# Generalization of Brouwer Fixed Point Theorem for infinite disk [duplicate]

I was thinking about Brower's fixed point theorem. Every continuous function $$f:D^n\rightarrow D^n$$ has a fixed point where $$D^n$$ is the closed unit ball in $$\mathbb{R}^n$$. I started thinking of a generalization. If we consider $$D^\infty\subseteq\ell^2$$ where $$D^\infty=\{(x_0,x_1,...)\vert\sum_{i=0}^\infty x_i^2\leq 1\}$$. Must it be true that every continuous function $$f:D^\infty\rightarrow D^\infty$$ contains a fixed point. When looking for an answer, I discovered the Schauder fixed point theorem, but I am not sure if it is applicable here (I am primarily worried about the condition that requires $$f(D^\infty)$$ to be contained in a compact subset of $$D^\infty$$).

My intuition tells me that such a function with no fixed points exists; however, it isn't clear why. If someone knows of such a function with no fixed points, or of a proof of why no such function exists, I would be interested to learn.

• @GeorgeDewhirst $D^\infty$ is not compact. A closed ball is only compact in a finite dimensional TVS. – Henno Brandsma Dec 19 '19 at 6:48
• @GeorgeDewhirst $D^{\infty}$ is only weakly compact, and not every continuous map is weakly continuous. – Conifold Dec 19 '19 at 6:51
• Wikipedia gives an explicit counterexample. – Conifold Dec 19 '19 at 6:59

Schauder does indeed not apply, as $$D^\infty$$ is not compact. The subspace $$H=\{(x_n) \in \ell^2\mid |x_n| \le \frac{1}{n}\}$$
known as the Hilbert cube (homeomorphic to $$[0,1]^{\Bbb N}$$ in the product topology) does have the fixed point property (FPP) in the sense that every continuous $$f: H \to H$$ has a fixed point. Not due to Schauder's theorem, but to the standard Brouwer fixed point theorem, in essence.
Wikipedia has a concrete counterexample to $$D^\infty$$ not having the FPP.