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.

I come across this problem in my functional analysis exercise book:


Let X be complete metric space and g be a function, If $g^q$ contracts then $g$ has a fixed point.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

HINT: Let $x_0\in X$. Given $x_n$, let $x_{n+1}=f^p(x_n)$. Use the hypothesis that $f^p$ is a contraction to show that the sequence of distances $\langle\rho(x_n,x_{n+1}):n\in\Bbb N\rangle$ shrinks exponentially fast towards $0$. Deduce from this that $\langle x_n:n\in\Bbb N\rangle$ is Cauchy; since $X$ is complete, it converges to some point $y$. Show that $f(y)=y$.

Now let $x_0'$ be another point of $X$, and construct the sequence $\langle x_n':n\in\Bbb N\rangle$ similarly. Show that $\langle\rho(x_n,x_n'):n\in\Bbb N\rangle$ coverges to $0$, and conclude that $\langle x_n':n\in\Bbb N\rangle$ also converges to $y$. Since $x_0'$ was arbitrary, conclude that $y$ is the unique fixed point of $f$.

share|improve this answer
    
How to know f(y)=y.It is the key,thanks. –  Joe Berg Oct 14 '12 at 16:35
2  
$y$ and $f(y)$ are fixed points of $f^p$. Thus $d(y,f(y)) = d(f^p(y),f^p(f(y)))$. Assume that $f(y)\neq y$. Since $f^p$ is a contraction $d(f^p(y),f^p(f(y))) < d(y,f(y))$. We get a contradiction: $d(y,f(y)) = d(f^p(y),f^p(f(y))) < d(y,f(y))$. –  Yury Oct 14 '12 at 16:42
    
@89085731: See Yury’s comment just above this. (He used $d$ instead of $\rho$ for the metric, but that shouldn’t pose any problem.) –  Brian M. Scott Oct 14 '12 at 16:45
    
@Yury Thanks a lot. –  Joe Berg Oct 14 '12 at 16:45

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.