Spectral extrema of graphs: Forbidden star-path forests

TL;DR

This paper determines the maximum spectral radius of large graphs that exclude specific star-path forest subgraphs, advancing spectral graph theory by identifying extremal structures under these constraints.

Contribution

It introduces new extremal spectral bounds for graphs forbidding certain star-path forest configurations, extending previous results in spectral extremal graph theory.

Findings

Identifies maximum spectral radius for graphs avoiding $kS_{\ell-1}\cup P_{\ell}$

Determines spectral bounds for graphs excluding $k_1S_{2\ell -1}\cup k_2P_{2\ell}$

Establishes extremal graphs for forbidden star-path forests

Abstract

A path of order $n$ is denoted by $P_n$, and a star of order $n$ is denoted by $S_{n-1}$. A star-path forest is a forest whose connected components are paths and stars. In this paper we determine the maximum spectral radius of graphs that contain no copy of $kS_{\ell-1}\cup P_{\ell}$, $k_1S_{2\ell -1}\cup k_2P_{2\ell}$ or $kS_{4}\cup 2P_{5}$ for $n$ appropriately large.

Equations

$$e(G)\leq{\rm{max}}\;\bigg{\{}\binom{k-1}{2}+(n-k+1),\;\binom{\lceil(k+1)/2\rceil}{2}+\bigg{\lfloor}\frac{k-1}{2}\bigg{\rfloor}\left(n-\bigg{\lceil}\frac{k+1}{2}\bigg{\rceil}\right)\bigg{\}}.$$

$${\rm{ex}}(n,\,S_{\ell})\leq\bigg{\lfloor}\frac{(\ell-1)n}{2}\bigg{\rfloor},$$

$${\rm{ex}}(n,\,kS_{\ell})=\left\{\begin{aligned} &\binom{n}{2},\quad\quad\quad\quad\quad\quad\;\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\;\;\;if\;n<k(\ell+1),\\ &\binom{k\ell+k-1}{2}+\binom{n-k\ell-k+1}{2},\quad\quad\;\quad\quad\quad\quad\quad\quad\;if\;k(\ell+1)\leq n\leq(k+1)\ell+k-1,\\ &\binom{k\ell+k-1}{2}+\bigg{\lfloor}\frac{(\ell-1)(n-k\ell-k+1)}{2}\bigg{\rfloor},\quad\quad\quad\quad\quad if\;(k+1)\ell+k\leq n<\frac{k\ell^{2}+2k\ell+2k-2}{2},\\ &\binom{k-1}{2}+(n-k+1)(k-1)+\bigg{\lfloor}\frac{(\ell-1)(n-k+1)}{2}\bigg{\rfloor},\;if\;n\geq\frac{k\ell^{2}+2k\ell+2k-2}{2}.\end{aligned}\right.$$

$$\mathrm{ex}\left(n,k{S}_{\ell -1}\cup P_{\ell }\right)=\left(k+\frac{\ell }{2}-1\right)n-\frac{k^2+(\ell -1)(k+r)-r^2}{2}.$$

$$\mathrm{Ex}(n,k{S}_{\ell -1}\cup P_{\ell })=\{\begin{array}{rl}& \{K_k\vee (d{K}_{\ell -1}\cup K_r),S_{n,k+\frac{\ell }{2}-1}\},\ell iseven,andr=\frac{\ell }{2}orr=\frac{\ell -2}{2},\\ & \{K_k\vee (d{K}_{\ell -1}\cup K_r)\},otherwise.\end{array}$$

$${\rm{ex}}(n,\,kS_{\ell-1}\cup P_{\ell})\leq\left(k+\bigg{\lfloor}\frac{\ell}{2}\bigg{\rfloor}-\frac{1}{2}\right)(n-1).$$

$$\mathrm{Ex}(n,k_1{S}_{2\ell -1}\cup k_2{P}_{2\ell })=\{S_{n,k_1+\ell k_2-1}\}.$$

$$\mathrm{Ex}(n,k{S}_4\cup 2P_5)=\{S_{n,k+3}^{+}\}.$$

$${\mathrm{Ex}}_{sp}(n,k_1{S}_{2\ell -1}\cup k_2{P}_{2\ell })=\{S_{n,k_1+\ell k_2-1}\}.$$

$${\mathrm{Ex}}_{sp}(n,k{S}_4\cup 2P_5)=\{S_{n,k+3}^{+}\}.$$

$${\mathrm{Ex}}_{sp}(n,k{P}_3)=\{F_{n,k}\}.$$

$$\rho (G)\le \frac{\delta -1+\sqrt{8e-4\delta n+(\delta +1{)}^2}}{2}.$$

$$\rho (G)\ge \frac{p-1+\sqrt{4pn-4p^2+c}}{2},$$

$$\rho (S_{n,p})=\frac{p-1+\sqrt{4pn-4p^2+(p-1{)}^2}}{2}.$$

$$\rho (G)\ge \rho (S_{n,p})=\frac{p-1+\sqrt{4pn-4p^2+(p-1{)}^2}}{2}.$$

$$\rho (G)\le \frac{\delta (G)-1+\sqrt{8e(G)-4\delta (G)n+(\delta (G)+1{)}^2}}{2}\le \frac{p-1+\sqrt{8e(G)-4pn+(p+1{)}^2}}{2}.$$

$$2e(H)=tr\left(A^2(H)\right)\ge {\rho }^2(H)>(2p+1)q>2e(S_{q,p}).$$

$$\rho (H)\le \frac{p-1+\sqrt{8e(H)-4pq+(p+1{)}^2}}{2}.$$

$$e(G)\leq\left(\bigg{\lfloor}\frac{\ell}{2}\bigg{\rfloor}-1\right)n.$$

$$e(G)\leq\bigg{\lfloor}\frac{n}{2}\bigg{\rfloor}\bigg{\lceil}\frac{n}{2}\bigg{\rceil}-1\leq\left(\bigg{\lfloor}\frac{\ell}{2}\bigg{\rfloor}-1\right)n.$$

$$e(G)\leq{\rm{max}}\left\{\binom{\ell-2}{2}+(n-\ell+2),\,\binom{\lceil\frac{\ell}{2}\rceil}{2}+\bigg{\lfloor}\frac{\ell-2}{2}\bigg{\rfloor}\left(n-\bigg{\lceil}\frac{\ell}{2}\bigg{\rceil}\right)\right\}\leq\left(\bigg{\lfloor}\frac{\ell}{2}\bigg{\rfloor}-1\right)n.$$

$$e(G)=e(G_{1}\cup\cdots\cup G_{s})\leq\sum_{i=1}^{s}\left(\bigg{\lfloor}\frac{\ell}{2}\bigg{\rfloor}-1\right)n_{i}\leq\left(\bigg{\lfloor}\frac{\ell}{2}\bigg{\rfloor}-1\right)n.$$

