One Turán Type Problem on Uniform Hypergraphs

Abstract

Let $n,m,p,r\in N$ with $p\ge n\ge r$. For a hypergraph, if each edge has r vertices, then the hypergraph is called an r-graph. Define $e_r(n,m;p)$ to be the maximum number of edges of an r-graph with p vertices in which every subgraph of n vertices has at most m edges. Researching this function constitutes a Turán type problem. In this paper, on the one hand, for fixed p, we present some results about the exact values of $e_r(n,m;p)$ for small m compared to n; on the other hand, for sufficient large p, we use the combinatorial technique of double counting to give an upper bound of $e(n,m;p)$ and obtain a lower bound of $e_r(n,m;p)$ by applying the lower bound of the independent set of a hypergraph.

1. Introduction

All graphs considered in this article are undirected and contain no loops or multiple edges. Let $r\in \mathbb{N}$ with $r\ge 2$, an r-uniform hypergraph G is a pair $\left(V\right(G),E(G\left)\right)$ consisting of vertex set $V\left(G\right)$ and edge set $E\left(G\right)$ of r-element subsets of $V\left(G\right)$. When $r=2$, G is a simple graph; when $r\ge 3$, G is also called an r-graph. We call $\left|V\right(G\left)\right|$ the order of G and $\left|E\right(G\left)\right|$ the size of G and let $e\left(G\right)=\left|E\right(G\left)\right|$.

For a vertex $v\in V\left(G\right)$, the degree of v is the number of edges in $E\left(G\right)$ containing v and is denoted by $d\left(v\right)$. We denote by $\Delta \left(G\right)$ the maximum degree of the vertices of G and by $d\left(G\right)$ their average degree, ${\displaystyle \frac{1}{\left|V\right(G\left)\right|}}$${\sum }_{v\in V\left(G\right)}d\left(v\right)$. Let $n\in \mathbb{N}$ and $\left|V\right(G\left)\right|=n$, sometimes we use $\left({\displaystyle \genfrac{}{}{0pt}{}{\left[n\right]}{k}}\right)$ to denote $\{K:K\subset V(G),|K|=k\}.$ If $\left|V\right(G\left)\right|=n$ and $E\left(G\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{\left[n\right]}{r}}\right)$, then G is called a complete r-uniform hypergraph and denoted by $K_n^{\left(r\right)}$; when $r=2$, we omit r and denote the complete graph by $K_n$.

For two r-graphs F and G, if any subhypergraph of G is not isomorphic to F, then we say G is F-free; otherwise, G contains at least one copy of F. Let $\mathcal{F}$ be a nonempty set of r-graphs; if G is F-free for any $F\in \mathcal{F}$, then we say G is $\mathcal{F}$-free. For a positive p and r-graph set $\mathcal{F}$, define $e{x}_r(p,\mathcal{F})$ as the maximum number of edges of an r-graph with p vertices which is $\mathcal{F}$-free; this is called the Turán number [1] of $\mathcal{F}$. When the set $\mathcal{F}$ only contains one element F, we write it as $e{x}_r(p,F)$ by simplicity. Moreover, when $r=2$, we usually use $ex(p,F)$ rather than $e{x}_2(p,F)$. Meanwhile the Turán density of an r-graph F is defined as

$$\pi \left(F\right)=li{m}_{p\to \infty }{\displaystyle \frac{e{x}_r(p,F)}{\left(\genfrac{}{}{0pt}{}{p}{r}\right)}}.$$

In the year 1941, Turán [2] firstly did research about the function $ex(p,F)$ for a general simple graph F. At present, there have been plenty of results about this function for the case $r=2$, most of which focus on determining the bounds of the Turán number for different graphs. It seems difficult to determine the exact values of Turán numbers even for 2-graphs, except for several classes of graphs containing complete graphs, roads, etc. One of the classic results is the Turán number with respect to a complete graph.

