The Maximum Spectral Radius of Graphs Without Friendship Subgraphs

Sebastian Cioab˘a∗ Lihua Feng† Department of Mathematical Sciences School of Mathematics and Statistics University of Delaware Central South University Newark, DE, U.S.A. New Campus, Changsha, Hunan, P.R. China [email protected] [email protected] Michael Tait‡ Department of Mathematics and Statistics Villanova University Villanova, PA, U.S.A. [email protected] Xiao-Dong Zhang§ School of Mathematical Sciences, MOE-LSC, SHL-MAC Shanghai Jiao Tong University Shanghai, P. R. China [email protected]

Submitted: Nov 29, 2019; Accepted: Oct 16, 2020; Published: Oct 30, 2020 c The authors. Released under the CC BY-ND license (International 4.0).

Abstract A graph on 2k + 1 vertices consisting of k triangles which intersect in exactly one common vertex is called a k−friendship graph and denoted by Fk. This paper determines the graphs of order n that have the maximum (adjacency) spectral radius among all graphs containing no Fk, for n suﬃciently large. Mathematics Subject Classiﬁcations: 05C50; 05C38 ∗S. M. Cioab˘ais partially supported by NSF grants DMS-1600768 and CIF-1815922 and a JSPS Fellowship. †L. Feng is partially supported by NSFC (Nos. 11871479, 11671402), Hunan Provincial Natural Science Foundation (2016JJ2138, 2018JJ2479) and Mathematics and Interdisciplinary Sciences Project of CSU. ‡M. Tait is partially supported by NSF grant DMS-2011553. §X.-D. Zhang is partially supported by NSFC (Nos. 11971311, 11531001); the Montenegrin-Chinese Science and Technology Cooperation Project (No.3-12) (Corresponding author). the electronic journal of combinatorics 27(4) (2020), #P4.22 https://doi.org/10.37236/9179 1 Introduction

In this paper, we consider only simple and undirected graphs. Let G be a simple connected graph with vertex set V (G) = {v1, . . . , vn} and edge set E(G) = {e1, . . . , em}. Let d(vi) (or dG(vi)) be the degree of a vertex v in G. The adjacency matrix of G is A(G) = (aij)n×n with aij = 1 if two vertices vi and vj are adjacent in G, and aij = 0 otherwise. The largest eigenvalue of A(G), denoted by λ(G) or λ1(G), is called the spectral radius of G. The spectral radius of a graph gives some information about how dense the graph is. For example, it is well-known that the average degree of G is at most λ1(G) which is at most the maximum degree of G. In spectral graph theory, Brualdi and Solheid [5] proposed the following problem: Given a set of graphs, try to find a tight upper bound for the spectral radius in this set and characterize all extremal graphs. This problem is widely studied in the literature for many classes of graphs, such as graphs with a given number of cut vertices [3], diameter [8, 18], radius [18], domination number [29], size [28], Euler genus [9], and clique or independence number [30], and additionally for subgraphs of the hypercube [4]. The following problem regarding the adjacency spectral radius was proposed in [25]: What is the maximum spectral radius of a graph G on n vertices without a subgraph isomorphic to a given graph F ? Fiedler and Nikiforov [13] obtained tight sufficient conditions for graphs to be Hamiltonian or traceable. Additionally, Nikiforov obtained spectral strengthenings of Turán’stheorem [24] and the K˝ovari-Sós-Turántheorem [22] when the forbidden graphs are complete or complete bipartite respectively. This motivates further study for such question, see [12, 13, 20, 23, 25]. The Turánnumber of a graph F is the maximum number of edges that may be in an n-vertex graph without a subgraph isomorphic to F , and is denoted by ex(n, F ). A graph on n vertices with no subgraph F and with ex(n, F ) edges is called an extremal graph for F and we denote by Ex(n, F ) the set of all extremal graphs on n vertices for F . Understanding ex(n, F ) and Ex(n, F ) for various graphs F is a cornerstone of extremal graph theory (see [2,7, 11, 14, 15, 17, 27] for surveys). A graph on 2k + 1 vertices consisting of k triangles which intersect in exactly one common vertex is called a k−friendship graph (also known as a k-fan) and denoted by Fk. In [10], the following result is proved.

