A stability result for $C_{2k+1}$ -free graphs

Sijie Ren¹¹1Department of Mathematics, Taiyuan University of Technology, Taiyuan 030024, P. R. China. E-mail:[email protected]. Jian Wang²²2Department of Mathematics, Taiyuan University of Technology, Taiyuan 030024, P. R. China. E-mail:[email protected]. Research supported by Natural Science Foundation of Shanxi Province No. RD2200004810. Shipeng Wang³³3Department of Mathematics, Jiangsu University, Zhenjiang, Jiangsu 212013, P. R. China. E-mail:[email protected]. Research supported by NSFC No.12001242. Weihua Yang⁴⁴4Department of Mathematics, Taiyuan University of Technology, Taiyuan 030024, P. R. China. E-mail:[email protected]. Research supported by NSFC No.11671296.

Abstract

A graph $G$ is called $C_{2k+1}$ -free if it does not contain any cycle of length $2k+1$ . In 1981, Ha̋ggkvist, Faudree and Schelp showed that every $n$ -vertex triangle-free graph with more than $\frac{(n-1)^{2}}{4}+1$ edges is bipartite. In this paper, we extend their result and show that for $1\leq t\leq 2k-2$ and $n\geq 318t^{2}k$ , every $n$ -vertex $C_{2k+1}$ -free graph with more than $\frac{(n-t-1)^{2}}{4}+\binom{t+2}{2}$ edges can be made bipartite by either deleting at most $t-1$ vertices or deleting at most $\binom{\lfloor\frac{t+2}{2}\rfloor}{2}+\binom{\lceil\frac{t+2}{2}\rceil}{2}-1$ edges. The construction shows that this is best possible.

Keywords: stability; odd cycles; non-bipartite graphs.

1 Introduction

We consider only simple graphs. For a graph $G$ , we use $V(G)$ and $E(G)$ to denote its vertex set and edge set, respectively. Let $e(G)$ denote the number of edges of $G$ . For $S\subseteq{V(G)}$ , let $G[S]$ denote the subgraph of $G$ induced by $S$ . Let $G-S$ denote the subgraph induced by $V(G)\backslash{S}$ . For simplicity, we write $E(S)$ and $e(S)$ for $E(G[S])$ and $e(G[S])$ , respectively. Let $N_{G}(v)$ denote the set of neighbors of $v$ in $G$ . For $S\subseteq V(G)$ , let $N_{G}(v,S)$ denote the set of neighbors of $v$ in $S$ . Let $\deg_{G}(v)=|N_{G}(v)|$ and $\deg(v,S)=|N_{G}(v,S)|$ . The minimum degree of $G$ is defined as the minimum of $\deg_{G}(v)$ over all $v\in V(G)$ . For a subgraph $H$ of $G$ , we also write $N_{G}(v,H)$ , $\deg_{G}(v,H)$ for $N_{G}(v,V(H))$ , $\deg_{G}(v,V(H))$ , respectively. For $E\subseteq E(G)$ , let $G-E$ denote the graph obtained from $G$ by deleting edges in $E$ . For disjoint sets $S,T\subseteq V(G)$ , let $G[S,T]$ denote the subgraph of $G$ with the vertex set $S\cup T$ and the edge set $\left\{xy\in E(G)\colon x\in S,\ y\in T\right\}$ . Let $e_{G}(S,T)=e(G[S,T])$ . If the context is clear, we often omit the subscript $G$ .

Given a graph $F$ , we say that $G$ is $F$ -free if it does not contain $F$ as a subgraph. The Turán number ${\rm ex}(n,F)$ is defined as the maximum number of edges in an $F$ -free graph on $n$ vertices. Let $T_{r}(n)$ denote the Turán graph, the complete $r$ -partite graph on $n$ vertices with $r$ partition classes, each of size $\lfloor\frac{n}{r}\rfloor$ or $\lceil\frac{n}{r}\rceil$ . Let $t_{r}(n)=e(T_{r}(n))$ . The classic Turán theorem [15] tells us that $T_{r}(n)$ is the unique graph attaining the maximum number of edges among all $K_{r+1}$ -free graphs, where the $r=2$ case is Mantel’s theorem [13].

An edge $e\in E(H)$ is called a color-critical edge of $H$ if $\chi(H-e)<\chi(H)$ . Let $H$ be a graph with $\chi(H)=r+1$ that contains a color-critical edge. Simonovits [14] proved that there exists $n_{0}(H)$ such that if $n>n_{0}(H)$ , then ${\rm ex}(n,H)=t_{r}(n)$ and $T_{r}(n)$ is the unique extremal graph. Since $\chi(C_{2k+1})=3$ and $C_{2k+1}$ contains a color-critical edge, ${\rm ex}(n,C_{2k+1})$ is implied by Simonovits’ result for sufficiently large $n$ . Dzido [6] observed that ${\rm ex}(n,C_{2k+1})$ can be read out from the works of Bondy [3, 4], Woodall [16], and Bollobás [1] (pp. 147–156). Later, Füredi and Gunderson [9] determined the $ex(n,C_{2k+1})$ for all $n$ and gave a complete description of the extremal graphs.

Theorem 1.1 ([6], [9]).

For $k\geq 2$ and $n\geq 4k-2$ ,

{\rm ex}(n,C_{2k+1})=\left\lfloor\frac{n^{2}}{4}\right\rfloor.

(1.1)

In 1981, Ha̋ggkvist, Faudree and Schelp [10] proved a stability result for Mantel’s Theorem.

Theorem 1.2 ([10]).

Let $G$ be a non-bipartite triangle-free graph on $n$ vertices. Then

e(G)\leq\frac{(n-1)^{2}}{4}+1.

Let $H_{0}$ be a graph obtained from $T_{2}(n-1)$ by replacing an edge $xy$ with a path $xzy$ , where $z$ is a new vertex. It is easy to see that $H_{0}$ is a triangle-free graph with $\frac{(n-1)^{2}}{4}+1$ edges, which shows that Theorem 1.2 is sharp.

To state our results, we need the following two parameters. Define

d_{2}(G)=\min\left\{|T|\colon T\subseteq V(G),\ G-T\mbox{ is bipartite}\right\}

and

\gamma_{2}(G)=\min\left\{|E|\colon E\subseteq E(G),\ G-E\mbox{ is bipartite}\right\}.

Let $H(n,t)$ be a graph obtained from graphs $T_{2}(n-t-1)$ and $K_{t+2}$ by sharing a vertex. In this paper, we extend Theorem 1.2 to the following two results.

Theorem 1.3.

Let $n,k,t$ be integers with $1\leq t\leq 2k-2$ and $n\geq 318t^{2}k$ . If $G$ is a $C_{2k+1}$ -free ( $k\geq 2$ ) graph on $n$ vertices with $d_{2}(G)\geq t$ , then

e(G)\leq\left\lfloor\frac{(n-t-1)^{2}}{4}\right\rfloor+\binom{t+2}{2},

with equality if and only if $G=H(n,t)$ up to isomorphism.

Theorem 1.4.

Let $n,k,t$ be integers with $1\leq t\leq 2k-2$ and $n\geq 252t^{2}k$ . If $G$ is a $C_{2k+1}$ -free ( $k\geq 2$ ) graph on $n$ vertices with $\gamma_{2}(G)\geq\binom{\lfloor\frac{t+2}{2}\rfloor}{2}+\binom{\lceil\frac{t+2}{2}\rceil}{2}$ , then

\displaystyle e(G)\leq\left\lfloor\frac{(n-t-1)^{2}}{4}\right\rfloor+\binom{t+2}{2},

(1.2)

with equality if and only if $G=H(n,t)$ up to isomorphism.

The circumference $c(G)$ is defined as the length of a longest cycle of $G$ . The girth $g(G)$ is defined as the length of a shortest cycle of $G$ . We say $G$ is weakly pancyclic if it contains cycles of all lengths from $g(G)$ to $c(G)$ . We need the following result due to Brandt, Faudree and Goddard [5].

Theorem 1.5 ([5]).

Every non-bipartite graph $G$ of order $n$ with minimum degree $\delta(G)\geq\frac{n+2}{3}$ is weakly pancyclic with girth 3 or 4.

Let $P_{\ell+1}$ be a path of length $\ell$ and let $\mathcal{C}_{\geq\ell}$ be the family of cycles with lengths at least $\ell$ . We need the Erdős-Gallai Theorem for paths and cycles.

Theorem 1.6 ([8]).

For $n\geq\ell\geq 1$ ,

{\rm ex}(n,P_{\ell+1})\leq\frac{\ell-1}{2}n,\ {\rm ex}(n,\mathcal{C}_{\geq\ell+1})\leq\frac{\ell}{2}(n-1).

2 Some useful facts and inequalities

In this section, we prove some facts and inequalities that are needed.

Fact 2.1.

Let $C$ be a shortest odd cycle of $G$ . If $C$ has length at least 5, then every vertex in $V(G)\setminus V(C)$ has at most two neighbors on $C$ .

Proof.

Let $C=x_{0}x_{1}\ldots x_{2m}x_{0}$ be a shortest odd cycle of $G$ , $m\geq 2$ . Fix an arbitrary vertex $v\in V(G)\setminus V(C)$ . If $v$ has at least three neighbors on $C$ , there exist neighbors $x_{i}$ , $x_{j}$ of $v$ ( $0\leq i<j\leq 2m$ ) such that the distance between $x_{i}$ and $x_{j}$ on $C$ does not equal 2. Then one of $x_{i}x_{i+1}\ldots x_{j}vx_{i}$ and $x_{0}x_{1}\ldots x_{i}vx_{j}x_{j+1}\ldots x_{2m}x_{0}$ is an odd cycle shorter than $C$ , a contradiction. ∎

Fact 2.2.

Let $G$ be a graph with $e(G)>\frac{(n-t-1)^{2}}{4}$ , $t\geq 1$ . Let $C$ be a shortest odd cycle of $G$ . If $n\geq 6(t+1)$ , $C$ has length at most $\frac{2n}{3}$ .

Proof.

Let $S=V(C)$ and $s=|S|$ . Suppose for contradiction that $s\geq\frac{2n}{3}\geq 5$ . By Fact 2.1, each vertex in $G-S$ has at most two neighbors on $C$ . Then

	$\displaystyle e(G)$	$\displaystyle=e(S)+e(G-S)+e(S,G-S)$
		$\displaystyle\leq s+\binom{n-s}{2}+2(n-s)$
		$\displaystyle=2n-s+\frac{(n-s)(n-s-1)}{2}=:f(s).$

Note that $f(s)$ is a decreasing function for $\frac{2n}{3}\leq s\leq n$ . Since $n\geq 6(t+1)\geq 12$ , it follows that

e(G)\leq f\left(\frac{2n}{3}\right)=\frac{4n}{3}+\frac{n}{6}\left(\frac{n}{3}-1\right)=\frac{n}{6}\left(\frac{n}{3}+7\right)\leq\frac{n^{2}}{6}.

Since $n\geq 6(t+1)$ implies $\frac{n}{12}-\frac{t+1}{2}\geq 0$ ,

\displaystyle e(G)\leq\frac{n^{2}}{6}\leq\left(\frac{n}{6}+\frac{n}{12}-\frac{t+1}{2}\right)n=\frac{(n-2(t+1))n}{4}<\frac{(n-t-1)^{2}}{4},

a contradiction. ∎

Fact 2.3.

Let $G$ be a $C_{2k+1}$ -free graph and $C$ be an odd cycle of length $2\ell+1$ in $G$ . If $\ell\geq k+1$ then $\deg(x,C)\leq\ell$ for every $x\in V(G)\setminus V(C)$ .

Proof.

Assume that $C=v_{0}v_{1}v_{2}\ldots v_{2\ell}v_{0}$ . Let $x\in V(G)\setminus V(C)$ . Define $I_{i}$ as an indicator function by setting $I_{i}=1$ for $xv_{i}\in E(G)$ and setting $I_{i}=0$ otherwise. Since $G$ is $C_{2k+1}$ -free, one cannot have both $xv_{i}$ and $xv_{i+2k-1}\in E(G)$ (subscript modulo $2\ell$ ) for each $i\in\{0,1,\ldots,2\ell\}$ . Hence,

2\deg(x,C)=2\sum_{0\leq i\leq 2\ell}I_{i}=\sum_{0\leq i\leq 2\ell}(I_{i}+I_{i+2k-1})\leq 2\ell+1.

Therefore, $\deg(x,C)\leq\lfloor\frac{2\ell+1}{2}\rfloor=\ell$ . ∎

Let $a,b\geq 0$ be integers. We need the following equalities and inequalities.

	$\displaystyle\left\lceil\frac{a}{2}\right\rceil\left\lfloor\frac{b}{2}\right\rfloor+\left\lfloor\frac{a}{2}\right\rfloor\left\lceil\frac{b}{2}\right\rceil=\left\lfloor\frac{ab}{2}\right\rfloor,$		(2.1)
	$\displaystyle\left\lceil\frac{a}{2}\right\rceil\left\lceil\frac{b}{2}\right\rceil+\left\lfloor\frac{a}{2}\right\rfloor\left\lfloor\frac{b}{2}\right\rfloor=\left\lceil\frac{ab}{2}\right\rceil,$		(2.2)
	$\displaystyle\left\lfloor\frac{a}{2}\right\rfloor b\leq\left\lfloor\frac{ab}{2}\right\rfloor,$		(2.3)
	$\displaystyle\left\lceil\frac{a}{2}\right\rceil b\leq\left\lceil\frac{ab}{2}\right\rceil+\left\lfloor\frac{b}{2}\right\rfloor.$		(2.4)

Note that these equalities and inequalities can be easily verified by checking four cases: $a=2i,b=2j$ ; $a=2i,b=2j+1$ ; $a=2i+1,b=2j$ and $a=2i+1,b=2j+1$ .

Let $a_{1},a_{2},\ldots,a_{m}\geq 0$ be integers. By induction on $m$ , one can also obtain the following inequality.

\displaystyle\left\lceil\frac{a_{1}+a_{2}+\cdots+a_{m}}{2}\right\rceil\geq\left\lceil\frac{a_{1}}{2}\right\rceil+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{m}}{2}\right\rfloor.

(2.5)

Define $f(x):=\binom{x}{2}-\left\lfloor\frac{x^{2}}{4}\right\rfloor+\left\lfloor\frac{x}{2}\right\rfloor$ .

Lemma 2.4.

