Let $\alpha:I\rightarrow \mathbb R^2$ be a regular curve with arc length parametrization, $s_0\in I$ and $\kappa(s_0)\neq0.$
Show that it exists $\epsilon>0$ such that for all $s\neq s_0\in(s_0-\epsilon,s_0+\epsilon)$ $$f(s):=\langle N(s_0),T(s) \rangle\neq0$$
I know that the problem in the title is stated differently, but I strongly believe, it's equivalent to the problem above.
My attempt: We know $f(s_0)=0$ and $f$ is not constant, because otherwise $$0 \neq \kappa (s_0)=\langle N(s_0),T'(s_0) \rangle=f'(s_0)=0$$ Since $f$ is continuous it is guaranteed that $$ \epsilon := \begin{cases} \frac{1}{2}d(s_0,f^{-1}(\{0\})\setminus\{s_0\} & \text{if } f^{-1}(0)\neq \emptyset \\ 1 & \text{otherwise} \end{cases} $$ will do the job.
Notations:
- $T(s)$ is the tangent vector at the point $\alpha(s)$.
- $N(s)$ is the normal vector at the point $\alpha(s)$.
- $\kappa$ signed curvature.
- $d(s,A)$ is the shortest distance from point $s$ to set $A$.
- $f^{-1}(A)$ preimage of set $A$
Is my proof correct?
I am not sure how to rigorously proof the second part.
Any suggestions and clever ideas are appreciated!