Spectral Extrema of Bipartite Graphs: Forbidden an Even Cycle of Specified Length

Abstract

A graph is called H-free if the graph has no subgraph being isomorphic to H. The classical Turán problem aims to determine the maximum size of an H-free graph with given order. In 2010, Nikiforov introduced the spectral Turán-type problem: determine the maximum spectral radius of an H-free graph with given order. He also presented the following conjecture: for sufficiently large $\nu$ and for $k\ge 2$, the graph $K_k\vee (\overline{K_{\nu -k-2}}\cup K_2)$ is the unique $\nu$-vertex $\{C_{2k+1},C_{2k+2}\}$-free graph having the largest spectral radius, and $K_k\vee \overline{K_{\nu -k}}$ is the unique $\nu$-vertex $C_{2k+2}$-free graph having the largest spectral radius. Recently, this conjecture was confirmed by Cioabă, Desai, and Tait. In this contribution, we resolve Nikiforov’s even cycle conjecture for bipartite graphs. Let $\nu$ and k be two integers satisfying $\nu \ge 1$ if $k=1,$ and $\nu \ge {\left({\displaystyle \frac{7^2{2}^{24}k{\left(2k-1\right)}^2{(k+1)}^{20}}{{(2k-3)}^2}}\right)}^{{\displaystyle \frac{k+1}{k-1}}}$ if $k\ge 2$. Then, we show that the complete bipartite graph $K_{k,\nu -k}$ uniquely maximizes the spectral radius over the set of $C_{2k+2}$-free bipartite graphs of order $\nu$.

1. Introduction

Consider a simple graph $G=(V_G,E_G),$ where $V_G$ is the vertex set, and $E_G$ is the edge set. As usual, $\nu :=|V_G|$ is the order, and $m\left(G\right):=|E_G|$ is the size of $G,$ respectively. We use $c\left(G\right)$ to denote the number of components of G. And $K_{a,\nu -a},K_{\nu },C_{\nu }$, and $P_{\nu }$ denote the complete bipartite graph, the complete graph, the cycle, and the path on $\nu$ vertices, respectively. Assume that $\mathcal{H}$ is a family of graphs. Then, a graph is $\mathcal{H}$-free if this graph has no subgraph isomorphic to any element in $\mathcal{H}$. When the set $\mathcal{H}$ is a singleton, say $\left\{H\right\}$, then we write H-free instead of $\mathcal{H}$-free.

The following is the Füredi-Simonovits type problem [1]:

Problem 1.

Let $\mathbb{G}$ be a set of graphs with $G\in \mathbb{G}.$ Given a pair of parameters $(\sigma ,\iota )$ on G, one aims to maximize the second parameter ι under the condition that its first parameter σ is fixed and G is $\mathcal{H}$-free.

If $(\sigma ,\iota )=(\nu ,m(G\left)\right)$, then Problem 1 is the well-known Turán type problem: determine the maximum size, $\mathrm{ex}(\nu ,\mathcal{H})$, of an $\nu$-vertex $\mathcal{H}$-free graph. One calls $\mathrm{ex}(\nu ,\mathcal{F})$ the Turán number. In 1941, Turán [2] determined the exact value of $\mathrm{ex}(\nu ,K_{r+1})$ and the unique corresponding extremal graph, written as $T_r\left(\nu \right)$. This result was extended by the Erdős–Stone–Simonovits theorem [3,4] to some other graphs with given chromatic number, and the exact asymptotics for the Turán numbers of graphs with a chromatic number of at least three were given. One sees that

$$\mathrm{ex}(\nu ,H)=\left(1-{\displaystyle \frac{1}{\chi \left(H\right)-1}}+o\left(1\right)\right){\displaystyle \frac{{\nu }^2}{2}},$$

where $\chi \left(H\right)$ is the chromatic number of graph H. Notice that when H is bipartite, one merely obtains $\mathrm{ex}(\nu ,F)=o\left({\nu }^2\right)$. Indeed, although the even cycle $C_{2k}$ is very simple, it is still very hard to determine the exact asymptotics of $\mathrm{ex}(\nu ,C_{2k})$, which is wide open in the Turán-type problem. From [1], one only knows the order of magnitude for $k\in \{2,3,5\}$.

Given a family of graphs $\mathcal{H},$ the Zarankiewicz number $z(\nu ,\mathcal{H})$ of $\mathcal{H}$ is defined to be the maximum number of edges in an $\nu$-vertex $\mathcal{H}$-free bipartite graph. This study of cycles of bipartite graphs attracts much attention. Let ${\mathcal{C}}_k=\{C_l:3\le l\le k,l\equiv k\left(\mathrm{mod}2\right)\}$. Erdős and Simonovits [5] conjectured that for any family of bipartite graphs, say $\mathcal{H},$ $\mathrm{ex}(\nu ,\mathcal{H}\cup {\mathcal{C}}_k)\sim z(\nu ,\mathcal{H})$ holds for some odd integer k. Keevash, Sudakov, and Verstraëte [6] confirmed this conjecture when $\mathcal{H}={\mathcal{C}}_{2l},l=2,3,5$. It shows that the study of the Zarankiewicz number is an effective approach to investigate the Turán number. Furthermore, we know from [7,8] that development of cycles for bipartite graphs are important to the research on cycles of hypergraphs. For more details on this topic, one may consult the survey paper [1] and references [9,10,11,12,13,14].

Let $A\left(G\right)$ be the adjacency matrix of G. Then, its largest eigenvalue, written by $\lambda \left(G\right),$ is called the spectral radius of G. In Problem 1, if one lets $(\sigma ,\iota )=(\nu ,\lambda (G\left)\right)$, then it is just the spectral Turán-type problem, which is the well-known Brualdi–Solheid–Turán type problem (see [15] for details): determine the maximum spectral radius of an H-free graph of order $\nu .$ Nikiforov [15] was the first to study this problem systematically. Nikiforov [16] showed that $T_r\left(\nu \right)$ is the unique one having the maximum spectral radius over all $\nu$-vertex $K_{r+1}$-free graphs. The Brualdi–Solheid–Turán type problem was studied extensively in the past decade. The references [17,18,19,20,21] can be consulted for more details.

Given two graphs $G_1$ and $G_2,$ one defines $G_1\cup G_2$ to be their disjoint union. The graph $G_1\vee G_2=G_1\cup G_2+\{xy:x\in V_{G_1},y\in V_{G_2}\}$ is the join of $G_1$ and $G_2$. Let $S_{\nu ,k}:=K_k\vee \overline{K_{\nu -k}}$, and $S_{\nu ,k}^{+}:=K_k\vee (\overline{K_{\nu -k-2}}\cup K_2)$. In 2010, Nikiforov [15] posed the spectral even cycle conjecture as follows.

Conjecture 1

([15]). Let $k\ge 2$ and G be a graph of sufficiently large order $\nu$.

(i)

If G is $\{C_{2k+1},C_{2k+2}\}$-free, then $\lambda \left(G\right)\le \lambda \left(S_{\nu ,k}^{+}\right)$ with equality if and only if $G\cong S_{\nu ,k}^{+};$

(ii)

If G is $C_{2k+2}$-free, then $\lambda \left(G\right)\le \lambda \left(S_{\nu ,k}\right)$ with equality if and only if $G\cong S_{\nu ,k}$.

Very recently, Cioabă, Desai, and Tait [22] confirmed Conjecture 1 completely. Notice that $S_{\nu ,k}$ is a non-bipartite graph when $k\ge 2$. Motivated by Conjecture 1 and the studies on Zarankiewicz number, it is natural and interesting to investigate the bipartite analogous version of the Nikiforov’s even cycle conjecture.

