Proof. The following Hardy-type inequality due to Herbst [Reference Herbst22]:

\begin{equation*} 2^{2s}\frac{\Gamma^2\left(\frac{N+2s}{4}\right)}{\Gamma^2\left(\frac{N-2s}{4}\right)}\int_{\mathbb{R}^{N}} \frac{v^2(x)}{|x|^{2s}} \, dx \le \int_{\mathbb{R}^N} |\xi|^{2s} |\hat{u}(\xi)|^2 d \xi, \end{equation*}

where $\hat{u}$ is the Fourier transform of u, holds for any $v \in H^s(\mathbb{R}^N)$. Then (11) follows from (10).

Proof. Let $\phi \in C^\infty_c(\Omega \times [0,+\infty) )$ and $k \in \{3,\dots,N\}$. Then $\phi(\cdot,y)\in C_c^{\infty}(\Omega)$ for any $ y \in [0,\infty)$ and so multiplying by $y^{1-2s}$ and integrating over $(0,\infty)$ we deduce (14) from (4).

Proof of Theorem 2.7

We follow the proof of [Reference Capella, Dávila, Dupaigne and Sire8, Proposition 2.1]. Let $v \in \mathbb{H}^s(\Omega)$ and consider:

(27)\begin{equation} V(x,y):=\sum_{n=1}^\infty v_nY_{\alpha,k,n}(x)h_n(y), \quad \text{where } v_n=\int_{\Omega} v Y_{\alpha,k,n} \, dx, \end{equation}

and $h_n:(0,+\infty) \to \mathbb{R}$ is a solution to the problem:

(28)\begin{equation} \begin{cases} h_n''+\frac{1-2s}{y}h_n'-\mu_{\alpha,k,n}h_n=1,\quad \text{on } (0,+\infty),\\ h_n(0)=1, \\ \lim_{y \to \infty } h_n(y)=0. \end{cases} \end{equation}

From the proof of [Reference Capella, Dávila, Dupaigne and Sire8, Proposition 2.1], (28) admits a unique solution h_n for any $n \in \mathbb{N}\setminus\{0\}$ and:

(29)\begin{equation} -\lim_{y \to 0^+}y^{1-2s}h'_n(y)= c_{N,s}\mu^s_{\alpha,k,n}, \end{equation}

for some positive constant $c_{N,s} \gt 0$ depending only on N and s. Furthermore for any $y \in [0+,\infty)$ by (27) and Remark 2.1.

(30)\begin{align} &\int_{\Omega} \left|\frac{\partial V}{\partial y}(x,y)\right|^2 \, dx+ \int_{\Omega} |\nabla_x V(x,y)|^2 \, dx -\int_{\Omega} \frac{\alpha}{|x|^2_k}V^2(x,y) \, dx\nonumber\\ &\quad\quad=\sum_{n=1}^{\infty} v_n^2(h'_n(y))^2+\mu_{\alpha,k,n}v_n^2h_n(y)^2. \end{align}

Proceeding exactly as in [Reference Capella, Dávila, Dupaigne and Sire8, Proposition 2.1] we can show that (18) holds. Hence, in view of (14), $V \in H^1(C,y^{1-2s})$ and $\sum_{n=1}^j v_nY_{\alpha,k,n}(x)h_n(y) \to V$ in $H^1(C,y^{1-2s})$ as $j \to \infty$. In conclusion $V \in H^1_{0,L}(C,y^{1-2s})$ since $\sum_{n=1}^j v_nY_{\alpha,k,n}(x)h_n(y) \in H_{0,L}^1(C,y^{1-2s})$ for any $j \in \mathbb{N}, j \ge 1$.

