0
$\begingroup$

If we consider the closed unit ball $\overline{\mathcal{B}}$ of a normed vector space (infinite dimension) and a continuous map $\phi:\overline{\mathcal{B}}\to \overline{\mathcal{B}}$ does it have at least one fixed point ?

Notice that the closed unit ball in infinite dimension is not a compact set (according to Riesz's theorem) but could we apply the theorem even if the set is not compact ?

Is the compacity an important hypothesis ?

Thanks in advance !

$\endgroup$
2
0
$\begingroup$

Consider the following normed vector space:

$$V=\mathbb{R}\oplus \mathbb{R}\oplus\cdots$$ $$|v|=|(a_1,a_2,\ldots)|=\max(|a_1|, |a_2|, \ldots)$$

Note that $V$ is a direct sum, i.e. $V$ consists of sequences which are $0$ everywhere except for the finite number of indexes. Thus this is well defined and $V$ is a normed space.

Now let's define

$$\Phi:\mathcal{B}\to\mathcal{B}$$ $$\Phi(a_1, a_2, \ldots)=(1, a_1, a_2, \ldots)$$

The function is well defined (i.e. the image is in $\mathcal{B}$ because $|1|=1$) and continous but it does not have a fixed point. Indeed if $\Phi(v)=v$ for some $v=(a_1,a_2,\ldots)$ then by definition

$$a_1 = 1$$ $$a_2 = a_1 = 1$$ $$a_3 = a_2 = 1$$ $$\cdots$$ $$a_i = 1$$

In particular $v=(1, 1, \ldots)$ is a constant infinite sequence. It's a contradiction since sequences in $V$ have to be "finite" (in the sense having zeros almost everywhere).

$\endgroup$
5
  • $\begingroup$ What is the closed unit ball of your space ? $\endgroup$ – Maman Oct 26 '16 at 23:10
  • $\begingroup$ @Maman These are all sequences $(a_1, a_2, \ldots)$ such that $-1 \leq a_i \leq 1$ for any $i$ and with $a_j=0$ for almost all $j$. Similar argument is valid for classical norms (e.g. euclidean) as well, it's just that computations are way too long for such a lazy person as me. :P $\endgroup$ – freakish Oct 26 '16 at 23:21
  • $\begingroup$ Ok so the compacity of $\mathcal{B}$ does not count to apply the theorem ? $\endgroup$ – Maman Oct 26 '16 at 23:23
  • $\begingroup$ @Maman Well, $\mathcal{B}$ in my example is not compact. It's extremely hard to determine what exactly is needed for a space to have a fixed point property. For example Brouwer speculated that every compact and contractible space has FPP. But that was disproved, there are compact and contractible spaces without FPP. $\endgroup$ – freakish Oct 26 '16 at 23:27
  • $\begingroup$ Ok I start to understand $\endgroup$ – Maman Oct 26 '16 at 23:30

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.