5

A subset $E$ contained in $\mathbb{R}^n$ is such that the function $x \mapsto \left\Vert x\right\Vert^2$ is uniformly continuous on $E$. For $r > 0$, let $E_r$ denote the union of all open balls of radius $r$ contained in $E$. Prove that $E_r$ is bounded for all $r > 0$. Find an example showing that $E$ itself does not have to be bounded.

I have been working on this one for a while and I seem to be stumped. I know what the definitions are, but I'm having trouble getting started on this problem.

Thanks

Julien
  • 45,674
randi
  • 201

2 Answers2

2

The function $f$ is uniformly continuous on the subset $E_r$ as well.

If contrariwise $E_r$ is unbounded for some $r>0$, then there is a sequence of vectors $x_n$ such that $\Vert x_n\Vert\to\infty$, and $B(x_n,r)\subseteq E$. For all $\delta\in(0,r)$ both $x_n$ and $x'_n(\delta)=x_n(1+\delta/(2\Vert x_n\Vert))$ are then in $E_r$, because $\Vert x_n-x'_n(\delta)\Vert<\delta<r$. But $$ f(x'_n(\delta))=f(x_n)\left(1+\frac{\delta}{2\Vert x_n\vert}\right)^2, $$ so $$ f(x'_n(\delta))-f(x_n)\ge f(x_n)\frac{\delta}{2\Vert x_n\Vert}=\frac{\delta}2 \Vert x_n\Vert. $$ This set of differences of values of $f$ is unbounded in spite of the arguments being within $\delta$ of each other violating the assumption that the restriction of $f$ to $E_r$ is uniformly continuous.

For an example of unbounded $E$ I proffer $E=\mathbb{Z}\subset\mathbb{R}$ (map this on the $x$-axis, if you want this to work for any $n$). There are no distinct points within distance $<1$ of each other, so uniform continuity of any function is automatic. Yet $E$ is unbounded.

Jyrki Lahtonen
  • 140,891
  • If you want $E$ to be open, then you can `fatten' my example $E$ by surrounding each point $n\neq0$ with a ball of radius $1/4n^3$ or some other rapidly enough shrinking sequence of radii. – Jyrki Lahtonen Apr 05 '13 at 09:09
  • 1
    But is there a connected $E$ with this property? Some kind of a curve slowly spiralling out? I can't make anything work in 2D, but may be somebody else can (or move to higher dimensions). – Jyrki Lahtonen Apr 05 '13 at 12:07
1

Given $r>0$ there is a positive $\delta<r$ such that $x$, $y\in E$ with $|y-x|<\delta$ implies $|y|^2-|x|^2<1$. Let $m$ be the center of an $r$-ball $B\subset E_r$. Then $m$ and $y:=m+\delta{m\over2|m|}$ are both in $E$; furthermore $|y-m|<\delta$. It follows that $$|m|\delta+{\delta^2\over 4}=|y|^2-|m|^2<1\ ,$$ whence $|m|<{1\over\delta}$. This implies that all points of $B$ have a distance $<{1\over\delta}+r$ from $0$, and since $E_r$ is the union of such balls it follows that $E_r$ is bounded.

The set $E:=\{x\in{\mathbb R}^n\ |\ |x|\in {\mathbb N}\}$ is an unbounded set on which the function $x\to|x|^2$ is uniformly continuous.

Christian Blatter
  • 232,144