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In this paper we have the following situation on page 60. $E$ is a compact subset of $\mathbb{R}^\tau\cup\{\infty\}$ (one point compactification) for $\tau\geq2$, $M_0$ is a point in the boundary of $E$ and $u$ is subharmonic on an open neighborhood $V$ of $M_0$. M. Brelot says that we can extend $u$ to a function that is subharmonic on a neighborhood of$E$, that he steel calls $u$, by "modifying properly $u$ outside $V$". Can someone explains the detail of this extension?

By the way a function $v$ on an open set $W$ of $\mathbb{R}^\tau\cup\{\infty\}$ is called subharmonic if: either $W$ does not contain $\infty$ and in this case $v$ is subharmonic on $W$ in the usual sense, or $W$ contains $\infty$ and then 1) $v$ is upper semicontinuous at $\infty$ and 2) $v(\infty)\leq$ the mean integral value of $u$ over any ball $B$ that is relatively compact in $W$.

Recall that the extension of a subharmonic function to $\mathbb{R}^\tau$ is classic. If $u$ is subharmonic on a neighborhood of $\overline{V}$ (closure), where $V$ is a bounded open set in $\mathbb{R}^\tau$ such that $\mathbb{R}^\tau\setminus\overline{V}$ is connected, then there exists a subharmonic function $\bar{u}$ on $\mathbb{R}^\tau$ that coincides with $u$ on $\overline{V}$ (see Armitage and Gardiner, "classical potential theory", pg192).

But Brelot extends $u$ to an open set of $\mathbb{R}^\tau\cup\{\infty\}$ containg $E$. How??

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  • $\begingroup$ If $E$ does not contain $\infty$, you know how to extend to $R^n$. If $E$ contains $\infty$, suppose that $x_0\not\in E$, and perform the Kelvin transform which sends $x_0$ to $\infty$. If $E=R^n\cup\{\infty\}$ then there is no boundary point and no argument is needed. $\endgroup$ – Alexandre Eremenko Feb 23 at 12:37
  • $\begingroup$ Consider the case of $E$ containing $\infty$. Suppose the ball $B(x_0,r)\subset \complement E\cap V$ ($\complement E$=complement of $E$) and $u^*$ is the Kelvin transfrom of $u$ relative to the sphere $\partial B(x_0,r)$. If I am following correctely, $u^*$ is subharmonic on a neighborhood of $E$, but the restriction of $u^*$ to $V$ does not coincide with $u$. I need the restriction of the extension to coincide with $u$ on $V$ or $\overline{V}\cap E$ (assume $u$ is subharmonic on a neighborhood of $\overline{V}$) . $\endgroup$ – M. Rahmat Feb 23 at 21:50
  • $\begingroup$ If I understand correctly (unfortunately I do not speak French), Brelot refers here to one of his previous papers: Sur le rôle du point à l'infini dans la théorie des fonctions harmoniques, Ann. Éc. Norm. sup., t. 61, 1944. I did not follow this reference, but I guess the argument is not much different from the classical one. $\endgroup$ – Mateusz Kwaśnicki Feb 26 at 8:13
  • $\begingroup$ Yes, he refers to part 20, "potential representation", of the paper you cited. But I only found there a generalization of the classic result that a subharmonic function can be locally represented as the sum of a potential and a harmonic function. I didn't get what does it have to do with the extension of subharmonic functions... $\endgroup$ – M. Rahmat Feb 27 at 0:52

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