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LemmaStatement: AI-generatedProof: AI-generated
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The original shifted roots impose a minimum degree above log n

Statement

Let G=(V,E) be a finite undirected graph on n=∣V∣≥2 vertices with symmetric binary adjacency matrix A, allowing self-loops and counting each loop once in its vertex's degree. Write d(x)=∑y∈VAxy, D=diag⁡(d(x)), dmin⁡(G)=min⁡x∈Vd(x) and dmax⁡(G)=max⁡x∈Vd(x). Let I and J be the identity and all-ones matrices on the vertex space.

Take the admissible original parameters γ=2, ε=1/2, and let c−<0<c+ be the original shifted roots, so 2h(c−−1/2)=2h(c++1/2)=3/2, where h(x)=(1+x)log⁡(1+x)−x for x>−1. Then −1/2<c−<0. Put L=log⁡n, d∗=2L, α=8/2L and p=2L/n, so pn=d∗.

The degree target (6.9) for G is the condition (1+c−+α)pn≤dmin⁡(G)≤dmax⁡(G)≤(1+c+−α)pn. For every sufficiently large n, each of this degree target and the original (n,d∗,α,c−,c+)-expander predicate for G implies dmin⁡(G)>L. The latter predicate means all three inequalities ∥A−pJ∥≤αd∗,c−d∗I⪯ΔL,ΔL⪯c+d∗I, where ΔL=D−A−d∗I+(d∗/n)J, ∥⋅∥ is the operator norm on CV, and ⪯ is Hermitian quadratic-form order.

Facts & Assumptions

[F1]

The shifted-root definition requires positive logarithm arguments and the stated signs of the roots (Shifted rate roots for quantitative expansion).

[F2]

Two-sided expansion is exactly the norm inequality and the two centered-Laplacian order inequalities (Two-sided spectral expander — Definition 1.9).

[F3]

Every such expander has dmin⁡(G)≥(1+c−)d∗−d∗/n (Degree bounds from two-sided expansion — Lemma 3.1).

[F4]

The graph conventions give binary symmetric A, degrees as row sums with each loop counted once, the diagonal degree matrix D, and the degree extrema of the same finite nonempty graph G (Adjacency, degrees, graph Laplacian, and spectral notation).

[F5]

The natural logarithm is strictly increasing on (0,∞) and satisfies log⁡1=0 (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

Proof

Given: The graph, parameters and shifted roots in the Statement; in each implication assume its stated condition on G.

Goal: Prove −1/2<c−<0 and that each condition forces dmin⁡(G)>L for sufficiently large n.

1.1F1F4F5givenalgebra

The graph and its degree extrema have the conventions in [F4]. Since n≥2>1, [F5] gives L=log⁡n>log⁡1=0, so d∗=2L>0 and α=8/2L>0. The parameter inequalities are 0<1/2<1 and 1+1/2<2<1+(1/2)−2=5. Positivity of 1+c−−1/2 in [F1] gives c−>−1/2, and [F1] also gives c−<0. Therefore δ=1+2c−>0 is a fixed positive number. If the full displayed degree target holds, its lower inequality and pn=d∗=2L give dmin⁡(G)≥(1+c−+α)2L=L+(δ+2α)L>L, using L>0 and α>0.

2.1F2F3F4step 1.1algebra∎

If G satisfies the expander predicate, [F2] gives precisely the displayed norm and order inequalities. The graph hypotheses in [F4] allow [F3] to be applied at d∗=2L, giving dmin⁡(G)≥L+(δ−2/n)L>L whenever n>2/δ. Together with step 1.1, this proves both implications for every sufficiently large n.

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