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I am studying for an exam of convex analysis. One of the exercises that I am doing has the following request:

Let $\Omega$ be a nonempty, open and convex subset of $R^n$, and denote by $\Omega_\epsilon$ the set $$\Omega_\epsilon=\{ x \in \Omega: dist(x,∂\Omega)≥\epsilon\}$$ Show that $\Omega_\epsilon$ is convex.

I redefine the distance with $\inf_{y \in ∂\Omega} ||y-x|| $, writing that $$\Omega_\epsilon=\{ x \in \Omega: \inf_{y \in ∂\Omega} ||y-x|| ≥\epsilon\}$$.

I would like to use the definition: a set is convex if $$\forall x,z \in \Omega_\epsilon, \text{then } (1-t)x+tz \in \Omega_\epsilon \forall t \in [0,1].$$ I always get stuck and I do not know what to do (in particular due to the presence of the Infimum of the set).

Thanks in advance.

NotNow11
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    Possible strategy I haven't tried. If $P$ and $Q$ are in the enlarged set each is near a $P'$ and $Q'$ in the original set. Perhaps the points on the paramterized segment $PQ$ are near the corresponding points on $P'Q'$. – Ethan Bolker Jan 08 '23 at 00:26
  • Oh wait, I thought the "strategy I haven't tried yet" was from the OP. Do what Ethan suggests; it will work. – user7530 Jan 08 '23 at 00:54
  • The supporting hyperplane theorem gives you a painless argument. Take each supporting hyperplane and translate it parallel to itself, in the direction of the convex set a distance $\epsilon$. The points of $\Omega$ left out are points of $\Omega\setminus\Omega_\epsilon$, leaving $\Omega_\epsilon$ inside the half-space of the translated hyperplane. Hence, $\Omega_\epsilon$ is an intersection of half-spaces and therefore convex. – plop Jan 08 '23 at 02:38
  • You can get proper norm bars by using \| instead of ||. – joriki Jan 08 '23 at 12:15

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