Does there exist a function $f$ from $R^2$ to $R$ and constant $c_0>0$ such that for all $x,y\in R^2$, $|f(x)-f(y)| \ge c_0|x-y|^2$?

Question

I know the function must be injective as for $x \not= y$, $c_0|x-y|^2>0$. Thus the function cannot be continuous. I'm feeling that no such function exists but I'm not sure how I would justify this.

Answer 1

In this answer, it is proved that the Hilbert space-filling curve is Hölder continuous with exponent $\frac{1}{2}$. Althouth the original curve is a mapping $[0,1]\to[0,1]^2$, it is easy to extend this function to a surjection $g : \mathbb{R} \to \mathbb{R}^2$ which is also Hölder continuous: there exists $C > 0$ such that

$$ \| g(t) - g(s) \| \leq C |t - s|^{1/2}. $$

Since $g$ is a surjective, we can find an injective function $f : \mathbb{R}^2 \to \mathbb{R}$ such that $g\circ f = \operatorname{id}_{\mathbb{R}^2} $. Then this function satisfies

$$ \forall x, y \in \mathbb{R}^2 \ : \quad \| x - y \|^2 \leq C^2 |f(x) - f(y)|. $$

So the condition is satisfied with $c_0 = C^{-2} > 0$.