Let $\widehat{\mathbb R^ d}=\mathbb R^ d\cup \{ \infty \} $ have the one-point compactification topology. We use Helms's convention: superharmonicity at an interior infinity means lower semicontinuity there and $u(\infty )\ge M_{\partial B}u$ for every ball whose closed exterior lies in the domain. Harmonicity means the corresponding equality and both semicontinuities, in addition to ordinary harmonicity at finite points. This extra mean condition is part of the definition, and is not implied by ordinary bounded exterior harmonicity in dimension at least three.
Theorem 4.4.11.E16C09608A00 Let $B=B(a,R)$ and let $f\in L^1(\partial B,d\sigma )$. Put $m=(s_{d-1}R^{d-1})^{-1}\int f d\sigma $ and $r=|x-a|$. The exterior solution in this compactified convention is
\[ H^{\rm ext}_ Bf(x)=\frac1{s_{d-1}R}\int _{\partial B} \left[\frac{r^2-R^2}{|x-z|^ d}+\frac1{R^{d-2}}-\frac1{r^{d-2}}\right] f(z)d\sigma (z),\qquad r>R, \]
with $H^{\rm ext}_ Bf(\infty )=m$. It is harmonic including at infinity, positive when $f\ge 0$, sends one to one, and tends to $f$ at every continuity point of $f$ on the sphere. More generally an ordinary exterior harmonic $h$, continuous to the sphere with finite limit $\alpha $ at infinity, satisfies
\[ h(x)=K\bigl (H_ B(h|_{\partial B})\bigr )(x) +\alpha \bigl [1-(R/r)^{d-2}\bigr ]. \]
For $d=2$ necessarily $\alpha =m$; for $d\ge 3$, it is harmonic in Helms's sense at infinity exactly when $\alpha =m$.
Proof.
Invert about the sphere and apply Kelvin covariance. For $d\ge 3$, the Kelvin transform of bounded $h$ is bounded in absolute value by a multiple of the fundamental pole near $a$. Adding a sufficiently large multiple of that pole makes it nonnegative, so the proved Bôcher lemma applies. Its singular coefficient is $\alpha R^{d-2}$, by taking the limit back at infinity. Subtracting this singularity leaves a harmonic function on $B$; its continuous boundary data identify it by the ball Poisson formula, giving the displayed representation. In dimension two the Kelvin transform is bounded near the center and extends harmonically there, with center value $\alpha =m$.
For integrable $f$, define the right side with $\alpha =m$. The inverted ball Poisson integral is harmonic away from $a$ and its derivatives are controlled on compact subsets by the integrable boundary density. The inversion distance identity gives precisely the displayed kernel. Its first term is positive; for $d\ge 3$ so is the correction $R^{2-d}-r^{2-d}$, and for $d=2$ that correction is zero. Evaluating at $f=1$ shows the total kernel mass is one. Its limit at infinity is $m$, and the ball Poisson continuity-point estimate, together with the vanishing correction at $r=R$, proves the boundary assertion.
The Kelvin identity for concentric spherical means gives, for a general finite-limit $h$,
\[ M_{\partial B(a,t)}h=\alpha +(m-\alpha )(R/t)^{d-2}\quad (d\ge 3), \]
and the constant $m$ if $d=2$. For the chosen $\alpha =m$ these means are constant, so their radial flux is zero. The divergence theorem on the region between any large sphere containing $\overline B$ and a larger concentric sphere transfers this zero flux to the arbitrary sphere. The annular mean derivative formula then makes its subsequent spherical means constant; their limit is the limit at infinity, namely $m$. Thus every exterior-ball mean equals $h(\infty )$, as required. Conversely Helms's mean condition for the concentric spheres forces $\alpha =m$.
$\square$
Theorem 4.4.12.8FA95EB71F47 On a connected compactified open set a superharmonic function attaining its infimum is constant. On a proper open subset, nonnegative boundary lower limits imply nonnegativity. Entire superharmonic functions on $\widehat{\mathbb R^ d}$ are constant. Positive scaling, sums and minima preserve superharmonicity. Harmonic lowering is valid on interior balls and on exterior balls, directed harmonic convergence holds, and a nonempty saturated family under both kinds of lowering has harmonic or $-\infty $ infimum on each component. The point infinity is polar for $d=2$ and is not polar for $d\ge 3$.
Proof.
At finite points the minimum-set argument is the checked Euclidean one. If the minimum is attained at infinity, each large sphere mean is at least the minimum and at most the value there. Lower semicontinuity and positivity of the surface measure of open patches force equality at every point of each such sphere. Thus the minimum set contains an exterior neighborhood and is open and closed. For a proper open set each component has a nonempty boundary in the connected compact ambient space. Extend by boundary liminf; compactness forces a negative minimum to be attained inside if any negative value exists. The strong minimum principle would make the component a negative constant, contradicting its nonnegative boundary limits. For the entire compact space a minimum is attained and the function is constant; one cannot infer a nonnegative constant from an empty boundary.
For algebraic operations use the ordinary local result and the linear sphere inequalities at infinity; nontriviality follows from finite almost everywhere values on each Euclidean component. For exterior lowering, increasing continuous minorants of the lsc sphere data and the positive exterior Poisson kernel give a harmonic minorant bounded above by $u$ via compactified comparison. Finite interior values exclude an infinite limit. The same boundary-liminf/pasting proof as for interior lowering makes the replacement superharmonic at finite points. Near infinity it is the exterior harmonic solution with constant large-sphere means equal to its value there. For every other allowed sphere center, the finite-point superharmonic mean is concave as a function of $r^{2-d}$ (or $-\log r$ in the plane), and is constant at that value for all sufficiently large radii, by the zero-flux argument above. Concavity bounds its earlier means by the same value, giving the required inequality at infinity. Directed harmonic limits reduce to the Euclidean result away from infinity and the positive exterior kernel on a fixed sphere near infinity; directed monotone integration identifies the value and harmonicity there. The saturated-family proof then applies in both kinds of neighborhoods, exactly as for finite balls, with nonemptiness explicitly required.
For $d=2$, $\log |x-a|$ on an exterior neighborhood, with value infinity at infinity, is a superharmonic pole function there. For $d\ge 3$, concentric sphere means of a superharmonic function on an exterior are finite and concave functions of $t=r^{2-d}$. A supporting affine upper bound at an interior $t$ bounds them above as $t\downarrow 0$. If the value at infinity were infinite, lsc would instead force all sufficiently distant sphere means to tend to infinity. This contradiction proves nonpolarity.
$\square$
Lemma 4.4.13.58BE7FB3C948 Kelvin transformation preserves superharmonicity on domains avoiding its center. If $u$ is superharmonic on $B(a,R)$, harmonic near $a$, and $u(a)=0$, its Kelvin transform, extended by zero at infinity, is superharmonic on the compactified exterior.
Proof.
On compact subdomains avoiding the center, transform harmonic comparison functions, use positivity of the Kelvin multiplier and its involutivity, and transform the comparison back. This proves the superharmonic comparison property for the transformed lsc function. In the normalized case the transform is harmonic near infinity with zero large-sphere means: the original function is harmonic near $a$ with zero center value, so the concentric mean identity and the flux transfer in the exterior theorem apply. In particular its value at infinity is zero and it is continuous there. For any other exterior ball in the domain, its sphere means are concave in $r^{2-d}$ (or $-\log r$ in the plane), and are zero for all sufficiently large radii. Concavity makes them nonpositive at the smaller allowed radii. This is exactly the supermean condition at infinity.
$\square$
Pome
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