For integers $a,b,t\geq 0$ ,

	$\displaystyle f(t+1)=\binom{\lfloor\frac{t+2}{2}\rfloor}{2}+\binom{\lceil\frac{t+2}{2}\rceil}{2},$		(2.6)
	$\displaystyle f(a+b)=f(a)+f(b)+\left\lceil\frac{ab}{2}\right\rceil.$		(2.7)

Proof.

Note that $\binom{t+1}{2}-\lfloor\frac{(t+1)^{2}}{4}\rfloor$ can be viewed as the number of edges in $K_{\lfloor(t+1)/2\rfloor}\cup K_{\lceil(t+1)/2\rceil}$ . By adding a vertex to $K_{\lfloor(t+1)/2\rfloor}$ , we get a graph with $\binom{t+1}{2}-\lfloor\frac{(t+1)^{2}}{4}\rfloor+\lfloor(t+1)/2\rfloor=f(t+1)$ edges, which is exactly a copy of $K_{\lfloor(t+2)/2\rfloor}\cup K_{\lceil(t+2)/2\rceil}$ . Thus $\eqref{equalities2.6}$ follows.

Next we prove (2.7). Let $A_{1},A_{2}$ be two disjoint vertex sets with $|A_{1}|=\left\lceil\frac{a}{2}\right\rceil$ , $|A_{2}|=\left\lfloor\frac{a}{2}\right\rfloor$ and $B_{1},B_{2}$ be two disjoint sets with $|B_{1}|=\left\lfloor\frac{b}{2}\right\rfloor$ , $|B_{2}|=\left\lceil\frac{b}{2}\right\rceil$ . Clearly $(A_{1}\cup B_{1},A_{2}\cup B_{2})$ is an almost equal partition of $a+b$ vertices. For a set $S$ , let $K(S)$ denote the complete graph with vertex set $S$ . Note that $\binom{a+b}{2}-\left\lfloor\frac{(a+b)^{2}}{4}\right\rfloor$ can be viewed as the number of edges in $K(A_{1}\cup B_{1})\cup K(A_{2}\cup B_{2})$ . Similarly, $\binom{a}{2}-\left\lfloor\frac{a^{2}}{4}\right\rfloor$ can be viewed as the number of edges in $K(A_{1})\cup K(A_{2})$ , $\binom{b}{2}-\left\lfloor\frac{b^{2}}{4}\right\rfloor$ can be viewed as the number of edges in $K(B_{1})\cup K(B_{2})$ . Thus,

	$\displaystyle f(a+b)-f(a)-f(b)$	$\displaystyle=\|A_{1}\|\|B_{1}\|+\|A_{2}\|\|B_{2}\|+\left\lfloor\frac{a+b}{2}\right\rfloor-\left\lfloor\frac{a}{2}\right\rfloor-\left\lfloor\frac{b}{2}\right\rfloor$
		$\displaystyle=\left\lceil\frac{a}{2}\right\rceil\left\lfloor\frac{b}{2}\right\rfloor+\left\lfloor\frac{a}{2}\right\rfloor\left\lceil\frac{b}{2}\right\rceil+\left\lfloor\frac{a+b}{2}\right\rfloor-\left\lfloor\frac{a}{2}\right\rfloor-\left\lfloor\frac{b}{2}\right\rfloor.$

Define $\delta(a,b):=\lfloor\frac{a+b}{2}\rfloor-\lfloor\frac{a}{2}\rfloor-\lfloor\frac{b}{2}\rfloor$ . Note that $\delta(a,b)$ equals 1 if $a,b$ are both odd and equals 0 otherwise. By (2.1),

f(a+b)-f(a)-f(b)=\left\lfloor\frac{ab}{2}\right\rfloor+\delta(a,b)=\left\lceil\frac{ab}{2}\right\rceil.

∎

3 Two structure lemmas

Lemma 3.1.

Let $G$ be a $C_{2k+1}$ -free ( $k\geq 2$ ) graph on $n$ vertices with $e(G)\geq\lfloor\frac{(n-t-1)^{2}}{4}\rfloor+\binom{t+2}{2}$ , $t\geq 1$ . If $n\geq 78t^{2}k$ , then there exists $T\subseteq V(G)$ with $|T|\leq 3(t+1)$ such that $G-T$ is a bipartite graph on partite sets $X,Y$ and (i), (ii), (iii), (iv) hold.

(i)

$\delta(G-T)\geq\frac{n-3t-1}{3}$ ;
(ii)

$\frac{n}{2}-2\sqrt{tn}\leq|X|,|Y|\leq\frac{n}{2}+\sqrt{3tn}$ ;
(iii)

$|\left\{u\in V(G)\setminus T\colon\deg_{G-T}(u)\leq\frac{n}{2}-\sqrt{3tn}\right\}|\leq\sqrt{tn}$ ;
(iv)

For $v\in T$ , either $N(v,X\cup Y)\subset X$ or $N(v,X\cup Y)\subset Y$ holds.

Proof.

Set $G_{0}:=G$ . We obtain a bipartite subgraph from $G_{0}$ by a vertex-deletion procedure. Let us start with $i=0$ and obtain a sequence of graphs $G_{1},G_{2},\ldots,G_{\ell},\ldots$ as follows: In the $i$ th step, if there exists a vertex $v_{i+1}\in V(G_{i})$ with $\deg_{G_{i}}(v_{i+1})<\frac{(n-i)+2}{3}$ then let $G_{i+1}$ be the graph obtained from $G_{i}$ by deleting $v_{i+1}$ and go to the $(i+1)$ th step. Otherwise we stop and set $G^{*}=G_{i}$ , $T=\{v_{1},v_{2},\ldots,v_{i}\}$ . Clearly, the procedure will end in finite steps.

Let $n^{\prime}=|V(G^{*})|$ . We claim that $n^{\prime}\geq 4k$ . Indeed, otherwise

	$\displaystyle e(G)<\sum_{i=0}^{n-4k-1}\frac{(n-i)+2}{3}+\binom{4k}{2}$	$\displaystyle=\frac{(n+4k+5)(n-4k)}{6}+2k(4k-1)$
		$\displaystyle<\frac{(n+7k)(n-4k)}{6}+8k^{2}$
		$\displaystyle<\frac{(n-t-1)^{2}}{4}+\frac{(t+1)n}{2}-\frac{n^{2}}{12}+\frac{kn}{2}+4k^{2}.$

By $t+1<2k$ , we have $e(G)<\frac{(n-t-1)^{2}}{4}-\frac{n^{2}}{12}+\frac{3}{2}kn+4k^{2}$ . Since $n\geq 78t^{2}k>36k$ implies that $\frac{3}{2}kn<\frac{n^{2}}{24}$ , we have $\frac{n^{2}}{12}>\frac{3}{2}kn+4k^{2}$ . Thus $e(G)<\frac{(n-t-1)^{2}}{4}$ , a contradiction. Thus $n^{\prime}\geq 4k$ . Suppose to the contrary that $|T|\geq 3t+4$ . Since $G^{*}$ is $C_{2k+1}$ -free, by Theorem 1.1 we have $e(G_{3t+4})\leq\frac{(n-3t-4)^{2}}{4}$ . Thus,

	$\displaystyle e(G)$	$\displaystyle<\sum_{i=0}^{3t+3}\frac{(n-i)+2}{3}+\frac{(n-3t-4)^{2}}{4}$
		$\displaystyle\leq\frac{(3t+4)n}{3}+\frac{(n-3t-4)^{2}}{4}$
		$\displaystyle=\frac{(3t+4)n}{3}+\frac{(n-t-1)^{2}}{4}+\frac{(t+1)n-(3t+4)n}{2}+\frac{(3t+4)^{2}-(t+1)^{2}}{4}$
		$\displaystyle=\frac{(n-t-1)^{2}}{4}-\frac{n}{6}+\frac{(2t+3)(4t+5)}{4}$
		$\displaystyle<\frac{(n-t-1)^{2}}{4}-\frac{n}{6}+2(t+2)^{2}.$

As $n\geq 78t^{2}k\geq 12(t+2)^{2}$ , we get $e(G)<\frac{(n-t-1)^{2}}{4}$ , a contradiction. Thus $|T|\leq 3t+3$ and (i) follows from $\delta(G^{*})\geq\frac{n^{\prime}+2}{3}$ .

Suppose that $G^{*}$ is non-bipartite. As $\delta(G^{*})\geq\frac{n^{\prime}+2}{3}$ , by Theorem 1.5 we infer that $G^{*}$ is weakly pancyclic with girth 3 or 4. By (i) and $n\geq 78t^{2}k>6k+3t+1$ , $G^{*}$ contains a cycle of length $\delta(G^{*})+1\geq\frac{n-3t+2}{3}\geq 2k+1$ . Then the weakly pancyclicity implies that $G^{*}$ contains a cycle of length $2k+1$ , a contradiction. Thus $G^{*}$ is bipartite.

Let $X$ , $Y$ be two partite sets of $G^{*}$ . We are left to show (ii), (iii) and (iv). Note that

	$\displaystyle\|X\|\|Y\|\geq e(G^{*})$	$\displaystyle\geq e(G)-\sum_{i=0}^{3t+2}\frac{(n-i)+2}{3}$
		$\displaystyle\geq\frac{(n-t-1)^{2}}{4}-\frac{(3t+3)n}{3}$
		$\displaystyle\geq\frac{n^{2}}{4}-\frac{3}{2}(t+1)n$
		$\displaystyle\geq\frac{n^{2}}{4}-3tn.$

Note that $|X|+|Y|=n-|T|$ . Set $|X|=\frac{n-|T|}{2}+d$ and $|Y|=\frac{n-|T|}{2}-d$ . Then

\frac{(n-|T|)^{2}}{4}-d^{2}=|X||Y|\geq\frac{n^{2}}{4}-3tn\geq\frac{(n-|T|)^{2}}{4}-3tn.

It follows that $d\leq\sqrt{3tn}$ . Thus,

\frac{n-|T|}{2}-\sqrt{3tn}\leq|X|,|Y|\leq\frac{n-|T|}{2}+\sqrt{3tn}\leq\frac{n}{2}+\sqrt{3tn}.

By $n\geq 78t^{2}k>\frac{9}{(2-\sqrt{3})^{2}}t$ we have $(2-\sqrt{3})\sqrt{tn}\geq 3t$ . Then

|X|,|Y|\geq\frac{n-3t-3}{2}-\sqrt{3tn}\geq\frac{n}{2}-3t-\sqrt{3tn}\geq\frac{n}{2}-2\sqrt{tn}.

Thus (ii) holds.

To show (iii), suppose indirectly that there are at least $\sqrt{tn}$ vertices in $G^{*}$ with degree less than $\frac{n}{2}-\sqrt{3tn}$ . Let $U\subseteq V(G)$ with $|U|=\lfloor\sqrt{tn}\rfloor$ such that $\deg_{G^{*}}(u)\leq\frac{n}{2}-\sqrt{3tn}$ for all $u\in U$ . Then $\deg_{G}(u)\leq\frac{n}{2}-\sqrt{3tn}+|T|$ for all $u\in U$ . Note that $n\geq 78t^{2}k>4t$ implies $\sqrt{tn}\leq\frac{n}{2}$ and thereby $n-|U|\geq n-\sqrt{tn}\geq\frac{n}{2}\geq 4k$ . Since $G-U$ is $C_{2k+1}$ -free, using Theorem 1.1 we get $e(G-U)\leq\frac{(n-|U|)^{2}}{4}$ . Thus, by $|T|\leq 3(t+1)$ we obtain that

	$\displaystyle e(G)$	$\displaystyle\leq e(G-U)+\|U\|\left(\frac{n}{2}-\sqrt{3tn}+\|T\|\right)$
		$\displaystyle=\frac{n^{2}}{4}+\frac{\|U\|^{2}}{4}-\sqrt{3tn}\|U\|+\|T\|\|U\|$
		$\displaystyle\leq\frac{n^{2}}{4}+\frac{\|U\|^{2}}{4}-\sqrt{3}\|U\|^{2}+3(t+1)\sqrt{tn}$
		$\displaystyle\leq\frac{n^{2}}{4}-\left(\sqrt{3}-\frac{1}{4}\right)tn+3(t+1)\sqrt{tn}$
		$\displaystyle\leq\frac{(n-t-1)^{2}}{4}-\left(\sqrt{3}-\frac{5}{4}\right)tn+3(t+1)\sqrt{tn}.$

By $n\geq 78t^{2}k\geq\frac{36}{(\sqrt{3}-\frac{5}{4})^{2}}t$ we have $\left(\sqrt{3}-\frac{5}{4}\right)tn\geq 6t\sqrt{tn}\geq 3(t+1)\sqrt{tn}$ . Hence $e(G)<\frac{(n-t-1)^{2}}{4}$ , a contradiction. Therefore (iii) holds.

To show (iv), suppose for contradiction that for some $v\in T$ there exist $x\in N(v,X)$ and $y\in N(v,Y)$ . Let $X_{1}=N(y,X)\setminus\{x\}$ and $Y_{1}=N(x,Y)\setminus\{y\}$ . Note that

|X_{1}|,|Y_{1}|\geq\delta(G-T)-1\geq\frac{n-3t-1}{3}-1>\frac{n}{3}-3t.

