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Statement

Theorem XI (p. 17). Let $f$ be a real additive function with $f(m+1)\ge f(m)$ for every $m$. Then $f(m)=c\log m$ for a constant $c$.

Proof pointer

Pp. 17--18. For odd $m<n<2m$, monotonicity gives $f(m)\le f(n)\le f(2m)=f(2)+f(m)$, so $f$ is finitely distributed and Theorem V gives $f(m)=c\log m+\varphi(m)$ with $\sum(\varphi'(p))^2/p<\infty$. If $\varphi$ is not identically 0, Theorem X (p. 17, stated without proof) gives $\varphi(m+1)-\varphi(m)<-\delta$ for infinitely many $m$, contradicting monotonicity.

Read depth

Claims checked: the statement and its proof on pp. 17--18 read on the page images. Theorem X, on which the proof rests, is stated in the paper without proof (its proof is said to be similar to one in an earlier paper). Nothing here is independently reviewed.

Dependencies

Theorem V; Theorem X (p. 17), stated without proof.

Source. P. Erdős, On the distribution function of additive functions, Ann. of Math. (2) 47 (1946), 1--20, doi:10.2307/1969031; the edition read is named on the source card.

Bears on

• Problem 491: the theorem gives $f(n)=c\log n$, the problem's conclusion with error term 0, for additive $f$ that are nondecreasing, a hypothesis different from the problem's bounded differences.
• Problem 1122: the theorem is the problem's case in which the set $\{n: f(n+1)<f(n)\}$ is empty.