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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

Setting. As in Theorem 1: SA=A+A\mathcal S_{\mathcal A}=\mathcal A+\mathcal A, B(X,d)={x∈X:x−d∉X}\mathcal B(\mathcal X,d)=\{x\in\mathcal X:x-d\notin\mathcal X\}, and B(X,d,n)B(\mathcal X,d,n) is the counting function of B(X,d)\mathcal B(\mathcal X,d); A(N)A(N) is the number of elements of A\mathcal A up to NN (p. 329).

Theorem 2 (p. 331, quoted). "There is a positive absolute constant c2c_2 such that for every infinite Sidon set A\mathcal A and all d∈Nd\in\mathbb N we have"

lim sup⁡N→+∞B(SA,d,N)(A(N))−2>c2.(3.2)\limsup_{N\to+\infty}B(\mathcal S_{\mathcal A},d,N)(A(N))^{-2}>c_2. \qquad(3.2)

The paper adds that c2=10−7c_2=10^{-7} can be taken (p. 331).

Remarks on p. 331. For every infinite set of positive integers, B(SA,d,N)(A(N))−2<c3B(\mathcal S_{\mathcal A},d,N)(A(N))^{-2}<c_3 for all NN. The limsup in (3.2) cannot be replaced by a liminf: the paper sketches an infinite Sidon set built by the greedy algorithm along the rapidly growing scale N1=1000N_1=1000, Nk+1=NkNkN_{k+1}=N_k^{N_k}, adding at each stage a block of ≫Nk+11/10\gg N_{k+1}^{1/10} elements inside [Nk+1−[Nk+11/2],Nk+1][N_{k+1}-[N_{k+1}^{1/2}],N_{k+1}], and bounds B(SA,d,Nk)≪A(Nk−1)log⁡A(Nk−1)B(\mathcal S_{\mathcal A},d,N_k)\ll A(N_{k-1})\log A(N_{k-1}), far below A(Nk)2A(N_k)^2 on that scale. The authors conclude that Theorem 2 is best possible apart from the value of c2c_2.

Source. P. Erdős, A. Sárközy, V. T. Sós, On Sum Sets of Sidon Sets, I, J. Number Theory 47 (1994), 329--347, doi:10.1006/jnth.1994.1040; the statement and remarks on p. 331, the proof in Sections 4--8, pp. 331--337. The edition read is identified on the source card.

Read depth. Claims checked: the statement and the remarks were read clause by clause on the page images of the journal print. The proof was read but not checked step by step.

Proof pointer

Sections 4--8, pp. 331--337, by contradiction from the assumption that the limsup is below a small δ\delta (4.1). First, every infinite set has infinitely many NN with A(N+j)/A(N)<((N+j)/N)2A(N+j)/A(N)<((N+j)/N)^2 for all jj (4.2). For such NN, with r=e−1/Nr=e^{-1/N} and f(z)=∑a∈Azaf(z)=\sum_{a\in\mathcal A}z^a, the proof bounds J=∫01∣(1−zd)f2(z)∣2 dα\mathcal J=\int_0^1|(1-z^d)f^2(z)|^2\,d\alpha on the circle ∣z∣=r|z|=r from both sides.

From below (Section 6), Cauchy--Schwarz, Parseval and the Sidon property, which makes a−a′=da-a'=d have at most one solution, give J>10−4A2(N)\mathcal J>10^{-4}A^2(N) (6.4).

From above (Section 7), Parseval is applied to the coefficients v(n)v(n) of (1−zd)f2(z)(1-z^d)f^2(z). Each ∣v(n)∣|v(n)| is at most 22, and v(n)≠0v(n)\ne0 only when n∈B(SA,d)n\in\mathcal B(\mathcal S_{\mathcal A},d), or when n∉SAn\notin\mathcal S_{\mathcal A} and n−d∈SAn-d\in\mathcal S_{\mathcal A} (such nn up to mm are no more numerous than the elements of B(SA,d)\mathcal B(\mathcal S_{\mathcal A},d) up to mm, (7.7)), or when nn or n−dn-d is 2a2a for some a∈Aa\in\mathcal A. With (4.1) and (4.2) this gives J<500δA2(N)\mathcal J<500\delta A^2(N) (7.13).

The two bounds contradict each other for δ=10−7\delta=10^{-7} (Section 8, pp. 336--337).

Dependencies

None beyond the facts proved in the paper.

Bears on

  • Problem 864: the paper records in its Problem 5 (p. 346) that the method of this proof cannot be adapted to nearly Sidon sets, a class that by that page's note contains the sets of Problem 864 of growing size. The theorem concerns infinite Sidon sets and gives no bound for that problem.