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Statement

Theorem VI (pp. 3--4). Let $f$ be additive with $|f(p)|<c$ and $\sum_p(f(p))^2/p=\infty$, and put

$$A_n=\sum_{p\le n}\frac{f(p)}{p},\qquad B_n=\sum_{p\le n}\frac{(f(p))^2}{p}.$$

Let $N_\epsilon^+(d,n)$ count the $m\le n$ for which some $u>d$ has $\sum_{p\mid m,\,p\le u}f(p)>A_u+(1+\epsilon)\sqrt{2B_u\log\log B_u}$, and put $U^+(d)=\limsup_{n\to\infty}N_\epsilon^+(d,n)/n$. Then $\lim_{d\to\infty}U^+(d)=0$. If instead $N_\epsilon^-(d,n)$ counts the $m\le n$ for which some $u>d$ has $\sum_{p\mid m,\,p<u}f(p)>A_u+(1-\epsilon)\sqrt{2B_u\log\log B_u}$, then $\lim_{n\to\infty}N_\epsilon^-(d,n)/n=1$ for every $d$.

The print writes the divergence hypothesis as $\sum_p(f(p)/p)^2=\infty$, which fails for every bounded $f$; the form above matches the normalisation $B_n$ the theorem uses. For $f(p)=1$ the paper deduces (p. 4) that almost all $m$ have no large divisor $d$ with $\nu(d)>\log\log d+(1+\epsilon)\sqrt{2\log\log d\log\log\log\log d}$, while almost all have one with $1-\epsilon$ in place of $1+\epsilon$, $\nu(n)$ being the number of distinct prime factors of $n$.

Proof pointer

Not given: the paper omits the proof as very similar to that of the Erdős--Kac paper (p. 4).

Read depth

Claims checked: the statement read on the page images of pp. 3--4. No proof is given in the paper. Nothing here is independently reviewed.

Dependencies

None in the corpus.

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

None directly.