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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?

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1 Answer 1

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  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 $b$.


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. 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.

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  • $\begingroup$ I am sorry, but I didn't get where the integrals and the equal signs come from? Could you please explain? $\endgroup$
    – M. Rahmat
    Mar 10, 2021 at 16:22
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    $\begingroup$ You mean in item 3? I can elaborate if you like, but unfortunately that has to wait a bit, most likely until the end of this week. $\endgroup$ Mar 10, 2021 at 16:34
  • $\begingroup$ Don't worry about this, I will wait. Thanks. $\endgroup$
    – M. Rahmat
    Mar 10, 2021 at 18:24
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    $\begingroup$ 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. $\endgroup$ Mar 13, 2021 at 14:02

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