# Superharmonicity at infinity

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, 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?

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 $$b$$.

(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. 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.
• I added some details, but this is still not complete. Also, there were some terrible typos in the previous version ($b \geqslant 0$ instead of $b \leqslant 0$, $u$ instead of $u^\star$), sorry. Mar 13, 2021 at 14:02