Lagrangian densities of short 3-uniform linear paths and Tur\'an numbers of their extensions

TL;DR

This paper investigates the Lagrangian densities and Turán numbers of specific 3-uniform hypergraphs, proving certain paths are perfect and confirming a conjecture about their properties, with implications for extremal hypergraph theory.

Contribution

It establishes that the linear 3-uniform paths of lengths 3 and 4 are perfect, supporting a broader conjecture about all such paths, and determines Turán numbers for their extensions.

Findings

P_3 and P_4 are perfect hypergraphs.

Supports the conjecture that all 3-uniform linear paths are perfect.

Determines Turán numbers for extensions of these paths.

Abstract

For a fixed positive integer $n$ and an $r$-uniform hypergraph $H$, the Tur\'an number $ex(n,H)$ is the maximum number of edges in an $H$-free $r$-uniform hypergraph on $n$ vertices, and the Lagrangian density of $H$ is defined as $\pi_{\lambda}(H)=\sup \{r! \lambda(G) : G \;\text{is an}\; H\text{-free} \;r\text{-uniform hypergraph}\}$, where $\lambda(G)$ is the Lagrangian of $G$. For an $r$-uniform hypergraph $H$ on $t$ vertices, it is clear that $\pi_{\lambda}(H)\ge r!\lambda{(K_{t-1}^r)}$. We say that an $r$-uniform hypergraph $H$ on $t$ vertices is perfect if $\pi_{\lambda}(H)= r!\lambda{(K_{t-1}^r)}$. Let $P_t=\{e_1, e_2, \dots, e_t\}$ be the linear $3$-uniform path of length $t$, that is, $|e_i|=3$, $|e_i \cap e_{i+1}|=1$ and $e_i \cap e_j=\emptyset$ if $|i-j|\ge 2$. We show that $P_3$ and $P_4$ are perfect, this supports a conjecture in \cite{yanpeng} proposing that all $3$-uniform linear hypergraphs are perfect. Applying the results on Lagrangian densities, we determine the Tur\'an numbers of their extensions.

Equations

$$\lambda (G,\vec{x})=\sum _{e\in G}\prod _{i\in e}{x}_i.$$

$${\Delta }_n=\{\vec{x}=(x_1,x_2,\dots ,x_n)\in [0,\infty {)}^n:x_1+x_2+\cdots +x_n=1\}.$$

$$\lambda (G)=\max \{\lambda (G,\vec{x}):\vec{x}\in {\Delta }_n\}.$$

$$\lambda (G)=\lambda (K_t^2)=\frac{1}{2}(1-\frac{1}{t}).$$

$${\pi }_{\lambda }(F)=\sup \{r!\lambda (G):G\text{is an}F\text{-free}r\text{-graph}\}.$$

$$L_G(j\setminus i)=\left\{f\in \left(\genfrac{}{}{0ex}{}{V(G)\setminus \{i,j\}}{r-1}\right):f\cup \{j\}\in E(G)\mathrm{and}f\cup \{i\}\notin E(G)\right\},$$

$${\pi }_{ij}(G)=\left(E(G)\setminus \{f\cup \{j\}:f\in L_G(j\setminus i)\}\right)\cup \{f\cup \{i\}:f\in L_G(j\setminus i)\}.$$

$$\frac{\partial \lambda (G,\vec{x})}{\partial x_i}=\sum _{i\in e\in E(G)}\prod _{j\in e\setminus \{i\}}{x}_j.$$

$$\frac{\partial \lambda (G,\vec{x})}{\partial x_i}=r\lambda (G)$$

$$\lambda (G,\vec{y})-\lambda (G,\vec{x})=\sum _{\{i,j\}\subseteq e\in G}\left[\frac{(x_i+x_j{)}^2}{4}-x_i{x}_j\right]\prod _{k\in e\setminus \{i,j\}}{x}_k\ge 0.$$

$$\sum _{i=1}^n\frac{\partial \lambda (\mathcal{F},\vec{x})}{\partial x_i}\le \sum _{1\le i<j\le n}{x}_i{x}_j\le \left(\genfrac{}{}{0ex}{}{n}{2}\right)\frac{1}{n^2}.$$

$${\pi }_{\lambda }(F)=3!\lambda (K_{t-1}^3).$$

$${\pi }_{\lambda }(P_t)=3!\lambda (K_{2t}^3).$$

$$\lambda (G)=\lambda (G,\vec{x})=\lambda (L(1),\vec{x})+\lambda (G[[2,n]],\overrightarrow{x^{\prime }}),$$

$$\lambda (G)=\lambda (L(1),\vec{x})+\lambda (G[[2,n]],\vec{x})\le \frac{1}{2}{x}_1(1-x_1{)}^2\le \frac{1}{4}\left(\frac{2x_1+1-x_1+1-x_1}{3}\right)=\frac{1}{4}{\left(\frac{2}{3}\right)}^3=\frac{2}{27}.$$

$$ϵ(3)=\min \left\{\frac{5}{54}-0.0877,\frac{5}{54}-\frac{2}{27}\right\}>0.0048$$

$$\lambda (G)$$

$$ϵ(4)=\min \left\{\frac{7}{64}-0.1077,\frac{7}{64}-\frac{1250}{11907}\right\}\ge 0.0016$$

$$\lambda (\mathcal{F})\le \lambda (\mathcal{G})\le 0.1035<7/64=\lambda (K_8^3),$$

$$|G^{\ast }|\le \lambda (G^{\ast }[S])n^r\le \left(\frac{1}{r!}\frac{[m{]}_r}{m^r}-ϵ\right)n^r<\left(\frac{[m{]}_r}{m^r}-ϵ\right)\frac{n^r}{r!}.$$

$$|G^{\ast }|>|G|-\gamma n\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)\ge \left(\frac{[m{]}_r}{m^r}-\delta -r\gamma \right)\left(\genfrac{}{}{0ex}{}{n}{r}\right)>\left(\frac{[m{]}_r}{m^r}-ϵ\right)\frac{n^r}{r!},$$

Full text

Key Words: Lagrangian of hypegraphs, Turán number

1 Introduction

For a positive integer $n$, let $[n]$ denote $\{1,2,\ldots,n\}$. An $r$-uniform hypergraph or $r$-graph $G$ consists of a set $V(G)$ of vertices and a set $E(G)\subseteq V(G)^{(r)}$ of edges. A $2$-graph is called a simple graph. We write $G$ for $E(G)$ sometimes. An edge $e=\{a_{1},a_{2},\ldots,a_{r}\}$ will be simply denoted by $a_{1}a_{2}\ldots a_{r}$. An $r$-graph $F$ is a subgraph of an $r$-graph $G$, denoted by $F\subseteq G$, if $V(F)\subseteq V(G)$ and $E(F)\subseteq E(G)$. Given an $r$-graph $G$ and $U\subseteq V(G)$, the induced subgraph $G[U]$ is the $r$-graph with vertex set $U$ and edge set $\{e\in G:e\subseteq U\}$. Let $K^{r}_{t}$ denote the complete $r$-graph on $t$ vertices, and $K^{r-}_{t}$ be removing one edge from $K^{r}_{t}$. A hypergprah $H$ covers pairs if every pair of vertices is contained in some edge of $H$. The extension of an $r$-graph $F$, denoted by $H^{F}$, is defined as follows. For each pair of vertices $v_{i},v_{j}\in V(F)$ not covered in $F$, we add a set $B_{ij}$ of $r-2$ new vertices and the edge $\{v_{i},v_{j}\}\cup B_{ij}$, where all $B_{ij}$ are pairwise disjoint over all such pairs $\{i,j\}$.

Given an $r$-graph $F$, an $r$-graph $G$ is called $F$-free if it does not contain a copy of $F$ as a subgraph. For a fixed positive integer $n$ and an $r$-graph $F$, the Turán number of $F$, denoted by $ex(n,F)$, is the maximum number of edges in an $F$-free $r$-graph on $n$ vertices. Determining the value $ex(n,F)$ for a general $r$-graph $F$ is a challenging problem in extremal combinatorics. For simple graphs, Erdős, Stone and Simonovits determined the asymptotic value of Turán numbers of all graphs except bipartite graphs. Very few results are known for hypergraphs and a survey on this topic can be found in Keevash’s survey paper [11]. Lagrangian method has been a helpful tool for hypergraph Turán problem. We now proceed to define the Lagrangian of an $r$-graph.

Definition 1.1

Let $G$ be an $r$-graph on $[n]$ and let $\vec{x}=(x_{1},\ldots,x_{n})\in[0,\infty)^{n}$, define

[TABLE]

Denote

[TABLE]

The Lagrangian of $G$, denoted by $\lambda(G)$, is defined as

[TABLE]

The value $x_{i}$ is called the weight of the vertex $i$ and a vector $\vec{x}\in{\Delta_{n}}$ is called a feasible weight vector on $G$. A vector $\vec{y}\in{\Delta_{n}}$ is called an optimum weight vector on $G$ if $\lambda(G,\vec{y})=\lambda(G)$.

In [13], Motzkin and Straus established a connection between the Lagrangian of any given $2$-graph and it’s maximum complete subgraphs.

Theorem 1.2

([13])* If $G$ is a $2$-graph in which a maximum complete subgraph has $t$ vertices, then*

[TABLE]

Given an $r$-graph $F$, the *Lagrangian density * $\pi_{\lambda}(F)$ of $F$ is defined as

[TABLE]

The Lagrangian density of an $r$-graph is closely related to its Turán density.

Proposition 1.3

([18, 17])* Let $F$ be an $r$-graph. Then

$(i)$ $\pi(F)\leq\pi_{\lambda}(F);$

$(ii)$ $\pi(H^{F})=\pi_{\lambda}(F).$ In particular, if $F$ covers pairs, then $\pi(F)=\pi_{\lambda}(F).$*

Earlier applications of Lagrangians of hypergraphs include that Frankl and Rödl [5] applied it in disproving the long standing jumping constant conjecture of Erdős. Sidorenko [19], and Frankl and Füredi [4] applied Lagrangians of hypergraphs in finding Turán densities of hypergraphs, generalizing work of Motzkin and Straus [13], and Zykov [25]. More recent developments of the method were obtained by Pikhurko [17] and in the papers [7, 14, 2, 15, 9]. In addition to its applications, it is interesting in its own right to determine the maximum Lagrangian of $r$-graphs with certain properties. For example, a challenging conjecture of Frankl and Füredi [4] considers the question of determining the maximum Lagrangian among all $r$-graphs with the fixed number of edges. Talbot [20] made some breakthrough in confirming this conjecture for some cases. Subsequent progress in this conjecture were made in the papers of Tang, Peng, Zhang and Zhao [21], Tyomkyn [22], and Lei, Lu and Peng [12]. Recently, Gruslys, Letzter and Morrison [6] confirmed this conjecture for $r=3$ and the number of edges is sufficiently large. In this paper, we focus on the Lagrangian density of an $r$-graph $F$.

