Lower bound of the distance between the endpoints of an average-curvature-constrained path of a given length Let $\gamma$ be a continuously differentiable path in the plane parametrized by its arclength. We require that the average curvature of $\gamma$ is at most 1, meaning that for all $t < s < t + \pi$ the angle between $\gamma'(t)$ and $\gamma'(s)$ is at most $s-t$.
Suppose $\gamma$ has length $l$. We seek a lower bound on the euclidean distance $\|\gamma(l) - \gamma(0)\|$ between its endpoints.
In the paper Discrete Dubins Paths the authors give in Lemma 6.1, page 15, the lower bound $$\|\gamma(0) - \gamma(l)\| \geq 2 \sin (l/2),$$ whenever $l < \pi$.
My question is: Is this bound still valid for $l \leq 2\pi$? It seems to me that it should be true, but I am unable to find a reference. Does anyone else agree, or have I missed something?
 A: I was able to come up with a simple proof. Thanks to Sylvester Eriksson-Bique for helping me simplify it.
Theorem
Let $\gamma$ be a continuously differentiable path in the plane, parametrized by its arclength, with length $l$, and with average curvature at most 1. We have $$|| \gamma(0) - \gamma(l) || \geq 2 \sin(l/2)$$ whenever $l \leq 2\pi$.
Proof
The case when $l \leq \pi$ is covered by Lemma 6.1 of [1]. We will show that it holds for $l \leq 2 \pi$. Let $x,y,z = \gamma(0), \gamma(l/2), \gamma(l)$. Since $l/2 \leq \pi$ we get $||x - y|| \geq 2 \sin(l/4)$ and $||y - z|| \geq 2 \sin(l/2)$. Note that we have $\angle x y z \geq \pi - l/2$, which we get by Lemma 6.2 of [1]. Using these facts, and the law of cosines, we get
\begin{align*}
    ||x - z||^2 &= ||x - y||^2 + ||y - z||^2 - 2 ||x - y|| \cdot ||y - z|| \cdot \cos(\angle x y z) \\
                &= (||x - y|| - ||y - z||)^2 + 2 ||x - y|| \cdot ||y - z|| \cdot (1 - \cos(\angle x y z)) \\
                &\geq 2 ||x - y|| \cdot ||y - z|| \cdot (1 - \cos(\angle x y z)) \\
                &\geq 2 \cdot (2 \sin(l/2)) \cdot (2 \sin(l/2)) \cdot (1 - \cos(\angle x y z)) \\
                &=    8 \sin^2(l/2) \cdot (1 - \cos(\angle x y z)) \\
                &\geq 8 \sin^2(l/2) \cdot (1 - \cos(\pi - l/2)) \\
                &= 4 \sin^2(l/2)
  \end{align*}
which gives us $||x - z|| \geq 2 \sin(l/2)$.
References: [1] Discrete Dubins Paths
