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Source. Laczkovich (1984), printed pp. 109–110 (PDF pp. 1–2). These are definitions and scope conventions, not an additional theorem.

For an additive subgroup $G\subseteq\mathbb R$, the inequality is

$$2f(x)\le f(x+h)+f(x+2h) \qquad(x,h\in G,\ h>0). \tag{K}$$

All functions are finite real-valued. Throughout, nondecreasing means $f(a)\le f(b)$ whenever $a<b$ are in the domain. This is the meaning of the source's word “increasing”; strict monotonicity is not asserted. Constant functions satisfy (K).

For an irrational real number $\alpha$, write

$$G_\alpha=\mathbb Z\alpha+\mathbb Z =\{n\alpha+k:n,k\in\mathbb Z\}.$$

The source denotes this group by $I(\alpha)$. Irrationality makes its coefficient pair $(n,k)$ unique: two representations with distinct $n$ would express $\alpha$ as a rational number. The group is countable, contains $0$ and $1$, and is closed under integer linear combinations. Its density and the required positive-step decomposition are proved in Positive increments.

For a positive integer $n$, let $\mathcal F_n$ consist of all $f:\{0,\ldots,n\}\to\mathbb R$ such that

$$2f(i)\le f(i+h)+f(i+2h) \quad\text{whenever }i,h\in\mathbb Z,\quad 0\le i<i+h<i+2h\le n.$$

For $n=1$ this restriction is vacuous. The estimate in Lemma 1 still holds, but its expression $10K/n$ is not a statement at $n=0$. Restriction to any consecutive subinterval, followed by translation of its left endpoint to zero, preserves this condition.

The stronger inequality considered separately is

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

For nonnegative functions, (K*) implies (K). This implication uses nonnegativity: in general the maximum of two real numbers need not be at most their sum.

Bears on. Problem 1125.