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Suppose $U$ is an open set in $\mathbb{R}^{n}$ ($n\geq2$) whose complementary is not polar, and $f$ is a real-valued function defined at least on the boundary of $U$. We know that the generalized Dirichlet problem has a solution, i.e., if $f$ is continuous and all boundary points of $U$ are regular, there is a function $H_{f}^{U}(x)$ that is harmonic on $U$ and tends to $f(y)$, as $x\to y$, for all $y$ in the boundary of $U$. The same is also true if $f$ is superharmonic, but not necessarily continuous, on a neighborhood of the boundary of $U$.

Suppose now that $U$ is an open set in $\mathbb{R}^{n}\cup \{\infty\}$, the one point compacification of $\mathbb{R}^{n}$. Do the same results hold for $f$ continue , or $f$ superharmonic (not necessarily continuous)? Do you know reference for that stuff?

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    $\begingroup$ Not sure if I understand correctly, but I bet the answer is yes: if one looks at harmonic functions, the one-point compactification of $\mathbb{R}^n$ is just the unit sphere in $\mathbb{R}^{n+1}$ (via stereographic projection), and so $\infty$ is no different from any other point of $\mathbb{R}^n$. $\endgroup$ – Mateusz Kwaśnicki Jan 27 at 8:17
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    $\begingroup$ @Mateusz Kwasinski: $\infty$ is different when $n\geq 3$. $\endgroup$ – Alexandre Eremenko Jan 27 at 17:47
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    $\begingroup$ @AlexandreEremenko: In what sense? Kelvin transform allows one to transform "problems at infinity" into "problems at the origin". $\endgroup$ – Mateusz Kwaśnicki Jan 27 at 18:42
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    $\begingroup$ @Mateusz Kwasnicki: Yes, but Kelvin's transform looks somewhat different in $n=2$ and $n\geq 3$. $\endgroup$ – Alexandre Eremenko Jan 27 at 23:45
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I found a book that talks about the problem of Dirichlet for unbounded regions. This is Lester L. Helms' book on " potential theory", Springer, 2009, Chapter 5. According to this book, the answer to the first question is yes; i.e. if $f$ is continuous on the boundary of $U$ then $H^{U}_{f}(x)$ is harmonic and tends to $f(y)$, as $x\to y$, for all regular point $y$ in the boundary of $U$. But steel I do not know the answer to the second question.

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