On Brouwer fixed point theorem in infinite dimension

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 ?

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).

• What is the closed unit ball of your space ? – Maman Oct 26 '16 at 23:10
• @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 – freakish Oct 26 '16 at 23:21
• Ok so the compacity of $\mathcal{B}$ does not count to apply the theorem ? – Maman Oct 26 '16 at 23:23
• @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. – freakish Oct 26 '16 at 23:27
• Ok I start to understand – Maman Oct 26 '16 at 23:30