Theorem 1

([2]). Let $p\ge n\in \mathbb{N},$ then

$$ex(p,K_n)={\displaystyle \frac{(n-2)(p^2-r^2)}{2(n-1)}}+{\displaystyle \frac{r(r-1)}{2}},$$

where $p\equiv r(\mathit{mod}n-1)(r\in \{0,1,\dots ,n-2\left\}\right).$

For convenience, we set $t_{n-1}\left(p\right)=ex(p,K_n)$ and let $Ex(p,K_n)$ be the set of graphs which is $K_n$-free with $ex(p,K_n)$ edges. For any natural numbers $p,n$ with $p\ge n$, partition p vertices into $n-1$ parts $V_1,\dots ,V_{n-1}$ of nearly equal size, i.e., $|\left|V_i\right|-\left|V_j\right||\le 1$ for different $i,j.$ Then, this complete $(n-1)$-partite graph is called a Turán graph, denoted by $T_{n-1}\left(p\right)$. It is not hard to check that $e\left(T_{n-1}\left(p\right)\right)=t_{n-1}\left(p\right).$ So $T_{n-1}\left(p\right)$ is an extremal graph, i.e., $T_{n-1}\left(p\right)\in Ex(p,K_n)$. In addition, it was proved that the extremal graph is unique (up to isomorphism).

The exact values of Turán numbers for other types of graphs can be found in the literature [3,4,5,6,7,8,9]. For the graphs whose Turán number is difficult to calculate, it makes sense to know the asymptotic order of their Turán numbers. In fact, determining the Turán density of a given graph is also a classical extremal graph problem. Let $\chi \left(F\right)$ be the chromatic number of F, which means the minimum numbers of colors making any neighbors in F colored by distinct colors. Erdős, Stone [10], and Simonovits [11] proved the foundational theorem of extremal graph theory, known as the Erdős–Stone–Simonovits theorem for short.

Theorem 2

([10,11]). Let $p\in \mathbb{N}$ and $\mathcal{F}$ be a family of graphs, when p is large enough,

$$ex(p,\mathcal{F})={\displaystyle \frac{\chi \left(\mathcal{F}\right)-2}{\chi \left(\mathcal{F}\right)-1}}·{\displaystyle \frac{p^2}{2}}+o\left(p^2\right),$$

where $\chi \left(\mathcal{F}\right)=\mathit{min}\left\{\chi \right(F):F\in \mathcal{F}\}$.

The Erdős–Stone–Simonovits theorem states that the Turán density of any $\chi$-chromatic graph F is equal to ${\displaystyle \frac{\chi \left(\mathcal{F}\right)-2}{\chi \left(\mathcal{F}\right)-1}}.$ For any r-graph F, a well-known fact (proved in 1964 [12]) is that the Turán density of F always exists.

In this paper, we focused on one Turán type problem for r-graphs; however, we do not consider the concrete structure of an r-graph and we are concerned about the edges of the subhypergraphs. We want to find the maximum edges or the minimum density of r-graphs only by adding a limit to the number of edges of the subhypergraphs. Now, we give the definition of a function that we mainly consider in this article.

Definition 1.

Let $p,n,m,r\in \mathbb{N}$ with $p\ge n\ge r$, define $e_r(n,m;p)$ as the maximum number of edges of an r-graph with p vertices in which n vertices induce at most m edges, and just denoted by $e(n,m;p)$ when $r=2$.

Obviously, $e_r(n,\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right);p)=\left({\displaystyle \genfrac{}{}{0pt}{}{p}{r}}\right),$$e_r(n,\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right)-1;p)=e{x}_r(p,K_n^r).$

When $0\le m\le \left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right)$, we have $0\le e_r(n,m;p)\le \left({\displaystyle \genfrac{}{}{0pt}{}{p}{2}}\right).$

For convenience, we call G an $(n,m)$ graph if it has the property that any n vertices induce at most m edges.