One may see that [23] is relevant as a recent application of adjacency-matrix/graph-theoretic consistency analysis, illustrating a further use of the adjacency-matrix formalism central to the present paper. Note that [24] is relevant as a recent study of matrix/tensor representations of network (graph) adjacency structure, connecting to the matrix-analytic (Perron–Frobenius, adjacency-spectrum) tools used throughout this manuscript. Similarly, ref. [25] is relevant as a recent extremal-combinatorics result on edge-labeled trees obtaining Ramsey-type maximum/minimum bounds, methodologically in the same broad family of extremal/threshold problems as Theorem 1 below.

The following is our main result in this contribution, which establishes a spectral sufficient condition to ensure a bipartite graph with sufficiently large order to contain an even cycle of given length, i.e., we determine the unique bipartite graph forbidding an even cycle of described length having maximum spectral radius and sufficiently large order.

Theorem 1.

Let ν and k be two integers satisfying $\nu \ge 1$ if $k=1,$ and $\nu \ge {\left({\displaystyle \frac{7^2{2}^{24}k{\left(2k-1\right)}^2{(k+1)}^{20}}{{(2k-3)}^2}}\right)}^{{\displaystyle \frac{k+1}{k-1}}}$ if $k\ge 2$. Assume that G is a ν-vertex bipartite graph with $\lambda \left(G\right)\ge \sqrt{k(\nu -k)}$. Then, G contains $C_{2k+2}$ as a subgraph, unless $G\cong K_{k,\nu -k}$.

Further on we need the following lemmas.

Lemma 1

([22]). Let M be a non-negative symmetric matrix, $\mathbf{y}$ be a non-negative non-zero vector, and c be a positive constant. If $M\mathbf{y}\ge c\mathbf{y}$ entrywise, then $\lambda \left(M\right)\ge c$.

Lemma 2

([26]). Assume that $G=(X,Y)$ is a bipartite graph satisfying $\left|X\right|=a$ and $\left|Y\right|=b$ with $a\le b$. If G is $P_{2k+3}$-free for some integer $k\ge 2$, then 

$$m\left(G\right)\le \left\{\begin{array}{cc}ab,\hfill & if a\le k;\hfill \\ {(k+1)}^2,\hfill & if a=b=k+1;\hfill \\ a+(b-1)k,\hfill & if b>a=k+1 or k+1<a<2k+1;\hfill \\ 2{(k+1)}^2,\hfill & if a=b=2(k+1);\hfill \\ (a+b-2k)k,\hfill & if b>a=2(k+1) or a>2(k+1).\hfill \end{array}\right.$$

Furthermore, if $a=b=k+1$, then $m\left(G\right)={(k+1)}^2$ holds if and only if $G\cong K_{k+1,k+1}$.

Lemma 3

([27]). Let $G=(X,Y)$ be a bipartite graph with $\left|X\right|=a$ and $\left|Y\right|=b$. If G is $C_{2k+2}$-free with $k\ge 1$, then

$$m\left(G\right)\le \left\{\begin{array}{cc}(2k-1)\left({\left(ab\right)}^{{\displaystyle \frac{k+2}{2k+2}}}+a+b\right),\hfill & if k is even;\hfill \\ (2k-1)\left({\left(ab\right)}^{{\displaystyle \frac{1}{2}}}{a}^{{\displaystyle \frac{1}{k+1}}}+a+b\right),\hfill & if k is odd.\hfill \end{array}\right.$$

The remainder of this contribution is organized as follows. In Section 2, we give the necessary preliminaries involving some structure characterization. In Section 3, we present the proof of our main result (i.e., Theorem 1). Some concluding remarks are given in the Section 4.

2. Some Structural Characterizations

We give some preliminaries in this section, which are indeed some structure characterizations. Given a graph G with $v\in V_G$, denote the neighborhood of v in G by $N_G\left(v\right)$. Then, $d_G\left(v\right):=\left|N_G\left(v\right)\right|$ is the degree of v (in G). As usual, the subindex referring to the underlying graph is dropped if it is clear in the context. The notation $N_i\left(v\right)$ denotes the set of vertices at distance i from $v,i\ge 0$.

For a bipartite graph $H=(X,Y)$, assume $A\subseteq X$ and $B\subseteq Y$, let $H[A,B]$ be the subgraph of H, whose vertex set is $A\cup B$ and edge set consists of all the edges between A and B in H. Let $m_H(A,B)$ (or $m(A,B)$) be the size of $H[A,B]$.

Our first two lemmas are direct consequences of Lemmas 2 and 3, which give an upper boundary for the Zarankiewicz number of $P_{2k+3}$ and $C_{2k+2}$, respectively.

Lemma 4.

$z(\nu ,P_{2k+3})\le k(\nu -k)+1$ holds for $k\ge 2$, and the corresponding extremal graph is just $K_{k+1,k+1}$.

Lemma 5.

$z(\nu ,C_{2k+2})<\left(k-{\displaystyle \frac{1}{2}}\right)\left({\nu }^{{\displaystyle \frac{k+2}{k+1}}}+2\nu \right)$ holds for $k\ge 1$.

The subsequent consequence gives a sufficient condition for a bipartite graph containing a path of specified length with both ends in the same part of the graph.

Theorem 2.

Let $G=(X,Y)$ be a bipartite graph, and let k be a positive integer. If

$$m\left(G\right)>(k-1)\left|X\right|+k\left|Y\right|,$$

then there exists a path of length $2k$ with both ends in X.

Proof. 

Denote $\nu :=|V_G|$. We prove our result by induction on $\nu$. Firstly, we consider $\nu \le 2k$. Then, $\left|X\right|\le k$ or $\left|Y\right|\le k$ holds. Notice that

$$\left|Y\right|\left|X\right|\ge m\left(G\right)>(k-1)\left|X\right|+k\left|Y\right|.$$

(1)

If $\left|X\right|\le k$, then $\left|Y\right|\left|X\right|\le k\left|Y\right|$, a contradiction. Hence, $\left|Y\right|\le k$ and $\left|X\right|>k$. Then, $\left|Y\right|\left|X\right|\le k\left|X\right|$. Together with (1), one has $\left|X\right|>k\left|Y\right|$. Notice that $\left|X\right|<n\le 2k$. Then, $\left|Y\right|=1$, and so $k+(k-1)\left|X\right|<\left|X\right|$, which implies $k=1$. Thus, G is a star and possibly together with some isolated vertices. Applying (1) again yields $m\left(G\right)\ge 2$. Hence, there exists a path of length 2 with both ends in X, as desired.

Next, assume it is true for each bipartite graph of order less than $\nu$. We are to show our result is true for an $\nu$-vertex bipartite graph, say G, where $\nu \ge 2k+1$. We proceed by considering whether G is connected or not.

If G is disconnected, then let $G_1,G_2,\dots ,G_c$ be all the components of G, where $G_i=(X_i,Y_i)$ with $X_i=X\cap V_{G_i}$, and $Y_i=Y\cap V_{G_i},i=1,\dots ,c$. Suppose our result does not hold, i.e., there does not exist a path of length $2k$ with both ends in X. Thus, for each $i=1,\dots ,c$, $G_i$ does not contain any path $P_{2k+1}$ with both ends in $X_i$. By induction, one has for $1\le i\le c$,

$$m\left(G_i\right)\le (k-1)|X_i|+k|Y_i|.$$

(2)

Summing (2), for $i=1,2,\dots ,c$, gives us

$$m\left(G\right)\le (k-1)\left|X\right|+k\left|Y\right|,$$

a contradiction. Thus, there exists a path of length $2k$ with both ends in X.

