2. Proof of Theorem 1.4

In this section, we give a proof of Theorem 1.4. Let G be a $K_{2,s}$ -free graph with n vertices, m edges and $\delta (G)\ge 2s-1$ . It suffices to show that for any small real $\varepsilon> 0$ , there exists an integer $n_0> 0$ such that if $n\ge n_0$ , then G has a bisection $(V_1,V_2)$ such that $e(V_i)\le (1/4+\varepsilon )m$ for $i = 1,2$ . Throughout the proof, we tacitly assume that the number of vertices n is large enough. Since $\delta (G)\ge 2s-1$ , we have $m\ge (2s-1)n/2$ , which indicates that m is also large enough.

As a starting point for Problem 1.3, Bollobás and Scott [Reference Bollobás and Scott3] (see also [Reference Scott11]) suggested that one of $\Delta (G)=o(n)$ or $\delta (G)\rightarrow \infty $ might suffice. This was confirmed independently by several authors [Reference Lee, Loh and Sudakov9, Reference Xu, Yan and Yu14, Reference Xu, Yan and Zhang16]. We use the following result of Lee et al. [Reference Lee, Loh and Sudakov9].

It follows from Lemma 2.1 that to prove Theorem 1.4, we need only consider sparse graphs with large maximum degree. More formally, we may assume that

$$ \begin{align*}m< \frac n {\varepsilon^{2}} \quad\text{and}\quad \Delta(G)> \frac{\varepsilon^2 n}2.\end{align*} $$

In fact, for sparse graphs with small maximum degree, Lee et al. [Reference Lee, Loh and Sudakov9] gave the following strengthening of Lemma 2.1. The key benefit is its parametrisation in terms of the number of tight components.

Combining Lemmas 2.1 and 2.2, we see that the main obstacle for Problem 1.3 is the maximum degree condition. To work around this, we use the natural idea of Lee et al. [Reference Lee, Loh and Sudakov9], which was first used by Bollobás and Scott [Reference Bollobás and Scott1] and then by several others. First, partition $V(G)$ into A and $\overline {A}$ , where A consists of certain high degree vertices; then partition A into $A_1$ and $A_2$ with certain properties and partition $\overline {A}$ by Lemma 2.2; finally appropriately combine the vertex subsets of the two partitions. This leads to the following result.

Lemma 2.3 (Lee et al. [Reference Lee, Loh and Sudakov9]).

Given any real constants $C, \varepsilon> 0$ , there exist $\gamma ,n_0> 0$ for which the following holds. Let G be a given graph with $n\ge n_0$ vertices and at most $Cn$ edges, and let $A\subseteq V(G)$ be a set of $\leq \gamma n$ vertices which has already been partitioned into $A_1$ and $A_2$ . Let $\overline {A} =V(G)\setminus A$ , and suppose that every vertex in $\overline {A}$ has degree at most $\gamma n$ (with respect to the full G). Let $\tau $ be the number of tight components in $G[\overline {A}]$ . Then there is a bisection $(V_1,V_2)$ with $A_1\subseteq V_1$ and $A_2\subseteq V_2$ , such that for $i = 1,2$ ,

$$ \begin{align*}e(V_i )\le e(A_i )+\frac{e(A_i,\overline{A})}2+\frac{e(\overline{A})}4-\frac{n-\tau}8+\varepsilon n.\end{align*} $$

Now we use Lemma 2.3 and some additional ideas to prove Theorem 1.4. Let

$$ \begin{align*}A = \{v \in V(G) : d_G (v)\ge n^{3/4} \} \quad\text{and}\quad \overline{A} =V(G)\setminus A.\end{align*} $$

Suppose $A\neq \emptyset $ , otherwise we are already done by Lemma 2.2. Note that

$$ \begin{align*}2m =\sum_{v\in V(G)}d(v)\ge \sum_{v\in A}d(v)\ge |A|n^{3/4},\end{align*} $$

which, together with $m< n/\varepsilon ^{2}$ , yields

(2.1) $$ \begin{align} |A|< \frac{2 n^{1/4}}{\varepsilon^{2}}, \end{align} $$

and hence

$$ \begin{align*}e(A)\le{{|A|}\choose{2}}= O(n^{1/2}).\end{align*} $$

Partition A into $(A_1,A_2)$ such that $e(A_1 ,\overline {A})\ge e(A_2 ,\overline {A})$ and, subject to this,

