Introduction

Chemical compounds (like hydrocarbons) can be represented by means of graphs. A topological index T is just a number that encapsulates some property of graphs and such that T correlates with a certain molecular characteristic; therefore, it can be employed to grasp chemical properties and physical properties of chemical substances. They play an important role in mathematical chemistry, in particular, in the QSPR/QSAR (quantitative structure-property relationship/quantitative structure-activity relationship) investigations. Computational and mathematical properties of topological indices have been studied in depth for more than 50 years (see, e.g., [1,2,3,4,5,6,7,8] and the references therein). In particular, a main subject on this field is to obtain sharp bounds of topological indices.

See [9] for a review in a dialog manner discussing relevance of topological descriptors to chemical/physical properties.

One of the main topological indices is the following index, called Randić index, defined in [1] by

$$R\left(G\right)=\sum _{uv\in E\left(G\right)}\frac{1}{\sqrt{d_u{d}_v}},$$

where G is a graph and $d_w$ is the degree of the vertex $w\in V\left(G\right)$.

Along the paper, $G=\left(V\right(G),E(G\left)\right)$ will denote a (non-oriented) simple (without loops and multiple edges loops) finite graph without isolated vertices. Hence, every vertex has degree at least 1.

Two of the most popular alternatives to the Randić index are the first Zagreb and second Zagreb indices, denoted by $M_1$ and $M_2$, respectively, and defined by

$$M_1\left(G\right)=\sum _{uv\in E\left(G\right)}(d_u+d_v)=\sum _{u\in V\left(G\right)}{d}_u^2,M_2\left(G\right)=\sum _{uv\in E\left(G\right)}{d}_u{d}_v.$$

These two indexes are very useful in mathematical chemistry and so, they have been extensively studied, see [10,11,12,13] and the references therein. Further development of Zagreb-type indices deals with the applications to more complex chemical objects, e.g., large carbon-based species with regular structures such as polycyclic aromatic hydrocarbons [14] and carbon nanostructures [15].

Please note that there are topological indices of different types. They may treat only vertices, only edges, or both edges and vertices of the graph to calculate an index. Thus, the first Zagreb index belongs to the first class (every index in the first class, as the first Zagreb index, also belongs to the second class).

Along this paper we obtain results for topological indices in this first class.

The so-called harmonic index, is defined in [16] by

$$H\left(G\right)=\sum _{uv\in E\left(G\right)}\frac{2}{d_u+d_v}.$$

For more information about the properties of that index we refer to [17,18,19,20,21,22], and the book [23].

The inverse degree index $ID$ is defined as

$$ID\left(G\right)=\sum _{u\in V\left(G\right)}\frac{1}{d_u}=\sum _{uv\in E\left(G\right)}\left(\frac{1}{d_u^2}+\frac{1}{d_v^2}\right).$$

This index first attracted attention through many conjectures obtained by the computer programme called Graffiti [16]. Since then, several authors have studied its connections with other parameters of graphs: diameter, matching number, edge-connectivity, Wiener index, etc.; also, its chemical applications have been studied by many researchers (see [24,25,26,27,28,29]).

In [30,31,32] the general first Zagreb and general second Zagreb indices are defined by

$$M_1^{\alpha }\left(G\right)=\sum _{u\in V\left(G\right)}{d}_u^{\alpha },M_2^{\alpha }\left(G\right)=\sum _{uv\in E\left(G\right)}{\left(d_u{d}_v\right)}^{\alpha }.$$

Please note that the first Zagreb index $M_1$ is $M_1^2$, the inverse degree index $ID$ is $M_1^{-1}$, the forgotten index F is $M_1^3$, …; moreover, the Randić index R is $M_2^{-1/2}$, the second Zagreb index $M_2$ is $M_2^1$, the modified Zagreb index is $M_2^{-1}$, ….

The variable topological indices were introduced as a new way of characterizing heteroatoms (see [33,34]), and to assess the structural differences (see [35]). The idea behind the variable topological indices is that the parameter is determined during the regression in such a way that the error of estimate for a fixed property is minimized.

The sum lordeg index was introduced in [36]. It is defined as

$$SL\left(G\right)=\sum _{u\in V\left(G\right)}{d}_u\sqrt{log{d}_u}.$$

This index is interesting from an applied viewpoint since it correlates very well with the octanol-water partition coefficient for octane isomers [36], and so, it appears in numerical packages for the computation of topological indices [37]. For these reasons, in [38] is stated the open problem of finding appropriate bounds for this index.

In [39] the harmonic polynomial is introduced as

$$H(G,x)=\sum _{uv\in E\left(G\right)}{x}^{d_u+d_v-1}.$$

In [40,41,42] several properties of the harmonic polynomial are obtained. Please note that $2{\int }_0^{1}H(G,x)dx=H\left(G\right)$.

In [41] the inverse degree polynomial is introduced as

$$ID(G,x)=\sum _{u\in V\left(G\right)}{x}^{d_u-1}.$$

It should be noticed that ${\int }_0^{1}ID(G,x)dx=ID\left(G\right)$. Thus, the inverse degree polynomial $ID(G,x)$ can be used to obtain information about the inverse degree index $ID\left(G\right)$ of a graph G.

Given any function $f:{\mathbb{Z}}^{+}\to {\mathbb{R}}^{+}$, let us define the f-index

$$I_f\left(G\right)=\sum _{u\in V\left(G\right)}f\left(d_u\right),$$

and the f-polynomial of G by

$$P_f(G,x)=\sum _{u\in V\left(G\right)}{x}^{1/f\left(d_u\right)-1},$$

if $x>0$. Also, let us define $P_f(G,0)={lim}_{x\to 0^{+}}{P}_f(G,x)$. In particular, $P_f(G,x)=ID(G,x)$ if $f\left(t\right)=1/t$. It is clear that ${\int }_0^1{P}_f(G,x)dx=I_f\left(G\right)$.

Please note that many important indices can be obtained from $I_f$ by choosing appropriate functions f: if $f\left(t\right)=t^2$, then $I_f$ is the first Zagreb index; if $f\left(t\right)=t^{-1}$, then $I_f$ is the inverse degree index $ID$; if $f\left(t\right)=t^3$, then $I_f$ is the forgotten index F; in general, if $f\left(t\right)=t^{\alpha }$, then $I_f$ is the general first Zagreb index $M_1^{\alpha }$; if $f\left(t\right)=t\sqrt{logt}$, then $I_f$ is the sum lordeg index $SL$. Thus, each theorem in this paper about $I_f$ is a result for each one of these indices.

The f-polynomial of other graph operations (e.g., join, corona product, Mycielskian and line) is studied in [43].

Operations on graphs play an important role in Mathematical Chemistry, see e.g., [44,45], since many chemical structures appear as operations of graphs: The crystal structure of sodium chloride is the Cartesian product of two path graphs. The kernel of the iron crystal structure is the join of the cube graph $Q_3$ and an isolated vertex. The cyclobutane is the Cartesian product of two path graphs $P_2$. The alkane $C_3{H}_6$ can be represented as the corona product of the path graph $P_3$ and the null graph $N_2$. The cyclohexane $C_6{H}_{12}$ can be represented as the corona product of the cycle graph $C_6$ and the null graph $N_2$. The carbon nanotube $TU{C}_4(m,n)$ can be seen as the Cartesian product of the path graph $P_3$ and the cycle graph $C_5$. The fence (respectively, the closed fence) is the lexicographic product of $P_5$ and $P_2$ (respectively, $C_5$ and $P_2$). The zigzag polyhex nanotube $TUH{C}_6[2n,2]$ is the generalized hierarchical product of the path graph $P_2$ and the cycle graph $C_{2n}$. See [46,47].

Polynomials, in general, have lately proved to be useful in graph theory and, in particular, in Mathematical Chemistry (see [41,43,45,48,49,50,51,52,53,54,55]).

The main goal in this paper is to obtain information of many topological indices (each case of $I_f$ for some particular choice of f) of several graph products, from the information on topological indices of these factors, which are much easier to calculate than the products. Our approach is to obtain information about the corresponding f-polynomials, which are easy to calculate (see for instance Theorems 1, 11 and 17); then, we can deduce information on the $I_f$ index by using the formula ${\int }_0^1{P}_f(G,x)dx=I_f\left(G\right)$ (see for instance Theorems 4, 13 and 18). This is a good approach since the bounds of the f-polynomial of a product of two graphs allow use of analytic tools to bound the $I_f$ index of such a graph product, simplifying the proofs.

2. Background

The study of the effect of graph operations on topological indices is an active topic of research (see, e.g., [41,54,55,56]). We study in this section the f-polynomial of several graph products: lexicographic product, Cartesian sum and Cartesian product.

