Classical subharmonic mean value inequalities

Write $m(t)$ for the sphere average at $a$. The radial identity gives $m\mathrm{\prime }(t)=\frac{t}{n}\frac{1}{\mathrm{\mid }{B}_t\mathrm{\mid }}{\int }_{B_t(a)}\mathrm{\Delta }u\ge 0$. Continuity gives $m(t)\to u(a)$ as $t\downarrow 0$, so $m(r)\ge u(a)$.

For the converse, two integrations of the one-variable fundamental theorem along $a+t\theta$ give $u(a+r\theta )=u(a)+r\mathrm{\nabla }u(a)\cdot \theta +r^2{\int }_0^1(1-s){\theta }^T{D}^{2}u(a+sr\theta )\theta ds$. The continuity of $D^{2}u$ makes its remainder after replacing the Hessian by $D^{2}u(a)$ uniformly $o(r^2)$ for unit $\theta$. Reflection and permutation symmetry give sphere averages of ${\theta }_i$ and ${\theta }_i{\theta }_j$ equal to zero for $i\mathrm{\ne }j$, and of ${\theta }_i^2$ equal to $1\mathrm{/}n$. Thus $m(r)=u(a)+r^2\mathrm{\Delta }u(a)\mathrm{/}(2n)+o(r^2)$.

Polar coordinates give the ball average $\frac{n}{r^n}{\int }_0^r{t}^{n-1}m(t)dt\ge u(a)$. For signed $u$, apply the nonnegative polar formula to its positive and negative parts on the bounded ball; both are bounded and integrable. Replacing $u$ by $-u$ proves both reversed inequalities.

Integrating the expansion with the radial weights gives ball average $u(a)+r^2\mathrm{\Delta }u(a)\mathrm{/}(2(n+2))+o(r^2)$. Either assumed mean inequality, divided by $r^2>0$ and followed by $r\downarrow 0$, forces $\mathrm{\Delta }u(a)\ge 0$. Negation gives the superharmonic converse.