TITLE: Is there a function with this property?
QUESTION [25 upvotes]: Is there a real function over the real numbers with this property $\ \sqrt{|x-y|} \leq |f(x)-f(y)|$ ?
My guess is no but can anyone tell me why?
This came up as  a question of one of my collegues and i cant give an answer.

REPLY [3 votes]: Whenever $f(x_0)$ and $f(x_1)$ fall in an interval $[a, b]$, we must have $$|x_0 - x_1| \le (f(x_0) - f(x_1))^2 \le (a - b)^{2}.$$  Therefore, the pre-image of any interval $[a,b]$ under $f$ is contained in some interval of length $(a-b)^2$.  Now let $a_0=0$ and $a_k=\sum_{n=1}^{k}n^{-1}$ for $k=1,2,3...$  Since $a_k$ diverges, we have $\cup_k [a_k, a_{k+1}] = [0, \infty)$.  So we can write
$$
\begin{eqnarray}
f^{-1}\left([0, \infty)\right) &=& f^{-1}\left(\cup_k [a_k, a_{k+1}]\right) \\
&=& \cup_k f^{-1}\left([a_k, a_{k+1}]\right) \\
&\subseteq& \cup_k I_k,
\end{eqnarray}
$$
where each $I_k$ is an interval of length $(a_{k+1} - a_{k})^2 = 1/(k+1)^2$.  This countable union of intervals is measurable, and has measure $\le \sum_{k=0}^{\infty} 1/(k+1)^2 = \pi^2/6$.  Similarly, we can show that $f^{-1}\left((-\infty, 0]\right)$ is also contained in a measurable set with measure $\le \pi^2/6$.  Combining these results, we have $f^{-1}(\mathbb{R}) \subseteq Q$, where $Q$ is measurable with measure $\pi^2/3$.  Clearly $Q \neq \mathbb{R}$ (since $\mathbb{R}$ has infinite measure); hence $f^{-1}(\mathbb{R}) \neq \mathbb{R}$, and $f$ cannot be defined over the entire real line.
The same argument with $a_k$ replaced by $\sum_{n=1}^{k}(n+\alpha)^{-1}$ can be used to make the measure of $Q$ arbitrarily small (by choosing $\alpha$ sufficiently large), and we conclude that the domain of $f$ cannot contain any set of positive measure.