Spectral extrema of graphs with fixed size: cycles and complete bipartite graphs
aa r X i v : . [ m a t h . C O ] F e b Spectral extrema of graphs with fixed size: cyclesand complete bipartite graphs * Mingqing Zhai a , Huiqiu Lin b † , Jinlong Shu c a School of Mathematics and Finance, Chuzhou University, Chuzhou, Anhui 239012, China b Department of Mathematics, East China University of Science and Technology, Shanghai 200237, China c Department of Computer Science and Technology, East China Normal University, Shanghai 200237, China
Abstract:
Nikiforov [Some inequalities for the largest eigenvalue of a graph, Combin.Probab. Comput. 179–189] showed that if G is K r + -free then the spectral radius ρ ( G ) ≤√ m (1 − / r ), which implies that G contains C if ρ ( G ) > √ m . In this paper, we followthis direction on determining which subgraphs will be contained in G if ρ ( G ) > f ( m ),where f ( m ) ∼ √ m as m → ∞ . We first show that if ρ ( G ) ≥ √ m , then G contains K , r + unless G is a star; and G contains either C + or C + unless G is a complete bipartite graph,where C + t denotes the graph obtained from C t and C by identifying an edge. Secondly,we prove that if ρ ( G ) ≥ + q m − , then G contains pentagon and hexagon unless G isa book; and if ρ ( G ) > ( k − ) + q m + ( k − ) , then G contains C t for every t ≤ k + Keywords:
Complete bipartite subgraph; Cycle; Forbidden subgraph; Spectral radius;Adjacency matrix
AMS Classification:
Let F be a family of graphs. We say a graph G is F -free if it does not contain any F ∈F as a subgraph. The Tur´an number ex ( n , F ) is the maximum possible number of edgesin an F -free graph with n vertices. Tur´an type problems aim to study the Tur´an numberof fixed graphs. So far, there is a large volume of literature on Tur´an type problems, seethe survey paper [5].For a graph G , we use A ( G ) to denote the adjacency matrix. The spectral radius ρ ( G ) is the largest modulus of eigenvalues of A ( G ). Nikiforov proposed a spectral Tur´an * Supported by the National Natural Science Foundation of China (Nos. 11971445, 11771141 and12011530064). † Corresponding author. E-mail addresses: [email protected] (M. Zhai); [email protected] (H.Lin); [email protected] (J. Shu). F -free graph with n vertices. This can be viewed as the spectral analogue of Tur´an type problem. Thespectral Tur´an problem has received a great deal of attention in the past decades. Themaximum spectral radius of various graphs have been determined for large enough n , forexample, K r -free graphs [21], K s , t -free graphs [1], C k + -free graphs [8], induced K s , t -freegraphs [14], { K , , K } -minor free and { K , , K } -minor free graphs [20]. For more resultsin this direction, readers are referred to a survey by Nikiforov [13].The problem of characterizing graphs of given size with maximal spectral radii wasinitially posed by Brualdi and Ho ff man [3] as a conjecture, and solved by Rowlinson [18].Nosal [16] showed that every triangle-free graph G on m edges satisfies that ρ ( G ) ≤ √ m .Very recently, Lin, Ning and Wu [7] slightly improved the bound to ρ ( G ) ≤ √ m − G is non-bipartite and triangle-free.If we vary the spectral Tur´an problem by fixing the number of edges instead of thenumber of vertices, then the problem asks to determine the maximum spectral radius ofan F -free graph with m edges. Nikiforov [9–11] focused his attention on this kind ofproblem. For a family of graph F and an integer m , let G ( m , F ) denote the family of F -free graphs with m edges and without isolated vertices. In particular, if F = { F } , thenwe write G ( m , F ) for G ( m , F ). Nikiforov solved this problem for F = K r + and F = C . Theorem 1.1. ( [9, 10]) Let G ∈ G ( m , K r + ) . Then ρ ( G ) ≤ q m (1 − r ) , with equality ifand only if G is a complete bipartite graph for r = , and G is a complete regular r-partitegraph for r ≥ . Theorem 1.2. ( [11]) Let G ∈ G ( m , C ) , where m ≥ . Then ρ ( G ) ≤ √ m, with equalityif and only if G is a star. Theorems 1.1 and 1.2 assert that G contains C and C if ρ ( G ) > √ m . Based on thisobservation, we wish to know which graphs will be a subgraph of G if ρ ( G ) > f ( m ),where f ( m ) ∼ √ m as m → ∞ . We obtain several results in this direction. Let S kn denotethe graph obtained from K , n − by adding k disjoint edges within its independent set. Let C + t denote the graph obtained from C t and C by identifying an edge. The first result isabout K , r + -free graphs and { C + , C + } -free graphs. Theorem 1.3.
Let G be a graph with size m. Then the following two statements hold.(i) If G ∈ G ( m , K , r + ) with r ≥ and m ≥ r , then ρ ( G ) ≤ √ m, and equality holds ifand only if G is a star.(ii) If G ∈ G ( m , { C + , C + } ) with m ≥ , then ρ ( G ) ≤ √ m, and equality holds if and only ifG is a complete bipartite graph or one of S , S and S . It is clear that G ( m , C ) ⊆ G ( m , K , r + ) for any positive integer r , since C (cid:27) K , .Moreover, note that each C -free or C -free graph is also { C + , C + } -free. Thus we have G ( m , C ) ∪ G ( m , C ) ⊆ G ( m , { C + , C + } ) . Therefore, Theorem 1.3 generalizes related resultson C -free graphs [16] and C -free graphs [11].Let us turn our focus to graphs without given cycles. To state our results, we needthe following definition. Let S n , k be the complete split graph , i.e., the graph obtained byjoining each vertex of K k to n − k isolated vertices. Clearly, S n , is a star with n − S n , is a book with n − Theorem 1.4.
If either G ∈ G ( m , C ) with m ≥ or G ∈ G ( m , C ) with m ≥ , then ρ ( G ) ≤ + √ m − , and equality holds if and only if G (cid:27) S m + , . The extremal question on cycles with consecutive lengths is also an important topic inextremal graph theory. From [2], we know that if G is a graph of order n with m ( G ) > ⌊ n ⌋ ,then G contains a cycle of length t for every t ≤ ⌊ n + ⌋ . Our last result follows the re-search of Nikiforov on the spectral conditions for the existence of cycles with consecutivelengths. Nikiforov [8] proved that if G is a graph of order n and ρ ( G ) > p ⌊ n / ⌋ , then G contains a cycle of length t for every t ≤ n . Recently, Ning and Peng [15] improvedNikiforov’s result by showing G contains a cycle of length t for every t ≤ n . We provean edge analogue result as follows. Theorem 1.5.
Let G be a graph of size m. If ρ ( G ) > k − + √ m + ( k − ) , then G contains acycle of length t for every t ≤ k + . Interestingly, Nikiforov [12] proved that if G is a graph of order n with ρ ( G ) > k + √ kn + k − k , then G contains an even cycle C t for every even number t ≤ k + We need introduce some notations. For a graph G and a subset S ⊆ V ( G ), let G [ S ]denote the subgraph of G induced by S . Let e ( G ) denote the size of G . For two vertexsubsets S and T of G (where S ∩ T may not be empty), let e ( S , T ) denote the numberof edges with one endpoint in S and the other in T . e ( S , S ) is simplified by e ( S ). For avertex v ∈ V ( G ), let N ( v ) be the neighborhood of v , N [ v ] = N ( v ) ∪ { v } and N ( v ) be the setof vertices of distance two to v . In particular, let N S ( v ) = N ( v ) ∩ S and d S ( v ) = | N S ( v ) | .It is known that A ( G ) is irreducible nonnegative for a connected graph G . From thePerron-Frobenius Theorem, there is a unique positive unit eigenvector corresponding to ρ ( G ), which is called the Perron vector of G . Lemma 2.1. ( [11])
Let A and A ′ be the adjacency matrices of two connected graphs Gand G ′ on the same vertex set. Suppose that N G ( u ) & N G ′ ( u ) for some vertex u. If thePerron vector X of G satisfies X T A ′ X ≥ X T AX, then ρ ( G ′ ) > ρ ( G ) . A graph is called , if it is a connected graph without cut vertex. A block is a maximal 2-connected subgraph of a graph G . In particular, an end-block is a blockcontaining at most one cut vertex of G . Throughout the paper, let G ∗ denote an extremalgraph with maximal spectral radius in G ( m , F ) for each fixed F . Let ρ ∗ = ρ ( G ∗ ) and let X ∗ be the Perron vector of G ∗ with coordinate x v corresponding to the vertex v ∈ V ( G ∗ ).A vertex u in G ∗ is said to be an extremal vertex and denoted by u ∗ , if x u = max { x v | v ∈ V ( G ∗ ) } . Let N i ( u ∗ ) = { v | v ∈ N ( u ∗ ) , d N ( u ∗ ) ( v ) = i } . Lemma 2.2.
