# A question on the problem of Dirichlet

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?

• 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 $\mathbb{R}^n$. Jan 27, 2020 at 8:17
• @Mateusz Kwasinski: $\infty$ is different when $n\geq 3$. Jan 27, 2020 at 17:47
• @AlexandreEremenko: In what sense? Kelvin transform allows one to transform "problems at infinity" into "problems at the origin". Jan 27, 2020 at 18:42
• @Mateusz Kwasnicki: Yes, but Kelvin's transform looks somewhat different in $n=2$ and $n\geq 3$. Jan 27, 2020 at 23:45

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.