By K. A. Broughan

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**Example text**

It is Harnack’s inequality, applied to the ﬁrst derivatives of w, that supplies what is needed. In this context Harnack’s inequality says: if the distance between the surfaces S and St(en +γτ ) is of order t at one point, then it is of order t in a neighborhood of that point. 1. THE MAIN THEOREM. HEURISTIC CONSIDERATIONS AND STRATEGY 53 if γ = 1/λ, one of the two surfaces St(en ±γτ ) separates from S by a distance of order t, in at least half of the points. If we suppose that the “good” surface is St(en +γτ ) , then, by Harnack’s inequality, St(en +γτ ) stays ct-away from S in B1/2 .

In fact, the family vt (x) = sup u Bt (x) with t constant, can only detect a uniform enlargement of the monotonicity cone, and, as such, one cannot exploit the interior gain. For this purpose we ask the question: what are the conditions on a variable radius t(x) so that for any harmonic function u, vt will be always subharmonic. Here is the fundamental lemma. 7. 12) for C = C(n) large enough. Let u be continuous, deﬁned in a domain Ω so large that the function w(x) = sup u(x + ϕ(x)ν) |σ|=1 is well deﬁned in B1 .

9): B ⎧ Δwt = 0 in Ω+ (vϕt ) ∩ Ω ≡ Ωt ⎪ ⎪ ⎪ ⎨w ≡ 0 in Ω ¯ Ωt t ⎪ wt = 0 on ∂B9/10 ⎪ ⎪ ⎩ wt = u(x0 ) on ∂B1/8 (x0 ) Then for a small constant c, h and any ε > 0 small enough, Vt = vεϕt + cεwt (0 ≤ t ≤ 1) is a family of subsolution. Proof. 7. We have to check that Vt has the correct asymptotic behavior. Notice that F (Vt ) = F (vεϕt ). 17) with β¯ α ¯ ≥G 1 − ε|∇ϕt | 1 + ε|∇ϕt | 4. LIPSCHITZ FREE BOUNDARIES ARE C 1,γ 70 Since |∇ϕt | ≡ 0 outside B7/8 the right inequality is satisﬁed by vεϕt and hence by Vt since wt is positive.