On extremal results of multiplicative Zagreb indices of trees with given distance $k$-domination number

TL;DR

This paper establishes extremal bounds for the first and second multiplicative Zagreb indices of trees with a specified distance k-domination number, and characterizes the trees that attain these bounds.

Contribution

It provides sharp bounds for Zagreb indices of trees with a given distance k-domination number and characterizes the extremal trees.

Findings

Sharp lower bound for $_1$ of trees with given distance k-domination number.

Sharp upper bound for $_2$ of trees with given distance k-domination number.

Characterization of trees attaining the bounds.

Abstract

The first multiplicative Zagreb index $\Pi_1$ of a graph $G$ is the product of the square of every vertex degree, while the second multiplicative Zagreb index $\Pi_2$ is the product of the products of degrees of pairs of adjacent vertices. In this paper, we give sharp lower bound for $\Pi_1$ and upper bound for $\Pi_2$ of trees with given distance $k$-domination number, and characterize those trees attaining the bounds.

Equations

$$M_1(G)=\sum _{u\in V(G)}d(u{)}^2\text{mbox}and{M}_2(G)=\sum _{uv\in E(G)}d(u)d(v).$$

$${\Pi }_1(G)=\prod _{u\in V(G)}d(u{)}^2\text{mbox}and{\Pi }_2(G)=\prod _{uv\in E(G)}d(u)d(v).$$

$${\Pi }_1(S_n)<{\Pi }_1(T)<{\Pi }_1(P_n)\text{mbox}and{\Pi }_2(P_n)<{\Pi }_2(T)<{\Pi }_2(S_n).$$

$${\Pi }_1(T_{uv})<{\Pi }_1(T)\text{mbox}and{\Pi }_2(T_{uv})>{\Pi }_2(T).$$

$$\max \{{\Pi }_1(G^{\prime }),{\Pi }_1(G^{\prime \prime })\}<{\Pi }_1(G)$$

$$\min \{{\Pi }_1(G^{\prime }),{\Pi }_1(G^{\prime \prime })\}>{\Pi }_2(G).$$

$$f(T)=\prod _{w\in V(T)}(d_T(w)+1).$$

$$f(T-u)\ge 2^{n-2}(n-1)$$

$$d_T(v)\prod _{w\in V(T)\setminus \{u,v\}}(d_T(w)+1)\ge 2^{n-2}(n-1)$$

$$f(T)$$

$$h(T)=\prod _{w\in V(T)}(d_T(w)+1{)}^{d_T(w)+1}.$$

$$h(T-u)\le 4^{n-2}(n-1{)}^{n-1}$$

$$(d_T(v){)}^{d_T(v)}\prod _{w\in V(T)\setminus \{u,v\}}(d_T(w)+1{)}^{d_T(w)+1}\le 4^{n-2}(n-1{)}^{n-1}$$

$$h(T)$$

$$B_T=\{x\in V(T)|d_T(w)=1\text{mbox}and{\gamma }_k(T-w)={\gamma }_k(T)\}.$$

$$\max \{{\Pi }_1(T^{\prime }),{\Pi }_1(T^{\prime \prime })\}<{\Pi }_1(T)$$

$$\min \{{\Pi }_1(T^{\prime }),{\Pi }_1(T^{\prime \prime })\}>{\Pi }_2(T).$$

$${\Pi }_1(T_{n,k,s})=4^{ks-1}(n-ks{)}^2\text{mbox}and{\Pi }_2(T_{n,k,s})=4^{ks-1}(n-ks{)}^{n-ks}.$$

$${\Pi }_1(T)\ge 4^{2k-1}(n-2k{)}^2\text{mbox}and{\Pi }_2(T)\le 4^{2k-1}(n-2k{)}^{n-2k}.$$

$${\Pi }_1(T)\ge 4^{k{\gamma }_k-1}({\gamma }_k{)}^2\text{mbox}and{\Pi }_2(T)\le 4^{k{\gamma }_k-1}({\gamma }_k{)}^{{\gamma }_k}.$$

$${\Pi }_1(T)$$

$${\Pi }_1(T)\ge 4^{k{\gamma }_k-1}({\gamma }_k{)}^2$$

$${\Pi }_2(T)$$

$${\Pi }_2(T)\le 4^{k{\gamma }_k-1}({\gamma }_k{)}^{{\gamma }_k}$$

$${\Pi }_1(T)\ge 4^{3k-1}(n-3k{)}^2\text{mbox}and{\Pi }_2(T)\le 4^{3k-1}(n-3k{)}^{n-3k}.$$

