  A classical result states that every lower bounded superharmonic function on
$\Bbb R^2$ is constant. In this paper the following (stronger) one-circle
version is proven. If $f\colon \Bbb R^2\to (-\infty,\infty]$ is lower
semicontinuous, $\liminf_{|x|\to\infty} f(x)/\ln|x|\ge 0$, and, for every
$x\in\Bbb R^2$, $1/(2\pi) \int_0^{2\pi} f(x+r(x)e^{it}) dt\le f(x)$, where
$r\colon \Bbb R^2\to (0,\infty)$ 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 $r\le c|\cdot|+M$
on $\Bbb R^d$, there is a striking difference between the restricted volume
mean property for the cases $d=1$ and $d=2$.
