$f:\mathbb R^2 \to \mathbb R$ be a function , $|f(x)-f(y)|\ge 3\|x-y\| , \forall x,y \in \mathbb R^2$ ; is $f(\mathbb R^2)$ open in $\mathbb R$?

Question

Let $f:\mathbb R^2 \to \mathbb R$ be a function such that $|f(x)-f(y)|\ge 3\|x-y\| , \forall x,y \in \mathbb R^2$ , then is it true that $f$ maps open sets of $\mathbb R^2$ to open sets of $\mathbb R$ ? I can show that $f$ maps open sets to open sets in $f(\mathbb R^2)$ , so to prove the claim we only need to show that $f(\mathbb R^2)$ is open in $\mathbb R$ . Please help . Thanks in advance

Answer 1

In a sense, your claim is true. For every such map, $f(\Bbb{R}^2) \subset \Bbb{R}$ is open, simply because there is no such map.

To see this, note that $f$ is injective, so that $f^{-1} : M \to \Bbb{R}^2$ is well-defined and surjective, with $M := f(\Bbb{R}^2)$. But for $x,y \in M$, we have $x = f(x')$ and $y = f(y')$ for certain $x',y' \in \Bbb{R}^2$. Hence, $$ \|f^{-1}(x) - f^{-1}(y)\| = \| x' - y'\| \leq \frac{1}{3} |f(x') -f( y')| = \frac{1}{3} |x-y|, $$ so that $f^{-1} : M \to \Bbb{R}^2$ is Lipschitz continuous and surjective, where $M \subset \Bbb{R}$ (already suspicious).

But by Kirszbraun's Theorem (https://en.wikipedia.org/wiki/Kirszbraun_theorem, in the current special case, it is not hard to prove it, since we can consider each component individually), we can find a Lipschitz-continuous extension $g : \Bbb{R} \to \Bbb{R}^2$ of $f^{-1}$.

Consider this as a map $g : \Bbb{R} \times \{0\} \subset \Bbb{R}^2 \to \Bbb{R}^2$ (and, using Kirszbraun's theorem, extend it to a Lipschitz continuous map $h : \Bbb{R}^2 \to \Bbb{R}^2$).

It is well-known that Lipschitz maps from $\Bbb{R}^n$ to $\Bbb{R}^n$ map null-sets to null-sets (this is also not hard to see directly, using the fact that a null-set can be covered by countably many cubes $Q_n$ with $\sum_n vol(Q_n) < \varepsilon$ for arbitrary $\varepsilon > 0$ and that the image of a cube under a Lipschitz map $\Phi$ satisfies $\lambda(\Phi(Q)) \leq (10 \cdot C)^n \lambda(Q)$, where $C$ is the Lipschitz-consatnt and $\lambda$ is Lebesgue measure).

Hence, we conclude that $\Bbb{R}^2 = f^{-1}(M) = g(M) = h(M \times \{0\}) \subset h(\Bbb{R}\times \{0\})$ is a null-set in $\Bbb{R}^2$, a contradiction.