Suppose $f:(a,b)\to(a,b)$ satisfies $|f(x)-f(y)|<|x-y|,\forall\ x\neq y\in(a,b)$, $x_1\in(a,b),x_{n+1}=f(x_n)$, prove that $\{x_n\}$ is convergent. Here $-\infty<a<b<\infty$.
Any comments and other methods will welcome.Two goals for this post: (1) check my proof and give comments (2) better alternative solutions.
This question comes from the 2025 Graduate Entrance Examination for the Department of Mathematics of Nankai University.
The following is my proof:
(1) It is easy to see that $f$ is uniformly continuous on $(a,b)$, so the both the right limit $f(a+0)=\lim\limits_{x\to a^+}f(x)$ and the left limit $f(b-0)=\lim\limits_{x\to b^-}f(x)$ exist and $$f(a+0)\geq a,\quad f(b-0)\leq b.$$ Let $f(a)=f(a+0),f(b)=f(b-0)$, then $f:[a,b]\to[a,b]$ is continuous on $[a,b]$.
(2) For $x<y\in(a,b)$, we know that $$f(y)-f(x)\leq|f(y)-f(x)|<y-x\implies f(y)-y<f(x)-x;$$ $$f(x)-f(y)\leq|f(y)-f(x)|<y-x\implies f(x)+x<f(y)+y,$$ we get that the function $f(x)-x$ is strictly decreasing on $(a,b)$ and the function $f(x)+x$ is strictly increasing on $(a,b)$. Due to the continuity of $f(x)-x$ and $f(x)+x$ on $[a,b]$, we know that the function $f(x)-x$ is strictly decreasing on $[a,b]$ and the function $f(x)+x$ is strictly increasing on $[a,b]$.
(3) Claim: $$|f(x)-f(y)|<|x-y|,\quad\forall\ x\neq y\in[a,b].$$ If $|f(a)-f(x)|=|a-x|$, then $f(a)-f(x)=a-x$ or $f(a)-f(x)=x-a$. In both cases we can get $x=a$ by the facts of $(2)$. Similarly, if $|f(b)-f(x)|=|b-x|$, we can get $x=b$.
(4) By $(1),(3)$, $f$ has a unique fixed point $x_0\in[a,b]$(for proof, we can refer unique fixed point or note that: $f(x)-x$ satisfies $f(a)-a\geq0,f(b)-b\leq0$ and is is strictly decreasing on $[a,b]$, those imply that $f(x)-x$ has a unique zero, so $f$ has a unique fixed point. ). we will prove that the sequence $x_1\in(a,b),x_{n+1}=f(x_n)$ converges to $x_0$. The following idea is from @copper.hat's Proof. It is clear that $|x_{n+1}-x_0|\leq|x_n-x_0|$ for all $n$. Let $\delta=\inf\limits_n|x_n-x_0|$. If $\delta=0$, we are finished.
Suppose $\delta>0$. Let $$\lambda=\max_{\substack{|x-x_0|\ge\delta \\x\in[a,b]}}\frac{|f(x)-x_0|}{|x-x_0|}.$$ (the $\max$ exists because it is the $\sup$ of a continuous function over a compact domain), and note that $\lambda<1$ due to the claim of $(3)$. Since $|x_n-x_0|\ge\delta$ for all $n$, we have $|x_{n+1}-x_0|\le\lambda|x_n-x_0|$ which implies $\delta=0$, this contradicts $\delta>0$. Hence $\delta=0$ and $\lim\limits_{n\to\infty}x_n=x_0$.
