Spherical mean ($M_{r}f(x)$)

The spherical mean of a function $f$ on ${\mathbb{R}}^n$ averages it over a sphere of radius $r>0$ about $x$:Here $d\omega$ is surface measure on the unit sphere and $|S^{n-1}|$ is its total area. Using the unit sphere absorbs the physical surface-area factor $r^{n-1}$ into the change of variables. The expression requires an integrable restriction to that sphere; continuity near it is one sufficient condition.

For $f(y)=|y{|}^2$, expanding $|x+r\omega {|}^2$ and using the zero average of $\omega$ gives $M_{r}f(x)=|x{|}^2+r^2$. By contrast, the mean value property for harmonic functions gives $M_{r}u(x)=u(x)$ when $u$ is harmonic on a neighborhood of the closed ball. Averaging over a boundary sphere and averaging over the whole ball are different operations, even though both have a mean-value property for harmonic functions.

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 $-4\pi s^2{\bar{v}}^{\prime }(s)=M$ 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 $\Vert v{\Vert }_2^2\ge M^2(4\pi {)}^{-1}{\int }_a^R(1-s/R{)}^{2}ds$. 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.