12.07.2015 Views

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

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

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

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

comes almost (but not quite) <str<strong>on</strong>g>for</str<strong>on</strong>g> free.We have, <str<strong>on</strong>g>for</str<strong>on</strong>g> any x/∈ Ω δ ,∫|∇u(x)| ≤C∂Ω|f| dω,where C depends <strong>on</strong> 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 c<strong>on</strong>stant depending <strong>on</strong> Ω, δ, andX 0 .A moment’s thought will show that this is just a “global” versi<strong>on</strong> of the sufficient c<strong>on</strong>diti<strong>on</strong>from 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 functi<strong>on</strong> being integrated against σ is just χ ∂Ω . It has no “tail”because there is no room <str<strong>on</strong>g>for</str<strong>on</strong>g> <strong>on</strong>e:the “cube” Q b fills up all of ∂Ω.To fit the pattern of Theorem 3.1, the c<strong>on</strong>stant 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 c<strong>on</strong>sists in finding an appropriate c<strong>on</strong>diti<strong>on</strong> <strong>on</strong> the measure µ 2 which ensuresthat(∫(∫) 1/q|∇u| q dµ 2 ≤Ω|f| p vdω∂Ωholds <str<strong>on</strong>g>for</str<strong>on</strong>g> 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 rotati<strong>on</strong>, we canassume that the support of µ 2 —which we will hence<str<strong>on</strong>g>for</str<strong>on</strong>g>th call µ—lies in a set of the <str<strong>on</strong>g>for</str<strong>on</strong>g>m) 1/pR≡{(x, y) : ψ(x)

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!