Now, we consider that G is connected. When G is Hamiltonian, i.e., there exists a spanning cycle $C_{\nu }$ in G, then one sees clearly that G contains a path of length $2k$ with both ends in X. So, we may assume that G is non-Hamiltonian. We distinguish the proof into three possible cases.

Case 1. There is a vertex u in X satisfying $d\left(u\right)\le k-1$. Let $G^{\prime }:=G-u$. Then,

$$m\left(G^{\prime }\right)=m\left(G\right)-d\left(u\right)>(k-1)\left|X\right|+k\left|Y\right|-d\left(u\right)\ge (k-1)\left(\right|X|-1)+k\left|Y\right|.$$

By the inductive assumption, there is a path of length $2k$ with both ends in $X\backslash \left\{u\right\}\subseteq X$, as desired.

Case 2. There is a vertex v in Y satisfying $d\left(v\right)\le k$. By a similar argumentation as that in Case 1, one may show that there is a path of length $2k$ with both ends in X.

Case 3. $d\left(u\right)\ge k$ for all $u\in X$, and $d\left(v\right)\ge k+1$ for all $v\in Y$. Choose a maximal path $P=v_1{v}_2\dots v_p$ in G. Then, $v_1{v}_p\notin E_G$, and $N_G\left(v_1\right)\cup N_G\left(v_p\right)\subseteq V_P$. One considers the following possible subcases.

Subcase 3.1. $v_1,v_p\in X$. Notice that $N_G\left(v_1\right)\subseteq Y$. Then, $|V_P\cap Y|\ge d\left(v_1\right)\ge k$. Therefore, $|V_P\cap X|\ge k+1$. Thereby, $|V_P|\ge 2k+1$; that is, there is a path of length at least $2k$ with both ends in X. Obviously, G also contains a path of length $2k$ with both ends in X.

Subcase 3.2. $v_1\in X$ and $v_p\in Y$ or $v_1\in Y$ and $v_p\in X$. Assume, without loss of generality, that $v_1\in X$, and $v_p\in Y$. Then, $|V_P\cap Y|\ge d\left(v_1\right)+1\ge k+1$, and $|V_P\cap X|\ge d\left(v_p\right)+1\ge k+2$. Therefore, $|V_P|\ge 2k+3$. Hence, there is a path of length at least $2k+2$ with one end in X and the other in Y. Thus, there is a path of length $2k$ with both ends in X.

Subcase 3.3. $v_1,v_p\in Y$. Notice that $N_G\left(v_1\right)\subseteq X$. Then, $|V_P\cap X|\ge d\left(v_1\right)\ge 1+k$. Consequently, $|V_P\cap Y|\ge k+2$. Hence, $|V_P|\ge 2k+3$; that is, there is a path of length at least $2k+2$ with both ends in Y. It follows that there is a path of length $2k$ with both ends in X.

By Cases 1–3, we complete our proof. □

Lemma 6.

Let $G=(X_1,X_2)$ be a bipartite graph on ν vertices. If G is $C_{2k+2}$-free with $k\ge 1,$ then

$$\sum _{u\in V_G}{d}^2\left(u\right)\le k\nu (\nu -1).$$

Proof. 

Consider a vertex u of G. Assume $u\in X_i$ for some $i\in \{1,2\}$. Put $A:=N_G\left(u\right)$ and $B:=X_i\backslash \left\{u\right\}$. Notice that in $G[A\cup B]$, there does not exist a path of length $2k$ with both ends in A. By Lemma 2, one sees

$$m(A,B)\le (k-1)\left|A\right|+k\left|B\right|\le (k-1)d\left(u\right)+k(\nu -d(u)-1)=k(\nu -1)-d\left(u\right).$$

Obviously, $m(A,B)={\sum }_{v\in N_G\left(u\right)}(d\left(v\right)-1)$. Thus,

$$\sum _{v\in N_G\left(u\right)}d\left(v\right)\le k(\nu -1).$$

Consequently,

$$\sum _{u\in V_G}\sum _{v\in N_G\left(u\right)}d\left(v\right)\le \sum _{u\in V_G}k(\nu -1)=k\nu (\nu -1).$$

Note that, for each $v\in V_G$, the term $d\left(v\right)$ appears in the left-hand sum exactly $d\left(v\right)$ times, and so

$$\sum _{u\in V_G}{d}^2\left(u\right)=\sum _{u\in V_G}\sum _{v\in N_G\left(u\right)}d\left(v\right)\le k\nu (\nu -1).$$

This completes the proof. □

3. Proof of Theorem 1

We show Theorem 1 in this section. At first we recall some definitions. The distance between two vertices (in a graph) is the length of a shortest path connecting them. We use the notation $d(v,u)$ to denote the distance between vertices v and u. The diameter of a connected graph G is defined as $\mathrm{diam}\left(G\right)=max\{d(v,u):v,u\in V_G\}.$ We call a shortest path connecting any two vertices v and u satisfying $d(v,u)=\mathrm{diam}\left(G\right)$ a diametrical path. A vertex u of a graph G is called a cut vertex if $c(G-u)>c\left(G\right)$. A block of a graph is a subgraph which is nonseparable and is maximal with respect to this property. If G is separable, the block of G which contains just one cut vertex of G is referred to as its pendant block. The double star $D_{i,j}$ is formed from $K_{1,i}$ and $K_{1,j}$ by adding an edge to join their largest degree vertices.

From ([28], Corollary 1.3.8), one sees that a graph is connected if and only if its spectral radius is a simple eigenvalue with an eigenvector whose coordinates are non-zero and of the same sign. Any positive eigenvector $\mathbf{x}={({\mathbf{x}}_1,\dots ,{\mathbf{x}}_{\nu })}^T$ corresponding to the spectral radius is called a Perron vector. As usual, for a vector $\mathbf{y}={({\mathbf{y}}_1,\dots ,{\mathbf{y}}_{\nu })}^T$ it is assumed that the coordinate ${\mathbf{y}}_i$ (which is also called the weight) corresponds to the vertex $v_i(1\le i\le n)$.

Proof of Theorem 1.

Choose a $C_{2k+2}$-free bipartite graph, say $H_k=(X,Y),$ on $\nu$ vertices such that $\lambda \left(H_k\right)$ is maximum, where $k\ge 1$. Clearly, $\lambda \left(K_{k,\nu -k}\right)=\sqrt{k(\nu -k)}$, and so, it is sufficient to prove $H_k\cong K_{k,\nu -k}$.

For convenience, put $\lambda :=\lambda \left(H_k\right)$. One may see that $H_k$ must be connected. Otherwise, add an edge e to connect two disconnected components of $H_k$. The resultant graph $H_k+e$ does not create any cycles, and $\lambda \left(H_k\right)<\lambda (H_k+e)$, a contradiction. Assume that $\mathbf{x}$ is the Perron vector of $H_k$ whose maximum entry ${\mathbf{x}}_{u^{*}}$ equals 1. Assume, without loss of generality, that $u^{*}\in X$.

Firstly, we consider $k=1$. That is, $H_1$ forbids $C_4$. Choose a diametrical path, say $v_0{v}_1{v}_2\dots v_d$, of $H_1$; here, $d=\mathrm{diam}\left(H_1\right)$. We show the following claim.

Claim 1.

$d\le 4$, and there exists at most one cut vertex in $H_1$.

Proof of Claim 1.

We show Claim 1 by contradiction. Suppose $d\ge 5$. Let $G_1=H_1+v_0{v}_5$. Hence, $G_1$ is bipartite, and $\lambda \left(G_1\right)>\lambda \left(H_1\right)$. By the choice of $H_1$, we know that $G_1$ contains $C_4$ as a subgraph with $v_0{v}_5\in E_{C_4}$. Hence, there exists a path of length 3 that connects $v_0$ and $v_5$ in $H_1$. It follows that the distance between $v_0$ and $v_d$ is less than d, a contradiction.