$$\frac{d_T(z)}{d_T(z)-1}\ge \frac{n-k{\gamma }_k}{n-1-k{\gamma }_k}$$

$${\Pi }_1(T)={\Pi }_1(T-w)\cdot \frac{d_T(z{)}^2}{(d_T(z)-1{)}^2}.$$

$${\Pi }_1(T)$$

$$g(t)=\frac{x^x}{(x-1{)}^{x-1}}.$$

$${\Pi }_2(T)={\Pi }_2(T-w)\cdot \frac{(d_T(z){)}^{d_T(z)}}{(d_T(z)-1{)}^{d_T(z)-1}}={\Pi }_2(T-x)g(d_T(z)),$$

Full text

Key Words: first multiplicative Zagreb index, second multiplicative Zagreb index, trees, distance $k$-domination number.

Mathematics Subject Classification (2010): 05C05, 05C35, 05C69

1 Introduction

In this article we consider only simple, undirected and connected graphs. Let $G$ be a graph with vertex set $V(G)$ and edge set $E(G)$. The degree of $v\in V(G)$, denoted by $d_{G}(v)$, is the number of vertices in $G$ adjacent to $v$, and the neighborhood of $v$ is the set $N_{G}(v)=\{w\in V(G):vw\in E(G)\}$. Evidently, $|N_{G}(v)|=d_{G}(v)$. A vertex with degree one is called pendent vertex. The distance between any two vertices $u$ and $v$ of a graph $G$ is denoted by $d_{G}(u,v)$. The maximum distance from a vertex $v\in V(G)$ to all other vertices of $G$ is called eccentricity of $v$ in $G$. The diameter of a graph $G$ is the maximum eccentricity of all vertices of $G$.

A graph $G$ that has $n$ vertices and $n-1$ edges is called a tree. As usual, by $P_{n}$ and $S_{n}$ we denote the path and the star on $n$ vertices, respectively.

The first and the second Zagreb indices are among the oldest topological molecular descriptors, see [5]. They are defined as follows:

[TABLE]

Many interesting properties of them may be found in [2, 6, 7, 15, 20, 21].

In 2010, Todeschini et al. [4, 13] put forward the multiplicative Zagreb indices as follows:

[TABLE]

It is easily seen that $\Pi_{2}(G)=\prod\limits_{uv\in E(G)}d(u){d(v)}=\prod\limits_{u\in V(G)}d(u)^{d(u)}$. Some properties for the multiplicative Zagreb indices have been established, see [3, 8, 9, 10, 16, 17].

For a positive integer $k$, a set $D\subseteq V(G)$ is said to be distance $k$-dominating set of $G$ if for every vertex $u\in V(G)\setminus D$, $d_{G}(u,v)\leq k$ for some vertex $v\in D$. The minimum cardinality among all distance $k$-dominating set of $G$ is called the distance $k$-domination number of $G$, denoted by $\gamma_{k}(G)$. A distance $1$-dominating set of $G$ is known as a dominating set of $G$ and the distance $1$-domination number of $G$ is just the classical domination number of $G$.

Borovicanin and Furtula [1] presented sharp lower and upper bounds on Zagreb indices of trees in terms of domination number, and Wang et al. [18] found sharp lower and upper bounds on multiplicative Zagreb indices of trees in terms of domination number. Recently, Pei and Pan [12] investigated the connection between the Zagreb indices and the distance $k$-domination number of trees.

Motivated by the above results, in this paper, we study the multiplicative Zagreb indices of trees in terms of distance $k$-domination number. We provide sharp lower bound for $\Pi_{1}$ and upper bound for $\Pi_{2}$ in terms of distance $k$-domination number of a tree, and characterize those trees for which the bounds are attained.

2 Preliminaries

In this section, we present some propositions, definitions and lemmas which are helpful in our main results.

Lemma 2.1**.**

[3]** Let $T$ be a tree of order $n>5$ such that $T\ncong P_{n},S_{n}$. Then

[TABLE]

Let $T$ be a tree and $uv\in E(T)$ a non-pendent edge of $T$. Assume that $T-uv=T_{1}\cup T_{2}$ with vertex $u\in V(T_{1})$ and $v\in V(T_{2})$. Let $T_{uv}$ be the tree obtained by identifying the vertex $u$ of $T_{1}$ and the vertex $v$ of $T_{2}$ and attaching a pendent vertex $w$ to this vertex.