Since $G$ is $C_{2k+1}$ -free and $k\geq 2$ , $G[X_{1}\cup Y_{1}]$ contains no paths of length $2k-3$ . By Theorem 1.6 we infer $e(X_{1},Y_{1})\leq\frac{(2k-3)-1}{2}(|X_{1}|+|Y_{1}|)\leq(k-2)n$ . Thus,

	$\displaystyle e(G)$	$\displaystyle\leq e(G-T)+\sum_{0\leq i<\|T\|}\frac{n-i+2}{3}$
		$\displaystyle\leq\|X\|\|Y\|-\|X_{1}\|\|Y_{1}\|+e(X_{1},Y_{1})+\frac{n+2}{3}\|T\|$
		$\displaystyle\leq\frac{(n-\|T\|)^{2}}{4}-\left(\frac{n}{3}-3t\right)^{2}+(k-2)n+\frac{n+2}{3}\|T\|.$

Let $f(x):=\frac{(n-x)^{2}}{4}+\frac{n+2}{3}x$ . It is easy to check that $f(x)$ is a decreasing function on $[1,\frac{n}{3}-2]$ . Since $n\geq 78t^{2}k>9t+15$ implies $|T|\leq 3(t+1)\leq\frac{n}{3}-2$ , $f(|T|)\leq f(1)=\frac{(n-1)^{2}}{4}+\frac{n+2}{3}$ . Hence,

	$\displaystyle e(G)$	$\displaystyle\leq\frac{(n-1)^{2}}{4}-\left(\frac{n}{3}-3t\right)^{2}+(k-2)n+\frac{n+2}{3}$
		$\displaystyle\leq\frac{(n-t-1)^{2}}{4}+\frac{tn}{2}-\frac{n^{2}}{9}+2tn+kn$
		$\displaystyle\leq\frac{(n-t-1)^{2}}{4}-\left(\frac{n^{2}}{9}-\left(k+\frac{5}{2}t\right)n\right).$

By $n\geq 78t^{2}k>9k+13t$ we see that $\frac{n^{2}}{9}\geq(k+\frac{5}{2}t)n$ . Then $e(G)\leq\frac{(n-t-1)^{2}}{4}$ , a contradiction. Thus (iv) holds and the lemma is proven. ∎

A vertex $u$ is called a high-degree vertex if $\deg(u)\geq 2\sqrt{tn}$ , otherwise it is called a low-degree vertex. Let $B(G)$ be the set of all low-degree vertices in $G$ , that is,

B(G)=\left\{u\in V(G)\colon\deg(u)<2\sqrt{tn}\right\}.

Lemma 3.2.

Let $G$ be a $C_{2k+1}$ -free ( $k\geq 2$ ) graph on $n$ vertices with $e(G)\geq\lfloor\frac{(n-t-1)^{2}}{4}\rfloor+\binom{t+2}{2}$ , $t\geq 1$ and let $B=B(G)$ . If $n\geq 252t^{2}k$ , then $G-B$ is a bipartite graph on bipartite sets $X,Y$ with (i)-(vi) hold.

(i)

$|B|\leq t+1$ .
(ii)

$\frac{n}{2}-2\sqrt{tn}\leq|X|,|Y|\leq\frac{n}{2}+2\sqrt{tn}$ .
(iii)

$|\left\{v\in V(G)\setminus B\colon\deg_{G-B}(v)\leq\frac{n}{2}-\sqrt{3tn}\right\}|\leq\frac{3\sqrt{tn}}{2}$ .
(iv)

For $v\in B$ , either $N(v,X\cup Y)\subset X$ or $N(v,X\cup Y)\subset Y$ holds.
(v)

For $uv\in E(B)$ , there does not exist $x_{1},x_{2}\in X$ such that $ux_{1},vx_{2}\in E(G)$ . Similarly, there does not exist $y_{1},y_{2}\in Y$ such that $uy_{1},vy_{2}\in E(G)$ .
(vi)

Let $uwv$ be a path in $B$ . If $\deg(w,X\cup Y)\leq 1$ , then there does not exist $x\in X,y\in Y$ such that $uy,vx\in E(G)$ or $ux,vy\in E(G)$ .

Proof.

By Lemma 3.1 there exists $T\subseteq V(G)$ with $|T|\leq 3(t+1)$ such that $G-T$ is a bipartite graph and (i)-(iv) of Lemma 3.1 hold. Let $X_{0}$ , $Y_{0}$ be two partite sets of $G-T$ . Since $n\geq 252t^{2}k>144t$ , by Lemma 3.1 (i) we have $\delta(G-T)\geq\frac{n-t-1}{3}>\frac{n}{6}\geq 2\sqrt{tn}$ . Thus $B\subset T$ .

Suppose indirectly that $G-B$ is not bipartite. Let $C=v_{0}v_{1}v_{2}\ldots v_{2\ell}v_{0}$ be a shortest odd cycle in $G-B$ and let $S=V(C)$ . Clearly, $v_{0},v_{1},\ldots,v_{2\ell}$ are all high-degree vertices. Since $C$ is a shortest odd cycle, $\deg(v_{i},C)=2$ for each $v_{i}\in S$ , that is, $C$ has no chord. Since $n\geq 252t^{2}k>\frac{(3t+7)^{2}}{t}$ implies $\sqrt{tn}\geq 3t+7\geq|T|+4$ ,

|N(v_{i},(X_{0}\cup Y_{0})\setminus S)|=\deg(v_{i})-|T|-\deg(v_{i},C)\geq 2\sqrt{tn}-|T|-2\geq\sqrt{tn}+2.

By Lemma 3.1 (iii), each $v_{i}$ in $S$ has at least two neighbors in $X_{0}\cup Y_{0}\setminus S$ with degree at least $\frac{n}{2}-\sqrt{3tn}$ . By Lemma 3.1 (iv), for each $v_{i}$ in $S$ either $N(v_{i},X_{0}\cup Y_{0}\setminus S)\subset X_{0}\setminus S$ or $N(v_{i},X_{0}\cup Y_{0}\setminus S)\subset Y_{0}\setminus S$ holds. As $C$ is an odd cycle, there exist $v_{i}$ , $v_{i+1}$ on $C$ such that $N(v_{i},X_{0}\cup Y_{0}\setminus S)$ and $N(v_{i+1},X_{0}\cup Y_{0}\setminus S)$ are contained in the same one of $X_{0}\setminus S$ and $Y_{0}\setminus S$ . Without loss of generality, let $x_{1},x_{2}$ be two distinct vertices in $X_{0}\setminus S$ such that $v_{i}x_{1},v_{i+1}x_{2}\in E(G)$ and $\deg_{G-T}(x_{1}),\deg_{G-T}(x_{2})\geq\frac{n}{2}-\sqrt{3tn}$ . Let $Y_{1}=N(x_{1},Y_{0})\cap N(x_{2},Y_{0})\setminus\{v_{i},v_{i+1}\}$ . By Lemma 3.1 (ii) we know $|Y_{0}|\leq\frac{n}{2}+\sqrt{3tn}$ . Then

	$\displaystyle\|Y_{1}\|$	$\displaystyle\geq\|N(x_{1},Y_{0})\|+\|N(x_{2},Y_{0})\|-\|N(x_{1},Y_{0})\cup N(x_{2},Y_{0})\|-2$
		$\displaystyle\geq\deg_{G-T}(x_{1})+\deg_{G-T}(x_{2})-\|Y_{0}\|-2$
		$\displaystyle\geq\frac{n}{2}-3\sqrt{3tn}-2.$

Let $X_{1}=X_{0}\setminus\{x_{1},x_{2},v_{i},v_{i+1}\}$ . Note that

	$\displaystyle\delta(G[X_{1}\cup Y_{1}])\geq\delta(G-T)-(\|Y_{0}\|-\|Y_{1}\|)$	$\displaystyle\geq\frac{n-3t-1}{3}-4\sqrt{3tn}-2$
		$\displaystyle\geq\frac{n}{3}-t-4\sqrt{3tn}-3.$

By $n\geq 252t^{2}k\geq 504t$ we infer that

\frac{n}{3}-t-4\sqrt{3tn}\geq\frac{n}{3}-\frac{n}{504}-4\sqrt{3tn}\geq\frac{167n}{504}-4\sqrt{3tn}.

As $n\geq 504t$ implies $4\sqrt{3tn}\leq\frac{4n}{\sqrt{168}}<\frac{163n}{504}$ , we have $\frac{167n}{504}-4\sqrt{3tn}\geq\frac{n}{126}\geq 2k$ . Hence $\delta(G[X_{1}\cup Y_{1}])\geq 2k-3$ . Then there is a path of length at least $\delta(G[X_{1}\cup Y_{1}])+1\geq 2k-2$ . For $k\geq 3$ we choose a path $P$ of length $2k-4$ with both ends in $Y_{1}$ . For $k=2$ we simply choose $P$ as a vertex in $Y_{1}$ . Then $P$ together with $x_{1}v_{i}v_{i+1}x_{2}$ form a cycle of length $2k+1$ , a contradiction. Thus $G-B$ is bipartite.

Now we consider (i) and suppose indirectly that $|B|\geq t+2$ . Since $G-B$ is bipartite, $e(G-B)\leq\frac{(n-|B|)^{2}}{4}$ . It follows that

e(G)<e(G-B)+2\sqrt{tn}|B|\leq\frac{(n-|B|)^{2}}{4}+2\sqrt{tn}|B|.

Let $f(x)=\frac{(n-x)^{2}}{4}+2\sqrt{tn}x$ . Since $f(x)$ is convex and $t+2\leq|B|\leq n$ ,

e(G)=f(|B|)\leq\max\{f(t+2),f(n)\}.

Note that

	$\displaystyle f(t+2)=$	$\displaystyle\frac{(n-t-2)^{2}}{4}+2(t+2)\sqrt{tn}$
		$\displaystyle\leq\frac{(n-t-1)^{2}}{4}-\frac{n-t-2}{2}+2(t+2)\sqrt{tn}$
		$\displaystyle<\frac{(n-t-1)^{2}}{4}+\binom{t+2}{2}-1-\frac{n}{2}+2(t+2)\sqrt{tn}.$

Using $n\geq 252t^{2}k>16(t+2)^{2}t$ , we have $\frac{n}{2}\geq 2(t+2)\sqrt{tn}$ . Thus $f(t+2)<\frac{(n-t-1)^{2}}{4}+\binom{t+2}{2}-1$ . By $n\geq 252t^{2}k>144t$ we also have $2\sqrt{tn}\leq\frac{n}{6}\leq\frac{n-4t}{4}\leq\frac{n-2(t+1)}{4}$ . It implies that

\displaystyle f(n)=2n\sqrt{tn}\leq\frac{(n-2(t+1))n}{4}<\frac{(n-t-1)^{2}}{4}.

Thus $e(G)<\frac{(n-t-1)^{2}}{4}+\binom{t+2}{2}-1$ , a contradiction. Therefore (i) holds.

Let $X$ , $Y$ be two partite sets of $G-B$ . By $B\subseteq T$ we infer $X_{0}\subset X$ and $Y_{0}\subset Y$ . By Lemma 3.1 (ii), $|X|\geq|X_{0}|\geq\frac{n}{2}-2\sqrt{tn}$ and $|Y|\geq|Y_{0}|\geq\frac{n}{2}-2\sqrt{tn}$ . By $|X|\leq\frac{n}{2}+\sqrt{3tn}$ and $|T\setminus B|\leq|T|\leq 3t+3$ we have

|X|\leq|X_{0}|+|T\setminus B|\leq\frac{n}{2}+\sqrt{3tn}+3t+3\leq\frac{n}{2}+\sqrt{3tn}+6t.

By $n\geq 252t^{2}k\geq\frac{36}{(2-\sqrt{3})^{2}}t$ , we have $(2-\sqrt{3})\sqrt{tn}\geq 6t$ . It implies that $|X|\leq\frac{n}{2}+2\sqrt{tn}$ . Similarly, $|Y|\leq\frac{n}{2}+2\sqrt{tn}$ . Thus (ii) holds.

Since $B\subseteq T$ , $G-T$ is a subgraph of $G-B$ . For $n\geq 144t$ we have $\frac{\sqrt{tn}}{2}\geq 6t\geq 3t+3$ . By Lemma 3.1 (iii) we infer that

\displaystyle\left|\left\{v\in V(G)\setminus B\colon\deg_{G-B}(v)\leq\frac{n}{2}-\sqrt{3tn}\right\}\right|\leq\sqrt{tn}+|T\setminus B|\leq\sqrt{tn}+3t+3\leq\frac{3\sqrt{tn}}{2}.

Thus (iii) holds.

We are left to show (iv), (v) and (vi). Suppose indirectly that there exists $v\in B$ such that $v$ has two neighbors $x\in X$ and $y\in Y$ . By the definition of $B$ , we have $\deg(x),\deg(y)\geq 2\sqrt{tn}$ . Let $X_{1}=N(y,X)\setminus\{x\}$ and $Y_{1}=N(x,Y)\setminus\{y\}$ . Then

|X_{1}|\geq\deg(x)-|B|-1\geq 2\sqrt{tn}-t-2\geq 2\sqrt{tn}-3t.

Since $n\geq 252t^{2}k>64t$ implies that $\frac{\sqrt{tn}}{2}\geq 4t\geq 3t+1$ , we have $|X_{1}|\geq\frac{3\sqrt{tn}}{2}+1$ . Similarly $|Y_{1}|\geq\frac{3\sqrt{tn}}{2}+1$ .

If $k=2$ , then the $C_{5}$ -free property implies that there is no edge between $X_{1}$ and $Y_{1}$ . It follows that $e(G-B)\leq|X||Y|-|X_{1}||Y_{1}|\leq\frac{(n-|B|)^{2}}{4}-tn$ . Then

\displaystyle e(G)

\displaystyle\leq e(G-B)+2\sqrt{tn}|B|\leq\frac{(n-|B|)^{2}}{4}-tn+2\sqrt{tn}|B|.

Let $f(x):=\frac{(n-x)^{2}}{4}+2\sqrt{tn}x$ . It is easy to check that $f(x)$ is decreasing on $[1,n-4\sqrt{tn}]$ . Since $n\geq 252t^{2}k>25t$ implies $4\sqrt{tn}\leq\frac{4n}{5}$ , we have $|B|\leq t+1\leq n-4\sqrt{tn}$ . Then

\displaystyle e(G)\leq f(|B|)-tn\leq f(1)-tn=\frac{(n-1)^{2}}{4}+2\sqrt{tn}-tn\leq\frac{(n-t-1)^{2}}{4}-\frac{tn}{2}+2\sqrt{tn}.

By $n\geq 252t^{2}k$ we have $\frac{tn}{2}\geq 2\sqrt{tn}$ . It follows that $e(G)\leq\frac{(n-t-1)^{2}}{4}$ , a contradiction. Thus we may assume that $k\geq 3$ .

Since $|X_{1}|$ , $|Y_{1}|\geq\frac{3\sqrt{tn}}{2}+1$ , by (iii) there are $x_{1}\in X_{1}$ and $y_{1}\in Y_{1}$ such that $\deg(x_{1},Y),\deg(y_{1},X)\geq\frac{n}{2}-\sqrt{3tn}$ . Let $X_{2}=N(y_{1},X)\setminus\{x,x_{1}\}$ , $Y_{2}=N(x_{1},Y)\setminus\{y,y_{1}\}$ . Then

|X_{2}|\geq\deg(y_{1},X)-2\geq\frac{n}{2}-\sqrt{3tn}-2,\quad|Y_{2}|\geq\deg(x_{1},Y)-2\geq\frac{n}{2}-\sqrt{3tn}-2.

