One-radius results for supermedian functions on ${\mathbb{R}}^d$, $d\le 2$

A classical result states that every lower bounded superharmonic function on ${\mathbb{R}}^2$ is constant. In this paper the following (stronger) one-circle version is proven. If $f\text{\hspace{0.17em}⁣}:{\mathbb{R}}^2\to (-\mathrm{\infty },\mathrm{\infty }]$ is lower semicontinuous, ${\mathrm{lim\hspace{0.17em}inf}}_{\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }}f(x)\mathrm{/}\ln \mathrm{\mid }x\mathrm{\mid }\ge 0$, and, for every $x\in {\mathbb{R}}^2$, $1\mathrm{/}(2\pi ){\int }_0^{2\pi }f(x+r(x)e^{it})dt\le f(x)$, where $r\text{\hspace{0.17em}⁣}:{\mathbb{R}}^2\to (0,\mathrm{\infty })$ is continuous, $\sup_{x\in\Bbb R^2} (r(x)-|x|)<\infty$,and$\inf_{x\in\Bbb R^2} (r(x)-|x|)=-\infty$,then$f$ is constant. Moreover, it is shown that, with respect to the assumption $r\le c\mathrm{\mid }\cdot \mathrm{\mid }+M$ on ${\mathbb{R}}^d$, there is a striking difference between the restricted volume mean property for the cases $d=1$ and $d=2$.