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Source. Laczkovich (1984), printed p. 110 (PDF p. 2), the answer to Lawrence's Question 7.

Statement

If $f:G_{\sqrt2}\to[0,\infty)$ satisfies

$$2f(x)\le\max\{f(x+h),f(x+2h)\} \qquad(x,h\in G_{\sqrt2},\ h>0), \tag{1}$$

then $f$ vanishes identically.

Dependencies. Theorem 2 and density of the subgroup.

Source scope. The source first passes to one-sided limits of the nondecreasing function and then doubles its values along a finite chain. The argument below supplies the same doubling comparison directly with a small positive subgroup step, so no undeclared continuity or existence of a finite real-line extension is needed.

Bears on. Problem 1125, as a distinct stronger-inequality consequence. Compare the rational counterexample, which does not extend to this subgroup.

Proof

Nonnegativity makes the maximum in (1) at most the sum of the two later values. Thus $f$ satisfies (K), and Theorem 2 makes it nondecreasing on $G_{\sqrt2}$.

For any $x<y$ in this subgroup, choose a positive subgroup element $h<(y-x)/2$, using density. Both $x+h$ and $x+2h$ are below $y$, so monotonicity and (1) give

$$2f(x)\le\max\{f(x+h),f(x+2h)\}\le f(y). \tag{2}$$

Fix $x$. For every positive integer $m$, density permits a chain of $m$ points

$$x<x_1<\cdots<x_m<x+1$$

in the subgroup, for example by choosing one in each of $m$ disjoint ordered subintervals of $(x,x+1)$. Applying (2) successively, and then monotonicity at the last point, gives

$$0\le f(x)\le2^{-m}f(x_m)\le2^{-m}f(x+1).$$

The value $f(x+1)$ is finite and independent of $m$. Letting $m$ tend to infinity proves $f(x)=0$.