Distance function

[Rigel1]

Since the distance function from a set is considered in a number of posts, I propose here the following (standard) exercise.

Let $S$ be nonempty closed subset of $RR^n$; let us define the distance function from $S$:
$d_S(x) := "inf"_{y\in S} |x-y|$, $x\in RR^n$.

1) Prove that, since $S$ is closed, then the infimum in the definition of $d_S$ is attained, i.e.
$d_S(x) = "min"_{y\in S} |x-y|$, $x\in RR^n$.

We would like to use Weierstrass' theorem, but $S$ is not necessarily bounded.
Neverthless, we can restrict to a bounded portion of $S$ with the following standard trick.
Let $y_0\in S$ be any point of $S$. Let $R := |x-y_0|$, and define $S_R := \{y\in S: |x-y|\le R\}$.
It is clear that $S_R$ is a nonempty (it contains $y_0$) bounded closed set (being the intersection of $S$ with the closed ball $\bar{B}_R(x)$).
Since the function $y\mapsto |x-y|$ is continuous, then it attains its minimum on $S_R$ at a point $y_1\in S_R$.
Moreover, $|x-y| \ge R$ for every $y\in S\setminus S_R$, so that the point $y_1$ is a global minimum over $S$.

2) Prove that $d_S$ is a $1$-Lipschitz function, i.e.
$|d_S(x_0) - d_S(x_1)| \le |x_0-x_1|$ for every $x_0, x_1\in RR^n$.

Let $x_0, x_1\in RR^n$, and let $y_0, y_1\in S$ be such that $d_S(x_i) = |x-y_i|$, $i=0,1$.
Since $d(x_0) = "min"_{y\in S} |x_0-y| \le |x_0-y_1|$, we have that
$d(x_0) - d(x_1) \le |x_0 - y_1| - |x_1 - y_1| \le |x_0-x_1|$.
In a similar way, interchanging the role of $x_0$ and $x_1$ we obtain
$d(x_1) - d(x_0) \le |x_1-x_0|$,
so that 2) is proved.

Now we define the distance function from the boundary of a set, which is a special case of the distance function defined above.
Let $\Omega$ be a nonempty open subset of $RR^n$, with $\Omega\ne RR^n$.
Let $d:\bar{\Omega}\to RR$ be the distance function from the boundary of $\Omega$, defined by
$d(x) := \min_{y\in\partial\Omega} |x-y|$, $x\in \bar{\Omega}$.
(Note that, setting $S := RR^n\setminus\Omega$, $d$ coincides with the restriction of $d_S$ to $\bar{\Omega}$.)

3) Let $\Omega$ be as above, and assume in addition that $\Omega$ is a convex set.
Prove that $d$ is a concave function in $\bar{\Omega}$.

By 2), $d$ is continuous in $\bar{\Omega}$. Then, in order to prove 3), it is enough to show that
(*) $d((x_0+x_1)/2) \ge 1/2 d(x_0) + 1/2 d(x_1)$ for every $x_0, x_1\in\Omega$.
[Additional exercise: prove this claim.]

So, let $x_0,x_1\in \Omega$ and let us denote by $d_i$ the (positive) distances $d(x_i)$, $i=0,1$.
Let $z = (x_0+x_1)/2$; by convexity it is clear that $z\in \Omega$.
In order to prove the inequality in (*) we have to show that
(**) $B_r(z) \subset \Omega$, with $r := (d_0+d_1)/2$.
Let $x\in B_r(z)$; w.l.o.g. we can assume that $x\ne z$, so that $\rho := |x-z| > 0$.
Let us define the unit vector $\xi := (x-z)/\rho$, and let
$\rho_i := (2 d_i \rho)/(d_0+d_1)$, $\bar{x}_i := x_0 + \rho_i \xi$, $i=0,1$.
We have that
$|\bar{x}_i - x_i| = \rho_i = (2 d_i\rho)/(d_0+d_1) < (2 d_i r)/(d_0+d_1) = d_i$, $i=0,1$,
hence $\bar{x}_i \in B_{d_i}(x_i) \subset\Omega$ for every $i=0,1$.
By convexity, we have that
$(\bar{x}_0 + \bar{x}_1)/2 = (x_0+\rho_0\xi + x_1+\rho_1\xi)/2 = z + \rho\xi = x\in \Omega$,
hence (**) is proved.

Tentative picture:
[asvg]xmin = -2; xmax = 11; ymin = -3; ymax = 3; noaxes();
circle( [0, 0] , 1);
circle( [8, 0] , 2);
dot([0,0]);
dot([8,0]);
line([0,0],[0.3,0.3]);
line([8,0], [8.6, 0.6]);
stroke="red";
circle([4,0], 1.5);
dot([4,0]);
line([4,0], [4.45, 0.45]);
text([0,0], "x0", left);
text([8,0], "x1", left);
text([4,0], "z", left);
stroke="green";
path([[-1,-1], [-1,2], [2,2], [6,3], [10,2], [10,0], [10.5,-2], [7,-3], [2,-2], [-1,-1]]);
marker="dot";
path([[0.3,0.3], [4.45, 0.45], [8.6,0.6]]);[/asvg]