Regarding iterated maps $\mathbb{R}^2 \rightarrow \mathbb{R}^2$, and strictly decreasing norm

Question

Consider a map $g:\mathbb{R}^2 \rightarrow \mathbb{R}^2$, and the iteration $(x,y) \mapsto g(x,y)$. Say that the origin is an asymptotically stable fixed point and there is a region $A$ in the first quadrant where we can show that for all $(x_0,y_0) \in A$,

1. $g(x,y) \in A$
2. $\|g(x,y)\| < \|(x,y)\|$
3. $g(x,y)$ continuous

Now, my goal is to show that all $(x,y) \in A$ will eventually converge to the origin, or equivalently $$\lim_{k \to \infty} \|g^k(x,y)\| = 0.$$

I initially thought that the conditions listed above were sufficient, but now I have doubts. What if, even though the norm was strictly decreasing, it only approached some nonzero limit, with $$\lim_{k \to \infty} g^k(x,y) = p \ne (0,0) \in A.$$

My question is: could such a $p$ exist? For it cannot be a fixed point under the iteration due to (2) above. This means that $$\lim_{k \to \infty} g^k(p) \ne p$$

Basically I am pretty sure there is a contradiction in combining $$\lim_{k \to \infty} g^k(x,y) = p$$ with $$g(p) \ne p,$$ but I can't exactly pin it down.

Summed up, is it possible that there could be some sequence $\{(x,y),g(x,y), g^2(x,y), \ldots g^k(x,y)\}$ that never reaches the origin, even though there are no fixed points in $A$ besides the origin? And why not precisely? Thank you.

Answer 1

If $g$ is not continuous, then there is such a sequence, even in 1D, take for example $$g(x) = \begin{cases}1/4+x/2 & \mbox{ for } x > 1/2\\ 0 & \mbox{ else}\end{cases}$$

However if $g$ needs to be continous then there is no such function. Assume there is a sequence of points $p_k = g^k(x_0,y_0)$ not converging to 0.

Since $g$ and the norm are continuous, $f:(x,y) \mapsto \|g(x,y)\|$ is a continous function. But clearly $f(p_k)$ is a decreasing sequence and bounded from below, so $f(p_k)$ converges to some $r > 0$.

Since $\|p_k\| < \|p_0\|$ for all $k\in\mathbb{N}$, $p_k$ is a bounded sequence, so by Bolzano-Weierstrass, there is some converging subsequence $p_{k_l} \rightarrow p \neq 0$. Now on one hand by continuity of $f$

$$ r= \lim_{l\rightarrow \infty} f(p_{k_l}) = f(p) = \|g(p)\|$$

and on the other hand

$$ r = \lim_{l\rightarrow \infty} f(p_{k_l-1}) = \lim_{l\rightarrow \infty} \|g(p_{k_l-1})\| = \lim_{l\rightarrow \infty} \|p_{k_l}\| = \|p\|$$

and thus $\|p\|=r=\|g(p)\|$ which is a contradiction.

Edit: also I just noted, I need that A is closed, otherwise $p$ may not lie in $A$ and there probably again is a counterexample.

Answer 2

Here is a counterexample. Let the region $A$ be $$\{(x,y) \, | \, x > 0, y > 0, xy > 1\} $$ Let $\overline A$ be the closure of $A$: $$\{(x,y) \, | \, x>0, y>0 ,xy\ge 1\} $$ Define the map restricted to $\overline A$ as follows: on each ray $R$ through the origin in the first quadrant, map the subray $R \cap \overline A$ to itself by linearly shrinking $R \cap \overline A$ towards its base point on the hyperbola $xy=1$. Note that iteration of a point of $A$ converges to a point on the hyperbola. You can choose the linear shrinking factors on different rays to be equal, so that this produces a continuous map on $\overline A$.

Meanwhile, take a tiny closed ball $B$ around the origin and let the map restrict on $B$ to scalar multipliciation by a factor of $1/2$, so that origin is an "asymptotically stable fixed point".

Finally, extend continuously to all of $\mathbb{R}^2$.