2 Theorem 1. [10] For every k > 1, and for every n > 50k , if a graph G of order n satisﬁes e(G) > ex(n, Fk), then G contains a copy of a k−friendship graph, where n2 k2 − k if k is odd, ex(n, Fk) = + 2 3 4 k − 2 k if k is even. i The extremal graphs Gn,k (i = 1, 2) of Theorem1 are as follows. For odd k (where 1 n > 4k − 1) Gn,k is constructed by taking a complete bipartite graph with color classes n n of size d 2 e and b 2 c and embedding two vertex disjoint copies of Kk in one side. For even 2 k (where now n > 4k − 3) Gn,k is constructed by taking a complete equi-bipartite graph 2 3 and embedding a graph with 2k − 1 vertices, k − 2 k edges with maximum degree k − 1 1 2 in one side. The graphs Gn,k is unique up to isomorphism, but Gn,k is not.

the electronic journal of combinatorics 27(4) (2020), #P4.22 2 Our goal is to give the spectral counterpart of Theorem1. As the case k = 1 is just Mantel’s theorem, whose spectral version is also known (see [21]) so we consider k > 2. The main result of this paper is the following. Theorem 2. Let G be a graph of order n that does not contain a copy of a k−friendship graph, k > 2. For suﬃciently large n, if G has the maximal spectral radius, then

G ∈ Ex(n, Fk).

We note that one may form an equitable partition of a graph in Ex(n, Fk) and deter- mine its spectral radius as the root of a degree 3 (if k is odd) or degree 4 (if k is even) polynomial. We at last point out that, during our proof, we use the triangle removal lemma, so it is diﬃcult to present exactly how large we need our n to be.

2 Some Lemmas

Let G be a simple graph with matching number β(G) and maximum degree ∆(G). For given two integers β and ∆, deﬁne f(β, ∆) = max{|E(G)| : β(G) 6 β, ∆(G) 6 ∆}. Chv´ataland Hanson [6] obtained the following result.

Theorem 3 (Chv´ataland Hanson [6]). For every two positive integers β > 1 and ∆ > 1, we have ∆ β f(β, ∆) = ∆β + ∆β + β. 2 d∆/2e 6 We will frequently use the following special case proved by Abbott, Hanson and Sauer [1]: k2 − k if k is odd, f(k − 1, k − 1) = 2 3 k − 2 k if k is even.

The extremal graphs are exactly those we embedded into the Tur´angraph Tn,2 to obtain i the extremal Fk-free graph Gn,k (i = 1, 2). Essential to our proof are the following two lemmas: the triangle removal lemma and a stability result of F¨uredi. Lemma 4 (Triangle Removal Lemma [10, 14, 26]). For each ε > 0, there exists an 3 N = N(ε) and δ > 0 such that every graph G on n vertices with n > N with at most δn triangles can be made triangle-free by removing at most εn2 edges.

Lemma 5 (F¨uredi[16]). Suppose that K3 * G, |V (G)| = n, s > 0 and e(G) = e(Tn,2)−s. Then there exists a bipartite subgraph H, E(H) ⊆ E(G) such that e(H) > e(G) − s. The following lemma is needed in the sequel. Lemma 6. For any positive integer n, we have

n rlnm jnk 1 − < . 2 2 2 n

the electronic journal of combinatorics 27(4) (2020), #P4.22 3 3 The Proof of Theorem2

Let Gn,k be the set of all Fk-free graphs of order n. Let G ∈ Gn,k be a graph on n vertices with maximum spectral radius. The aim of this section is to prove that e(G) = ex(n, Fk) for n large enough. First note that G must be connected. Let λ1 be the spectral radius of G and let x be a positive eigenvector for it. We may normalize x so that it has maximum entry equal to 1, and let z be a vertex such that xz = 1. We prove the theorem iteratively, giving successively better lower bounds on both e(G) and the eigenvector entries of all of the other vertices, until ﬁnally we can show that e(G) = ex(n, Fk). Let H ∈ Ex(n, Fk). Then since G is the graph maximizing the spectral radius over all Fk-free graphs, in view of Theorem1, we must have

1T A(H)1 n n + f(k − 1, k − 1) n λ (G) λ (H) = 2 2 2 > . (1) 1 > 1 > 1T 1 n 2 The proof of Theorem2 is outlined as follows.

• We give a lower bound on e(G) as a function of λ1 and the number t of triangles in n2 G, which on ﬁrst approximation gives a bound of roughly 4 − O(kn). • Using the triangle removal lemma and F¨uredi’s stability result, we show that G has a very large maximum cut.

