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By Dahlberg B.E.J., Kenig C.E.

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3 (DeGiorgi 2], Nash 3]). If u is a solution of Lu = 0 in D, then u is Hlder continuous. 4 (Harnack's inequality). If u 0 and Lu = 0 in D and if K D is a compact set, then ess supK u C ess infK u where C = C (n; ; K; D). Remark: Harnack's inequality is a quantitative version of the maximum principle. Remark: For notational convenience we let min and max denote ess inf and ess sup, resp. 3. This is done using Harnack's inequality. Assume Lu = 0 in D = fx 2 Rn : jxj < 2g. Set M (r) = maxjxj r u m(r) = minjxj r u r < 1: Then M (r) ?

4 (John 6]). Assume f : R ! R measurable. Assume there exists an > 0 and a continuous function C : ! R, where = f(a; b) 2 R2 : a < bg, such that jfx 2 I : jf (x) ? C (I )j < gj > 31 jI j for each interval I = (a; b). Then f 2 BMO (R) and kf k C where C is independent of ; f and the function C . Proof. It is enough to prove the lemma for = 1. The arguments are similar to those one uses to prove John-Nirenberg's inequality once we have proved the following Claim: If I J R are intervals such that jJ j = 2jI j, then jC (I ) ?

2. If f 2 L2(@D), then there exists a u such that u=0 @u @n j@D = f in D on @D where the boundary values are taken in the sense np ru(Q) ! f (P ) as Q ! e. @D and M (ru) 2 L2 (@D) with kM (ru)k2 C kf k2 where C only depends on > 1 and k'0k1 . We recall that M u(P ) = supfju(Q)j : jP ? e. on @D. 1. 1 is valid if L2 replaced by Lp for 2 p 1. 51 This is a straightforward consequence of the maximum principle and interpolation between L2 and L1 , but apart from this result we discuss the Lp-theory for the Dirichlet problem and Neumann problem in Chapter 7.