Spectral extremal results on trees

Let ${\rm spex}(n,F)$ be the maximum spectral radius over all $F$-free graphs of order $n$, and ${\rm SPEX}(n,F)$ be the family of $F$-free graphs of order $n$ with spectral radius equal to ${\rm spex}(n,F)$. Given integers $n,k,p$ with $n>k>0$ and $0\leq p\leq \lfloor(n-k)/2\rfloor$, let $S_{n,k}^{p}$ be the graph obtained from $K_k\nabla(n-k)K_1$ by embedding $p$ independent edges within its independent set, where `$\nabla$' means the join product. For $n\geq\ell\geq 4$, let $G_{n,\ell}=S_{n,(\ell-2)/2}^{0}$ if $\ell$ is even, and $G_{n,\ell}=S_{n,(\ell-3)/2}^{1}$ if $\ell$ is odd. Cioab\u{a}, Desai and Tait [SIAM J. Discrete Math. 37 (3) (2023) 2228--2239] showed that for $\ell\geq 6$ and sufficiently large $n$, if $\rho(G)\geq \rho(G_{n,\ell})$, then $G$ contains all trees of order $\ell$ unless $G=G_{n,\ell}$. They further posed a problem to study ${\rm spex}(n,F)$ for various specific trees $F$. Fix a tree $F$ of order $\ell\geq 6$, let $A$ and $B$ be two partite sets of $F$ with $|A|\leq |B|$, and set $q=|A|-1$. We first show that any graph in ${\rm SPEX}(n,F)$ contains a spanning subgraph $K_{q,n-q}$ for $q\geq 1$ and sufficiently large $n$. Consequently, $\rho(K_{q,n-q})\leq {\rm spex}(n,F)\leq \rho(G_{n,\ell})$, we further respectively characterize all trees $F$ with these two equalities holding. Secondly, we characterize the spectral extremal graphs for some specific trees and provide asymptotic spectral extremal values of the remaining trees. In particular, we characterize the spectral extremal graphs for all spiders, surprisingly, the extremal graphs are not always the spanning subgraph of $G_{n,\ell}$.


Introduction
Given a graph G, let A(G) be its adjacency matrix, and ρ(G) or ρ(A(G)) be its spectral radius (i.e., the largest eigenvalue of A(G)).Given a graph family F , a graph is said to be F -free if it does not contain any copy of F ∈ F .For convenience, we write F-free instead of F -free if F = {F}.In 2010, Nikiforov [20] proposed the following Brualdi-Soheid-Turán type problem: What is the maximum spectral radius in any F-free graph of order n?The aforementioned value is called the spectral extremal value of F and denoted by spex(n, F).An F-free graph G is said to be extremal for spex(n, F), if |V (G)| = n and ρ(G) = spex(n, F).Denote by SPEX(n, F) the family of extremal graphs for spex(n, F).In the past decades, the Brualdi-Soheid-Turán type problem has been studied by many researchers for many specific graphs, such as complete graphs [18,24], odd cycles [19], even cycles [2,18,26,27], paths [20] and wheels [3,28].For more information, we refer the reader to [4,9,12,13,14,15,21,22,23].
Fix a tree F of order ℓ ≥ 4, let A and B be two partite sets of F with |A| ≤ |B|, and set q = |A| − 1.If q = 0, then we can see that F is a star, and the spectral extremal result is trivial.It remains the case q ≥ 1. Obviously, K q,n−q is F-free.Then it is natural to consider the following result, which will be frequently used in the following.
Theorem 1.1.For q ≥ 1 and sufficiently large n, any graph in SPEX(n, F) contains a spanning subgraph K q,n−q .
The validity of Conjecture 1.1 for P ℓ was proved by Nikiforov [20], for all brooms was proved by Liu, Broersma and Wang [16], for the family of all ℓ-vertex trees with diameter at most 4 was proved by Hou, Liu, Wang, Gao and Lv [10] when ℓ is even and Liu, Broersma and Wang [17] when ℓ is odd.Very recently, Cioabȃ, Desai and Tait [1] completely solved Conjecture 1.1.Thus, Conjecture 1.1 for the family of all ℓ-vertex trees with given diameter is true.Now we give a slightly stronger result.Theorem 1.3.Let n be sufficiently large, and F be a tree of order ℓ ≥ 4.
In [1], Cioabȃ, Desai and Tait also proposed the following question.
Question 1.2.( [1]) For sufficiently large n, what is the exact value of spex(n, F) for a tree F of order ℓ ≥ 6?
