Let $f\colon \mathbb{R}^2 \to \mathbb{R}^2$ be a $C^2$ uniformly expanding map that fixes the origin: that is, $f(0)=0$ and there is $\lambda>1$ such that $\|D_x f(v)\| \geq \lambda\|v\|$ for every $x$ and $v$. 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?

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.