Suppose $B$ is a ball in $\mathbb{R}^{m}$ and $u$ and $s$ are subharmonic on $B$. Suppose there is a closed subset $F$ of the closure of $B$ with no interior such that $v=u-s$ is subharmonic on $B\setminus F$. Is there a way to prove that $v$ is subharmonic on $B$? Notice that if $u$ and $s$ are $C^{2}$-smooth, the solution is easy: for each $x\in F$ we may take a sequence $(x_{n})$ in $B\setminus F$ that converges to $x$. Then by taking the limit of $\Delta v(x_{n})\geq0$ we can prove the case. But what if $u$ and $s$ are not smooth?

  • $\begingroup$ Something is missing in the statement. What if $F=B$, for example? $\endgroup$ Jul 6 '19 at 21:14
  • $\begingroup$ Sorry! I corrected. $\endgroup$
    – M. Rahmat
    Jul 6 '19 at 21:45
  • 2
    $\begingroup$ This is not correct: take $u(x)=0,s(x)=|x_1|$, where $x=(x_1,x_2)$ both subharmonic, $v(x)=-|x_1|$ is harmonic away from the line $F=\{ x:|x_1|=0\}$. $\endgroup$ Jul 7 '19 at 5:05

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