Suppose $D$ is an open, convex and bounded set in $R^3$ with a smooth boundary. I want to shrink $D$ a bit but preserve the same 'shape'. What I'd like to do is to take inward normals at each point of the boundary, move by $\epsilon$ on each of them and then consider the region bounded by this new 'boundary'. There is a theorem (I think it's called mattress theorem) which says that if the boundary is smooth and $\epsilon$ is small enough then the little normal intervals of length $\epsilon$ are pairwise disjoint. My question is - does this new set (boundary translated by $\epsilon$) form in fact a boundary of some open and convex subset of $D$? It seems intuitive that it should, but I have no idea on how to show that - a reference or at least a confirmation would be most appreciated. Easy to see that it works for balls for example, but that's a pretty special case.
(also I wasn't sure what tags to add, I'd be grateful if someone verified that, I've put the "pde" because this question arose when considering some partial differential equation)
It is easier to define the shrinking process differently, and then show it's equivalent to yours. Namely, let $$\begin{split}D_\epsilon&=\{x\in D:\operatorname{dist}(x,\partial D)>\epsilon\} \\ &= \{x\in D: \exists r>\epsilon \text{ such that } B_r(x)\subset D \} \end{split} \tag{1}$$ The second form shows that $D$ is convex: given $x_1,x_2\in D_\epsilon$, let $r$ be the smaller of $r_1,r_2$ and observe that for any $\lambda\in (0,1)$ the ball $B_r(\lambda x_1+(1-\lambda)x_2)$ is contained in $D$. Openness is clear from either form.
If $\partial D$ is $C^2$ smooth and $\epsilon$ is small, then the translation along normal indeed produces $\partial D_\epsilon$ (this fact does not rely on convexity). Indeed, consider the line segment from $x$ to the nearest boundary point: this segment is normal to the boundary. Pushing the boundary by $\epsilon$ along the normal is the same as taking the points at distance $\epsilon $ from the boundary.