Let $f: \mathbb{R}^{n^2} \times \mathbb{Z}^{n} \longrightarrow \mathbb{R}$ defined by
$$
f(X,z) = \prod\_{i=1}^{n} |x\_i z|,
$$
where $x\_i$ is the $i$-th row of $X$ and $x\_iz$ is a dot product of $x\_i$ and $z$.
My question is:
Is it true that
$$
\inf\_{z \neq 0} \lim\_{X \rightarrow 0} f(e^{X+C}, z) = \lim\_{X \rightarrow 0} \inf\_{z \neq 0} f(e^{X+C}, z) ?
$$
Some observations that I deduce about this:


1. $e^{X} = \displaystyle\sum\_{k=0}^{\infty} \dfrac{X^{k}}{k!}$ is infinitely differentiable and therefore $e^{X+C}$ is continuous.
2. I think that you don´t need to know about the constant $C$, but $C = \log b$ and the definition is that $e^{C}=b$. $C$ is a matrix, i.e., $C \in \mathbb{R}^{n^2}$.
3. Actually, still I believe that we don´t need to know about the function $f$ the same way that I wrote above. I think that the problem can be formulated as follows:


Given $f: \mathbb{R}^{n^2} \times \mathbb{Z}^{n} \longrightarrow \mathbb{R}$ continuous, is it true that $\inf\_{z \neq 0} \lim\_{X \rightarrow 0} f(X, z) = \lim\_{X \rightarrow 0} \inf\_{z \neq 0} f(X, z)$?


So I can´t do how prove it. I tested with some examples in Matlab and Wolfram Mathematica and the equality has always been valid.


One of the ways which I thought was to "open" the expressions of both sides of equality and try to reach the same result. To reach the infimum, I thought that $z$ should have all coordinates equal to zero, except 1 coordinate. In this case, I have $z \neq 0$ and maybe this is a vector that works to prove it. But I didn´t get anywhere.



Problem of analysis about matrix exponential, infimum and limit
Define $I(X)=\inf\_{z\neq0}f(e^X,z)$. Because $f$ is continuous, your question is precisely: is $I$ continuous? I claim it isn't always the case.



> 
> Call $\mathcal X$ the set of matrices $X$ such that $I(X)=0$. Then
> 
> 
> 1. $I$ is continuous precisely on $\mathcal X$;
> 2. $\mathcal X$ is a countable intersection of dense open sets (hence it is itself dense);
> 3. If $n=2$, the complement of $\mathcal X$ has measure zero but is dense.
> 
> 
> 


I do not know whether the complement of $\mathcal X$ is “large” (or even non-empty) in higher dimensions, although I believe that people from the Diophantine approximation realm would be able to use the ideas in 3. to settle the question. It looks similar to what is described at the [Subspace theorem](https://en.wikipedia.org/wiki/Subspace_theorem) page on Wikipedia, but I wasn't able to apply it directly to your situation.


Points 1. and 2. above really just hinge on the fact that $\mathcal X$ is dense. Suppose for a moment it is. Then $I$ cannot be continuous at $X\notin\mathcal X$. If however $I(X)=0$, then for all $\varepsilon>0$ there exists $z^\*\neq0$ such that $f(e^X,z^\*)<\varepsilon/2$, so $I(X+X')\leq f(\exp(X+X'),z^\*)<\varepsilon$ for all $X'$ small enough, which means that $I$ is continuous at $X$. Finally, $\mathcal X$ is the intersection of the open sets $\lbrace X:f(X,z)<2^{-k}\rbrace$, where $(z,k)$ ranges over a countable set of values.


### Density of $\mathcal X$


Here is a hands-on proof of the fact that $\mathcal X$ is dense, but I will give another proof of it later that I will actually need for point 3. Denoting by $\pi\_1:\mathbb R^n\to\mathbb R$ the first coordinate, it is clear that if $\ker (\pi\_1\circ X)\cap\mathbb Q^n\neq\langle0\rangle$, then $\inf\_{z\neq0}f(X,z)=0$. Indeed, any $w\neq0$ in the rational kernel will have some multiple $z$ with integer coordinates such that $|x\_1z| = |(\pi\_1\circ X)(z)| = 0$.


Fix a matrix $X$, and let us prove that $X$ is a limit of elements of $\mathcal X$. Using the previous observation, it suffices to show that it is a limit of matrices $X(k)$ such that $\pi\circ e^{X(k)}$ has non-trivial rational kernel. Choose $u\in\ker(\pi\_1\circ e^X)$ non zero. There exists a sequence of diagonal matrices $P(k)$ such that $P(k)u\in\mathbb Q^n$ and $P(k)$ converges to the identity. We see that $P(k)e^XP(k)^{-1}$ has such a non-trivial rational kernel; indeed,
$$e^XP(k)^{-1}(P(k)u) = e^Xu\in\ker(\pi\_1) = \ker(\pi\_1\circ P(k)),$$
so $\pi\_1\circ(P(k)e^XP(k)^{-1})(P(k)u)=0$ with $P(k)u\in\mathbb Q^n$ non zero. Setting $X(k)=P(k)XP(k)^{-1}$, this concludes because $e^{X(k)} = P(k)e^XP(k)^{-1}$.


### Preimage of dense sets


Another proof of the density of $\mathcal X$, that I will use for 3., relies on the following result.



