Notes on Schauder Estimates
نویسنده
چکیده
Proof. Let g(x) = u(x) − sup∂Br(y) u − r2 − |x − y|2 2n supBr(y) f . We have ∆g = ∆u + supBr(y) f ≥ − f + supBr(y) f ≥ 0, that is, g is subharmonic in Br(y). Then supBr(y) g = sup∂Br(y) g = 0, so g ≤ 0 in Br(y) and the lemma follows. Lemma 2. If u is a solution to ∆u = f in Br(y) and v solves ∆v = 0 and v = u on ∂Br(y), then r2 − |x − y|2 2n inf Br(y) f ≤ v(x) − u(x) ≤ r 2 − |x − y|2 2n sup Br(y) f ,
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