# If a bounded function $f:\Bbb R\to \Bbb R$ and $\left|\,f(x)-f(y)\right|<\left|x-y\right|$ for $x\ne y$, then there is an $a$ s.t. $f(a)=a$.

If a bounded function $f:\Bbb R\to \Bbb R$ and $\left|\,f(x)-f(y)\right|<\left|x-y\right|$ for $x\ne y$, then there is an $a$ s.t. $f(a)=a$.

What I know is $f$ should be uniformly continuous. As far as I know, for the fixed point theorem to holds, the above condition is not strong enough to imply a fixed point exists as there should be a contraction,i.e.$\left|\,f(x)-f(y)\right|<\alpha\,\left|x-y\right|$ where $\alpha<1$.

This is a question I saw from a book but now I wonder if it is true .

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it is true if f is bounded(note that f is automatically continuous) – Mehdi Mar 27 '13 at 4:18
you can prove it by finding a sequence ${f^n(b)}$ with a limit, say $l$, and prove that indeed $f(l)=l$ – Mehdi Mar 27 '13 at 4:24

I think I have a simpler proof without usage of Banach fixed point theorem. Let's assume that the range of $f$ is inside of some closed interval $X=[-M,M]$, which is compact. Let's view $f$ as a funcion from $X$ to $X$. Then consider the function on $X$, $g(x)=|f(x)-x|$, which obtains its minimum value, say $\delta$, at $x_{0}$ on $X$. If $\delta=0$, then $x_{0}$ is the fixed point. If $\delta>0$, by the assumption, we have $$|f(f(x_{0}))-f(x_{0})|<|f(x_{0})-x_{0}|=\delta.$$ This is a contradiction. So this proof applies equally well to any compact metric space.

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Nicely done, +1. – 1015 Mar 27 '13 at 11:59

Every bounded and continuous function has a fixed point. This implies the desired result.

Proof: Assume that $f$ is bounded. Then $f(x)\lt x$ for some $x$ close enough to $+\infty$, say $x=c$, and $f(x)\gt x$ for some $x$ close enough to $-\infty$, say $x=b$. Assume furthermore that $f$ is continuous. The intermediate value theorem guarantees the existence of some $a$ in $(b,c)$ such that $f(a)=a$.

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That's the way, +1. – 1015 Mar 27 '13 at 12:29

Edit: Did's and Yunfeng's proofs make mine utterly ridiculous...Oh well, I'll leave it like this.

Take $n\geq 1$ and consider the function $f_n(x):=f(\alpha_nx)$ with $\alpha_n=1-\frac{1}{n}<1$. Then $$|f_n(x)-f_n(y)|\leq\alpha_n|x-y|\qquad\forall x,y\in\mathbb{R}.$$ By Banach fixed point theorem, there exists a unique $x_n\in \mathbb{R}$ such that $$f_n(x_n)=x_n.$$ Snce $f$ is bounded, we obtain a bounded sequence $x_n$ from which we can extract a converging subsequence $x_{n_k}\longrightarrow x$ as $k\rightarrow +\infty$. Now $$|f_{n_k}(x_{n_k})-f(x)|\leq|f_{n_k}(x_{n_k})-f_{n_k}(x)|+|f_{n_k}(x)-f(x)|.$$ The lhs term tends to $0$ by the fact that each $f_{n_k}$ is $1$ Lipschitz. The rhs term tends to $0$ by continuity of $f$ at $x$. Hence $$f(x)=\lim f_{n_k}(x_{n_k})=\lim x_{n_k}=x$$ is a fixed point of $f$. Note that it is unique, as $f$ is a contraction.

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