Let $\mathcal{L}$ denote all circle maps of degree one with nondecreasing liftings (a map $f \in \mathcal{L}$ is of degree one if its lifting $F$ satisfies $F(x+1)=F(x)+1$) . I need to prove that if $g, f_1, f_2 , f_3 , \cdots \in \mathcal{L}$ have liftings $G, F_1, F_2 , F_3 , \cdots $ respectively and the sequence $\{F_i\}^{\infty}_{i=1}$ converges uniformly to $G$, then $$\lim_{i \to \infty} \rho(F_i) = \rho(G)$$ Where $\rho(F)$ is the rotation number of the lifting $F$ and is defined by $$\rho(F)= \lim_{n \to \infty} \frac{F^n(x)-x}{n}$$ $F^n(x)$ denotes the n-th iterate of the function $F$ (not the n-th power). If $F$ is the lifting of a map $f \in \mathcal{L}$, It is known that the above limit exists for all $x \in \mathbb{R}$ and Moreover, it is independent of $x$.
My unsuccessful Try
I need to show that for each $\epsilon > 0$, there exists $M \in \mathbb{N}$ such that for all $i \geq M$, $$|\rho(F_i)- \rho(G)| = | \lim_{n \to \infty} \frac{F_i^n(x)-x}{n} - \lim_{n \to \infty} \frac{G^n(x)-x}{n} | < \epsilon$$ Because rotation number is independent of $x$, we can let $x=0$, so it is enough to show that $$ | \lim_{n \to \infty} \frac{F_i^n(0)-G^n(0)}{n} | < \epsilon $$ I can show that for each iterate $n$, the sequence $\{F^n_i(x)\}^{\infty}_{i=1}$ converges to $G^n(x)$ pointwise (It is quite easy by an induction argument and using the continuity of $G$ )
If I could show that for all n there exists $N$, such that for all $i \geq N$ , $|F_i^n(0)-G^n(0)|<K$ (for some $K$) then the problem is solved, but pointwise convergence of $\{F^n_i(0)\}^{\infty}_{i=1}$ to $G^n(0)$, does not imply that.
This problem is really proving that the map $\rho : F \mapsto \rho(F)$ is continuous. and one can try to prove that the preimage of open sets is open. But that seems more involved, I mean , what topology should one consider on the space of liftings? compact-open topology ...?
I would really appreciate any help on this problem.
A proof on the lines of what you are attempting can essentially be found in [Devaney, An Introduction to Chaotic Dynamical Systems, p.105-107].
Here is what Devaney does, slightly rewritten.
We want to show for some $M$, $|\rho(F_i) - \rho(G)| < \epsilon$ for all $i \geq M$.
Choose $n$ such that $2/n < \epsilon$. Then there exists some $r$ such that $$r-1 < G^n(x) - x < r+1$$ for all $x$. As you say, $F^n_i$ converges to $G^n$ uniformly as $i \to \infty$ so for all $x$ and for any $\delta$ we may choose $M$ large enough so that $|F^n_i(x) - G^n(x)| < \delta$ for all $i \geq M$. Taking $\delta$ to be small enough, we can force $$r-1 < F^n_i(x) - x< r+1$$ for all $x$ as well.
Next we apply the first fact above repeatedly to $0, G^n(0), G^{2n}(0), ...G^{(m-1)n}(0)$:
$$r-1 < G^{n}(0) - 0< r+1$$
$$r-1 < G^{2n}(0) - G^n(0) < r+1$$
$$...$$
$$r-1 < G^{mn}(0) - G^{(m-1)n}(0) < r+1$$
Adding these all together, we have $$m(r-1) < G^{mn}(0) < m(r+1)$$ The same argument shows $$m(r-1) < F_i^{mn}(0) < m(r+1)$$
Then subtracting and dividing by $nm$ we have $$\left|\frac{F_i^{mn}(0)- G^{mn}(0) }{mn}\right| < 2/n < \epsilon$$ for all $m$. If we take the limit as $m$ approaches infinity, since $$\lim_{m\to \infty} \frac{F_i^{mn}(0)}{mn} = \rho(F_i)$$ and $$\lim_{m\to \infty} \frac{G^{mn}(0)}{mn} = \rho(G)$$ we have $|\rho(F_i)- \rho(G)| < \epsilon$ for all $i \geq M$, proving continuity of $\rho$.