# 'Shrunken Version' of a convex set is also convex

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)

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 $$K$$.