On a conjecture of Nikiforov involving a spectral radius condition for a graph to contain all trees

We partly confirm a Brualdi-Solheid-Tur\'{a}n type conjecture due to Nikiforov, which is a spectral radius analogue of the well-known Erd\H{o}s-S\'os Conjecture that any tree of order $t$ is contained in a graph of average degree greater than $t-2$. We confirm Nikiforov's Conjecture for all brooms and for a larger class of spiders. For our proofs we also obtain a new Tur\'{a}n type result which might turn out to be of independent interest.


Introduction
A central problem in extremal graph theory is the following Turán-type problem: for a given graph H, what is the maximum number of edges in an H-free graph with a given order? In the past decades, much attention has been paid to a spectral version of this question, that is, what is the maximum spectral radius of an H-free graph with a given order? The latter type of problem is called a Brualdi-Solheid-Turán type problem in [16] by Nikiforov. Examples of such problems are numerous since every Turán type problem gives rise to a corresponding Brualdi-Solheid-Turán type problem. As argued in [16], "the study of Brualdi-Solheid-Turán type problems is an important topic in spectral graph theory". Several groups of researchers have studied the relationship between the spectral radius and forbidden subgraphs (such as cliques, paths, cycles and complete bipartite subgraphs). We refer to [1,9,13,14,16,18,19,20] for more information.
Motivated by these problems and earlier works, we study a conjecture due to Nikiforov [16], which is a spectral radius analogue of the well-known Erdős-Sós Conjecture that a graph of average degree greater than t − 2 admits any tree of order t. Before we give more details concerning our work, we start by giving some essential definitions and introducing some useful notation.
Let G = (V (G), E(G)) be a simple undirected graph with vertex set V (G) and edge set E(G). We use |G| := |V (G)| and e(G) := |E(G)| to denote the order and size of G, respectively. Let µ(G) be the largest eigenvalue of the adjacency matrix A(G) of G. We call µ(G) the spectral radius of G. A graph G is said to be H-free if H is not a subgraph of G. In order to avoid confusion, please note that we mean subgraph here and not induced subgraph. The Turán number of H is the maximum number of edges in an H-free graph of order n, and denoted by ex(n, H). Given two disjoint graphs G and H, the disjoint union of G and H, denoted by G ∪ H, is the graph with vertex set V (G) ∪ V (H) and edge set E(G) ∪ E(H). We use mG to denote the disjoint union of m copies of G. The join of G and H, denoted by G ∨ H, is the graph obtained from G ∪ H by adding edges joining every vertex of G to every vertex of H.
Adopting the notation of [16], let S n,k denote the graph obtained by joining every vertex of a complete graph K k to every vertex of an independent set of order n − k, that is, S n,k = K k ∨ K n−k . Let S + n,k be the graph obtained from S n,k by adding a single edge joining two vertices of the independent set of S n,k . In addition, from [16] we know and µ(S n,k ) < µ(S + n,k ) < µ(S n,k ) + 1 n − k − 2 (n − k)/k .
In [16], Nikiforov proved that Conjecture 1.1 holds for paths. 16]) Let k ≥ 2 and let G be a graph of sufficiently large order n.
Recently, Hou et al. [10] proved that Conjecture 1.1 (a) holds for all trees of diameter at most four. Liu, Broersma and Wang [12] proved that Conjecture 1.1 (b) holds for all trees of diameter at most four, except for the subdivision of K 1,k+1 in which every edge is subdivided precisely once.
In this paper, we study Conjecture 1.1 for spiders. A spider is a tree with at most one vertex of degree more than 2. The vertex of degree more than 2 is called the center of the spider (if all vertices have degree 1 or 2, so if the spider is a path, then any designated vertex can act as its center). A leg of a spider is a path from the center to a vertex of degree 1. The length of a leg is the number of edges of the leg. We use S(t 1 , t 2 , . . . , t m ) to denote a spider consisting of one designated center and m legs with lengths t 1 , t 2 , . . . , t m ; see Figure 1 for an example. Thus S(t 1 , t 2 , . . . , t m ) has 1 + m i=1 t i vertices and The aforementioned Erdős-Sós Conjecture has been confirmed for several classes of spiders in a series of papers; see [6,7,8]. Our first contribution is the following spectral radius result on the existence of a class of spiders. By imposing two restrictions on the number of odd legs, i.e., of odd length, we can prove the following theorem involving spiders. Theorem 1.3. Let k ≥ 2 and let S be a spider of order 2k + 3 with r odd legs and s legs of length 1. If r ≥ 3, 2s − r ≥ 2 and n is sufficiently large, then every graph G of order n with µ(G) ≥ µ(S n,k ) contains S as a subgraph.
