# Is a continuous function on compact convex set where the boundary is mapped to the set a self mapping?

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.

• As you can see, there are counterexamples. I do wonder, however, if there are injective counterexamples? May 22 at 9:45
• @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.
– Bill
May 22 at 13:01
• @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 $g\circ f$. May 22 at 13:46
• @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? May 22 at 14:46

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