If we let $\mathcal{H}=\left\{H\mathrm{is}\mathrm{an}\mathrm{r}\text{-}\mathrm{graph}\right|$ any subgraph $F\subset H$ induced by n vertices contains at least $m+1$ edges}, then we have $e_r(n,m;p)=e{x}_r(p,\mathcal{H}),$ so this is one Turán type problem. In fact, as early as 1963, G. Dirac [13] began to study this function for general simple graphs, and later there were several authors [14,15,16] who also did some work with this function and generalized this problem from general simple graphs to hypergraphs. It is worth mentioning that there is a similar function with respect to hypergraphs investigated by Brown, Erdős, and T. Sós [17,18], where the function becomes much more involved and sometimes extremely deep. For more about Turán type hypergraph results, we referred to [19,20] and the surveys by Füredi [21] and Sidorenko [22].

Here, let us consider the function $e_r(n,m;p)$ from two perspectives: its exact values for fixed r and p; its asymptotic values for fixed n and m and sufficiently large n.

Now, we list all the known and new results about this function from the above two perspectives.

Theorem 3

([16]). Let $p,n,k\in \mathbb{N}$ with $p\ge n\ge k,$

1. 

$e(n,n-1;p)=ex(p,\{C_3,\dots ,C_n\});$

2. 

$e(n,t_{k-1}\left(n\right);p)=t_{k-1}\left(p\right),$ where $k\ge 3$;

3. 

$e(n,\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2}}\right)-k;p)=t_{n-k}\left(p\right),$ where $n\ge 2k$;

4. 

$t_{k-1}\left(p\right)\le e(n,m;p)\le t_k\left(p\right)$, where $t_{k-1}\left(n\right)\le m<t_k\left(n\right).$

For the case $m\le n-2$, the value of $e(n,m;p)$ was obtained by different authors (refer to [14,15,23]), but the original papers are not easily found, so we give a proof of this in Section 2.

Theorem 4.

Let $m,n,p\in \mathbb{N}$ with $2\le m\le n\le p.$ If $n-1=k(m-1)+r,k\in \mathbb{N},r\in \{0,1,\dots ,m-2\}$, then

$$e(n,n-m;p)=\left[{\displaystyle \frac{(k-1)p+r}{k}}\right]=p-m+1+\left[{\displaystyle \frac{n-1-p}{\left[\frac{n-1}{m-1}\right]}}\right].$$

For a given r-graph G, if a subset S of $V\left(G\right)$ does not contain any edge in $E\left(G\right)$, i.e., $\left({\displaystyle \genfrac{}{}{0pt}{}{S}{r}}\right)\cap E\left(G\right)=\varnothing$, then S is called an independent set and the independence number of G is

$$\alpha \left(G\right):=max\left\{\right|S|:S\subset V(G\left)\mathrm{is}\mathrm{an}\mathrm{independent}\mathrm{set}\right\}.$$

By the following lemma, we can give a lower bound of $e_r(n,m;p)$ in Section 3.

Lemma 1

([24]). For a given positive number $r\ge 2$, there exists a constant C which only depends on the value of r such that for any r-graph H

$$\alpha \left(H\right)\ge C{\displaystyle \frac{\left|V\right(H\left)\right|}{d{\left(H\right)}^{\frac{1}{r-1}}}}.$$

For sufficiently large p, we use the combinatorial technique of double counting to give an upper bound of $e_r(n,m;p)$, so we have the following result.

Theorem 5.

Let $p,n,m,r\in \mathbb{N}$ with $p\ge n\ge r$. If $e_r(n,m;p)\ge m+1$, then

$$C_1{p}^{{\displaystyle \frac{(m+1)r-n}{m}}}\le e_r(n,m;p)\le C_2{p}^{\lceil {\displaystyle \frac{(m+1)r-n}{m}}\rceil }$$

where $C_1,C_2$ are constants which depend only on the values of $n,m,r$.

2. Proof of Theorem 4

In this section, all the graphs are simple, and $T_n$ denotes a tree with n vertices without any specific structural constraints.

Lemma 2.

Let $m,n,p\in \mathbb{N}$ with $2\le m\le n\le p.$ If $n-1=k(m-1)+r,k\in \mathbb{N},r\in \{0,1,\dots ,m-2\}$, then

$$e(n,n-m;p)\ge \left[{\displaystyle \frac{(k-1)p+r}{k}}\right].$$

Proof. 

