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edited Aug 19, 2019 at 1:58

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This conclusion is almost certainly true, and the argument is morally true, but I don’t understand the precise argument given. In particular, what variables are the supremum and infimum being taken over? Also the argument given fails if all of the maps are the identity map (so the argument is not correct in the generality claimed).

What you’re looking at is an example of an iterated function system: you have $x_{n+1}$ is $h_{\omega_n}(x_n)$. I haven’t looked in detail, but I assume that there is an interval $J$ such that $h_\omega(x)\in J$ for all $\omega \in[-1,1]$? Assuming this, what remains is to show (ideally) that each $h_\omega$ is contracting on all of $J$. This would allow one to complete the argument given. Failing this, it would suffice to show that there exists a sequence $\delta_n\to 0$ such that $h_{\omega_1}\circ\ldots h_{\omega_n}(J)$ is of length at most $\delta_n$ for each choice of $\omega_1,\ldots\omega_n$. Failing this, I would try for an argument based on negativity of the Lyapunov ExponentsExponent.

Assuming that one of these conditions holds, you can find the unique invariant measure by looking at the distribution of $\lim_{n\to\infty}h_{\omega_{-1}}\circ\ldots\circ h_{\omega_{-n}}(0)$. The conditions above are to ensure that the limit exists.

This conclusion is almost certainly true, and the argument is morally true, but I don’t understand the precise argument given. In particular, what variables are the supremum and infimum being taken over? Also the argument given fails if all of the maps are the identity map (so the argument is not correct in the generality claimed).

What you’re looking at is an example of an iterated function system: you have $x_{n+1}$ is $h_{\omega_n}(x_n)$. I haven’t looked in detail, but I assume that there is an interval $J$ such that $h_\omega(x)\in J$ for all $\omega \in[-1,1]$? Assuming this, what remains is to show (ideally) that each $h_\omega$ is contracting on all of $J$. This would allow one to complete the argument given. Failing this, it would suffice to show that there exists a sequence $\delta_n\to 0$ such that $h_{\omega_1}\circ\ldots h_{\omega_n}(J)$ is of length at most $\delta_n$ for each choice of $\omega_1,\ldots\omega_n$. Failing this, I would try for an argument based on Lyapunov Exponents.

This conclusion is almost certainly true, and the argument is morally true, but I don’t understand the precise argument given. In particular, what variables are the supremum and infimum being taken over? Also the argument given fails if all of the maps are the identity map (so the argument is not correct in the generality claimed).

What you’re looking at is an example of an iterated function system: you have $x_{n+1}$ is $h_{\omega_n}(x_n)$. I haven’t looked in detail, but I assume that there is an interval $J$ such that $h_\omega(x)\in J$ for all $\omega \in[-1,1]$? Assuming this, what remains is to show (ideally) that each $h_\omega$ is contracting on all of $J$. This would allow one to complete the argument given. Failing this, it would suffice to show that there exists a sequence $\delta_n\to 0$ such that $h_{\omega_1}\circ\ldots h_{\omega_n}(J)$ is of length at most $\delta_n$ for each choice of $\omega_1,\ldots\omega_n$. Failing this, I would try for an argument based on negativity of the Lyapunov Exponent.

Assuming that one of these conditions holds, you can find the unique invariant measure by looking at the distribution of $\lim_{n\to\infty}h_{\omega_{-1}}\circ\ldots\circ h_{\omega_{-n}}(0)$. The conditions above are to ensure that the limit exists.

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created Aug 19, 2019 at 1:37

23.2k
5
63
98

This conclusion is almost certainly true, and the argument is morally true, but I don’t understand the precise argument given. In particular, what variables are the supremum and infimum being taken over? Also the argument given fails if all of the maps are the identity map (so the argument is not correct in the generality claimed).

What you’re looking at is an example of an iterated function system: you have $x_{n+1}$ is $h_{\omega_n}(x_n)$. I haven’t looked in detail, but I assume that there is an interval $J$ such that $h_\omega(x)\in J$ for all $\omega \in[-1,1]$? Assuming this, what remains is to show (ideally) that each $h_\omega$ is contracting on all of $J$. This would allow one to complete the argument given. Failing this, it would suffice to show that there exists a sequence $\delta_n\to 0$ such that $h_{\omega_1}\circ\ldots h_{\omega_n}(J)$ is of length at most $\delta_n$ for each choice of $\omega_1,\ldots\omega_n$. Failing this, I would try for an argument based on Lyapunov Exponents.