3.3.15 Harmonic reflection across a sphereA9175B0B7BA5

Proof. This is a short consequence of the checked local Riesz decomposition. On a sufficiently small ball define $u=h+\Phi (\cdot -a)$ off $a$ and $u(a)=+\infty $. The fundamental function tends to infinity, and $h\ge 0$; thus $u$ is lsc and is superharmonic, using locality off $a$ and the trivial supermean inequality at $a$. It is locally integrable by the checked local integrability theorem. Its positive Riesz measure $-\Delta u$ is supported at $a$, so it equals $b\delta _ a$ with $b\ge 0$. Since $-\Delta \Phi =c_ d\delta _ a$, where $c_2=2\pi $ and $c_ d=(d-2)s_{d-1}$ otherwise, Weyl's lemma gives $u=(b/c_ d)\Phi +v$ almost everywhere. Equality holds everywhere off $a$ by continuity. Hence $h=(b/c_ d-1)\Phi +v$ there, and its nonnegativity near $a$ forces $c=b/c_ d-1\ge 0$. This also proves removability when $h$ is bounded, or when a one-sided bounded harmonic function has a finite limit along one approach to $a$. $\square$

Proof. First, odd reflection across a hyperplane preserves harmonicity for a harmonic function continuous and zero at that plane. The reflected function is continuous; at a center on the plane its spherical means vanish by the reflection symmetry, and away from the plane sufficiently small spheres have the usual harmonic mean property. The checked local mean characterization therefore proves harmonicity across the plane.

Translate $a=0$. Add a sufficiently large positive multiple of $\log (R/|x|)$ if $d=2$, or of $|x|^{2-d}-R^{2-d}$ otherwise, so that the result is nonnegative on the inner sphere. The minimum principle makes it nonnegative throughout the annulus. These added functions already extend harmonically across the outer sphere, so it suffices to treat $h\ge 0$. Choose $b$ on the outer sphere and invert about a sphere centered at $b$. The outer sphere minus $b$ becomes a hyperplane, and the inner closed ball becomes a compact ball strictly in one of its half-spaces. The Kelvin transform is harmonic outside that compact ball in the half-space, continuous and zero at its planar boundary. Reflect it oddly. Applying the inverse Kelvin transform gives a harmonic extension on a neighborhood of the original closed annulus, except possibly at $b$. On its exterior side this extension is nonpositive, since the reflected function is nonpositive and the Kelvin multiplier is positive. On the interior side it tends to zero at $b$, so it is bounded above on a punctured neighborhood of $b$. Apply the preceding lemma to a constant minus this extension. Its singular coefficient must vanish because the extension has limit zero along the interior approach. Thus the singularity at $b$ is removable. Compactness of the outer sphere gives one $\varepsilon >0$ valid around the whole sphere. Finally subtract the added radial harmonic function. $\square$