We postpone all proofs to later sections. Theorem 1.3 confirms Conjecture 1.1 (b) for all spiders satisfying the condition in the statement of the theorem. Since µ(S n,k ) < µ(S + n,k ), Theorem 1.3 is in fact a stronger result. Let S be a spider of order 2k + 2 with r odd legs and s legs of length 1 such that r ≥ 2 and 2s − r ≥ 1. Let S ′ be the graph obtained from S by adding an extra pendant edge at the center of S. Then S ′ is a spider of order 2k + 3 with r + 1 odd legs and s + 1 legs of length 1 such that r + 1 ≥ 3 and 2(s + 1) − (r + 1) ≥ 2. Applying Theorem 1.3, we immediately derive the following result which confirms Conjecture 1.1 (a) for a class of spiders. Corollary 1.4. Let k ≥ 2 and let S be a spider of order 2k + 2 with r odd legs and s legs of length 1. If r ≥ 2, 2s − r ≥ 1 and n is sufficiently large, then every graph G of order n with µ(G) ≥ µ(S n,k ) contains S as a subgraph.
Our next result deals with the special subclass of spiders called brooms. For s, t ≥ 1, a broom B s,t is a tree on s + t vertices obtained by identifying the center of a star K 1,s and an end-vertex of a path P t ; see Figure 2. Note that the broom B s,t can be viewed as a spider S(t 1 , t 2 , . . . , t s+1 ), where t 1 = · · · = t s = 1 and t s+1 = t − 1. Moreover, B 1,t = P t+1 and B s,1 = K 1,s . It is easy to check that if n is large enough, then S n,k contains all brooms of order 2k + 3 except for B 1,2k+2 and B 2,2k+1 , and S + n,k contains all brooms of order 2k + 3 except for B 1,2k+2 .  (a) If µ(G) ≥ µ(S n,k ), then G contains all brooms of order 2k + 2, unless G = S n,k .
(b) If µ(G) ≥ µ(S + n,k ), then G contains all brooms of order 2k + 3, unless G = S + n,k . In order to prove Theorem 1.5, we shall use Theorems 1.2, 1.3 and the following Turán type result for connected graphs involving the broom B 2,2k+1 . Theorem 1.6 below might be of independent interest. Theorem 1.6. For k ≥ 2 and n sufficiently large, let G be a connected graph of order n. If e(G) ≥ e(S + n,k ) = kn − k(k+1) 2 + 1, then G contains B 2,2k+1 as a subgraph.
The remainder of this paper is organized as follows. In the next section, we provide some auxiliary results that will be used in our proofs. In Section 3, we prove Theorem 1.3. Section 4 is devoted to our proof of Theorem 1.5. In Section 5, we prove Theorem 1.6. Finally, we conclude this paper with some remarks on generalized brooms and by presenting some open problems in Section 6.

Preliminaries
In this section, we provide some additional terminology and lemmas that we will use. Let G be a connected graph. For any vertex we denote the sub-path v i v i+1 · · · v j by v i P v j . All logarithms in this paper are to the base 2. We use the standard Bachmann-Landau notation to indicate asymptotic growth rates of functions.
We will use the following known lemma on matrices in the set-up of the proof of Theorem 1.3. Given an n × n matrix A, let A ij be the (i, j)-th entry of A for 1 ≤ i, j ≤ n.
) Given a, b ∈ Z + and an n × n nonnegative symmetric irreducible matrix A, let µ be the largest eigenvalue of A and µ ′ be the largest root of A linear forest is a forest all whose components are paths. We shall use the following known results on the Turán numbers of paths and linear forests in the proof of Theorem 1.3.

Lemma 2.2. ([5])
For any positive integers t and n, we have ex(n, P t ) ≤ t−2 2 n. Lemma 2.3. (see [4,Theorem 2.2]) For any integer ℓ ≥ 2 and sufficiently large n, we have ex(n, ℓP 3 ) < ℓ − 1 2 n. Lemma 2.4. (see [11,Theorem 2]) For any integer ℓ ≥ 2, let F = 1≤i≤ℓ P a i be a linear forest with a i ≥ 2 for all i ∈ [ℓ]. If at least one a i is not 3, then for n sufficiently large, Next we introduce two lemmas that will be used in the proof of Theorem 1.6. Lemma 2.6. ( [3]) For k ≥ 2 and n sufficiently large, let G be a connected graph of order n with G = S + n,k . If e(G) ≥ e(S + n,k ) = kn − k(k+1) 2 + 1, then G contains P 2k+3 as a subgraph.