If F is a 2-connected graph and u ∗ is an extremal vertex of G ∗ , then thefollowing claims hold. ( i ) G ∗ is connected. ( ii ) There exists no cut vertex in V ( G ∗ ) \{ u ∗ } , and hence d ( u ) ≥ for any u ∈ V ( G ∗ ) \ N [ u ∗ ] . ( iii ) If F is C -free, then N ( v ) = N ( v ) for any non-adjacent vertices v , v of degree two. Proof. (i) Otherwise, let H be a component of G ∗ with ρ ( H ) = ρ ∗ and H be anothercomponent. Selecting a vertex v ∈ V ( H ) and an edge w w ∈ E ( H ), let G be the graphobtained from G ∗ by adding an edge vw and deleting an edge w w with possible isolatedvertex. Note that vw is a cut edge in G . Then G ∈ G ( m , F ). Moreover, ρ ( G ) > ρ ( H ) = ρ ∗ ,a contradiction.(ii) Suppose to the contrary that there exists a cut vertex in V ( G ∗ ) \ { u ∗ } . Then G ∗ hasat least two end-blocks. We may assume that B is an end-block of G ∗ with u ∗ < V ( B ) anda cut vertex v ∈ V ( B ). Let G ′ = G ∗ + { u ∗ w | w ∈ N ( v ) ∩ V ( B ) } − { vw | w ∈ N ( v ) ∩ V ( B ) } .Note that F is 2-connected. Then G ′ is F -free, and X ∗ T ( A ( G ′ ) − A ( G ∗ )) X ∗ = X u i u j ∈ E ( G ′ ) x u i x u j − X u i u j ∈ E ( G ∗ ) x u i x u j = X w ∈ N ( v ) ∩ V ( B ) x u ∗ − x v ) x w ≥ . Note that N G ∗ ( u ∗ ) & N G ′ ( u ∗ ). By Lemma 2.1, ρ ( G ′ ) > ρ ∗ , a contradiction.(iii) Let v , v be a pair of non-adjacent vertices of G ∗ with N ( v ) = { u , u } and N ( v ) = { w , w } , where | N ( v ) ∩ N ( v ) | ≤
1. We may assume that x u + x u ≥ x w + x w .Define G ′′ = G ∗ + v u + v u − v w − v w . Notice that F is C -free and v , v are a pairof vertices of degree two with the same neighbourhood in G ′′ . Thus, G ′′ is still F -free.However, by Lemma 2.1, we have ρ ( G ′′ ) > ρ ∗ , a contradiction. (cid:3) We simplify A ( G ∗ ) by A . Then ρ ∗ x u ∗ = ( AX ∗ ) u ∗ = X u ∈ N ( u ∗ ) x u + X u ∈ N ( u ∗ ) \ N ( u ∗ ) x u . (1)Notice that ρ ∗ is the spectral radius of A . Thus, ρ ∗ x u ∗ = X u ∈ V ( G ∗ ) a (2) u ∗ u x u = d ( u ∗ ) x u ∗ + X u ∈ N ( u ∗ ) \ N ( u ∗ ) d N ( u ∗ ) ( u ) x u + X w ∈ N ( u ∗ ) d N ( u ∗ ) ( w ) x w , (2)where a (2) u ∗ u is the number of walks of length two from u ∗ to the vertex u . Equalities (1)and (2) will be frequently used in the following proof. We first consider K , r + -free graphs. Let G ∗ be an extremal graph for G ( m , K , r + ).Note that K , m ∈ G ( m , K , r + ). We have ρ ∗ ≥ ρ ( K , m ) = √ m . For convenience, let U = N ( u ∗ ) and W = V ( G ∗ ) \ N [ u ∗ ], where u ∗ is an extremal vertex of G ∗ . Then (2)becomes ρ ∗ x u ∗ = | U | x u ∗ + X u ∈ U d U ( u ) x u + X w ∈ W d U ( w ) x w . (3)Notice that G ∗ is K , r + -free. Then we have the following claim. Claim 3.1. d U ( u ) ≤ r for any vertex u ∈ U ∪ W . Furthermore, we give the following claims.
Claim 3.2.
If e ( U ) = , then G ∗ (cid:27) K , m . Proof.
Note that ρ ∗ ≥ √ m . By (3), we have mx u ∗ ≤ ρ ∗ x u ∗ ≤ ( | U | + e ( U ) + e ( U , W )) x u ∗ . This implies that e ( W ) ≤ e ( U ) . If e ( U ) =
0, then e ( W ) =
0. Therefore, G ∗ is a bipartitegraph and thus G ∗ is triangle-free. By Theorem 1.1, G ∗ (cid:27) K s , t for some s and t with s ≤ t .Note that m ≥ r . If s ≥
2, then G ∗ contains K , r + as a subgraph. Hence s =
1, asdesired. (cid:3)
Now let U = { u ∈ U | P w ∈ N W ( u ) d W ( w ) > r } and U = U \ U . Claim 3.3.
If U , ∅ , then P u ∈ U d U ( u ) < e ( W ) . Proof.
Denote N W ( U ) = ∪ u ∈ U N W ( u ). Given w ∈ N W ( U ). By Claim 3.1, d U ( w ) ≤ r ,and hence d W ( w ) is counted at most r times in P u ∈ U P w ∈ N W ( u ) d W ( w ). Thus we have X w ∈ N W ( U ) d W ( w ) ≥ r X u ∈ U X w ∈ N W ( u ) d W ( w ) > r X u ∈ U r = r | U | . (4)Note that N W ( U ) ⊆ W . Thus by (4), e ( W ) = X w ∈ W d W ( w ) ≥ X w ∈ N W ( U ) d W ( w ) > r | U | . It follows from Claim 3.1 that 12 X u ∈ U d U ( u ) ≤ r | U | < e ( W ) , as desired. (cid:3) Now let U ′ = { u ∈ U | d W ( u ) > r ( ρ ∗ + r )2( ρ ∗ − r ) } and U ′′ = U \ U ′ . Claim 3.4.
For any u ∈ U ′ , d U ( u ) x u + P w ∈ N W ( u ) x w < d W ( u ) x u ∗ . Proof.
Let u ∈ U ′ . By Claim 3.1, d U ( u ) ≤ r and | N U ( w ) | ≤ r for any w ∈ N W ( u ). By thedefinition of U , P w ∈ N W ( u ) | N W ( w ) | ≤ r . Thus, ρ ∗ X w ∈ N W ( u ) x w = X w ∈ N W ( u ) ρ ∗ x w = X w ∈ N W ( u ) X v ∈ N W ( w ) x v + X v ∈ N U ( w ) x v ≤ ( r + d W ( u ) · r ) x u ∗ . (5)Note that d W ( u ) > r ( ρ ∗ + r )2( ρ ∗ − r ) . By (5) we have12 d U ( u ) x u + X w ∈ N W ( u ) x w ≤ r x u ∗ + r ρ ∗ ( r + d W ( u )) x u ∗ < d W ( u ) x u ∗ , as claimed. (cid:3) Claim 3.5.
For any u ∈ U ′′ , x u < x u ∗ . Proof.
Let u ∈ U ′′ . By (5), we have ρ ∗ x u = x u ∗ + X v ∈ N U ( u ) x v + X w ∈ N W ( u ) x w ≤ + r + r ρ ∗ ( r + d W ( u )) ! x u ∗ . (6)By the definition of U ′′ , we have d W ( u ) ≤ r ( ρ ∗ + r )2( ρ ∗ − r ) . Thus by (6), we have ρ ∗ x u ≤ + r + r ρ ∗ − r ) ! x u ∗ . (7)In order to show that x u < x u ∗ , it su ffi ces to show that 1 + r + r ρ ∗ − r ) < ρ ∗ , or equivalently, ρ ∗ − (3 r + ρ ∗ − ( r − r ) > . Since r ≥ ρ ∗ ≥ √ m ≥ r , we have ρ ∗ − (3 r + ρ ∗ − ( r − r ) ≥ r ( r − ≥ . (8)And if equality in (8) holds, then all the above inequalities will be equalities. In particular,(5) becomes an equality. This implies that x v = x u ∗ for each w ∈ N W ( u ) and v ∈ N U ( w ).Note that u ∈ N U ( w ) for each w ∈ N W ( u ). Then x u = x u ∗ . However, (7) and (8) imply that x u ≤ x u ∗ , a contradiction. Hence, (8) is a strict inequality. The claim follows. (cid:3) In the following, we give the proof of Theorem 1.3 (i).
Proof.
