# An irrational Lagrangian density of a single hypergraph

Zilong Yan School of Mathematics, Hunan University, Changsha 410082, P.R.
China. Email<EMAIL_ADDRESS>Yuejian Peng  Corresponding author. School
of Mathematics, Hunan University, Changsha, 410082, P.R. China. Email:
<EMAIL_ADDRESS>Supported in part by National Natural Science Foundation of
China (No. 11931002).

###### Abstract

The Turán number of an $r$-uniform graph $F$, denoted by $ex(n,F)$, is the
maximum number of edges in an $F$-free $r$-uniform graph on $n$ vertices. The
Turán density of $F$ is defined as
$\pi(F)=\underset{{n\rightarrow\infty}}{\lim}{ex(n,F)\over{n\choose r}}.$
Denote $\Pi_{\infty}^{(r)}=\\{\pi(\mathcal{F}):\mathcal{F}{\rm\ is\ a\ family\
of}\ r{\rm-uniform\ graphs}\\},$
$\Pi_{fin}^{(r)}=\\{\pi(\mathcal{F}):\mathcal{F}{\rm\ is\ a\ finite\ family\
of}\ r{\rm-uniform\ graphs}\\}$ and
$\Pi_{t}^{(r)}=\\{\pi(\mathcal{F}):\mathcal{F}{\rm\ is\ a\ family\ of}\ r{\rm-
uniform\ graphs\ and}\\\ \ |\mathcal{F}|\leq t\\}.$ For graphs, Erdős-Stone-
Simonovits ([7], [8]) showed that
$\Pi_{\infty}^{(2)}=\Pi_{fin}^{(2)}=\Pi_{1}^{(2)}=\\{0,{1\over 2},{2\over
3},...,\\\ {l-1\over l},...\\}.$ We know quite few about the Turán density of
an $r$-uniform graph for $r\geq 3$. Baber and Talbot [2], and Pikhurko [27]
showed that there is an irrational number in $\Pi_{3}^{(3)}$ and
$\Pi_{fin}^{(3)}$ respectively, disproving a conjecture of Chung and Graham
[5]. Baber and Talbot [2] asked whether $\Pi_{1}^{(r)}$ contains an irrational
number. The Lagrangian of a hypergraph has been a useful tool in hypergraph
extremal problems. The Lagrangian density  of an $r$-uniform graph $F$ is
$\pi_{\lambda}(F)=\sup\\{r!\lambda(G):G\;is\;F\text{-}free\\}$, where
$\lambda(G)$ is the Lagrangian of an $r$-uniform graph $G$. Sidorenko [31]
showed that the Lagrangian density of an $r$-uniform hypergraph $F$ is the
same as the Turán density of the extension of $F$. In this paper, we show that
the Lagrangian density of $F=\\{123,124,134,234,567\\}$ (the disjoint union of
$K_{4}^{3}$ and an edge) is ${\sqrt{3}\over 3}$, consequently, the Turán
density of the extension of $F$ is an irrational number, answering the
question of Baber and Talbot.

Keywords: Hypergraph Lagrangian, Lagrangian density, Turán density

##  1 Introduction

For a set $V$ and a positive integer $r$, let $V^{(r)}$ denote the family of
all $r$-subsets of $V$. An $r$-uniform graph or $r$-graph $G$ consists of a
set $V(G)$ of vertices and a set $E(G)\subseteq V(G)^{(r)}$ of edges. Let
$e(G)$ denote the number of edges of $G$. 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 $H$ is a subgraph of an $r$-graph $G$, denoted by
$H\subseteq G$, if $V(H)\subseteq V(G)$ and $E(H)\subseteq E(G)$. A subgraph
of $G$ induced by $V^{\prime}\subseteq V$, denoted as $G[V^{\prime}]$, is the
$r$-graph with vertex set $V^{\prime}$ and edge set $E^{\prime}=\\{e\in
E(G):e\subseteq V^{\prime}\\}$. For $S\subseteq V(G)$, let $G-S$ denote the
subgraph of $G$ induced by $V(G)\setminus S$. Let $G^{c}$ denote the
complement $r$-graph of an $r$-graph $G$ with $V(G^{c})=V(G)$ and
$E(G^{c})=\\{e:e\in V(G)^{r}\setminus E(G)\\}$. Let $K^{r}_{t}$ denote the
complete $r$-graph on $t$ vertices. Let $K^{r-}_{t}$ be obtained by removing
one edge from $K_{t}^{r}$. For a positive integer $n$, let $[n]$ denote
$\\{1,2,3,\ldots,n\\}$.

For a family ${\mathcal{F}}$ of $r$-graphs, an $r$-graph $G$ is called
$\mathcal{F}$-free if it does not contain an isomorphic copy of any $r$-graph
of $\mathcal{F}$. For a fixed positive integer $n$ and a family of $r$-graphs
$\mathcal{F}$, the Turán number of $\mathcal{F}$, denoted by
$ex(n,\mathcal{F})$, is the maximum number of edges in an $\mathcal{F}$-free
$r$-graph on $n$ vertices. An averaging argument of Katona, Nemetz and
Simonovits [17] shows that the sequence ${ex(n,\mathcal{F})\over{n\choose r}}$
is non-increasing. Hence
$\underset{n\rightarrow\infty}{\lim}{ex(n,\mathcal{F})\over{n\choose r}}$
exists. The Turán density of $\mathcal{F}$ is defined as

$\pi(\mathcal{F})=\lim_{n\rightarrow\infty}{ex(n,\mathcal{F})\over{n\choose
r}}.$

If $\mathcal{F}$ consists of an single $r$-graph $F$, we simply write
$ex(n,\\{F\\})$ and $\pi(\\{F\\})$ as $ex(n,F)$ and $\pi(F)$. Denote

$\Pi_{\infty}^{(r)}=\\{\pi(\mathcal{F}):\mathcal{F}{\rm\ is\ a\ family\ of\
}r{\rm-uniform\ graphs}\\},$
$\Pi_{fin}^{(r)}=\\{\pi(\mathcal{F}):\mathcal{F}{\rm\ is\ a\ finite\ family\
of\ }r{\rm-uniform\ graphs}\\}$

and

$\Pi_{t}^{(r)}=\\{\pi(\mathcal{F}):\mathcal{F}{\rm\ is\ a\ family\ of\ }r{\rm-
uniform\ graphs\ and}\ |\mathcal{F}|\leq t\\}.$

Clearly,

$\Pi_{1}^{(r)}\subseteq\Pi_{2}^{(r)}\subseteq\cdots\subseteq\Pi_{fin}^{(r)}\subseteq\Pi_{\infty}^{(r)}.$

For 2-graphs, Erdős-Stone-Simonovits ([7], [8]) determined the Turán numbers
of all non-bipartite graphs asymptotically. Their result implies that

$\Pi_{\infty}^{(2)}=\Pi_{fin}^{(2)}=\Pi_{1}^{(2)}=\\{0,{1\over 2},{2\over
3},...,{l-1\over l},...\\}.$

Very few results are known for $r\geq 3$. In [5] Chung and Graham proposed the
conjecture that every element in $\Pi_{fin}^{(r)}$ is a rational number. Baber
and Talbot [2], and Pikhurko [27] disproved this conjecture by showing that
there is an irrational number in $\Pi_{3}^{(3)}$ and $\Pi_{fin}^{(3)}$,
respectively. Baber and Talbot [2] asked whether $\Pi_{1}^{(r)}$ contains an
irrational number. In this paper, we answer this question by showing that the
Lagrangian density of the disjoint union of $K_{4}^{3}$ and an edge is an
irrational number.

The hypergraph Lagrangian method has been helpful in hypergraph extremal
problems.

###### 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 the Lagrangian
function

$\lambda(G,\vec{x})=\sum_{e\in E(G)}\prod\limits_{i\in e}x_{i}.$

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

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

where

$\Delta=\\{\vec{x}=(x_{1},x_{2},\ldots,x_{n}):\sum_{i=1}^{n}x_{i}=1,x_{i}\geq
0\;for\;every\;i\in[n]\\}.$

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

In [22], Motzkin and Straus established a connection between the Lagrangian of
a $2$-graph and it’s maximum complete subgraphs.

###### Theorem 1.2

([22]) If $G$ is a $2$-graph in which a maximum complete subgraph has $t$
vertices, then $\lambda(G)=\lambda(K_{t}^{2})={1\over 2}(1-{1\over t})$.

They also applied this connection to give another proof of the theorem of
Turán on the Turán density of complete graphs. Since then the Lagrangian
method has been a useful tool in hypergraph extremal problems. Earlier
applications include that Frankl and Rödl [11] applied it in disproving the
long standing jumping constant conjecture of Erdős. Sidorenko [31] applied
Lagrangians of hypergraphs to first find infinitely many Turán densities of
hypergraphs. More recent developments of this method were obtained in [26, 2,
13, 24, 3, 25, 15, 16, 14, 37]. Determining the Lagrangian of a hypergraph is
much more difficult than graphs and there is no conclusion similar to Theorem
1.2 for hypergraphs. It is of great interests to estimate Lagrangians of
hypergraphs that have some certain properties. In 1980’s, Frankl and Füredi
[9] asked the question that for a given integer $m$, what is the maximum
Lagrangian among all $r$-graphs with $m$ edges? They conjectured that the
$r$-graph with $m$ edges formed by taking the first $m$ sets in the colex
ordering of ${\mathbb{N}}^{(r)}$ has the largest Lagrangian of all $r$-graphs
with $m$ edges. For distinct $A,B\in{\mathbb{N}}^{r}$ we say that $A$ is less
than $B$ in the colex ordering if $max(A\triangle B)\in B$, where $A\triangle
B=(A\setminus B)\cup(B\setminus A)$. By Theorem 1.2, this conjecture is true
when $r=2$. For hypergraphs, Talbot [32] first proved the conjecture for $r=3$
and ${\ell\choose 3}\leq m\leq{\ell\choose 3}+{{\ell-1}\choose 2}-\ell$, where
$\ell>0$ is an integer. Subsequent progress in this conjecture were made in
the papers of Tang, Peng, Zhang and Zhao [33, 34], Tyomkyn [36], Lei, Lu and
Peng [20], Nikiforov[23], Lei and Lu[19], and Lu[21]. Recently, Gruslys,
Letzter and Morrison [12] confirmed this conjecture for $r$ and ${\ell\choose
r}\leq m\leq{\ell\choose r}+{{\ell-1}\choose r-1}$ if $\ell$ is sufficiently
large. They also found infinitely many counterexamples for all $r\geq 4$. As
remarked in [12], it would be interesting to find the maximisers for other
values of $m$ though it might be a very hard problem. In this paper, we will
apply the connection of the Lagrangian density and the Turán density of an
$r$-graph to answer the question of Baber and Talbot. Our proof relies heavily
on the estimation of Lagrangians of $3$-graphs.

The Lagrangian density  $\pi_{\lambda}(F)$ of an $r$-graph $F$ is defined to
be

$\pi_{\lambda}(F)=\sup\\{r!\lambda(G):G\>\>{\rm is}\>\>F{\text{-}}{\rm
free}\\}.$

A pair of vertices $\\{i,j\\}$ is covered in a hypergraph $F$ if there exists
an edge $e$ in $F$ such that $\\{i,j\\}\subseteq e$. We say that $F$ covers
pairs if every pair of vertices in $F$ is covered. Let $r\geq 3$ and $F$ be an
$r$-graph. The extension of $F$, denoted by $H^{F}$ is obtained as follows:
For each pair of vertices $v_{i}$ and $v_{j}$ 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 the $B_{ij}$’s are pairwisely disjoint over all such pairs $\\{i,j\\}$.

The Lagrangian density is closely related to the Turán density. The following
proposition is implied by Theorem $2.6$ in [30] (see Proposition $5.6$ in [3]
and Corollary $1.8$ in [31] for the explicit statement).

###### Proposition 1.3

([30, 3, 31]) 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
$\pi(F)=\pi_{\lambda}(F).$

To answer the question of Baber and Talbot, we show that the Lagrangian
density of $\\{123,124,134,234,\\\ 567\\}$, the disjoint union of $K_{4}^{3}$
and an edge, denoted as $K_{4}^{3}\cup e$, is ${\sqrt{3}\over 3}$. The
following is our main Theorem.

###### Theorem 1.4

$\pi_{\lambda}(K_{4}^{3}\cup e)=\frac{\sqrt{3}}{3}$.

Applying Theorem 1.4 and Proposition 1.3, we see that the Turán density of the
extension of $K_{4}^{3}\cup e$ is ${\sqrt{3}\over 3}$.

For an $r$-graph $H$ on $t$ vertices, it is clear that $\pi_{\lambda}(H)\geq
r!\lambda{(K_{t-1}^{r})}$. An $r$-graph $H$ on $t$ vertices is
$\lambda$-perfect if $\pi_{\lambda}(H)=r!\lambda{(K_{t-1}^{r})}$. Theorem 1.2
implies that all $2$-graphs are $\lambda$-perfect. Theorem 1.4 indicates that
$K_{4}^{3}\cup e$ is not $\lambda$-perfect. We can show however that
$K_{4}^{3}\cup k\cdot e$, the disjoint union of $K_{4}^{3}$ and $k$ disjoint
edges is $\lambda$-perfect for $k\geq 2$.

###### Theorem 1.5

$K_{4}^{3}\cup k\cdot e$ is $\lambda$-perfect for $k\geq 2$.

In Section 2, we give a sketch of the proof of Theorem 1.4. In Section 3, we
will give the proof of Theorem 1.5. In Section 4, we give some preliminaries
on KKT conditions for continuous optimization problems and properties of
hypergraph Lagrangians. In Section 5, we prove the main Lemmas needed in the
proof of Theorem 1.4.

##  2 Sketch of the proof of Theorem 1.4

The following three 3-graphs are to be used throughout the paper.

B(2, n-2): the 3-graph with vertex set $[n]$ and edge set
$E(B(2,n-2))=\\{e\in{[n]\choose 3}:e\cap\\{1,2\\}\not=\emptyset\\}$, i.e.,
every edge in $B(2,n-2)$ contains vertex $1$ or $2$ or both. Note that
$B(2,n-2)$ is $K_{4}^{3}\cup e$-free, we will show that it is an extremal
$3$-graph for $K_{4}^{3}\cup e$ (in terms of Lagrangian density).

$\mathbf{X_{i}:}$ the 3-graph with vertex set $[2i+2]$ such that
$\\{1,2,2j+1,2j+2\\}$ form $K_{4}^{3}$ for all $j$, $1\leq j\leq i$, i.e., it
consists of $i$ copies of $K_{4}^{3}$ all sharing vertices $\\{1,2\\}$.

$\mathbf{Y_{i}:}$ the 3-graph with vertex set $[i+3]$ such that
$\\{1,2,3,j+3\\}$ form $K_{4}^{3}$ for all $j$, $1\leq j\leq i$, i.e., it
consists of $i$ copies of $K_{4}^{3}$ all sharing vertices $\\{1,2,3\\}$.

An $r$-graph $G$ is dense if $\lambda(G^{\prime})<\lambda(G)$ for every proper
subgraph $G^{\prime}$ of $G$.

Sketch of the proof of Theorem 1.4: For the lower bound, note that $B(2,n-2)$
is $K_{4}^{3}\cup e$-free, we shall prove
$\underset{n\to\infty}{\lim}\lambda(B(2,n))={\sqrt{3}\over 18}$ in Lemma 4.10.
So $\pi_{\lambda}(K_{4}^{3}\cup e)\geq
3!\underset{{n\rightarrow\infty}}{\lim}\lambda(B(2,n-2))={\sqrt{3}\over 3}$.

For the upper bound, let $G$ be a $K_{4}^{3}\cup e$-free 3-graph, our goal is
to show that $\lambda(G)\leq\frac{\sqrt{3}}{18}$. If $G$ is not dense, then
there exists a proper subgraph $G^{\prime}$ of $G$ such that
$\lambda{(G^{\prime})}=\lambda{(G)}$ and $|V(G^{\prime})|<|V(G)|.$ If
$G^{\prime}$ is dense, then we stop. Otherwise, we continue this process until
we find a dense subgraph $G^{\prime\prime}$ such that
$\lambda(G^{\prime\prime})=\lambda(G)$. This process terminates since the
number of vertices is reduced by at least one in each step. To show
$\lambda(G)\leq\frac{\sqrt{3}}{18}$, it’s sufficient to show that
$\lambda(G^{\prime\prime})\leq\frac{\sqrt{3}}{18}$. So we may assume that $G$
is a dense $K_{4}^{3}\cup e$-free 3-graph. Suppose that
$\lambda(G)>{\sqrt{3}\over 18}$, we will prove the following lemmas in Section
5.

###### Lemma 2.1

Let $G$ be a dense $K_{4}^{3}\cup e$-free 3-graph with
$\lambda(G)>{\sqrt{3}\over 18}$. Then $G$ is $X_{2}$-free.

###### Lemma 2.2

Let $G$ be a dense $K_{4}^{3}\cup e$-free 3-graph with
$\lambda(G)>{\sqrt{3}\over 18}$. Then $G$ contains at least two copies of
$K_{4}^{3}$.

###### Lemma 2.3

Let $G$ be a dense $K_{4}^{3}\cup e$-free 3-graph with
$\lambda(G)>{\sqrt{3}\over 18}$. Then $G$ is $Y_{2}$-free.

By Lemma 2.2, $G$ contains two copies of $K_{4}^{3}$. Since $G$ is
$K_{4}^{3}\cup e$-free, these two copies of $K_{4}^{3}$ must have $2$ or $3$
vertices in common. So $G$ contains a copy of $X_{2}$ or $Y_{2}$, a
contradiction to Lemmas 2.1 and 2.3.

To complete the proof of Theorem 1.4, what remains is to show those three main
lemmas. They will be given in Section 5.

##  3 The proof of Theorem 1.5

In order to prove Theorem 1.5, we need some lemmas form [38]. Let $S_{2,t}$
denote the 3-graph with vertex set $\\{v_{1},v_{2},u_{1},u_{2},...,u_{t}\\}$
and edge set $\\{v_{1}v_{2}u_{1},v_{1}v_{2}u_{2},...,v_{1}v_{2}u_{t}\\}$. A
result of Sidorenko in [31] implies that $S_{2,t}$ is $\lambda$-perfect.

###### Theorem 3.1

([38]) If $H$ is $\lambda$-perfect, then $H\cup S_{2,t}$ is $\lambda$-perfect
for any $t\geq 1$.

###### Claim 3.2

([38]) Let $G$ be a 3-graph with $\lambda(G)>\lambda(K_{k+1}^{3})$ and let
$\vec{x}$ be an optimal weight vector. Then for any $v\in V(G)$, its weight
$x_{v}$ satisfies that $x_{v}<1-\frac{\sqrt{k(k-1)}}{k+1}$.

###### Claim 3.3

([38]) Let $v$ be a vertex in a 3-graph $G$ and $x_{v}$ be the weight of v in
an optimal weight vector $\vec{x}$ of $G$. If $G-\\{v\\}$ is $H$-free, then
$\lambda(G)\leq\frac{\pi_{\lambda}(H)(1-x_{v})^{3}}{6(1-3x_{v})}$.

###### Remark 3.4

([38]) $f(x)=\frac{(1-x)^{3}}{1-3x}$ is increasing in $(0,\frac{1}{3}).$

###### Definition 3.5

For $v\in V(G)$, the link graph of $v$ in $G$, denote by $G_{v}$, is the graph
on vertex set $V(G)$ and the edge set $\\{e\setminus\\{v\\}:v\in e\in
E(G)\\}$. Let $\omega(G_{v})$ be the number of vertices in a maximum complete
subgraph in $G_{v}$.

###### Claim 3.6

([38]) Let a 3-graph $G$ be $H\cup S_{2,t}$-free, where H is a 3-graph with s
vertices. Let $v\in V(H)$. If $H\subseteq G-\\{v\\}$, then $\omega(G_{v})\leq
s+t$.

###### Claim 3.7

([38]) Let a 3-graph G be $H\cup S_{2,t}$-free, where H is a 3-graph with s
vertices. Let
$S_{2,s+t}=\\{v_{1}v_{2}b_{1},v_{1}v_{2}b_{2},...,v_{1}v_{2}b_{s+t}\\}\subseteq
G$. Then $G-\\{v_{1},v_{2}\\}$ is H-free.

###### Claim 3.8

([38]) Let a 3-graph G be $H\cup S_{2,t}$-free, where H is a 3-graph with s
vertices. If $H\subseteq G-\\{v_{1}\\}$ and $H\nsubseteq G-\\{v_{1},v_{2}\\}$,
then $\omega((G-\\{v_{2}\\})_{v_{1}})\leq s+t-1.$

_Proof of Theorem 1.5._ By Theorem 3.1, it’s sufficient to show that
$K_{4}^{3}\cup 2\cdot e$ is $\lambda$-perfect. Note that $K_{4}^{3}\cup 2\cdot
e$ has 10 vertices. It’s sufficient to show that if $G$ is $K_{4}^{3}\cup
2\cdot e$-free dense 3-graph then $\lambda(G)\leq\lambda(K_{9}^{3})$. Suppose
on the contrary that $\lambda(G)>\lambda(K_{9}^{3})={28\over 243}$. Let
$\vec{x}$ be an optimal weight vector of $G$.

_Case 1._ There exists $v\in V(G)$ with weight $x_{v}$ such that $G-\\{v\\}$
is $K_{4}^{3}\cup e$-free.

By Claim 3.2, $x_{v}<1-\frac{2\sqrt{14}}{9}$. By Theorem 1.4 and Claim 3.3,

$\lambda(G)\leq\frac{\frac{\sqrt{3}}{18}(1-x_{v})^{3}}{1-3x_{v}}=f(x_{v}).$
$None$

Since $f(x_{v})$ is increasing in $[0,1-\frac{2\sqrt{14}}{9}]$, then

$\displaystyle\lambda(G)$ $\displaystyle\leq$ $\displaystyle
f(1-\frac{2\sqrt{14}}{9})$ $\displaystyle=$
$\displaystyle\frac{28\sqrt{42}}{729(3\sqrt{14}-9)}$ $\displaystyle\leq$
$\displaystyle\frac{28}{243},$

a contradiction.

