Vertex degree sums for matchings in 3-uniform hypergraphs

TL;DR

This paper establishes a sharp degree sum condition in 3-uniform hypergraphs that guarantees the existence of a large matching, confirming a previously conjectured bound and extending prior results.

Contribution

The paper proves a new optimal degree sum condition for matchings in 3-uniform hypergraphs, generalizing earlier specific cases and confirming a conjecture by the authors.

Findings

Degree sum condition guarantees matchings of size s

Condition is proven to be best possible

Extends previous case when s = n/3

Abstract

Let $n, s$ be positive integers such that $n$ is sufficiently large and $s\le n/3$. Suppose $H$ is a 3-uniform hypergraph of order $n$. If $H$ contains no isolated vertex and $deg(u)+ deg(v) > 2(s-1)(n-1)$ for any two vertices $u$ and $v$ that are contained in some edge of $H$, then $H$ contains a matching of size $s$. This degree sum condition is best possible and confirms a conjecture of the authors [Electron. J. Combin. 25 (3), 2018], who proved the case when $s= n/3$.

Equations

$${\sigma }_2(H_{n,s}^3)$$

$${\sigma }_2(H_{n,s}^2)$$

$$0<\epsilon \ll {\epsilon }^{\prime }\ll {\epsilon }^{\prime \prime }\ll {\eta }_1\ll {\eta }_2\ll \gamma \ll {\gamma }^{\prime }\ll \tau \ll 1.$$

$$\sum _{i=1}^3\left(\sum _{j=1}^2{\mathrm{deg}}_{G_i}(u_j)+{\mathrm{deg}}_{G_i}(v_1)\right)\le \max \{4a+7,3a+2b+5\}.$$

$${\mathrm{deg}}_{G_i}(v_1)\le 2\text{for }i=2,3$$

$$\mathrm{deg}(v_0)+\mathrm{deg}(u)$$

$$\text{deg}(u)+\text{deg}(v_0)\le (n-1)(2s+{\epsilon }^{\prime }n-1)-\left(\genfrac{}{}{0ex}{}{2{\epsilon }^{\prime \prime }n}{2}\right),$$

$$\text{deg}(u)\le \left(\genfrac{}{}{0ex}{}{|U|-1}{2}\right)<\left(\genfrac{}{}{0ex}{}{2s+{\epsilon }^{\prime }n}{2}\right),$$

$$\mathrm{deg}(u_1)+\mathrm{deg}(u_2)+\mathrm{deg}(v_0)\le 2\left(\left(\genfrac{}{}{0ex}{}{|U|-1}{2}\right)+|W|(U|-1)\right)+\left(\genfrac{}{}{0ex}{}{|U|}{2}\right)-\left(\genfrac{}{}{0ex}{}{s-{\epsilon }^{\prime \prime }n}{2}\right).$$

$$\mathrm{deg}(u_1)+\mathrm{deg}(u_2)+\mathrm{deg}(v_0)>3\left(2sn-\epsilon n^2\right)-2\left(\genfrac{}{}{0ex}{}{|U|}{2}\right)-\left(\genfrac{}{}{0ex}{}{|U|-1}{2}\right)-|W|(|U|-1).$$

$$3\left(\left(\genfrac{}{}{0ex}{}{|U|-1}{2}\right)+|W|(|U|-1)+\left(\genfrac{}{}{0ex}{}{|U|}{2}\right)\right)-\left(\genfrac{}{}{0ex}{}{s-{\epsilon }^{\prime \prime }n}{2}\right)$$

$$\text{or}(|U|-1)(n-1)-\frac{1}{3}\left(\genfrac{}{}{0ex}{}{s-{\epsilon }^{\prime \prime }n}{2}\right)$$

$${\sigma }_2(H)\le \mathrm{deg}(v_0)+\mathrm{deg}(u_0)$$

$$\le (2n-|U|)|M|+\frac{|U{|}^2}{2}.$$

$${\sigma }_2(H)$$

$$=2sn-2{\epsilon }^{\prime }{n}^2+3s{\epsilon }^{\prime }n+\frac{3}{2}{\epsilon }^{\prime 2}{n}^2\le 2sn-{\epsilon }^{\prime }{n}^2+\frac{3}{2}{\epsilon }^{\prime 2}{n}^2$$

$$N(u_0,v_1)\cap U_2=\varnothing .$$

$$\mathrm{deg}(v_0)\le \left(\genfrac{}{}{0ex}{}{|U_1|}{2}\right)+2|M^{\prime }||U_2|+2(|U_1|-2|M^{\prime }|)=\left(\genfrac{}{}{0ex}{}{|U_1|}{2}\right)+2|U_1|+|M^{\prime }|(2|U_2|-4).$$

$$\mathrm{deg}(u_0)$$

$$\mathrm{deg}(v_0)+\mathrm{deg}(u_0)\le \left(\genfrac{}{}{0ex}{}{|U_1|}{2}\right)+\left(\genfrac{}{}{0ex}{}{|U|-1}{2}\right)+|U_1|(|W|+2)+(|U_2|-1)|W_1|+|M^{\prime }|(|U_2|-3).$$

$$\mathrm{deg}(v_0)+\mathrm{deg}(u_0)$$

$$=\left(\genfrac{}{}{0ex}{}{|U|}{2}\right)-\left(\genfrac{}{}{0ex}{}{|U_2|}{2}\right)+\left(\genfrac{}{}{0ex}{}{|U|-1}{2}\right)+|U_1||W|={\left(|U|-1\right)}^2-\left(\genfrac{}{}{0ex}{}{|U_2|}{2}\right)+2|M|\left(n-|U|\right).$$

$$\text{deg}(v_0)+\text{deg}(u_0)$$

$${\sigma }_2(H)\le \text{deg}(v_0)+\text{deg}(u_0)$$

$$\sum _{i=1}^3\mathrm{deg}(u_i)>3sn-\frac{3}{2}\epsilon n^2.$$

$$\sum _{i=1}^2\mathrm{deg}(u_i)+\mathrm{deg}(v_1)\ge {\sigma }_2(H)+\mathrm{deg}(u_2)>3sn-\frac{3}{2}\epsilon n^2.$$

$$\sum _{i=1}^2|L_{u_i}(e,U_3\cup W_2)|+|L_{v_1}(e,U_3\cup W_2)|=\sum _{i=1}^3\left(\sum _{j=1}^2{\mathrm{deg}}_{G_i}(u_j)+{\mathrm{deg}}_{G_i}(v_1)\right).$$

$$\sum _{i=1}^3\mathrm{deg}(u_i)$$

$$=18\left(\genfrac{}{}{0ex}{}{|M|}{2}\right)+9|M|+6\left(3s-|U|\right)(3s-3|M|)+3(|M|+|U|-3s)(3s-3|M|+3)$$

$$=(9|U|-18s+9)|M|+(3s-|U|)(9s-9).$$

$$\sum _{i=1}^3\mathrm{deg}(u_i)$$

$$3s^2-sn=s(3s-n)\le \max \left\{-{\eta }_{1}n(n-3{\eta }_{1}n),-\tau n(n-3\tau n)\right\}=-{\eta }_{1}n(n-3{\eta }_{1}n).$$

Full text

Yi Zhang and Mei Lu are supported by the National Natural Science Foundation of China (Grant 11771247). Yi Zhao is partially supported by NSF grant DMS-1700622.