Note that ρ ∗ ≥ √ m . According to (3), we have( m − | U | ) x u ∗ ≤ X u ∈ U d U ( u ) x u + X w ∈ W d U ( w ) x w , (9)where X u ∈ U d U ( u ) x u = X u ∈ U ∪ U ′ d U ( u ) x u + X u ∈ U ′′ d U ( u ) x u + X u ∈ U ∪ U ′ d U ( u ) x u . On one hand, by Claim 3.5, x u < x u ∗ for u ∈ U ′′ . Thus,12 X u ∈ U ∪ U ′ d U ( u ) x u + X u ∈ U ′′ d U ( u ) x u ≤ X u ∈ U d U ( u ) x u ∗ = e ( U ) x u ∗ , (10)where inequality is strict if U ′′ , ∅ and e ( U ′′ , U ) ,
0. On the other hand, by Claim 3.3and Claim 3.4,12 X u ∈ U ∪ U ′ d U ( u ) x u ≤ e ( W ) x u ∗ + X u ∈ U ′ d W ( u ) x u ∗ − X w ∈ N W ( u ) x w , (11)where inequality is strict if U ∪ U ′ , ∅ . Moreover, X u ∈ U ′ d W ( u ) x u ∗ − X w ∈ N W ( u ) x w ≤ X u ∈ U d W ( u ) x u ∗ − X w ∈ N W ( u ) x w = e ( U , W ) x u ∗ − X w ∈ W d U ( w ) x w . Combining with (11), we have12 X u ∈ U ∪ U ′ d U ( u ) x u ≤ ( e ( W ) + e ( U , W )) x u ∗ − X w ∈ W d U ( w ) x w . (12)Combining (10), (12) with (9), we have( m − | U | ) x u ∗ ≤ ( e ( U ) + e ( W ) + e ( U , W )) x u ∗ = ( m − | U | ) x u ∗ . This implies that, if m , | U | , then both (10) and (11) are equalities. Recall that if (11) isan equality then U ∪ U ′ = ∅ (i.e., U = U ′′ ), and if (10) is an equality then either U ′′ = ∅ or e ( U ′′ , U ) =
0. Note that U , ∅ . It follows that U ′′ = U , ∅ and thus e ( U ) =
0. ByClaim 3.2, we have G ∗ (cid:27) K , m . If m = | U | , we also have G ∗ (cid:27) K , m . This completes theproof. (cid:3) In what follows, we consider { C + , C + } -free graphs. Before the final proof, we shallgive some lemmas on characterizing the extremal graph G ∗ . Recall that S kn is the graphobtained from K , n − by adding k disjoint edges within its independent set. Lemma 3.1. ( [11]) ρ ( S kn ) is the largest root of the polynomialf ( x ) = x − x − ( n − x + n − − k . Moreover, ρ ( S n ) > √ n for ≤ n ≤ . Remark 3.1.
Note that S m is a { C + , C + } -free graph of size m. Lemma 3.1 implies that thecondition m ≥ in Theorem 1.3 (ii) is best possible. Lemma 3.1 also implies the following result.
Lemma 3.2. ρ ( S kn ) = p e ( S kn ) = for ( n , k ) ∈ { (9 , , (8 , , (7 , } . Proof.
For ( n , k ) ∈ { (9 , , (8 , , (7 , } , we have e ( S kn ) = k = − n . Let ρ = ρ ( S kn ).By Lemma 3.1, we have ρ − ρ − ( n − ρ + n − = ( ρ − ρ + ρ − n + = . Note that n ≤
9. The largest root of above polynomial is ρ =
3, as desired. (cid:3)
Recall that N i ( u ∗ ) = { u | u ∈ N ( u ∗ ) , d N ( u ∗ ) ( u ) = i } . Since G ∗ is C + -free, we cansee that G ∗ [ N ( u ∗ )] may consist of isolated vertices and isolated edges. Thus, N ( u ∗ ) = N ( u ∗ ) ∪ N ( u ∗ ). Let N i ( u ∗ ) = { w | w ∈ N ( u ∗ ) , d N i ( u ∗ ) ( w ) ≥ } for i ∈ { , } . Note that G ∗ is { C + , C + } -free. Then N ( u ∗ ) ∩ N ( u ∗ ) = ∅ and d N ( u ∗ ) ( w ) = w ∈ N ( u ∗ ) . Since G ∗ is an extremal graph, ρ ∗ ≥ ρ ( K , m ) = √ m . Combining with (2), we have( m − d ( u ∗ )) x u ∗ ≤ X u ∈ N ( u ∗ ) x u + X w ∈ N ( u ∗ ) x w + X w ∈ N ( u ∗ ) d N ( u ∗ ) ( w ) x w . (13)By (13) and the definition of u ∗ , we have( m − d ( u ∗ )) x u ∗ ≤ (2 e ( N ( u ∗ )) + e ( N ( u ∗ ) , N ( u ∗ ))) x u ∗ . (14)Let W = V ( G ∗ ) \ N [ u ∗ ]. Note that e ( N ( u ∗ )) = e ( N ( u ∗ )) . (13) and (14) imply that e ( W ) ≤ e ( N ( u ∗ )) , (15)and if e ( W ) = e ( N ( u ∗ )), then x u = x u ∗ for any vertex u ∈ N ( u ∗ ) ∪ N ( u ∗ ). ① ① ①✛✚ ✘✙① ① u ∗ u i − u i ① ① ①✛✚ ✘✙ u ∗ u i − u i w i w ′ i Fig. 1:
The two cases when i ∈ A . Lemma 3.3.
If m ≥ , then e ( W ) = . Proof. If N ( u ∗ ) = ∅ , then by (15), e ( W ) =
0. In the following, let N ( u ∗ ) , ∅ and E ( G ∗ [ N ( u ∗ )]) = { u i − u i | i = , , . . . , t } . Recall that G ∗ [ N ( u ∗ )] consists of isolatededges. For each vertex u i ∈ N ( u ∗ ), let φ ( u i ) be the number of paths u i ww ′ with w , w ′ ∈ W .Clearly, φ ( u i ) = P w ∈ N W ( u i ) d W ( w ) . Furthermore, we can divide { , , . . . , t } into two subsets A and A , where A = { i | φ ( u i − ) + φ ( u i ) ≤ } , A = { i | φ ( u i − ) + φ ( u i ) ≥ } . Note that G ∗ is C + -free. Thus, N W ( u i − ) ∩ N W ( u i ) = ∅ for i ∈ { , , . . . , t } , and hence e ( W ) ≥ X i ∈ A ( φ ( u i − ) + φ ( u i )) ≥ | A | . (16)Moreover, we have the following claim. Claim 3.6. x u i − + x u i < x u ∗ for each i ∈ A . Proof.
By Lemma 2.2, d ( w ) ≥ w ∈ W . Recall that d N ( u ∗ ) ( w ) = w ∈ N ( u ∗ ). Thus, d W ( w ) ≥ w ∈ N ( u ∗ ). Given i ∈ A . If φ ( u i − ) + φ ( u i ) =
1, then | N W ( u i − ) ∪ N W ( u i ) | =
1. Moreover, d W ( w i ) = w i ∈ N W ( u i − ) ∪ N W ( u i ). We may assume without loss of generality that N W ( u i − ) = { w i } and N W ( w i ) = { w ′ i } (see Fig.1). Then ρ ∗ x u i = x u ∗ + x u i − and ρ ∗ x w i = x w ′ i + x u i − ≤ ρ ∗ x u i . Thus, ρ ∗ x u i − = ρ ∗ ( x u ∗ + x u i + x w i ) ≤ ( ρ ∗ + x u ∗ + x u i − . It follows that x u i − ≤ ρ ∗ + ρ ∗ − x u ∗ and x u i = ρ ∗ ( x u ∗ + x u i − ) ≤ ρ ∗ + ρ ∗ − x u ∗ . Since ρ ∗ ≥ √ m ≥ √
13, we have 2 ρ ∗ + < ρ ∗ −
2. Thus, x u i − + x u i < x u ∗ , as desired.If φ ( u i − ) + φ ( u i ) =
0, then | N W ( u i − ) ∪ N W ( u i ) | = x u i − = x u i and ρ ∗ x u i − = x u ∗ + x u i . Hence, x u i − = ρ ∗ − x u ∗ and x u i − + x u i = ρ ∗ − x u ∗ < x u ∗ . (cid:3) ③③③ ③③③③③③ u ∗ Fig. 2:
The extremal graph G ∗ with e ( G ∗ ) = t = e ( W ) = e ( N ( u ∗ ) , N ( u ∗ )) = | N ( u ∗ ) | = According to Claim 3.6, if A , ∅ , then X u ∈ N ( u ∗ ) x u = X i ∈ A ∪ A ( x u i − + x u i ) < ( | A | + | A | ) x u ∗ = ( e ( N ( u ∗ )) + | A | ) x u ∗ . Thus by (13), we have( m − d ( u ∗ )) x u ∗ < ( e ( N ( u ∗ )) + | A | + e ( N ( u ∗ ) , N ( u ∗ ))) x u ∗ . This implies that e ( W ) < | A | , which contradicts (16). Hence, A = ∅ , that is, N ( u ∗ ) = ∪ i ∈ A { u i − , u i } . By (13), we have( m − d ( u ∗ )) x u ∗ ≤ X i ∈ A ( x u i − + x u i ) + X w ∈ N ( u ∗ ) d N ( u ∗ ) ( w ) x w ≤ (2 | A | + e ( N ( u ∗ ) , N ( u ∗ ))) x u ∗ . (17)Notice that | A | = e ( N ( u ∗ )). (17) implies that e ( W ) ≤ | A | . Combining with (16), we have e ( W ) = | A | . Now, both (16) and (17) become equalities. Therefore, φ ( u i − ) + φ ( u i ) = i ∈ A , and x u = x u ∗ for any vertex u ∈ N ( u ∗ ) ∪ N ( u ∗ ). Note that x u ≤ ρ ∗ x u ∗ < x u ∗ for any vertex u of degree two. It follows that d ( u ) ≥ u ∈ N ( u ∗ ) ∪ N ( u ∗ ).If A , ∅ , then for any i ∈ A , both u i − and u i have neighbors in W and each of theirneighbors is of degree at least 3. Hence φ ( u i − ) + φ ( u i ) ≥
4, a contradiction. Therefore, A = ∅ and e ( W ) = | A | = (cid:3) Lemma 3.4. If ≤ m ≤ , then e ( W ) = . Proof.