_Case 2._ For any $v\in V(G)$, $K_{4}^{3}\cup e\subseteq G-\\{v\\}$.

Since $\lambda(G)>\lambda(K_{9}^{3})$ and $S_{2,8}$ is $\lambda$-perfect, then
$S_{2,8}=\\{v_{1}v_{2}b_{1},v_{1}v_{2}b_{2},...,v_{1}v_{2}b_{8}\\}\subseteq
G$. By Claim 3.7, $G-\\{v_{1},v_{2}\\}$ is $K_{4}^{3}\cup e$-free. Applying
Claim 3.6 ($s=7,t=1$), we have

$\omega(G_{v_{1}})\leq 8\quad{\rm and}\quad\omega(G_{v_{2}})\leq 8.$

Applying Claim 3.8 ($s=7,t=1$), we have $\omega((G-\\{v_{2}\\})_{v_{1}})\leq
7$ and $\omega((G-\\{v_{1}\\})_{v_{2}})\leq 7$.

Assume the weight of $v_{1}$ and $v_{2}$ are $a_{1}$ and $a_{2}$ respectively,
and $a_{1}+a_{2}=2a$. Since $G-\\{v_{1},v_{2}\\}$ is $K_{4}^{3}\cup e$-free
and by Theorem 1.4, the contribution of edges containing neither $v_{1}$ nor
$v_{2}$ to $\lambda(G,\vec{x})$ is at most $\frac{\sqrt{3}}{18}(1-2a)^{3}$.
Since $\omega((G-\\{v_{2}\\})_{v_{1}})\leq 7$ and
$\omega((G-\\{v_{1}\\})_{v_{2}})\leq 7$, by Theorem 1.2, the contribution of
edges containing either $v_{1}$ or $v_{2}$ to $\lambda(G,\vec{x})$ is at most
$2\times\frac{1}{2}a(1-\frac{1}{7})(1-2a)^{2}$. The contribution of edges
containing both $v_{1}$ and $v_{2}$ to $\lambda(G,\vec{x})$ is at most
$a^{2}(1-2a)$. Therefore

$\displaystyle\lambda(G)$ $\displaystyle\leq$
$\displaystyle\frac{\sqrt{3}}{18}(1-2a)^{3}+a^{2}(1-2a)+\frac{6}{7}a(1-2a)^{2}$
$\displaystyle\leq$
$\displaystyle\frac{1}{10}(1-2a)^{3}+a^{2}(1-2a)+\frac{6}{7}a(1-2a)^{2}=f(a)$
$\displaystyle f^{\prime}(a)$ $\displaystyle=$
$\displaystyle\frac{66a^{2}-86a+9}{35}.$

Since $f(a)$ is increasing in $[0,\frac{43-\sqrt{1255}}{66}]$ and is
decreasing in $[\frac{43-\sqrt{1255}}{66},1]$, then $\lambda(G)\leq
f(\frac{43-\sqrt{1255}}{66})<\frac{28}{243}$. $\Box$

##  4 Preliminaries

###  4.1 Karush-Kuhn-Tucker Conditions

Let us consider the optimisation problem:

maximise $f(x)$

subject to $g_{i}(x)\leq 0$, $i=1,\dots,m,$ (3.1)

where $x\in\mathbb{R}^{n}$ and $f$ and $g_{i}$ are differentiable functions
from $\mathbb{R}^{n}$ to $\mathbb{R}$ for all $i$. Let $\nabla{f(x)}$ be the
gradient of $f$ at $x$ i.e. the vector in $\mathbb{R}^{n}$ whose $i$th
coordinate is ${\partial\over\partial{x_{i}}}f(x)$. We say that KKT conditions
hold at $x^{*}\in\mathbb{R}^{n}$ if there exist
$\lambda_{1},\dots\lambda_{m}\in\mathbb{R}$ such that

  1. (i)

$\nabla{f(x^{*})}=\sum_{i=1}^{m}\lambda_{i}\nabla{g_{i}(x^{*})},$

  2. (ii)

$\lambda_{i}\geq 0,i=1,\dots,m,$

  3. (iii)

$\lambda_{i}g_{i}(x^{*})=0,i=1,\dots,m.$

We call the constraints linear if $g_{1},\dots,g_{m}$ are all affine
functions.

###### Theorem 4.1

([4],[15]) If the constraints of (3.1) are linear, then any optimum point of
(3.1) must satisfy the KKT conditions.

###  4.2 Properties of the Lagrangian function

The following fact follows immediately from the definition of the Lagrangian.

###### Fact 4.2

Let $G_{1}$, $G_{2}$ be $r$-graphs and $G_{1}\subseteq G_{2}$. Then
$\lambda(G_{1})\leq\lambda(G_{2}).$

###### Fact 4.3

([11]) 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

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

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

Given an $r$-graph $G$, and $i,j\in V(G),$ define

$L_{G}(j\setminus i)=\\{e:i\notin e,e\cup\\{j\\}\in
E(G)\>and\>e\cup\\{i\\}\notin E(G)\\}.$

###### Fact 4.4

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 $i,j\in[n]$, $i\neq j$ satisfying
$L_{G}(i\setminus j)=L_{G}(j\setminus i)=\emptyset$. 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 $y_{i}=y_{j}={1\over
2}(x_{i}+x_{j})$. Then $\lambda(G,\vec{y})\geq\lambda(G,\vec{x})$.
Furthermore, if the pair $\\{i,j\\}$ is contained in an edge of $G$, $x_{i}>0$
for each $1\leq i\leq n$, and $\lambda(G,\vec{y})=\lambda(G,\vec{x})$, then
$x_{i}=x_{j}$.

Proof of Fact 4.4. Since $L_{G}(i\setminus j)=L_{G}(j\setminus i)=\emptyset$,
then

$\lambda(G,\vec{y})-\lambda(G,\vec{x})=\sum_{\\{i,j\\}\subseteq e\in
G}\left({(x_{i}+x_{j})^{2}\over 4}-x_{i}x_{j}\right)\prod\limits_{k\in
e\setminus\\{i,j\\}}x_{k}\geq 0.$

If the pair $\\{i,j\\}$ is contained in an edge of $G$ and $x_{i}>0$ for each
$1\leq i\leq n$, then the equality holds only if $x_{i}=x_{j}$. $\Box$

###### Fact 4.5

Let $\vec{x}=(x_{1},x_{2},\dots,x_{n})$ be an optimum vector for an $r$-graph
$G$ on $[n]$. If $L_{G}(j\setminus i)=\emptyset$, then we may assume that
$x_{i}\geq x_{j}$.

Proof of Fact 4.5. If $x_{i}<x_{j}$, then let $\epsilon={x_{j}-x_{i}\over 2}$
and
$\vec{x^{\prime}}=(x_{1},x_{2},\dots,x_{i}+\epsilon,\dots,x_{j}-\epsilon,\dots,x_{n})$.
Since $L_{G}(j\setminus i)=\emptyset$, then

$\lambda(G,\vec{x^{\prime}})-\lambda(G,\vec{x})\geq\sum_{\\{i,j\\}\subseteq
e\in G}((x_{i}+\epsilon)(x_{j}-\epsilon)-x_{i}x_{j})\prod\limits_{k\in
e\setminus\\{i,j\\}}x_{k}\geq{(x_{j}-x_{i})^{2}\over
4}\sum_{\\{i,j\\}\subseteq e\in G}\prod\limits_{k\in
e\setminus\\{i,j\\}}x_{k}\geq 0.$

$\Box$

###### Fact 4.6

([11]) Let $G=(V,E)$ be a dense $r$-graph. Then $G$ covers pairs.

Let $C_{r,m}$ denote the $r$-graph with $m$ edges formed by taking the first
$m$ sets in the colex ordering of $\mathbb{N}^{r}$. The following two results
was given in [35] and [32], respectively.

###### Lemma 4.7

([35]) Let $m$ and $t$ be positive integers satisfying the condition that
${t\choose 3}-6\leq m\leq{t\choose 3}-3$. Let $G$ be a 3-graph with $m$ edges.
Then $\lambda(G)\leq\lambda(C_{3,m})$.

###### Lemma 4.8

([32]) For any integers $m$, $t$ and $r$ satisfying the condition that
${t-1\choose r}\leq m\leq{t-1\choose r}+{t-2\choose r-1}$, then we have
$\lambda(C_{r,m})=\lambda(K_{t-1}^{r})$.

###### Fact 4.9

Let $f(x)=\frac{x^{2}(1-x)}{4}+\frac{x(1-x)^{2}}{2}$, where $0\leq x\leq 1$.
Then $f(x)\leq\frac{\sqrt{3}}{18}$ and equality holds only if
$x=\frac{3-\sqrt{3}}{3}$.

_Proof of Fact 4.9._ Since $f^{\prime}(x)=\frac{3x^{2}-6x+2}{4},$ then $f(x)$
is increasing when $x\in[0,\frac{3-\sqrt{3}}{3}]$ and decreasing when
$x\in[\frac{3-\sqrt{3}}{3},1]$. Therefore $f(x)\leq
f(\frac{3-\sqrt{3}}{3})=\frac{\sqrt{3}}{18}.$ $\Box$

###### Lemma 4.10

$\lambda(B(2,n-2))\leq\frac{\sqrt{3}}{18}$ and
$\underset{n\to+\infty}{\lim}\lambda(B(2,n-2))={\sqrt{3}\over 18}.$

_Proof of Lemma 4.10._ Let $\vec{x}=\\{x_{1},x_{2},\dots,x_{n}\\}$ be an
optimum vector of $\lambda(B(2,n-2))$. Let $x_{1}+x_{2}=a$ and $b=1-a$. Then

$\displaystyle\lambda(B(2,n-2))$ $\displaystyle\leq$
$\displaystyle\frac{a^{2}(1-a)}{4}+a\bigg{(}\frac{1-a}{n-2}\bigg{)}^{2}{n-2\choose
2}$ $\displaystyle\leq$
$\displaystyle\frac{a^{2}(1-a)}{4}+\frac{a(1-a)^{2}}{2}.$

By Fact 4.9, $\lambda(B(2,n-2))\leq\frac{\sqrt{3}}{18}$. Note that
${}^{\prime}=^{\prime}$ holds only if $a=\frac{3-\sqrt{3}}{3}$ and
$n\to\infty$. $\Box$

###  4.3 Preliminaries for the main Lemmas

In this section we introduce two hypergraphs and show their Lagrangian are
less than ${\sqrt{3}\over 18}$. When giving some proofs in Section 5, we will
change our hypergraph by replacing some edges such that the new hypergraph has
non-decreasing Lagrangian and is isomorphic with the subgraph of the following
two hypergraphs.

$\mathbf{H_{1}:}$ the 3-graph with vertex set $[n]$ and edge set
$E(H_{1})=E(B(2,n-2))\setminus\\{2ij:i,j\in[n]\setminus\\{1,2,3,4,5,6\\}\\}\cup\\{345,346\\}$.

$\mathbf{H_{2}:}$ a 3-graph with vertex set $[n]$ and edge set
$E(H_{2})=E(B(2,n-2))\setminus\\{2ij:i,j\in D\\}\cup\\{34i:i\in D\\}$, where
$D$ is a subset of $[n]\setminus\\{1,2,3,4\\}$ with $|D|\geq 2$.

###### Lemma 4.11

$\lambda(H_{1})<\frac{\sqrt{3}}{18}$.

_Proof of Lemma 4.11._ Let $\vec{x}=(x_{1},x_{2},...,x_{n})$ be an optimum
vector of $\lambda(H_{1})$. Assume that $x_{1}=a$, $x_{2}=b$, $x_{3}+x_{4}=c$,
$x_{5}+x_{6}=d$, $e=1-a-b-c-d$. By Fact 4.4, we may assume that
$x_{5}=x_{6}=\frac{d}{2}$ and $x_{3}=x_{4}=\frac{c}{2}$. By Fact 4.5, we may
assume that $a\geq b$. If $c=0$ or $d=0$, then
$\lambda(H_{1})<\lambda(B(2,n-2))\leq\frac{\sqrt{3}}{18}$. If $b=0$, then

$\displaystyle\lambda(H_{1})$ $\displaystyle\leq$ $\displaystyle
a(\frac{c^{2}}{4}+\frac{d^{2}}{4}+\frac{e^{2}}{2}+cd+ce+de)+\frac{c^{2}d}{4}$
$\displaystyle\leq$
$\displaystyle\frac{a(c+d+e)^{2}}{2}+\frac{(\frac{c}{2}+\frac{c}{2}+d+e)^{3}}{27}=\frac{a(1-a)^{2}}{2}+\frac{(1-a)^{3}}{27}=f(a).$
$\displaystyle f^{\prime}(a)$ $\displaystyle=$
$\displaystyle\frac{25a^{2}}{18}-\frac{16a}{9}+\frac{7}{18}.$

Note that $f(a)$ is increasing in $[0,\frac{7}{25}]$ and decreasing in
$[{7\over 25},1]$, then $\lambda\leq f(\frac{7}{25})<\frac{\sqrt{3}}{18}$. So
we may assume that $a,b,c,d>0$. By Fact 4.3,

$\frac{\partial\lambda}{\partial x_{1}}+\frac{\partial\lambda}{\partial
x_{2}}=\frac{\partial\lambda}{\partial x_{5}}+\frac{\partial\lambda}{\partial
x_{6}}.$

Note that

$\frac{\partial\lambda}{\partial x_{1}}+\frac{\partial\lambda}{\partial
x_{2}}\geq
bc+bd+be+cd+ce+de+\frac{c^{2}}{4}+\frac{d^{2}}{4}+ac+ad+ae+cd+ce+de+\frac{c^{2}}{4}+\frac{d^{2}}{4},$

and

$\frac{\partial\lambda}{\partial x_{5}}+\frac{\partial\lambda}{\partial
x_{6}}=ab+ac+ae+bc+be+(a+b)\frac{d}{2}+\frac{c^{2}}{4}+ab+ac+ae+bc+be+(a+b)\frac{d}{2}+\frac{c^{2}}{4}.$

Therefore

$2cd+2ce+2de+\frac{d^{2}}{2}\leq 2ab+ac+ae+bc+be.$

If $a+b\leq d$, then $2ab\leq\frac{d^{2}}{2}$, $ac+bc\leq cd$ and $ae+be\leq
de$. So we have $a=b$ and $e=c=0$, a contradiction to $c>0$. So we may assume
that $a+b>d$, then

$\displaystyle\lambda(H_{1})$ $\displaystyle\leq$ $\displaystyle
ab(c+d+e)+a(\frac{c^{2}}{4}+\frac{d^{2}}{4}+\frac{e^{2}}{2}+cd+ce+de)+b(\frac{c^{2}}{4}+\frac{d^{2}}{4}+cd+ce+de)+\frac{c^{2}d}{4}$
$\displaystyle\leq$ $\displaystyle
ab(1-a-b)+\frac{(a+b)(c+d+e)^{2}}{2}+\frac{c^{2}d}{4}-\frac{c^{2}a}{4}-\frac{c^{2}b}{4}$
$\displaystyle<$ $\displaystyle ab(1-a-b)+\frac{(a+b)(1-a-b)^{2}}{2}$
$\displaystyle\leq$
$\displaystyle\frac{(a+b)^{2}(1-a-b)}{4}+\frac{(a+b)(1-a-b)^{2}}{2}.$

By Fact 4.9, then $\lambda(H_{1})<\frac{\sqrt{3}}{18}.$ $\Box$

###### Lemma 4.12

$\lambda(H_{2})\leq\frac{\sqrt{3}}{18}$.

_Proof of Lemma 4.12._ Let $\vec{x}=(x_{1},x_{2},...,x_{n})$ be an optimum
vector of $\lambda(H_{2})$. Let $x_{1}=a$, $x_{2}=b$, $x_{3}+x_{4}=c$,
$\sum_{v\in D}x_{v}=d$, $\sum_{v\in E}x_{v}=e$. We have

$\displaystyle\lambda(H_{2})$ $\displaystyle\leq$ $\displaystyle
ab(c+d+e)+a(\frac{c^{2}}{2}+\frac{d^{2}}{2}+\frac{e^{2}}{2}+cd+ce+de)+b(\frac{c^{2}}{4}+\frac{e^{2}}{2}+cd+ce+de)+\frac{c^{2}d}{4}$
$\displaystyle=$ $\displaystyle\lambda(a,b,c,d,e)=\lambda$

under the constraint

$\left\\{\begin{array}[]{lr}a+b+c+d+e=1,&\\\ a\geq 0,b\geq 0,c\geq 0,d\geq 0\
and\ e\geq 0.&\end{array}\right.$ (1)

Note that if $c=0$ or $d=0$, then

$\displaystyle\lambda$ $\displaystyle<$
$\displaystyle\frac{(a+b)^{2}(1-a-b)}{4}+\frac{(a+b)(1-a-b)^{2}}{2},$

by Fact 4.9, then $\lambda\leq\frac{\sqrt{3}}{18}$. So we may assume that
$c,d>0$. By Theorem 4.1, then $\frac{\partial\lambda}{\partial
c}=\frac{\partial\lambda}{\partial d}$. By direct calculation,

$\displaystyle\frac{\partial\lambda}{\partial c}$ $\displaystyle=$
$\displaystyle ab+a(c+d+e)+b(\frac{c}{2}+d+e)+\frac{cd}{2}$
$\displaystyle\frac{\partial\lambda}{\partial d}$ $\displaystyle=$
$\displaystyle ab+a(c+d+e)+b(c+e)+\frac{c^{2}}{4},$

then $\frac{c}{2}(\frac{c}{2}+b)=d(\frac{c}{2}+b)$, so $c=2d.$

By Fact 4.5, we may assume that $a\geq b$. We claim that $b>0$. If $b=0$, then

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
a(\frac{c^{2}}{2}+\frac{d^{2}}{2}+\frac{e^{2}}{2}+cd+ce+de)+\frac{c^{2}d}{4}$
$\displaystyle\leq$
$\displaystyle\frac{a(c+d+e)^{2}}{2}+\bigg{(}\frac{\frac{c}{2}+\frac{c}{2}+d+e}{3}\bigg{)}^{3}$
$\displaystyle=$
$\displaystyle\frac{a(1-a)^{2}}{2}+\frac{(1-a)^{3}}{27}=f(a),$ $\displaystyle
f^{\prime}(a)$ $\displaystyle=$
$\displaystyle\frac{25a^{2}}{18}-\frac{16a}{9}+\frac{7}{18}.$

Note that $f(a)$ is increasing in $[0,\frac{7}{25}]$ and decreasing in
$[\frac{7}{25},1]$, then $\lambda\leq f(\frac{7}{25})\leq\frac{\sqrt{3}}{18}$.
So we may assume that $a,b>0$. By Theorem 4.1, then
$\frac{\partial\lambda}{\partial a}=\frac{\partial\lambda}{\partial b}$.
Therefore $(a-b)(c+d+e)=\frac{c^{2}}{4}+\frac{d^{2}}{2}.$

We claim that $e>0$. If $e=0$, recall that we have shown that $c=2d$ and
$(a-b)(c+d)=\frac{c^{2}}{4}+\frac{d^{2}}{2}$, then $a=b+\frac{d}{2}$. Since
$a+b+c+d=1$, then $a=\frac{1}{2}-\frac{5d}{4}$ and
$b=\frac{1}{2}-\frac{7d}{4}$. So

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{-53d^{3}-12d^{2}+12d}{16}$ $\displaystyle\lambda^{\prime}$
$\displaystyle=$ $\displaystyle\frac{-159d^{2}}{16}-\frac{3d}{2}+\frac{3}{4}.$

Note that $\lambda$ is increasing in $[0,\frac{2\sqrt{57}-4}{53}]$ and
decreasing in $[\frac{2\sqrt{57}-4}{53},1]$, so $\lambda\leq 0.094.$

Therefore we may assume that $a,b,c,d,e>0$. Recall that $c=2d$. By Theorem
4.1, then $\frac{\partial\lambda}{\partial d}=\frac{\partial\lambda}{\partial
e}$. Since

$\displaystyle\frac{\partial\lambda}{\partial d}$ $\displaystyle=$
$\displaystyle ab+a(c+d+e)+b(c+e)+\frac{c^{2}}{4}$
$\displaystyle\frac{\partial\lambda}{\partial e}$ $\displaystyle=$
$\displaystyle ab+a(c+d+e)+b(c+d+e),$

then $d=b$. So

$\displaystyle\lambda$ $\displaystyle<$
$\displaystyle\frac{(a+b)^{2}(1-a-b)}{4}+\frac{(a+b)(1-a-b)^{2}}{2}+\frac{c^{2}d}{4}-\frac{bc^{2}}{4}-\frac{bd^{2}}{2}$
$\displaystyle<$
$\displaystyle\frac{(a+b)^{2}(1-a-b)}{4}+\frac{(a+b)(1-a-b)^{2}}{2}.$

By Fact 4.9, then $\lambda\leq\frac{\sqrt{3}}{18}.$ $\Box$

##  5 Proofs of the main Lemmas

To complete the proof of Theorem 1.4, what remains is to show Lemma 2.1 to
2.3. In this section, we prove these lemmas. Throughout this section, let $G$
be a dense $K_{4}^{3}\cup e$-free 3-graph on vertex set $[n]$ with Lagrangian
$\lambda(G)>\frac{\sqrt{3}}{18}$. Let $\vec{x}=\\{x_{1},x_{2},\dots,x_{n}\\}$
be an optimum vector for $\lambda(G)$. Since
$\lambda(K_{6}^{3})=\frac{5}{54}\leq\frac{\sqrt{3}}{18}$, then $n\geq 7$.

###### Lemma 5.1

([1],[28]) $\pi(K_{4}^{3})\leq 0.5615$.

By Fact 4.6, Proposition 1.3 and Lemma 5.1,
$\pi_{\lambda}(K_{4}^{3})=\pi(K_{4}^{3})\leq
0.5615<3!\frac{\sqrt{3}}{18}<3!\lambda(G)$. So $K_{4}^{3}\subseteq G$. Without
loss of generality, assume that $\\{1,2,3,4\\}$ forms a $K_{4}^{3}$ in $G$.

