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Source. Laczkovich (1984), printed pp. 112 and 114 (PDF pp. 4–6). The proof expands the source's finite-witness characterization and the use of that characterization in Theorem 2.

Statement

Fix an integer $N\ge2$ and $H\subseteq\mathbb R$. A set $U$ is $N$-closed if

$$\bigl(x+h,x+2h,\ldots,x+Nh\in U,\ h>0\bigr) \ \Longrightarrow\ x\in U \tag{1}$$

for all real $x,h$. Let $H^{(N)}$ be the intersection of all $N$-closed sets containing $H$.

Then $x\in H^{(N)}$ if and only if there is a finite list $x_0,\ldots,x_t=x$ such that each $x_j$ is either in $H$, or has $x_j+ih$ among earlier entries for all $1\le i\le N$, for some $h>0$.

If $H$ lies in an additive subgroup $G$, then $H^{(N)}\subseteq G$, and every step $h$ in such a witness also lies in $G$. Consequently, if $f:G\to\mathbb R$ satisfies (K) and $f\le M$ on $H$, then $f\le M$ on $H^{(N)}$.

There is also a useful finite-seed observation. If $H$ contains $N$ consecutive points

$$z,z+1,\ldots,z+N-1,$$

then $H^{(N)}$ contains every $z-j$ for integers $j\ge0$.

Dependencies. The group and inequality conventions are in Definitions. Only finite induction is needed.

Bears on. Problem 1125, through Lemma 2 and Theorem 2.

Proof

Put $H_0=H$, and let $H_{j+1}$ be $H_j$ together with every point whose $N$ later equally spaced points all lie in $H_j$. The union $V=\bigcup_{j\ge0}H_j$ is $N$-closed. Indeed, the finitely many later points in (1) belong to finitely many stages; their largest stage contains all of them, so the next stage contains $x$.

Every $N$-closed set containing $H$ contains every $H_j$, by induction. Thus $V=H^{(N)}$. A point in $H_j$ has a finite witness list: for a new point, concatenate finite witness lists for its $N$ parents and append the point. Repetition of entries is harmless. Conversely, any such list lies in every $N$-closed superset of $H$, by induction along the list. This proves the characterization, including $H=\varnothing$, whose closure is empty.

An additive subgroup is $N$-closed because the first two later points give

$$h=(x+2h)-(x+h)\in G,\qquad x=2(x+h)-(x+2h)\in G.$$

Hence $H^{(N)}\subseteq G$. The same two identities show that the steps in a finite witness list lie in $G$. Starting with $f\le M$ on $H$, induction along the list now gives

$$f(x_j)\le\frac{f(x_j+h)+f(x_j+2h)}2\le M$$

at every new point. This uses only the first two parents, even when $N>2$.

Finally, the $N$ displayed consecutive seed points yield $z-1$ by taking $h=1$. The last $N$ known consecutive points then yield $z-2$, and induction yields every point claimed. The restriction $N\ge2$ is essential for the subgroup assertion: one later point alone does not determine a step in $G$.