Weighted inequalities for gradients on non-smooth domains ...

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Weighted inequalities for gradients on non-smooth domains ...

comes almost (but not quite) ong>forong> free.We have, ong>forong> any x/∈ Ω δ ,∫|∇u(x)| ≤C∂Ω|f| dω,where C depends on Ω, δ, and our choice of X 0 (which, without loss of generality, we may assumelies in Ω \ Ω δ ). This inequality comes from the fact thatlinked to the inequality:By Hölder’s Inequality,sup |∇u(x)| ≤Cδ −1 sup |u(x)|x/∈Ω δ x/∈Ω δ/2sup |u(x)| ≤C Ω,δ |f| dω.x/∈Ω δ/2∫∂Ω∫∂Ω(∫ ) 1/p (∫|f| dω ≤ |f| p vdω∂Ω∂Ω) 1/p′σdω .Thus, to get (2.1), it is sufficient to haveµ(Ω \ Ω δ ) 1/q (∫∂Ω) 1/p′σdω ≤ c ′ ,where c ′ is small positive constant depending on Ω, δ, andX 0 .A moment’s thought will show that this is just a “global” version of the sufficient conditionfrom Theorem 3.1, with ∂Ω (a bounded set, recall) playing the role of a cube Q b ,andΩ\ Ω δpretending to be T (Q b ). The bump function being integrated against σ is just χ ∂Ω . It has no “tail”because there is no room ong>forong> one:the “cube” Q b fills up all of ∂Ω.To fit the pattern of Theorem 3.1, the constant c ′ should really be cdiam(Ω)ω(∂Ω) = cdiam(Ω).But, of course, it is—assuming we choose c properly!That is (almost) all we will say about the parts of Ω lying far from the boundary.Our problem now consists in finding an appropriate condition on the measure µ 2 which ensuresthat(∫(∫) 1/q|∇u| q dµ 2 ≤Ω|f| p vdω∂Ωholds ong>forong> all f ∈ L ∞ , where we remember that µ 2 is supported in Ω δ .By cutting the integral up, taking δ small enough, rescaling, and doing a rotation, we canassume that the support of µ 2 —which we will henceong>forong>th call µ—lies in a set of the ong>forong>m) 1/pR≡{(x, y) : ψ(x)

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