Let t be the number of triangles in G ∗ [ N [ u ∗ ]]. It follows from (15) that e ( W ) ≤ e ( N ( u ∗ )) = t , and if e ( W ) = e ( N ( u ∗ )), then x u = x u ∗ for any vertex u ∈ N ( u ∗ ) ∪ N ( u ∗ ). By Lemma 2.2, d ( u ) ≥ u ∈ N ( u ∗ ) ∪ N ( u ∗ ). Assume that W , ∅ . First, we show that e ( W ) ≤ d ( u ) = u ∈ N ( u ∗ ) ∪ N ( u ∗ ), then ρ ∗ x u = P u ∈ N ( u ) x u ≤ x u ∗ and x u ≤ ρ ∗ x u ∗ < x u ∗ (since ρ ∗ ≥ √ m ≥ e ( W ) < e ( N ( u ∗ )), that is, e ( N ( u ∗ )) ≥ e ( W ) +
1. Therefore, e ( G ∗ ) ≥ t + e ( W ) + e ( N ( u ∗ ) , N ( u ∗ )) ≥ e ( W ) + + | N ( u ∗ ) | . ★★★◗◗◗ ✇✇ ✇ ✇✇ ✇✇ ✇✇✇✇ ✇✇ ✇✇ ✇✇ ✇ ✇✇✇✇✇✇ ✇✇ ✇ ✇✇✇ ✇ u ∗ v v v wu ∗ u ∗ u ∗ S S S S ∗ Fig. 3:
Possible extremal graphs for m = Note that e ( G ∗ ) ≤
12. Thus e ( W ) ≤
2, and if e ( W ) = | N ( u ∗ ) | =
1. Now we canfind pendant vertices in W , which contradicts Lemma 2.2. Hence, e ( W ) ≤ d ( u ) ≥ u ∈ N ( u ∗ ) ∪ N ( u ∗ ). However, d N [ u ∗ ] ( u ) = u ∈ N ( u ∗ ) and d N ( u ∗ ) ( w ) = w ∈ N ( u ∗ ) . Hence, | N ( u ∗ ) | ≥ | N ( u ∗ ) | = t .Furthermore, e ( G ∗ ) ≥ t + e ( W ) + e ( N ( u ∗ ) , N ( u ∗ )) ≥ e ( W ) + | N ( u ∗ ) | ≥ e ( W ) . (18)If e ( W ) ≥
2, then all the inequalities in (18) must be equalities. This implies that e ( G ∗ ) = t = e ( W ) = e ( N ( u ∗ ) , N ( u ∗ )) = | N ( u ∗ ) | = N ( u ∗ ) are of degree two, a contradiction. Thus we also have e ( W ) ≤ e ( W ) = w w is the unique edge of G ∗ [ W ]. Recall that d ( w ) ≥ w ∈ W and d N ( u ∗ ) ( w ) = w ∈ N ( u ∗ ) ⊆ W . Hence, W \ { w , w } ⊆ N ( u ∗ ). If w , w < N ( u ∗ ), then d ( u ) = u ∈ N ( u ∗ ). If w i ∈ N ( u ∗ ) for some i ∈ { , } , then d ( w i ) =
2, and d ( u ) , u ∈ N ( u ∗ ). This implies thatthere are at least | N ( u ∗ ) | vertices of degree two in N ( u ∗ ) ∪ { w , w } . Note that x u ≤ ρ ∗ x u ∗ for any vertex u of degree two. Thus by (13), we have( m − d ( u ∗ )) x u ∗ ≤ e ( N ( u ∗ )) + e ( N ( u ∗ ) , N ( u ∗ )) − (1 − ρ ∗ ) | N ( u ∗ ) | ! x u ∗ . (19)Notice that | N ( u ∗ ) | = e ( N ( u ∗ )) and ρ ∗ ≥
3. (19) implies that e ( W ) ≤ e ( N ( u ∗ )) = t . It follows that m > t + e ( W ) ≥ e ( W ) =
10 and hence ρ ∗ ≥ √ m >
3. Again by (19)we have e ( W ) < e ( N ( u ∗ )) . Thus, t ≥ e ( W ) + m > t + e ( W ) >
12, acontradiction. Therefore, e ( W ) =
0, as desired. (cid:3)
In the following, we give the proof of Theorem 1.3 (ii).
Proof.
From Lemmas 3.3 and 3.4, we have e ( W ) =
0. If N ( u ∗ ) = ∅ , then G ∗ is triangle-free. By Theorem 1.1, G ∗ is a complete bipartite graph, as desired.Next, assume that N ( u ∗ ) , ∅ . Then e ( N ( u ∗ )) ≥
1. Note that d ( w ) ≥ d N ( u ∗ ) ( w ) = w ∈ N ( u ∗ ). Then d W ( w ) ≥ w ∈ N ( u ∗ ). However e ( W ) =
0, thus, N ( u ∗ ) = ∅ and W = N ( u ∗ ) . Hence, d ( u i − ) = d ( u i ) = u i − u i of G ∗ [ N ( u ∗ )]. Similarly as above, we have x u i − = x u i = ρ ∗ − x u ∗ . By (13), we have( m − d ( u ∗ )) x u ∗ ≤ X u ∈ N ( u ∗ ) x u + X w ∈ N ( u ∗ ) d N ( u ∗ ) ( w ) x w ≤ e ( N ( u ∗ )) ρ ∗ − + e ( N ( u ∗ ) , N ( u ∗ )) ! x u ∗ . (20)1It follows that e ( W ) ≤
0, since m ≥ ρ ∗ ≥ √ m ≥
3. This implies that e ( W ) = m = x w = x u ∗ for any vertex w ∈ N ( u ∗ ).Given any w ∈ N ( u ∗ ), ρ ∗ x u ∗ = ρ ∗ x w = P u ∈ N ( w ) x u ≤ d ( w ) x u ∗ . It follows that d ( w ) ≥ ρ ∗ ≥
3. Hence | N ( u ∗ ) | ≥ e ( G ∗ ) ≥ e ( N ( u ∗ )) + | N ( u ∗ ) | + | N ( u ∗ ) | . Since e ( G ∗ ) = | N ( u ∗ ) | ≤
1. If | N ( u ∗ ) | =
1, then e ( N ( u ∗ )) = | N ( u ∗ ) | = G ∗ (cid:27) S ∗ (seeFig.3). Note that d ( v i ) = v i of w . Then, x v i ≤ ρ ∗ x u ∗ < x u ∗ and hence x w = ρ ∗ ( x v + x v + x v ) < x u ∗ . This implies that (20) is a strict inequality, that is, e ( W ) < N ( u ∗ ) = ∅ , i.e., W = ∅ . Then G ∗ is isomorphic to one of S , S and S (see Fig.3). This completes the proof. (cid:3) In this section, we characterize the extremal graph G ∗ for C -free or C -free graphs.The following first lemma was proposed as a conjecture by Guo, Wang and Li [6] andconfirmed by Sun and Das [19] recently. Lemma 4.1. ( [19])
Let v ∈ V ( G ) with d G ( v ) ≥ . Then ρ ( G ) ≤ p ρ ( G − v ) + d G ( v ) − .The equality holds if and only if either G (cid:27) K n or G (cid:27) K , n − with d G ( v ) = . Lemma 4.2. ( [17])
Let G be a connected graph, and X = ( x u , x u , . . . , x u n ) T be its eigen-vector corresponding to ρ = ρ ( G ) with k X k p = . Then for ≤ p < ∞ , the maximumcoordinate x u ≤ ( ρ p − + ρ p − ) p . Now we come back to consider the desired extremal graph S m + , in Theorem 1.4. Lemma 4.3.
