0
$\begingroup$

Let $\Omega$ = closed ball $B_1(0)$ in $\mathbb{R}^n$ with metric d induced by the Euclidean norm. Suppose the mapping $T: \Omega \to \Omega$ satisfies

$d(Tx,Ty) \leq d(x,y)$ for all $x,y \in \Omega$

Prove that there exists at least on fixed point of T. Hint consider the map $T_k = (1-\frac{1}{k})T$

So I first start off with proving T is a contraction, nonetheless, consider

$|T_k(x) - T_k(y)| = |(1-\frac{1}{k})T(x) - (1-\frac{1}{k})T(y)|$

$=|(1-\frac{1}{k})(T(x)-T(y))|$

$\leq |1-\frac{1}{k}||T(x)-T(y)$

However, the answer says that

$|T_k(x) - T_k(y)| \leq (1-\frac{1}{k})^2|x-y|$

Where did the square come from is it because we're dealing with the Euclidean norm or did I do something wrong? Any help would be greatly appreciated

$\endgroup$
  • $\begingroup$ I guess it is a mistake in the "answer". $\endgroup$ – Peter Melech Nov 10 '18 at 9:55
  • $\begingroup$ But, see if it is a mistake then $(1-\frac{1}{k})$ wouldn't satisfy a constant, and this would not be contraction map, however, if it is squared then it would be a contraction because $c=(1-\frac{1}{k})^2$ is a constant between $0\leq c < 1$ $\endgroup$ – lastgunslinger Nov 10 '18 at 10:00
  • $\begingroup$ For $k>1$ You have an appropriate constant $0<1-\frac{1}{k}<1$ that gives You a contraction, what is different if it is squared? $\endgroup$ – Peter Melech Nov 10 '18 at 10:14
  • $\begingroup$ It must be strictly less than 1, cannot be equal to 1, since as k -> infinity this will tend to one and won't satisfy my constant, however if it is squared, when you expand it, then and then take k -> infinity for the $\frac{1}{k^2} - \frac{2}{k} +1$, you always get a value < 1, which satisfies the condition $\endgroup$ – lastgunslinger Nov 10 '18 at 10:28
  • $\begingroup$ For every $k$ it $\textbf{is}$ strictly less than one (besides $\lim_{k\rightarrow\infty}(1-\frac{1}{k})^2=1$ too). So every $T_k$ has a fixpoint. Now You need to argue via convergence. $\endgroup$ – Peter Melech Nov 10 '18 at 10:37
1
$\begingroup$

So the inequality condition gives the continuity of $T $. Now use Brouwer fixed-point theorem to get your result.

$\endgroup$
  • $\begingroup$ I think the OP is asked to prove this using Banach's fixpoint theorem. Brouwer's theorem works without the sequence $T_k$ though,+1 $\endgroup$ – Peter Melech Nov 10 '18 at 10:47
  • $\begingroup$ Hmm..but it's not a easy theorem to prove. As of now, I can't think of any proof using Banach's fixed point theorem. Proving it only using fixed point theorem will be a big deal. $\endgroup$ – Larsson Nov 10 '18 at 10:59
  • $\begingroup$ Why "a big deal"? Every $T_k$ has a fixpoint by Banach and $T_k\rightarrow T$ in the strong operator topology yields the desird result $\endgroup$ – Peter Melech Nov 10 '18 at 11:03
  • $\begingroup$ I see... you have to use Strong Operator Topology. $\endgroup$ – Larsson Nov 10 '18 at 11:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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