For $x,y\in V(G)$, let $N^{*}(x,y)=\\{v:vxy\in E(G)\\}$.

###### Claim 5.2

$K_{5}^{3}\nsubseteq G$.

Proof of Claim 5.2. Assume that $K_{5}^{3}\subseteq G$. Since $n\geq 7$, then
there are $x,y\in[n]\setminus V(K_{5}^{3})$, since $G$ is dense, then there
exists $z\in N^{*}(x,y)$. The edge $xyz$ together with the $K_{5}^{3}$
contains a $K_{4}^{3}\cup e$, a contradiction. $\Box$

###### Claim 5.3

For $v\in[n]$, $\omega(G_{v})\geq 3$ and
$x_{v}<1-\sqrt{\frac{\sqrt{3}}{3}\frac{\omega(G_{v})}{\omega(G_{v})-1}}$.
Furthermore, if $v\in[n]\setminus\\{1,2,3,4\\}$, then $3\leq\omega(G_{v})\leq
4$. Therefore $x_{v}<0.0694$ if $\omega(G_{v})=3$, and $x_{v}<0.12262$ if
$\omega(G_{v})=4$.

Proof of Claim 5.3. For any $v\in[n]$, applying Fact 4.3 and Theorem 1.2, we
have

$\frac{\sqrt{3}}{6}<3\lambda=\frac{\partial\lambda}{\partial
x_{v}}\leq\bigg{(}\frac{1-x_{v}}{\omega(G_{v})}\bigg{)}^{2}{\omega(G_{v})\choose
2}.$

Then

$x_{v}<1-\sqrt{\frac{\sqrt{3}}{3}\frac{\omega(G_{v})}{\omega(G_{v})-1}}.$

If $v\in\\{1,2,3,4\\}$, then it’s clear that $\omega(G_{v})\geq 3$. Let
$v\notin[n]\setminus\\{1,2,3,4\\}$. If $\omega(G_{v})\leq 2$, then $x_{v}<0$,
a contradiction. Therefore $\omega(G_{v})\geq 3.$ If $\omega(G_{v})\geq 5$,
then $\omega(G_{v})$ contains at least 1 vertex in
$[n]\setminus\\{1,2,3,4\\}$, therefore $G$ contains either a $K_{5}^{3}$ (if
$\omega(G_{v})$ contains only 1 vertex in $[n]\setminus\\{1,2,3,4\\}$) or a
$K_{4}^{3}\cup e$ (if $\omega(G_{v})$ contains at least 2 vertices in
$[n]\setminus\\{1,2,3,4\\}$), a contradiction. Therefore
$3\leq\omega(G_{v})\leq 4$ for all $v\in[n]\setminus\\{1,2,3,4\\}$. $\Box$

###### Claim 5.4

For 2 vertices $a$ and $b$ in $[n]$, $x_{a}+x_{b}\leq\frac{3-\sqrt{3}}{3}$.

Proof of Claim 5.4. Applying Fact 4.3, we have

$\displaystyle\frac{\sqrt{3}}{3}$ $\displaystyle\leq$ $\displaystyle
6\lambda(G)=\frac{\partial\lambda(G)}{\partial
x_{a}}+\frac{\partial\lambda(G)}{\partial x_{b}}$ $\displaystyle\leq$
$\displaystyle
x_{b}(1-x_{a}-x_{b})+\bigg{(}\frac{1-x_{a}-x_{b}}{n-2}\bigg{)}^{2}{{n-2}\choose
2}+x_{a}(1-x_{a}-x_{b})+\bigg{(}\frac{1-x_{a}-x_{b}}{n-2}\bigg{)}^{2}{{n-2}\choose
2}$ $\displaystyle\leq$
$\displaystyle(x_{a}+x_{b})(1-x_{a}-x_{b})+(1-x_{a}-x_{b})^{2}$
$\displaystyle=$ $\displaystyle 1-(x_{a}+x_{b}).$

Therefore $x_{a}+x_{b}\leq\frac{3-\sqrt{3}}{3}.$ $\Box$

###### Claim 5.5

If $G-\\{v\\}$ is $K_{4}^{3}$-free for some $v\in[n]$, then $x_{v}>0.0848$.

Proof of Claim 5.5. Applying Lemma 5.1 and Proposition 1.3, we have

$\lambda(G-\\{v\\},\vec{x})\leq(1-x_{v})^{3}\frac{0.5615}{6}.$

Therefore

$\displaystyle\lambda(G)$ $\displaystyle\leq$
$\displaystyle(1-x_{v})^{3}\frac{0.5615}{6}+x_{v}\frac{\partial\lambda(G)}{\partial
x_{v}}$ $\displaystyle=$
$\displaystyle(1-x_{v})^{3}\frac{0.5615}{6}+3x_{v}\lambda(G),$

the last equality follows from Fact 4.3. So

$\displaystyle\lambda(G)$ $\displaystyle\leq$
$\displaystyle\frac{0.5615}{6}\frac{(1-x_{v})^{3}}{1-3x_{v}}.$

Note that $\frac{(1-x_{v})^{3}}{1-3x_{v}}$ is increasing in $[0,\frac{1}{3})$.
If $x_{v}\leq 0.0848$, then $\lambda(G)\leq 0.09622\leq\frac{\sqrt{3}}{18}$, a
contradiction. $\Box$

Let $M^{r}_{t}$ be an $r$-graph on $tr$ vertices with $t$ disjoint edges.

###### Claim 5.6

$n\geq 8$.

Proof of Claim 5.6. Assume that and $V(G)=[7]$. Recall that $\\{1,2,3,4\\}$
forms $K_{4}^{3}$. By Claim 5.3, then
$\omega(G_{5}),\omega(G_{6}),\omega(G_{7})\leq 4$ and $x_{5}+x_{6}+x_{7}\leq
3\times 0.12262$. Therefore $x_{1}+x_{2}+x_{3}+x_{4}\geq 0.63214>4\times
0.158,$ then, without loss of generality, let $x_{1}\geq 0.158$. By Claim 5.3,
$x_{v}<0.151$ if $\omega(G_{v})\leq 5$. Therefore $\omega(G_{1})=6$. Since $G$
is $K_{4}^{3}\cup e$-free, then $G-\\{1\\}$ is $M_{2}^{3}$-free. Hefetz and
Keevash ([13]) proved that $\pi_{\lambda}(M_{2}^{3})\leq\frac{12}{25}$. So
$\lambda(G-\\{1\\})\leq\frac{2}{25}$. Therefore

$\displaystyle\lambda(G)$ $\displaystyle\leq$ $\displaystyle
x_{1}\bigg{(}\frac{1-x_{1}}{6}\bigg{)}^{2}{6\choose
2}+\frac{2}{25}(1-x_{1})^{3}$ $\displaystyle=$
$\displaystyle\frac{5}{12}x_{1}(1-x_{1})^{2}+\frac{2}{25}(1-x_{1})^{3}=f(x_{1}).$
$\displaystyle f^{\prime}(x_{1})$ $\displaystyle=$
$\displaystyle\frac{5}{12}(1-x_{1})^{2}-\frac{5}{6}x_{1}(1-x_{1})-\frac{6}{25}(1-x_{1})^{2}$
$\displaystyle=$ $\displaystyle\frac{(1-x_{1})(53-303x_{1})}{300}.$

So $f(x_{1})$ is increasing in $[0,\frac{53}{303}]$ and decreasing in
$[\frac{53}{303},1]$, then $f(x_{1})\leq
f(\frac{53}{303})<0.095<\frac{\sqrt{3}}{18}.$ $\Box$

###  5.1 $G$ is $K_{5}^{3-}$-free

###### Lemma 5.7

$G$ is $K_{5}^{3-}$-free.

Proof of Lemma 5.7. Assume that $K_{5}^{3-}\subseteq G$ with vertex set
$\\{1,2,3,4,5\\}$ and $345\notin E(G)$. Since $G$ is $K_{4}^{3}\cup e$-free,
then for any $x,y\in[n]\setminus\\{1,2,3,4,5\\}$, we have
$N^{*}(x,y)\subseteq\\{1,2\\}$. By Claim 5.6, $n\geq 8$. Recall that
$\vec{x}=(x_{1},x_{2},\dots,x_{n})$ is an optimal vector for $G$.

Case 1. $x_{1},x_{2}\geq x_{3},x_{4},x_{5}$.

We say that a vertex $v\in[n]\setminus\\{1,2,3,4,5\\}$ is a good vertex if for
the set of edges in

$B_{v}=\\{vij\in E(G):\\{i,j\\}\in\\{3,4,5\\}^{(2)}\\},$

there exist the same number of triples in

$A_{v}=\\{vij\in
E(G^{c}):\\{i,j\\}\cap\\{1,2\\}\not=\emptyset,\\{i,j\\}\subset\\{1,2,3,4,5\\}\\}$

such that $\sum_{vij\in B_{v}}x_{v}x_{i}x_{j}\leq\sum_{vij\in
A_{v}}x_{v}x_{i}x_{j}$. In this case, we say that $B_{v}$ can be replaced by
$A_{v}$. Otherwise, we call $v$ a bad vertex.

We call $v34,v35,v45$ bad edges for $v\in[n]\setminus\\{1,2,3,4,5\\}$. Note
that for a good vertex $v\in[n]\setminus\\{1,2,3,4,5\\}$, replacing $B_{v}$ by
$A_{v}$ in $G$ does not reduce the Lagrangian. Let $B$ be the set of all bad
vertices. If $B=\emptyset$, then we can replace $B_{v}$ by $A_{v}$ in $G$ for
each $v\in[n]\setminus\\{1,2,3,4,5\\}$, obtain $G^{0}$, and all edges in
$E(G^{0})$ are incident to 1 or 2. So $G^{0}\subseteq B(2,n-2)$. Hence
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(B(2,n-2))=\frac{\sqrt{3}}{18}$, a
contradiction.

Case 1.1. There exists $v\in B$ such that exactly 1 of $v34,v35,v45$ is in
$E(G)$.

Without loss of generality, let $v34\in E(G)$. Since $G$ is $K_{5}^{3}$-free,
then at least 1 of $v12,v13,v23,v14,v24\notin E(G)$. Since $x_{1},x_{2}\geq
x_{3},x_{4}$, then $v34$ can be replaced by that missing edge, so $v\notin B$,
a contradiction.

Case 1.2. There exists $v\in B$ such that exactly 2 of $v34,v35,v45$ in
$E(G)$.

Without loss of generality, let $v34,v35\in E(G)$. Since $\\{v,1,2,3,4\\}$
can’t form a $K_{5}^{3}$, then at least 1 of $v12,v13,v23,v14,v24$ is not in
$E(G)$.

If $v12\notin E(G)$, then $v13,v14,v23,v24\in E(G)$. Otherwise we can replace
$v34,v35$ by that missing edge and $v12$ with $\lambda(G)$ non-decreasing.
Contradict to $v\in B$. Therefore $\\{v,1,3,4\\},\\{v,2,3,4\\}$ form
$K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup e$-free, then $N^{*}(x,y)=\emptyset$
for $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, a contradiction to that $G$ is
dense.

If $v14\notin E(G)$ (or $v24\notin E(G)$), then $\\{v,1,3,5\\},\\{v,2,3,5\\}$
form $K_{4}^{3}$. Otherwise we can replace $v34,v35$ by $v14$ and one of the
missing edges, a contradiction to $v\in B$. Since $G$ is $K_{4}^{3}\cup
e$-free, then $N^{*}(x,y)=\emptyset$ for
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, a contradiction to that $G$ is dense.

If $v23\notin E(G)$ (or $v13\notin E(G)$), then
$\\{v,1,3,4\\},\\{v,1,3,5\\},\\{v,1,2,4\\},\\{v,1,2,5\\}$ form $K_{4}^{3}$.
Since $G$ is $K_{4}^{3}\cup e$-free, then for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, we have $N^{*}(x,y)=\\{1\\}$ and
$x34,x35\notin E(G)$. So for any $x\in[n]\setminus\\{v,1,2,3,4,5\\}$ only,
possibly, $x45\in E(G)$ is a bad edge incident to $x$. By the proof of Case
1.1, $x\notin B$, so $B=\\{v\\}$. Replacing $v35$ by $v23,$ adding $345$ and
replacing $B_{x}$ by $A_{x}$ for all $x\in[n]\setminus\\{v,1,2,3,4,5\\}$, we
obtain $G^{0}$. Note that $G^{0}$ is contained in an isomorphic copy of
$H_{1}$ (view $v$ in $G^{0}$ as 6 in $H_{1}$), then
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(H_{1})\leq\frac{\sqrt{3}}{18}$ by
Lemma 4.11, a contradiction to $\lambda(G)>\frac{\sqrt{3}}{18}.$

Case 1.3. There exists $v\in B$ such that $v34,v35,v45\in E(G)$.

Since $v\in B$, then at most two of $\\{v12,v13,v23,v14,v24,v15,v25\\}$ are
not in $E(G)$, otherwise we can replace $B_{v}$ by the three missing edges. We
claim that there are 2 of those edges not in $E(G)$. Suppose that there is
only 1 of those edges missing in $E(G)$. If only $v12\notin E(G)$, then
$\\{v,1,3,4\\},\\{v,2,3,4\\}$ form $K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup
e$-free and $|G|\geq 8$, then there are $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$
such that $N^{*}(x,y)=\emptyset$, a contradiction to that $G$ is dense. If
only $vij\notin E(G)$ for some $i\in\\{1,2\\}$ and some $j\in\\{3,4,5\\}$,
then $\\{v,1,2,k,l\\}$ form a $K_{5}^{3}$, where
$\\{k,l\\}=\\{3,4,5\\}\setminus\\{j\\}$, a contradiction to Claim 5.2. So we
may assume that there are two of those edges not in $E(G)$.

If $v12\notin E(G)$ and $vij\notin E(G)$, where $i\in\\{1,2\\}$ and
$j\in\\{3,4,5\\}$, then $\\{v,1,k,l\\}$ and $\\{v,2,k,l\\}$
($\\{k,l\\}=\\{3,4,5\\}\setminus\\{j\\}$) form $K_{4}^{3}$. Since $G$ is
$K_{4}^{3}\cup e$-free, then $N^{*}(x,y)=\emptyset$ for
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, contradicting to $G$ being dense.

If $v1i,v2j\notin E(G)$, where $i,j\in\\{3,4,5\\}$, then $\\{v,1,p,q\\}$ and
$\\{v,2,s,t\\}$ form $K_{4}^{3}$, where
$\\{p,q\\}=\\{3,4,5\\}\setminus\\{i\\}$ and
$\\{s,t\\}=\\{3,4,5\\}\setminus\\{j\\}$. Since $G$ is $K_{4}^{3}\cup e$-free,
then $N^{*}(x,y)=\emptyset$ for $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$,
contradicting to $G$ being dense.

If $vij,vik\notin E(G)$, where $i\in\\{1,2\\}$ and
$\\{j,k\\}\in\\{3,4,5\\}^{(2)}$, without loss of generality, assume that
$v13,v14\notin E(G)$, then
$\\{v,2,3,4\\},\\{v,2,3,5\\},\\{v,2,4,5\\},\\{v,1,2,5\\}$ form $K_{4}^{3}$.
Since $G$ is $K_{4}^{3}\cup e$-free, then for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, we have $N^{*}(x,y)=\\{2\\}$ and
$x34\notin E(G)$. By the proof of Case 1.2, $x\notin B$, so $B=\\{v\\}$.
Replacing $B_{x}$ by $A_{x}$ for all $x\in[n]\setminus\\{v,1,2,3,4,5\\}$,
replacing $v35,v45$ by $v13,v14$ and adding 345, we obtain $G^{0}$ which is
contained in an isomorphic copy of $H_{1}$ (view $v$ in $G^{0}$ as 6 in
$H_{1}$). So
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(H_{1})\leq\frac{\sqrt{3}}{18}$ by
Lemma 4.11, a contradiction to $\lambda(G)>\frac{\sqrt{3}}{18}.$

Case 2. $x_{1},x_{3}\geq x_{2},x_{4},x_{5}.$

A vertex $v\in[n]\setminus\\{1,2,3,4,5\\}$ is a good vertex if for the edges
in

$B_{v}=\\{vij\in E(G):ij\in\\{2,4,5\\}^{(2)}\\},$

there exist the same number of triples in

$A_{v}=\\{vi^{\prime}j^{\prime}\in
E(G^{c}):\\{i^{\prime},j^{\prime}\\}\cap\\{1,3\\}\not=\emptyset,\\{i^{\prime},j^{\prime}\\}\subset\\{1,2,3,4,5\\}\\}$

such that the substitute $vi^{\prime}j^{\prime}$ for $vij\in B_{v}$ satisfies
$|\\{i^{\prime},j^{\prime}\\}\cap\\{i,j\\}|=1$ or
$\\{i^{\prime},j^{\prime}\\}=\\{1,3\\}$. Note that $\sum_{vij\in
B_{v}}x_{v}x_{i}x_{j}\leq\sum_{vij\in A_{v}}x_{v}x_{i}x_{j}$. In this case, we
say that $B_{v}$ can be replaced by $A_{v}$. Otherwise, we call $v$ a bad
vertex. We call $v24,v25,v45$ bad edges for $v\in[n]\setminus\\{1,2,3,4,5\\}$.

Let $B$ be the vertex set containing all bad vertices.

Case 2.1. There exists $v\in B$ and exactly one of $\\{v24,v25,v45\\}$ is in
$E(G)$.

Case 2.1.1. $v45\in E(G)$.

Since $v\in B$, then $v13,v14,v34,v15,v35\in E(G)$. Therefore
$\\{v,1,3,4\\},\\{v,1,3,5\\},\\{v,1,4,5\\}$ form $K_{4}^{3}$, then for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$ we have $N^{*}(x,y)=\\{1\\}$ and
$x23,x24,x25\notin E(G)$. Otherwise $K_{4}^{3}\cup e\subseteq G$. For any
$u\in B$, since $u24,u25\notin E(G)$, then $u45\in E(G)$ and
$\\{u,1,3,4\\},\\{u,1,3,5\\},\\\ \\{u,1,4,5\\}$ form $K_{4}^{3}$.

If $B=\\{v\\}$, then let $E=[n]\setminus\\{1,2,3,4,5,v\\}$. Note that
$N^{*}(x,y)=1$ for $x,y\in E$. Replacing $B_{x}$ by $A_{x}$ for $x\in E$, we
obtain $G^{0}$. Then $G^{0}$ is contained in an isomorphic copy of $H_{1}$
(view 3 in $G^{0}$ as 2 in $H_{1}$, view $245,v45$ in $G^{0}$ as $345,346$ in
$H_{1}$). So
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(H_{1})\leq\frac{\sqrt{3}}{18}$ by
Lemma 4.11, a contradiction.

If $B=\\{v,u\\}$, then let $E=[n]\setminus(B\cup\\{1,3,4,5\\})$. Note that
$2\in E$. Since $x$ is a good vertex for $x\in E\setminus\\{2\\}$, then $x45$
can be replaced by 1 of $\\{x13,x14,x15,x34,x35\\}$. Note that $N^{*}(x,y)=1$
for $x,y\in E\cup\\{2\\}$. Replace $uv2$ by $uv3$, then the obtained $G^{0}$
is contained in an isomorphic copy of $H_{1}$ (view 3 in $G^{0}$ as 2 in
$H_{1}$, view $v45,u45$ in $G^{0}$ as $345,346$ in $H_{1}$). So
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(H_{1})\leq\frac{\sqrt{3}}{18}$ by
Lemma 4.11, a contradiction.

If $|B|\geq 3$, then $N^{*}(x,y)=\\{1\\}$ for $x,y\in B$. Otherwise if $xy2\in
E(G)$, then $\\{z,1,4,5\\}\cup\\{x,y,2\\}$ forms $K_{4}^{3}\cup e$ in $G$ for
$x,y,z\in B$. Let $E=[n]\setminus(B\cup\\{1,3,4,5\\})$. Replacing $B_{x}$ by
$A_{x}$ for each $x\in E$, we obtain $G^{0}$ which is contained in an
isomorphic copy of $H_{2}$ (view $B$ in $G^{0}$ as $D$ in $H_{2}$, view 3 in
$G^{0}$ as 2 in $H_{2}$, view $i45(i\in B)$ in $G^{0}$ as $i34(i\in D)$ in
$H_{2}$). So
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(H_{2})\leq\frac{\sqrt{3}}{18}$ by
Lemma 4.12, a contradiction.

Case 2.1.2. $v2i\in E(G)$, where $i$ is 4 or 5.

Since $v\in B$, then $v12,v23,v1i,v3i,v13\in E(G)$, so $\\{v,1,2,3,i\\}$ forms
a $K_{5}^{3}$, a contradiction to Claim 5.2.

Case 2.2. There exists $v\in B$ and exactly two of $\\{v24,v25,v45\\}$ are in
$E(G)$.

Case 2.2.1. $v24,v25\in E(G)$.

Since $\\{v,1,2,3,4\\}$ can’t form a $K_{5}^{3}$, then at least one of
$\\{v12,v23,v14,v34,v13\\}$ is not in $E(G)$.

If $v12\notin E(G)$, then $v13,v14,v23,v34\in E(G)$. Otherwise we can replace
$v24,v25$ by $v12$ and that missing edge, a contradiction to $v\notin B$.
Therefore $\\{v,1,3,4\\},\\{v,2,3,4\\}$ form $K_{4}^{3}$. Since $G$ is
$K_{4}^{3}\cup e$-free, then $N^{*}(x,y)=\emptyset$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, a contradiction to that $G$ is dense.

If $v23\notin E(G)$, then $v12,v13,v14,v15,v34,v35\in E(G)$. Therefore
$\\{v,1,2,4\\},\\{v,1,2,5\\},\\{v,1,3,4\\},\\\ \\{v,1,3,5\\}$ form
$K_{4}^{3}$, then for all $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$ we have
$N^{*}(x,y)=\\{1\\}$ and $x24,x25\notin E(G)$, by the proof of Case 2.1,
$x\notin B$, so $B=\\{v\\}$. Let $E=[n]\setminus\\{1,2,3,4,5,v\\}$. Deleting
$v25$, adding $v23$, and replacing $B_{x}$ by $A_{x}$ for $x\in E$, we obtain
$G^{0}$ which is contained in an isomorphic copy of $H_{1}$ (view 3, 2 in
$G^{0}$ as 2, 3 in $H_{1}$ respectively, view $v24$ in $G^{0}$ as $346$ in
$H_{1}$). So
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(H_{1})\leq\frac{\sqrt{3}}{18}$ by
Lemma 4.11, a contradiction.

