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Source. Jin's sixteen-page author manuscript, Section 4, definition and Lemma 1 on p. 14, with the proof continued on p. 15. Page numbers are also PDF page numbers.

Write C(a,b)=∣C∩[a,b]∣C(a,b)=|C\cap[a,b]| for integer endpoints a≤ba\leq b.

Statement. Let A,B⊆N0A,B\subseteq\mathbb N_0, and let h≥1h\geq1 be an integer such that hB=N0hB=\mathbb N_0. Suppose 0≤a≤b0\leq a\leq b and

γ=A(a,b)b−a+1=min⁡a≤t≤bA(a,t)t−a+1>0,\gamma=\frac{A(a,b)}{b-a+1} =\min_{a\leq t\leq b}\frac{A(a,t)}{t-a+1}>0,

where the minimum is over integers (Jin's term for the equality is that AA has a minimal forward ratio on [a,b][a,b]). Then

(A+B)(a,b)≥(b−a+1)γ1−1/h.(A+B)(a,b)\geq(b-a+1)\gamma^{1-1/h}.

For h≥2h\geq2 the assertion also holds when γ=0\gamma=0, with right side zero. For h=1h=1 and γ=0\gamma=0, only the trivial nonnegative bound is asserted; the expression 000^0 is not assigned a value.

Proof. First the minimal-prefix condition gives a bound for every tail. For a<z≤ba<z\leq b,

A(a,z−1)≥γ(z−a),A(a,z-1)\geq\gamma(z-a),

so subtracting from A(a,b)=γ(b−a+1)A(a,b)=\gamma(b-a+1) gives

A(z,b)≤γ(b−z+1).A(z,b)\leq\gamma(b-z+1).

The same inequality is an equality when z=az=a.

Set m=b−am=b-a and

F=(A∩[a,b])−a⊆[0,m].F=(A\cap[a,b])-a\subseteq[0,m].

This is a nonempty finite set with ∣F∣=γ(m+1)|F|=\gamma(m+1). Translation of the tail bound gives

∣F∩[z,m]∣≤γ(m−z+1)(0≤z≤m).|F\cap[z,m]|\leq\gamma(m-z+1)\qquad(0\leq z\leq m).

Take any nonempty A′⊆FA'\subseteq F and put z=min⁡A′z=\min A'. Because hB=N0hB=\mathbb N_0, all integers from zz to mm belong to A′+hBA'+hB. No smaller integer belongs to this sumset, since its summands are nonnegative and every element of A′A' is at least zz. Therefore

(A′+hB)∩[0,m]=[z,m],(A'+hB)\cap[0,m]=[z,m],

where the right side denotes the integer interval. Also

0<∣A′∣≤∣F∩[z,m]∣≤γ(m−z+1).0<|A'|\leq|F\cap[z,m]|\leq\gamma(m-z+1).

Consequently every subset in the minimum defining Dm,hD_{m,h} for A0=FA_0=F satisfies

∣(A′+hB)∩[0,m]∣∣A′∣=m−z+1∣A′∣≥1γ.\frac{|(A'+hB)\cap[0,m]|}{|A'|} =\frac{m-z+1}{|A'|}\geq\frac1\gamma.

It follows that Dm,h≥γ−1D_{m,h}\geq\gamma^{-1}. The external [[additive_bases/jin_2014_density_versions_plunnecke_inequality/theorem_3|Theorem 3]] now gives

∣(F+B)∩[0,m]∣∣F∣≥Dm,1≥Dm,h1/h≥γ−1/h.\frac{|(F+B)\cap[0,m]|}{|F|} \geq D_{m,1}\geq D_{m,h}^{1/h}\geq\gamma^{-1/h}.

Every element counted in (F+B)∩[0,m](F+B)\cap[0,m], after adding aa, belongs to (A+B)∩[a,b](A+B)\cap[a,b]. Multiplying the last display by ∣F∣=γ(m+1)|F|=\gamma(m+1) proves

(A+B)(a,b)≥(m+1)γ1−1/h.(A+B)(a,b)\geq(m+1)\gamma^{1-1/h}.

If h≥2h\geq2 and γ=0\gamma=0, nonnegativity proves the separately stated zero bound. This completes the proof.

Source normalization. The manuscript writes A0=A−aA_0=A-a before invoking Theorem 3. The finite set FF above explicitly removes irrelevant elements outside [a,b][a,b], ensuring that the translated set is nonnegative as that theorem requires. The printed sentence on p. 15 says z=min⁡A0z=\min A_0; the correct minimum in its subset argument is min⁡A′\min A'. The proof above checks the ratio for every nonempty A′A', so neither a choice of the wrong minimum nor an unjustified equality of minima is needed. No zero is adjoined to the original set AA.

Dependencies. Theorem 3 is an external Plünnecke graph inequality; the tail estimate and all other steps are included here.

Bears on. #35, through [[additive_bases/jin_2014_density_versions_plunnecke_inequality/theorem_2|Theorem 2]].