line segment in $\mathcal{G}_0$ with both end points on $\partial\mathcal{G}_0$. By substracting a linear function, we assume that $u_0\geq0$ and $\ell$ lies in $\{x_3=0\}$. From the definition of $\Omega'$, we also have $\ell\subset\Omega'$ with both end points on $\partial\Omega'\cap\partial\Omega$. By a traslation and a dilation of the coordiantes, we may assume furthermore that $$\label{e418}
   \ell=\{(0,x_2,0)\,:\,-1\leq x_2\leq 1\}$$ with the endpoints $(0,\pm1)\in\partial\Omega'\cap\partial\Omega$.

Since $\varphi$ is smooth, uniformly convex in a neighborhood of $\Omega$ and $u_0=\varphi$ on $\partial\Omega$, it follows $$\label{e419}
   u_0(x)=\varphi(x)\leq\frac{C}{2}|x_1|^2,\quad x\in\partial\Omega'\cap\partial\Omega.$$ By the convexity of $u_0$, $$\label{e420}
   u_0(x)\leq\frac{C}{2}|x_1|^2,\quad x\in\Omega'.$$

Consider the Legendre transform $u_0^*$ of $u_0$ in $\Omega^*=D\varphi(\Omega)$, given by $$\label{e421}
   u_0^*(y)=\sup\{x\cdot y-u_0(x),\ \ x\in\Omega\},\quad y\in\Omega^*.$$ Since both endpoints $(0,\pm1)\in\partial\Omega'\cap\partial\Omega$, by the uniform convexity of $\varphi$, $0\notin D\varphi(\partial\Omega)$. Hence $0\in\Omega^*$ is an interior point. By , we have $$\begin{aligned}
   u_0^*(0,y_2)\!\!&\geq&\!\!|y_2|, \label{e422} \\
   u_0^*(y)\!\!&\geq&\!\!\frac{1}{2C}y_1^2. \label{e423}
  \end{aligned}$$ Therefore, ${\mbox{det}}\,D^2u_0^*$ is not bounded from above near the origin by Lemma \[l401\].

But on the other hand, by the a priori estmiate in Lemma \[l203\], ${\mbox{det}}\,D^2u_0^*$ must be bounded. Indeed, consider the Legendre transform $u_m^*$ of $u_m$. By the approximations , , and , $u_m^*$ satisfies the equation $$\label{e424}
   U^{*ij}w_{ij}^*=-f_m(Du^*){\mbox{det}}\,D^2u^*\quad\mbox{in }\Omega^*_{\varepsilon_m},$$ where $f_m=f+\beta_m\chi_{D_m}$ and $$\Omega^*_{\varepsilon_m}=\{y\in\Omega^*\,:\,{\mbox{dist}}(y,\partial\Omega^*)>\varepsilon_m\}$$ with $\varepsilon_m\to 0$ as $m\to\infty$. By the growth estimates 