Now we give partial answers to Question 1.2 in Theorems 1.4 and 1.5.
Theorem 1.4.If q ≥ 1 and δ ≥ 2, then S 1 n,q is F-free.Moreover, for sufficiently large n, Denote by ex(n, A ) the maximum size in any A -free graph of order n, and EX(n, A ) the family of n-vertex A -free graphs with ex(n, A ) edges.Now we give the characterization of the spectral extremal graphs for spex(n, F) when δ = 1.
Theorem 1.5.For q ≥ 1 and sufficiently large n, SPEX(n, F) ⊆ H (n, q, A ) if and only if δ = 1, where Particularly, we shall show that SPEX(n, S a+1,b+1 ) = {K a,n−a }, where the double star S a+1,b+1 is obtained from K 1,a and K 1,b by joining the centers with a new edge for a ≤ b.If F = S a+1,b+1 , then q = a, δ = 1 and β (F) = 2.By the definition of A , we can see that EX(k, A ) = {aK 1 }.By Theorem 1.5 (i), SPEX(n, F) = {K a,n−a } for sufficiently large n.
However, it seems difficult to determine SPEX(n, F) when δ ≥ 2, and so we leave this as a problem.In the following, we provide asymptotic spectral extremal values of all trees.Note that ρ(K q,n−q ) = q(n − q).From [20] we know ρ(S 0 n,q ) = q−1 2 + qn − A tree of order ℓ ≥ 4 is said to be a spider if it contains at most one vertex of degree at least 3.The vertex of degree at least 3 is called the center of the spider (if any vertex is of degree 1 or 2, then the spider is a path and any vertex of degree two can be taken to be the center).A leg of a spider is a path from the center to a leaf, and the length of a leg is the number of its edges.Let k ≥ 2 and let F be a spider of order 2k + 3 with r legs of odd length and s legs of length 1.If r ≥ 3 and s ≥ 1, then for sufficiently large n.This means that every graph G of order n with ρ(G) ≥ ρ(S 0 n,k ) contains F as a subgraph.Then we can derive the following result on spiders, which was originally proved by Liu, Broersma and Wang [16].
Corollary 1.1.( [16]) Let k ≥ 2 and let F be a spider of order 2k + 3 with r legs of odd length 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 ρ(G) ≥ ρ(S 0 n,k ) contains F as a subgraph.The Erdős-Sós Conjecture has been confirmed for some special families of spiders (see [5,6,7,25]).Recently, Fan, Hong and Liu [8] has resolved this conjecture for all spiders.The spectral Erdős-Sós Conjecture has also been confirmed for several classes of spiders (see [16]).In this paper, we completely characterize SPEX(n, F) for all spiders F with q ≥ 1.
Theorem 1.6.Let r 1 , r 2 , r 3 , r, s and ℓ be non-negative integers with r = r 1 + r 2 + r 3 and ℓ ≥ 4, and let F be a spider of order ℓ with r 1 legs of odd length at least 5, r 2 legs of length 3, r 3 legs of length 1 and s legs of even length.Let n be sufficiently large.Then 2 Proof of Theorem 1.1 Before beginning our proof, we first give some notations not defined previously.Let G be a simple graph.We use V (G) to denote the vertex set, E(G) the edge set, |V (G)| the number of vertices, e(G) the number of edges, ν(G) the maximum number of independent edges, respectively.Given a vertex v ∈ V (G) and a vertex subset S ⊆ V (G), we denote N G (v) the set of neighbors of v in G, and let (2) By the definition of q, we obtain A standard graph theory exercise shows that for any tree F with ℓ ≥ 2 vertices, In this section, we always assume that n is sufficiently large and G ⋆ is an extremal graph to spex P (n, F), and let ρ ⋆ denote its spectral radius.By Perron-Frobenius theorem, there exists a non-negative eigenvector We also choose a positive constant ε and a positive integer φ satisfying which will be frequently used later.First, we give a rough estimation on ρ ⋆ .
Proof.By (3), K q,n−q is F-free.Hence, ρ ⋆ ≥ ρ(K q,n−q ) = q(n − q) as G ⋆ is an extremal graph to spex P (n, F).Given a vertex u ∈ V (G ⋆ ), denote by N i (u) the set of vertices at distance i from u. Now we prove the upper bound.Note that Combining (4) gives (ρ ⋆ ) 2 ≤ 2e(G ⋆ ) ≤ 2ℓn, which leads to ρ ⋆ ≤ √ 2ℓn, as desired. Set for some positive integer η.We shall constantly give an upper bound of |L η | and a lower bound for degrees of vertices in L η (see Lemmas 2.2-2.5).
