---

Statement

Setting (p. 51). $f$ satisfies $f(x+1)=f(x)$, $\int_0^1f(x)\,dx=0$ and $\int_0^1f(x)^2\,dx=1$, and $n_1<n_2<\cdots$ satisfies $n_{k+1}/n_k>c>1$.

(6) (p. 52). The paper states that an easy modification of the construction of Theorem 1 shows the existence of an $f$ and a sequence $n_k$ such that for almost all $x$

$$\limsup_{N\to\infty}\frac{1}{N(\log\log N)^{1/2-\epsilon}}\Big(\sum_{k=1}^Nf(n_kx)\Big)=\infty.$$

The print does not say whether one $f$ and $n_k$ serve every $\epsilon>0$ or whether they depend on $\epsilon$.

(7) (p. 52). The paper states that it can show that for almost all $x$

$$\lim_{N\to\infty}\frac{1}{N(\log N)^{1/2+\epsilon}}\Big(\sum_{k=1}^Nf(n_kx)\Big)=0,$$

which by the setting is asserted for every such $f$ and $n_k$. The print does not quantify $\epsilon$ in (7) either.

The paper says there is again a gap between (6) and (7), that (6) seems to give the right order of magnitude, and that it cannot prove this. It also records (p. 52) that the $f$ of Theorem 1 is unbounded and that whether the strong law (2) holds for every bounded $f$ remains open.

Proof pointer

None in the paper: neither (6) nor (7) is proved there.

Read depth

Claims checked: (6), (7) and the surrounding remarks were read clause by clause on the page image of p. 52. There is no proof to check. A second reader checked the statements, hypotheses, labels and page against the print.

Dependencies

None.

Source. P. Erdős, On the strong law of large numbers, Trans. Amer. Math. Soc. 67 (1949), 51--56; the edition read is named on the source card.

Bears on

• Problem 995: (6) and (7) bound the almost-everywhere growth the problem asks to estimate, for $f$ normalized as in the paper, from below by $N(\log\log N)^{1/2-\epsilon}$ for some $f$ and $n_k$ and from above by $o(N(\log N)^{1/2+\epsilon})$ for all, both stated without proof. Since the lower exponent is below $1/2$, (6) does not answer the problem's $o(N\sqrt{\log\log N})$ question.