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

TL;DR

This paper proves a stronger one-circle version of a classical result, showing that certain lower semicontinuous functions on b^2 with specific mean value properties must be constant, highlighting differences between dimensions 1 and 2.

Contribution

It establishes a new one-circle mean value characterization for supermedian functions on b^2, extending classical harmonic function results and revealing dimension-dependent behaviors.

Findings

Functions satisfying the mean value inequality are constant under given conditions.

A dimensional difference exists in the volume mean property between b^1 and b^2.

The result generalizes classical superharmonic function properties to a one-radius setting.

Abstract

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 $d=2$.

Equations

$$\underset{|x|\to \infty }{\mathrm{liminf}}f(x)/\ln |x|\ge 0$$

$${\sigma }_{x,r(x)}(f):=\frac{1}{2\pi }{\int }_0^{2\pi }f(x+r(x)e^{it})dt\le f(x)(x\in \text{mathbbm}{R}^2).$$

$$|r(x)-r(y)|\le L|x-y|.$$

$${\sigma }_{x,r(x)}(f)=f(x)(x\in \text{mathbbm}{R}^2).$$

$$(t+1)\ln (t+1)+(t-1)\ln (t-1)=2t$$

$$r(x)\le c_0|x|+M\text{mbox}forallx\in \text{mathbbm}R\setminus [-M,M].$$

$$r\le |\cdot |+M$$

$$B(x,\rho ):=\{y\in \text{mathbbm}{R}^2:|y-x|\le \rho \}\text{mbox}andS(x,\rho ):=\{y\in \text{mathbbm}{R}^2:|y-x|=\rho \}.$$

$$r(x)\le |x|+M\text{mbox}forallx\in B(0,M{)}^c.$$

$$f(x_0)\le f\text{mbox}forsome{x}_0\in B(0,M+2).$$

$$f=0\text{mbox}onS(x,r(x)),\text{mbox}wheneverx\in \text{mathbbm}{R}^2\text{mbox}suchthatf(x)=0.$$

$${\alpha }_n(u):={\alpha }_{n-1}(u)+r({\alpha }_{n-1}(u)u)(n\in \text{mathbbm}N).$$

$$f({\alpha }_n(u)u)=0.$$

$${\alpha }_n>M\text{mbox}onS.$$

$$f(\alpha u)=0\text{mbox}forevery\alpha \ge {\alpha }_n(u).$$

$$\{\alpha u:a\ge \alpha \ge {\alpha }_n(u)\}\cup \{{\alpha }_n(v)v:v\in S\}.$$

$$f=0\text{mbox}onB(x,r(x)).$$

$$t:=\sup \{s\in [0,1]:f(sx+(1-s)y)>0\}\text{mbox}andz:=tx+(1-t)y.$$

$$f(x):=\min \{1,|x_1{|}^{-1}\}\text{mbox}andr(x):=6\max \{1,x_1^2\}.$$

$$a:=\max \{|x_1|,1\},A:=\{y\in \text{mathbbm}{R}^2:|y_1|\le 2a\}.$$

$${\sigma }_{x,r(x)}(A)\le a^{-1}/2$$

$${\sigma }_{x,r(x)}(f)\le {\sigma }_{x,r(x)}(A)+\sup f(\text{mathbbm}{R}^2\setminus A)\le a^{-1}/2+a^{-1}/2=f(x).$$

$${\sigma }_{x,r(x)}(A)\le 4\frac{\alpha }{2\pi }=\frac{2}{\pi }\alpha \le \sin \alpha \le \frac{3a}{r(x)}.$$

$$\underset{t\to \infty }{\lim }\frac{\tilde{r}(t,0)}{t\ln t}=\frac{2}{\pi }.$$

$${\int }_{\text{mathbbm}{R}^2\setminus \tilde{A}}fd{\sigma }_{x,\rho }\le \sup f(\text{mathbbm}{R}^2\setminus \tilde{A})\le \frac{1}{k}f(x)\text{mbox}forevery\rho >0,$$