Now choose two cut vertices u and v such that $d(u,v)$ is as large as possible. Assume, without loss of generality, that $x_u\ge x_v$. Then, move a pendant block of $H_1$ from v to u, and the resultant graph is denoted by $G_1^{\prime }$. One sees that there does not exist odd cycle or $C_4$ in $G_1^{\prime }.$ That is to say, $G_1^{\prime }$ is bipartite, forbidding $C_4$ and having $\nu$ vertices. Another fact should be considered, that is, with the help of the Rayleigh quotient, we obtain $\lambda \left(G_1^{\prime }\right)>\lambda \left(H_1\right)$, a contradiction. □

If $d=3$, then the double star $D_{d_{H_1}\left(v_1\right)-1,d_{H_1}\left(v_2\right)-1}$ is a spanning subgraph of $H_1$. Recall that $H_1$ is a $C_4$-free bipartite graph. Then, $H_1\cong D_{d_{H_1}\left(v_1\right)-1,d_{H_1}\left(v_2\right)-1}$, and so $H_1$ has two cut vertices, which contradicts Claim 1.

If $d=4$, then $H_1^{\prime }$ is a spanning subgraph of $H_1$ (see Figure 1). Since $H_1$ is a $C_4$-free bipartite graph, each vertex in $(N_{H_1}\left(v_1\right)\cup N_{H_1}\left(v_3\right))\backslash \left\{v_2\right\}$ is a pendant vertex in $H_1$. It follows that $H_1$ has at least two cut vertices, which also contradicts Claim 1. Applying Claim 1 again yields $d\le 2$, and, therefore, $H_1\cong K_{1,\nu -1}$, as desired.

Figure 1. Graph $H_1^{\prime }$ with labelled vertices $v_0,v_1,v_2,v_3$ and $v_4$.

In the remainder of our proof, we need only consider $k\ge 2$ and $\nu \ge {\left({\displaystyle \frac{7^2{2}^{24}k{\left(2k-1\right)}^2{(k+1)}^{20}}{{(2k-3)}^2}}\right)}^{{\displaystyle \frac{k+1}{k-1}}}$. We proceed by showing the following claims.

Claim 2.

$\sqrt{k(\nu -k)}\le \lambda \left(H_k\right)\le \sqrt{k(\nu -1)}$ and $min\left\{\right|X|,|Y\left|\right\}\ge k$.

Proof of Claim 2.

Clearly, $K_{k,\nu -k}$ is $C_{2k+2}$-free. Hence, $\lambda \left(H_k\right)\ge \lambda \left(K_{k,\nu -k}\right)=\sqrt{k(\nu -k)}$.

Suppose that $min\left\{\right|X|,|Y\left|\right\}\le k-1$. Since $H_k$ is a subgraph of $K_{\left|X\right|,\left|Y\right|}$, one has $\lambda \left(H_k\right)\le \lambda \left(K_{\left|X\right|,\left|Y\right|}\right)=\sqrt{\left|Y\right|\left|X\right|}<\sqrt{k(\nu -k)}$, a contradiction. Thus, $min\left\{\right|X|,|Y\left|\right\}\ge k$ and so $max\left\{\right|X|,|Y\left|\right\}\le \nu -k$.

Let $u\in V_{H_k}$. Since $H_k$ is $C_{2k+2}$-free, in the graph $H_k[N\left(u\right),N_2\left(u\right)]$, there is no paths of length $2k$ with both ends in $N\left(u\right)$. In view of Lemma 2, one has

$$m(N\left(u\right),N_2\left(u\right))\le (k-1)\left|N\left(u\right)\right|+k|N_2\left(u\right)|\le (k-1)d\left(u\right)+k(\nu -d\left(u\right)-1)=k(\nu -1)-d\left(u\right).$$

(3)

Notice that $A^2\left(H_k\right)\mathbf{x}={\lambda }^2\mathbf{x}$. Together with (3), we obtain

$$\begin{array}{cc}{\lambda }^2{\mathbf{x}}_u\hfill & =\sum _{v\in N\left(u\right)}\lambda {\mathbf{x}}_v=\sum _{v\in N\left(u\right)}\sum _{w\in N\left(v\right)}{\mathbf{x}}_w\hfill \\ & =d\left(u\right){\mathbf{x}}_u+\sum _{v\in N\left(u\right)}\sum _{w\in N_2\left(u\right)\cap N\left(v\right)}{\mathbf{x}}_w\hfill \\ & \le d\left(u\right)+m(N\left(u\right),N_2\left(u\right))\le k(\nu -1).\hfill \end{array}$$

(4)

It follows from ${\mathbf{x}}_{u^{*}}=1$ that ${\lambda }^2={\lambda }^2{\mathbf{x}}_{u^{*}}\le k(\nu -1)$, that is, $\lambda \left(H_k\right)\le \sqrt{k(\nu -1)}$, as desired. □

With foresight, we define $\eta ,ϵ$ and $\alpha$ as follows:

$$\begin{array}{cc}\hfill & \eta :={\displaystyle \frac{1}{2k}}=min\left\{{\displaystyle \frac{1}{2k}},1-{\displaystyle \frac{1}{16k^3}}\right\},\hfill \\ \hfill & ϵ:={\displaystyle \frac{1}{16{(1+k)}^5}}<min\left\{{\displaystyle \frac{\eta }{3}},{\displaystyle \frac{1}{16k^3}},{\displaystyle \frac{\eta }{8k^3(k+2)+2}}\right\},\hfill \\ \hfill & \alpha :={\displaystyle \frac{1}{32{(1+k)}^5}}\le min\left\{{\displaystyle \frac{ϵ}{2}},{\displaystyle \frac{1}{4k}}\right\}.\hfill \end{array}$$

Define L to be the set of vertices whose weight is greater than $\alpha$, and let S be the set of vertices whose weight is no greater than $\alpha$, i.e.,

$$L:=\{u\in V_{H_k}:{\mathbf{x}}_u>\alpha \}\mathrm{and}S:=V_{H_k}\backslash L=\{u\in V_{H_k}:{\mathbf{x}}_u\le \alpha \}.$$

Define

$$M:=\left\{u\in V_{H_k}:{\mathbf{x}}_u\ge {\displaystyle \frac{\alpha }{3}}\right\}\mathrm{and}{L}^{\prime }:=\left\{u\in V_{H_k}:{\mathbf{x}}_u\ge \eta \right\}.$$

Let u be in $V_{H_k}$. Then, put

$$L_i\left(u\right):=N_i\left(u\right)\cap L,S_i\left(u\right):=N_i\left(u\right)\cap S,\mathrm{and}{M}_i\left(u\right):=N_i\left(u\right)\cap M.$$

Claim 3.

$\left|L\right|<{\displaystyle \frac{\left(2k-1\right)\left({\nu }^{\frac{k+2}{k+1}}+2\nu \right)}{\sqrt{k(\nu -k)}\alpha }}$ and $\left|M\right|<{\displaystyle \frac{3\left(2k-1\right)\left({\nu }^{\frac{k+2}{k+1}}+2\nu \right)}{\sqrt{k(\nu -k)}\alpha }}$.

Proof of Claim 3.

