Making images arbitrarily dense under an expanding map

Question

Let $f\colon \mathbb{R}^2 \to \mathbb{R}^2$ be a $C^2$ uniformly expanding diffeomorphism that fixes the origin: that is, $f(0)=0$ and there is $\lambda>1$ such that $d(f(x),f(y)) \geq \lambda d(x,y)$ for all $x,y\in \mathbb{R}^2$. [The original question just asked for a locally expanding map; I've clarified that it should be a globally expanding diffeomorphism.]

Let $X=\{(t,0) : t\in \mathbb{R}\}$ be the $x$-axis in $\mathbb{R}^2$. Is it possible that the images $f^n(X)$ become arbitrarily dense in the unit ball? Or do they satisfy some sort of "uniformly nowhere dense" condition?

More precisely, my first instinct is to expect that the following result is true: for every $f$ as above, there is $\delta>0$ such that for every $n\in \mathbb{N}$, there is some $y\in B(0,1)$ such that $B(y,\delta) \cap f^n(X) = \emptyset$.

After some effort I've been unable to prove this statement. On the other hand, playing around with candidate counterexamples hasn't gotten me anywhere either: the closest I've come is to consider the maps \begin{align*} g(x,y) &= (x, y + A \sin(Rx)), \\ h(x,y) &= (x + A\sin(Ry), y) \end{align*} for some choice of the parameters $A$ and $R$, then choose $c>0$ large enough that $f(x,y) = ch(g(x,y))$ is uniformly expanding. Taking $A=.06$ and $R=100$ gave some interesting pictures, but numerically it seems that I can only make the images $f^n(X)$ continue to get denser in the unit ball if I take $c$ small enough that $f$ is not expanding everywhere.

Which leads me to the question: does every expanding map $f$ as in the first paragraph admit a $\delta$ satisfying the condition in the second paragraph? Or is there a clever counterexample hiding out there somewhere?

Edit: As suggested in the comments, another natural class of maps to consider take the form $f(z) = c e^{ig(|z|)} z$ for $z\in \mathbb{C}$, where $g\colon [0,\infty) \to \mathbb{R}$ must be a $C^2$ function with $g'(0)=0$ to make $f$ be $C^2$. Then $f^n(X)$ spirals around the origin, but we can control the total amount of spiraling by a bounded distortion result: Writing $X^+$ for the positive $x$-axis, then given $r>0$, the point on $f^n(X^+)$ with modulus $r$ has argument given by $h(r) := \sum_{k=0}^{n-1} g(c^{-k} r)$, and we have $$ |h(r) - h(t)| \leq \sum_{k=0}^{n-1} |g(c^{-k}r) - g(c^{-k}t)| \leq \sum_{k=0}^\infty |g|_{\mathrm{Lip}} c^{-k} |r-t| = C|r-t|, $$ which means that $f^n(X^+)$ is the graph in polar coordinates of a function $\theta(r)$ that is $C$-Lipschitz. Then it is not hard to show that there is $\delta>0$ satisfying the condition above.

Answer 1

You are looking at things from a totally wrong perspective, i.e., you try to construct a complicated mapping for a simple curve while it is much easier to construct complicated curves for simple mappings. Also expansion is something that grows and gets more complicated with every step and you do not want to fight monsters unless absolutely forced to do it. Contractions, on the other hand, are tame and beautiful. So let's construct the inverse map instead.

Next note that Hartman-Grobman tells you that (unless you try something really exotic), all maps near the origin after a change of variable are just linear, so there is no point in stretching your imagination here. Take some linear contraction $T$ you know (say $(x,y)\mapsto (x/2,y/32)$) and notice that if you have any horizontal interval of some fixed length $\delta$ in the unit disk at the distance at least $\delta$ from the vertical axis, then its image under the $n$-th iteration is a horizontal interval of length $\delta\ell$ at the distance at most $\ell=2^{-n}$ from the origin lying between the curves $y=\pm x^4$. Now we can easily construct a $C^1$ curve $y=\varphi(x)$ with the first derivative bounded by a small number and tending to $0$ that jumps between these curves so that the distance between jumps is o-small of the jump location as we approach the origin. Then it will intersect any of your intervals if $n$ is large enough in terms of $\delta$. Let $\Phi$ be the $C^1$ automorphism of the plane moving this curve to the horizontal line (say $(x,y)\mapsto(x,y-\varphi(x))$). Now just put $f^{-1}=\Phi\circ T\circ\Phi^{-1}$.

Answer 2

Edited Answer (with minor corrections from comments below)

I think this answer is morally the same as @fedja's above who beat me to it, but maybe slightly more explicit. Set $f(x)=x^5\sin(\pi/x)/(1+|x|^5)$, so that it looks like $x^5\sin(\pi/x)$ near the origin and define $h(x,y)=(x,y+\epsilon f(x))$. This is a global $C^2$ diffeomorphism and satisfies $\|D_xh(v)\|\ge c_\epsilon\|v\|$ for all $x$ and $v$ for a uniform $c_\epsilon$. Choose $\epsilon$ such that $c_\epsilon>1/\sqrt 2$. Now set $T(x,y)=(2x,2048y)$ and consider $\tilde T=h^{-1}\circ T\circ h$ (so that $\tilde T$ is globally expanding). Iterates of the $x$-axis under $\tilde T$ are conjugate by $h$ to iterates of the graph of $f$ under $T$, so it suffices to show that iterates of $\Gamma_f$, the graph of $f$, under $T$ become dense in the unit ball.

Let $2^n>2/\epsilon$ and consider the image of $\Gamma_f$ under $T^n$. Let $k$ be in the range $n$ to $2n-1$. $\Gamma_f$ crosses the $x$ axis $2^k$ times in $(2^{-(k+1)},2^{-k}]$ with crossings roughly $2^{-2k}$-dense. The height of the graph is approximately $\pm 2^{-5k}\epsilon$ between each pair of zeros. Iterating $n$ times, the image of the graph reaches height $2^{11n}\times \pm 2^{-5k}\epsilon$ between each pair of zeros, so that the $n$th iterate of the graph crosses $2^k$ times between roughly $\pm 2^{11n-5k}\epsilon$ (since $2^{11n-5k}>2^n$, the $n$th iterate of the part of $\Gamma_f$ with $x$ values between $2^{-(k+1)}$ and $2^{-k}$ covers $[-1,1]$) for $x$ values that are $2^{n-2k}$-dense (so at least $2^{-n}$ dense) in each range $[2^{n-(k+1)},2^{n-k}]$ with $k=n,\ldots,2n-1$. That is the image of $\Gamma_f$ under $T^n$ contains "branches" that cover $[-1,1]$ and that are $2^{-n}$ dense in the range $\pm[2^{-n},1]$. In particular, the $T$ iterates of $\Gamma_f$ become dense in the unit ball, so the $\tilde T$ iterates of the $x$-axis become dense in the unit ball.