$$e(G)\leq\left(k+\bigg{\lfloor}\frac{\ell}{2}\bigg{\rfloor}-1\right)n.$$

$$\begin{split}e(H)&\geq h(n-h)>\left(k+\frac{\ell}{2}-2\right)n-\frac{(k-1)^{2}}{2}-\frac{(\ell-2)(k-1)}{2}\\ &\geq\binom{k-1}{2}+(n-k+1)(k-1)+\bigg{\lfloor}\frac{(\ell-2)(n-k+1)}{2}\bigg{\rfloor}\\ &={\rm{ex}}(n,\,kS_{\ell-1}).\end{split}$$

$$m_0=e(H)-e(H-S_{\ell -1})\ge h(n-h)-(h-1)(n-\ell )=n-h^2+(h-1)\ell .$$

$$d_{H[W_0]}(u)\ge \frac{n-h^2+(h-1)\ell }{\ell }-(k-1).$$

$$|W_1|\ge |W_0|-\ell \ge \frac{n-h^2+(h-1)\ell }{\ell }-(k-1)-\ell \ge (\ell -1)k.$$

$$\begin{split}e(H)&\leq e\left(H[\overline{U}]\right)+k(n-k)+\frac{k^{2}}{4}\\ &\leq\left(\bigg{\lfloor}\frac{\ell}{2}\bigg{\rfloor}-1\right)(n-k)+k(n-k)+\frac{k^{2}}{4}\\ &\leq\left(k+\bigg{\lfloor}\frac{\ell}{2}\bigg{\rfloor}-1\right)n.\end{split}$$

$$|S_1\cap \cdots \cap S_k|\ge \sum _{i=1}^k|S_i|-(k-1)|{\cup }_{i=1}^k{S}_i|.$$

$$e(A,B)=e(A,B\setminus A)+e(A,A\cap B)=e(A,B\setminus A)+2e(G[A\cap B])+e(A\setminus B,A\cap B).$$

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1 Introduction

In this paper, we consider undirected graphs without loops or multiedges. The order of a graph $G=\left(V(G),\,E(G)\right)$ is the number of its vertices, and the size of a graph $G$ is the number of its edges, denoted by $e(G)$. The adjacent matrix $A(G)=(a_{ij})$ of $G$ is a matrix, where $a_{ij}=1$ if $v_{i}$ is adjacent to $v_{j}$, and 0 otherwise. The spectral radius of $G$ is the largest eigenvalue of $A(G)$, denoted by $\rho(G)$. For a vertex $v\in V(G)$, the neighborhood of $v$ in $G$ is denoted by $N_{G}(v)=\{u\in V(G):uv\in E(G)\}$ or simply $N(v)$ and $N[v]=N(v)\cup\{v\}$. Denote $N_{G}^{2}(v)$ or simply $N^{2}(v)$ by the set of vertices at distance two from $v$ in $G$.

Given two vertex-disjoint graphs $G$ and $H$, the union of graphs $G$ and $H$ is the graph $G\cup H$ with vertex set $V(G)\cup V(H)$ and edge set $E(G)\cup E(H)$. In particular, $G=kH$ is the vertex-disjoint union of $k$ copies of $H$. The join of $G$ and $H$, denoted by $G\vee H$, is the graph obtained from $G\cup H$ by adding all edges between $V(G)$ and $V(H)$. For a graph $G$ and its subgraph $H$, $G-H$ denotes the subgraph induced by $V(G)\setminus V(H)$. For $U\subseteq V(G)$, let $G[U]$ be the subgraph of $G$ induced by $U$. A path of order $n$ is denoted by $P_{n}$, and a star of order $n$ is denoted by $S_{n-1}$. Let $S_{n,\,h}$ be the graph of order $n$ obtained by joining a clique of order $h$ with an independent set of order $n-h$, i.e., $S_{n,\,h}=K_{h}\vee\overline{K}_{n-h}$. Let $S^{+}_{n,\,h}$ be the graph obtained from $S_{n,\,h}$ by adding one edge, i.e., $S^{+}_{n,\,h}=K_{h}\vee\left(K_{2}\cup\overline{K}_{n-h-2}\right)$ .

A graph is $H$-free if it does not contain a copy of $H$ as a subgraph. The Turán number ex$(n,\,H)$ is the maximum number of edges in a graph of order $n$ which is $H$-free. We denote by Ex$(n,\,H)$ the set of $H$-free graphs of order $n$ with ex$(n,\,H)$ edges, and call a graph in Ex$(n,\,H)$ an extremal graph for $H$.

In 1956, Erdős and Gallai [9] studied Turán numbers of paths. Then in 2008, Balister and Győri [4] gave an improvement of the extremal graph theorem for $P_{k}$. In 2019, Lan et al. [18] determined the Turán number of a star. A linear forest is a forest whose connected components are paths. A star forest is a forest whose connected components are stars. A star-path forest is a forest whose connected components are paths and stars. In 2011, Bushaw and Kettle [3] determined the Turán numbers of $kP_{\ell}$ for sufficiently large $n$, which was extended by Lidiciký, Liu and Palmer [16]. Yuan and Zhang [25, 24] determined the Turán numbers of linear forests containing at most one odd path for all $n$. In 2022, Li, Yin and Li [17] determined the Turán numbers of star forests. Very recently, Fang and Yuan [10] determined the Turán numbers of $kS_{\ell-1}\cup P_{\ell}$, $k_{1}S_{2\ell-1}\cup k_{2}P_{2\ell}$ and $kS_{4}\cup 2P_{5}$ for $n$ appropriately large.

Theorem 1.1** (Balister and Győri [4]).**

Let $G$ be a connected graph of order $n$ containing no path of order $k+1$, $n>k\geq 3$. Then

[TABLE]

Theorem 1.2** (Lan, Li, Shi and Tu [18]).**

If $\ell\geq 3$ and $n\geq\ell+1$, then

[TABLE]

with one extremal graph is the $(\ell-1)$-regular graph of order $n$.

Theorem 1.3** (Li, Yin and Li [17]).**

*If $k\geq 2$ and $\ell\geq 3$, then *

[TABLE]

Theorem 1.4** (Fang and Yuan [10]).**

Suppose $n=k+d(\ell-1)+r\geq(\ell^{2}-\ell+1)k+\frac{\ell^{2}+3\ell-2}{2}$, where $k\geq 1$, $\ell\geq 4$ and $0\leq r<\ell-1$. Then

[TABLE]

Moreover,

[TABLE]

Based on the result of Theorem 1.4, we have

[TABLE]

Theorem 1.5** (Fang and Yuan [10]).**

Suppose $n\geq(4\ell^{2}-2\ell+1)k_{1}+(2\ell^{2}+3\ell-4)k_{2}+3$, where $k_{1}\geq 1$, $k_{2}\geq 2$ and $\ell\geq 2$. Then

[TABLE]

Theorem 1.6** (Fang and Yuan [10]).**

Suppose $n\geq 21k+38$ where $k\geq 1$. Then

[TABLE]

In 2010, Nikiforov [20] proposed the spectral counterpart of Turán type extremal problem.