For an $r$-graph $H$ on $t$ vertices, it is clear that $\pi_{\lambda}(H)\geq r!\lambda{(K_{t-1}^{r})}$. Let us say that an $r$-graph $H$ on $t$ vertices is perfect if $\pi_{\lambda}(H)=r!\lambda{(K_{t-1}^{r})}$. Theorem 1.2 implies that all $2$-graphs are perfect. It is interesting to explore what kind of hypergraphs are perfect. Sidorenko [19] showed that the $(r-2)$-fold enlargement of a tree with order greater than some number $A_{r}$ is perfect. Hefetz and Keevash [7] showed that a $3$-uniform matching of size 2 is perfect. Jiang, Peng and Wu [10] verified that any $3$-uniform matching is perfect. Pikhurko [17], and Norin and Yepremyan [15] showed that an $r$-uniform tight path of length 2 is perfect for $r=4$ and $r=5$ or $6$ respectively. Jenssen [9] showed that a path of length 2 formed by two edges intersecting at $r-2$ vertices is perfect for $r=3,4,5,6,7$. An $r$-graph is linear if any two edges have at most 1 vertex in common. Hu, Peng and Wu [8], and Chen, Liang and Peng [3] showed that the disjoint union of a $3$-uniform linear path of length $2$ or $3$ and a $3$-uniform matching, and the disjoint union of a $3$-uniform tight path of length $2$ and a $3$-uniform matching are perfect. Yan and Peng [24] showed that the $3$-uniform linear cycle of length 3 ({123, 345, 561}) is perfect, and $F_{5}$ ({123, 124, 345}) is not perfect (by determining its Lagrangian density). Bene Watts, Norin and Yepremyan [1] showed that an $r$-uniform matching of size 2 is not perfect for $r\geq 4$ confirming a conjecture of Hefetz and Keevash [7]. Wu, Peng and Chen [23] showed the same result for $r=4$ independently. Though an $r$-uniform matching of size 2 is not perfect for $r\geq 4$, we think that an $r$-uniform matching with large enough size is perfect. Yan and Peng proposed the following conjecture in [24].

Conjecture 1.4

([24])* For $r\geq 3$, there exists $n$ such that a linear $r$-graph with at least $n$ vertices is perfect.*

A natural and interesting question is whether a linear hyperpath perfect? Let $P_{t}=\{e_{1},e_{2},\dots,e_{t}\}$ be the $3$-uniform linear path of length $t$, that is, $|e_{i}|=3$, $|e_{i}\cap e_{i+1}|=1$ and $e_{i}\cap e_{j}=\emptyset$ if $|i-j|\geq 2$. We show that $P_{3}$ and $P_{4}$ are perfect in this paper. In the joint work with Jiang [10], we applied the fact that left-compressing an $M^{r}_{t}$-free $r$-graph yields an $M^{r}_{t}$-free $r$-graph, where $M^{r}_{t}$ is an $r$-uniform matching of size $t$. In general, left-compressing a $P_{t}$-free $3$-graph may not result in a $P_{t}$-free $3$-graph. However, we manage to prove that left-compressing a dense $P_{t}$-free $3$-graph will result in $P_{t}$-free $3$-graph for $t=3$ or $4$ by structural analysis, and determine the Lagrangian density of $P_{3}$, and $P_{4}$.

In the next section, we give some useful properties of the Lagrangian function. In Section 3, we prove that left-compressing a $P_{3}$-free 3-graph ($P_{4}$-free 3-graph) that covers pairs results in a $P_{3}$-free ($P_{4}$-free) $3$-graph, and show that $P_{t}$ is perfect for $t=3$ or $4$. In Section 4, we give the Turán numbers of their extensions by using a similar stability argument for lager enough $n$ as in [17] and several other papers.

2 Some properties of the Lagrangian function

In this section, we develop some useful properties of Lagrangian functions. The following fact follows immediately from the definition of the Lagrangian.

Fact 2.1

Let $F$, $G$ be $r$-graphs and $F\subseteq G$. Then $\lambda(F)\leq\lambda(G).$

Given an $r$-graph $G$ and a vertex $i\in V(G)$, the link of $i$ in $G$, denoted by $L_{G}(i)$, is the $(r-1)$-graph with edge set $\left\{e\in{V(G)\setminus\{i\}\choose r-1}:e\cup\{i\}\in E(G)\right\}$. We will drop the subscript $G$ when there is no confusion. Given $i,j\in V(G)$, define

[TABLE]

and define the compression of $j$ to $i$ as

[TABLE]

We say $G$ on vertex set $[n]$ is left-compressed if for every $i,j$, $1\leq i<j\leq n$, $L_{G}(j\setminus i)=\emptyset$. By the definition of $\pi_{ij}(G)$, it’s straightforward to verify the following fact.

Fact 2.2

Let $G$ be an $r$-graph on the vertex set $[n]$. Let $\vec{x}=(x_{1},x_{2},\dots,x_{n})$ be a feasible weight vector on $G$. If $x_{i}\geq x_{j}$, then $\lambda(\pi_{ij}(G),\vec{x})\geq\lambda(G,\vec{x})$.

An $r$-graph $G$ is dense if $\lambda(G^{\prime})<\lambda(G)$ for every subgraph $G^{\prime}$ of $G$ with $|V(G^{\prime})|<|V(G)|$. This is equivalent to that no coordinate in all optimum weight vector is zero.

Fact 2.3

([5])* If $G$ is a dense $r$-graph then $G$ covers pairs.*

Let $G$ be an $r$-graph on $[n]$ and $\vec{x}=(x_{1},x_{2},\dots,x_{n})$ be a weight vector on $G$. If we view $\lambda(G,\vec{x})$ as a function in variables $x_{1},\dots,x_{n}$, then

[TABLE]

Lemma 2.4

([5])* Let $G$ be an $r$-graph on $[n]$. Let $\vec{x}=(x_{1},x_{2},\dots,x_{n})$ be an optimum weight vector on $G$. Then*

[TABLE]

for every $i\in[n]$ with $x_{i}>0$.

Lemma 2.5

Let $G$ be an $r$-graph on $[n]$. Let $\vec{x}=(x_{1},x_{2},\dots,x_{n})$ be a feasible weight vector on $G$. Let $i,j\in[n]$, where $i\neq j$. Suppose that $L_{G}(i\setminus j)=L_{G}(j\setminus i)=\emptyset$. Let $\vec{y}=(y_{1},y_{2},\dots,y_{n})$ be defined by letting $y_{\ell}=x_{\ell}$ for every $\ell\in[n]\setminus\{i,j\}$ and letting $y_{i}=y_{j}=(x_{i}+x_{j})/2$. Then $\lambda(G,\vec{y})\geq\lambda(G,\vec{x})$. Furthermore, if the pair $\{i,j\}$ is contained in some edge of $G$ and $\lambda(G,\vec{y})=\lambda(G,\vec{x})$, then $x_{i}=x_{j}$.

Proof.

Since $L_{G}(i\setminus j)=L_{G}(j\setminus i)=\emptyset$, we have

[TABLE]

If the pair $\{i,j\}$ is contained in some edge of $G$, then equality holds only if $x_{i}=x_{j}$.

The following facts are consequences of Lemma 2.5.

Fact 2.6

$\lambda(K_{t}^{r})={t\choose r}{1\over t^{r}}.$**

Recall that $K^{r-}_{t}$ is the $r$-graph by removing one edge from $K^{r}_{t}$.

Fact 2.7

$\lambda(K_{4}^{3-})={4\over 81}<0.0494.$**

Proof.

Let $\vec{x}=(x_{1},x_{2},x_{3},x_{4})$ be an optimum weighting of $K_{4}^{3-}$. By Lemma 2.5, we can assume that $x_{1}=a$ and $x_{2}=x_{3}=x_{4}=b$. So $a+3b=1$. Then $\lambda(K_{4}^{3-})=3ab^{2}=3(1-3b)b^{2}\leq{4\over 3}\left(1-3b+1.5b+1.5b\over 2\right)^{3}={4\over 81}<0.0494.$

Fact 2.8

$\lambda(K_{6}^{3-})<0.0887.$**

Proof.

Let $\vec{x}=(x_{1},\dots,x_{6})$ be an optimum weighting of $K_{6}^{3-}$. By Lemma 2.5, we can assume that $x_{1}=x_{2}=x_{3}=a$ and $x_{4}=x_{5}=x_{6}=b$. So $3a+3b=1$. Then $\lambda(K_{6}^{3-})=a^{3}+9a^{2}b+9ab^{2}=a^{3}+3a^{2}(1-3a)+a(1-3a)^{2}=a^{3}-3a^{2}+a.$ Let $f(a)=a^{3}-3a^{2}+a$, we have $f^{\prime}(a)=3a^{2}-6a+1$ and $f^{\prime\prime}(a)=6a-6$. Since $f^{\prime\prime}(a)<0$, $f(a)$ reaches the maximaum on interval $[0,1/3]$ at $a$ satisfying $f^{\prime}(a)=0$. Then direct calculation $f(a)\leq f({3-\sqrt{6}\over 3})<0.0887.$

Fact 2.9

$\lambda(K_{8}^{3-})<0.1077.$**

Proof.

Let $\vec{x}=(x_{1},\dots,x_{8})$ be an optimum weighting of $K_{8}^{3-}$. By Lemma 2.5, we can assume that $x_{1}=\dots=x_{5}=a$ and $x_{6}=x_{7}=x_{8}=b$. So $5a+3b=1$. Then $\lambda(K_{8}^{3-})=10a^{3}+30a^{2}b+15ab^{2}=(5a^{3}-20a^{2}+5a)/3.$

Let $f(a)=(5a^{3}-20a^{2}+5a)/3$. Then $f^{\prime}(a)=(15a^{2}-40a+5)/3$ and $f^{\prime}(a)=0$ implies that $a=(4\pm\sqrt{13})/3$. It’s easy to see that $\max_{0<a<1}f(a)=f((4-\sqrt{13})/3)<0.1077$.

The following result in [16] is useful for determining the Lagrangian of some hypergraph containing a large clique.

Theorem 2.10

( [16])* Let $m$ and $l$ be positive integers satisfying ${l-1\choose 3}\leq m\leq{l-1\choose 3}+{l-2\choose 2}$. Let $G$ be a $3$-graph with $m$ edges and $G$ contains a complete subgraph of order $l-1$. Then $\lambda(G)=\lambda([l-1]^{(3)})$.*

3 The Lagrangian densities of $P_{3}$ and $P_{4}$

We first study a property of dense $3$-graphs.

Lemma 3.1

Let $i=1$ or $2$. Let $\mathcal{F}$ be a dense $3$-graph with $n\geq 6-i$ vertices. Then there are $e_{1},e_{2}\in\mathcal{F}$ such that $|e_{1}\cap e_{2}|=i$.

Proof.

