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Suppose that $\mathbf a = (a_1,...,a_d)$ is a constant in $\mathbb R_{+}^d \setminus \{\mathbf 0\}$ (some $a_i$ can be $0$ but not all, all are $\ge 0$).

Define the set $$S(\mathbf a) = \left\{\mathbf x \in \mathbb C^d:\; \sum_{i=1}^d \frac{x_i +1}{x_i -1}a_i = 1 \right\}.$$

Define the distance of the set to zero as :

$$d(\mathbf a) = \min\limits_{\mathbf x \in S(a)} \sum_{i=1}^d \lvert x_i \rvert$$

How can I caracterize this distance w.r.t the constant $\mathbf a$ ? I am specifically interested in conditions on $\mathbf a$ that will garentie $d(\mathbf a)$ to be smaller or greater than $1$.

I tried to consider the function $f(x) = \frac{x+1}{x-1}$, which is it's own inverse, and apply it componentwise on the frontier of the $\mathbf a$-simplex : $\Delta(\mathbf a) = \left\{\mathbf x:\; \langle \mathbf x, \mathbf a \rangle = 1\right\}$. But this did not help me as my complex geometry knowledge are not as good as i'd like.

Please re-tag the question if you feel the tags are not rights.

Edit: cross-posted on mathematica stack exchange to eventually find aa numerical/analyticial solution

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  • $\begingroup$ If I understand it right, $x\mapsto \frac{x+1}{x-1}$ maps the unit disk in $\mathbb C$ to the left half-plane $\Re z\le 0$, so you cannot even make all $|x_i|\le 1$, forget about $d(a)$. Perhaps you meant something else? $\endgroup$
    – fedja
    Feb 4, 2021 at 15:16
  • $\begingroup$ Are you saying that $d(a)$ will always be greater than one ? If yes, this is exactly what i want and i did not manage to proove it. if no, please clarify your remark I did not understand :/ $\endgroup$
    – lrnv
    Feb 4, 2021 at 15:33
  • $\begingroup$ Yes, this is exactly what I'm saying: $d(a)\ge\max|x_i|> 1$. $\endgroup$
    – fedja
    Feb 4, 2021 at 16:49
  • $\begingroup$ This is a very good news for me, but sadly I'm not getting why it is so.. would you mind developing a little more the argument ? Your max is taken over every $\mathbf x \in S(\mathbf a)$, or over every $x_i$ dimension wise ? $\endgroup$
    – lrnv
    Feb 4, 2021 at 21:56
  • $\begingroup$ I posted the details as an answer. If there are more questions, feel free to ask them. My formula in the comment was rather sloppily written, indeed. It should be $d(a)\ge \min_{x\in S(a)}\max|x_i|>1$. $\endgroup$
    – fedja
    Feb 4, 2021 at 22:54

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OK, here are full details. Hope that I don't say any nonsense.

The first observation is that $x\mapsto \frac{x+1}{x-1}$ maps the unit disk $|x|\le 1$ to the left half-plane $z\le 0$. You can see it by the general linear fractional nonsense or just observe that $$ 2\Re \frac{x+1}{x-1}=\frac{x+1}{x-1}+\frac{\bar x+1}{\bar x-1}=2\frac{|x|^2-1}{|x-1|^2}\le 0. $$ Thus, if all $|x_i|\le 1$, your sum in the definition of $S(a)$ has a non-positive real part, so it has no chance to be $1$. This immediately shows that $d(a)\ge 1$.

To show that $d(a)>1$, we just remove all $a_i$ that are $0$ (they are of no help anyway), so we can assume that all $a_i>0$, and observe that then $S(a)\cap\{x:|x_i|\le 2, |x_i-1|>\delta\ \forall i\}$ is a compact set for all $\delta>0$, so the only chance to get $d(a)=1$ (and, thereby, not attained) is to make one of $x_i$ approach $1$ and the rest tend to $0$ in the minimizing sequence. But this creates a huge term $a_i\frac{x_i+1}{x_i-1}$ that cannot be compensated by other terms, so this case is ruled out.

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  • $\begingroup$ Thanks a lot for this proof, this is now a lot more clearer for me, nd it corresponds to my empirical studies. I do have a last followup question: Given $\mathbf a$, is there a way to characterise a lower bound greater than $1$ for $d(\mathbf a)$ ? In the sense $d(\mathbf a) > f(\mathbf a) > 1$ ? Stated differently, is there a way of choosing the $\mathbf a$ parameter to make $d(a)$ approach $1$ as close as we want or not ? $\endgroup$
    – lrnv
    Feb 5, 2021 at 9:02
  • $\begingroup$ If you want, when $d=1$, $S(a)$ has just one element $x = \frac{1+a}{1-a}$, and therefore $d(a) = \lvert \frac{1+a}{1-a} \rvert$. Since this is one-dimensional, by plotting the function I found out that $d(a) > \min(a, \frac{1}{a})$ for all $a \in ]0;+\infty[$, which is perfect for me as it allows to take a minimum easily over several different $a$. This bound is not tight at all, but it simplifies a lot the function. I was hoping for something similar in more dimensions. $\endgroup$
    – lrnv
    Feb 5, 2021 at 9:34
  • $\begingroup$ @Irnv Given $a$, is there a way to characterise a lower bound greater than $1$ for $d(a)$. Yes, of course. If you want, I can think of what could be a decent bound. Is there a way of choosing the $a$ parameter to make $d(a)$ approach $1$ as close as we want or not? Yes. Just take $a_1=\delta>0$ very small, $x_1=1+\delta$, $a_2=1+\delta$, $x_2=0$ in dimension $2$, for instance. $\endgroup$
    – fedja
    Feb 5, 2021 at 16:12
  • $\begingroup$ Very nice example on how $d(a)$ might approach zero. If you find a decent bound, It'll help. Btw maybe the general problem could be solved: for $d=2$ and specific values of $a$, wolfram alpha seems to find a minimum: wolframalpha.com/input/… and maybe with a more powerfull software an expression for $d(a)$ could be found. $\endgroup$
    – lrnv
    Feb 5, 2021 at 17:08
  • $\begingroup$ @Irnv A pretty cheap but reasonably accurate (up to a factor of 5 or so) effective bound for the case when all $a_i>0$ and $d\ge 2$ is $d(a)-1\ge \frac{\min_i a_i}{1+\sum_i a_i}$. If it is sufficient for your purposes, I'll post the derivation. For the general problem it looks like you can get a closed form expression sometimes but I'm not sure about "always": there are two cases to consider and if the second one can dominate, then the explicit form may be impossible. $\endgroup$
    – fedja
    Feb 6, 2021 at 2:22

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