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I was working in this problem for a long time and I didn't have success.

Someone could help me, please?

The problem:

Let $f: \mathbb{R}^{n^2} \times \mathbb{Z}^{n} \longrightarrow \mathbb{R}$ defined by $$ f(X,z) = \prod_{i=1}^{n} |x_i^{T} z|, $$ where $x_i^{T}$ is the $i$-th row of $X$.

My question is: If $f(e^{X}B, z) = 0 \iff z = 0 \;(B \in \mathbb{R}^{n^2})$ is it true that $$ \inf_{z \neq 0} \lim_{X \rightarrow 0} f(e^{X}B, z) = \lim_{X \rightarrow 0} \inf_{z \neq 0} f(e^{X}B, z) ? $$ Remark:

  1. $e^{X} = \exp(X)= \displaystyle\sum_{k=0}^{\infty} \dfrac{X^{k}}{k!}$ is infinitely differentiable.

  2. $B$ is a full rank matrix. In my problem this is a matrix of lattice for me.

  3. $X$ is an antisymmetric matrix. I can see with the hypothesis of the problem the matrix $X$ converges to zero. For my propose I used it because the exponential of antisymmetric is an orthogonal matrix. I think that I don't add this hypothesis in this question. But I worked with matrices $X$ antysymetrics.

I had a question like this two or three months ago, but the answered don't help me.

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    $\begingroup$ Cross-posted: math.stackexchange.com/questions/4188872/… $\endgroup$
    – Matthew Daws
    Commented Aug 10, 2021 at 12:17
  • $\begingroup$ @AlexandreEremenko This was also my first idea, but it seems that the infimum is taken over $z\in\mathbb Z^n\setminus \{0\}$. $\endgroup$
    – Jochen Wengenroth
    Commented Aug 10, 2021 at 18:16
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    $\begingroup$ Can you be clear about the quantifiers in your question? Is $X$ a fixed matrix? Is your hypothesis that for all $B\in M_n(\mathbb R)$, $f(e^XB,z)=0$ if and only if $z=0$? And the same quantifiers on the conclusion? $\endgroup$
    – Anthony Quas
    Commented Aug 10, 2021 at 18:39
  • $\begingroup$ @AnthonyQuas $X$ is an any matrix and $B$ is a full rank matrix. In my hypothesis, when $f(e^XB, z) = 0 \iff z = 0$ (1) implies that the matrix $X$ converges to zero. So my question is if the equality that I've written is true for all $B$ and $z$ with the hypothesis (1). But I think that don't need used it (1) for prove my question. $\endgroup$
    – Thiago Alexandre
    Commented Aug 12, 2021 at 14:40
  • $\begingroup$ @MatthewDaws yes, I've tried to get a solution one month ago on mathexchange, but anybody could help me. And my friend told me that this question is more appropriate here. So I put here this question. I am a beginning reasearch and don't use internet too much. So I didn't know this site some years ago. $\endgroup$
    – Thiago Alexandre
    Commented Aug 12, 2021 at 14:43

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