I'm having some trouble wrapping my head around the Browder fixed-point theorem. The statement of the theorem is:
- If $X$ is a uniformly convex Banach space, and
- If $K \subset X$ is nonempty, convex, closed, and bounded, and
- If $f: K \to K$ is a function which does not increase distances (i.e. $|f(x) - f(y)| \leq |x - y|$, for all $x, y \in K$), then
- $f$ has a fixed point $x \in K$.
But it seems easy to produce counterexamples. For instance, $\ell^2(\Bbb{N})$, the vector space of square-summable sequences, is uniformly convex. Let $K$ be the convex hull of the standard basis vectors $\{ \vec{e}_i \}_{i \geq 1}$, then $K$ is also nonempty, closed, and bounded. If we take the unilateral shift operator $R: \ell^2(\Bbb{N})\to \ell^2(\Bbb{N})$ given by $$R(a_1, a_2, a_3, ...) = R(0, a_1, a_2, a_3, ...)$$ then $R(K) \subset K$ and $R$ does not increase distances (since it is, in fact, an isometry onto its image). But $R$ clearly can't have a fixed point in $K$: the only fixed point of $R$ is the zero sequence, which is not in the convex hull of the standard basis vectors.
Am I misunderstanding the conditions to apply this fixed-point theorem? Or, does my example not satisfy those conditions in a way that I'm not seeing? What am I missing here?