The Maximum Spectral Radius of Graphs Without Friendship Subgraphs

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. Cioabă is 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


Introduction
In this paper, we consider only simple and undirected graphs. Let G be a simple connected graph with vertex set V (G) = {v 1 , . . . , v n } and edge set E(G) = {e 1 , . . . , e m }. Let d(v i ) (or d G (v i )) be the degree of a vertex v in G. The adjacency matrix of G is A(G) = (a ij ) n×n with a ij = 1 if two vertices v i and v j are adjacent in G, and a ij = 0 otherwise. The largest eigenvalue of A(G), denoted by λ(G) or λ 1 (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 λ 1 (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's theorem [24] and the Kővari-Sós-Turán theorem [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 Turán number 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 F k .
[10] For every k 1, and for every n 50k 2 , if a graph G of order n satisfies e(G) > ex(n, F k ), then G contains a copy of a k−friendship graph, where The extremal graphs G i n,k (i = 1, 2) of Theorem 1 are as follows. For odd k (where n 4k − 1) G 1 n,k is constructed by taking a complete bipartite graph with color classes of size n 2 and n 2 and embedding two vertex disjoint copies of K k in one side. For even k (where now n 4k − 3) G 2 n,k is constructed by taking a complete equi-bipartite graph and embedding a graph with 2k − 1 vertices, k 2 − 3 2 k edges with maximum degree k − 1 in one side. The graphs G 1 n,k is unique up to isomorphism, but G 2 n,k is not.
Our goal is to give the spectral counterpart of Theorem 1. 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, F k ).
We note that one may form an equitable partition of a graph in Ex(n, F k ) and determine 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.
Theorem 3 (Chvátal and Hanson [6]). For every two positive integers β 1 and ∆ 1, we have We will frequently use the following special case proved by Abbott, Hanson and Sauer [1]: The extremal graphs are exactly those we embedded into the Turán graph T n,2 to obtain the extremal F k -free graph G i n,k (i = 1, 2). Essential to our proof are the following two lemmas: the triangle removal lemma and a stability result of Füredi.
Lemma 4 (Triangle Removal Lemma [10,14,26]). For each ε > 0, there exists an N = N (ε) and δ > 0 such that every graph G on n vertices with n N with at most δn 3 triangles can be made triangle-free by removing at most εn 2 edges.
The following lemma is needed in the sequel.