Let ρ ⋆ = ρ ( S m + , ) . Then ρ ⋆ is the largest root of ρ − ρ − ( m − = . Andhence, ρ ⋆ = + √ m − . Proof.
Let X be the Perron vector of S m + , . Let u , u be the dominating vertices, and v , v , . . . , v m − be the remaining vertices of degree two in S m + , . By symmetry, we have ρ ⋆ x u = x u + m − x v , ρ ⋆ x v = x u . Clearly, ρ ⋆ − ρ ⋆ − ( m − = (cid:3) Let R k denote the graph obtained from k copies of K by sharing a vertex and H t , s = h T , S i be a bipartite graph with | T | = t and | S | = s (note that H t , s is not necessarilycomplete bipartite). Let H t , s ◦ R k denote the graph obtained by joining the dominatingvertex u of R k with the independent set T of H t , s , for k ≥ t , s ≥ Lemma 4.4.
If k ≥ and m = k + t ≥ , then ρ ( H t , ◦ R k ) < + √ m − . Proof.
Let ρ = ρ ( H t , ◦ R k ), and X be the Perron vector of H t , ◦ R k . Let u be the dominatingvertex, u , u , . . . , u k be the remaining vertices of R k , and v , v , . . . , v t be the vertices in H t , . By symmetry, we have ρ x v = x u , ρ x u = kx u + tx v , ρ x u = x u + x u . ✈✈✈✈✈✈✈ v v v v t w w w s ✈ ✈ ✈✈✈✈ ✈ u u u u ... ... ... Fig. 4: The graph H t , s ◦ R k , where k ≥ t , s ≥ f ( ρ ) = ρ − ρ − (3 k + t ) ρ + t = . (21)Let ρ ⋆ = + √ m − . Then ρ ⋆ = ρ ⋆ + ( m − = ρ ⋆ + (6 k + t − . (22)Substituting (22) to (21), we have f ( ρ ⋆ ) = (3 k − ρ ⋆ − (6 k − t − ρ ⋆ > m ≥ f ( ρ ⋆ ) > k + t − ≥
0. Moreover, (22) implies that f ′ ( ρ ) = ρ − ρ − (3 k + t ) > ρ ≥ ρ ⋆ . Hence, f ( ρ ) > ρ ≥ ρ ⋆ . Thus, ρ ( H t , ◦ R k ) < ρ ⋆ . (cid:3) Remark 4.1.
If m = , then t = k = , ρ ⋆ = and f ( ρ ⋆ ) = k + t − = − . Hence, ρ ( H , ◦ R ) > ρ ⋆ . Notice that H , ◦ R is C -free. The condition m ≥ is best possiblefor Theorem 1.4 (i). We now characterize the extremal graph G ∗ for C -free graphs. Note that G ∗ [ N ( u ∗ )]contains no paths of length 3 for forbidding C . Thus, we have the following result on thelocal structure of G ∗ . Lemma 4.5.
Each connected component of G ∗ [ N ( u ∗ )] is either a triangle or a star K , r for some r ≥ , where K , is a singleton component. Notice that S m + , is C t -free for t ≥
5. By Lemma 4.3, we have ρ ∗ ≥ ρ ( S m + , ) = + √ m − . Recall that N ( u ∗ ) = { u | u ∈ N ( u ∗ ) , d N ( u ∗ ) ( u ) = } and W = V ( G ∗ ) \ N [ u ∗ ].For convenience, let W = N W ( N ( u ∗ )), i.e., ∪ u ∈ N ( u ∗ ) N W ( u ). The following lemma gives aclearer local structure of the extremal graph G ∗ . Lemma 4.6. e ( W ) = and W = W . Proof.
According to (1) and (2), we have( ρ ∗ − ρ ∗ ) x u ∗ = | N ( u ∗ ) | x u ∗ + X u ∈ N ( u ∗ ) \ N ( u ∗ ) ( d N ( u ∗ ) ( u ) − x u − X u ∈ N ( u ∗ ) x u + X w ∈ W d N ( u ∗ ) ( w ) x w . (23)3Let N + ( u ∗ ) = N ( u ∗ ) \ N ( u ∗ ). Then( ρ ∗ − ρ ∗ ) x u ∗ ≤ | N ( u ∗ ) | + e ( N + ( u ∗ )) − | N + ( u ∗ ) | − X u ∈ N ( u ∗ ) x u x u ∗ + e ( W , N ( u ∗ )) x u ∗ (24)Note that ρ ∗ ≥ + √ m − , that is, ρ ∗ − ρ ∗ ≥ m −
1. Combining with (24), we have e ( W ) ≤ e ( N + ( u ∗ )) − | N + ( u ∗ ) | − X u ∈ N ( u ∗ ) x u x u ∗ + , (25)since e ( N ( u ∗ )) = e ( N + ( u ∗ )). According to (23) and (24), the equality in (25) holds if andonly if ρ ∗ − ρ ∗ = m − x w = x u ∗ for any w ∈ W and u ∈ N + ( u ∗ ) with d N ( u ∗ ) ( u ) ≥
2. ByLemma 4.5, e ( N + ( u ∗ )) ≤ | N + ( u ∗ ) | and hence e ( W ) ≤ . Assume that e ( W ) =
1. Let w i w j be the unique edge in G ∗ [ W ]. By Lemma 2.2, d ( w ) ≥ w ∈ W . This implies that both w i and w j have neighbors in N ( u ∗ ). Notethat G ∗ is C -free. Thus, w i , w j must have the same and the unique neighbor u in N ( u ∗ ).Now u is a cut vertex, which contradicts Lemma 2.2. Hence, e ( W ) = . Suppose that W \ W , ∅ . For the reason of C -free, we can see that d ( w ) ≤ d ( w ) =
2) for each w ∈ W \ W . By Lemma 2.2, all the vertices in W \ W havethe same neighborhood, say, { u , u } . For avoiding pentagons, u and u are in the samecomponent H of G ∗ [ N ( u ∗ )] and H can only be a copy of K , . Let G = G ∗ − { wu | w ∈ W \ W } + { wu ∗ | w ∈ W \ W } . Clearly, G is C -free. Note that x u ∗ ≥ x u and N G ∗ ( u ∗ ) & N G ( u ∗ ).By Lemma 2.1, ρ ( G ) > ρ ∗ , a contradiction. Hence, W = W . (cid:3) Lemma 4.7.
Either G ∗ (cid:27) S m + , or G ∗ (cid:27) H t , s ◦ R k for some k ≥ and some bipartitegraph H t , s of size m − k − t. Proof.
By Lemma 4.6 and (25), we have e ( N + ( u ∗ )) ≥ | N + ( u ∗ ) | + X u ∈ N ( u ∗ ) x u x u ∗ − . (26)Firstly, assume that N ( u ∗ ) , ∅ . Then P u ∈ N ( u ∗ ) x u x ∗ u > e ( N + ( u ∗ )) ≥ | N + ( u ∗ ) | .It follows from Lemma 4.5 that G ∗ [ N + ( u ∗ )] consists of disjoint union of k triangles. If k =
0, then G ∗ is triangle-free. By Theorem 1.1, ρ ∗ ≤ √ m < + √ m − , a contradiction.Thus, k ≥
1. By Lemma 4.6, G ∗ (cid:27) H | N ( u ∗ ) | , | W | ◦ R k for some bipartite graph H | N ( u ∗ ) | , | W | .Secondly, assume that N ( u ∗ ) = ∅ . Then W = ∅ , and by Lemma 4.6, W = ∅ . Thus,(25) becomes e ( N + ( u ∗ )) − | N + ( u ∗ ) | ≥ −
1. Let c be the number of star-components of G ∗ [ N + ( u ∗ )]. Then e ( N + ( u ∗ )) − | N + ( u ∗ ) | = − c . It follows that c ≤
1. If c =
0, then G ∗ (cid:27) R k for k = m . By Lemma 4.4, ρ ∗ < + √ m − , a contradiction. Thus, c =
1. Let k be the numberof triangle-components of G ∗ [ N + ( u ∗ )]. If k ≥
1, then G ∗ is isomorphic to K , r • R k for some r ≥ ρ ∗ x u = x u + x u ∗ . Since m ≥ ρ ∗ ≥ + √ m − > x u = x u ∗ ρ ∗ − < x u ∗ . This implies that (25) is strict. Correspondingly, e ( W ) < e ( N + ( u ∗ )) − | N + ( u ∗ ) | + = − c + = , a contradiction. Hence, k =
0. Now G ∗ [ N ( u ∗ )] (cid:27) K , r and G ∗ is obtained by joining u ∗ with each vertex of K , r . Thus, G ∗ (cid:27) S m + , . This completes the proof. (cid:3) ✈ v ✈ ✈ ✈✈✈✈ ✈ u ∗ u u u ... v r ✈ v ✈ ... w ✈ Fig. 5: K , r • R k for r ≥ k ≥ Proof.