Since

$$\begin{array}{cc}\hfill p-r(k+1)& \ge n-r(k+1)\hfill \\ & =k(m-1)+r+1-r(k+1)\hfill \\ & \ge k+1.\hfill \end{array}$$

we let $p-r(k+1)=ak+b,a\in \mathbb{N},b\in \{0,1,\dots ,k-1\}$. Set $G=r{T}_{k+1}\cup a{T}_k\cup T_b.$ Then, we have

$$\begin{array}{cc}\hfill e\left(G\right)& =e\left(r{T}_{k+1}\right)+e\left(a{T}_k\right)+e\left(T_b\right)\hfill \\ & =rk+a(k-1)+b-1\hfill \\ & =rk+a(k-1)+[b-b/k]\hfill \\ & =[rk+(ak+b\left)\right(1-1/k\left)\right]\hfill \\ & =[rk+(p-r(k+1)){\displaystyle \frac{k-1}{k}}]\hfill \\ & =\left[{\displaystyle \frac{(k-1)p+r}{k}}\right].\hfill \end{array}$$

Now, we just need to prove that this graph is an $(n,n-m)$ graph to complete the proof of the lemma.

Let $G_0$ be an arbitrary subgraph of G induced by n vertices. Obviously, $G_0$ is a forest, and then we suppose that $G_0$ has t components, among which there are c trees with $k+1$ vertices; thereby $c\le r.$ If $t\le m-1$, then $n=v\left(G_0\right)\le c(k+1)+(t-c)k=kt+c\le k(m-1)+r=n-1,$ which is impossible! Hence, $t\ge m$ and $e\left(G_0\right)=n-t\le n-m,$ which means that G is an $(n,n-m)$ graph. □

Lemma 3.

Let $m,n,p\in \mathbb{N}$ with $2\le m\le n\le p.$ If $n-1=k(m-1)+r,k\in \mathbb{N},r\in \{0,1,\dots ,m-2\}$, G is an $(n,n-m)$ graph with p vertices, and G does not contain subtree with at least $k+1$ vertices, then

$$e(n,n-m;p)\le \left[{\displaystyle \frac{(k-1)p+r}{k}}\right].$$

Proof. 

Case 1: G is a forest.

Assume that G has t subtrees, then by the condition, we know each component contains at most k vertices; thereby, $p\le kt$. Hence,

$$e\left(G\right)=p-t\le p-p/k={\displaystyle \frac{(k-1)p}{k}}\le \left[{\displaystyle \frac{(k-1)p+r}{k}}\right].$$

Case 2: G is not a forest.

Suppose that G has t components, and let $G={\cup }_{i=1}^t{G}_i,$ where $G_1,\dots ,G_s$ are trees and $G_{s+1},\dots ,G_t$ are not; obviously, $0\le s<t$. Let $v\left(G_i\right)=p_i,e\left(G_i\right)=q_i,$ for $i=1,\dots ,t.$

When $p_i\le q_i$ for $s+1\le i\le t$, without loss of generality, we may assume that $p_1\le p_2\le \cdots p_s\le p_{s+1}\le \cdots \le p_t$.

If ${\sum }_{i=s+1}^t{p}_i\ge n,$ then denote the minimal integer by u such that $u\in \{s+1,\cdots ,t\}$ and ${\sum }_{i=s+1}^u{p}_i\ge n.$ Let T be a subtree of $G_u$, and $v\left(T\right)=n-{\sum }_{i=s+1}^{u-1}{p}_i$. Then, $G_{s+1}\cup \cdots \cup G_{u-1}\cup T$ induces a subgraph with vertices ${\sum }_{i=s+1}^{u-1}{p}_i+n-{\sum }_{i=s+1}^{u-1}{p}_i=n$, and edges

$$e(G_{s+1}\cup \cdots G_{u-1})+e\left(T\right)\ge \sum _{i=s+1}^{u-1}{p}_i+n-\sum _{i=s+1}^{u-1}{p}_i-1=n-1>n-m.$$

This contradicts that G is an $(n,n-m)$ graph. So, we have ${\sum }_{i=s+1}^t{p}_i\le n-1.$ Hence, we have $s\ge 1.$

