Why is the Ruelle Transfer Operator a bounded operator?

Question

Let $X^+$ be the space of one sided sequences of a finite alphabet. (Analogous to the space of binary sequences, but where we permit $k$ symbols.)

Moreover, define the metric for $x,y\in X^+$ for some $\theta \in (0,1)$ by:

$$ d(x,y) = \theta ^N $$ where $N$ is the largest integer such that $x_i = y_i$ for all $i < N$. Also let $\sigma$ be the left-sided shift map.

In the literature on subshifts of finite types we then commonly encounter the space $F_\theta(X^+) = \Big\{ f : f \text{ is continuous, } var_n f \leq C \theta^n \Big\}$ which denotes the space of $d_\theta$-Lipschitz continuous functions. The norm here is $||f||_\theta = |f|_\infty + |f|_\theta$ where $|f|_\theta = \sup_n \{\frac{var_n f}{\theta^n} \}$ and $ var_n f = \sup \{ |f(x) - f(y)| : x_i = y_i, i < n \}. $

I then wish to show that the operator $L_f:F_\theta(X^+) \to F_\theta(X^+) $ defined by:

$$L_f w(x) = \sum_{y \in \sigma^{-1}(x)} e^{f(y)} w(y) $$

is a bounded linear operator. Linearity is trivial, and apparently boundedness is too. Unfortunately, I can't seem to see why $L_f$ is bounded. I can't see boundedness because I believe that the sum on the right is infinite and need not converge - though I may be wrong.

Thanks in advance for any help.

Answer 1

As you stated, $X^+$ has a finite alphabet so the pre-image of any $x\in X^+$ under the shift map must be a finite set.

You may want to take a look at this book by Parry and Pollicott. Another reference I've been using describes this in some depth is the book Positive Transfer Operators and Decay of Correlations by Viviane Baladi.

Answer 2

Observe that,

$|L_fw(x)|=|\sum_{y \in \sigma^{-1}(x)} e^{f(y)} w(y)|\leqslant \sum_{y \in \sigma^{-1}(x)}| e^{f(y)} w(y)|\leqslant e^{-|f|_\infty}|w|_\infty$.

Dividing in booth sides by $|w|_\infty \neq 0$ and taking the sup for $w\neq 0$ it follows that $L_f$ act as a bounded operator on the space of the continuous functions.

To see that $L_f$ acts as a bounded linear operator on $F_\theta(X^+)$ we just need to show that the Lipschitz constant of $L_fw$ exists. This is also straightforward.

First, each pre image of the set $\sigma^{-1}y$ is the form, $ax=(a, x_1, x_2, \ldots)$ where $a\in \{1, \ldots, k\}$. Secondly, observe that by the Intermediate Value theorem there exists a constant $\gamma$ such that, $|e^{f(x)}-e^{f(y)}|\leqslant \gamma Lip(f)d_\theta(x,y).$ Therefore,

$$ |L_fw(x)-L_fw(y)|=|\sum_{a\in \{1, \ldots, k\}}w(ax)e^{f(ax)}-\sum_{a\in \{1, \ldots, k\}}w(ay)e^{f(ay)}| \\ = \sum_{a\in \{1, \ldots, k\}} \big[|w(ax)||e^{f(ax)}-e^{f(ay)}|-|e^{f(ay}||w(ax)-w(ay)|\big] \\ \leqslant \gamma Lip(f)|w|_\infty \theta d_\theta(x,y)+ e^{|f|_\infty}\theta Lip(w) d_\theta(x,y) \\ = \big(\gamma Lip(f)|w|_\infty \theta + e^{|f|_\infty}\theta Lip(w) \big)d_\theta(x,y) $$ wich implies that $L_fw$ is Lipschitz.