# Equicontinuity of $\{f_{2n}\circ f_{2n-1}\}$

Let $(X,D)$ be a compact metric space and $\{f_n\}_{n\in\mathbb{N}}$ be a sequence of homeomorphisms of $(X,d)$. It is easy to see that if $\{f_n\}$ is uniformly convergent then $\{g_n\}$ defined by $g_n=f_{2n}\circ f_{2n-1}$ is also uniformly convergent. Thus, for every $\epsilon>0$ there is $\delta>0$ such that if $d(x,y)<\delta$ then $d(g_n(x), g_n(y))<\epsilon$. This means that uniform convergence of $\{f_n\}$ implies that the family $\{f_{2n}\circ f_{2n-1}\}$ is equicontinuous.

Are there other conditions which imply that the sequence $\{f_{2n}\circ f_{2n-1}\}$ is equicontinuous?

• Take an equicontinuous family $\{g_n\}$ and choose $\{f_{2n-1}\}$ arbitrarily. Then set $f_{2n}= g_n\circ \left(f_{2n-1}\right)^{-1}$. So, what is your question? – ifw Apr 17 '15 at 12:48
• Dear Andrade thanks a lot for your reply. I explain my question in more details: Let $(X,d)$ be a compact metric space and $f_n:X\rightarrow X$ be continuous for every $n\in\mathbb{N}$. Are there conditions which imply that the sequence $\{f_{2n}of_{2n-1}\}$ is equicontinuous? For example if $\{f_n\}$ is uniform convergence then $\{f_{2n}of_{2n-1}$ is equicontinuous. – Ali Barzanouni Apr 17 '15 at 18:55