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I am working on a control theory problem that either just caught me on a blind spot or is beyond me. I guess it's not a new question, but I couldn't find anything about it.

Let $p(s), s\in \mathbb{R}, p(s) \in \mathbb{R}^n$ be a parametrized curve. Let $p'=\frac{dp}{ds}$, and denote the (common) scalar product by $\langle\cdot,\cdot\rangle$. For a given $s^*$, the set $N(s^*) = \{x | \langle p(s^*)-x,p'(s^*)\rangle=0\}$ defines all points on the normal plane defined by the point $p(s^*)$ and the vector $p'(s^*)$.

My hypothesis is that

$$p'\neq 0 \implies \bigcup_{s\in\mathbb{R}}{N(s)} = \mathbb{R}^n$$

i.e. the union of all normal planes covers the entire space. It is pretty obvious for $n=1,2$, yet I struggle at giving a formal proof on it. I am also unsure on how it is in the general case $n>2$. Any ideas would be so kind.

Edit: Let $p(s)$ be a continuous, piecewise defined polynomial. Is it true for this case?

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  • $\begingroup$ For n=2, think to the evolvent of a circle p(s). The normal lines to it are the tangents to the circle, which only cover the exterior of it (and if you approximate $C^1$ the curve p(s) by polynomials, the covered set stay close) $\endgroup$
    – Pietro Majer
    Commented Dec 8, 2023 at 22:05
  • $\begingroup$ The hypothesis is true in the case that $p$ is periodic, i.e. factors through a map $p:S^1 \to \mathbb{R}^n$. In this case, given a point $x \in \mathbb{R}^n$, the closest point (and furthest point) to $x$ on the curve will have the desired property. $\endgroup$
    – Ethan Dlugie
    Commented Dec 8, 2023 at 22:42
  • $\begingroup$ For a $C^1$ curve $p:\mathbb R\to\mathbb R^n$, and $x\in\mathbb R^n$, there is a minimizing sequence $(s_k)_k$ for $d(s):=\|p(s)-x\|^2$ with $p(s_k)$ converging to some $x^*$ and $d'(s_k)=o(1)$. Assuming also $\|p'(s)\|\ge\epsilon>0$ for all $s$, this implies that the union of normal planes is dense. $\endgroup$
    – Pietro Majer
    Commented Dec 9, 2023 at 10:32

2 Answers 2

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$\newcommand\R{\mathbb R}$This is not true in general. E.g., let $p(s):=(\arctan s,0,\dots,0)$. Then $$\bigcup_{s\in\R}N_s=(-\pi/2,\pi/2)\times\R^{n-1}\ne\R^n.$$

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  • $\begingroup$ As $arctan(0) = 0$, the above condition $p'\neq0$ is not met. $\endgroup$
    – user517853
    Commented Dec 8, 2023 at 19:32
  • $\begingroup$ @user517853 : $p'(s)=(\frac1{1+s^2},0,\dots,0)\ne(0,\dots,0)$ for any $s$. Anyhow, if you for some reason don't like $p(s)=0$, replace $\arctan s$ by $\pi/2+\arctan s$. $\endgroup$
    – Iosif Pinelis
    Commented Dec 8, 2023 at 19:38
  • $\begingroup$ My mistake, I shouldn't respond late at night. Anyhow: Can you provide a counterexample if $p(s)$ is a piecewise polynomial, as in my edit above? $\endgroup$
    – user517853
    Commented Dec 8, 2023 at 19:40
  • $\begingroup$ @user517853 : One should not edit the post so as to invalidate a valid answer. So, please roll back your edit. $\endgroup$
    – Iosif Pinelis
    Commented Dec 8, 2023 at 19:48
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For a point $x \in \mathbb{R}^n$, saying that $x$ lies on the normal plane at $p(s^*)$ is equivalent to saying that $s^*$ is a critical point for the distance function between $x$ and $p(s)$.

If your curve happens to be periodic, i.e. factors through a map $p:S^1 \to \mathbb{R}^n$, then a distance minimizer (and a distance maximizer) is guaranteed to exist. This condition is rare for a randomly given curve.

If your curve goes off to infinity in both directions, i.e. $\lim_{s \to \pm \infty} |p(s)| = \infty$, then a distance minimizer is guaranteed to exist. This condition should be easy to establish for any given parametrized curve.

One can manufacture counterexamples when these conditions fail. For instance, let $p(t)=\exp(it)/t$ in $\mathbb{C} \approx \mathbb{R}^2$. The magnitude of $p(t)$ has no critical points, so the origin lies on no normal line. It would be interesting to understand what other conditions can be imposed on $p$ when it is neither periodic nor escaping to $\infty$ so that the normal hyperplanes cover the space. Maybe something about closedness of the image?

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  • $\begingroup$ Yes, the existence of a minimum distance is more or less the background of my question. The periodic and infinite conditions are general enough for what I had in mind, thank you. $\endgroup$
    – user517853
    Commented Dec 10, 2023 at 10:54

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