Problem 1.1**.**

Given a graph $H$, what is the maximum $\rho(G)$ of a graph $G$ of order $n$ without $H$ as a subgraph?

This problem has been intensively investigated in the literature for many classes of graphs. Guiduli [12] and Nikiforov [21] independently studied the case $H=K_{r}$. In 2010, Nikiforov [20] studied the case $H$ is a path or cycle of specified length. Chen, Liu and Zhang in 2019 [5] studied the case $H$ is a linear forest, and in 2021 they [6] studied the case $H$ is a star forest. For other classes of graphs, the readers may be referred to [7, 13, 23, 26, 27]. Motivated by Problem 1.1, we will give the maximum value of the spectral radius and characterize corresponding extremal graphs for some kinds of star-path forests. Denote by Ex${}_{sp}(n,\,H)$ the set of $H$-free graphs of order $n$ with maximum spectral radius.

For any graph $F$, if its extremal graph is $S_{n,\,p}$ (resp. $S^{+}_{n,\,p}$), then we will prove that for appropriate $n$, its spectral extremal graph is also $S_{n,\,p}$ (resp. $S^{+}_{n,\,p}$).

Theorem 1.7**.**

For any graph $F$, suppose $n\geq m$ and Ex$(n,\,F)=\{S_{n,\,p}\}$. Then when $n\geq\,$max$\,\{2^{4p},\,m^{2}\}$, we have Ex${}_{sp}(n,\,F)=\{S_{n,\,p}\}$. 2. 2.

For any graph $F$, suppose $n\geq m$ and Ex$(n,\,F)=\{S^{+}_{n,\,p}\}$. Then when $n\geq{\rm{max}}\,\{2^{4p},\,m^{2}\}$, we have EX${}_{sp}(n,\,F)=\{S^{+}_{n,\,p}\}$.

The spectral extremal graphs for $k_{1}S_{2\ell-1}\cup k_{2}P_{2\ell}$ and $kS_{4}\cup 2P_{5}$ can be determined based on the results of Theorems 1.5 – 1.7.

Lemma 1.1**.**

Suppose $k_{1}\geq 1$, $k_{2}\geq 2$, $\ell\geq 2$ and $n\geq$ max $\{2^{4(k_{1}+\ell k_{2}-1)},\,[(4\ell^{2}-2\ell+1)k_{1}+(2\ell^{2}+3\ell-4)k_{2}+3]^{2}\}$. Then

[TABLE]

Lemma 1.2**.**

Suppose $k\geq 1$ and $n\geq$ max $\{2^{4(k+3)},\,(21k+38)^{2}\}$. Then

[TABLE]

Now we will consider spectral extremal graphs for $kS_{\ell-1}\cup P_{\ell}$. If $\ell=2$, then $kS_{\ell-1}\cup P_{\ell}=(k+1)P_{2}$ holds. In 2007, Feng, Yu and Zhang [11] gave the spectral extremal graph for $kP_{2}$.

Theorem 1.8** (Feng, Yu and Zhang [11]).**

Suppose $k\geq 1$ and $n\geq 2k$.

If $n=2k$ or $2k+1$, then ${\rm{Ex}}_{sp}(n,\,kP_{2})=\{K_{n}\}$ holds. 2. 2.

If $2k+2\leq n<3k+2$, then ${\rm{Ex}}_{sp}(n,\,kP_{2})=\{K_{2k+1}\cup\overline{K_{n-2k-1}}\}$ holds. 3. 3.

If $n=3k+2$, then ${\rm{Ex}}_{sp}(n,\,kP_{2})=\{K_{2k+1}\cup\overline{K_{n-2k-1}},\,K_{k}\vee\overline{K_{n-k}}\}$ holds. 4. 4.

If $n>3k+2$, then ${\rm{Ex}}_{sp}(n,\,kP_{2})=\{K_{k}\vee\overline{K_{n-k}}\}$ holds.

If $\ell=3$, then $kS_{\ell-1}\cup P_{\ell}=(k+1)P_{3}$ holds. In 2019, Chen, Liu and Zhang [5] gave the spectral extremal graph for $kP_{3}$. Set $F_{n,\,k}=K_{k-1}\vee(dK_{2}\cup K_{s})$, where $n-(k-1)=2d+s$ and $0\leq s<2$.

Theorem 1.9** (Chen, Liu and Zhang [5]).**

Suppose $k\geq 2$ and $n\geq 8k^{2}-3k$. Then

[TABLE]

For $\ell\geq 4$, we give the spectral extremal graphs for $kS_{\ell-1}\cup P_{\ell}$.

Theorem 1.10**.**

Suppose $k\geq 1$, $\ell\geq 4$ and $n\geq 8\left(k+\lfloor\frac{\ell}{2}\rfloor\right)^{3}t^{8}$ where $t=(\ell^{2}-\ell+1)k+\frac{\ell^{2}+3\ell-2}{2}$.

If $\ell$ is even, then ${\rm{Ex}}_{sp}(n,\,kS_{\ell-1}\cup P_{\ell})=S_{n,\,\frac{2k+\ell-2}{2}}$ holds. 2. 2.

If $\ell$ is odd, then ${\rm{Ex}}_{sp}(n,\,kS_{\ell-1}\cup P_{\ell})=S^{+}_{n,\,\frac{2k+\ell-3}{2}}$ holds.

2 Proof of Theorem 1.7

We would like to point out that the proof of Theorem 1.7 is inspired by [5], and the following two lemmas are very useful to prove Theorem 1.7.

Theorem 2.1** (Hong, Shu and Fang [14], Nikiforov [19]).**

Let $G$ be a graph of order $n$ with the minimum degree $\delta=\delta(G)$ and $e=e(G)$. Then

[TABLE]

Theorem 2.2** (Nikiforov [20]).**

Suppose $c\geq 0$, $p\geq 2$, $n\geq 2^{4p}$. If $\delta(G)<p$ and

[TABLE]

then there exists a subgraph $H$ of order $q\geq\lfloor\sqrt{n}\rfloor$ satisfying one of the following conditions:

$q=\lfloor\sqrt{n}\rfloor$* and $\rho(H)>\sqrt{(2p+1)q}$;* 2. 2.

$q>\lfloor\sqrt{n}\rfloor$, $\delta(H)\geq p$ and $\rho(H)>\frac{p-1+\sqrt{4pq-4p^{2}+c+2}}{2}$.