Lemma 2.2**.**

[19]** Let $T$ be a tree with a non-pendent edge $uv$. Then

[TABLE]

Lemma 2.3**.**

[19]** Let $u$ and $v$ be two distinct vertices in a graph $G$. Let $u_{1},\dots,u_{r}$ be pendent neighbors of $u$ and $v_{1},\dots,v_{t}$ pendent neighbors of $v$. Define $G^{\prime}=G-\{vv_{1},\dots,vv_{t}\}+\{uv_{1},\dots,uv_{t}\}$ and $G^{\prime\prime}=G-\{uu_{1},\dots,uu_{r}\}+\{vu_{1},\dots,vu_{r}\}$. Then

[TABLE]

and

[TABLE]

Lemma 2.4**.**

[11]** Let $G$ be a connected graph of order $n$ with $n\geq k+1$. Then $\gamma_{k}(G)\leq\lfloor\frac{n}{k+1}\rfloor$.

Lemma 2.5**.**

[12]** Let $T$ be a tree of order $n$ with maximum degree $\Delta$ and distance $k$-domination number $\gamma_{k}\geq 2$. Then $\Delta\leq n-k\gamma_{k}$.

Lemma 2.6**.**

[14*]** Let $T$ be a tree on $(k+1)n$ vertices. Then $\gamma_{k}(T)=n$ if and only if one of the following conditions holds:

$(1)$ $T$ is any tree on $k+1$ vertices;

$(2)$ $T=R\circ k$ for some tree $R$ on $n\geq 1$ vertices, where $R\circ k$ is the graph obtained by taking one copy of $R$ and $|V(R)|$ copies of the path $P_{k-1}$ of length $k-1$ and then joining the $ith$ vertex of $R$ to exactly one end vertex in the $ith$ copy of $P_{k-1}$.*

Lemma 2.7**.**

Let $T$ be a tree of order $n$. Let

[TABLE]

Then $f(T)\geq 2^{n-1}n$ with equality if and only if $T\cong S_{n}$.

Proof.

If $n=1$, it is obviously. Suppose that $n\geq 2$ and that the result is true for a tree of order $n-1$. Let $u$ be a pendent vertex of $T$, being adjacent to vertex $v$. By induction assumption,

[TABLE]

i.e.,

[TABLE]

with equality if and only if $T-u\cong S_{n-1}$. Now we have

[TABLE]

with equalities if and only if $d_{T}(v)=n-1$ and $T-u\cong S_{n-1}$, i.e., $T\cong S_{n}$. ∎

Lemma 2.8**.**

Let $T$ be a tree of order $n$. Let

[TABLE]

Then $h(T)\leq 4^{n-1}n^{n}$ with equality if and only if $T\cong S_{n}$.

Proof.

If $n=1$, it is obviously. Suppose that $n\geq 2$ and that the result is true for a tree of order $n-1$. Let $u$ be a pendent vertex of $T$, being adjacent to vertex $v$. By induction assumption,

[TABLE]

i.e.,

[TABLE]

with equality if and only if $T-u\cong S_{n-1}$. Now we have

[TABLE]

with equalities if and only if $d_{T}(v)=n-1$ and $T-u\cong S_{n-1}$, i.e., $T\cong S_{n}$. ∎

For a graph $G$ with $S\subset V(G)$, let $N_{G}(S)=\cup_{v\in S}N_{G}(v)$.

Lemma 2.9**.**

Let T be a tree with minimum value of first multiplicative Zagreb index or maximum value of second multiplicative zagreb index among all $n$-vertex trees with distance $k$-domination number $\gamma_{k}$. Let

[TABLE]

If $B_{T}\neq\emptyset$, then $|N_{T}(B_{T})|=1$.

Proof.