For any vertex $u\in L$, based on Claim 2, one has

$$\sqrt{k(\nu -k)}\alpha <\sqrt{k(\nu -k)}{\mathbf{x}}_u\le \lambda {\mathbf{x}}_u=\sum _{w\in N\left(u\right)}{\mathbf{x}}_w\le d\left(u\right).$$

Summing the above inequalities for all $u\in L$, together with Lemma 5, gives us

$$\sqrt{k(\nu -k)}\alpha \left|L\right|<\sum _{u\in L}\lambda {\mathbf{x}}_u\le \sum _{u\in L}d\left(u\right)\le 2m\left(H_k\right)\le \left(2k-1\right)\left({\nu }^{{\displaystyle \frac{k+2}{k+1}}}+2\nu \right).$$

Consequently, $\left|L\right|<{\displaystyle \frac{\left(2k-1\right)\left({\nu }^{\frac{k+2}{k+1}}+2\nu \right)}{\sqrt{k(\nu -k)}\alpha }}$.

Similarly, if one replaces $\alpha$ with ${\displaystyle \frac{\alpha }{3}}$ throughout the deduction process as above, we may obtain the upper bound for $\left|M\right|$, which is omitted. □

Claim 4.

$d\left(u\right)\ge {\displaystyle \frac{\alpha (\nu -k)}{k}}$ for all $u\in L$ and $\left|L\right|\le {\displaystyle \frac{2k^3}{{\alpha }^2}}$.

Proof of Claim 4.

Firstly, we show the first part by contradiction if there exists a vertex u in L satisfying $d\left(u\right)<{\displaystyle \frac{\alpha (\nu -k)}{k}}$. For convenience, assume ${\mathbf{x}}_u=c$. Based on Claim 2 and (4), one has

$$k(\nu -k)c\le {\lambda }^2{\mathbf{x}}_u=d\left(u\right)c+\sum _{v\in N\left(u\right)}\sum _{w\in N_2\left(u\right)\cap N\left(v\right)}{\mathbf{x}}_w<{\displaystyle \frac{\alpha (\nu -k)c}{k}}+\sum _{v\in N\left(u\right)}\sum _{w\in N_2\left(u\right)\cap N\left(v\right)}{\mathbf{x}}_w,$$

which implies that

$$\left(k-{\displaystyle \frac{\alpha }{k}}\right)(\nu -k)c\le \sum _{v\in N\left(u\right)}\sum _{w\in N_2\left(u\right)\cap N\left(v\right)}{\mathbf{x}}_w.$$

Put $Q:={\sum }_{v\in N\left(u\right)}{\sum }_{w\in N_2\left(u\right)\cap N\left(v\right)}{\mathbf{x}}_w$. Next, we show the subsequent fact.

Fact 1.

There exist at least $\left({\displaystyle \frac{2k}{3}}-{\displaystyle \frac{1}{k}}\right)(\nu -k)c$ terms ${\mathbf{x}}_w$ with $w\in M_2\left(u\right)$ in the sum Q.

Proof of Fact 1.

Suppose to the contrary that there are less than $\left({\displaystyle \frac{2k}{3}}-{\displaystyle \frac{1}{k}}\right)(\nu -k)c$ terms ${\mathbf{x}}_w$ with $w\in M_2\left(u\right)$ in the sum Q. Therefore,

$$\begin{array}{cc}\left(k-{\displaystyle \frac{\alpha }{k}}\right)(\nu -k)c\hfill & \le Q=\sum _{v\in N\left(u\right)}\sum _{w\in M_2\left(u\right)\cap N\left(v\right)}{\mathbf{x}}_w+\sum _{v\in N\left(u\right)}\sum _{w\in (S_2\left(u\right)\backslash M_2\left(u\right))\cap N\left(v\right)}{\mathbf{x}}_w\hfill \\ & <\left({\displaystyle \frac{2k}{3}}-{\displaystyle \frac{1}{k}}\right)(\nu -k)c+m(N\left(u\right),S_2\left(u\right)\backslash M_2\left(u\right)){\displaystyle \frac{\alpha }{3}}.\hfill \end{array}$$

(5)

Notice that $H_k[N\left(u\right),S_2\left(u\right)\backslash M_2\left(u\right)]$ is an induced subgraph of $H_k$ and that there is no path of length $2k+2$ in $H_k[N\left(u\right),S_2\left(u\right)\backslash M_2\left(u\right)]$. By Lemma 4, one has $m(N\left(u\right),S_2\left(u\right)\backslash M_2\left(u\right))\le k(\nu -1-k)+1\le k(\nu -k)$. Together with (5) and $\alpha <c$, one has ${\displaystyle \frac{1}{k}}-{\displaystyle \frac{\alpha }{k}}<0$, i.e., $\alpha >1$, a contradiction.

We finish the proof of Fact 1. □

By Fact 1, one sees

$$m(N\left(u\right),M_2\left(u\right))\ge \left({\displaystyle \frac{2k}{3}}-{\displaystyle \frac{1}{k}}\right)(\nu -k)c>\left({\displaystyle \frac{2k}{3}}-{\displaystyle \frac{1}{k}}\right)(\nu -k)\alpha .$$

(6)

Another fact should be considered. In graph $H_k[N\left(u\right),M_2\left(u\right)]$, there is no path of length $2k$ with both ends in $N\left(u\right)$. Based on Lemma 2, one has

$$m(N\left(u\right),M_2\left(u\right))\le (k-1)\left|N\left(u\right)\right|+k|M_2\left(u\right)|<{\displaystyle \frac{\alpha (\nu -k)(k-1)}{k}}+k\left|M_2\left(u\right)\right|.$$

(7)

Combining (6) and (7), we get

$$\begin{array}{c}\hfill |M_2\left(u\right)|>\left({\displaystyle \frac{2}{3}}-{\displaystyle \frac{1}{k^2}}\right)(\nu -k)\alpha -{\displaystyle \frac{\alpha (\nu -k)(k-1)}{k^2}}=\left({\displaystyle \frac{2}{3}}-{\displaystyle \frac{1}{k}}\right)(\nu -k)\alpha .\end{array}$$

Then, together with Claim 3, one has

$${\displaystyle \frac{3\left(2k-1\right)\left({\nu }^{\frac{k+2}{k+1}}+2\nu \right)}{\sqrt{k(\nu -k)}\alpha }}>\left({\displaystyle \frac{2}{3}}-{\displaystyle \frac{1}{k}}\right)(\nu -k)\alpha ,$$

it is equivalent to

$${\nu }^{{\displaystyle \frac{k+2}{k+1}}}+2\nu >\left({\displaystyle \frac{2}{3}}-{\displaystyle \frac{1}{k}}\right){\displaystyle \frac{{\alpha }^2\sqrt{k}}{3\left(2k-1\right)}}{(\nu -k)}^{{\displaystyle \frac{3}{2}}},$$

which implies

$$3{\nu }^{{\displaystyle \frac{k+2}{k+1}}}>{\displaystyle \frac{{\alpha }^2\sqrt{k}(2k-3)}{9k\left(2k-1\right)}}{(\nu -k)}^{{\displaystyle \frac{3}{2}}}.$$

That is,

$${\nu }^{{\displaystyle \frac{1-k}{2(k+1)}}}>{\displaystyle \frac{{\alpha }^2(2k-3)}{27\sqrt{k}\left(2k-1\right)}}{\left(1-{\displaystyle \frac{k}{\nu }}\right)}^{{\displaystyle \frac{3}{2}}}.$$

(8)

Note that $\nu >50k>{\displaystyle \frac{k}{1-{\left(\frac{27}{28}\right)}^{\frac{2}{3}}}}$. By (8), we obtain

$${\nu }^{{\displaystyle \frac{1-k}{2(k+1)}}}>{\displaystyle \frac{{\alpha }^2(2k-3)}{28\sqrt{k}\left(2k-1\right)}}.$$

Consequently,

$$\nu <{\left({\displaystyle \frac{784k{\left(2k-1\right)}^2}{{(2k-3)}^2{\alpha }^4}}\right)}^{{\displaystyle \frac{k+1}{k-1}}}={\left({\displaystyle \frac{7^2{2}^{24}k{\left(2k-1\right)}^2{(k+1)}^{20}}{{(2k-3)}^2}}\right)}^{{\displaystyle \frac{k+1}{k-1}}},$$

which contradicts our assumption. Thus, $d\left(u\right)\ge {\displaystyle \frac{\alpha (\nu -k)}{k}}$ for each $u\in L$.

