Does every diffeomorphism between $\mathbb{B}^n$ and $\mathbb{R}^n$ satisfies that $\lim_{\|x\| \to 1} \|f(x)\| = \infty$

Question

In this question: Diffeomorphism: Unit Ball vs. Euclidean Space a number of diffeomorphisms between $\mathbb{B}^n$ and $\mathbb{R}^n$ are presented. They all have in common that the points closer to the border are sent "further". This is: $$\lim_{\|x\| \to 1} \|f(x)\| = \infty$$

Intuitively these diffeomorphisms "extend" the ball more or less uniformly, as to fill the space. Is there any diffeomorphism that doesn't have this property, so a sucession of points that converge to the border remain at a bounded distance from the origin?

Answer 1

Nice problem! We can prove that any homeomorphism (i.e. only preserving the topology) between $\mathbb{R}^n$ and $\mathbb{B}^n$ satisfies $\lim_{\lVert x \rVert \to 1} \lVert f(x) \rVert = \infty$.

To show this, let me recall the following fact.

Let $C \subset \mathbb{B}^n=\{x \in \mathbb{R}^n \mid \lVert x \rVert <1 \}$ be a compact subset. Then, there is some $0<c<1$ such that $\lVert x \rVert \leq c$ for any $x \in C$.

(It is a nice problem if you haven't done it before!)

Keeping this in mind, assume that $f\colon \mathbb{B}^n \to \mathbb{R}^n$ is a homeomorphism, and let $M>0$ be given. Consider $C=\{x \in \mathbb{R}^n \mid \lVert x \rVert \leq M\}$, which is a compact subset of $\mathbb{R}^n$. Take its preimage $f^{-1}(C) \subset \mathbb{B}^n$, which is a compact subset as $f$ is an isomorphism. Take the value $c$ from the fact above, and let $\delta=1-c \in (0, 1)$. This way, if $x \in f^{-1}(C)$, then $\lVert x \rVert \leq c$. Conversely, if $\lVert x \rVert >c$, then $c \notin f^{-1}(C)$.

If $\lvert \lVert x \rVert - 1 \rvert < \delta$, then $-(\lVert x \rVert -1) < \delta$, and so $\lVert x \rVert > 1-\delta=c$. By the preceding remark, we get $x \notin f^{-1}(C)$. Then $f(x) \notin C$, i.e. $\lVert f(x) \rVert > M$. This shows that $\lVert f(x) \rVert \to \infty$, as required.