• We show that no vertex has many neighbors on its side of the partition, and then we reﬁne this by considering eigenvector entries to show that in fact no vertex has more than a constant number of neighbors on its side of the partition.

n • We show that no vertices have degree much smaller than 2 , and this allows us to reﬁne our lower bound on both e(G) and on the eigenvector entry of each vertex.

• Once we know that all vertices have eigenvector entry very close to 1, we may show that the partition is balanced. This shows that G can be converted to a graph in Ex(n, Fk) by adding or removing a constant number of edges, and this allows us to show that e(G) = ex(n, Fk).

We now proceed with the details. First we prove a lemma which gives a lower bound on e(G) in terms of λ1 and the number of triangles in G. Lemma 7. If G has t triangles, then

2 3t e(G) > λ1 − . λ1

Proof. Let λ1 be the spectral radius of G and let x be a positive eigenvector scaled such that it has maximum entry equal to 1, and let z be a vertex with maximum eigenvector

the electronic journal of combinatorics 27(4) (2020), #P4.22 4 P 2 P P P entry i.e., xz = 1. Then λ1xu = v∼u xv and λ1xu = v∼u λ1xv = v∼u w∼v xw. We consider the following triple sum:

X 2 X X X λ1xu = xw. u∈V u∈V v∼u w∼v The sum counts over all ordered walks on three vertices (with possible repetition), and is weighted by the eigenvector entry of the last vertex. Instead of summing over ordered triples of vertices, we count by considering the ﬁrst edge in the walk. If a given walk has ﬁrst edge uv, then xw will be counted by this edge exactly once if w is adjacent to exactly one of u or v and exactly twice if {u, v, w} forms a triangle. Therefore, the sum is equal to     X  X X X  X  X X X  2 xw + xw + xw =  xw + xw + xw uv∈E wu∈E wu∈E wv∈E uv∈E wu∈E wu∈E wv∈E wv∈E wv6∈E wu6∈E wv∈E wv6∈E   X  X X X  6  xw + xw + xw uv∈E wu∈E wv6∈E wv∈E wv∈E   X  X X  6  xw + xw uv∈E wu∈E w∈V wv∈E   X  X X  6  1 + xw uv∈E wu∈E w∈V wv∈E X = 3t + e(G) xw. w∈V Hence 2 3t e(G) > λ1 − P . w∈V xw On the other hand, X X λ1 = λ1xz = xw 6 xw. w∼z w∈V Therefore 2 3t 2 3t e(G) > λ1 − P > λ1 − . w∈V xw λ1 So the assertion holds. Corollary 8. If the number of triangles of G is t, then 6t e(G) λ2 − . > 1 n

the electronic journal of combinatorics 27(4) (2020), #P4.22 5 2 3t Proof. In view of inequality (1), and the function f(x) = λ1 − x is strictly increasing with respect to x, the assertion follows.

Lemma 9. Suppose the matching number of a graph H of order n is at most k − 1. Then e(H) 6 kn, i.e., ex(n, Mk) 6 kn, where Mk is a matching of size k.

Proof. By Theorem3, e(H) 6 f(k − 1, n − 1) 6 (k − 1)(n − 1 + 1) < kn. 1 δ2 Lemma 10. Let ε and δ be ﬁxed positive constants with δ < 10(k+1)2 , ε < 16 . There exists an N(ε, δ, k) such that G has a partition V = S ∪ T which gives a maximum cut, and

1 e(S,T ) − ε n2 > 4

for n > N(ε, δ, k). Furthermore 1 √ 1 √ − ε n |S|, |T | + ε n. 2 6 6 2

Proof. Since G is Fk-free, the neighborhood of any vertex does not have Mk (a matching of size k) as a subgraph. Thus by Lemma9, we can obtain the following upper bound for the number of triangles,

X X X X 2 3t = e(G[N(v)]) 6 ex(d(v) + 1,Mk) 6 ex(n, Mk) 6 kn = kn . v∈V (G) v∈V (G) v∈V (G) v∈V (G)

kn2 k 3 3 k This gives t 6 3 . So t 6 3n n 6 δn for n > N2 > 3δ . From Corollary8, we obtain n2 e(G) − 2kn. (2) > 4

By Lemma4, there exists an N1(ε, k) such that the graph G1 obtained from G by deleting 1 2 at most 10 εn edges is K3-free. For N = max{N1,N2}, the size of the graph G1 of order 1 2 n > N satisfies e(G1) > e(G) − 10 εn . Note that e(G1) 6 e(Tn,2) by Turán’sTheorem. Define s , e(Tn,2) − e(G1) > 0.