Since $G$ is $C_{2k+1}$ -free and $k\geq 3$ , $G[X_{2}\cup Y_{2}]$ contains no paths of length $2k-5$ . By Theorem 1.6 we infer $e(X_{2},Y_{2})\leq\frac{(2k-5)-1}{2}(|X_{2}|+|Y_{2}|)\leq(k-3)n$ . Thus,

	$\displaystyle e(G)$	$\displaystyle\leq e(G-B)+2\sqrt{tn}\|B\|$
		$\displaystyle\leq\|X\|\|Y\|-\|X_{2}\|\|Y_{2}\|+e(X_{2},Y_{2})+2\sqrt{tn}\|B\|$
		$\displaystyle\leq\frac{(n-\|B\|)^{2}}{4}-\left(\frac{n}{2}-\sqrt{3tn}-2\right)^{2}+(k-3)n+2\sqrt{tn}\|B\|.$

Let $f(x):=\frac{(n-x)^{2}}{4}+2\sqrt{tn}x$ . Recall that $f(x)$ is decreasing on $[1,n-4\sqrt{tn}]$ . Since $n\geq 252t^{2}k>25t$ implies $|B|\leq t+1\leq n-4\sqrt{tn}$ , $f(|B|)\leq f(1)=\frac{(n-1)^{2}}{4}+2\sqrt{tn}$ . Thus,

	$\displaystyle e(G)$	$\displaystyle\leq f(\|B\|)-\left(\frac{n}{2}-\sqrt{3tn}-2\right)^{2}+(k-3)n$
		$\displaystyle\leq\frac{(n-1)^{2}}{4}+2\sqrt{tn}-\left(\frac{n}{2}-\sqrt{3tn}-2\right)^{2}+(k-3)n$
		$\displaystyle\leq 2\sqrt{tn}+(\sqrt{3tn}+2)n-3tn-4\sqrt{3tn}+(k-3)n$
		$\displaystyle\leq(k+\sqrt{3tn})n-\frac{t+1}{2}n.$

By $n\geq 252t^{2}k>20k+75t$ we have $(k+\sqrt{3tn})n\leq(\frac{n}{20}+\frac{n}{5})n=\frac{n^{2}}{4}$ . It implies that $e(G)\leq\frac{n^{2}}{4}-\frac{t+1}{2}n<\frac{(n-t-1)^{2}}{4}$ , a contradiction. Thus (iv) holds.

To show $(v)$ , suppose for contradiction that there exist $uv\in E(B)$ and $x_{1},x_{2}\in X$ such that $ux_{1},vx_{2}\in E(G)$ . We distinguish two cases.

Case 1. $k=2$ .

Then $|B|\leq t+1\leq 2k-1=3$ . By $u,v\in B$ we have $2\leq|B|\leq 3$ . Since $G$ is $C_{5}$ -free, for every $ux^{\prime},vx^{\prime\prime}\in E(G)$ with $x^{\prime},x^{\prime\prime}\in X$ we have $N(x^{\prime},Y)\cap N(x^{\prime\prime},Y)=\emptyset$ . It follows that

\displaystyle\deg(x^{\prime},Y)+\deg(x^{\prime\prime},Y)\leq|Y|.

(3.1)

If $|B|=3$ , then $e(B)+e(B,X\cup Y)\leq 2\sqrt{tn}|B|=6\sqrt{tn}$ . By (3.1), $\deg(x_{1},Y)+\deg(x_{2},Y)\leq|Y|$ . Since $|X|+|Y|-1=n-4$ , it implies that

e(X,Y)\leq|X\setminus\{x_{1},x_{2}\}||Y|+\deg(x_{1},Y)+\deg(x_{2},Y)\leq(|X|-1)|Y|\leq\left\lfloor\frac{(n-4)^{2}}{4}\right\rfloor.

Then

	$\displaystyle e(G)=e(B)+e(B,X\cup Y)+e(X,Y)$	$\displaystyle\leq 6\sqrt{tn}+\left\lfloor\frac{(n-4)^{2}}{4}\right\rfloor$
		$\displaystyle=\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor-\left\lfloor\frac{n-3}{2}\right\rfloor+6\sqrt{tn}.$

By $n\geq 252t^{2}k>144t$ we have $\left\lfloor\frac{n-3}{2}\right\rfloor+6>\frac{n}{2}\geq 6\sqrt{tn}$ . It implies that $e(G)<\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor+6$ , a contradiction. Thus $|B|=2$ , that is, $B=\{u,v\}$ .

By (iv), $N(u,X\cup Y)\subset X$ and $N(v,X\cup Y)\subset X$ . Applying (3.1) with $x^{\prime}=x_{1}$ and $x^{\prime\prime}=x_{2}$ ,

\displaystyle\deg(x_{1},Y)+\deg(x_{2},Y)\leq|Y|.

(3.2)

If $\deg(u,X)+\deg(v,X)\leq 4$ , then $e(B,X\cup Y)=\deg(u,X)+\deg(v,X)\leq 4$ . As $|X|+|Y|-1=n-3$ ,

\ e(X,Y)\leq|X\setminus\{x_{1},x_{2}\}||Y|+\deg(x_{1},Y)+\deg(x_{2},Y)\leq(|X|-1)|Y|\leq\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor.

Then

\displaystyle e(G)

\displaystyle=e(B)+e(B,X\cup Y)+e(X,Y)\leq 1+4+\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor<\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor+6,

a contradiction. Thus we may assume that $\deg(u,X)+\deg(v,X)\geq 5$ .

Without loss of generality, we further assume that $\deg(u,X)\geq\deg(v,X)$ . Then $\deg(u,X)\geq 3$ and $e(B,X\cup Y)\leq 2\deg(u,X)$ . Let $X_{u}=N(u,X)\setminus\{x_{1},x_{2}\}$ . Note that $x_{2}$ is a high-degree vertex. It follows that $\deg(x_{2},Y)\geq 2\sqrt{tn}-|\{u,v\}|=2\sqrt{tn}-2$ . For every $x\in X_{u}$ , by applying (3.1) with $x^{\prime}=x$ and $x^{\prime\prime}=x_{2}$ we get $\deg(x,Y)\leq|Y|-\deg(x_{2},Y)\leq|Y|-2\sqrt{tn}+2$ . By (3.2) we have

	$\displaystyle\ e(X,Y)$	$\displaystyle\leq\|X\setminus(X_{u}\cup\{x_{1},x_{2}\})\|\|Y\|+\deg(x_{1},Y)+\deg(x_{2},Y)+\|X_{u}\|(\|Y\|-2\sqrt{tn}+2)$
		$\displaystyle\leq(\|X\|-1)\|Y\|-\|X_{u}\|(2\sqrt{tn}-2)$
		$\displaystyle\leq\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor-(2\sqrt{tn}-2)(\deg(u,X)-2).$

Thus

	$\displaystyle e(G)$	$\displaystyle=e(B)+e(B,X\cup Y)+e(X,Y)$
		$\displaystyle\leq 1+2\deg(u,X)+\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor-(2\sqrt{tn}-2)(\deg(u,X)-2)$
		$\displaystyle\leq 1-(2\sqrt{tn}-4)\deg(u,X)+4\sqrt{tn}-4+\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor,$

Since $\deg(u,X)\geq 3$ , we have

e(G)\leq\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor-3(2\sqrt{tn}-4)+4\sqrt{tn}-3=\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor-2\sqrt{tn}+9<\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor+6,

a contradiction.

Case 2. $k\geq 3$ .

Since $\deg(x_{1}),\deg(x_{2})\geq 2\sqrt{tn}-t-1\geq\frac{3\sqrt{tn}}{2}+2$ and by (iii), there exist distinct vertices $y_{1}\in N(x_{1})\cap Y,y_{2}\in N(x_{2})\cap Y$ such that $\deg(y_{1},X)\geq\frac{n}{2}-\sqrt{3tn}>\frac{n}{2}-2\sqrt{tn}$ and $\deg(y_{2},X)\geq\frac{n}{2}-\sqrt{3tn}>\frac{n}{2}-2\sqrt{tn}$ . Let $X_{1}=N(y_{1},X)\cap N(y_{2},X)\setminus\{x_{1},x_{2}\}$ , $Y_{1}=Y\setminus\{y_{1},y_{2}\}$ . Since $|X|\leq\frac{n}{2}+2\sqrt{tn}$ and $|Y|\geq\frac{n}{2}-2\sqrt{tn}$ ,

	$\displaystyle\|X_{1}\|$	$\displaystyle=\|N(y_{1},X)\|+\|N(y_{2},X)\|-\|N(y_{1},X)\cup N(y_{2},X)\|-2$
		$\displaystyle\geq 2\left(\frac{n}{2}-2\sqrt{tn}\right)-\|X\|-2$
		$\displaystyle\geq\frac{n}{2}-6\sqrt{tn}-2.$

Note also that $|Y_{1}|=|Y|-2\geq\frac{n}{2}-2\sqrt{tn}-2$ . If there exists a path of length $2k-6$ with both ends in $X_{1}$ , then together with $y_{1}x_{1}uvx_{2}y_{2}$ we obtain a cycle of length $2k+1$ , a contradiction. Thus, there does not exist a path of length $2k-6$ with both ends in $X_{1}$ . This implies that $G[X_{1},Y_{1}]$ is $P_{2k-3}$ -free. By Theorem 1.6,

e(X_{1},Y_{1})\leq\frac{(2k-3)-1}{2}(|X_{1}|+|Y_{1}|)\leq(k-2)n.

Then

	$\displaystyle e(G)$	$\displaystyle\leq\|X\|\|Y\|-\|X_{1}\|\|Y_{1}\|+e(X_{1},Y_{1})+(t+1)\cdot 2\sqrt{tn}$
		$\displaystyle\leq\frac{(n-2)^{2}}{4}-\left(\frac{n}{2}-6\sqrt{tn}-2\right)\left(\frac{n}{2}-2\sqrt{tn}-2\right)+(k-2)n+2(t+1)\sqrt{tn}$
		$\displaystyle\leq\frac{(n-2)^{2}}{4}-\frac{n^{2}}{4}+4n\sqrt{tn}-12tn+2n+(k-2)n+2(t+1)\sqrt{tn}$
		$\displaystyle\leq(4\sqrt{tn}+k)n-12tn+2(t+1)\sqrt{tn}$
		$\displaystyle\leq(4\sqrt{tn}+k)n-\frac{t+1}{2}n-10tn+4t\sqrt{tn}.$

It is easy to verify that $10tn\geq 4t\sqrt{tn}$ . By $n\geq 252t^{2}k>20k+400t$ we have $(4\sqrt{tn}+k)n\leq(\frac{n}{5}+\frac{n}{20})n=\frac{n^{2}}{4}$ . It implies that $e(G)\leq\frac{n^{2}}{4}-\frac{t+1}{2}n<\frac{(n-t-1)^{2}}{4}$ , a contradiction. Thus (v) holds.

To show $(vi)$ , let $uwv$ be a path in $B$ and $\deg(w,X\cup Y)\leq 1$ , assume that there exist $x\in X,y\in Y$ such that $uy,vx\in E(G)$ . We distinguish two cases.

Case 1. $k=2$ .

Since $|B|\leq t+1\leq 2k-1=3$ , $B=\{u,v,w\}$ . By (iv), we have $N(u,X\cup Y)\subset Y$ and $N(v,X\cup Y)\subset X$ . It follows that

e(B,X\cup Y)\leq\deg(u,Y)+\deg(v,{X})+\deg(w,X\cup Y)\leq\deg(u,{Y})+\deg(v,{X})+1.

Since $G$ is $C_{5}$ -free, there is no edge between $N(u,Y)$ and $N(v,X)$ . That is,

\displaystyle e(X,Y)\leq|X||Y|

\displaystyle-\deg(u,{Y})\deg(v,{X}).

It follows that

	$\displaystyle e(G)$	$\displaystyle=e(B)+e(B,X\cup Y)+e(X,Y)$
		$\displaystyle\leq 3+\deg(u,{Y})+\deg(v,{X})+1+\|X\|\|Y\|-\deg(u,{Y})\deg(v,{X})$
		$\displaystyle=5+\|X\|\|Y\|-(\deg(u,{Y})-1)(\deg(v,{X})-1).$

Since $|X|+|Y|=n-3$ , $|X||Y|\leq\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor$ . Moreover, $\deg(u,{Y})\geq 1$ and $\deg(v,{X})\geq 1$ , we have $e(G)\leq 5+\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor$ . It follows that $e(G)<\left\lfloor\frac{(n-2)^{2}}{4}\right\rfloor+3$ for $t=1$ and $e(G)<\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor+6$ for $t=2$ , a contradiction.

Case 2. $k\geq 3$ .

Let $X_{1}=N(y,X)\setminus\{x\}$ and $Y_{1}=N(x,Y)\setminus\{y\}$ . Recall that $x,y$ are both high-degree vertices. Since $n\geq 252t^{2}k>64t$ implies $|B|\leq t+1\leq 2t\leq\frac{\sqrt{tn}}{2}-2$ , we infer that

|X_{1}|,|Y_{1}|\geq 2\sqrt{tn}-|B|-1\geq\frac{3\sqrt{tn}}{2}+1>\sqrt{tn}.

It follows that $|X_{1}||Y_{1}|>tn$ .

If $k=3$ then by the $C_{7}$ -free property there is no edge between $X_{1}$ and $Y_{1}$ . Thus,

	$\displaystyle e(G)$	$\displaystyle=e(B)+e(B,X\cup Y)+e(X,Y)$
		$\displaystyle\leq(t+1)\cdot 2\sqrt{tn}+\|X\|\|Y\|-\|X_{1}\|\|Y_{1}\|$
		$\displaystyle\leq\frac{(n-3)^{2}}{4}-tn+2(t+1)\sqrt{tn}$
		$\displaystyle\leq\frac{(n-t-1)^{2}}{4}-\frac{t+1}{2}n+2(t+1)\sqrt{tn}.$

By $n\geq 252t^{2}k\geq 16t$ we have $\frac{t+1}{2}n\geq 2(t+1)\sqrt{tn}$ . Thus $e(G)\leq\frac{(n-t-1)^{2}}{4}$ , a contradiction.

