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Small total induced-copy expectation forces many homogeneous k-sets

Statement

Let 1kt be integers, and let F be a finite family of graphs, each with at least one vertex, such that every F-free graph has the (t,k)-homogeneous property. Let G be a finite graph on n2t vertices. Choose X uniformly from [V(G)]2t and define

Y(X):=HFindH(G[X]).

If E[Y]t/2, then G has at least

12(n2t)k

homogeneous vertex sets of size k.

Facts & Assumptions

Given: Positive integers 1kt, a finite family F of graphs, each with at least one vertex, a finite graph G on n2t vertices, the uniform choice of X[V(G)]2t, and the hypothesis E[Y]t/2.

[L1]

A graph is F-free exactly when it is H-free for every HF (H-free and F-free graphs under the induced-subgraph convention).

[L2]

Y(X) is a nonnegative real random variable on the uniform probability space on [V(G)]2t, and its expectation is the average value over that finite outcome set (The uniform probability space on a nonempty finite set, Expectation of a real random variable on a finite probability space, The induced-embedding count indH(G)).

[L3]

If a nonnegative random variable has expectation at most t/2, then the probability that it is at least t is at most 1/2 (Markov's inequality on a finite probability space).

[L4]

If at least half of the 2t-element subsets of V(G) contain a t-element induced subgraph in a class with the (t,k)-homogeneous property, then G has at least 12(n/(2t))k homogeneous k-sets (Many good 2t-vertex subsets force many homogeneous k-sets).

Proof

technique · direct
1.1

Since Y is nonnegative and E[Y]t/2, [L3] gives P(Yt)1/2, so with probability at least 1/2 one has Y<t.

L2L3
2.1

Fix a set X[V(G)]2t with Y(X)<t. For each induced embedding counted by Y(X) choose one vertex from its image; this is possible because every graph in F has at least one vertex. Delete from X every chosen vertex. Since fewer than t embeddings were counted, fewer than t vertices are deleted, so at least t vertices remain.

step 1.1choose
3.1

Let T be any t-element subset of the remaining vertices. If some HF had an induced embedding into G[T], then that same embedding would already have been counted in Y(X), so step 2.1 would have deleted a vertex from its image. Because the image lies in T, this contradicts the choice of T. Thus G[T] is F-free by [L1].

step 2.1L1choose
4.1

Steps 2.1 and 3.1 show that with probability at least 1/2, a uniformly random 2t-element subset of V(G) contains a t-element induced subgraph that is F-free. Applying [L4] to the class of F-free graphs proves the claimed lower bound on homogeneous k-sets.

step 1.1step 3.1L4

Depends on

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