v_\xi} dx_0 \\
& = -\sum_{\ell=0}^{s-1} \po^\ell f (0,\xi) \overline{B_{s,\ell}
v_\xi (0)} +\sum_{j=0}^{s} (-1)^j \bn{s}{j} |\xi|^{2(s-j)}
\int_0^\infty f(x_0,\xi) \overline{\po^{2j} v_\xi } dx_0 \\
& = -
\sum_{\ell=0}^{s-1} \po^\ell f (0,\xi) \overline{B_{s,\ell}
v_\xi (0)} + |\xi|^{2s} \int_0^\infty f(x_0,\xi) 
    \overline{v_\xi} dx_0 \\ 
& \qquad -\sum_{j=1}^{s} (-1)^j \bn{s}{j} |\xi|^{2(s-j)} \bigl(
|\xi|^{2(s-j)} f(0,\xi)\po^{2j-1} v_x (0) 
+ \int_0^\infty \po f(x_0,\xi) \overline{\po^{2j-1} v_\xi } dx_0 
\bigr) \\
& =  -\sum_{\ell=1}^{s-1} \po^\ell f (0,\xi) \overline{B_{s,\ell} 
v_\xi (0)} 
+ |\xi|^{2s} \int_0^\infty f(x_0,\xi) \overline{v_\xi} dx_0 \\
& \qquad +\sum_{k=0}^{s-1} (-1)^k \bn{s}{k+
1} |\xi|^{2(s-k-1)}
\int_0^\infty \po f(x_0,\xi) \overline{\po^{2k+1} v_\xi } dx_0 .\end{aligned}$$ By applying integration by parts $(s-1)$ more times to the last term in the right hand side above, we obtain that $$\label{v-xi}
0= \sum_{j=0}^{s} \bn{s}{j} |\xi|^{2(s-j)}\int_0^\infty \po^j
f(x_0,\xi) \overline{\po^j v_\xi} dx_0$$ for all $\xi\neq0$.

Now, for each $\xi\neq0$ we can pick $f$ so that $f(\cdot,\xi) =v_\xi$. Substituting in (\[v-xi\]) we obtain that $$\sum_{j
=0}^{s} \bn{s}{j} |\xi|^{2(s-j)} \int_0^\infty |\po^j v_\xi
(x_0)|^2 dx_0 =0,$$ that is, $v_\xi=0$.

Thus, we only need to prove the claim. If, for $p\ge m$ we set $F_k (p,m):=\sum_{j=0}^{m} \binom{k+j}{j}\binom{p-j}{m-j}$, we wish to show that $$\label{claim}
F_k (p,m) =\bn{p+k+1}{m} .$$ Observe that (\[claim\]) holds true for $m=0,1$ and $p\ge1$, and for $p=m$, by direct computation and 