We shall also apply the following partial solution of the Erdős-Sós Conjecture.
If G is a graph on n vertices with e(G) > (t−2)n 2 , then G contains every t-vertex spider with three legs.
We end this section with some known results about µ(G).
Lemma 2.8. ( [15]) If G is a graph with n vertices, m edges and δ(G) = δ, then Lemma 2.10. ( [16]) Let the numbers c ≥ 0, k ≥ 2 and n ≥ 2 4k , and let G be a graph of then there exists a subgraph H of G satisfying one of the following conditions: 3 Proof of Theorem 1.3 Let G be a graph with V (G) = [n] and µ(G) ≥ µ(S n,k ), where n is sufficiently large. For a contradiction, suppose that G contains no copy of S. Without loss of generality, we may Since the (i, u)-entry of A(G) 2 is the number of walks of length 2 between i and u, we have We complete the proof by first proving the following claim and then distinguishing two cases based on the degree of u.
Then G contains S n,k as a subgraph. Since S is a bipartite graph, we may assume that Thus S is a subgraph of S n,k . This contradicts the assumption that G contains no copy of S.
We divide the rest of the proof into two cases: In this case, since B u ≥ 0 and by equality (3), we have Thus (1)) kn log n . We claim that there are n 2 log n vertices in N 2 (u) each of which has at least k neighbors in N 1 (u). Otherwise, we have e(N 1 (u), n log k+1 n vertices in N 2 (u) which have k common neighbors in N 1 (u). Without loss of generality, let X ⊆ N 1 (u) and Y ⊆ N 2 (u) with |X| = k and |Y | = n log k+1 n ≫ 2k + 2 such that X is completely joined to Y . Similarly as in the last paragraph of the proof of Claim 3.1, we derive that G[X ∪ Y ] contains a copy of S, a contradiction.
We consider two subcases based on the number of edges between N 1 (u) and N 2 (u).
We next claim that L u contains the following linear forest.
There is a s+1≤i≤m P t i in L u such that each P t i has an end-vertex in N 1 (u).
Proof. If t i = 2 for all s + 1 ≤ i ≤ m, then s = r ≥ 3. Since 2k + 3 is odd, we have that r is even. Hence, we further have s = r ≥ 4. Thus m − s = (2k + 3 − 1 − s)/2 ≤ k − 1. By Lemmas 2.2 and 2.4, for sufficiently large N , we have ex (N, (m − s)P 2 ) ≤ (k − 2)N . By inequality (5), there is an (m − s)P 2 in L u . By the definition of L u , each P 2 has an end-vertex in N 1 (u). Next, we assume that at least one t i (s + 1 ≤ i ≤ m) is not 2. We now show that there is a s+1≤i≤m P t i +1 in L u . If r = s and m − s = m − r = 1, then t m + 1 ≤ 2k + 3 − 1 − r + 1 ≤ 2k since r ≥ 3. By Lemma 2.2 and inequality (5), there is a P tm+1 in L u . If m = r and m − s = r − s = 1, then s ≥ 3 since 2 ≤ 2s − r = 2s − (s + 1). In this case, we also have t m + 1 ≤ 2k + 3 − 1 − s + 1 ≤ 2k. Thus there is a P tm+1 in L u by Lemma 2.2 and inequality (5). If m − s ≥ 2, then by Lemma 2.4 and since 2s − r ≥ 2, for sufficiently large N , we have ex   N, be the first and second vertex along the path P t i +1 , respectively. By the definition of L u , at least one of v ′ i and v ′′ i is contained in N 1 (u). Hence, there is a s+1≤i≤m P t i in L u such that each P t i has an end-vertex in N 1 (u).
By Claim 3.2, there is a s+1≤i≤m P t i in L u such that each P t i has an end-vertex in N 1 (u). Together with vertex u and s additional vertices in N 1 (u), this s+1≤i≤m P t i forms a copy of S in G, a contradiction.
By Claim 3.3, there is a s+1≤i≤m P t i in G[N 1 (u)]. Together with vertex u and s additional vertices in N 1 (u), this s+1≤i≤m P t i forms a copy of S in G, a contradiction. This contradiction completes our proof of Theorem 1.3.

Proof of Theorem 1.5
In this section, we give our proof of Theorem 1.5, which confirms Conjecture 1.1 for all brooms. We first prove that the result holds for B 2,2k+1 .
