Estimation optimale du gradient du semi-groupe de la chaleur sur le ...

D. Brydges, A. Talarczyk / Journal of Functional Analysis 236 (2006) 682–711 699
Proposition 4.1. Assume that L>0, d 1, B =∇, and Λ is some bounded open set in R d ,
such that the set Λ−2L ={x ∈ Λ: dist(x, ∂Λ) > 2L} is nonempty. If y ∈ Λ−2L then ALϕ(y) =
d
R hL(y − x)ϕ(x)dx, where
hL(x) =
1
|B|L 2 |x| d
∂BL
2x · z −|x| 2 1|x−z|LσL(dz), (4.2)
where σL denotes the surface measure (normalized to 1) on a sphere of radius L. In the case
d = 1, σL(dx) = 1
2 (δ−L(dx) + δL(dx)) where δt(dx) denotes a unit point mass at x = t.
Notice that since σL is a uniform measure on the sphere ∂BL the value of hL(x) depends only
on |x|.
Proof. From Theorem 2.8 and Lemma 2.11, for ϕ ∈ H+(Λ) and y ∈ Λ−2L we have
By translation invariance,
ALϕ(y) = 1
|BL|
ALϕ(y) = 1
|BL|
y+BL
y+BL
where ϕx(z) = ϕ(x + z).
Recall the well-known formula for the Poisson kernel of the ball BL
PBLϕ(y) =
∂BL
Px+BL ϕ(y)dx. (4.3)
PBL ϕx(y − x)dx, (4.4)
Ld−2 (L2 −|y| 2 )
|y − z| d
ϕ(z)σL(dz).
Note that this is valid also for d = 1. Applying this to (4.4), then changing the order of integration
and substituting x = x ′ − z we obtain
ALϕ(y) = 1
|BL|
∂BL
= 1
|BL|
= 1
|BL|
=
R d
∂BL
∂BL
1x∈y+BL
1x∈z+y+BL
1z∈x−y+BL
hL(y − x)ϕ(x)dx
Ld−2 (L2 −|y − x| 2 )
|y − x − z| d
ϕ(x + z) dx σL(dz)
Ld−2 (L2 −|y − x + z| 2 )
|y − x| d
ϕ(x)dx σL(dz)
Ld−2 (L2 −|y − x + z| 2 )
|y − x| d
σL(dz)ϕ(x) dx