which there exists $\phi\in C^k(\overline{B'_r})$ such that $${\mathcal{T}}u(x'):=
\lim_{x_n\searrow0} \frac{u(x',x_n)}{x_n^\sigma}=\phi(x')\qquad{\mbox{ for all }}
x'\in B_r'.$$ Also, given $r>0$ and $\phi\in C^k(\overline{B'_r})$, we consider the array $$\label{mul}{\mathcal{D}}^k\phi(0):=\big(
D^\gamma \phi(0)
\big)_{{\gamma\in{\ensuremath{\mathbb{N}}}^{n-1}}\atop{|\gamma|{\leqslant}k}}\;.$$ Namely, the array ${\mathcal{D}}^k\phi(0)$ contains all the derivatives of $\phi$ at the origin, up to order $k$. Fixing some order in the components of the multiindex, we can consider ${\mathcal{D}}^k\phi(0)$ as a vector in ${\ensuremath{\mathbb{R}}}^{N_k}$, with $$\label{ennekappa} N_k:=\sum_{j=0}^k (n-1)^j.$$ The following result states that the linear space produced in this way is “as large as possible”:

\[fan\] We have that $$\label{SP} \Big\{ {\mathcal{D}}^k\phi(0), {\mbox{ with }}\phi={\mathcal{T}}u
{\mbox{ and }}u\in {\mathcal{H}}_k\Big\}={\ensuremath{\mathbb{R}}}^{N_k}.$$

Suppose, by contradiction, that the linear space in the left hand side of  does not exhaust the whole of ${\ensuremath{\mathbb{R}}}^{N_k}$. Then, there would exist $$\label{0876tg456}
\omega\in {\ensuremath{\mathbb{R}}}^{N_k}\setminus\{0\}$$ such that the linear space in the left hand side of  lies in the orthogonal space of $\omega$. Namely, for every $u\in {\mathcal{H}}_k$ with $\phi={\mathcal{T}}u$, $$\label{mul2}
{\mathcal{D}}^k\phi(0)\cdot\omega=0.$$ Recalling the notation in , we can write $\omega=(\omega_\gamma)_{{\gamma\in{\ensuremath{\mathbb{N}}}^{n-1}}\atop{|\gamma|{\leqslant}k}}$, and then  takes the form $$\label{mul3}
\sum_{{\gamma\in{\ensuremath{\mathbb{N}}}^{n-1}}\atop{|\gamma|{\leqslant}k}}\omega_\gamma\,
D^\gamma\phi(0)=0,\qquad
{\mbox{ for all~$u\in {\mathcal{H}}_k$ with~$\phi={\mathcal{T}}u$. }}$$ Now, we exploit Lemma \[BAN\] with $\vartheta:=\sigma$. In the notation of Lemma \[BAN\], we take $P=(P_1,P_2)\in{\mathcal{C}}_\alpha\subset{\ensuremath{\mathbb{R}}}^2$ such that $v(P)>0$ and define $$\tilde v(x_1,x_2):=\frac{v(P)\;x_2^\sigma}{P_2^\sigma}.$$ We 