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A general theorem in spectral extremal graph theory

John Byrne University of Delaware, Department of Mathematical Sciences, [email protected]    Dheer Noal Desai University of Wyoming, Department of Mathematics and Statistics. The University of Memphis, Department of Mathematical Sciences, [email protected]    Michael Tait Villanova University, Department of Mathematics & Statistics, [email protected]. Research partially supported by NSF grant DMS-2245556.
Abstract

The extremal graphs EX(n,)\mathrm{EX}(n,\mathcal{F}) and spectral extremal graphs SPEX(n,)\mathrm{SPEX}(n,\mathcal{F}) are the sets of graphs on nn vertices with maximum number of edges and maximum spectral radius, respectively, with no subgraph in \mathcal{F}. We prove a general theorem which allows us to characterize the spectral extremal graphs for a wide range of forbidden families \mathcal{F} and implies several new and existing results. In particular, whenever EX(n,)\mathrm{EX}(n,\mathcal{F}) contains the complete bipartite graph Kk,nkK_{k,n-k} (or certain similar graphs) then SPEX(n,)\mathrm{SPEX}(n,\mathcal{F}) contains the same graph when nn is sufficiently large. We prove a similar theorem which relates SPEX(n,)\mathrm{SPEX}(n,\mathcal{F}) and SPEXα(n,)\mathrm{SPEX}_{\alpha}(n,\mathcal{F}), the set of \mathcal{F}-free graphs which maximize the spectral radius of the matrix Aα=αD+(1α)AA_{\alpha}=\alpha D+(1-\alpha)A, where AA is the adjacency matrix and DD is the diagonal degree matrix.

1 Introduction

For a family of graphs \mathcal{F} and α[0,1)\alpha\in[0,1), we define ex(n,)\mathrm{ex}(n,\mathcal{F}), spex(n,)\mathrm{spex}(n,\mathcal{F}), and spexα(n,)\mathrm{spex}_{\alpha}(n,\mathcal{F}) to be the maximum number of edges, maximum spectral radius of the adjacency matrix, and maximum spectral radius of the matrix AαA_{\alpha} respectively over all nn-vertex graphs which do not contain any FF\in\mathcal{F} as a subgraph. Here Aα=αD+(1α)AA_{\alpha}=\alpha D+(1-\alpha)A where DD is the diagonal degree matrix and AA is the adjacency matrix. So spex(n,)=spex0(n,)\mathrm{spex}(n,\mathcal{F})=\mathrm{spex}_{0}(n,\mathcal{F}) and spex1/2(n,)\mathrm{spex}_{1/2}(n,\mathcal{F}) is one half of the maximum spectral radius of the signless Laplacian over \mathcal{F}-free graphs. As is standard, we define EX(n,)\mathrm{EX}(n,\mathcal{F}), SPEX(n,)\mathrm{SPEX}(n,\mathcal{F}), and SPEXα(n,)\mathrm{SPEX}_{\alpha}(n,\mathcal{F}) as the set of extremal graphs for each respective function, and we call the graphs in each set (edge) extremal, spectral extremal, and alpha spectral extremal, respectively. When ={F}\mathcal{F}=\{F\} is a single graph we use the standard notation ex(n,F)\mathrm{ex}(n,F) instead of ex(n,{F})\mathrm{ex}(n,\{F\}) and similarly with the other functions. If the family of extremal graphs consists of a single graph HH we will write EX(n,)=H\mathrm{EX}(n,\mathcal{F})=H instead of EX(n,)={H}\mathrm{EX}(n,\mathcal{F})=\{H\}, and similarly with the other functions. Our goal in this paper is to prove general theorems which capture many previous and new results on these parameters.

Spectral extremal graph theory has become very popular recently. Several sporadic results appeared over the years, for example results of Wilf [59] and Nosal [54], before the field was studied systematically. All results in the area which regard the adjacency matrix can be understood under the general framework of Brualdi-Solheid problems [4], where one seeks to maximize the spectral radius of the adjacency matrix over a family of graphs. When the family is defined by a forbidden subgraph, this problem has a Turán-type flavor, and the spectral Turán problem was introduced systematically by Nikiforov [50]. Since then many papers have been written, and we refer to the excellent recent survey [41]. We also note that there are hypergraph versions of these problems (see for example [37, 52]), but we only consider graphs in this paper.

Researchers have also frequently considered other matrices besides the adjacency matrix. Signless Laplacian versions of Brualdi-Solheid problems were studied extensively and Nikiforov introduced the AαA_{\alpha} matrix as a way to merge these two theories [53].

With so much recent activity in this area, we believe that it is valuable to have general theorems that apply to families of forbidden graphs. One question that naturally presents itself when reviewing previous results is: when are the edge-extremal graphs the same as the spectral-extremal graphs? That is, when do EX(n,F)\mathrm{EX}(n,F) and SPEXα(n,F)\mathrm{SPEX}_{\alpha}(n,F) either coincide or overlap?

Some of the first results in the area suggest that perhaps the problems are the same. For example, the spectral Turán theorem [47] shows EX(n,Kr+1)=SPEX(n,Kr+1)\mathrm{EX}(n,K_{r+1})=\mathrm{SPEX}(n,K_{r+1}), and [48] shows that EX(n,C2k+1)=SPEX(n,C2k+1)=Kn/2,n/2\mathrm{EX}(n,C_{2k+1})=\mathrm{SPEX}(n,C_{2k+1})=K_{n/2,n/2} for large enough nn. Furthermore, for any non-bipartite FF, the theorems in [21, 22, 49] show that the asymptotics of the two functions coincide. However, the two problems can also be vastly different, for example when FF is an even cycle [13, 50, 61, 62]. One might then guess that there is a difference between FF bipartite and non-bipartite, but it is a bit more subtle than that. In [15], it was shown that when forbidding large enough odd wheels the two extremal families have similar structure but are in fact disjoint.

Hence we seek general criteria which can determine whether the extremal families overlap or not. One such very nice result is in [58], proving a conjecture in [15], where it was shown that for any graph FF such that any graph in EX(n,F)\mathrm{EX}(n,F) is a Turán graph plus O(1)O(1) edges, then for large enough nn one has that SPEX(n,F)EX(n,F)\mathrm{SPEX}(n,F)\subseteq\mathrm{EX}(n,F). Another very nice paper proving general results about trees was written [23].

In this paper we prove some general results which apply to many forbidden graphs or families. As special cases of our theorems, we recover results from [7, 8, 9, 10, 11, 12, 14, 16, 23, 27, 28, 35, 39, 40, 43, 44, 50, 55, 56, 60, 64, 65] and additionally we can apply the theorems to prove several new results. More details about specific applications are given after statements of our theorems.

When studying a classical Turán-type problem, every edge increases the objective function by the same amount. But when trying to determine spex(n,F)\mathrm{spex}(n,F), some edges will contribute more to the spectral radius than others, and so vertices of large degree are incentivized. When α>0\alpha>0, the presence of the diagonal degree matrix incentivizes large degree vertices even further. This paper is an attempt to formalize this phenomenon.

The rest of this paper is organized as follows. In the rest of Section 1 we define our notation and state our main results. In Section 2 we prove results relating EX(n,)\mathrm{EX}(n,\mathcal{F}) and SPEX(n,)\mathrm{SPEX}(n,\mathcal{F}). In Section 3, we prove results relating SPEX(n,)\mathrm{SPEX}(n,\mathcal{F}) and SPEXα(n,)\mathrm{SPEX}_{\alpha}(n,\mathcal{F}). In Section 4 we give concluding remarks.

1.1 Notation

For graphs GG and HH, G¯\overline{G} denotes the complement of GG, G+HG+H denotes the join of GG and HH, GHG\cup H denotes their disjoint union, and aGa\cdot G denotes the disjoint union of aa copies of GG; if a=a=\infty then this means countably many copies. We write HGH\subseteq G if HH is a subgraph of GG. We write Δ(G)\Delta(G) for the maximum degree of GG. We write A=A(G)A=A(G) for the adjacency matrix, D=D(G)D=D(G) for the diagonal degree matrix, and Aα=Aα(G)=αD(G)+(1α)A(G)A_{\alpha}=A_{\alpha}(G)=\alpha D(G)+(1-\alpha)A(G). For a symmetric matrix MM we use λ1(M)\lambda_{1}(M) to mean the largest eigenvalue of the matrix, and for a graph GG we write λ(G)\lambda(G) and λα(G)\lambda_{\alpha}(G) for λ1(A(G))\lambda_{1}(A(G)) and λ1(Aα(G))\lambda_{1}(A_{\alpha}(G)), respectively. For vV(G)v\in V(G) and UV(G)U\subseteq V(G), we write N(v)N(v) for the neighborhood of vv, NU(v)N_{U}(v) for N(v)UN(v)\cap U, and dU(v)d_{U}(v) for |NU(v)||N_{U}(v)|. For ii\in\mathbb{N}, we write Ni(v)N_{i}(v) for the set of vertices at distance ii from vv. For U,WV(G)U,W\subseteq V(G), we write e(U,W)e(U,W) for |{(u,w):uU,wW,uw}||\{(u,w):u\in U,w\in W,u\sim w\}|, and e(U)e(U) for e(U,U)/2e(U,U)/2; we write E(U,W)E(U,W) and E(U)E(U) for the corresponding sets of edges. We write G[U]G[U] for the subgraph induced by UU and if UW=U\cap W=\emptyset we write G[U,W]G[U,W] for the bipartite subgraph on vertices UWU\sqcup W and with all edges in GG between UU and WW. Our notation for specific classes of graphs is standard; we note that PnP_{n} denotes the path of order nn, MnM_{n} is a maximum matching on nn vertices, and Km,K_{m,\infty} means that the larger partite set has a countable set of vertices. Sometimes we will consider the graph on nn vertices which is the disjoint union of a maximum number of copies of GG, plus isolated vertices in the remainder, and we will call this the maximal union of GG.

Let \mathcal{F} be a family of graphs and let GG be a graph. We denote by exG(n,)\mathrm{ex}^{G}(n,\mathcal{F}) the maximum number of edges in an \mathcal{F}-free graph which contains a copy of GG, and we write EXG(n,)\mathrm{EX}^{G}(n,\mathcal{F}) for the corresponding extremal graphs. We write exc(n,)\mathrm{ex}_{c}(n,\mathcal{F}) for the connected extremal number of \mathcal{F}, the maximum number of edges in a connected \mathcal{F}-free graph, and EXc(n,)\mathrm{EX}_{c}(n,\mathcal{F}) for the set of corresponding extremal graphs.

1.2 Main results

Our first goal is to relate the extremal graphs and spectral extremal graphs. In [63] it was proven that, if EX(n,F)=Kk+Knk¯\mathrm{EX}(n,F)=K_{k}+\overline{K_{n-k}} or EX(n,F)=Kk+(K2Knk2¯)\mathrm{EX}(n,F)=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}), then SPEX(n,F)=EX(n,F)\mathrm{SPEX}(n,F)=\mathrm{EX}(n,F) if nn is large enough. The following theorem generalizes this result (see the discussion following Proposition 1.2.)

Theorem 1.1.

Let \mathcal{F} be a family of graphs. Suppose that ex(n,)=O(n)\mathrm{ex}(n,\mathcal{F})=O(n), Kk+1,K_{k+1,\infty} is not \mathcal{F}-free, and for nn large enough EXKk,nk(n,)H\mathrm{EX}^{K_{k,n-k}}(n,\mathcal{F})\ni H, where one of the following holds:

  • (a)

    H=Kk,nkH=K_{k,n-k}

  • (b)

    H=Kk+Knk¯H=K_{k}+\overline{K_{n-k}}

  • (c)

    H=Kk+(K2Knk2¯)H=K_{k}+(K_{2}\cup\overline{K_{n-k-2}})

  • (d)

    H=Kk+XH=K_{k}+X where e(X)Qn+O(1)e(X)\leq Qn+O(1) for some Q[0,3/4)Q\in[0,3/4) and \mathcal{F} is finite

  • (e)

    H=Kk+XH=K_{k}+X where e(X)Qn+O(1)e(X)\leq Qn+O(1) for some Q[3/4,)Q\in[3/4,\infty) and \mathcal{F} is finite

  • (f)

    (e) holds, all but a bounded number of vertices of XX have constant degree dd, Kk+(K1,d+1)K_{k}+(\infty\cdot K_{1,d+1}) is not \mathcal{F}-free.

If (a), (b), or (c) holds then for nn large enough, SPEX(n,)=H\mathrm{SPEX}(n,\mathcal{F})=H.

If (d) or (f) holds then for nn large enough, SPEX(n,)EXKk+Knk¯(n,)\mathrm{SPEX}(n,\mathcal{F})\subseteq\mathrm{EX}^{K_{k}+\overline{K_{n-k}}}(n,\mathcal{F}).

If (e) holds then for nn large enough all graphs in SPEX(n,)\mathrm{SPEX}(n,\mathcal{F}) are of the form Kk+YK_{k}+Y, where e(Y)e(X)O(n1/2)e(Y)\geq e(X)-O(n^{1/2}) and Δ(Y)\Delta(Y) is bounded in terms of \mathcal{F}.

We make two remarks about Theorem 1.1. First, we show in Section 2 any graph satisfying (d) must have e(X)=Qn+O(1)e(X)=Qn+O(1) for some Q{0,1/2,2/3}Q\in\{0,1/2,2/3\}. Secondly, Theorem 1.1 (d) is sharp in that it fails if Q=3/4Q=3/4. The reason is that there are two non-isomorphic trees of order 44.

Proposition 1.2.

There exists a finite family of graphs \mathcal{F} such that EX(n,)=K2+X\mathrm{EX}(n,\mathcal{F})=K_{2}+X, where e(X)=3n/4+O(1)e(X)=3n/4+O(1), and for infinitely many nn, SPEX(n,)EX(n,)=\mathrm{SPEX}(n,\mathcal{F})\cap\mathrm{EX}(n,\mathcal{F})=\emptyset.

We now turn to applications. We would like to make use of the existing literature on the extremal functions ex(n,)\mathrm{ex}(n,\mathcal{F}) and exc(n,)\mathrm{ex}_{c}(n,\mathcal{F}). Thus, note that if EX(n,)HKk,nk\mathrm{EX}(n,\mathcal{F})\ni H\supseteq K_{k,n-k} or EXc(n,)HKk,nk\mathrm{EX}_{c}(n,\mathcal{F})\ni H\supseteq K_{k,n-k} then EXKk,nk(n,)H\mathrm{EX}^{K_{k,n-k}}(n,\mathcal{F})\ni H, and moreover in cases (a)-(d) or else if Q<1Q<1, we have that Kk+1,K_{k+1,\infty} is not \mathcal{F}-free. Furthermore it is not too difficult to show that if exc(n,)=O(n)\mathrm{ex}_{c}(n,\mathcal{F})=O(n) then ex(n,)=O(n)\mathrm{ex}(n,\mathcal{F})=O(n). This implies that in such cases, if we replace the assumption “EXKk,nk(n,)H\mathrm{EX}^{K_{k,n-k}}(n,\mathcal{F})\ni H and ex(n,)=O(n)\mathrm{ex}(n,\mathcal{F})=O(n)” in Theorem 1.1 with either “EX(n,)H\mathrm{EX}(n,\mathcal{F})\ni H” or “EXc(n,)H\mathrm{EX}_{c}(n,\mathcal{F})\ni H,” then we obtain a special case of the result, which generalizes Theorem 1.7 of [63].

Corollary 1.3.

Suppose that \mathcal{F} is a family of graphs such that, for nn large enough, EX(n,)H\mathrm{EX}(n,\mathcal{F})\ni H (or EXc(n,)H\mathrm{EX}_{c}(n,\mathcal{F})\ni H), where HH is one of the graphs in (a)-(d) or (f) of Theorem 1.1, and moreover that \mathcal{F} is finite in cases (d) and (f) and Q<1Q<1 in case (f). Then for nn large enough, SPEX(n,)EX(n,)\mathrm{SPEX}(n,\mathcal{F})\subseteq\mathrm{EX}(n,\mathcal{F}) (or SPEX(n,)EXc(n,)\mathrm{SPEX}(n,\mathcal{F})\subseteq\mathrm{EX}_{c}(n,\mathcal{F})).

We list some applications of Theorem 1.1 and Corollary 1.3. All of our results only apply when nn is large enough, and this condition is assumed everywhere in the list below. The literature for specific families \mathcal{F} contains many results with a specific lower bound on the ‘large enough’ nn, and thus our results do not always imply the quantitative part of the results that we mention.

