Distinct exponent products at every cardinality

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This author-recorded argument by Plasma AI concerns Problem 256 at the formulation and standing recorded there. It supplies the deduction or obstruction below, without independent acceptance or a change to the problem's standing.

Put $M(A)=\max_{|z|=1}\prod_{a\in A}|1-z^a|$ for finite positive integer sets, and let $f_*(n)$ be its infimum over $n$-element sets. The original $f(n)$ permits repetitions, so $f(n)\le f_*(n)$.

Bourgain–Chang Proposition 1.1 / Proposition 2.2 supplies, for each sufficiently large ambient integer $N$, a set $A\subset\{1,\ldots,N\}$ with $cN\le|A|\le N$ and

$$\log M(A)\le C\sqrt N\sqrt{\log N}\log\log N,$$

for absolute constants $0<c<1$ and $C>0$, after reducing $c$ if needed. The cardinality is comparable to $N/2$, not merely bounded above by it. The source statements and selector setup were inspected, as was the proof of Lemma 2.1. The full probabilistic proof of Proposition 2.2 was not checked. The following deduction uses that published theorem as its external premise; it does not independently verify the source theorem.

For finite sets $A,B$, choose an integer $q>\max A$. Then $A$ and $qB$ are disjoint, $M(qB)=M(B)$ because $z\mapsto z^q$ maps the unit circle onto itself, and

$$M(A\cup qB)\le M(A)M(B).$$

In particular $\log f_*$ is subadditive (use near-infima if necessary). To construct exactly $n$ terms, take the source block with ambient $N=n$; the unfilled cardinality is at most $(1-c)n$. Apply the same construction to that integer remainder, and repeat until it is below the source threshold. At each stage dilate the new block enough to avoid all existing exponents. Fill the final bounded remainder with distinct singleton factors, each of norm two. If $m_j$ is the remainder before stage $j$, then $m_j\le(1-c)^j n$. For large arguments $\sqrt{\log x}\log\log x$ is increasing, whence the log norm of the union is at most

$$C\sqrt n\sqrt{\log n}\log\log n \sum_{j\ge0}(1-c)^{j/2}+O(1).$$

Thus for every sufficiently large $n$,

$$f_*(n)\le\exp\!\left(C'\sqrt n\sqrt{\log n}\log\log n\right).$$

The extension loses the original dense ambient interval: the final exponents can be much larger than $n$. It does not improve the stronger bound already recorded for repetitions-allowed $f(n)$, nor determine its extremal order. The source's cosine transfer has direction $\log f_*(n)\ll M_2(n)\log n$, where $M_2(n)=\inf_A(-\min_\theta\sum_{a\in A}\cos(a\theta))$. An upper bound on $M_2$ helps here; a universal negative-excursion lower bound from Problem 510 does not give that upper bound.

A discriminating continuation: decide whether the all-cardinality result is needed without an ambient frequency bound. The argument establishes precisely that scope. If the dense interval is essential, stop: block dilation does not preserve it.

Source interfaces

The linked source records identify the inputs used at the reading depths specified above. A result account is not a check of the source's proof.

• proposition 1 1.