If $k\geq 4$ , by (iii) there exist $x_{1}\in X_{1}$ , $y_{1}\in Y_{1}$ such that $\deg(x_{1},{Y})>\frac{n}{2}-2\sqrt{tn}$ and $\deg(y_{1},{X})>\frac{n}{2}-2\sqrt{tn}$ . Let $X_{2}=N(y_{1},X)\setminus\{x,x_{1}\}$ and $Y_{2}=N(x_{1},Y)\setminus\{y,y_{1}\}$ . By the $C_{2k+1}$ -free property there is no a path of length $2k-7$ with one end in $X_{2}$ and the other one in $Y_{2}$ . By Theorem 1.6,

e(X_{2},Y_{2})\leq\frac{(2k-7)-1}{2}(|X_{1}|+|Y_{1}|)\leq(k-4)n.

Using $|X_{2}|,|Y_{2}|\geq\frac{n}{2}-2\sqrt{tn}-2$ , we have

	$\displaystyle e(G)$	$\displaystyle=e(B)+e(B,X\cup Y)+e(X,Y)$
		$\displaystyle\leq(t+1)\cdot 2\sqrt{tn}+\|X\|\|Y\|-\|X_{2}\|\|Y_{2}\|+e(X_{2},Y_{2})$
		$\displaystyle\leq\left\lfloor\frac{(n-3)^{2}}{4}\right\rfloor-\left(\frac{n}{2}-2\sqrt{tn}-2\right)^{2}+(k-4)n+2(t+1)\sqrt{tn}$
		$\displaystyle\leq(2\sqrt{tn}+2)n-4tn+(k-4)n+4t\sqrt{tn}$
		$\displaystyle\leq(k+2\sqrt{tn})n-\frac{t+1}{2}n-2tn+4t\sqrt{tn}.$

It is easy to verify that $2tn\geq 4t\sqrt{tn}$ . By $n\geq 252t^{2}k>10k+64t$ we have $(k+2\sqrt{tn})n\leq(\frac{n}{10}+\frac{n}{8})n<\frac{n^{2}}{4}$ . It implies that $e(G)\leq\frac{n^{2}}{4}-\frac{t+1}{2}n<\frac{(n-t-1)^{2}}{4}$ , a contradiction. Thus (vi) holds and the lemma is proven. ∎

4 The Proof of Theorem 1.3

Proof of Theorem 1.3.

Let $G$ be a $C_{2k+1}$ -free graph on $n$ vertices with $d_{2}(G)\geq t$ . If $e(G)<\lfloor\frac{(n-t-1)^{2}}{4}\rfloor+\binom{t+2}{2}$ then we have nothing to do. Thus we assume that

\displaystyle e(G)\geq\frac{(n-t-1)^{2}}{4}+\binom{t+2}{2}

(4.1)

and we are left to show that $G=H(n,t)$ up to isomorphism. By Lemma 3.2, there exists $B\subseteq V(G)$ with $t\leq|B|\leq t+1$ such that $G-B$ is a bipartite graph on partite sets $X,Y$ with (i)-(vi) of Lemma 3.2 hold.

Claim 1.

Let $B^{\prime}\subseteq B$ . If $G-B^{\prime}$ is non-bipartite, then each shortest odd cycle in $G-B^{\prime}$ has length at most $2k-1$ .

Proof.

Since $G-B$ is bipartite and $G-B^{\prime}$ is non-bipartite, $|B^{\prime}|\leq|B|-1\leq t$ . Let $C$ be a shortest cycle in $G-B^{\prime}$ , let $S=V(C)$ . Suppose for contradiction that $|S|=2\ell+1\geq 2k+3$ . By Fact 2.3 each $v\in B^{\prime}$ has at most $\ell$ neighbors on $C$ . By Fact 2.1 each $v\in V(G)\setminus(B^{\prime}\cup S)$ has at most two neighbors on $C$ . It follows that

	$\displaystyle e(S)+e(G-S,S)$	$\displaystyle\leq 2\ell+1+{2\left(n-\|B^{\prime}\|-\|S\|\right)+\|B^{\prime}\|\ell}$
		$\displaystyle\leq 2\ell+1+{2\left(n-t-2\ell-1\right)+t\ell}$
		$\displaystyle=2n+(t-2)\ell-2t-1.$

By Fact 2.2 we have $2\ell+1\leq\frac{2n}{3}$ . It follows that $n-2\ell-1\geq\frac{n}{3}\geq 4k$ . Since $G-S$ is $C_{2k+1}$ -free, by Theorem 1.1 we have $e(G-S)\leq{\frac{(n-2\ell-1)^{2}}{4}}$ . Thus,

\displaystyle e(G)

\displaystyle=e(S)+e(G-S,S)+e(G-S)\leq 2n+(t-2)\ell-2t-1+\frac{(n-2\ell-1)^{2}}{4}=:f(\ell).

Since $f(\ell)$ is a convex function of $\ell$ with $k+1\leq\ell\leq\frac{n-1}{2}$ , $e(G)\leq\max\{f(\frac{n-1}{2}),f(k+1)\}$ . By $n\geq 318t^{2}k>14t+2$ we have $\frac{n-2(t+1)}{4}\geq 3t$ . It follows that

f\left(\frac{n-1}{2}\right)=2n+\frac{(t-2)(n-1)}{2}-2t-1\leq 3tn\leq\frac{(n-2(t+1))n}{4}<\frac{(n-t-1)^{2}}{4}.

For $t\leq 2k-3$ ,

	$\displaystyle f(k+1)$	$\displaystyle=2n+(t-2)(k+1)-2t-1+\frac{(n-2k-3)^{2}}{4}$
		$\displaystyle=2n+(t-2)(k+1)-2t-1+\frac{(n-2k+2)^{2}}{4}-\frac{5}{2}n+\frac{(2k+3)^{2}-(2k-2)^{2}}{4}$
		$\displaystyle<\frac{(n-2k+2)^{2}}{4}-\frac{n}{2}+(t-2)k-2+\frac{5(4k+1)}{4}$
		$\displaystyle<\frac{(n-2k+2)^{2}}{4}-\frac{n}{2}+(t+3)k.$

Since $n\geq 318t^{2}k$ implies $\frac{n}{2}\geq(t+3)k$ , we infer that $f(k+1)<\frac{(n-t-1)^{2}}{4}$ . For $t=2k-2$ ,

	$\displaystyle f(k+1)$	$\displaystyle=2n+2(k-2)(k+1)-4k+3+\frac{(n-2k-3)^{2}}{4}$
		$\displaystyle=\frac{(n-2k+1)^{2}}{4}+2k^{2}-2k+1$
		$\displaystyle\leq\frac{(n-2k+1)^{2}}{4}+\binom{2k}{2}-1$
		$\displaystyle=\frac{(n-t-1)^{2}}{4}+\binom{t+2}{2}-1.$

Thus $e(G)\leq\max\{f(\frac{n-1}{2}),f(k+1)\}\leq\frac{(n-t-1)^{2}}{4}+\binom{t+2}{2}-1$ , contradicting (4.1). ∎

Claim 2.

Let $B^{\prime}\subseteq B$ such that $G-B^{\prime}$ is non-bipartite and let $C$ be a shortest odd cycle in $G-B^{\prime}$ . Then there is at most one high-degree vertex on $C$ .

Proof.

Suppose that $C=v_{0}v_{1}\ldots v_{2\ell}v_{0}$ and there are two high-degree vertices on $C$ , say $v_{i},v_{j}$ with $0\leq i<j\leq 2\ell$ . Set $Q_{1}=v_{i}v_{i+1}\ldots v_{j}$ and $Q_{2}=v_{j}v_{j+1}\ldots v_{2\ell}v_{0}\ldots v_{i}$ . Let $|V(Q_{1})|=\ell_{1}+1$ and $|V(Q_{2})|=\ell_{2}+1$ . By Claim 1 we know that $\ell\leq k-1$ . It follows that $\ell_{1},\ell_{2}\in\{1,2,\ldots,2k-2\}$ . Without loss of generality, assume that $\ell_{1}$ is odd and $\ell_{2}$ is even.

Note that $C$ is an induced cycle and each $v_{i}$ has at most $2$ neighbors on $C$ . Let $X_{1}=X\setminus S$ and $Y_{1}=Y\setminus S$ . Since $n\geq 318t^{2}k>144t$ implies $\deg(v_{i})\geq 2\sqrt{tn}\geq\frac{3\sqrt{tn}}{2}+6t$ , we have

\deg(v_{i},X_{1}\cup Y_{1})\geq\deg(v_{i})-|B|-\deg(v_{i},S)\geq\frac{3\sqrt{tn}}{2}+6t-(t+1)-2\geq\frac{3\sqrt{tn}}{2}+2.

Similarly, $\deg(v_{j},X_{1}\cup Y_{1})\geq\frac{3\sqrt{tn}}{2}+2$ . By Lemma 3.2 (iii), there exist distinct vertices $x_{1}\in N(v_{i},X_{1}\cup Y_{1}),x_{2}\in N(v_{j},X_{1}\cup Y_{1})$ such that $\deg(x_{1},X\cup Y)>\frac{n}{2}-2\sqrt{tn}$ and $\deg(x_{2},X\cup Y)>\frac{n}{2}-2\sqrt{tn}$ . Then by Claim 1

\deg(x_{1},X_{1}\cup Y_{1})\geq\deg(x_{1},X\cup Y)-|S|\geq\frac{n}{2}-2\sqrt{tn}-2k+1.

Similarly, $\deg(x_{2},X_{1}\cup Y_{1})\geq\frac{n}{2}-2\sqrt{tn}-2k+1$ . Now we distinguish two cases.

Case 1. $x_{1}$ and $x_{2}$ belong to the same one of $X_{1},Y_{1}$ .

Without loss of generality, assume $x_{1},x_{2}\in X_{1}$ . Let $Y_{2}=N(x_{1},Y_{1})\cap N(x_{2},Y_{1})$ and $X_{2}=X_{1}\setminus\{x_{1},x_{2}\}$ . By Lemma 3.2 (ii) $|X_{2}|\geq|X|-|S|-2\geq\frac{n}{2}-2\sqrt{tn}-2k-1$ . By Lemma 3.2 (ii) $|Y_{1}|\leq|Y|\leq\frac{n}{2}+2\sqrt{tn}$ . Then

	$\displaystyle\|Y_{2}\|=\|N(x_{1},Y_{1})\cap N(x_{2},Y_{1})\|$	$\displaystyle\geq\deg(x_{1},Y_{1})+\deg(x_{2},Y_{1})-\|Y_{1}\|$
		$\displaystyle\geq 2\left(\frac{n}{2}-2\sqrt{tn}-2k\right)-\|Y_{1}\|$
		$\displaystyle\geq\frac{n}{2}-6\sqrt{tn}-4k.$

Let

L=\left\{v\in V(G)\setminus B\colon\deg(v,X\cup Y)\leq\frac{n}{2}-\sqrt{3tn}\right\}.

By Lemma 3.2 (iii) we have $|L|\leq\frac{3\sqrt{tn}}{2}$ . It follows that $G[X\setminus L,Y\setminus L]$ is a subgraph with minimum degree at least $\frac{n}{2}-\sqrt{3tn}-|L|>\frac{n}{2}-4\sqrt{tn}$ . Note that

|Y\setminus Y_{2}|\leq\left(\frac{n}{2}+2\sqrt{tn}\right)-\left(\frac{n}{2}-6\sqrt{tn}-4k\right)=8\sqrt{tn}+4k.

Then $G^{\prime}:=G[X_{2}\setminus L,Y_{2}\setminus L]$ is a graph on $|X_{2}|+|Y_{2}|-|L|$ vertices with

\delta(G^{\prime})\geq\frac{n}{2}-4\sqrt{tn}-|Y\setminus Y_{2}|\geq\frac{n}{2}-12\sqrt{tn}-4k.

Since $n\geq 318t^{2}k\geq\frac{318^{2}}{153^{2}}\times 12^{2}t$ implies $\sqrt{tn}\leq\frac{1}{12}\times\frac{153}{318}n$ , we have $\frac{n}{2}-12\sqrt{tn}\geq\frac{6}{318}n\geq 6k$ . It follows that $\delta(G^{\prime})\geq 2k$ . Thus there exists a path of length at least $\delta(G^{\prime})+1\geq 2k+1$ . For $\ell_{1}\leq 2k-4$ , we choose a sub-path $Q$ of length $2k-3-\ell_{1}$ with both ends in $Y_{2}$ . For $\ell_{1}=2k-3$ , we choose $Q$ as an arbitrary vertex in $Y_{1}$ . Then $v_{j}x_{2}Qx_{1}Q_{1}$ forms a cycle of length $2k+1$ , a contradiction.

Case 2. $x_{1}$ and $x_{2}$ belong to different ones of $X_{1}$ , $Y_{1}$ .

Without loss of generality, assume that $x_{1}\in X_{1}$ and $x_{2}\in Y_{1}$ . Let $X_{2}=N(x_{2},X_{1})\setminus\{x_{1}\}$ and $Y_{2}=N(x_{1},Y_{1})\setminus\{x_{2}\}$ . Then $|X_{2}|\geq\deg(x_{2},X_{1})-1\geq\frac{n}{2}-2\sqrt{tn}-2k$ . Similarly, $|Y_{2}|\geq\frac{n}{2}-2\sqrt{tn}-2k$ . Let

L=\left\{v\in V(G)\setminus B\colon\deg(v,X\cup Y)\leq\frac{n}{2}-\sqrt{3tn}\right\}.

|Y\setminus Y_{2}|\leq\left(\frac{n}{2}+2\sqrt{tn}\right)-\left(\frac{n}{2}-2\sqrt{tn}-2k\right)=4\sqrt{tn}+2k.

Then $G^{\prime}:=G[X_{2}\setminus L,Y_{2}\setminus L]$ is a graph on $|X_{2}|+|Y_{2}|-|L|$ vertices with

\delta(G^{\prime})\geq\frac{n}{2}-4\sqrt{tn}-|Y\setminus Y_{2}|\geq\frac{n}{2}-8\sqrt{tn}-2k.

Since $n\geq 318t^{2}k>\frac{318^{2}}{155^{2}}\times 64t$ implies $\sqrt{tn}\leq\frac{1}{8}\times\frac{155}{318}n$ , we have $\frac{n}{2}-8\sqrt{tn}\geq\frac{4}{318}n\geq 4k$ . It implies that $\delta(G^{\prime})\geq 2k$ . Then there exists a path of length at least $\delta(G^{\prime})+1\geq 2k+1$ . For $\ell_{2}\leq 2k-4$ , we choose a sub-path $Q$ of length $2k-3-\ell_{2}$ with one end in $X_{1}$ and the other one in $Y_{1}$ . Then $v_{i}x_{1}Qx_{2}Q_{2}$ forms a cycle of length $2k+1$ , a contradiction.

