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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Nikiforov: for every H and every ϵ∈(0,12) there is δ>0 such that every graph G with ind⁡H(G)<(δ∣V(G)∣)∣V(H)∣ has an ϵ-restricted vertex set of size at least δ∣V(G)∣

Statement

Fix a graph H with h=∣V(H)∣ and a real ϵ∈(0,12). Then there exists δ>0 such that every nonempty finite simple graph G with

ind⁡H(G)<(δ∣V(G)∣)h

has an ϵ-restricted vertex set of size at least δ∣V(G)∣.

Facts & Assumptions

Given: A graph H with h vertices and a real ϵ∈(0,12).

[L1]

For each 0<γ<1, the self-regular-subset theorem gives a constant δ1(γ)>0 such that every nonempty graph on n vertices has a subset W with ∣W∣≥δ1(γ)n and (W,W) γ-regular (Every finite graph has a linearly large ϵ-self-regular vertex subset, ϵ-regular pairs and self-regular vertex sets).

[L2]

The induced counting constants include N=N(H,η), and if 0<γ≤γ(H,η), ∣W∣≥N, (W,W) is γ-regular, and its self-density lies between η and 1−η, then ind⁡H(G)≥c(H,η)∣W∣h (A large γ-self-regular set whose density lies between η and 1−η forces at least c∣W∣∣V(H)∣ induced copies of H).

[L4]

Every nonempty set of at most two vertices is 0-restricted, hence ϵ-restricted (Sparsity is preserved when the parameter grows, and every nonempty set of at most two vertices is 0-restricted).

Proof

technique · direct
1.1L2L3algebra

Set η:=ϵ/4 and choose γ>0 so small that γ≤ϵ/8 and γ≤γ(H,η) from [L2]. Then (η+γ)/(1−γ)≤ϵ, and the dense trimming parameter from [L3] is also at most ϵ, because η=ϵ/4, γ≤ϵ/8, and ϵ<1/2.

2.1step 1.1L1L2choose

If h=0, the induced-copy hypothesis is never satisfied: both sides of its displayed inequality are 1. Thus suppose h≥1. Let δ1:=δ1(γ) be the constant of [L1], let c:=c(H,η) and N:=N(H,η) be the constants of [L2], put N′:=max⁡{N,1}, and set δ:=min⁡{δ1c1/h, (1−γ)δ1, δ1/N′, 1}. Then δ>0 and depends only on H and ϵ.

3.1step 2.1L1L4choosealgebra

Now let G be a nonempty graph on n vertices with ind⁡H(G)<(δn)h. If n<N′/δ1, then any singleton X is 0-restricted by [L4] and satisfies ∣X∣=1>δn, because δ≤δ1/N′. Hence suppose n≥N′/δ1. By [L1] choose W⊆V(G) with ∣W∣≥δ1n≥N′≥N such that (W,W) is γ-regular.

4.1step 1.1step 2.1step 3.1L2algebra

If η≤dG(W,W)≤1−η, then [L2] gives ind⁡H(G)≥c∣W∣h≥cδ1hnh≥(δn)h, contrary to the hypothesis on G. Therefore either dG(W,W)<η or dG(W,W)>1−η.

5.1step 1.1step 2.1step 3.1step 4.1L3

In the first case, the low-density trimming lemma in [L3] yields a subset W′⊆W with ∣W′∣>(1−γ)∣W∣≥δn that is ϵ-sparse by step 1.1. In the second case, the high-density trimming lemma in [L3] yields a subset W′⊆W with the same size bound that is ϵ-dense. In either case W′ is ϵ-restricted.

6.1step 3.1step 4.1step 5.1∎

Step 3.1 handles small n, and steps 4.1 and 5.1 handle all remaining cases, so every graph satisfying the induced-copy bound has an ϵ-restricted set of size at least δ∣V(G)∣.

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