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Let $n\geq3$ and $A\subset\mathbb{R}^n$ be a discrete subset. How to characterize all the positive harmonic function on $\mathbb{R}^n\setminus A$ ?

Rmk: This is from Exercise 17 in Chapter 3 of the textbook named 'Harmonic Function Theory' (GTM 137) by Axler, Bourdon, and Ramey. If $A = \{p_1,\dotsc,p_m\}$ is finite, It's easy to show the positive harmonic functions $V$ can only be of the form $$V = \sum_{i = 1}^m \frac{a_i}{|x - p_i|^{n-2}} + C$$ where $C>0$ is constant and $a_i >0$ (You could find one proof at Sheldon Axler's answer to Characterize a positive harmonic functions with multi-singularities). As Noam commented, if the sum $$V = \sum_{i = 1}^\infty \frac{a_i}{|x - p_i|^{n-2}} + C $$ converges for $a_i\geq0$, then $V$ is a solution. But it's harder to show they are the only solutions with singularities at $A$. Moreover, since Bôcher's theorem classifies positive harmonic function on punctured ball, I also want to know is there any possibility to classify positive harmonic function on any open set in $\mathbb{R}^n$ ?

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    $\begingroup$ I'd think that $C + \sum_{p\in A} a_p / \left|x-p\right|^{n-2}$ would still work as long as the sum converges $-$ and the harder part would be proving that these are the only solutions. $\endgroup$ Commented Feb 22 at 23:32
  • $\begingroup$ Do you want $U:=\Bbb R^b\setminus A$ to be open? Else what will "harmonic" mean at a point $x\in U$ that is not interior to $U$? $\endgroup$ Commented Feb 22 at 23:51
  • $\begingroup$ @JohnDawkins $A$ is discrete by assumption, hence closed. $\endgroup$ Commented Feb 23 at 2:26
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    $\begingroup$ @Alexandre Eremenko, you are right, thank you to point out my mistakes. I have corrected them. $\endgroup$ Commented Feb 23 at 15:31
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    $\begingroup$ Matemáticos Chibchas: Usually, discrete means that each point of $A$ is isolated. If $x\in\Bbb R^n$ with $\|x\|\not=0$, then $A:=\{n^{-1}\cdot x: n=1,2,\ldots\}$ is discrete, but not closed. $\endgroup$ Commented Feb 23 at 17:54

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Let $n$ be the dimension of the space. When $n=2$ such functions do not exist (except constants). When $n\geq 3$ every positive harmonic function in $R^n\backslash \{ p_k\}$ is of the form $$u(x)=C+\sum_k\frac{a_k}{|x-p_k|^{n-2}},\quad a_k\geq 0$$ where the series is convergent, that is $$ \sum \frac{a_k}{|p_k|^{n-2}}<\infty. \label{488337_1}\tag{1} $$ For the proof, first notice that in a neighborhood $V_k$ of each $p_k$ we must have $$u(x)=v_k(x)+a_k|x-p_k|^{2-n},\quad x\in V_k,$$ where $v_k$ is harmonic, and $a_k>0$. This follows from the local theorem that you cite. So $u(p_k)=+\infty$ (or $u$ is harmonic in $V_k$). Therefore the function $-u$ is subharmonic in the whole space and bounded from above (in fact $<0$). The structure theorem for bounded subharmonic function is given in Theorem 3.20 of the book

W. Hayman, Subharmonic functions, I, Academic Press, 1976.

and it gives us the desired representation. The main point of the proof is convergence of \eqref{488337_1}, and this is derived from Jensen's inequality.

When $n=2$ the same argument shows that all such functions must be constant, no matter whether $\{ p_k\}$ is finite or infinite, since all subharmonic functions in $R^2$ bounded from above are constant.

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  • $\begingroup$ @John Dawkins: nothing. It is harmonic at $0=(0,0,0)$. $\endgroup$ Commented Feb 27 at 4:31
  • $\begingroup$ My real question is this: In this instance, what does "harmonic at $(0,0,0)$" mean? $\endgroup$ Commented Feb 27 at 17:36
  • $\begingroup$ I do not understand what's special about $(0,0,0)$. "Harmonic" means "satisfies Laplace equation" as in any other point. $\endgroup$ Commented Feb 28 at 14:34
  • $\begingroup$ Doesn't $\Delta u(0,0,0) =0$ require $u$ to be defined in a neighborhood of $(0,0,0)$? $\endgroup$ Commented Feb 28 at 22:47
  • $\begingroup$ In your example in the first comment it is defined at $(0,0,0)$: your points $p_k$ do not include $(0,0,0)$. $\endgroup$ Commented Mar 1 at 13:56

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