Suppose that $|N_{T}(B_{T})|\geq 2$, say $u$, $v\in N_{T}(B_{T})$. If $u^{\prime}\notin D$ for some pendent neighbor $u^{\prime}$ of $u$, then $D\setminus\{u^{\prime}\}\cup\{u\}$ is a distance $k$-dominating set of $T$. So we may assume that no pendent neighbor of $u$ and $v$ is in $D$. Define $T^{\prime}=T-vv^{\prime}+uv^{\prime}$ and $T^{\prime\prime}=T-uu^{\prime}+vu^{\prime}$, where $u^{\prime}$ ($v^{\prime}$, respectively) is a pendent neighbor of $u$ ($v$, respectively). Then $\gamma_{k}(T)=\gamma_{k}(T^{\prime})=\gamma_{k}(T^{\prime\prime})$. By Lemma 2.3,

[TABLE]

and

[TABLE]

a contradiction. Hence $|N_{T}(B_{T})|=1$. ∎

3 Main results

In this section, we present sharp lower bounds of first multiplicative Zagreb index and upper bounds for second multiplicative zagreb index of a tree of order $n$ with distance $k$-domination number $\gamma_{k}$.

A tree is starlike if it contains at most one vertex of degree at least three. Obviously, a starlike tree is either a path or a tree with exactly one vertex of degree at least three. In the latter case, it consists of pendent paths at common vertex.

Definition 3.1**.**

For positive integers $n$, $k$ and $s$ with $n\geq(k+1)s$, define $T_{n,k,s}$ to be a starlike tree with maximum degree $n-ks$, and if it is not a path, then it has one pendent path of length $k-1$, $s-1$ pendent paths of length $k$ and $n-(k+1)s$ paths of length $1$.

Note that

[TABLE]

As mentioned earlier, sharp lower bounds on first multiplicative Zagreb index and upper bounds on the second multiplicative Zagreb index of an $n$-vertex tree with distance $1$-domination number have been given in [18], so we only consider $k\geq 2$.

Definition 3.2**.**

If $P=v_{0}v_{1}\dots v_{d}$ is a diametric path of tree $T$ of order $n$, then denote by $T_{i}$ the component of $T-v_{i-1}v_{i}-v_{i}v_{i+1}$ containing $v_{i}$ for $i=1,\dots,d-1$.

Definition 3.3**.**

Denoted by $T^{a}_{n,k,2}$ the tree formed from the path $P_{2k+2}=v_{0}v_{1}\dots v_{2k+1}$ by joining $n-2(k+1)$ pendent vertices to $v_{a}$, where $a\in\{1,\dots,k\}$.

For a graph $G$, it is obvious that $\gamma_{k}(G)\leq\gamma_{1}(G)$ for $k\geq 2$. Note also that $\gamma_{1}(S_{n})=1$. Thus, by Lemma 2.1, we have the following result.

Theorem 3.1**.**

Let $T$ be an $n$-vertex tree and $\gamma_{k}(T)=1$. Then $\Pi_{1}(T)\geq n^{2}$ and $\Pi_{2}(T)\leq n^{n}$. Either equality holds if and only if $T\cong S_{n}$.

Theorem 3.2**.**

Let $T$ be a tree of order $n$ with $\gamma_{k}(T)=2$. Then

[TABLE]

Either equality holds if and only if $T\cong T^{a}_{n,k,2}$ with $a\in\{1,\dots,k\}$.

Proof.

Let $T$ be a tree of order $n$ with distance $k$-domination number $2$ that minimize the first multiplicative Zagreb index and maximize the second multiplicative Zagreb index respectively.

Let $P=v_{0}\dots v_{d}$ be a diametric path of $T$. If $d\leq 2k$, then $\{v_{\lfloor\frac{d}{2}\rfloor}\}$ is a distance $k$-dominating set of $T$, a contradiction. If $d\geq 2k+2$, for $i\in\{1,\dots,d-2\}$, $T_{v_{i}v_{i+1}}$ is a tree of order $n$ with distance $k$-domination number $2$, by Lemma 2.2, we have $\Pi_{1}(T_{v_{i}v_{i+1}})<\Pi_{1}(T)$, and $\Pi_{2}(T_{v_{i}v_{i+1}})>\Pi_{2}(T)$, also a contradiction. Hence $d=2k+1$.

If $T_{i}$ is not a star for some $i\in\{1,\dots,2k\}$, then as above, $T_{v_{i}v_{i+1}}$ is a tree of order $n$ with distance $k$-domination number $2$ such that $\Pi_{1}(T_{v_{i}v_{i+1}})<\Pi_{1}(T)$, and $\Pi_{2}(T_{v_{i}v_{i+1}})>\Pi_{2}(T)$, a contradiction. Thus each $T_{i}$ for $i\in\{1,\dots,d-1\}$ is a star with center $v_{i}$. Now by Lemma 2.3, for some $a\in\{1,\dots,k\}$, $T\cong T^{a}_{n,k,2}$.

