1
$\begingroup$

Let $X_1,\ldots,X_m$ be Lipchitz continuous vector fields in an open set $\Omega \subset \Re^n$, Let $d(\cdot, \cdot)$ denote the control distance associated to $X$. With respect to this control distance we can define balls etc.

Let us also define the Sobolev spaces $W^{1,p} (\Omega) = [u \in L^p (\Omega) | Xu := (X_1 u, \ldots, X_m u) \in L^p(\Omega) ]$ (in a weak sense), let also $W_0^{1,p}$ be the closure of $C_0^\infty (\Omega)$ under the norm $||u||_{1,p}^p = ||u||_p + ||Xu||_p$. Let us also assume that for every compact set $K \subset \Omega$ there exists positive constants $C_D = C_D(X,K), C_L = C_L(X,K), C_P = C_P (X,K), R = R(X,K)$ such that for every $x \in K$ and $0 < r < R$ one has

(D) $|B(x,r)| \geq C_D |B(x,2r)|$

(L) $d(\cdot, x)$ is differentiable a.e. in $\Omega$ and $||Xd(\cdot,x)||_{L^\infty (\Omega)} \leq C_L$ for every $x \in K$.

(P) $\int_{B(x,r)} |u - u_B|^2 dx \leq C_P \int_{B(x,2r)} |Xu|^2 dx$.

As such we have a doubling metric measure space $(K,d,dx)$, with Poincare' inequality.

I am currently reading a paper stating that in the above setting we obtain the existence of cutoff functions in metric balls satisfying the common $|X u| < C/r$, and $u = 1$ on $B(x,r)$ and $u = 0$ outside $B(x,2r)$, and $u \in W_0^{1,\infty}(\Omega)$.

I have never seen this result, and to me it seems strange. Why would one otherwise mess about with balls defined from level surfaces of the fundamental solution to the real part of the Kohn-Laplacian in Hörmander vector-fields?

$\endgroup$
0

1 Answer 1

2
$\begingroup$

Apparently one can compose a 1-d cutoff function with the metric to obtain this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.