Proof of Theorem 1.7 (1).  In [20], it is pointed out that

[TABLE]

Suppose $G\in{\rm{Ex}}_{sp}(n,\,F)$. Since $S_{n,\,p}$ is $F$-free, we have

[TABLE]

Case 1. $\delta(G)\geq p$.

It is easy to see the function $f(x)=\frac{x-1+\sqrt{8e-4xn+(x+1)^{2}}}{2}$ is decreasing with respect to $x$. Then by Theorem 2.1 we have

[TABLE]

Together with (2.2), we have $e(G)\geq pn-\frac{p^{2}+p}{2}=e(S_{n,\,p})$. Then $G=S_{n,\,p}$ holds.

Case 2. $\delta(G)<p$.

If we take $c=(p-1)^{2}$ in Theorem 2.2, then there is a graph $H\subseteq G$ of order $q\geq\lfloor\sqrt{n}\rfloor$ satisfying one of the following conditions:

$q=\lfloor\sqrt{n}\rfloor$ and $\rho(H)>\sqrt{(2p+1)q}$; 2. 2.

$q>\lfloor\sqrt{n}\rfloor$, $\delta(H)\geq p$ and $\rho(H)>\frac{p-1+\sqrt{4pq-4p^{2}+c+2}}{2}>\frac{p-1+\sqrt{4pq-4p^{2}+(p-1)^{2}}}{2}$.

When $q=\lfloor\sqrt{n}\rfloor$ and $\rho(H)>\sqrt{(2p+1)q}$, then

[TABLE]

Noting that $q=\lfloor\sqrt{n}\rfloor\geq m$ and Ex$(q,\,F)=\{S_{q,\,p}\}$ hold, then $H$ contains a copy of $F$ and then $G$ contains a copy of $F$. This is a contradiction.

When $q>\lfloor\sqrt{n}\rfloor$, $\delta(H)\geq p$ and $\rho(H)>\frac{p-1+\sqrt{4pq-4p^{2}+(p-1)^{2}}}{2}$, applying Theorem 2.1, we have

[TABLE]

Combining with $\rho(H)>\frac{p-1+\sqrt{4pq-4p^{2}+(p-1)^{2}}}{2}$, we have $e(H)>qp-\frac{p^{2}+p}{2}=e(S_{q,\,p})$. Hence $H$ contains a copy of $F$ and then $G$ contains a copy of $F$. That is a contradiction.

The proof of Theorem 1.7 (2) is similar to that of (1). $\square$

3 Some auxiliary results

To prove Theorem 1.10, we need a rough bound on the size of $kS_{\ell-1}\cup P_{\ell}$-free bipartite graphs.

Lemma 3.1**.**

Let $\ell\geq 4$ and $n$ be positive integers. If $G$ is a $P_{\ell}$-free bipartite graph of order $n$, then

[TABLE]

Proof.

When $n<\ell$, we have $e(G)\leq\lfloor\frac{n}{2}\rfloor\lceil\frac{n}{2}\rceil\leq\left(\lfloor\frac{\ell}{2}\rfloor-1\right)n.$ When $n\geq\ell$, we show the inequality by using induction on $n$. When $n=\ell$, since $G$ is bipartite and $P_{\ell}$-free, we have

[TABLE]

Suppose that $n>\ell$ and the inequality holds for all $\ell\leq n^{\prime}<n$. If $G$ is connected, then by Theorem 1.1, we have

[TABLE]

Now we suppose $G_{1},\,G_{2},\,\cdots,\,G_{s}$ are the connected components of $G$ with $\lvert V(G_{i})\rvert=n_{i}$. If $n_{i}\geq\ell$, then by induction hypothesis $e(G_{i})\leq\left(\lfloor\frac{\ell}{2}\rfloor-1\right)n_{i}$ holds. If $n_{i}<\ell$, then we have $e(G_{i})\leq\lfloor\frac{n_{i}}{2}\rfloor\lceil\frac{n_{i}}{2}\rceil\leq\left(\lfloor\frac{\ell}{2}\rfloor-1\right)n_{i}$. Therefore, we obtain

[TABLE]

∎

Lemma 3.2**.**

Let $G$ be a $kS_{\ell-1}\cup P_{\ell}$-free bipartite graph of order $n\geq\left(k+\lfloor\frac{\ell}{2}\rfloor-1\right)^{2}-(k+\lfloor\frac{\ell}{2}\rfloor-2)\ell+\ell^{2}k+\ell^{2}-\ell$ with $k\geq 0$ and $\ell\geq 4$. Then

[TABLE]

Proof.

We show the inequality by using induction on $k$. When $k=0$, the result holds by Lemma 3.1. Suppose that $k\geq 1$ and the conclusion holds for all $k^{\prime}<k$. Write $h=k+\lfloor\frac{\ell}{2}\rfloor-1$. Let $H$ be a $kS_{\ell-1}\cup P_{\ell}$-free bipartite graph of order $n$ with maximum size. Since the complete bipartite graph $K_{h,\,n-h}$ is $kS_{\ell-1}\cup P_{\ell}$-free, we have

[TABLE]

Hence $H$ contains a copy of $kS_{\ell-1}$. The fact that $H$ is $kS_{\ell-1}\cup P_{\ell}$-free implies $H-S_{\ell-1}$ is $(k-1)S_{\ell-1}\cup P_{\ell}$-free. By induction hypothesis we have $e(H-S_{\ell-1})\leq(h-1)(n-\ell)$. Let $m_{0}$ be the number of edges incident with the vertices of a $S_{\ell-1}$ in $H$. Then we have

[TABLE]

Then each copy of $S_{\ell-1}$ in $H$ contains a vertex of degree at least $\frac{n-h^{2}+(h-1)\ell}{\ell}$. Let $U\subseteq V(H)$ with $\lvert U\rvert=k$ and each vertex in $U$ belongs to distinct $S_{\ell-1}$ with degree at least $\frac{n-h^{2}+(h-1)\ell}{\ell}$. Set $\overline{U}=V(H)\setminus U$, $W=\cup_{u\in U}N(u)$ and $W_{0}=W\cap\overline{U}$ . For any $u\in U$, we have

[TABLE]

Then $\lvert W_{0}\rvert\geq\frac{n-h^{2}+(h-1)\ell}{\ell}-(k-1)$ holds. If $H[\overline{U}]$ contains a copy of $P_{\ell}$, we set $W_{1}=W_{0}\setminus V(P_{\ell})$, then we have

[TABLE]

Then we may find $k$ copies of $S_{\ell-1}$ in $H-P_{\ell}$ with $k$ center vertices in $U$ and leaves vertices in $W_{1}$, then $H$ contains a copy of $kS_{\ell-1}\cup P_{\ell}$, a contradiction. Therefore, $H[\overline{U}]$ is $P_{\ell}$-free and bipartite. By Lemma 3.1, we have $e(H[\overline{U}])\leq\left(\lfloor\frac{\ell}{2}\rfloor-1\right)(n-k)$. Then we deduce

[TABLE]

Therefore we have $e(G)\leq e(H)\leq\left(k+\lfloor\frac{\ell}{2}\rfloor-1\right)n$. ∎

Suppose $S_{1},\,\cdots,\,S_{k}$ are $k$ finite sets, and the following inequality was proved by [7].

