On heat kernels on Lie groups.
Let G be a homogeneous Lie group. We prove that for every closed, homogeneous subset Γ of G* which is invariant under the coadjoint action, there exists a regular kernel P such that P goes to 0 in any representation from Γ and P satisfies the Rockland condition outside Γ. We prove a subelliptic estimate as an application.
Let G be a Lie group, Xj right invariant vector fields on G, which generate (as a Lie algebra) the Lie algebra of G, L = -Σ Xj 2. (...) In this paper we consider L1(G) boundedness of F(L) for (some) metabelian G and a distinguished L on G. Of the main interest is that the group is of exponential growth, and possibly higher rank. Previously positive results about higher rank groups...
Let be a symmetric α-stable semigroup of probability measures on a homogeneous group N, where 0 < α < 2. Assume that are absolutely continuous with respect to Haar measure and denote by the corresponding densities. We show that the estimate , x≠0, holds true with some integrable function Ω on the unit sphere Σ if and only if the density of the Lévy measure of the semigroup belongs locally to the Zygmund class LlogL(N╲e). The problem turns out to be related to the properties of the maximal...
We prove that on a product of generalized Heisenberg groups, a Hörmander type multiplier theorem for Rockland operators is true with the critical index n/2 + ϵ, ϵ>0, where n is the euclidean (topological) dimension of the group.
Let L be a positive Rockland operator of homogeneous degree d on a graded homogeneous group G and let be the convolution kernels of the semigroup generated by L. We prove that if τ(x) is a Riemannian distance of x from the unit element, then there are constants c>0 and C such that . Moreover, if G is not stratified, more precise estimates of at infinity are given.
We show that, if a certain Sobolev inequality holds, then a scale-invariant elliptic Harnack inequality suffices to imply its a priori stronger parabolic counterpart. Neither the relative Sobolev inequality nor the elliptic Harnack inequality alone suffices to imply the parabolic Harnack inequality in question; both are necessary conditions. As an application, we show the equivalence between parabolic Harnack inequality for on , (i.e., for ) and elliptic Harnack inequality for on .
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