# Entropy of $f^{m(x)+n}$ of full shift

Let $$(X,\mu,f)$$ be a two-sided full shift system. Assume that there is $$t \in \mathbb{N}$$ such that for every $$n \in \mathbb{N}$$ and $$x \in X$$, we can define $$T(x)=f^{n+m(x)}(x)$$, where $$m(x) \leq t;$$ $$m(x) \in \mathbb{N}$$ and that depends on $$x$$. I also assume that $$T$$ is measure preserving.

It is well-known that $$h_{\mu}(f^n)=nh_{\mu}(f)$$, where $$h_{\mu}$$ is the measure-theoretic entropy.

Can we use the above fact to say $$h_{\mu}(T) \leq (n+t)h_{\mu}(f)$$?

Edit: As it was mentioned in comments, $$T$$ is not necessarily measure preserving. I add the assumption that $$T$$ is measure preserving.

• To begin with, your new transformation need not be measure preserving.
– R W
Oct 18, 2022 at 11:24
• @RW Thanks for your comment. Even, isn't it a measure preserving when $(X, \mu, f)$ is a full shift?
Oct 18, 2022 at 11:37
• What is $\mathbb{N}$? Oct 18, 2022 at 12:58
• @VilleSalo Natural numbers
Oct 18, 2022 at 13:18
• @IlkkaTörmä The reason that I asked questions is: I have a preserving function like $T$, where it is some iteration of $f$, and I want to see whether the above entropy relation holds or not. Unfortunately, I can not exactly write the function as it is so complicated, but I knew $f$, which is some iteration of $T$ is preserving.
If I understand right, your function $$m$$ (for a fixed $$n$$) takes on only finitely many values, which are all measurable sets. You can then define the partition $$\{m^{-1}(i)\}$$ and the associated finite $$\sigma$$-algebra $$\mathcal{B}$$.
If we define $$\mathcal{A}$$ to be any finite sub-$$\sigma$$-algebra of your original $$\sigma$$-algebra of measurable sets, then it's well-known (Corollary 4.10 of Walters) that $$k^{-1} H_\mu\left(\bigvee_{i=0}^{k-1} T^{-i} \mathcal{A} \right)$$ converges (in fact decreases) to $$h_\mu(T, \mathcal{A})$$.
In your case, I think by definition $$\bigvee_{i = 0}^k T^{-i} \mathcal{A}$$ is contained in $$\bigvee_{i = 0}^{k(n+t)-1} f^{-i} (\mathcal{A} \vee \mathcal{B})$$. Therefore, $$k^{-1} H_\mu\left(\bigvee_{i=0}^{k-1} T^{-i} \mathcal{A} \right) \leq k^{-1} H_\mu\left(\bigvee_{i = 0}^{k(n+t)-1} f^{-i} (\mathcal{A} \vee \mathcal{B}) \right) = (n+t) (k(n+t))^{-1} H_\mu\left(\bigvee_{i = 0}^{k(n+t)-1} f^{-i} (\mathcal{A} \vee \mathcal{B}) \right).$$
But by the above, the first quantity approaches $$h_\mu(T, \mathcal{A})$$ and the final quantity approaches $$(n+t) h_\mu(f, \mathcal{A} \vee \mathcal{B})$$. Therefore, $$h_\mu(T, \mathcal{A}) \leq (n+t) h_\mu(f, \mathcal{A} \vee \mathcal{B})$$, and then taking the supremum over $$\mathcal{A}$$ yields $$h_\mu(T) \leq (n+t) h_\mu(f)$$.