$f(x+1)=f(x)+1$ for monotone continuous $f$ defined on $\Bbb R$, $g_n(x)=f^n(x)-x$, show $\lim_{n\to\infty}\frac{g_n(x)}{n}$ is independent of $x$. $f(x+1)=f(x)+1$ for monotone continuous $f$ defined on $\Bbb R$, $g_n(x)=f^n(x)-x$, where $f^n(x)=\underbrace{f\circ \cdots\circ f}_n(x)$. Clearly, $g_n(x)$ is periodice with period $1$. How to show $\lim_{n\to\infty}\frac{g_n(x)}{n}$ exists and is independent of $x$?
Periodicity of $g_n$ implies the boundedness of $g_n$, but the bound may be depend on $n$! How to use the condition $f$ is monotone?
 A: I am using the notation $f_n(x)=\underbrace{f\circ \cdots\circ f}_n(x)$.
Note that for $0 \le x < y < 1$ by monotonicity of $f_n$ we have $$f_n(y) - 1 = f_n(y-1) \le f_n(x) \le f_n(y) \le f_n(y+1) = f_n(y) + 1.$$ Therefore, by $1$-periodicity of $g_n$, for any $x,y \in \mathbb{R}$ we have
$$|g_n(x) - g_n(y)| \le |f_n(\{x\}) - f_n(\{y\})| + |\{x\}-\{y\}| \le 2$$ where, $\{x\} = x - [x]$ denotes the fractional part of $x$. This means in particular that if the limit exists for some $x \in [0,1)$ then it is independent of $x$.
Now, if $\exists \, x_0 \in [0,1)$ and a $n_0 \in \mathbb{N}$ such that $g_{n_0}(x_0) = m \in \mathbb{Z}$ then note that $f_{qn_0}(x_0) = x_0 + qm$ for any $q \in \mathbb{N}$. Then writing $n = qn_0 + r$ for $0 \le r < n_0$ we see that $$g_n(x_0) = f_{n_0q + r}(x_0) - x_0 = f_r(f_{n_0q}(x_0)) - x_0 = f_r(x_0) + qm - x_0 = g_r(x_0) + qm$$
thus, the sequence $\left\{\frac{g_n(x_0)}{n}\right\}_{n \in \mathbb{N}}$ converges to $\frac{m}{n_0}$. Otherwise, we are left with the case $g_n(x) \not\in \mathbb{Z}$  for all $n \in \mathbb{N}$ and $x \in \mathbb{R}$. In particular by continuity and $1$-periodicity of $g_n$ we may infer $\exists \, N_n = N(n) \in \mathbb{Z}$ such that $$N_n-1 < g_n(x) = f_n(x) - x < N_n$$ for all $x \in \mathbb{R}$. Therefore, writing $\displaystyle g_{mn}(x) = f_{mn}(x) - x = \sum\limits_{j=1}^{m} f_{jn}(x) - f_{(j-1)n}(x) = \sum\limits_{j=1}^{m} g_{n}(f_{(j-1)n}(x))$ we see that $$m(N_n-1) < g_{mn}(x) < mN_n$$ or equivalently $\displaystyle \left|\frac{g_{mn}(x)}{mn} - \frac{N_n}{n}\right| < \frac{1}{n}$. Interchanging the role of $m$ with $n$ we get $\displaystyle \left|\frac{g_{mn}(x)}{mn} - \frac{N_m}{m}\right| < \frac{1}{m}$. Then by triangle inequality we have $$\left|\frac{N_n}{n} - \frac{N_m}{m}\right| < \frac{1}{n} + \frac{1}{m}$$ that is the sequence $\left\{\frac{N_n}{n}\right\}_{n \in \mathbb{N}}$ and consequently $\left\{\frac{g_n(x)}{n}\right\}_{n \in \mathbb{N}}$ are Cauchy sequence which converges.
