What I've done so far:

Let $B=\{x\in\mathbb{R}^n : \|x\| \leq 1\}$ be the closed unit ball in $\mathbb{R}^n$ equipped with the standard Euclidean metric. Let $f \colon B \to B$ be a function such that $\|f(x)-f(y)\|\leq\|x-y\|$ for all $x,y \in B$.

I have already shown $f$ has a fixed point.

Next the question asks me to:

Consider $B$ equal to the closed unit ball in $l^2$. $B = \left\{ (x_n) \in \mathbb R^\omega : \sum_{n=1}^\infty x_n^2 \leq 1 \right\}$ with the $l^2$-norm $\|x\|= \left( \sum_{n=1}^\infty x_n^2 \right)^{1/2}$, where $x=(x_n)$.

How do I prove this case does not have a fixed point?

  • 1
    $\begingroup$ What operators on $\ell^2$? do you know? There's a really standard example that is Lipschitz with norm $1$ and no fixed point. $\endgroup$ – T. Bongers Apr 14 '16 at 2:29
  • $\begingroup$ Let the operator be some function that satisfies the contraction mapping theorem? I'm not 100% sure what you're asking. Also what is the example? Maybe I can work with that? $\endgroup$ – LB2015 Apr 14 '16 at 2:50
  • $\begingroup$ What operators do you know on $\ell^2$? As in, specific operators you've worked with. $\endgroup$ – T. Bongers Apr 14 '16 at 2:51
  • $\begingroup$ This is the first time I have seen $l^2$ $\endgroup$ – LB2015 Apr 14 '16 at 3:04
  • $\begingroup$ What if $f$ is the zero map? Then $f$ is 1-Lipschitz and has $0$ as a fixed point. Do you require something more, like surjectivity of $f$? $\endgroup$ – Jon Warneke Apr 14 '16 at 4:53

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