Let $l={\sum }_{i=s+1}^t(q_i-p_i)$. For $s+1\le i\le t$, we have $q_i\ge p_i,$ so $l\ge 0.$ Because ${\sum }_{i=1}^s{p}_i=p-{\sum }_{i=s+1}^t{p}_i\ge n-{\sum }_{i=s+1}^t{p}_i\ge 1,$ we can choose one subgraph denoted by $G_0$ among the components $G_1\cup \cdots G_s$ such that $v\left(G_0\right)=n-{\sum }_{i=s+1}^t{p}_i$ and $e\left(G_0\right)=n-{\sum }_{i=s+1}^t{p}_i-s$. Now, $G_0\cup G_{s+1}\cup \cdots G_t$ is a subgraph of G with vertices n and edges at least $n-{\sum }_{i=s+1}^t{p}_i-s+{\sum }_{i=s+1}^t{q}_i=n-s+l.$ Because G is an $(n,n-m)$ graph, we have $n-s+l\le n-m$ that is $l\le s-m.$

If ${\sum }_{i=s+1}^t{p}_i+{\sum }_{i=1}^{l+m-1}{p}_i\ge n,$ then, similar to above, we can find a subgraph also denoted by $G_0$ among the components ${\cup }_{i=1}^{l+m-1}{G}_i$ with vertices $\left|V\left(G_0\right)\right|=n-{\sum }_{i=s+1}^t{p}_i$ and edges $e\left(G_0\right)\ge n-{\sum }_{i=s+1}^t{p}_i-(l+m-1).$ Now, we have that $G_0\cup G_{s+1}\cup \cdots \cup G_t$ is a subgraph of G with vertices n and edges at least

$$n-\sum _{i=s+1}^t{p}_i-(l+m-1)+\sum _{i=s+1}^t{q}_i=n-m+1>n-m,$$

which contradicts the property of G. Hence,

$$\sum _{i=s+1}^t{p}_i+\sum _{i=1}^{l+m-1}{p}_i\le n-1.$$

So

$$\sum _{i=l+m}^s{p}_i=p-\sum _{i=1}^t{p}_i-\sum _{i=1}^{l+m-1}{p}_i\ge p-(n-1).$$

On the other hand,

$$\sum _{i=l+m}^s{p}_i\le \sum _{i=l+m}^{s}k=k(s-l-m+1).$$

Combining the above two inequalities, we have that

$$p-(n-1)\le k(s-l-m+1).$$

Therefore, $l\le s-m+1-{\displaystyle \frac{p-n+1}{k}}.$ Now, we can estimate the edges of G like this,

$$\begin{array}{cc}\hfill e\left(G\right)& =\sum _{i=s+1}^t{q}_i+\sum _{i=1}^s(p_i-1)\hfill \\ & =\sum _{i=s+1}^t{p}_i+l+\sum _{i=1}^s{p}_i-s=p+l-s\hfill \\ & \le p+s-m+1-{\displaystyle \frac{p-n+1}{k}}-s\hfill \\ & =p-m+1-{\displaystyle \frac{p-n+1}{k}}={\displaystyle \frac{(k-1)p+r}{k}}.\hfill \end{array}$$

Therefore, we obtain that $e\left(G\right)\le \left[{\displaystyle \frac{(k-1)p+r}{k}}\right].$ □

Now, we prove Theorem 4.

Proof. 

Since $k=\left[{\displaystyle \frac{n-1}{m-1}}\right],$ by simple derivation, we have that

$$\begin{array}{cc}\hfill \left[{\displaystyle \frac{(k-1)p+r}{k}}\right]& =p+\left[{\displaystyle \frac{r-p}{k}}\right]=p+\left[{\displaystyle \frac{n-1-k(m-1)-p}{k}}\right]\hfill \\ & =p-m+1+\left[{\displaystyle \frac{n-1-p}{\left[\frac{n-1}{m-1}\right]}}\right].\hfill \end{array}$$

Now, we prove the theorem by induction on m. When $m=2$, $n-1=(n-1)·(2-1)+0$ and so $k=n-1,r=0.$ Then, by Lemma 2, we know that $e(n,n-2;p)\ge \left[{\displaystyle \frac{(n-2)p}{n-1}}\right].$ Furthermore, G is an $(n,n-2)$ graph, so G cannot contain one subtree with vertices $n=k+1$; this, together with Lemma 3, implies that

$$e(n,n-2;p)\le \left[{\displaystyle \frac{(n-2)p}{n-1}}\right].$$

So, the theorem is true for $m=2$.

