Let $\Omega \subset \mathbb{R}^n$ be an open open subset. Let $u,v\colon \Omega\to \mathbb{R}$ be two functions such that at least one of them is compactly supported. Assume each of $u$ and $v$ can be presented as a difference of two bounded subharmonic functions in $\Omega$. Thus in particular the distributional Laplacians $\Delta u,\Delta v$ are well defined as signed measures on $\Omega$.
Question. Is it true that
$$\int_\Omega u(x)\Delta v(x) dx=\int_\Omega v(x)\Delta u(x) dx?$$
Remark. (1) The expressions under the both integrals are well defined as signed measures with compact support. Thus both sides make sense.
(2) The simplest unknown to me case is $n=2$.
Best Answer
Without loss of generality, $u$ has compact support $K\subset\Omega$. Therefore the (signed) measure $\Delta u$ is supported in $K$ as well. Let $(\phi_k)$ be a sequence of smooth (radial) mollifiers such that $\phi_k*u$ is supported in $K^\delta$ (the closed $\delta$ neighborhood of $K$, with $\delta>0$ so small that $K^\delta\subset\Omega$). Additionally, suppose $0\le\phi_k$, $\int_{\Bbb R^n}\phi_k(x)\phantom{!}dx=1$, and $\phi_k$ is supported in the ball $B(0,\delta/k)$, for each $k\ge 1$. Then $\lim_k \phi_k*u=u$ pointwise and boundedly, because (for example) $u$ is finely continuous. Likewise $\lim_k\phi_k*v=v$. Then $$ \eqalign{ \int_\Omega u(x)\cdot\Delta v(dx) &=\lim_k\int_\Omega (\phi_k*u)(x)\cdot\Delta v(dx)\cr &=\lim_k\int_\Omega \Delta(\phi_k*u)(x)\cdot v(x) \phantom{b}dx\cr &=\lim_k\int_\Omega (\phi_k*\Delta u)(x)\cdot v(x) \phantom{b}dx\cr &=\lim_k\int_\Omega (\phi_k*v )(x)\phantom{b}\Delta u(dx)\cr &=\int_\Omega v(x)\cdot\Delta u(dx).\cr } $$ Here the second equality is just the definition of $\Delta v$ as a distribution; the third likewise; the fourth is Fubini.
(Edited per suggestion of makt, to fix the vacuity noticed by Mateusz Kwaśnicki.)