Consider a 1-periodic function $\varphi \in C^1(\mathbb R;\mathbb R^n)$ 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 $\varphi(0)$.
Question: Is there a $\delta>0$ such that for all $x\in N \cap B_\delta(\varphi(0))$ we have $$|x-\varphi(0)|\le C\cdot |x-\varphi|,$$ where the constant $C>0$ is independent of $x$ and where $|x-\varphi| :=\inf_{t\in[0,1]}|x-\varphi(t)|$ is the distance between $x$ and the curve?
It is clear that as $x$ approaches the curve, $|x-\varphi(0)|$ will converge to $|x-\varphi|$, but can we even establish such a local bound? If $\varphi'$ is constant in a neighborhood of 0, we even have $|x-\varphi(0)|= |x-\varphi|$. So if $\varphi'$ is continuous, the above shouldn't be too far-fetched, or am I missing something?
I suppose that people from differential geometry (which I am hardly familiar with at all) might know more about this kind of problem than me...
Thanks a lot in advance!