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Motivation

The motivation for this question arises from the following problem: Is there a closed, simple, and infinitely differentiable path on the (complex) plane such that every straight line has a parallel line that intersects the path at infinitely many points?

To approach this, I attempt to construct such a path by first defining $\gamma: [-1, 1] \to \mathbb{C}$ as follows:

$$ \gamma(t) = \begin{cases} \cos\left(\frac{1}{t^2}\right)\exp\left(-\frac{1}{t^2}\right) + \sin\left(\frac{1}{t^2}\right)\exp\left(-\frac{1}{t^2}\right)i, & \text{if } -1 \leq t < 0, \\ 0, & \text{if } t = 0, \\ \cos\left(\frac{1}{t^2}\right)\exp\left(-\frac{1}{t^2} + \varepsilon t^2\right) + \sin\left(\frac{1}{t^2}\right)\exp\left(-\frac{1}{t^2} + \varepsilon t^2\right)i, & \text{if } 0 < t \leq 1. \end{cases} $$

This path is infinitely differentiable ($C^{\infty}$, defined below), simple (injective) if $0 < \varepsilon < 2\pi$, and intersects every straight line through $0$ infinitely many times. However, it is not closed. To create a closed path, I need a path $\eta : [0, 1] \to \mathbb{C}$ connecting $\gamma(1)$ to $\gamma(-1)$ such that $\gamma \eta$, the concatenation of $\gamma$ and $\eta$ (defined below), forms an infinitely differentiable, closed simple path.


The Problem

The above discussion leads to the following general question:

If a path $\gamma$ on euclidean space is $C^k$ (resp. $C^k$-smooth), injective, is there a path $\eta$ such that $\gamma \eta$ is closed, simple, and $C^k$ (resp. $C^k$-smooth)?

Here $C^k$ (resp. $C^k$-smooth) could be either $C^{\infty}$ (resp. $C^{\infty}$-smooth) or $C^k$ (resp. $C^k$-smooth) for some positive integer $k$.


Relevant Definitions

Concatenation

For two paths $f, g: [0, 1] \to X$ in a topological space $X$, if $f(1) = g(0)$, their concatenation $f g$ is defined as:

$$ fg(t) = \begin{cases} f(2t), & \text{if } 0 \leq t \leq 1/2, \\ g(2t - 1), & \text{if } 1/2 \leq t \leq 1. \end{cases} $$

"Smoothness" conditions

  • A path $\gamma: [0, 1] \to \mathbb{R}^n$ is $C^k$ if it is $k$-times differentiable on $[0, 1]$ (using one-sided derivatives at $t = 0$ and $t = 1$). If $\gamma$ forms a closed loop (i.e., $\gamma(0) = \gamma(1)$), the one-sided derivatives at the endpoints must coincide: $\gamma^{(n)}(0^+) = \gamma^{(n)}(1^-)$ for $1 \leq n \leq k$.

  • A path $\gamma$ is $C^k$-smooth if it is $C^k$ and $\gamma'(t) \neq 0$ for all $t \in [0, 1]$.

  • A path is $C^\infty$ (resp. $C^\infty$-smooth) if it is $C^k$ (resp. $C^k$-smooth) for all $k \geq 1$.


Edit Summary

  • Added what it means for a path to be $C^k$ and $C^k$-smooth.
  • The problem originally consisted of a two-part question. First, I asked: "Given an infinitely differentiable path $\gamma: [0, 1] \to \mathbb{R}^n$ such that $\gamma(0) \neq \gamma(1)$, does there exist a path $\eta: [0, 1] \to \mathbb{R}^n$ connecting $\gamma(1)$ to $\gamma(0)$ such that the concatenated path $\gamma \eta$ is closed and infinitely differentiable?" This part was removed, as it suffices to concatenate $\gamma(t)$ with $\gamma(1-t)$.
  • Generalized the problem to various "smoothness" conditions.
  • Removed the possible extensions of the problem.
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    $\begingroup$ Please specify what you mean by "infinitely-differentiable path" in the original problem: there exists an infinitely differentiable parametrization $\gamma(t)$ or there exists such a parametrization with $\gamma'(t)\neq 0$? $\endgroup$ Commented Nov 18 at 12:29
  • $\begingroup$ Thank you for pointing that out @Alexandre, I will clarify that. In the motivating problem, it would be interesting to come up with smooth paths ($\gamma'(t)\neq 0$), and it is possible to construct a $C^1$-smooth path that has the desired property. Also, I just realized that the first part of my problem is trivial, it would suffice to concatenate $\gamma(t)$ with $\gamma(1 - t)$ if the resulting path is not required to be simple. It is the second part of the problem that is interesting, and it can be asked of many kinds of paths ($C^1$, $C^1$-smooth, etc). $\endgroup$ Commented Nov 18 at 14:40
  • $\begingroup$ All problems are doable, but it think to answer the question in the motivation you don't need the constrictions in the subsequeent problems. $\endgroup$ Commented Nov 18 at 16:18
  • $\begingroup$ I heavily revised my question, I hope it is clearer now. @Pietro, could you please specify which priors in my question are unnecessary? They would all be necessary as far as I understand. $\endgroup$ Commented Nov 18 at 19:45
  • $\begingroup$ ops, my comment above has been made obscure by the auto-correct. I just meant that the question in the Motivation may be easier than the subsequent problems (as a side note: it should be better to ask one question per post) $\endgroup$ Commented Nov 18 at 20:06

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Solution of your original question: A smooth curve cannot intersect infinitely many times lines in all directions.

Let $\gamma: T\to \Gamma$ be the parametrization of your curve, where $T$ is the unit circle, $\gamma\in C^\infty$ and $\gamma'(t)\neq 0$ for all $t$. Then $$f(t)=\frac{\gamma'(t)}{|\gamma'(t)|}$$ is a $C^\infty$ map $T\to T$. Applying Sard's Lemma, we obtain that almost every value of $f$ is regular, therefore, the $f$-preimage of almost every value is finite.

On the other hand, if the curve intersects a line in direction $\theta\in T$ infinitely many times, then it must have infinitely many points where the direction of the tangent line to $\Gamma$ is $\theta$.

Indeed, suppose that $t_k$ are the points where $\gamma(t_k)$ intersects some line $\ell$ with direction $\theta$, wlog $\theta$ is the horizontal direction and $\ell$ is the $x$-axis. Let $t^*$ be an accumulation point of $t_k$. Then in a neighborhood of $t_k$, $\Gamma$ is a graph of a smooth function, $\phi(x)$ which intersect the horizontal axis infinitely many times in any neighborhood of $\gamma(t^*)$,
so by Rolle's theorem, $\phi'(t)=0$ infinitely many times in a neighborhood of $t^*$.

The best reference for Sard's Lemma is J. Milnor, Topology from the differentiab;e viewpoint, UP of Virginia, Charlottesville, 1965,

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