# Comparing integrals of bounded subharmonic functions

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 $$n=2$$.

• @Christian Remling. You are right. But then it is always true, by approximation: choose shooth $v_n$ converging to $v$ in $D'$. Aug 7 at 14:14
• @AlexandreEremenko: I had the same thought, but is it clear: if $v_n\to v$ in $\mathcal D'$, then I only know that $\int u\Delta v_n\to\int u\Delta v$ if $u$ is also smooth. Aug 7 at 14:16
• @Christian Remling: I think justifying the limit in your second comment is a purely technical problem, not very hard. See Landkof, Introduction to modern potential theory, Ch. II, Potentials with finite energy. Aug 7 at 14:34
• @makt: But $\Delta v$ can have a point mass, while $u$ can be infinite at this point. Moreover, differences of subharmonic functions are in general not everywhere defined: in $u=u_1-u_2$, both $u_j$ can be $-\infty$ at some point. And $\Delta v$ can have a point mass at this point. So there are problems with definition of these integrals. The probems disappear if you consider "potentials of finite energy" instead of differences of subharmonic functions. Aug 7 at 14:57
• @AlexandreEremenko : I explicitly stated that both functions are bounded.
– makt
Aug 7 at 15:31

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.
• Good point. Strike the first sentence of my answer (save the stipulation that $u$ be of compact support), and remove "positive" from the second sentence. Aug 8 at 16:13