The Cartesian product $G_1\times G_2$ of the graphs $G_1$ and $G_2$ has the vertex set $V(G_1\times G_2)=V\left(G_1\right)\times V\left(G_2\right)$ and $(u_i,v_j)(u_k,v_l)$ is an edge of $G_1\times G_2$ if $u_i=u_k$ and $v_j{v}_l\in E\left(G_2\right)$, or $u_i{u}_k\in E\left(G_1\right)$ and $v_j=v_l$.

The lexicographic product $G_1\odot G_2$ of the graphs $G_1$ and $G_2$ has $V\left(G_1\right)\times V\left(G_2\right)$ as vertex set, so that two distinct vertices $(u_i,v_j),(u_k,v_l)$ of $V(G_1\odot G_2)$ are adjacent if either $u_i{u}_k\in E\left(G_1\right)$, or $u_i=u_k$ and $v_j{v}_l\in E\left(G_2\right)$.

The Cartesian sum $G_1\oplus G_2$ of the graphs $G_1$ and $G_2$ has the vertex set $V(G_1\oplus G_2)=V\left(G_1\right)\times V\left(G_2\right)$ and $(u_i,v_j)(u_k,v_l)$ is an edge of $G_1\oplus G_2$ if $u_i{u}_k\in E\left(G_1\right)$ or $v_j{v}_l\in E\left(G_2\right)$.

In [57] is defined the following Zagreb polynomial

$$M_1^{*}(G,x):=\sum _{u\in V\left(G\right)}{d}_u{x}^{d_u}.$$

Please note that $x{\left(xID(G,x)\right)}^{\prime }=M_1^{*}(G,x)$.

In ([43], Propositions 1, 2 and 3) appear the following useful results.

Proposition 1.

If G is a graph of order n and $f:{\mathbb{Z}}^{+}\to {\mathbb{R}}^{+}$, then:

• $P_f(G,x)$ is a polynomial if and only if $1/f\left(d_u\right)\in {\mathbb{Z}}^{+}$ for every $u\in V\left(G\right)$,
• $P_f(G,x)$ is a positive $C^{\infty }$ function on $(0,\infty )$,
• $P_f(G,x)$ is a continuous function on $[0,\infty )$ if and only if $P_f(G,0)<\infty$,
• $P_f(G,x)$ is a continuous function on $[0,\infty )$ if and only if $f\left(d_u\right)\le 1$ for every $u\in V\left(G\right)$,
• $P_f(G,x)$ is an integrable function on $[0,A]$ for every $A>0$, and ${\int }_0^1{P}_f(G,x)dx=I_f\left(G\right)$,
• $P_f(G,x)$ is increasing on $(0,\infty )$ if and only if $f\left(d_u\right)\le 1$ for every $u\in V\left(G\right)$,
• $P_f(G,x)$ is strictly increasing on $(0,\infty )$ if and only if $f\left(d_u\right)\le 1$ for every $u\in V\left(G\right)$, and $f\left(d_v\right)\ne 1$ for some $v\in V\left(G\right)$,
• $P_f(G,x)$ is convex on $(0,\infty )$ if $f\left(d_u\right)\in (0,1/2]\cup [1,\infty )$ for every $u\in V\left(G\right)$,
• $P_f(G,x)$ is strictly convex on $(0,\infty )$ if $f\left(d_u\right)\in (0,1/2]\cup [1,\infty )$ for every $u\in V\left(G\right)$, and $f\left(d_v\right)\notin \{1/2,1\}$ for some $v\in V\left(G\right)$,
• $P_f(G,x)$ is concave on $(0,\infty )$ if $f\left(d_u\right)\in [1/2,1]$ for every $u\in V\left(G\right)$,
• $P_f(G,x)$ is strictly concave on $(0,\infty )$ if $f\left(d_u\right)\in [1/2,1]$ for every $u\in V\left(G\right)$, and $f\left(d_v\right)\notin \{1/2,1\}$ for some $v\in V\left(G\right)$,
• $P_f(G,1)=n$.

If $\alpha \in \mathbb{R}$ and $f\left(t\right)=t^{\alpha }$, then Proposition 1 gives ${\int }_0^1{P}_f(G,x)dx=M_1^{\alpha }\left(G\right)$.

Proposition 2.

If G is a k-regular graph with n vertices and $f:{\mathbb{Z}}^{+}\to {\mathbb{R}}^{+}$, then $P_f(G,x)=n{x}^{1/f\left(k\right)-1}.$

Next, we show the polynomial $P_f$ for some important graphs: $K_n$ (complete graph), $C_n$ (cycle graph), $Q_n$ (hypercube graph), $K_{n_1,n_2}$ (complete bipartite graph), $S_n$ (star graph), $P_n$ (path graph), and $W_n$ (wheel graph).

Proposition 3.

If $f:{\mathbb{Z}}^{+}\to {\mathbb{R}}^{+}$, then

$$\begin{array}{cc}\hfill P_f(K_n,x)=n{x}^{1/f(n-1)-1},& P_f(C_n,x)=n{x}^{1/f\left(2\right)-1},\hfill \\ \hfill P_f(Q_n,x)=2^n{x}^{1/f\left(n\right)-1},& P_f(K_{n_1,n_2},x)=n_1{x}^{1/f\left(n_2\right)-1}+n_2{x}^{1/f\left(n_1\right)-1},\hfill \\ \hfill P_f(S_n,x)=x^{1/f(n-1)-1}+(n-1)x^{1/f\left(1\right)-1},& P_f(P_n,x)=(n-2)x^{1/f\left(2\right)-1}+2x^{1/f\left(1\right)-1},\hfill \\ \hfill P_f(W_n,x)=x^{1/f(n-1)-1}+(n-1)x^{1/f\left(3\right)-1}.\end{array}$$

Choose $\delta \in {\mathbb{Z}}^{+}$ and $f:{\mathbb{Z}}^{+}\to {\mathbb{R}}^{+}$. f satisfies the $\delta$-additive property 1 ($f\in A{P}_1\left(\delta \right)$) if

$$\frac{1}{f(a+b)}\ge \frac{1}{f\left(a\right)}+\frac{1}{f\left(b\right)}$$

for every $a,b\in {\mathbb{Z}}^{+}$ with $a,b\ge \delta$.

f satisfies the $\delta$-additive property 2 ($f\in A{P}_2\left(\delta \right)$) if

$$\frac{1}{f(a+b)}\le \frac{1}{f\left(a\right)}+\frac{1}{f\left(b\right)}$$

for every $a,b\in {\mathbb{Z}}^{+}$ with $a,b\ge \delta$.

f satisfies the $\delta$-additive property 3 ($f\in A{P}_3\left(\delta \right)$) if

$$\frac{1}{f(a+b)}\le min\left\{\frac{1}{f\left(a\right)},\frac{1}{f\left(b\right)}\right\}$$

for every $a,b\in {\mathbb{Z}}^{+}$ with $a,b\ge \delta$.

3. Inequalities for Cartesian Products

First, we prove pointwise inequalities of $P_f(G_1\times G_2,x)$ involving $P_f(G_1,x)$ and $P_f(G_2,x)$.

Theorem 1.

Let $\delta \in {\mathbb{Z}}^{+}$, and let $G_1$ and $G_2$ be two graphs with minimum degree at least δ. For $x\in (0,1]$, the f-polynomial of the Cartesian product $G_1\times G_2$ satisfies.

$\left(1\right)$ If $f\in A{P}_1\left(\delta \right)$, then

$$P_f(G_1\times G_2,x)\le x{P}_f(G_1,x)P_f(G_2,x).$$

$\left(2\right)$ If $f\in A{P}_2\left(\delta \right)$, then

$$P_f(G_1\times G_2,x)\ge x{P}_f(G_1,x)P_f(G_2,x).$$

$\left(3\right)$ If $f\in A{P}_3\left(\delta \right)$ and $G_1$ and $G_2$ have $n_1$ and $n_2$ vertices, respectively, then

$$P_f(G_1\times G_2,x)\ge max\left\{n_2{P}_f(G_1,x),n_1{P}_f(G_2,x)\right\}.$$

Proof. 

If $(u,v)\in V(G_1\times G_2)$, we have $d_{(u,v)}=d_u+d_v$ and its corresponding monomial of the f-polynomial is

$$x^{1/f(d_u+d_v)-1}.$$

Assume that $f\in A{P}_1\left(\delta \right)$. Since $d_u\ge \delta$ for every $u\in V\left(G_1\right)\cup V\left(G_2\right)$, $f\in A{P}_1\left(\delta \right)$ and $x\in (0,1]$,

$$\begin{array}{cc}\hfill P_f(G_1\times G_2,x)& =\sum _{u\in V\left(G_1\right)}\sum _{v\in V\left(G_2\right)}{x}^{1/f(d_u+d_v)-1}\le \sum _{u\in V\left(G_1\right)}\sum _{v\in V\left(G_2\right)}{x}^{1/f\left(d_u\right)+1/f\left(d_v\right)-1}\hfill \\ & =x\sum _{u\in V\left(G_1\right)}{x}^{1/f\left(d_u\right)-1}\sum _{v\in V\left(G_2\right)}{x}^{1/f\left(d_v\right)-1}=x{P}_f(G_1,x)P_f(G_2,x).\hfill \end{array}$$

If $f\in A{P}_2\left(\delta \right)$, then by a quite similar argument the result is obtained.