By direct calculation , $\Pi_{1}(T)=\Pi_{1}(T^{a}_{n,k,2})=4^{2k-1}(n-2k)^{2}$ and $\Pi_{2}(T)=\Pi_{2}(T^{a}_{n,k,2})=4^{2k-1}(n-2k)^{n-2k}$ for $a\in\{1,\dots,k\}$. ∎

Lemma 3.1**.**

Let $T$ be a tree of order $n$ with distance $k$-domination number $\gamma_{k}\geq 3$. If $n=(k+1)\gamma_{k}$, then

[TABLE]

Either equality holds if and only if $T\cong T_{n,k,\gamma_{k}}$.

Proof.

By Lemma 2.6, we have $T=R\circ k$ for some tree $R$ on $\gamma_{k}$ vertices. For $w\in V(R)$, $d_{R}(w)=d_{T}(w)-1$. Thus

[TABLE]

where $f(R)=\prod\limits_{w\in V(R)}(d_{R}(w)+1)$. By Lemma 2.7, $f^{2}(R)\geq\left(2^{\gamma_{k}-1}\cdot\gamma_{k}\right)^{2}$ with equality if and only if $R\cong S_{\gamma_{k}}$. Therefore

[TABLE]

with equality if and only if $T=R\circ k$ with $R\cong S_{\gamma_{k}}$, i.e., $T\cong T_{n,k,\gamma_{k}}$.

Similarly,

[TABLE]

where $h(T)=\prod\limits_{w\in V(R)}(d_{R}(w)+1)^{d_{R}(w)+1}$. By Lemma 2.8, $h(T)\leq 4^{\gamma_{k}-1}(\gamma_{k})^{\gamma_{k}}$ with equality if and only if $R\cong S_{\gamma_{k}}$. Therefore

[TABLE]

with equality if and only if $T=R\circ k$ with $R\cong S_{\gamma_{k}}$, i.e., $T\cong T_{n,k,\gamma_{k}}$.

∎

Lemma 3.2**.**

Let $T$ be a tree of order $n$ with $\gamma_{k}(T)=3$, then

[TABLE]

Either equality holds if and only if $T\cong T_{n,k,3}$.

Proof.

By Lemmas 2.4 and 3.1, we have $n\geq(k+1)\gamma_{k}$, and the result holds for $n=(k+1)\gamma_{k}$. We present our proof by induction on $n$. Suppose that $n>3(k+1)$ and the result is true for $n-1$.

Let $P=v_{0}\dots v_{d}$ be a diametric path and $D$ be a minimum distance $k$-dominating set of $T$. We claim that $d\geq 2k+2$, for otherwise, $\{v_{k},v_{k+1}\}$ is a distance $k$-dominating set, a contradiction. We may choose distance $k$-dominating set $D$ of cardinality $\gamma_{k}(G)$ with $\{v_{k},v_{d-k}\}\subseteq D$ such that $(\cup_{a=0}^{k}V(T_{a})\setminus\{v_{k}\})\cap D=\emptyset$ and $(\cup_{a=d-k}^{d}V(T_{a})\setminus\{v_{d-k}\})\cap D=\emptyset$.

Let $v_{0}=w_{1},v_{d}=w_{2},\dots,w_{m}$ be all the pendent vertices of $T$ and $B_{T}=\{w_{i}|1\leq i\leq m,\gamma_{k}(T-w_{i})=\gamma_{k}(T)\}$.

We claim that $|B_{T}|\geq 1$.

Suppose that $B_{T}=\emptyset$. Then $\gamma_{k}(T-w_{i})=\gamma_{k}(T)-1$ for $1\leq i\leq m$.

For some $i\in\{1,\dots,k,d-k,\dots,d-1\}$, if $d_{T}(v_{i})\geq 3$, then $V(T_{i})\cap\{w_{3},\dots,w_{m}\}\neq\emptyset$. As $\{v_{k},v_{d-k}\}\in D$, we have $\gamma_{k}(T-z)=\gamma_{k}(T)$ for $z\in V(T_{i})\cap\{w_{3},\dots,w_{m}\}$, a contradiction. It follows that $d_{T}(v_{i})=2$ for $i\in\{1,\dots,k,d-k,\dots,d-1\}$.