Now, we assume that $m\ge 3$ and the theorem is true for $m-1$.

Suppose that G is an $(n,n-m)$ graph and $v\left(G\right)=p$. Let $G_1,\dots ,G_t$ be all components of G and $v\left(G_i\right)=p_i$ for $i\in \{1,\dots ,t\}.$ By Lemma 2, it suffices to prove that $e\left(G\right)\le \left[{\displaystyle \frac{(k-1)p+r}{k}}\right].$ If G does not contain a subtree with at least $k+1$ vertices, then, by Lemma 3, we have completed it. So, from now, we assume G contains some component denoted by $G_i$ such that $v\left(G_i\right)=p_i\ge k+1$ and $e\left(G_i\right)\ge p_i-1$; then, we claim that $p_i\le n-m+1$; otherwise, $p_i\ge n-m+2$ and $e\left(G_i\right)\ge n-m+1$; then we can find a subgraph in $G_i$ denoted by H satisfying that $v\left(H\right)=n-m+2$ and $e\left(H\right)=n-m+1$. Again, this contradicts the property of G.

Now, we have that $G-G_i$ is an $(n-p_i,n-p_i-m+1)$ graph. Since $n-p_i-1\ge m-2,$ we may assume that $n-p_i-1=k^{\prime }(m-2)+r^{\prime },r^{\prime }\in \mathbb{N},r^{\prime }\in \{0,1,\dots ,m-3\}.$ Note that $2\le m-1\le n-p_i\le p-p_i$; by the induction, we have

$$e(n-p_i,n-p_i-m+1;p-p_i)=\left[{\displaystyle \frac{(k^{\prime }-1)(p-p_i)+r^{\prime }}{k^{\prime }}}\right].$$

As $p_i\ge k+1=\left[{\displaystyle \frac{n-1}{m-1}}\right]+1>{\displaystyle \frac{n-1}{m-1}}$, we have that ${\displaystyle \frac{n-p_i-1}{m-2}}-{\displaystyle \frac{n-1}{m-1}}={\displaystyle \frac{n-1-(m-1)p_i}{(m-2)(m-1)}}<0$, and so

$$k^{\prime }=\left[{\displaystyle \frac{n-p_i-1}{m-2}}\right]\le {\displaystyle \frac{n-p_i-1}{m-2}}<{\displaystyle \frac{n-1}{m-1}}.$$

Therefore, $k^{\prime }\le \left[{\displaystyle \frac{n-1}{m-1}}\right]=k.$ Since

$$\begin{array}{cc}\hfill & p_i-1+\left[{\displaystyle \frac{(k^{\prime }-1)(p-p_i)+r^{\prime }}{k^{\prime }}}\right]-{\displaystyle \frac{(k-1)p+r}{k}}\hfill \\ & \le p_i-1+{\displaystyle \frac{(k^{\prime }-1)(p-p_i)+r^{\prime }}{k^{\prime }}}-{\displaystyle \frac{(k-1)p+r}{k}}\hfill \\ & ={\displaystyle \frac{(k^{\prime }-k)p+k{p}_i+k{r}^{\prime }-k^{\prime }r-k{k}^{\prime }}{k{k}^{\prime }}}\hfill \\ & ={\displaystyle \frac{(k^{\prime }+k)p+k{p}_i+k(n-p_i-1-k^{\prime }(m-2))-k^{\prime }(n-1-k(m-1))-k{k}^{\prime }}{k{k}^{\prime }}}\hfill \\ & ={\displaystyle \frac{(k-k^{\prime })(p-n+1)}{k{k}^{\prime }}}\le 0,\hfill \end{array}$$

we have that

$$\begin{array}{cc}\hfill e\left(G\right)& =e\left(G_i\right)+e(G-G_i)\hfill \\ & \le p_i-1+e(n-p_i,n-p_i-m+1;p-p_i)\hfill \\ & =p_i-1+\left[{\displaystyle \frac{(k^{\prime }-1)(p-p_i)+r^{\prime }}{k^{\prime }}}\right]\hfill \\ & \le {\displaystyle \frac{(k-1)p+r}{k}}.\hfill \end{array}$$

