Let $X=(X_1,\cdots,X_m)$ be a system of Hörmander vector fields defined on $\mathbb{R}^n$. The Sobolev space $W_{X}^{1,p}(\Omega)$ is defined by $$W_{X}^{1,p}(\Omega):=\{u\in L^p(\Omega)|X_iu\in L^p(\Omega),~i=1,...,m\}$$ endowed with the norm $$\|u\|_{1,p}=\|u\|_{L^p(\Omega)}+\|Xu\|_{L^p(\Omega)}:=\|u\|_{L^p(\Omega)}+\sum_{i=1}^m\|X_iu\|_{L^p(\Omega)}.$$ Here $Xu$ is understood in the weak sense. We denote $W_{X,0}^{1,p}(\Omega)$ the closure of $C_0^\infty(\Omega)$ under the norm $\|\cdot\|_{1,p}$.
Question: Suppose $\Omega\subset\mathbb{R}^n$ is an open bounded domain. Let $v\in W_{X,0}^{1,p}(\Omega)$ with $\|Xv\|_{L^p(\Omega)}=0$, can we prove $v\equiv0$ in $\Omega$? (If not, what conditions can be added to obtain $v\equiv 0$?)
Attempt: We first have $Xv=0$ a.e. in $\Omega$, then we can extends $v$ with zero outside $\Omega$. One can check $v\in W_{X,0}^{1,p}(\mathbb{R}^n)$ with $Xv=0$ a.e. in $\mathbb{R}^n$. Next we have $$\int_{\mathbb{R}^n}Xv\cdot X\phi dx=0,~~~\forall \phi\in C_0^\infty(\mathbb{R}^n).$$$$\int_{\mathbb{R}^n}v \sum_{i=1}^mX_i^*X_i\phi dx=-\int_{\mathbb{R}^n}Xv\cdot X\phi dx=0,~~~\forall \phi\in C_0^\infty(\mathbb{R}^n).$$ Then I want to use the hypoellipticity of $\triangle_X:=\sum_{i=1}^mX_i^*X_i$$\triangle_X:=-\sum_{i=1}^mX_i^*X_i$, which implies $u\in C^\infty(\mathbb{R}^n)$ then things become easy. However I am not sure in this case one can treat $u$ a weak solution of $\triangle_Xu=0$ or not, though $u$ is a distribution.