$(x_-,y)\in
{\ensuremath{\mathbb{R}}}^{n+1}\setminus E_{00}$ for all $y\in{\ensuremath{\mathbb{R}}}$, while $(x_+,y_+)\in E_{00}$ for some $y_+\in{\ensuremath{\mathbb{R}}}$.

This and  give that $(x_+,y)\in E_{00}$ for all $y{\leqslant}y_+$. In particular, for all $k\in{\ensuremath{\mathbb{N}}}$ sufficiently large, we have that $(x_+,-k)\in E_{00}$, with $(x_-,-k)\in {\ensuremath{\mathbb{R}}}^{n+1}\setminus E_{00}$. Consequently, by , we see that $
\left(\frac{x_+}k,-1\right)\in E_{00}$ and $\left(\frac{x_-}k,-1\right)\in {\ensuremath{\mathbb{R}}}^{n+1}\setminus E_{00}$. As a result, by taking the limit as $k\to+\infty$, we conclude that $-e_{n+1}\in
\overline{(\partial E_{00})\cap\{x_n>0\}}$. But this is in contradiction with , and therefore necessarily $\omega_-=\varnothing$.

Similarly, one proves that $\omega_+=\varnothing$, and this completes the proof of .

The proof of  is thus completed, and we now focus on the proof of . For this, assume the converse: then, there exists a sequence $x^{(k)}=(x^{(k)}_1,\dots,x^{(k)}_n)$ such that $x^{(k)}_n>0$, $x^{(k)}\to0$ as $k\to+\infty$ and $|u(x^{(k)})|{\geqslant}a_0$, for some $a_0>0$. Up to a sign change, we can suppose that $
u(x^{(k)}){\geqslant}a_0$. Hence, we have that $( x^{(k)}, u(x^{(k)}))\in
(\partial E_{00})\cap\{x_n>0\}$ and then, recalling , we deduce that $
\left(\frac{ x^{(k)}}{ u(x^{(k)})},1\right)\in
(\partial E_{00})\cap\{x_n>0\}$.

Accordingly, taking the limit as $k\to+\infty$, we find that $e_{n+1}\in\overline{
(\partial E_{00})\cap\{x_n>0\} }$. This is in contradiction with  and, as a result, the proof of  is complete.

[F]{}rom Lemma \[3243-237gndlsm\] we also deduce a regularity result of the following type:

\[lemma33\] Let 