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Source. Laczkovich (1984), printed p. 109 (PDF p. 1), crediting J. Lawrence, reference [3]. Laczkovich supplies the formula and states that it satisfies the stronger inequality. The verification below is expanded here. Lawrence's separately cited paper is not used as an independently inspected source.

Statement

For positive rational $r$, let $\nu(r)$ be the least positive integer $j$ for which $rj!\in\mathbb Z$. Define $F:\mathbb Q\to[0,\infty)$ by

$$F(r)= \begin{cases} 0,&r\le0,\\ 2^{\,r\nu(r)!},&r>0. \end{cases}$$

Then for every $x\in\mathbb Q$ and positive $h\in\mathbb Q$,

$$2F(x)\le\max\{F(x+h),F(x+2h)\}. \tag{1}$$

In particular $F$ satisfies (K), but it is neither nondecreasing nor nonincreasing. It is unbounded on every rational interval $(u,v)$ with $0<u<v$.

Bears on. Problem 1125: this separates the rational domain from the real-line theorem. It is not a counterexample to that theorem.

Proof

The level $\nu(r)$ exists because the positive denominator of a reduced fraction divides a sufficiently large factorial. Its exponent $r\nu(r)!$ is a positive integer.

If $x\le0$, (1) follows from nonnegativity. Suppose $x>0$, and write $n=\nu(x)$. At least one of $x+h$ and $x+2h$ has level at least $n$. This is automatic when $n=1$. For $n\ge2$, if both levels were at most $n-1$, their products with $(n-1)!$ would be integers. The identity

$$x=2(x+h)-(x+2h)$$

would make $x(n-1)!$ an integer, contradicting minimality of $n$.

Choose such a later point $y$, and put $m=\nu(y)\ge n$. Since $y>x>0$ and $m!\ge n!$,

$$ym!>xn!.$$

Both sides are integers, so $ym!\ge xn!+1$. Therefore $F(y)\ge2F(x)$, proving (1). As $F\ge0$, its maximum at the two later points is at most their sum, so (K) follows.

For explicit failures of the two monotonicity directions, observe

$$F(0)=0<F(1)=2,\qquad F(2/3)=2^4=16>F(1).$$

Here $\nu(2/3)=3$. Thus $F$ is neither nonincreasing nor nondecreasing.

Finally, fix $0<u<v$. For every sufficiently large integer $m$, the interval $(2^m u,2^m v)$ has length greater than $2$, and so contains an odd integer $k_m$. Put $r_m=k_m/2^m\in(u,v)$. Its reduced denominator is $2^m$. For each fixed integer $J$, this denominator fails to divide $J!$ for all sufficiently large $m$, so $\nu(r_m)\to\infty$. Consequently

$$F(r_m)=2^{r_m\nu(r_m)!}\ge2^{u\nu(r_m)!}\longrightarrow\infty.$$

This proves the stated local unboundedness.