1. Introduction

A $k$-uniform hypergraph $H$ (in short, $k$-graph) is a pair $(V,E)$, where $V$ is a finite set of vertices and $E$ is a family of $k$-element subsets of $V$. Note that a 2-graph is simply a graph. Let $V(H)$ and $E(H)$ denote the vertex set and edge set of $H$, respectively. A matching of size $s$ in $H$ is a family of $s$ pairwise disjoint edges of $H$. If the matching covers all the vertices of $H$, then we call it a perfect matching. Given a set $S\subseteq V$, the degree $\deg_{H}(S)$ of $S$ is the number of the edges of $H$ containing $S$. We simply write $\deg(S)$ when $H$ is obvious from the context. Further, let $\delta_{\ell}(H)=\min\{\deg(S):S\subseteq V(H),|S|=\ell\}$.

Given integers $\ell<k\leq n$ such that $k$ divides $n$, let $m_{\ell}(k,n)$ denote the smallest integer $m$ such that every $k$-graph $H$ on $n$ vertices with $\delta_{\ell}(H)\geq m$ contains a perfect matching. In recent years the problem of determining $m_{\ell}(k,n)$ has received much attention (see [2, 5, 6, 7, 8, 9, 10, 12, 14, 17, 16, 18, 20, 21, 22]). In particular, Rödl, Ruciński and Szemerédi [18] determined $m_{k-1}(k,n)$ for all $k\geq 3$ and sufficiently large $n$. Treglown and Zhao [20, 21] determined $m_{\ell}(k,n)$ for all $\ell\geq k/2$ and sufficiently large $n$. More Dirac-type results on hypergraphs can be found in surveys [15, 27].

A well-known result of Ore [13] extended Dirac’s theorem by determining the smallest degree sum of two non-adjacent vertices that guarantees a Hamilton cycle in graphs. Ore-type problems for hypergraphs have been studied recently. For example, Tang and Yan [19] studied the degree sum of two $(k-1)$-sets that guarantees a tight Hamilton cycle in $k$-graphs. Zhang and Lu [23] studied the degree sum of two $(k-1)$-sets that guarantees a perfect matching in $k$-graphs. Zhang, Zhao and Lu [26] determined the minimum degree sum of two adjacent vertices that guarantees a perfect matching in 3-graphs without isolated vertices, see Theorem 2 (two vertices in a hypergraph are adjacent if there exists an edge containing both of them). Note that one may study the minimum degree sum of two arbitrary vertices and that of two non-adjacent vertices that guarantees a perfect matching instead. In fact, it was mentioned in [26] that the former equals to $2m_{1}(3,n)-1$ while the latter does not exist.

Let us define (potential) extremal 3-graphs for the matching problem. For $1\leq\ell\leq 3$, let $H^{\ell}_{n,s}$ denote the 3-graph of order $n$, whose vertex set is partitioned into two sets $S$ and $T$ of size $n-s\ell+1$ and $s\ell-1$, respectively, and whose edge set consists of all triples with at least $\ell$ vertices in $T$. A well-known conjecture of Erdős  [3], recently verified for $3$-graphs [4, 11] , implies that $H_{n,s}^{1}$ or $H_{n,s}^{3}$ is the densest $3$-graph on $n$ vertices not containing a matching of size $s$. On the other hand, Kühn, Osthus and Treglown [10] showed that for sufficiently large $n$, $H_{n,s}^{1}$ has the largest minimum vertex degree among all 3-graphs on $n$ vertices not containing a matching of size $s$.

Theorem 1**.**

[10]** There exists $n_{0}\in\mathbb{N}$ such that if $H$ is a $3$-graph of order $n\geq n_{0}$ with $\delta_{1}(H)>\delta_{1}(H^{1}_{n,s})=\binom{n-1}{2}-\binom{n-s}{2}$ and $n\geq 3s$, then $H$ contains a matching of size $s$.

Given a 3-graph $H$, let $\sigma_{2}(H)$ denote the minimum $\deg(u)+\deg(v)$ among all adjacent vertices $u$ and $v$. It is easy to see that

[TABLE]

The following is [26, Theorem 1], which implies that, when $n$ is divisible by 3 and sufficiently large, $H^{2}_{n,n/3}$ has the largest $\sigma_{2}(H)$ among all $n$-vertex 3-graphs $H$ containing no isolated vertex or perfect matching.

Theorem 2**.**

[26]** There exists $n_{0}\in\mathbb{N}$ such that the following holds for all integers $n\geq n_{0}$ that are divisible by $3$. Let $H$ be a $3$-graph of order $n$ without isolated vertex. If $\sigma_{2}(H)>\sigma_{2}(H^{2}_{n,n/3})=\frac{2}{3}n^{2}-\frac{8}{3}n+2$, then $H$ contains a perfect matching.

Zhang, Zhao and Lu [26, Conjecture 12] further conjectured that for sufficiently large $n$ and any $s<n/3$, $H^{2}_{n,s}$ has the largest $\sigma_{2}(H)$ among all $n$-vertex 3-graphs $H$ containing no isolated vertex or matching of size $s$. In this paper we verify this conjecture.

Theorem 3**.**

There exists $n_{1}\in\mathbb{N}$ such that the following holds for all integers $n\geq n_{1}$ and $s\leq n/3$. If $H$ is a $3$-graph of order $n$ without isolated vertex and $\sigma_{2}(H)>\sigma_{2}(H_{n,s}^{2})=2(s-1)(n-1)$, then $H$ contains a matching of size $s$.

Since two theorems have different extremal hypergraphs, Theorem 3 does not imply Theorem 1 (analogously Theorem 1 does not imply Erdős’ matching conjecture for 3-graphs). On the other hand, one may wonder why we assume that $H$ contains no isolated vertex in Theorem 3 (especially when $s<n/3$). In fact, as shown in the concluding remarks of [26], Theorem 3 implies another conjecture [26, Conjecture 13], which determines the largest $\sigma_{2}(H)$ among all 3-graphs containing no matching of size $s$. Note that $\sigma_{2}(H_{n,s}^{2})\geq\sigma_{2}(H_{n,s}^{3})$ if and only if $s\leq(2n+4)/9$.

Corollary 4**.**

There exists $n_{2}\in\mathbb{N}$ such that the following holds. Suppose that $H$ is a $3$-graph of order $n\geq n_{2}$ and $2\leq s\leq n/3$. If $\sigma_{2}(H)>\sigma_{2}(H_{n,s}^{2})$ and $s\leq(2n+4)/9$ or $\sigma_{2}(H)>\sigma_{2}(H_{n,s}^{3})$ and $s>(2n+4)/9$, then $H$ contains a matching of size $s$.

