I am reading Galdi's Introduction to the mathematical theory of Navier Stokes equations and there is an argument which comes up quite often that I really don't understand.
In many theorems of Chapter $3$, we prove the existence of solutions of a problem defined on an open set $\Omega \subset \mathbb R^n$ that statisfies the following condition:
$$\Omega = \bigcup_{k = 1}^N \Omega_k$$
where each $\Omega_k$ is star shaped with respect to some open ball $B_k$ with $\overline{B}_k \subset \Omega_k$.
Then, in the next chapters, he uses these results but for $\Omega$ bounded and locally Lipschitz. Therefore, it seems to me that a bounded locally Lipschitz open set should satisfy the above condition, but I really have no idea how to show that.. Clearly $\overline{\Omega}$ is compact, so we can cover it by a finite number of balls, but then how do we prove that they are star-shaped with respect to some other balls $B_k$ ? Any idea ?
Best Answer
The full, rigorous argument for this is tedious and not something one should do in public :) Instead, let me give you an example that hopefully gives the right idea.
Basically it all boils down to choosing your cover in a smart way, ensuring that you can fit a sufficiently tall cylinder centered at the boundary at every point. Why does this help? Well, imagine the graph is given locally in the usual first $n-1$ coordinates, i.e. $$ \partial\Omega= \{ (x',x_n)\in \mathbb{R}^n: x_n=f(x'),\, |x'|<R\}, $$ for a Lipschitz function $f$ with constant $M$ (we may assume $f(0)=0$). Now notice that if we consider the cylinder $$ \mathcal{C}:=\{ (x',x_n): |x'|<R, |x_n|<30MR\}, $$ then $\mathcal{C}\cap \Omega$ satisfies your starshaped condition. One way to look at this is that the cylinder is "tall" enough to guarantee that the top of it lies well above $f$, actually far enough so that any cone with "aperture" $\cot^{-1} M$ and vertex on $\partial\Omega\cap \mathcal{C}$, contains the whole top of $\mathcal{C}$; and recall that the Lipschitz condition means that all these (open) cones are disjoint from the graph of $f$.
Edit:
Let $X=(0, 30MR)$ be the center of the top of our cylinder $\mathcal{C}$, then we show that the line segments joining $X$ and any point on the boundary are all above the graph of $f$. From the argument you'll see that there's some wiggle room so that the same is true for points in a neighborhood of $X$.
First notice that the set $$ \Gamma(x'):= \{ (y',y_n)\in \mathbb{R}^{n-1}\times \mathbb{R}: M|y'-x'|<(y_n-f(x'))\}\subset \mathbb{R}^n, $$ is the convex right cone with aperture $\cot^{-1} M$, with vertex $(x',f(x'))$ and opening upward. I claim that for any $x'\in \mathbb{R}^{n-1}$, we have $\Gamma(x')\cap \text{Graph}(f)=\emptyset$.
The claim is actually a characterization of $f$ being a Lipschitz function with constant $M$ (which we assume defined on the whole space $\mathbb{R}^{n-1}$). Let $(y',y_n)\in \Gamma(x')\cap \text{Graph}(f)$ for some $x'\in \mathbb{R}^{n-1}$. By definition of the graph, this means $y_n=f(y')$, and by the Lipschitz condition on $f$, $$ |f(y')-f(x')|\leq M|x'-y'|, $$ however, since $(y',f(y'))\in \Gamma(x')$, $$ M|x'-y'|<(f(y')-f(x')), $$ which is clearly a contradiction, and the claim is proved.
To finish we notice that for any $|x'|<R$ we have $X\in \Gamma(x')$: Since $f(0)=0$, $|f(x')|\leq M|x'|<MR $, so that $(30MR-f(x')) \geq 20MR >M|x'|$ and this is exactly the definition of $X\in \Gamma(x')$.