Now we show the second part of Claim 4. By Lemma 6, we obtain

$$\left|L\right|{\displaystyle \frac{{\alpha }^2{(\nu -k)}^2}{k^2}}\le \sum _{u\in L}{d}^2\left(u\right)\le k\nu (\nu -1).$$

Therefore,

$$\left|L\right|\le {\displaystyle \frac{k^{3}n(\nu -1)}{{\alpha }^2{(\nu -k)}^2}}\le {\displaystyle \frac{2k^3}{{\alpha }^2}}.$$

(9)

The last inequality in (9) follows by $\nu >4k-1>{\displaystyle \frac{4k-1+\sqrt{8k^2-8k+1}}{2}}$. □

Indeed, we may refine the lower bound on the degree of vertices in L.

Claim 5.

$d\left(u\right)\ge (\nu -k)({\mathbf{x}}_u-ϵ)$ for all u in L.

Proof of Claim 5.

We prove Claim 5 by contradiction. Suppose that there is a vertex $u\in L$ such that $d\left(u\right)<(\nu -k)({\mathbf{x}}_u-ϵ)$. For convenience, put ${\mathbf{x}}_u=c$. Then, $c>ϵ$. In view of (4), one has

$$k(\nu -k)c\le {\lambda }^2{\mathbf{x}}_u\le d\left(u\right)c+m(N\left(u\right),L_2\left(u\right))+m(N\left(u\right),S_2\left(u\right))\alpha .$$

Notice that $H_k[N\left(u\right),L_2\left(u\right)]$ and $H_k[N\left(u\right),S_2\left(u\right)]$ do not contain $P_{2k+1}$ with both ends in $N\left(u\right)$. Based on Lemma 2, we have

$$m(N\left(u\right),L_2\left(u\right))\le (k-1)\left|N\left(u\right)\right|+k|L_2\left(u\right)|$$

and

$$m(N\left(u\right),S_2\left(u\right))\le (k-1)\left|N\left(u\right)\right|+k|S_2\left(u\right)|\le (k-1)|N\left(u\right)|+k(\nu -\left|N\left(u\right)\right|-|L_2\left(u\right)\left|\right)=k(\nu -|L_2\left(u\right)\left|\right)-\left|N\left(u\right)\right|.$$

Consequently, by Claim 4 and the fact that $|L_2\left(u\right)|\le |L|$, we get

$$\begin{array}{cc}\hfill k(\nu -k)c& <d\left(u\right)c+(k-1)\left|N\left(u\right)\right|+k|L_2\left(u\right)|+(k(\nu -|L_2\left(u\right)\left|\right)-\left|N\left(u\right)\right|)\alpha \hfill \\ \hfill & <(\nu -k)(c-ϵ)(k-1+c-\alpha )+k\nu \alpha +k(1-\alpha )|L_2\left(u\right)|\hfill \\ \hfill & \le (\nu -k)(c-ϵ)(k-1+c-\alpha )+k(\nu -k)\alpha +k^2\alpha +{\displaystyle \frac{2k^4(1-\alpha )}{{\alpha }^2}},\hfill \end{array}$$

that is,

$$(\nu -k)[(c-ϵ)(1-c+\alpha )+k(ϵ-\alpha )]<k^2\alpha +{\displaystyle \frac{2k^4(1-\alpha )}{{\alpha }^2}},$$

which implies

$$(\nu -k)(ϵ-\alpha )<k\alpha +{\displaystyle \frac{2k^3(1-\alpha )}{{\alpha }^2}}.$$

Recall that $\alpha ={\displaystyle \frac{ϵ}{2}}$. Hence,

$$\nu <k+{\displaystyle \frac{2k^3(1-\alpha )}{{\alpha }^3}}<{\left({\displaystyle \frac{784k{\left(2k-1\right)}^2}{{(2k-3)}^2{\alpha }^4}}\right)}^{{\displaystyle \frac{k+1}{k-1}}},$$

which contradicts our assumption. □

Claim 6.

$L^{\prime }\subseteq X$ and $L^{\prime }\backslash \left\{u^{*}\right\}\subseteq L_2\left(u^{*}\right)$.

Proof of Claim 6.

Let u be any vertex in $L^{\prime }\backslash \left\{u^{*}\right\}$. Based on Claim 5, one has

$$d\left(u^{*}\right)\ge (\nu -k)(1-ϵ)\mathrm{and}d\left(u\right)\ge (\nu -k)(\eta -ϵ).$$

(10)

Suppose that $u\in L^{\prime }\cap Y$. Then, $N\left(u^{*}\right)\cap N\left(u\right)=\varnothing$. Recall that $\eta >3ϵ$ and $ϵ=2\alpha$, one has

$$n\ge d\left(u\right)+d\left(u^{*}\right)\ge (\nu -k)(1+\eta -2ϵ)>(\nu -k)(1+ϵ)=(\nu -k)(1+2\alpha ),$$

which implies $\nu <{\displaystyle \frac{k}{2\alpha }}+k<{\left({\displaystyle \frac{784k{\left(2k-1\right)}^2}{{(2k-3)}^2{\alpha }^4}}\right)}^{{\displaystyle \frac{k+1}{k-1}}}$. This contradicts our assumption. Thereby, $L^{\prime }\subseteq X$.

In view of Claim 2 and (10), we know $|Y\backslash N\left(u^{*}\right)|\le ϵ(\nu -k)$, and, for every vertex z in $L^{\prime }$, one has $d\left(z\right)\ge (\nu -k)(\eta -ϵ)>2ϵ(\nu -k)$. Recall that $L^{\prime }\subseteq X$. Hence, $L^{\prime }\backslash \left\{u^{*}\right\}\subseteq L_2\left(u^{*}\right)$, as desired. □

Claim 7.

$m\left(N\right(u),L)\le (k+ϵ)(\nu -k)$ for all vertices $u\in L^{\prime }$.

Proof of Claim 7.

Notice that there does not exist a path of length $2k$ in $H_k[N\left(u\right),L\backslash \left\{u\right\}]$ such that both ends of the path lie in $N\left(u\right)$. Based on Lemma 4 and Claim 4, we have

$$m(N\left(u\right),L)=m(N\left(u\right),L\backslash \left\{u\right\})+\left|N\left(u\right)\right|\le (k-1)\left|N\left(u\right)\right|+k\left|L\right|+\left|N\left(u\right)\right|\le k(\nu -k)+{\displaystyle \frac{2k^4}{{\alpha }^2}}<(k+ϵ)(\nu -k).$$

(11)

The last inequality in (11) follows by $ϵ=2\alpha$ and $\nu \ge {\left({\displaystyle \frac{784k{\left(2k-1\right)}^2}{{(2k-3)}^2{\alpha }^4}}\right)}^{{\displaystyle \frac{k+1}{k-1}}}>{\displaystyle \frac{k^4}{{\alpha }^3}}+k={\displaystyle \frac{2k^4}{{\alpha }^2ϵ}}+k$. □

Claim 8.

${\mathbf{x}}_u\ge 1-{\displaystyle \frac{1}{16k^3}}$ and $d\left(u\right)\ge (\nu -k)\left(1-{\displaystyle \frac{1}{8k^3}}\right)$ for all vertices $u\in L^{\prime }$. Moreover, $|L^{\prime }|=k$.