If $v34\notin E(G)$ (or $v14\notin E(G)$ or both), then
$\\{v12,v13,v15,v23,v35\\}\subset E(G)$, so $\\{v,1,3,5\\},\\{v,2,3,\\\ 5\\}$
form $K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup e$-free, then
$N^{*}(x,y)=\emptyset$ for all $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, a
contradiction to that $G$ is dense.

If $v13\notin E(G)$, then $\\{v12,v14,v15,v23,v34,v35\\}\subseteq E(G)$,
otherwise we can replace $v24,v25$ by that missing edge and $v13$, so
$\\{v,1,2,4\\},\\{v,1,2,5\\},\\{v,2,3,4\\},\\{v,2,3,5\\}$ form $K_{4}^{3}$.
Since $G$ is $K_{4}^{3}\cup e$-free, then $N^{*}(x,y)=\\{2\\}$ and
$x34,x35,x14\notin E(G)$ for all $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$. Note
that $x\notin B$ for $x\in[n]\setminus\\{v,1,2,3,4,5\\}$, otherwise we can
replace the all bad edges incident to $x$ by $x34,x35,x14$. Deleting $v25$,
adding $v13$, replacing $B_{x}$ by $A_{x}$ and replacing $xv2,xy2$ by
$xv3,xy1$ for $x,y\in[n]\setminus\\{1,2,3,4,5,v\\}$, we obtain $G^{0}$ which
is contained in an isomorphic copy of $H_{1}$ (view 3, 2 in $G^{0}$ as 2, 3 in
$H_{1}$ respectively, view $v24$ in $G^{0}$ as $346$ in $H_{1}$). So
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(H_{1})\leq\frac{\sqrt{3}}{18}$ by
Lemma 4.11, a contradiction.

Case 2.2.2. $v24,v45\in E(G)$ (the proof for $v25,v45\in E(G)$ is identical).

Since $\\{v,1,2,3,4\\}$ can’t form a $K_{5}^{3}$, then at least 1 of
$\\{v23,v12,v13,v14,v34\\}$ is not in $E(G)$.

If $v23\notin E(G)$, since $v\in B$, then $v13,v14,v15,v34,v35\in E(G)$. So
$\\{v,1,3,4\\},\\{v,1,3,5\\},\\{v,1,4,5\\}$ form $K_{4}^{3}.$ Since $G$ is
$K_{4}^{3}\cup e$-free, then $N^{*}(x,y)=\\{1\\}$ and $x23,x24,x25\notin E(G)$
for all $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$. By the proof of case 2.1, then
$x\notin B$, so $B=\\{v\\}$. Let $E=[n]\setminus\\{1,2,3,4,5,v\\}$. Deleting
$v24$, adding $v23$, deleting $xv2$, adding $xv3$ for $x\in E$, and replacing
$B_{x}$ by $A_{x}$ for $x\in E$, we obtain $G^{0}$ which is contained in an
isomorphic copy of $H_{1}$ (view 3 in $G^{0}$ as 2 in $H_{1}$, view
$\\{245,v45\\}$ in $G^{0}$ as $\\{345,346\\}$ in $H_{1}$). So
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(H_{1})\leq\frac{\sqrt{3}}{18}$ by
Lemma 4.11, a contradiction. So we may assume that $v23\in E(G)$.

If $v12\notin E(G)$, since $v\in B$, then $v13,v14,v23,v34\in E(G)$, so
$\\{v,1,3,4\\},\\{v,2,3,4\\}$ form $K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup
e$-free, then $N^{*}(x,y)=\emptyset$ for
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, contradicting to $G$ being dense. So we
may assume that $v12\in E(G)$.

If $v13\notin E(G)$, since $v\in B$, then $v14,v15,v23,v34\in E(G)$, so
$\\{v,1,4,5\\},\\{v,2,3,4\\}$ form $K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup
e$-free, then $N^{*}(x,y)=\emptyset$ for
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, contradicting to $G$ being dense. So we
may assume that $v13\in E(G)$.

If $v14\notin E(G)$, since $v\in B$, then $v13,v15,v23,v34,v35\in E(G)$, so
$\\{v,1,3,5\\},\\{v,2,3,4\\}$ form $K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup
e$-free, then $N^{*}(x,y)=\emptyset$ for
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, contradicting to $G$ being dense. So we
may assume that $v14\in E(G)$.

If $v34\notin E(G)$, since $v\in B$, then $v12,v13,v14,v15,v23,v35\in E(G)$,
so $\\{v,1,2,3\\},\\{v,1,2,4\\},\\{v,1,3,\\\ 5\\},\\{v,1,4,5\\}$ form
$K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup e$-free, then $N^{*}(x,y)=\\{1\\}$
and $x24,x45\notin E(G)$ for all $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, which
means, by the proof of Case 2.1., $x\notin B$ and $B=\\{v\\}$. Let
$E=[n]\setminus\\{1,2,3,4,5,v\\}$. Deleting $v24$, adding $v34$, replacing
$B_{x}$ by $A_{x}$, deleting $xv2$ and adding $xv3$ for $x\in E$, we obtain
$G^{0}$ which is contained in an isomorphic copy of $H_{1}$ (view 3 in $G^{0}$
as 2 in $H_{1}$, view $\\{245,v45\\}$ in $G^{0}$ as $\\{345,346\\}$ in
$H_{1}$). So
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(H_{1})\leq\frac{\sqrt{3}}{18}$ by
Lemma 4.11, a contradiction.

Case 2.3. There exists $v\in B$ and $v24,v25,v45\in E(G)$.

Since $v\in B$, then at most 2 of $\\{v12,v13,v14,v15,v23,v34,v35\\}$ are not
in $E(G)$, otherwise we can replace $B_{v}=\\{v24,v25,v45\\}$ by those 3
missing edges in $A_{v}$. We claim that exactly 2 of
$\\{v12,v13,v14,v15,v23,\\\ v34,v35\\}$ are not in $E(G)$. Otherwise there is
only 1 of those edges not in $E(G)$. If only $v12\notin E(G)$ (or only
$v23\notin E(G)$), then $\\{v,1,4,5\\}$ , $\\{v,2,4,5\\}$ form $K_{4}^{3}$.
Since $G$ is $K_{4}^{3}\cup e$-free, then $N^{*}(x,y)=\emptyset$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, a contradiction to $G$ being dense. If
only $v13$ (or one of $\\{v14,v15,v34,v35\\})$ is not in $E(G)$, then
$\\{v,1,2,4,5\\}$ forms an $K_{5}^{3}$, a contradiction. So we can assume that
there are exactly two of $\\{v12,v13,v14,v15,v23,v34,v35\\}$ not in $E(G)$.

If $v12\notin E(G)$, and $v23\notin E(G)$ (the discussion for $v23$ replaced
by 1 of $\\{v13,v14,v15,v34,v35\\}$ is similar), then
$\\{v,1,4,5\\},\\{v,2,4,5\\}$ form $K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup
e$-free, then $N^{*}(x,y)=\emptyset$ for
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, contradicting to $G$ being dense. So we
may assume that $v12\in E(G)$.

If $v13\notin E(G)$, and $v23\notin E(G)$ (or 1 of $\\{v34,v35\\}\notin
E(G)$), then $\\{v,1,4,5\\},\\{v,2,4,5\\}$ form $K_{4}^{3}$. Therefore
$N^{*}(x,y)=\emptyset$ for all $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, a
contradiction. If $v13\notin E(G)$ and $v14\notin E(G)$ (or $v15\notin E(G)$,
since 4 and 5 are symmetric, we only discuss $v14$ here), then
$\\{v,2,3,4\\},\\{v,2,3,5\\},\\{v,2,4,5\\},\\{v,1,2,5\\}$ form $K_{4}^{3}$.
Therefore $N^{*}(x,y)=\\{2\\}$ and $x13,x14,x15,xv3\notin E(G)$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$. Since we can replace $B_{x}$ by
$x13,x14,x15$ for $x\in[n]\setminus\\{v,1,2,3,4,5\\}$, then $x\notin B$, so
$B=\\{v\\}$. Let $E=[n]\setminus\\{1,2,3,4,5,v\\}$. Deleting $v24$ and $v25$,
adding $v13$ and $v14$, deleting $xv2$, adding $xv3$, deleting $xy2$, adding
$xy1$ for $x,y\in E$, and replacing $B_{x}$ by $A_{x}$ for $x\in E$, we obtain
$G^{0}$. Note that $G^{0}$ is contained in an isomorphic copy of $H_{1}$ (view
3 in $G^{0}$ as 2 in $H_{1}$, view $\\{245,v45\\}$ in $G^{0}$ as
$\\{345,346\\}$ in $H_{1}$). So
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(H_{1})\leq\frac{\sqrt{3}}{18}$ by
Lemma 4.11, a contradiction. So we may assume that $v13\in E(G)$.

If $v14\notin E(G)$, and $v23\notin E(G)$ (or $v34\notin E(G)$), then
$\\{v,1,3,5\\}$ and $\\{v,2,4,5\\}$ form $K_{4}^{3}$. Therefore
$N^{*}(x,y)=\emptyset$ for all $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, a
contradiction. If $v14,v15\notin E(G)$ ($v14,v35\notin E(G)$ is similar), then
$\\{v,1,2,3\\},\\{v,2,3,4\\},\\{v,2,3,5\\},\\{v,2,4,5\\}$ form $K_{4}^{3}$.
Therefore $N^{*}(x,y)=\\{2\\}$ and $x13,x14,x15,x45\notin E(G)$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$. So we can replace $B_{x}$ by
$x13,x14,x15$ for $x\in[n]\setminus\\{v,1,2,3,4,5\\}$, then $x\notin B$, and
$B=\\{v\\}$. Let $E=[n]\setminus\\{1,2,3,4,5,v\\}$. Deleting $v24$ and $v25$,
adding $v14$ and $v15$, deleting $xv2$, adding $xv3$, deleting $xy2$, adding
$xy1$ for $x,y\in E$, and replacing $B_{x}$ by $A_{x}$ for $x\in E$, we obtain
$G^{0}$ which is contained in an isomorphic copy of $H_{1}$ (view 3 in $G^{0}$
as 2 in $H_{1}$, view $\\{245,v45\\}$ in $G^{0}$ as $\\{345,346\\}$ in
$H_{1}$). By Lemma 4.11, then
$\lambda(G)\leq\lambda(G^{0})\leq\frac{\sqrt{3}}{18}$, a contradiction. So we
may assume that $v14\in E(G)$.

Similar to $v14$, we may assume that $v15\in E(G)$, but then $\\{v,1,2,4,5\\}$
forms a $K_{5}^{3}$, a contradiction to Claim 5.2.

Case 3. $x_{3},x_{4}\geq x_{1},x_{2},x_{5}$.

A vertex $v\in[n]\setminus\\{1,2,3,4,5\\}$ is a good vertex if for the edges
in

$B_{v}=\\{vij\in E(G):ij\in\\{1,2,5\\}^{(2)}\\},$

there exist the same number of triples in

$A_{v}=\\{vi^{\prime}j^{\prime}\in
E(G^{c}):\\{i^{\prime},j^{\prime}\\}\cap\\{3,4\\}\not=\emptyset,\\{i^{\prime},j^{\prime}\\}\subset\\{1,2,3,4,5\\}\\}$

such that the substitute $vi^{\prime}j^{\prime}$ for $vij\in B_{v}$ satisfies
$|\\{i^{\prime},j^{\prime}\\}\cap\\{i,j\\}|=1$ or
$\\{i^{\prime},j^{\prime}\\}=\\{3,4\\}$. Note that $\sum_{vij\in
B_{v}}x_{v}x_{i}x_{j}\leq\sum_{vij\in A_{v}}x_{v}x_{i}x_{j}$. In this case, we
say that $B_{v}$ can be replaced by $A_{v}$. Otherwise we call $v$ a bad
vertex. We call $v12,v15,v25$ bad edges for $v\in[n]\setminus\\{1,2,3,4,5\\}$.

Let $B$ be the set of all bad vertices. Let
$E=[n]\setminus(B\cup\\{1,2,3,4,5\\})$.

Case 3.1. There exists $v\in B$ and there is exactly 1 of $v12,v15,v25\in
E(G)$.

Case 3.1.1. $v12\in E(G)$.

Since $v\in B$, then $\\{v13,v14,v23,v24,v34\\}\subseteq E(G)$, so
$\\{v,1,2,3,4\\}$ forms a $K_{5}^{3}$, a contradiction to Claim 5.2.

Case 3.1.2. $v15\in E(G)$ (the case $v25\in E(G)$ is similar).

Since $v\in B$, then $v13,v14,v34,v35,v45\in E(G)$. So
$\\{v,1,3,4\\},\\{v,1,3,5\\},\\{v,1,4,5\\}$ form $K_{4}^{3}$. Since $G$ is
$K_{4}^{3}\cup e$-free, then $N^{*}(x,y)=\\{1\\}$ and
$v12,v25,x23,x24,x25\notin E(G)$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$. Therefore for all $u\in B$, only
possibly, $u12,u15\in E(G)$. Since we can replace $u12$ by $u23$ and $u\in B$,
then $u15$ can’t be replaced. So $\\{u,1,3,4\\},\\{u,1,3,5\\},\\{u,1,4,5\\}$
form $K_{4}^{3}$.

If $|B|=1$, i.e. $B=\\{v\\}$, then replacing $B_{x}$ by $A_{x}$, deleting
$xv1,xv2,xy1$, adding $xv3,xv4,xy3$ for all $x,y\in E$, we obtain $G^{0}$
which is contained in an isomorphic copy of $H_{1}$ (view 3, 4 in $G^{0}$ as
1, 2 in $H_{1}$ respectively, view $\\{v15,125\\}$ in $G^{0}$ as
$\\{345,346\\}$ in $H_{1}$). Then
$\lambda(G)\leq\lambda(G^{0})\leq\frac{\sqrt{3}}{18}$ by Lemma 4.11, a
contradiction.

If $|B|=2$, i.e. $B=\\{v,v^{\prime}\\}$, then $v^{\prime}15\in E(G)$ (or
$v^{\prime}12,v^{\prime}15\in E(G)$ but we can replace $v^{\prime}12$ by
$v^{\prime}23$). Deleting $vv^{\prime}1,vv^{\prime}2$, adding
$vv^{\prime}3,vv^{\prime}4$, and replacing $xy1,xv1,xv^{\prime}1,x12,125$ by
$xy3,xv3,xv^{\prime}3,x23,345$ respectively for all $x,y\in E$, we obtain
$G^{0}$. View $3,4,1,5,v,v^{\prime}$ in $G^{0}$ as $1,2,3,4,5,6$ in $H_{1}$,
respectively. Note that $N^{*}(x,2)=\\{3\\}$ in $G^{0}$ for $x\in E$, so
$G^{0}$ is contained in an isomorphic copy of $H_{1}$. Hence
$\lambda(G)\leq\lambda(G^{0})\leq\frac{\sqrt{3}}{18}$ by Lemma 4.11, a
contradiction.

If $|B|\geq 3$, then $vv^{\prime}2,xv2\notin E(G)$ for any $v,v^{\prime}\in B$
and $x\in E$, since otherwise
$\\{v^{\prime\prime},1,3,4\\}\cup\\{v,v^{\prime},2\\}$ forms $K_{4}^{3}\cup e$
for $v^{\prime\prime}\in B$ and $x\in E$. Replacing $vv^{\prime}1$ by
$vv^{\prime}3$, deleting $xy1,xv1$, adding $xy3,xv3$ for all $x,y\in E$ and
$v,v^{\prime}\in B$, and replacing $B_{x}$ by $A_{x}$ for $x\in E$, we obtain
$G^{0}$. So $N^{*}(v,v^{\prime})=\\{3\\}$ for $v,v^{\prime}\in B$ in $G^{0}$.
View $\\{3,4\\}$ in $G^{0}$ as $\\{1,2\\}$ in $H_{2}$, view $v15$ in $G^{0}$
($v\in B$) as $i34$ ($i\in D$) in $H_{2}$, then $G^{0}$ is contained in an
isomorphic copy of $H_{2}$. So
$\lambda(G)\leq\lambda(G^{0})\leq\lambda(H_{2})\leq\frac{\sqrt{3}}{18}$ by
Lemma 4.12, a contradiction.

Case 3.2. $v\in B$ and exactly 2 of $v12,v15,v25\in E(G)$.

Case 3.2.1. $v12,v15\in E(G)$. (The case that $v12,v25\in E(G)$ is similar.)

Since $\\{v,1,2,3,4\\}$ can’t form a $K_{5}^{3}$, then there is at least 1 of
$\\{v13,v14,v23,v24,v34\\}$ not in $E(G)$. Since $v\in B$, then we may assume
that $v35,v45\in E(G)$. Otherwise we may replace $v12,v15$ by 1 of
$\\{v13,v14,v23,v24,v34\\}$ and 1 of $v35,v45\in E(G)$, a contradiction. If
$v13\notin E(G)$ (or $v14\notin E(G)$), since $v\in B$, then $\\{v,2,3,4\\}$
and $\\{v,1,4,5\\}$ (or $\\{v,1,3,5\\}$) form $K_{4}^{3}$, then
$N^{*}(x,y)=\emptyset$ for all $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$. So we
may assume that $v13,v14\in E(G)$.

If $v23\notin E(G)$ (or $v24\notin E(G)$ or both), since $v\in B$, then
$v34\in E(G)$. So $\\{v,1,3,4\\},\\{v,1,3,5\\},\\{v,1,\\\ 4,5\\}$ form
$K_{4}^{3}$. So $N^{*}(x,y)=\\{1\\}$ and $x23,x24,x25\notin E(G)$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$. If $v^{\prime}\in B\setminus\\{v\\}$,
note that $v^{\prime}25\notin E(G)$, only possibly,
$v^{\prime}12,v^{\prime}15\in E(G)$. And $v^{\prime}12$ can be replaced by
$v^{\prime}23$, so $v^{\prime}15$ can’t be replaced, therefore
$\\{v^{\prime},1,3,4\\},\\{v^{\prime},1,3,5\\},\\{v^{\prime},1,4,5\\}$ form
$K_{4}^{3}$.

If $B=\\{v\\}$, then replacing $B_{x}$ by $A_{x}$ for $x\in E$, deleting
$xy1,xv1,xv2$, and adding $xy3,xv3,xv4$ for $x,y\in E$, we obtain $G^{0}$
which is contained in an isomorphic copy of $H_{1}$ (view $3,4,v$ in $G^{0}$
as $1,2,6$ in $H_{1}$ respectively, view $\\{v15,125\\}$ in $G^{0}$ as
$\\{345,346\\}$ in $H_{1}$), a contradiction.

If $B=\\{v,v^{\prime}\\}$, then replace $v12$ by $v23$. Delete
$125,1vv^{\prime},2vv^{\prime},xy1$ and add
$345,3vv^{\prime},4vv^{\prime},xy3$ for $x,y\in E$, respectively. Replace
$x12$ by $x23$. Since $x$ is a good vertex for $x\in E$, then $x15$ can be
replaced by 1 of the missing edges in $\\{x13,x14,x34,x35,x45\\}$. Let $G^{0}$
be the resulting 3-graph, note that $N^{*}(x,y)=\\{3\\}$ for $x,y\in
E\cup\\{2\\}$ in $G^{0}$ (view $3,4,1,5,v,v^{\prime}$ in $G^{0}$ as
$1,2,3,4,5,6$ in $H_{1}$ respectively). Then $G^{0}$ is contained in an
isomorphic copy of $H_{1}$, a contradiction.

If $|B|\geq 3$, then replace $v12$ by $v23$ for $v\in B$. Delete $125$ and add
$345$ for $x\in E$. Replace $B_{x}$ by $A_{x}$ for $x\in E$. Since $G$ is
$K_{4}^{3}\cup e$-free, then $N^{*}(v,v^{\prime})=\\{1\\}$ for any
$v,v^{\prime}\in B$. Replace $vv^{\prime}1$ by $vv^{\prime}3$ for
$v,v^{\prime}\in B$. Let $G^{0}$ be the resulting 3-graph. Note that
$N^{*}(x,y)=\\{3\\}$ for $x,y\in B$ in $G^{0}$. View $\\{3,4\\}$ in $G^{0}$ as
$\\{1,2\\}$ in $H_{2}$, view $v15$ in $G^{0}$ ($v\in B$) as $i34$ ($i\in D$)
in $H_{2}$. Then $G^{0}$ is contained in an isomorphic copy of $H_{2}$, a
contradiction.

So we may assume that $v23,v24\in E(G)$, then $v34\notin E(G)$. Therefore
$\\{v,1,2,3\\},\\{v,1,2,4\\},\\{v,1,3,\\\ 5\\},\\{v,1,4,5\\}$ form
$K_{4}^{3}$. So $N^{*}(x,y)=\\{1\\}$ and $x23,x24,x35,x45\notin E(G)$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, so $B=\\{v\\}$. Deleting
$v12,1xv,2xv,xy1$, adding $v34,3xv,4xv,xy3$ for $x,y\in E$, replacing $B_{x}$
by $A_{x}$ for $x\in E$, we obtain $G^{0}$. View $\\{3,4\\}$ in $G^{0}$ as
$\\{1,2\\}$ in $H_{1}$, view $\\{v15,125\\}$ in $G^{0}$ as $\\{345,346\\}$ in
$H_{1}$. Then $G^{0}$ is contained in an isomorphic copy of $H_{1}$, a
contradiction.

Case 3.2.2. $v15,v25\in E(G)$.

If $v34\notin E(G)$, since $v\in B$, then $\\{v,1,4,5\\},\\{v,2,4,5\\}$ form
$K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup e$-free, then $N^{*}(x,y)=\emptyset$
for $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, contradicting to $G$ being dense.
Therefore $v34\in E(G)$.

