claimroutineinferred_dep

Marginal atom lower bounds by vertex removal

Claim·§2.3 Degree control and a maximizing rectangle·inside propIncidence piece with degree and expansion control

Statement (paper)

For all sufficiently large tt, every p∈Pp\in P and q∈Qq\in Q satisfy μ(p)≥γ/∣P∣\mu(p)\ge\gamma/|P| and κ(q)≥γ/∣Q∣\kappa(q)\ge\gamma/|Q|, where γ=27/50\gamma=27/50; that is, holds, and multiplying by ∣E∣|E| gives the corresponding lower bounds on the degrees d(p)d(p) and d(q)d(q) in .

In plain English

For all large tt, every p∈Pp\in P and q∈Qq\in Q has marginal mass μ(p)≥γ/∣P∣\mu(p)\ge\gamma/|P| and κ(q)≥γ/∣Q∣\kappa(q)\ge\gamma/|Q| with γ=27/50\gamma=27/50. Equivalently, after multiplying by ∣E∣|E|, every degree d(p)d(p) and d(q)d(q) meets the corresponding lower bound.

Why it's here

This gives the vertex-wise lower bounds on marginals, which are the atom condition in the proposition. The likely route is vertex removal: compare the density-maximizing rectangle from the previous claim with the rectangle that omits a single low-degree vertex. Since that rectangle keeps a t−o(a)t^{-o(a)} fraction of the vertices and edges, a vertex with too small a marginal could be removed to raise the ratio w/(uv)γw/(uv)^\gamma, contradicting maximality.
Classified routine:
A standard vertex-removal argument applied to the density-maximizing rectangle of claim-inc-piece-5. If some vertex had marginal below gamma/|P|, deleting it would increase the ratio w/(uv)^gamma, contradicting maximality; this is the usual technique for obtaining a minimum-degree subgraph.

Sources

PaperOpen
Section
§2.3 Degree control and a maximizing rectangle
Page
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sections/01-incidence.tex:327
Leannot aligned

Phase 2 aligns this node to a Lean declaration.

Friction

  • info
    inferred_dep
    outline: "Marginal atom lower bounds by vertex removal" rests on claim-inc-piece-5