Therefore, we can obtain that $e\left(G\right)\le \left[{\displaystyle \frac{(k-1)p+r}{k}}\right]$. □

Next, we give several corollaries of Theorem 4, which mainly give a simpler expression of $e(n,m;p)$ for small m.

Corollary 1.

Let $k,n,p\in \mathbb{N}$ and $k+1\le n\le p.$ If a graph G satisfies $v\left(G\right)=p,e\left(G\right)\ge \left[{\displaystyle \frac{(k-1)p}{n-1}}\right],$ then G must contain a subgraph induced by n vertices with at least k edges.

Proof. 

Let $m=n+1-k,$ then $n\ge m\ge 2.$ By Lemma 2, we can deduce that

$$\begin{array}{cc}\hfill e(n,k-1;p)& =e(n,n-m;p)\hfill \\ & =p-(m-1)+\left[-{\displaystyle \frac{p-(n-1)}{\left[\frac{n-1}{m-1}\right]}}\right]\hfill \\ & \le p-(m-1)+[-{\displaystyle \frac{p-(n-1)}{\frac{n-1}{m-1}}}]\hfill \\ & =\left[{\displaystyle \frac{(n-m)p}{n-1}}\right],\hfill \end{array}$$

this, together with the given condition, implies the result. □

Corollary 2.

Let $n,p\in \mathbb{N}$ with $p\ge n$ and $v\left(G\right)=p$. If $e\left(G\right)=q\le p,$ then G must contain a subgraph induced by n vertices with edges large than $\lceil {\displaystyle \frac{(n-1)q}{p}}\rceil .$

Proof. 

As $\lceil {\displaystyle \frac{(n-1)q}{p}}\rceil <1+{\displaystyle \frac{(n-1)q}{p}}$, we deduce that $(\lceil {\displaystyle \frac{(n-1)q}{p}}\rceil -1)p<(n-1)q$; thereby, $q>{\displaystyle \frac{(\lceil \frac{(n-1)q}{p}\rceil -1)p}{n-1}}$. In Corollary 1, let $k=\left[{\displaystyle \frac{(n-1)q}{p}}\right]$, then we can obtain the desired result. □

Corollary 3.

Let $p,m,n,r\in \mathbb{N}$ with $r(m+1)\le n\le p.$ Then $e_r(n,m;p)=m.$

Proof. 

Suppose that G is an $(n,m)$ graph and $e\left(G\right)=e_r(n,m;p)$. Assume to the contrary that $e\left(G\right)\ge m+1$. Since $m+1$ edges were at most induced by $r(m+1)(\le n)$ vertices, then we can find a subgraph induced by n vertices with at least $m+1$ edges, which is a contradiction.

For the case $r=2$, the result also follows from Theorem 4. □

At the end of this section, we give a formula that is easier to use for the case $m=n-3$, and the proof is simple and so omitted.

Corollary 4.

Let $p,n\in \mathbb{N}$ with $p\ge n\ge 3.$ Then

$$e(n,n-3;p)=\left\{\begin{array}{cc}\left[{\displaystyle \frac{(n-3)p}{n-1}}\right],\hfill & 2\nmid n\hfill \\ {\displaystyle \frac{(n-4)p+2}{n-2}}],\hfill & 2\mid n.\hfill \end{array}\right.$$

3. Proof of Theorem 5

Proof. 

