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 $|f(P)-f(Q)| > ||P-Q||$


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!
 A: Consider the domain $U \subset \mathbb{R}^2$ obtained from the annulus $1 <r <2$ 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 $<1$. 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.  
A: Consider $$h(t)=\begin{cases}0&\text{if }t\le 0\\e^{-1/t}&\text{if }t>0\end{cases}$$
which is a smooth(!) function. Then 
$$ g(t)=\frac{h(t+1)\cdot t+h(1-t)\cdot 0}{h(t+1)+h(1-t)}$$
is a function that smoothly switches from constant $g(t)=0$ (for $t<-1$) to $g(t)=t$ (for $t>1$). You may also check that $|g'(t)|\le 1$. Now let
$$f(x,y)= \operatorname{sgn}(y)\cdot g(x)$$ 
on the open set  $$U=\mathbb R^2\setminus\bigl([-1,\infty)\times \{0\}\bigr).$$
Then check points $P=(x,y)$ and $Q=(x-y)$ with $x\gg 0$ and $0<y\ll x$:
$$ |f(P)-f(Q)|=2x\quad \gg\quad  \|P-Q\|=2y.$$
A: 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.