[TABLE]

The following formulas are introduced in [27]. For a graph $G$ and any two vertex subsets $A$ and $B$ of $V(G)$, $e(A,\,B)$ denotes the number of the edges of $G$ with one end vertex in $A$ and the other in $B$.

[TABLE]

[TABLE]

[TABLE]

4 Some characterizations of the graphs in EX${}_{sp}(n,\,kS_{\ell-1}\cup P_{\ell})$

Suppose $G\in{\rm{Ex}}_{sp}(n,\,kS_{\ell-1}\cup P_{\ell})$. For $V_{1},\,V_{2}\subseteq V(G)$ and $V_{1}\cap V_{2}=\emptyset$, $G[V_{1},\,V_{2}]$ denotes the induced bipartite graph with one partite set $V_{1}$ and the other partite set $V_{2}$. In this section we always suppose $k\geq 1$, $\ell\geq 4$, $h=k+\lfloor\frac{\ell}{2}\rfloor-1$, $t=(\ell^{2}-\ell+1)k+\frac{\ell^{2}+3\ell-2}{2}$, $\alpha=\frac{1}{2(h+1)t^{2}}$ and $n\geq\frac{t^{2}}{\alpha^{3}}$. The fact that $S_{n,\,h}$ is $kS_{\ell-1}\cup P_{\ell}$-free implies

[TABLE]

Firstly we will show $G$ is connected. Suppose to the contrary that $G$ is disconnected and $G_{1}$ is a component of $G$ with $\rho(G_{1})=\rho(G)\geq\sqrt{hn}$. Let $u\in V(G_{1})$ with the maximum degree in $G_{1}$, $G^{\prime}$ be the graph obtained from $G_{1}$ by attaching a pendent edge at $u$ and $n-\lvert V(G_{1})\rvert-1$ isolated vertices. Then $G^{\prime}$ is $kS_{\ell-1}\cup P_{\ell}$-free. Otherwise there is a copy of $kS_{\ell-1}\cup P_{\ell}$ in $G_{1}$ as $d_{G_{1}}(u)\geq\rho(G_{1})\geq\sqrt{hn}\geq\ell-1$. However, the fact $\rho(G^{\prime})>\rho(G_{1})=\rho(G)$ contradicts $G\in{\rm{Ex}}_{sp}(n,\,kS_{\ell-1}\cup P_{\ell})$. Hence $G$ is connected. Let x be the Perron vector of $G$, i.e., $A(G)\textbf{x}=\rho(G)\textbf{x}$ and $\|\textbf{x}\|_{2}=1$. Set $x_{z}=$max$\{x_{v}:\,v\in V(G)\}$.

Write $\rho=\rho(G)$ for convenience. Let $R=\{v\in V(G):x_{v}>\alpha x_{z}\}$, and $\overline{R}=V(G)\setminus R$. We will evaluate the cardinality of $R$.

Lemma 4.1**.**

$\lvert R\rvert\leq 2\sqrt{hn}$.

Proof.

For each vertex $v\in R$, we have

[TABLE]

Summing the inequality (4) for all $v\in R$, we have

[TABLE]

If $\lvert R\rvert>2\sqrt{hn}$, we have $\lvert R\rvert>t$ as $n\geq\frac{t^{2}}{\alpha^{3}}$. Since $G[R]$ is $kS_{\ell-1}\cup P_{\ell}$-free, by (1.1) we have $e(G[R])\leq(h+\frac{1}{2})\lvert R\rvert$. Furthermore, by Lemma 3.2 and the fact that the bipartite graph $G[R,\,\overline{R}]$ is $kS_{\ell-1}\cup P_{\ell}$-free, $e(R,\,\overline{R})\leq hn$ holds. Hence we obtain

[TABLE]

Since $n\geq\frac{t^{2}}{\alpha^{3}}\geq\frac{(4h+2)^{2}}{\alpha^{2}}$, we have $\sqrt{hn}\alpha-2\left(h+\frac{1}{2}\right)\geq\frac{1}{2}\sqrt{hn}\alpha$. Then from (4.3) we have

[TABLE]

which is a contradiction. Therefore $\lvert R\rvert\leq 2\sqrt{hn}$ holds. ∎

Furthermore, we set $R^{\prime}=\{v\in V(G):x_{v}>4\alpha x_{z}\}$.

Lemma 4.2**.**

For each $v$ in $R^{\prime}$, $d_{G}(v)>\frac{1}{3}\alpha n$ holds.

Proof.

First we prove a claim.

Claim. For any vertex $v$ in $R$, if $d_{G}(v)\leq\frac{1}{3}\alpha n$, then $e\left(G[(N(v)\cup R)\setminus\{v\}]\right)\leq\frac{(5h-2)\alpha n}{6}$ holds.

When $\lvert(N(v)\cup R)\setminus\{v\}\rvert<t$, we have

[TABLE]

where the last inequality holds as $n\geq\frac{t^{2}}{\alpha^{3}}$ and $\alpha=\frac{1}{2(h+1)t^{2}}$.

Now we suppose $\lvert(N(v)\cup R)\setminus\{v\}\rvert\geq t$. Since $G[N(v)\cup R]$ is $kS_{\ell-1}\cup P_{\ell}$-free, by (1.1), we have

[TABLE]

which implies

[TABLE]

where the last second inequality follows from $n\geq\frac{t^{2}}{\alpha^{3}}\geq\frac{(12h+6)^{2}h}{(3h-1)^{2}\alpha^{2}}$.

Suppose to the contrary that there is a vertex $v\in R^{\prime}$ with $d_{G}(v)\leq\frac{1}{3}\alpha n$. By the fact

[TABLE]

and the above claim, we obtain

[TABLE]

From

[TABLE]

we have $d_{G}(v)\geq 4\sqrt{hn}\alpha\geq t$. Since G\big{[}N\left[v\right]\big{]} is $kS_{\ell-1}\cup P_{\ell}$-free, we obtain e\left(G\big{[}N\left[v\right]\big{]}\right)\leq(h+\frac{1}{2})d_{G}(v). Therefore, $e(G[N(v)])\leq\left(h-\frac{1}{2}\right)d_{G}(v)$ holds. It follows

[TABLE]

Moreover, from Lemma 3.2, we have $e(N(v),\,N^{2}(v)\setminus R)\leq hn$, which implies

[TABLE]

Combining (4.4) –  (4.6), we have

[TABLE]

On the other hand, by (4.1) we have

[TABLE]

Then we obtain the following inequality

[TABLE]

Simplifying the inequality, we obtain $h\leq-\frac{2}{9}$, which is a contradiction. Therefore, each vertex in $R^{\prime}$ has degree larger than $\frac{1}{3}\alpha n$. ∎

Lemma 4.3**.**

$\lvert R^{\prime}\rvert\leq\frac{3(h+1)}{\alpha}$.