If $f\in A{P}_3\left(\delta \right)$, then

$$\begin{array}{cc}\hfill P_f(G_1\times G_2,x)& =\sum _{u\in V\left(G_1\right)}\sum _{v\in V\left(G_2\right)}{x}^{1/f(d_u+d_v)-1}\ge \sum _{u\in V\left(G_1\right)}\sum _{v\in V\left(G_2\right)}{x}^{1/f\left(d_u\right)-1}\hfill \\ & =\sum _{u\in V\left(G_1\right)}{x}^{1/f\left(d_u\right)-1}\sum _{v\in V\left(G_2\right)}1=n_2{P}_f(G_1,x).\hfill \end{array}$$

The inequality involving $P_f(G_2,x)$ is obtained in a similar way. □

Remark 1.

If $\delta =1$, then the condition $G_1$ and $G_2$ have minimum degree at least δ, is satisfied for every graph $G_1,G_2$.

Theorem 1 has the following consequence when we consider $f\left(t\right)=t^{\alpha }$.

Theorem 2.

Let $G_1$ and $G_2$ be two graphs of order $n_1$ and $n_2$, respectively, $\alpha \in \mathbb{R}$ and $f\left(t\right)=t^{\alpha }$. For $x\in (0,1]$, the f-polynomial of the Cartesian product $G_1\times G_2$ satisfies.

$\left(1\right)$ If $\alpha \le -1$, then $f\in A{P}_1\left(1\right)$ and

$$P_f(G_1\times G_2,x)\le x{P}_f(G_1,x)P_f(G_2,x).$$

$\left(2\right)$ If $\alpha \in [-1,0]$, then $f\in A{P}_2\left(1\right)$ and

$$P_f(G_1\times G_2,x)\ge x{P}_f(G_1,x)P_f(G_2,x).$$

$\left(3\right)$ If $\alpha \ge 0$, then $f\in A{P}_3\left(1\right)$ and

$$P_f(G_1\times G_2,x)\ge max\left\{n_2{P}_f(G_1,x),n_1{P}_f(G_2,x)\right\}.$$

Proof. 

The inequalities are consequences of the following facts and Theorem 1.

If $\alpha \le -1$, then $-\alpha \ge 1$ and

$$\frac{1}{f(x+y)}={(x+y)}^{-\alpha }\ge x^{-\alpha }+y^{-\alpha }=\frac{1}{f\left(x\right)}+\frac{1}{f\left(y\right)}$$

for every $x,y>0$, and so, $f\in A{P}_1\left(1\right)$.

If $\alpha \in [-1,0]$, then $-\alpha \in [0,1]$ and

$$\frac{1}{f(x+y)}={(x+y)}^{-\alpha }\le x^{-\alpha }+y^{-\alpha }=\frac{1}{f\left(x\right)}+\frac{1}{f\left(y\right)}$$

for every $x,y>0$, and so, $f\in A{P}_2\left(1\right)$.

If $\alpha \ge 0$, then $1/t^{\alpha }$ is a decreasing function on $(0,\infty )$ and

$$\frac{1}{f(x+y)}=\frac{1}{{(x+y)}^{\alpha }}\le min\left\{\frac{1}{x^{\alpha }},\frac{1}{y^{\alpha }}\right\}=min\left\{\frac{1}{f\left(x\right)},\frac{1}{f\left(y\right)}\right\}$$

for every $x,y>0$, and so, $f\in A{P}_3\left(1\right)$. □

Theorem 2 yields for the inverse degree polynomial:

Corollary 1.

Let $G_1$ and $G_2$ be two graphs, the ID polynomial of the Cartesian product $G_1\times G_2$ is

$$ID(G_1\times G_2,x)=xID(G_1,x)ID(G_2,x).$$

Since $f\left(t\right)=t\sqrt{logt}$ is an increasing function on $[1,\infty )$, $f\in A{P}_3\left(2\right)$ and Theorem 1 has the following consequence. Recall that a vertex with degree 1 is called pendant vertex. Please note that $f\left(t\right)=t\sqrt{logt}$ is a positive function on ${\mathbb{Z}}^{+}\backslash \left\{1\right\}$, and so, $P_f(G,x)$ is well-defined for every graph G without pendant vertices.

Please note that a graph has minimum degree at least 2 if and only if it does not have pendant vertices.

Theorem 3.

Let $G_1$ and $G_2$ be two graphs with minimum degree at least 2 and of order $n_1$ and $n_2$, respectively. If $f\left(t\right)=t\sqrt{logt}$, then $f\in A{P}_3\left(2\right)$ and the f-polynomial of the Cartesian product $G_1\times G_2$ satisfies for $x\in (0,1]$

$$P_f(G_1\times G_2,x)\ge max\left\{n_2{P}_f(G_1,x),n_1{P}_f(G_2,x)\right\}.$$

Next, we obtain bounds for $I_f(G_1\times G_2)$ by using the previous inequalities for $P_f(G_1\times G_2,x)$. This is a good approach since the pointwise bounds of $P_f(G_1\times G_2,x)$ allow use of analytic tools to bound $I_f(G_1\times G_2)$. We start with the case $f\in A{P}_3\left(\delta \right)$.

Theorem 4.

Let $\delta \in {\mathbb{Z}}^{+}$, $G_1$ and $G_2$ be two graphs of order $n_1$ and $n_2$, respectively, and minimum degree at least δ. If $f\in A{P}_3\left(\delta \right)$, then

$$I_f(G_1\times G_2)\ge max\left\{n_2{I}_f\left(G_1\right),n_1{I}_f\left(G_2\right)\right\}.$$

Proof. 

Theorem 1 gives

$$P_f(G_1\times G_2,x)\ge max\left\{n_2{P}_f(G_1,x),n_1{P}_f(G_2,x)\right\},$$

for every $x\in (0,1]$. Thus, Proposition 1 leads to

$$\begin{array}{cc}\hfill I_f(G_1\times G_2)& ={\int }_0^1{P}_f(G_1\times G_2,x)dx\ge n_2{\int }_0^1{P}_f(G_1,x)dx=n_2{I}_f\left(G_1\right).\hfill \end{array}$$

The same argumentation gives $I_f(G_1\times G_2)\ge n_1{I}_f\left(G_2\right)$. □

To deal with the cases $f\in A{P}_1\left(\delta \right)$ and $f\in A{P}_2\left(\delta \right)$, we need some technical results.

Lemma 1.

[58] If $f_1,\dots ,f_k$ are non-negative convex functions on $[a,b]$, then

$$\frac{1}{b-a}{\int }_a^b{\displaystyle \prod _{i=1}^k}{f}_i\left(x\right)dx\ge \frac{2^k}{k+1}{\displaystyle \prod _{i=1}^k}\frac{1}{b-a}{\int }_a^b{f}_i\left(x\right)dx.$$

Lemma 1 can be slightly improved as follows.

Proposition 4.

If $f_1,\dots ,f_k$ are convex non-negative functions on $(a,b)$, then

$$\frac{1}{b-a}{\int }_a^b{\displaystyle \prod _{i=1}^k}{f}_i\left(x\right)dx\ge \frac{2^k}{k+1}{\displaystyle \prod _{i=1}^k}\frac{1}{b-a}{\int }_a^b{f}_i\left(x\right)dx.$$

Proof. 

If $0<\epsilon <(b-a)/2$, then $f_1,\dots ,f_k$ are convex non-negative functions on $[a+\epsilon ,b-\epsilon ]$, and Lemma 1 gives

$$\frac{1}{b-a-2\epsilon }{\int }_{a+\epsilon }^{b-\epsilon }{\displaystyle \prod _{i=1}^k}{f}_i\left(x\right)dx\ge \frac{2^k}{k+1}{\displaystyle \prod _{i=1}^k}\frac{1}{b-a-2\epsilon }{\int }_{a+\epsilon }^{b-\epsilon }{f}_i\left(x\right)dx.$$

Since $f_1,\dots ,f_k$ are non-negative on $(a,b)$, the functions $F,G:(0,(b-a)/2)\to [0,\infty )$ defined by

$$F\left(\epsilon \right)={\int }_{a+\epsilon }^{b-\epsilon }{\displaystyle \prod _{i=1}^k}{f}_i\left(x\right)dx,G\left(\epsilon \right)={\displaystyle \prod _{i=1}^k}{\int }_{a+\epsilon }^{b-\epsilon }{f}_i\left(x\right)dx,$$

are decreasing, and so, there exist their limits as $\epsilon \to 0^{+}$ (although they can be ∞). By taking $\epsilon \to 0^{+}$ in the above inequality, we obtain the result. □

Lemma 2.