By Lemma5, G1 contains a bipartite subgraph G2 such that e(G2) > e(G1) − s. Hence, for n suﬃciently large, we have

e(G2) > e(G1) − s = 2e(G1) − e(Tn,2) 1 2e(G) − e(T ) − εn2 > n,2 5 n2 n2 1 2 − 2kn − − εn2 > 4 4 5

the electronic journal of combinatorics 27(4) (2020), #P4.22 6 1 − ε n2. > 4 Therefore, G has a partition V = S ∪ T which gives a maximum cut such that 1 e(S,T ) e(G ) − ε n2. (3) > 2 > 4

1 √ Furthermore, without loss of generality, we may assume that |S| 6 |T |. If |S| < ( 2 − ε)n, 1 √ then |T | = n − |S| > ( 2 + ε)n. So 1 √ 1 √ 1 e(S,T ) |S||T | < − ε n + ε n = − ε n2, 6 2 2 4 which contradicts to Eq. (3). Therefore it follows that 1 √ 1 √ − ε n |S|, |T | + ε n. 2 6 6 2 Hence the assertion holds.

Lemma 11. Let k > 2. Denote by 1 1 L := v : d(v) − n . 6 2 4(k + 1) Then 2 |L| 6 16k . Proof. Suppose that |L| > 16k2. Then let L0 ⊆ L with |L0| = 16k2. Then it follows that

0 X e(G − L ) > e(G) − d(v) v∈L0 n2 1 1 − 2kn − 16k2 − n > 4 2 4(k + 1) (n − 16k2)2 > + k2. 4 for n a suﬃciently large constant depending only on k, where the second inequality is by 0 (2). Hence by Theorem1, G − L contains Fk, which implies that G contains Fk. So the assertion holds. We will also need the following lemma which can be proved by induction or double counting.

Lemma 12. Let A1, ··· ,Ap be p ﬁnite sets. Then

p p

X [ |A1 ∩ A2 ∩ · · · ∩ Ap| > |Ai| − (p − 1) Ai . (4) i=1 i=1

the electronic journal of combinatorics 27(4) (2020), #P4.22 7 For a vertex v, let dS(v) = |N(v) ∩ S| and dT (v) = |N(v) ∩ T |, and let

W := {v ∈ S : dS(v) > δn} ∪ {v ∈ T : dT (v) > δn} be the set of vertices that have many neighbors which are not in the cut. Let L be as in Lemma 11, that is 1 1 L = v : d(v) − n . 6 2 4(k + 1) Next we show that actually W and L are empty.

Lemma 13. For the above W , we have

2ε 2k2 |W | < n + , δ δn and W \ L is empty.

1 2 n2 2 Proof. Note that e(S,T ) > 4 − ε n by Lemma 10, and e(G) 6 ex(n, Fk) 6 4 + k by Theorem1. Hence n2 1 e(S) + e(T ) = e(G) − e(S,T ) + k2 − − ε n2 = εn2 + k2. (5) 6 4 4

On the other hand, if we let W1 = W ∩ S and W2 = W ∩ T , then we deduce X X X X 2e(S) = dS(u) > dS(u) > |W1|δn, 2e(T ) = dT (u) > dT (u) > |W2|δn. u∈S u∈W1 u∈T u∈W2 So δn |W |δn e(S) + e(T ) (|W | + |W |) = . (6) > 1 2 2 2 By (5) and (6), we get |W |δn εn2 + k2, 2 6 i.e., 2(εn2 + k2) |W | . (7) 6 δn Suppose that W \ L 6= ∅. We now prove that this is impossible. Let L1 = L ∩ S and L2 = L ∩ T . Without loss of generality, there exists a vertex 1 u ∈ W1 \ L1. Since S and T form a maximum cut, dT (u) > 2 d(u). On the other hand, 1 1 u 6∈ L because u ∈ W1 \ L1. Therefore d(u) > 2 − 4(k+1) n. So

1 1 1 d (u) d(u) − n. T > 2 > 4 8(k + 1)

the electronic journal of combinatorics 27(4) (2020), #P4.22 8 2 1 δ2 On the other hand, |L| 6 16k . Hence, for ﬁxed δ < 10(k+1)2 , ε < 16 and suﬃciently large n, we have 1 √ 2k2 1 √ |S \ (W ∪ L)| − ε n − δn − − 16k2 − ε − δ n − 18k2 k. > 2 δn > 2 >

