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Let $f: [0,1]^n \rightarrow \mathbb{R}_+^n$ be a vector-field where the function $f_i$ for each dimension $i$ is defined as:

$$ f_i(\vec{x}) = \sqrt{ \sum_{j=1}^n x_i \cdot x_j \cdot C_{i,j}} $$

Where we are given the $n \times n$ matrix denoted $C$ with elements $C_{i,j} \in [0,1]$ for all $i$ and $j$, and the diagonal equals 1 so $C_{i,i}=1$, and the other elements are symmetrical so $C_{i,j} = C_{j,i}$.

We are also given $\vec{a} = [a_1, a_2, .., a_n]$ with $a_i \in [0,1]$ for all $i$.

Question 1: Does a solution $\vec{x} \in [0,1]^n$ exist that causes $f(\vec{x})$ to equal $\vec{a}$?

Question 2: Is $\vec{x}$ unique?

Question 3: How can we find $\vec{x}$?

Please explain in a manner that can be understood by amateur mathematicians like myself.

Note that I have found a couple of numerical algorithms to solve this problem, one is shown in another question, and another way is to derive the inverse of $f_i(\vec{x})$ and use that to find each $x_i$ that causes $f_i(\vec{x}) = a_i$, but this causes $f_j(\vec{x}) \neq a_j$ for the other dimensions $j$, so we need to do this for several iterations and then it converges to the correct solution. However, I am having trouble formally proving the convergence of those algorithms, so I would like to know if we can at least claim that a unique solution even exists, using only mathematical arguments?

Thanks!

questiondude
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  • Your $f$ does not map to $[0,1]^n$. A counterexample is given in the case $n=2$, where $C$ consists of all ones and $\vec x = (1,1)$. Maybe you want to multiply the sum in the definition of $f$ with $\frac 1n$. Then indeed, $f : [0,1]^n\to[0,1]^n$. For simplicity, I would also leave away the square root because the desired statement $[0,1]^n\subset f([0,1]^n)$ is independent of it. – amsmath Sep 06 '21 at 07:23
  • I can at least show that, when you define $f$ as I suggested (with $\frac 1n$ and without square root), it is locally a diffeomorphism on $(0,1)^n$. To see this, find first that for $x\in (0,1)^n$ we have$$n\operatorname{diag}(x_1,\ldots,x_n)^{-1}\cdot f'(x) = C+D,\quad\text{where }D=\sum_{j=1}^nx_j\operatorname{diag}(c_{1j}x_1^{-1},\ldots,c_{nj}x_n^{-1}).$$Now, let $v\in\mathbb R^n$ such that $f'(x)v=0$. Then $(C+D)v=0$ and in particular, $v^T(C+D)v = 0$. The latter means that$$2\sum_k v_k^2 + \sum_k\sum_{j>k}c_{jk}(a_{jk}v_k + a_{jk}^{-1}v_j)^2 = 0,$$where $a_{jk}=x_j/x_k$, hence $v=0$. – amsmath Sep 06 '21 at 09:18
  • Oops! I made a mistake. The co-domain should be $\mathbb{R}_+^n$ and not $[0,1]^n$. My actual problem is a bit more complicated than what I have written here and I also have an additional normalization step at the end to bring the solution for $x_i$ back to the $[0,1]$ range. I made a mistake when translating my actual problem to a concise question here. However, I do need the square-root and I cannot simply normalize by taking $1/n$. Does that affect your solution? And thanks very much for taking a crack at this problem! – questiondude Sep 06 '21 at 09:29
  • "However, I do need the square-root" - No. For the questions you pose, you don't. The answer would be the same without the square root. – amsmath Sep 06 '21 at 09:32
  • I had to lookup "diffeomorphism" as I am not familiar with that term. I am trying to understand the implications of what you wrote. Have you given a proof that $\vec{x}$ exists and it is a unique solution to $f(\vec{x})=\vec{a}$ because $f$ is a diffeomorphism? Have I understood correctly? – questiondude Sep 06 '21 at 10:01
  • I didn't prove it's a diffeomorphism. Then you would (almost) be done. ;-) I only proved that it is locally a diffeomorphism, meaning that for every $x\in (0,1)^n$ there exists an open neighborhood $U$ around $x$ such that the restriction $f : U\to f(U)$ is a diffeomorphism. In particular, $f(U)$ is open. But that doesn't answer any of your questions, yet. – amsmath Sep 06 '21 at 10:04
  • Hehe! OK! Well, I look forward to seeing if you can prove it and I greatly appreciate the time and effort you are putting into this! I also find it very interesting that you and the other poster are attacking the problem from seemingly very different angles. – questiondude Sep 06 '21 at 10:49

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