> 
> **Fact.** The matrix exponential is a local diffeomorphism, except over a singular set included in a countable union of algebraic subsets. In particular, the singular set is a countable union of closed sets of measure zero and empty interior.
> 
> 
> 


This is a fairly immediate consequence of the fact that the differential of the exponential is invertible at $X$ if the differences between the eigenvalues of $X$ are not multiples of $2i\pi$ (the [Wikipedia page](https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map) for the derivative of the exponential cites Rossmann, *Lie Groups...* section 1.2, Proposition 7 for this fact). The important thing for us is that the singular set has empty interior, so the preimage of a dense set by the exponential is still a dense set. Indeed, for any non-empty open set $\mathcal U\subset\mathbb R^{n^2}$, we can find a matrix $X\in\mathcal U$ that is not in the singular set for the exponential. This makes it a local diffeomorphism around $X$, so the dense set must intersect the image of a small neighbourhood of $X$, small enough that it is contained in $\mathcal U$, which precisely means that the preimage of the dense set intersects $\mathcal U$. A similar reasoning shows that the preimage of a set of measure zero still has measure zero.


Now it is clear that $\inf\_{z\neq0}f(X,z)$ must be zero for a matrix $X$ with rational coefficients. So $\mathcal X$ contains the preimage of the set of such matrices. But it is also clear that the set of matrices with rational coefficients is dense, so $\mathcal X$ contains a dense subset, which gives another proof of the density of $\mathcal X$.


### The case $n=2$


For point 3., we prove first that the complement is dense. According to the remark above, it suffices to show that the set of matrices $X$ with $\inf\_{z\neq0}f(X,z)>0$ is dense. I claim that it contains the following matrices, which clearly constitute a dense set:
$$ X = \begin{pmatrix}
 a & -\alpha \\
 b & -\beta
 \end{pmatrix} $$
with $a,b\in\mathbb Q^\*$, $\alpha$ and $\beta$ quadratic irrational numbers (i.e. $[\mathbb Q(\alpha):\mathbb Q]=[\mathbb Q(\beta):\mathbb Q]=2$), and $\alpha/a\neq\beta/b$. For instance, $\alpha$ and $\beta$ could be elements of $\sqrt2+\mathbb Q$, which is already enough to ensure the density.


The important feature of quadratic irrational numbers is that they have finite [Markov constant](https://en.wikipedia.org/wiki/Markov_constant). In other words, there exists $\varepsilon\_\alpha>0$ such that
$$ \left|\frac pq-\alpha\right|\geq\frac{\varepsilon\_\alpha}{q^2} $$
for all $p\in\mathbb Z$, $q\in\mathbb N^\*$, and similarly for $\beta$. Now let us consider $X$ a matrix as above, and show that it must satisfy $\inf\_{z\neq0}f(X,z)>0$.


Let $d>0$ is the distance between $\alpha/a$ and $\beta/b$, and $z=(p,q)$ a non-zero element of $\mathbb Z^2$. There are 4 (overlapping) cases to consider:


* If $q=0$, then $f(X,z)=|abp^2|\geq|ab|>0$.
* If $q\neq0$ and $|p/q-\alpha/a|\leq d/2$, then first of all $|pa-q\alpha|>|q|\varepsilon\_\alpha/|q\_aq|^2$, where $q\_a$ is the denominator of $a$. Moreover, $|pb-q\beta|=|bq|\cdot|p/q-\beta/b|\geq|bq|d/2$. All in all,
$$ f(X,z)
\geq \frac{\varepsilon\_\alpha}{|q\_a|^2|q|}\cdot|b|\cdot|q|\frac d2
 = \frac{\varepsilon\_\alpha|b|d}{2|q\_a|}>0. $$
* The case $q\neq0$, $|p/q-\beta/b|\leq d/2$ is similar.
* If $q\neq0$, $|p/q-\alpha/a|>d/2$ and $|p/q-\beta/b|>d/2$, then we see easily that $f(X,z)>|abq^2|\cdot(d/2)^2\geq|ab|d^2/4>0$.


All the inequalities $f(X,z)>0$ hold uniformly in $z$, so we have $\inf\_{z\neq0}f(X,z)>0$ as announced.


Last but not least, it remains to show that $\mathcal X$ has full measure. We will show that $\mathcal X$ contains the preimage under $\exp$ of a set of full measure; in other words, we look for a big set of matrices $X$ such that $\inf\_{z\neq0}f(X,z)=0$.


Using [Khinchin's theorem](https://en.wikipedia.org/wiki/Diophantine_approximation#Khinchin%27s_theorem_on_metric_Diophantine_approximation_and_extensions) on Diophantine approximations, the set of reals $r$ such that $|q\_n|^2\cdot|p\_n/q\_n-r|\to0$ along a good rational approximation $p\_n/q\_n\to r$ has full measure; for instance, choose $\psi(x)=1/(x\ln(x)^{1/2})$ in the statement on the Wikipedia page. We call these numbers *(mildly) well approximated.* From this, we can deduce that the set of matrices
$$ X = \begin{pmatrix}
 a & -b \\
 c & -d
 \end{pmatrix}, $$
with $b/a$ (well-defined and) well-approximated, has full measure. Indeed, $(a,b)\mapsto(a,b/a)$ is a local diffeomorphism almost everywhere, and the reasoning above (with the exponential) together with Fubini's theorem are enough to conclude.


I claim that these matrices satisfy $\inf\_{z\neq0}f(X,z)=0$. Indeed, let $X$ be such a matrix, and $p\_n/q\_n$ a sequence of rationals that approximate $a/b$ in the sense described above. Set $z\_n=(p\_n,q\_n)$. Then, for $n$ large enough,
$$ f(X,z\_n)
 = |acq\_n^2|\cdot\left|\frac{p\_n}{q\_n}-\frac ba\right|\cdot\left|\frac{p\_n}{q\_n}-\frac dc\right|
\leq |ac|\cdot|q\_n|^2\cdot\left|\frac{p\_n}{q\_n}-\frac ba\right|\cdot2\left|\frac ba-\frac dc\right|, $$
which tends to zero as $n$ goes to infinity. This shows that the complement of $\mathcal X$ is dense, and concludes the proof of point 3.


