since these rescalings approach $E_{00}$ in the Hausdorff distance (see [@2019arXiv190405393D]), the claim in  follows.

Moreover, by , $$\label{ADX-EE-568-029-3984112}
E_{00}\cap\{x_n<0\}=\{x_{n+1}<0\}\cap\{x_n<0\},$$ and, by Lemma 2.2 in [@2019arXiv190405393D], we have that $$\label{ADX-EE-568-029-3984112-NIA}
{\mbox{$tE_{00}=E_{00}$ for every~$t>0$.
}}$$ [F]{}rom this and , we are in the position of using Lemma \[3243-237gndlsm\], and thus deduce from  that $$\label{6BOu-1BIS-TRIS}
\min\Big\{ {\rm dist}\big(e_{n+1},(\partial E_{00})\cap\{x_n>0\}\big),\;
{\rm dist}\big(-e_{n+1},(\partial E_{00})\cap\{x_n>0\}\big)\Big\}>0.$$ Then, for all $x\in{\ensuremath{\mathbb{R}}}^{n-1}\times(0,+\infty)$, we set $$u_{00}(x):=\sup\big\{
y {\mbox{ s.t. }}(x,y)\in E_{00}
\big\}.$$ By extending $u_{00}$ to vanish in ${\ensuremath{\mathbb{R}}}^{n-1}\times(-\infty,0)$, we find that $E_{00}$ is the subgraph of $u_{00}$, as desired. To this aim, it remains to prove that the image of $u_{00}$ is ${\ensuremath{\mathbb{R}}}$, namely that for every $x\in{\ensuremath{\mathbb{R}}}^{n-1}\times(0,+\infty)$, $$\label{71-02-23884oe}
u_{00}(x)\not\in\{-\infty,+\infty\}.$$ As a matter of fact, in light of , it is sufficient to prove  for every $x\in S^{n-1}_+:=\{x=(x_1,\dots,x_n)\in{\ensuremath{\mathbb{R}}}^n
{\mbox{ s.t. $|x|=1$ and~$x_n>0$}}\}$. Hence, we set $$\begin{aligned}
&&
\omega :=\big\{ x\in S^{n-1}_+{\mbox{ s.t. }}u_{00}(x)\not\in\{-\infty,+\infty\}\big\},\\
&&
\omega_+ :=\big\{ x\in S^{n-1}_+{\mbox{ s.t. }}u_{00}(x)=+\infty\big\}\\
{\mbox{and }}&&
\omega_-:=\big\{ x\in S^{n-1}_+{\mbox{ s.t. }}u_{00}(x)=-\infty\big\},\end{aligned}$$ and, to prove , we want to show that $\omega_+=\varnothing=\omega_-$.

For a contradiction, assume that $\omega_-\ne\varnothing$. By , we also know that $\omega_-\ne S^{n-1}_+$. Hence, we can take $x_-\in\omega_-$ and $x_+\in\omega\cup\omega_+$. By construction, we have that $u_{00}(x_-)=-\infty$ and $u_{00}(x_+)\in{\ensuremath{\mathbb{R}}}\cup\{+\infty\}$, and therefore 