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

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


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

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

  • $\begingroup$ 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

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

  • $\begingroup$ 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

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