= Spherical mean of a Poisson solution outside its source
{title2=$\bar v(s)=\frac{M}{4\pi}(s^{-1}-R^{-1})$}
In three dimensions let $v\in H_0^1(B_R)$ solve $-\Delta v=f$, where $f\in L^{6/5}(B_R)$ is supported in $B_a$ and $0<a<R$. Write $M=\int f$ and $\bar v(s)=(4\pi)^{-1}\int_{S^2}v(s\omega)\,d\omega$. Outside $B_a$, the <weak solution> is a <harmonic function>. The <weak formulation> against radial <test functions> gives $-4\pi s^2\bar v'(s)=M$ for $a<s<R$. The zero <Sobolev trace> gives $\bar v(R)=0$; integration proves the displayed formula, without assuming the source is a <radial function>.
<Cauchy-Schwarz inequality> on spheres gives $\|v\|_2^2\geq M^2(4\pi)^{-1}\int_a^R(1-s/R)^2\,ds$. For fixed nonzero $M$ and fixed $a$, this grows linearly in $R$. Such solutions on expanding <balls> cannot remain bounded in the full $H^1(\mathbb R^3)$ <norm> after <zero extension of H01>.
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