  • Paths. Balister, Győri, Lehel, and Schelp [1] showed that EXc(n,P2+2)=K+Kn¯\mathrm{EX_{c}}(n,P_{2\ell+2})=K_{\ell}+\overline{K_{n-\ell}} and EXc(n,P2+3)=K+(K2Kn2¯)\mathrm{EX_{c}}(n,P_{2\ell+3})=K_{\ell}+(K_{2}\cup\overline{K_{n-\ell-2}}) for any 1\ell\geq 1. Thus, Corollary 1.3 and Theorem 1.1 (b) and (c) give SPEX(n,P2+2)=K+Kn¯\mathrm{SPEX}(n,P_{2\ell+2})=K_{\ell}+\overline{K_{n-\ell}} and SPEX(n,P2+3)=K+(K2Kn2¯)\mathrm{SPEX}(n,P_{2\ell+3})=K_{\ell}+(K_{2}\cup\overline{K_{n-\ell-2}}). This result was shown by Nikiforov [51].

  • Matchings. Erdős and Gallai [18] showed that EX(n,M2k+2)=Kk+Knk¯\mathrm{EX}(n,M_{2k+2})=K_{k}+\overline{K_{n-k}}. Thus, SPEX(n,M2k+2)=Kk+Knk¯\mathrm{SPEX}(n,M_{2k+2})=K_{k}+\overline{K_{n-k}}. This result was shown by Feng, Yu, and Zhang [27] (and also proved in a different way by Csikvári [16]).

  • Copies of P3P_{3}. Bushaw and Kettle [5] showed that EX(n,kP3)=Kk1+Mnk+1\mathrm{EX}(n,k\cdot P_{3})=K_{k-1}+M_{n-k+1}. Thus, Theorem 1.1 (d) gives SPEX(n,kP3)=Kk1+Mnk+1\mathrm{SPEX}(n,k\cdot P_{3})=K_{k-1}+M_{n-k+1} This was shown by Chen, Liu, and Zhang [8].

  • Linear forests. Lidicky, Liu, and Palmer [42] generalized the result of [18]: if F=i=1jPviF=\bigcup_{i=1}^{j}P_{v_{i}} for vi2v_{i}\geq 2, j2j\geq 2, and at least one vi3v_{i}\neq 3, then EX(n,F)=Kk+Knk¯\mathrm{EX}(n,F)=K_{k}+\overline{K_{n-k}} if some viv_{i} is even and EX(n,F)=Kk+(K2Knk2¯)\mathrm{EX}(n,F)=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}) if all viv_{i} are odd, where k=(i=1jvi/2)1k=\left(\sum_{i=1}^{j}\lfloor v_{i}/2\rfloor\right)-1. Thus, SPEX(n,F)=Kk+Knk¯\mathrm{SPEX}(n,F)=K_{k}+\overline{K_{n-k}} if some viv_{i} is even and SPEX(n,F)=Kk+(K2Knk2¯)\mathrm{SPEX}(n,F)=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}) if all viv_{i} are odd. This result was shown by Chen, Liu, and Zhang [8].

  • Star forests. Lidicky, Liu, and Palmer [42] investigated the extremal number of F=i=1kK1,diF=\bigcup_{i=1}^{k}K_{1,d_{i}}, where d1dkd_{1}\geq\cdots\geq d_{k} and k2k\geq 2. In this case it is possible that EX(n,)SPEX(n,)=.\mathrm{EX}(n,\mathcal{F})\cap\mathrm{SPEX}(n,\mathcal{F})=\emptyset. Note that EXKk1,nk+1(n,)=Kk1+X\mathrm{EX}^{K_{k-1,n-k+1}}(n,\mathcal{F})=K_{k-1}+X, where XX is an almost (dk1)(d_{k}-1)-regular graph; Kk,FK_{k,\infty}\supseteq F; and Kk1+(K1,dk)K_{k-1}+(\infty\cdot K_{1,d_{k}}) is not \mathcal{F}-free. Thus Theorem 1.1 (f) gives that SPEX(n,)EXKk+Knk¯(n,)\mathrm{SPEX}(n,\mathcal{F})\subseteq\mathrm{EX}^{K_{k}+\overline{K_{n-k}}}(n,\mathcal{F}). Then one can show using equitable partitions that if nk+1n-k+1 is even, we actually have SPEX(n,)={Kk+X:X is (dk1)-regular}\mathrm{SPEX}(n,\mathcal{F})=\{K_{k}+X:X\text{ is }(d_{k}-1)\text{-regular}\}. This was shown by Chen, Liu, and Zhang [9].

  • Trees. At the same time that we were finishing this paper Fang, Lin, Shu, and Zhang [23] proved several theorems studying trees. These include the results listed below about “certain small trees”. Our result below about “almost all trees” is complementary to the results in [23]; their results determine for which trees TT SPEX(n,T)=Kk,nk\mathrm{SPEX}(n,T)=K_{k,n-k} but the characterization is phrased in terms of coverings and the extremal number of a decomposition-like family.

    • Certain small trees. Caro, Patkós and Tuza [6] determined EXc(n,T)\mathrm{EX}_{c}(n,T) for many small trees TT. Let Sa1,,ajS_{a_{1},\ldots,a_{j}} denote the spider obtained by identifying paths of orders a1+1,,aj+1a_{1}+1,\ldots,a_{j}+1 at one endpoint, and let Da,bD_{a,b} denote the double star on a+b+2a+b+2 vertices whose two non-leaf vertices have degrees a+1a+1 and b+1b+1. Let D2,2D_{2,2}^{*} be the graph obtained by attaching a leaf to a leaf of D2,2D_{2,2}. Among their results there are some to which we can apply Theorem 1.1 which are not already mentioned above: (i) EXc(n,S2,2,1)=K2+Kn2¯\mathrm{EX}_{c}(n,S_{2,2,1})=K_{2}+\overline{K_{n-2}}; (ii) EXc(n,S3,1,1)=K1+Mn1\mathrm{EX}_{c}(n,S_{3,1,1})=K_{1}+M_{n-1}; (iii) EXc(n,D2,2)=K2,n2\mathrm{EX}_{c}(n,D_{2,2})=K_{2,n-2}; (iv) EXc(n,D2,2)=K2+Kn2¯\mathrm{EX}_{c}(n,D_{2,2}^{*})=K_{2}+\overline{K_{n-2}}; (v) EXc(n,S3,2,1)=K2+Kn2¯\mathrm{EX}_{c}(n,S_{3,2,1})=K_{2}+\overline{K_{n-2}}; (vi) EXc(n,D2,3)=K2,n2\mathrm{EX}_{c}(n,D_{2,3})=K_{2,n-2}.

    • Almost all trees. Let TT be a tree with proper coloring V=ABV=A\sqcup B, where k+1=|A||B|k+1=|A|\leq|B|, such that BB contains a vertex of degree at least 3, exactly one of whose neighbors is not a leaf. (In the proof of Proposition 1.4 we will show that ‘almost all trees’ indeed have this property.) It is well-known that ex(n,T)=O(n)\mathrm{ex}(n,T)=O(n). Observe that Kk,K_{k,\infty} is TT-free. To show it is TT-saturated, it remains to show that adding any edge to Kk,K_{k,\infty} creates a copy of TT. Let L,RL,R be the partite sets with |L|=k|L|=k. Since AA has a leaf, e(R)=0e(R)=0, or else we could embed TT in GG by placing this leaf incident to an edge in RR and placing the rest of AA in LL. To show e(L)=0e(L)=0, suppose LL has an edge. Then we can embed TT in GG by placing at least 2 leaves of AA in RR and the rest of AA in LL, a contradiction. Thus, Theorem 1.1 (a) gives SPEX(n,T)=Kk,nk\mathrm{SPEX}(n,T)=K_{k,n-k}.

  • Spectral Erdős-Sós theorem. Note that GG contains all trees on 2k+22k+2 vertices if and only if GG is not \mathcal{F}-free, where \mathcal{F} is the set of all graphs which contain all trees on 2k+22k+2 vertices. Note that EXKk,nk(n,)=Kk+Knk¯\mathrm{EX}^{K_{k,n-k}}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}} (see e.g. Lemma 4.6 of [14]). Thus, we have SPEX(n,)=Kk+Knk¯\mathrm{SPEX}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}}. Similarly, if \mathcal{F}^{\prime} is the set of all graphs which contain all trees on 2k+32k+3 vertices, then SPEX(n,)=Kk+(K2Knk2¯)\mathrm{SPEX}(n,\mathcal{F}^{\prime})=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}). These results were shown by Cioabă and the second and third authors [14].

  • Long cycles. Let ={C,C+1,}\mathcal{F}=\{C_{\ell},C_{\ell+1},\ldots\} be the set of cycles of length at least 5\ell\geq 5. Note EXKk,nk(n,)=Kk+Knk¯\mathrm{EX}^{K_{k,n-k}}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}} if \ell is odd and EXKk,nk(n,)=Kk+(K2Knk2¯)\mathrm{EX}^{K_{k,n-k}}(n,\mathcal{F})=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}) if \ell is even, where k=(1)/2k=\lfloor(\ell-1)/2\rfloor; this can be viewed as a special case of a result of Kopylov [38]. Erdős and Gallai [18] showed that ex(n,)12(k1)(n1)\mathrm{ex}(n,\mathcal{F})\leq\frac{1}{2}(k-1)(n-1). Thus, SPEX(n,)=Kk+Knk¯\mathrm{SPEX}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}} if \ell is odd and SPEX(n,)=Kk+(K2Knk2¯)\mathrm{SPEX}(n,\mathcal{F})=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}) if \ell is even. This result was shown by Gao and Hou [28]. In fact, the result also follows from the spectral even cycle theorem [13], but we have included it to demonstrate the versatility of Theorem 1.1.

  • Arithmetic progression of cycles. Let \mathcal{F} be the set of cycles of length \ell modulo rr, <r\ell<r. Bollobás [3] showed that ex(n,)<r1((r+1)r1)n\mathrm{ex}(n,\mathcal{F})<r^{-1}((r+1)^{r}-1)n. Thus, if \ell is even and at least 5, then SPEX(n,)=K/21+(K2Kn/21¯)\mathrm{SPEX}(n,\mathcal{F})=K_{\ell/2-1}+(K_{2}\cup\overline{K_{n-\ell/2-1}}) and if ,r\ell,r are both odd then SPEX(n,)=K(r+)/21,n(r+)/2+1\mathrm{SPEX}(n,\mathcal{F})=K_{(r+\ell)/2-1,n-(r+\ell)/2+1}. The first result follows from the spectral even cycle theorem, while the second one appears to be new.

  • Interval of even cycles. Note that GG contains kk consecutive even cycle lengths if and only if GG is not \mathcal{F}-free, where \mathcal{F} is the set of all graphs which contain kk consecutive even cycle lengths. Verstraëte [57] proved that ex(n,)<3kn\mathrm{ex}(n,\mathcal{F})<3kn. Thus, SPEX(n,)=Kk+(K2Knk2¯)\mathrm{SPEX}(n,\mathcal{F})=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}). This result appears to be new.

  • Disjoint cycles. Let \mathcal{F} be the set of all disjoint unions of kk cycles (of possibly different lengths), where k2k\geq 2. Erdős and Pósa [19] showed that EX(n,)=K2k1+Kn2k+1¯\mathrm{EX}(n,\mathcal{F})=K_{2k-1}+\overline{K_{n-2k+1}}. Thus, SPEX(n,)=K2k1+Kn2k+1¯.\mathrm{SPEX}(n,\mathcal{F})=K_{2k-1}+\overline{K_{n-2k+1}}. This result was shown by Liu and Zhai [43].

  • Disjoint long cycles. Let \mathcal{F} be the set of all disjoint unions of kk cycles, each of length at least 5\ell\geq 5. Harvey and Wood [34] showed that ex(n,)<23kn\mathrm{ex}(n,\mathcal{F})<\frac{2}{3}k\ell n. Thus, if \ell is even then SPEX(n,)=Kk/21+(K2Knk/21¯)\mathrm{SPEX}(n,\mathcal{F})=K_{k\ell/2-1}+(K_{2}\cup\overline{K_{n-k\ell/2-1}}) and if \ell is odd then SPEX(n,)=Kk(+1)/21+Knk(+1)/2+1¯\mathrm{SPEX}(n,\mathcal{F})=K_{k(\ell+1)/2-1}+\overline{K_{n-k(\ell+1)/2+1}}. The first result follows from the spectral extremal result for tCt\cdot C_{\ell} [24], while the second result appears to be new.

  • Disjoint equicardinal cycles. Let \mathcal{F} be the set of all disjoint unions of two cycles of the same length. Häggkvist [33] showed the ex(n,)<6n\mathrm{ex}(n,\mathcal{F})<6n. Thus, SPEX(n,)=K3+Kn3¯\mathrm{SPEX}(n,\mathcal{F})=K_{3}+\overline{K_{n-3}}. This result appears to be new.

  • Minors. Note that GG is FF-minor free if and only if GG is \mathcal{F}-free, where \mathcal{F} is the set of all graphs which have an FF-minor. Suppose that Kk+Knk¯K_{k}+\overline{K_{n-k}} is \mathcal{F}-saturated; then EXKk,nk(n,)=Kk+Knk¯\mathrm{EX}^{K_{k,n-k}}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}}, because Kk¯+(K2K¯)\overline{K_{k}}+(K_{2}\cup\overline{K_{\infty}}) has a Kk+(K2K¯)K_{k}+(K_{2}\cup\overline{K_{\infty}})-minor. Mader [46] showed that ex(n,)<(k1)2(k12)1n\mathrm{ex}(n,\mathcal{F})<(k-1)2^{{k-1\choose 2}-1}n. Thus SPEX(n,)=Kk+Knk¯\mathrm{SPEX}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}}. We list below some special cases.

    • The KkK_{k}-minor free spectral extremal graph is Kk2+Knk+2¯K_{k-2}+\overline{K_{n-k+2}}. This was shown by the third author [55].

    • If FF is obtained by deleting from KkK_{k} the edges of disjoint paths, not all of order 3, then the FF-minor free spectral extremal graph is Kk3+Knk+3¯K_{k-3}+\overline{K_{n-k+3}}. This was shown recently by Chen, Liu, and Zhang [12].

    • If FkF_{k} is the friendship graph with kk triangles then the FkF_{k}-minor free spectral extremal graph is Kk+Knk¯K_{k}+\overline{K_{n-k}}. This was shown by He, Li, and Feng [35].

    If SPEX(n,F)=Kk+Knk¯\mathrm{SPEX}(n,F)=K_{k}+\overline{K_{n-k}} then SPEX(n,)=Kk+Knk¯\mathrm{SPEX}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}} holds automatically, without making use of Theorem 1.1. The same applies to SPEXα(n,F)\mathrm{SPEX}_{\alpha}(n,F).

  • Topological subdivisions. The extremal result of [46] is actually stronger than as stated in the application above; it applies even when \mathcal{F} is the family of all topological subdivisions of KkK_{k}. A similar argument to the minor case then proves the following: if \mathcal{F} is the family of all topological subdivisions of FF, and Kk+Knk¯K_{k}+\overline{K_{n-k}} is \mathcal{F}-saturated then SPEX(n,)=Kk+Knk¯\mathrm{SPEX}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}}.

  • Chorded cycles. Let \mathcal{F} be the family of all chorded cycles. Pósa (see e.g. in [32]) showed that ex(n,)<2n3\mathrm{ex}(n,\mathcal{F})<2n-3. Thus SPEX(n,)=K2,n2\mathrm{SPEX}(n,\mathcal{F})=K_{2,n-2}. This was shown recently by Zheng, Huang, and Wang [65] and gives an answer to a question posed by Gould (Question 3 in [32]).

  • Two disjoint chorded cycles. Let 1\mathcal{F}_{1} be the family of all graphs made of two disjoint cycles, one of which has a chord, and let 2\mathcal{F}_{2} be the family of all graphs made of two disjoint chorded cycles. Bialostocki, Finkel, and Gyárfás [2] showed that EX(n,1)K4,n4\mathrm{EX}(n,\mathcal{F}_{1})\ni K_{4,n-4} and EX(n,2)K5,n5\mathrm{EX}(n,\mathcal{F}_{2})\ni K_{5,n-5}. Thus SPEX(n,1)=K4,n4\mathrm{SPEX}(n,\mathcal{F}_{1})=K_{4,n-4} and SPEX(n,2)=K5,n5\mathrm{SPEX}(n,\mathcal{F}_{2})=K_{5,n-5}. This result appears to be new.