Theorem 4.1. For integers k ≥ 2 and n sufficiently large, every graph G of order n with µ(G) ≥ µ(S + n,k ) contains B 2,2k+1 as a subgraph.
If G is disconnected, then G contains a component G ′ with µ(G ′ ) = µ(G). Then µ(G ′ ) ≥ µ(S + n,k ) ≥ µ S + |G ′ |,k . Thus we can apply the above arguments to G ′ and complete the proof.
Now we have all ingredients to present our proof of Theorem 1.5.
Proof of Theorem 1.5. For positive integers s and t, let B s,t be a broom. Note that B 1,t is in fact a path. In this case, the result follows from Theorem 1.2. Hence, we may assume that s ≥ 2 in the following arguments. The broom B s,t can be viewed as a spider S(t 1 , t 2 , . . . , t s+1 ), where t 1 = · · · = t s = 1 and t s+1 = t − 1. Such a spider has at least s legs of length 1. Let r be the number of odd legs. So s ≤ r ≤ s + 1. We first prove Theorem 1.5 (a). Since s ≥ 2, we have 2s − r ≥ 2s − (s + 1) ≥ 1. By Corollary 1.4, the result holds. We next prove Theorem 1.5 (b). The case for B 2,2k+1 follows from Theorem 4.1. For the case s ≥ 3, we have 2s − r ≥ 2s − (s + 1) ≥ 2. By Theorem 1.3, the result holds. 9 5 Proof of Theorem 1.6 Let G be a connected graph of order n with e(G) ≥ e(S + n,k ) = kn − k(k + 1)/2 + 1. We first consider the case G = S + n,k . Let v 1 , v 2 , . . . , v k denote the k vertices with degree n − 1, u 1 , u 2 , . . . , u n−k−2 the n − k − 2 vertices with degree k, and w 1 , w 2 the two vertices with degree k + 1. Then w 1 w 2 v 1 u 1 v 2 u 2 · · · v k u k is a path of order 2k + 2. Together with the edge v k u k+1 , this path forms a B 2,2k+1 in G.
Next, we consider the case G = S + n,k . For a contradiction, suppose that G contains no B 2,2k+1 . Let P = v 1 v 2 · · · v ℓ be a longest path in G. By Lemma 2.6, we have ℓ ≥ 2k + 3. Let X = V (P ) and Y = V (G) \ V (P ). We next state and prove two claims.
Proof. We first suppose 4k ≤ ℓ ≤ n − 1. Since P is a longest path, we have uv 1 / ∈ E(G) and uv ℓ / ∈ E(G) for any vertex u ∈ Y . Since G is connected, there is an i ∈ {2, 3, . . . , ℓ − 1} and a vertex w ∈ Y such that v i w ∈ E(G). Since ℓ ≥ 4k, at least one of the paths v 1 · · · v i and v i · · · v ℓ has order at least 2k + 1, say v 1 · · · v i . Together with the edge v i w, the path v 1 · · · v i v i+1 forms a graph that contains B 2,2k+1 as a subgraph, a contradiction.
If e(X, V (H)) + e(H) ≤ (k − 1/2)|H| for each component H in G − X, then e(G) ≤ (k − 1/2)(n − ℓ) + ℓ 2 < kn − k(k + 1)/2 + 1 when n is sufficiently large, a contradiction. Thus there exists a component H in G−X with e(X, V (H))+e(H) > (k−1/2)|H|. In the remainder of this proof, let H denote such a component. Our aim is to show that e(X, V (H)) + e(H) ≤ (k − 1/2)|H|, contradicting the above. For any vertex v ∈ V (H), let s v be the number of neighbors of v in X, and let p v be the length of a longest path in H starting at v (so this longest path has p v + 1 vertices). We next state and prove three claims. Proof. By symmetry, we only prove vv ℓ−2k+1 / ∈ E(G) for any v ∈ V (H). For a contradiction, suppose that there is a vertex u ∈ V (H) with uv ℓ−2k+1 ∈ E(G). Note that |H| ≥ 2; otherwise if |H| = 1, then e(X, V (H)) + e(H) = s u ≤ k − 1 ≤ (k − 1/2)|H| by Claims 5.1 and 5.2 (ii), a contradiction. Before we derive at a contradiction, we first make some other observations.