Lemma 2.2.For every positive integer µ, we have Proof.By Lemma 2.1, we get for each u ∈ L η .Summing this inequality over all vertices in L η , we obtain Consequently, |L η | ≤ n 0.6 for sufficiently large n.
Given an arbitrary vertex u ∈ V (G ⋆ ).For simplicity, we use N i , L η i and L η i instead of N i (u), N i (u) ∩ L η and N i (u) \ L η , respectively.By Lemma 2.1, we have Since x w ≤ 2e(L x w . Since Now we deal with the case w ∈ L where e(L η 4).Combining ( 6)-( 9), we obtain Now we show that d G ⋆ (u) ≥ n (2ℓ) µ+1 for any u ∈ L µ .By (4), we have where the last inequality holds as 6 and n is sufficiently large.Combining (10) and (11), we obtain as u ∈ L µ .Combining these with η = µ + 2 we obtain where the first inequality holds as q ≥ 1 and ℓ ≥ 4. Consequently, d G ⋆ (u) ≥ n (2ℓ) µ+1 .Summing this inequality over all vertices in L µ , we obtain , completing the proof.
Lemma 2.3.For every positive integer µ and every u ∈ L µ , we have d be the subset of L η 1 in which each vertex has at least q neighbors in L is empty, as desired.Now we deal with the case |L η 1 ∪ L η 2 | ≥ q.Suppose to the contrary that |L . Moreover, one can observe that Hence, G ⋆ contains a copy of K q+1,ℓ , and so contains a copy of F by (3), which gives a contradiction.The claim holds.Thus, where the last inequality holds because both 10) and ( 12), we have Proof.We first show the lower bounds of x u and |N 1 (u)| for any u ∈ L 1 .Suppose to the contrary that there exists a vertex . By Lemma 2.3, we get For convenience, we set From ( 13) we can see that u 0 has a neighbor in L η 1 , which is also a neighbor of u ⋆ .Thus, where x u 0 − 1 < −ε by the previous assumption.Combining this with (12) and setting η = φ , we have (5), contradicting (13).Therefore, Finally, we prove that |L 1 | = q.We first suppose |L 1 | ≥ q + 1.Note that every vertex u ∈ L 1 has at most 2εn non-neighbors.It follows that any q vertices in L 1 have at least n − 2qεn ≥ n 2 common neighbors by (5).Hence, G ⋆ contains a copy of K q+1,ℓ , and so contains a copy of F by (3), which gives a contradiction.Hence, which gives a contradiction.Therefore, For convenience, we use L, L i and L i instead of L 1 , N i (u) ∩ L 1 and N i (u) \ L 1 , respectively.Now, let R 1 be the subset of V (G ⋆ ) \ L in which every vertex is a non-neighbor of some vertex in L and R = V (G ⋆ ) \ (L ∪ R 1 ).Thus, |R 1 | ≤ 2εn|L| ≤ n (2ℓ) 3 by ( 5), and so 2 .Now, we prove that the eigenvector entries of vertices in R ∪ R 1 are small.
Proof of Theorem 1.1.From (4) we know that e(R 1 ) ≤ ℓ|R 1 |.Then there exists a vertex We modify the graph G ⋆ by deleting all edges incident to v 1 and joining v 1 to all vertices in L to obtain the graph G ⋆⋆ .We first claim that G ⋆⋆ is F-free.Suppose to the contrary, then G ⋆⋆ contains a subgraph F ′ isomorphic to F. From the modification, we can see that This indicates that a copy of F is already present in G ⋆ , which gives a contradiction.The claim holds.Now we claim that ρ(G ⋆⋆ ) > ρ ⋆ .By (14) and Lemma 2.5, we have x w ≤ (q − 1) + ∑ w∈N R (v 1 ) x w ≤ (q − 1) By Lemma 2.4, ∑ w∈L x w ≥ q(1 − ε).Combining this with ( 15) and ( 5), we have x w   ≥ 0.
3 Proofs of Theorems 1.2-1.6 In this section, we first record several technique lemmas that we will use.
Lemma 3.1.( [21]) Let H 1 be a graph on n 0 vertices with maximum degree d and H 2 be a graph on n − n 0 vertices with maximum degree d ′ .H 1 and H 2 may have loops or multiple edges, where loops add 1 to the degree.