Let us explain our approach towards Theorem 3. The case when $s\leq n/13$ was already solved by Zhang and Lu [24] in a stronger form. Note that $\sigma_{2}(H_{n,s}^{2})>\sigma_{2}(H_{n,s}^{1})$. The following theorem shows that, when $n\geq 13s$, not only $H_{n,s}^{2}$ is the (unique) 3-graph with the largest $\sigma_{2}(H)$ among all $H$ containing no isolated vertex or a matching of size $s$, but also $H^{1}_{n,s}$ is the sub-extremal 3-graph for this problem. (In fact, Zhang and Lu [24] conjectured that Theorem 5 holds for all $n\geq 3s$. If true, this strengthens Theorem 1 and actually provides a link between Ore’s and Dirac’s problems.)

Theorem 5**.**

[25]** Let $n,s$ be positive integers and $H$ be a 3-graph of order $n\geq 13s$ without isolated vertex. If $\sigma_{2}(H)>\sigma_{2}(H^{1}_{n,s})=2\left(\binom{n-1}{2}-\binom{n-s}{2}\right)$, then either $H$ contains a matching of size $s$ or $H$ is a subgraph of $H_{n,s}^{2}$.

Therefore it suffices to prove Theorem 3 for reasonably large $s$. For such $s$, we actually prove a (stronger) stability theorem.

Theorem 6**.**

Given $0<\varepsilon\ll\tau\ll 1$, let $n$ be sufficiently large and $\tau n<s\leq n/3$. If $H$ is a $3$-graph of order $n$ without isolated vertex such that $\sigma_{2}(H)>2sn-\varepsilon n^{2}$, then either $H$ is a subgraph of $H_{n,s}^{2}$ or $H$ contains a matching of size $s$.

Theorem 3 follows from Theorem 6 immediately. Indeed, if $\sigma_{2}(H)>\sigma_{2}(H^{2}_{n,s})$, then it is easy to see that $H$ is not a subgraph of $H_{n,s}^{2}$.111Unfortunately $\sigma_{2}$ is not a monotone function: for example, adding an edge to $H^{2}_{n,s}$ indeed reduces the value of $\sigma_{2}$ because two vertices in $S$ now become adjacent and their degree sum is smaller than $\sigma_{2}(H^{2}_{n,s})$. Suppose instead, that $V(H)$ can be partitioned $S\cup T$ such that $|S|=n-2s+1$, $|T|=2s-1$, and every edge of $H$ contains at least two vertices of $T$. Since $H$ contains no isolated vertices, every vertex of $S$ is adjacent to some vertex of $T$. Thus $\sigma_{2}(H)\leq\deg(u)+\deg(v)$ for some $u\in S$ and $v\in T$. Consequently $\sigma_{2}(H)\leq\sigma_{2}(H^{2}_{n,s})$, a contradiction. We therefore apply Theorem 6 to derive that $H$ contains a matching of size $s$. Furthermore, Theorem 6 implies that $H_{n,s}^{2}$ is the unique extremal 3-graph for Theorem 3 because all proper subgraphs $H$ of $H_{n,s}^{2}$ satisfy $\sigma_{2}(H)<\sigma_{2}(H_{n,s}^{2})$.

In order to prove Theorem 6, we follow the same approach as in [26]: using the condition on $\sigma_{2}(H)$, we greedily extend a matching of $H$ until it has $s$ edges. An important intermediate step is finding a matching that covers a certain number of low-degree vertices (see Lemma 7). Nevertheless, the proof of Theorem 6 does require new ideas: in particular, the meaning of an optimal matching is more complicated (see Definition 8); we proceed differently depending on whether the number of low-degree vertices in the optimal matching is at the threshold. In one case we reduce the problem to that of finding a perfect matching in a subgraph of $H$ and apply the main result of [26] (see Theorem 9).

This paper is organized as follows. In Section 2, we give an outline of the proof along with some preliminary results. We prove Lemma 7 in Section 3 and complete the proof in Section 4.

Notation Given a graph $G$ and a vertex $u$ in $G$, $N_{G}(u)$ is the set of neighbors of $u$ in $G$. Suppose $H$ is a 3-uniform hypergraph. For $u\neq v\in V(H)$, let $N_{H}(u,v)=\{w\in V(H):\{u,v,w\}\in E(H)\}$ (the subscript is often omitted when $H$ is clear from the context). Given three subsets $V_{1},V_{2},V_{3}$ of $V(H)$, we say that an edge $\{v_{1},v_{2},v_{3}\}\in E(H)$ is a type of $V_{1}V_{2}V_{3}$ if $v_{i}\in V_{i}$ for $1\leq i\leq 3$. Given a vertex $v\in V(H)$ and a subset $A\subseteq V(H)$, we define the link $L_{v}(A)=\{uw:u,w\in A\mbox{~{}and~{}}\{u,v,w\}\in E(H)\}$. When $A$ and $B$ are two disjoint subsets of $V(H)$, we let $L_{v}(A,B)=\{uw:u\in A,~{}w\in B\mbox{~{}and~{}}\{u,v,w\}\in E(H)\}$.

We write $0<a_{1}\ll a_{2}\ll a_{3}$ if we can choose the constants $a_{1},a_{2},a_{3}$ from right to left. More precisely there are increasing functions $f$ and $g$ such that given $a_{3}$, whenever we choose some $a_{2}\leq f(a_{3})$ and $a_{1}\leq g(a_{2})$, all calculations needed in our proof are valid.

2. Outline of the proof and preliminaries

Let $n$ be sufficiently large and $\tau n<s\leq n/3$. Suppose $H$ is a $3$-graph of order $n$ without isolated vertex and $\sigma_{2}(H)>2sn-\varepsilon n^{2}$. Let $U=\{u\in V(H):\deg(u)>sn-\frac{\varepsilon}{2}n^{2}\}$ and $W=V\setminus U$. Then any two vertices of $W$ are not adjacent – otherwise $\sigma_{2}(H)\leq 2sn-\varepsilon n^{2}$, a contradiction. If $|U|\leq 2s-1$, then $H$ is a subgraph of $H_{n,s}^{2}$ and we are done. We thus assume that $|U|\geq 2s$.

Throughout the proof we use small constants

[TABLE]

We first prove the following lemma, which is an extension of [26, Lemma 4].

Lemma 7**.**

Given $0<\varepsilon\ll\tau\ll 1$, let $n$ be sufficiently large and $\tau n<s\leq n/3$. Suppose $H$ is a $3$-graph of order $n$ without isolated vertex and $\sigma_{2}(H)>2sn-\varepsilon n^{2}$. Let $U=\{u\in V(H):\deg(u)>sn-\varepsilon n^{2}/2\}$ and $W=V\setminus U$. If $2s\leq|U|\leq 3s$, then $H$ contains a matching of size $3s-|U|$, each of which contains exactly one vertex of $W$.