By Fact 2.3, $\mathcal{F}$ being dense implies that $\mathcal{F}$ covers pairs. For $i=1$, since $n\geq 5$ and $\mathcal{F}$ covers pairs, it is easy to see that $\mathcal{F}$ has at least two edges. Let $e,f\in\mathcal{F}$. Let $a\in e\setminus f$ and $b\in f\setminus e$. Since $\mathcal{F}$ covers pairs, there exists one an edge $g\in\mathcal{F}$ such that $a,b\in g$. Hence either $|e\cap g|=1$ or $|f\cap g|=1$ (or both).

For $i=2$. Let $\vec{x}=(x_{1},x_{2},\dots,x_{n})$ be an optimum weighting of $\mathcal{F}$. Note that $x_{i}>0$ for all $i\in[n]$. Suppose for any edge pair $e_{1},e_{2}\in\mathcal{F}$, $|e_{1}\cap e_{2}|=0$ or $1$. This implies that for every pair $i,j\in V(\mathcal{F})$, $\{i,j\}$ is contained in at most one edge of $\mathcal{F}$. Then for every pair $i,j\in V(\mathcal{F})$, $x_{i}x_{j}$ appears in at most one of $\frac{\partial\lambda(\mathcal{F},\vec{x})}{\partial x_{i}}$, $i\in[n]$. Hence

[TABLE]

By Fact 2.4, $\frac{\partial\lambda(\mathcal{F},\vec{x})}{\partial x_{i}}=3\lambda(\mathcal{F})$ for every $i\in[n]$. Hence $3n\lambda(\mathcal{F})\leq{n\choose 2}{1\over n^{2}}$, so $\lambda(\mathcal{F})\leq{n-1\over 6n^{2}}\leq{1\over 32}$. Note that $\mathcal{F}$ contains at least one edge, so by Fact 2.1, $\lambda(\mathcal{F})\geq{1\over 27}>{1\over 32}$, a contradiction.

Denote $T_{2}=\{123,124\}$ and $F_{5}=\{123,124,345\}$. Note that $F_{5}=H^{T_{2}}$ and the following result in [19, 2] is a consequence of the above.

Lemma 3.2

([19, 2])* Let $F\in\{P_{2},T_{2}\}$ and $t=|V(F)|$. Then*

[TABLE]

Furthermore, for any $F$-free and $K_{t-1}^{3}$-free $3$-graph $G$, there exists $\epsilon=\epsilon(t)>0$ such that $\lambda(G)\leq\lambda(K_{t-1}^{3})-\epsilon$.

Proof.

Let $F\in\{P_{2},T_{2}\}$ and $t=|V(F)|$. Since $K_{t-1}^{3}$ is $F$-free, we have $\pi_{\lambda}(F)\geq 3!\lambda(K_{t-1}^{3}).$ Let $G$ be $F$-free and let $G^{*}$ be a subgraph of $G$ such that $G^{*}$ is dense and $\lambda(G^{*})=\lambda(G)$. By Lemma 3.1, $|V(G)|\leq t-1$ and hence $\lambda(G^{*})\leq\lambda(K_{t-1}^{3})$. Thus $\pi_{\lambda}(F)=3!\lambda(K_{t-1}^{3}).$ Furthermore, if $G$ is also $K_{t-1}^{3}$-free, then $G^{*}$ is a copy of some subgraph of $K_{t-1}^{3-}$. Note that $K_{3}^{3-}=\emptyset$. Hence $\lambda(G^{*})\leq\lambda(K_{4}^{3})-\epsilon$ for $t=5$, where $\epsilon=\lambda(K_{4}^{3})-\lambda(K_{4}^{3-})={1\over 16}-{4\over 81}={17\over 1296}$, and $\lambda(G^{*})=0$ for $t=4$.

3.1 Left-compressing $P_{3}$-free $3$-graphs covering pairs

Lemma 3.3

Let $\mathcal{F}$ be a $P_{3}$-free $3$-graph with $n\geq 6$ vertices that covers pairs. Let $i,j\in V(\mathcal{F})$. Then $\pi_{ij}(\mathcal{F})$ is also $P_{3}$-free. Furthermore, if $\mathcal{F}$ is also $K^{3}_{6}$-free, then $\pi_{ij}(\mathcal{F})$ is $K^{3}_{6}$-free.

Proof.

Suppose for the contrary that there is a copy of $P_{3}$, denoted by $P$, such that $P\subseteq\pi_{ij}(\mathcal{F})$. By the definition of $\pi_{ij}(\mathcal{F})$, for every edge $f\in\pi_{ij}(\mathcal{F})$ with $\{i,j\}\subseteq f$, $f\in\mathcal{F}$; for every edge $f\in\pi_{ij}(\mathcal{F})$ with $j\in f$ and $i\notin f$, $f$ and $(f\setminus\{j\})\cup\{i\}\in\mathcal{F}$. Since $\mathcal{F}$ is $P_{3}$-free, there is $e\in P$ such that $e\notin\mathcal{F}$ and $(e\setminus\{i\})\cup\{j\}\in\mathcal{F}$. There are two cases according to the degree of $i$ in $P$.

Case 1. $d_{P}(i)=1$. There are two subcases according to the degree of $j$ in $P$.

Subcase 1.1. $d_{P}(j)=0$. Then $(P\setminus e)\cup\{(e\setminus\{i\})\cup\{j\}\}$ is a copy of $P_{3}$ in $\mathcal{F}$, a contradiction.

Subcase 1.2. $d_{P}(j)=1$ or $2$. Then $\{f\cup\{i\}:f\in L_{P}(j\setminus i)\}\cup\{f\cup\{j\}:f\in L_{P}(i\setminus j)\}\cup\{e\in P:\text{ both}\>i,j\notin e\}$ (i.e., exchange $i$ and $j$ in $P$) is a copy of $P_{3}$ in $\mathcal{F}$, a contradiction.

Case 2. $d_{P}(i)=2$. There are three subcases according to the degree of $j$ in $P$.

Subcase 2.1. $d_{P}(j)=2$. We may assume $P=\{abi,icj,jdf\}$. Then $abj\in\mathcal{F}$ and $idf\in\mathcal{F}$. So $\{abj,jci,idf\}$ forms a copy of $P_{3}$ in $\mathcal{F}$, a contradiction.

Subcase 2.2. $d_{P}(j)=1$. If $ij$ is contained in some edge $f$ of $P$, then $\{(e\setminus\{i\})\cup\{j\},f,g\}$ forms a copy of $P$ in $\mathcal{F}$, where $g\in P\setminus\{e,f\}$, a contradiction. Now assume that $P=\{abi,icd,dfj\}$. If $abj\in\mathcal{F}$, then we get a contradiction that $\{abj,dfj,icd\}$ forms a copy of $P_{3}$ in $\mathcal{F}$. Otherwise $jcd\in\mathcal{F}$. Note that $dfi\in\mathcal{F}$. Then we get a contradiction that $\{abi,ifd,dcj\}$ forms a copy of $P_{3}$ in $\mathcal{F}$.

Subcase 2.3. $d_{P}(j)=0$. We can assume that $P=\{abi,icd,dgh\}$. If both $abj,jcd$ are in $\mathcal{F}$, then $\{abj,jcd,dgh\}$ forms a copy of $P_{3}$ in $\mathcal{F}$, a contradiction. Otherwise we get a copy of $\{a_{1}a_{2}b_{0},b_{0}c_{2}c_{1},b_{1}b_{2}b_{3}\}$ in $\mathcal{F}$. Relabel the vertices of $\mathcal{F}$ such that $\{a_{1}a_{2}b_{0},b_{0}c_{2}c_{1},b_{1}b_{2}b_{3}\}\subseteq\mathcal{F}$. Since $\mathcal{F}$ covers pairs, then for every $p\in[2]$ and $q\in[3]$, there exists $x_{pq}\in V(\mathcal{F})$ such that $a_{p}b_{q}x_{pq}\in\mathcal{F}$. Then $x_{pq}=a_{k}$ for $k\in\{1,2\}\setminus\{p\}$, i.e., $a_{1}a_{2}b_{1},a_{1}a_{2}b_{2},a_{1}a_{2}b_{3}\in\mathcal{F}$, since otherwise there exists $p\in[2]$ and $q\in[3]$ such that $\{b_{0}c_{2}c_{1},a_{p}b_{q}x_{pq},b_{1}b_{2}b_{3}\}$ ($x_{pq}\in\{c_{1},c_{2}\}$) or $\{a_{1}a_{2}b_{0},b_{0}c_{2}c_{1},a_{p}b_{q}x_{pq}\}$ ($x_{pq}\notin\{c_{1},c_{2}\}$) forms a copy of $P_{3}$ in $\mathcal{F}$. In a similar way, $c_{1}c_{2}b_{1},c_{1}c_{2}b_{2},c_{1}c_{2}b_{3}\in\mathcal{F}$. Then $\{a_{1}a_{2}b_{1},b_{1}b_{2}b_{3},b_{3}c_{1}c_{2}\}$ forms a copy of $P_{3}$ in $\mathcal{F}$, a contradiction.

Next, suppose that $\mathcal{F}$ is $K^{3}_{6}$-free. Since $\mathcal{F}$ covers pairs, $\{i,j\}$ is contained in some edge $g$ of $\mathcal{F}$. Suppose for contradiction that $\pi_{ij}(\mathcal{F})$ contains a copy $K$ of $K^{3}_{6}$. Clearly $V(K)$ must contain $i$. If $V(K)$ also contains $j$ then it is easy to see that $K$ also exists in $\mathcal{F}$, contradicting $\mathcal{F}$ being $K^{3}_{6}$-free. All the edges in $K$ not containing $i$ also exist in $\mathcal{F}$. Without loss of generality assume that $[4]\subseteq V(K)\setminus g$. Thus $\{123,34i,g\}$ or $\{123,34j,g\}$ forms a copy of $P_{3}$ in $\mathcal{F}$, a contradiction.

3.2 Left-compressing $P_{4}$-free $3$-graphs covering pairs

Denote $F_{1}$ as the $3$-graph with the union of two disjoint $P_{2}$’s, $F_{2}$ as the $3$-graph with the union of disjoint $P_{1}$ and $P_{3}$, and $F_{3}$ as the $3$-graph with vertex set $\{a_{1},a_{2},a_{3},b_{1},b_{2},c_{1},c_{2},d_{1},d_{2}\}$ and edge set $\{a_{1}a_{2}a_{3},a_{1}b_{1}b_{2},a_{2}c_{1}c_{2}$, $a_{3}d_{1}d_{2}\}$. The following lemma is vital for the proof of the Lagrangian density of $P_{4}$ and its proof is postponed to Subsection 3.4.

Lemma 3.4

If $\mathcal{F}$ be a $P_{4}$-free $3$-graph with $n\geq 9$ vertices that covers pairs, then $\mathcal{F}$ is $F_{1}$-free and $F_{2}$-free. Furthermore, if $\lambda(\mathcal{F})\geq\lambda(K_{8}^{3})-0.005$, then $\mathcal{F}$ is also $F_{3}$-free.