$${\int }_{\tilde{A}}fd{\sigma }_{x,\rho }\sim 4\cdot \frac{1}{2\pi \rho }{\int }_1^{kt}\frac{1}{\tau }d\tau =\frac{2\ln (kt)}{\pi \rho }\sim \frac{2}{\pi }\cdot \frac{t\ln t}{\rho }f(x)$$

$$r^{-\alpha }(x)=\min \{|cx{|}^{-\alpha },M^{-\alpha }\},x\in \text{mathbbm}{R}^2.$$

$$I(\alpha ):=\frac{1}{2\pi }{\int }_0^{2\pi }|1+c{e}^{it}{|}^{-\alpha }dt,\alpha \ge 0.$$

$$I^{\prime }(0)=-\frac{1}{2\pi }{\int }_0^{2\pi }\ln |1+c{e}^{it}|dt=-\ln c<0.$$

$${\sigma }_{x,r(x)}(r^{-\alpha })\le {\sigma }_{x,r(x)}(c^{-\alpha }|\cdot {|}^{-\alpha })=\frac{c^{-\alpha }}{2\pi }{\int }_0^{2\pi }|x+c|x|e^{it}{|}^{-\alpha }dt=|cx{|}^{-\alpha }I(\alpha )<r^{-\alpha }(x).$$

Full text

Moreover, it is shown that, with respect to the assumption $r\leq c|\cdot|+M$ on $\mathbbm{R}^{d}$, there is a striking difference between the restricted volume mean property for the cases $d=1$ and $d=2$.

2000 Mathematics Subject Classification: 31A05

Keywords: Superharmonic function, Liouville’s theorem, supermedian function, one-radius theorem

1 Introduction and results

It is a well-known fact that every lower bounded superharmonic function on $\mathbbm{R}^{2}$ is constant. We recall that superharmonic functions $u$ on $\mathbbm{R}^{2}$ are lower semicontinuous functions on $\mathbbm{R}^{2}$ such that $u>-\infty$, $u\not\equiv\infty$, and, for every circle $S(x,\rho)$ of center $x\in\mathbbm{R}^{2}$ and radius $\rho>0$, the average $\sigma_{x,\rho}(u)$ of $u$ on $S(x,\rho)$ is at most $u(x)$. In this note, we shall present the following stronger result (where, as usual, we do not distinguish between $\mathbbm{C}$ and $\mathbbm{R}^{2}$).

Theorem 1.1**.**

Let $r$ be a strictly positive real function on $\mathbbm{R}^{2}$ such that

• (i)

$r$* is continuous,*

• (ii)

$\limsup_{|x|\to\infty}\,(r(x)-|x|)<\infty$,

• (iii)

$\liminf_{|x|\to\infty}(r(x)-|x|)=-\infty$.111Having (i), properties (ii),(iii) are equivalent to $\sup_{x\in\mathbbm{R}^{2}}(r(x)-|x|)<\infty$, $\inf_{x\in\mathbbm{R}^{2}}(r(x)-|x|)=-\infty$, respectively.

Let $f>-\infty$ be a lower semicontinuous numerical function on $\mathbbm{R}^{2}$ such that

[TABLE]

and $f$ is $(\sigma,r)$-supermedian, that is,

[TABLE]

Then $f$ is constant.

Remarks 1.2**.**

1. Obviously, $r\colon\mathbbm{R}^{2}\to(0,\infty)$ has the properties (i) – (iii), if there exists $L\in(0,1)$ such that, for all $x,y\in\mathbbm{R}^{2}$,

[TABLE]

However, assuming only that $|r(x)-r(y)|<|x-y|$ (which implies (i) and (ii)), even the conclusion breaks down. Indeed, if $f:=(1-|x|)^{+}$ and $r:=|x|+2+(|x|+1)^{-1}$, then, for all $x,y\in\mathbbm{R}^{2}$, $\sigma_{x,r(x)}(f)=0\leq f(x)$ and the inequality $|r(x)-r(y)|<|x-y|$ holds, since $(|x|+1)^{-1}>(|y|+1)^{-1}$ if $|x|<|y|$. In fact, none of the properties (i), (ii), (iii) may be dropped (see Section 3). Moreover, since the function $-\ln(|\cdot|^{2}+1)$ is superharmonic, it is clear that (1.1) cannot be replaced by $\liminf_{|x|\to\infty}f(x)/\ln|x|>-\infty$.