  • Disjoint chorded cycles. Let \mathcal{F} be the family of all graphs made of kk disjoint chorded cycles. The result of Gao and Wang [30] implies that ex(n,)=O(n)\mathrm{ex}(n,\mathcal{F})=O(n) via a well-known bound on max cut. Thus, SPEX(n,)=K3k1,n3k+1\mathrm{SPEX}(n,\mathcal{F})=K_{3k-1,n-3k+1}. This result appears to be new.

  • Multiply chorded cycles. Let \mathcal{F} be the family of all cycles with k(k2)+1k(k-2)+1 chords, k2k\geq 2. Gould, Horn, and Magnant [31] showed that ex(n,)=O(n)\mathrm{ex}(n,\mathcal{F})=O(n). Thus, SPEX(n,)=Kk,nk\mathrm{SPEX}(n,\mathcal{F})=K_{k,n-k}. This result appears to be new.

  • Cycles with kk incident chords. Let \mathcal{F} be the family of cycles with kk chords all incident to the same vertex. Jiang [36] showed that EX(n,)Kk+1,nk1\mathrm{EX}(n,\mathcal{F})\ni K_{k+1,n-k-1}. Thus, SPEX(n,)=Kk+1,nk1\mathrm{SPEX}(n,\mathcal{F})=K_{k+1,n-k-1}. This result appears to be new.

Proposition 1.4.

The proportion of all labelled mm-vertex trees TT such that for some kk, SPEX(n,T)=Kk,nk\mathrm{SPEX}(n,T)=K_{k,n-k} for nn large enough, approaches 11 as mm\to\infty.

Our second goal is to relate the spectral extremal graphs and alpha spectral extremal graphs.

Theorem 1.5.

Let \mathcal{F} be a family of graphs containing some bipartite graph FF for which FvF-v is a forest, or such that ex(n,)=O(n)\mathrm{ex}(n,\mathcal{F})=O(n). Suppose that for nn large enough, SPEX(n,)H\mathrm{SPEX}(n,\mathcal{F})\ni H, where either:

  • (a)

    H=Kk+Knk¯H=K_{k}+\overline{K_{n-k}}

  • (b)

    H=Kk+(K2Knk2¯)H=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}).

Then for any α(0,1)\alpha\in(0,1), for any nn large enough, SPEXα(n,)=H\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=H.

We give a list of new and existing results implied by Theorem 1.5; if no reference is given for the required spectral result that means that it falls under the previous list of applications.

  • Paths. We have SPEXα(n,P2+2)=K+Kn¯\mathrm{SPEX}_{\alpha}(n,P_{2\ell+2})=K_{\ell}+\overline{K_{n-\ell}} and SPEXα(n,P2+3)=K+(K2Kn2¯)\mathrm{SPEX}_{\alpha}(n,P_{2\ell+3})=K_{\ell}+(K_{2}\cup\overline{K_{n-\ell-2}}). This was shown by Chen, Liu, and Zhang [11].

  • Matchings. We have SPEXα(n,M2k+2)=Kk+Knk¯\mathrm{SPEX}_{\alpha}(n,M_{2k+2})=K_{k}+\overline{K_{n-k}}. This was shown by Yuan and Shao [60].

  • Linear forests. Generalizing the previous two results: if F=i=1jPviF=\bigcup_{i=1}^{j}P_{v_{i}} where vi2v_{i}\geq 2, j2j\geq 2, and at least one vi3v_{i}\neq 3, then SPEXα(n,)=Kk+Knk¯\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}} if some viv_{i} is even and SPEXα(n,)=Kk+(K2Knk2¯)\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}) if all viv_{i} are odd, where k=(i=1jvi/2)1k=\left(\sum_{i=1}^{j}\lfloor v_{i}/2\rfloor\right)-1. This was shown by Chen, Liu, and Zhang [11].

  • Certain small trees. We have (i) SPEXα(n,S2,2,1)=Kk+Kn2¯\mathrm{SPEX}_{\alpha}(n,S_{2,2,1})=K_{k}+\overline{K_{n-2}}; (ii) SPEXα(n,D2,2)=K2+Kn2¯\mathrm{SPEX}_{\alpha}(n,D_{2,2}^{*})=K_{2}+\overline{K_{n-2}}; (iii) SPEXα(n,S3,2,1)=K2+Kn2¯\mathrm{SPEX}_{\alpha}(n,S_{3,2,1})=K_{2}+\overline{K_{n-2}}. These results appear to be new.

  • Spectral Erdős-Sós theorem. Let \mathcal{F} be the set of all graphs which contain all trees on 2k+22k+2 vertices, and let \mathcal{F}^{\prime} be the set of all graphs which contain all trees on 2k+32k+3 vertices. Then SPEXα(n,)=Kk+Knk¯\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}} and SPEXα(n,)=Kk+(K2Knk2¯)\mathrm{SPEX}_{\alpha}(n,\mathcal{F}^{\prime})=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}). These results were shown by Chen, Li, Li, Yu, and Zhang [7].

  • Even cycles and consecutive cycles. Cioabă and the second and third authors [13] showed that SPEX(n,C2+2)=K+(K2Kn2¯)\mathrm{SPEX}(n,C_{2\ell+2})=K_{\ell}+(K_{2}\cup\overline{K_{n-\ell-2}}) and SPEX(n,{C2+1,C2+2})=K+Kn¯\mathrm{SPEX}(n,\{C_{2\ell+1},C_{2\ell+2}\})=K_{\ell}+\overline{K_{n-\ell}}, for k2k\geq 2. Thus, SPEXα(n,C2+2)=Kk+(K2Kn2¯)\mathrm{SPEX}_{\alpha}(n,C_{2\ell+2})=K_{k}+(K_{2}\cup\overline{K_{n-\ell-2}}) and SPEXα(n,{C2+1,C2+2})=K+Kn¯\mathrm{SPEX}_{\alpha}(n,\{C_{2\ell+1},C_{2\ell+2}\})=K_{\ell}+\overline{K_{n-\ell}}. This was shown by Li and Yu [39].

  • Intersecting even cycles. Let C1,,tC_{\ell_{1},\ldots,\ell_{t}} be obtained by intersecting the cycles C1,CtC_{\ell_{1}}\ldots,C_{\ell_{t}} at a unique vertex. The second author [17] showed that if 1,,t2\ell_{1},\ldots,\ell_{t}\geq 2 and some t>2\ell_{t}>2, then SPEX(n,C21,,2t)=Kk+(K2Knk2¯)\mathrm{SPEX}(n,C_{2\ell_{1},\ldots,2\ell_{t}})=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}), where k=i=1t(i1)k=\sum_{i=1}^{t}(\ell_{i}-1). Thus, SPEXα(n,C21,,2t)=Kk+(K2Knk2¯)\mathrm{SPEX}_{\alpha}(n,C_{2\ell_{1},\ldots,2\ell_{t}})=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}). This result appears to be new.

  • Long cycles. If ={C,C+1,}\mathcal{F}=\{C_{\ell},C_{\ell+1},\ldots\} then SPEXα(n,)=Kk+Knk¯\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}} if \ell is odd and SPEXα(n,)=Kk+(K2Knk2¯)\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}) if \ell is even, where k=(1)/2k=\lfloor(\ell-1)/2\rfloor. This result is implied by the result forbidding even cycles or consecutive cycles.

  • Arithmetic progression of cycles. Let \mathcal{F} be the set of cycles of length \ell modulo rr, <r\ell<r. If \ell is even and at least 5, then SPEXα(n,)=K/21+(K2Kn/21¯)\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{\ell/2-1}+(K_{2}\cup\overline{K_{n-\ell/2-1}}). This result is implied by the result forbidding even cycles.

  • Interval of even cycles. Let \mathcal{F} be the family of all graphs which contain kk consecutive even cycle lengths. Then SPEXα(n,)=Kk+(K2Knk2¯)\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}). This result appears to be new.

  • Disjoint cycles. If \mathcal{F} is the set of all disjoint unions of kk cycles (of possibly different lengths) then SPEXα(n,)=K2k1+Kn2k+1¯\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{2k-1}+\overline{K_{n-2k+1}}. This was shown by Li, Yu, and Zhang [40].

  • Disjoint long cycles. Let \mathcal{F} be the set of all disjoint unions of kk cycles, each of length at least 5\ell\geq 5. If \ell is even then SPEXα(n,)=Kk/21+Knk/2+1¯\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k\ell/2-1}+\overline{K_{n-k\ell/2+1}} and if \ell is odd then SPEXα(n,)=Kk(+1)/21+(K2Knk(+1)/2+1¯)\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k(\ell+1)/2-1}+(K_{2}\cup\overline{K_{n-k(\ell+1)/2+1}}). This result seems to be new.

  • Disjoint equicardinal cycles. If \mathcal{F} is the family of all disjoint unions of two cycles of the same length, then SPEXα(n,)=K3+Kn3¯\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{3}+\overline{K_{n-3}} for nn large enough. This result seems to be new.

  • Minors. If \mathcal{F} is the family of all graphs which have an FF-minor and Kk+Knk¯K_{k}+\overline{K_{n-k}} is \mathcal{F}-saturated, then SPEXα(n,)=Kk+Knk¯\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}}.

    • If F=KkF=K_{k} then SPEXα(n,)=Kk2+Knk+2¯\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k-2}+\overline{K_{n-k+2}}. This was shown by Chen, Liu, and Zhang [10].

    • If FF is obtained by deleting from KkK_{k} the edges of disjoint paths, not all of length 3, then SPEXα(n,)=Kk3+Knk+3¯\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k-3}+\overline{K_{n-k+3}}. This result appears to be new.

    • If FF is the friendship graph with kk triangles SPEXα(n,)=Kk+Knk¯\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}}. This was shown by Wang and Zhang [64].

  • Topological subdivisions. If \mathcal{F} is the family of all topological subdivisions of FF and Kk+Knk¯K_{k}+\overline{K_{n-k}} is \mathcal{F}-saturated, then SPEXα(n,)=Kk+Knk¯\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k}+\overline{K_{n-k}}.

We observe that there are fewer options for the assumed extremal graph in Theorem 1.5 than in Theorem 1.1. In the case that H=Kk,nkH=K_{k,n-k} there is a simple counterexample that explains this: let FF be any graph such that SPEX(n,F)=Kk,nk\mathrm{SPEX}(n,F)=K_{k,n-k} for nn large enough, where k>1k>1. Then K1,nK_{1,n} is also FF-free, and if α\alpha is sufficiently close to 1 then λα(K1,n)α(n1)>λα(Kk,nk)\lambda_{\alpha}(K_{1,n})\geq\alpha(n-1)>\lambda_{\alpha}(K_{k,n-k}) (see e.g. [53]). Thus, one could only hope to extend Theorem 1.5 to this case if α\alpha is restricted to a certain range. When HH is of the form Kk+XK_{k}+X, then the situation may be more subtle and we do not attempt to address every case.

2 Spectral extremal results

In this section we prove results concerning SPEX(n,)\mathrm{SPEX}(n,\mathcal{F}). In subsections 2.1, 2.2, 2.3, and 2.4 we prove Theorem 1.1; in subsection 2.5 we prove Proposition 1.2; and in subsection 2.6 we prove Proposition 1.4. Unless stated otherwise, all claims are asserted only for nn large enough.

2.1 Structure of edge extremal graphs

Proposition 2.1.

Under the assumptions of Theorem 1.1 (d) XX can be obtained from a maximal union of P1P_{1}, of P2P_{2}, or of P3P_{3} by adding or deleting a bounded number of edges.

Proof.

Note that Kk+1,FK_{k+1,\infty}\supseteq F for some FF\in\mathcal{F}. Let δ\delta be the smallest degree in the partite set of FF of size k+1k+1. Note that Δ(X)δ1\Delta(X)\leq\delta-1, because FKk+(K1,δKnkδ1¯)F\subseteq K_{k}+(K_{1,\delta}\cup\overline{K_{n-k-\delta-1}}).

Consider the components of XX. We claim the number of components of order at least 4 is bounded; otherwise, the union of these components is a subgraph YXY\subseteq X with |V(Y)||V(Y)|\to\infty and |e(Y)|3|V(Y)|/4|e(Y)|\geq 3|V(Y)|/4, thus contradicting the fact that exKk,nk(n,)kn+Qn+O(1)\mathrm{ex}^{K_{k,n-k}}(n,\mathcal{F})\leq kn+Qn+O(1) for nn large enough. Moreover, the order of a component of XX is bounded; otherwise, by taking the largest component YY, we find a subgraph YXY\subseteq X with |V(Y)||V(Y)|\to\infty and |e(Y)||V(Y)|1|e(Y)|\geq|V(Y)|-1, again contradicting our bound on exKk,nk(n,)\mathrm{ex^{K_{k,n-k}}}(n,\mathcal{F}).

Now let ν=max{i:there exist an unbounded number of components of X of order i}{1,2,3}\nu=\max\{i:\text{there exist an unbounded number of components of $X$ of order }i\}\in\{1,2,3\}. If ν=1\nu=1 then e(X)e(X) is bounded, and we are done.

Suppose ν=2\nu=2. Then XX is the union of an unbounded number copies of P2P_{2}, a bounded number of graphs of bounded order, and isolated vertices. Suppose for contradiction there are 2 isolated vertices in XX; then adding an edge e0e_{0} between them creates a copy F0F_{0} of a graph from \mathcal{F}. Since there were already an unbounded number of disjoint edges in XX, one such edge e1e_{1} is disjoint from F0F_{0}, and the neighborhoods of its endpoints contain the neighborhoods of the endpoints of e0e_{0}, so we can replace e0e_{0} by e1e_{1} to find F1F0F_{1}\simeq F_{0} which already existed in Kk+XK_{k}+X, a contradiction. Thus there is at most 1 isolated vertex. Hence, XX is obtained from MnkM_{n-k} by deleting and adding a bounded number of edges.

Suppose ν=3\nu=3. Then XX is the union of an unbounded number of copies of P3P_{3}, a bounded number of graphs of bounded order, and graphs of order less than 3. We will show that the components smaller than P3P_{3} span at most 5 vertices; for otherwise, we can find either 3 P1P_{1} components, a P1P_{1} component and a P2P_{2} component, or three P2P_{2} components. In any case we can add and delete edges for a net increase in e(X)e(X) without creating any components on more than 3 vertices, and such that the copy of some F0F_{0}\in\mathcal{F} which must have been created uses a vertex from a newly created P3P_{3} component; but similarly to the previous case, we find another P3P_{3} component (or two disjoint P3P_{3} components if needed) which was not used in F0F_{0}, hence there was some F1F0F_{1}\simeq F_{0} which already existed, a contradiction. This shows that XX is obtained from a maximal union of P3P_{3} by deleting and adding a bounded number of edges. ∎

2.2 Common features of the spectral extremal graphs

The purpose of this subsection is to prove the following proposition.

Proposition 2.2.

Assume the setup of any of the cases (a)-(f) of Theorem 1.1. If GSPEX(n,)G\in\mathrm{SPEX}(n,\mathcal{F}) then GKk,nkG\supseteq K_{k,n-k}.

Let HH be the edge extremal graph; when we need to specify the number of vertices, we will write HnH_{n} instead of HH. Set λ=λ1(G)\lambda=\lambda_{1}(G), and let xx be a Perron eigenvector of GG, scaled so that its greatest entry is xz=1x_{z}=1. Let FF\in\mathcal{F} be a graph such that FKk+1,F\subseteq K_{k+1,\infty}, and let m=|V(F)|m=|V(F)|, m=sup{|V(F)|:F}m^{\prime}=\sup\{|V(F^{\prime})|:F^{\prime}\in\mathcal{F}\}. Let CC be chosen such that ex(n,)Cn\mathrm{ex}(n,\mathcal{F})\leq Cn. We choose positive constants η,ε,σ,β\eta,\varepsilon,\sigma,\beta to satisfy the following inequalities:

  • (i)

    β<k2C\beta<\frac{k}{2C}

  • (ii)

    10Cσ<ε210C\sigma<\varepsilon^{2}

  • (iii)

    ε<ηC\varepsilon<\eta\leq C

  • (iv)

    σ<ε230k\sigma<\frac{\varepsilon^{2}}{30k}

  • (v)

    (8k3+2)ε<η(8k^{3}+2)\varepsilon<\eta

  • (vi)

    ε<(2/51/4)/k3\varepsilon<(2/5-1/4)/k^{3}

  • (vii)

    (k+ε)η+ε2/2<1/2(k+\varepsilon)\eta+\varepsilon^{2}/2<1/2

  • (viii)

    (2C+m)η<114k3.(2C+m)\eta<1-\frac{1}{4k^{3}}.