We are left with the case $\ell_{2}=2k-2$ . The $C_{2k+1}$ -free property implies that there is no edge between $N(v_{i},X_{1})$ and $N(v_{j},Y_{1})$ . Since $n\geq 318t^{2}k\geq\frac{16}{(2-\frac{23}{25})^{2}}t$ implies $4t\leq\left(2-\frac{23}{25}\right)\sqrt{tn}$ ,

\deg(v_{i},X_{1})\geq\deg(v_{i})-|B|-\deg(v_{i},S)\geq 2\sqrt{tn}-t-3\geq 2\sqrt{tn}-4t\geq\frac{23}{25}\sqrt{tn}.

Similarly, $\deg(v_{j},Y_{1})\geq\frac{23}{25}\sqrt{tn}$ . Then

e(G-B)\leq|X||Y|-|N(v_{i},X_{1})||N(v_{j},Y_{1})|\leq\frac{(n-|B|)^{2}}{4}-\frac{5tn}{6}.

By Lemma 3.2 we have

\displaystyle e(G)=e(G-B)+2\sqrt{tn}|B|\leq\frac{(n-|B|)^{2}}{4}-\frac{5tn}{6}+2\sqrt{tn}|B|.

Let $f(x):=\frac{(n-x)^{2}}{4}+2\sqrt{tn}x$ . It is easy to verify that $f(x)$ is decreasing on $[1,n-4\sqrt{tn}]$ . Note that $n\geq 318t^{2}k>25t$ implies $|B|\leq t+1\leq n-4\sqrt{tn}$ . By $|B|\geq t$ we infer that

e(G)\leq f(t)-\frac{5tn}{6}=\frac{(n-t)^{2}}{4}+2t\sqrt{tn}-\frac{5tn}{6}.

Since $n\geq 36t$ we have $2t\sqrt{tn}\leq\frac{tn}{3}$ . Then $e(G)\leq\frac{(n-t)^{2}}{4}-\frac{tn}{2}\leq\frac{(n-t-1)^{2}}{4}+\frac{n-t}{2}-\frac{tn}{2}\leq\frac{(n-t-1)^{2}}{4}$ , contradicting (4.1). Thus there is an edge $xy$ between $N(v_{i},X_{1}\cup Y_{1})$ and $N(v_{j},X_{1}\cup Y_{1})$ . It follows that $xyQ_{2}x$ is a cycle of length $2k+1$ , a contradiction. Thus the claim holds. ∎

Now we choose $B_{0}\subset B$ such that $G-B_{0}$ is bipartite and $|B_{0}|$ is minimal. By $d_{2}(G)\geq t$ we have $t\leq|B_{0}|\leq t+1$ . By the minimality of $|B_{0}|$ , we infer that for each $u\in B_{0}$ , $G-(B_{0}\setminus\{u\})$ is not bipartite. Let $C_{u}$ be a shortest odd cycle in $G-(B_{0}\setminus\{u\})$ . By Claim 2, there is at most one high-degree vertex on $C_{u}$ . Since $|V(C_{u})|\geq 3$ , $C_{u}$ contains at least two low-degree vertices. As $V(C_{u})\cap B_{0}=\{u\}$ , we infer that $C_{u}$ contains a low-degree vertex that is not in $B_{0}$ , implying that $B\setminus B_{0}\neq\emptyset$ . By $t\leq|B_{0}|\leq|B|\leq t+1$ we have $|B|=t+1$ and $|B_{0}|=t$ . Then $C_{u}$ contains exactly two low-degree vertices and one high-degree vertex, that is, $|V(C_{u})|=3$ . Let $B\setminus B_{0}=\{u_{0}\}$ and let $x_{u}$ be a high-degree vertex on $C_{u}$ . Then $V(C_{u})=\{u,u_{0},x_{u}\}$ . Since $u$ is chosen from $B_{0}$ arbitrarily, we conclude that $u_{0}u\in E(G)$ for all $u\in B_{0}$ .

Assume that $B=\{u_{0},u_{1},u_{2},\ldots,u_{t}\}$ and let $B_{i}=B\setminus\{u_{i}\}$ , $i=1,2,\ldots,t$ . We claim that $G-B_{i}$ is bipartite. Indeed, otherwise let $C^{i}$ be a shortest odd cycle in $G-B_{i}$ . Note that $|B_{i}|=t$ . By applying Claim 2 to $C^{i}$ , $C^{i}$ contains at most one high-degree vertex. It follows that $C^{i}$ contains at least two low-degree vertices. But there is only one low-degree vertex $u_{i}$ in $G-B_{i}$ , a contradiction. Thus $G-B_{i}$ is bipartite. Let $B_{ij}=B\setminus\{u_{i},u_{j}\}$ for $0\leq i<j\leq t$ . By $d_{2}(G)\geq t$ and $|B_{ij}|=t-1$ , we see that $G-B_{ij}$ is non-bipartite.

Claim 3.

For $0\leq i<j\leq t$ , there exists a high-degree vertex $x_{ij}$ such that $u_{i}u_{j}x_{ij}$ forms a triangle in $G$ .

Proof.

Let $C$ be a shortest odd cycle in $G-B_{ij}$ . By Claim 2, $C$ contains at most one high-degree vertex. Since $G-B_{ij}$ contains exactly two low-degree vertices $u_{i},u_{j}$ , $C$ has to be a triangle on $\{u_{i},u_{j},x_{ij}\}$ for some high-degree vertex $x_{ij}$ . ∎

By Claim 3, $B$ spans a clique of size $t+1$ in $G$ . Moreover, for each $u_{i}u_{j}\in E(B)$ there exists a high-degree vertex $x_{ij}$ such that $C_{ij}=u_{i}u_{j}x_{ij}u_{i}$ is a triangle in $G$ .

Claim 4.

For each $u_{i}u_{j}\in E(B)$ , $N(u_{i},X\cup Y)=N(u_{j},X\cup Y)=\{x_{ij}\}$ .

Proof.

Otherwise there exists a high-degree vertex $y\neq x_{ij}$ such that $u_{i}x_{ij},u_{j}y\in E(G)$ . By Lemma 3.2 (iv), it implies that $x_{ij},y$ belong to same side of $X$ or $Y$ . By symmetry assume that $y,x_{ij}\in X$ , contradicting (v) of Lemma 3.2. ∎

Recall that $G[B]$ is a clique of size $t+1$ . By Claim 4, $x_{ij}=x_{i^{\prime}j^{\prime}}=:x^{*}$ for all $u_{i}u_{j},u_{i^{\prime}}u_{j^{\prime}}\in E(B)$ . That is, each vertex in $B$ is adjacent to the unique high-degree vertex $x^{*}$ . Hence $G$ is isomorphic to a subgraph of $H(n,t)$ . By (4.1) we conclude that $G=H(n,t)$ up to isomorphism and the theorem is proven. ∎

5 The Proof of Theorem 1.4

Proof of Theorem 1.4.

Let $G$ be a $C_{2k+1}$ -free on $n$ vertices with

\displaystyle\gamma_{2}(G)\geq\binom{\lfloor\frac{t+2}{2}\rfloor}{2}+\binom{\lceil\frac{t+2}{2}\rceil}{2}\overset{\eqref{equalities2.6}}{=}f(t+1).

(5.1)

If $e(G)<\left\lfloor\frac{(n-t-1)^{2}}{4}\right\rfloor+\binom{t+2}{2}$ then we are done. Thus we may assume that

\displaystyle e(G)\geq\left\lfloor\frac{(n-t-1)^{2}}{4}\right\rfloor+\binom{t+2}{2}

(5.2)

and we are left to show that $G=H(n,t)$ up to isomorphism. By Lemma 3.2, there exists $B\subseteq V(G)$ with $|B|\leq t+1$ such that $G-B$ is a bipartite graph on partite sets $X,Y$ with (i)-(vi) of Lemma 3.2 hold. Let

	$\displaystyle B_{11}=\{u\in B\colon\deg(u,Y)=1\},\quad B_{12}=\{u\in B\colon\deg(u,X)=1\},$
	$\displaystyle X_{2}=\{u\in B\colon\deg(u,Y)\geq 2\},\quad Y_{2}=\{u\in B\colon\deg(u,X)\geq 2\},$
	$\displaystyle B_{0}=\{u\in B\colon\deg(u,X\cup Y)=0\}.$

By Lemma 3.2 (iv), $B_{11}\cap B_{12}=\emptyset$ , $B_{11}\cap Y_{2}=\emptyset$ , $X_{2}\cap B_{12}=\emptyset$ and $X_{2}\cap Y_{2}=\emptyset$ . It follows that $(B_{0},B_{11},B_{12},X_{2},Y_{2})$ is partition of $B$ . Furthermore, we have

\displaystyle e(X_{2},X)=0,\ e(Y_{2},Y)=0,\ e(B_{11},X)=0,\ e(B_{12},Y)=0.

(5.3)

Claim 5.

$X_{2}$ and $Y_{2}$ are both independent sets of $G$ .

Proof.

Suppose not, by symmetry assume that $uv\in E(X_{2})$ . Since $\deg(u,Y)\geq 2$ and $\deg(v,Y)\geq 2$ , there are different vertices $y_{1},y_{2}\in Y$ such that $uy_{1},vy_{2}\in E(G)$ , contradicting (v) of Lemma 3.2. Thus $X_{2}$ is an independent set. Similarly, $Y_{2}$ is an independent set. ∎

Claim 6.

$e(B_{11},X_{2})=0$ and $e(B_{12},Y_{2})=0$ .

Proof.

Suppose that there exist $u\in B_{11}$ , $v\in X_{2}$ such that $uv\in E(G)$ . Since $\deg(u,Y)=1$ and $\deg(v,Y)\geq 2$ , one can find different vertices $y_{1},y_{2}\in Y$ such that $uy_{1},vy_{2}\in E(G)$ , contradicting Lemma 3.2 (v). Thus $e(B_{11},X_{2})=0$ . Similarly, $e(B_{12},Y_{2})=0$ . ∎

Set $|B_{0}|=z$ , $|B_{11}|=p$ , $|B_{12}|=q$ , $|X_{2}|=x$ and $|Y_{2}|=y$ . Clearly $x+y+z+p+q=|B|\leq t+1$ . Now we further partition $B_{0}$ to $X_{0}$ and $Y_{0}$ such that $|X_{0}|=\left\lfloor\frac{z}{2}\right\rfloor$ and $|Y_{0}|=\left\lceil\frac{z}{2}\right\rceil$ . Partition $B_{11}$ to $X_{11}$ , $Y_{11}$ such that $|X_{11}|=\left\lceil\frac{p}{2}\right\rceil$ , $|Y_{11}|=\left\lfloor\frac{p}{2}\right\rfloor$ and partition $B_{12}$ to $X_{12}$ , $Y_{12}$ such that $|X_{12}|=\left\lfloor\frac{q}{2}\right\rfloor$ , $|Y_{12}|=\left\lceil\frac{q}{2}\right\rceil$ . Let

\widetilde{X}=X_{0}\cup X_{11}\cup X_{12}\cup X_{2}\cup X,\ \widetilde{Y}=Y_{0}\cup Y_{11}\cup Y_{12}\cup Y_{2}\cup Y.

Clearly, $(\widetilde{X},\widetilde{Y})$ is a bi-partition of $V(G)$ (as shown in Figure 1) and $\gamma_{2}(G)\leq e(\widetilde{X})+e(\widetilde{Y})$ .

Refer to caption — Figure 1: The bi-partition $(\widetilde{X},\widetilde{Y})$ of $V(G)$ .

By Claim 6 we have $e(B_{11},X_{2})=0$ . Using $e(B_{0},X)=0$ and (5.3), we obtain that

e(\widetilde{X})=e(X_{0})+e(X_{11})+e(X_{12})+e(X_{0},X_{11}\cup X_{12}\cup X_{2})+e(X_{11},X_{12})+e(X_{12},X_{2})+e(X_{12},X).

Similarly,

e(\widetilde{Y})=e(Y_{0})+e(Y_{11})+e(Y_{12})+e(Y_{0},Y_{11}\cup Y_{12}\cup Y_{2})+e(Y_{11},Y_{12})+e(Y_{11},Y_{2})+e(Y_{11},Y).

Note that

	$\displaystyle e(X_{0})\leq\binom{\|X_{0}\|}{2},\ e(X_{11})\leq\binom{\|X_{11}\|}{2},\ e(X_{12})\leq\binom{\|X_{12}\|}{2},$
	$\displaystyle e(X_{0},X_{11}\cup X_{12}\cup X_{2})\leq\|X_{0}\|(\|X_{11}\|+\|X_{12}\|+\|X_{2}\|),$
	$\displaystyle e(X_{11},X_{12})\leq\|X_{11}\|\|X_{12}\|,\ e(X_{12},X_{2})\leq\|X_{12}\|\|X_{2}\|.$

Similarly,

	$\displaystyle e(Y_{0})\leq\binom{\|Y_{0}\|}{2},\ e(Y_{11})\leq\binom{\|Y_{11}\|}{2},\ e(Y_{12})\leq\binom{\|Y_{12}\|}{2},$
	$\displaystyle e(Y_{0},Y_{11}\cup Y_{12}\cup Y_{2})\leq\|Y_{0}\|(\|Y_{11}\|+\|Y_{12}\|+\|Y_{2}\|),$
	$\displaystyle e(Y_{11},Y_{12})\leq\|Y_{11}\|\|Y_{12}\|,\ e(Y_{11},Y_{2})\leq\|Y_{11}\|\|Y_{2}\|.$

By the definitions of $B_{12}$ and $B_{11}$ , $e(X_{12},X)\leq|X_{12}|$ and $e(Y_{11},Y)\leq|Y_{11}|$ .

