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Let $K ⊂ R^n$ be convex and compact with $0$ in the interior of $K$. Let $f ∈ C(K, R^n)$ with $f(∂K) ⊂ K$.

If this is the case, do we in fact have $f(K) \subset K$. It is probably not the case that the image of a convex set is convex as those are hard to prove, but we at least know it is compact and connected (also path-connected). But at the same time, just being continuous is not a strong enough property to make further conclusions. The image of boundary is not necessarily a boundary unless $f$ is a diffeomorphism, but here we don't even have a homeomorphism, but just continuity.

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    $\begingroup$ As you can see, there are counterexamples. I do wonder, however, if there are injective counterexamples? $\endgroup$ Commented May 22, 2022 at 9:45
  • $\begingroup$ @TheoBendit thanks. I wonder though if the conditions I gave are enough for f to have a fixed point on K even though it is not a self mapping. $\endgroup$
    – Bill
    Commented May 22, 2022 at 13:01
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    $\begingroup$ @Bill I think they are sufficient. Assume for simplicity that $K=\overline{B_1(0)}$ (in the general case one could replace $|x|$ by the Minkowski functional for $K$). Let $g:K\to K$ be defined by $g(x)=x$ if $|x|<1$ and $g(x)=x/|x|$ otherwise. Then apply Brouwer to $g\circ f$. $\endgroup$
    – leoli1
    Commented May 22, 2022 at 13:46
  • $\begingroup$ @TheoBendit I don't think there are any. Maybe one can argue as follows: Since $f$ is injective, $f(K)$ has to lie entirely in one component of $\Bbb R^n\setminus f(\partial K)$. By invariance of domain only $\partial K$ could map into the boundary of $f(K)$, hence it has to be the interior component and as $f(\partial K)\subset K$ we get $f(K)\subset K$. This relies on two very non-trivial facts (generalized Jordan curve theorem and IoD), perhaps there is a simpler argument? $\endgroup$
    – leoli1
    Commented May 22, 2022 at 14:46

2 Answers 2

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A simple counter-example: $K=[-1,1], f(x)=2(x^{2}-1)$. Note that $f(0)=-2 \notin K$.

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  • $\begingroup$ Is this wrong? Why is there a downvote? $\endgroup$ Commented May 22, 2022 at 9:27
  • $\begingroup$ @Bill Do you agree with this answer? Or do you find a mistake. please let me know. $\endgroup$ Commented May 22, 2022 at 23:15
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No, not necessarily. Let's consider the closed, convex, compact unit disc $K \subseteq \Bbb{R}^2$. Define the continuous function: $$f(x,y) = (2 - 2\|(x, y)\|, 0) = \left(2 - 2\sqrt{x^2 + y^2},0\right).$$ The boundary of $K$ consists of all points such that $\|(x, y)\| = 1$. Thus, $$f(\partial K) = \{(0,0)\} \subseteq K.$$ But, the point $(0, 0) \in K$ maps to $(2, 0) \notin K$, so $f$ is not a self-map on $K$.

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