Definition 8**.**

We call a matching $M$ optimal if (i) $M$ contains a submatching $M_{1}=\{e\in M:e\cap W\neq\emptyset\}$ of size at least $3s-|U|$; (ii) subject to (i), $|M|$ is as large as possible; (iii) subject to (i) and (ii), $|M_{1}|$ is as large as possible.

Lemma 7 shows that $H$ contains an optimal matching $M$. We separate the cases when $|M_{1}|=3s-|U|$ and when $|M_{1}|>3s-|U|$. When $|M_{1}|=3s-|U|$, we first consider the case when $s\leq n/3-\eta_{1}n$. If no vertex of $U_{3}:=U\setminus V(M)$ is adjacent to any vertex of $W_{2}:=W\setminus V(M)$, then the assumption $|M_{1}|=3s-|U|$ forces $\sum_{i=1}^{3}\deg(u_{i})$ to be smaller than $3sn-\frac{3}{2}\varepsilon n^{2}$ for any three vertices $u_{1},u_{2},u_{3}\in U_{3}$. If some vertex $u_{1}\in U_{3}$ is adjacent to $v_{1}\in W_{2}$, then the fact $v_{1}\in W$ reduces $\sum_{i=1}^{2}\deg(u_{i})+\deg(v_{1})$ to a number less than $3sn-\frac{3}{2}\varepsilon n^{2}$ (where $u_{2}$ is another vertex of $U_{3}$). When $s>n/3-\eta_{1}n$, we consider $H^{\prime}=H[V\setminus W_{2}]$. Since $|W_{2}|=n-3s$ is very small, we deduce that $\sigma_{2}(H^{\prime})$ is greater than $2sn-\eta_{2}n^{2}$. This allows us to apply the following theorem from [26] to obtain a perfect matching of $H^{\prime}$, which is also a matching of size $s$ of $H$.

Theorem 9**.**

[26]** There exist $\eta_{2}>0$ and $n_{0}\in\mathbb{N}$ such that the following holds for all integers $n\geq n_{0}$ that are divisible by $3$. Suppose that $H$ is a $3$-graph of order $n$ without isolated vertex and $\sigma_{2}(H)>2n^{2}/3-\eta_{2}n^{2}$, then either $H$ is a subgraph of $H^{2}_{n,n/3}$ or $H$ contains a perfect matching.

Now consider the case when $|M_{1}|>3s-|U|$. Let $W^{\prime}:=\{v\in W:\deg(v)\leq sn-s^{2}/2+\gamma^{\prime}n^{2}\}$. If $|W^{\prime}|$ is very small, then we can find a matching of size $s$ in $H[V\setminus W^{\prime}]$ by Theorem 1. When $|W^{\prime}|$ is not small, we consider $u_{1},u_{2},u_{3}\in U_{3}$. If one of $u_{1}$, $u_{2}$, $u_{3}$ is adjacent to one vertex from $W^{\prime}$, then $\sum_{i=1}^{3}\deg(u_{i})$ becomes much larger than $3sn$; otherwise we show that $\sum_{i=1}^{3}\deg(u_{i})<3sn-\frac{3}{2}\varepsilon n^{2}$ by proceeding with the cases when $|W^{\prime}\cap W_{1}|>\gamma n/2$ and when $|W^{\prime}\cap W_{2}|>\gamma n/2$ separately.

In the proof we need several (simple) extremal results on (hyper)graphs. Lemma 10 is Observation 1.8 of Aharoni and Howard [1]. Lemmas 11 and 12 are from [26]. A $k$-graph $H$ is called $k$-partite if $V(H)$ can be partitioned into $V_{1},\ldots,V_{k}$, such that each edge of $H$ meets every $V_{i}$ in precisely one vertex. If all parts are of the same size $n$, we call $H$ $n$-balanced.

Lemma 10**.**

[1*]**

Let $F$ be the edge set of an $n$-balanced $k$-partite $k$-graph. If $F$ does not contain $s$ disjoint edges, then $|F|\leq(s-1)n^{k-1}$.*

Lemma 11**.**

[26]** Let $G_{1},G_{2},G_{3}$ be three graphs on the same set $V$ of $n\geq 4$ vertices such that every edge of $G_{1}$ intersects every edge of $G_{i}$ for both $i=2,3$. Then $\sum_{i=1}^{3}\sum_{v\in A}\deg_{G_{i}}(v)\leq 6(n-1)$ for any set $A\subset V$ of size $3$.

Lemma 12**.**

[26]** Let $G_{1},G_{2},G_{3}$ be three graphs on the same set $V$ of $n\geq 5$ vertices such that for any $i\neq j$, every edge of $G_{i}$ intersects every edge of $G_{j}$. Then $\sum_{i=1}^{3}\sum_{v\in A}\deg_{G_{i}}(v)\leq 3(n+1)$ for any set $A\subset V$ of size $3$.

Following the same proof of Lemmas 11 and 12 from [26], we obtain another lemma and omit its proof.

Lemma 13**.**

Let $G_{1},\cdots,G_{k}$ be $k$ graphs on the same set $V$ of $n\geq 4$ vertices such that for any $1\leq i<j\leq k$, every edge of $G_{i}$ intersects every edge of $G_{j}$. Then $\sum_{i=1}^{k}\sum_{v\in A}\deg_{G_{i}}(v)\leq kn$ for any set $A\subset V$ of size $2$.∎

The following lemma needs slightly more work so we include a proof.

Lemma 14**.**

Given two disjoint vertex sets $A=\{u_{1},u_{2},\ldots,u_{a}\}$ and $B=\{v_{1},v_{2},\ldots,v_{b}\}$ with $a\geq 3$ and $b\geq 1$. Let $G_{i}$, $i=1,2,3$, be graphs on $A\cup B$ such that every vertex of $B$ is isolated vertex in $G_{1}$, and every edge of $G_{i}$ ($i=2,3$) contains at least one vertex of $A$. If there are no two disjoint edges (i) one from $G_{1}$ and the other from $G_{2}$ or $G_{3}$; or (ii) one from $G_{2}$ and the other from $G_{3}$, and at least one of them contains a vertex from $B$, then

[TABLE]

Proof.

For convenience, let $s_{i}=\sum_{j=1}^{2}\deg_{G_{i}}(u_{j})+\deg_{G_{i}}(v_{1})$ for $i=1,2,3$ and $y=s_{1}+s_{2}+s_{3}$. Below we show that $y\leq\max\{4a+7,3a+2b+5\}$.