To see that such a choice exists, we choose η\eta first with the additional constraint that η<k/4\eta<k/4; this allows that some ε>0\varepsilon>0 will satisfy (vii). Then we choose ε,σ,\varepsilon,\sigma, and β\beta in that order. With V=V(G)V=V(G), let L={vV:xvη}L^{\prime}=\{v\in V:x_{v}\geq\eta\}, L={vV:xvσ}L=\{v\in V:x_{v}\geq\sigma\}, M={vV:xvβσ}M=\{v\in V:x_{v}\geq\beta\sigma\}, and S=VLS=V-L. For vVv\in V, let Li(v)=LNi(v)L_{i}(v)=L\cap N_{i}(v), Mi(v)=MNi(v)M_{i}(v)=M\cap N_{i}(v), Si(v)=SNi(v)S_{i}(v)=S\cap N_{i}(v).

The proof of Proposition 2.2 is technical; a high-level outline is as follows.

  • 1.

    Show that |L||L| and |M||M| are not too large, i.e. there are not too many vertices with large eigenweight.

  • 2.

    Show that |L||L| is in fact bounded.

  • 3.

    Relate the eigenweight of vertices to their degrees.

  • 4.

    Show that eigenweights of vertices in LL^{\prime} are close to 1.

  • 5.

    Show that |L|=k|L^{\prime}|=k.

  • 6.

    Show that if uLu\not\in L^{\prime} then uvu\sim v for all vLv\in L^{\prime}.

Lemma 2.3.

We have

k(nk)λ2Cn.\sqrt{k(n-k)}\leq\lambda\leq\sqrt{2Cn}.
Proof.

Since Kk,nkK_{k,n-k} is \mathcal{F}-free we have

λλ1(Kk,nk)=k(nk).\lambda\geq\lambda_{1}(K_{k,n-k})=\sqrt{k(n-k)}.

For the upper bound,

λ2=λ2xz=uzwuxwuzd(u)uVd(u)=2e(G)2Cn\lambda^{2}=\lambda^{2}x_{z}=\sum_{u\sim z}\sum_{w\sim u}x_{w}\leq\sum_{u\sim z}d(u)\leq\sum_{u\in V}d(u)=2e(G)\leq 2Cn

Lemma 2.4.

We have

|L|2Cnσk(nk),|M|2Cnσβk(nk).|L|\leq\frac{2Cn}{\sigma\sqrt{k(n-k)}},\ \ |M|\leq\frac{2Cn}{\sigma\beta\sqrt{k(n-k)}}.
Proof.

For any vVv\in V we have λxv=uvxud(v)\lambda x_{v}=\sum_{u\sim v}x_{u}\leq d(v). Hence

k(nk)|L|σλ|L|σvLλxvvVλxvvVd(v)=2e(G)2Cn.\sqrt{k(n-k)}|L|\sigma\leq\lambda|L|\sigma\leq\sum_{v\in L}\lambda x_{v}\leq\sum_{v\in V}\lambda x_{v}\leq\sum_{v\in V}d(v)=2e(G)\leq 2Cn.

Solving for |L||L| proves the first claim. A similar argument applies to |M||M|. ∎

Lemma 2.5.

For any vLv\in L,

d(v)σk4(1+3C)n.d(v)\geq\frac{\sigma k}{4(1+3C)}n.
Proof.

Let vLv\in L. We have

σk(nk)λ2xv=uvwuxwd(v)+2e(N1)+e(N1,M2)+uvwuwN2M2xwB\sigma k(n-k)\leq\lambda^{2}x_{v}=\sum_{u\sim v}\sum_{w\sim u}x_{w}\leq d(v)+2e(N_{1})+e(N_{1},M_{2})+\underbrace{\sum_{u\sim v}\sum_{\begin{subarray}{c}w\sim u\\ w\in N_{2}-M_{2}\end{subarray}}x_{w}}_{B}

from the eigenvector equation, where in the first three terms in the right hand side we used the estimate xw1x_{w}\leq 1. We estimate e(N1)e(N_{1}), e(N1,M2)e(N_{1},M_{2}), and BB as follows. First, N1N_{1} induces a subgraph of order d(v)d(v) which does not contain any element of \mathcal{F}, so e(N1)Cd(v)e(N_{1})\leq Cd(v). Similarly we have e(N1,M2)C(2Cnσβk(nk)+d(v))e(N_{1},M_{2})\leq C\left(\frac{2Cn}{\sigma\beta\sqrt{k(n-k)}}+d(v)\right). For BB, we use xwσβx_{w}\leq\sigma\beta and so we have Bσβe(G)σβCnB\leq\sigma\beta e(G)\leq\sigma\beta Cn. Rearranging the inequality, we get

(σkσβC)nσk22C2nσβk(nk)(1+3C)d(v).\left(\sigma k-\sigma\beta C\right)n-\sigma k^{2}-\frac{2C^{2}n}{\sigma\beta\sqrt{k(n-k)}}\leq(1+3C)d(v).

By inequality (i), taking nn large enough gives

d(v)σk4(1+3C)n.d(v)\geq\frac{\sigma k}{4(1+3C)}n.

Lemma 2.6.

There is a constant C1=C1()C_{1}=C_{1}(\mathcal{F}) such that |L|C1|L|\leq C_{1}.

Proof.

We use the same argument from Lemma 2.4 with the updated information from Lemma 2.5. We have

|L|σk4(1+3C)nvLd(v)2e(G)2Cn|L|\frac{\sigma k}{4(1+3C)}n\leq\sum_{v\in L}d(v)\leq 2e(G)\leq 2Cn

so

|L|8C(1+3C)σk=:C1.|L|\leq\frac{8C(1+3C)}{\sigma k}=:C_{1}.

Lemma 2.7.

For any vertex vv, we have

k(nk)xvd(v)xv+e(S1,L1L2)+ε2n2.k(n-k)x_{v}\leq d(v)x_{v}+e(S_{1},L_{1}\cup L_{2})+\frac{\varepsilon^{2}n}{2}.
Proof.

Let c=xvc=x_{v}. We have

k(nk)c\displaystyle k(n-k)c λ2c=uvwuxw=d(v)c+uvwuwvxw\displaystyle\leq\lambda^{2}c=\sum_{u\sim v}\sum_{w\sim u}x_{w}=d(v)c+\sum_{u\sim v}\sum_{\begin{subarray}{c}w\sim u\\ w\neq v\end{subarray}}x_{w}
d(v)c+uS1(v)wuwL1(v)L2(v)xw+2e(S1)σ+2e(L)+e(L1,S1)σ+e(N1,S2)σ.\displaystyle\leq d(v)c+\sum_{u\in S_{1}(v)}\sum_{\begin{subarray}{c}w\sim u\\ w\in L_{1}(v)\cup L_{2}(v)\end{subarray}}x_{w}+2e(S_{1})\sigma+2e(L)+e(L_{1},S_{1})\sigma+e(N_{1},S_{2})\sigma.

We apply e(G)Cne(G)\leq Cn and |L|C1|L|\leq C_{1} to get

2e(S1)σ+2e(L)+e(L1,S1)σ+e(N1,S2)σ2Cnσ+2C12+Cnσ+Cnσ5Cnσ.2e(S_{1})\sigma+2e(L)+e(L_{1},S_{1})\sigma+e(N_{1},S_{2})\sigma\leq 2Cn\sigma+2C_{1}^{2}+Cn\sigma+Cn\sigma\leq 5Cn\sigma.

So

k(nk)c\displaystyle k(n-k)c d(v)c+uS1(v)wu,wL1(v)L2(v)xw+5Cnσd(v)c+e(S1,L1L2)+ε2n2\displaystyle\leq d(v)c+\sum_{u\in S_{1}(v)}\sum_{w\sim u,w\in L_{1}(v)\cup L_{2}(v)}x_{w}+5Cn\sigma\leq d(v)c+e(S_{1},L_{1}\cup L_{2})+\frac{\varepsilon^{2}n}{2}

where the last inequality holds by (ii). ∎

Lemma 2.8.

If vLv\in L^{\prime} and c=xvc=x_{v}, then d(v)(cε)nd(v)\geq(c-\varepsilon)n.

Proof.

Assume for a contradiction that d(v)<cnεnd(v)<cn-\varepsilon n. Thus Lemma 2.7 gives

k(nk)c(cnεn)c+e(S1,L1L2)+ε2n2.k(n-k)c\leq(cn-\varepsilon n)c+e(S_{1},L_{1}\cup L_{2})+\frac{\varepsilon^{2}n}{2}.

Hence

e(S1,L1L2)(kc+ε)ncck2ε2n2=(kc)nc+εncck2ε2n2(k1)nc+ε2n2,e(S_{1},L_{1}\cup L_{2})\geq(k-c+\varepsilon)nc-ck^{2}-\frac{\varepsilon^{2}n}{2}=(k-c)nc+\varepsilon nc-ck^{2}-\frac{\varepsilon^{2}n}{2}\geq(k-1)nc+\frac{\varepsilon^{2}n}{2},

using that vLv\in L^{\prime} implies that cη>εc\geq\eta>\varepsilon and c1c\leq 1.

Now we claim there are at least n\sqrt{n} vertices in S1S_{1} with at least kk neighbors in L1L2L_{1}\cup L_{2}. For if not, then e(S1,L1L2)<(k1)|S1|+|L|n<(k1)(cε)n+C1ne(S_{1},L_{1}\cup L_{2})<(k-1)|S_{1}|+|L|\sqrt{n}<(k-1)(c-\varepsilon)n+C_{1}\sqrt{n} because |S1|d(v)|S_{1}|\leq d(v) and |L|C1|L|\leq C_{1}. Thus (k1)cn(k1)εn+C1n>e(S1,L1L2)(k1)cn+ε2n2(k-1)cn-(k-1)\varepsilon n+C_{1}\sqrt{n}>e(S_{1},L_{1}\cup L_{2})\geq(k-1)cn+\frac{\varepsilon^{2}n}{2} which is a contradiction for nn large enough. Thus, let DD be a set of n\sqrt{n} vertices in S1S_{1} each with at least kk neighbors in LL. Since there are only (|L|k){|L|\choose k} options for these kk neighbors, there is some set of kk vertices in L1L2L_{1}\cup L_{2} with at least n/(|L|k)\sqrt{n}/{|L|\choose k} common neighbors in DD. Since |L|C1|L|\leq C_{1} is bounded, we have n/(|L|k)\sqrt{n}/{|L|\choose k}\to\infty as nn\to\infty, and so by adding the vertex vv we find that GKk+1,G\supseteq K_{k+1,\ell} for arbitrarily large \ell. However Kk+1,FK_{k+1,\ell}\supseteq F\in\mathcal{F} for \ell large enough, which is a contradiction. ∎

Lemma 2.9.

For the vertex zz with xz=1x_{z}=1, we have (1ε)kne(S1,{z}L1L2)(k+ε)n(1-\varepsilon)kn\leq e(S_{1},\{z\}\cup L_{1}\cup L_{2})\leq(k+\varepsilon)n.

Proof.

For the lower bound, we use Lemma 2.7:

k(nk)(1)d(z)+e(S1,L1L2)+ε2n2e(S1,{z}L1L2)+C1+ε2n2e(S1,{z}L1L2)+kεnk2k(n-k)(1)\leq d(z)+e(S_{1},L_{1}\cup L_{2})+\frac{\varepsilon^{2}n}{2}\leq e(S_{1},\{z\}\cup L_{1}\cup L_{2})+C_{1}+\frac{\varepsilon^{2}n}{2}\leq e(S_{1},\{z\}\cup L_{1}\cup L_{2})+k\varepsilon n-k^{2}

where the second inequality follows from e(S1,{z})d(z)C1e(S_{1},\{z\})\geq d(z)-C_{1} and the last inequality follows from ε<2k\varepsilon<2k. For the upper bound, suppose for a contradiction that e(S1,{z}L1L2)>(k+ε)ne(S_{1},\{z\}\cup L_{1}\cup L_{2})>(k+\varepsilon)n. Let γ=ε/C1\gamma=\varepsilon/C_{1}. We claim there are at least γn\gamma n vertices inside S1S_{1} with at least kk neighbors in L1L2L_{1}\cup L_{2}. If not then

e(S1,L1L2)<(k1)|S1|+|L|γn(k1)n+εn=(k+ε1)ne(S_{1},L_{1}\cup L_{2})<(k-1)|S_{1}|+|L|\gamma n\leq(k-1)n+\varepsilon n=(k+\varepsilon-1)n

because |S1|n|S_{1}|\leq n and |L|C1|L|\leq C_{1}. Since d(z)nd(z)\leq n, we then have e(S1,{z}L1L2)(k+ε)ne(S_{1},\{z\}\cup L_{1}\cup L_{2})\leq(k+\varepsilon)n, contradicting the assumption. Hence there is a subset DS1D\subseteq S_{1} with at least γn\gamma n vertices such that every vertex in DD has at least kk neighbors in L1L2L_{1}\cup L_{2}. Since there are at most (|L|k){|L|\choose k} options for these kk neighbors, there exists some set of kk vertices in L1L2L_{1}\cup L_{2} with at least γn/(|L|k)\gamma n/{|L|\choose k} common neighbors in S1S_{1}. Since |L|C1|L|\leq C_{1}, we have γn/(|L|k)\gamma n/{|L|\choose k}\to\infty as nn\to\infty, so adding in the vertex zz means that GKk+1,G\supseteq K_{k+1,\ell} for arbitrarily large \ell, hence contains FF, which is a contradiction. ∎

Lemma 2.10.

For all vLv\in L^{\prime}, d(v)(125k3)nd(v)\geq\left(1-\frac{2}{5k^{3}}\right)n and xv114k3x_{v}\geq 1-\frac{1}{4k^{3}}.

Proof.

We first show xv114k3x_{v}\geq 1-\frac{1}{4k^{3}}. Assume for a contradiction there is some vLv\in L^{\prime} with xv<114k3x_{v}<1-\frac{1}{4k^{3}}. Then the eigenvector equation for zz gives

k(nk)\displaystyle k(n-k) λ2<e(S1(z),{z}L1(z)L2(z){v})+|N1(z)N1(v)|xv+ε2n2\displaystyle\leq\lambda^{2}<e(S_{1}(z),\{z\}\cup L_{1}(z)\cup L_{2}(z)-\{v\})+|N_{1}(z)\cap N_{1}(v)|x_{v}+\frac{\varepsilon^{2}n}{2}
<(k+ε)n|S1(z)N1(v)|+|N1(z)N1(v)|(114k3)+ε2n2\displaystyle<(k+\varepsilon)n-|S_{1}(z)\cap N_{1}(v)|+|N_{1}(z)\cap N_{1}(v)|\left(1-\frac{1}{4k^{3}}\right)+\frac{\varepsilon^{2}n}{2}
=kn+εn+|L1(z)N1(v)||N1(z)N1(v)|14k3+ε2n2\displaystyle=kn+\varepsilon n+|L_{1}(z)\cap N_{1}(v)|-|N_{1}(z)\cap N_{1}(v)|\frac{1}{4k^{3}}+\frac{\varepsilon^{2}n}{2}

where in the first line, the ε2n/2\varepsilon^{2}n/2 term absorbs both the eigenweights of the vertices in SS (using inequality (iv)) and the weights from the edges in LL since |L|C1|L|\leq C_{1}. This implies

14k3|N1(z)N1(v)|<εn+ε2n2+|L|+k22εn.\frac{1}{4k^{3}}|N_{1}(z)\cap N_{1}(v)|<\varepsilon n+\frac{\varepsilon^{2}n}{2}+|L|+k^{2}\leq 2\varepsilon n.

However, since vLv\in L^{\prime} we have xvηx_{v}\geq\eta and so d(v)(ηε)nd(v)\geq(\eta-\varepsilon)n, so |N1(z)N1(v)|(η2ε)n.|N_{1}(z)\cap N_{1}(v)|\geq(\eta-2\varepsilon)n. This contradicts inequality (v). Now since xv114k3x_{v}\geq 1-\frac{1}{4k^{3}}, we have d(v)(xvε)n(114k3ε)n(125k3)nd(v)\geq(x_{v}-\varepsilon)n\geq\left(1-\frac{1}{4k^{3}}-\varepsilon\right)n\geq\left(1-\frac{2}{5k^{3}}\right)n using inequality (vi). ∎

Lemma 2.11.

We have |L|=k|L^{\prime}|=k.

Proof.

If |L|k+1|L^{\prime}|\geq k+1 then the common neighborhood of k+1k+1 vertices in LL^{\prime} would have at least n(k+1)25k3nn-(k+1)\frac{2}{5k^{3}}n vertices, so GG contains Kk+1,K_{k+1,\ell} for arbitrarily large \ell, a contradiction.

