Let $\gamma: [0,1] \to \mathbb R^2$ be a parameterized curve, where $\gamma(t) = (x(t),y(t))$ is continuously differentiable with $(x' (t), y'(t)) \ne (0,0)$ for all $t \in [0,1]$.
Let $N(t) = (-y'(t),x'(t))$ be the normal vector to the tangent vector $\gamma' (t)$.
Is it true that for all $t \in (0,1)$, there exists an open ball $B_r (t)$ with radius $r > 0$ such that for all $(x,y) \in B_r (t)$, there exists $s \in (0,1)$ and $\alpha \in \mathbb R$ with $\gamma (s) + \alpha N(s) = (x,y)$ ?
If we assume that $\gamma'$ also is continuously differentiable, we can use the Inverse Function Theorem: Define $f: \mathbb R^2 \to \mathbb R^2$ by $f( s, \alpha ) = \gamma (s) + \alpha N(s)$. Then the Jacobian matrix of $f$ is:
$\left[ \begin{array} ,x'(s) - \alpha y''(s) & y'(s)+ \alpha x''(s) \\ -y'(s) & x'(s) \end{array} \right]$
The determinant of the Jacobian evaluated at $t$ is:
$x'(t)^2 + y'(t)^2 \ne 0$
which shows that the matrix is invertible. It follows from the Inverse Function Theorem that $f$ is invertible in an open ball with center $t$, which proves the claim.
However, when $\gamma'$ is not differentiable, I am unsure how to create the proof. My attempt would be to fix $t_0,t_1 \in (0,1)$ with $t_0 < t < t_1$ and $K > 0$ such that the curve with image:
$\{ f( t_0, \alpha) : \alpha \in [-K,K] \} \cup \{ f( t_1, \alpha) : \alpha \in [-K,K] \} \cup \{ f( s, K ) : s \in [t_0,t_1] \} \cup \{ f( s, -K ) : s \in [t_0,t_1] \}$
has positive distance $d$ from $\gamma(t)$. Then we take the limit $t_0,t_1 \to t$ and $K \to 0$, so the curve contracts continuously to $\gamma(t)$. Then we can use tools from algebraic topology to show that for all $(x,y) \in B_d (t)$, the contraction must pass through the point $(x,y)$.
Is such an approach viable?