Consider a 1-periodic function $\varphi \in C^2(\mathbb R;\mathbb R^n)$ with non-vanishing derivative and such that $\varphi|_{[0,1)}$ is a bijection. Let
$$ N:=\{\varphi(0)+z\,|\,z\in\mathbb R^n, \, z^\top\varphi'(0)=0\}= \varphi(0) +\langle \varphi'(0)\rangle^\bot \subset \mathbb R^n$$
the hyperplane that is orthogonal to the closed curve $\varphi(\mathbb R)=\varphi([0,1])$ in the point $\varphi(0)$.
In the context of this (similar and still open) question, I was told that the following is true:
There is a $\delta>0$ such that for all $x\in N \cap B_\delta(\varphi(0))$ we have $$|x-\varphi(0)| = |x-\varphi|,$$ where $|x-\varphi| :=\inf_{t\in[0,1]}|x-\varphi(t)|$ is the distance between $x$ and the curve.
Apparently, this follows from the existence of a tubular neighborhood. From an intuitive point of view, I can believe this, but since I am hardly an expert on differential geometry, I have the following questions:
1.) Is there an elementary proof for this statement (i.e. one that does not incorporate more involved tools from differential geometry)?
2.) Even if not, is there a quotable reference where this is stated in a way which is not obscured by much greater generality?
3.) Is there an intuitively graspable reason why this statement fails for curves that are only $C^1$?
Thanks a lot in advance!
Let me give an informal view on the problem. I believe that this can also be transferred to a simple proof. At least, it gives some insight why the property fails for $C^1$ curves.
Let us fix $x \in N$ and define $r := | x - \varphi(0)|$. Then, the property $r = |x-\varphi|$ is equivalent to: "the open ball $U_r(x)$ at $x$ with radius $r$ does not intersect $\varphi$". Note that $\varphi(0)$ lies on the boundary of this ball. Now, we see that this is somehow related to the curvature $\kappa$ of the curve in $0$. As long as $r$ is smaller than $1/\kappa$, the open ball $U_r(x)$ cannot intersect $\varphi$. Hence, you can take $\delta < 1/\kappa$.
However, if $\varphi$ is merely $C^1$, you can have (formally) $\kappa = +\infty$ and therefore, the choice $0 < \delta < 1/\kappa$ is not possible.