We first observe that if $\deg_{G_{i}}(v_{1})\geq 3$ for some $i\in\{2,3\}$, then $E(G_{1})=\emptyset$ and $G_{i^{\prime}}$ is a star centered at $v_{1}$, where $i^{\prime}=5-i$. Indeed, if $G_{1}$ or $G_{i^{\prime}}$ contains an edge $e$ not incident to $v_{1}$, then $e$ is disjoint from some edge of $G_{i}$ that is incident to $v_{1}$ – this contradicts our assumption. The observation implies that if $\deg_{G_{i}}(v_{1})\geq 3$ for both $i=2,3$, then $E(G_{1})=\emptyset$ and both $G_{2}$ and $G_{3}$ are stars centered at $v_{1}$. In this case, $s_{i}\leq a+2$ for $i=2,3$ and thus $y\leq 2(a+2)$. If $\deg_{G_{2}}(v_{1})\geq 3$ and $\deg_{G_{3}}(v_{1})\leq 2$, then $E(G_{1})=\emptyset$ and $G_{3}$ consists of at most two edges incident to $v_{1}$. In this case, $y\leq 2(a+b-1)+a+4=3a+2b+2$. The case when $\deg_{G_{2}}(v_{1})\leq 2$ and $\deg_{G_{3}}(v_{1})\geq 3$ is analogous. We thus assume that

[TABLE]

for the rest of the proof.

Next, we observe that if $|N_{G_{i}}(u_{j})\cap B|\geq 2$ for some $i\in\{2,3\}$ and some $j\in\{1,2\}$, then $G_{i^{\prime}}$ is a star centered at $u_{j}$ for $i^{\prime}\in\{1,2,3\}\setminus\{i\}$. This is again due to our assumption on $G_{1},G_{2}$ and $G_{3}$. The observation implies that if $|N_{G_{i}}(u_{j})\cap B|\geq 2$ for both $j=1,2$, then $E(G_{i^{\prime}})\subseteq\{u_{1}u_{2}\}$ and consequently, $s_{i^{\prime}}\leq 2$ for $i^{\prime}\in\{1,2,3\}\setminus\{i\}$. By (2), we have $s_{i}\leq 2(a+b-1)+2$. Therefore, $y\leq 2(a+b-1)+2+4=2a+2b+4$. The observation also implies that if $|N_{G_{i}}(u_{j})\cap B|\geq 2$ for both $i=2,3$, then $G_{1},G_{2},G_{3}$ are all stars centered at $u_{j}$. In this case, $s_{1}\leq a$ and $s_{i}\leq a+b+1$ for $i=2,3$, which implies that $y\leq a+2(a+b+1)=3a+2b+2$. We now consider the case when $|N_{G_{2}}(u_{1})\cap B|\geq 2$, $|N_{G_{2}}(u_{2})\cap B|\leq 1$, and $|N_{G_{3}}(u_{1})\cap B|\leq 1$. Thus $G_{3}$ is a star (centered at $u_{1}$) of size at most $a$, which yields $s_{3}\leq a+2$. Now suppose $N_{G_{2}}(u_{2})\cap B\subseteq\{v_{p}\}$ for some $p$. Let $A^{\prime}:=A\cup\{v_{p}\}$ (note that $|A^{\prime}|=a+1\geq 4$). Since every edge of $G_{1}$ intersects every edge of $G_{2}$, we can apply Lemma 13 to $G_{1}[A^{\prime}]$ and $G_{2}[A^{\prime}]$ and obtain that $\sum_{i=1}^{2}\sum_{j=1}^{2}\deg_{G_{i}[A^{\prime}]}(u_{j})\leq 2a+2$. Since $|N_{G_{2}}(u_{1})\cap(B\setminus\{v_{p}\})|\leq b-1$ and $\deg_{G_{2}}(v_{1})\leq 2$, it follows that $s_{1}+s_{2}\leq 2a+2+b-1+2=2a+b+3$ and $y\leq 2a+b+3+a+2=3a+b+5$.

We thus assume that $|N_{G_{i}}(u_{j})\cap B|\leq 1$ for $i=2,3$ and $j=1,2$. Suppose $N_{G_{2}}(u_{2})\cap B\subseteq\{v_{p}\}$ for some $p$ and let $A^{\prime}:=A\cup\{v_{p}\}$. We apply Lemma 13 to $G_{1}[A^{\prime}]$ and $G_{2}[A^{\prime}]$ and obtain that $\sum_{i=1}^{2}\sum_{j=1}^{2}\deg_{G_{i}[A^{\prime}]}(u_{j})\leq 2a+2$. Since $|N_{G_{2}}(u_{1})\cap B|\leq 1$ and $\deg_{G_{2}}(v_{1})\leq 2$, it follows that $s_{1}+s_{2}\leq 2a+2+1+2$. On the other hand, we have $s_{3}\leq 2a+2$ because $\deg_{G_{3}}(u_{j})\leq a$ for $j=1,2$ and $\deg_{G_{3}}(v_{1})\leq 2$. Thus $y\leq 2a+5+2a+2=4a+7$. ∎

3. Proof of Lemma 7

The proof is similar to that of [26, Lemma 4]. Let $M$ be a largest matching of $H$ such that each edge of $M$ contains (exactly) one vertex of $W$. To the contrary, assume $|M|\leq 3s-|U|-1$. Let $U_{1}=V(M)\cap U$, $U_{2}=U\setminus U_{1}$, $W_{1}=V(M)\cap W$ and $W_{2}=W\setminus W_{1}$. Since $|U|\geq 2s$, we have $|U_{2}|=|U|-2|M|\geq 2$. Since $|W_{2}|=|W|-|M|$ and $|W|\geq 3s-|U|$, it follows that $W_{2}\neq\emptyset$.

Below is a sketch of the proof. We first assume $|U|<2s+\varepsilon^{\prime}n$. In this case every vertex in $U$ is adjacent to some vertex in $W$. If $|M|$ is not close to $s$, then we easily obtain a contradiction because $U_{2}$ is not small. When $|M|$ is close to $s$, we consider three vertices $u_{1}\neq u_{2}\in U_{2}$ and $v_{0}\in W_{2}$, and derive a contradiction on $\deg(u_{1})+\deg(u_{2})+\deg(v_{0})$. Next we assume $|U|\geq 2s+\varepsilon^{\prime}n$. In this case $U_{2}$ is not small. If no vertex of $W_{2}$ is adjacent to any vertex of $U_{2}$, then consider two adjacent vertices $v_{0}\in W_{2}$ and $u_{0}\in U_{1}$. We have $\deg(v_{0})\leq\binom{2|M|}{2}$, which eventually yields that $\deg(v_{0})+\deg(u_{0})<2sn-\varepsilon n^{2}$. Now assume $v_{0}\in W_{2}$ is adjacent to some vertex $u_{0}\in U_{2}$. In this case we define $M^{\prime}$ consisting of all $e\in M$ that contains a vertex $u^{\prime}\in U$ such that $|N(v_{0},u^{\prime})\cap U_{2}|\geq 3$. We show that if $|M^{\prime}|$ is small, then $\deg(v_{0})$ is small; otherwise $\deg(u_{0})$ is small. In either case we derive that $\deg(v_{0})+\deg(u_{0})<2sn-\varepsilon n^{2}$.

We now give the details of the proof.

**Case 1. ** $2s\leq|U|<2s+\varepsilon^{\prime}n$.

In this case we have the following two claims.