If $v35\notin E(G)$ (or $v45\notin E(G)$), then $\\{v,1,3,4\\},\\{v,2,3,4\\}$
form $K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup e$-free, then
$N^{*}(x,y)=\emptyset$ for $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$,
contradicting to $G$ being dense. Therefore $v35,v45\in E(G)$.

If $v13\notin E(G)$ (or $v14\notin E(G)$ or neither $v13$ nor $v14$ is in
$E(G)$ or $v23\notin E(G)$ or $v24\notin E(G)$ or neither $v23$ nor $v24$ is
in $E(G)$), then $\\{v,2,3,4\\},\\{v,2,3,5\\},\\{v,2,4,5\\}$ form $K_{4}^{3}$.
So $N^{*}(x,y)=\\{2\\}$ and $x13,x14,x15\notin E(G)$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$. By Case 3.2.1, $x\notin B$ for
$x\in[n]\setminus\\{v,1,2,3,4,5\\}$, so $B=\\{v\\}$. Replacing $B_{x}$ by
$A_{x}$ for $x\in E$, deleting $xy2,v15,1xv,2xv$, and adding
$xy3,v13,3xv,4xv$, respectively, we obtain$G^{0}$. Note that
$N^{*}(x,y)=\\{3\\}$ for $x,y\in E$ in $G^{0}$. View $\\{3,4\\}$ in $G^{0}$ as
$\\{1,2\\}$ in $H_{1}$, view $\\{v25,125\\}$ in $G^{0}$ as $\\{345,346\\}$ in
$H_{1}$. Then $G^{0}$ is contained in an isomorphic copy of $H_{1}$, a
contradiction.

Case 3.3. There exists $v\in B$ and $v12,v15,v25\in E(G)$.

Since $v\in B$, then there are at most 2 of
$\\{v13,v14,v23,v24,v34,v35,v45\\}$ not in $E(G)$. We claim that there are
exactly 2 of those edges not in $E(G)$. Otherwise if there is at most 1 of
$\\{v13,v14,v23,v24,v34,\\\ v35,v45\\}$ not in $E(G)$, without loss of
generality, say at most $v13\notin E(G)$, then $\\{v,1,4,5\\},\\{v,2,4,5\\}$
form $K_{4}^{3}$. So $N^{*}(x,y)=\emptyset$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, a contradiction. So we may assume that
there are exactly 2 of $v13,v14,v23,v24,v34,v35,v45$ not in $E(G)$.

Assume that $v34\notin E(G)$. If $v13\notin E(G)$ (the case that 1 of
$\\{v14,v23,v24\\}$ is not in $E(G)$ is similar), then
$\\{v,1,4,5\\},\\{v,2,3,5\\}$ form $K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup
e$-free, then $N^{*}(x,y)=\emptyset$ for
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, contradicting to $G$ being dense. If
$v35\notin E(G)$ (or $v45\notin E(G)$), then $\\{v,1,4,5\\},\\{v,2,4,5\\}$
form $K_{4}^{3}$. Since $G$ is $K_{4}^{3}\cup e$-free, then
$N^{*}(x,y)=\emptyset$ for $x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$,
contradicting to $G$ being dense. So we may assume that $v34\in E(G)$.

Assume that 1 of $\\{v13,v14,v23,v24\\}$ is not in $E(G)$. Without loss of
generality, let $v13\notin E(G)$. If $v14\notin E(G)$, then
$\\{v,1,2,5\\},\\{v,2,3,4\\},\\{v,2,3,5\\},\\{v,2,4,5\\}$ form $K_{4}^{3}$. So
$N^{*}(x,y)=\\{2\\}$ and $x13,x14,x34,x35\notin E(G)$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$. Therefore $B=\\{v\\}$. Replacing
$v12,v15$ by $v13,v14$, replacing $B_{x}$ by $A_{x}$ for $x\in E$, deleting
$1xv,2xv,xy2$ and adding $3xv,4xv,xy3$ for $x,y\in E$, respectively. Let
$G^{0}$ be the resulting 3-graph. Note that $N^{*}(x,y)=\\{3\\}$ for $x,y\in
E$ in $G^{0}$. View $\\{3,4\\}$ in $G^{0}$ as $\\{1,2\\}$ in $H_{1}$, view
$\\{v25,125\\}$ in $G^{0}$ as $\\{345,346\\}$ in $H_{1}$. Then $G^{0}$ is
contained in an isomorphic copy of $H_{1}$, a contradiction. If $v23\notin
E(G)$ (or $v35\notin E(G)$), then $\\{v,1,4,5\\},\\{v,2,4,5\\}$ form
$K_{4}^{3}$. So $N^{*}(x,y)=\emptyset$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, a contradiction. If $v24\notin E(G)$,
then $\\{v,1,4,5\\},\\{v,2,3,5\\}$ form $K_{4}^{3}$. Since $G$ is
$K_{4}^{3}\cup e$-free, then $N^{*}(x,y)=\emptyset$ for
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, contradicting to $G$ being dense. If
$v45\notin E(G)$, then
$\\{v,1,2,4\\},\\{v,1,2,5\\},\\{v,2,3,4\\},\\{v,2,3,5\\}$ form $K_{4}^{3}$. So
$N^{*}(x,y)=\\{2\\}$ and $x14,x15,x34,x35\notin E(G)$ for all
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$. Therefore $B=\\{v\\}$. Replacing
$v12,v25$ by $v13,v45$, replacing $B_{x}$ by $A_{x}$ for $x\in E$, deleting
$1xv,2xv,xy2$ and adding $3xv,4xv,xy3$ for $x,y\in E$, respectively. Let
$G^{0}$ be the resulting 3-graph. Note that $N^{*}(x,y)=\\{3\\}$ for $x,y\in
E$ in $G^{0}$. View $\\{3,4\\}$ in $G^{0}$ as $\\{1,2\\}$ in $H_{1}$, view
$\\{v15,125\\}$ in $G^{0}$ as $\\{345,346\\}$ in $H_{1}$. Then $G^{0}$ is
contained in an isomorphic copy of $H_{1}$, a contradiction. So it’s
sufficient to consider $v35,v45\notin E(G)$. However
$\\{v,1,3,4\\},\\{v,2,3,4\\}$ form $K_{4}^{3}$ in this situation. Since $G$ is
$K_{4}^{3}\cup e$-free, then $N^{*}(x,y)=\emptyset$ for
$x,y\in[n]\setminus\\{v,1,2,3,4,5\\}$, contradicting to $G$ being dense.
$\Box$

###  5.2 $G$ does not contain two copies of $K_{4}^{3}$ sharing two vertices

Before giving the proof of Lemma 2.1, we will prove the following Lemmas.

###### Lemma 5.8

$G$ is $X_{4}$-free.

Proof of Lemma 5.8. Assume that $G$ contains an $X_{4}$ with the vertex set
$\\{1,2,3,4,5,6,7,8,9,10\\}$, and $\\{1,2,3,4\\}$, $\\{1,2,5,6\\}$,
$\\{1,2,7,8\\}$, $\\{1,2,9,10\\}$ form $K_{4}^{3}$. Since $G$ is
$K_{4}^{3}\cup e$-free, then there is no edge in $V(G)\setminus\\{1,2\\}$.
Therefore $G$ is a subgraph of $B(2,n-2)$. By Lemma 4.10,
$\lambda(G)\leq\frac{\sqrt{3}}{18}$. $\Box$

###### Lemma 5.9

$G$ is $X_{3}$-free.

Proof of Lemma 5.9. Assume that $G$ contains an $X_{3}$ with vertex set
$\\{1,2,3,4,5,6,7,8\\}$ and $\\{1,2,3,4\\}$, $\\{1,2,5,6\\}$, $\\{1,2,7,8\\}$
form $K_{4}^{3}$. Denote $C=\\{3,4,5,6,7,8\\}$,
$D=\\{x\in[n]\setminus\\{1,2,3,4,5,6,7,8\\}:x12\in E(G)\\}$ and
$E=[n]\setminus(\\{1,2\\}\cup C\cup D)$. Note that $x12\notin E(G)$ for $x\in
E$. Since $G$ is $K_{4}^{3}\cup e$-free and $X_{4}$-free, then
$N^{*}(x,y)=\\{1\\}$ or $\\{2\\}$ for $x,y\in D$, and there is no edge between
$C$ and $D\cup E$. And if there is an edge $e_{1}$ in $C$, then
$|e_{1}\cap\\{3,4\\}|$=$|e_{1}\cap\\{5,6\\}|$=$|e_{1}\cap\\{7,8\\}|=1$, i.e.
$G[C]$ is a 3-partite 3-graph and $\lambda(G[C])\leq\frac{1}{27}$. Set

$x_{1}=a,\ x_{2}=b,\ \sum_{v\in C}x_{v}=c,\ \sum_{v\in D}x_{v}=d\ and\
\sum_{v\in E}x_{v}=e.$

Without loss of generality, assume that $a\geq b$, then replacing $xy2$ by
$xy1$ for all $x,y\in D$ does not decrease the Lagrangian. So

$\displaystyle\lambda(G)$ $\displaystyle\leq$ $\displaystyle
ab(c+d)+a(\bigg{(}\frac{c}{6}\bigg{)}^{2}{6\choose
2}+cd+ce+de+\frac{d^{2}}{2}+\frac{e^{2}}{2})+b(\bigg{(}\frac{c}{6}\bigg{)}^{2}{6\choose
2}+cd+ce+de+\frac{e^{2}}{2})+\frac{c^{3}}{27}$ $\displaystyle\leq$
$\displaystyle
ab(c+d)+a\bigg{(}\frac{5c^{2}}{12}+c(d+e)+\frac{(d+e)^{2}}{2}\bigg{)}+b\bigg{(}\frac{5c^{2}}{12}+c(d+e)+\frac{(d+e)^{2}}{2}\bigg{)}+\frac{c^{3}}{27}.$

Let $\delta=d+e$. Then

$\lambda(G)\leq\lambda(a,b,c,\delta)=ab(c+\delta)+a(\frac{5c^{2}}{12}+c\delta+\frac{\delta^{2}}{2})+b(\frac{5c^{2}}{12}+c\delta+\frac{\delta^{2}}{2})+\frac{c^{3}}{27}\triangleq\lambda$

subject to

$\left\\{\begin{array}[]{lr}a+b+c+\delta=1,&\\\ a\geq 0,\ b\geq 0,\ c\geq 0\
and\ \delta\geq 0.&\end{array}\right.$ (2)

For simplicity of the notation, we assume that $\lambda$ reaches the maximum
at $(a,b,c,\delta)$.

If $c=0$, then
$\lambda\leq\frac{(a+b)^{2}[1-(a+b)]}{4}+\frac{(a+b)[1-(a+b)]^{2}}{2}$. By
Fact 4.9, $\lambda\leq\frac{\sqrt{3}}{18}.$

If $\delta=0$, then

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
abc+\frac{5(a+b)c^{2}}{12}+\frac{c^{3}}{27}$ $\displaystyle\leq$
$\displaystyle\frac{(a+b)^{2}c}{4}+\frac{5(a+b)c^{2}}{12}+\frac{c^{3}}{27}$
$\displaystyle=$
$\displaystyle\frac{(1-c)^{2}c}{4}+\frac{5(1-c)c^{2}}{12}+\frac{c^{3}}{27}=f(c).$
$\displaystyle f^{\prime}(c)$ $\displaystyle=$
$\displaystyle\frac{(1-c)^{2}-2c(1-c)}{4}+\frac{10c(1-c)-5c^{2}}{12}+\frac{c^{2}}{9}$
$\displaystyle=$ $\displaystyle\frac{-14c^{2}-6c+9}{36}.$

So $f(c)$ is increasing in $[0,\frac{3(\sqrt{15}-1)}{14}]$, then
$f_{max}=f(\frac{3(\sqrt{15}-1)}{14})<0.0921.$

If $c,\delta>0$, then by Theorem 4.1, we have $\frac{\partial\lambda}{\partial
c}=\frac{\partial\lambda}{\partial\delta},$ solving it, we obtain that
$\frac{ac+bc}{6}=\frac{c^{2}}{9}.$ Therefore $c=\frac{3(a+b)}{2}$. So

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
ab(1-a-b)+\frac{(a+b)(c+\delta)^{2}}{2}-\frac{c^{2}(a+b)}{12}+\frac{c^{3}}{27}$
$\displaystyle\leq$
$\displaystyle\frac{(a+b)^{2}[1-(a+b)]}{4}+\frac{(a+b)[1-(a+b)]^{2}}{2}+\frac{c^{3}}{27}-\frac{c^{3}}{18}$
$\displaystyle<$
$\displaystyle\frac{(a+b)^{2}[1-(a+b)]}{4}+\frac{(a+b)[1-(a+b)]^{2}}{2}$
$\displaystyle\leq$ $\displaystyle\frac{\sqrt{3}}{18},$

the last inequality follows from Fact 4.9. $\Box$

Proof of Lemma 2.1. Assume that $G$ contains an $X_{2}$ with vertex set
$\\{1,2,3,4,5,6\\}$ and $\\{1,2,3,4\\}$, $\\{1,2,5,6\\}$ form $K_{4}^{3}$. We
prove the following claim first.

###### Claim 5.10

$\lambda(G[\\{1,2,3,4,5,6\\}])\leq\frac{2}{25}.$

Proof of Claim 5.10. Denote $A=\\{1,2,3,4,5,6\\}$. By Lemma 5.7, $G$ is
$K_{5}^{3-}$-free, then $e(G[A^{\prime}])\leq 8$ for all $A^{\prime}\subseteq
A$ and $|A^{\prime}|=5$. By double counting, we have

${6-3\choose 2}e(G[A])\leq{6\choose 5}\times 8.$

Therefore $e(G[A])\leq 16={6-1\choose 3}+{6-2\choose 2}<{6\choose 3}-3=17$. By
Lemma 4.7 and Lemma 4.8, $\lambda(G[A])\leq\lambda(K_{5}^{3})=\frac{2}{25}.$
$\Box$

Denote $C=\\{3,4,5,6\\}$, $D=\\{x\in V(G)\setminus\\{1,2,3,4,5,6\\}:x12\in
E(G)\\}$ and $E=V(G)\setminus(\\{1,2\\}\cup C\cup D)$. Note that $x12\notin
E(G)$ for $x\in E$. Since $G$ is $K_{4}^{3}\cup e$-free, then $x34,x56\notin
E(G)$ and only possibly, $x35,x36,x45,x46\in E(G)$ for $x\in D\cup E$. Let

$x_{1}=a,x_{2}=b,\sum_{v\in C}x_{v}=c,\sum_{v\in D}x_{v}=d,\sum_{v\in
E}x_{v}=e.$

Without loss of generality, assume that $a\geq b$. Note that the contribution
of the edges between $C$ and $D\cup E$ to $\lambda(G)$ is at most
$(d+e)(x_{3}+x_{4})(x_{5}+x_{6})\leq\frac{(d+e)c^{2}}{4}$. Since $G$ is
$X_{3}$-free, then $N^{*}(x,y)=\\{1\\}$ or $\\{2\\}$ for $x,y\in D$. If
$xy2\in E(G)$, we delete $xy2$ and add $xy1$, this does not reduce the
Lagrangian. Hence

$\displaystyle\lambda(G)$ $\displaystyle\leq$ $\displaystyle
abd+a(cd+ce+de+\frac{d^{2}}{2}+\frac{e^{2}}{2})+b(cd+ce+de+\frac{e^{2}}{2})+\frac{2(a+b+c)^{3}}{25}+\frac{(d+e)c^{2}}{4}$
$\displaystyle=$ $\displaystyle\lambda(a,b,c,d,e)=\lambda$

under the constraints $a+b+c+d+e=1$, $a,b,c,d,e\geq 0$.

To simplify the notation, we assume that $\lambda$ reaches the maximum at
$(a,b,c,d,e)$, Note that $a\geq b$.

###### Claim 5.11

$\lambda(a,0,c,d,e)\leq\frac{\sqrt{3}}{18}$.

The proof of Claim 5.11 is given in Appendix.

Let us continue the proof of Lemma 2.1. We have shown that $a\geq b>0$ (Claim
5.11). If $d=0$, substitute it into (5.2), then

$\lambda(a,b,c,0,e)=a(ce+\frac{e^{2}}{2})+b(ce+\frac{e^{2}}{2})+\frac{2(a+b+c)^{3}}{25}+\frac{ec^{2}}{4}.$

So $\lambda(a+b,0,c,0,e)$ also gets the maximum value, a contradiction to
$a\geq b>0$ when $\lambda$ gets the maximum. So $a,b,d>0$. By Theorem 4.1,
$\frac{\partial\lambda}{\partial a}=\frac{\partial\lambda}{\partial b}$,
combining with

$\displaystyle\frac{\partial\lambda}{\partial a}$ $\displaystyle=$
$\displaystyle
bd+cd+ce+de+\frac{d^{2}}{2}+\frac{e^{2}}{2}+\frac{6(a+b+c)^{2}}{25},$
$\displaystyle\frac{\partial\lambda}{\partial b}$ $\displaystyle=$
$\displaystyle ad+cd+ce+de+\frac{e^{2}}{2}+\frac{6(a+b+c)^{2}}{25},$

we get

$\displaystyle a=b+\frac{d}{2}.$ (4)

###### Claim 5.12

$\lambda(a,b,c,d,0)\leq\frac{\sqrt{3}}{18}$.

The proof of Claim 5.12 is given in Appendix.

Let us continue the proof of Lemma 2.1. We have shown that $a,b,d,e>0$. By
Theorem 4.1, we have $\frac{\partial\lambda}{\partial
d}=\frac{\partial\lambda}{\partial e}$. Since

$\displaystyle\frac{\partial\lambda}{\partial d}$ $\displaystyle=$
$\displaystyle ab+ac+ae+ad+bc+be+\frac{c^{2}}{4},$
$\displaystyle\frac{\partial\lambda}{\partial e}$ $\displaystyle=$
$\displaystyle ac+ad+ae+bc+bd+be+\frac{c^{2}}{4},$

then $a=d$. Recall (4), so

$\displaystyle a=d=2b.$ (5)

###### Claim 5.13

$\lambda(a,b,0,d,e)\leq\frac{\sqrt{3}}{18}$.

Proof of Claim 5.13. Assume $c=0$. By (5), then $e=1-a-b-c-d=1-5b$.
Substituting it into (5.2), then

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
abd+a(de+\frac{d^{2}}{2}+\frac{e^{2}}{2})+b(de+\frac{e^{2}}{2})+\frac{2(a+b)^{3}}{25}$
$\displaystyle=$ $\displaystyle
4b^{3}+b\bigg{(}6b(1-5b)+4b^{2}+\frac{3(1-5b)^{2}}{2}\bigg{)}+\frac{2(3b)^{3}}{25}$
$\displaystyle=$ $\displaystyle\frac{883b^{3}}{50}-9b^{2}+\frac{3b}{2}=f(b),$
$\displaystyle f^{\prime}(b)$ $\displaystyle=$
$\displaystyle\frac{2649b^{2}}{50}-18b+\frac{3}{2}.$

So $f(b)$ is increasing in $[0,\frac{450-5\sqrt{153}}{2649}]$ or
$[\frac{450+5\sqrt{153}}{2649},1]$. Note that $a+b+d+e=1$, then $5b<1$, so
$\lambda\leq f(\frac{450-5\sqrt{153}}{2649})\leq 0.083$. $\Box$

Let us continue the proof of Lemma 2.1. We have shown that $a=d=2b$ and
$e=1-5b-c$ and $a,b,c,d,e>0$. Substituting them into (5.2), we have

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
4b^{3}+b(6bc+3ce+6be+4b^{2}+\frac{3e^{2}}{2})+\frac{2(3b+c)^{3}}{25}+\frac{(2b+e)c^{2}}{4}$
$\displaystyle=$
$\displaystyle\frac{31b^{3}}{2}-\frac{3bc^{2}}{2}-9b^{2}+\frac{3b}{2}+\frac{2(3b+c)^{3}}{25}+\frac{(1-3b-c)c^{2}}{4}$
$\displaystyle\leq$ $\displaystyle
16b^{3}-\frac{3bc^{2}}{2}-9b^{2}+\frac{3b}{2}+\frac{(3b+c)^{3}}{12}+\frac{(1-3b-c)c^{2}}{4}=\lambda_{0}(b,c)$

under the constraints $5b+c\leq 1,b,c\geq 0.$ Now we estimate the optimum
value of $\lambda_{0}$. For simplicity of the notation, let $\lambda_{0}$
reach the maximum value at $(b,c)$.

If $b=0$, then
$\lambda_{0}=\frac{c^{3}}{12}+\frac{(1-c)c^{2}}{4}\leq\frac{3c^{2}-2c^{3}}{12}=f(c)$.
Since $f^{\prime}(c)=\frac{c(1-c)}{2}$, then $\lambda$ is increasing in
$[0,1]$. Therefore $\lambda_{0}\leq f(1)=\frac{1}{12}<0.09.$

If $c=0$, then
$\lambda_{0}=16b^{3}-9b^{2}+\frac{3b}{2}+\frac{(3b)^{3}}{12}=\frac{73b^{3}}{4}-9b^{2}+\frac{3b}{2}\leq
19b^{3}-9b^{2}+\frac{3b}{2}=f(b)$. Since
$f^{\prime}(b)=57b^{2}-18b+\frac{3}{2}>0$, so $f(b)$ is increasing in $[0,1]$.
Therefore $\lambda_{0}\leq f(\frac{1}{5})=0.092.$

If $5b+c=1$, then

$\displaystyle\lambda_{0}$ $\displaystyle=$ $\displaystyle
16b^{3}-\frac{3b(1-5b)^{2}}{2}-9b^{2}+\frac{3b}{2}+\frac{(1-2b)^{3}}{12}+\frac{b(1-5b)^{2}}{2},$
$\displaystyle\lambda_{0}^{\prime}$ $\displaystyle=$
$\displaystyle-29b^{2}+4b.$

Then $\lambda_{0}\leq\lambda_{0}(4/29)\leq 0.0961<\frac{\sqrt{3}}{18}$.

