# 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 $$\mathbb{R}$$?

• Existence of a point with the dense orbit is one of the definitions of topological transitivity. To see some examples of such maps on $\mathbb{R}$, google "transitive map on R"
– erz
Commented Apr 29, 2020 at 2:03
• I found a topologically transitive map on \mathbb{R} with backward orbit of every point dense on \mathbb{R}, but not the forward orbits.(Reference:- researchgate.net/publication/…) Commented Apr 29, 2020 at 3:35
• @yogamat therefore the paper you link, + the one I provide (Proposition 2) seems to answer your question, except that the proof doesn't tell you which point has a dense forward orbit: it just tells you that every point in a dense $G_\delta$-dense subset $B$ has a dense orbit. This set $B$ is explicit but exhibiting a point in it doesn't seem easy in such an example as the piecewise affine map from your link.
– YCor
Commented Apr 29, 2020 at 5:20
• mathoverflow.net/a/363010/53155
– erz
Commented Jun 14, 2020 at 4:42
• A more challenging question: does there exist a Lipschitz map on ${\bf R}$ with dense orbit? Commented Sep 27, 2020 at 3:19

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.

• @Matt F.: No. If $x < 1/5$, then $T(x) = 1-5x$. Going back to the real number, we get $S(x) =1-5\{x\}+ \lfloor x \rfloor + F(\{x\}) = \lfloor x \rfloor + 1-5\{x\}-1 = \lfloor x \rfloor -5\{x\}$ whenever $\{x\} < 1/5$, so the sequence $(x_n)$ can decrease. Commented Sep 22, 2021 at 22:25
• Basically, the graph of this function starts from $(0,0)$, goes down to $(1/5, -1)$, goes up to $(4,/5, 2)$, goes down to $(6/5, 0)$, goes up to $(9/5, 3)$, etc. I will post a graph tomorrow to make the construction more intuitive (because it really is) ; it's a situation where a picture is worth a thousand words. Commented Sep 22, 2021 at 22:28
• @Matt F.: The images are now there. Commented Sep 23, 2021 at 9:21
• Now I agree that this works from a generic starting point, and can be tweaked to (apparently) work from $0$. I would describe your construction as: Let $$S_0(x) = \lfloor x \rfloor + \min(5(x- \lfloor x \rfloor), 6 - 5(x - \lfloor x \rfloor)),$$ $$S(x)=S_0(x-\frac15)-1.$$ Then it is easy to see that both $S_0$ and $S$ also define maps on $\mathbb{R}/\mathbb{N}$. Furthermore $$S(x)=S_0(x-\frac13)-\frac{13}{15}$$ seems to give a dense forward orbit of $0$, and (answering @TerryTao's question) with a Lipschitz function.
– user44143
Commented Sep 24, 2021 at 17:05

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.

• @D.Thomine's answer referenced above. Commented Sep 22, 2021 at 23:46
• It now looks to me like this is different from D. Thomine's answer, and his is probably right and easier to prove correct.
– user44143
Commented Sep 24, 2021 at 16:25