# Distance function has gradient magnitude equal to one

Suppose $\Omega$ is a subset of $\mathbb{R}^n$ with smooth boundary, $\partial\Omega$. Let’s define the function,

\[g(x) = \min_{y \in \partial\Omega}|x-y|\]which gives the distance from a point to the boundary of the region.

Let’s assume that the gradient of $g(x)$ is defined almost everywhere and show that it has magnitude one. This will also apply to the signed distance function which is positive inside $\Omega$ and negative outside (or the opposite, depending on convention). The signed distance function is used in the level set method and the PDE, $|Dg|=1$, is an instance of the eikonal equation.

First let’s choose some $x$ and suppose that $y^* = \text{arg}\min_{y \in \partial\Omega} |x-y|$ is unique.

Now let’s consider the line segment, $(1-t)x + ty^*$. Along this segment,

\[\begin{align*} |(1-t)x + ty - y^*| = |(1-t)x +ty - ty^* - (1-t)y^*)| &\leq (1-t)|x-y| + t|y-y^*| \\ \min_{y \in \partial\Omega} |(1-t)x + ty^*| &\leq \min_{y \in \partial\Omega} \biggl[ (1-t)|x-y| + t|y-y^*| \biggr] \\ g\left( (1-t)x + ty^*\right) &\leq (1-t)g(x) \\ \end{align*}\]but on the other hand, the inequality,

\[|(1-t)x + ty^* - y| = |x-y - t(x-y^*)| \geq |x-y| - t|x-y^*|\]yields the result, \(g\left( (1-t)x + ty^*\right) \geq (1-t)g(x)\)

So, we have that $g\left( (1-t)x + ty^*\right) = (1-t)g(x)$ where by taking derivatives with respect to $t$, we obtain the relation,

\[\langle Dg\left( (1-t)x + ty^*\right), y^*-x \rangle = -g(x)\]which says that \(|Dg\left( (1-t)x + ty^*\right)| |y^*-x| \cos\theta = -g(x)\)

… or since $|y^* - x| = g(x)$,

\[|Dg\left( (1-t)x + ty^*\right)| \cos\theta = -1\]Now note that when $t=0$, $Dg(x)$, points in the direction of maximal increase while \(y^*-x\) points from $x$ to the closest point on the boundary, \(\partial\Omega\). However, the direction, \(y^* - x\), is clearly in the direction of maximum decrease since otherwise, \(y^*\) would not be the minimizing point on the boundary. Therefore, $y^*-x = -Dg(x)$, so $\cos\theta = -1$, which must therefore hold for all $t \in [0,1]$.

So, we’ve shown that for any point, $x$, not on the boundary, if there is a unique closest point to the boundary, then we have,

\[|Dg| = 1\]