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For any positive integer $n$, let $$A_n=\idotsint\limits_{\substack{x_1+\cdots+x_n+y_1+\cdots+y_n\leq1\\x_1,\cdots,x_n,y_1,\cdots,y_n\geq0}}\prod_{i,j=1}^n(x_i-y_j)dx_1\cdots dx_ndy_1\cdots dy_n.$$ It is easy to prove that $A_n=0$ when $n$ is odd.

I conjecture that $A_n>0$ when $n$ is even.

To prove it, I constructed the following real polynomial: $$\varphi_n(X_1,X_2,\cdots,X_{2n})=\prod_{\substack{1\leq i\leq n\\n+1\leq j\leq 2n}}(X_i-X_j)$$ and let $$S_{\varphi_n}(X_1,X_2,\cdots,X_{2n})=\sum_{\sigma=[i_1,i_2,\cdots,i_{2n}]\in S_{2n}}f(X_{i_1},X_{i_2},\cdots,X_{i_{2n}}).$$ Here the sum is computed over all permutations $\sigma=[i_1,i_2,\cdots,i_{2n}](\sigma(1)=i_1,\sigma(2)=i_2,\cdots,\sigma(2n)=i_{2n})$ of the set $\{1,2,\cdots,2n\}$ and the set of all such permutations is denoted $S_{2n}.$

Then it is easy to see $$S_{\varphi_n}(a_1,a_2,\cdots,a_{2n})\geq0\ \text{for every}\ (a_1,a_2,\cdots,a_{2n})\in \mathbb{R}^{2n},\text{when}\ n\ \text{is even}$$ $$\Longrightarrow$$ $$A_n=\dfrac{1}{\left(2n\right)!}\idotsint\limits_{\substack{x_1+x_2+\cdots+x_{2n}\leq1\\x_1,x_2,\cdots,x_{2n}\geq0}}S_{\varphi_n}(x_1,x_2,\cdots,x_{2n})dx_1dx_2\cdots dx_{2n}>0,\text{when}\ n\ \text{is even}$$ I believed the former is right, but I can not prove it. So I asked another question here. Unfortunately, it has been proved to be wrong:

When $n$ is even and $n\geq6$, there exists $(a_1,a_2,\cdots,a_{2n})\in \mathbb{R}^{2n}$ such that $S_{\varphi_n}(a_1,a_2,\cdots,a_{2n})<0$.

Now I have no idea how to prove or disprove my original conjecture: $$A_n>0\quad\text{when}\ n\ \text{is even}.$$ So I ask for some help.

Community
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user173856
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  • If I make some generalizations:

    For any positive integers $m,n$, let $$A_{m,n}=\idotsint\limits_{\substack{x_1+\cdots+x_m+y_1+\cdots+y_n\leq1\x_1, \cdots,x_m,y_1,\cdots,y_n\geq0}}\prod_{i=1}^m\prod_{j=1}^n(x_i-y_j)dx_1\cdots dx_mdy_1\cdots dy_n,$$ then we have

    $(1)A_{n,n}=0$ when $n$ is odd.

    $(2)$Follow the answer given by fedja, we can similarly prove that $A_{m,n}>0$ when one of $m,n$ is even.

    What about the sign of $A_{m,n}$ when $m,n$ are different odd numbers?

     – user173856 Dec 24 '15 at 02:31

1 Answers1

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Let's prove a bit more, namely that the integral over $x_i>0, y_j>0, \sum_i x_i+\sum_j y_j=S$ with respect to the $2n-1$-dimensional Lebesgue measure is positive. Note that due to the obvious scaling $x_i,y_j\mapsto tx_i,ty_j$, this integral depends on $S$ in a trivial way (as some pure power function of $S$). Hence, we can just as well introduce any other fast decaying positive weight $w(S)$ and prove that $\idotsint_{(0,+\infty)^{2n}}\prod\dots w(x_1+\dots+x_n+y_1+\dots+\dots y_n)dx_1\dots dx_n dy_1\dots dy_n>0$ instead. Obviously the easiest weight to deal with is just $\exp(-x_1-\dots x_n-y_1-\dots-y_n)$, in which case the integration in $x$ gives $\left(\int_0^\infty P_y(x)e^{-x}\right)^n$ where $P_y(x)=\prod_j(x-y_j)$. For even $n$ this is never negative and certainly not zero for at least some $y$ (say, all $y_j$ equal to each other). Thus the new weighted integral is positive and, thereby, the claim follows.

fedja
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