We next observe that there is at most one vertex in H which is adjacent to ⌈(4k − ℓ)/2⌉ vertices in X. Otherwise, suppose that there are two vertices v ′ , v ′′ ∈ V (H) with s v ′ = s v ′′ = ⌈(4k − ℓ)/2⌉. This implies that v ′ (resp., v ′′ ) is either adjacent to both v ℓ−2k+1 and v ℓ−2k+3 or adjacent to both v ℓ−2k+2 and v ℓ−2k+4 . Since H is connected, there is a path Q in H connecting v ′ and v ′′ . It is easy to check that G[X ∪ V (Q)] contains a path longer than P , a contradiction.
Therefore, by Lemma 2.5 and Claim 5.2 (ii), we have Proof. By Claim 5.2 (i) and Claim 5.3, we have and thus completes the proof. Let v i 1 , . . . , v is v be all the neighbors of v in X, where i 1 < · · · < i sv . In order to avoid a P ℓ+1 in G, we have i sv ≤ ℓ−(p v +1) and This implies that 2(s v + p v ) ≤ ℓ − 1.
Proof. The case k = 2 follows from Claim 5.3. We next consider the case k = 3. In this case, for any vertex v ∈ V (H), v 5 is the only possible neighbor of v in X.
In order to avoid a B 2,2k+1 , each vertex v ∈ V ′ (resp., v ∈ V ′′ ) has at most two neighbors in V ′ (resp., V (H)). Thus e(X, V (H)) + e(H) ≤ e(X, In the following arguments, we may assume that k ≥ 4. For a contradiction, suppose that there is a vertex u ∈ V (H) with uv k+2 ∈ E(G). Let H be the set of all components of H − u and let H ′ ∈ H. For any vertex v ∈ V (H ′ ), let p ′ v be the length of a longest path in H ′ starting at v. Then p ′ v ≤ p v . In the following, we show that Otherwise, suppose that Q is a path in H ′ of length 2k − 1 starting at v. Let w be the other end-vertex of Q. Since H is connected and H ′ is a component of H − u, there is a path Q ′ connecting u and Q. We choose such a path Q ′ with the minimum length, that is, Then one of the paths vQx and xQw has length at least k, say vQx. Then v 1 · · · v k+2 uQ ′ xQv is a path of length at least 2k + 3. This contradiction implies p ′ v ≤ 2k − 2. In order to avoid a P 2k+4 , if p ′ v ≥ k, then vu / ∈ E(G). Thus  Claim 5.6. If C = ∅, then for any vertex v ∈ C, the degree of v in H is at most k − 1.
Proof. By the definition of C, we may assume that w 0 w 1 w 2 · · · w 2k is a path in H, where w 0 := v. Let W = {w 1 , w 2 , . . . , w 2k }. Since G is connected and s v = 0, there is a path Q connecting a vertex of X and a vertex of W . We choose such a path with minimal length, that is, Q has exactly one common vertex with X and exactly one common vertex with W . We may assume that Q = u 0 u 1 · · · u t for some t ≥ 1, where u 0 ∈ X and u t = w i for some 1 ≤ i ≤ 2k.
Recall that for any vertex v ∈ A ∪ B, we have s v + p v /2 ≤ k − 1/2. By Lemma 2.
This contradiction completes the proof of Theorem 1.6.

Concluding remarks
In this paper, we proved that a Brualdi-Solheid-Turán type conjecture due to Nikiforov (Conjecture 1.1) holds for a class of spiders. In particular, we also confirmed Conjecture 1.1 for all brooms. For integers s ≥ 1 and t ≥ ℓ ≥ 1, a generalized broom B ℓ s,t is a tree on s + t vertices obtained from a path P t by attaching s pendant edges at the ℓ-th vertex along the path. Note that B ℓ s,t = B t+1−ℓ s,t and B 1 s,t = B t s,t = B s,t . Using Theorem 1.3, it is easy to derive the following result for generalized brooms.
Corollary 6.1. For integers k ≥ 2, let T (resp., T ′ ) be the set of all generalized brooms B ℓ s,t of order 2k + 3 with s ≥ 3 (resp., of order 2k + 2 with s ≥ 2). Then every graph G of sufficiently large order n with µ(G) ≥ µ(S n,k ) contains all graphs in T and T ′ .
An interesting and natural question is to study Conjecture 1.1 for generalized brooms B ℓ s,t of order 2k + 3 with s ≤ 2 (resp., of order 2k + 2 with s = 1) and 2 ≤ ℓ ≤ t − 1. Another direction is to study Conjecture 1.1 for other classes of spiders. Hopefully this will also lead to new ideas and approaches for resolving Conjecture 1.1 for general trees.