# Volume doubling, uniform Poincaré, counterexample

The Poincaré inequality and the volume doubling property are important notions related to heat kernel estimates.

Pavel Gyrya and Laurent Saloff-Coste obtain the two sided heat kernel estimate of Neumann and Dirichlet heat kernels on inner uniform domains by showing the Poincaré inequality and the volume doubling property for the canonical Dirichlet forms on inner uniform domains.

Namely, for inner uniform domains, we can check the Poincaré inequality and the volume doubling property.

It is easy to check that domains with cusps are not inner uniform domains. For example, $$D=\{(x,y) \in \mathbb{R}^2 \mid x>0,\ y<1,\ y>x^{1/2}\}$$ is not inner uniform.

The closure $$\bar{D}$$ of $$D$$ is regarded as a metric space endowed with the shortest path metric $$\rho$$.

I am concerned with whether the Poincaré inequality holds on $$\bar{D}$$. Namely, there exists $$P_0$$ such that for any $$x \in \bar{D}$$, $$r>0$$, and smooth $$f$$, \begin{align*} (\ast)\quad \inf_{\xi \in \mathbb{R}} \int_{B(x,r)}|f-\xi|^2\,dm \le P_o r^2 \int_{B(x,r)}|\nabla f|^2\,dm, \end{align*} where $$B(x,r)=\{y \in \bar{D} \mid \rho(y,x) and $$\nabla f$$ denotes the distributional derivative of $$f$$. $$m$$ is the Lebesgue measure on $$D$$

My question

Does $$(\ast)$$ hold? The volume doubling property holds. Let $$x$$ be the origin and let $$f$$ be a smooth bump function with $$\nabla f=1$$, $$f=1$$ on $$B(x,r/2)$$, and $$f=0$$ outside $$B(x,r)$$. Then, RHS of $$(\ast)=P_0 r^2 m(B(x,r)\setminus B(x,r/2)) \sim r^2 \times r\sqrt{r}=r^{7/2}$$. What is the order of LHS of $$(\ast)$$?

The answer is yes. The inequality (*) is true. Eriksson-Bique, et al.  proved that on power cusp domains $$M^{1,p}=W^{1,p}$$, where $$W^{1,p}$$ is the classical Sobolev space and $$M^{1,p}$$ is the space of all $$u\in L^p$$ such that $$(1)\qquad |u(x)-u(y)|\leq d(x,y)(g(x)+g(y)) \quad a.e.$$ for some $$0\leq g\in L^p$$.

The spaces $$M^{1,p}$$ has been introduced in , see also  for a more detailed exposition.

If $$u\in W^{1,p}$$ on $$D$$, then by the result of , (1) is satisfied. Integrating (1) we obtain $$\int_B |u-u_B|\leq Cr\int_B g$$ Now Lemma 6 in  yields that $$\int_B |u-u_B|\leq C'r\int_B |\nabla u|$$ which proves inequality (*).

 P. Hajlasz, Sobolev spaces on metric-measure spaces. (Heat kernels and analysis on manifolds, graphs, and metric spaces (Paris, 2002)), 173--218, Contemp. Math. , 338, Amer. Math. Soc., Providence, RI, 2003.

 P. Hajlasz, A new characterization of the Sobolev space. (Dedicated to Professor Aleksander Pelczynski on the occasion of his 70th birthday.) Studia Math. 159 (2003), 263--275.

 P. Hajlasz, Sobolev spaces on an arbitrary metric space, Potential Analysis , 5 (1996), 403--415.

 S. Eriksson-Bique, P. Koskela, J. Maly, Z. Zhu, Pointwise inequalities for Sobolev functions on outward cuspidal domains. arXiv:1912.04555.

• The content of the question has changed and I am sorry, but is $M^{1,p}=W^{1,p}$ holds for the following domain $D$? $D=S \cup \bigcup_{n=2}^{\infty}R_n$, where $S=\{u+iv \mid |u|<1,\ |v|<1\}$ and $R_n=\{u+iv \mid 0 \le u-1 <n2^{-n} \log 2,\ |v-v_n|<2^{-n}\}$. This is also a typical example of non-uniform domain. Jul 23, 2019 at 12:02