Lemma 3.5

Let $\mathcal{F}$ be a $P_{4}$-free $3$-graph with $n\geq 9$ vertices that covers pairs. If $\lambda(\mathcal{F})\geq\lambda(K_{8}^{3})-0.005$, then $\pi_{ij}(\mathcal{F})$ is also $P_{4}$-free. Furthermore, if $\mathcal{F}$ is also $K^{3}_{8}$-free, then $\pi_{ij}(\mathcal{F})$ is $K^{3}_{8}$-free.

Proof.

The proof is similar to Lemma 3.3. Suppose for the contrary that there is a copy of $P_{4}$, denoted by $P^{\prime}$, such that $P^{\prime}\subseteq\pi_{ij}(\mathcal{F})$. Since $\mathcal{F}$ is $P_{4}$-free then there is $e^{*}\in P^{\prime}$ such that $i\in e^{*}$, $e^{*}\notin\mathcal{F}$ and $(e^{*}\setminus\{i\})\cup\{j\}\in\mathcal{F}$.

Case 1. $d_{P^{\prime}}(i)=1$. If $d_{P^{\prime}}(j)=0$. Then $(P^{\prime}\setminus e^{*})\cup\{(e^{*}\setminus\{i\})\cup\{j\}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Otherwise $d_{P^{\prime}}(j)=1$ or $2$. Then $\{f^{*}\cup\{i\}:f^{*}\in L_{P^{\prime}}(j\setminus i)\})\cup\{f^{*}\cup\{j\}:f^{*}\in L_{P^{\prime}}(i\setminus j)\})\cup\{f^{*}\in P^{\prime}:\{i,j\}\subseteq f^{*}\>{\text{o}r\>both}\>i,j\notin f^{*}\}$ contains a copy of $P_{4}$ in $\mathcal{F}$, a contradiction.

Case 2. $d_{P^{\prime}}(i)=2$.

Subcase 1. $d_{P^{\prime}}(j)=0$. There are two subcases: $P^{\prime}=\{abi,icd,def,fgh\}$ or $P^{\prime}=\{abc,cdi,ief,fgh\}$. First we consider $P^{\prime}=\{abi,icd,def,fgh\}$. If both $abj,jcd\in\mathcal{F}$, then $\{abj,jcd,def,fgh\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Otherwise we get $\{abi,jcd,def,fgh\}$ or $\{abj,icd,def,fgh\}$ in $\mathcal{F}$, which is isomorphic to $F_{2}$. By Lemma 3.4, this is a contradiction. Suppose $P^{\prime}=\{abc,cdi,ief,fgh\}$. If both $cdj,jef\in\mathcal{F}$, then $\{abc,cdj,jef,fgh\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Otherwise we get $\{abc,cdi,jef,fgh\}$ or $\{abc,cdj,ief,fgh\}$ in $\mathcal{F}$, which is isomorphic to $F_{1}$. By Lemma 3.4, this is a contradiction.

Subcase 2. $d_{P^{\prime}}(j)=1$. If $ij$ is contained in some edge $g^{*}$ of $P^{\prime}$, then $(h^{*}\setminus\{i\})\cup\{j\}\in\mathcal{F}$ for some $h^{*}\in P^{\prime}$ with $i\in h^{*}$ and $j\notin h^{*}$. Hence $\{(h^{*}\setminus\{i\})\cup\{j\},g^{*}\})\cup(\{P\setminus\{g^{*},h^{*}\})$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Otherwise then $P^{\prime}=\{abi,icd,djf,fgh\}$ or $\{abi,icd,dfg,gjh\}$ or $\{jab,bci,idf,fgh\}$. Denote the edges $e_{1},e_{2},e_{3}\in P^{\prime}$ such that $i\in e_{1},e_{2}$ and $j\in e_{3}$. If both $(e_{1}\setminus\{i\})\cup\{j\},(e_{2}\setminus\{i\})\cup\{j\}\in\mathcal{F}$, then $\{(e_{1}\setminus\{i\})\cup\{j\},(e_{2}\setminus\{i\})\cup\{j\},(e_{3}\setminus\{j\})\cup\{i\}\}\cup(P\setminus\{e_{1},e_{2},e_{3}\})$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. So assume exactly one of $(e_{1}\setminus\{i\})\cup\{j\},(e_{2}\setminus\{i\})\cup\{j\}$ is not in $\mathcal{F}$.

For $P^{\prime}=\{abi,icd,dfj,fgh\}$, if $abj\in\mathcal{F}$ then $\{abj,icd,dfj,fgh\}$ forms a copy of $F_{3}$ in $\mathcal{F}$, by Lemma 3.4, this is a contradiction. If $jcd\in\mathcal{F}$ then $\{dfi,jcd,fgh,abi\}$ forms a copy of $F_{3}$ in $\mathcal{F}$, a contradiction.

For $P^{\prime}=\{abi,icd,dfg,jgh\}$, if $abj\in\mathcal{F}$ then $\{icd,dfg,jgh,abj\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. If $jcd\in\mathcal{F}$ then $\{jcd,dfg,ghi,iab\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction.

For $P^{\prime}=\{abj,ibc,idf,fgh\}$, if $jbc\in\mathcal{F}$ then $\{jbc,abi,idf,fgh\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. If $jdf\in\mathcal{F}$ then $\{abj,bci,jdf,fgh\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction.

Subcase 3. $d_{P^{\prime}}(j)=2$. If $\{i,j\}$ is contained in some edge $e_{4}$ of $P^{\prime}$, denote the edge of $P^{\prime}$ containing $i$ but not $j$ as $e_{5}$, the edge of $P^{\prime}$ containing $j$ but not $i$ as $e_{6}$, and the edge of $P^{\prime}$ containing neither $i$ nor $j$ as $e_{7}$, then $\{(e_{6}\setminus\{j\})\cup\{i\},e_{7},e_{4},(e_{5}\setminus\{i\})\cup\{j\}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Otherwise we can assume that $P^{\prime}=\{abi,icd,dej,jfg\}$. If both $abj,jcd$ in $\mathcal{F}$, then $\{abj,jcd,dei,ifg\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. If $abj\notin\mathcal{F}$ and $jcd\in\mathcal{F}$, then $\{abi,dei,jcd,jgh\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Otherwise $abj\in\mathcal{F}$ and $jcd\notin\mathcal{F}$, then $\{abj,dej,icd,ifg\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction.

Next, suppose that $\mathcal{F}$ is $K^{3}_{8}$-free. Since $\mathcal{F}$ covers pairs, $\{i,j\}$ is contained in some edge $g^{*}$ of $\mathcal{F}$. Suppose for contradiction that $\pi_{ij}(\mathcal{F})$ contains a copy $K$ of $K^{3}_{8}$. Clearly, $V(K)$ must contain $i$. If $V(K)$ also contains $j$ then it is easy to see that $K$ also exists in $\mathcal{F}$, contradicting $\mathcal{F}$ being $K^{3}_{8}$-free. By our assumption, $V(K)$ contains at least $6$ vertices outside $g^{*}$. Without loss of generality assume that $[6]\subseteq V(K)\setminus g^{*}$. Thus $\{123,345,56i,g^{*}\}$ or $\{123,345,56j,g^{*}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction.

3.3 Lagrangian densities of $P_{3}$ and $P_{4}$

We will perform the following algorithm in the proof of Theorem 3.7. In the algorithm, $t=3$ or $4$.

Algorithem 3.6

(Dense and left-compressed $3$-graph)**

Input:* A $P_{t}$-free $3$-graph $G$ on vertex set $[n]$ ( with $\lambda(G)\geq\lambda(K_{8}^{3})-0.005$ for $t=4$).*

Output:* A $P_{t}$-free $3$-graph $G^{\prime}$ that is dense and left-compressed, and satisfies that $\lambda(G^{\prime})\geq\lambda(G)$.*

Step 1.* If $G$ is dense then let $G^{\prime}=G$ and go to Step 2. Otherwise replace $G$ by a dense subgraph $G^{\prime}$ with the same Lagrangian and go to Step 2.*

Step 2.* Let $\vec{x}=(x_{1},\dots,x_{n})$ be an optimum weighting of $G^{\prime}$. Assume that $x_{1}\geq x_{2}\geq...\geq x_{|V(G^{\prime})|}$ since otherwise we can relabel the vertices. If $G^{\prime}$ is left-compressed then terminate. Otherwise there exist vertices $i,j$, where $i<j$, such that $L_{G^{\prime}}(j\setminus i)\neq\emptyset$, then replace $G^{\prime}$ by $\pi_{ij}(G^{\prime})$ and go to step 1.*

Note that the algorithm terminates after finite many steps. By Lemma 3.3 or 3.5 and Fact 2.2. The output of the algorithm $G^{\prime}$ is a dense and left-compressed $P_{t}$-free $3$-graph with Lagrangian at lest $\lambda(G)$.

Theorem 3.7

Let $t=3$ or $4$. Then

[TABLE]

Furthermore, for any $P_{t}$-free and $K_{2t}^{3}$-free $3$-graph $G$, there exists $\epsilon=\epsilon(t)>0$ such that $\lambda(G)\leq\lambda(K_{2t}^{3})-\epsilon$.

Proof.

Let $t=3$ or $4$. Let $F$ be a $P_{t}$-free $3$-graph with with $\lambda(F)\geq\lambda(K_{8}^{3})-0.005$. Applying Algorithm 3.6 to $F$ and let $G$ be the final hypergraph obtained. Thus $G$ is a $P_{t}$-free $3$-graph that is dense and left-compressed, and $\lambda(G)\geq\lambda(K_{8}^{3})-0.005$. Assume that $V(G)=[n]$. If $n\leq 2t$ then $\lambda(G)\leq\lambda(K_{2t}^{3})$. Furthermore, if $F$ is $K_{2t}^{3}$-free then $G$ is $K_{2t}^{3}$-free by Lemma 3.3 and Lemma 3.5, and hence $\lambda(G)\leq\lambda(K_{2t}^{3-})<\lambda(K_{2t}^{3})$. Now suppose that $n\geq 2t+1$. We use induction on $t\geq 2$. The base case $t=2$ is guaranteed by Lemma 3.2. Suppose that the result holds for $t-1$. Let $\vec{x}=(x_{1},\dots,x_{n})$ be an optimum weighting of $G$. Denote $L(1)=\{e\in G:1\in e\}$. Then

[TABLE]

where $\vec{x^{\prime}}=(x_{2},\dots,x_{n})$. Clearly, $\lambda(L(1),\vec{x})\leq x_{1}\sum_{2\leq i<j\leq n}x_{i}x_{j}\leq{1\over 2}x_{1}(1-x_{1})^{2}$. For the second term, we divide it into two cases according to $t=3$ or $4$.

Case $t=3$. We claim that $G[[2,n]]=\emptyset$. Otherwise since $G$ is left-compressed and so does $G[[2,n]]$ (on vertex set $[2,n]$), then $234\in G[[2,n]]$. $G$ covers pairs and $n\geq 7$ implies that $167\in G[[2,n]]$ and $125\in G[[2,n]]$. Then $\{167,125,234\}$ forms a copy of $P_{3}$, a contradiction. Hence

[TABLE]

Note that by Fact 2.6 and 2.8, we have $\lambda(K_{6}^{3})={5\over 54}$ and $\lambda(K_{6}^{3-})=0.0887$. Let

[TABLE]

and we are done for the case $t=3$.

