7
$\begingroup$

I have a question regarding the construction of a barrier frequently used in PDE. The barrier used is the following:

Let $\Omega$ be a uniformly convex domain in $\mathbb{R}^n$ with $C^2$ boundary. Here uniformly convex means there exists some $r>0$ such that every point in $\partial \Omega$ satisfies an interior sphere condition for a sphere with radius $r$. Then there exists a uniformly convex defining function $h \in C^2(\Omega)$, that is a function satisfying $$ h < 0 \text{ in } \Omega, \quad h = 0 \text{ on }\partial\Omega$$ and $$ |Dh| = 1 \text{ on }\partial\Omega, \quad D^2h \geq \delta I \text{ in } \Omega,$$ for some $\delta >0$.

My question is the following: Whilst I understand how such a function can be constructed in some neighborhood of the boundary by taking for example $h(x) = -\text{dist}(x,\partial \Omega)+\text{dist}(x,\partial \Omega)^2$ (as outlined in Gilbarg and Trudinger $\S$14.6 or the footnote on page 40 of Figalli's Monge–Ampère book), how does one extend this function to the entire domain? That is how does one explicitly construct $h$?

$\endgroup$
3
  • 2
    $\begingroup$ Would it work if you let $h(x) = (m - \mathrm{dist}(x,\partial\Omega))^2 - m^2$, where $m = \inf\mathrm{dist}(x,\partial\Omega)$? The term being squared is convex and positive, so its square is too. I'm not sure how to check the uniform convexity condition though. $\endgroup$
    – user856
    May 24 '19 at 10:48
  • $\begingroup$ Rahul, thanks for the help, do you possibly mean sup instead of inf? As far as I can tell if inf is correct, and the inf is taken over $\Omega$, we will have $m=0$. Either way, since your function is $h(x) = -2m\text{dist}(x,\partial\Omega) + \text{dist}(x,\partial \Omega)^2$ I believe the same difficulties which arise for the function I mentioned in my question arise for this new $h$, which is that I only know how to control the second derivatives of $\text{dist}(x,\partial \Omega)$ in a neighbourhood of $\partial \Omega.$ $\endgroup$ May 24 '19 at 22:13
  • $\begingroup$ does such function exist when $\Omega$ is only convex ? Could you please mention a precise reference for the existence of $h$ as above (I can't find it in the mentioned references). Many thanks. $\endgroup$
    – Migalobe
    Mar 2 at 12:49
4
+500
$\begingroup$

You start with your function $h$ defined in a neighborhood $V$ of $\partial\Omega$ and you extend it to $\Omega$ by taking the convex envelope of $h$, \begin{align} h^{\ast\ast}(x)=\inf\left\{ \lambda_{1}h(x_{1})+\cdots+\lambda_{n+1}% h(x_{n+1}):\,\sum_{i=1}^{n+1}\lambda_{i}=1,\\\,\lambda_{i}\geq0,\,\sum _{i=1}^{n+1}x_{i}=x,\,x_{i}\in\Omega\right\} . \end{align} Then you consider a standard mollifier $\varphi\in C_{c}^{\infty}% (\mathbb{R}^{n})$, with $\operatorname*{supp}\varphi\subseteq B(0,1)$, $\varphi\geq0$ and $\int_{\mathbb{R}^{n}}\varphi\,dx=1$ and you define $$ h_{\varepsilon}(x)=(\varphi_{\varepsilon}\ast h^{\ast\ast})(x)=\int% _{\mathbb{R}^{n}}\varphi_{\varepsilon}(x-y)h^{\ast\ast}(y)\,dy $$ for $x$ defined in $\Omega^{\varepsilon}:=\{x\in\mathbb{R}^{n}% :\,\operatorname*{dist}(x,\Omega)<\varepsilon\}$, where $0<\varepsilon \leq\varepsilon_{0}$ and $\varepsilon_{0}>0$ is so small that $\Omega ^{\varepsilon_{0}}\subset V\cup\Omega$. The function $h_{\varepsilon}$ is still convex and of of class $C^{\infty}$ but it no longer coincide with $h$ on $V$. Then you consider a cut-off function $\phi\in C_{c}^{\infty }(\mathbb{R}^{n})$ such that $\phi=1$ in $\Omega^{\varepsilon_{0}/2}$ and $\phi=0$ outside $\Omega^{\varepsilon_{0}}$ and finally you take $$ f:=\phi h+(1-\phi)(h_{\varepsilon}+\delta|x|^{2}), $$ where $\delta>0$ is very small. With some work you can check that this does the job.

