I am self-studying some PDE stuff, and the following inequality arose in a proof of the theorem, without any further explanations.
Let $U$ be a bounded Lipschitz domain. By a Lipschitz domain, I mean the following.
There exist constants $r_0 , M>0$ such that for each $z \in \partial U$, there exists a Lipschitz map $\phi: \mathbb{R}^{n-1} \to \mathbb R$ with $|\nabla \phi|_{L^\infty} \leq M$ with $$ B(z,r_0) \cap U = B(z,r_0) \cap \big\{ \big(x,\phi(x)\big) : x \in \mathbb{R}^{n-1} \big\}.$$
The inequality says that for any harmonic polynomial $f$ of degree $\leq d$,
$$\int_{U_{2r}} | \nabla f(x)| ^2 dx \leq C \int_{U} | \nabla f(x)| ^2 dx$$
for a constant $C>0$ depending on $d$, $n$, $r$, $N$, and $M$. Here, $U_{2r} $ is the set of points whose distance from $U$ is less than $2r$. $N$ denotes the number of open balls (as in the definition of "Lipschitz domain") which are required to cover $\partial U$.
I am lost since I do not see any reason for the inequality to be true. Could someone help me, please? Any bits of help would be appreciated. Thank you!
Note: the statement in the question was used as the final step for a proof of the following inequality.
$$|\nabla f|_{L^2 (\partial U)} \leq C_{d, n, r, N, M} |\nabla f|_{L^2(U)}$$