Some authors define superharmonicity at infinity in the following way. A function $u$ is superharmonic on an open set $V\subset\mathbb{R}^m\cup\{\infty\}$ (one point compactification), containing infinity, if it is superharmonic on $V\setminus\{\infty\}$ in the regular way, and at infinity, $u$ is lower semicontinuous and $u(\infty)$ is bounded below by the average of $u$ over any sphere $S(y,r)$, if the complement of the ball $B(y,r)$ lies in $V$ (see *Introduction to potential theory*, by Helms, pg. 198).


Some other authors give a different definition. In this new definition, a function $u$ is superharmonic on $V$ if it is superharmonic on $V\setminus\{\infty\}$ and if the Kelvin transform $u^\*$ of $u$, that is already superharmonic on $V^\*\setminus\{\infty\}$, can be defined at $0$ so that it is superharmonic on $V$ (see *[Classical Potential Theory](http://fourier.math.uoc.gr/%7Epapadim/potential_theory.pdf)*, by Papadimitrakis, available freely on the web, pg. 154). Here,
$$u^\*(x)=\left(\frac{1}{|x|}\right)^{m-2}u(x^\*)$$ and $x^\*=\frac{1}{|x|^2}x.$


It seems that these two defintions are not equivalent for $m\geq3$, because constant functions are superharmonic everywher according to the first definition but according to the second definition the only constant function that is superharmonic is $\equiv0$.


My questions are: suppose $m\geq3$.


1. Are there some conditions under which these two definitions are equivalent?
2. Does the second definition implies the first?



Superharmonicity at infinity
1. None of the definitions implies the other:


	* The function $|x|^{2-m}$ is superharmonic (in fact: harmonic) in $\mathbb R^m \cup \{\infty\} \setminus \{0\}$ according to the second definition, but it is not according to the first one.
	* Conversely, the function $-1$ is superharmonic (in fact: harmonic) in $\mathbb R^m \cup \{\infty\} \setminus \{0\}$ in the sense of the first definition and it is not according to the second one.
2. If we assume that $u \geqslant 0$, then every $u$ superharmonic in $\mathbb R^m \setminus K$ for a compact $K$ is superharmonic in $\mathbb R^m \cup \{\infty\} \setminus K$ according to the second definition, so for *positive* superharmonic function, the first definition clearly implies the second one (but not vice versa).
3. I believe things become clearer if one writes both definitions in terms of the Kelvin transform $u^\*(x) = |x|^{2 - m} u(|x|^{-2} x)$.


	* The second definition requires that $u^\*$ is superharmonic in a neighbourhood of $0$, so in a sufficiently small ball $B\_r$ we have
	$$ u^\*(x) = \int\_{B\_r \setminus \{0\}} (|x - y|^{2 - m} - |y|^{2 - m}) \mu(dy) + (h(x) - h(0)) + a |x|^{2 - m} + b$$
	for some $\mu \geqslant 0$, some $h$ harmonic in $B\_r$, some $a \geqslant 0$ and some $b \in \mathbb R$.
	* On the other hand, the first definition requires that $u^\*$ is superharmonic in a punctured neighbourhood of $0$, and $|x|^{m - 2} u^\*(x)$ satisfies the "super-mean-value property at $0$", so in a sufficiently small ball $B\_r$ we *should* have
	$$ u^\*(x) = \int\_{B\_r \setminus \{0\}} (|x - y|^{2 - m} - |y|^{2 - m}) \mu(dy) + (h(x) - h(0)) + a |x|^{2 - m} + b$$
	for some $\mu \geqslant 0$, some $h$ harmonic in $B\_r$, some $a \in \mathbb R$ and some $b \leqslant 0$. Note: I might be wrong here, I did not have time to think about this carefully.The difference is of course in the admissible range of $a$ and $b$.




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*Edit: Some additional comments to the above item 3.*


(A) If $u^\*$ is superharmonic in a ball $B\_R$, then in a smaller ball $B\_r$ we have
$$ u^\*(x) = \int\_{B\_r} |x - y|^{2 - m} + h(x) $$
for some harmonic $h$; this is the [Riesz decomposition theorem](https://encyclopediaofmath.org/wiki/Riesz_theorem(2)). If we take out the atom at $0$, then we can write
$$ u^\*(x) = \int\_{B\_r \setminus \{0\}} |x - y|^{2 - m} \mu(dy) + h(x) + a |x|^{2 - m} $$
for some $a \geqslant 0$, and this is equivalent to what I wrote above.


(B) Suppose now that $u^\*$ is the Kelvin transform of a function superharmonic at infinity in the sense of the first definition. Then $u^\*$ is superharmonic in $B\_R \setminus \{0\}$, and since $u$ is bounded from below by a constant $-M$ in a neighbourhood of $\infty$, we have
$$ u^\*(x) + M |x|^{2 - m} \ge 0 $$
in some ball $B\_R$. This implies that $u^\*(x) + M |x|^{2 - m}$ is superharmonic in $B\_R$, not just $B\_R \setminus \{0\}$. By the same argument as in (A) we find that
$$ u^\*(x) = \int\_{B\_r \setminus \{0\}} |x - y|^{2 - m} \mu(dy) + h(x) + a |x|^{2 - m} $$
for some harmonic $h$, but this time $a$ need not be nonnegative (we only know that $a \ge -M$, but $M$ can be arbitrarily large). This leads us to the expression given in the original answer, with an arbitrary real $b$.


Now why in fact we need $b \leqslant 0$? I did not check this carefully, but the average of $|x|^{m - 2} u^\*(x)$ over a small ball $B\_s$ seems to be equal to $a + c\_m b s^{m - 2} + o(s^{m - 2})$ for an appropriate constant $c\_m > 0$, and this is no greater than the limit $a$ only if $b \leqslant 0$.


The above shows that $b \leqslant 0$ is necessary. The next question is whether is is also sufficient. I guess it is, and this should be fairly easy to verify, but I have to stop here for the time being.