Recall that $f(x):=\binom{x}{2}-\left\lfloor\frac{x^{2}}{4}\right\rfloor+\left\lfloor\frac{x}{2}\right\rfloor$ . Then

	$\displaystyle e(X_{0})+e(Y_{0})\leq\binom{\|X_{0}\|}{2}+\binom{\|Y_{0}\|}{2}=\binom{z}{2}-\left\lfloor\frac{z^{2}}{4}\right\rfloor=f(z)-\left\lfloor\frac{z}{2}\right\rfloor,$
	$\displaystyle e(X_{11})+e(Y_{11})+e(Y_{11},Y)\leq\binom{\|X_{11}\|}{2}+\binom{\|Y_{11}\|}{2}+\|Y_{11}\|=\binom{p}{2}-\left\lfloor\frac{p^{2}}{4}\right\rfloor+\left\lfloor\frac{p}{2}\right\rfloor=f(p),$
	$\displaystyle e(X_{12})+e(Y_{12})+e(X_{12},X)\leq\binom{\|X_{12}\|}{2}+\binom{\|Y_{12}\|}{2}+\|X_{12}\|=\binom{q}{2}-\left\lfloor\frac{q^{2}}{4}\right\rfloor+\left\lfloor\frac{q}{2}\right\rfloor=f(q).$

Moreover,

	$\displaystyle e(X_{0},X_{11}\cup X_{12}\cup X_{2})+e(X_{11},X_{12})\leq\left\lfloor\frac{z}{2}\right\rfloor\left(\left\lceil\frac{p}{2}\right\rceil+\left\lfloor\frac{q}{2}\right\rfloor+x\right)+\left\lceil\frac{p}{2}\right\rceil\left\lfloor\frac{q}{2}\right\rfloor,$
	$\displaystyle e(Y_{0},Y_{11}\cup Y_{12}\cup Y_{2})+e(Y_{11},Y_{12})\leq\left\lceil\frac{z}{2}\right\rceil\left(\left\lfloor\frac{p}{2}\right\rfloor+\left\lceil\frac{q}{2}\right\rceil+y\right)+\left\lfloor\frac{p}{2}\right\rfloor\left\lceil\frac{q}{2}\right\rceil,$

and

\displaystyle e(X_{12},X_{2})\leq\left\lfloor\frac{q}{2}\right\rfloor x,\ e(Y_{11},Y_{2})\leq\left\lfloor\frac{p}{2}\right\rfloor y.

Thus,

	$\displaystyle e(\widetilde{X})+e(\widetilde{Y})$	$\displaystyle\leq f(p)+f(q)+f(z)-\left\lfloor\frac{z}{2}\right\rfloor+\left\lfloor\frac{z}{2}\right\rfloor\left(\left\lceil\frac{p}{2}\right\rceil+\left\lfloor\frac{q}{2}\right\rfloor+x\right)$
		$\displaystyle+\left\lceil\frac{z}{2}\right\rceil\left(\left\lfloor\frac{p}{2}\right\rfloor+\left\lceil\frac{q}{2}\right\rceil+y\right)+\left\lceil\frac{p}{2}\right\rceil\left\lfloor\frac{q}{2}\right\rfloor+\left\lfloor\frac{p}{2}\right\rfloor\left\lceil\frac{q}{2}\right\rceil+\left\lfloor\frac{q}{2}\right\rfloor x+\left\lfloor\frac{p}{2}\right\rfloor y.$

Using (2.1) and (2.2), we obtain that

	$\displaystyle e(\widetilde{X})+e(\widetilde{Y})\leq$	$\displaystyle f(p)+f(q)+f(z)-\left\lfloor\frac{z}{2}\right\rfloor+\left\lfloor\frac{zp}{2}\right\rfloor+\left\lceil\frac{zq}{2}\right\rceil+\left\lfloor\frac{z}{2}\right\rfloor x$
		$\displaystyle+\left\lceil\frac{z}{2}\right\rceil y+\left\lfloor\frac{pq}{2}\right\rfloor+\left\lfloor\frac{q}{2}\right\rfloor x+\left\lfloor\frac{p}{2}\right\rfloor y.$

By (2.3) and (2.4), we arrive at

	$\displaystyle e(\widetilde{X})+e(\widetilde{Y})\leq$	$\displaystyle f(p)+f(q)+f(z)-\left\lfloor\frac{z}{2}\right\rfloor+\left\lfloor\frac{zp}{2}\right\rfloor+\left\lceil\frac{zq}{2}\right\rceil+\left\lfloor\frac{zx}{2}\right\rfloor$
		$\displaystyle+\left\lceil\frac{zy}{2}\right\rceil+\left\lfloor\frac{y}{2}\right\rfloor+\left\lfloor\frac{pq}{2}\right\rfloor+\left\lfloor\frac{qx}{2}\right\rfloor+\left\lfloor\frac{py}{2}\right\rfloor.$

Since

	$\displaystyle f(p+q+z)-f(p)-f(q)-f(z)$	$\displaystyle\overset{\eqref{equalities2.5}}{=}f(p+q)-f(p)-f(q)+\left\lceil\frac{z(p+q)}{2}\right\rceil$
		$\displaystyle\overset{\eqref{equalities2.5}}{=}\left\lceil\frac{z(p+q)}{2}\right\rceil+\left\lceil\frac{pq}{2}\right\rceil$
		$\displaystyle\overset{\eqref{equalities2.3}}{\geq}\left\lfloor\frac{zp}{2}\right\rfloor+\left\lceil\frac{zq}{2}\right\rceil+\left\lceil\frac{pq}{2}\right\rceil,$

we have

\displaystyle e(\widetilde{X})+e(\widetilde{Y})

\displaystyle\leq f(p+q+z)-\left\lfloor\frac{z}{2}\right\rfloor+\left\lfloor\frac{zx}{2}\right\rfloor+\left\lceil\frac{zy}{2}\right\rceil+\left\lfloor\frac{y}{2}\right\rfloor+\left\lfloor\frac{qx}{2}\right\rfloor+\left\lfloor\frac{py}{2}\right\rfloor.

Then,

$\displaystyle e(\widetilde{X})+e(\widetilde{Y})$	$\displaystyle\leq f(p+q+z)+\left\lfloor\frac{zx}{2}\right\rfloor+\left\lceil\frac{zy}{2}\right\rceil+\left\lfloor\frac{y}{2}\right\rfloor+\left\lfloor\frac{qx}{2}\right\rfloor+\left\lfloor\frac{py}{2}\right\rfloor$	(5.4)
	$\displaystyle\overset{\eqref{equalities2.5}}{=}f(p+q+z+x)-f(x)-\left\lceil\frac{x(p+q+z)}{2}\right\rceil+\left\lfloor\frac{zx}{2}\right\rfloor+\left\lceil\frac{zy}{2}\right\rceil+\left\lfloor\frac{y}{2}\right\rfloor$
	$\displaystyle\qquad+\left\lfloor\frac{qx}{2}\right\rfloor+\left\lfloor\frac{py}{2}\right\rfloor$
	$\displaystyle\overset{\eqref{equalities2.3}}{\leq}f(p+q+z+x)+\left\lceil\frac{zy}{2}\right\rceil+\left\lfloor\frac{y}{2}\right\rfloor+\left\lfloor\frac{py}{2}\right\rfloor$	(5.5)
	$\displaystyle\overset{\eqref{equalities2.5}}{\leq}f(p+q+z+x+y)-f(y)-\left\lceil\frac{y(p+q+z+x)}{2}\right\rceil+\left\lceil\frac{zy}{2}\right\rceil+\left\lfloor\frac{y}{2}\right\rfloor+\left\lfloor\frac{py}{2}\right\rfloor$
	$\displaystyle\leq f(p+q+z+x+y)-\left(\left\lceil\frac{y(p+q+z+x)}{2}\right\rceil-\left\lceil\frac{zy}{2}\right\rceil-\left\lfloor\frac{py}{2}\right\rfloor\right)$	(5.6)
	$\displaystyle\overset{\eqref{equalities2.3}}{\leq}f(p+q+z+x+y).$

Since $f(x)$ is an increasing function and $p+q+z+x+y\leq t+1$ ,

\binom{\lfloor\frac{t+2}{2}\rfloor}{2}+\binom{\lceil\frac{t+2}{2}\rceil}{2}\leq\gamma_{2}(G)\leq e(\tilde{X})+e(\tilde{Y})\leq f(t+1)\overset{\eqref{equalities2.6}}{=}\binom{\lfloor\frac{t+2}{2}\rfloor}{2}+\binom{\lceil\frac{t+2}{2}\rceil}{2}.

Then we have equality in (5.4), (5.5), (5.6) and

\displaystyle p+q+z+x+y=t+1.

(5.7)

It follows that

\left\lfloor\frac{z}{2}\right\rfloor=0,\ f(x)=0,\ f(y)=\left\lfloor\frac{y}{2}\right\rfloor.

Hence $z\leq 1$ , $x\leq 1$ and $y\leq 2$ .

Recall that $(B_{0},B_{11},B_{12},X_{2},Y_{2})$ is partition of $B$ and $|B_{0}|=z\leq 1$ . Let us define a different partition of $B_{11}$ and $B_{12}$ . Let $X_{11}^{*}=\cup_{v\in Y_{2}}N(v,B_{11})$ and $Y_{12}^{*}=\cup_{u\in X_{2}}N(u,B_{12})$ . Set $|B_{11}\setminus X_{11}^{*}|=p_{1}$ , $|X_{11}^{*}|=p_{2}$ , $|B_{12}\setminus Y_{12}^{*}|=q_{1}$ and $|Y_{12}^{*}|=q_{2}$ . Clearly, $p=p_{1}+p_{2}$ and $q=q_{1}+q_{2}$ . We partition $B_{11}\setminus X_{11}^{*}$ into $X_{11}^{\prime}$ and $Y_{11}^{\prime}$ such that $|X_{11}^{\prime}|=\left\lceil\frac{p_{1}}{2}\right\rceil$ and $|X_{12}^{\prime}|=\left\lfloor\frac{p_{1}}{2}\right\rfloor$ . Similarly, we partition $B_{12}\setminus Y_{12}^{*}$ into $X_{12}^{\prime}$ and $Y_{12}^{\prime}$ such that $|X_{12}^{\prime}|=\left\lfloor\frac{q_{1}}{2}\right\rfloor$ and $|Y_{12}^{\prime}|=\left\lceil\frac{q_{1}}{2}\right\rceil$ . Let

X^{\prime}=X_{11}^{\prime}\cup X_{11}^{*}\cup X_{12}^{\prime}\cup X_{2}\cup X\cup B_{0},\ Y^{\prime}=Y_{11}^{\prime}\cup Y_{12}^{*}\cup Y_{12}^{\prime}\cup Y_{2}\cup Y.

Clearly, $(X^{\prime},Y^{\prime})$ is also a bi-partition of $V(G)$ (as shown in Figure 2) and $\gamma_{2}(G)\leq e(X^{\prime})+e(Y^{\prime})$ .

Claim 7.

If $X_{11}^{*}\neq\emptyset$ , then $X_{11}^{*}$ is an independent set and $e(X_{11}^{\prime},X_{11}^{*})=0$ . Similarly, if $Y_{12}^{*}\neq\emptyset$ , then $Y_{12}^{*}$ is an independent set and $e(Y_{12}^{*},Y_{12}^{\prime})=0$ .

Proof.

Suppose that there is an edge $wv$ with $w\in X_{11}^{*}$ and $v\in X_{11}^{\prime}\cup X_{11}^{*}$ . By the definition of $X_{11}^{*}$ , there exists $u\in Y_{2}$ such that $uw\in E(G)$ . Then $uwv$ be a path in $B$ . Note that $\deg(w,Y)=1$ , $\deg(v,Y)=1$ and $\deg(u,X)\geq 2$ . It leads to a contradiction with Lemma 3.2 (vi). Thus the claim holds. ∎

Claim 8.

$p_{2}+q_{2}+x+y+z\leq 1$ .

Proof.

If $z=1$ , let $B_{0}=\{w\}$ , then by Lemma 3.2 (vi) we infer that $e(\{w\},B_{11}\cup X_{2})=0$ or $e(\{w\},B_{12}\cup Y_{2})=0$ . If $e(\{w\},B_{12}\cup Y_{2})=0$ then one can interchange $X$ and $Y$ , $X_{2}$ and $Y_{2}$ , $X_{11}^{\prime}$ and $Y_{12}^{\prime}$ , $X_{11}^{*}$ and $Y_{12}^{*}$ , $X_{12}^{\prime}$ and $Y_{11}^{\prime}$ . Thus by symmetry we may assume that $e(\{w\},B_{11}\cup X_{2})=0$ . If $z=0$ then $e(B_{0},B_{11}\cup X_{2})=0$ and $e(B_{0},B_{12}\cup Y_{2})=0$ are obvious. Thus in either case we may assume that $e(B_{0},B_{11}\cup X_{2})=0$ .

By Claim 7, $e(X_{11}^{\prime},X_{11}^{*})=0$ . By Claim 6 we have $e(B_{11},X_{2})=0$ . By the definition of $Y_{12}^{*}$ we have $e(X_{12}^{\prime},X_{2})=0$ . Using $e(B_{0},X)=0$ , (5.3) and by Claim 5, Claim 7, we arrive at

e(X^{\prime})=e(X_{11}^{\prime})+e(X_{12}^{\prime})+e(X_{11}^{\prime},X_{12}^{\prime})+e(X_{11}^{*},X_{12}^{\prime})+e(X_{12}^{\prime},X)+e(B_{0},X_{12}^{\prime}).

Similarly, $e(Y^{\prime})=e(Y_{11}^{\prime})+e(Y_{12}^{\prime})+e(Y_{11}^{\prime},Y_{12}^{*})+e(Y_{11}^{\prime},Y_{12}^{\prime})+e(Y_{11}^{\prime},Y)$ .

Since $X_{11}^{\prime},Y_{11}^{\prime}$ is almost balanced partition of $p_{1}$ vertices, we infer that $e(X_{11}^{\prime})+e(Y_{11}^{\prime})\leq\binom{p_{1}}{2}-\lfloor\frac{p_{1}^{2}}{4}\rfloor$ . Note that each vertex in $Y_{11}^{\prime}$ has exactly one neighbor in $Y$ . Thus,

\displaystyle e(X_{11}^{\prime})+e(Y_{11}^{\prime})+e(Y_{11}^{\prime},Y)\leq\binom{p_{1}}{2}-\left\lfloor\frac{p_{1}^{2}}{4}\right\rfloor+\left\lfloor\frac{p_{1}}{2}\right\rfloor=f(p_{1}).

(5.8)

Similarly,

\displaystyle e(X_{12}^{\prime})+e(Y_{12}^{\prime})+e(X_{12}^{\prime},X)\leq f(q_{1}).