([59] Corollary 5.2) If $f_1,\dots ,f_k$ are convex non-negative functions on $[a,b]$, then

$${\int }_a^b{\displaystyle \prod _{i=1}^k}{f}_i\left(x\right)dx\le \frac{2}{k+1}{\left({\displaystyle \prod _{i=1}^k}{\int }_a^b{f}_i\left(x\right)dx\right)}^{1/k}{\left({\displaystyle \prod _{i=1}^k}\left(f_i\left(a\right)+f_i\left(b\right)\right)\right)}^{1-1/k}.$$

Since any non-negative concave function on $(a,b)$ has finite lateral limits at a and b, ([59], Corollary 4.3) can be stated as follows.

Lemma 3.

If $f_1,f_2$ are non-negative concave functions on $(a,b)$, then

$$\begin{array}{cc}\hfill \frac{2}{3}\frac{1}{b-a}{\int }_a^b{f}_1\left(x\right)dx\frac{1}{b-a}{\int }_a^b{f}_2\left(x\right)dx& \le \frac{1}{b-a}{\int }_a^b{f}_1\left(x\right)f_2\left(x\right)dx\hfill \\ & \le \frac{4}{3}\frac{1}{b-a}{\int }_a^b{f}_1\left(x\right)dx\frac{1}{b-a}{\int }_a^b{f}_2\left(x\right)dx.\hfill \end{array}$$

With these technical results, we can deal now with the cases $f\in A{P}_1\left(\delta \right)$ and $f\in A{P}_2\left(\delta \right)$.

Theorem 5.

Let $\delta \in {\mathbb{Z}}^{+}$, $G_1$ and $G_2$ be graphs of orders $n_1$ and $n_2$, respectively, and minimum degree at least δ. If the f-polynomials of $G_1$ and $G_2$ are convex functions on $(0,1)$, then

$\left(1\right)$ If $f\in A{P}_1\left(\delta \right)$, then

$$I_f(G_1\times G_2)\le \frac{1}{2}{\left(\frac{1}{2}{\left(n_1+P_f(G_1,0)\right)}^2{\left(n_2+P_f(G_2,0)\right)}^2{I}_f\left(G_1\right)I_f\left(G_2\right)\right)}^{1/3}.$$

$\left(2\right)$ If $f\in A{P}_2\left(\delta \right)$, then

$$I_f(G_1\times G_2)\ge I_f\left(G_1\right)I_f\left(G_2\right).$$

Proof. 

Assume first that $f\in A{P}_1\left(\delta \right)$. Theorem 1 gives

$$P_f(G_1\times G_2,x)\le x{P}_f(G_1,x)P_f(G_2,x),$$

for every $x\in (0,1]$.

If $P_f(G_1,0)=\infty$ or $P_f(G_2,0)=\infty$, then $\left(1\right)$ trivially holds.

If $P_f(G_1,0)<\infty$ and $P_f(G_2,0)<\infty$, then $P_f(G_1,x)$ and $P_f(G_2,x)$ are continuous functions on $[0,1]$ by Proposition 1, and so, they are convex on $[0,1]$. Proposition 1 and Lemma 2 give

$$\begin{array}{cc}\hfill I_f(G_1\times G_2)& ={\int }_0^1{P}_f(G_1\times G_2,x)dx\le {\int }_0^{1}x{P}_f(G_1,x)P_f(G_2,x)dx\hfill \\ & \le \frac{1}{2}{\left({\int }_0^{1}xdx{\int }_0^1{P}_f(G_1,x)dx{\int }_0^1{P}_f(G_2,x)dx\right)}^{1/3}\hfill \\ & ·{\left(((0+1)\left(P_f(G_1,0)+P_f(G_1,1)\right)\left(P_f(G_2,0)+P_f(G_2,1)\right)\right)}^{2/3}\hfill \\ & =\frac{1}{2}{\left(\frac{1}{2}{\left(n_1+P_f(G_1,0)\right)}^2{\left(n_2+P_f(G_2,0)\right)}^2{I}_f\left(G_1\right)I_f\left(G_2\right)\right)}^{1/3}.\hfill \end{array}$$

Assume now that $f\in A{P}_2\left(\delta \right)$. Theorem 1 gives

$$P_f(G_1\times G_2,x)\ge x{P}_f(G_1,x)P_f(G_2,x),$$

for every $x\in (0,1]$. Since $P_f(G_1,x)$ and $P_f(G_2,x)$ are convex functions on $(0,1)$, Proposition 1 and Proposition 4 give

$$\begin{array}{cc}\hfill I_f(G_1\times G_2)& ={\int }_0^1{P}_f(G_1\times G_2,x)dx\ge {\int }_0^{1}x{P}_f(G_1,x)P_f(G_2,x)dx\hfill \\ & \ge 2{\int }_0^{1}xdx{\int }_0^1{P}_f(G_1,x)dx{\int }_0^1{P}_f(G_2,x)dx\hfill \\ & =I_f\left(G_1\right)I_f\left(G_2\right).\hfill \end{array}$$

 □

If the f-polynomials of $G_1$ and $G_2$ are concave functions on $(0,1)$, we can obtain simpler bounds for $I_f(G_1\times G_2)$.

Theorem 6.

Let $\delta \in {\mathbb{Z}}^{+}$, $G_1$ and $G_2$ be two graphs with minimum degree at least δ. If $f\in A{P}_1\left(\delta \right)$ and the f-polynomials of $G_1$ and $G_2$ are concave functions on $(0,1)$, then

$$I_f(G_1\times G_2)\le \frac{4}{3}{I}_f\left(G_1\right)I_f\left(G_2\right).$$

Proof. 

Since $f\in A{P}_1\left(\delta \right)$, Theorem 1 gives

$$P_f(G_1\times G_2,x)\le x{P}_f(G_1,x)P_f(G_2,x).$$

for every $x\in (0,1]$. Since $P_f(G_1,x)$ and $P_f(G_2,x)$ are concave functions on $(0,1)$, a simple combination of Proposition 1 and Lemma 3 yields

$$\begin{array}{cc}\hfill I_f(G_1\times G_2)& ={\int }_0^1{P}_f(G_1\times G_2,x)dx\le {\int }_0^{1}x{P}_f(G_1,x)P_f(G_2,x)dx\hfill \\ & \le {\int }_0^1{P}_f(G_1,x)P_f(G_2,x)dx\le \frac{4}{3}{\int }_0^1{P}_f(G_1,x)dx{\int }_0^1{P}_f(G_2,x)dx\hfill \\ & =\frac{4}{3}{I}_f\left(G_1\right)I_f\left(G_2\right).\hfill \end{array}$$

 □

We can obtain more bounds for $I_f(G_1\times G_2)$ if the image of f does not intersect $(a/2,a)$ for some constant $a>0$.

Theorem 7.

Let $\delta \in {\mathbb{Z}}^{+}$, $G_1$ and $G_2$ be two graphs of orders $n_1$ and $n_2$, respectively, and minimum degree at least δ, and $a>0$. If $f:{\mathbb{Z}}^{+}\cap [\delta ,\infty )\to (0,a/2]\cup [a,\infty )$, then

$\left(1\right)$ If $f\in A{P}_1\left(\delta \right)$, then

$$I_f(G_1\times G_2)\le \frac{1}{2}{\left(\frac{a}{2}{\left(n_1+P_{f/a}(G_1,0)\right)}^2{\left(n_2+P_{f/a}(G_2,0)\right)}^2{I}_f\left(G_1\right)I_f\left(G_2\right)\right)}^{1/3}.$$

$\left(2\right)$ If $f\in A{P}_2\left(\delta \right)$, then

$$I_f(G_1\times G_2)\ge \frac{1}{a}{I}_f\left(G_1\right)I_f\left(G_2\right).$$

Proof. 

Let be the function $g=f/a$. Hence, $g:{\mathbb{Z}}^{+}\cap [\delta ,\infty )\to (0,1/2]\cup [1,\infty )$ and Proposition 1 gives that $P_g(G_1,x)$ and $P_g(G_2,x)$ are convex functions on $(0,\infty )$.

If $f\in A{P}_1\left(\delta \right)$, then $f/a\in A{P}_1\left(\delta \right)$ and Theorem 5 gives

$$\begin{array}{cc}\hfill \frac{1}{a}{I}_f(G_1\times G_2)& =I_{f/a}(G_1\times G_2)\hfill \\ & \le \frac{1}{2}{\left(\frac{1}{2}{\left(n_1+P_{f/a}(G_1,0)\right)}^2{\left(n_2+P_{f/a}(G_2,0)\right)}^2{I}_{f/a}\left(G_1\right)I_{f/a}\left(G_2\right)\right)}^{1/3}\hfill \\ & =\frac{1}{2}{\left(\frac{1}{2}{\left(n_1+P_{f/a}(G_1,0)\right)}^2{\left(n_2+P_{f/a}(G_2,0)\right)}^2\frac{1}{a}{I}_f\left(G_1\right)\frac{1}{a}{I}_f\left(G_2\right)\right)}^{1/3}.\hfill \end{array}$$

If $f\in A{P}_2\left(\delta \right)$, then $f/a\in A{P}_2\left(\delta \right)$ and Theorem 5 gives

$$\begin{array}{c}\hfill \frac{1}{a}{I}_f(G_1\times G_2)=I_{f/a}(G_1\times G_2)\ge I_{f/a}\left(G_1\right)I_{f/a}\left(G_2\right)=\frac{1}{a}{I}_f\left(G_1\right)\frac{1}{a}{I}_f\left(G_2\right).\end{array}$$

 □

The first inequality in Theorem 7 can be improved as follows.