Suppose that u is adjacent to k vertices u1, . . . , uk in S \ (W ∪ L). Since ui 6∈ L, 1 1 we have d(ui) > 2 − 4(k+1) n. On the other hand, dS(ui) 6 δn by ui ∈/ W . So 1 1 dT (ui) = d(ui) − dS(ui) > 2 − 4(k+1) − δ n. By Lemma 12, we have

|NT (u) ∩ NT (u1) ∩ · · · ∩ NT (uk)| > |NT (u)| + |NT (u1)| + ... + |NT (uk)| − k|NT (u) ∪ NT (u1) ∪ · · · ∪ NT (uk)| > dT (u) + dT (u1) + ··· + dT (uk) − k|T | 1 1 1 1 1 √ − n + − − δ n · k − k + ε n > 4 8(k + 1) 2 4(k + 1) 2 1 √ = − kδ − k ε n > k 8(k + 1)

1 δ2 for suﬃciently large n, where the last inequality is from δ < 10(k+1)2 and ε < 16 . So there exist k vertices v1, . . . vk in T such that the induced subgraph by two partitions {u1, . . . , uk} and {v1, . . . , vk} is complete bipartite. It follows that G contains Fk, this is a contradiction. Therefore u is adjacent to at most k − 1 vertices in S \ (W ∪ L). Hence, δ2 in view of ε < 16 , we have

dS(u) 6 |W | + |L| + k − 1 2ε 2k2 < n + + 16k2 + k − 1 δ δn 2δ 2k2 < n + + 17k2 3 δn < δn for suﬃciently large n. This is a contradiction to the fact that u ∈ W . Similarly, there is no vertex u ∈ W2 \ L2, Hence W \ L = ∅.

Lemma 14. L is empty, and both G[S] and G[T ] are K1,k and Mk-free. Proof. We will prove the result from the following two claims. Claim 1: There exist independent sets IS ⊆ S and IT ⊆ T such that

2 2 |IS| > |S| − 18k , and |IT | > |T | − 18k .

Indeed, let u1, . . . , u2k ∈ S \ L. Then ui ∈/ L which implies 1 1 d(u ) − n. i > 2 4(k + 1)

the electronic journal of combinatorics 27(4) (2020), #P4.22 9 By Lemma 13, dS(ui) 6 δn. Hence 1 1 d (u ) = d(u ) − d (u ) − − δ n. T i i S i > 2 4(k + 1) Furthermore, by Lemma 12, we have

2k 2k 2k \ X [ NT (ui) > |NT (ui)| − (2k − 1) NT (ui) i=1 i=1 i=1 1 1 1 √ − − δ n · 2k − (2k − 1) + ε n > 2 4(k + 1) 2 1 √ = − 2kδ − (2k − 1) ε n 2(k + 1) > k

for suﬃciently large n. Hence there exist k vertices v1, . . . , vk such that the induced subgraph by two partitions {u1, . . . , u2k} and {v1, . . . , vk} is a complete bipartite graph. So G[S \ L] is both K1,k and Mk-free, otherwise G contains Fk, i.e., uu1v1, . . . uukvk for d(u) > k, or v1u1u2, . . . , v1u2k−1u2k for {u1u2, . . . u2k−1u2k} being a matching of size k. Hence the maximum degree and the maximum matching number of G[S \ L] are at most k − 1, respectively. By Theorem3,

e(G[S \ L]) 6 f(k − 1, k − 1). The same argument gives

e(G[T \ L]) 6 f(k − 1, k − 1). Since G[S \L] has at most f(k −1, k −1) edges, then the subgraph obtained from G[S \L] by deleting one vertex of each edge in G[S \L] contains no edges, which is an independent set of G[S \ L]. So there exists an independent set IS ⊆ S such that 3 |I | |S \ L| − f(k − 1, k − 1) |S| − k k − − 16k2 |S| − 18k2. S > > 2 >