Suppose that G ∗ (cid:27) H t , s ◦ R k for some k ≥ H t , s = < T , S > (see Fig.4). By the symmetry, ρ ∗ x u i = x u i + x u for each vertex u i in any K -block of G ∗ .Recall that ρ ∗ >
3. Thus, x u i = x u ρ ∗ − < x u . If x v ≥ x u for some vertex v , u , then v ∈ T ∪ S .Thus we can select a block B (cid:27) K and define G = G ∗ + { v j | j ∈ N B ( u ) } − { u j | j ∈ N B ( u ) } .Clearly, G is C -free. And by Lemma 2.1, we have ρ ( G ) > ρ ∗ , a contradiction. It followsthat x v < x u for any vertex v , u .Notice that ρ ∗ ≥ + √ m − . Lemma 4.4 implies that s ≥
1. If s =
1, say S = { w } and v ∈ N ( w ), then we define G ′ = G ∗ + uw − v w . Clearly, G ′ is C -free. Since x v < x u ,we have ρ ( G ′ ) > ρ ∗ , a contradiction. Thus, s ≥
2. By Lemma 2.2, d ( w i ) ≥ w i ∈ S . Hence, t ≥ e ( T , S ) ≥ s .Without loss of generality, we may assume that x v , x v attain the largest two coordi-nates among all vertices in T and w attains the maximum degree among all vertices in S (see Fig.4). Since G ∗ is an extremal graph, all the vertices in S are adjacent to v and v .This implies that N ( v ) = N ( v ) and hence x v = x v . If e ( T , S ) = s , then d ( w ) = ρ ∗ x w = x v . If e ( T , S ) = s +
1, then d ( w ) = ρ ∗ x w = x v + x v + x v ≤ x v , where v ∈ N ( w ) \ { v , v } . In both cases, x w < x v , since ρ ∗ >
3. Define G ′′ = G ∗ + v v − w v if e ( T , S ) = s or G ′′ = G ∗ + v v − w v if e ( T , S ) = s +
1. One can see that G ′′ is C -free and ρ ( G ′′ ) > ρ ∗ , a contradiction. Thus, e ( T , S ) ≥ s +
2. It follows that t ≥ m = k + t + e ( T , S ) ≥ k + s + . Let { u , u , u , u } be a 4-clique of G ∗ . Next, we useinduction on m to show ρ ∗ < + √ m − .Firstly, assume that k = . Recall that s ≥
2. Thus, m ≥
15. By Lemma 4.1, we have ρ ∗ ≤ ρ ( G ∗ − u ) + ≤ ρ ( G ∗ − u − u ) + ≤ ρ ( G ∗ − u − u − u ) + . Note that G ∗ − u − u − u is a bipartite graph of size m −
6. By Theorem 1.1, we have ρ ( G ∗ − u − u − u ) ≤ m −
6. Hence, ρ ∗ ≤ m + < ( + √ m − ) , since m ≥ k ≥ . Then m ≥
21. Note that x u = x u = x u (see Fig.4).Thus ρ ∗ x u = x u + x u . By Lemma 4.2, x u ≤ √ and thus x u = ( x u ρ ∗ − ≤ ρ ∗ − . (27)Let G ′′′ = G ∗ − { u i u j | ≤ i < j ≤ } . We have ρ ( G ′′′ ) ≥ X ∗ T ( A ( G ′′′ )) X ∗ = X ∗ T ( A ( G ∗ )) X ∗ − x u i x u j = ρ ∗ − x u ρ ( G ′′′ )( ρ ( G ′′′ ) − ≥ ( ρ ∗ − x u )( ρ ∗ − x u − > ρ ∗ − ρ ∗ − ρ ∗ − x u . Combing with (27), we have ρ ( G ′′′ )( ρ ( G ′′′ ) − > ρ ∗ − ρ ∗ − ρ ∗ − ρ ∗ − . (28)Note that e ( G ′′′ ) = m − G ′′′ belongs to the class of H t + , s ◦ R k − . On one hand, byinduction hypothesis, ρ ( G ′′′ ) < + √ m − , that is, ρ ( G ′′′ )( ρ ( G ′′′ ) − < m − . On the otherhand, recall that ρ ∗ ≥ + √ m − and m ≥
21. We have ρ ∗ ≥ ρ ∗ − ρ ∗ − ≤
3. Thusby (28), ρ ∗ − ρ ∗ < m − , that is, ρ ∗ < + √ m − .In both cases, we get a contradiction: ρ ∗ < + √ m − . Thus, G ∗ (cid:29) H t , s ◦ R k . Accordingto Lemma 4.7, G ∗ (cid:27) S m + , . This completes the proof. (cid:3) In what follows, we consider C -free graphs. Recall that W = V ( G ∗ ) \ N [ u ∗ ]. Let W H = N W ( V ( H )) for any component H of G ∗ [ N ( u ∗ )]. Since G ∗ is C -free, U H i ∩ U H j = ∅ for any two component H i and H j of G ∗ [ N ( u ∗ )], unless one is an isolated vertex and theother is a star. Note that ρ ∗ ≥ ρ ( S m + , ) = + √ m − , that is, ρ ∗ − ρ ∗ ≥ m − . And P w ∈ W d N ( u ∗ ) ( w ) x w ≤ e ( W , N ( u ∗ )) x u ∗ . Combining with (23), we have m − − | N ( u ∗ ) | − e ( W , N ( u ∗ )) + X u ∈ N ( u ∗ ) x u x u ∗ x u ∗ ≤ X u ∈ N ( u ∗ ) \ N ( u ∗ ) (cid:0) d N ( u ∗ ) ( u ) − (cid:1) x u . Define γ ( H ) = P u ∈ V ( H ) ( d H ( u ) − x u for any non-trivial component H of G ∗ [ N ( u ∗ )]. Then e ( N ( u ∗ )) + e ( W ) + X u ∈ N ( u ∗ ) x u x u ∗ − x u ∗ ≤ X H γ ( H ) , (29)where H takes over all non-trivial components of G ∗ [ N ( u ∗ )]. Lemma 4.8. G ∗ [ N ( u ∗ )] contains no any cycle of length four or more. Proof.
