I Dense orbit and fixed point question

1. Jan 2, 2017

Zafa Pi

Let f be a continuous function of a metric space, M, to itself with a dense orbit and a fixed point.
I.e. there exists z such that the set {f(n)(z)} for all n ∊ N (where f(n) is the nth iterate of f) is dense in M, and there exists p such that f(p) = p.

Does this imply that f spreads?
I.e. does there exists δ > 0 such that for x ≠ y there is an n0 (depending on x and y) with |f(n0)(x) - f(n0)(y)| > δ.

I can prove this for M = R or S1, but I don't know in general, or even in the case where M = the orbit and p.

Last edited: Jan 2, 2017
2. Jan 3, 2017

andrewkirk

I don't think so. The function $f:\mathbb R\to\mathbb R$ given by $f(x)=x/2$ has a dense orbit and a fixed point at 0. But given $\delta>0$, for any two points $x,y$ separated by a distance less than $\delta$ there is no $n$ such that $|f^{(n)}(x)-f^{(n)}(y)|<\delta$.

Also, the identity function on any metric space has a dense orbit, and every point is a fixed point. But again it does not satisfy the above definition of 'spreads'.

3. Jan 3, 2017

Zafa Pi

I suspect your last < should have been >. But that doesn't matter since f(x) = x/2 doesn't have a dense orbit in R. The orbit of z converges to 0 for any z. The orbit of z for the identity function is just z, hardly dense.

4. Jan 3, 2017

andrewkirk

You seem to be using the concept of 'orbit' in a non-standard way. I have only ever seen the orbit of a function defined in relation to a particular point. See for instance here. Since you have not specified a point, guesswork is needed as to what you mean when you write the 'orbit of a function' rather than the 'orbit of a function at a point'.

Hence, when you write that f has a dense orbit in M, I interpret that to mean that the set $\{f^{(n)}(x)\ :\ x\in M,\ n\in\mathbb N\}$ is dense in M. It sounds like you are interpreting it to mean that, for every $x\in X$ the orbit of $f$ at $x$ is dense in $X$.

The last < was intended as written but, given that your question is different from what it appeared to be, the point is moot.

5. Jan 3, 2017

Zafa Pi

The definition you've given (x is fixed at one point and n runs over all members of N) is exactly the one I gave, except I gave a particular point z in M.
For a fixed x the orbit orbit of x for f(x) = x/2 is {f(x), f(f(x)), f(f(f(x))), ...} = {x/2, x/4, x/8, ...} which converges to 0 in R, which is not dense in R.

6. Jan 3, 2017

The Bill

This thread seems to be using the idea of dense periodic orbits either wrong, or in a way different from what I've seen in dynamical systems theory. I've seen it said that a system has dense periodic orbits in its phase space, meaning that the set of points which generate a periodic orbit is dense in the phase space. The individual orbits aren't dense, the points which generate orbits are.

Unless(as I implied may be the case?) I'm confused and the terminology is being used differently here.

7. Jan 3, 2017

Zafa Pi

The notion of dense orbits (or topological transitivity) frequently shows up in chaotic dynamical systems. A periodic orbit is finite and can't be dense unless it is the entire space.
For example the logistic function 4x(1-x) on [0,1] has a dense orbit and two fixed points and spreads (post #1). It also has a dense set of periodic points, but I don't care.

8. Jan 3, 2017

Zafa Pi

I would be grateful and delighted if someone could decide whether f spreads when M (metric) = {x0, x1, x2, ...} U {p}, and has no isolated points, while f is continuous, and f(xn) = xn+1, f(p) = p.

9. Jan 4, 2017

The Bill

I think you might want to rework your definition of spread, since it is always trivially true for any function you pick if you iterate the function 0 times. You have the condition only that there need exist a single iteration where the distance between the iterates for $x$ and $y$ is nonzero. The only functions that never spread under your definition are the constant functions.

10. Jan 5, 2017

Zafa Pi

The definition I use for the orbit of a point is the same as everyone's. The definition I use for being dense is the same as everyone's. the definition I use for for "spread" may be new (I don't know), but is almost the same as the one for "sensitivity to initial conditions". I believe you are having difficulty with the quantifiers. Here is the definition that f spreads:
There exists δ > 0 such that for x ≠ y there is an n0 (depending on x and y) with |f(n0)(x) - f(n0)(y)| > δ.

It means that no matter how close together x and y are there is some iterate of f when applied to x and y separates them to more than δ apart.
So for example if f is the identity function notice that |x - y| = |f(n0)(x) - f(n0)(y)|. Hence if |x - y| < δ no iterate separates them to more than δ apart.
If f(x) = x2 on {1,∞) then f spreads using δ = 1. It's irrelevant, but f has a fixed point at 1, and no dense orbit.

11. Jan 5, 2017

The Bill

For the identity function: If x ≠ y then there always exists δ less than |x - y| and greater than 0. Such a δ is going to be less than |x - y| for every iteration. Therefore the identity spreads according to your definition.

12. Jan 5, 2017

Zafa Pi

The existence of δ precedes the x and y. This is what I meant when I said, "I believe you are having difficulty with the quantifiers." I suggest you look up the definition of "sensitivity to initial conditions".

13. Jan 6, 2017

The Bill

So, the maximal δ if it exists could then be a measurement of the function's spread according to your notion?

14. Jan 6, 2017

Zafa Pi

Indeed. There will always be a sup. I don't think it is necessarily easy to find. What do you think it is for the logistic function, post #7?

I don't know if my definition of spread is different than Devaney's definition of "sensitive to initial conditions", i.e. there exists δ > 0 such that for any x and any neighborhood U containing x there is y in U and no with |f(n0))(x) - f(n0))(y)| > δ. I would like to know.

15. Jan 6, 2017

The Bill

Honestly, I have no idea what you meant by the metric equaling a countable set of points plus a fixed point.

16. Jan 6, 2017

Zafa Pi

M is a countable dense in itself metric space as given in post #8. Thanks for your interest, but I think I'm done with this thread.

17. Jan 6, 2017

The Bill

Sorry, I guess when I studied nonlinear dynamical systems I jumped to continuous systems too quickly. My reaction to this has been affected by that, I suppose.