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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 $n=2$.

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  • $\begingroup$ @Christian Remling. You are right. But then it is always true, by approximation: choose shooth $v_n$ converging to $v$ in $D'$. $\endgroup$
    – Alexandre Eremenko
    Aug 7, 2021 at 14:14
  • $\begingroup$ @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. $\endgroup$
    – Christian Remling
    Aug 7, 2021 at 14:16
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    $\begingroup$ @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. $\endgroup$
    – Alexandre Eremenko
    Aug 7, 2021 at 14:34
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    $\begingroup$ @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. $\endgroup$
    – Alexandre Eremenko
    Aug 7, 2021 at 14:57
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    $\begingroup$ @AlexandreEremenko : I explicitly stated that both functions are bounded. $\endgroup$
    – asv
    Aug 7, 2021 at 15:31

1 Answer 1

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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.)

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    $\begingroup$ The only bounded, subharmonic and compactly supported function is constant zero. :-) $\endgroup$
    – Mateusz Kwaśnicki
    Aug 7, 2021 at 19:57
  • $\begingroup$ 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. $\endgroup$
    – John Dawkins
    Aug 8, 2021 at 16:13
  • $\begingroup$ @JohnDawkins : Would you like to edit your answer? $\endgroup$
    – asv
    Aug 9, 2021 at 8:00

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