for all large s > 0. Now, we absorb the first term on the right-hand side into the left-hand side by choosing s > 0 large, we get for all large s > 0. Replacing the integration domain on the left-hand side by D × { y ; | y − y$_{0}$ | < L ρ } ⊂ Ω$_{0}$ and using the facts that χ$_{0}$ ( | y − y$_{0}$ | ) = 1 in D × { y ; | y − y$_{0}$ | < L ρ } and we obtain, for all s ≥ s$_{0}$ , where s$_{0}$ is some constant. Since κ$_{2}$ > κ$_{1}$, the last inequality implies for all s ≥ s$_{0}$ , where κ = κ$_{2}$ − κ$_{1}$ > 0. We separately consider the two cases: Case 1. Let M ≥ d. Choosing s ≥ 0 such that we obtain Case 2. Let M < d. Then setting s = 0 in (4.23) we have ∫ Ω$_{0}$ χ 2 $_{0}$( | y − y$_{0}$ | ) | f | $^{2}$e 2 sϕ ( x,y, $^{0)}$dxdy ≤ CM $^{2}$e2 sκ$_{1}$ + Ce $^{cs}$d 2 e 2 sϕ ( x,y, 0) = exp ( 2 se γψ ( x,y, 0) ) > exp (2 se $^{γϵ}$) = e 2 $^{sκ$_{2}$}$, e 2 sκ$_{2}$ ∫ D × { y ; | y − y$_{0}$ | < L ρ } | f | $^{2}$dxdy ≤ CM $^{2}$e2 sκ$_{1}$ + Ce $^{cs}$d 2 ∫ D × { y ; | y − y$_{0}$ | < L ρ } | f | $^{2}$dxdy ≤ CM $^{2}$e − 2 sκ + Ce $^{cs}$d 2 (4.23) M $^{2}$e − 2 sκ = e $^{Cs}$d $^{2}$, that is , s = 2 C + 2 κ log M d ≥ 0 , ∫ D × { y ; | y − y$_{0}$ | < L ρ } | f | $^{2}$dxdy ≤ 2 M 2 C C +2 κ d 4 κ C +2 κ . ∫ D × { y ; | y − y$_{0}$ | < L ρ } | f | $^{2}$dxdy ≤ 2 Cd $^{2}$. 30