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Zhi-Wei Sun
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On the set $\{\sum_{k=1}^n \lambda_ka_k:\ a_1,\ldots,a_k\ \text{are distinct elements of}\ A\}|$

For a field $F$ let $p(F)=p$ if the characteristic of $F$ is a prime $p$, and $p(F)=+\infty$ if $F$ is of characteristic zero.

In 2007 I considered the linear extension of the Erdos-Heilbronn conjecture, and conjectured that (cf. http://arXiv.org/abs/0810.0467) for any nonzero elements $\lambda_1,\ldots,\lambda_n$ of a field $F$ with $p(F)\not=n+1$ and a finite subset $A$ of $F$ we have \begin{align}&|\{\lambda_1a_1+\ldots+\lambda_n a_n:\ a_1,\ldots,a_n\ \text{are distinct elements of}\ A\}| \\&\qquad \qquad \ge\min\{p(F)-\delta,\, n(|A|-n)+1\},\end{align} where $\delta$ is $1$ if $n=2$ and $a_1+a_2=0$, and $\delta=0$ otherwise.

Motivated by the above as well as Question 316142 of mine, here I ask the following question.

QUESTION: Is my following conjecture true?

Conjecture. Let $\lambda_1,\ldots,\lambda_n\ (n\ge3)$ be positive integers with $\lambda_1\le\ldots\le\lambda_n\le n$ and $\gcd(\lambda_1,\ldots,\lambda_n)=1$. Let $F$ be a field with $p(F)>n+1$. Then, for any finite subset $A$ of $F$ with $|A|\ge n+\delta_{n,3}$ we have \begin{align}&\bigg|\bigg\{\sum_{k=1}^n\lambda_ka_k:\ a_1,\ldots,a_n\ \text{are distinct elements of}\ A\bigg\}\bigg| \\\ge&\min\bigg\{p(F),\ (\lambda_1+\ldots+\lambda_n)(|A|-n)+\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)+1\bigg\}.\end{align}

Now let me explain where the lower bound comes from. Suppose that $A$ is just the subset $\{1,\ldots,m\}$ of the rational field $\mathbb Q$. For the set $$S=\{\lambda_1a_1+\ldots+\lambda_na_n:\ a_1,\ldots,a_n\ \text{are distinct elements of}\ A\},$$ its minimal element should be $\sum_{k=1}^n\lambda_k(n+1-k)$ while its maximal element should be $\sum_{k=1}^n\lambda_k(m-n+k)$. Note that \begin{align}&\bigg|\bigg\{\sum_{k=1}^n\lambda_k(n+1-k),、 \ldots,\ \sum_{k=1}^n\lambda_k(m-n+k)\bigg\}\bigg| \\=&(\lambda_1+\ldots+\lambda_n)(m-n)+\sum_{k=1}^{\lfloor n/2\rfloor}(n+1-2k)(\lambda_{n+1-k}-\lambda_k)+1. \end{align}

Any comments are welcome!

Zhi-Wei Sun
  • 15.6k
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  • 20
  • 67