As $d_{T}(v_{1})=2$, we have $\gamma_{k}(T-v_{0})=\gamma_{k}(T)-1$. Note that $d_{T}(v_{1},v_{k+1})=k$ and $(\cup_{a=0}^{k}V(T_{a})\setminus\{v_{k}\})\cap D=\emptyset$. Thus $v_{k+1}\in D$. Similarly, $v_{d-k-1}\in D$. If $d>2k+2$, then $v_{k},v_{k+1},v_{d-k-1},v_{d-k}$ are all distinct, a contradiction to the fact that $\gamma_{k}(T)=3$. Hence $d=2k+2$ and $D=\{v_{k},v_{k+1},v_{d-k}\}$.

If $d_{T}(v_{k+1})=2$, then $T\cong P_{2k+3}$ and $\{v_{k},v_{d-k}\}$ is a distance $k$-dominating set, a contradiction. Then $d_{T}(v_{k+1})\geq 3$ and thus $m\geq 3$. If $m>3$, then $\gamma_{k}(T-w_{i})=\gamma_{k}(T)$ for some $i\in\{1,\dots,m\}$, which is impossible. Hence $m=3$. Thus $T_{k+1}$ is a path with end vertices $v_{k+1}$ and $w_{3}$. By the definition of distance $k$-domination number, we have $d(v_{k+1},w_{3})=k$. It follows that $|V(T)|=3(k+1)$, which is contradiction. This proves our Claim.

Now by our claim and Lemma 2.9, we have $|N_{T}(B_{T})|=1$.

Let $w$ be a pendent vertex such that $\gamma_{k}(T-w)=\gamma_{k}(T)$ and $z$ being a unique vertex adjacent to $x$. Then by lemma 2.5, we have $d_{T}(z)\leq n-k\gamma_{k}$.

[TABLE]

with equality if and only if $d_{T}(z)=n-k\gamma_{k}$. Note that

[TABLE]

By induction hypothesis, we have

[TABLE]

with equalities if and only if $T-w\cong T_{n-1,k,\gamma_{k}}$ and $d_{T}(z)=\Delta=n-k\gamma_{k}$, i.e., $T\cong T_{n,k,\gamma_{k}}$.

Let

[TABLE]

Obviously, $g^{\prime}(x)=g(x)\log\frac{x}{x-1}>0$ for $x>1$. Thus $g(x)$ is strictly increasing for $x>1$, implying that $g(d_{T}(z))\leq g(n-k\gamma_{k})$ with equality if and only if $d_{T}(z)=n-k\gamma_{k}$. Similarly as above,

[TABLE]

and by induction hypothesis, we have

[TABLE]

with equalities holds if and only if $T-w\cong T_{n-1,k,\gamma_{k}}$ and $d_{T}(z)=\Delta=n-k\gamma_{k}$, i.e., $T\cong T_{n,k,\gamma_{k}}$. ∎

Let $D$ be a distance $k$-dominating set of a graph $G$. Let $N^{a}_{G}(v)$ be the set of vertices with distance $a$ from $v$. A vertex $v\in V(G)$ is called a private $k$-neighbor of $u$ with respect to $D$ if $\cup_{a=0}^{k}N^{a}_{G}(v)\cap D=\{u\}$, that is $d_{G}(v,u)\leq k$ and $d_{G}(v,x)\geq k+1$, for any vertex $x\in D\setminus\{u\}$.

Theorem 3.3**.**

Let $T$ be a tree of order $n$ with distance $k$-domination number $\gamma_{k}\geq 3$. Then

[TABLE]

Either equality holds if and only if $T\cong T_{n,k,\gamma_{k}}$.

Proof.

Let $P=v_{0}\dots v_{d}$ be a diametric path of $T$. Define $B_{T}=\{w\in V(T)|d_{T}(w)=1\mbox{ and }\gamma_{k}(T-w)=\gamma_{k}(T)\}$. We may choose distance $k$-dominating set $D$ of cardinality $\gamma_{k}$ with $\{v_{k},v_{d-k}\}\subseteq D$ such that $(\cup_{a=0}^{k}V(T_{a})\setminus\{v_{k}\})\cap D=\emptyset$ and $(\cup_{a=d-k}^{d}V(T_{a})\setminus\{v_{d-k}\})\cap D=\emptyset$.