2. In Theorem 1.1, we may just as well assume that $f$ does not attain the value $\infty$ (it suffices to consider the functions $f_{n}:=\min\{f,n\}$, $n\in\mathbbm{N}$, which are $(\sigma,r)$-supermedian provided $f$ is $(\sigma,r)$-supermedian; if these functions are constant, then $f:=\lim_{n\to\infty}f_{n}$ is constant). **

Let us assume, for a moment, that $f\colon\mathbbm{R}^{2}\to\mathbbm{R}$ is continuous and $(\sigma,r)$-median, that is, such that

[TABLE]

P.C. Fenton [2] showed that $f$ has to be constant provided $f$ is lower bounded, $r$ is continuous and, for some $x_{0}\in\mathbbm{R}^{2}$, the set $\{x\in\mathbbm{R}^{2}\colon r(x)>|x-x_{0}|\}$ is bounded, a requirement which may be replaced by the weaker property (ii) (see Remark 2.1). If $f$ is bounded, then (ii) alone (without any further assumption on $r$) is sufficient to conclude that $f$ is constant ([7, Theorem 1.1], cf. also [5]). On the other hand, there exist $r\colon\mathbbm{R}^{2}\to(0,\infty)$ and a continuous $(\sigma,r)$-median function $f$ on $\mathbbm{R}^{2}$ such that $r\leq 4(|\cdot|+1)$ and $\min f(\mathbbm{R}^{2})=0$, $\max f(\mathbbm{R}^{2})=1$ (see [5, Proposition 6.1] or [7, Section 5]).

An essential step for the strong version [7, Theorem 1.1] of Liouville’s theorem consists in proving that, assuming (ii), every lower semicontinuous $(\sigma,r)$-supermedian function $f\geq 0$ on $\mathbbm{R}^{2}$ attains a minimum. It immediately implies that constant functions are the only lower semicontinuous, lower bounded functions $f$ on $\mathbbm{R}^{2}$ which are $(\lambda,r)$-supermedian, that is, which have the property that, for every $x\in\mathbbm{R}^{2}$, the average $\lambda_{x,r(x)}(f)$ of $f$ on the (closed) disk $B(x,r(x))$ is at most $f(x)$ (see [7, Corollary 6.1]). We recall that $(\lambda,r)$-supermedian functions are $(\sigma,\tilde{r})$-supermedian for some function $0<\tilde{r}\leq r$ (cf. [7, Section 6]). The following result shows that the existence of a minimum fails, if (ii) is replaced by an inequality $r\leq c|\cdot|+M$, where $c>1$.

Proposition 1.3**.**

Let $c>1$, $M>0$, and $r(x):=\max\{c|x|,M\}$, $x\in\mathbbm{R}^{2}$. Then the functions $r^{-\alpha}$ on $\mathbbm{R}^{2}$ are $(\sigma,r)$-supermedian (and $(\lambda,r)$-supermedian) provided $\alpha>0$ is sufficiently small.

So, assuming that $r\leq c|\cdot|+M$, where $c,M\in(0,\infty)$, a result of Liouville type for $(\lambda,r)$-supermedian functions on $\mathbbm{R}^{2}$ holds if and only if $c\leq 1$.

On the real line, this will turn out to be strikingly different. By the following proposition, such a result of Liouville type on $\mathbbm{R}$ holds if and only if $c\leq c_{0}$, where $c_{0}\in[2.50,2.51]$ is the unique solution to the equation

[TABLE]

in $(1,\infty)$ (see Section 5).

Proposition 1.4**.**

Let $r\colon\mathbbm{R}\to(0,\infty)$ and $M>0$ such that

[TABLE]

*Then every lower semicontinuous $(\lambda,r)$-supermedian function $f>-\infty$ on $\mathbbm{R}$ which is lower bounded *(​or satisfies $\liminf_{|x|\to\infty}f(x)/\ln|x|\geq 0$) is constant. 2. 2.