Theorem 8.

Let $\delta \in {\mathbb{Z}}^{+}$, $G_1$ and $G_2$ be two graphs of orders $n_1$ and $n_2$, respectively, and minimum degree at least δ, and $a>0$. If $f:{\mathbb{Z}}^{+}\cap [\delta ,\infty )\to (0,a/2]$, and $f\in A{P}_1\left(\delta \right)$, then

$$I_f(G_1\times G_2)\le \frac{1}{2}{\left(\frac{a}{2}{n}_1^2{n}_2^2{I}_f\left(G_1\right)I_f\left(G_2\right)\right)}^{1/3}.$$

Proof. 

Theorem 7 gives

$$I_f(G_1\times G_2)\le \frac{1}{2}{\left(\frac{a}{2}{\left(n_1+P_{f/a}(G_1,0)\right)}^2{\left(n_2+P_{f/a}(G_2,0)\right)}^2{I}_f\left(G_1\right)I_f\left(G_2\right)\right)}^{1/3}.$$

Since $f\le a/2$ on ${\mathbb{Z}}^{+}\cap [\delta ,\infty )$, we have $a/f-1\ge 1$ and $P_{f/a}(G_1,0)=P_{f/a}(G_2,0)=0$, and so, we obtain the result. □

Theorems 2, 8 (with $a=2$), 7 (with $a=2$) and 4 have the following consequence for the general first Zagreb index.

Theorem 9.

Let $G_1$ and $G_2$ be two graphs of orders $n_1$ and $n_2$, respectively, and $\alpha \in \mathbb{R}$.

$\left(1\right)$ If $\alpha \le -1$, then

$$M_1^{\alpha }(G_1\times G_2)\le \frac{1}{2}{\left(n_1^2{n}_2^2{M}_1^{\alpha }\left(G_1\right)M_1^{\alpha }\left(G_2\right)\right)}^{1/3}.$$

$\left(2\right)$ If $\alpha \in [-1,0]$, then

$$M_1^{\alpha }(G_1\times G_2)\ge \frac{1}{2}{M}_1^{\alpha }\left(G_1\right)M_1^{\alpha }\left(G_2\right).$$

$\left(3\right)$ If $\alpha \ge 0$, then

$$M_1^{\alpha }(G_1\times G_2)\ge max\left\{n_2{M}_1^{\alpha }\left(G_1\right),n_1{M}_1^{\alpha }\left(G_2\right)\right\}.$$

As a rather simple consequence of the above theorem, we have for the first Zagreb, forgotten and inverse degree indices:

Corollary 2.

If $G_1$ and $G_2$ are two graphs with $n_1$ and $n_2$ vertices, respectively, then

$$\begin{array}{cc}\hfill M_1(G_1\times G_2)& \ge max\left\{n_2{M}_1\left(G_1\right),n_1{M}_1\left(G_2\right)\right\},\hfill \\ \hfill F(G_1\times G_2)& \ge max\left\{n_{2}F\left(G_1\right),n_{1}F\left(G_2\right)\right\},\hfill \\ \hfill \frac{1}{2}ID\left(G_1\right)ID\left(G_2\right)& \le ID(G_1\times G_2)\le \frac{1}{2}{\left(n_1^2{n}_2^{2}ID\left(G_1\right)ID\left(G_2\right)\right)}^{1/3}.\hfill \end{array}$$

Since $f\left(t\right)=t\sqrt{logt}\in A{P}_3\left(2\right)$, Theorem 4 gives the following result for the $SL$ index.

Theorem 10.

If $G_1$ and $G_2$ are graphs without pendant vertices and with $n_1$ and $n_2$ vertices, respectively, then

$$SL(G_1\times G_2)\ge max\left\{n_{2}SL\left(G_1\right),n_{1}SL\left(G_2\right)\right\}.$$

A particular consequence of Theorems 9 and 10 is the following.

Corollary 3.

If $C_{n_1}$ and $C_{n_1}$ are the cycle graphs with $n_1$ and $n_2$ vertices, respectively, then

$$\begin{array}{cc}\hfill M_1(C_{n_1}\times C_{n_2})& \ge 4n_1{n}_2,\hfill \\ \hfill F(C_{n_1}\times C_{n_2})& \ge 8n_1{n}_2,\hfill \\ \hfill \frac{1}{8}{n}_1{n}_2\le ID(C_{n_1}\times C_{n_2})& \le \frac{1}{2^{5/3}}{n}_1{n}_2,\hfill \\ \hfill SL(C_{n_1}\times C_{n_2})& \ge n_1{n}_{2}2log2.\hfill \end{array}$$

4. Inequalities for Lexicographic Products

We start this section by proving pointwise inequalities of $P_f(G_1\odot G_2,x)$ involving the f-polynomials of $G_1$ and $G_2$.

Theorem 11.

Let $\delta \in {\mathbb{Z}}^{+}$, $G_1$ and $G_2$ be two graphs of orders $n_1$ and $n_2$, respectively, and minimum degree at least δ. The f-polynomial of the lexicographic product $G_1\odot G_2$ satisfies the following inequalities for $x\in (0,1]$.

$\left(1\right)$ If $f\in A{P}_1\left(\delta \right)$, then

$$P_f(G_1\odot G_2,x)\le x^{n_2}{P}_f(G_1,x^{n_2})P_f(G_2,x).$$

$\left(2\right)$ If $f\in A{P}_2\left(\delta \right)$, then

$$P_f(G_1\odot G_2,x)\ge x^{n_2}{P}_f(G_1,x^{n_2})P_f(G_2,x).$$

$\left(3\right)$ If $f\in A{P}_3\left(\delta \right)$, then

$$P_f(G_1\odot G_2,x)\ge max\left\{n_2{P}_f(G_1,x),n_1{P}_f(G_2,x)\right\}.$$

Proof. 

If $(u,v)\in V(G_1\odot G_2)$, then $d_{(u,v)}=n_2{d}_u+d_v$.

Assume that $f\in A{P}_1\left(\delta \right)$. Since $d_u\ge \delta$ for every $u\in V\left(G_1\right)\cup V\left(G_2\right)$, one can prove by induction that

$$\frac{1}{f(n_2{d}_u+d_v)}\ge \frac{n_2}{f\left(d_u\right)}+\frac{1}{f\left(d_v\right)}.$$

Since $x\in (0,1]$,

$$\begin{array}{cc}\hfill P_f(G_1\odot G_2,x)& =\sum _{u\in V\left(G_1\right)}\sum _{v\in V\left(G_2\right)}{x}^{1/f(n_2{d}_u+d_v)-1}\hfill \\ & \le \sum _{u\in V\left(G_1\right)}{x}^{n_2/f\left(d_u\right)}\sum _{v\in V\left(G_2\right)}{x}^{1/f\left(d_v\right)-1}\hfill \\ & =\sum _{u\in V\left(G_1\right)}{\left(x^{n_2}\right)}^{1/f\left(d_u\right)-1}{x}^{n_2}{P}_f(G_2,x)\hfill \\ & =x^{n_2}{P}_f(G_1,x^{n_2})P_f(G_2,x).\hfill \end{array}$$

If $f\in A{P}_2\left(\delta \right)$, then same argument allows obtaining of the result.

If $f\in A{P}_3\left(\delta \right)$, then

$$\begin{array}{cc}\hfill P_f(G_1\odot G_2,x)& =\sum _{u\in V\left(G_1\right)}\sum _{v\in V\left(G_2\right)}{x}^{1/f(n_2{d}_u+d_v)-1}\hfill \\ & \ge \sum _{u\in V\left(G_1\right)}\sum _{v\in V\left(G_2\right)}{x}^{1/f\left(d_u\right)-1}=n_2{P}_f(G_1,x).\hfill \end{array}$$

A similar argument gives $P_f(G_1\odot G_2,x)\ge n_1{P}_f(G_2,x)$. □

Theorems 2 and 11 have the following consequence when $f\left(t\right)=t^{\alpha }$.

Proposition 5.

Let $G_1$ and $G_2$ be two graphs with $n_1$ and $n_2$ vertices, respectively, $\alpha \in \mathbb{R}$ and $f\left(t\right)=t^{\alpha }$. The f-polynomial of the lexicographic product $G_1\odot G_2$ satisfies the following inequalities for $x\in (0,1]$.