If $B_{T}=\emptyset$ then for $i=0,d,\gamma_{k}(T-v_{i})=\gamma_{k}(T)-1$. If $B_{T}\neq\emptyset$, then by Lemma 2.9, $|N_{T}(B_{T})|=1$. If $v_{0},v_{d}\in B_{T}$, then since $d-1>1$, we have $\{v_{1},v_{d-1}\}\subseteq|N_{T}(B_{T})|$ and thus $|N_{T}(B_{T})|>1$, a contradiction. Thus, in either case, we may assume that $\gamma_{k}(T-v_{0})=\gamma_{k}(T)-1$, and thus $\{v_{k},v_{k+1},v_{d-k}\}\subseteq D$.

For $i=1,\dots,k$, if some $T_{i}$ is not a star with center $v_{i}$, then applying Lemma 2.2 for a non-pendent edge $e$ of $T_{i}$ to obtain a tree $T_{e}$, we have $\Pi_{1}(T_{e})<\Pi_{1}(T)$ and $\Pi_{2}(T_{e})>\Pi_{2}(T)$. Thus, we may assume that $T_{i}$ is a star with center $v_{i}$ for all $i=1,\dots,k$.

If there are at least two vertices, say $v_{i}$ and $v_{j}$ with $1\leq i<j\leq k$, with degree at least $3$ in $T$, then by Lemma 2.3 we may find a tree $T^{\prime}$ by moving the pendent edges at $v_{i}$ to $v_{j}$ or via such that $\Pi_{1}(T^{\prime})<\Pi_{1}(T)$ and $\Pi_{2}(T^{\prime})>\Pi_{2}(T)$. So we may assume that there is at most one vertex among vertices $v_{1},\dots,v_{k}$ with degree at least $3$. That is, among vertices $v_{1},\dots,v_{k}$, either each vertex has degree $2$ or exactly one vertex, say $v_{i_{0}}$ has at least one pendent neighbor, where $1\leq i_{0}\leq k$. Let

[TABLE]

Let $s$ is number of pendent edges at $v_{i_{0}}$. If $s=0$, then $T=T^{\prime\prime}$.

Suppose that $s\geq 1$. Then for $i=1,\dots,k,d_{T^{\prime\prime}}(v_{i})=2$ and $D$ is minimum distance $k$-dominating set of $T^{\prime\prime}$. Let $PN_{k,D}(z)$ be the set of all private $k$-neighbors of $z$ with respect to $D$ in $T^{\prime\prime}$. Then for any $z\in\bigcup_{a=0}^{k}N^{a}_{T^{\prime\prime}}(v_{k})\setminus\{v_{0},\dots,v_{k}\}$, $d_{T^{\prime\prime}}(z,v_{k+1})\leq k$. It follows that $D\setminus\{v_{k}\}$ is a distance $k$-dominating set of the tree $T^{\prime\prime}-\{v_{0},\dots,v_{k}\}$. Also, $PN_{k,D}(v_{k+1})\subseteq V(T^{\prime\prime})\setminus\{v_{0},\dots,v_{k}\}$. It shows that $D\setminus\{v_{k}\}$ is a minimum distance $k$-dominating set of $T^{\prime\prime}-\{v_{0},\dots,v_{k}\}$. Thus $\gamma_{k}(T^{\prime\prime}-\{v_{0},\dots,v_{k}\})=\gamma_{k}-1$ and $\gamma_{k}(T^{\prime\prime}-\{v_{0},\dots,v_{k-1}\})=\gamma_{k}-1$.

Let $\Gamma=\prod_{u\in V(T)\setminus\{v_{i_{0}},v_{k+1}\}}d^{2}_{T}(u)$ and $\Phi=\prod_{u\in V(T)\setminus\{v_{i_{0}},v_{k+1}\}}(d_{T}(u))^{d_{T}(u)}$. Then

[TABLE]

and thus $\Pi_{1}(T)\geq\Pi_{1}(T^{\prime\prime})$ with equality if and only if $d_{T}(v_{k+1})=2$. Also

[TABLE]

where $F(s)=4\left(d_{T}(v_{k+1})+s\right)^{d_{T}(v_{k+1})+s}-(s+2)^{s+2}(d_{T}(v_{k+1}))^{d_{T}(v_{k+1})}$. It is easy to check that $F(s)$ is an increasing function for $s\geq 0$. Thus $\Pi_{2}(T^{\prime\prime})\geq\Pi_{2}(T)$ with equality if and only if $d_{T}(v_{k+1})=2$.

