Can $n+1$ distinct positive vectors in $\mathbb{R}^n_{>0}$ agree on $n$ distinct weighted $p$-norms?

Question

Consider $n$ distinct positive numbers $\{p^{(1)},...,p^{(n)}\}\subset [1,\infty)$ along with weights $\{w^{(1)},...,w^{(n)}\}\subset\Delta^{n-1}$ and scalars $\{q^{(1)},...,q^{(n)}\}\subset(0,\infty)$.

Let $X$ be the set of positive vectors defined by the $n$ weighted weighted $p$-norms, $$X=\left\{\vec x\in (0,\infty)^n\ \middle|\ \left(\sum_{i=1}^n w^{(k)}_i (x_i)^{p^{(k)}}\right)^{1/p^{(k)}}\!\!\!\!=q^{(k)}\text{ for all } k=1,...,n\right\}.$$

Is it true that $|X|\leq n$?

Update: Connor points out that a simple permutation over components always maintains all $p$-norms, so obviously the statement is wrong as written for $n\geq 3$. But are there less `special' counterexamples?

More formally, if you consider all the sets $X\subset (0,\infty)^n$ of cardinality $|X|=n+1$ that satisfy the definition above for some sets of $w$, $p$ and $q$, and you put them all into a `superset' $Y$... does $Y$ have zero Lebesgue measure as a subset of $((0,\infty)^n)^{n+1}$?

My ultimate goal is to establish the following: If I take a random countable collection of positive $n$-dimensional vectors (in the Lebesgue sense), then for almost all $p>0$, at most $n$ of those vectors will lie in the same weighted $l_p$-sphere (for any weights $w$, and any radius $q$).

Answer 1

I haven't found an answer to your question, but here is a different spin to your question, that might help finding an answer: if we define the weighted p-norm $$ \lVert x \rVert_{w,p} = \left( \sum_{i=1}^n w_i x_i^p \right)^{1/p},$$ and $$\mathbb{R}^n_{+,o} = \{ x \in (0, \infty)^n: x_1 \leq x_2 \leq \cdots \leq x_n \},$$ then if I understand correctly, you would like to show that for any choice choice of weights $\{w^{(1)},...,w^{(n)}\}\subset\Delta^{n-1}$, norms $\{p^{(1)},...,p^{(n)}\}\subset [1,\infty)$, and scalars $\{q^{(1)},...,q^{(n)}\}\subset(0,\infty)$, the set $$ X = \bigcap_{k=1}^n \left\{ x \in \mathbb{R}^n_{+,o} : \lVert x \rVert_{w^{(k)}, p^{(k)}} = q^{(k)} \right\} $$ has cardinality at most $n$. Of course there are some choices of weights, norms, and scalars that trivially yield $|X| > n$, but we rule these out. This invites a new way of viewing the problem: you are interested in the cardinality of the intersection of weighted $l_p$ spheres, and looking into this literature might lead to a solution (can any analyst help?). Another way at looking at the problem is that you are essentially looking at solutions to a system of polynomial equations, and this could also lead to a solution.

(I apologize that this is perhaps more of a comment than an answer, but since I am new to stackexchange I do not have the reputation to post comments yet.)

Answer 2

After having gathered some help from various sources on several related sub-questions, I'm going to give this a shot myself. Comments are very much appreciated!

Let us define the mapping $g:\mathbb{R}^n\times [1,\infty)\to\mathbb R^n$ component-wise by $g_i(x,p)=x_i^p$. Note first that the collection of vectors $X=(x^1,...,x^{n+1})\in\mathbb (R^n)^{n+1}$ share the same weighted $p$-norm if and only if the vectors $\{g(x^1,p),...,g(x^{n+1},p)\}$ are affinely dependent, which in turn is equivalent to requiring that $$G(X,p)=\textsf{Det}\left(\ \left[\begin{array}{ccc} (x_1^1)^p &\cdots &(x_1^{n+1})^p\\ \vdots &\ddots &\vdots\\ (x_1^n)^p &\cdots &(x_n^{n+1})^p\\ 1 &\cdots&1 \end{array} \right]\ \right)=0.$$ $G$ is an analytic function (meaning it can be written as a convergent power series). Its pre-image $G^{-1}(0)$ is therefore bounded and Lebesgue measurable.

I now use Fubini's theorem to show that $G^{-1}(0)$ has measure zero in $(\mathbb R^n)^{n+1}\times [1,\infty)$. Indeed, consider first what happens when $p=1$. Then it can be shown by induction that $n+1$ vectors are almost always affinely independent ($G\neq 0$) since any affine subspace over $\leq n$ points is of dimensionality $\leq n-1$. So there exists a set $E\subseteq(\mathbb R^n)^{n+1}$ of measure zero such that for all collections $X\notin E$, $G(X,\cdot)$ is a nonzero generalized polynomial. But in that case, it admits at most a finite number of roots, which we refer to as the zero-measure set $D(X)\subset[1,\infty)$.

We know that $$G^{-1}(0)\subseteq \big(E\times[0,\infty)\big)\cup \left(\bigcup_{X\in\mathbb R^{n\times (n+1)}} X\times D(X)\right),$$ so by Fubini's theorem, its measure is bounded above by $$\int_1^\infty \mu_X(E)d\mu_\lambda+\int_X \mu_\lambda(D(X))d\mu_X=0.$$

I'm not very well versed with measure theory anymore, so comments are more than welcome!!