Suppose that H is an $(n,m)$ graph with p vertices. To prove the lower bound, firstly, construct an $m+1$-graph denoted by F as follows:

$V\left(F\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{\left[p\right]}{r}}\right)$, and $m+1$ different vertices $A_1,A_2,\dots ,A_{m+1}\in \left({\displaystyle \genfrac{}{}{0pt}{}{\left[p\right]}{r}}\right)$ form an edge of F if and only if $\left|{\cup }_{i=1}^{m+1}{A}_i\right|\le n.$

Because H is an $(n,m)$ graph, n vertices induce at most m edges, then we claim that

$$\mathrm{any}m+1\mathrm{edges}\mathrm{contain}\mathrm{at}\mathrm{least}n+1\mathrm{vertices},$$

otherwise, there exist $m+1$ edges denoted by $A_1,A_2,\dots ,A_{m+1}$ such that $\left|{\cup }_{i=1}^{m+1}{A}_i\right|\le n,$ which contradicts the definition of H. The claim together with the construction of F shows that $A_1,\dots ,A_{m+1}$ do not form an edge of F. Therefore, $\alpha \left(F\right)=e\left(H\right)$.

On the other hand, for an edge $\{A_1,\dots ,A_{m+1}\}$ of F, it follows that ${\cup }_{i=1}^{m+1}{A}_i$ contains at most n vertices of H. Then, for a vertex $A_i$ of F, since $\left|A_i\right|=r$, there are at most $\left({\displaystyle \genfrac{}{}{0pt}{}{p-r}{n-r}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right)}^m$ edges of F containing $A_i$. Therefore, we have

$$d\left(F\right)\le \Delta \left(F\right)\le \left({\displaystyle \genfrac{}{}{0pt}{}{p-r}{n-r}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right)}^m.$$

By Lemma 1, we have

$$\begin{array}{cc}\hfill e\left(H\right)& =\alpha \left(F\right)\ge C_1{\displaystyle \frac{\left|V\right(F\left)\right|}{d^{\frac{1}{m}}\left(F\right)}}\hfill \\ & =C_1\left({\displaystyle \genfrac{}{}{0pt}{}{p}{r}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{p-r}{n-r}}\right)}^{-1/m}·{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{r}}\right)}^{-1}\hfill \\ & =\Omega \left(p^{{\displaystyle \frac{(m+1)r-n}{m}}}\right),\hfill \end{array}$$

where $C_1$ is a constant which depends only on the values of $n,m,r$.

Now, we show the upper bound of $e_r(n,m;p)$. Suppose that H is an $(n,m)$ graph with p vertices.

Setting $t=\lceil {\displaystyle \frac{(m+1)r-n}{m}}\rceil$, for any subset $T\in \left({\displaystyle \genfrac{}{}{0pt}{}{\left[p\right]}{t}}\right)$, there at most exist m edges of H which contain T simultaneously; otherwise, suppose $T\in \left({\displaystyle \genfrac{}{}{0pt}{}{\left[p\right]}{t}}\right)$ and $A_1,A_2,\dots ,A_{m+1}\in E\left(H\right)$ such that $T\subset {\cap }_{i=1}^{m+1}{A}_i$, then

$$\left|{\cup }_{i=1}^{m+1}{A}_i\right|\le \left|T\right|+\sum _{i=1}^{m+1}|A_i\setminus T|=t+(m+1)(r-t)\le n,$$

which contradicts the above claim. Now, we consider the set $\{(T,A):T\in \left({\displaystyle \genfrac{}{}{0pt}{}{\left[p\right]}{t}}\right),A\in E\left(H\right),T\subset A\}$. By counting its size in different ways, we obtain that

$$e\left(H\right)·\left({\displaystyle \genfrac{}{}{0pt}{}{r}{t}}\right)\le \left({\displaystyle \genfrac{}{}{0pt}{}{p}{t}}\right)·m.$$

Then

$$e_r(n,m;p)\le C_2{p}^{\lceil {\displaystyle \frac{(m+1)r-n}{m}}\rceil },$$

where $C_2$ is a constant which depends only on the values of $n,m,r$. □

Remark 1.

From the bounds of $e_r(n,m;p)$, we can evaluate the Turán density of an $(n,m)$ graph, so maybe it is interesting to search some concrete graphs which are $(n,m)$ graphs.