Let $D (\mathbb R^d)$ is the space of test functions and $D' (\mathbb R^d)$ that of distributions. For $u \in D' (\mathbb R^d)$, its distributional Laplacian $\Delta u \in D' (\mathbb R^d)$ is defined by $$ _{D'} \langle \Delta u, \varphi \rangle_D := {_{D'}} \langle u, \Delta \varphi \rangle_D , \quad \varphi \in D (\mathbb R^d). $$
Assume that $\Delta u$ has a representation in $L^2 (\mathbb R^d)$, i.e., there is $f \in L^2 (\mathbb R^d)$ such that $$ _{D'} \langle \Delta u, \varphi \rangle_D := \int_{\mathbb R^d} f (\Delta \varphi), \quad \varphi \in D (\mathbb R^d). $$
Does $u$ have a representation in $H^1 (\mathbb R^d)$?
Thank you so much for your help!
Not necessarily. One easy way to construct a counterexample is to take $v \in H^1(\mathbb{R}^d)$ and then consider the distribution $u = 1+ v$. Then the distributional Laplacian of $u$ coincides with distributional Laplacian of $v$ since $\Delta 1 = 0$. As a result, $\Delta u$ has such a representation but we clearly do not have $u \in L^2(\mathbb{R}^d)$.