The well-known König-Egerváry Theorem is as follows.
By the proof of Theorem 1.1, we can see that We then give three lemmas to characterize G ⋆ [L] and G ⋆ [R], which help us to present an approach to prove the remaining theorems.Lemma 3.3.Let n be sufficiently large and H be a graph of order q.Then H∇(n Proof.Suppose first that H is A -free.Then we show that H∇(n − q)K 1 is F-free.Otherwise, embed F into H∇(n − q)K 1 , where S = V (F) ∩V (H).Then F[S] ⊆ H[S] and S is a covering set of F. By the definition of A , F[S] ∈ A , which contradicts that H is A -free.Hence, H∇(n − q)K 1 is F-free.Suppose then that H is not A -free.By the definition of A , there exists a covering set S of F such that |S| ≤ q and F[S] ⊆ H.We can further find that H∇(n − q)K 1 contains a copy of F. Therefore, H∇(n − q)K 1 is F-free if and only if This completes the proof.Proof.Since δ = 1, there exists a vertex v ∈ A of degree 1 in F. Let A ′ = A \ {v} and B ′ = B ∪{v}.Obviously, |A ′ | = q and F[B ′ ] consists of an edge and some isolated vertices, which implies that F ⊆ K 1 q,ℓ−q .If e(G ⋆ [R]) ≥ 1, then G ⋆ must contain a copy of K 1 q,n−q , and so contains a copy of F, a contradiction.Thus, e(G ⋆ [R]) = 0.By Lemma 3.3, G ⋆ [L] is A -free, which implies that e(G ⋆ [L]) ≤ ex(q, A ). Now we prove that e(G ⋆ [L]) = ex(q, A ). Suppose to the contrary, then e(G ⋆ [L]) < e(Q A ), where Q A ∈ EX(q, A ). Clearly, e(Q A ) ≤ e(K q ) = q 2 .By Lemma 2.4 and ( 5), we have Lemma 3.5.Let n be sufficiently large and δ ≥ 2. Then S 1 n,q is F-free and e(G ⋆ [R]) ≥ 1.
Proof.We first prove that S 1 n,q is F-free, where Y 1 is the set of dominating vertices of S 1 n,q and In graph F, let B 1 be the set of vertices in Y 1 adjacent to at least one vertex in A 2 .Then, B 1 ⊆ B, and thus Hence, e(G ⋆ [R]) ≥ 1.This completes the proof.
Combining Lemmas 3.4 and 3.5, we can directly get Theorem 1.5.Having Lemmas 3.1-3.5,we are ready to complete the proofs of the remaining theorems.
Conversely, suppose β (F) = (ℓ − 1)/2 and δ ≥ 2. Combining (2) gives |A| = q + 1 = (ℓ − 1)/2.We first claim that G ⋆ [R] is 2K 2 -free.Otherwise, G ⋆ contains a copy of K 2 q,n−q .Let v 1 , v 2 be two endpoints of a longest path P in F. Since F is not a star, the path P is of length at least 3, which implies that v 1 , v 2 have no common neighbors.Since δ ≥ 2, we have q, and F[B ′ ] consists of two independent edges and some isolated vertices.This indicates that F ⊆ K 2 q,ℓ−q .However, G ⋆ contains a copy of K 2 q,n−q , and so contains a copy of F, a contradiction.Hence, G ⋆ [R] is 2K 2 -free.
We then claim that G ⋆ [R] is P 3 -free.Since δ ≥ 2, we have This indicates that all vertices in A are of degree 2. Choose an arbitrary vertex v 0 ∈ A. Set A ′′ = A \ {v 0 } and B ′′ = B ∪ {v 0 }.Then A ′′ is an independent set of F with |A ′′ | = (ℓ − 3)/2, and F[B ′′ ] consists of a path of length 2 with center v 0 and some isolated vertices.This implies that G ⋆ [R] is P 3 -free.

4
. This, together with Lemma 3.5, gives that It remains the upper bound.We shall prove that ∆ ≤ δ − 1, where ∆ is the maximum degree of G ⋆ [R].Suppose to the contrary that there exists a vertex u ∈ R with d R ( u) ≥ δ .Choose a vertex u 0 ∈ A with d F (u 0 ) = δ .Then we can embed F into G ⋆ by embedding A \ {u 0 } into L, and embedding B ∪ {u 0 } into R such that u = u 0 .This contradicts that G ⋆ is F-free.The claim holds.Applying d = q − 1, n 0 = q and d ′ = ∆ with Lemma 3.1, we have ρ ⋆ ≤ ρ(J ⋆ ).By direct computation, we have Since n is sufficiently large and ∆ ≤ δ − 1, we obtain that This completes the proof.
