# Interesting and intuitive affirmation involving convex sets

Let $\Omega_1$ and $\Omega_2$ two open, bounded and convex domains in $R^n$ with $\Omega_1 \supset \overline{\Omega_2}$ and $0 \in \Omega_2.$

Suppose that for each $x_0 \in \partial (\Omega_1 \setminus \overline{\Omega_2})$ exists a ball $B(z, \delta) \subset (\Omega_1 \setminus \overline{\Omega_2} )$ with $x_0 \in \overline{B(z, \delta) }$, where $\delta >0$ is independent of $x_0.$ (*)

A paper that I am reading says :

Then exists $\alpha > 0$, independent of $x_0,$ and $z$ (as in (*)) such that $|(z-x_0).x_0 |\geq \alpha |z - x_0| |x_0|$

Drawing pictures I cans see that the above affirmation is ocurring, but I have no idea to a proof of the affirmation. Someone could help me with a proof?

First observe that there exist $0 < a < b < \infty$ such that $a < \vert x_0 \vert < b$ for all $x_0 \in \partial (\Omega_1 \setminus \Omega_2)$, since $\Omega_i$ is bounded, $0 \in \Omega_2$ and $\Omega_2$ is open.
Now let $x_0$ and $z$ be arbitrary, satisfying $(*)$. Then $\vert z - x_0 \vert = \delta$, since $x_0 \in \partial B(z,\delta)$. It follows that $$\vert z - x_0 \vert \vert x_0 \vert = \delta \vert x_0 \vert \leq \delta b$$ Therefore, it suffices to show that there exists $0 < m$ (independent of $z$ and $x_0$) such that $$\vert (z - x_0).x_0\vert > m$$ as one just needs to choose $\alpha > 0$ with $\alpha \delta b < m$:
Let us proof this claim for all $x_0 \in \partial \Omega_2 \subset \partial (\Omega_1 \setminus \Omega_2)$. The proof for $x_0 \in \partial \Omega_1$ is more or less the same (note that $\partial (\Omega_1 \setminus \Omega_2) = \partial \Omega_1 \cup \partial \Omega_2$).
Let $x_0 \in \partial \Omega_2$. A supporting half space $H$ at $x_0$ is defined as a closed halfspace containing $\Omega_2$ and with $x_0$ in its boundary. It is not hard to see that there exists a supporting halfspace $H_0$ at $x_0$ which contains $B(z,\delta)$, in fact one can choose the halfspace tangent to $B(z,\delta)$ at $x_0$.
The following is a rather well known general fact: Since $\Omega_2$ is compact with $0$ in its interior, there exists $c > 0$ such that for each $x_0 \in \partial \Omega_2$ and for each supporting halfspace $H$ at $x_0$ the angle between the segment from $x_0$ to $0$ and $H$ is bigger than $c$. (A way to proof this is to assume on the contrary and consider a sequence of supporting half spaces where this angle goes to $0$. Then one can consider a limit of this half spaces which is a supporting halfspace itself containing $0$ in its boundary, contradicting the fact that $0$ is an interior point of $\Omega_2$)
Now, since $0$ is an interior point of $\overline \Omega_2$, it follows that the angle between the segment from $x_0$ to $0$ and $H_0$ is bounded below by $c$, independent of $x_0$ and $z$. Since $H_0$ is tangent to $B(z,\delta)$ at $x_0$ it follows that the angle between the segments from $x_0$ to $0$ and from $x_0$ to $z$ is bigger than $\pi/2 + c$. Translating this means that the angle between $x_0$ and $z - x_0$, say $\omega$, is smaller that $\pi/2 - \tilde c$ for some $0 < \tilde c < \pi/2$ independent of $x_0$ and $z$. Therefore, $$(z - x_0).x_0 = \cos(\omega)\vert z - x_0 \vert \vert x_0 \vert > \epsilon \vert z - x_0 \vert \vert x_0\vert = \epsilon \delta \vert x_0 \vert > \epsilon \delta a,$$ independet of $x_0$ and $z$, for some $\epsilon > 0$. Setting $m = \epsilon \delta a$ we are done.