Claim 15**.**

$|M|\geq s-\varepsilon^{\prime\prime}n$.

Proof.

To the contrary, assume that $|M|<s-\varepsilon^{\prime\prime}n$. Fix $v_{0}\in W_{2}$. Then $\deg(v_{0})\leq\binom{|U|}{2}-\binom{|U_{2}|}{2}$ because there is no edge of type $U_{2}U_{2}W_{2}$. Since $v_{0}$ is not an isolated vertices, $v_{0}$ is adjacent to some vertex $u\in U$. Trivially $\deg(u)\leq\binom{|U|-1}{2}+(|U|-1)|W|$. Thus

[TABLE]

Since $|U|\geq 2s$ and $|M|<s-\varepsilon^{\prime\prime}n$, it follows that $|U_{2}|=|U|-2|M|>2\varepsilon^{\prime\prime}n$. As a result,

[TABLE]

which contradicts the condition that $\text{deg}(u)+\text{deg}(v_{0})>2sn-\varepsilon n^{2}$ because $\varepsilon\ll\varepsilon^{\prime}\ll\varepsilon^{\prime\prime}$. ∎

Claim 16**.**

Every vertex in $U$ is adjacent to one vertex in $W$.

Proof.

To the contrary, assume that $u\in U$ is not adjacent to any vertex in $W$. Then

[TABLE]

which contradicts the condition that $\text{deg}(u)>sn-\frac{1}{2}\varepsilon n^{2}$ because $\tau n<s\leq n/3$ and $\varepsilon\ll\varepsilon^{\prime}\ll\tau$. ∎

Fix $u_{1}\neq u_{2}\in U_{2}$ and $v_{0}\in W_{2}$. Trivially $\deg(w)\leq\binom{|U|}{2}$ for any vertex $w\in W$ and $\deg(u)\leq\binom{|U|-1}{2}+|W|(|U|-1)$ for any vertex $u\in U$. Furthermore, for any two distinct edges $e_{1},e_{2}\in M$, we observe that at least one triple of type $UUW$ with one vertex in $e_{1}$, one vertex in $e_{2}$ and one vertex in $\{u_{1},u_{2},v_{0}\}$ is not an edge by the choice of $M$. By Claim 15, $|M|\geq s-\varepsilon^{\prime\prime}n$. Thus,

[TABLE]

On the other hand, Claim 16 implies that $u_{i}$ is adjacent to some vertex in $W$ for $i=1,2$. We know that $v_{0}$ is adjacent to some vertex in $U$. Therefore, $\deg(u_{i})>\left(2sn-\varepsilon n^{2}\right)-\binom{|U|}{2}$ for $i=1,2$, and $\deg(v_{0})>\left(2sn-\varepsilon n^{2}\right)-\left(\binom{|U|-1}{2}+|W|(|U|-1)\right)$. It follows that

[TABLE]

The upper and lower bounds for $\deg(u_{1})+\deg(u_{2})+\deg(v_{0})$ together imply that

[TABLE]

which is impossible because $|U|<2s+\varepsilon^{\prime}n$, $\tau n<s\leq n/3$, and $\varepsilon\ll\varepsilon^{\prime}\ll\varepsilon^{\prime\prime}\ll\tau$.

**Case 2. ** $2s+\varepsilon^{\prime}n\leq|U|\leq 3s$.

We consider the following two subcases.

**Subcase 2.1. ** No vertex in $U_{2}$ is adjacent to any vertex in $W_{2}$.

Fix $v_{0}\in W_{2}$. Then $\deg(v_{0})\leq\binom{|U_{1}|}{2}=\binom{2|M|}{2}$. Since $v_{0}$ is not an isolated vertex, $v_{0}$ is adjacent to some vertex $u_{0}\in U_{1}$. We know that $\deg(u_{0})\leq\binom{|U|-1}{2}+(|U|-1)|W|-|U_{2}||W_{2}|$ because no vertex in $U_{2}$ is adjacent to any vertex in $W_{2}$. Since $|W|=n-|U|$, $|U_{2}|=|U|-2|M|$ and $|W_{2}|=n-|U|-|M|$, we derive that

[TABLE]

Since $|M|<3s-|U|$, it follows that

[TABLE]

Note that the quadratic function $\frac{3}{2}x^{2}-(3s+2n)x$ is minimized at $x=s+\frac{2}{3}n$. Since $2s+\varepsilon^{\prime}n\leq|U|\leq 3s\leq s+\frac{2}{3}n$, we derive that

[TABLE]

because $s\leq n/3$. Since $\varepsilon\ll\varepsilon^{\prime}$, this contradicts the assumption that $\sigma_{2}(H)>2sn-\varepsilon n$.

**Subcase 2.2. ** Two vertices $u_{0}\in U_{2}$ and $v_{0}\in W_{2}$ are adjacent.

Let $M^{\prime}=\{e\in M:\exists\,u^{\prime}\in e,|N(v_{0},u^{\prime})\cap U_{2}|\geq 3\}$. Assume $\{u_{1},u_{2},v_{1}\}\in M^{\prime}$ such that $u_{1},u_{2}\in U_{1}$, $v_{1}\in W_{1}$ and $|N(v_{0},u_{1})\cap U_{2}|\geq 3$. We claim that

[TABLE]

Indeed, if $\{u_{0},v_{1},u_{3}\}\in E(H)$ for some $u_{3}\in U_{2}$, then we can find $u_{4}\in U_{2}\setminus\{u_{0},u_{3}\}$ such that $\{v_{0},u_{1},u_{4}\}\in E(H)$. Replacing $\{u_{1},u_{2},v_{1}\}$ by $\{u_{0},v_{1},u_{3}\}$ and $\{v_{0},u_{1},u_{4}\}$ gives a larger matching than $M$, a contradiction.

By the definition of $M^{\prime}$, we have

[TABLE]

By (3), we have

[TABLE]

and consequently

[TABLE]

Since $|M^{\prime}|\leq|M|=|W_{1}|=\frac{|U_{1}|}{2}$, it follows that

[TABLE]

Since $|M|\leq 3s-|U|$ and $|U_{2}|=|U|-2|M|\geq 3|U|-6s$, we have

[TABLE]

Note that the quadratic function $-\frac{3}{2}x^{2}+(12s-2n)x$ is maximized at $x=4s-\frac{2}{3}n$. Since $3s\geq|U|\geq 2s+\varepsilon^{\prime}n\geq 4s-\frac{2}{3}n$, we have

[TABLE]

because $s\leq n/3$. Since $\varepsilon\ll\varepsilon^{\prime}$, this contradicts the assumption that $\sigma_{2}(H)>2sn-\varepsilon n$.