Proof of Claim 8.

Let u be a vertex in $L^{\prime }$. Suppose that ${\mathbf{x}}_u<1-{\displaystyle \frac{1}{16k^3}}$. In view of Claim 6, one has $u\in L_2\left(u^{*}\right)$. Notice that $H_k[N\left(u^{*}\right),N_2\left(u^{*}\right)]$ contains no path of length $2k+2$. Combining with Lemma 4 gives us

$$m(N\left(u^{*}\right),S_2\left(u^{*}\right))\le m(N\left(u^{*}\right),N_2\left(u^{*}\right))\le k(\nu -1-k)+1<k(\nu -k).$$

Refining (4) applied to the vertex $u^{*}$ one has

$$\begin{array}{cc}k(\nu -k)\hfill & \le d\left(u^{*}\right)+m(N\left(u^{*}\right),L_2\left(u^{*}\right)\backslash \left\{u\right\})+|N\left(u^{*}\right)\cap N\left(u\right)|{\mathbf{x}}_u+m(N\left(u^{*}\right),S_2\left(u^{*}\right))\alpha \hfill \\ & <m(N\left(u^{*}\right),L)-|N\left(u^{*}\right)\cap N\left(u\right)|+|N\left(u^{*}\right)\cap N\left(u\right)|\left(1-{\displaystyle \frac{1}{16k^3}}\right)+\alpha k(\nu -k)\hfill \\ & \le (k+ϵ)(\nu -k)-{\displaystyle \frac{|N\left(u^{*}\right)\cap N\left(u\right)|}{16k^3}}+\alpha k(\nu -k),\hfill \end{array}$$

(12)

where (12) follows by Claim 7. Notice that $\alpha ={\displaystyle \frac{ϵ}{2}}$. Then,

$$|N\left(u^{*}\right)\cap N\left(u\right)|<16k^3(\nu -k)(ϵ+\alpha k)=8k^3ϵ(\nu -k)(k+2).$$

On the other hand, since $\{u^{*},u\}\subseteq L^{\prime }\subseteq X$, by Claims 2 and 5, one has

$$|N\left(u\right)\cap N\left(u^{*}\right)|=|N\left(u\right)|+|N\left(u^{*}\right)|-|N\left(u\right)\cup N\left(u^{*}\right)|\ge (\nu -k)(1+\eta -2ϵ)-(\nu -k)=(\nu -k)(\eta -2ϵ).$$

Consequently,

$$8k^3ϵ(\nu -k)(k+2)>(\nu -k)(\eta -2ϵ),$$

i.e., $ϵ\ge {\displaystyle \frac{\eta }{8k^3(k+2)+2}}$, this contradicts our choice for $ϵ$. Therefore, ${\mathbf{x}}_u\ge 1-{\displaystyle \frac{1}{16k^3}}$ holds for each $u\in L^{\prime }$.

Since $ϵ\le {\displaystyle \frac{1}{16k^3}}$, by Claim 5, one has

$$d\left(u\right)\ge (\nu -k)({\mathbf{x}}_u-ϵ)\ge (\nu -k)\left(1-{\displaystyle \frac{1}{16k^3}}-ϵ\right)\ge (\nu -k)\left(1-{\displaystyle \frac{1}{8k^3}}\right),$$

as desired.

Now, we prove $|L^{\prime }|=k$. If this is not true, then either $|L^{\prime }|\ge k+1$, or $|L^{\prime }|\le k-1$. For the former, there is a vertex subset $L^{″}$ of $L^{\prime }$ satisfying $|L^{″}|=k+1$. Note that, for each vertex x in $L^{″}$, one has $d\left(x\right)\ge (\nu -k)(1-{\displaystyle \frac{1}{8k^3}})$. Then there are at most ${\displaystyle \frac{k+1}{8k^3}}(\nu -k)$ vertices in Y each of which is not adjacent to some vertices in $L^{″}$. Clearly, $ϵ<{\displaystyle \frac{1}{2}}$. Based on Claim 5, one sees

$$\begin{array}{cc}\left|Y\right|-{\displaystyle \frac{(\nu -k)(k+1)}{8k^3}}\hfill & \ge d\left(u^{*}\right)-{\displaystyle \frac{(\nu -k)(k+1)}{8k^3}}\hfill \\ & \ge (\nu -k)\left[(1-ϵ)-{\displaystyle \frac{k+1}{8k^3}}\right]\hfill \\ & \ge {\displaystyle \frac{1}{4}}(\nu -k)\ge k+1.\hfill \end{array}$$

(13)

The last inequality in (13) follows by $\nu \ge 5k+4$. That is to say, there is a vertex subset $\check{Y}$ of Y such that $|\check{Y}|\ge k+1$, and each vertex in $\check{Y}$ is adjacent to each one in $L^{″}$. Therefore, $K_{k+1,k+1}$ is a subgraph of $H_k$; this is impossible.

In the remainder of our proof, we consider $|L^{\prime }|\le k-1$. Based on Claim 7 and (4), one has

$$\begin{array}{cc}\hfill k(\nu -k)& \le d\left(u^{*}\right)+m(N\left(u^{*}\right),L^{\prime }\backslash \left\{u^{*}\right\})+m(N\left(u^{*}\right),L\backslash L^{\prime })\eta +m(N\left(u^{*}\right),S_2\left(u^{*}\right))\alpha \hfill \\ \hfill & \le m(N\left(u^{*}\right),L^{\prime })+m(N\left(u^{*}\right),L\backslash L^{\prime })\eta +\alpha k(\nu -k)\hfill \\ \hfill & \le (k-1)(\nu -k)+(k+ϵ)(\nu -k)\eta +\alpha k(\nu -k).\hfill \end{array}$$

Together with $\alpha <{\displaystyle \frac{1}{4k}}$ and $\eta ={\displaystyle \frac{1}{2k}}$, we get

$$1\le k\alpha +\eta (k+ϵ)<{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2k}}\le 1,$$

which is impossible. Thus, $|L^{\prime }|=k$.

We finish the proof of Claim 8. □

By Claim 8, the cardinality of $L^{\prime }$ equals k, and every vertex, say z, in $L^{\prime }$ satisfies $d\left(z\right)\ge (\nu -k)\left(1-{\displaystyle \frac{1}{8k^3}}\right)$. Thus, it follows that the cardinality of the common neighborhood of $L^{\prime }$ is at least $(\nu -k)\left(1-{\displaystyle \frac{1}{8k^2}}\right)$. That is, at most $(\nu -k){\displaystyle \frac{1}{8k^2}}$ vertices in Y are not adjacent to all of $L^{\prime }$. Let

$$R=\left\{r\in V_{H_k}\backslash L^{\prime }:|N\left(r\right)\cap L^{\prime }|\le k-1\right\}\mathrm{and}W=V_{H_k}\backslash (L^{\prime }\cup R).$$

Clearly, $V_{H_k}=L^{\prime }\cup R\cup W$. Furthermore,

$$\left|R\right|\le \nu -k-(\nu -k)\left(1-{\displaystyle \frac{1}{8k^2}}\right)={\displaystyle \frac{\nu -k}{8k^2}}$$

(14)

and

$$\left|W\right|\ge (\nu -k)\left(1-{\displaystyle \frac{1}{8k^2}}\right)>{\displaystyle \frac{1}{2}}(\nu -k)\ge 2k+2.$$

(15)

The last inequality in (15) follows by $\nu \ge 5k+4.$

By (15), if one adds a new vertex and connects it with each vertex in $L^{\prime }$ by an edge, no cycle of length $2k+2$ is created, otherwise such cycle already exists in $H_k$.