Therefore we may assume that $\lambda_{0}$ gets maximum when $b,c>0$ and
$5b+c<1$. By Theorem 4.1,

$\displaystyle\frac{\partial\lambda_{0}}{\partial b}$ $\displaystyle=$
$\displaystyle
48b^{2}-\frac{9c^{2}}{4}-18b+\frac{3}{2}+\frac{3(3b+c)^{2}}{4}=0$
$\displaystyle\frac{\partial\lambda_{0}}{\partial c}$ $\displaystyle=$
$\displaystyle-3bc+\frac{(3b+c)^{2}}{4}+\frac{2c-6bc-3c^{2}}{4}=0.$

Equivalently, $\frac{\partial\lambda_{0}}{\partial
b}-3\times\frac{\partial\lambda_{0}}{\partial c}=0$ and
$\frac{\partial\lambda_{0}}{\partial c}=0$. Solving these two equations, we
have

$c=\frac{1-12b+32b^{2}}{1-9b}=\frac{(1-8b)(1-4b)}{1-9b}\quad{\rm and}\quad
9b^{2}-12bc+2c-2c^{2}=0.$

Recall that $c>0$ and $b<\frac{1}{5}$. So $0<b<\frac{1}{9}$ or
$\frac{1}{8}\leq b\leq\frac{1}{5}$. Combining the above equations, we have
$b(2137b^{3}-882b^{2}+125b-6)=0$. Let $f(b)=2137b^{3}-882b^{2}+125b-6$. Since
$f^{\prime}(b)=6411b^{2}-1764b+125>0$, then $f(b)$ is increasing in
$[0,\frac{1}{4}]$. However $f(0),f(\frac{1}{9})<0$ and $f(\frac{1}{8})>0$, so
$f(b)=0$ has no solution in $0<b<\frac{1}{9}$ and $\frac{1}{8}\leq
b\leq\frac{1}{5}$, a contradiction. This completes the proof of Lemma 2.1.
$\Box$

###  5.3 $G$ contains at least two copies of $K_{4}^{3}$

In this section, we give the proof of Lemma 2.2.

Proof of Lemma 2.2. Recall that $\\{1,2,3,4\\}$ forms a $K_{4}^{3}$. Assume
that $G$ contains no other $K_{4}^{3}$, in other words, $v$ does not belong to
any $K_{4}^{3}$ for any $v\in[n]\setminus\\{1,2,3,4\\}$. We claim that
$|G_{v}\cap\\{12,13,14,23,24,34\\}|\leq 4$. Since otherwise
$G_{v}[\\{1,2,3,4\\}]$ contains a triangle and $v$ is contained in the
$K_{4}^{3}$ formed by $v$ and the vertices in this triangle. Since $G$ is
$K_{4}^{3}\cup e$-free, then $G_{v}$ does not contain an edge in
$[n]\setminus\\{1,2,3,4\\}$. By Claim 5.3, $\omega(G_{v})\geq 3$, so the
maximum clique of $G_{v}$ contains at least 2 vertices in $\\{1,2,3,4\\}$,
therefore $|G_{v}\cap\\{12,13,14,23,24,34\\}|\geq 1$. Let

$A=\\{1,2,3,4\\},$ $A_{1}=\\{v\in[n]\setminus
A:|G_{v}\cap\\{12,13,14,23,24,34\\}|=1\\},$ $A_{2}=\\{v\in[n]\setminus
A:|G_{v}\cap\\{12,13,14,23,24,34\\}|=2\\},$ $A_{3}=\\{v\in[n]\setminus
A:|G_{v}\cap\\{12,13,14,23,24,34\\}|=3\\},$ $A_{4}=\\{v\in[n]\setminus
A:|G_{v}\cap\\{12,13,14,23,24,34\\}|=4\\}.$

Without loss of generality, let’s assume that $x_{1}\geq x_{2}\geq x_{3}\geq
x_{4}$. Then $x_{1}x_{2}\geq x_{1}x_{3}\geq x_{2}x_{3},x_{1}x_{4}\geq
x_{2}x_{4}\geq x_{3}x_{4}.$ We aim to give an upper bound of
$\lambda(G,\vec{x})$, therefore we can assume that $v12,v13\in E(G)$ for $v\in
A_{2}$, $v12\in E(G)$ for $v\in A_{1}$. Set

$x_{1}=a,x_{2}=b,x_{3}=c,x_{4}=d,\sum_{v\in A_{1}}x_{v}=h,\sum_{v\in
A_{2}}x_{v}=g,\sum_{v\in A_{3}}x_{v}=f,\sum_{v\in A_{4}}x_{v}=e.$

Since $G$ doesn’t contain two copies of $K_{4}^{3}$, then the deletion of any
1 of $\\{123,124,134,234\\}$ of $E(G)$ makes $G$ $K_{4}^{3}$-free. So
$abc,abd,acd,bcd>0.00264$ since otherwise $\lambda(G)\leq
0.00264+\frac{0.5615}{6}$(in view of Lemma 5.1)$\leq\frac{\sqrt{3}}{18}.$ So
$(a+b)cd>2\times 0.00264$. By Claim 5.4, $a+b\leq\frac{3-\sqrt{3}}{3}$, then
$cd\geq\frac{6\times 0.00264}{3-\sqrt{3}}$. Therefore $c+d\geq
2\sqrt{cd}>0.22354.$ If $d<0.11177$, then $b+c\geq 2\sqrt{bc}>0.307.$

To complete the proof, we show the following three claims in Appendix.

###### Claim 5.14

$\lambda(G[A\cup A_{4}],\vec{x})\leq 0.0789(a+b+c+d+e)^{3}.$

###### Claim 5.15

$\lambda(G[A\cup A_{4}\cup A_{3}],\vec{x})\leq 0.092(a+b+c+d+e+f)^{3}.$

###### Claim 5.16

$\lambda(G)\leq\frac{\sqrt{3}}{18}.$

###  5.4 $G$ does not contain two copies of $K_{4}^{3}$ sharing three vertices

Proof of Lemma 2.3. Assume that $G$ contains an $Y_{2}$ with the vertex set
$\\{1,2,3,4,5\\}$, where $\\{1,2,3,4\\}$ and $\\{1,2,3,5\\}$ form $K_{4}^{3}$.
Since $G$ is $K_{4}^{3}\cup e$-free, then any two $K_{4}^{3}$ in $G$ must
intersect 2 or 3 vertices. Since $G$ is $X_{2}$-free (Lemma 2.1), then any two
$K_{4}^{3}$ in $G$ must intersect 3 vertices. Therefore $G-\\{3\\}$ cannot
contain a $K_{4}^{3}$ since it cannot intersect with $\\{1,2,3,4\\}$ and
$\\{1,2,3,5\\}$ at three vertices in the same time. Let $x_{1}=a$, $x_{2}=b$,
$x_{3}=c$, $x_{4}+x_{5}=d$ and assume that $a\geq b\geq c$. By Claim 5.5,
$c>0.08$. Let $D=\\{4,5\\}$, and

$E_{0}=\\{v:v\in[n]\setminus\\{1,2,3,4,5\\}\ and\
|G_{v}\cap\\{12,13,23\\}|=0\\},$ $E_{1}=\\{v:v\in[n]\setminus\\{1,2,3,4,5\\}\
and\ |G_{v}\cap\\{12,13,23\\}|=1\\},$
$E_{2}=\\{v:v\in[n]\setminus\\{1,2,3,4,5\\}\ and\
|G_{v}\cap\\{12,13,23\\}|\geq 2\\}.$

Set $\sum_{v\in E_{0}}x_{v}=g$, $\sum_{v\in E_{1}}x_{v}=f$ and $\sum_{v\in
E_{2}}x_{v}=e$. Since $G$ is $K_{4}^{3}\cup e$-free, then for
$x,y\in[n]\setminus\\{1,2,3,4,5\\}$, $N^{*}(x,y)\subseteq\\{1,2,3\\}$. If
$x\in D$ and $y\in E_{2}$, then $N^{*}(x,y)\subseteq\\{1,2,3,4,5\\}$. We claim
that $|N^{*}(x,y)\cap\\{1,2,3\\}|\leq 2$ for $x,y\in E_{2}$ or $x\in D$ and
$y\in E_{2}$, otherwise there exists two vertices $z,w\in
N^{*}(x,y)\cap\\{1,2,3\\}$ such that $zw\in G_{x}\cap G_{y}$. So
$\\{x,y,z,w\\}$ forms a $K_{4}^{3}$, which forms an $X_{2}$ with
$G[\\{1,2,3,4\\}]$, a contradiction. Therefore we may let
$N^{*}(x,y)\cap\\{1,2,3\\}\subseteq\\{1,2\\}$ for those $x,y$ with Lagrangian
non-decreasing. Since $G$ is $K_{4}^{3}\cup e$-free, then all edges in
$G[V-\\{1,2,3\\}]$ must contain $\\{4,5\\}$. Since $G$ is $K_{5}^{3}$-free,
then at least one of $\\{145,245,345\\}$ is not in $E(G)$. (Indeed, $G$ is
$K_{5}^{3-}$-free, at least two of $\\{145,245,345\\}$ are not in $E(G)$. But
it seems to be easier to estimate the Lagrangian below if we relax it to be
one.) We may assume that 345 is not in $E(G)$. Therefore

$\displaystyle\lambda(G)$ $\displaystyle\leq$ $\displaystyle
ab(c+d+e+f)+(ac+bc)(d+e)+(a+b)(de+\frac{e^{2}}{2})+(a+b+c)(df+dg+eg+fg+ef$
$\displaystyle+$
$\displaystyle\frac{f^{2}}{2}+\frac{g^{2}}{2})+\frac{d^{2}(a+b+e+f+g)}{4}$
$\displaystyle=$ $\displaystyle
ab(c+d+e+f)+(ac+bc)(d+e)+(a+b)(de+\frac{e^{2}}{2})+(a+b+c)\bigg{(}d(g+f)+e(g+f)$
$\displaystyle+$
$\displaystyle\frac{(f+g)^{2}}{2}\bigg{)}+\frac{d^{2}(a+b+e+f+g)}{4}$
$\displaystyle=$ $\displaystyle\lambda(a,b,c,d,e,f,g)=\lambda.$

Note that
$\lambda(a,b,c,d,e,f,g)\leq\lambda(\frac{a+b}{2},\frac{a+b}{2},c,d,e,f+g,0)$,
then we may assume that $g=0$ and $a=b$. Then let $\alpha=a+b$, so

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}(c+d+e+f)}{4}+\alpha
c(d+e)+\alpha(de+\frac{e^{2}}{2})+(\alpha+c)(df+ef+\frac{f^{2}}{2})+\frac{d^{2}(\alpha+e+f)}{4}$

subject to

$\left\\{\begin{array}[]{lr}\alpha+c+d+e+f=1,&\\\ c\geq 0.08,\alpha,d,e,f\geq
0.&\end{array}\right.$ (6)

Note that

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}(c+d+e+f)}{4}+\alpha
c(d+e)+\alpha(\frac{d^{2}}{2}+de+\frac{e^{2}}{2})+(\alpha+c)(df+ef+\frac{f^{2}}{2})+\frac{d^{2}(e+f-\alpha)}{4}.$

If $e+f-\alpha<\frac{d}{2}$, then
$\lambda\leq\lambda(a,c,d-\epsilon,e+\epsilon,f)$ for $\epsilon>0$ small
enough, a contradiction. So we may assume that $e+f\geq\alpha+\frac{d}{2}$ or
$d=0$.

###### Claim 5.17

$\lambda(0,c,d,e,f)\leq\frac{\sqrt{3}}{18}$.

Proof of Claim 5.17. If $\alpha=0$, then $c+d+e+f=1$ and

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
c(df+ef+\frac{f^{2}}{2})+\frac{d^{2}(e+f)}{4}$ $\displaystyle\leq$
$\displaystyle
c\frac{(d+e+f)^{2}}{2}+\bigg{(}\frac{\frac{d}{2}+\frac{d}{2}+e+f}{3}\bigg{)}^{3}$
$\displaystyle=$
$\displaystyle\frac{c(1-c)^{2}}{2}+\frac{(1-c)^{3}}{27}=\frac{25c^{3}-48c^{2}+21c+2}{54}=f(c),$
$\displaystyle f^{\prime}(c)$ $\displaystyle=$
$\displaystyle\frac{25c^{2}-32c+7}{18}=\frac{(25c-7)(c-1)}{18}.$

Therefore $\lambda\leq f(\frac{7}{25})=0.0864.$ $\Box$

Let’s continue the proof of Lemma 2.3. We claim that if $c>0.08$, then
$\alpha\geq c$. If $\alpha<c$, then $\alpha\leq\frac{1}{2}$, and for
$0<\epsilon<\frac{a}{3}$,

$\displaystyle\lambda(\alpha+\epsilon,c-\epsilon,d,e,f)-\lambda(\alpha,c,d,e,f)$
$\displaystyle\geq$
$\displaystyle\frac{(\alpha+\epsilon)^{2}(c-\epsilon+d+e+f)-(\alpha)^{2}(c+d+e+f)}{4}$
$\displaystyle>$ $\displaystyle\epsilon(2\alpha-3\alpha^{2}-3\epsilon\alpha)$
$\displaystyle>$ $\displaystyle\epsilon(2\alpha-4\alpha^{2})\geq 0,$

a contradiction.

###### Claim 5.18

$\lambda(\alpha,c,d,e,0)\leq\frac{\sqrt{3}}{18}$.

The proof of Claim 5.18 is given in Appendix.

###### Claim 5.19

$\lambda(\alpha,c,d,0,f)\leq\frac{\sqrt{3}}{18}$.

The proof of Claim 5.19 is given in Appendix.

By the above claims, we may assume that $\alpha,e,f>0$, so by Theorem 4.1, we
have $\frac{\partial\lambda}{\partial e}=\frac{\partial\lambda}{\partial f}$.
In view of (LABEL:eq6),

$\displaystyle\frac{\partial\lambda}{\partial e}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+\alpha c+\alpha d+\alpha
e+(\alpha+c)f+\frac{d^{2}}{4}$ $\displaystyle\frac{\partial\lambda}{\partial
f}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+(\alpha+c)(d+e+f)+\frac{d^{2}}{4}.$

So $\alpha=d+e.$

###### Claim 5.20

$\lambda(\alpha,c,0,e,f)\leq\frac{\sqrt{3}}{18}$.

The proof of Claim 5.20 is given in Appendix.

Let’s continue the proof of Lemma 2.3. The above claims indicate that we may
assume $d,e>0$, then by Theorem 4.1, we have $\frac{\partial\lambda}{\partial
d}=\frac{\partial\lambda}{\partial e}$. Since

$\displaystyle\frac{\partial\lambda}{\partial d}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+\alpha c+\alpha
e+(\alpha+c)f+\frac{d(\alpha+e+f)}{2}$
$\displaystyle\frac{\partial\lambda}{\partial e}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+\alpha c+\alpha d+\alpha
e+(\alpha+c)f+\frac{d^{2}}{4},$

then $\alpha+\frac{d}{2}=e+f$. Since $\alpha=d+e$, then $f=\frac{3d}{2}$. Note
that $\alpha+c+d+e+f=1$, then $f=1-2\alpha-c$ and
$d=\frac{2(1-2\alpha-c)}{3}>0$ and $e=\frac{7\alpha+2c-2}{3}$. Substituting
these into (LABEL:eq6), we have

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}(1-\alpha)}{4}+\alpha^{2}c+\alpha((\frac{2(1-2\alpha-c)}{3})(\frac{7\alpha+2c-2}{3})+\frac{(\frac{7\alpha+2c-2}{3})^{2}}{2})$
$\displaystyle+$
$\displaystyle(\alpha+c)(\alpha(1-2\alpha-c)+\frac{(1-2\alpha-c)^{2}}{2})+\frac{(1-2\alpha-c)^{2}}{9}(1-c-\frac{2(1-2\alpha-c)}{3})$
$\displaystyle=$
$\displaystyle\frac{-5\alpha^{3}}{108}+\frac{14\alpha^{2}c}{9}-\frac{11\alpha^{2}}{36}+\frac{23\alpha
c^{2}}{18}-\frac{14\alpha
c}{9}+\frac{5\alpha}{18}+\frac{25c^{3}}{54}-\frac{8c^{2}}{9}+\frac{7c}{18}+\frac{1}{27}$
$\displaystyle=$ $\displaystyle\lambda(\alpha,c).$

If $c=0.08$, then

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{-5\alpha^{3}}{108}-\frac{1.63\alpha^{2}}{9}+\frac{1.4536\alpha}{9}+\frac{1.6928}{27},$
$\displaystyle\lambda^{\prime}$ $\displaystyle=$
$\displaystyle\frac{-5\alpha^{2}}{36}-\frac{3.26\alpha}{9}+\frac{1.4536}{9}.$

Note that $\lambda$ is increasing in $[0,\frac{3\sqrt{4971}-163}{125}]$, then
$\lambda\leq 0.096.$ So we may assume that $c>0.08$. Recall that $\alpha\geq
c$ when $c>0.08$ and $e=\frac{7\alpha+2c-2}{3},f=1-2\alpha-c>0$, so
$\alpha>\frac{2}{9}$ and $2\alpha+c<1$. So we can maximize $\lambda(\alpha,c)$
subject to

$\left\\{\begin{array}[]{lr}2\alpha+c\leq 1,&\\\ \alpha\geq\frac{2}{9},&\\\
c\geq 0.08.&\end{array}\right.$ (7)

Consider

$\displaystyle\lambda$ $\displaystyle(\alpha+{c-0.08\over
2},0.08)-\lambda(\alpha,c)$ $\displaystyle=$
$\displaystyle{(2-25c)(32500\alpha^{2}+26250\alpha
c-25500\alpha+9375c^{2}-16150c+4728)\over 500000}=\lambda_{0},$
$\displaystyle\lambda_{0}^{\prime}|_{\alpha}$
$\displaystyle=\frac{(2-25c)(260\alpha+105c-102)}{2000}.$

If $260\alpha+105c-102>0$, then $\lambda_{0}^{\prime}|_{\alpha}<0$. So
$\lambda_{0}\geq\lambda_{0}(\frac{1-c}{2},c)$. Therefore

$\displaystyle\lambda_{0}$ $\displaystyle\geq$
$\displaystyle\frac{-7c^{3}}{32}+\frac{11c^{2}}{32}-\frac{c}{32}+\frac{103}{250000}=\lambda_{1},$
$\displaystyle\lambda_{1}^{\prime}$ $\displaystyle=$
$\displaystyle\frac{-21c^{2}}{32}+\frac{11c}{16}-\frac{1}{32}.$

Note that $\lambda_{1}$ is increasing in $[0.08,1]$, so
$\lambda_{1}\geq\lambda_{1}(0.08)=0$. So $\lambda$ gets maximum when $c=0.08$,
a contradiction.

If $260\alpha+105c-102<0$, then $\lambda_{0}^{\prime}|_{a}>0$. So
$\lambda_{0}\geq\lambda_{0}(\frac{2}{9},c)$. Therefore

$\displaystyle\lambda_{0}$ $\displaystyle\geq$
$\displaystyle{(2-25c)(9375c^{2}-{30950c\over 3}+{53968\over 81})\over
500000}=\lambda_{2}$ $\displaystyle\lambda_{2}^{\prime}$ $\displaystyle=$
$\displaystyle\frac{83c}{75}-\frac{45c^{2}}{32}-{6041\over 81000}.$

Note that $\lambda_{2}$ is increasing in $[0.08,0.7]$, then
$\lambda_{2}\geq\lambda_{2}(0.08)=0$. So $\lambda$ gets maximum when $c=0.08$,
a contradiction. The proof of Lemma 2.3 is completed. $\Box$

##  6 Remark

Let $\Lambda_{t}^{(r)}=\\{\pi_{\lambda}(\mathcal{F}):\mathcal{F}{\rm\ is\ a\
family\ of}\ r{\rm-uniform\ graphs\ and}\ |\mathcal{F}|\leq t\\}.$ Proposition
1.3 (Proposition 1.3 can be generalized to a family of $r$-graphs) implies
that $\Lambda_{t}^{(r)}\subseteq\Pi_{t}^{(r)}$.

###### Question 1

Is $\Lambda_{t}^{(r)}$ the same as $\Pi_{t}^{(r)}$?

Let us propose the following conjecture implying that there exists an
$r$-graph whose Turán density is an irrational number.

###### Conjecture 6.1

If $c\cdot{r!\over r^{r}}$ is in $\Pi_{1}^{(r)}$ for $r\geq 2$, then
$c\cdot{p!\over p^{p}}$ is in $\Pi_{1}^{(p)}$ for $p\geq r$.

##  7 Appendix

We give theoretical proofs for Claims 5.11, 5.12, 5.14-5.16 and 5.18-5.20 in
this section, we have also used Lingo to run the optimization problems. The
outcome by Lingo is consistent with the expected optimum values. We can
provide the programming upon request.

###  7.1 Proof of Claim 5.11

Substitute $b=0$ into (5.2), then

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
a(cd+ce+de+\frac{d^{2}}{2}+\frac{e^{2}}{2})+\frac{2(a+c)^{3}}{25}+\frac{(d+e)c^{2}}{4},$
$\displaystyle=$ $\displaystyle
a[c(d+e)+\frac{(d+e)^{2}}{2}]+\frac{2(a+c)^{3}}{25}+\frac{(d+e)c^{2}}{4},$
$\displaystyle=$ $\displaystyle
a[c\delta+\frac{\delta^{2}}{2}]+\frac{2(a+c)^{3}}{25}+\frac{\delta c^{2}}{4},$

where $\delta=d+e$, then $a+c+\delta=1$.