Case $t=4$. We claim that $G[[2,n]]$ is $P_{3}$-free. Otherwise suppose that there is a copy of $P_{3}$, denoted by $P^{\prime}$, in $G[[2,n]]$. Since $n\geq 9$ and $|V(P^{\prime})|=7$, there is $v\in\{2,\dots,n\}\setminus V(P^{\prime})$. Let $u\in V(P^{\prime})$ such that $d_{P^{\prime}}(u)=1$. Since $G$ covers pairs and $G$ is left-compressed, we have $1(n-1)n\in G$. Then $1uv\in G$. Hence $P^{\prime}\cup\{1uv\}$ forms a copy of $P_{4}$ in $G$, a contradiction.

Let $H$ be a dense subgraph of $G[[2,n]]$ with $\lambda(H)=\lambda(G[[2,n]])$. Note that $H$ must be an induced subgraph of $G[[2,n]]$. Every weight in an optimum weighting $\vec{y}$ of $H$ is positive. If $H$ is not an induced subgraph of $G[[2,n]]$, let $H^{\prime}$ be an induced subgraph of $G[[2,n]]$ on $V(H)$, then $\lambda(G[[2,n]])\geq\lambda(H^{\prime},\vec{y})>\lambda(H,\vec{y})=\lambda(G[[2,n]])$, a contradiction. Note that $\lambda(G[[2,n]])$ is left-compressed on $[2,n]$. Suppose $V(H)=\{i_{1},\dots,i_{s-1}\}$, where $i_{1}<\dots<i_{s-1}$ and $s$ is a positive integer. For each $i_{j_{1}}i_{j_{2}}i_{j_{3}}\in H$ with $j_{1}<j_{2}<j_{3}$, since $G[[2,n]]$ is left-compressed on $[2,n]$ and $H$ is an induced subgraph of $G[[2,n]]$, $i_{k_{1}}i_{k_{2}}i_{k_{3}}\in H$ for every $\{i_{k_{1}},i_{k_{2}},i_{k_{3}}\}\subseteq V(H)$ satisfying $j_{l}\geq k_{l}$ for every $l\in[3]$. Relabel the vertex $i_{j}$ of $H$ with $j+1$ for each $j\in[s-1]$, then $H$ is left-compressed on vertex set $[2,s]$.

If $s\geq 8$, then $\lambda(H)\leq{2\over 27}$ following from the case of $t=3$ with the number of vertices not less than $7$. Otherwise $s\leq 7$. When $s\leq 6$, then $\lambda(H)\leq\lambda(K_{5}^{3})={2\over 25}$. Now assume that $s=7$, we claim that $346\notin H$. Otherwise suppose that $346\in H$. Since $H$ is left-compressed, $\{346,235,127,189\}$ forms a copy of $P_{4}$ in $G$, a contradiction. Thus all $ijk$ with $i\geq 3$, $j\geq 4$, $k\geq 6$ and $i+j+k>13$ are not in $H$. Then $H\subseteq\{234,235,245,345,236,246,256,345,346,356,456,237,247,257,267\}:=H^{*}$. Since the maximum clique of $H^{*}$ is on $[2,6]$ and the number of edges of $H^{*}$ is $15$, then by Theorem 2.10, we have $\lambda(H^{*})=\lambda(K_{5}^{3})={2\over 25}$. Hence $\lambda(G[[2,n]])\leq\max\left\{{2\over 27},{2\over 25}\right\}={2\over 25}.$ Then

[TABLE]

Note that by Fact 2.6 and 2.9, we have $\lambda(K_{8}^{3})={7\over 64}$ and $\lambda(K_{8}^{3-})=0.1077$. Let

[TABLE]

and we are done for the case $t=4$.

3.4 Proof of Lemma 3.4

Given a $3$-graph $\mathcal{F}$ that covers pairs and $a,b\in V(\mathcal{F})$, let Cover$(a,b)$ denote the property that $\{a,b\}$ is covered by $\mathcal{F}$ and $N_{ab}=\{v\in V(\mathcal{F}):abv\in\mathcal{F}\}$. We repeatly use the property that $\mathcal{F}$ covers pairs to prove Lemma 3.4.

*Lemma 3.4. Let $\mathcal{F}$ be a $P_{4}$-free $3$-graph with $n\geq 9$ vertices that covers pairs. Then $\mathcal{F}$ is $F_{1}$-free and $F_{2}$-free. Furthermore, if $\lambda(\mathcal{F})\geq\lambda(K_{8}^{3})-0.005$, then $\mathcal{F}$ is also $F_{3}$-free. *

Proof.

Note that since $\mathcal{F}$ covers pairs, we have $N_{uv}\neq\emptyset$ for every pair of vertices $u,v\in V(\mathcal{F})$.

(i) First suppose that $\mathcal{F}$ contains a copy of $F_{1}$, denoted as $F$. Denote $V(F)=\{a,b_{1},b_{2},b_{3},b_{4},$ $c_{1},c_{2},c_{3},c_{4},d\}$ and $E(F)=\{c_{1}c_{2}a,ac_{3}c_{4},b_{1}b_{2}d,db_{3}b_{4}\}$.

Claim 1. $N_{b_{i}c_{j}}\subseteq\{b_{i^{\prime}},c_{j^{\prime}}\}$, where $i,j,i^{\prime},j^{\prime}$ satisfy $\{i,i^{\prime}\}=\{1,2\}$ or $\{3,4\}$ and $\{j,j^{\prime}\}=\{1,2\}$ or $\{3,4\}$. By symmetry, we only prove that $N_{b_{1}c_{1}}\subseteq\{b_{2},c_{2}\}$. If $a\in N_{b_{1}c_{1}}$, then $\{c_{3}c_{4}a,ac_{1}b_{1},b_{1}b_{2}d,$ $db_{3}b_{4}\}$ forms a copy of $P_{4}$, a contradiction. Similarly, all $d,c_{3},c_{4},b_{3},b_{4}$ and vertices in $V(\mathcal{F})\setminus V(F)$ are not in $N_{b_{1}c_{1}}$. Thus we complete the proof of the claim.

By Claim 1, $N_{b_{1}c_{1}}\neq\emptyset$ and $N_{b_{1}c_{1}}\subseteq\{b_{2},c_{2}\}$, then $b_{1}b_{2}c_{1}\in\mathcal{F}$ or $b_{1}c_{1}c_{2}\in\mathcal{F}$. By symmetry, assume that $b_{1}b_{2}c_{1}\in\mathcal{F}$. Then $b_{3}b_{4}c_{3}\notin\mathcal{F}$. Otherwise $\{b_{1}b_{2}c_{1},c_{1}c_{2}a,ac_{3}c_{4},b_{3}b_{4}c_{3}\}$ forms a copy of $P_{4}$, a contradiction. So by Claim 1, Cover$(b_{3},c_{3})$ implies that $b_{3}c_{3}c_{4}\in\mathcal{F}$. Now we show that $b_{3}b_{4}c_{2}\notin\mathcal{F}$. Otherwise $\{b_{1}b_{2}c_{1},c_{1}c_{2}a,b_{3}b_{4}c_{2},b_{3}c_{3}c_{4}\}$ forms a copy of $P_{4}$, a contradiction. Thus by Claim 1, Cover$(b_{4},c_{2})$ implies that $b_{4}c_{1}c_{2}\in\mathcal{F}$. Then $\{b_{3}c_{3}c_{4},db_{3}b_{4},b_{4}c_{1}c_{2},b_{1}b_{2}c_{1}\}$ forms a copy of $P_{4}$, a contradiction.

(ii) Suppose that $\mathcal{F}$ contains a copy of $F_{2}$, denoted as $F^{\prime}$. Denote $V(F^{\prime})=\{a_{1},a_{2},a_{3},c_{1},c_{2},c_{3},c_{4},$ $b_{1},b_{0},b_{2}\}$ and $E(F^{\prime})=\{a_{1}a_{2}a_{3},c_{1}c_{2}b_{1},b_{1}b_{0}b_{2},b_{2}c_{3}c_{4}\}$.

Case 1. $c_{1}c_{2}a_{i},c_{3}c_{4}a_{i}\in\mathcal{F}$ for all $i\in[3]$. Cover$(c_{2},c_{3})$ implies that either $c_{1}c_{2}c_{3}\in\mathcal{F}$ or $c_{2}c_{3}c_{4}\in\mathcal{F}$. Indeed, $a_{i}\notin N_{c_{2}c_{3}}$ for each $i\in[3]$, otherwise without loss of generality suppose that $c_{2}c_{3}a_{1}\in\mathcal{F}$, then $\{a_{1}a_{2}a_{3},a_{1}c_{2}c_{3},c_{3}c_{4}b_{2},b_{2}b_{0}b_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $b_{1},b_{2}\notin N_{c_{2}c_{3}}$, otherwise without loss of generality assume that $c_{2}c_{3}b_{1}\in\mathcal{F}$, then $\{b_{0}b_{1}b_{2},b_{1}c_{2}c_{3},c_{3}c_{4}a_{3},a_{3}a_{2}a_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $b_{0}\notin N_{c_{2}c_{3}}$, otherwise $\{a_{3}a_{2}a_{1},$ $c_{1}c_{2}a_{1},c_{2}c_{3}b_{0},b_{0}b_{1}b_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $x\notin N_{c_{2}c_{3}}$ for some vertex $x\in V(\mathcal{F})\setminus V(F)$, otherwise $\{b_{0}b_{1}b_{2},b_{2}c_{3}c_{4},c_{2}c_{3}x,c_{1}c_{2}a_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Therefore either $c_{1}c_{2}c_{3}\in\mathcal{F}$ or $c_{2}c_{3}c_{4}\in\mathcal{F}$. Without loss of generality assume that $c_{1}c_{2}c_{3}\in\mathcal{F}$.

Consider the property Cover$(a_{1},b_{0})$. $a_{2},a_{3}\notin N_{a_{1}b_{0}}$, otherwise by symmetry, assume that $a_{1}b_{0}a_{2}\in\mathcal{F}$, then $\{a_{1}b_{0}a_{2},b_{0}b_{1}b_{2},b_{2}c_{3}c_{4},c_{1}c_{2}c_{3}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $b_{1},b_{2}\notin N_{a_{1}b_{0}}$, otherwise by symmetry, assume that $a_{1}b_{0}b_{1}\in\mathcal{F}$, then $\{c_{3}c_{4}a_{3},a_{3}a_{2}a_{1},a_{1}b_{0}b_{1},b_{1}c_{1}c_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $c_{i}\notin N_{a_{1}b_{0}}$ for each $i\in[4]$, otherwise by symmetry, assume that $a_{1}b_{0}c_{1}\in\mathcal{F}$, then $\{c_{3}c_{4}a_{3},a_{3}a_{2}a_{1},$ $a_{1}b_{0}c_{1},b_{1}c_{1}c_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $x\notin N_{a_{1}b_{0}}$ for some vertex $x\in V(\mathcal{F})\setminus V(F)$, otherwise $\{a_{3}a_{2}a_{1},a_{1}b_{0}x,b_{0}b_{1}b_{2},b_{2}c_{3}c_{4}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Hence $\{a_{1},b_{0}\}$ is not covered by any edge of $\mathcal{F}$, which contradicts that $\mathcal{F}$ covers pairs.

