where $u(x):=U(x,0)$ and $v(x)=V(x,0)$.

Identity follows by testing the equation for $U$ with $U$ and the equation for $V$ with $V$ and by integrating by parts over $B_r^+$.

To prove we first observe that $U,V\in
H^2(B_r^+)$ for all $r\in(0,R)$. Indeed, since $\frac{\partial U}{\partial \nu}= 0$ on $B_R'$, the function $$\widetilde U(x,t)=
\begin{cases}
  U(x,t),&\text{if }t>0,\\
  U(x,-t),&\text{if }t<0,
\end{cases}$$ satisfies the equation $\Delta \widetilde U=\widetilde V$, where $\widetilde V(x,t)=
  V(x,t)$ if $t>0$ and $\widetilde V(x,t)=V(x,-t)$ if $t<0$. Since $\widetilde V\in L^2(B_R)$, by classical elliptic regularity we have that $\widetilde U\in H^2(B_r)$ and hence $U\in
H^2(B_r^+)$ for all $r\in(0,R)$. By the Gagliardo Trace Theorem we have that $u=\mathop{\rm Tr}U\in H^{1/2}(B_r')$ for all $r\in(0,R)$ . Since $h\in
C^1(B_R')$ we have that $hu \in  H^{1/2}(B_r')$ for all $r\in(0,R)$. Therefore, for all $r\in(0,R)$, $V$ satisfies $$\begin{cases}
  \Delta V=0,&\text{in }B_r^+,\\
\frac{\partial V}{\partial \nu}\in H^{1/2}(B_r').&
\end{cases}$$ From elliptic regularity under Neumann boundary conditions (see in particular [@Salsa Theorem 8.13]) we conclude that $V\in H^2(B_r^+)$ for all $r\in(0,R)$.

Since, for every $r\in(0,R)$, $U,V\in
H^2(B_r^+)$, we can test the equation for $U$ with $\nabla U\cdot z$ (which belongs to $H^1(B_r^+)$) and the equation for $V$ with $\nabla V\cdot z$ (which belongs to $H^1(B_r^+)$), thus obtaining .

\[l:Hpos\] Let $(U,V)\in H^1(B_R^+)\times 