Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 .

share|improve this question
    
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
add comment

3 Answers 3

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.

share|improve this answer
1  
Nicely done, +1. –  1015 Mar 27 '13 at 11:59
add comment

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 $f(a)=a$.

share|improve this answer
1  
That's the way, +1. –  1015 Mar 27 '13 at 12:29
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.