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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.

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

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