If, however, $c>c_{0}$, $M>0$, and $r:=\max(c|\cdot|,M)$, then the function $r^{-\alpha}$ is $(\lambda,r)$-supermedian provided $\alpha>0$ is sufficiently small.

Finally, let us recall that every bounded harmonic function on $\mathbbm{R}^{d}$, $d\geq 1$, is constant. Hence the results of Liouville type, which we discussed until now, are special cases of one-radius results for harmonic functions on open sets $U$. The trivial requirement that $r$ be at most the distance to $U^{c}$, if $U\neq\mathbbm{R}^{d}$, implies the existence of a real $M>0$ such that

[TABLE]

(which may justify considering (1.5) as a natural assumption on $r$ in the case $U=\mathbbm{R}^{d}$).

One radius-results for harmonic functions have a long history (see the survey papers [11, 3] and the references therein). If $U$ is an arbitrary open set in $\mathbbm{R}^{d}$, then every continuous $(\lambda,r)$-median functions $f\colon U\to\mathbbm{R}$ admitting a (sub)harmonic minorant and a (super)harmonic majorant is harmonic (provided (1.5) holds, if $U=\mathbbm{R}^{d}$).

Let us now return to continuous bounded $(\sigma,r)$-median functions. If $d=1$, the corresponding result fails almost trivially both for $\mathbbm{R}$ and $(-1,1)$ (in the first case consider $f(x):=\sin x$ and $r(x):=2\pi$, for the interval see e.g. [1, Section IV.3], cf. also [9]). We already mentioned the positive result for $U=\mathbbm{R}^{2}$. On the unit disk, however, there exists a continuous function $0\leq f\leq 1$ having the one-circle property which is not harmonic (see [8] and [6]). The corresponding general problems for $\mathbbm{R}^{d}$, $d\geq 3$, are unsolved, both for the open unit ball and the entire space (if, however, $r$ is Lipschitz with constant $L\in(0,1)$, the answer is positive [10, Theorem 2]).

2 Proof of Theorem 1.1

Let $f\colon\mathbbm{R}^{2}\to\mathbbm{R}$ be lower semicontinuous such that (1.1) holds and $f$ is $(\sigma,r)$-supermedian, where $r\colon\mathbbm{R}^{2}\to(0,\infty)$ satisfies (i), (ii), and (iii). For all $x\in\mathbbm{R}^{2}$ and $\rho>0$, let

[TABLE]

By (ii), there is a real $M>0$ such that

[TABLE]

If $f$ is lower bounded, then, by [5, Proposition 2.1] (see also [7]),

[TABLE]

In fact, a short look at the proof for (2.1) reveals that it is valid as well under our weaker assumption (1.1). So we may suppose without loss of generality that $f(0)=0$ and $f\geq 0$ (we can replace $f$ by the function $x\mapsto f(x_{0}+x)-f(x_{0})$). Since $f$ is lower semicontinuous, we know, by (1.2), that

[TABLE]

Let us use the technique developed in [2]. We define an increasing sequence $(\alpha_{n})$ of continuous real functions on the unit circle $S:=S(0,1)$ by $\alpha_{0}:=r(0)$ and

[TABLE]

By induction, we conclude from (2.2) that, for all $u\in S$ and $n\in\mathbbm{N}$,

[TABLE]

Since $r$ is continuous and strictly positive, we obtain immediately that $\lim\alpha_{n}=\infty$. So there exists $n\in\mathbbm{N}$ such that

[TABLE]

For the moment, let us fix $u\in S$. We claim that

[TABLE]

Indeed, suppose that (2.4) does not hold. Then there exists a maximal real $a$ such that $a\geq\alpha_{n}(u)$ and $f(\alpha u)=0$ for every $\alpha\in[\alpha_{n}(u),a]$. We may join the points $y_{0}:=au$ and $y_{1}:=-\alpha_{n}(-u)u$ continuously by an arc $\gamma:[0,1]\to\mathbbm{R}^{2}$ contained in the set

[TABLE]

In particular, $f\circ\gamma=0$ by (2.3). Fix $0<\beta\leq r(y_{0})$ and let $z:=y_{0}+\beta u$ so that $|z-y_{0}|\leq r(y_{0})$. Clearly, $|z-y_{1}|\geq r(y_{1})$, since the origin is contained in the line segment from $y_{1}$ to $y_{0}$, $r(y_{1})\leq|y_{1}|+M$, and $a+\beta\geq\alpha_{n}(u)>M$. By continuity of $r$, there exists $s\in[0,1]$ such that $|z-\gamma(s)|=r(\gamma(s))$. Since $f(\gamma(s))=0$, we conclude by (2.2) that $f(z)=0$. Thus $a\geq a+r(y_{0})$, a contradiction proving (2.4).

