Dense properties of weighted Sobolev space define on $\mathbb{R}^n$

Question

Consider the usual Sobolev space $H^1(\mathbb{R}^n)$ and $H^1_0(\mathbb{R}^n)$, where $H^1_0(\mathbb{R}^n)$ is the closure of $C_0^\infty(\mathbb{R}^n)$ with respect to the norm of $H^1(\mathbb{R}^n)$. We know $H^1(\mathbb{R}^n)=H^1_0(\mathbb{R}^n)$. Now let $X=(X_1,\cdots,X_m)$ be a system of H"{o}rmander vector fields on $\mathbb{R}^n$. Consider the weighted Sobolev space: $$H_{X}^1(\mathbb{R}^n):=\{u\in L^2(\mathbb{R}^n)|X_iu\in L^2(\mathbb{R}^n),~i=1,\cdots,m\},$$ where $X_iu$ is understood in the disbitrution sense. I want to know whether $H_X^1(\mathbb{R}^n)=H^1_{X,0}(\mathbb{R}^n)$ or not? So far, I know for the vector fields with bounded coefficients this fact still holds, or for the vector fields with some homogeneous conditions. But what about general case? Could one give a counterexample that indicates $H_X^1(\mathbb{R}^n)\neq H^1_{X,0}(\mathbb{R}^n)$?

Here I suggest consider a very simple case: $\bar{X}=e^{x^2}\partial_x$ in one-dimension case. But I stuck in constructing a function $u$ in $H^1_{\bar{X},0}(\mathbb{R}^n)$ but $u\notin H^1_{\bar{X}}(\mathbb{R}^n)$.

Answer 1

I think in dimension 1 you won't be able to produce a counterexample, see §4 of

V. V. Zhikov, "Weighted Sobolev spaces", Mat. Sb., 189:8 (1998), 27–58; Sb. Math., 189:8 (1998), 1139–1170

For some counterexamples, see for instance this paper :

V., Chiado Piat; Serra Cassano, Francesco, "Some remarks about the density of smooth functions in weighted Sobolev spaces", Journal of Convex Analysis v. 1, n. 2 (1994), p. 135-142