Case 2. $c_{1}c_{2}a_{i}$ or $c_{3}c_{4}a_{i}\notin\mathcal{F}$ for some $i\in[3]$. Without loss of generality assume that $c_{1}c_{2}a_{1}\notin\mathcal{F}$. Consider the property Cover$(a_{1},c_{1})$. Then $N_{a_{1}c_{1}}=\{b_{2}\}$. Since if $a_{2}\in N_{a_{1}c_{1}}$, then $\{a_{1}a_{2}c_{1},c_{1}c_{2}b_{1},b_{1}b_{0}b_{2},$ $b_{2}c_{3}c_{4}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, $a_{3}\notin N_{a_{1}c_{1}}$. If $b_{1}\in N_{a_{1}c_{1}}$, then $\{a_{3}a_{2}a_{1},$ $a_{1}c_{1}b_{1},b_{1}b_{0}b_{2},b_{2}c_{3}c_{4}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, $b_{0},c_{3},c_{4}\notin N_{a_{1}c_{1}}$. If $x\in N_{a_{1}c_{1}}$ for some vertex $x\in V(\mathcal{F})\setminus V(F^{\prime})$, then $\{a_{3}a_{2}a_{1},a_{1}xc_{1},c_{1}c_{2}b_{1},b_{1}b_{0}b_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, $N_{a_{1}c_{2}}=\{b_{2}\}$. We claim that $c_{3}c_{4}a_{2},c_{3}c_{4}a_{3}\notin\mathcal{F}$; otherwise without loss of generality assume that $c_{3}c_{4}a_{2}\in\mathcal{F}$, then $\{c_{3}c_{4}a_{2},a_{3}a_{2}a_{1},a_{1}c_{1}b_{2},b_{1}b_{2}b_{3}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction.

Hence $c_{1}c_{2}a_{1}\notin\mathcal{F}$ implies that $a_{1}c_{1}b_{2},a_{1}c_{2}b_{2}\in\mathcal{F}$ and $c_{3}c_{4}a_{2},c_{3}c_{4}a_{3}\notin\mathcal{F}$. Similarly, $c_{3}c_{4}a_{2},c_{3}c_{4}a_{3}\notin\mathcal{F}$ imply $c_{1}c_{2}a_{3},c_{1}c_{2}a_{2}\notin\mathcal{F}$ and $c_{1}c_{2}a_{2}\notin\mathcal{F}$ implies that $c_{3}c_{4}a_{1}\notin\mathcal{F}$. Thus $c_{1}c_{2}a_{i},c_{3}c_{4}a_{i}\notin\mathcal{F}$ for each $i\in[3]$. Then $N_{a_{i}c_{j}}=\{b_{k_{j}}\}$ for every $i\in[3],j\in[4]$, where $k_{1},k_{2}=2$ and $k_{3},k_{4}=1$. Thus $\{a_{3}b_{1}c_{3},a_{2}b_{1}c_{4},a_{2}b_{2}c_{1},a_{1}b_{2}c_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. So $\mathcal{F}$ is $F_{2}$-free.

(iii) Let $\mathcal{F}$ satisfy $\lambda(\mathcal{F})\geq\lambda(K_{8}^{3})-0.005$. Suppose that $\mathcal{F}$ contains a copy of $F_{3}$, denoted as $F^{\prime\prime}$. Let $V(F^{\prime\prime})=\{a_{1},a_{2},a_{3},c_{1},c_{2},b_{1},b_{2},d_{1},d_{2}\}$ and $E(F^{\prime\prime})=\{a_{1}a_{2}a_{3},a_{1}b_{1}b_{2},a_{2}c_{1}c_{2},a_{3}d_{1}d_{2}\}$. We first prove that $V(\mathcal{F})=V(F^{\prime\prime})$. Otherwise suppose that $V(\mathcal{F})\setminus V(F_{3})\neq\emptyset$, let $x\in V(\mathcal{F})\setminus V(F_{3})$. The following claim is a simple consequence of $\mathcal{F}$ being $P_{4}$-free.

We claim that $xb_{1}b_{2},xc_{1}c_{2},xd_{1}d_{2}\notin\mathcal{F}$. To prove this, we first assume that there are at least two of $xb_{1}b_{2},xc_{1}c_{2},xd_{1}d_{2}$ in $\mathcal{F}$. Suppose $xb_{1}b_{2},xc_{1}c_{2}\in\mathcal{F}$. If $b_{1}\in N_{xd_{2}}$, then $\{xb_{1}d_{2},d_{2}d_{1}a_{3},a_{3}a_{1}a_{2},a_{2}c_{1}c_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, all $b_{2},c_{1},c_{2}$ and the vertices in $V(\mathcal{F})\setminus(V(F_{3})\cup\{x\})$ (if there exists) are not in $N_{xd_{2}}$. If $a_{3}\in N_{xd_{2}}$, then $\{b_{1}b_{2}x,xd_{2}a_{3},a_{1}a_{2}a_{3},a_{2}c_{1}c_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. If $a_{1}\in N_{xd_{2}}$, then $\{b_{1}b_{2}x,xd_{2}a_{1},a_{1}a_{2}a_{3},a_{2}c_{1}c_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, $a_{2}\notin N_{xd_{2}}$. Hence $N_{xd_{2}}=\{d_{1}\}$.

Now we show that $\{b_{2},c_{2}\}$ is not covered by any edge of $\mathcal{F}$ and we get a contradiction. $b_{1}\notin N_{b_{2}c_{2}}$since otherwise $\{d_{2}d_{1}a_{3},a_{3}a_{1}a_{2},a_{2}c_{1}c_{2},b_{1}b_{2}c_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, all $c_{1}$ and the vertices in $V(\mathcal{F})\setminus(V(F_{3})\cup\{x\})$ (if there exists) are not in $N_{b_{2}c_{2}}$. $a_{1}\notin N_{b_{2}c_{2}}$ since otherwise $\{b_{1}b_{2}x,b_{2}c_{2}a_{1},a_{1}a_{2}a_{3},a_{3}d_{1}d_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, $a_{2}\notin N_{b_{2}c_{2}}$. $a_{3}\notin N_{b_{2}c_{2}}$ since otherwise $\{d_{1}d_{2}x,b_{1}b_{2}x,b_{2}c_{2}a_{3},a_{1}a_{2}a_{3}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $d_{1}\notin N_{b_{2}c_{2}}$ since otherwise $\{b_{1}b_{2}x,b_{2}c_{2}d_{1},a_{3}d_{1}d_{2},a_{1}a_{2}a_{3}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, $d_{2}\notin N_{b_{2}c_{2}}$. Thus, $\{b_{2},c_{2}\}$ is not covered by any edge of $\mathcal{F}$, a contradiction.

Now assume that there is exactly one of $xb_{1}b_{2},xc_{1}c_{2},xd_{1}d_{2}$ in $\mathcal{F}$. Without loss of generality assume that $xb_{1}b_{2}\in\mathcal{F}$ and $xc_{1}c_{2},xd_{1}d_{2}\notin\mathcal{F}$. Consider the property Cover$\{x,c_{2}\}$. $b_{1}\notin N_{xc_{2}}$since otherwise $\{d_{2}d_{1}a_{3},a_{3}a_{1}a_{2},a_{2}c_{1}c_{2},c_{2}b_{1}x\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, all $b_{2},d_{1},d_{2},c_{1}$ and the vertices in $V(\mathcal{F})\setminus(V(F_{3})\cup\{x\})$ (if there exists) are not in $N_{xc_{2}}$. $a_{1}\notin N_{xc_{2}}$since otherwise $\{b_{1}b_{2}x,xc_{2}a_{i},a_{1}a_{2}a_{3},a_{3}d_{1}d_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Hence $N_{xc_{2}}\neq\emptyset$ implies that $y_{xc_{2}}=a_{3}$, that is, $xc_{2}a_{3}\in\mathcal{F}$. Similarly, $a_{2}\notin N_{xc_{2}}$. Then $N_{xc_{2}}=\{a_{3}\}$. Similarly, $N_{xd_{2}}=\{a_{2}\}$.

Consider the property Cover$\{b_{2},c_{2}\}$. $b_{1}\notin N_{b_{2}c_{2}}$since otherwise $\{d_{2}d_{1}a_{3},a_{3}a_{1}a_{2},a_{2}c_{1}c_{2},c_{2}b_{1}b_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, all $c_{1},x$ and the vertices in $V(\mathcal{F})\setminus(V(F_{3})\cup\{x\})$ (if there exists) are not in $N_{b_{2}c_{2}}$. $a_{1}\notin N_{xc_{2}}$since otherwise $\{b_{1}b_{2}x,b_{2}c_{2}a_{1},a_{1}a_{2}a_{3},a_{3}d_{1}d_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, $a_{2}\notin N_{b_{2}c_{2}}$. $d_{1}\notin N_{xc_{2}}$since otherwise $\{xb_{1}b_{2},b_{2}c_{2}d_{1},d_{1}d_{2}a_{3},a_{1}a_{2}a_{3}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, $d_{2}\notin N_{b_{2}c_{2}}$. Hence $N_{b_{2}c_{2}}=\{a_{3}\}$. Then $\{a_{1}b_{1}b_{2},b_{2}c_{2}a_{3},a_{3}d_{1}d_{2},d_{2}a_{2}x\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Hence $xb_{1}b_{2},xc_{1}c_{2},xd_{1}d_{2}\notin\mathcal{F}$.