Fixing $R>0$ such that $\alpha_{n}\leq R$ on $S$, we therefore know that $f=0$ on $\mathbbm{R}^{2}\setminus B(0,R)$. Since $\liminf_{|x|\to\infty}(r(x)-|x|)=-\infty$, there exists $x\in\mathbbm{R}^{2}$ such that $|x|-r(x)>R$, and hence

[TABLE]

Suppose that there is a point $y\in\mathbbm{R}^{2}$ such that $f(y)>0$. We define

[TABLE]

Since $r(z)>0$, there exists a point $\tilde{y}\in[y,z]$ such that $|\tilde{y}-z|<r(z)$ and $f(\tilde{y})>0$. By (2.6), $|\tilde{y}-x|>r(x)$. By continuity of $r$, we conclude that there exists $\tilde{z}\in(z,x)$ such that $|\tilde{y}-\tilde{z}|=r(\tilde{z})$. So $f(\tilde{z})>0$, by (2.2). However, by definition of $z$, $f=0$ on $(z,x)$. Thus there is no point $y\in\mathbbm{R}^{2}$ such that $f(y)>0$, that is, $f$ is identically zero, and the proof of Theorem 1.2 is finished.

Remark 2.1**.**

If $f$ is even continuous and $(\sigma,r)$-median, then (iii) is not needed to conclude that $f$ is constant.

Indeed, it suffices to observe that (iii) has not been used to obtain that $f=\inf f(\mathbbm{R}^{2})$ outside a compact set, and hence $\gamma:=\sup f(\mathbbm{R}^{2})<\infty$. Then, just using (i) and (ii), we get as well that $\gamma-f=\inf(\gamma-f)(\mathbbm{R}^{2})$ outside a compact set, that is, $f=\sup f(\mathbbm{R}^{2})$ outside a compact set. Thus $\inf f(\mathbbm{R}^{2})=\sup f(\mathbbm{R}^{2})$, $f$ is constant. **

3 Examples

Simple examples show that a continuous bounded $(\sigma,r)$-supermedian function $f$ on $\mathbbm{R}^{2}$ may be non-constant, if any of the properties (i), (ii), or (iii) of $r$ is violated.

1. Let $f:=(1-|x|)^{+}$, $x\in\mathbbm{R}^{2}$. Taking $r(x):=3$, if $|x|<2$, and $r(x):=1$, if $|x|\geq 2$, we observe that, of course, property (i), that is, the continuity of $r$, cannot be omitted (or replaced by lower semicontinuity). Considering $r:=|x|+2+(|x|+1)^{-1}$ we already noted in Remark 1.2.1 that (iii) cannot be dropped (of course, for this purpose, it would be sufficient to take $r(x):=|x|+2$).

2. Finally, let us prove that the conclusion of Theorem 1.1 fails, if property (ii) is omitted. For $x\in\mathbbm{R}^{2}$, let

[TABLE]

Clearly, $0\leq f\leq 1$, $f$ and $r$ are continuous functions, and $r$ satisfies (iii), since $r(0,t)=6$ for every $t\in\mathbbm{R}$. To prove that (1.2) holds, we fix $x\in\mathbbm{R}^{2}$ and define

[TABLE]

Then $f(x)=a^{-1}$. We shall see that

[TABLE]

and hence

[TABLE]

To prove (3.1) let $\alpha$ denote the maximal angle between the $x_{2}$-axis and the lines connecting $x$ with one of the four points $y\in S(x,r(x))$ satisfying $|y_{1}|=2a$. Then

