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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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Correct finite-time bound for partial-homogeneity zero crossing

Statement

Let M≥0, η>0, L≥0, ϵ>0, and let w:[t0,t0+2M/η]→[−M,0] and θ:[t0,t0+2M/η]→Rq be absolutely continuous until the first time w reaches zero. Assume w(t0)∈[−M,0], θ(t0)=θ0, and, almost everywhere before the crossing,

w′(t)≥η/2,q′(t)≤∣w(t)∣(ϵ+Lq(t)),q(t)=∣θ(t)−θ0∣.

Then w reaches zero at some τ≤t0+2M/η. At the first such time,

q(τ)≤{ϵL(eLM2/η−1),L>0,ϵM2η,L=0.

Thus the crossing remains in ∣θ−θ0∣<r0 whenever the displayed bound is less than r0. The crossing-time bound depends on the initial negative-weight magnitude M.

Facts & Assumptions

Given: The differential inequalities and initial data in the Statement.

[A1]

The functions w and θ are absolutely continuous on each interval before the first zero of w.

Proof

Given: M,η,L,ϵ,w,θ as in the Statement.

1.1givenA1algebra

Suppose first that M>0 and w has not reached zero by t0+2M/η. Integrating w′≥η/2 gives w(t0+2M/η)≥w(t0)+M≥0, contradicting the assumed strict negativity before the first crossing. Hence the first crossing time τ exists and satisfies the stated bound. If M=0, then w(t0)=0 and τ=t0.

2.1A1step 1.1algebra∎

On [t0,τ], w is strictly increasing with w′≥η/2, so it is an admissible absolutely continuous time change. Its inverse satisfies dt/dw≤2/η. If L>0, set p=ϵ+Lq. The inequality for q gives dp/dw≤(2L/η)∣w∣p, and integration from w(t0) to 0 yields p(τ)≤ϵeLM2/η. Rearranging gives q(τ)≤ϵ(eLM2/η−1)/L. If L=0, direct integration gives q(τ)≤(2ϵ/η)∫w(t0)0∣s∣ds=ϵw(t0)2/η≤ϵM2/η. The same bounds hold with zero right-hand side when M=0.

Source note

Chizat and Bach's printed proof claims a common crossing time T<2/η for every w0∈[−M,0]. Integrating w′≥η/2 gives the correct worst-case bound 2M/η. The estimate above proves hitting at an actual time τ and the corresponding θ-displacement bound; it does not assert w(T)=0 at one fixed time for all initial weights.

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Sources