$r\in(0,1)$}.$$ For any $r\in (0,1)$ and $k\in {{\mathbb N}}$, let us define the functions $$\begin{aligned}
&D_k(r)=r^{-N+1} \bigg[\int_{B_r^+} \left(|\nabla
  U_{\lambda_{n_k}}|^2+|\nabla V_{\lambda_{n_k}}|^2+\lambda_{n_k}^2 U_{\lambda_{n_k}}\!
  V_{\lambda_{n_k}}
\right)\,dz\\
&\hskip6cm-\lambda_{n_k}\int_{B_r'} h(\lambda_{n_k} x)u_{\lambda_{n_k}}\!(x)v_{\lambda_{n_k}}\!(x)\, dx  \bigg],\\
&H_k(r)=r^{-N}\int_{S_r^+} (U_{\lambda_{n_k}}^2+V_{\lambda_{n_k}}^2)\, dS,\end{aligned}$$ where we have set $v_\lambda=\mathop{\rm Tr}V_\lambda$. By direct calculations we have $$\label{NkNw}
{\mathcal
    N}_k(r):=\frac{D_k(r)}{H_k(r)}=\frac{D(\lambda_{n_k}r)}{H(\lambda_{n_k}r)}
  ={\mathcal N}(\lambda_{n_k}r) \quad \text{for all } r\in (0,1).$$ From it follows that, for any fixed $r\in (0,1)$, $$\label{convDk}
 D_k(r)\to \widetilde D(r)\quad\text{and}\quad H_k(r)\to \widetilde
 H(r)\quad\text{as }k\to+\infty$$ where $$\label{Dw(r)}
  \widetilde D (r)=
r^{-N+1} \int_{B_r^+} \left(|\nabla
  \widetilde U|^2+|\nabla \widetilde V|^2
\right)dz\quad\text{and}\quad
\widetilde H (r)=
r^{-N}\int_{S_r^+} (\widetilde U^2+\widetilde V^2)\, dS$$ for all $r\in(0,1)$. We observe that $\widetilde H (r)>0$ for all $r\in(0,1)$; indeed, if $\widetilde H (\bar r)=0$ for some $\bar r\in(0,1)$, the fact that $\widetilde
U,\widetilde V$ (and their even extension for $t<0$) are harmonic would imply that $\widetilde
U\equiv\widetilde V\equiv 0$ in $B_{\bar r}^+$, thus contradicting the classical unique continuation principle. Therefore the function $$\widetilde {\mathcal N}(r):=\frac{\widetilde  D(r)}{\widetilde  H(r)}$$ is well defined for $r\in (0,1)$. From (\[NkNw\]), (\[convDk\]), and Lemma \[l:limitN\], we deduce that $$\label{Nw(r)}
\widetilde {\mathcal    N}(r)=\lim_{k\to \infty} {\mathcal N}(\lambda_{n_k}r)=\gamma$$ for all $r\in (0,1)$. Therefore $\widetilde {\mathcal N}$ is constant in $(0,1)$ and hence $\widetilde {\mathcal
  N}'(r)=0$ for any $r\in (0,1)$. Arguing as in the proof 