Proof.

If $\lvert R^{\prime}\rvert\leq t$, then $\lvert R^{\prime}\rvert\leq\frac{3(h+1)}{\alpha}$ holds as $\alpha=\frac{1}{2(h+1)t^{2}}$. Now we suppose $\lvert R^{\prime}\rvert>t$, then by (1.1) we have $e(G[R^{\prime}])\leq\left(h+\frac{1}{2}\right)\lvert R^{\prime}\rvert$. By Lemma 4.2, $\sum_{v\in R^{\prime}}d_{G}(v)\geq\frac{1}{3}\alpha n\lvert R^{\prime}\rvert$ holds, and we have

[TABLE]

Therefore we have

[TABLE]

Noting $n\geq\frac{t^{2}}{\alpha^{3}}$, we have $\alpha n\geq(6h+3)(h+1)$, together with $\left(h+\frac{1}{2}\right)n\geq\frac{1}{3}\alpha n\lvert R^{\prime}\rvert-\left(h+\frac{1}{2}\right)\lvert R^{\prime}\rvert$, we obtain

[TABLE]

∎

Lemma 4.4**.**

If $v$ is a vertex with $x_{v}=mx_{z}$ and $\frac{1}{2(h+1)}\leq m\leq 1$, then $d_{G}(v)>\left(m-\frac{1}{6(h+1)}\right)n$.

Proof.

Suppose to the contrary that there is a vertex $v$ with $x_{v}=mx_{z}$ $\left(\frac{1}{2(h+1)}\leq m\leq 1\right)$ and $d_{G}(v)\leq\left(m-\frac{1}{6(h+1)}\right)n.$ By the definition of $R^{\prime}$, $v\in R^{\prime}$ holds. Let $M=N(v)\cup N^{2}(v)$. From (1.1) and (3.3), we obtain

[TABLE]

Since $v\in R^{\prime}$, we have $d_{G}(v)>\frac{1}{3}\alpha n$ from Lemma 4.2. Then $G[R^{\prime}\setminus\{v\},\,N(v)\setminus R^{\prime}]$ is $(k-1)S_{\ell-1}\cup P_{\ell}$-free. Otherwise we may add a copy of $S_{\ell-1}$ with $v$ as center vertex to obtain a copy of $kS_{\ell-1}\cup P_{\ell}$ in $G$. Then by Lemma 3.2 we have

[TABLE]

Then from Lemma 3.2 and (3.2) we have

[TABLE]

If $\lvert R^{\prime}\rvert\geq t$, then by (1.1), $e(G[R^{\prime}])\leq(h+\frac{1}{2})\lvert R^{\prime}\rvert$ holds. From Lemma 4.3, $\lvert R^{\prime}\rvert\leq\frac{3(h+1)}{\alpha}$ holds, then we have

[TABLE]

If $\lvert R^{\prime}\rvert<t$, then $e(G[R^{\prime}])<\frac{t(t-1)}{2}$ holds. We have

[TABLE]

the last inequality holds as $\alpha=\frac{1}{2(h+1)t^{2}}$.

For each vertex $u\in M\setminus R^{\prime}$, the definition of $R^{\prime}$ implies $x_{u}\leq 4\alpha x_{z}$. Therefore, combining (4.7) – (4.10), we have

[TABLE]

On the other hand, by (4.1), we have

[TABLE]

Therefore, we obtain

[TABLE]

Since $d_{G}(v)\leq\left(m-\frac{1}{6(h+1)}\right)n$, we have

[TABLE]

Let

[TABLE]

Then $g(x)\geq g(1)$ holds when $x\in[\frac{1}{2(h+1)},\,1]$. So $g(m)\geq g(1)$ holds and we have

[TABLE]

Then we get $n\leq 54h(h+1)^{4}t^{2}$, while it contradicts the fact $n\geq 8(h+1)^{3}t^{8}$. ∎

Let $R^{\prime\prime}=\{v\in V(G):\,x_{v}\geq\frac{1}{2(h+1)}x_{z}\}$. Clearly $R^{\prime\prime}\subseteq R^{\prime}\subseteq R$ holds. To prove the cardinality of $R^{\prime\prime}$, we prove a lower bound of the degree of vertices in $R^{\prime\prime}$.

Lemma 4.5**.**

For each $v$ in $R^{\prime\prime}$, we have $d_{G}(v)\geq\left(1-\frac{5}{6(h+1)}\right)n$.

Proof.

By Lemma 4.4, it suffices to prove $x_{v}\geq\left(1-\frac{2}{3(h+1)}\right)x_{z}$ for every $v\in R^{\prime\prime}$. Suppose to the contrary that there is a vertex $v\in R^{\prime\prime}$ such that $x_{v}=mx_{z}$, where $\frac{1}{2(h+1)}\leq m<1-\frac{2}{3(h+1)}$. Let $R_{1}=N(z)\cap R^{\prime}$, $R_{2}=N^{2}(z)\cap R^{\prime}$, $S_{1}=N(z)\setminus R_{1}$ and $S_{2}=N^{2}(z)\setminus R_{2}$. By (1.1) and Lemma 3.2, we have

[TABLE]

From Lemma 4.3 we have $\lvert R^{\prime}\rvert\leq\frac{3(h+1)}{\alpha}$, therefore we deduce

[TABLE]

Noting that

[TABLE]

holds. By Lemma 3.2, we have $e(S_{1},\,R_{1}\cup R_{2}\cup\{z\})\leq hn.$ Then

[TABLE]

Together with the fact that $hnx_{z}\leq\rho^{2}x_{z}$ and $\lvert R^{\prime}\rvert\leq\frac{3(h+1)}{\alpha}$, we obtain

[TABLE]

Since $\alpha=\frac{1}{2(h+1)t^{2}}$ and $n\geq\frac{t^{2}}{\alpha^{3}}$, we get

[TABLE]

Since $z,\,v\in R^{\prime}$, by Lemma 4.4 we have $d_{G}(v)>\left(m-\frac{1}{6(h+1)}\right)n>\frac{n}{3(h+1)}$ and $d_{G}(z)>\left(1-\frac{1}{6(h+1)}\right)n$. So

[TABLE]

holds, a contradiction to (4.12). Hence we have $d_{G}(v)\geq(1-\frac{5}{6(h+1)})n$ for each $v$ in $R^{\prime\prime}$. ∎

Now we prove the exact value of $\lvert R^{\prime\prime}\rvert$.

