Existence of convex defining functions for convex domains 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 $h$? 
 A: 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.
