Problem:
Given a domain $\mathcal{D}\subset\mathbb{R}^{l}$, we can find $l$ points $\boldsymbol{v}_{i}\in\mathcal{D}$, $i=1,\cdots,l$. Each point is a column vector with dimension $l\times1$. They form a matrix $\mathbf{V}$, $ \mathbf{V}=\left[\begin{array}{ccc} \boldsymbol{v}_{1}, & \cdots & ,\boldsymbol{v}_{l}\end{array}\right] $. Denote the set of all such $\mathbf{V}$ that is invertible as $\mathcal{V}$. That is, $$ \mathcal{V}=\{\mathbf{V}=\left[\begin{array}{ccc} \boldsymbol{v}_{1}, & \cdots & ,\boldsymbol{v}_{l}\end{array}\right]:\boldsymbol{v}_{i}\in\mathcal{D},\mathbf{V}\text{ is invertible}\}. $$
Consider a given vector $\boldsymbol{y}$. For a $\mathbf{V}\in\mathcal{V}$, define $$ \boldsymbol{x}=\mathbf{V}^{-1}\boldsymbol{y}. $$
Denote $\mathbf{V}^{*}$ as the invertible matrix that minimizes the L1 norm of $\boldsymbol{x}$. That is \begin{align*} \mathbf{V}^{*} & =\arg\min_{\mathbf{V}\in\mathcal{V}}\left|\left|\boldsymbol{x}\right|\right|_{1}\\ & =\arg\min_{\mathbf{V}\in\mathcal{V}}\left|\left|\mathbf{V}^{-1}\boldsymbol{y}\right|\right|_{1}. \end{align*} Denote $\boldsymbol{x}^{*}=\mathbf{V}^{*-1}\boldsymbol{y}$.
For an arbitrary $\boldsymbol{u}\in\mathcal{D}$, denote $\boldsymbol{z}=\mathbf{V}^{*-1}\boldsymbol{u}$. If for each component of $\boldsymbol{x}^{*}$, we have $\left|x_{i}^{*}\right|\neq0$, $i=1,\cdots,l$, prove that
\begin{align*} \left|\sum_{i=1}^{l}z_{i}\frac{x_{i}^{*}}{\left|x_{i}^{*}\right|}\right| & \leq1. \end{align*}
Things I have tried:
I have also post this problem here to reach out to more audience.
The case when $l=1$ is easy to prove. Denote the domain $\mathcal{D}$ as $[a,b]$. For a given $y$, denote the optimal value of $v$ that minimize $\left|\frac{y}{v}\right|$ as $v^{*}$. One can easily get that $v^{*}$ is either $a$ or $b$.
The optimal condition means that , $\forall u\in[a,b]$, we have \begin{align*} \left|x^{*}\right| & =\left|\frac{y}{v^{*}}\right|\\ & \leq\left|\frac{y}{u}\right|. \end{align*} So $$ \left|\frac{1}{v^{*}}u\right|\leq1, $$ which is exactly the inequality we want to prove. I also try some special cases when $l=2$, and the conclusion holds. But I cannot find a general proof for $l\geq2$.
I also asked ChatGPT 4 for help, but found no valid proof from it. However, it gave some inspiration. Sherman–Morrison formula might be useful when considering replacing the $i$ column of $\mathbf{V}^{*}$ with $\boldsymbol{u}$ ($\boldsymbol{y}$) and multiplying the resulted matrix with $\boldsymbol{y}$ ($\boldsymbol{u}$). But I got no further progress.
About the domain $\mathcal{D}$:
In the problem $\mathcal{D}$ should be compact, i.e., closed and bounded.
But it seems that $\mathcal{D}$ is not necessarily convex. Consider the following toy and random example when $l=2$ and $\mathcal{D}$ is non-convex.
The domain $\mathcal{D}$ is given by $$ \mathcal{D}=\left\{ \left(v_{1},v_{2}\right)^{\top}:v_{1}\in\left[-6,6\right],\left|v_{1}\right|\leq v_{2}\leq\frac{1}{2}\left|v_{1}\right|+3\right\} . $$ One can easily see that it is non-convex in this plot of $\mathcal{D}$.
The vector $\boldsymbol{y}$ is given by $\boldsymbol{y}=\left(9,-2\right)^{\top}$. After some efforts, one can get that \begin{align*} \mathbf{V}^{*} & =\left[\begin{array}{cc} 6 & -6\\ 6 & 6 \end{array}\right],\\ \boldsymbol{x}^{*} & =\left(\frac{7}{12},-\frac{11}{12}\right)^{\top}. \end{align*}
We then have \begin{align*} \boldsymbol{z} & =\mathbf{V}^{*-1}\left(v_{1},v_{2}\right)^{\top}\\ & =\left(\frac{v_{1}}{12}+\frac{v_{2}}{12},\frac{v_{2}}{12}-\frac{v_{1}}{12}\right)^{\top}. \end{align*}
It is then $$ |z_{1}-z_{2}|=\left|\frac{v_{1}}{6}\right|\leq1. $$
By the way, the vector $\boldsymbol{y}$ can not be $\boldsymbol{0}$, and I think we can assume that the first component of $\boldsymbol{y}$ is non-zero and scale it to 1.
In this way, $\mathbf{V}^{*}$ remains the same, $\left|\left|\boldsymbol{x}^{*}\right|\right|_{1}$ changes up to a constant, $\text{sgn}(\boldsymbol{x}^{*})$ either remains the same or is reversed. The conclusion will still holds if it is correct.
But I do not know whether this assumption helps.
