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 $h$?

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    $\begingroup$ 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

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.

  • $\begingroup$ 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
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    $\begingroup$ I added more details. Hope it answers your questions. $\endgroup$
    – Gio67
    May 28 '19 at 5:12

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