Find function $f: U \subset \mathbb{R}^2 \to \mathbb{R}$ with $||Df(x,y)|| \leq 1$ such that $P,Q \in U$ exist with $|f(P) - f(Q)| > || P-Q||$ I am looking for a continuously-differentiable function $f: U \to \mathbb{R}, U \subset \mathbb{R}^2$ which satisfies the following requirements:


$U$ is open set
$U$ is connected (right word?), i.e. $\forall P,Q \in U\ \ \exists$ continuous $\gamma: [0,1] \to U$ with $\gamma(0)=P$ and $\gamma(1)=Q$
$||Df(x)|| \leq 1$ for all $x \in U$ ($Df(x)$ is Jacobi Matrix)
There are points $P,Q \in U$ with $|f(P)-f(Q)| &gt; ||P-Q||$


Essentially I am looking for a counter-example for the mean value theorem if U is not convex.

I had been thinking about trying to construct an (open) spiral like this: http://www.mathematische-basteleien.de/spiral22.gif
, but can't really get it work.

Any ideas / hints / other examples?

Thanks a lot in advance!
Consider the domain $U \subset \mathbb{R}^2$ obtained from the annulus $1 &lt;r &lt;2$ by removing the real half axis $[0, \infty)$. Consider the function $f(x,y) = \text{arg}( x + i y)= \phi$ in polar coordinates. $f$ maps $U$ onto $(0,2 \pi)$. The gradient of $f$ is $\nabla \arctan(y/x) = (- \frac{y}{x^2 + y^2}, \frac{x}{x^2 + y^2})$ so its norm is $&lt;1$. Check to see that points very close by in the distance from $\mathbb{R}^2$ have values of $f$ differing by about $2 \pi$. 

The point is that the distance from $\mathbb{R}^2$ is in general not the intrinsic distance ( the infimum of lengths of paths ) if $U$ is not convex. However, for that intrinsic distance the inequality will still hold, in fact you can check on this example.
Answer was provided before edit required continuity.

F(x,y)=0 elsewhere

F(y,y^2)=1

Now consider the origin. All gateaux derivatives exist but the function isn't even continuous.