Clearly, (14) tells us an upper bound on $\left|R\right|$. In fact, we hope to show that $\left|R\right|=0$. In order to obtain this fact, we need the following claim.

Claim 9.

$\lambda {\mathbf{x}}_u\ge k-{\displaystyle \frac{1}{16k^2}}$ for each u in $V_{H_k}$.

Proof of Claim 9.

We prove Claim 9 by contradiction. That is, for some $u\in V_{H_k}$, $\lambda {\mathbf{x}}_u<k-{\displaystyle \frac{1}{16k^2}}$ happens, then together with Claim 2, one has $\sqrt{k(\nu -k)}{\mathbf{x}}_u<k-{\displaystyle \frac{1}{16k^2}}$, which implies ${\mathbf{x}}_u<{\displaystyle \frac{k-\frac{1}{16k^2}}{\sqrt{k(\nu -k)}}}=\sqrt{{\displaystyle \frac{k}{\nu -k}}}\left(1-{\displaystyle \frac{1}{16k^3}}\right)<1-{\displaystyle \frac{1}{16k^3}}$. By Claim 8, $u\notin L^{\prime }$. Define $G_k=H_k-\{wu:w\in N_{H_k}\left(u\right)\}+\{wu:w\in L^{\prime }\}$. One sees that $G_k$ is bipartite, and it is $C_{2k+2}$-free. Clearly, the order of $G_k$ is $\nu$. Combining with Rayleigh quotient and Claim 8 yields

$$\begin{array}{cc}\hfill \lambda \left(G_k\right)-\lambda \left(H_k\right)& \ge {\mathbf{x}}^T(A\left(G_k\right)-A\left(H_k\right))\mathbf{x}=2{\mathbf{x}}_u\left(\sum _{w\in L^{\prime }}{\mathbf{x}}_w-\sum _{w\in N_{H_k}\left(u\right)}{\mathbf{x}}_w\right)\hfill \\ \hfill & =2{\mathbf{x}}_u\left(\sum _{w\in L^{\prime }}{\mathbf{x}}_w-\lambda {\mathbf{x}}_u\right)>2{\mathbf{x}}_u\left(k\left(1-{\displaystyle \frac{1}{16k^3}}\right)-\left(k-{\displaystyle \frac{1}{16k^2}}\right)\right)=0.\hfill \end{array}$$

That is, $\lambda \left(G_k\right)>\lambda \left(H_k\right)$, a contradiction to the choice of $H_k$. □

Now we are ready to show R is an empty set.

Claim 10.

$R=\varnothing$.

Proof of Claim 10.

We prove Claim 10 by reduction to absurdity. Suppose that $R\ne \varnothing$. Choose a vertex $u\in R$. If $u\in R\cap X$, then $\left|N\right(u)\cap R|\le 1$. Otherwise, there exists a cycle $C_{2k+2}$ in $H_k$, which will not happen. Notice that ${\mathbf{x}}_v<\eta$ for any vertex $v\in R$. Then,

$$\lambda {\mathbf{x}}_u=\sum _{z\in N\left(u\right)}{\mathbf{x}}_z=\sum _{z\in N\left(u\right)\cap R}{\mathbf{x}}_z+\sum _{z\in N\left(u\right)\cap R\cap Y}{\mathbf{x}}_z<\eta +\sum _{z\in N\left(u\right)\cap R\cap Y}{\mathbf{x}}_z.$$

Recall that $\eta <1-{\displaystyle \frac{1}{16k^3}}$. Together with Claim 9, one has

$${\displaystyle \frac{1}{\lambda {\mathbf{x}}_u}}\sum _{z\in N\left(u\right)\cap R\cap Y}{\mathbf{x}}_z>1-{\displaystyle \frac{\eta }{\lambda {\mathbf{x}}_u}}\ge 1-{\displaystyle \frac{\eta }{k-\frac{1}{16k^2}}}=1-{\displaystyle \frac{16k^2\eta }{16k^3-1}}>1-{\displaystyle \frac{1}{k}}>{\displaystyle \frac{7}{8k}}.$$

(16)

Now, we consider $u\in R\cap Y$. Notice that u has at most $k-1$ neighbors in $L^{\prime }$. Then,

$$\lambda {\mathbf{x}}_u=\sum _{z\in N\left(u\right)}{\mathbf{x}}_z=\sum _{z\in N\left(u\right)\cap L^{\prime }}{\mathbf{x}}_z+\sum _{z\in N\left(u\right)\cap R\cap X}{\mathbf{x}}_z\le k-1+\sum _{z\in N\left(u\right)\cap R\cap X}{\mathbf{x}}_z.$$

It follows from Claim 9 that

$${\displaystyle \frac{1}{\lambda {\mathbf{x}}_u}}\sum _{z\in N\left(u\right)\cap R\cap X}{\mathbf{x}}_z\ge 1+{\displaystyle \frac{1-k}{\lambda {\mathbf{x}}_u}}\ge 1+{\displaystyle \frac{1-k}{k-\frac{1}{16k^2}}}=1+{\displaystyle \frac{16k^2(1-k)}{16k^3-1}}={\displaystyle \frac{16k^2-1}{16k^3-1}}>{\displaystyle \frac{7}{8k}}.$$

(17)

Define ${\tilde{H}}_k=H_k[R\cap X,R\cap Y]$ and ${\mathbf{y}=\mathbf{x}|}_R$ (the restriction of $\mathbf{x}$ on the set R). By (16) and (17), one has

$$M{\mathbf{y}}_u=\sum _{z\in N_{{\tilde{H}}_k}\left(u\right)}{\mathbf{x}}_z=\sum _{z\in N_{H_k}\left(u\right)\cap R}{\mathbf{x}}_z\ge {\displaystyle \frac{7}{8k}}\lambda {\mathbf{x}}_u={\displaystyle \frac{7}{8k}}\lambda {\mathbf{y}}_u.$$

Applying Lemma 1 and Claim 2, we get $\lambda \left({\tilde{H}}_k\right)\ge {\displaystyle \frac{7}{8k}}\lambda \ge {\displaystyle \frac{7}{8}}\sqrt{{\displaystyle \frac{\nu -k}{k}}}$. We also notice that there does not exist a cycle $C_{2k+2}$ in ${\tilde{H}}_k$. Hence, Claim 2 implies

$$\lambda \left({\tilde{H}}_k\right)\le \sqrt{k\left(\right|R|-1)}<\sqrt{{\displaystyle \frac{\nu -k}{8k}}}<{\displaystyle \frac{7}{8}}\sqrt{{\displaystyle \frac{\nu -k}{k}}},$$

a contradiction. Thus, $R=\varnothing$. □

By Claim 10, R is an empty set, and so $V_{H_k}=L^{\prime }\cup W$. Together with Claim 8, we deduce $H_k\cong K_{k,\nu -k}$. Therefore, we finish the proof for Theorem 1. □

4. Further Research Problems

In this contribution, our research concerns the spectral Turán-type problem of bipartite graphs. We show that the $C_{2k+2}$-free bipartite graph with sufficient large order $\nu$ having maximum spectral radius is $K_{k,\nu -k}$.

Denote the family of $\nu$-vertex bipartite graphs by ${\mathcal{B}}_{\nu }$, and let $\mathcal{B}(p,q)$ be the family of all bipartite graphs with two parts of sizes p and q, respectively. We propose two further research problems as follows.

Problem 2.

How can we characterize the $C_{2k+2}$-free bipartite graph among ${\mathcal{B}}_{\nu }\backslash \left\{K_{k,\nu -k}\right\}$ having the largest spectral radius?

Problem 3.

How can we characterize the $C_{2k+2}$-free bipartite graph among $\mathcal{B}(p,q)$ having the largest spectral radius?