I'm trying to prove the following claim:
Let $\mathbf x _1,...,\mathbf x_n\in \mathbf R ^p$ and $f:\mathbf R ^p \times ... \times \mathbf R ^p \ \ \text{(n times!)}\rightarrow \mathbf R.$ Suppose that $$f(\mathbf x _1 + \mathbf c ,...,\mathbf x _n + \mathbf c)=f(\mathbf x _1 ,...,\mathbf x _n)$$ for each $\mathbf c \in \mathbf R ^p$ and each set of $\mathbf x _i$'s. Then there exist a function $g:\mathbf R ^p \times ...\times \mathbf R ^p \ \ (n-1 \text { times!}) \rightarrow \mathbf R$ such that:$$f(\mathbf x _1,...,\mathbf x _n)=g(\mathbf x _1-\mathbf x _2,\mathbf x _2-\mathbf x _3,...,\mathbf x _{n-1}-\mathbf x _n).$$
So to speak (and clarify last notation), that means that if $f$ is invariant by uniform translation of points $\mathbf x _i$'s, then it can depend only on relative coordinates $\mathbf x _i - \mathbf x _j$. The idea seems simple, yet I can't find a rigorous proof of why this is true. Any hints?
Observation. There's no need to use all couples ($i,j$) to write $g$, but only the $n-1$ couples $(1,2),(2,3),(3,4),...$. In fact, if $i< j $, $x_i-x_j=\sum _{k=i} ^{k=j-1} x_k-x_{k+1}$.
Try $g(y_1,y_2,\ldots,y_{n-1})=f(y_1+y_2+\cdots+y_{n-1},y_2+y_3+\cdots+y_{n-1},\ldots,y_{n-1},0)$.