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Statement

Setting (first text page, unnumbered, and p. -2-): 0<a1<a2<⋯0<a_1<a_2<\cdots is an infinite sequence of integers and f(n)f(n) is the number of solutions of ai+aj=na_i+a_j=n, read in any of the paper's three conventions: (I) i≠ji\ne j counted twice and i=ji=j once, (II) i≠ji\ne j counted once and i=ji=j once, (III) i≠ji\ne j counted once and i=ji=j excluded.

Theorem 2 (p. -2-, quoted). "If c>0c>0 or c=0c=0 and ak<Ak2a_k<Ak^2, then lim‾⁡n→∞1n∑k=0n(f(k)−c)2>0\varlimsup_{n\to\infty}\frac1n\sum_{k=0}^{n}(f(k)-c)^2>0."

The paper states it for all three conventions. The print does not say how AA is quantified; read as a positive constant, the hypothesis bounds aka_k above by a multiple of k2k^2. The proof treats only the case in which Ak2≤ak≤Bk2Ak^2\le a_k\le Bk^2 holds for large kk with positive constants A,BA,B (the paper's (2)).

Context (p. -2-). The paper recalls that in conventions (I) and (II) Dirac and Newman proved that f(n)f(n) cannot be constant for n>n0n>n_0. If f(n)=cf(n)=c for all large nn, the mean in Theorem 2 tends to 00; and c=0c=0 is impossible for an infinite sequence, since f(a1+ak)≥1f(a_1+a_k)\ge1 for every k≥2k\ge2 in each convention. So Theorem 2 contains that result and extends it to convention (III) (a filing derivation).

Source. P. Erdős and W. H. J. Fuchs, On a problem of additive number theory, J. London Math. Soc. 31 (1956), 67--73, doi:10.1112/jlms/s1-31.1.67, read in the August 1954 Cornell University technical report printing (Report No. 11, OSR-TN-54-216) identified in the source card: the setting on the first text page (PDF p. 5), the conventions and Theorem 2 on p. -2- (PDF p. 7), the proof on p. -8- (PDF p. 19), read on the page images. The journal version's statement was not compared.

Read depth. Claims checked: the setting, the conventions and the statement were read clause by clause on the page images. The proof was read on the page image but not checked step by step. Nothing here is independently reviewed.

Proof pointer

P. -8-. As for Theorem 1, the paper treats only the case Ak2≤ak≤Bk2Ak^2\le a_k\le Bk^2 for large kk, calling the others trivial. With g(z)=∑kzakg(z)=\sum_kz^{a_k}, the generating function of ff is g(z)2g(z)^2 or 12(g(z)2±g(z2))\frac12\bigl(g(z)^2\pm g(z^2)\bigr) according to the convention. On the circle z=reiθz=re^{i\theta}, Parseval's formula and the Schwarz inequality bound (∑n(f(n)−c)2r2n)1/2\bigl(\sum_n(f(n)-c)^2r^{2n}\bigr)^{1/2} below by a multiple of the integral of the absolute difference between that generating function and c(1−z)−1c(1-z)^{-1}. The integral of ∣g(z)∣2\lvert g(z)\rvert^2 is of order (1−r)−1/2(1-r)^{-1/2} under the growth condition, while those of ∣g(z2)∣\lvert g(z^2)\rvert and ∣1−z∣−1\lvert1-z\rvert^{-1} are O((1−r)−1/4+log⁡11−r)O((1-r)^{-1/4}+\log\frac1{1-r}). Hence ∑n(f(n)−c)2r2n>K(1−r)−1\sum_n(f(n)-c)^2r^{2n}>K(1-r)^{-1}, so the partial sums tn=∑k≤n(f(k)−c)2t_n=\sum_{k\le n}(f(k)-c)^2 satisfy ∑ntnr2n>K(1−r)−2\sum_nt_nr^{2n}>K(1-r)^{-2}, which gives lim‾⁡tn/n>0\varlimsup t_n/n>0.

Bears on

No problem page in the corpus cites this theorem, and none is recorded here.