$\left(1\right)$ If $\alpha \le -1$, then

$$P_f(G_1\odot G_2,x)\le x^{n_2}{P}_f(G_1,x^{n_2})P_f(G_2,x).$$

$\left(2\right)$ If $\alpha \in [-1,0]$, then

$$P_f(G_1\odot G_2,x)\ge x^{n_2}{P}_f(G_1,x^{n_2})P_f(G_2,x).$$

$\left(3\right)$ If $\alpha \ge 0$, then

$$P_f(G_1\odot G_2,x)\ge max\left\{n_2{P}_f(G_1,x),n_1{P}_f(G_2,x)\right\}.$$

Theorem 5 gives the following equality for the inverse degree polynomial.

Corollary 4.

Given two graphs $G_1$ and $G_2$, of order $n_1$ and $n_2$, respectively, the ID polynomial of the lexicographic product $G_1\odot G_2$ is

$$ID(G_1\odot G_2,x)=x^{n_2}ID(G_1,x^{n_2})ID(G_2,x).$$

Since $f\left(t\right)=t\sqrt{logt}\in A{P}_3\left(2\right)$, Theorem 11 has the following consequence.

Theorem 12.

Let $G_1$ and $G_2$ be two graphs with minimum degree two and of order $n_1$ and $n_2$, respectively. If $f\left(t\right)=t\sqrt{logt}$, then the f-polynomial of the lexicographic product $G_1\odot G_2$ satisfies for $x\in (0,1]$

$$P_f(G_1\odot G_2,x)\ge max\left\{n_2{P}_f(G_1,x),n_1{P}_f(G_2,x)\right\}.$$

Next, we obtain bounds for $I_f(G_1\odot G_2)$ by using the previous inequalities for $P_f(G_1\odot G_2,x)$. We start with $f\in A{P}_3\left(\delta \right)$.

Theorem 13.

Let $\delta \in {\mathbb{Z}}^{+}$, $G_1$ and $G_2$ be two graphs of order $n_1$ and $n_2$, respectively, and minimum degree at least δ. If $f\in A{P}_3\left(\delta \right)$, then

$$I_f(G_1\odot G_2)\ge max\left\{n_2{I}_f\left(G_1\right),n_1{I}_f\left(G_2\right)\right\}.$$

Proof. 

Theorem 11 gives

$$P_f(G_1\odot G_2,x)\ge n_2{P}_f(G_1,x).$$

for every $0\le x\le 1$. Thus, Proposition 1 leads to $I_f(G_1\odot G_2)\ge n_2{I}_f\left(G_1\right)$. A similar argument gives the inequality $I_f(G_1\odot G_2)\ge n_1{I}_f\left(G_2\right)$. □

We deal now with $f\in A{P}_1\left(\delta \right)\cup A{P}_2\left(\delta \right)$.

Theorem 14.

Let $\delta \in {\mathbb{Z}}^{+}$, $G_1$ and $G_2$ be two graphs of orders $n_1$ and $n_2$, respectively, and minimum degree at least δ, and $a>0$. If $f:{\mathbb{Z}}^{+}\cap [\delta ,\infty )\to (0,a/2]$, then the following inequalities hold.

$\left(1\right)$ If $f\in A{P}_1\left(\delta \right)$, then

$$I_f(G_1\odot G_2)\le \frac{1}{2}{\left(\frac{n_1^2{n}_{2}a}{2}{I}_f\left(G_1\right)I_f\left(G_2\right)\right)}^{1/3}.$$

$\left(2\right)$ If $f\in A{P}_2\left(\delta \right)$, then

$$I_f(G_1\odot G_2)\ge \frac{1}{n_{2}a}{I}_f\left(G_1\right)I_f\left(G_2\right).$$

Proof. 

Consider the function $g=f/a$. We have $g:{\mathbb{Z}}^{+}\cap [\delta ,\infty )\to (0,1/2]$ and so, Proposition 1 implies that $P_g(G_1,x)$ and $P_g(G_2,x)$ are convex functions on the open interval $(0,\infty )$ and continuous on the closed interval $[0,\infty )$; hence, they are convex when $x\in [0,1]$.

If $f\in A{P}_1\left(\delta \right)$, then $f/a\in A{P}_1\left(\delta \right)$ and Theorem 11 implies

$$P_{f/a}(G_1\odot G_2,x)\le x^{n_2}{P}_{f/a}(G_1,x^{n_2})P_{f/a}(G_2,x).$$

Notice that $f/a\le 1/2$ implies $a/f-1\ge 1$, and thus, $P_{f/a}(G_i,0)=0$ and $P_{f/a}(G_i,1)=n_i$ for $i=1,2$. Since x is a convex function when $x\in [0,1]$, Lemma 2 implies

$$\begin{array}{cc}& \frac{1}{a}{I}_f(G_1\odot G_2)=I_{f/a}(G_1\odot G_2)={\int }_0^1{P}_{f/a}(G_1\odot G_2,x)dx\hfill \\ & \le {\int }_0^1{x}^{n_2}{P}_{f/a}(G_1,x^{n_2})P_{f/a}(G_2,x)dx\hfill \\ & \le \frac{1}{2}{\left({\int }_0^{1}xdx{\int }_0^1{x}^{n_2-1}{P}_{f/a}(G_1,x^{n_2})dx{\int }_0^1{P}_{f/a}(G_2,x)dx\right)}^{1/3}{\left(1·n_1·n_2\right)}^{2/3}\hfill \\ & =\frac{1}{2}{\left(\frac{1}{2}\frac{1}{n_2}{\int }_0^1{P}_{f/a}(G_1,t)dt\frac{1}{a}{I}_f\left(G_2\right)n_1^2{n}_2^2\right)}^{1/3}\hfill \\ & =\frac{1}{2}{\left(\frac{n_1^2{n}_2}{2a^2}{I}_f\left(G_1\right)I_f\left(G_2\right)\right)}^{1/3}.\hfill \end{array}$$

If $f\in A{P}_2\left(\delta \right)$, thus $f/a\in A{P}_2\left(\delta \right)$ and Theorem 11 implies

$$P_{f/a}(G_1\odot G_2,x)\ge x^{n_2}{P}_{f/a}(G_1,x^{n_2})P_{f/a}(G_2,x).$$

Thus, Lemma 1 gives

$$\begin{array}{cc}\hfill \frac{1}{a}{I}_f(G_1\odot G_2)& =I_{f/a}(G_1\odot G_2)={\int }_0^1{P}_{f/a}(G_1\odot G_2,x)dx\hfill \\ & \ge {\int }_0^1{x}^{n_2}{P}_{f/a}(G_1,x^{n_2})P_{f/a}(G_2,x)dx\hfill \\ & \ge 2{\int }_0^{1}xdx{\int }_0^1{x}^{n_2-1}{P}_{f/a}(G_1,x^{n_2})dx{\int }_0^1{P}_{f/a}(G_2,x)dx\hfill \\ & =2\frac{1}{2}\frac{1}{n_{2}a}{I}_f\left(G_1\right)\frac{1}{a}{I}_f\left(G_2\right)=\frac{1}{n_2{a}^2}{I}_f\left(G_1\right)I_f\left(G_2\right).\hfill \end{array}$$

 □

Now, we deduce several inequalities for many topological indices of lexicographic products.

Theorems 2, 14 (with $a=2$) and 13 have the following consequence for the variable first Zagreb index.

Theorem 15.

Let $G_1$ and $G_2$ be two graphs of orders $n_1$ and $n_2$, respectively, and $\alpha \in \mathbb{R}$.

$\left(1\right)$ If $\alpha \le -1$, then

$$M_1^{\alpha }(G_1\odot G_2)\le \frac{1}{2}{\left(n_1^2{n}_2{M}_1^{\alpha }\left(G_1\right)M_1^{\alpha }\left(G_2\right)\right)}^{1/3}.$$

$\left(2\right)$ If $\alpha \in [-1,0]$, then

$$M_1^{\alpha }(G_1\odot G_2)\ge \frac{1}{2n_2}{M}_1^{\alpha }\left(G_1\right)M_1^{\alpha }\left(G_2\right).$$

$\left(3\right)$ If $\alpha \ge 0$, then

$$M_1^{\alpha }(G_1\odot G_2)\ge max\left\{n_2{M}_1^{\alpha }\left(G_1\right),n_1{M}_1^{\alpha }\left(G_2\right)\right\}.$$

Theorem 15 has the following consequence for the first Zagreb, forgotten and inverse degree.

Corollary 5.