Since G ∗ is C -free, G ∗ [ N ( u ∗ )] does not contain any path of length 4 and any cycleof length more than 4. Moreover, if G ∗ [ N ( u ∗ )] is of circumference 4, then componentscontaining C are isomorphic to C , C + e or K ; if G ∗ [ N ( u ∗ )] is of circumference 3, thencomponents containing C are isomorphic to K , r + e for r ≥
2. Let H be the family ofcomponents of G ∗ [ N ( u ∗ )] which contain C and H ′ be the family of other components of G ∗ [ N ( u ∗ )]. Then H ′ consists of trees or unicyclic components. Hence, γ ( H ) ≤ (2 e ( H ) − | H | ) x u ∗ ≤ e ( H ) x u ∗ , (30)for any H ∈ H ′ . Moreover, note that G ∗ is C -free. Thus, W H ∩ W H = ∅ and e ( W H , W H ) = H , H ∈ H with H , H . We can see that e ( W ) ≥ X H ∈H e ( W H , W ) . (31)6Combining (30), (31) with (29) and abandoning the item P u ∈ N ( u ∗ ) x u x u ∗ , we have X H ∈H ( e ( H ) + e ( W H , W )) − x u ∗ ≤ X H ∈H γ ( H ) . Suppose that H , ∅ . In order to get a contradiction, it su ffi ces to show γ ( H ) < ( e ( H ) + e ( W H , W ) − x u ∗ for each H ∈ H . Let H ∗ ∈ H with V ( H ∗ ) = { u , u , u , u } . First, assume that W H ∗ = ∅ .Let x u i ∗ = max { x u i | ≤ i ≤ } . Then ρ ∗ x u i ∗ = P u ∈ N ( u i ∗ ) x u ≤ x u ∗ + x u i ∗ . Note that ρ ∗ ≥ + √ m − >
5, since m >
21. Hence x u i ∗ ≤ x u ∗ ρ ∗ − < x u ∗ and γ ( H ∗ ) ≤ (2 e ( H ∗ ) −| H ∗ | ) x u i ∗ < ( e ( H ∗ ) − x u ∗ , as desired. Next assume that W H ∗ , ∅ . We consider two cases.Firstly, there are k ≥ W H ∗ with mutual distinct neighbor in V ( H ∗ ). Since G ∗ is C -free, d N ( u ∗ ) ( w ) = w ∈ W H ∗ , and N H ∗ ( w ) = N H ∗ ( w ′ ) for each pair ofadjacent vertices w , w ′ ∈ W H ∗ . Let w , w , . . . , w k ∈ W H ∗ with mutual distinct neighborin V ( H ∗ ). Then { w , w , . . . , w k } is an independent set. By Lemma 2.2, d ( w i ) ≥ i ∈ { , , . . . , k } and d ( w i ) = w i . Thus, P ≤ i ≤ k d ( w i ) ≥ k −
1. Thisimplies that e ( W H ∗ , W ) ≥ k − e ( H ∗ ) ≤ e ( K ) ≤ e ( W H ∗ , W ) − k + ≤ e ( W H ∗ , W ) + . (32)Note that | H ∗ | =
4. By (32), we have γ ( H ∗ ) = X u i ∈ V ( H ∗ ) ( d H ∗ ( u i ) − x u i ≤ (2 e ( H ∗ ) − | H ∗ | ) x u ∗ ≤ ( e ( H ∗ ) + e ( W H ∗ , W ) − x u ∗ . (33)If (33) holds in equality, then (32) also holds in equality. This implies that k = H ∗ (cid:27) K and x u i = x u ∗ for each u i ∈ V ( H ∗ ). Since k =
2, we can find a vertex u i ∈ V ( H ∗ ) with N W ( u i ) = ∅ . Thus, ρ ∗ x u i = P u ∈ N ( u i ) x u ≤ x u ∗ . That is, x u i ≤ ρ ∗ x u ∗ < x u ∗ . Thus, (33) isstrict, as desired.Secondly, all vertices in W H ∗ have a common neighbor, say u , in V ( H ∗ ). Now N W ( u i ) = ∅ for i ∈ { , , } . Let x u = max { x u i | i = , , } . Thus, ρ ∗ x u ≤ ( x u ∗ + x u ) + ( x u + x u ) ≤ x u ∗ + x u ), i.e., x u ≤ ρ ∗ − x u ∗ < x u ∗ . Note that d H ∗ ( u ) ≤
3. By thedefinition of γ ( H ∗ ), we have γ ( H ∗ ) ≤ ( d H ∗ ( u ) − x u + (2 e ( H ∗ ) − d H ∗ ( u ) − x u < ( d H ∗ ( u ) − + ( 43 e ( H ∗ ) − d H ∗ ( u ) − ! x u ∗ ≤ ( 43 e ( H ∗ ) − x u ∗ . It follows that γ ( H ∗ ) < e ( H ∗ ) x u ∗ , since e ( H ∗ ) ≤
2. Moreover, note that d ( w ) ≥ w ∈ W H ∗ . Thus, e ( W H ∗ , W ) ≥ γ ( H ∗ ) < ( e ( H ∗ ) + e ( W H ∗ , W ) − x u ∗ . (cid:3) Lemma 4.8 implies that each component of G ∗ [ N ( u ∗ )] is a tree or a unicyclic graph K , r + e . Let c be the number of non-trivial tree-components of G ∗ [ N ( u ∗ )]. Then X H γ ( H ) ≤ X H (2 e ( H ) − | H | ) x u ∗ = ( e ( N ( u ∗ )) − c ) x u ∗ , H takes over all non-trivial components of G ∗ [ N ( u ∗ )]. Thus, (29) becomes e ( W ) + X u ∈ N ( u ∗ ) x u x u ∗ − x u ∗ ≤ − cx u ∗ . Equivalently, e ( W ) ≤ − c − X u ∈ N ( u ∗ ) x u x u ∗ . (34) Lemma 4.9. e ( W ) = and G ∗ [ N ( u ∗ )] contains no triangle. Proof.
By (34), we have e ( W ) ≤
1, and if e ( W ) =
1, then c = N ( u ∗ ) = ∅ . Supposethat e ( W ) =
1. Then, each component H of G ∗ [ N ( u ∗ )] is isomorphic to K , r + e for some r ≥
2. Let w w be the unique edge in G ∗ [ W ]. For avoiding hexagons, w , w belong tothe same W H and they must have a unique and common neighbor u ∈ V ( H ). This leads toa pendant triangle and a cut vertex u , which contradicts Lemma 2.2. Hence, e ( W ) = G ∗ [ N ( u ∗ )] contains triangles, that is, G ∗ [ N ( u ∗ )] contains acomponent H ∗ (cid:27) K , r + e . We have the following claims. Claim 4.1. H ∗ (cid:29) C . Proof.
Assume that H ∗ (cid:27) C and V ( H ∗ ) = { u , u , u } . If W H ∗ = ∅ , then x u = x u = x u and ρ ∗ x u = x u ∗ + x u . Thus, x u = x u ∗ ρ ∗ − < x u ∗ , since ρ ∗ >
5. Therefore, γ ( H ∗ ) = X ≤ i ≤ ( d H ∗ ( u i ) − x u i = x u < ( e ( H ∗ ) − x u ∗ . Note that e ( W ) =
0, and by (30), γ ( H ) ≤ e ( H ) x u ∗ for any other component H . Thus, (29)becomes an impossible inequality: e ( N ( u ∗ )) + X u ∈ N ( u ∗ ) x u x u ∗ − x u ∗ ≤ X H γ ( H ) < ( e ( N ( u ∗ )) − x u ∗ . Hence, W H ∗ , ∅ . Clearly, 2 ≤ d ( w ) ≤ | H ∗ | = w ∈ W H ∗ . If there exists w ∈ W H ∗ with d ( w ) =
3, then W H ∗ = { w } (otherwise we will get a hexagon). If d ( w ) = w ∈ W H ∗ , then by Lemma 2.2, all the vertices in W H ∗ have common neighborhood, say { u , u } . In both cases, we define G = G ∗ + { wu ∗ | w ∈ W H ∗ } − { wu | w ∈ W H ∗ } . Clearly, G is C -free. Moreover, since x u ∗ ≥ x u , by Lemma 2.1, ρ ( G ) > ρ ∗ , a contradiction. Thus,the claim holds. (cid:3) Claim 4.2. W H ∗ = ∅ and H ∗ is the unique non-trivial component of G ∗ [ N ( u ∗ )] . Proof.
Claim 4.1 implies that H ∗ (cid:27) K , r + e for some r ≥
3. Since G ∗ is C -free, d N ( u ∗ ) ( w ) = w ∈ W H ∗ . Note that e ( W ) =
0. Thus, d ( w ) = w ∈ W H ∗ ,which contradicts Lemma 2.2. Hence, W H ∗ = ∅ .Now V ( H ∗ ) contains r − H ∗ . Hence, H ∗ is the unique component which contains triangles.8Suppose that G ∗ [ N ( u ∗ )] contains a non-trivial tree-component H , i.e., c ≥
1. By (34), N ( u ∗ ) = ∅ and c =
1. Since no vertices of degree two out of H ∗ , we have W H , ∅ (otherwise, each leaf of H is of degree two in G ∗ ), and d ( w ) ≥ w ∈ W H . Notethat e ( W ) = G ∗ is C -free. We can see that H (cid:27) K , and | W H | =
1. Let W H = { w } and u be the central vertex of H . Clearly, G ∗ + wu ∗ − wu is C -free. And by Lemma 2.1, ρ ( G ∗ + uu ∗ − uv ) > ρ ∗ , a contradiction. The claim holds. (cid:3) Claim 4.3.
Let W = N W ( N ( u ∗ )) . Then W = ∅ . Proof.
Suppose that W , ∅ . Then N ( u ∗ ) , ∅ . By Lemma 2.2, d ( w ) , w ∈ W .And since no vertices of degree two out of H ∗ , d ( w ) , w ∈ N ( u ∗ ) ∪ W .If d ( u ) ≥ u ∈ N ( u ∗ ), then u has at least two neighbors w , w ∈ W with min { d ( w ) , d ( w ) } ≥
3. Note that e ( W ) = e ( W , V ( H ∗ )) = u , u ∈ N ( u ∗ ) \ { u } with u i ∈ N ( w i ). Then u ∗ u w uw u u ∗ is a C (see Fig.6), a contradiction. Thus, d ( u ) = u ∈ N ( u ∗ ),that is, W = ∅ . (cid:3) ✛✚✘✙✛✚✘✙✇✇ ✇✇✇ ✫✪✬✩✇ ❝❝❝❝ w w u u u u ∗ H ∗ W N ( u ∗ )Fig. 6: The extremal graph G ∗ in Claim 3.According to Claims 4.1-4.3, W = ∅ , and G ∗ is obtained by joining u ∗ with H ∗ (cid:27) K , r + e and | N ( u ∗ ) | isolated vertices. Let u , u , u ∈ V ( H ∗ ) with d H ∗ ( u ) = d H ∗ ( u ) = d H ∗ ( u ) = r . Then x u = x u and ρ ∗ x u = x u + x u + x u ∗ ≤ x u + x u ∗ . Hence, x u ≤ ρ ∗ − x u ∗ < x u ∗ , since ρ ∗ > . Thus, γ ( H ∗ ) = ( r − x u + x u + x u ≤ ( r − x u ∗ + x u < rx u ∗ = ( e ( H ∗ ) − x u ∗ . Then, (29) becomes an impossible inequality: e ( H ∗ ) + X u ∈ N ( u ∗ ) x u x u ∗ − x u ∗ < ( e ( H ∗ ) − x u ∗ . This completes the proof. (cid:3)
In the following, we give the proof of Theorem 1.4 (ii).