If $a=0$, then

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{2c^{3}}{25}+\frac{\delta
c^{2}}{4}\leq\frac{c^{3}}{12}+\frac{(1-c)c^{2}}{4}=\frac{3c^{2}-2c^{3}}{12}=f(c)$
$\displaystyle f^{\prime}(c)$ $\displaystyle=$
$\displaystyle\frac{c-c^{2}}{2}=\frac{c(1-c)}{2}.$

So $f(c)$ is increasing in $[0,1]$, then $\lambda\leq
f(1)=\frac{1}{12}<\frac{\sqrt{3}}{18}.$

If $\delta=0$, then

$\lambda=\frac{2(a+c)^{3}}{25}\leq\frac{2}{25}.$

If $c=0$, then

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{a\delta^{2}}{2}+\frac{2a^{3}}{25}\leq\frac{a(1-a)^{2}}{2}+\frac{a^{3}}{12}=\frac{7a^{3}-12a^{2}+6a}{12}=f(a)$
$\displaystyle f^{\prime}(a)$ $\displaystyle=$
$\displaystyle\frac{7a^{2}-8a+2}{4}.$

So $f(a)$ is increasing in $[0,\frac{4-\sqrt{2}}{7}]$ or
$[\frac{4+\sqrt{2}}{7},1]$, then $\lambda\leq
max\\{f(1),f(\frac{4-\sqrt{2}}{7})\\}.$ Note that $f(\frac{4-\sqrt{2}}{7})\leq
0.078$, then $\lambda\leq f(1)=\frac{1}{12}\leq\frac{\sqrt{3}}{18}.$

Therefore we may assume that $a,c,\delta>0$. Substituting $c=1-a-\delta$ into
(7.1), then

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
a[(1-a-\delta)\delta+\frac{\delta^{2}}{2}]+\frac{2(1-\delta)^{3}}{25}+\frac{\delta(1-a-\delta)^{2}}{4}$

gets its maximum inside interior points. By Theorem 4.1,

$\displaystyle\frac{\partial\lambda}{\partial a}$ $\displaystyle=$
$\displaystyle\frac{\delta}{2}-\frac{3a\delta}{2}=0,$
$\displaystyle\frac{\partial\lambda}{\partial\delta}$ $\displaystyle=$
$\displaystyle\frac{(1-a-\delta)(1+3a-3\delta)}{4}-\frac{6(1-\delta)^{2}}{25}=0.$

Then $a=\frac{1}{3}$ and $\delta=\frac{26-10\sqrt{2}}{51}$, and
$\lambda<0.09.$ $\Box$

###  7.2 Proof of Claim 5.12

Substitute $e=0$ into (5.2), then

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
abd+a(cd+\frac{d^{2}}{2})+bcd+\frac{2(a+b+c)^{3}}{25}+\frac{dc^{2}}{4}.$ (9)

If $c=0$, applying (4), then $b=3a-1$ and $d=2-4a$. By Theorem 4.1,
$\frac{\partial\lambda}{\partial b}=\frac{\partial\lambda}{\partial d}$.
Combining with

$\displaystyle\frac{\partial\lambda}{\partial b}$ $\displaystyle=$
$\displaystyle ad+\frac{6(a+b)^{2}}{25},$
$\displaystyle\frac{\partial\lambda}{\partial d}$ $\displaystyle=$
$\displaystyle ab+ad,$

we get $ab=\frac{6(a+b)^{2}}{25}$. Substituting $b=3a-1$, we have
$21a^{2}-23a+6=0$, then $a=\frac{2}{3}$ or $\frac{3}{7}$. Note that if
$a=\frac{2}{3}$, then $d<0$, a contradiction. So $a=\frac{3}{7}$,
$b=\frac{2}{7}$ and $d=\frac{2}{7}$. Therefore $\lambda=\frac{4}{49}\leq
0.082.$

If $c\not=0$, substitute $c=1-a-b-d$ into (9),

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
abd+a[(1-a-b-d)d+\frac{d^{2}}{2}]+b(1-a-b-d)d+\frac{2(1-d)^{3}}{25}+\frac{d(1-a-b-d)^{2}}{4}$
$\displaystyle=$ $\displaystyle
ad-a^{2}d-abd-\frac{ad^{2}}{2}+bd-b^{2}d-bd^{2}+\frac{2(1-d)^{3}}{25}+\frac{d(1-a-b-d)^{2}}{4}.$

subject to $a+b+d\leq 1$, and $a.b,d\geq 0$. Since $a+b+d<1$ and we have
showned that $a,b,d>0$, then by Theorem 4.1,

$\displaystyle\frac{\partial\lambda}{\partial a}$ $\displaystyle=$
$\displaystyle d-2ad-bd-\frac{d^{2}}{2}-\frac{d(1-a-b-d)}{2}=0,$
$\displaystyle\frac{\partial\lambda}{\partial b}$ $\displaystyle=$
$\displaystyle-ad+d-2bd-d^{2}-\frac{d(1-a-b-d)}{2}=0,$
$\displaystyle\frac{\partial\lambda}{\partial d}$ $\displaystyle=$
$\displaystyle a-a^{2}-ab-
ad+b-b^{2}-2bd-\frac{6(1-d)^{2}}{25}+\frac{(1-a-b-d)^{2}}{4}-\frac{d(1-a-b-d)}{2}=0.$

Then we have $b=1-3a$ (by solving $\frac{\partial\lambda}{\partial a}=0$) and
$d=8a-2$ (by solving $\frac{\partial\lambda}{\partial b}=0$), substitute this
into $\frac{\partial\lambda}{\partial d}=0$, we have $1266a^{2}-787a+121=0$.
Therefore $a=\frac{787-5\sqrt{265}}{2532}$ and $\lambda\leq 0.0939$. $\Box$

###  7.3 Proof of Claim 5.14

Since $G$ contains no other $K_{4}^{3}$, then for any $v\in A_{4}$, there is
no $K_{3}$ in $G_{v}[A]$, so $G_{v}[A]$ forms a $C_{4}$. Hence

$\lambda(G_{v}[A],\vec{x})\leq
max\\{(a+b)(c+d),(a+c)(b+d),(a+d)(b+c)\\}\leq\frac{(a+b+c+d)^{2}}{4}.$

Since $G$ is $K_{4}^{3}\cup\\{e\\}$-free, then $N^{*}(x,y)\subseteq A$ for
$x,y\in[n]\setminus A$. Since $G$ contains only one $K_{4}^{3}$, we claim that
$|N^{*}(x,y)|\leq 2$ for all $x,y\in A_{4}$. Recall that $G_{x}$ and $G_{y}$
are $C_{4}$’s, then $G_{x}[A]\cap G_{y}[A]$ must be two vertex disjoint edges
or $G_{x}[A]=G_{y}[A]$. If $|N^{*}(x,y)|\geq 3$, then there are $z,w\in
N^{*}(x,y)\subseteq A$ such that $zw\in G_{x}\cap G_{y}$. Then $\\{x,y,z,w\\}$
forms a $K_{4}^{3}$, a contradiction. Recall that $a\geq b\geq c\geq d$, then
we may assume that $N^{*}(x,y)=\\{1,2\\}$ for $x,y\in A_{4}$ with the
Lagrangian non-decreasing. Therefore

$\displaystyle\lambda(G[A\cup A_{4}],\vec{x})$ $\displaystyle\leq$
$\displaystyle abc+abd+acd+bcd+\frac{(a+b)e^{2}}{2}+e\frac{(a+b+c+d)^{2}}{4},$

subject to $a+b+c+d+e=1$ and $c+d>0.22354$ and $a+b>c+d$. Let
$\alpha={a+b\over a+b+c+d+e}$, $\gamma={c+d\over a+b+c+d+e}\geq 0.22354$ and
$\eta={e\over a+b+c+d+e}$. So

$\displaystyle{\lambda(G[A\cup A_{4}],\vec{x})\over(a+b+c+d+e)^{3}}$
$\displaystyle\leq$
$\displaystyle\frac{\alpha^{2}\gamma}{4}+\frac{\alpha\gamma^{2}}{4}+\frac{\alpha\eta^{2}}{2}+\eta\frac{(\alpha+\gamma)^{2}}{4}=\lambda(\alpha,\gamma,\eta)\triangleq\lambda$

subject to

$\left\\{\begin{array}[]{lr}\alpha+\gamma+\eta=1,&\\\ \gamma\geq 0.22354,&\\\
\alpha\geq\gamma.&\end{array}\right.$ (10)

If $\eta=0$, then
$\lambda\leq\lambda(\frac{1}{2},\frac{1}{2},0)=\frac{1}{16}.$

If $\alpha=\gamma$, then $\eta=1-2\alpha$. So

$\lambda=\frac{\alpha^{3}+\alpha\eta^{2}}{2}+\eta\alpha^{2}=\frac{\alpha(\alpha+\eta)^{2}}{2}=\frac{\alpha(1-\alpha)^{2}}{2}\leq\frac{(\frac{2\alpha+1-\alpha+1-\alpha}{3})^{3}}{4}=\frac{2}{27}<0.075.$

If $\gamma=0.22354$, then $\eta=0.77646-\alpha$

$\displaystyle\lambda$ $\displaystyle\leq$
$\displaystyle\frac{\alpha^{3}}{4}-0.6382\alpha^{2}+0.38823\alpha+0.0097=\lambda_{0}$
$\displaystyle\lambda_{0}^{\prime}$ $\displaystyle=$
$\displaystyle\frac{3\alpha^{2}}{4}-1.2764\alpha+0.38823.$

So $\lambda_{0}$ is increasing in $[0,0.3965684]$. Then $\lambda\leq 0.0789$.

Therefore we may assume that $\lambda$ gets the maximum in its interior
points. By Theorem 4.1, then
$\frac{\partial\lambda}{\partial\gamma}=\frac{\partial\lambda}{\partial\eta}$.
Combining with

$\displaystyle\frac{\partial\lambda}{\partial\gamma}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+\frac{\alpha\gamma}{2}+\eta\frac{(\alpha+\gamma)}{2}$
$\displaystyle\frac{\partial\lambda}{\partial\eta}$ $\displaystyle=$
$\displaystyle\alpha\eta+\frac{(\alpha+\gamma)^{2}}{4},$

we get $\gamma^{2}=2\eta(\gamma-\alpha)$, contradicting to $\alpha>\gamma$.
This completes the proof of Claim 5.14. $\Box$

###  7.4 Proof of Claim 5.15.

Note that for $x,y\in A_{3}$, if $|N^{*}(x,y)|=4$ (i.e. $N^{*}(x,y)=A$) and
there exists $zw\in G_{x}[A]\cap G_{y}[A]$, then $\\{x,y,z,w\\}$ forms a
$K_{4}^{3}$, a contradiction. So for $x,y\in A_{3}$, if $|N^{*}(x,y)|=4$, then
$G_{x}[A]\cap G_{y}[A]=\emptyset$ and $G_{x}[A]\cup G_{y}[A]={A\choose 2}$.
Such pairs $\\{x,y\\}$ are partitioned into at most ${6\choose 3}/2$ groups
$\\{B_{0i},B_{1i}\\}$ such that $G_{x}[A]$ are the same for all $x\in B_{0i}$,
$G_{y}[A]$ are the same for all $y\in B_{1i}$, and $G_{x}[A]\cup G_{y}[A]$ is
a partition of ${A\choose 2}$ for all $x\in B_{0i}$ and $y\in B_{1i}$ for each
$i$. Assume that there are $s$ such groups. So $|N^{*}(x,y)|\leq 3$ for
$x,y\in A_{4}\cup A_{3}$ except $x\in B_{0i}$ and $y\in B_{1i}$, then we may
assume that $N^{*}(x,y)=\\{1,2,3\\}$ for such $x,y$ with the Lagrangian non-
decreasing. Note that $|N^{*}(x,y)|\leq 2$ for $x,y\in B_{0i}$ or $x,y\in
B_{1i}$ for $1\leq i\leq s$. Therefore we may assume that
$N^{*}(x,y)=\\{1,2\\}$ for such $x,y$ with the Lagrangian non-decreasing. And
for all $x\in A_{3}$, if $x23\notin E(G)$, then we may assume that
$x12,x13,x14\in E(G)$. If $x23\in E(G)$, since $\\{x,1,2,3\\}$ doesn’t span
$K_{4}^{3}$, then one of $x12,x13$ does not belong to $E(G)$, we can replace
$x23$ by that edge and replace other 2 edges $xij$, $ij\in A^{(2)}$ from
$\\{x12,x13,x14\\}$ with the Lagrangian non-decreasing. Let $f_{0i}=\sum_{v\in
B_{0i}}x_{v}$ and $f_{1i}=\sum_{v\in B_{1i}}x_{v}$ and $f_{i}=f_{0i}+f_{1i}$
for $1\leq i\leq s$. Let $f^{\prime}=f-\sum_{i=1}^{s}f_{i}$. By Claim 5.14 and
the above analysis, we have

$\displaystyle\lambda(G[A\cup A_{4}\cup A_{3}],\vec{x})$ $\displaystyle\leq$
$\displaystyle
0.0789(a+b+c+d+e)^{3}+(ab+ac+ad)f+(ef+f^{\prime}\sum_{i=1}^{s}f_{i}+\sum_{1\leq
i\not=j\leq s}f_{i}f_{j}$ $\displaystyle+$ $\displaystyle\frac{f^{\prime
2}}{2})(a+b+c)+\sum_{i=1}^{s}\frac{f_{0i}^{2}+f_{1i}^{2}}{2}(a+b)+\sum_{i=1}^{s}f_{0i}f_{1i}(a+b+c+d)$
$\displaystyle\leq$ $\displaystyle
0.0789(a+b+c+d+e)^{3}+(ab+ac+ad)f+ef(a+b+c)+{(a+b)f^{2}\over 2}$
$\displaystyle+$ $\displaystyle c(f^{\prime}\sum_{i=1}^{s}f_{i}+\sum_{1\leq
i\not=j\leq s}f_{i}f_{j}+\frac{f^{\prime
2}}{2})+(c+d)\sum_{i=1}^{s}f_{0i}f_{1i}$ $\displaystyle\leq$ $\displaystyle
0.0789(a+b+c+d+e)^{3}+(ab+ac+ad)f+ef(a+b+c)+{(a+b)f^{2}\over 2}$
$\displaystyle+$ $\displaystyle c(f^{\prime}\sum_{i=1}^{s}f_{i}+\sum_{1\leq
i\not=j\leq s}f_{i}f_{j}+\frac{f^{\prime
2}}{2})+(c+d)\sum_{i=1}^{s}{f_{i}^{2}\over 4}$ $\displaystyle=$
$\displaystyle\lambda(f^{\prime},f_{1},\dots,f_{s}).$

Note that

$\displaystyle\lambda(f,0,\dots,0)$ $\displaystyle-$
$\displaystyle\lambda(f^{\prime},f_{1},\dots,f_{s})$ $\displaystyle=$
$\displaystyle\frac{cf^{2}}{2}-c(f^{\prime}\sum_{i=1}^{s}f_{i}+\sum_{1\leq
i\not=j\leq s}f_{i}f_{j}+\frac{f^{\prime
2}}{2})-(c+d)\sum_{i=1}^{s}{f_{i}^{2}\over 4}$ $\displaystyle=$ $\displaystyle
c{\sum_{i=1}^{s}f_{i}^{2}\over 2}-(c+d)\sum_{i=1}^{s}{f_{i}^{2}\over 4}$
$\displaystyle\geq$ $\displaystyle 0$

since $c\geq d$. So we may assume that $f^{\prime}=f$ and $f_{i}=0$ for $1\leq
i\leq s$. So

$\displaystyle\lambda(G[A\cup A_{4}\cup A_{3}],\vec{x})$ $\displaystyle\leq$
$\displaystyle 0.0789(a+b+c+d+e)^{3}+(ab+ac+ad)f+(ef+\frac{f^{2}}{2})(a+b+c)$
(11) $\displaystyle=$ $\displaystyle
0.0789(a+(b+c)+d+e)^{3}+(a(b+c)+ad)f+(ef+\frac{f^{2}}{2})(a+(b+c))$
$\displaystyle=$ $\displaystyle\lambda.$

Let $\alpha={a\over(a+b+c+d+e+f)}$, $\beta={b+c\over a+b+c+d+e+f}$,
$\delta={d\over a+b+c+d+e+f}$, $\eta={e\over a+b+c+d+e+f}$ and $\rho={f\over
a+b+c+d+e+f}$, and let $\tau={\lambda\over(a+b+c+d+e+f)^{3}}$. Then

$\tau=0.0789(\alpha+\beta+\delta+\eta)^{3}+(\alpha\beta+\alpha\delta)\rho+(\eta\rho+\frac{\rho^{2}}{2})(\alpha+\beta),$

and it’s sufficient to prove that $\tau\leq 0.092.$

Case 1. $d\geq 0.11177$.

In this case, note that $\delta\geq d\geq 0.11177$. Recall that $b+c\geq 2d$,
then $\beta\geq b+c\geq 0.22354$. Note that $\tau$ is non-decreasing if we
change $(\beta,\delta)$ to $(\beta+\epsilon,\delta-\epsilon)$ for
$\epsilon>0$. So, we may assume that $\delta=0.11177$. By Claim 5.4,
$a\leq{3-\sqrt{3}\over 3}-d\leq 3\times 0.11177\leq b+c+d$, so
$\alpha\leq\beta+\delta$. Therefore $\tau$ is non-decreasing if we change
$(\alpha,\beta)$ to $(\alpha+\epsilon,\beta-\epsilon)$ for small $\epsilon$,
we may assume that $\beta=0.22354$. Therefore

$\displaystyle\tau$ $\displaystyle=$ $\displaystyle
0.0789(\alpha+0.33531+\eta)^{3}+0.33531\alpha\rho+(\eta\rho+\frac{\rho^{2}}{2})(\alpha+0.22354)$
(12)

subject to $\alpha+0.33531+\eta+\rho=1$, $\alpha,\eta,\rho\geq 0$. If
$\rho=0,$ then $\tau=0.0789$. So we may assume that $\rho>0$.

If $\alpha=0$, then $\eta=0.66469-\rho$. So

$\displaystyle\tau$ $\displaystyle\leq$
$\displaystyle-0.0789\rho^{3}+0.125\rho^{2}-0.08\rho+0.0789=f(\rho),$
$\displaystyle f^{\prime}(\rho)$ $\displaystyle=$
$\displaystyle-0.2367\rho^{2}+0.25\rho-0.08<0.$

Note that $f(\rho)$ is decreasing in $[0,1]$, then $\tau_{0}\leq f(0)\leq
0.0789.$ So we may assume that $\alpha>0$.

If $\eta>0$, then by Theorem 4.1, we have
$\frac{\partial\tau}{\partial\alpha}=\frac{\partial\tau}{\partial\eta}$, so
$\alpha=0.11177+\eta+\frac{\rho}{2}$. Therefore $\rho=1.55292-4\alpha$ and
$\eta=3\alpha-0.88823$. So

$\displaystyle\tau$ $\displaystyle\leq$ $\displaystyle
1.05\alpha^{3}-3.55\alpha^{2}+1.55\alpha-0.094=\tau_{0}$
$\displaystyle\tau_{0}^{\prime}$ $\displaystyle=$ $\displaystyle
3.15\alpha^{2}-7.1\alpha+1.55.$

Note that $\tau_{0}$ is increasing in $[0,\frac{71-4\sqrt{193}}{63}]$, then
$\tau_{0}\leq 0.09.$

If $\eta=0$, then $\rho=0.66469-\alpha>0$. So

$\displaystyle\tau$ $\displaystyle=$ $\displaystyle
0.0789(\alpha+0.33531)^{3}+0.33531\alpha(0.66469-\alpha)+\frac{(0.66469-\alpha)^{2}}{2}(\alpha+0.22354)$
$\displaystyle\leq$ $\displaystyle
0.5789\alpha^{3}-0.8088\alpha^{2}+0.3219\alpha+0.0524=\tau_{0}$
$\displaystyle\tau_{0}^{\prime}$ $\displaystyle=$ $\displaystyle
1.7367\alpha^{2}-1.6176\alpha+0.3219.$

Note that $\tau_{0}$ is increasing in $[0,\frac{2696-\sqrt{1056819}}{5789}]$
or $[\frac{2696+\sqrt{1056819}}{5789},0.66469]$, then $\tau_{0}\leq 0.092.$

Case 2. $d<0.11177.$

In this case, we know that $b+c\geq 0.307$, so $\beta\geq 0.307$. By Claim
5.5, we know that $d\geq 0.0848$, so $\delta\geq 0.0848$. Note that $\tau$ is
non-decreasing if we change $(\beta,\delta)$ to
$(\beta+\epsilon,\delta-\epsilon)$ for $\epsilon>0$, so we may let
$\delta=0.0848$ in $\tau$. Therefore

$\displaystyle\tau$ $\displaystyle=$ $\displaystyle
0.0789(\alpha+\beta+0.0848+\eta)^{3}+(\alpha\beta+0.0848\alpha)\rho+(\eta\rho+\frac{\rho^{2}}{2})(\alpha+\beta)$

subject to

$\left\\{\begin{array}[]{lr}\alpha+\beta+0.0848+\eta+\rho=1,&\\\ \beta\geq
0.307,&\\\ \alpha,\rho\geq 0&\end{array}\right.$ (13)

Note that $\rho>0$.

If $\alpha=0$, then we claim that $\beta=0.307$, this is because that $\tau$
is non-decreasing if we change $(\alpha,\beta)$ to
$(\alpha+\epsilon,\beta-\epsilon)$ for small $\epsilon$. So
$\eta=0.6082-\rho$. Substituting these into (LABEL:9), we have

$\displaystyle\tau$ $\displaystyle\leq$
$\displaystyle-0.0789\rho^{3}+0.042\rho^{2}+0.002\rho+0.0789=f(\rho),$
$\displaystyle f^{\prime}(\rho)$ $\displaystyle=$
$\displaystyle-0.2367\rho^{2}+0.084\rho+0.002.$

Note that $f(\rho)$ is increasing in $[0,{2\sqrt{6215}+140\over 789}]$ and
decreasing in $[{2\sqrt{6215}+140\over 789},1]$, then $\tau\leq
f({2\sqrt{6215}+140\over 789})<0.085.$ So we may assume that $\alpha>0$.