On the other hand if |L|k1|L^{\prime}|\leq k-1 then a refinement of Lemma 2.7 applied to zz along with Lemma 2.9 gives the following contradiction,

k(nk)d(z)+e(S1,L{z})+e(S1,(L1L2)L)η+ε2n2(k1)n+(k+ε)nη+ε2n2<(k1/2)n,k(n-k)\leq d(z)+e(S_{1},L^{\prime}-\{z\})+e(S_{1},(L_{1}\cup L_{2})-L^{\prime})\eta+\frac{\varepsilon^{2}n}{2}\leq(k-1)n+(k+\varepsilon)n\eta+\frac{\varepsilon^{2}n}{2}<(k-1/2)n,

where the last inequality follows from (vii). ∎

Since |L|=k|L^{\prime}|=k and every vertex in LL^{\prime} has degree at least (125k3)n\left(1-\frac{2}{5k^{3}}\right)n, the common neighborhood of LL^{\prime} has at least (125k2)n(1-\frac{2}{5k^{2}})n vertices. Let RR be this common neighborhood and E=VLRE=V-L^{\prime}-R. Note that |E|25k2n|E|\leq\frac{2}{5k^{2}}n. The following lemma completes the proof of Proposition 2.2.

Lemma 2.12.

We have E=E=\emptyset.

Proof.

Suppose otherwise. Note that

vEdER(v)2e(E)+e(E,R)2C|E|+m|E|.\sum_{v\in E}d_{E\cup R}(v)\leq 2e(E)+e(E,R)\leq 2C|E|+m|E|.

Hence, there exists some vEv\in E such that dER(v)2C+md_{E\cup R}(v)\leq 2C+m. Consider the graph GG^{\prime} obtained by deleting all edges between vv and ERE\cup R, and adding any missing edges between vv and LL^{\prime}. The decrease in xTAxx^{T}Ax is at most (2C+m)xvη(2C+m)x_{v}\eta, and the increase is at least xv(114k3)x_{v}\left(1-\frac{1}{4k^{3}}\right), hence λ(G)>λ(G)\lambda(G^{\prime})>\lambda(G) by inequality (viii).

Case 1: \mathcal{F} is finite. Since λ(G)>λ(G)\lambda(G^{\prime})>\lambda(G), GF0G^{\prime}\supseteq F_{0} for some F0F_{0}\in\mathcal{F}, and F0F_{0} must use the vertex vv. Since |V(F0)|m<|V(F_{0})|\leq m^{\prime}<\infty, there is some uRV(F0)u\in R-V(F_{0}), so by replacing vv with uu we find some F1GF_{1}\subseteq G which is isomorphic to F0F_{0}, a contradiction.

Case 2: \mathcal{F} is infinite. Let E1=EvE_{1}=E-v, and note that G[E1]G^{\prime}[E_{1}] is \mathcal{F}-free. Furthermore, any vertex in E1E_{1} has neighbors only in LL^{\prime}, E1E_{1}, and RR. Then similarly to the argument above we find v2E1v_{2}\in E_{1} such that dVL(v2)2C+md_{V-L^{\prime}}(v_{2})\leq 2C+m. Delete all edges between v2v_{2} and VLV-L^{\prime} and add all edges between v2v_{2} and LL^{\prime} to obtain G′′G^{\prime\prime}, and note that xTAxx^{T}Ax again increases, so that λ(G′′)>λ(G)\lambda(G^{\prime\prime})>\lambda(G). Set E2=E1v2E_{2}=E_{1}-v_{2}, and continue in this way until we obtain Ei=E_{i}=\emptyset. Since the resulting graph G(i)G^{(i)} satisfies λ(G(i))>λ(G)\lambda(G^{(i)})>\lambda(G), G(i)G^{(i)} is not \mathcal{F}-free. We now consider whether we are in case (a), (b), or (c) in the statement of Theorem 1.1.

Subcase (a). Since G(i)Kk,nkG^{(i)}\supseteq K_{k,n-k} and is not \mathcal{F}-free, G(i)G^{(i)} must have an edge either in LL^{\prime} or RR; but since |R||R|\to\infty, this implies that G(i)[LR]G^{(i)}[L^{\prime}\cup R] is not \mathcal{F}-free. Since no edges in LRL^{\prime}\cup R were modified to get from GG to G(i)G^{(i)}, this implies G[LR]G[L^{\prime}\cup R] is not \mathcal{F}-free, a contradiction.

Subcase (b). Note that e(G(i)[R])=e(G[R])(k2)e(G[L])(k2)e(G^{(i)}[R])=e(G[R])\leq{k\choose 2}-e(G[L^{\prime}])\leq{k\choose 2}. Thus we can delete all edges in E(G(i)[R])E(G^{(i)}[R]) and add the same number of edges in LL^{\prime} to obtain a subgraph of HH (which is therefore \mathcal{F}-free). However, it is easy to see that this operation further increases xTAxx^{T}Ax so that the spectral radius of the resulting graph exceeds λ(G)\lambda(G), which is a contradiction.

Subcase (c). Noting e(G(i)[R])(k2)e(G[L])+1(k2)+1e(G^{(i)}[R])\leq{k\choose 2}-e(G[L^{\prime}])+1\leq{k\choose 2}+1, the same argument, where we now delete all but one edge in E(G(i)[R])E(G^{(i)}[R]), gives the required contradiction.

We observe that our arguments for the case where \mathcal{F} is infinite apply when H=Kk+Knk¯H=K_{k}+\overline{K_{n-k}} or H=Kk+(K2Knk2)H=K_{k}+(K_{2}\cup K_{n-k-2}), but not when H=Kk+XH=K_{k}+X where e(X)2e(X)\geq 2. This is because, after deleting the edges in RR and adding the same number of edges in LL^{\prime}, there is more than one possible graph spanned by the remaining edges in RR, hence we do not necessarily end up with a subgraph of HH.

2.3 Eigenweight estimates

We now know that there is a set LL^{\prime} of kk vertices which are adjacent to every vertex in R=V(G)LR=V(G)\setminus L^{\prime}. Any other vertex may have at most m1m-1 neighbors in V(G)LV(G)\setminus L^{\prime}, otherwise GG contains a Kk+1,mK_{k+1,m} and is not FF-free. In this subsection we give more refined estimates on the eigenvector entries.

Lemma 2.13.

For vRv\in R we have xvkλmx_{v}\leq\frac{k}{\lambda-m}.

Proof.

Let v=argmax{xv:vR}v=\arg\max\{x_{v}:v\in R\}. Then we have

λxv=uvuLxu+uvuRxuk+mxv.\lambda x_{v}=\sum_{\begin{subarray}{c}u\sim v\\ u\in L\end{subarray}}x_{u}+\sum_{\begin{subarray}{c}u\sim v\\ u\in R\end{subarray}}x_{u}\leq k+mx_{v}.

Lemma 2.14.

For vRv\in R we have xvk(1λ+dR(v)λ(λm)).x_{v}\leq k\left(\frac{1}{\lambda}+\frac{d_{R}(v)}{\lambda(\lambda-m)}\right).

Proof.

We have

λxvk+wNR(v)xwk+dR(v)kλm\lambda x_{v}\leq k+\sum_{w\in N_{R}(v)}x_{w}\leq k+d_{R}(v)\frac{k}{\lambda-m}

by Lemma 2.13. ∎

Lemma 2.15.

For vRv\in R, we have xvkλO(n1).x_{v}\geq\frac{k}{\lambda}-O(n^{-1}).

Proof.

First we find a lower bound on eigenweights for vertices in LL^{\prime}. Note that

λ=λxzk+vRxvvRxvλk.\lambda=\lambda x_{z}\leq k+\sum_{v\in R}x_{v}\Longrightarrow\sum_{v\in R}x_{v}\geq\lambda-k.

Thus, for any wLw\in L^{\prime} we have

λxwvRxvλkxw1O(n1/2).\lambda x_{w}\geq\sum_{v\in R}x_{v}\geq\lambda-k\Longrightarrow x_{w}\geq 1-O(n^{-1/2}).

Hence, for vRv\in R we have

λxvwLxwk(1O(n1/2))xvkλO(n1).\lambda x_{v}\geq\sum_{w\in L^{\prime}}x_{w}\geq k\left(1-O(n^{-1/2})\right)\Longrightarrow x_{v}\geq\frac{k}{\lambda}-O(n^{-1}).

Combining the estimates of Lemma 2.14 and Lemma 2.15, we see that for vRv\in R, xv=kλ+Θ(n1)x_{v}=\frac{k}{\lambda}+\Theta(n^{-1}).

2.4 Proof of Theorem 1.1

Note that Proposition 2.2 already proves Theorem 1.1 (a). Thus, let GG be the spectral extremal graph for any remaining case (b)-(f). For notational convenience, let X=Knk¯X=\overline{K_{n-k}} and Q=0Q=0 in case (b), and let X=K2Knk2¯X=K_{2}\cup\overline{K_{n-k-2}} and Q=0Q=0 in case (c). This way, we can write H=Kk+XH=K_{k}+X in all cases. Note that, since we now know that GKk,nkG\supseteq K_{k,n-k}, we have e(G)exKk,nk(n,)=e(Kk+X)e(G)\leq\mathrm{ex}^{K_{k,n-k}}(n,\mathcal{F})=e(K_{k}+X).

Lemma 2.16.

The set LL^{\prime} induces a clique.

Proof.

Suppose not. We transform GG into Kk+XK_{k}+X by deleting all edges in RR and then adding all edges in XX and in G[L]G[L^{\prime}]. Note e(R)e(X)+(k2)e(R)\leq e(X)+\binom{k}{2}, since GKk,nkG\supseteq K_{k,n-k} and HEXKk,nk(n,)H\in\mathrm{EX}^{K_{k,n-k}}(n,\mathcal{F}). At least one added edge was in LL^{\prime} and thus increased xTAxx^{T}Ax by at least (114k3)2>12\left(1-\frac{1}{4k^{3}}\right)^{2}>\frac{1}{2}. Thus the net change in xTAxx^{T}Ax is at least

12+e(X)(kλO(n1))2(e(X)+(k2))(kλ+O(n1))2\frac{1}{2}+e(X)\left(\frac{k}{\lambda}-O(n^{-1})\right)^{2}-\left(e(X)+{k\choose 2}\right)\left(\frac{k}{\lambda}+O(n^{-1})\right)^{2}
12e(X)O((λn)1)(k2)(kλ+O(n1))2>12QO(n1/2)(k2)O(n1)>0\geq\frac{1}{2}-e(X)\cdot O\left((\lambda n)^{-1}\right)-{k\choose 2}\left(\frac{k}{\lambda}+O(n^{-1})\right)^{2}>\frac{1}{2}-Q\cdot O(n^{-1/2})-{k\choose 2}\cdot O(n^{-1})>0

which is a contradiction. ∎

Lemma 2.16 now gives that all vertices in LL^{\prime} are dominating and hence all have eigenvector entry equal to 11. This means that the error term in Lemma 2.15 can be removed, and

xvkλx_{v}\geq\frac{k}{\lambda} (1)

Armed with this we may now prove the theorem.

Proof Theorem 1.1 (e), (f).

We first only assume only (e). We already know that G=Kk+YG=K_{k}+Y for some YY. Now, we can delete all edges in RR and add the edges in XX to transform GG into XX. The change in xTAxx^{T}Ax is at least

e(Y)(kλ+O(n1))2+e(X)(kλ)2\displaystyle-e(Y)\left(\frac{k}{\lambda}+O(n^{-1})\right)^{2}+e(X)\left(\frac{k}{\lambda}\right)^{2} (e(X)e(Y))k2λ2e(Y)O(n3/2))\displaystyle\geq(e(X)-e(Y))\frac{k^{2}}{\lambda^{2}}-e(Y)O(n^{-3/2}))
(e(X)e(Y))k2λ2O(n)O(n3/2)\displaystyle\geq(e(X)-e(Y))\frac{k^{2}}{\lambda^{2}}-O(n)\cdot O(n^{-3/2})

The last quantity above cannot be positive, and so e(X)e(Y)=O(n1/2)e(X)-e(Y)=O(n^{1/2}), which proves the first claim. We already know that Δ(Y)\Delta(Y) is bounded (by mm).

Now assume (f). We first improve bounds on xvx_{v} for vYv\in Y. Note that all but a bounded number of vR=V(Y)v\in R=V(Y) have dY(v)dd_{Y}(v)\leq d. By the above, e(X)e(Y)=O(n1/2)e(X)-e(Y)=O(n^{1/2}), so the number of vertices in YY with degree less than dd is O(n1/2)O(n^{1/2}). Call a vertex vRv\in R bad if dR(v)dd_{R}(v)\neq d or vuv\sim u and dR(u)dd_{R}(u)\neq d for some uYu\in Y, and good otherwise. Since Δ(R)\Delta(R) is bounded, the number of bad vertices in RR is O(n1/2)O(n^{1/2}). Now for any good vertex vv,

λ2xv\displaystyle\lambda^{2}x_{v} =uLλxu+dk+uNR(v)wNR(u)xw\displaystyle=\sum_{u\in L^{\prime}}\lambda x_{u}+dk+\sum_{u\in N_{R}(v)}\sum_{w\in N_{R}(u)}x_{w}
kλ+dk+d2kλ\displaystyle\geq k\lambda+dk+d^{2}\frac{k}{\lambda}
xv\displaystyle x_{v} kλ+kdλ2+kd2λ3.\displaystyle\geq\frac{k}{\lambda}+\frac{kd}{\lambda^{2}}+\frac{kd^{2}}{\lambda^{3}}.

On the other hand,

λ2xv\displaystyle\lambda^{2}x_{v} =uLλxu+dk+uNR(v)wNR(u)xw\displaystyle=\sum_{u\in L^{\prime}}\lambda x_{u}+dk+\sum_{u\in N_{R}(v)}\sum_{w\in N_{R}(u)}x_{w}
kλ+dk+d2kλm\displaystyle\leq k\lambda+dk+d^{2}\frac{k}{\lambda-m}
xv\displaystyle x_{v} kλ+kdλ2+kd2λ2(λm).\displaystyle\leq\frac{k}{\lambda}+\frac{kd}{\lambda^{2}}+\frac{kd^{2}}{\lambda^{2}(\lambda-m).}

Combining the two estimates one obtains that xv=kλ+kdλ2+kd2λ3+O(n2)x_{v}=\frac{k}{\lambda}+\frac{kd}{\lambda^{2}}+\frac{kd^{2}}{\lambda^{3}}+O(n^{-2}). Now we delete all edges in YY and add all edges in XX. Note that at most mO(n1/2)=O(n1/2)m\cdot O(n^{1/2})=O(n^{1/2}) edges of YY are incident to a bad vertex. It follows that for some N=O(n1/2)N=O(n^{1/2}) , there is a set of NN edges of XX and a set of NN edges of YY which contain all edges of XX or YY, respectively, incident to a bad vertex. Then the change in xTAxx^{T}Ax is at least

(e(Y)N)(kλ+kdλ2+kd2λ3+O(n2))2+(e(X)N)(kλ+kdλ2+kd2λ3)2\displaystyle-(e(Y)-N)\left(\frac{k}{\lambda}+\frac{kd}{\lambda^{2}}+\frac{kd^{2}}{\lambda^{3}}+O(n^{-2})\right)^{2}+(e(X)-N)\left(\frac{k}{\lambda}+\frac{kd}{\lambda^{2}}+\frac{kd^{2}}{\lambda^{3}}\right)^{2}
N(kλ+O(n1))2+N(kλ)2\displaystyle-N\left(\frac{k}{\lambda}+O(n^{-1})\right)^{2}+N\left(\frac{k}{\lambda}\right)^{2}
(e(X)e(Y))k2λ2e(Y)kλO(n2)NkλO(n1)\displaystyle\geq(e(X)-e(Y))\frac{k^{2}}{\lambda^{2}}-e(Y)\frac{k}{\lambda}\cdot O(n^{-2})-N\frac{k}{\lambda}\cdot O(n^{-1})
(e(X)e(Y))O(n1)QnO(n5/2)O(n1/2)O(n3/2)\displaystyle\geq(e(X)-e(Y))\cdot O(n^{-1})-Qn\cdot O(n^{-5/2})-O(n^{1/2})\cdot O(n^{-3/2})

If e(X)e(Y)=ω(1)e(X)-e(Y)=\omega(1) then the quantity above is positive, a contradiction. Hence e(X)e(Y)=O(1)e(X)-e(Y)=O(1), which implies that the number of bad vertices is in fact O(1)O(1). Therefore we can take N=O(1)N=O(1) and repeat the previous argument to show that when we transform Kk+YK_{k}+Y to KK+XK_{K}+X the change in xTAxx^{T}Ax is at least

(e(X)e(Y))O(n1)QnO(n5/2)O(1)O(n3/2)(e(X)-e(Y))\cdot O(n^{-1})-Qn\cdot O(n^{-5/2})-O(1)\cdot O(n^{-3/2})

which implies e(X)e(Y)=0e(X)-e(Y)=0; hence GEXKk+Knk¯(n,)G\in\mathrm{EX}^{K_{k}+\overline{K_{n-k}}}(n,\mathcal{F}). ∎

Proof of Theorem 1.1 (b), (c), (d).

