‘Shrunken Version’ of a convex set is also convex

Question

I'm trying to show that for a convex set $K$ in $\mathbb{R}^n $ (possibly bounded, if that makes things easier), the set $K_{\epsilon}:= \{x\in K: \text{dist}(x,\partial K)>\epsilon\}$ is also convex (I don't really care whether we consider open or closed sets since I only have to integrate over the set). How could I prove that?
I've tried the following: For any boundary point $p$, we can find a hyperplane $H_p$ s.t. $p\in H_p$ and $H_p$ separates $K$. Now my idea was to shift all hyperplanes by $\epsilon$, then we can write $K_{\epsilon}$ as the intersection over all these shifted hyperplanes and hence it would be convex as an intersection of convex sets. But I don't see why exactly we can actually write $K_{\epsilon}$ as this intersection, it's just intuitively clear to me. Is this a good approach? Is it even correct? How can I go about proving it? (Is the statement even true? If it helps, I might also assume we are in $\mathbb{R}^2$ and that we have a convex bounded lipschitz domain)

Answer 1

If $p$ and $q$ are in $K_\epsilon$, the closed balls $B_\epsilon(p)$ and $B_\epsilon(q)$ of radius $\epsilon$ centred at $p$ and $q$ are contained in $K$. If $0 \le \lambda \le 1$, we need to show that the ball of radius $\epsilon$ centred at $\lambda p + (1-\lambda) q$ is contained in $K$. But any member of this ball is $\lambda p + (1-\lambda) q + v$ where $\|v\| \le \epsilon$, and it is thus $\lambda (p+v) + (1-\lambda) (q+v)$ where $p+v \in B_\epsilon(p)$ and $q+v \in B_\epsilon(q)$ are both in $K$.