Existence of continuous map on real numbers with dense orbit? Does there exist a continuous map $f:\mathbb{R}\rightarrow \mathbb{R}$ such that the forward orbit of 0 is dense in $\mathbb{R}$?
 A: As in this previous answer of mine, emulating random walks works pretty well for this kind of question.
Consider the map $T : \mathbb{S}_1 \to \mathbb{S}_1$ defined by
$$T(x) = 5x [1] \ \text{ if } \ 1/5 \leq x < 4/5,$$
$$T(x) = -5x [1] \ \text{ otherwise.}$$
Its graph is as follows:

The map $T$ is continuous, preserves the Lebesgue measure, is ergodic, and much more.
Now, let me introduce the $\mathbb{Z}$-extension $\widetilde{T} : \mathbb{S}_1 \times \mathbb{Z} \to \mathbb{S}_1 \times \mathbb{Z}$ of $T$ defined by
$$\widetilde{T} (x, p) := (T(x), p+F(x)),$$
where $F(x) = -1$ for $x \in [0,2/5)$, then $F(x) = 0$ for $x \in [2/5,3/5)$ and $F(x) = +1$ for $x \in [3/5,1)$. Note that $\widetilde{T}$ preserves the uniform ($\sigma$-finite) measure on $\mathbb{S}_1 \times \mathbb{Z}$.
The second coordinate of $\widetilde{T}^n (x, p)$ is $p+S_n F(x) := p+\sum_{k=0}^{n-1} F (T^k (x))$. Under the Lebesgue measure on $\mathbb{S}_1$, the sequence $(F \circ T^k)_{k \geq 0}$ is a sequence of i.i.d. random variables of symmetric distribution $2/5\cdot \delta_{-1} + 1/5 \cdot \delta_0+ 2/5\cdot \delta_{+1}$, so that the process $(S_n F)_{n \geq 0}$ is ergodic and recurrent.
A bit more work (but not that much, given the simplicity of the model) gives that $\widetilde{T}$ is ergodic and recurrent for the uniform measure on $\mathbb{S}_1 \times \mathbb{Z}$. This is very much folklore, although I have to admit it can get annoying to pinpoint the best reference and fill the gaps. As a consequence, almost every point has a dense orbit.
Now, everything is on $\mathbb{S}_1 \times \mathbb{Z}$; however, is we identify $\mathbb{S}_1$ with $[0,1)$ and then $\mathbb{S}_1 \times \mathbb{Z}$ with $\mathbb{R}$, we get a map $S$ from $\mathbb{R}$ to $\mathbb{R}$. The specific choice of $T$ gives that $S$ is continuous, actually 5-Lipschitz, with a sawtooth-like graph (in black, the line with equation $y=x$):

Again, Lebesgue-almost every point has a dense orbit. Here is the picture of an orbit from a random (uniform in $[0,1]$) starting point:

The orbit of $0$ is not dense, as it is a fixed point; however, conjugating by a Lebesgue-generic translation gives the map we want.
The same construction work on $\mathbb{R}^2$ (just use $S \times S$), where Lebesgue almost every orbit will be dense. Things get more annoying in higher dimension, since the random walks are no longer recurrent. This can be solved by getting a tweak $R$ of $S$ favouring orbits recurring quickly to zero, ensuring that $R$ preserves a unique absolutely continuous invariant measure, with respect to which it is mixing; then $(R, R, \ldots, R)$ is mixing with respect to the product measure on $\mathbb{R}^n$, and from there topologically mixing.
A: Here is a possibility, inspired by D. Thomine's answer (without deciding whether the answers are the same or whether either is right). Let
$$S(x)=(-1)^{\lfloor x\rfloor}(\pi+|x|)(1-2x+2\lfloor x \rfloor))$$
Then $S$ looks like

and the first $10000$ iterates of $0$ look like

That graph goes from $-14.2$ to $14.6$, and after a million iterates it goes from $-26.0$ to $26.6$. The first four moments and the extremes of these iterates are roughly what one would expect from a normal distribution with standard deviation $4$. So I suspect that the infinite set of iterates is roughly normal and dense in $\mathbb{R}$; perhaps someone will see how to use ergodic theory to prove that.