Lemma 4.6**.**

$\lvert R^{\prime\prime}\rvert=h$.

Proof.

Firstly, we prove $\lvert R^{\prime\prime}\rvert>h-1$. Suppose to the contrary that $\lvert R^{\prime\prime}\rvert\leq h-1$. Let $M=N(z)\cup N^{2}(z)$. By (1.1) and (3.3), we obtain

[TABLE]

The fact $M\cap R^{\prime\prime}\subseteq R^{\prime\prime}\setminus\{z\}$ and (3.4) imply

[TABLE]

Combining (4.13) and (4.14), we have

[TABLE]

Together with the fact $hn\leq\rho^{2}(G)$, it follows

[TABLE]

which may deduce $h\leq 0$, a contradiction. Hence, $\lvert R^{\prime\prime}\rvert>h-1$.

If $\lvert R^{\prime\prime}\rvert\geq h+1$, suppose $\{v_{1},\,v_{2},\cdots,v_{h+1}\}\subseteq R^{\prime\prime}$ and $W=N(v_{1})\cap\cdots\cap N(v_{h+1})$. By (3.1), it follows

[TABLE]

the last inequality holds as $n\geq\frac{t^{2}}{\alpha^{3}}$. When $\ell$ is even, we may find a path $P_{\ell}=u_{1}v_{1}\cdots u_{\frac{\ell}{2}}v_{\frac{\ell}{2}}$, where $u_{i}\in W$ ($i=1,\,2,\,\cdots,\frac{\ell}{2}$). As $d_{G}(v_{i})\geq\left(1-\frac{5}{6(h+1)}\right)n$ ($i=\frac{\ell}{2}+1,\cdots,h+1$), we may find a copy of $kS_{\ell-1}$ in $G-P_{\ell}$ with $v_{\frac{\ell}{2}+1},\cdots,v_{h+1}$ as the $k$ center vertices. Then there is a copy of $kS_{\ell-1}\cup P_{\ell}$ in $G$, a contradiction. When $\ell$ is odd, we may find a path $P_{\ell}=u_{1}v_{1}\cdots u_{\frac{\ell-1}{2}}v_{\frac{\ell-1}{2}}u_{\frac{\ell+1}{2}}$ where $u_{i}\in W$ ($i=1,\,2,\,\cdots,\frac{\ell+1}{2}$) and a copy of $kS_{\ell-1}$ in $G-P_{\ell}$ with $v_{\frac{\ell+1}{2}},\cdots,v_{h+1}$ as the $k$ center vertices. Then there is a copy of $kS_{\ell-1}\cup P_{\ell}$ in $G$, a contradiction. Therefore, $\lvert R^{\prime\prime}\rvert=h$ holds. ∎

5 Proof of Theorem 1.10

Suppose $k\geq 1$, $\ell\geq 4$, $h=k+\lfloor\frac{\ell}{2}\rfloor-1$, $t=(\ell^{2}-\ell+1)k+\frac{\ell^{2}+3\ell-2}{2}$, $\alpha=\frac{1}{2(h+1)t^{2}}$ and $n\geq\frac{t^{2}}{\alpha^{3}}$. Let $G\in{\rm Ex}_{sp}(n,\,kS_{\ell-1}\cup P_{\ell})$, $x_{z}={\rm max}\{x_{v}:v\in V(G)\}$ and $R^{\prime\prime}=\{v\in V(G):\,x_{v}\geq\frac{1}{2(h+1)}x_{z}\}$. By Lemma 4.6, $\lvert R^{\prime\prime}\rvert=h$ holds. Write $R^{\prime\prime}=\{v_{1},\,v_{2},\cdots,v_{h-1},\,z\}$ and $W=\{v\in V(G):R^{\prime\prime}\subseteq N_{G}(v)\}$. By Lemma 4.5, we have $d_{G}(v)\geq n-\frac{5n}{6(h+1)}$ for each $v\in R^{\prime\prime}$. Hence when $n\geq 8(h+1)^{3}t^{8}$, we have

[TABLE]

Claim A. $G[\overline{R^{\prime\prime}}]$ is $S_{\ell-1}$-free.

Suppose to the contrary that $G[\overline{R^{\prime\prime}}]$ contains a copy of $S_{\ell-1}$. Let $R_{1}=\{v_{1},\,v_{2},\cdots,v_{k-1}\}\subseteq R^{\prime\prime}$. Since each vertex in $R^{\prime\prime}$ has degree at least $\left(1-\frac{5}{6(h+1)}\right)n$, we may find a $(k-1)S_{\ell-1}$ with $v_{1},\cdots,v_{k-1}$ as the $k-1$ center vertices and leaves vertices in $V(G)\setminus(R^{\prime\prime}\cup V(S_{\ell-1}))$. Then there is a copy of $kS_{\ell-1}$ in $G$. When $\ell$ is even, we have $\lvert R^{\prime\prime}\setminus R_{1}\rvert=\frac{\ell}{2}$. Since $\lvert W\rvert-k\ell>\ell$ when $n\geq\frac{t^{2}}{\alpha^{3}}$, we may find a copy of $P_{\ell}=v_{k}u_{1}\cdots zu_{\frac{\ell}{2}}$, where $u_{i}\in W\setminus V\left(kS_{\ell-1}\right)$ ($i=1,\cdots,\frac{\ell}{2}$). Then $G$ contains a copy of $kS_{\ell-1}\cup P_{\ell}$, a contradiction. When $\ell$ is odd, we have $\lvert R^{\prime\prime}\setminus R_{1}\rvert=\frac{\ell-1}{2}$. We may find a copy of $P_{\ell}=u_{1}v_{k}\cdots u_{\frac{\ell-1}{2}}zu_{\frac{\ell+1}{2}}$, where $u_{i}\in W\setminus V\left(kS_{\ell-1}\right)$ ($i=1,\,\cdots,\,\frac{\ell+1}{2}$), a contradiction. Therefore, $G[\overline{R^{\prime\prime}}]$ is $S_{\ell-1}$-free.

Claim B. $z$ is adjacent to each vertex of $\overline{R^{\prime\prime}}$.

Suppose to the contrary that there is a vertex $v$ in $\overline{R^{\prime\prime}}$ not adjacent to $z$. Let $G_{1}$ be a graph obtained from $G$ by deleting all edges incident with $v$ in $G[\overline{R^{\prime\prime}}]$ and adding an edge $zv$. If $G_{1}$ contains a copy of $kS_{\ell-1}\cup P_{\ell}$, we have $zv\in E(kS_{\ell-1}\cup P_{\ell})$. Since $zv$ is a pendent edge in $G_{1}$ and $\lvert W\rvert>(k+1)\ell$, we may find a vertex $w\in W$ and use $zw$ as a replacement of $zv$, then we obtain a copy of $kS_{\ell-1}\cup P_{\ell}$ in $G$, a contradiction. Therefore, $G_{1}$ is $kS_{\ell-1}\cup P_{\ell}$-free. The fact that $G[\overline{R^{\prime\prime}}]$ is $S_{\ell-1}$-free implies $d_{G[\overline{R^{\prime\prime}}]}(v)<\ell-1$. When $w\notin R^{\prime\prime}$, we have $x_{w}<\frac{1}{2(h+1)}x_{z}$. It follows

[TABLE]

Then we have

[TABLE]

the inequality deduces that $z$ is adjacent to each vertex of $\overline{R^{\prime\prime}}$.

