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A classical result states that every lower bounded superharmonic function on$\Bbb R^2$ is constant. In this paper the following (stronger) one-circleversion is proven. If $f\colon \Bbb R^2\to (-\infty,\infty]$ is lowersemicontinuous, $\liminf_{|x|\to\infty} f(x)/\ln|x|\ge 0$, 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 $f$ isconstant.  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 volumemean property for the cases $d=1$ and $d=2$.

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