If $\beta>0.307$, then by Theorem 4.1,
$\frac{\partial\tau}{\partial\alpha}=\frac{\partial\tau}{\partial\beta}$,
simplifying it, we have $\alpha=\beta+0.0848>0.3918$. If $\eta>0$, then by
Theorem 4.1, we have
$\frac{\partial\tau}{\partial\alpha}=\frac{\partial\tau}{\partial\eta}$,
simplyfying it, we get $\alpha=0.0848+\eta+\frac{\rho}{2}.$ Hence
$1.1754<3\alpha=\alpha+\beta+2\times
0.0848+\eta+\frac{\rho}{2}<1+0.0848=1.0848$, a contradiction. So $\eta=0$.
Then $\rho=1-\alpha-\beta-0.0848=1-2\alpha>0$ and $\beta=\alpha-0.0848.$ So
$0.39<\alpha<0.5$. And

$\displaystyle\tau$ $\displaystyle=$ $\displaystyle
0.0789(2\alpha)^{3}+\alpha^{2}(1-2\alpha)+\frac{(1-2\alpha)^{2}}{2}(2\alpha-0.0848)$
$\displaystyle\leq$ $\displaystyle
2.65\alpha^{3}-3.1695\alpha^{2}+1.1695\alpha-0.0424=\tau_{0},$
$\displaystyle\tau_{0}^{\prime}$ $\displaystyle=$ $\displaystyle
7.95\alpha^{2}-6.339\alpha+1.1695.$

Note that $\tau_{0}$ is decreasing in $[0.39,0.5]$, then
$\tau_{0}\leq\tau_{0}(0.39)\leq 0.089.$

So we may assume that $\beta=0.307$. If $\eta>0$, then by Theorem 4.1,
$\frac{\partial\tau}{\partial\alpha}=\frac{\partial\tau}{\partial\eta}$,
simplifying it, we have $\alpha=0.0848+\eta+\frac{\rho}{2}.$ Since
$\alpha+\eta+\rho=1-\beta-0.0848=0.6082$, then $\eta=3\alpha-0.7778$ and
$\rho=1.386-4\alpha>0.$ Since
$\frac{\partial\tau}{\partial\eta}=\frac{\partial\tau}{\partial\rho}$, then
$0.2367(\alpha+\beta+0.0848+\eta)^{2}=\alpha\beta+0.0848\alpha+\eta(\alpha+\beta)$.
By direct calculation,
$\frac{492\alpha^{2}}{625}-\frac{395603\alpha}{312500}+\frac{685129883}{2500000000}=0$,
and $\alpha=\frac{395603-20\sqrt{180576895}}{492000}$. But $3\alpha<0.774$,
then $\eta=3\alpha-0.7778<0$, a contradiction. So $\eta=0$, then
$\rho=1-\alpha-\beta-0.0848=0.6082-\alpha>0.$ So

$\displaystyle\tau$ $\displaystyle=$ $\displaystyle
0.0789(\alpha+0.3918)^{3}+0.3918\alpha(0.6082-\alpha)+\frac{(0.6082-\alpha)^{2}}{2}(\alpha+0.307)$
$\displaystyle\leq$ $\displaystyle
0.5789\alpha^{3}-0.75375\alpha^{2}+0.27286\alpha+0.06153=\tau_{0},$
$\displaystyle\tau_{0}^{\prime}$ $\displaystyle=$ $\displaystyle
1.7367\alpha^{2}-1.5075\alpha+0.27286.$

Note that $\tau_{0}$ is increasing in
$[0,\frac{5025}{11578}-\frac{\sqrt{942631005}}{173670}]$, then $\tau_{0}\leq
0.092.$ This complete the proof of Claim 5.15. $\Box$

###  7.5 Proof of Claim 5.16

By Claim 5.15, we have

$\displaystyle\lambda(G)$ $\displaystyle\leq$ $\displaystyle
0.092(a+b+c+d+e+f)^{3}+(ab+ac)g+abh+(eg+eh+fg+fh+\frac{g^{2}}{2}+gh$
$\displaystyle+$ $\displaystyle\frac{h^{2}}{2})(a+b+c+d)$ $\displaystyle\leq$
$\displaystyle
0.092(a+\beta+d+\zeta)^{3}+a\beta\eta+(\zeta\eta+\frac{\eta^{2}}{2})(a+\beta+d)$
$\displaystyle=$ $\displaystyle\lambda,$

where $\eta=g+h$, $\beta=b+c$ and $\zeta=e+f$.

Case 1. $d\geq 0.11177.$

Since replacing $(a,d)$ by $(a+\epsilon,d-\epsilon)$ for $\epsilon>0$ will not
decrease $\lambda$, so we may assume that $d=0.11177$. Let $\alpha=a+\beta$.
Then

$\displaystyle\lambda$ $\displaystyle\leq$ $\displaystyle
0.092(\alpha+\delta+\zeta)^{3}+\frac{\alpha^{2}\eta}{4}+(\zeta\eta+\frac{\eta^{2}}{2})(\alpha+\delta),$

subject to

$\left\\{\begin{array}[]{lr}\alpha+\delta+\zeta+\eta=1,&\\\
\delta=0.11177,\alpha,\zeta,\eta\geq 0.&\end{array}\right.$ (14)

If $\eta=0$, then $\lambda=0.092$, we are done. So assume that $\eta>0.$

If $\alpha=0$, then $\zeta=0.88823-\eta$.

$\displaystyle\lambda$ $\displaystyle\leq$
$\displaystyle-0.092\eta^{3}+0.221\eta^{2}-0.1767\eta+0.092=\lambda_{0}$
$\displaystyle\lambda_{0}^{\prime}$ $\displaystyle=$
$\displaystyle-0.276\eta^{2}+0.442\eta-0.1767.$

Note that $\lambda_{0}$ is decreasing in $[0,0.7700236]$ or $[0.8314257,1]$.
Therefore $\lambda_{0}\leq 0.095.$ So assume that $\alpha>0.$

If $\zeta=0$, then $\eta=0.88823-\alpha>0$. So

$\displaystyle\lambda$ $\displaystyle\leq$ $\displaystyle
0.342\alpha^{3}-0.58\alpha^{2}+0.3\alpha+0.045=\lambda_{0}$
$\displaystyle\lambda_{0}^{\prime}$ $\displaystyle=$ $\displaystyle
1.026\alpha^{2}-1.16\alpha+0.3.$

Note that $\lambda_{0}$ is increasing in $[0,\frac{290-5\sqrt{286}}{513}]$ or
$[\frac{290+5\sqrt{286}}{513},0.88823]$. Therefore $\lambda_{0}\leq 0.095.$

If $\zeta>0$, then by Theorem 4.1, we have
$\frac{\partial\lambda}{\partial\alpha}=\frac{\partial\lambda}{\partial\zeta}$,
i.e. $2\zeta+\eta=\alpha+2\delta$. Since $\alpha+\delta+\zeta+\eta=1$, then
$\zeta=2\alpha-0.66469$ and $\eta=1.55292-3\alpha>0$. So
$0.33234<\alpha<0.51764$. And

$\displaystyle\lambda$ $\displaystyle\leq$ $\displaystyle
0.234\alpha^{3}-0.7115\alpha^{2}+0.4765\alpha+0.00385=\lambda_{0}$
$\displaystyle\lambda_{0}^{\prime}$ $\displaystyle=$ $\displaystyle
0.702\alpha^{2}-1.423\alpha+0.4765.$

Note that $\lambda_{0}$ is increasing in
$[0,\frac{1423-11\sqrt{5677}}{1404}]$. Therefore $\lambda_{0}\leq 0.096.$

Case 2. $d<0.11177.$

In this case, we have $\beta\geq 0.307$. By Claim 5.5, then $d\geq 0.0848$.
Since replacing $(a,d)$ by $(a+\epsilon,d-\epsilon)$ for $\epsilon>0$ will not
decrease $\lambda$, so we may assume $d=0.0848$. So

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
0.092(a+\beta+\delta+\zeta)^{3}+a\beta\eta+(\zeta\eta+\frac{\eta^{2}}{2})(a+\beta+\delta),$

subject to

$\left\\{\begin{array}[]{lr}a+\beta+\delta+\zeta+\eta=1,&\\\ \beta\geq
0.307,&\\\ \delta=0.0848.&\end{array}\right.$ (15)

If $a=0$, we claim that $\beta=0.307$ since $\lambda$ is non-decreasing if we
change $(a,\beta)$ to $(a+\epsilon,\beta-\epsilon)$ for small $\epsilon$. So
$\zeta=0.6082-\eta$. Then

$\displaystyle\lambda$ $\displaystyle\leq$
$\displaystyle-0.092\eta^{3}+0.0801\eta^{2}-0.037\eta+0.092=f(\eta),$
$\displaystyle f^{\prime}(\eta)$ $\displaystyle=$
$\displaystyle-0.276\eta^{2}+0.1602\eta-0.037<0.$

Therefore $\lambda_{0}\leq f(0)=0.092$. So we may assume that $a>0$.

If $\beta>0.307$, then by Theorem 4.1, $\frac{\partial\lambda}{\partial
a}=\frac{\partial\lambda}{\partial\beta}$, simplifying it, we get
$a=\beta>0.307$. If $\zeta>0$, then we have $\frac{\partial\lambda}{\partial
a}=\frac{\partial\lambda}{\partial\zeta}$, i.e.
$a+\delta=\zeta+\frac{\eta}{2}.$ So
$1<3a+2\delta=a+\beta+\delta+\zeta+\frac{\eta}{2}<1$, a contradiction. Then
$\zeta=0$. So $\eta=1-a-\beta-\delta=1-2a-\delta=0.9152-2a>0$, then
$a<0.4576$. Recall that $\delta=0.0848$. So

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
0.092(2a+0.0848)^{3}+(0.9152-2a)a^{2}+\frac{(0.9152-2a)^{2}}{2}(2a+0.0848).$

By a direct calculation on the derivative of $\lambda_{0}$, we obtain that
$\lambda_{0}$ is increasing in $[0,0.2138466]$. Therefore $\lambda_{0}\leq
0.096.$

So we may assume that $\beta=0.307$ and $\delta=0.0848$. If $\zeta=0$, then
$\eta=1-a-\beta-\delta=0.6082-a$. By Theorem 4.1, then
$\frac{\partial\lambda}{\partial a}=\frac{\partial\lambda}{\partial\eta}$.
Combining them, we get
$0.276(a+0.3918)^{2}+\frac{(0.6082-a)^{2}}{2}=(0.6082-a)(a+0.0848)+0.307a$.
Solving the equation for $a$ and substituting the values into $\lambda$, we
obtain that $\lambda\leq 0.095.$ So we may assume that $\zeta>0$. By Theorem
4.1, then $\frac{\partial\lambda}{\partial
a}=\frac{\partial\lambda}{\partial\zeta}$, i.e.
$a+\delta=\zeta+\frac{\eta}{2}$. So $\eta=1.0468-4a$ and $\zeta=3a-0.4386.$
Then $0.1462<a<0.2617$ and

$\displaystyle\lambda$ $\displaystyle=$ $\displaystyle
0.092(4a-0.0468)^{3}+0.307(1.0468-4a)a+((3a-0.4386)(1.0468-4a)$
$\displaystyle+$ $\displaystyle\frac{(1.0468-4a)^{2}}{2})(a+0.3918).$

By a direct calculation on the derivative of $\lambda$, we obtain that
$\lambda$ is increasing in
$[0,\frac{716959}{1770000}-\frac{\sqrt{211982461}}{70800}]$ or
$[\frac{716959}{1770000}+\frac{\sqrt{211982461}}{70800},1]$. Recall that
$0.1462<a<0.2617$, then $\lambda<0.0961$. $\Box$

###  7.6 Proof of Claim 5.18

If $f=0$, then

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}(c+d+e)}{4}+\alpha
c(d+e)+\alpha(de+\frac{e^{2}}{2})+\frac{d^{2}(\alpha+e)}{4}$

subject to

$\left\\{\begin{array}[]{lr}\alpha+c+d+e=1,&\\\ c\geq
0.08.&\end{array}\right.$ (16)

If $d=0$, then $\lambda=\frac{\alpha^{2}(c+e)}{4}+\alpha
ce+\alpha\frac{e^{2}}{2}\leq\lim_{n\to\infty}B(2,n-2)=\frac{\sqrt{3}}{18}$. So
we are done.

If $e=0$, then $\lambda=\frac{\alpha^{2}(c+d)}{4}+\alpha cd+\frac{\alpha
d^{2}}{4}\leq\lambda(K_{5}^{3})<\frac{\sqrt{3}}{18}$. So we are done.

So we may assume that $d,e>0$. By direct calculation,

$\displaystyle\frac{\partial\lambda}{\partial\alpha}$ $\displaystyle=$
$\displaystyle\frac{\alpha(c+d+e)}{2}+cd+ce+de+\frac{e^{2}}{2}+\frac{d^{2}}{4}$
$\displaystyle\frac{\partial\lambda}{\partial c}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+\alpha d+\alpha e$
$\displaystyle\frac{\partial\lambda}{\partial d}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+\alpha c+\alpha e+\frac{d(\alpha+e)}{2}$
$\displaystyle\frac{\partial\lambda}{\partial e}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+\alpha c+\alpha d+\alpha
e+\frac{d^{2}}{4}.$

By Theorem 4.1, $\frac{\partial\lambda}{\partial
d}=\frac{\partial\lambda}{\partial e}$, combining with the above equations, we
get $d=2e-2\alpha.$ If $c>0.08$, then by Theorem 4.1,
$\frac{\partial\lambda}{\partial c}=\frac{\partial\lambda}{\partial e}$,
combining the equations, we get $\alpha c+{d^{2}\over 4}=0$, a contradiction.
So $c=0.08$, therefore $\alpha+d+e=0.92$, combining with $d=2e-2\alpha$, we
get $d=\frac{1.84-4\alpha}{3}$ and $e=\frac{\alpha+0.92}{3}$. Since
$\frac{\partial\lambda}{\partial\alpha}=\frac{\partial\lambda}{\partial e}$,
then $\frac{5\alpha^{2}}{36}+\frac{307\alpha}{450}-\frac{3473}{11250}=0$. Then
$\alpha=\frac{3\sqrt{14331}-307}{125}$. By direct calculation, $\lambda\leq
0.096<\frac{\sqrt{3}}{18}$, a contradiction. $\Box$

###  7.7 Proof of Claim 5.19

If $e=0$, in view of (LABEL:eq6), then

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}(c+d+f)}{4}+\alpha
cd+(\alpha+c)(df+\frac{f^{2}}{2})+\frac{d^{2}(\alpha+f)}{4},$

subject to

$\left\\{\begin{array}[]{lr}\alpha+c+d+f=1,&\\\ c\geq 0.08;\alpha,d,f\geq
0.&\end{array}\right.$ (17)

If $d=0$, then we have

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}(c+f)}{4}+(\alpha+c)\frac{f^{2}}{2}.$

We relax the constraint $c\geq 0.08$ to $c\geq 0$ and $\alpha+c+f=1$. If
$c=0$, then
$\lambda=\frac{\alpha^{2}(1-\alpha)}{4}+\alpha\frac{(1-\alpha)^{2}}{2}\leq\frac{\sqrt{3}}{18}$
by Fact 4.9. If $c>0$, recall that $f>0$, then by Theorem 4.1, we have
$\frac{\partial\lambda}{\partial c}=\frac{\partial\lambda}{\partial f}$, then
$f=2(\alpha+c)$, combining with $\alpha+c+f=1$, we have $f=\frac{2}{3}$ and
$c=\frac{1}{3}-\alpha$. So
$\lambda=\frac{\alpha^{2}(1-\alpha)}{4}+\frac{2}{27}$, where
$\alpha<\frac{1}{3}$. Since $\lambda^{\prime}=\frac{\alpha(2-3\alpha)}{4}$,
then $\lambda$ is increasing in $[0,\frac{1}{3}]$, i.e.
$\lambda\leq\frac{(\frac{1}{3})^{2}\frac{2}{3}}{4}+\frac{2}{27}=\frac{5}{54}\leq\frac{\sqrt{3}}{18}.$

So we may assume that $d>0$. Recall that $e+f>\alpha+\frac{d}{2}$ and $e=0$,
so $f>\alpha+\frac{d}{2}$. By direct calculation,

$\displaystyle\frac{\partial\lambda}{\partial\alpha}$ $\displaystyle=$
$\displaystyle\frac{\alpha(c+d+f)}{2}+cd+df+\frac{f^{2}}{2}+\frac{d^{2}}{4}$
$\displaystyle\frac{\partial\lambda}{\partial c}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+\alpha d+df+\frac{f^{2}}{2}$
$\displaystyle\frac{\partial\lambda}{\partial d}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+\alpha c+(\alpha+c)f+\frac{d(\alpha+f)}{2}$
$\displaystyle\frac{\partial\lambda}{\partial f}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+(\alpha+c)(d+f)+\frac{d^{2}}{4}.$

By Theorem 4.1, we have $\frac{\partial\lambda}{\partial
d}=\frac{\partial\lambda}{\partial f}$, i.e.

$\frac{d^{2}}{4}+(\alpha+c)d=\alpha c+\frac{d(\alpha+f)}{2}>\alpha
c+\frac{d(2\alpha+\frac{d}{2})}{2},$

then $d>\alpha.$ If $c>0.08$, then
$\frac{\partial\lambda}{\partial\alpha}=\frac{\partial\lambda}{\partial c}$.
Therefore $\frac{\alpha^{2}}{4}+\alpha
d=\frac{\alpha(c+d+f)}{2}+cd+\frac{d^{2}}{4}$. Since $f>\alpha+\frac{d}{2},$
then $\frac{\alpha^{2}}{4}+\alpha
d>\frac{\alpha(c+\alpha+\frac{3d}{2})}{2}+cd+\frac{d^{2}}{4}>\frac{\alpha^{2}}{2}+\frac{d^{2}+3\alpha
d}{4}$. So $\alpha d>\alpha^{2}+d^{2}$, a contradiction. So $c=0.08$. Since
$\frac{\partial\lambda}{\partial\alpha}=\frac{\partial\lambda}{\partial f}$
and $f=0.92-\alpha-d$, then

$\frac{\alpha^{2}}{4}+\alpha d+\alpha
f+cf-\frac{\alpha(1-\alpha)}{2}-df-\frac{f^{2}}{2}=0.$

Substituting $c=0.08$ and $f=0.92-\alpha-d$ into it, we have

$\frac{\alpha^{2}}{4}+\alpha
d+\alpha(0.92-\alpha-d)+0.08(0.92-\alpha-d)-\frac{\alpha(1-\alpha)}{2}-d(0.92-\alpha-d)-\frac{(0.92-\alpha-d)^{2}}{2}=0.$

Simplifying it, we have
$\frac{d^{2}}{2}-\frac{2d}{25}+\frac{63\alpha}{50}-\frac{3\alpha^{2}}{4}-\frac{437}{1250}=0$,
then
$d=\sqrt{\frac{3\alpha^{2}}{2}-\frac{63\alpha}{25}+\frac{441}{625}}+\frac{2}{25}$.
Note that $0.92=\alpha+d+f>\alpha+\alpha+\frac{3\alpha}{2}=\frac{7\alpha}{2}$,
then $0<\alpha<0.2629$. Therefore
$\phi(\alpha)=\frac{3\alpha^{2}}{2}-\frac{63\alpha}{25}+\frac{441}{625}>\phi(0.2629)>0.1467$.
So $d>\sqrt{0.1467}+\frac{2}{25}>0.46$ and $f>0.23+\alpha$, then
$\alpha+d+f>\alpha+\sqrt{\frac{3\alpha^{2}}{2}-\frac{63\alpha}{25}+\frac{441}{625}}+\frac{2}{25}+0.23+\alpha\triangleq\psi(\alpha)>\psi(0)=1.15>1$,
a contradiction. $\Box$

###  7.8 Proof of Claim 5.20

If $d=0$, then we have $\alpha+c+e+f=1$ and $\alpha=e$. In view of
(LABEL:eq6),

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}(c+e+f)}{4}+\alpha
ce+\alpha\frac{e^{2}}{2}+(\alpha+c)(ef+\frac{f^{2}}{2}).$

Then

$\displaystyle\frac{\partial\lambda}{\partial\alpha}$ $\displaystyle=$
$\displaystyle\frac{\alpha(c+e+f)}{2}+ce+\frac{e^{2}}{2}+ef+\frac{f^{2}}{2}$
$\displaystyle\frac{\partial\lambda}{\partial c}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+\alpha e+ef+\frac{f^{2}}{2}$
$\displaystyle\frac{\partial\lambda}{\partial e}$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}}{4}+\alpha c+\alpha e+(\alpha+c)f,$

If $c>0.08$, then by Theorem 4.1, we have
$\frac{\partial\lambda}{\partial\alpha}=\frac{\partial\lambda}{\partial c}$,
so $\frac{\alpha(1-\alpha)}{2}+\alpha
c+\frac{\alpha^{2}}{2}=\frac{\alpha^{2}}{4}+\alpha^{2}$, then
$0.08<c=\frac{5\alpha-2}{4}$ and $0<f=\frac{6-13\alpha}{4}$. So
$0.464<\alpha<\frac{6}{13}<0.462$, a contradiction.

If $c=0.08$, then $f=0.92-2\alpha$. So

$\displaystyle\lambda$ $\displaystyle=$
$\displaystyle\frac{\alpha^{2}(1-\alpha)}{4}+0.08\alpha^{2}+\frac{\alpha^{3}}{2}+(\alpha+0.08)(\alpha(0.92-2\alpha)+\frac{(0.92-2\alpha)^{2}}{2})$
$\displaystyle=$
$\displaystyle\frac{\alpha^{3}}{4}-\frac{59\alpha^{2}}{100}+\frac{437\alpha}{1250}+\frac{529}{15625}.$
$\displaystyle\lambda^{\prime}$ $\displaystyle=$
$\displaystyle\frac{3\alpha^{2}}{4}-\frac{59\alpha}{50}+\frac{437}{1250}.$

Note that $\lambda$ is increasing in $[0,\frac{59-\sqrt{859}}{75}]$, then
$\lambda\leq 0.0955$. $\Box$

Acknowledgement We are grateful to two reviewers for checking all the details
and giving us valuable comments to help improve the presentation. The research
is supported in part by National Natural Science Foundation of China (No.
11931002).

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