Fix constants T0∈(0,1/8) and κ,cg,εg,δg>0 witnessing Assumption 29.1. Throughout the proof the stopping width is exactly
η=T01/3∈(0,1/2). Choose once and for all
ε=21min{1,εg,T01/3}>0. The published discharge Corollary 26.2 supplies Assumption 26.1 with universal constants c0,C1>0 and t1(n)=c0(logn)−2 for n≥3. Since t1(n)→0, choose an integer N≥3 such that t1(n)≤T0 for every n≥N. Enlarge N if necessary so that
Ln:=1+loglogn≥1(n≥N). Let CL>0 be the universal constant in the supply inequality (33.11). Set
a=min{4CLκT0,2cg}>0. Fix n≥N and suppose, for a contradiction, that hn⋆≤a/Ln. The established finite-dimensional bound Theorem 26.2 gives hn⋆>0. Also hn⋆<∞: the standard Gaussian is an admissible isotropic law and a balanced coordinate halfspace has finite perimeter. The definition (0.5) of the infimum therefore supplies an isotropic log-concave probability μ on Rn with
hμ≤(1+ε)hn⋆≤2hn⋆. This is simultaneously within the near-worst class of the completion and the class of Theorem 33.1, with ε≤T01/3.
By Lemma 33.1, Iμ(1/2)=hμ/2. By the definition of Iμ(1/2) as an infimum over balanced measurable cuts, for every integer j≥1 there is such a cut Ej with
0≤ej:=μ+(Ej)−Iμ(1/2)<min{δg,j−1}. In particular, every Ej has finite initial lower outer Minkowski perimeter, is admissible for the completion, and satisfies e0(Ej)=eˉ0(Ej)=ej. No existence of an exactly minimizing cut is needed.
All time, dimension, and near-worst hypotheses of Corollary 33.1 now hold. Apply its supply estimate separately to each fixed Ej, using its own mass process and stopping time τη(Ej). Since T04/3≤1≤Ln, it gives
∫0T0E[eˉt(Ej)1{t<τη(Ej)}]dt≤T0ej+CLhμ(T04/3+Ln)≤T0ej+4CLhn⋆Ln≤T0ej+κT0. Thus Assumption 29.1 implies μ+(Ej)≥cg for every j. Passing to the limit in the scalar perimeter values, which converge to Iμ(1/2) by construction, gives
hμ=2Iμ(1/2)≥2cg. On the other hand, the hypothesized smallness and the choice of a give
hμ≤2hn⋆≤Ln2a≤cg, a contradiction. We have proved hn⋆>a/Ln, which implies the claimed weak inequality. Finally, once loglogn≥1, one has 1+loglogn≤2loglogn, proving the last assertion.