From Proposition 2.1, we have e(X)=Qn+O(1)e(X)=Qn+O(1) for some Q{0,1/2,2/3}Q\in\{0,1/2,2/3\}. The first case is Q=0Q=0, in other words e(X)=O(1)e(X)=O(1) and consequently e(R)=O(1)e(R)=O(1). If GEXKk+Knk¯(n,)G\not\in\mathrm{EX}^{K_{k}+\overline{K_{n-k}}}(n,\mathcal{F}), then delete all edges in RR and add all edges in XX to make the graph isomorphic to Kk+XK_{k}+X. Let s,ts,t be the number of edges deleted and added, respectively, which satisfy s<t=O(1)s<t=O(1). Then the change in xTAxx^{T}Ax is at least

s(kλ+O(n1))2+t(kλ)2(ts)k2λ2O((λn)1)O(n2)>0-s\left(\frac{k}{\lambda}+O(n^{-1})\right)^{2}+t\left(\frac{k}{\lambda}\right)^{2}\geq(t-s)\frac{k^{2}}{\lambda^{2}}-O\left((\lambda n)^{-1}\right)-O(n^{-2})>0

which is a contradiction. Actually, this argument proves the following fact: whenever Kk+XK_{k}+X can be obtained from Kk+YK_{k}+Y by deleting a bounded number of edges and adding more edges than were deleted, we have λ1(Kk+X)>λ1(Kk+Y)\lambda_{1}(K_{k}+X)>\lambda_{1}(K_{k}+Y).

The second case is Q=1/2Q=1/2. We first note that since GG is \mathcal{F}-saturated, only a bounded number of vertices vRv\in R have dR(v)2d_{R}(v)\geq 2, by the same argument as Proposition 2.1. On the other hand, e(R)e(R) must be unbounded, or the same argument as in the case Q=0Q=0 would show λ1(G)<λ1(Kk+X)\lambda_{1}(G)<\lambda_{1}(K_{k}+X). Therefore, the argument in Proposition 2.1 shows that G[R]G[R] has at most one isolated vertex. Thus GG is obtained from a maximal union of P2P_{2} by deleting and adding a bounded number of edges. Since the same is true of Kk+XK_{k}+X, we can delete and add a bounded number of edges to transform GG into a graph isomorphic to Kk+XK_{k}+X. By the fact mentioned above, this implies that GEXKk+Knk¯(n,)G\in\mathrm{EX}^{K_{k}+\overline{K_{n-k}}}(n,\mathcal{F}).

The final case Q=2/3Q=2/3 is similar. We find that YY is the union of copies of P3P_{3} and smaller graphs, plus a bounded number of graphs of bounded order. The number of copies of P3P_{3} must be unbounded; otherwise, we have e(Y)<23ne(Y)<\frac{2}{3}n, and we may substitute ts=Ω(n)t-s=\Omega(n) in the argument above, and find

s(kλ+O(n1))2+t(kλ)2Ω(n)k2λ2O(n)((λn)1)O(n)O(n2)>0-s\left(\frac{k}{\lambda}+O(n^{-1})\right)^{2}+t\left(\frac{k}{\lambda}\right)^{2}\geq\Omega(n)\frac{k^{2}}{\lambda^{2}}-O(n)\cdot((\lambda n)^{-1})-O(n)\cdot O(n^{-2})>0

which gives a contradiction. The argument of Proposition 2.1 then shows that GG is obtained by deleting and adding a bounded number of edges to a maximal union of P3P_{3}. By the fact mentioned above, this implies that GEXKk+Knk¯(n,)G\in\mathrm{EX}^{K_{k}+\overline{K_{n-k}}}(n,\mathcal{F}). ∎

Note that the conclusions stated in (b) and (c) follow because in these cases EXKk+Kk,nk¯(n,)\mathrm{EX}^{K_{k}+\overline{K_{k,n-k}}}(n,\mathcal{F}) contains a single graph up to isomorphism.

2.5 A counterexample

Proof of Proposition 1.2.

Let \mathcal{F} consist of the all the following graphs:

  1. 1.

    3K1,43\cdot K_{1,4};

  2. 2.

    K2,6X5Y5K_{2,6}\cup X_{5}\cup Y_{5}, for all connected graphs X5,Y5X_{5},Y_{5} on 5 vertices;

  3. 3.

    K2,6K1,3X5K_{2,6}\cup K_{1,3}\cup X_{5}, for all connected graphs X5X_{5} on 5 vertices;

  4. 4.

    K2,6X9K_{2,6}\cup X_{9}, for all connected graphs X9X_{9} on 9 vertices;

  5. 5.

    K2,6X8K_{2,6}\cup X_{8}, for all connected graphs X8X_{8} on 8 vertices except for P8P_{8};

  6. 6.

    X5Y5Z5W5X_{5}\cup Y_{5}\cup Z_{5}\cup W_{5}, for all connected graphs X5,Y5,Z5,W5X_{5},Y_{5},Z_{5},W_{5} on 5 vertices;

  7. 7.

    CiCjCkC_{i}\cup C_{j}\cup C_{k} for 3i,j,k73\leq i,j,k\leq 7.

We have chosen a very large family \mathcal{F}, which makes the extremal problem for \mathcal{F} easy. We suspect that a similar counterexample could be obtained with an even smaller family \mathcal{F}.

Lemma 2.17.

If n2(mod4)n\equiv 2\pmod{4} is large enough, then we have that EX(n,)=K2+(P8n104P4)\mathrm{EX}(n,\mathcal{F})=K_{2}+\left(P_{8}\cup\frac{n-10}{4}\cdot P_{4}\right).

Proof.

One can check that H=K2+(P8n104P4)H=K_{2}+\left(P_{8}\cup\frac{n-10}{4}\cdot P_{4}\right) is \mathcal{F}-free. Now let HH^{\prime} be the extremal graph, so that

e(H)2(n2)+1+7+n1043=114n4+1+7152=114n72.e(H^{\prime})\geq 2(n-2)+1+7+\frac{n-10}{4}\cdot 3=\frac{11}{4}n-4+1+7-\frac{15}{2}=\frac{11}{4}n-\frac{7}{2}.

By forbidding (1), we may choose vertices x,yx,y such that any other vertex zz has at most 1111 neighbors in V(G){x,y}V(G)\setminus\{x,y\}. Consider the components of G{x,y}G-\{x,y\}; these are either graphs on 3 or 4 vertices containing a C3C_{3} or C4C_{4}, respectively (and by (7) there are at most 2 of each type), or connected graph of order at least 55 (and by (6) there are at most 3 of these), or trees on at most 44 vertices. We claim that the order of any graphs of the second type is at most 51195\cdot 11^{9}. Let KK be such a component with |V(K)|5119|V(K)|\geq 5\cdot 11^{9}; let Br(v)B_{r}(v) denote {uV(K):dK(u,v)r}\{u\in V(K):d_{K}(u,v)\leq r\}. Then for any vv, |B8(v)|1+11++118119|B_{8}(v)|\leq 1+11+\cdots+11^{8}\leq 11^{9} since Δ(K)11\Delta(K)\leq 11, and consequently we can find vertices v1,,v5v_{1},\ldots,v_{5} such that for iji\neq j, d(vi,vj)9d(v_{i},v_{j})\geq 9. This implies that B4(vi)B_{4}(v_{i}), 1i51\leq i\leq 5, induce 5 disjoint connected graphs on at least 5 vertices, which contradicts (6). Putting this all together, we find a partition V(H)=UBV(H^{\prime})=U\sqcup B where UU is a union of trees of order at most 4, |B|2+23+24+351191111|B|\leq 2+2\cdot 3+2\cdot 4+3\cdot 5\cdot 11^{9}\leq 11^{11}, and x,yBx,y\in B. Hence, since Δ(G[Bxy])11\Delta(G[B-x-y])\leq 11, we have

114n72d(x)+d(y)+111111+n43d(x)+d(y)2n7211122n1113.\frac{11}{4}n-\frac{7}{2}\leq d(x)+d(y)+11^{11}\cdot 11+\frac{n}{4}\cdot 3\Longrightarrow d(x)+d(y)\geq 2n-\frac{7}{2}-11^{12}\geq 2n-11^{13}.

So d(x)n1113d(x)\geq n-11^{13}, d(y)n1113d(y)\geq n-11^{13}, and |N(x)N(y)|n21113|N(x)\cap N(y)|\geq n-2\cdot 11^{13}. Note that xx and yy are contained in disjoint copies of K1,4K_{1,4} so by (1), every other vertex of HH^{\prime} has at most 3 neighbors outside {x,y}\{x,y\}. Now we have

e(N(x)N(y))114n722(n1)211135=34nO(1)e(N(x)\cap N(y))\geq\frac{11}{4}n-\frac{7}{2}-2(n-1)-2\cdot 11^{13}\cdot 5=\frac{3}{4}n-O(1)

which implies we can find 2 disjoint edges in N(x)N(y)N(x)\cap N(y). Hence xx and yy are contained in disjoint copies of C3C_{3}. Thus by (5) and (7), all components of GxyG-x-y, except possibly one (which by (4) can be of order at most 88), are trees of order at most 4. Such a graph cannot have more edges than HH. If e(H)=e(H)e(H^{\prime})=e(H), then Gxy=aP4bK1,3KG-x-y=a\cdot P_{4}\cup b\cdot K_{1,3}\cup K, where KK is a connected graph on 8 vertices (using (4)). By (5), we have K=P8K=P_{8}; but then by (3), b=0b=0. It follows that HHH^{\prime}\subseteq H, hence HHH^{\prime}\simeq H. ∎

As in the proof above, let H=K2+(P8n104P4)H=K_{2}+\left(P_{8}\cup\frac{n-10}{4}\cdot P_{4}\right). Let G=K2+n24K1,3G=K_{2}+\frac{n-2}{4}\cdot K_{1,3}. One can check that GG is also \mathcal{F}-free. We now show that λ1(G)>λ1(H)\lambda_{1}(G)>\lambda_{1}(H). Note that GG and HH have equitable partitions with quotient matrices

BG=[114(n2)34(n2)203210],BH=[112(n10)12(n10)2222211000021000002001100200101020001012000010].B_{G}=\begin{bmatrix}1&\frac{1}{4}(n-2)&\frac{3}{4}(n-2)\\ 2&0&3\\ 2&1&0\\ \end{bmatrix},\ B_{H}=\begin{bmatrix}1&\frac{1}{2}(n-10)&\frac{1}{2}(n-10)&2&2&2&2\\ 2&1&1&0&0&0&0\\ 2&1&0&0&0&0&0\\ 2&0&0&1&1&0&0\\ 2&0&0&1&0&1&0\\ 2&0&0&0&1&0&1\\ 2&0&0&0&0&1&0\end{bmatrix}.

The characteristic polynomials are

pG(x)\displaystyle p_{G}(x) =x3x2+(12n)x+93n\displaystyle=x^{3}-x^{2}+(1-2n)x+9-3n
pH(x)\displaystyle p_{H}(x) =x73x6+(32n)x5+(3+n)x4+(7n22)x3+(1n)x2+(104n)x+3n.\displaystyle=x^{7}-3x^{6}+(3-2n)x^{5}+(3+n)x^{4}+(7n-22)x^{3}+(1-n)x^{2}+(10-4n)x+3-n.

By the Weyl inequalities, λ1(G),λ1(H)2(n2)\lambda_{1}(G),\lambda_{1}(H)\geq\sqrt{2(n-2)} and λ2(G),λ2(H)<2(n2)\lambda_{2}(G),\lambda_{2}(H)<\sqrt{2(n-2)}. Thus pGp_{G} is negative on [2(n2),λ1(G))[\sqrt{2(n-2)},\lambda_{1}(G)) and positive on (λ1(G),)(\lambda_{1}(G),\infty), and pHp_{H} is negative on [2(n2),λ1(H))[\sqrt{2(n-2)},\lambda_{1}(H)) and positive on (λ1(H),)(\lambda_{1}(H),\infty). So, to show λ1(G)>λ1(H)\lambda_{1}(G)>\lambda_{1}(H) it suffices to find some x>2(n2)x>\sqrt{2(n-2)} such that pG(x)<0<pH(x)p_{G}(x)<0<p_{H}(x). For a constant C>0C>0, we compute

pG(2(n2)+5/4+Cn1/2)\displaystyle p_{G}(\sqrt{2(n-2)}+5/4+Cn^{-1/2}) =4Cn1382n2+o(n)\displaystyle=4C\sqrt{n}-\frac{13}{8\sqrt{2}}\sqrt{n-2}+o(\sqrt{n})
pH(2(n2)+5/4+Cn1/2)\displaystyle p_{H}(\sqrt{2(n-2)}+5/4+Cn^{-1/2}) =16Cn5/2522n2n2+o(n5/2)\displaystyle=16Cn^{5/2}-\frac{5}{2\sqrt{2}}n^{2}\sqrt{n-2}+o(n^{5/2})

as nn\to\infty. Choosing 13322<C<5322\frac{13}{32\sqrt{2}}<C<\frac{5}{32\sqrt{2}}, we find that pG(2(n2)+5/4+Cn1/2)p_{G}(\sqrt{2(n-2)}+5/4+Cn^{-1/2})\to-\infty while pH(2(n2)+5/4+Cn1/2)p_{H}(\sqrt{2(n-2)}+5/4+Cn^{-1/2})\to\infty. ∎

2.6 Forbidden trees

Proof of Proposition 1.4.

We will assume that all labelled trees have vertex set [n][n] and let TT denote a uniformly random labelled tree on [n][n]. For iji\neq j in [n][n], let AijA_{ij} be the event that iji\sim j, ii has a neighbor different from jj which is adjacent precisely to ii and 2 leaves, and jj has a neighbor different from ii which is adjacent precisely to jj and 2 leaves; let XijX_{ij} be the indicator function of AijA_{ij}. Observe that if AijA_{ij} occurs then TT is of the form described in the application ‘Almost all trees’ of Theorem 1.1, for some k4k\geq 4. Therefore, defining X=i<jXijX=\sum_{i<j}X_{ij}, it suffices to show that [X=0]0\mathbb{P}[X=0]\to 0. Our argument is a variation on [20] in which it was shown that almost all labelled trees have a vertex with two leaves.

For i,j,k,[n]i,j,k,\ell\in[n] (not necessarily distinct), we define Nn(ij)N_{n}(ij) to be the number of labelled trees on [n][n] which contain the edge ijij, and we define Nn(ij,k)N_{n}(ij,k\ell) to be the number of labelled trees on [n][n] which contain both edges ijij and kk\ell. From Lemma 6 of [45], we have the following formulas.

Lemma 2.18.

For any i,j,k,i,j,k,\ell distinct we have:

  • (a)

    Nn(ij)=2nn3N_{n}(ij)=2n^{n-3}

  • (b)

    Nn(ij,ik)=3nn4N_{n}(ij,ik)=3n^{n-4}

  • (c)

    Nn(ij,k)=4nn4N_{n}(ij,k\ell)=4n^{n-4}.

Next we estimate 𝔼[X]\mathbb{E}[X] and 𝔼[X2]\mathbb{E}[X^{2}]. To specify a tree TT such that AijA_{ij} occurs, we choose the neighbor of ii and its pair of leaves, then we choose the neighbor of jj and its pair of leaves, and then attach a tree containing the edge ijij and the unused vertices in [n][n]. For a constant cc we have that (nc)nc=nncec+Θ(nnc2)(n-c)^{n-c}=n^{n-c}e^{-c}+\Theta(n^{n-c-2}). This gives:

[Aij]\displaystyle\mathbb{P}[A_{ij}] =1nn2(n2)(n32)(n5)(n62)Nn6(ij)=12e6n+o(n1)\displaystyle=\frac{1}{n^{n-2}}(n-2){n-3\choose 2}(n-5){n-6\choose 2}N_{n-6}(ij)=\frac{1}{2e^{6}n}+o(n^{-1})
𝔼[X]\displaystyle\mathbb{E}[X] =(n2)(12e6n+o(n1))=n4e6+o(n)\displaystyle={n\choose 2}\left(\frac{1}{2e^{6}n}+o(n^{-1})\right)=\frac{n}{4e^{6}}+o(n)
𝔼[X]2\displaystyle\mathbb{E}[X]^{2} =n216e12+o(n2).\displaystyle=\frac{n^{2}}{16e^{12}}+o(n^{2}).