Proof of Theorem 1.6.Let v ⋆ be the center of the spider F, and let C denote the set of vertices at odd distance from v ⋆ in F. Then C ∈ {A, B}.Combining Lemma 3.2, we can observe that Note that S r n,(ℓ−r−1)/2 contains a copy of F. Then r ′ ≤ r, where r ′ is the minimum integer such that S r ′ n,(ℓ−r−1)/2 contains a copy of F. By ( 16), δ ≥ 2.Then, from Lemma 3.5 we know that S 1 n,(ℓ−r−1)/2 is F-free, which implies that r ′ ≥ 2. Now we shall prove r ′ = r.Otherwise, r ′ < r.Embed F into S r ′ n,(ℓ−r−1)/2 , where Y 1 is the set of dominating vertices of S r ′ n,(ℓ−r−1)/2 and Y 2 = V (S r ′ n,(ℓ−r−1)/2 ) \Y 1 .Set V (F) ∩V (Y 1 ) = A ′ and V (F) ∩ V (Y 2 ) = B ′ .By the definition of r ′ , F[B ′ ] contains exactly r ′ independent edges, say e 1 , e 2 , . . ., e r ′ , and some isolated vertices.Contracting e i as a vertex for each i ∈ {1, . .., r ′ } in F and S r ′ n,(ℓ−r−1)/2 , we obtain a corresponding spider F ′ and a corresponding graph S 0 n−r ′ ,(ℓ−r−1)/2 .Then, F ′ ⊆ S 0 n−r ′ ,(ℓ−r−1)/2 as F ⊆ S r ′ n,(ℓ−r−1)/2 .Now we shall prove that S 0 n−r ′ ,(ℓ−r−1)/2 is F ′ -free, which gives a contradiction.By Claim 3.1, any leg of F has at most one of these independent edges.If r 3 = 0, then F ′ has exactly r − r ′ ≥ 1 legs of length 3 and r ′ legs of length 2. If r 3 = 1, then either F ′ has exactly r − 1 − r ′ legs of length 3, r ′ legs of length 2 and one leg of length 1, or F ′ has exactly r − r ′ ≥ 1 legs of length 3 and r ′ − 1 legs of length 2. Let A ′ and B ′ be two partite sets of F ′ with |A ′ | ≤ |B ′ |.In all situations, we can observe that |V (F ′ )| = ℓ − r ′ and F ′ has exactly r − r ′ ≥ 1 legs of odd length.By a similar discussion of ( 17 is F-free.This completes the proof of Theorem 1.6.

Theorem 1 . 2 .
Let ℓ ≥ 6 and d ∈ {4, . . ., ℓ − 1}, and let G be a graph of sufficiently large order n.(i) If at least one of ℓ and d is even, then there exists a tree F of order ℓ and diameter d such that SPEX(n, F) = {G n,ℓ }. (ii) If both ℓ and d are odd and ρ(G) ≥ ρ(S 0 n,(ℓ−3)/2 ), then G contains all trees of order ℓ and diameter d unless G = S 0 n,(ℓ−3)/2 .It is interesting to find all those trees F satisfying SPEX(n, F) = {G n,ℓ }.Question 1.1.For sufficiently large n, which tree F of order ℓ ≥ 6 can satisfy SPEX(n, F) = {G n,ℓ }?A covering of a graph is a set of vertices which meets all edges of the graph.Let β (G) denote the minimum number of vertices in a covering of G. Set δ := min{d F (x) : x ∈ A}.Inspired by the work of Cioabȃ, Desai and Tait, we provide an answer to Question 1.1.

Lemma 3 . 4 .
Given a non-nagative integer p ≤ b/2, let K p a,b be the graph obtained from aK 1 ∇bK 1 by embedding p independent edges into the partite set of size b.Let n be sufficiently large and δ = 1.Then e(G ⋆ [R]) = 0 and G ⋆ [L] ∈ EX(q, A ).

2 .
For non-negative integers a, b, c with a ≥ b + 1 and c ≥ 1, let S(a, b, c) be the spider with a − b − 1 legs of length 1, b legs of length 2 and one leg of length c.Clearly, |V (S(a, b, c))| = (a − b − 1) + 2b + c + 1 = a + b + c.