EDIT: Added more details and corrected misprints. In the region where $0<\phi<1$ write $$ f=h+(1-\phi)(h_{\varepsilon}-h+\delta|x|^{2}). $$ Then \begin{align*} \partial_{ij}f & =\partial_{ij}h+(1-\phi)(\partial_{ij}(h_{\varepsilon }-h)+2\delta\delta_{i,j})\\ & -\partial_{i}\phi(\partial_{j}(h_{\varepsilon}-h)+2\delta x_{j}% )-\partial_{j}\phi(\partial_{i}(h_{\varepsilon}-h)+2\delta x_{j})\\ & -\partial_{ij}\phi(h_{\varepsilon}-h+\delta|x|^{2}). \end{align*} By taking $\varepsilon_{0}$ sufficiently small you can assume that $h^{\ast\ast}=h$ in $\Omega^{\varepsilon_{0}}$ and that the Hessian matrix satisfies the inequality $H_{h}(x)\geq c_{0}I_{n}$ for some $c_{0}>0$. Then in $\Omega^{\varepsilon_{0}}$, \begin{align*} \partial_{ij}(h_{\varepsilon}-h)(x) & =\int_{\mathbb{R}^{n}}\varphi _{\varepsilon}(y)\partial_{ij}h(x-y)\,dy-\partial_{ij}h(x)\\ & =\int_{\mathbb{R}^{n}}\varphi_{\varepsilon}(y)(\partial_{ij}h(x-y)-\partial _{ij}h(x))\,dy\\ & =\int_{B(0,\varepsilon)}\varphi_{\varepsilon}(y)(\partial_{ij}% h(x-y)-\partial_{ij}h(x))\,dy \end{align*} and similarly $$ \partial_{i}(h_{\varepsilon}-h)(x)=\int_{B(0,\varepsilon)}\varphi _{\varepsilon}(y)(\partial_{i}h(x-y)-\partial_{i}h(x))\,dy $$ and the same for $h_{\varepsilon}-h$. Now you have that $\Vert\partial_{i}% \phi\Vert_{\infty}\leq C/\varepsilon_{0}$ and $\Vert\partial_{ij}\phi \Vert_{\infty}\leq C/\varepsilon_{0}^{2}$. Using uniform continuity, given $\eta>0$ you can find $\varepsilon$ so small that $|h(x-y)-h(x)|\leq \eta\varepsilon_{0}^{2}$, $|\partial_{i}h(x-y)-\partial_{i}h(x)|\leq \eta\varepsilon_{0}$, and $|h(x-y)-h(x)|\leq\eta\varepsilon_{0}$ for all $y\in B(0,\varepsilon)$. You can also take $\delta$ smaller that $\eta \varepsilon_{0}^{2}$. Hence, you can estimate% \begin{align*} |\partial_{ij}\phi(h_{\varepsilon}-h+\delta|x|^{2})| & \leq\Vert\partial _{ij}\phi\Vert_{\infty}(|h_{\varepsilon}-h|+\delta^{2}R)\\ & \leq C\varepsilon_{0}^{-2}(\eta\varepsilon_{0}^{2}+\eta\varepsilon_{0}% ^{2}R)\leq C\eta(1+R). \end{align*} You will have similar estimates for the other terms. Using $H_{h}(x)\geq c_{0}I_{n}$ you should get that $$ H_{f}(x)\geq\frac{1}{2}c_{0}I_{n}% $$ provided $\eta$ is small enough.

$\endgroup$
3
  • $\begingroup$ Gio67, do you have any advice for checking the lower bound on the Hessian. Obtaining the lower bound is clear on $\{x; \text{dist}(x,\partial \Omega) < \frac{\varepsilon}{2}\}$ and $\{x; \text{dist}(x,\partial \Omega) > \varepsilon\}$ where $\phi$ is $1$ and $0$ respectively. However on $\{x; \frac{\varepsilon}{2} < \text{dist}(x,\partial \Omega) < \varepsilon \}$ the only expression I have been able to work out for the Hessian is $$D^2f = \phi D^2h + (1-\phi)D^2v+[Dh \otimes D\phi+D\phi \otimes Dh-D\phi \otimes Dv - Dv \otimes D\phi]+(h-v)D^2\phi,$$ where $v = h_{\varepsilon}+\delta|x|^2$. $\endgroup$ May 28 '19 at 1:40
  • $\begingroup$ Provided $\phi$ is $C^2$ I believe it must have points of concavity and convexity where $D^2\phi$ is comparable in size to $\varepsilon^{-2}$ and I am not sure how to obtain a lower bound for the Hessian at these points. Thanks for your help so far. $\endgroup$ May 28 '19 at 1:58
  • 1
    $\begingroup$ I added more details. Hope it answers your questions. $\endgroup$
    – Gio67
    May 28 '19 at 5:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.