$u$ be an $s$-minimal graph in ${\ensuremath{\mathbb{R}}}^{n-1}\times(0,+\infty)$. Assume that $$\label{pppADX}
{\mbox{$u(x_1,\dots,x_n)=0$ if~$x_n<0$.}}$$ Assume also that $u$ is positively homogeneous of degree $1$, i.e. $$\label{pppADX6}
u(tx)=tu(x)\qquad{\mbox{ for all $x\in{\ensuremath{\mathbb{R}}}^n$ and~$t>0$.}}$$ Then, we have that $$\label{ADX0}
u\in 
L^\infty_{\rm loc}({\ensuremath{\mathbb{R}}}^n)\cap C([-1,1]^{n-1}\times[0,1])\cap
C^\infty( {\ensuremath{\mathbb{R}}}^{n-1}\times(0,+\infty)).$$

We claim that $$\label{6BOu} u\in
L^\infty_{\rm loc}({\ensuremath{\mathbb{R}}}^n).$$ Suppose not. Then there exist $R>0$ and $x^{(k)}=(x^{(k)}_1,\dots,x^{(k)}_n)\in B_R$ such that $|u(x^{(k)})|{\geqslant}k$. Without loss of generality, we can suppose that $u(x^{(k)}){\geqslant}k$. Then, by , we see that $x^{(k)}_n>0$, and, recalling , we have that $\big( t x^{(k)},t u(x^{(k)})\big)\in
\partial E_u$ for every $t>0$ (being $E_u$ defined in ), and then, in particular, $$\left( \frac{x^{(k)}}{u(x^{(k)})},1\right)\in\partial E_u.$$ Accordingly, taking the limit as $k\to+\infty$, we find that $e_{n+1}\in\overline{\partial E_u\cap\{x_n>0\}}$. This is in contradiction with , whence the proof of  is complete.

As a result, by Theorem 1.1 in [@MR3934589] we obtain that $u$ is smooth in $\{x_n>0\}$, and the continuity up to $\{x_n=0\}$ follows from Theorem 1.1 of [@MR3516886]. This proves the claim in .

Now, we show that if the second blow-up is either empty or full in a halfspace, then the original $s$-minimal graph is necessarily boundary discontinuous:

\[554327h:AJSJD2374859ty\] Let $u$ be an $s$-minimal graph in $(-2,2)\times(0,4)$. Assume that there exists $h>0$ such that 