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

Question

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!

"if not all requirements of the theorem are met" <-- specifically the requirement is that $U$ is convex. — Martin Brandenburg, Jun 09 '15 at 09:45
After seeing RowanS's non-fitting answer, perhaps you should also add that $f$ is differentiable. — Martin Brandenburg, Jun 09 '15 at 10:02
@MartinBrandenburg Thanks for the hint, I completely forgot about that! It even should be continously-differentiable. — johnnycrab, Jun 09 '15 at 10:05

score 0 · Answer 1 · answered Jun 09 '15 at 10:09

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.

score 0 · Accepted Answer · answered Jun 09 '15 at 10:09

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 $0<y\ll x$: $$ |f(P)-f(Q)|=2x\quad \gg\quad \|P-Q\|=2y.$$

RowanS · Answer 3 · 2015-06-09T10:11:11.097

-1

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.

edited Jun 09 '15 at 10:11

answered Jun 09 '15 at 09:47

RowanS

1,086

Your example is not even continuous, while the OP asked for a "continuously-differentiable function". – Alex M. Jun 09 '15 at 10:14
As stated in his answer he answered before the edit required differentiability. – johnnycrab Jun 09 '15 at 10:23
He added the continuity requirement after I game my answer. – RowanS Jun 09 '15 at 10:24

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

3 Answers3