[TABLE]

If $|x_{1}|\leq 1$, then $a=1$, $r(x)=6$, and hence $3a/r(x)=1/2=1/(2a)$. If $|x_{1}|>1$, then $a=|x_{1}|$, $r(x)=6x_{1}^{2}$, and hence $3a/r(x)=1/(2|x_{1}|)=1/(2a)$. Thus (3.1) holds

A closer look would reveal that $f$ is $\tilde{r}$-superharmonic with respect to a continuous function $0<\tilde{r}\leq r$ satisfying $\tilde{r}(x)=\tilde{r}(|x_{1}|,0)$ and

[TABLE]

This follows from the fact that, given $t\geq 1$ and $k\in\mathbbm{N}$, the point $x:=(t,0)$ and the set $\tilde{A}:=\{y\in\mathbbm{R}^{2}\colon|y_{1}|\leq kt\}$ satisfy

[TABLE]

and, for large $\rho$,

[TABLE]

(which, incidentally, shows that the limit behavior in (3.2) is optimal for our function $f$).

4 Proof of Proposition 1.3

Let $c>1$, $M>0$, and $r:=\max\{c|\cdot|,M\}$ so that, for every $\alpha>0$,

[TABLE]

We define

[TABLE]

Then $I(0)=1$ and

[TABLE]

So there exists $\alpha_{0}>0$ such that $I<1$ on $(0,\alpha_{0}]$. Let us fix $\alpha\in(0,\alpha_{0}]$ and $x\in\mathbbm{R}^{2}$. If $c|x|>M$, then $r(x)=c|x|$ and hence

[TABLE]

If $c|x|\leq M$, then $r(x)=M$, and hence $\sigma_{x,r(x)}(r^{-\alpha})\leq M^{-\alpha}=r^{-\alpha}(x)$. Thus $r^{-\alpha}$ is $(\sigma,r)$-supermedian.

Since $\lambda_{x,\rho}=2\rho^{-2}\int_{0}^{\rho}\sigma_{x,s}\,sds$ (and $\int_{0}^{2\pi}\ln|1+ce^{is}|\,ds=2\pi\ln 1=0$, if $s\in(0,1)$), we obtain similarly that $r^{-\alpha}$ is $(\lambda,r)$-supermedian provided $\alpha>0$ is sufficiently small.

5 Proof of Proposition 1.4

Let us define

[TABLE]

Then $\psi$ is continuous, $\lim_{t\to 1}\psi(t)=2\ln 2-2<0$, $\lim_{t\to\infty}\psi(t)=\infty$. Moreover,

[TABLE]

hence $\psi$ is strictly decreasing on $(0,\sqrt{2}\,)$ and strictly increasing on $(\sqrt{2},\infty)$. So there exists $c_{0}\in(1,\infty)$ such that $\psi(c_{0})=0$,

[TABLE]

In fact, $2,50<c_{0}<2,51$ (since $\psi(2.50)<0$ and $\psi(2.51)>0$).

1. Let $r\colon\mathbbm{R}\to(0,\infty)$ and $M>0$ such that

[TABLE]

Let $\varphi:=\ln^{+}(|\cdot|-M)$ (so that $\varphi(x)=0$, if $-(M+1)\leq x\leq M+1$). We claim that there exists $\tilde{M}>0$ such that, for every $x\in\mathbbm{R}\setminus[\tilde{M},-\tilde{M}]$,

[TABLE]

Since \lim_{x\to\infty}(x/M)\ln\bigl{(}1-(M/x)\bigr{)}=-\ln^{\prime}1=-1, there exists $\tilde{M}\geq 1+c_{0}+2M$ such that

[TABLE]

For a while, let us fix $x\in\mathbbm{R}\setminus[-\tilde{M},\tilde{M}]$. To prove (5.1) we may assume, by symmetry, that $x$ is positive. Let $y\in(0,c_{0}x+M)$. If $x-y\geq M+1$, then

[TABLE]

since $\varphi$ is concave on $(M+1,\infty)$. If $M+1>x-y\geq-(M+1)$, then (5.3) holds, since $\varphi(x-y)=0$ and $x+y-M\leq x(1+c_{0})\leq(x-M)^{2}$. Therefore (5.1) holds, if $x-r(x)\geq-(M+1)$.

