While working on a research problem on fuzzy metric spaces, I came across a special symmetric function $F_n:X^n\times (0,\infty)\to [0,1]$ i.e.
\begin{equation*} F_n(x_1,x_2,\dots,x_n,t)=F_n(x_{\pi(1)},x_{\pi(2)},...,x_{\pi(n)},t) \end{equation*} for every permutation $\pi$ of $\{1,2,...,n\}$ such that

\begin{equation*} [F_n(x_1,x_2,...,x_n,t)]^{n-2}=\prod_{1\le i_1<i_2<\dots<i_{n-1}\le n} F_{n-1}(x_{i_1},x_{i_2},\dots x_{i_{n-1}},t) \end{equation*} Where $[F_n]^m=F_n\ast F_n\ast\dots(m \quad\text{times})$, $\ast$ being continuous $t$-norm.

I am interested in finding a relation between $F_n(x_1,x_2,\dots,x_n,t)$ and $F_2(x_i,x_j,t), (1\le i<j\le n$). Any suggestions on how to approch the problem?

Note: Here is a somewhat similar situation in graph theory. if we take $F_n$ as the sum of all distances $d(x_i,x_j),1\le i<j\le n$ between different pairs of vertices $x_i$ and $x_j$ of a complete graph $K_n$ with vertex set $X=\{x_1,x_2,\dots,x_n\}$. Then $F_n:X^n\to \mathbb{R}$ represents a symmetric function in variables $x_1,x_2,\dots,x_n\in X$ such that \begin{equation} (n-2)F_n(x_1,x_2,...,x_n)=\sum_{1\le i_1<i_2<\dots<i_{n-1}\le n} F_{n-1}(x_{i_1},x_{i_2},\dots x_{i_{n-1}}) \end{equation} And We have \begin{equation} F_n(x_1,x_2,...,x_n)=\sum_{1\le i<j\le n} F_2(x_i,x_j) \end{equation}

  • 2
    $\begingroup$ The last equality which your wrote holds for the general case too. It can be proved by so straight induction that I am even shy to write it as an answer. Moreover, this proof can be easier seen than read. $\endgroup$ Dec 30 '15 at 5:36
  • $\begingroup$ @AlexRavsky Please write your answer for fuzzy metric $F_n$. $\endgroup$
    – K A Khan
    Jan 1 '16 at 12:52
  • 1
    $\begingroup$ I see the question has been changed. I don’t know what is a fuzzy metric, $t$-norm and what “*” means. But if “*” is the usual multiplication then it seems that for each $t$ the proof is essentially the same, and it holds for an arbitrary (not necessarily symmetric) non-negative function. Do you wish me to write an answer for this case? $\endgroup$ Jan 1 '16 at 16:09
  • $\begingroup$ For fuzzy metric space a good reference is emis.ams.org/journals/IJOPCM/Vol/10/IJOPCM(vol.3.2.2.J.10).pdf (See definition 2.3) and for t-norm please see en.wikipedia.org/wiki/T-norm. Actually I want to prove the relation $F_n(x_1,x_2,...,x_n,t)=\prod_{1\le i<j\le n} F_2(x_i,x_j,t)$. $\endgroup$
    – K A Khan
    Jan 2 '16 at 6:15

We prove that for each $t$

$\mbox{(1) } F_n(x_1,x_2,...,x_n,t)]^{(n-2)!}=\left[\prod_{1\le i<j\le n} F_2(x_i,x_j,t)\right]^{(n-2)!}$

even without the symmetry assumption by straight induction by $n$, starting from $n=3$. For $n=3$ Claim (1) is the same as the condition connecting functions $F_{n}$ and $F_{n-1}$ (I understand the products with respect to the binary operation $*$). Assume that we already have proved the Claim (1) for a number $n-1$. Then

$$[F_n(x_1,x_2,...,x_n,t)]^{(n-2)!}=\left[\prod_{1\le i_1<i_2<\dots<i_{n-1}\le n} F_{n-1}(x_{i_1},x_{i_2},\dots x_{i_{n-1}},t)\right]^{(n-3)!}= \prod_{1\le i<j\le n} [F_2(x_i,x_j,t)]^{ (n-2)!}$$ (this follows from the commutativity and associativity of the operation $*$) and a remark that each pair $(i,j)$ ($i<j$) is in exactly $n-2$ sequences $(i_1,\dots,i_{n-1})$ where $1\le i_1<\dots<i_{n-1}\le n$, that is in all such sequences except $(1,\dots,\hat{i},\dots,j,\dots,n-1,n)$ and $(1,\dots, i,\dots, \hat{j},\dots,n-1,n)$, where $\hat{k}$ means that the number $k$ is skipped from the sequence.

If for each $n>0$ a function $a\mapsto [a]^n$ is strictly monotone then we can drop the powers and obtain that

$$F_n(x_1,x_2,...,x_n,t)=\prod_{1\le i<j\le n} F_2(x_i,x_j,t).$$


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