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 $\alpha =m$.
$\square$
Pome
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