4. Proof of Theorem 6

Suppose $H$ is a $3$-graph of order $n$ without isolated vertex and $\sigma_{2}(H)>2sn-\varepsilon n^{2}$. Let $U=\{u\in V(H):\deg(u)>sn-\varepsilon n^{2}/2\}$ and $W=V\setminus U$. We know that no two vertices in $W$ are adjacent and $|U|\geq 2s$. Let $M$ be an optimal matching as in Definition 8. By Lemma 7, such $M$ exists. Let $M_{2}=M\setminus M_{1}$, $U_{1}=V(M_{1})\cap U$, $U_{2}=V(M_{2})$, $U_{3}=U\setminus V(M)$, $W_{1}=V(M_{1})\cap W$ and $W_{2}=W\setminus W_{1}$. Since $M$ is optimal, no edge of $H$ is of type $W_{2}U_{3}U_{3}$ or $W_{2}U_{2}U_{3}$. In addition, for any $e\in M_{1}$, there are no two disjoint edges $e_{1},e_{2}\in e\cup W_{2}\cup U_{3}$ such that $(e_{1}\cup e_{2})\cap W_{2}\neq\emptyset$.

Suppose to the contrary, that $|M|\leq s-1$. We know that $|U_{3}|=|U|+|M_{1}|-3|M|\geq 3+|M_{1}|-(3s-|U|)\geq 3$. Let $u_{1},u_{2},u_{3}\in U_{3}$. Since $u_{i}\in U$ for $i=1,2,3,$ we have

[TABLE]

On the other hand, if $u_{1}$ is adjacent to some $v_{1}\in W_{2}$, then

[TABLE]

Claim 17**.**

For any two distinct edges $e_{1}$, $e_{2}$ from $M$, we have $\sum_{i=1}^{3}|L_{u_{i}}(e_{1},e_{2})|\leq 18$ and $\sum_{i=1}^{2}|L_{u_{i}}(e_{1},e_{2})|$ $+|L_{v_{1}}(e_{1},e_{2})|\leq 18.$

Proof.

Let $H_{1}$ be the 3-partite subgraph of $H$ induced on three parts $e_{1}$, $e_{2}$, and $\{u_{1},u_{2},u_{3}\}$. We observe that $H_{1}$ does not contain a perfect matching by the choice of $M$. By Lemma 10, we obtain that $|E(H_{1})|=\sum_{i=1}^{3}|L_{u_{i}}(e_{1},e_{2})|\leq 18$. Similarly, we have $\sum_{i=1}^{2}|L_{u_{i}}(e_{1},e_{2})|+|L_{v_{1}}(e_{1},e_{2})|\leq 18.$ ∎

We proceed in two cases.

Case 1. $|M_{1}|=3s-|U|$. In this case, we have $|M_{2}|=|M|+|U|-3s$, $|U_{3}|=3s-3|M|$ and $|W_{2}|=n-3s$.

Claim 18**.**

For any $e\in M_{1}$, we have

(i) $\sum_{i=1}^{2}|L_{u_{i}}(e,U_{3}\cup W_{2})|+|L_{v_{1}}(e,U_{3}\cup W_{2})|\leq\max\{4|U_{3}|+7,3|U_{3}|+2|W_{2}|+5\}$,where $v_{1}\in W_{2}$;

(ii) $\sum_{i=1}^{3}|L_{u_{i}}(e,U_{3})|\leq 6|U_{3}|$.

Proof.

Assume $e=\{u_{1}^{\prime},u_{2}^{\prime},u_{3}^{\prime}\}\in M_{1}$ with $u_{1}^{\prime}\in W_{1}$ and $u_{2}^{\prime},u_{3}^{\prime}\in U_{1}$.

(i) Let $A=U_{3}$, $B=W_{2}$, and $E(G_{i})=L_{u_{i}^{\prime}}(U_{3}\cup W_{2})$ for $i=1,2,3$. By the choice of $M$, there are not two disjoint edges, one from $G_{1}$ and the other from $G_{2}$ or $G_{3}$; or one from $G_{2}$ and the other from $G_{3}$, and at least one of them contains one vertex from $B$. Furthermore, it is easy to see that

[TABLE]

The desired inequality thus follows from Lemma 14.

(ii) For $i=1,2,3$, let $G_{i}$ be the graph obtained from $L_{u_{i}^{\prime}}(U_{3})$ after adding an isolated vertex $u^{*}$. Then $|V(G_{i})|=|U_{3}|+1\geq 4$. By the choice of $M$, every edge of $G_{1}$ intersects every edge of $G_{2}$ and $G_{3}$. The desired inequality thus follows from Lemma 11. ∎

Claim 19**.**

For any $e\in M_{2}$, we have

(i) $\sum_{i=1}^{3}|L_{u_{i}}(e,U_{3})|\leq 3(|U_{3}|+3)$;

(ii) $\sum_{i=1}^{2}|L_{u_{i}}(e,U_{3})|\leq 3(|U_{3}|+1)$.

Proof.

Assume $e=\{u_{1}^{\prime},u_{2}^{\prime},u_{3}^{\prime}\}\in M_{2}$ with $u_{1}^{\prime},u_{2}^{\prime},u_{3}^{\prime}\in U_{2}$.

(i) For $i=1,2,3$, let $G_{i}$ be the graph obtained from $L_{u_{i}^{\prime}}(U_{3})$ after adding two isolated vertices $u^{\prime}$ and $u^{\prime\prime}$. Then $|V(G_{i})|=|U_{3}|+2\geq 5$. Since $M$ is optimal, the desired inequality follows from Lemma 12.

(ii) For $i=1,2,3$, let $G_{i}$ be the graph obtained from $L_{u_{i}^{\prime}}(U_{3})$ after adding an isolated vertex $u^{*}$. Then $|V(G_{i})|=|U_{3}|+1\geq 4$. Since $M$ is optimal, the desired inequality follows from Lemma 13. ∎

Claim 20**.**

$s>n/3-\eta_{1}n$.

Proof.

Suppose $s\leq n/3-\eta_{1}n$. We first consider the case that $u_{1},u_{2},u_{3}$ are not adjacent to any vertex of $W_{2}$.

Following Claim 17, we have

[TABLE]

Furthermore, by Claims 18 (ii) and 19 (i), we obtain that

[TABLE]

Since $|M|\leq s-1$, it follows that

[TABLE]

Since $\tau n<s\leq n/3-\eta_{1}n$ and $\eta_{1}<\tau$, we know that

[TABLE]

Consequently, $\sum_{i=1}^{3}\deg(u_{i})<9s^{2}\leq 3sn-3\eta_{1}n(n-3\eta_{1}n)$. Since $\varepsilon\ll\eta_{1}$, this contradicts (4).

