If $0<r\neq 1$ and $U$ is convex then $\partial U\cap\partial (rU)=\emptyset$

Question

Let $U\subseteq\mathbb{R}^n$ be convex and closed. I'm trying to find conditions for which the following happens:

If $0<r\neq 1$ then $\partial U\cap\partial (rU)=\emptyset$.

It seems to me that this is not true if $0$ is in the frontier of $U$, for instance $U=\{(0,y):y\in\mathbb{R}\}$ or $U=\{(x,y):y\ge |x|\}$. In fact, in both examples $U=rU$.

So, let us suppose that $0\in\text{Int} (U)$. Does this suffice to prove that $\partial U\cap\partial (rU)=\emptyset$? I couldn't find counterexamples so far.

What do you think?

Thank you.

Answer 1

WLOG assume that $r>1$ (if $0<r<1$ then let $U' = rU$ and $r' = \frac{1}{r}$, so $r'U' = U$ and $r'>1$). Suppose that the intersection is nonempty. Then, there is some $x\in \partial (rU)$ such that $x\in\partial U$. Since $x\in \partial (rU)$, $y = r^{-1}x\in \partial U$. Thus, both $x$ and $y$ are on the boundary of $U$. We can draw a line segment from $0$ through $x$, which will also pass through $y$. If $0\in\text{Int}(U)$ there is an open neighborhood of $0$ also in $\text{Int}(U)$ since the interior is always open. Then, take the collection of line segments (not including endpoints) from points in that neighborhood to $x$. The union $I$ of all these line segments is an open set containing $y$. Since $U$ is convex, $I\subset U$. Thus, there is an open subset of $U$ containing $y$, contradicting that $y\in\partial U$. Thus, if $0\in\text{Int}(U)$ then the intersection must be empty.

If $0\notin\text{Int}(U)$ then $0 \in \partial U$ or $0 \in \text{Ext}(U)$. If $0\in\partial U$ then obviously $0 \in \partial (rU)$ so the intersection will be nonempty. If $0 \in \text{Ext}(U)$, then let the shortest distance between $0$ and $U$ be $R$ and the largest distance be $M$. If $0<r<\frac{R}{M}$, then $rU\cap U = \emptyset$. However, if $U$ is a line segment not through the origin, then $rU\cap U = \emptyset$ for all $r\neq 1$.

Thus, $\partial U \cap \partial (rU) = \emptyset$ for all $r$ if $0\in\text{Int}(U)$ and $\partial U \cap \partial (rU) \neq \emptyset$ for all $r$ if $0\in \partial U$.

Answer 2

I have an approach that seems promising to me but I can't push it through. I'll enter it here in the hope that someone else can complete it.

$\partial U$ can be described as $(t, f(t))$ for $0 \le t \lt 2\pi$.

Suppose $(0, 0) \in U$ \ $\partial U$.

Suppose $\partial U \cap \partial(rU) \ne \emptyset$ where $0 < r < 1$.

Then there are $a, b \in [0, 2\pi)$ such that $(a, f(a)) = r(b, f(b))$ so that $a = rb$ and $f(a) = rf(b)$.

Consider the line from $(b, f(b))$ to $(rb, rf(b))$. This has equation $c(b, f(b))+(1-c)(rb, rf(b))$ for $c \in \mathbb{R}$ or

$c(b-rb, f(b)-rf(b))+(rb, rf(b))\\ =c(1-r)(b, f(b))+r(b, f(b))\\ =c(1-r)(b, f(b))+(a, f(a)) $.

Frustrating. I can't see where this leads to a contradiction. What I would like to do is show that this line which passes through two points on $\partial U$ also passes through a third point, which would be a contradiction.