The same argument gives that there is an independent set IT ⊆ T with

2 |IT | > |T | − 18k . So Claim 1 holds. Recall that z is a vertex with maximum eigenvector entry. Since xz = 1, and X n d(z) x = λ x = λ . > w 1 z 1 > 2 w∼z Hence z∈ / L. the electronic journal of combinatorics 27(4) (2020), #P4.22 10 Without loss of generality, we may assume that z ∈ S. Since the maximum degree in the induced subgraph G[S \ L] is at most k − 1 (containing no K1,k), from Lemma 11, we 2 have |L| 6 16k and 2 2 dS(z) = dS∩L(z) + dS\L(z) 6 k − 1 + 16k 6 17k . Therefore, by Claim 1, we have X λ1 = λ1xz = xv v∼z X X = xv + xv v∼z,v∈S v∼z,v∈T X X X = xv + xv + xv

v∼z,v∈S v∼z,v∈IT v∼z,v∈T \IT X X 6 dS(z) + xv + 1 v∈IT v∈T \IT 2 X 6 17k + xv + |T | − |IT | v∈IT X 2 2 6 xv + 17k + 18k v∈IT X 2 6 xv + 35k . v∈IT So X 2 xv > λ1 − 35k . (8) v∈IT Claim 2: L = ∅. 1 1 By way of contradiction, assume that there is a vertex v ∈ L, i.e., d(v) 6 ( 2 − 4(k+1) )n. Consider the graph G+ with vertex set V (G) and edge set E(G+) = E(G \{v}) ∪ {vw : w ∈ IT }. Note that adding a vertex incident with vertices in IT does not create any + triangles, and so G is Fk-free. By (8), we have that   T + x (A(G ) − A(G)) x 2xv X X λ (G+) − λ (G) = x − x 1 1 > xTx xT x  w u w∈IT uv∈E(G) 2x v λ − 35k2 − d (v) > xT x 1 G 2x 1 1 v λ − 35k2 − − n > xT x 1 2 4(k + 1) 2x n 1 1 v − 35k2 − − n > xT x 2 2 4(k + 1) 2x n = v − 35k2 > 0, xT x 4(k + 1) the electronic journal of combinatorics 27(4) (2020), #P4.22 11 1 1 where the last step uses n large enough and that if v ∈ L, then dG(v) 6 2 − 4k+4 n. This contradicts G has the largest spectral radius over all Fk-free graphs and so L must be empty. Next we may reﬁne the structure of G.

Lemma 15. For n and k as before, we have n n − 4k |S|, |T | + 4k, (9) 2 6 6 2

n2 e(G) − 12k2, (10) > 4 and n n − 14k2 δ(G) λ ∆(G) + 5k. (11) 2 6 6 1 6 6 2

Proof. From Lemma 14, both G[S] and G[T ] are K1,k and Mk-free, so we have e(S) + 2 e(T ) 6 2f(k − 1, k − 1) < 2k . This means that the number of triangles in G is bounded above by 2k2n since any triangle contains an edge of E(S) ∪ E(T ). By Corollary8, we have 6t n2 12k2n n2 e(G) λ2 − − = − 12k2. > 1 n > 4 n 4 n n Suppose that |S| 6 2 − 4k, then |T | = n − |S| > 2 + 4k. Hence n n n2 e(G) = e(S) + e(T ) + e(S,T ) 2k2 + |S||T | 2k2 + − 4k + 4k = − 14k2, 6 6 2 2 4

n2 2 which contradicts to e(G) > 4 − 12k . So we have n n − 4k |S|, |T | + 4k. 2 6 6 2 Moreover, by Lemma 14, the maximum degree of G[S] is at most k − 1. This implies that n n ∆(G) + 4k + k − 1 + 5k. 6 2 6 2 So n λ ∆(G) + 5k. 1 6 6 2 n 2 Furthermore, we claim that the minimum degree of G is at least 2 − 14k . Otherwise, removing a vertex v of minimum degree d(v), we have

e(G − v) = e(G) − d(v) n2 n − 12k2 − − 14k2 > 4 2 n2 n = − + 2k2 4 2

the electronic journal of combinatorics 27(4) (2020), #P4.22 12 (n − 1)2 1 = + k2 + k2 − 4 4 (n − 1)2 > + k2, 4

which implies G − v contains Fk by Theorem1.

