Let $D\subset\mathbb{R}^n$ be a non-empty open set and $C$ be a cone function with vertex $x_o$ and slope $b$. Then $L(C, D)=|b|$. Moreover, if $D$ is bounded and $x_o\notin D$, then $L(C,\partial D)=|b|$.
Some notations. For cone function with vertex $x_o$ we mean a function $C$ of the form $C(x)=a+b|x-x_o|$ where $a$ and $b$ are the height and the slope respectively. Moreover, the half-line $$ t\longmapsto x_o+t(x-x_o),\quad x\neq x_o $$ is called the ray of $C$ through $x$.
Now, proving that if $D$ contains two distinct points on the same ray of a cone function $C$ with vertex $x_o$ and slope $b$, it holds $L(C, D)=|b|$, is simple. Indeed if $y\neq w$ are two points on the same ray of $C$, we have, for a certain $x^*$, $y=x_o+\alpha(x^*-x_o)$ and $w=x_o+\beta(x^*-x_o)$ with $\alpha,\beta\geq0$, $\alpha\neq\beta$. Then $$ \frac{|C(y)-C(w)|}{|y-w|}=\ldots=|b|. $$
But, how can I prove the statements before? Some helps?
Thank You
Let $z\in D$, let $\nu := \frac{z-x_0}{\|z-x_0\|}$ and define $$ t^- := \inf\{t<0:\ z+s\nu\in D\ \forall s\in [t,0]\}, \quad t^+ := \sup\{t>0:\ z+s\nu\in D\ \forall s\in [0,t]\}. $$ (The two sets are nonempty since, being $D$ open, $z$ is an interior point of $D$.) Then the points $y^\pm := z + t^\pm \nu$ belong to $\partial D$ and to the same ray.