A problem about convex domains

Question

Let $\Omega\subseteq\mathbb{R}^{n}$ be an open convex domain and let $A$ be a compact subset in $\Omega$. Prove that, there is $c>0$ such that $$ x+\left(1-c\left|x-y\right|\right)\left(z-y\right)\in\Omega,\quad\forall z\in\Omega,\forall x,y\in A, \left|x-y\right|<\dfrac{1}{c}. $$

Answer 1

I think that the statement is incorrect.

Example: $\Omega = (0,4)$, $A = [\epsilon, 2]$, with $\epsilon >0$ and $\epsilon \ll 1$.

Let $x = \epsilon$, $y = 2\epsilon$. Then the constraint implies that $\forall z \in \Omega$: $$\epsilon + (1 - c \epsilon)(z-2\epsilon) > 0$$ Taking $z \to 0$, we must have $$\epsilon + (1-c\epsilon)(-2\epsilon) \geq 0 \Leftrightarrow \epsilon(2c\epsilon -1) \geq 0 \Rightarrow c \geq \frac{1}{2\epsilon}.$$ Now take $x=1$, $y =2$, $z=3$. Then $$x + (1 - c|x-y|)(z-y) = 1 + (1 - c) = 2 - c \leq 2 - \frac{1}{2\epsilon},$$ which is negative for $\epsilon$ small enough.

Edit 1: When I wrote this, the OP had forgotten to mention that the condition must be satisfied only for $|x-y|\leq 1/c$.

Edit 2: Here is a proof of the weaker result. Rearranging, we have $$x + (1 - c|x-y|)(z-y) = (1 - c|x-y|) z + c|x-y|\,\underbrace{\left(\frac{1}{c|x-y|}\, x + \left(1 - \frac{1}{c|x-y|}\right)y\right)}_{:=w}.$$ Since $\Omega$ is convex, it suffices to show that there exists $c$ such that $w \in \Omega$ for all $x,y \in A$. For this, let us compute $$|w-x| = \left(\frac{1}{c|x-y|} - 1 \right)|x-y| = \frac{1}{c}(1 - c|x-y|) \leq \frac{1}{c}.$$ Since $A$ is compact the distance from $A$ to $\mathbb R^n \backslash \Omega$ is strictly positive; let us call it $d(A,\mathbb R^n \backslash \Omega)$. By choosing $c > 1/d(A,\mathbb R^n \backslash \Omega)$, $|w - x| \leq d(A,\mathbb R^n \backslash \Omega)$, so $w\in \Omega$ and the result is proved.