120k2 Lemma 16. For all u ∈ V (G), we have that xu > 1 − n . Proof. Without loss of generality, we may assume that z ∈ S. We consider the following two cases. 2 2 Case 1: u ∈ S. Then dS(u) 6 k as e(G[S]) 6 k . Hence we obtain n |N (u)| = d (u) = d(u) − d (u) δ(G) − d (u) − 14k2 − k2 T T S > S > 2 n − 15k2. > 2

|NT (u) ∩ NT (z)| = |NT (u)| + |NT (z)| − |NT (u) ∪ NT (z)| > 2δT (G) − |T | n n n 2 − 15k2 − + 4k − 34k2. > 2 2 > 2 X X X λ1xu − λ1xz = xv + xv + xv v∼u,v∈T,v∼z v∼u,v∈T,v6∼z v∼u,v∈S X X X − xv − xv − xv v∼z,v∈T,v∼u v∼z,v∈T,v6∼u v∼z,v∈S X X > − xv − xv v∼z,v∈T,v6∼u v∼z,v∈S X X > − 1 − 1 v∼z,v∈T,v6∼u v∼z,v∈S > −(dT (z) − |NT (u) ∩ NT (z)|) − dS(z) n n − + 5k − − 34k2 − k2 > 2 2 2 > −40k . Therefore, for any u ∈ S, we have 40k2 40k2 80k2 xu > 1 − > 1 − n = 1 − . (12) λ1 2 n Case 2: u ∈ T . By (12),

X X 80k2 λ x = x x 1 − d (u). 1 u v > v > n S v∼u v∼u,v∈S Since n − 14k2 δ(G) d(u) = d (u) + d (u), 2 6 6 S T

the electronic journal of combinatorics 27(4) (2020), #P4.22 13 and dT (u) 6 k as the maximum degree in G[T ] is at most k − 1, we have dS(u) > n 2 n 2 2 − 14k − k > 2 − 15k . Hence

80k2 80k2 n 2 (1 − n )dS(u) (1 − n )( 2 − 15k ) xu > > n λ1 2 + 5k n 2 1200k4 2 − 55k + n = n 2 + 5k 2 1200k4 55k + 5k − n = 1 − n 2 + 5k 120k2 > 1 − . n From the above two cases, the result follows. Using this reﬁned bound on the eigenvector entries, we may show that the partition V = S ∪ T is balanced. Lemma 17. The sets S and T have sizes as close as possible. That is

|S| − |T | 6 1.

Proof. Without loss of generality, we may assume that |T | > |S|. Denote S0 := {v ∈ S : N(v) ⊆ T }, T 0 := {v ∈ T : N(v) ⊆ S}.

2 2 Since e(G[S]) 6 k , there exist at most 2k vertices in S having a neighbor in S. Hence 0 2 |S | > |S| − 2k . Similarly, 0 2 |T | > |T | − 2k . 0 Let C ⊆ T be a set having |T | − |S| vertices, which exists since, from (9), |T | − |S| 6 8k 0 2 n 2 and |T | > |T | − 2k > 2 − 4k − 2k > 8k. Then G \ C is a graph on 2|S| vertices such that (2|S|)2 e(G) − e(C,S) = e(G \ C) ex(2|S|,F ) + f(k − 1, k − 1). 6 k 6 4 Hence 2 e(G) 6 |S| + |C||S| + f(k − 1, k − 1) = |S||T | + f(k − 1, k − 1).

Let B = K|S|,|T | be the complete bipartite graph with partite sets S and T , and let G1 = G[S] ∪ G[T ] and G2 be the graph with edges E(B) \ E(G). Note that e(G) = e(B) + e(G1) − e(G2) and so e(G1) − e(G2) = e(G) − e(B) 6 f(k − 1, k − 1). By Lemma 16 we have, 120k2 2 240k2 xT x n 1 − > n 1 − = n − 240k2, > n n

the electronic journal of combinatorics 27(4) (2020), #P4.22 14 p 2 and that λ1(B) = |S||T |. By Lemma 15, also note that e(G1) 6 2k , we obtain n2 n2 e(S,T ) = e(G) − e(G ) − 12k2 − 2k2 = − 14k2, 1 > 4 4 which implies that n2 e(G ) = e(B) − e(S,T ) |S||T | − − 14k2 14k2. 2 6 4 6 So, bearing in mind the inequality (1), we have 2 jnk lnm 2f(k − 1, k − 1) xT (A(B) + A(G ) − A(G ))x + λ = 1 2 n 2 2 n 6 1 xT x xT A(B)x xT A(G )x xT A(G )x = + 1 − 2 . xT x xT x xT x Note that by Lemma 16,

