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Can the Kakutani's fixed point theorem's be extended to say that there exists a fixed point inside the set (not on boundary)(I am not sure how to formally state this). For a $n$-dimensional compact, convex subset $X$ of $\mathbb{R}^n$, that would mean that there exists a fixed point $x \in X$ such that $B_{\epsilon}(x) \in X$. I think that the usual proofs could be extended to show this. Is that so?

Edit : It seems the assumption that $f(x)$ is not a point may be required.

Edit 2 : As Mike pointed out the above assumption is not good enough.

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  • $\begingroup$ One obvious condition I can think of is: any point from the boundary is mapped to the inside. $\endgroup$ – Ilya Apr 8 '16 at 12:58
  • $\begingroup$ Yes that's true, but I believe it might be true in general as I think the proofs by themselves produce an interior point. $\endgroup$ – nikhil_vyas Apr 8 '16 at 13:17
  • $\begingroup$ It is definitely not true that "Any continuous map $B^n \to B^n$ has a fixed point in the interior". Take a constant map $f(x)=c$, where $c$ is in the boundary. You need some (probably mild) assumption. $\endgroup$ – user98602 Apr 8 '16 at 14:12
  • $\begingroup$ @MikeMiller See the edit. $\endgroup$ – nikhil_vyas Apr 8 '16 at 14:23
  • $\begingroup$ No, that's not good enough. There are homeomorphisms of the ball that have only one fixed point (on the boundary). These homeomorphisms necessarily send the boundary to itself; I'm thinking of a condition along the lines of "Some boundary point is mapped to the interior". (Of course, if every boundary point is mapped to the interior, the result you want follows trivially from Brouwer.) $\endgroup$ – user98602 Apr 8 '16 at 14:26

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