Hi everyone!
Let $\Omega$ a bounded open set in $ \mathbf{R}^n$ ($n \geq 2$) provided with a boundary $\partial\Omega\in\mathcal{C}^1$.
$$\exists \delta>0,\, \forall x\in \partial\Omega : x-\varepsilon n(x) \in\Omega,\quad \varepsilon\in ]0,\delta],$$ where $n(x)$ is the outer normal to $\partial\Omega$ at point $x$
My question: what is the geometric interpretation of the condition above?
Thank you for your answers :)
For the point $\;x\;$ in the boundary $\;\partial\Omega\;$ consider $\;n(x),\;$ the outer normal to $\;\partial\Omega\;$ at the point $\;x$. Geometrically, this means that you are considering a normal vector which initial point is $\;x\;$ and which direction goes "outside" $\;\Omega$.
Now, if you take $\;\varepsilon\in ]0,\delta]\;$ for the given $\;\delta\;$ (which is claimed that always exists), the vector $\;-\varepsilon n(x)\;$ is an scaling of $\;n(x)\;$ which is point now "inside" $\;\Omega,\;$ because of the minus sign. Then, $\;x-\varepsilon n(x)\;$ means "moving" the boundary point $\;x\;$ in the direction given by $\;-\varepsilon n(x),\;$ that is, inside $\;\Omega$.
Since $\;x\;$ is a point in the boundary, the transformation $\;x-\varepsilon n(x)\;$ necessarily needs to be in $\;\Omega$.
Thus, the claim is stating that is always possible to find a positive $\delta$ such that if we take values in $\;]0,\delta]\;$ and move from the points of the boundary inwards, we will get inside $\;\Omega$.