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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 $$\int_{\mathbb{R}^n}v \sum_{i=1}^mX_i^*X_i\phi dx=-\int_{\mathbb{R}^n}Xv\cdot X\phi dx=0,~~~\forall \phi\in C_0^\infty(\mathbb{R}^n).$$ Then I want to use the hypoellipticity of $\triangle_X:=-\sum_{i=1}^mX_i^*X_i$, 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.

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There exist $c>0$ and $s>0$, such that for all smooth functions $v$ compactly supported in $\Omega$, $$ \sum_{1\le j\le m}\Vert X_jv\Vert_{L^2}\ge c\Vert v\Vert_{W^{s,2}}. $$ The largest (i.e. the best) possible $s$ is related to the number of brackets of the $X_j$ necessary to generate the tangent bundle. In fact $$ s=\frac1{k+1}, $$ $\bullet$ $k=0$ in the elliptic case (the $X_j$ generate the tangent bundle),

$\bullet$ $k=1$ if the $X_j, [X_j,X_l]$ generate the tangent bundle,

$\bullet$ $k=2$ if the $X_j, [X_{j_1},X_{j_2}], [X_{j_1},[X_{j_2},X_{j_3}] ]$ generate the tangent bundle. Etc.

I believe that this inequality and some Friedrichs regularization argument should suffice to prove your sought result.

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  • $\begingroup$ Thanks for your answers. But what I worry about is whether $v$ can be a weak solution of $-\triangle_Xu=0$ in the distribution sense, i.e., $\langle v,-\triangle_X\phi\rangle$ for $\phi\in C_0^\infty(\mathbb{R}^n)$. I think this should be correct since the condition $Xv=0$ a.e. is too strong. By the definition of weak derivative we obtain $-\triangle_Xu=0$ in the distribution sense. Then $v$ smooth and we can prove the conclusion. $\endgroup$
    – Houa
    Commented Feb 21 at 8:39

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