Is there a proof that the double pendulum path reaches every point in a closed region eventually? The simplest double pendulum are just two rods attached to each other. The main rod may or may not flip depending on the initial conditions.
If it flips, the path that the end of pendulum passes covers the space between two concentric circles. Is it true that all points are covered if we let pendulum spinning to infinity?
If it does not ever flip, there is the max height the path will reach on both sides, but apart from that it is still a space between two concentric circles. The same question in this case, all points that are bellow this height and between the two circles will be visited?
The simulations are strongly confirming this, making the path if we take it as a fractal having dimension $2$.
Any proof of this in the literature?
If you take it like a probability of the end of pendulum being somewhere like in this post
https://physics.stackexchange.com/questions/349915/probability-density-to-find-the-end-of-a-double-pendulum
I need only the entire space of probability being different from $0$. I am somewhat sure it has a higher dimension than $1$, but not certain if it has dimension $2$ most of the time, all of the time, sometimes or never.
Regarding the dynamics and complexity it would be very natural to assume that, at least, the end of pendulum reaches almost every point meaning the number of points it does not reach has measure $0$.
But it is just me. (I am not sure if this question is that difficult that I cannot find even a hint about it. It is still just a question about fractal dimension.)
This is an illustration of day or two the simulation running:

 A: As far as I understand your question, you are asking this: Given some initial conditions of the double pendulum, does the trajectory of the pendulum's end-point visit every point in some "permissible region"? You also guess that this region is a section of the area between two concentric circles which lies below a certain horizontal line. Please correct me in case I misunderstood.
I believe the answer to this question is no. First an intuitive explanation: The trajectory is a "well-behaved" (continuous, differentiable with respect to time, etc.) curve and as such has a zero area. So it obviously cannot cover the whole region whose area is non-zero. Here you could of course object, pointing to the existence of space-filling curves. The key point is that space-filling curves have to be "irregular" in a way that the trajectory cannot be.
To make this argument more rigorous, note that the path $\vec{x}(t)$ taken by the end-point of the pendulum is a Lipschitz function (since the velocity is bounded by the finite energy of the system). The trajectory $T$ is the image of $\mathbb{R}$ under $\mathbf{x}(t)$. There is apparently a theorem (cited for example here; I didn't check this by going to the original source) which says that transformation by a Lipschitz function increases the Hausdorff measure of a set by at most a constant multiplicative factor. Since $\mathcal{H}^2(\mathbb{R}) = 0$, this gives us $\mathcal{H}^2(T) = 0$, so the trajectory cannot fill your region of non-zero area.
That being said, I do think that your intuition is correct in some sense. As the double pendulum is a chaotic system (given sufficiently high energy), it is quite reasonable to believe that its trajectory will eventually get within an arbitrarily small distance $\varepsilon>0$ of every point in the system's phase space which is consistent with energy conservation. This last property would follow easily from ergodicity but I could not find any results showing that the double pendulum is (not) ergodic.