Let us assume next that $x-r(x)<-(M+1)$. Then $t:=(r(x)-M)/x\in(1,c_{0}]$,

[TABLE]

Since $\psi(t)\leq\psi(c_{0})=0$, we hence see that

[TABLE]

that is, $r(x)\lambda_{x,r(x)}(\varphi)\leq tx\ln x+1$. Thus

[TABLE]

where the right side is positive by (5.2), since $t\leq c_{0}$. This finishes the proof of (5.1).

Now let $f>-\infty$ be a lower semicontinuous $(\lambda,r)$-supermedian function on $\mathbbm{R}$ such that $\liminf_{|x|\to\infty}f(x)/\ln|x|\geq 0$. To prove that $f$ is constant, we use ideas from [4] and [7]. There exists $x_{0}\in[-\tilde{M},\tilde{M}]$ such that

[TABLE]

We intend to show that $f\geq f(x_{0})$ on $\mathbbm{R}$. Then the lower semicontinuity of $f$ and the inequalities $\lambda_{x,r(x)}(f)\leq f(x)$, $x\in\mathbbm{R}$, will imply that the set $A:=\{x\in\mathbbm{R}\colon f(x)=f(x_{0})\}$ is both closed and open, hence $A=\mathbbm{R}$, $f=f(x_{0})$.

Fixing $\varepsilon>0$, it suffices to prove that

[TABLE]

Obviously, $\tilde{f}$ is lower semicontinuous, lower bounded, and $\lim_{|x|\to\infty}\tilde{f}(x)=\infty$. Therefore $\tilde{f}$ attains a minimum on $\mathbbm{R}$, and the non-empty set

[TABLE]

is closed. Let $z\in\tilde{A}$ with minimal absolute value. If $|z|>\tilde{M}$, then $\lambda_{z,r(z)}(\tilde{f})\leq\tilde{f}(z)$, and hence $[z-r(z),z+r(z)]\subset\tilde{A}$. This is impossible, by our choice of $z$. Thus $z\in[-\tilde{M},\tilde{M}]$, and $\tilde{f}\geq\tilde{f}(z)\geq f(z)\geq f(x_{0})$.

2. Finally, let $c>c_{0}$, $M>0$, and $r:=\max(c|\cdot|,M)$. We have to show that the function $r^{-\alpha}$ is $(\lambda,r)$-supermedian provided $\alpha>0$ is sufficiently small. To that end we define

[TABLE]

Then $\Psi^{\prime}(\beta)=(c+1)^{\beta}\ln(c+1)+(c-1)^{\beta}\ln(c-1)-2c$. In particular, $\Psi^{\prime}(1)=\psi(c)>0$. So there exists $\alpha\in(0,1)$ such that $\Psi(1-\alpha)<0$.

Let us now fix $x\in\mathbbm{R}$. If $c|x|\leq M$, then $r(x)=M$, and hence $\lambda_{x,r(x)}(r^{-\alpha})\leq M^{-\alpha}=r^{-\alpha}(x)$. So let us assume that $c|x|<M$ and hence $r(x)=c|x|$. Then

[TABLE]

Since $\Psi(1-\alpha)<0$, we conclude that $\lambda_{x,r(x)}(|\cdot|^{-\alpha})\leq|x|^{-\alpha}$ and hence

[TABLE]

Thus $r^{-\alpha}$ is $(\lambda,r)$-supermedian.

Remark 5.1**.**

If even $r\leq c|\cdot|+M$ for some $c\in(0,c_{0})$, then the conclusion in (1) of Proposition 1.4 is still valid, if the condition $\liminf_{|x|\to\infty}f(x)/\ln|x|\geq 0$ is replaced by the weaker assumption $\liminf_{|x|\to\infty}f(x)/\ln|x|>-\infty$. Indeed, by means of the function $\Psi$, we may then prove that, for some $\alpha>0$, the function $|\cdot|^{\alpha}$ (which will replace $\varphi$) is $(\lambda,r)$-supermedian. **