If $G_1$ and $G_2$ are two graphs with $n_1$ and $n_2$ vertices, respectively, then

$$\begin{array}{cc}\hfill M_1(G_1\odot G_2)& \ge max\left\{n_2{M}_1\left(G_1\right),n_1{M}_1\left(G_2\right)\right\},\hfill \\ \hfill F(G_1\odot G_2)& \ge max\left\{n_{2}F\left(G_1\right),n_{1}F\left(G_2\right)\right\},\hfill \\ \hfill \frac{1}{2n_2}ID\left(G_1\right)ID\left(G_2\right)\le ID(G_1\odot G_2)& \le \frac{1}{2}{\left(n_1^2{n}_{2}ID\left(G_1\right)ID\left(G_2\right)\right)}^{1/3}.\hfill \end{array}$$

Since $f\left(t\right)=t\sqrt{logt}\in A{P}_3\left(2\right)$, Theorem 13 allows deduction of the following result for the $SL$ index.

Theorem 16.

If $G_1$ and $G_2$ are graphs without pendant vertices and with $n_1$ and $n_2$ vertices, respectively, then

$$SL(G_1\odot G_2)\ge max\left\{n_{2}SL\left(G_1\right),n_{1}SL\left(G_2\right)\right\}.$$

5. Inequalities for Cartesian Sums

We start this last section by proving pointwise inequalities of $P_f(G_1\oplus G_2,x)$ involving the f-polynomials of $G_1$ and $G_2$.

Theorem 17.

Let $\delta \in {\mathbb{Z}}^{+}$, $G_1$ and $G_2$ be two graphs of orders $n_1$ and $n_2$, respectively, and minimum degree at least δ. For $x\in (0,1]$, the f-polynomial of the Cartesian sum $G_1\oplus G_2$ satisfies.

$\left(1\right)$ If $f\in A{P}_1\left(\delta \right)$, then

$$P_f(G_1\oplus G_2,x)\le x^{n_1+n_2-1}{P}_f(G_1,x^{n_2})P_f(G_2,x^{n_1}).$$

$\left(2\right)$ If $f\in A{P}_2\left(\delta \right)$, then

$$P_f(G_1\oplus G_2,x)\ge x^{n_1+n_2-1}{P}_f(G_1,x^{n_2})P_f(G_2,x^{n_1}).$$

$\left(3\right)$ If $f\in A{P}_3\left(\delta \right)$, then

$$P_f(G_1\oplus G_2,x)\ge max\left\{n_2{P}_f(G_1,x),n_1{P}_f(G_2,x)\right\}.$$

Proof. 

If $(u,v)\in V(G_1\oplus G_2)$, then $d_{(u,v)}=n_2{d}_u+n_1{d}_v$.

Suppose that $f\in A{P}_1\left(\delta \right)$. Since $d_u\ge \delta$ for every $u\in V\left(G_1\right)\cup V\left(G_2\right)$, one can prove by induction that

$$\frac{1}{f(n_2{d}_u+n_1{d}_v)}\ge \frac{n_2}{f\left(d_u\right)}+\frac{n_1}{f\left(d_v\right)}.$$

Since $x\in (0,1]$,

$$\begin{array}{cc}\hfill P_f(G_1\oplus G_2,x)& =\sum _{u\in V\left(G_1\right)}\sum _{v\in V\left(G_2\right)}{x}^{1/f(n_2{d}_u+d_v)-1}\hfill \\ & \le \sum _{u\in V\left(G_1\right)}{x}^{n_2/f\left(d_u\right)}\sum _{v\in V\left(G_2\right)}{x}^{n_1/f\left(d_v\right)}{x}^{-1}\hfill \\ & =\sum _{u\in V\left(G_1\right)}{\left(x^{n_2}\right)}^{1/f\left(d_u\right)-1}{x}^{n_2}\sum _{v\in V\left(G_2\right)}{\left(x^{n_1}\right)}^{1/f\left(d_v\right)-1}{x}^{n_1}{x}^{-1}\hfill \\ & =x^{n_1+n_2-1}{P}_f(G_1,x^{n_2})P_f(G_2,x^{n_1}).\hfill \end{array}$$

If $f\in A{P}_2\left(\delta \right)$, then a similar argument allows obtaining of the corresponding inequality.

Suppose that $f\in A{P}_3\left(\delta \right)$. We deduce

$$\begin{array}{cc}\hfill P_f(G_1\oplus G_2,x)& =\sum _{u\in V\left(G_1\right)}\sum _{v\in V\left(G_2\right)}{x}^{1/f(n_2{d}_u+n_1{d}_v)-1}\hfill \\ & \ge \sum _{u\in V\left(G_1\right)}\sum _{v\in V\left(G_2\right)}{x}^{1/f\left(d_u\right)-1}=n_2{P}_f(G_1,x).\hfill \end{array}$$

A similar argument gives $P_f(G_1\oplus G_2,x)\ge n_1{P}_f(G_2,x)$. □

Theorems 2 and 17 have the following consequence for $f\left(t\right)=t^{\alpha }$.

Proposition 6.

Let $G_1$ and $G_2$ be two graphs of orders $n_1$ and $n_2$, respectively, $\alpha \in \mathbb{R}$ and $f\left(t\right)=t^{\alpha }$. For $x\in (0,1]$, the f-polynomial of the Cartesian sum $G_1\oplus G_2$ satisfies the following inequalities for.

$\left(1\right)$ If $\alpha \le -1$, then

$$P_f(G_1\oplus G_2,x)\le x^{n_1+n_2-1}{P}_f(G_1,x^{n_2})P_f(G_2,x^{n_1}).$$

$\left(2\right)$ If $\alpha \in [-1,0]$, then

$$P_f(G_1\oplus G_2,x)\ge x^{n_1+n_2-1}{P}_f(G_1,x^{n_2})P_f(G_2,x^{n_1}).$$

$\left(3\right)$ If $\alpha \ge 0$, then

$$P_f(G_1\oplus G_2,x)\ge max\left\{n_2{P}_f(G_1,x),n_1{P}_f(G_2,x)\right\}.$$

Proposition 6 allows deduction of the following equality for the inverse degree polynomial.

Corollary 6.

Given two graphs $G_1$ and $G_2$, or order $n_1$ and $n_2$, respectively, the ID polynomial of the Cartesian sum $G_1\oplus G_2$ is

$$ID(G_1\oplus G_2,x)=x^{n_1+n_2-1}ID(G_1,x^{n_2})ID(G_2,x^{n_1}).$$

Since $f\left(t\right)=t\sqrt{logt}\in A{P}_3\left(2\right)$, Theorem 17 has the following consequence.

Corollary 7.

Let $G_1$ and $G_2$ be two graphs with minimum degree two and of order $n_1$ and $n_2$, respectively. If $f\left(t\right)=t\sqrt{logt}$, then the f-polynomial of the Cartesian sum $G_1\oplus G_2$ satisfies for $x\in (0,1]$

$$P_f(G_1\oplus G_2,x)\ge max\left\{n_2{P}_f(G_1,x),n_1{P}_f(G_2,x)\right\}.$$

Next, we obtain bounds for $I_f(G_1\oplus G_2)$ by using the previous inequalities for $P_f(G_1\oplus G_2,x)$. We start when $f\in A{P}_3\left(\delta \right)$.

Theorem 18.

Let $\delta \in {\mathbb{Z}}^{+}$, $G_1$ and $G_2$ be two graphs of orders $n_1$ and $n_2$, respectively, and minimum degree at least δ. If $f\in A{P}_3\left(\delta \right)$, then

$$I_f(G_1\oplus G_2)\ge max\left\{n_2{I}_f\left(G_1\right),n_1{I}_f\left(G_2\right)\right\}.$$

Proof. 

Theorem 17 gives

$$P_f(G_1\oplus G_2,x)\ge n_2{P}_f(G_1,x).$$

for every $0<x\le 1$. Hence, $I_f(G_1\oplus G_2)\ge n_2{I}_f\left(G_1\right)$ by Proposition 1. A similar argument gives the inequality $I_f(G_1\oplus G_2)\ge n_1{I}_f\left(G_2\right)$. □

We consider now the case $f\in A{P}_1\left(\delta \right)\cup A{P}_2\left(\delta \right)$.

Theorem 19.

Let $\delta \in {\mathbb{Z}}^{+}$, $G_1$ and $G_2$ be two graphs of orders $n_1$ and $n_2$, respectively, and minimum degree at least δ, and $a>0$. If $f:{\mathbb{Z}}^{+}\cap [\delta ,\infty )\to (0,a/2]$, then the following inequalities hold.

$\left(1\right)$ If $f\in A{P}_1\left(\delta \right)$, then

$$I_f(G_1\oplus G_2)\le \frac{1}{2}{\left(\frac{n_1{n}_{2}a}{2}{I}_f\left(G_1\right)I_f\left(G_2\right)\right)}^{1/3}.$$

$\left(2\right)$ If $f\in A{P}_2\left(\delta \right)$, then

$$I_f(G_1\oplus G_2)\ge \frac{1}{n_1{n}_{2}a}{I}_f\left(G_1\right)I_f\left(G_2\right).$$

Proof. 