(2.2) $$ \begin{align} \theta: = e(A_1 ,\overline{A})-e(A_2 ,\overline{A}) \end{align} $$

is minimised. Since $e(A_1,\overline {A})+e(A_2,\overline {A})=e(A,\overline {A})$ , from (2.2), we see that

$$ \begin{align*}e(A_2 ,\overline{A})\le e(A_1 ,\overline{A}) = \frac{e(A,\overline{A})+\theta}2.\end{align*} $$

By (2.1), $|A| =O(n^{1/4})$ . For each $v\in \overline {A}$ , we have $d_G (v)< n^{3/4}$ . Since n is sufficiently large (by choosing $n_0$ large), it follows from Lemma 2.3 (with $C=1/\varepsilon ^2$ ) that G has a bisection $(V_1,V_2)$ with $A_1\subseteq V_1$ and $A_2\subseteq V_2$ , such that for $i = 1,2$ ,

$$ \begin{align*} e(V_i )\le e(A_i )+\frac{e(A,\overline{A})+\theta}4+\frac{e(\overline{A})}4-\frac{n-\tau}8+\varepsilon n \le\frac14\bigg(\theta+\frac{\tau}2-\frac n2\bigg)+\frac{m}{4}+\frac{3\varepsilon}{4} m, \end{align*} $$

where $\tau $ is the number of tight components in $G[\overline {A}]$ . The last inequality holds as ${e(A_i ) = O(n^{1/2} )}$ and $m\ge (2s-1)n/2$ . Then, to prove $e(V_i)\leq (1/4+\varepsilon )m$ , it suffices to show

(2.3) $$ \begin{align} \theta+\frac \tau 2\leq\frac {n} 2+\varepsilon m. \end{align} $$

Now we prove (2.3) through carefully bounding $\theta $ and $\tau $ . Consider the partition $(T,K)$ of $\overline {A}$ , where T consists of all vertices of the tight components in $G[\overline {A}]$ and $K:=\overline {A}\setminus T$ . Let $T_0$ be the set of isolated vertices in $G[\overline {A}]$ and denote $T_1=T\setminus T_0$ . By Lemma 1.1, each component of $G[T_1]$ has at least three vertices. Therefore,

$$ \begin{align*}\tau\le |T_0|+\frac{|T_1|}{3}.\end{align*} $$

Let

$$ \begin{align*}S = \{v\in \overline{A }: d_A(v)\ge2\}.\end{align*} $$

Then $T_0\subset S$ since $\delta (G)\ge 2s-1\ge 3$ . To give a reasonable bound for $\tau $ , we bound $|S|$ by using the condition that G is $K_{2,s}$ -free.

Since G is $K_{2,s}$ -free, any pair of vertices in A has at most $s-1$ common neighbours in G (and thus in S). Through (double) counting the number of $K_{1,2}$ with the 2-degree vertex in S and the two pendent vertices in A, we have

$$ \begin{align*}(s-1){{|A|}\choose{2}}\ge\sum_{v\in S}{{d_A(v)}\choose{2}}\ge |S|.\end{align*} $$

Since $|A|= O(n^{1/4})$ by (2.1), we see that $|S| = O(n^{1/2})$ . This proves Claim 1.

For $s\ge 3$ , we give a better bound for $\tau $ .

Clearly, each vertex in S falls in at most one tight component in $G[\overline {A}]$ . Now we show that each tight component in $G[\overline {A}]$ has a vertex in S, which implies that $\tau \le |S|= O(n^{1/2})$ . Suppose in contrast that $T'$ is a tight component in $G[\overline {A}]$ which does not contain a vertex of S. This means each vertex of $T'$ has at most one neighbour in A. Considering one of the endblocks of $T'$ , by Lemma 1.1, it is an odd clique with minimum degree at least $2s-2$ , and hence contains a $K_{2s-1}$ . Since $s\ge 3$ , it also contains a $K_{2,s}$ , which gives a contradiction. This proves Claim 2.