Now we claim that for every $A\in\{xb_{i}a_{2},xb_{i}a_{3},xc_{i}a_{1},xc_{i}a_{3},xd_{i}a_{1},xd_{i}a_{2}:i\in[2]\}$, we have $A\notin\mathcal{F}$. By symmetry, suppose that $xb_{2}a_{2}\in\mathcal{F}$. Consider the property Cover$\{b_{1},c_{1}\}$. We first show that all $d_{1},d_{2},a_{3}$ are not in $N_{b_{1}c_{1}}$; otherwise, denote such an edge by $e$, then $\{xb_{2}a_{2},a_{2}c_{1}c_{2},e,d_{1}d_{2}a_{3}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similar to the above $b_{1}\notin N_{b_{2}c_{2}}$, we have $b_{2}\notin N_{b_{1}c_{1}}$. Similarly, $c_{2}$ and the vertices in $V(\mathcal{F})\setminus(V(F_{3}))$ are not in $N_{b_{1}c_{1}}$. Hence we have $b_{1}c_{1}a_{1}\in\mathcal{F}$ or $b_{1}c_{1}a_{2}\in\mathcal{F}$. By symmetry, $b_{1}c_{2}a_{1}\in\mathcal{F}$ or $b_{1}c_{2}a_{2}\in\mathcal{F}$. We claim that $b_{1}c_{1}a_{1}\notin\mathcal{F}$ or $b_{1}c_{2}a_{2}\notin\mathcal{F}$. Otherwise let $e^{\prime}\in{\{b_{2},x\}\choose 1}\times{\{a_{3},d_{1},d_{2}\}\choose 2}$, if $e^{\prime}\in\mathcal{F}$, then $\{e^{\prime},xa_{2}b_{2},b_{1}c_{2}a_{2},b_{1}c_{1}a_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Hence $\left({\{b_{2},x\}\choose 1}\times{\{a_{3},d_{1},d_{2}\}\choose 2}\right)\cap\mathcal{F}=\emptyset$. Then Cover($b_{2},d_{1}$) implies that $b_{2}d_{1}a_{1}\in\mathcal{F}$ or $b_{2}d_{1}a_{2}\in\mathcal{F}$. Then $\{d_{1}d_{2}a_{3},b_{2}d_{1}a_{1},a_{1}c_{1}b_{1},b_{1}c_{2}a_{2}\}$ or $\{d_{2}a_{3}d_{1},d_{1}b_{2}a_{2},a_{2}c_{2}b_{1},b_{1}a_{1}c_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, $b_{1}c_{1}a_{2}\notin\mathcal{F}$ or $b_{1}c_{2}a_{1}\notin\mathcal{F}$. Hence $b_{1}c_{1}a_{1},b_{1}c_{2}a_{1}\in\mathcal{F}$ or $b_{1}c_{1}a_{2},b_{1}c_{2}a_{2}\in\mathcal{F}$.

Case 1. $b_{1}c_{1}a_{1},b_{1}c_{2}a_{1}\in\mathcal{F}$. Consider the property Cover$\{x,d_{1}\}$. $b_{1}\notin N_{xd_{1}}$since otherwise $\{xb_{1}d_{1},d_{1}d_{2}a_{3},a_{3}a_{1}a_{2},a_{2}c_{1}c_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, all $b_{2},c_{1},c_{2}$ and the vertices not in $V(F^{\prime\prime})$ are not in $N_{xd_{1}}$. $a_{1}\notin N_{xd_{1}}$since otherwise $\{d_{1}d_{2}a_{3},d_{1}xa_{1},a_{1}c_{1}b_{1},c_{1}c_{2}a_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $a_{2}\notin N_{xd_{1}}$since otherwise $\{d_{1}d_{2}a_{3},d_{1}xa_{2},a_{2}c_{1}c_{2},b_{1}c_{2}a_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $a_{3}\notin N_{xd_{1}}$since otherwise $\{xd_{1}a_{3},xb_{2}a_{2},a_{2}c_{1}c_{2},b_{1}c_{2}a_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Thus $\{x,d_{1}\}$ is not covered in any edge of $\mathcal{F}$, a contradiction.

Case 2. $b_{1}c_{1}a_{2},b_{1}c_{2}a_{2}\in\mathcal{F}$. Consider the property Cover$\{x,d_{1}\}$. The same as the above in Case 1, all $b_{1},b_{2},c_{1},c_{2}$ and other vertices not in $V(F^{\prime\prime})$ are not in $N_{xd_{1}}$. $a_{1}\notin N_{xd_{1}}$since otherwise $\{d_{1}d_{2}a_{3},d_{1}xa_{1},a_{1}b_{1}b_{2},b_{1}c_{1}a_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $a_{2}\notin N_{xd_{1}}$since otherwise $\{d_{1}d_{2}a_{3},d_{1}xa_{2},a_{2}c_{1}b_{1},b_{1}b_{2}a_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Suppose $a_{3}\in N_{xd_{1}}$. Now consider the property Cover$\{c_{1},d_{2}\}$. $b_{1}\notin N_{c_{1}d_{2}}$since otherwise $\{xb_{2}a_{2},a_{1}a_{2}a_{3},a_{3}d_{1}d_{2},c_{1}d_{2}b_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $b_{2}\notin N_{c_{1}d_{2}}$since otherwise $\{b_{1}c_{2}a_{2},a_{1}a_{2}a_{3},a_{3}d_{1}d_{2},c_{1}d_{2}b_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $d_{1}\notin N_{c_{1}d_{2}}$since otherwise $\{b_{1}b_{2}a_{1},a_{1}a_{3}a_{2},a_{2}c_{2}c_{1},c_{1}d_{2}d_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Similarly, $c_{2}\notin N_{c_{1}d_{2}}$. Then $N_{c_{1}d_{2}}\subseteq\{a_{1},a_{2},a_{3}\}$. If $a_{1}\notin N_{c_{1}d_{2}}$, then $\{xd_{1}a_{3},xb_{2}a_{2},a_{2}c_{1}c_{2},c_{1}d_{2}a_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. If $a_{2}\notin N_{c_{1}d_{2}}$, then $\{a_{1}b_{1}b_{2},xb_{2}a_{2},a_{2}c_{1}d_{2},d_{2}d_{1}a_{3}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. If $a_{3}\notin N_{c_{1}d_{2}}$, then $\{c_{1}d_{2}a_{3},xd_{1}a_{3},xb_{2}a_{2},a_{1}b_{1}b_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Thus $\{x,d_{1}\}$ is not covered in any edge of $\mathcal{F}$, a contradiction.

Hence for every $A\in\{xb_{i}a_{2},xb_{i}a_{3},xc_{i}a_{1},xc_{i}a_{3},xd_{i}a_{1},xd_{i}a_{2}:i\in[2]\}$, we have $A\notin\mathcal{F}$. Let $y\in\{b_{1},b_{2},c_{1},c_{2},d_{1},d_{2}\}$, consider the property Cover$(x,y)$. We have proved that $N_{xy}\cap(\{b_{1},b_{2},c_{1},c_{2},d_{1},d_{2},$ $a_{1},a_{2},a_{3}\}\setminus\{y\})=\emptyset$. $z\notin N_{xy}$ for some vertex not in $V(F^{\prime\prime})\cup\{x\}$since otherwise $xyz$ connects to the endpoint of a linear path of 3 in $F^{\prime\prime}$ and we get a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. Hence $xb_{i}a_{1},xc_{i}a_{2},xd_{i}a_{3}\in\mathcal{F}$ for every $i\in[2]$. Consider the property Cover($b_{2},c_{2}$). We have proved that $b_{1},c_{1}$ and vertex not in $V(F^{\prime\prime})$ are not in $N_{b_{2}c_{2}}$. Let $i=1,2$, we claim that $b_{2}c_{2}d_{i}\notin\mathcal{F}$; otherwise $\{b_{1}b_{2}a_{1},a_{1}a_{3}a_{2},a_{2}c_{2}c_{1},c_{1}d_{2}d_{1}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. $\{xb_{1}a_{1},a_{1}a_{2}a_{3},a_{3}d_{1}d_{2},b_{2}c_{2}d_{i}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. If $b_{2}c_{2}a_{1}\in\mathcal{F}$, then $\{d_{1}a_{3}x,xb_{1}a_{1},a_{1}b_{2}c_{2},c_{2}c_{1}a_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. If $b_{2}c_{2}a_{2}\in\mathcal{F}$, then $\{a_{1}b_{1}b_{2},b_{2}c_{2}a_{2},$ $a_{2}c_{1}x,xd_{1}a_{3}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. If $b_{2}c_{2}a_{3}\in\mathcal{F}$, then $\{d_{1}d_{2}a_{3},a_{3}b_{2}c_{2},a_{1}b_{2}x,xc_{1}a_{2}\}$ forms a copy of $P_{4}$ in $\mathcal{F}$, a contradiction. So $\{b_{2},c_{2}\}$ is not covered by any edge of $\mathcal{F}$, a contradiction. Hence $V(\mathcal{F})=V(F_{3})$.

We deduce the value $\lambda(\mathcal{F})$. Denote $B=\{b_{1},b_{2},c_{1},c_{2},d_{1},d_{2}\}$, $E^{\prime}=\{b_{1}b_{2}x,c_{1}c_{2}y,d_{1}d_{2}z:x\in B\setminus\{b_{1},b_{2}\},y\in B\setminus\{c_{1},c_{2}\},z\in B\setminus\{d_{1},d_{2}\}\}$. Since for every $e\in E^{\prime}$, $e\notin\mathcal{F}$, then $\mathcal{F}\subseteq{V(\mathcal{F})\choose 3}\setminus E^{\prime}:=\mathcal{G}$. Then $\lambda(\mathcal{F})\leq\lambda(\mathcal{G})$. Let $\vec{x}$ be an optimum weighting of $\mathcal{G}$. By Lemma 2.5, we can assume that $x_{a_{1}}=x_{a_{2}}=x_{a_{3}}=x$, $x_{b_{1}}=x_{b_{2}}=x_{c_{1}}=x_{c_{2}}=x_{d_{1}}=x_{d_{2}}=y$. So $3x+6y=1$. Then $\lambda(\mathcal{G})=x^{3}+8y^{3}+18x^{2}y+45xy^{2}$. Let $L(x,y,\gamma)=x^{3}+8y^{3}+18x^{2}y+45xy^{2}-\gamma(3x+6y-1)$. Then ${\partial L\over\partial x}={\partial L\over\partial y}={\partial L\over\partial\gamma}=0$ implies that $y={\sqrt{873}-15\over 162}$. So

[TABLE]

which contradicts that $\lambda(\mathcal{F})\geq\lambda(K_{8}^{3})-0.005$. This completes the proof.

4 Turán numbers of the extensions

If $L$ is a hypergraph on $[t]$, then a blowup of $L$ is a hypergraph $G$ whose vertex set can be partitioned into $V_{1},\ldots,V_{t}$ such that $E(G)=\bigcup_{e\in L}\prod_{i\in e}V_{i}$. Let $T_{t}^{r}(n)$ be the balanced blowup of $K_{t}^{r}$ on $n$ vertices, i.e., $V(T_{t}^{r}(n))=V_{1}\cup V_{2}\cup\dots\cup V_{t}$ such that $V_{i}\cap V_{j}=\emptyset$ for every $i\neq j$ and $|V_{1}|\leq|V_{2}|\leq\dots\leq|V_{t}|\leq|V_{1}|+1$, and $E(T_{t}^{r}(n))=\{S\in\binom{[n]}{r}:\forall i\in[t],|S\cap V_{i}|\leq 1\}$. Let $t_{t}^{r}(n)=|T_{t}^{r}(n)|$. The main result in this section is as follows.

Theorem 4.1

Let $F\in\{P_{3},P_{4}\}$. Then $ex(n,H^{F})=t_{|V(F)|-1}^{3}(n)$ for sufficiently large $n$. Moreover, if $n$ is sufficiently large and $G$ is an $H^{F}$-free $3$-graph on $[n]$ with $|G|=t_{|V(F)|-1}^{3}(n)$, then $G\cong T_{|V(F)|-1}^{3}(n)$.