Set $R_{2}=\{v_{1},\cdots,v_{k}\}\subseteq R^{\prime\prime}$.

Claim C. When $\ell$ is even, there is no edge in $G[\overline{R^{\prime\prime}}]$.

Suppose to the contrary that there is an edge in $G[\overline{R^{\prime\prime}}]$, denoted by $uv$. By Claim B, we have $uz$, $vz\in E(G)$. Since each vertex in $R^{\prime\prime}$ has degree at least $\left(1-\frac{5}{6(h+1)}\right)n$, we may find a copy of $kS_{\ell-1}$ in $G$ with the $k$ center vertices in $R_{2}$ and leaves vertices in $\overline{R^{\prime\prime}}$. Then $\lvert R^{\prime\prime}\setminus R_{2}\rvert=\frac{\ell}{2}-1$ holds. As $\lvert W\rvert>(k+1)\ell$, we may find a copy of $P_{\ell}=u_{1}v_{k+1}\cdots u_{\frac{\ell}{2}-2}v_{h-1}u_{\frac{\ell}{2}-1}zvu$, where $u_{i}\in W\setminus V(kS_{\ell-1})$ ($i=1,\cdots,\frac{\ell}{2}-1$). Then $G$ contains a copy of $kS_{\ell-1}\cup P_{\ell}$, a contradiction.

Claim D. When $\ell$ is odd, there is at most one edge in $G[\overline{R^{\prime\prime}}]$.

Suppose to the contrary that there is a $P_{3}\subseteq G[\overline{R^{\prime\prime}}]$, denoted by $uvw$. By Claim B, we have $wz\in E(G)$. Since each vertex in $R^{\prime\prime}$ has degree at least $\left(1-\frac{5}{6(h+1)}\right)n$, we may find a copy of $kS_{\ell-1}$ in $G$ with the $k$ center vertices in $R_{2}$ and leaves vertices in $\overline{R^{\prime\prime}}$. Then $\lvert R^{\prime\prime}\setminus R_{2}\rvert=\frac{\ell-3}{2}$ holds. As $\lvert W\rvert>(k+1)\ell$, we may find a path $P_{\ell}=u_{1}v_{k+1}u_{2}\cdots v_{h-1}u_{\frac{\ell-3}{2}}zwvu$ where $u_{i}\in W\setminus V(kS_{\ell-1})$ ($i=1,\cdots,\frac{\ell-3}{2}$). Then there is a copy of $kS_{\ell-1}\cup P_{\ell}$ in $G$, a contradiction. Hence $G$ is $P_{3}$-free.

Suppose to the contrary $2P_{2}\subseteq G[\overline{R^{\prime\prime}}]$, denoted by $w_{1}w_{2}$, $w_{3}w_{4}$. If one of $\{w_{1},\,w_{2},\,w_{3},\,w_{4}\}$ is adjacent to a vertex in $R^{\prime\prime}\setminus\{z\}$, without loss of generality, say $v_{\frac{\ell-5}{2}}w_{1}\in E(G)$. We may find a $P_{\ell}$ with $\frac{\ell-3}{2}$ vertices in $R^{\prime\prime}$, $P_{\ell}=u_{1}v_{1}\cdots u_{\frac{\ell-5}{2}}v_{\frac{\ell-5}{2}}w_{1}w_{2}zw_{3}w_{4}$, $u_{i}\in W$ ($i=1,\,\cdots,\,\frac{\ell-5}{2}$). Moreover, we may find a $kS_{\ell-1}$ with the remaining $k$ vertices of $R^{\prime\prime}$ as the $k$ center vertices and leaves vertices in $\overline{R^{\prime\prime}}\setminus P_{\ell}$, a contradiction. Hence we have $N_{G[R^{\prime\prime}]}(w_{i})=\{z\}$ ($i=1,\,2,\,3,\,4$). Let $G_{2}$ be a graph obtained from $G$ by deleting edge $w_{1}w_{2}$ and adding edge $w_{1}v_{1}$. If $G_{2}$ contains a copy of $kS_{\ell-1}\cup P_{\ell}$, we have $w_{1}v_{1}\in E(kS_{\ell-1}\cup P_{\ell})$. Since $w_{1}v_{1}$ is a pendent edge in $G_{2}$ and $\lvert W\rvert>(k+1)\ell$, we may find a vertex $w_{5}\in W$ and use $w_{5}v_{1}$ as a replacement of $w_{1}v_{1}$. Then we obtain a copy of $kS_{\ell-1}\cup P_{\ell}$ in $G$, a contradiction. Hence $G_{2}$ is $kS_{\ell-1}\cup P_{\ell}$-free. Since $v_{1}\in R^{\prime\prime}$, $w_{2}\in\overline{R^{\prime\prime}}$, we have $x_{w_{2}}<\frac{1}{2(h+1)}x_{z}\leq x_{v_{1}}.$ Then

[TABLE]

this contradiction implies that $G[\overline{R^{\prime\prime}}]$ is 2$P_{2}$-free. Hence there is at most one edge in $G[\overline{R^{\prime\prime}}]$.

When $\ell$ is even, $G[\overline{R^{\prime\prime}}]=\overline{K}_{n-h}$ holds from Claim C. By the facts that the spectral radius of a graph does not decrease by adding an edge and $S_{n,\,h}$ is $kS_{\ell-1}\cup P_{\ell}$-free, each vertex in $R^{\prime\prime}$ is adjacent to each vertex in $\overline{R^{\prime\prime}}$ and $G[R^{\prime\prime}]=\{K_{n}\}$ holds. Then we have $G=S_{n,\,h}$. When $\ell$ is odd, $G[\overline{R^{\prime\prime}}]=K_{2}\cup\overline{K}_{n-h-2}$ holds from Claim D. Since $S^{+}_{n,\,h}$ is $kS_{\ell-1}\cup P_{\ell}$-free, each vertex in $R^{\prime\prime}$ is adjacent to each vertex in $\overline{R^{\prime\prime}}$ and $G[R^{\prime\prime}]=\{K_{n}\}$ holds. Then we have $G=S^{+}_{n,\,h}$.

Declaration

The authors have declared that no competing interest exists.