Two contractions with the same fixed point

Question

We have $f,g: \mathbb{R} \to \mathbb{R}$ two contractions with the same fixed point and $(x_{n})_{n\geq1}$ a sequence with real numbers with the propriety that $x_{n+1}\in \{f(x_{n}),g(x_{n})\}$, for any $n\geq1$. Show that $(x_{n})_{n\geq1}$ is convergent.

I was thinking to solve in 3 cases. First case when $x_{n+1} = f(x_{n})$ only, the second case when $x_{n+1} = g(x_{n})$ only and the last case when $x_{n+1} = f(x_{n})$ or $x_{n+1} = g(x_{n})$, when the sequence has alternating recursion with unknown periodicity. I was thinking breaking the sequence in 2 subsequences $x_{k_{m}}$ and $x_{q_{p}}$, where the sequence $k_{m}$ is the sequence of the positions (when the function f is in the recursion) and $q_{p}$ is the sequence of the positions (when the function $g$ is in the recursion) and for a sequence with "$n$ terms" we will have "$m$ terms" for the first subsequence + "$p$ terms" for the second subsequence. For the first two cases the problem is easily solved using Banach Theorem for fixed point. What should I use for the third case? The definition of contraction?

Answer 1

Let $L<1$ be a common Lipschitz constant of $f$ and $g$ and let $a$ be the common fixed point of $f$ and $g$. As observed in the comments by Hagen von Eitzen $n \mapsto |x_n-a|$ is decreasing: $$ x_{n+1}=f(x_n) ~ \Rightarrow ~ |x_{n+1}-a| = |f(x_n)-f(a)| \le L|x_n-a| \le |x_n-a|, $$ $$ x_{n+1}=g(x_n) ~ \Rightarrow ~ |x_{n+1}-a| = |g(x_n)-g(a)| \le L|x_n-a| \le |x_n-a|. $$ Hence $\gamma:= \lim_{n \to \infty} |x_n-a|$ exists. We have $\gamma \ge 0$ and want to show $\gamma =0$: Since $(|x_n-a|)$ is convergent $(x_n)$ is bounded and therefore has a convergent subsequence $(x_{n_k})$ with limit $b$, say. In the iteration steps $n_k \to n_k+1$ at least one of the functions $f$ or $g$ is chosen infinite often, w.l.o.g. let it be $f$. Thus there is a subsequence $(x_{n_{k_j}})$ of $(x_{n_k})$ such that $$ x_{n_{k_j}+1}=f(x_{n_{k_j}}) \to f(b) \quad (j \to \infty). $$ Thus $$ \gamma = \lim_{k \to \infty} |x_{n_k}-a| = |b-a|, $$ and $$ \gamma = \lim_{j \to \infty} |x_{n_{k_j}+1}-a| = |f(b)-a|. $$ Now $|b-a|=|f(b)-a| =|f(b)-f(a)| \le L|b-a|$ which yields $a=b$. Therefore $\gamma=0$, which means $x_n \to a$ $(n \to \infty)$.