(5.9)

It is easy to see that

	$\displaystyle e(X_{11}^{\prime},X_{12}^{\prime})+e(Y_{11}^{\prime},Y_{12}^{\prime})\leq\left\lceil\frac{p_{1}}{2}\right\rceil\left\lfloor\frac{q_{1}}{2}\right\rfloor+\left\lfloor\frac{p_{1}}{2}\right\rfloor\left\lceil\frac{q_{1}}{2}\right\rceil\overset{\eqref{equalities2.1-1}}{\leq}\left\lfloor\frac{p_{1}q_{1}}{2}\right\rfloor,$		(5.10)
	$\displaystyle e(X_{11}^{},X_{12}^{\prime})\leq p_{2}\left\lfloor\frac{q_{1}}{2}\right\rfloor,\ e(Y_{11}^{\prime},Y_{12}^{})\leq\left\lfloor\frac{p_{1}}{2}\right\rfloor q_{2}$		(5.11)

and $e(\{w\},X_{12}^{\prime})\leq z\left\lfloor\frac{q_{1}}{2}\right\rfloor$ . Adding (5.8), (5.9),(5.10) and (5.11), we get

$\displaystyle\gamma_{2}(G)$	$\displaystyle\leq e(X_{11}^{\prime})+e(Y_{11}^{\prime})+e(Y_{11}^{\prime},Y)+e(X_{12}^{\prime})+e(Y_{12}^{\prime})+e(X_{12}^{\prime},X)+e(X_{11}^{\prime},X_{12}^{\prime})$
	$\displaystyle+e(Y_{11}^{\prime},Y_{12}^{\prime})+e(X_{11}^{},X_{12}^{\prime})+e(Y_{11}^{\prime},Y_{12}^{})+e(\{w\},X_{12}^{\prime})$
	$\displaystyle\leq f(p_{1})+f(q_{1})+\left\lfloor\frac{p_{1}q_{1}}{2}\right\rfloor+p_{2}\left\lfloor\frac{q_{1}}{2}\right\rfloor+\left\lfloor\frac{p_{1}}{2}\right\rfloor q_{2}+z\left\lfloor\frac{q_{1}}{2}\right\rfloor$
	$\displaystyle\overset{\eqref{equalities2.5}}{\leq}f(p_{1}+q_{1})+p_{2}\left\lfloor\frac{q_{1}}{2}\right\rfloor+\left\lfloor\frac{p_{1}}{2}\right\rfloor q_{2}+z\left\lfloor\frac{q_{1}}{2}\right\rfloor$	(5.12)
	$\displaystyle\overset{\eqref{equalities2.5}}{\leq}f(p_{1}+q_{1}+p_{2}+q_{2}+x+y+z)-f(p_{2}+q_{2}+x+y+z)$
	$\displaystyle-\left\lceil\frac{(p_{1}+q_{1})(p_{2}+q_{2}+x+y+z)}{2}\right\rceil+p_{2}\left\lfloor\frac{q_{1}}{2}\right\rfloor+\left\lfloor\frac{p_{1}}{2}\right\rfloor q_{2}+z\left\lfloor\frac{q_{1}}{2}\right\rfloor$

Since

\left\lceil\frac{(p_{1}+q_{1})(p_{2}+q_{2}+x+y+z)}{2}\right\rceil\overset{\eqref{equalities2.3}}{\geq}\left\lfloor\frac{q_{1}p_{2}}{2}\right\rfloor+\left\lfloor\frac{p_{1}q_{2}}{2}\right\rfloor+\left\lfloor\frac{q_{1}z}{2}\right\rfloor\overset{\eqref{equalities2.2}}{\geq}p_{2}\left\lfloor\frac{q_{1}}{2}\right\rfloor+\left\lfloor\frac{p_{1}}{2}\right\rfloor q_{2}+z\left\lfloor\frac{q_{1}}{2}\right\rfloor,

we conclude that $\gamma_{2}(G)\leq f(t+1)-f(p_{2}+q_{2}+x+y+z)\leq f(t+1)$ and equality holds if and only if $p_{2}+q_{2}+x+y+z\leq 1$ . ∎

Claim 9.

$z=0$ .

Proof.

By Claim 8, $p_{2}+q_{2}+x+y+z\leq 1$ . Suppose $z=1$ . Then $p_{2}+q_{2}+x+y=0$ . Since $G-B$ is $C_{2k+1}$ -free, $e(G-B)\leq\left\lfloor\frac{(n-t-1)^{2}}{4}\right\rfloor$ . Moreover, by (5.7) we infer $e(B,G-B)=p_{1}+q_{1}=t$ and $e(B)\leq\binom{t+1}{2}$ . Thus

e(G)\leq\left\lfloor\frac{(n-t-1)^{2}}{4}\right\rfloor+\binom{t+1}{2}+t<\left\lfloor\frac{(n-t-1)^{2}}{4}\right\rfloor+\binom{t+2}{2},

contradicting (5.2). ∎

Claim 10.

$x+y=0$ .

Proof.

By Claim 8 we have $p_{2}+q_{2}+x+y+z\leq 1$ . Suppose that $x+y=1$ . Then $p_{2}+q_{2}+z=0$ . By (5.7), $p_{1}+q_{1}=t$ . Using (5.12) we have

\displaystyle\gamma_{2}(G)\leq f(p_{1}+q_{1})=f(t)<f(t+1),

contradicting (5.1). ∎

By Claims 9 and 10, $x=y=z=0$ . Then $X_{11}^{*}=\cup_{v\in Y_{2}}N(v,B_{11})=\emptyset$ and $Y_{12}^{*}=\cup_{u\in X_{2}}N(u,B_{12})=\emptyset$ , that is, $p_{2}=q_{2}=0$ . Therefore $B=B_{11}\cup B_{12}$ .

Claim 11.

$G[B]$ is a clique.

Proof.

Since $e(B,X\cup Y)=e(B_{11},Y)+e(B_{12},X)\leq|B_{11}|+|B_{12}|=t+1$ . Moreover, $G[X\cup Y]$ is bipartite and $|X|+|Y|=n-t-1$ , we have $e(X\cup Y)\leq\left\lfloor\frac{(n-t-1)^{2}}{4}\right\rfloor$ . By $e(G)\geq\left\lfloor\frac{(n-t-1)^{2}}{4}\right\rfloor+\binom{t+2}{2}$ , we have

e(B)\geq e(G)-e(B,X\cup Y)-e(X\cup Y)\geq\binom{t+2}{2}-(t+1)=\binom{t+1}{2}.

Thus $B$ forms a clique in $G$ . ∎

Claim 12.

Either $p=0$ or $q=0$ holds.

Proof.

Suppose that $p\geq 1$ and $q\geq 1$ . Then we show that $|B|\geq 3$ . Otherwise, $|B_{11}|=|B_{12}|=1$ and it follows that $G$ is a bipartite. That is, $\gamma_{2}(G)=0<f(t+1)$ , contradicting (5.1). By Claim 11 and $p\geq 1$ , $q\geq 1$ , we can find a path $uwv$ with $u\in B_{11}$ , $v\in B_{12}$ and $w\in B_{11}\cup B_{12}$ . Since $\deg(u,Y)=1$ , $\deg(v,X)=1$ and $\deg(w,X\cup Y)=1$ , there exist $x\in X$ and $y\in Y$ such that $vx,uy\in E(G)$ , contradicting Lemma 3.2 (vi). Thus the claim holds. ∎

Without loss generality, assume that $p=0$ . By (5.7) and Claims 9, 10 we obtain that $q=t+1$ . Similarly, if $q=0$ , then $p=t+1$ . That is, the equality (5.2) holds if and only if $p=0$ , $q=t+1$ or $p=t+1$ , $q=0$ . That is, $B=B_{11}$ or $B=B_{12}$ . Recall that $B_{11}\cap B_{12}=\emptyset$ . For each $uv\in E(B)$ , there exists a unique vertex $w_{uv}\in G-B$ such that $uw_{uv},vw_{uv}\in E(G)$ . Otherwise, one can find distinct vertices $w_{1},w_{2}$ in $X$ (or $Y$ ) such that $uw_{1},vw_{2}\in E(G)$ , contradicting Lemma 3.2 (v). By Claim 11, $B$ forms a clique in $G$ . Thus, each vertex in $B$ is adjacent to the unique vertex in $G-B$ . Hence $G$ is isomorphic to a subgraph of $H(n,t)$ . By the assumption (5.2) we infer that $G=H(n,t)$ up to isomorphism and the theorem is proven. ∎

6 Concluding remarks

In this paper, we obtain an edge-stability result and a vertex-stability result for $C_{2k+1}$ -free graphs with at least $\frac{(n-2k+1)^{2}}{4}+\binom{2k}{2}$ edges. It should be mention that Korándi, Roberts and Scott [12] proposed a challenging conjecture concerning $C_{2k+1}$ -free graphs with $\frac{n^{2}}{4}-\delta n^{2}$ edges.

For a graph $F$ , an $F$ -blowup is a graph obtained from $F$ by replacing each vertex with an independent set and replacing each edge by a complete bipartite graph. Let $\mathcal{B}(n,k)$ be a class of graphs consisting of all $C_{2k+3}$ -blowups on $n$ vertices.

Conjecture 6.1 ([12]).

Fixed $k\geq 2$ and let $\delta>0$ be small enough. Then for any $\delta>\delta_{0}>0$ and large enough $n$ , the following holds. For every $C_{2k+1}$ -free graph $G$ on $n$ vertices with $(\frac{1}{4}-\delta_{0})n^{2}\geq e(G)\geq(\frac{1}{4}-\delta)n^{2}$ edges, there is a graph $G^{*}\in\mathcal{B}(n,k)$ satisfying $e(G^{*})\geq e(G)$ and $\gamma_{2}(G^{*})\geq\gamma_{2}(G)$ .

We define a class of graphs $\mathcal{G}(n,k)$ on $n$ vertices as follows: a graph $G$ on $n$ vertices is in $\mathcal{G}(n,k)$ if $G$ has exactly one block (2-connected component) being complete bipartite and all the other blocks are cliques of size at most $2k$ .

Motivated by Theorem 1.4 and Conjecture 6.1, we make the following conjecture.

Conjecture 6.2.

Fixed $k\geq 2$ and let $\delta>0$ be small enough. Then for any $\delta>0$ and large enough $n$ , the following holds. For every $C_{2k+1}$ -free graph $G$ on $n$ vertices with $e(G)\geq(\frac{1}{4}-\delta)n^{2}$ edges, there is a graph $G^{*}\in\mathcal{B}(n,k)\cup\mathcal{G}(n,k)$ satisfying $e(G^{*})\geq e(G)$ and $\gamma_{2}(G^{*})\geq\gamma_{2}(G)$ .

References

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	$\displaystyle e(G)$	$\displaystyle\leq e(G-T)+\sum_{0\leq i<\|T\|}\frac{n-i+2}{3}$
		$\displaystyle\leq\|X\|\|Y\|-\|X_{1}\|\|Y_{1}\|+e(X_{1},Y_{1})+\frac{n+2}{3}\|T\|$
		$\displaystyle\leq\frac{(n-\|T\|)^{2}}{4}-\left(\frac{n}{3}-3t\right)^{2}+(k-2)n+\frac{n+2}{3}\|T\|.$

	$\displaystyle\|Y_{1}\|$	$\displaystyle\geq\|N(x_{1},Y_{0})\|+\|N(x_{2},Y_{0})\|-\|N(x_{1},Y_{0})\cup N(x_{2},Y_{0})\|-2$
		$\displaystyle\geq\deg_{G-T}(x_{1})+\deg_{G-T}(x_{2})-\|Y_{0}\|-2$
		$\displaystyle\geq\frac{n}{2}-3\sqrt{3tn}-2.$

	$\displaystyle e(G)$	$\displaystyle\leq e(G-B)+2\sqrt{tn}\|B\|$
		$\displaystyle\leq\|X\|\|Y\|-\|X_{2}\|\|Y_{2}\|+e(X_{2},Y_{2})+2\sqrt{tn}\|B\|$
		$\displaystyle\leq\frac{(n-\|B\|)^{2}}{4}-\left(\frac{n}{2}-\sqrt{3tn}-2\right)^{2}+(k-3)n+2\sqrt{tn}\|B\|.$

	$\displaystyle\|X_{1}\|$	$\displaystyle=\|N(y_{1},X)\|+\|N(y_{2},X)\|-\|N(y_{1},X)\cup N(y_{2},X)\|-2$
		$\displaystyle\geq 2\left(\frac{n}{2}-2\sqrt{tn}\right)-\|X\|-2$
		$\displaystyle\geq\frac{n}{2}-6\sqrt{tn}-2.$

	$\displaystyle\|Y_{2}\|=\|N(x_{1},Y_{1})\cap N(x_{2},Y_{1})\|$	$\displaystyle\geq\deg(x_{1},Y_{1})+\deg(x_{2},Y_{1})-\|Y_{1}\|$
		$\displaystyle\geq 2\left(\frac{n}{2}-2\sqrt{tn}-2k\right)-\|Y_{1}\|$
		$\displaystyle\geq\frac{n}{2}-6\sqrt{tn}-4k.$

A stability result for C2​k+1C_{2k+1}-free graphs

Abstract

1 Introduction

Theorem 1.1 ([6], [9]).

Theorem 1.2 ([10]).

Theorem 1.3.

Theorem 1.4.

Theorem 1.5 ([5]).

Theorem 1.6 ([8]).

2 Some useful facts and inequalities

Fact 2.1.

Proof.

Fact 2.2.

Proof.

Fact 2.3.

Proof.

Lemma 2.4.

Proof.

3 Two structure lemmas

Lemma 3.1.

Proof.

Lemma 3.2.

Proof.

4 The Proof of Theorem 1.3

Proof of Theorem 1.3.

Claim 1.

Proof.

Claim 2.

Proof.

Claim 3.

Proof.

Claim 4.

Proof.

5 The Proof of Theorem 1.4

Proof of Theorem 1.4.

Claim 5.

Proof.

Claim 6.

Proof.

Claim 7.

Proof.

Claim 8.

Proof.

Claim 9.

Proof.

Claim 10.

Proof.

Claim 11.

Proof.

Claim 12.

Proof.

6 Concluding remarks

Conjecture 6.1 ([12]).

Conjecture 6.2.

References

A stability result for $C_{2k+1}$ -free graphs