To prove the theorem, we need several results from [2]. Similar results are obtained independently in [15]. Let ${\mathcal{K}}_{p}^{F}$ denote the family of $r$-graphs $H$ that contains a set $C$ of $p$ vertices such that the subgraph of $H$ induced by $C$ contains a copy of $F$ and such that every pair in $C$ is covered in $H$. Let $[m]_{r}=m\times(m-1)\times\dots\times(m-r+1)$.

Definition 4.2

([2]) Let $m,r\geq 2$ be positive integers. Let $F$ be an $r$-graph that has at most $m+1$ vertices satisfying $\pi_{\lambda}(F)\leq{[m]_{r}\over m^{r}}$. We say that ${\mathcal{K}}_{m+1}^{F}$ is $m$-stable if for every real $\varepsilon>0$ there are a real $\delta>0$ and an integer $n_{1}$ such that if $G$ is a ${\mathcal{K}}_{m+1}^{F}$-free $r$-graph with at least $n\geq n_{1}$ vertices and more than $({[m]_{r}\over m^{r}}-\delta){n\choose r}$ edges, then $G$ can be made $m$-partite by deleting at most $\varepsilon n$ vertices.**

Given an $r$-graph $G$ and a real $\alpha$ with $0<\alpha\leq 1$, we say that $G$ is $\alpha$-dense if $G$ has minimum degree at least $\alpha{|V(G)|-1\choose r-1}$. Let $i,j\in V(G)$, we say $i$ and $j$ are nonadjacent if $\{i,j\}$ is not contained in any edge of $G$. Given a set $U\subseteq V(G)$, we say $U$ is an equivalence class of $G$ if for every two vertices $u,v\in U$, $L_{G}(u)=L_{G}(v)$. Given two nonadjacent nonequivalent vertices $u,v\in V(G)$, symmetrizing $u$ to $v$ refers to the operation of deleting all edges containing $u$ of $G$ and adding all the edges $\{u\}\cup A,A\in L_{G}(v)$ to $G$. We use the following algorithm from [2].

Algorithem 4.3

(Symmetrization and cleaning with threshold $\alpha$ [2])*

Input: An $r$-graph $G$.

Output: An $r$-graph $G^{*}$.

Initiation: Let $G_{0}=H_{0}=G.$ Set $i=0$.

Iteration: For each vertex $u$ in $H_{i}$, let $A_{i}(u)$ denote the equivalence class that $u$ is in. If either $H_{i}$ is empty or $H_{i}$ contains no two nonadjacent nonequivalent vertices, then let $G^{*}=H_{i}$ and terminate. Otherwise let $u,v$ be two nonadjacent nonequivalent vertices in $H_{i}$, where $d_{H_{i}}(u)\geq d_{H_{i}}(v)$. We symmetrize each vertex in $A_{i}(v)$ to $u$. Let $G_{i+1}$ denote the resulting graph. If $G_{i+1}$ is $\alpha$-dense, then let $H_{i+1}=G_{i+1}$. Otherwise let $L=G_{i+1}$ and repeat the following: let $z$ be any vertex of minimum degree in $L$. Redefine $L=L-z$ unless in forming $G_{i+1}$ from $H_{i}$ we symmetrized the equivalence class of some vertex $v$ in $H_{i}$ to some vertex in the equivalence class of $z$ in $H_{i}$. In that case, we redefine $L=L-v$ instead. Repeat the process until $L$ becomes either $\alpha$-dense or empty. Let $H_{i+1}=L$. We call the process of forming $H_{i+1}$ from $G_{i+1}$ “cleaning”. Let $Z_{i+1}$ denote the set of vertices removed, so $H_{i+1}=G_{i+1}-Z_{i+1}$. By our definition, if $H_{i+1}$ is nonempty then it is $\alpha$-dense. *

Theorem 4.4

([2])* Let $m,r\geq 2$ be positive integers. Let $F$ be an $r$-graph that has at most $m$ vertices or has $m+1$ vertices one of which has degree $1$. There exists a real $\gamma_{0}=\gamma_{0}(m,r)>0$ such that for every positive real $\gamma<\gamma_{0}$, there exist a real $\delta>0$ and an integer $n_{0}$ such that the following is true for all $n\geq n_{0}$. Let $G$ be an ${\mathcal{K}}_{m+1}^{F}$-free $r$-graph on $[n]$ with more than $({[m]_{r}\over m^{r}}-\delta){n\choose r}$ edges. Let $G^{*}$ be the final $r$-graph produced by Algorithm 4.3 with threshold ${[m]_{r}\over m^{r}}-\gamma$. Then $|V(G^{*})|\geq(1-\gamma)n$ and $G^{*}$ is $({[m]_{r}\over m^{r}}-\gamma)$-dense. Furthermore, if there is a set $W\subseteq V(G^{*})$ with $|W|\geq(1-\gamma_{0})|V(G^{*})|$ such that $W$ is the union of a collection of at most $m$ equivalence classes of $G^{*}$, then $G[W]$ is $m$-partite.       *

Theorem 4.5

([2])* Let $m,r\geq 2$ be positive integers. Let $F$ be an $r$-graph that either has at most $m$ vertices or has $m+1$ vertices one of which has degree $1$. Suppose either $\pi_{\lambda}(F)<{[m]_{r}\over m^{r}}$ or $\pi_{\lambda}(F)={[m]_{r}\over m^{r}}$ and ${\mathcal{K}}_{m+1}^{F}$ is $m$-stable. Then there exists a positive integer $n_{2}$ such that for all $n\geq n_{2}$ we have $ex(n,H_{m+1}^{F})=t_{m}^{r}(n)$ and the unique extremal $r$-graph is $T_{m}^{r}(n)$.*

The following proposition follows immediately from the definition and is implicit in many papers (see [11] for instance).

Proposition 4.6

*Let $r\geq 2$. Let $L$ be an $r$-graph and $G$ be a blowup of $L$. Suppose $|V(G)|=n$. Then $|G|\leq\lambda(L)n^{r}$.       *

Theorem 4.1 follows from the following Lemma, which is implicit in [2] and [15].

Lemma 4.7

Let $m,r\geq 2$ be positive integers. Let $F$ be an $r$-graph that has at most $m$ vertices or has $m+1$ vertices of which $r-1$ vertices of an edge has degree $1$ and $\pi_{\lambda}(F)\leq{[m]_{r}\over m^{r}}$. Suppose there is a constant $c>0$ such that for every $F$-free and $K_{m}^{r}$-free $r$-graph $L$, $\lambda(L)\leq\lambda(K_{m}^{r})-\epsilon$ holds. Then ${\mathcal{K}}_{m+1}^{F}$ is $m$-stable, consequently $ex(n,H_{m+1}^{F})=t_{m}^{r}(n)$ and the unique extremal $r$-graph is $T_{m}^{r}(n)$.

Proof. Let $\varepsilon>0$ be given. Let $\delta,n_{0}$ be the constants guaranteed by Theorem 4.4. We can assume that $\delta$ is small enough and $n_{0}$ is large enough. Let $\gamma>0$ satisfy $\gamma<\varepsilon$ and $\delta+r\gamma<\epsilon$. Let $G$ be a ${\mathcal{K}}_{m+1}^{F}$-free $r$-graph on $n>n_{0}$ vertices with more than $({[m]_{r}\over m^{r}}-\delta){n\choose r}$ edges. Let $G^{*}$ be the final $r$-graph produced by applying Algorithm 4.3 to $G$ with threshold ${[m]_{r}\over m^{r}}-\gamma$. By Algorithm 4.3, if $S$ consists of one vertex from each equivalence class of $G^{*}$, then $G^{*}[S]$ covers pairs and $G^{*}$ is a blowup of $G^{*}[S]$.

First, suppose that $|S|\geq m+1$. If $F\subseteq G^{*}[S]$, then since $G^{*}[S]$ covers pairs we can find a member of ${\mathcal{K}}_{m+1}^{F}$ in $G^{*}[S]$ by using any $(m+1)$-set that contains a copy of $F$ as the core, contradicting $G^{*}$ being ${\mathcal{K}}_{m+1}^{F}$-free. So $G^{*}[S]$ is $F$-free. We claim that $G^{*}[S]$ is $K^{r}_{m}$-free. Otherwise suppose $G^{*}[S]$ contains a copy of $K^{r}_{m}$. When $|V(F)|=m$, $K^{r}_{m}$ contains a copy of $F$ clearly. So suppose that $|V(F)|=m+1$ and $F$ has $r-1$ vertices of one edge of degree $1$. Let $e=\{v_{1},\dots,v_{r}\}\in F$ with $d_{F}(v_{1})=\dots=d_{F}(v_{r-1})=1$. Let $u_{1}\in S\setminus V(K^{r}_{m})$ since $|S|\geq m+1$, and let $u_{2}\in V(K^{r}_{m})$, since $G^{*}[S]$ covers pairs, there is an edge containing $\{u_{1},u_{2}\}$ in $G^{*}[S]$, denote as $\{u_{1},\dots,u_{r}\}$. Assume that $V(F)=\{v_{1},\dots,v_{m+1}\}$. We define an injective function $f$ from $V(F)$ to $S$ with $f(v_{i})=u_{i}$ for every $i\in[m+1]$, where $u_{r+1},\dots,u_{m+1}$ are arbitrary $m+1-r$ vertices in $V(K^{r}_{m})\setminus\{u_{2},\dots,u_{r}\}$. It’s clear that $f$ preserves edges and hence $G^{*}[S]$ contains a copy of $F$, a contradiction. Thus, by our assumption, $\lambda(G^{*}[S])\leq\frac{1}{r!}\frac{[m]_{r}}{m^{r}}-\epsilon$. By Proposition 4.6, we have

[TABLE]

Now, during the process of obtaining $G^{*}$ from $G$, symmetrization never decreases the number of edges. Since at most $\gamma n$ vertices are deleted in the process (see Theorem 4.4),

[TABLE]

contradicting (1). So $|S|\leq m$. Hence, $W=V(G^{*})$ is the union of at most $m$ equivalence classes of $G^{*}$. By Theorem 4.4, $|W|\geq(1-\gamma)n$ and $G[W]$ is $m$-partite. Hence, $G$ can be made $m$-partite by deleting at most $\gamma n<\varepsilon n$ vertices. Thus, $\mathcal{K}_{m+1}^{F}$ is $m$-stable. Then the result holds by Theorem 4.5.

5 Remark

If one could show that left-compressing a dense $P_{t}$-free $3$-graph will result in a $P_{t}$-free $3$-graph, then it would not be hard to show that $P_{t}$ is perfect. It seems to be hard for general $t$. All known cases regarding Lagrangian densities are listed in Section 1 (to the best of our knowledge). For $3$-uniform graphs spanned by 3 edges, there is still one remaining unsolved cases: $K_{4}^{3-}$ ({123, 124, 134}). Since the Lagrangian density and the Turán density of $K_{4}^{3-}$ equal, it would be very interesting in determining the Turán density of $K_{4}^{3-}$ by its Lagrangian density.

Acknowledgement We thank Tao Jiang for helpful discussions.