Proof.
By Lemma 4.8 and Lemma 4.9, G ∗ [ N ( u ∗ )] consists of tree-components. (34)implies that c ≤
1, and if c = N ( u ∗ ) = ∅ . Assume that c =
0. Then G ∗ is bipartite,since e ( W ) =
0. By Theorem 1.1, ρ ∗ ≤ √ m < + √ m − , a contradiction. Hence, c = N ( u ∗ ) = ∅ . Let H be the unique component of G ∗ [ N ( u ∗ )].9Since G ∗ is C -free, diam ( H ) ≤
3. If diam ( H ) =
3, then H is a double star. Forforbidding C , d N ( u ∗ ) ( w ) = w ∈ W H . This implies W H = ∅ (otherwise, d ( w ) = e ( W ) = H contains two non-adjacent vertices u , u with d G ∗ ( u ) = d G ∗ ( u ) = N ( u ) , N ( u ). It contradicts Lemma 2.2. Thus, diam ( H ) ≤
2, i.e., H is astar K , r . Let V ( H ) = { u , u , . . . , u r } , where u is the center vertex of H . Note that ρ ∗ > ρ ∗ x u ∗ = P ≤ i ≤ r x u i ≤ ( r + x u ∗ . Hence, r > w ∈ W H with at least two neighbors, say u , u , in V ( H ) \ { u } , then u ∗ u wu u u u ∗ is a hexagon, a contradiction. It follows that d ( w ) = u ∈ N ( w ) forany w ∈ W H . By Lemma 2.2, all the vertices of W H have common neighbors, say u , u .However, N ( u ) = { u ∗ , u } , N ( w ) for w ∈ W H . This contradicts Lemma 2.2. Hence, W H = ∅ and G ∗ is the join of u ∗ with a star H . Clearly, G ∗ (cid:27) S m + , . This completes theproof. (cid:3)
In this section, we first consider the existence of cycles with consecutive lengths. Weneed introduce a well-known result due to Erd˝os and Gallai.
Lemma 5.1. ( [4]) For every k ≥ , ex ( n , P k + ) ≤ ( k − n, with equality if and only ifn = kt, the extremal graph is the disjoint union of K k ’s. In the following, we give the proof of Theorem 1.5.
Proof.
Let G be a graph of size m and Y be an eigenvector of A ( G ) corresponding to ρ = ρ ( G ) with P | G | i = y i =
1. Define f ( A ( G )) = A ( G ) − (2 k − A ( G ), where k is a positiveinteger. Then f ( A ( G )) Y = f ( ρ ) Y . Thus, f ( ρ ) = | G | X i = f ( ρ ) y i = | G | X i = ( f ( A ( G )) Y ) i = | G | X i = | G | X j = f i j y j = | G | X j = | G | X i = f i j y j = | G | X j = f j ( G ) y j , (35)where f i j is the ( i , j )-element of f ( A ( G )) and f j ( G ) is the sum of the j -th column of f ( A ( G )). For any j ∈ V ( G ), let U j = V ( G ) \ N [ j ], and g j ( G ) be the sum of the j -th columnof A ( G ). Clearly, g j ( G ) = X i ∈ N ( j ) d G ( i ) = d G ( j ) + e ( N ( j )) + e ( N ( j ) , U j ) . Furthermore, f j ( G ) = d G ( j ) + e ( N ( j )) + e ( N ( j ) , U j ) −
12 (2 k − d G ( j ) ≤ m + e ( N ( j )) −
12 (2 k − d G ( j ) . If ρ > k − + √ m + ( k − ) , then f ( ρ ) > m . (35) implies that there exists some j ∗ ∈ V ( G ) suchthat f j ∗ ( G ) > m . Thus, e ( N ( j ∗ )) > (2 k − d G ( j ∗ ). By Lemma 5.1, G [ N ( j ∗ )] contains acopy of P k + . It follows that G contains a cycle C t for every t ≤ k + (cid:3) Theorems 1.1-1.2 imply that if ρ ( G ) ≥ √ m , then G contains C and C unless G is star S m + , . From Theorem 1.4 we know that if ρ ( G ) ≥ + √ m − , then G contains C t for every t ≤ G is a book S m + , . These results inspire us to look for a more general spectralcondition for cycles with consecutive lengths, which is stated in the following conjecture.0 Conjecture 5.1.
Let k be a fixed positive integer and G be a graph of su ffi ciently largesize m without isolated vertices. If ρ ( G ) ≥ k − + √ m − k + , then G contains a cycle of lengtht for every t ≤ k + , unless G (cid:27) S mk + k + , k . On the other hand, we know that ρ ( G ) ≤ √ m for several kinds of graphs, such as,bipartite graphs, C -free graphs, C -free graphs, K , r + -free graphs, { C + , C + } -free graphs,and so on. A natural question: how large can a graph family be such that ρ ≤ √ m ? An r-book , denoted by B r , is the graph obtained from r triangles by sharing one edge. At theend of this paper, we pose the following conjecture. Conjecture 5.2.
Let G ∈ G ( m , B r + ) and m ≥ m ( r ) for large enough m ( r ) . Then ρ ( G ) ≤√ m, with equality if and only if G is a complete bipartite graph. Acknowledgements
We would like to show our great gratitude to anonymous refereesfor their valuable suggestions which largely improve the quality of this paper.
References [1] L. Babai, B. Guiduli, Spectral extrema for graphs: the Zarankiewicz problem,
Elec-tron. J. Combin. (1) (2009), Research Paper 123, 8 pp.[2] B. Bollob´as, Extremal graph theory, Academic Press, 1978.[3] R.A. Brualdi, A.J. Ho ff man, On the spectral radius of (0 ,
1) matrices,
Linear AlgebraAppl. (1985) 133–146.[4] P. Erd˝os, T. Gallai, On maximal paths and circuits in graphs, Acta Math. Acad. Sci.Hungar. (1959) 337–356.[5] Z. F¨uredi, M. Simonovits, The history of degenerate (bipartite) extremal graph prob-lems, Erd˝os centennial, Bolyai Soc. Math. Stud. (2013) 169–264.[6] J.M. Guo, Z.W. Wang, X. Li, Sharp upper bounds of the spectral radius of a graph, Discrete Math. (2019) 2559–2563.[7] H.Q. Lin, B. Ning, B. Wu, Eigenvalues and triangles in graphs,
Combin. Probab.Comput. (2020) doi:10.1017 / S0963548320000462.[8] V. Nikiforov, A spectral condition for odd cycles in graphs,
Linear Algebra Appl. (2008) 1492–1498.[9] V. Nikiforov, Some inequalities for the largest eigenvalue of a graph,
Combin. Probab.Comput. (2002) 179–189.[10] V. Nikiforov, Walks and the spectral radius of graphs, Linear Algebra Appl. (2006) 257–268.[11] V. Nikiforov, The maximum spectral radius of C -free graphs of given order andsize, Linear Algebra Appl. (2009) 2898–2905.1[12] V. Nikiforov, The spectral radius of graphs without paths and cycles of specifiedlength,
Linear Algebra Appl. (2010) 2243–2256.[13] V. Nikiforov, Some new results in extremal graph theory,
London Math. Soc. Lect.Note Ser. (2011) 141–182.[14] V. Nikiforov, M. Tait, C. Timmons, Degenerate Tur´an problems for hereditary prop-erties,
Electron. J. Combin. (4) (2018), Paper 4.39, 11 pp.[15] B. Ning, X. Peng, Extensions of the Erd˝os-Gallai theorem and Luo’s Theorem, Com-bin. Probab. Comput. (2020) 128–136.[16] E. Nosal, Eigenvalues of Graphs (Master’s Thesis), University of Calgary, 1970.[17] B. Papendieck, P. Recht, On maximal entries in the principal eigenvector of graphs, Linear Algebra Appl. (2000) 129–138.[18] P. Rowlinson, On the maximal index of graphs with a prescribed number of edges,
Linear Algebra Appl. (1988) 43–53.[19] S.W. Sun, K.C. Das, A conjecture on spectral radius of graphs,
Linear Algebra Appl. (2020) 74–80.[20] M. Tait, J. Tobin, Three conjectures in extremal spectral graph theory,
J. Combin.Theory Ser. B (2017) 137–161.[21] H. Wilf, Spectral bounds for the clique and independence numbers of graphs,
J.Combin. Theory Ser. B40