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Source. Jin's sixteen-page author manuscript, Theorem 2 on p. 3; the simplified proof is Section 4, pp. 14–15. This is Jin's proof of the theorem attributed there to Plünnecke (1970), not a transcription of the unacquired 1970 paper. All page numbers refer to the author manuscript.

Statement. Let $A,B\subseteq\mathbb N_0$ and let $h\geq1$ be an integer such that $hB=\mathbb N_0$. Define

$$\alpha=\sigma(A)=\inf_{N\geq1}\frac{|A\cap[1,N]|}{N}.$$

For $h\geq2$,

$$\sigma(A+B)\geq\alpha^{1-1/h}.$$

For $h=1$ and $\alpha>0$, $\sigma(A+B)=1$, which is the same formula. For $h=1$ and $\alpha=0$, the asserted bound is only $\sigma(A+B)\geq0$. The undefined expression $0^0$ is not used.

The hypothesis $hB=\mathbb N_0$ forces $0\in B$. It is equivalent to representation of every nonnegative integer by at most $h$ elements of $B$ when $0\in B$ is given, by padding with zeros. No hypothesis $0\in A$ is imposed.

Proof. If $\alpha=0$, the stated zero bound is immediate. If $h=1$, then $B=\mathbb N_0$; positive Schnirelmann density forces $1\in A$ by the cutoff $N=1$. Thus $1+\mathbb N_0\subseteq A+B$ and the density is one. If $\alpha=1$, then $A\subseteq A+B$ because $0\in B$, so again the density is one. It remains to treat $h\geq2$ and $0<\alpha<1$.

Fix any integer $N\geq1$. Write $A(a,b)=|A\cap[a,b]|$. We construct integers

$$1=n_0<n_1<\cdots<n_k=N+1$$

by the following finite procedure. If $n_{i-1}\leq N$, put $a=n_{i-1}$ and define

$$\alpha_i=\min_{a\leq t\leq N}\frac{A(a,t)}{t-a+1}.$$

There are finitely many nonempty prefixes, so a minimizer exists. Let $t_i$ be the greatest integer attaining this minimum and set $n_i=t_i+1$. Then $n_i>n_{i-1}$ and $n_i\leq N+1$. Each step consumes at least one integer, so the procedure stops after at most $N$ steps, necessarily at $n_k=N+1$. On each block $[n_{i-1},n_i-1]$ the density is $\alpha_i$ and is minimal among all prefixes of that block.

We next check the density monotonicity used by Jin. The first block starts at 1, so $\alpha_1\geq\sigma(A)=\alpha$. If a next block exists, its concatenation with the preceding block has density

$$\frac{(n_i-n_{i-1})\alpha_i+(n_{i+1}-n_i)\alpha_{i+1}} {n_{i+1}-n_{i-1}}.$$

Both lengths are positive. If $\alpha_{i+1}<\alpha_i$, this is less than $\alpha_i$, contradicting the minimum over prefixes ending at most $N$ from $n_{i-1}$. If $\alpha_{i+1}=\alpha_i$, the same minimum is attained at the larger endpoint $n_{i+1}-1>n_i-1$, contradicting the greatest-endpoint choice. Hence

$$0<\alpha\leq\alpha_1<\alpha_2<\cdots<\alpha_k\leq1.$$

Apply the fully proved [[additive_bases/jin_2014_density_versions_plunnecke_inequality/lemma_1|Lemma 1]] to each block. The output intervals are disjoint and partition $[1,N]$, so

$$\begin{aligned} |(A+B)\cap[1,N]| &=\sum_{i=1}^k (A+B)(n_{i-1},n_i-1)\\ &\geq\sum_{i=1}^k(n_i-n_{i-1})\alpha_i^{1-1/h}\\ &\geq\sum_{i=1}^k(n_i-n_{i-1})\alpha^{1-1/h} =N\alpha^{1-1/h}. \end{aligned}$$

The second inequality uses that $x^{1-1/h}$ is increasing for $x\geq0$. Dividing by $N$ and taking the infimum over all positive integer cutoffs proves the theorem. Every choice above was made inside a finite interval; there is no assumption that the global infimum defining $\sigma(A)$ is attained.

Implication for Problem 35. For $\alpha=0$ the requested increment bound is zero. For $0<\alpha\leq1$, put $x=1-\alpha\in[0,1)$ and $r=1/h$. The function

$$g(x)=(1-x)^{-r}-1-rx$$

satisfies $g(0)=0$ and $g'(x)=r((1-x)^{-r-1}-1)\geq0$. Therefore

$$\alpha^{1-1/h} =\alpha(1-x)^{-1/h} \geq\alpha\left(1+\frac{x}{h}\right) =\alpha+\frac{\alpha(1-\alpha)}h.$$

This applies also to the separately proved positive-density order-one case. Taking $h=k$ yields exactly the bound asked in Problem 35.

Dependencies and scope. Lemma 1 is the only additional same-paper lemma required for this proof, and its proof is included in full. Its external input is [[additive_bases/jin_2014_density_versions_plunnecke_inequality/theorem_3|Theorem 3]]. Jin's asymptotic and Banach density arguments are not prerequisites for Section 4. The order-one endpoint qualification matters: with $A=\{2\}$ and $B=\mathbb N_0$, both $\sigma(A)$ and $\sigma(A+B)$ vanish, so reading the printed formula using $0^0=1$ would be false.

Bears on.

• #35: with the elementary inequality above and $h=k$, the theorem implies the inequality the problem asks for, for every basis of order $k\geq1$ and every $A$.
• #37: every Schnirelmann basis (which contains 0) is an essential component, because the bound is strictly larger than $\alpha$ for $0<\alpha<1$; this says nothing about lacunary sets.
• #38: this controls $A+B$ for a basis $B$; it does not assert that one shift $b\in B$ gives a density increment, and the problem asks about sets that are not bases.