Next we consider AijAikA_{ij}\cap A_{ik}. There are 2 types of trees which realize this event: (i) those in which the special neighbor of ii in AijA_{ij} is the same as the special neighbor of ii in AikA_{ik} (the special neighbor of jj and kk must be distinct in a tree), and (ii) those in which they are different. So we have

[AijAik]\displaystyle\mathbb{P}[A_{ij}\cap A_{ik}] n93nn13+n123nn16nn2=O(n2).\displaystyle\leq\frac{n^{9}3n^{n-13}+n^{12}3n^{n-16}}{n^{n-2}}=O(n^{-2}).

Finally, if i,j,k,i,j,k,\ell are distinct, then we have

[AijAk]n4(n2)44(n12)n16nn2=14e12n2+o(n2).\begin{aligned} \mathbb{P}[A_{ij}\cap A_{k\ell}]&\leq\frac{n^{4}{n\choose 2}^{4}4(n-12)^{n-16}}{n^{n-2}}=\frac{1}{4e^{12}n^{2}}+o(n^{-2})\end{aligned}.

Therefore, we have

𝔼[X2]\displaystyle\mathbb{E}[X^{2}] =i<j[Aij]+i,j,k[AijAik]+{i,j},{k,} disjoint[AijAk]\displaystyle=\sum_{i<j}\mathbb{P}[A_{ij}]+\sum_{i,j,k}\mathbb{P}[A_{ij}\cap A_{ik}]+\sum_{\{i,j\},\{k,\ell\}\text{ disjoint}}\mathbb{P}[A_{ij}\cap A_{k\ell}]
(n2)1e6n+n3O(n2)+(n2)(n22)(14e12n2+o(n2))\displaystyle\leq{n\choose 2}\frac{1}{e^{6}n}+n^{3}\cdot O(n^{-2})+{n\choose 2}{n-2\choose 2}\left(\frac{1}{4e^{12}n^{2}}+o(n^{-2})\right)
=n216e12+o(n2).\displaystyle=\frac{n^{2}}{16e^{12}}+o(n^{2}).

Therefore Var(X)=o(n2)\mathrm{Var}(X)=o(n^{2}), and Chebyshev’s inequality gives

[X=0][|X𝔼[X]|n5e6]25e12o(n2)n20.\mathbb{P}[X=0]\leq\mathbb{P}\left[|X-\mathbb{E}[X]|\geq\frac{n}{5e^{6}}\right]\leq\frac{25e^{12}\cdot o(n^{2})}{n^{2}}\to 0.

3 Alpha spectral extremal results

Let \mathcal{F} be the family of graphs in Theorem 1.5. We first consider the case where \mathcal{F} contains some bipartite graph FF with a vertex vv such that FvF-v is a forest.

Lemma 3.1.

There exists c=c(F)c=c(F) such that, for any graph GG with V(G)=UWV(G)=U\sqcup W and

2e(U)+e(U,W)>3c|U|+c|W|,2e(U)+e(U,W)>3c|U|+c|W|,

there exists a copy of FvF-v in GG such that NF(v)N_{F}(v) embeds in UU.

Proof.

Let vv be the vertex such that FvF-v is a forest, and let Δ=Δ(Fv)\Delta=\Delta(F-v). Let V(F)=ABV(F)=A\sqcup B be a proper coloring with vBv\in B. Then FvF-v has a proper coloring A(B{v})A\sqcup(B\setminus\{v\}), and AA contains all the neighbors of vv in FF. Set a=|A|a=|A| and b=|B|b=|B|.

Let TT be a forest obtained from 2 copies of FvF-v by selecting a vertex from AA from each component (observe that every component of FvF-v must contain a vertex in AA) and making it adjacent to the corresponding vertex in the other copy of FvF-v. Now if t=|V(T)|=2(|V(F)|1)t=|V(T)|=2(|V(F)|-1) then it is known that ex(n,T)(t2)n\mathrm{ex}(n,T)\leq(t-2)n.

Now consider the graph GG with V(G)=UWV(G)=U\sqcup W, appearing in the statement of this lemma. If e(U)>(t/22)|U|e(U)>(t/2-2)|U| then G[U]FvG[U]\supseteq F-v and we are done. If e(U,W)>(t2)(|U|+|W|)e(U,W)>(t-2)(|U|+|W|) then G[U,W]TG[U,W]\supseteq T, and by the design of TT this means there is a copy of FvF-v in GG in which AA embeds in UU. Hence, to guarantee the conclusion of the lemma, it suffices that e(U)>(t2)|U|e(U)>(t-2)|U| or e(U,W)>(t2)(|U|+|W|)e(U,W)>(t-2)(|U|+|W|), and we may take e.g. c=tc=t. ∎

If GG is any FF-free graph, then for any uV(G)u\in V(G), there cannot be a copy of FvF-v in G[N1(u)N2(u)]G[N_{1}(u)\cup N_{2}(u)] such that all the neighbors of NF(v)N_{F}(v) embed in N1(u)N_{1}(u). Applying Lemma 3.1 gives

2e(N1(v))+e(N1(v),N2(v))3cd(v)+c(nd(v)1)=2cd(v)+c(n1)3cn.2e(N_{1}(v))+e(N_{1}(v),N_{2}(v))\leq 3cd(v)+c(n-d(v)-1)=2cd(v)+c(n-1)\leq 3cn. (2)

If \mathcal{F} does not contain a graph FF so that FvF-v is a forest, then we assume that ex(n,)=O(n)\mathrm{ex}(n,\mathcal{F})=O(n). By choosing cc large enough in Equation 2, we may assume that the equation holds in either case, for some c=c()c=c(\mathcal{F}), for any \mathcal{F}-free graph.

Lemma 3.2.

([53]) Let GG be a graph and 0α10\leq\alpha\leq 1, then

αΔ(G)λα(G)αΔ(G)+(1α)λ(G).\alpha\Delta(G)\leq\lambda_{\alpha}(G)\leq\alpha\Delta(G)+(1-\alpha)\lambda(G).
Lemma 3.3.

For any \mathcal{F}-free graph GG we have λ(G)k(nk)+k\lambda(G)\leq\sqrt{k(n-k)}+k.

Proof.

We have λ(G)spex(n,)λ(Kk+(K2Knk2¯))\lambda(G)\leq\mathrm{spex}(n,\mathcal{F})\leq\lambda(K_{k}+(K_{2}\cup\overline{K_{n-k-2}})), and by the Weyl inequalities we have

λ(Kk+(K2Knk2¯))λ(Kk,nk)+k=k(nk)+k.\lambda(K_{k}+(K_{2}\cup\overline{K_{n-k-2}}))\leq\lambda(K_{k,n-k})+k=\sqrt{k(n-k)}+k.

Lemma 3.4.

Let GG be a connected graph and let yy be a Perron vector of Aα(G)A_{\alpha}(G). Then for all vV(G)v\in V(G), we have

λα(G)yv\displaystyle\lambda_{\alpha}(G)y_{v} =αd(v)yv+(1α)uvyu\displaystyle=\alpha d(v)y_{v}+(1-\alpha)\sum_{u\sim v}y_{u}
λα2(G)yv\displaystyle\lambda_{\alpha}^{2}(G)y_{v} =αd(v)λα(G)yv+α(1α)uvd(u)yu+(1α)2wvuwyu\displaystyle=\alpha d(v)\lambda_{\alpha}(G)y_{v}+\alpha(1-\alpha)\sum_{u\sim v}d(u)y_{u}+(1-\alpha)^{2}\sum_{w\sim v}\sum_{u\sim w}y_{u}
λα(G)\displaystyle\lambda_{\alpha}(G) =1yTy(αvV(G)d(v)yv2+(1α)uvE(G)yuyv).\displaystyle=\frac{1}{y^{T}y}\left(\alpha\sum_{v\in V(G)}d(v)y_{v}^{2}+(1-\alpha)\sum_{uv\in E(G)}y_{u}y_{v}\right).

Now let GG be a graph in SPEXα(n,)\mathrm{SPEX}_{\alpha}(n,\mathcal{F}) and let xx be its Perron vector, scaled so that its largest entry is xz=1x_{z}=1. Let σ\sigma be a positive constant satisfying 3cσ<1α3c\sigma<1-\alpha and (3c+1)σ<α(1α)(3c+1)\sigma<\alpha(1-\alpha). With V=V(G)V=V(G), let L={vV:xv>σ}L=\{v\in V:x_{v}>\sigma\} and let S=VLS=V-L. Also, write Li(v)=LNi(v)L_{i}(v)=L\cap N_{i}(v) and Si(v)=SNi(v)S_{i}(v)=S\cap N_{i}(v). As in Section 2, all claims below are only asserted for nn large enough unless stated otherwise.

Lemma 3.5.

For nn sufficiently large, we have that

spexα(n,)max{αn+kαk12k(k+1)α3nα2(k+1+α)+αk,αn+kαk1α,α(n1)}\mathrm{spex}_{\alpha}(n,\mathcal{F})\geq\max\left\{\alpha n+\frac{k}{\alpha}-k-1-\frac{2k(k+1)}{\alpha^{3}n-\alpha^{2}(k+1+\alpha)+\alpha k},\alpha n+\frac{k}{\alpha}-k-1-\alpha,\alpha(n-1)\right\}
Proof.

To obtain the lower bound, observe that we have assumed Kk+Knk¯K_{k}+\overline{K_{n-k}} is \mathcal{F}-free. Thus, spexα(n,)λα(Kk+Knk¯)\mathrm{spex}_{\alpha}(n,\mathcal{F})\geq\lambda_{\alpha}(K_{k}+\overline{K_{n-k}}) and

λα(Kk+Knk¯)max{αn+kαk12k(k+1)α3nα2(k+1+α)+αk,αn+kαk1α,α(n1)},\lambda_{\alpha}(K_{k}+\overline{K_{n-k}})\geq\max\left\{\alpha n+\frac{k}{\alpha}-k-1-\frac{2k(k+1)}{\alpha^{3}n-\alpha^{2}(k+1+\alpha)+\alpha k},\alpha n+\frac{k}{\alpha}-k-1-\alpha,\alpha(n-1)\right\},

see for example [7]. ∎

Lemma 3.6.

For all vLv\in L, we have d(v)(11(k+1)n)nd(v)\geq\left(1-\dfrac{1}{(k+1)\sqrt{n}}\right)n.

Proof.

Suppose to the contrary that there is some vLv\in L with d(v)<(11(k+1)n)nd(v)<\left(1-\dfrac{1}{(k+1)\sqrt{n}}\right)n. Then,

λα2xv\displaystyle\lambda_{\alpha}^{2}x_{v} =αd(v)λαxv+α(1α)uvd(u)xu+(1α)2wvuwxu\displaystyle=\alpha d(v)\lambda_{\alpha}x_{v}+\alpha(1-\alpha)\sum_{u\sim v}d(u)x_{u}+(1-\alpha)^{2}\sum_{w\sim v}\sum_{u\sim w}x_{u}
αd(v)λαxv+α(1α)(d(v)+2e(N1(v))+e(N1(v),N2(v)))\displaystyle\leq\alpha d(v)\lambda_{\alpha}x_{v}+\alpha(1-\alpha)(d(v)+2e(N_{1}(v))+e(N_{1}(v),N_{2}(v)))
+(1α)2(d(v)+2e(N1(v))+e(N1(v),N2(v)))\displaystyle+(1-\alpha)^{2}(d(v)+2e(N_{1}(v))+e(N_{1}(v),N_{2}(v)))
<αd(v)λαxv+α(1α)(d(v)+3cn)+(1α)2(d(v)+3cn)\displaystyle<\alpha d(v)\lambda_{\alpha}x_{v}+\alpha(1-\alpha)(d(v)+3cn)+(1-\alpha)^{2}(d(v)+3cn)
<αd(v)λαxv+(1α)(3c+1)n.\displaystyle<\alpha d(v)\lambda_{\alpha}x_{v}+(1-\alpha)(3c+1)n.

Then

λα(λααd(v))xv<(1α)(3c+1)n.\lambda_{\alpha}(\lambda_{\alpha}-\alpha d(v))x_{v}<(1-\alpha)(3c+1)n.

On the other hand, from Lemma 3.5 and d(v)<(11(k+1)n)nd(v)<\left(1-\dfrac{1}{(k+1)\sqrt{n}}\right)n we have

λα(λααd(v))xv>α(n1)[α(n1)α(11(k+1)n)n]σ=Ω(n3/2).\lambda_{\alpha}(\lambda_{\alpha}-\alpha d(v))x_{v}>\alpha(n-1)\left[\alpha(n-1)-\alpha\left(1-\dfrac{1}{(k+1)\sqrt{n}}\right)n\right]\sigma=\Omega(n^{3/2}).

which gives a contradiction. ∎

Lemma 3.7.

There is some FF^{\prime}\in\mathcal{F} such that FKk+1,F^{\prime}\subseteq K_{k+1,\infty}. Consequently, FKk+1,mF^{\prime}\subseteq K_{k+1,m} for some fixed mm.

Proof.

If not, then Kk+1,nk1K_{k+1,n-k-1} is \mathcal{F}-free. Since λ(Kk+1,nk1)>λ(Kk+(K2Knk2¯)\lambda(K_{k+1,n-k-1})>\lambda(K_{k}+(K_{2}\cup\overline{K_{n-k-2}}), this contradicts the fact that spex(n,)λ(Kk+(K2Knk2¯)\mathrm{spex}(n,\mathcal{F})\leq\lambda(K_{k}+(K_{2}\cup\overline{K_{n-k-2}}). ∎

Lemma 3.8.

The size of LL is kk.

Proof.

First assume for a contradiction that |L|k+1|L|\geq k+1. Then there is a set LLL^{\prime}\subseteq L with L=k+1L^{\prime}=k+1. Therefore, every vertex vLv\in L^{\prime} has d(v)(11(k+1)n)nd(v)\geq\left(1-\dfrac{1}{(k+1)\sqrt{n}}\right)n, and the common neighborhood of the vertices in LL^{\prime} has at least nnn-\sqrt{n} vertices. This implies that for nn large enough, GKk+1,nnG\supseteq K_{k+1,n-\sqrt{n}} which contradicts Lemma 3.7.

To show |L|k|L|\geq k, assume to the contrary that |L|k1|L|\leq k-1. In cases (b) and (c), we have

λα2\displaystyle\lambda_{\alpha}^{2} =λα2xz=αd(z)λαxz+α(1α)uzd(u)xu+(1α)2wzuwxu\displaystyle=\lambda_{\alpha}^{2}x_{z}=\alpha d(z)\lambda_{\alpha}x_{z}+\alpha(1-\alpha)\sum_{u\sim z}d(u)x_{u}+(1-\alpha)^{2}\sum_{w\sim z}\sum_{u\sim w}x_{u}
αd(z)λα+α(1α)(uL1(z)d(u)xu+uS1(z)d(u)xu)\displaystyle\leq\alpha d(z)\lambda_{\alpha}+\alpha(1-\alpha)\left(\sum_{u\in L_{1}(z)}d(u)x_{u}+\sum_{u\in S_{1}(z)}d(u)x_{u}\right)
+(1α)2(wzuL1(w)xu+wzuS1(w)xu)\displaystyle+(1-\alpha)^{2}\left(\sum_{w\sim z}\sum_{u\in L_{1}(w)}x_{u}+\sum_{w\sim z}\sum_{u\in S_{1}(w)}x_{u}\right)
<αd(z)λα+α(1α)((k2)n+(d(z)+2e(N1(z))+e(N1(z),N2(z)))σ)\displaystyle<\alpha d(z)\lambda_{\alpha}+\alpha(1-\alpha)\left((k-2)n+(d(z)+2e(N_{1}(z))+e(N_{1}(z),N_{2}(z)))\sigma\right)
+(1α)2((k1)n+(2e(N1(z))+e(N1(z),N2(z)))σ)\displaystyle+(1-\alpha)^{2}\left((k-1)n+(2e(N_{1}(z))+e(N_{1}(z),N_{2}(z)))\sigma\right)
<αd(z)λα+α(1α)((k2)n+(d(z)+3cn)σ)+(1α)2(3cnσ+(k1)n)\displaystyle<\alpha d(z)\lambda_{\alpha}+\alpha(1-\alpha)\left((k-2)n+(d(z)+3cn)\sigma\right)+(1-\alpha)^{2}\left(3cn\sigma+(k-1)n\right)
αd(z)λα+(1α)((k1)n+3cnσ).\displaystyle\leq\alpha d(z)\lambda_{\alpha}+(1-\alpha)((k-1)n+3cn\sigma).

