Let $U$ be an open set in $\mathbb{R}^{n}$ with $n\geq2$ and $V$ an open set containing the boundary $\partial U$ of $U$. Suppose $u$ is subharmonic on $V$. We know that the generalized solution of Dirichlet problem $H^{U}_{u}(x)$ exists, i.e. is a harmonic function on $U$ that $\to u(y)$ as $x\to y$, for all regular point $y\in\partial U$. My question is: can we say that $$u(x)\leq H^{U}_{u}(x)$$ for all $x\in V\cap U$? Notice that the answer would be yes, if $u$ was subharmonic on $U$, and not on a neighborhood of the boundary of $U$ only.

## 1 Answer

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The answer is no. E.g., let $U:=\{x\in\mathbb R^n\colon|x|<1\}$, $V:=\{x\in\mathbb R^n\colon1/2<|x|<2\}$ (or $V:=\mathbb R^n\setminus\{0\}$), $u(x):=|x|^{2-n}-1$ for $x\in V$ if $n\ge3$, and $u(x):=-\ln|x|$ for $x\in V$ if $n=2$. Then $u$ is harmonic and hence subharmonic on $V$. However, $u>0=H_u^U$ on $V\cap U$.