The Proof of Theorem 2
Let G n,k be the set of all F k -free graphs of order n. Let G ∈ G n,k be a graph on n vertices with maximum spectral radius. The aim of this section is to prove that e(G) = ex(n, F k ) for n large enough. First note that G must be connected. Let λ 1 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 x z = 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, F k ).
Let H ∈ Ex(n, F k ). Then since G is the graph maximizing the spectral radius over all F k -free graphs, in view of Theorem 1, we must have The proof of Theorem 2 is outlined as follows.
• We give a lower bound on e(G) as a function of λ 1 and the number t of triangles in G, which on first approximation gives a bound of roughly n 2 4 − O(kn). • Using the triangle removal lemma and Füredi'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.
• We show that no vertices have degree much smaller than n 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, F k ) by adding or removing a constant number of edges, and this allows us to show that e(G) = ex(n, F k ).
We now proceed with the details. First we prove a lemma which gives a lower bound on e(G) in terms of λ 1 and the number of triangles in G.
Proof. Let λ 1 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 entry i.e., x z = 1. Then We consider the following triple sum: 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 x w 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 On the other hand, So the assertion holds.
Proof. In view of inequality (1), and the function f (x) = λ 2 1 − 3t 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) kn, i.e., ex(n, M k ) kn, where M k is a matching of size k.
Lemma 10. Let ε and δ be fixed positive constants with δ < 1 10(k+1) 2 , ε < δ 2 16 . There exists an N (ε, δ, k) such that G has a partition V = S ∪ T which gives a maximum cut, and for n N (ε, δ, k). Furthermore Proof. Since G is F k -free, the neighborhood of any vertex does not have M k (a matching of size k) as a subgraph. Thus by Lemma 9, we can obtain the following upper bound for the number of triangles, This gives t kn 2 3 . So t k 3n n 3 δn 3 for n N 2 k 3δ . From Corollary 8, we obtain By Lemma 4, there exists an N 1 (ε, k) such that the graph G 1 obtained from G by deleting at most 1 10 εn 2 edges is K 3 -free. For N = max{N 1 , N 2 }, the size of the graph G 1 of order n N satisfies e(G 1 ) e(G) − 1 10 εn 2 . Note that e(G 1 ) e(T n,2 ) by Turán's Theorem. Define s e(T n,2 ) − e(G 1 ) 0.
By Lemma 5, G 1 contains a bipartite subgraph G 2 such that e(G 2 ) e(G 1 ) − s. Hence, for n sufficiently large, we have the electronic journal of combinatorics 27(4) (2020), #P4.22 Therefore, G has a partition V = S ∪ T which gives a maximum cut such that Furthermore, without loss of generality, we may assume that |S| |T |.
which contradicts to Eq. (3). Therefore it follows that Hence the assertion holds.
Proof. Suppose that |L| > 16k 2 . Then let L ⊆ L with |L | = 16k 2 . Then it follows that for n a sufficiently large constant depending only on k, where the second inequality is by (2). Hence by Theorem 1, G − L contains F k , which implies that G contains F k . So the assertion holds.
We will also need the following lemma which can be proved by induction or double counting.
Lemma 12. Let A 1 , · · · , A p be p finite sets. Then be the set of vertices that have many neighbors which are not in the cut. Let L be as in Lemma 11, that is n .
Next we show that actually W and L are empty.
Lemma 13. For the above W , we have and W \ L is empty.
Suppose that W \ L = ∅. We now prove that this is impossible. Let L 1 = L ∩ S and L 2 = L ∩ T . Without loss of generality, there exists a vertex u ∈ W 1 \ L 1 . Since S and T form a maximum cut, d T (u) n.
the electronic journal of combinatorics 27(4) (2020), #P4.22 On the other hand, |L| 16k 2 . Hence, for fixed δ < 1 10(k+1) 2 , ε < δ 2 16 and sufficiently large n, we have Suppose that u is adjacent to k vertices u 1 , . . . , u k in S \ (W ∪ L). Since u i ∈ L, we have d(u i ) for sufficiently large n, where the last inequality is from δ < 1 10(k+1) 2 and ε < δ 2 16 . So there exist k vertices v 1 , . . . v k in T such that the induced subgraph by two partitions {u 1 , . . . , u k } and {v 1 , . . . , v k } is complete bipartite. It follows that G contains F k , this is a contradiction. Therefore u is adjacent to at most k − 1 vertices in S \ (W ∪ L). Hence, in view of ε < δ 2 16 , we have for sufficiently large n. This is a contradiction to the fact that u ∈ W . Similarly, there is no vertex u ∈ W 2 \ L 2 , Hence W \ L = ∅. Proof. We will prove the result from the following two claims. Claim 1: There exist independent sets I S ⊆ S and I T ⊆ T such that Indeed, let u 1 , . . . , u 2k ∈ S \ L. Then u i / ∈ L which implies n.
the electronic journal of combinatorics 27(4) (2020), #P4.22 By Lemma 13, d S (u i ) δn. Hence Furthermore, by Lemma 12, we have The same argument gives that there is an independent set I T ⊆ T with So Claim 1 holds. Recall that z is a vertex with maximum eigenvector entry. Since x z = 1, and Hence z / ∈ L.
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 K 1,k ), from Lemma 11, we have |L| 16k 2 and Therefore, by Claim 1, we have Claim 2: L = ∅. By way of contradiction, assume that there is a vertex v ∈ L, i.e., d(v) ( 1 2 − 1 4(k+1) )n. Consider the graph G + with vertex set V (G) and edge set E(G + ) = E(G \ {v}) ∪ {vw : w ∈ I T }. Note that adding a vertex incident with vertices in I T does not create any triangles, and so G + is F k -free. By (8), we have that where the last step uses n large enough and that if v ∈ L, then d G (v) 1 2 − 1 4k+4 n. This contradicts G has the largest spectral radius over all F k -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 and Proof. From Lemma 14, both G[S] and G[T ] are K 1,k and M k -free, so we have e(S) + e(T ) 2f (k − 1, k − 1) < 2k 2 . This means that the number of triangles in G is bounded above by 2k 2 n since any triangle contains an edge of E(S) ∪ E(T ). By Corollary 8, we have Suppose that |S| n 2 − 4k, then |T | = n − |S| n 2 + 4k. Hence which contradicts to e(G) n 2 4 − 12k 2 . So we have n 2 − 4k |S|, |T | n 2 + 4k.
Moreover, by Lemma 14, the maximum degree of G[S] is at most k − 1. This implies that ∆(G) n 2 + 4k + k − 1 n 2 + 5k. So Furthermore, we claim that the minimum degree of G is at least n 2 − 14k 2 . Otherwise, removing a vertex v of minimum degree d(v), we have which implies G − v contains F k by Theorem 1.
Lemma 16. For all u ∈ V (G), we have that x u 1 − 120k 2 n . Proof. Without loss of generality, we may assume that z ∈ S. We consider the following two cases.
Case 1: u ∈ S. Then d S (u) k 2 as e(G[S]) k 2 . Hence we obtain Therefore, for any u ∈ S, we have Case 2: u ∈ T . By (12), the electronic journal of combinatorics 27(4) (2020), #P4.22 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 Proof. Without loss of generality, we may assume that |T | |S|. Denote Since e(G[S]) k 2 , there exist at most 2k 2 vertices in S having a neighbor in S. Hence Similarly, |T | |T | − 2k 2 .
Assume to the contrary, so |T | |S| + 2. We consider two cases. Case 1: n is even. Since |S| + |T | = n, we have 2 n n 2 So by (13), we have 1 n < 2 n n 2 This is a contradiction for sufficiently large n. This is a contradiction for sufficiently large n. Therefore for n large enough we must have that ||S| − |T || 1.
Finally, we show that e(G) = ex(n, F k ).
Proof of Theorem 2. By way of contradiction, we assume that e(G) ex(n, F k ) − 1. Let H be an F k -free graph with ex(n, F k ) 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 Ex(n, F k ) has a maximum cut of size n 2 /4 ). 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.,  (10) we have that |E + | < 15k 2 . Now, by the Rayleigh quotient [19] and Lemma 16, we have that for sufficiently large n, where the last second inequality by |E − | < 2k 2 and |E + | |E − |+1 and the last inequality by |E + | < 15k 2 . Therefore we have that for n large enough, λ 1 (H) > λ 1 (G), a contradiction. Hence e(G) = e(H).
From the above discussion, we complete the proof of Theorem 2.