Now we assume, without loss of generality, that $u_{1}$ is adjacent to $v_{1}$. The choice of $M$ implies that $L_{v}(e,U_{3})=L_{u}(e,W_{2})=\emptyset$ for any $v\in W_{2}$, $u\in U_{3}$ and $e\in M_{2}$. By Claim 17, we have

[TABLE]

We know that $4|U_{3}|+7\geq 3|U_{3}|+2|W_{2}|+5$ if and only if $|U_{3}|\geq 2|W_{2}|-2$. If $|U_{3}|\geq 2|W_{2}|-2$, then by (8), Claim 18 (i) and Claim 19 (ii), we have

[TABLE]

Since $|M|\leq s-1$ and $|U|\geq 2s$, it follows that

[TABLE]

Following (7), we have $\sum_{i=1}^{2}\deg(u_{i})+\deg(v_{1})<3sn-3\eta_{1}n(n-3\eta_{1}n)+n/3-3$. Since $\varepsilon\ll\eta_{1}$ and $n$ is sufficiently large, this contradicts (5).

If $|U_{3}|<2|W_{2}|-2$, by (8), Claim 18 (i) and Claim 19 (ii), we have

[TABLE]

Since $|M|\leq s-1$ and $|U|\geq 2s$, it follows that

[TABLE]

Applying (7), we have $\sum_{i=1}^{2}\deg(u_{i})+\deg(v_{1})<3sn-\eta_{1}n(n-3\eta_{1}n)$, which contradicts (5) because $\varepsilon\ll\eta_{1}$. ∎

By Claim 20, we have $|W_{2}|=n-3s<3\eta_{1}n$. Let $H^{\prime}=H[V\setminus W_{2}]$. We claim that $\sigma_{2}(H^{\prime})>2n^{2}/3-\eta_{2}n^{2}$. Indeed, recall that $\text{deg}_{H}(u)+\text{deg}_{H}(v)\geq 2n^{2}/3-\varepsilon n^{2}$ for any two adjacent vertices $u$ and $v$ of $H^{\prime}$. Since $|W_{2}|<3\eta_{1}n$ and $\varepsilon\ll\eta_{1}\ll\eta_{2}$, it follows that

[TABLE]

Since $\eta_{2}\ll 1$, we may apply Theorem 9 and conclude that either $H^{\prime}$ is a subgraph of $H_{3s,s}^{2}$ or $H^{\prime}$ contains a perfect matching. In the former case, there is a partition of $V(H^{\prime})$ into two sets $|T|=2s-1$ and $|S|=s+1$ such that for every vertex $u\in S$,

[TABLE]

On the other hand, since $U\subseteq V(H^{\prime})$ and $|U|\geq 2s$, there exists a vertex $u\in U\cap S$ such that

[TABLE]

which is a contradiction. Therefore $H^{\prime}$ must contain a perfect matching, which is a matching of size $s$ in $H$.

Case 2. $|M_{1}|>3s-|U|$.

The difference from Case 1 is that, for any edge $e\in M$, we cannot find two disjoint edges $e_{1},e_{2}$ from $e\cup U_{3}\cup W_{2}$ – otherwise we can replace $M$ by $M\setminus\{e\}\cup\{e_{1},e_{2}\}$ contradicting the assumption that $M$ is an optimal matching.

Note that $|U_{3}|=|U|+|M_{1}|-3|M|\geq 3s+1-3|M|\geq 4$.

Claim 21**.**

For any $e\in M$, $\sum_{i=1}^{3}|L_{u_{i}}(e,U_{3}\cup W_{2})|\leq 3(|U_{3}|+|W_{2}|+2)$.

Proof.

Assume $e=\{u_{1}^{\prime},u_{2}^{\prime},u_{3}^{\prime}\}\in M$. For $i=1,2,3$, let $G_{i}$ be the graph obtained from $L_{u_{i}^{\prime}}(U_{3}\cup W_{2})$ after adding an isolated vertex $u^{*}$. Then $|V(G_{i})|=|U_{3}|+|W_{2}|+1\geq 5$. Since $H$ contains no two disjoint edges $e_{1},e_{2}$ from $e\cup U_{3}\cup W_{2}$, we know that for any $i\neq j$, every edge of $G_{i}$ intersects every edge of $G_{j}$. The desired inequality thus follows from Lemma 12. ∎

By Claims 17 and 21, we obtain that

[TABLE]

Let $W^{\prime}=\{v\in W:\deg(v)\leq sn-s^{2}/2+\gamma^{\prime}n^{2}\}$. If $|W^{\prime}|\leq\gamma n$, then we let $H^{\prime}:=H[V\setminus W^{\prime}]$. By the definition of $W^{\prime}$, $\deg_{H}(u)>sn-s^{2}/2+\gamma^{\prime}n^{2}$ for every $u\in V(H^{\prime})\cap W$. For any $u\in V(H^{\prime})\cap U$, $\deg_{H}(u)>sn-\varepsilon n^{2}/2>sn-s^{2}/2+\gamma^{\prime}n^{2}$ because $s>\tau n$ and $\varepsilon\ll\gamma^{\prime}\ll\tau$. Therefore every vertex $u\in V(H^{\prime})$ satisfies

[TABLE]

because $|W^{\prime}|\leq\gamma n$, $\gamma\ll\gamma^{\prime}$, and $n$ is sufficiently large. By Theorem 1, $H^{\prime}$ contains a matching of size $s$.

We thus assume that $|W^{\prime}|>\gamma n$ for the rest of the proof. If one of $u_{1},u_{2},u_{3}$ is adjacent to a vertex of $W^{\prime}$, then

[TABLE]

which contradicts (4) because $s>\tau n$ is sufficiently large and $\varepsilon\ll\gamma^{\prime}\ll\tau$.

If none of $u_{1},u_{2},u_{3}$ is adjacent to a vertex of $W^{\prime}$, then we distinguish the following two subcases.

Subcase 2.1. $|W^{\prime}\cap W_{1}|>\gamma n/2$.

Let $M^{\prime}=\{e\in M:e\cap W^{\prime}\neq\emptyset\}$, thus $|M^{\prime}|>\gamma n/2$. Since $u_{1},u_{2},u_{3}$ are not adjacent to any vertex in $W^{\prime}\cap W_{1}$, then for any distinct $e_{1}$, $e_{2}$ from $M^{\prime}$, we have

[TABLE]

By Claims 17, 21 and (10), we have

[TABLE]

Since $|M^{\prime}|>\gamma n/2$, it follows that

[TABLE]

which contradicts (4) because $s\leq n/3$ and $\varepsilon\ll\gamma$.

Subcase 2.2. $|W^{\prime}\cap W_{1}|\leq\gamma n/2$.

Since $|W^{\prime}|>\gamma n$, we have $|W^{\prime}\cap W_{2}|>\gamma n/2$. Let $W_{2}^{*}=W_{2}\setminus W^{\prime}$. Then $W_{2}\setminus W_{2}^{*}=W^{\prime}\cap W_{2}$. By Claim 21, we obtain that $\sum_{i=1}^{3}|L_{u_{i}}(V(M),U_{3}\cup W_{2}^{*})|\leq 3|M|\left(|U_{3}|+|W_{2}^{*}|+2\right)$. Therefore,

[TABLE]

which contradicts (4) because $|M|\leq s$, $\tau n<s$, and $\varepsilon\ll\gamma\ll\tau$. This completes the proof of Theorem 6.