How to prove that $n^{-2}[x+g(x)+g\circ g(x)+\cdots +g^{\circ n}(x)]$ converges when $n\to\infty$ [duplicate]

Let $f:\mathbb R\to\mathbb R$ be a periodic function with period $1$. We assume that $f$ is Lipschitz continuous, and in particular, we assume that there exists an $L\in (0,1)$, such that $$ |f(x)-f(y)| \le L|x-y|, \quad \text{for all $x,y\in\mathbb R$.} $$ Let also $g(x)=x+f(x).$ Show that the limit $$\lim_{n\to +\infty}\frac1{n^2}[x+g(x)+g(g(x))+\cdots +g^{\circ n}(x)]$$ exists and is independent of $x$, where, for every $n\ge1$, $g^{\circ n}$ is $g$ composed with itself $n$ times, thus $g^{\circ 1}=g$ and, for every $n\ge1$, $g^{\circ n+1}=g\circ g^{\circ n}$.

My attempt: Since $$f(x+1)=f(x),\forall x\in R$$ then $$g(g(x))=g(x+f(x))=x+f(x)+f(x+f(x))$$ $$g(g(g(x)))=g(x+f(x)+f(x+f(x)))=x+f(x)+f(x+f(x))+f(x+f(x)+f(x+f(x)))$$ $$\cdots\cdots $$ and I have $$|g(x)-g(y)|=|x-y+f(x)-f(y)|\le |x-y|+|f(x)-f(y)|<(L+1)|x-y|$$ where $L+1>1$.

So $g$ is also Lipschitz continuous, and that's all I can do.

Thank you for you help.

Solution 1:

First attempt. One needs to use the Stolz–Cesàro theorem. According to this theorem:

\begin{align} \lim_{n\to \infty}\dfrac{x+g(x)+g(g(x))+\cdots +\underbrace{g(g(\cdots(g(x))))}_{n-1}}{n^2}&=\lim_{n\to \infty}\frac{\underbrace{g(g(\cdots(g(x))))}_{n}}{2n+1}\\=&\lim_{n\to \infty}\frac{\underbrace{g(g(\cdots(g(x))))}_{n+1}-\underbrace{g(g(\cdots(g(x))))}_{n}}{2}, \end{align} provided that the last limit exists. Define $x_n=\underbrace{g(g(\cdots(g(x))))}_{n}$. Then $$ \underbrace{g(g(\cdots(g(x))))}_{n+1}=g(x_n)=f(x_n)+x_n, $$ and hence $$ \underbrace{g(g(\cdots(g(x))))}_{n+1}-\underbrace{g(g(\cdots(g(x))))}_{n}=f(x_n)=f\bigg(\underbrace{g(g(\cdots(g(x))))}_{n}\bigg) $$ We need show that the limit of $f(x_n)$ exists.