X 120k2 240k2 xT A(G )x = 2 x x 2e(G )(1 − )2 2e(G ) 1 − , 2 u v > 2 n > 2 n uv∈E(G2) therefore 2 jnk lnm 2f(k − 1, k − 1) + n 2 2 n 2 2e(G ) 2e(G )(1 − 240k ) λ (B) + 1 − 2 n 6 1 xT x xT x 2 2(e(G ) − e(G )) 2e(G ) 240k λ (B) + 1 2 + 2 n 6 1 xT x xT x 2 2f(k − 1, k − 1) 2 · 14k2 240k p|S||T | + + n 6 xT x xT x 2 2f(k − 1, k − 1) 1 1 2 · 14k2 240k p|S||T | + + 2f(k − 1, k − 1) − + n 6 n xT x n 240k2 n(1 − n ) ! 2 2f(k − 1, k − 1) 1 1 2 · 14k2 240k p|S||T | + + 2k2 − + n 6 n 240k2 n 240k2 n(1 − n ) n(1 − n ) 2f(k − 1, k − 1) 480k4 6720k4 = p|S||T | + + + n n(n − 240k2) n(n − 240k2) 2f(k − 1, k − 1) 7200k4 = p|S||T | + + . n n(n − 240k2)

Then 2 jnk lnm 7200k4 − p|S||T | . (13) n 2 2 6 n(n − 240k2)

the electronic journal of combinatorics 27(4) (2020), #P4.22 15 Assume to the contrary, so |T | > |S| + 2. We consider two cases. Case 1: n is even. Since |S| + |T | = n, we have

2 jnk lnm n rn n − p|S||T | − − 1 + 1 n 2 2 > 2 2 2 r n n2 1 1 = − − 1 = > . 2 4 n q n2 n 2 + 4 − 1 So by (13), we have 1 2 jnk lnm 7200k4 8000k4 < − p|S||T | . n n 2 2 6 n(n − 240k2) 6 n2 This is a contradiction for suﬃciently large n. Case 2: n is odd. Since |S| + |T | = n, we have s 2 jnk lnm n2 − 1 n − 3 n + 3 − p|S||T | − n 2 2 > 2n 2 2 √ n − 1 n2 − 9 (n − 1 )2 − (n2 − 9) = n − = n √ 2 2 1 2 2(n − n + n − 9) 1 7 + 2 1 = √n . 1 2 > n 2(n − n + n − 9) So by (13), we have 1 2 jnk lnm 7200k4 8000k4 < − p|S||T | . n n 2 2 6 n(n − 240k2) 6 n2 This is a contradiction for suﬃciently large n. Therefore for n large enough we must have that ||S| − |T || 6 1.

Finally, we show that e(G) = ex(n, Fk).

Proof of Theorem2 . By way of contradiction, we assume that e(G) 6 ex(n, Fk)−1. Let H be an Fk-free graph with ex(n, Fk) edges on the same vertex set as G, where S and T induce a complete bipartite graph in H (this is possible because every graph in 2 Ex(n, Fk) has a maximum cut of size bn /4c). Let E+ and E− be sets of edges such that E(G) ∪ E+ \ E− = E(H), and choose E+ and E− to be as small as possible, i.e., E+ = E(H) \ E(G) and E− = E(G) \ E(H). Since |E(G) ∩ E(H)| + |E−| = e(G) < e(H) = |E(G) ∩ E(H)| + |E+| which implies that |E+| > |E−| + 1. Furthermore, we have 2 2 that |E−| 6 e(G[S]) + e(G[T ]) < 2k . Finally, by (10) we have that |E+| < 15k . Now, by the Rayleigh quotient [19] and Lemma 16, we have that

xT A(H)x 2 X 2 X λ (H) = λ (G) + x x − x x 1 > xT x 1 xT x i j xT x i j ij∈E+ ij∈E− the electronic journal of combinatorics 27(4) (2020), #P4.22 16 ! 2 120k2 2 λ (G) + |E | 1 − − |E | > 1 xT x + n − 2 240k2 (120k2)2 λ (G) + |E | − |E | − |E | + |E | > 1 xT x + − n + n2 + 2 240k2 (120k2)2 λ (G) + 1 − |E | + |E | > 1 xT x n + n2 +

> λ1(G),

2 for suﬃciently large n, where the last second inequality by |E−| < 2k and |E+| > |E−|+1 2 and the last inequality by |E+| < 15k . Therefore we have that for n large enough, λ1(H) > λ1(G), a contradiction. Hence e(G) = e(H). From the above discussion, we complete the proof of Theorem2.

Acknowledgements The authors would like to express their sincere thanks to the two referees for their helpful comments and suggestions of the manuscript.

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