Now we bound $\theta $ . In the partition $(A_1,A_2)$ of A, since $A\neq \emptyset $ , we have $A_1\neq \emptyset $ .

For otherwise, through moving v from $A_1$ to $A_2$ ,

$$ \begin{align*} \theta'&=e(A_1\setminus \{v\},\overline{A})-e(A_2\cup \{v\},\overline{A}) \\ &=e(A_1,\overline{A})-d_{\overline{A}}(v)-e(A_2,\overline{A})-d_{\overline{A}}(v) \\ &=\theta-2d_{\overline{A}}(v) \\ &>-\theta. \end{align*} $$

However, $\theta '=\theta -2d_{\overline {A}}(v)<\theta $ . This implies that $|\theta '|<\theta $ , which is a contradiction to the optimality of the partition $(A_1,A_2)$ . This proves Claim 3.

For some fixed $v_0\in A_1$ , by Claim 3, $\theta \le d_{\overline {A}}(v_0)$ . We give a bound for $d_{\overline {A}}(v_0)$ .

Denote $X=N_{\overline {A}}(v_0)\setminus S$ and $Y=\overline {A}\setminus X$ . We show that

$$ \begin{align*}|X|\le |Y|.\end{align*} $$

Then the claim follows immediately.

For any connected component B of $G[\overline {A}]$ , no matter whether it belongs to $G[T]$ or $G[K]$ , let $B\cap X=C$ and $B\cap Y=D$ . To prove $|X|\le |Y|$ , it suffices to show $|C|\le |D|.$

Summing up the degrees of all vertices in C,

$$ \begin{align*}\sum_{v\in C}{d(v)}=2e(C)+e(C,A)+e(C,D).\end{align*} $$

Since G contains no $K_{2,s}$ , the maximum degree of $G[C]$ is no more than $s-1$ , which implies

$$ \begin{align*}e(C)\le \frac{(s-1)|C|}{2}.\end{align*} $$

Note that $(C\cap S)\subset (X\cap S)=\emptyset $ , so that $e(C,A)\le |C|$ . Therefore, on the one hand,

$$ \begin{align*} e(C,D) & = \sum_{v\in C}{d(v)}-2e(C)-e(C,A) \\ & \ge (2s-1)|C|-(s-1)|C|-|C| \\ & \ge (s-1)|C|. \end{align*} $$

On the other hand, for any vertex y of D, if $d_C(y)\ge s$ , then a $K_{2,s}$ can be found easily in $G[{v_0\cup y\cup N_C(y)}]$ . Hence, y has at most $s-1$ neighbours in C, which implies

$$ \begin{align*}e(C,D)\le (s-1)|D|.\end{align*} $$

We conclude that

$$ \begin{align*}|C|\le |D|.\end{align*} $$

For the second inequality, by Lemma 1.1, if G is $K_{2,2}$ -free, each block of a tight component in $G[\overline {A}]$ has three vertices. Therefore, $v_0$ has at most one neighbour in each such block. It is easy to see $d_{T}(v_0)\le |T_0|+\tfrac 13|T_1|$ . Through considering nontight components B (restrict B in $G[K]$ ), our proof above implies that ${d_{K}(v_0)\le \tfrac 12|K|+|S|}$ . Thus, $d_{\overline {A}}(v_0)=d_{T}(v_0)+d_{K}(v_0)\le |T_0|+\tfrac 13|T_1|+\tfrac 12|K|+|S|$ when $s=2$ . This proves Claim 4.

When $s\ge 3$ , combining Claims 2–4,

$$ \begin{align*}\theta+\frac {\tau}2\le \frac{|\overline{A}|}2+|S|+\frac{|S|}2\le \frac n2+\varepsilon m.\end{align*} $$

When $s=2$ , by Claims 1 and 3 and the second inequality of Claim 4,

$$ \begin{align*}\theta+\frac {\tau}2\le |T_0|+\frac{|T_1|}{3}+\frac{|K|}2+|S|+\frac{|T_1|/3+O(n^{1/2})}2\le \frac n2+\varepsilon m,\end{align*} $$

where the final inequality follows as $T_0\subset S$ . This completes the proof of (2.3).