Now we have shown that $\Pi_{1}(T)\geq\Pi_{1}(T^{\prime\prime})$ and $\Pi_{2}(T^{\prime\prime})\geq\Pi_{2}(T)$ with either equality if and only if $s=0$ (i.e., $T=T^{\prime\prime}$) or $d_{T}(v_{k+1})=2$.

In the following, we prove that, $\Pi_{1}(T^{\prime\prime})\geq 4^{k\gamma_{k}-1}(n-k\gamma_{k})^{2}$ and $\Pi_{2}(T^{\prime\prime})\leq 4^{k\gamma_{k}-1}(n-k\gamma_{k})^{n-k\gamma_{k}}$ with either equality if and only if $T^{\prime\prime}\cong T_{n,k,\gamma_{k}}$.

By Lemma 3.2, the result holds for $n\geq(k+1)\gamma_{k}$ and $\gamma_{k}=3$. Suppose that $\gamma_{k}\geq 4$, and the result is true for $n\geq(k+1)(\gamma_{k}-1)$.

Note that $\gamma_{k}(T^{\prime\prime}-\{v_{0},\dots,v_{k}\})=\gamma_{k}-1$ and $|V(T^{\prime\prime}-\{v_{0},\dots,v_{k}\})|=n-k-1>(k+1)(\gamma_{k}-1)$. Then

[TABLE]

with equalities if and only if $T^{\prime\prime}-\{v_{0},\dots,v_{k}\}\cong T^{\prime\prime}_{n-k-1,k,\gamma_{k}-1}$, and $d_{T^{\prime\prime}}(v_{k+1})=\Delta=n-k\gamma_{k}$. Recall that for $i=1,\dots,k$, $d_{T^{\prime\prime}}(v_{i})=2$. Thus $\Pi_{1}(T^{\prime\prime})\geq 4^{k\gamma_{k}-1}(n-k\gamma_{k})^{2}$ with equality if and only if $T^{\prime\prime}\cong T_{n,k,\gamma_{k}}$.

Similarly, we have

[TABLE]

with equalities if and only if $T^{\prime\prime}-\{v_{0},\dots,v_{k}\}\cong T^{\prime\prime}_{n-k-1,k,\gamma_{k}-1}$, and $d_{T^{\prime\prime}}(v_{k+1})=\Delta=n-k\gamma_{k}$. Also for $i=1,\dots,k$, $d_{T^{\prime\prime}}(v_{i})=2$. Thus $\Pi_{2}(T^{\prime\prime})\leq 4^{k\gamma_{k}-1}(n-k\gamma_{k})^{n-k\gamma_{k}}$ with equality if and only if $T\cong T_{n,k,\gamma_{k}}$.

Now we conclude that $\Pi_{1}(T)\geq\Pi_{1}(T^{\prime\prime})\geq 4^{k\gamma_{k}-1}(n-k\gamma_{k})^{2}$ with equality in the first inequality if and only if $T=T^{\prime\prime}$ or $T\not\cong T^{\prime\prime}$ and $d_{T}(v_{k+1})=2$, and with equality in the second inequality if and only if $T^{\prime\prime}\cong T_{n,k,\gamma_{k}}$. We show that if $T\not\cong T^{\prime\prime}$ and $d_{T}(v_{k+1})=2$, then $T^{\prime\prime}\not\cong T_{n,k,\gamma_{k}}$. Otherwise, say $T^{\prime\prime}=T_{n,k,\gamma_{k}}$. As $d_{T^{\prime\prime}}(v_{k+1})=n-k\gamma_{k}$, there are $n-(k+1)\gamma_{k}$ pendent edges at $v_{k+1}$ in $T^{\prime\prime}$. By the above argument, $d_{T}(v_{k+1})=\gamma_{k}>2$, a contradiction. Therefore $\Pi_{1}(T)\geq 4^{k\gamma_{k}-1}(n-k\gamma_{k})^{2}$ with equality if and only if $T\cong T_{n,k,\gamma_{k}}$. Similarly, $\Pi_{2}(T)\leq 4^{k\gamma_{k}-1}(n-k\gamma_{k})^{n-k\gamma_{k}}$ with equality if and only if $T\cong T_{n,k,\gamma_{k}}$. ∎