with $m = 0$, a locally correct matching is $(\sigma, \theta)$, where $\sigma(t) = 0$ and $\theta(t) = t \cdot n$.

Since $m = 0$, $P$ is just a single point, $p_0$. The [Fréchet distance]{} between a point and a curve is the maximal distance between the point and any point on the curve: $\delta_\text{F}(p_0, Q_\theta[a,b]) = d_\mu[a,b]$. This implies that the matching $\mu$ is locally correct.

\[lem:linesegments\] For two polygonal curves $P$ and $Q$ with $m = n = 1$, a locally correct matching is $(\sigma, \theta)$, where $\sigma(t) = \theta(t) = t$.

The free space diagram of $P$ and $Q$ is a single cell and thus the free space is a convex area for any value of $\varepsilon$. Since $\mu = (\sigma, \theta)$ is linear, we have that $d_\mu[a,b] = \max\left\{ d_\mu(a), d_\mu(b) \right\}$: if there would be a $t$ with $a < t < b$ such that $d_\mu(t) > \max\left\{ d_\mu(a), d_\mu(b) \right\}$, then the free space at $\varepsilon = \max\left\{ d_\mu(a), d_\mu(b) \right\}$ would not be convex. Since $d_\mu[a,b] = \max\left\{ d_\mu(a), d_\mu(b) \right\} \leq \delta_\text{F}(P_\sigma[a,b],Q_\theta[a,b]) \leq d_\mu[a,b]$, we conclude that $\mu$ is locally correct.

![(a) Curves with the free space diagram for $\varepsilon = \delta_\text{F}(P,Q)$ 