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I was reading the paper

Quelques remarques sur les problemes elliptiques quasilineaires du second ordre, P. L. Lions, Journal d’Analyse Mathématique volume 45, pages 234–254(1985)

and on page 251 he claims that (using it in the proof), the distance function near the boundary is concave for a general open bounded domain, which I don't think it is true. I can get this fact if the domain is convex, is there any result saying this for a non-convex domain? Say, interior ball condition is enough?


Edit: On page 251, here is the part I suspected that he used that fact: \begin{align} |u(\overline{x}-s\mathbf{n}(\overline{x}))-u(y)| &= |u(\overline{x}-s\mathbf{n}(\overline{x}))-u(\overline{y} - s\mathbf{n}(\overline{y}))|\\ &\leq \frac{C}{s^{1-\theta}}|\overline{x} - s\mathbf{n}(\overline{x}) - y|. \end{align} My reasoning is that, he has $|\nabla u(x)\leq \frac{C}{d(x)^{1-\theta}}$ earlier. Therefore to connect the line between $\overline{x}-s\mathbf{n}(\overline{x})$ and $y = \overline{y}-s\mathbf{n}(\overline{y})$ we need that every point on this line segment need to have distance bigger or equal to the distance from $\overline{x}-s\mathbf{n}(\overline{x})$ or $y$, in short, we want \begin{equation} d_{\partial \Omega}\Big(\lambda\big(\overline{x}-s\mathbf{n}(\overline{x})\big)+(1-\lambda )y\Big) \geq s \end{equation} provided that at $\lambda = 0$ or $\lambda = 1$ we have \begin{equation} d_{\partial\Omega}(y) = d_{\partial\Omega}(\overline{x}-s\mathbf{n}(\overline{x})) = s. \end{equation} We could say quasi-concave is enough, but I do believe he meant concave.

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    $\begingroup$ You're sure about the reference ? I checked page 251 and I don't see where this is stated or used. $\endgroup$ Nov 26, 2020 at 20:51
  • $\begingroup$ See edit. I just added some more details. $\endgroup$
    – Sean
    Nov 26, 2020 at 21:19

1 Answer 1

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The domain is assumed regular therefore at least Lipschitz in my opinion, and this is sufficient to ensure quasi-convexity of the domain. This means that for $z_1,z_2\in \Omega$ you have $\mathrm{d}_\Omega(z_1,z_2) \lesssim_\Omega |z_1-z_2|$ where $\mathrm{d}_\Omega$ is the geodesic distance (minimal length of a continuous $\Omega$-valued path between the two points). Here you have the assumption $s\geq |x-y|$ so WLOG $s>0$. In particular $\overline{x}-s \mathbf{\overline{x}}$ and $y$ are two points of $\Omega$ and you can therefore use your estimate on the gradient on the given path.

For a reference about quasi-convexity you can check Sections 2.5.1 and 2.5.2 of

Brudnyi, A.; Brudnyi, Y., Methods of geometric analysis in extension and trace problems. Volume 1. Monographs in Mathematics, 102. Birkh¨auser/Springer, 2012.

EDIT : sorry, I got confused in the previous answer. You invoke the quasi-convexity of $\partial\Omega$ to get the existence of a map $\gamma:[0,1]\rightarrow\partial\Omega$ linking $\overline{x}$ to $\overline{y}$. Then, $\gamma_s(\sigma):=\gamma(\sigma)-s\mathbf{n}(\gamma(\sigma))$ maps $\overline{x}-s\mathbf{n}(\overline{x})$ to $y$ and satisfies $\mathrm{d}_{\partial_\Omega}(\gamma_s)\geq s$ all the way. The length of $\gamma_s$ should be (I confess, I did not check it) comparable to the one of $\gamma$, which is, by quasi-convexity, comparable to $|\overline{x}-\overline{y}|$, itself comparable to $|x-y|$ (beginning of page 251).

That's just the idea that in $\Omega= \mathbf{R}^2\setminus \text{B}(0,1)$ to connect two points $x$ and $y$ of norm $1+\varepsilon$, you do not follow the straightline but a portion of arc (of radius $1+\varepsilon$).

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  • $\begingroup$ But I need a lower bound on the distance, i.e., quasi-convexity is not enough here, I want quasi-concavity of the distance, right? $\endgroup$
    – Sean
    Nov 29, 2020 at 19:12
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    $\begingroup$ You're right, I got confused, I corrected the answer, hope it works out for you ! $\endgroup$ Nov 29, 2020 at 19:58
  • $\begingroup$ Thank you! I am checking it out now. $\endgroup$
    – Sean
    Dec 1, 2020 at 2:22
  • $\begingroup$ I got it, thank you very much! $\endgroup$
    – Sean
    Dec 2, 2020 at 18:13

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