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

Question

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.

Answer 1

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 $K$.

Answer 2

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