Consider the function $g=f/a$. Therefore, $g:{\mathbb{Z}}^{+}\cap [\delta ,\infty )\to (0,1/2]$ and Proposition 1 implies that $P_g(G_1,x)$ and $P_g(G_2,x)$ are convex functions on the open interval $(0,\infty )$ and continuous on the closed interval $[0,\infty )$; hence, they are convex when $0\le x\le 1$.

If $f\in A{P}_1\left(\delta \right)$, then the function $f/a$ belongs to $A{P}_1\left(\delta \right)$ and Theorem 17 implies

$$P_{f/a}(G_1\oplus G_2,x)\le x^{n_1+n_2-1}{P}_{f/a}(G_1,x^{n_2})P_{f/a}(G_2,x^{n_1}).$$

Notice that $f/a\le 1/2$ implies $a/f-1\ge 1$, and thus, $P_{f/a}(G_i,0)=0$ and $P_{f/a}(G_i,1)=n_i$ for $i=1,2$. Since x is a convex function on the interval $[0,1]$, Lemma 2 implies

$$\begin{array}{cc}& \frac{1}{a}{I}_f(G_1\oplus G_2)=I_{f/a}(G_1\oplus G_2)={\int }_0^1{P}_{f/a}(G_1\oplus G_2,x)dx\hfill \\ & \le {\int }_0^1{x}^{n_1+n_2-1}{P}_{f/a}(G_1,x^{n_2})P_{f/a}(G_2,x^{n_1})dx\hfill \\ & \le \frac{1}{2}{\left({\int }_0^{1}xdx{\int }_0^1{x}^{n_2-1}{P}_{f/a}(G_1,x^{n_2})dx{\int }_0^1{x}^{n_1-1}{P}_{f/a}(G_2,x^{n_1})dx\right)}^{1/3}{\left(1·n_1·n_2\right)}^{2/3}\hfill \\ & =\frac{1}{2}{\left(\frac{1}{2}\frac{1}{n_2}{\int }_0^1{P}_{f/a}(G_1,t)dt\frac{1}{n_1}{\int }_0^1{P}_{f/a}(G_2,t)dt{n}_1^2{n}_2^2\right)}^{1/3}\hfill \\ & =\frac{1}{2}{\left(\frac{n_1{n}_2}{2a^2}{I}_f\left(G_1\right)I_f\left(G_2\right)\right)}^{1/3}.\hfill \end{array}$$

If $f\in A{P}_2\left(\delta \right)$, thus $f/a$ belongs to $A{P}_2\left(\delta \right)$ and Theorem 17 implies

$$P_{f/a}(G_1\oplus G_2,x)\ge x^{n_1+n_2-1}{P}_{f/a}(G_1,x^{n_2})P_{f/a}(G_2,x^{n_1}).$$

Thus, Lemma 1 gives

$$\begin{array}{cc}\hfill \frac{1}{a}{I}_f(G_1\oplus G_2)& =I_{f/a}(G_1\oplus G_2)={\int }_0^1{P}_{f/a}(G_1\oplus G_2,x)dx\hfill \\ & \ge {\int }_0^1{x}^{n_1+n_2-1}{P}_{f/a}(G_1,x^{n_2})P_{f/a}(G_2,x^{n_1}).dx\hfill \\ & \ge 2{\int }_0^{1}xdx{\int }_0^1{x}^{n_2-1}{P}_{f/a}(G_1,x^{n_2})dx{\int }_0^1{x}^{n_1-1}{P}_{f/a}(G_2,x^{n_1})dx\hfill \\ & =2\frac{1}{2}\frac{1}{n_{2}a}{I}_f\left(G_1\right)\frac{1}{n_{1}a}{I}_f\left(G_2\right)=\frac{1}{n_1{n}_2{a}^2}{I}_f\left(G_1\right)I_f\left(G_2\right).\hfill \end{array}$$

 □

Theorems 2, 19 (with $a=2$) and 18 have the following consequence for the general first Zagreb index.

Theorem 20.

Let $G_1$ and $G_2$ be two graphs of orders $n_1$ and $n_2$, respectively, and $\alpha \in \mathbb{R}$.

$\left(1\right)$ If $\alpha \le -1$, then

$$M_1^{\alpha }(G_1\oplus G_2)\le \frac{1}{2}{\left(n_1{n}_2{M}_1^{\alpha }\left(G_1\right)M_1^{\alpha }\left(G_2\right)\right)}^{1/3}.$$

$\left(2\right)$ If $\alpha \in [-1,0]$, then

$$M_1^{\alpha }(G_1\oplus G_2)\ge \frac{1}{2n_1{n}_2}{M}_1^{\alpha }\left(G_1\right)M_1^{\alpha }\left(G_2\right).$$

$\left(3\right)$ If $\alpha \ge 0$, then

$$M_1^{\alpha }(G_1\oplus G_2)\ge max\left\{n_2{M}_1^{\alpha }\left(G_1\right),n_1{M}_1^{\alpha }\left(G_2\right)\right\}.$$

Theorem 20 has the following consequence for the first Zagreb, forgotten and inverse degree indices.

Corollary 8.

If $G_1$ and $G_2$ are two graphs of orders $n_1$ and $n_2$, respectively, then

$$\begin{array}{cc}\hfill M_1(G_1\oplus G_2)& \ge max\left\{n_2{M}_1\left(G_1\right),n_1{M}_1\left(G_2\right)\right\}.\hfill \\ \hfill F(G_1\oplus G_2)& \ge max\left\{n_{2}F\left(G_1\right),n_{1}F\left(G_2\right)\right\}.\hfill \\ \hfill \frac{1}{2n_1{n}_2}ID\left(G_1\right)ID\left(G_2\right)\le ID(G_1\oplus G_2)& \le \frac{1}{2}{\left(n_1{n}_{2}ID\left(G_1\right)ID\left(G_2\right)\right)}^{1/3}.\hfill \end{array}$$

Since $f\left(t\right)=t\sqrt{logt}\in A{P}_3\left(2\right)$, Theorem 18 implies the following result for the $SL$ index.

Theorem 21.

If $G_1$ and $G_2$ are graphs without pendant vertices and of order $n_1$ and $n_2$, respectively, then

$$SL(G_1\oplus G_2)\ge max\left\{n_{2}SL\left(G_1\right),n_{1}SL\left(G_2\right)\right\}.$$

We obtain the following result by using the previous ideas.

Lemma 4.

Let G be a graph and Γ a subgraph of G, $f:{\mathbb{Z}}^{+}\to {\mathbb{R}}^{+}$ an increasing function, and $x\in (0,1]$. Then

$$P_f(\Gamma ,x)\le P_f(G,x),I_f(\Gamma )\le I_f\left(G\right).$$

If f is a decreasing function and $V(\Gamma )=V\left(G\right)$, then we obtain the converse inequalities.

Lemma 4 has the following consequence, relating the polynomials and indices of Cartesian products, lexicographic products and Cartesian sums.

Proposition 7.

Let $G_1$ and $G_2$ be two graphs, $f:{\mathbb{Z}}^{+}\to {\mathbb{R}}^{+}$ an increasing function, and $x\in (0,1]$. Then

$$\begin{array}{cc}\hfill P_f(G_1\times G_2,x)& \le P_f(G_1\odot G_2,x)\le P_f(G_1\oplus G_2,x),\hfill \\ \hfill I_f(G_1\times G_2)& \le I_f(G_1\odot G_2)\le I_f(G_1\oplus G_2).\hfill \end{array}$$

If f is a decreasing function, then we obtain the converse inequalities.

6. Conclusions

There are several graph products which play an important role in graph theory. Three of these products are Cartesian product, lexicographic product and Cartesian sum. Many topological indices can be written as $I_f\left(G\right)={\sum }_{u\in V\left(G\right)}f\left(d_u\right),$ for an appropriate choice of the function f (e.g., first Zagreb, inverse degree, forgotten, general first Zagreb and sum lordeg indices). By using the f-polynomial $P_f(G,x)$ introduced in [43], we obtain in this paper several inequalities of every topological index which can be written as $I_f$ for a function f in these classical graph products, from the information on topological indices of their factors, which are much easier to calculate than the products. These results are interesting from the theoretical viewpoint, and also from the point of view of applications since many chemical compounds can be represented by graph products (see the introduction). Our approach is to obtain information about the corresponding f-polynomials, which are easy to calculate (as in Theorems 1, 11 and 17); thus, we can deduce information on the $I_f$ index by using the formula ${\int }_0^1{P}_f(G,x)dx=I_f\left(G\right)$ (as in Theorems 4, 13 and 18). This is a good approach since the bounds of the f-polynomial of a product of two graphs allow the use of analytic tools to bound the $I_f$ index of such a product, simplifying the proofs.

In [43] appear similar results for corona product and join. Consequently, two natural open questions are to study this problem for strong product and tensor product of graphs.