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Source. Laczkovich (1984), Theorem 2, statement on printed p. 110 and proof on pp. 114–115 (PDF pp. 2 and 6–7).

Statement

Let $\alpha$ be irrational with bounded regular continued-fraction partial quotients. Suppose $f:G_\alpha\to\mathbb R$ satisfies

$$2f(x)\le f(x+h)+f(x+2h) \qquad(x,h\in G_\alpha,\ h>0).$$

Then $f$ is nondecreasing on $G_\alpha$.

Dependencies. Lemma 2, backward propagation, positive increments, and Lemma 1. The continued-fraction facts imported by Lemma 2 remain explicit. No regularity assumption on $f$ is used.

Bears on. Problem 1125, through Theorem 1.

Proof

Fix $a<b$ in $G_\alpha$. Apply Lemma 2 with closure parameter $2$ and endpoint $b$, obtaining a finite seed $H$. It is nonempty because its closure contains $b$ whereas the empty seed has empty closure. Let

$$M=\max_{x\in H}f(x).$$

Backward propagation shows $f(x)\le M$ for all $x\in G_\alpha$ with $x\le b$, in particular $f(b)\le M$.

Set $g(x)=\max\{f(x),f(b)\}$. This truncation also satisfies the inequality. If $g(x)=f(x)$, bound the two later $f$ values by the corresponding $g$ values; if $g(x)=f(b)$, both later $g$ values are at least $f(b)$. On $[a,b]\cap G_\alpha$ we therefore have

$$f(b)\le g(x)\le M,\qquad |g(x)|\le K:=\max\{|f(b)|,|M|\}. \tag{1}$$

This single $K$ is fixed before choosing any progression length.

Let $N\ge1$ be an arbitrary integer. The positive-increment decomposition supplies $c,d\in G_\alpha$, $c,d>0$, with

$$Nc+(N+1)d=b-a.$$

All the points from $a$ to $a+Nc$ in steps of $c$, followed by the points from $a+Nc$ to $b$ in steps of $d$, lie in the interval where (1) holds. Restricting $g$ to either finite progression gives a sequence satisfying the hypotheses of Lemma 1: every valid integer step corresponds to a positive integer multiple of $c$ or $d$ in $G_\alpha$.

Two applications of that lemma yield

$$g(a)\le g(a+Nc)+\frac{10K}{N} \le g(b)+10K\left(\frac1N+\frac1{N+1}\right).$$

Since $K$ is independent of $N$, letting $N$ tend to infinity gives $g(a)\le g(b)=f(b)$. Finally $f(a)\le g(a)$, so $f(a)\le f(b)$. The pair $a<b$ was arbitrary.