Therefore,

λα(λααd(z))\displaystyle\lambda_{\alpha}(\lambda_{\alpha}-\alpha d(z)) <(1α)(k1+3cσ)n.\displaystyle<(1-\alpha)(k-1+3c\sigma)n.

But from Lemma 3.5 we have

λα(λααd(z))\displaystyle\lambda_{\alpha}(\lambda_{\alpha}-\alpha d(z)) α(n1)(αn+kαk14k2α4nα(n1))=(1α)(kα)n+O(1)\displaystyle\geq\alpha(n-1)\left(\alpha n+\frac{k}{\alpha}-k-1-\frac{4k^{2}}{\alpha^{4}n}-\alpha(n-1)\right)=(1-\alpha)(k-\alpha)n+O(1)

using 2k(k+1)/(α3nα2(k+1+α)+ak)4k2/(α4n)2k(k+1)/(\alpha^{3}n-\alpha^{2}(k+1+\alpha)+ak)\leq 4k^{2}/(\alpha^{4}n). Since 3cσ<1α3c\sigma<1-\alpha, these two inequalities give a contradiction. ∎

It follows from Lemma 3.6 and Lemma 3.8 that L=kL=k and for all vL,v\in L, d(v)(11(k+1)n)nd(v)\geq\left(1-\dfrac{1}{(k+1)\sqrt{n}}\right)n. Let RR denote the common neighborhood of all vertices in LL. Then |R|(1k(k+1)n)n>nn|R|\geq\left(1-\dfrac{k}{(k+1)\sqrt{n}}\right)n>n-\sqrt{n}. Let E=SRE=S\setminus R, so that |E|n|E|\leq\sqrt{n}. We will show that E=E=\emptyset and so GKk,nkG\supseteq K_{k,n-k}.

Lemma 3.9.

For any vLv\in L, we have xv1αx_{v}\geq 1-\alpha.

Proof.

Assume to the contrary that there is some vLv\in L with xv<1αx_{v}<1-\alpha. Then the second-degree eigenvector equation with respect to zz gives

λα(λααd(z))\displaystyle\lambda_{\alpha}(\lambda_{\alpha}-\alpha d(z)) α(1α)uzd(u)xu+(1α)2wzuwxu\displaystyle\leq\alpha(1-\alpha)\sum_{u\sim z}d(u)x_{u}+(1-\alpha)^{2}\sum_{w\sim z}\sum_{u\sim w}x_{u}
α(1α)(uL1(z)d(u)xu+uS1(z)d(u)xu)\displaystyle\leq\alpha(1-\alpha)\left(\sum_{u\in L_{1}(z)}d(u)x_{u}+\sum_{u\in S_{1}(z)}d(u)x_{u}\right)
+(1α)2(wzuL1(w)xu+wzuS1(w)xu)\displaystyle\ +(1-\alpha)^{2}\left(\sum_{w\sim z}\sum_{u\in L_{1}(w)}x_{u}+\sum_{w\sim z}\sum_{u\in S_{1}(w)}x_{u}\right)
α(1α)((k1)n+(3c+1)σn)+(1α)2((k1+xv)n+(3c+1)σn)\displaystyle\leq\alpha(1-\alpha)\left((k-1)n+(3c+1)\sigma n\right)+(1-\alpha)^{2}\left((k-1+x_{v})n+(3c+1)\sigma n\right)
(α(k1)+(1α)(kα)+(3c+1)σ)(1α)n.\displaystyle\leq\left(\alpha(k-1)+(1-\alpha)(k-\alpha)+(3c+1)\sigma\right)(1-\alpha)n.

On the other hand, from Lemma 3.5 we have λα(λααd(z))(1α)(kα)n+O(1)\lambda_{\alpha}(\lambda_{\alpha}-\alpha d(z))\geq(1-\alpha)(k-\alpha)n+O(1) which, since α(k1)+(1α)(kα)+(3c+1)σ<(kα),\alpha(k-1)+(1-\alpha)(k-\alpha)+(3c+1)\sigma<(k-\alpha), gives a contradiction. ∎

Note that since FKk+1,mF^{\prime}\subseteq K_{k+1,m}, for any vLv\not\in L we have dR(v)<md_{R}(v)<m.

Lemma 3.10.

For any vSv\in S we have xv=O(n1)x_{v}=O(n^{-1}).

Proof.

Let v=argmaxuSxuv=\arg\max_{u\in S}x_{u}. Then we have

λαxv\displaystyle\lambda_{\alpha}x_{v} αd(v)xv+(1α)(k+(m+n)xv)\displaystyle\leq\alpha d(v)x_{v}+(1-\alpha)(k+(m+\sqrt{n})x_{v})
α(k+m+n)xv+(1α)(k+(m+n)xv)\displaystyle\leq\alpha(k+m+\sqrt{n})x_{v}+(1-\alpha)(k+(m+\sqrt{n})x_{v})
k+(m+n)xv\displaystyle\leq k+(m+\sqrt{n})x_{v}

so

(λαmn)xvkxvkλαmn=O(n1).(\lambda_{\alpha}-m-\sqrt{n})x_{v}\leq k\Longrightarrow x_{v}\leq\frac{k}{\lambda_{\alpha}-m-\sqrt{n}}=O(n^{-1}).

Lemma 3.11.

We have E=E=\emptyset.

Proof.

Delete all edges between EE and ERE\cup R and add all missing edges between EE and LL, to obtain a graph GG^{\prime}. The deletions decrease xTAαxx^{T}A_{\alpha}x by at most

α|E|(2m+n)O(n2)+(1α)|E|(m+n)O(n2)=o(1),\alpha|E|(2m+\sqrt{n})\cdot O(n^{-2})+(1-\alpha)|E|(m+\sqrt{n})\cdot O(n^{-2})=o(1),

while the additions increase xTAαxx^{T}A_{\alpha}x by at least α|E|(1α)2.\alpha|E|(1-\alpha)^{2}. If EE\neq\emptyset then the net change is positive. Hence, GF0G^{\prime}\supseteq F_{0} for some F0F_{0}\in\mathcal{F}. This implies that e(R)1e(R)\geq 1 in case (a) and e(R)2e(R)\geq 2 in case (b). Either way, it follows that e(L)<(k2)e(L)<{k\choose 2}, since GG is \mathcal{F}-free. Delete from GG^{\prime} all edges in RR and add an edge in LL, to obtain a subgraph of HH. The deletions decrease xTAαxx^{T}A_{\alpha}x by at most

α|R|O(n1/2)O(n2)+(1α)|R|O(n1/2)O(n2)=O(n1/2)\alpha|R|\cdot O(n^{1/2})\cdot O(n^{-2})+(1-\alpha)|R|\cdot O(n^{1/2})\cdot O(n^{-2})=O(n^{-1/2})

while the addition increases xTAαxx^{T}A_{\alpha}x by at least (1α)3(1-\alpha)^{3}, so the net change is again positive. Thus λα(G)<λα(G)<λα(H)\lambda_{\alpha}(G)<\lambda_{\alpha}(G^{\prime})<\lambda_{\alpha}(H) which is a contradiction. ∎

Lemma 3.12.

The set LL induces a clique.

Proof.

Suppose not. Then we transform GG into Kk+Knk¯K_{k}+\overline{K_{n-k}} by deleting all edges in RR and adding at least one edge in LL. The deletions decrease xTAαxx^{T}A_{\alpha}x by at most

αnmO(n2)+(1α)nm2O(n2)=O(n1).\alpha nm\cdot O(n^{-2})+(1-\alpha)\frac{nm}{2}\cdot O(n^{-2})=O(n^{-1}).

while the additions increase xTAαxx^{T}A_{\alpha}x by at least (1α)3(1-\alpha)^{3}, which is a contradiction. ∎

Proof of Theorem 1.5.

By Lemma 3.12, we know GKk+Knk¯.G\supseteq K_{k}+\overline{K_{n-k}}. In case (a), Kk+Knk¯K_{k}+\overline{K_{n-k}} is \mathcal{F}-saturated, hence G=Kk+Knk¯G=K_{k}+\overline{K_{n-k}}. In case (b), there is at most one edge in RR, which implies G=Kk+(K2Knk2¯)G=K_{k}+(K_{2}\cup\overline{K_{n-k-2}}). ∎

4 Concluding remarks

Our aim in this paper was to study situations where SPEX(n,)EX(n,)\mathrm{SPEX}(n,\mathcal{F})\subseteq\mathrm{EX}(n,\mathcal{F}) or SPEXα(n,)SPEX(n,)\mathrm{SPEX}_{\alpha}(n,\mathcal{F})\subseteq\mathrm{SPEX}(n,\mathcal{F}). In fact, we have given ‘recipes’ which allow one to determine these extremal graphs in more general situations. Many more applications of these results could be stated, and in our list of applications we have restricted ourselves to families for which the edge extremal graphs were studied in previous papers. There are also applications which follow easily from the methods of this paper but are not strictly consequences of our main theorems. We list some such cases below.

  • When one has a ‘nice characterization’ of the graphs XX such that Kk+XK_{k}+X is \mathcal{F}-free, stronger results can be proved. For example, suppose Kk+XK_{k}+X if \mathcal{F}-free if and only if XP3X\not\supseteq P_{3}; then one can remove the assumption that \mathcal{F} is finite from Theorem 1.1 (c) and prove an analagous result for the alpha spectrum. This allows one to prove the following results.

    • SPEX(n,{C4,C5,})=K1+Mn1.\mathrm{SPEX}(n,\{C_{4},C_{5},\ldots\})=K_{1}+M_{n-1}. This also follows from the spectral extremal result for C4C_{4} [47].

    • SPEXα(n,C4)=K1+Mn1\mathrm{SPEX}_{\alpha}(n,C_{4})=K_{1}+M_{n-1} [56].

    • K2k1+Mn2k+1K_{2k-1}+M_{n-2k+1} has the largest (alpha) spectral radius without kk disjoint even cycles.

    • K3+Mn3K_{3}+M_{n-3} has the largest (alpha) spectral radius without 2 disjoint theta graphs; see [29] for the definition and required extremal result.

    Another such ‘nice characterization’ occurs in the spectral result for star forests; here Kk+XK_{k}+X is FF-free if and only if Δ(X)<dk\Delta(X)<d_{k}. Thus we can obtain the corresponding alpha spectral result from [10].

  • In some cases our methods work for extremal graphs of the form W+Kn|W|¯W+\overline{K_{n-|W|}}, where WW is a fixed graph which is neither empty nor complete. For example, if \mathcal{F} is finite and every graph in EXK|W|,n|W|(n,)\mathrm{EX}^{K_{|W|,n-|W|}}(n,\mathcal{F}) contains ex(|W|,)\mathrm{ex}(|W|,\mathcal{F}) edges in WW and no edges in VWV-W, then under the other assumptions of Theorem 1.1, we have that any graphs in SPEX(n,)\mathrm{SPEX}(n,\mathcal{F}) are W+Kn|W|¯W+\overline{K_{n-|W|}} where WEX(n,|W|)W\in\mathrm{EX}(n,|W|). As consequence, we obtain the spectral analogues of Theorems 1.6 and 1.7 of [44].

  • As mentioned in the introduction, when SPEX(n,)=Kk,nk\mathrm{SPEX}(n,\mathcal{F})=K_{k,n-k} then under the assumptions of Theorem 1.5 we can show that SPEXα(n,)=Kk,nk\mathrm{SPEX}_{\alpha}(n,\mathcal{F})=K_{k,n-k} for a certain range of α\alpha. However, additional arguments would be needed to characterize SPEXα(n,)\mathrm{SPEX}_{\alpha}(n,\mathcal{F}) for every α[0,1)\alpha\in[0,1).

Additionally, when our methods do not determine the extremal graphs exactly, one can sometimes still obtain general structural results. For example, let \mathcal{F} be any finite family of graphs such that ex(n,)=O(n)\mathrm{ex}(n,\mathcal{F})=O(n). Let k=max{:K, is -free}<.k=\max\{\ell:K_{\ell,\infty}\text{ is }\mathcal{F}\text{-free}\}<\infty. Then the proof of Proposition 2.2 gives that any GSPEX(n,)G\in\mathrm{SPEX}(n,\mathcal{F}) has GKk,nkG\supseteq K_{k,n-k}. Moreover, any vertex in the partite set of GG of size nkn-k has degree at most k+mk+m, where mm is the smallest minimum degree of a vertex in the partite set of size k+1k+1 of any FF\in\mathcal{F} satisfying FKk+1,F\subseteq K_{k+1,\infty}.

We end by listing some open questions.

  • To prove Theorem 1.1 in its generality we require that ex(n,)=O(n)\mathrm{ex}(n,\mathcal{F})=O(n). However, many of the techniques in the proof were originally discovered in the proof of the spectral even-cycle theorem [13] where the Turán number is much larger. When considering a specific forbidden family, the structure of the forbidden graph(s) can be used and in some cases the condition on the Turán number can be relaxed. When α=0\alpha=0, perhaps the assumption that FvF-v is a forest for some bipartite FF\in\mathcal{F} could be useful. For example, it would be interesting to try to prove a spectral version of the general result of Faudree and Simonovits [26]. In any case, it seems possible that our assumption ex(n,)=O(n)\mathrm{ex}(n,\mathcal{F})=O(n) could be weakened or modified, and one could obtain a general result which implies, e.g., the spectral even cycle theorem.

  • For the case α>0\alpha>0, we do not know whether the assumption that FvF-v is a forest could be weakened. For example, if one may assume only that F{v1,,vk}F-\{v_{1},\ldots,v_{k}\} is a forest then perhaps one could determine the alpha spectral extremal graphs for kC2k\cdot C_{2\ell}.

  • We would like to obtain a counterexample similar to Proposition 1.2 using a smaller family \mathcal{F} or ideally a single graph FF. Alternatively, perhaps Theorem 1.3 be strengthened when \mathcal{F} consists of a single forest FF. It seems reasonable to guess that, for 3/4Q13/4\leq Q\leq 1, we have SPEX(n,F)EXKk+Knk¯(n,F)\mathrm{SPEX}(n,F)\subseteq\mathrm{EX}^{K_{k}+\overline{K_{n-k}}}(n,F) while for any Q>1Q>1 this is not necessarily the case. The reason is that, when Q1Q\leq 1 we can use the fact that XX is a union of bounded trees or of cycles; while for certain Q>1Q>1, certain path-star forests [25] provide counterexamples.

  • The only step in the proof of Theorem 1.1 which prevents us from allowing \mathcal{F} to be infinite in all cases is Lemma 2.12. We were unable to modify our arguments to work for all infinite graph families or to find a counterexample showing this cannot be done. In some cases, the nature of the graph family \mathcal{F} allows one to prove Lemma 2.12 without resorting to the argument we use for cases (a)-(c). For example, if ={C4,C5,}\mathcal{F}=\{C_{4},C_{5},\ldots\}, then the ‘nice characterization’ argument alluded to above achieves this, but alternatively one can argue instead of Lemma 2.12 that the existence of a long cycle in the modified graph GG^{\prime} implies the existence of a different long cycle in the original graph GG. Ad-hoc arguments of this kind may work for many other natural infinite families.

  • In cases (d) and (e) of Theorem 1.1, it is not too difficult to show that if Q<1Q<1, then there is some RR such that e(X)=Rn+O(1)e(X)=Rn+O(1). Can this be shown for Q1Q\geq 1?

  • As mentioned in the introduction, the extremal graphs covered by Theorem 1.5 are more restricted than those covered by Theorem 1.1. Our methods should give some information about SPEXα(n,)\mathrm{SPEX}_{\alpha}(n,\mathcal{F}) when \mathcal{F} is as in cases (a), (d), (e), or (f) of Theorem 1.1, but we have not attempted to determine the strongest possible statement one can prove using our methods.

  • When G=Kk,nkG=K_{k,n-k}, we have seen that the extremal function exG(n,)\mathrm{ex}^{G}(n,\mathcal{F}) is relevant to determining the spectral extrema for many families of forbidden graphs. Perhaps this motivates further study of this or other ‘restricted extremal problems’; in particular, it would be interesting to see if there are cases in which determination of EXKk,nk(n,)\mathrm{EX}^{K_{k,n-k}}(n,\mathcal{F}) is nontrivial while also providing an application of Theorem 1.1.

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