In contrast to [Reference Capella, Dávila, Dupaigne and Sire8, Proposition 2.1], V might not be smooth for y > 0 since the functions $Y_{\alpha,k,n}$ might not be smooth on Ω. Then we prove that V satisfies (16) in the weak sense given by (17). Let $\phi \in C_c^{\infty}(\Omega \times [0,+\infty))$. Then,

\begin{equation*} \phi(x,y)=\sum_{n=1}^\infty \phi_n(y)Y_{\alpha,k,n}(x), \quad \text{where } \phi_n(y):=\int_{\Omega} \phi(x,y) Y_{\alpha,k,n}(x) \, dx, \end{equation*}

and similarly to (30):

(31)\begin{equation} \int_{\Omega} |\nabla \phi(x,y)|^2 \, dx -\int_{\Omega} \frac{\alpha}{|x|^2_k}\phi^2(x,y) \, dx=\sum_{n=1}^{\infty}(\phi'_n(y))^2+\mu_{\alpha,k,n}\phi_n(y)^2. \end{equation}

Then by (27) and Remark 2.1

(32)\begin{align} & \int_{\Omega} \nabla V(x,y) \cdot \nabla \phi(x,y)\, dx-\int_{\Omega} \frac{\alpha}{|x|^2_k}V(x,y)\phi(x,y) \, dx\nonumber\\ &=\sum_{n=1}^{\infty} v_nh'_n(y)\phi'_n(y)+\mu_{\alpha,k,n}v_nh_n(y)\phi_n(y). \end{align}

Furthermore, for any $j \in \mathbb{N}$, by Hölder’s inequality:

\begin{align*} &\left|\int_{0}^{+\infty} y^{1-2s}\left(\sum_{n=j}^{\infty} v_nh'_n(y)\phi'_n(y)+\mu_{\alpha,k,n}v_nh_n(y)\phi_n(y) \right)\, dy\right| \\ &\quad\le \frac{1}{2} \int_{0}^{+\infty} y^{1-2s}\left(\sum_{n=j}^{\infty} v_n^2(h'_n(y))^2+\mu_{\alpha,k,n}v_n^2h_n(y)^2 \right)\, dy\\ &\quad\quad+\frac{1}{2} \int_{0}^{+\infty} y^{1-2s}\left(\sum_{n=j}^{\infty} (\phi'_n(y))^2+\mu_{\alpha,k,n}\phi_n(y)^2 \right)\, dy. \end{align*}

By (30), (31) and the Monotone Convergence Theorem we conclude that:

\begin{equation*} \lim_{j \to \infty}\int_{0}^{\infty} y^{1-2s}\left(\sum_{n=j}^{\infty} v_nh'_n(y)\phi'_n(y)+\mu_{\alpha,k,n}v_nh_n(y)\phi_n(y) \right)\, dy=0. \end{equation*}

Hence we may change the order of summation and integration in (32) obtaining:

\begin{align*} & \int_{C} y^{1-2s}\left(\nabla V\cdot\nabla \phi - \frac{\alpha}{|x|^2_k}V\phi\right) \, dz\\ & =\sum_{n=1}^{\infty}v_n\int_{0}^{+\infty}y^{1-2s}(h'_n(y)\phi'_n(y)+\mu_{\alpha,k,n}h_n(y)\phi_n(y)) \, dy. \end{align*}

An integration by parts, in view of (28) and (29), yields:

\begin{equation*} \int_{0}^{+\infty}y^{1-2s}(h'_n(y)\phi'_n(y)+\mu_{\alpha,k,n}h_n(y)\phi_n(y)) \, dy=c_{N,s}\mu_{\alpha,k,n}^s\phi_n(0). \end{equation*}

It follows that:

\begin{equation*} \int_{C} y^{1-2s}\nabla V\cdot\nabla \phi \, dz -\int_{C}y^{1-2s} \frac{\alpha}{|x|^2_k}V\phi \, dz=c_{N,s}\sum_{n=1}^{\infty}\mu_{\alpha,k,n}^s v_n\phi_n(0), \end{equation*}

and so we have proved (17). If V ₁ and V ₂ solve (16) then by (1), (17) and (14) we deduce that:

\begin{equation*} \int_{C} y^{1-2s}|\nabla (V_1-V_2)|^2 \, dz=0, \quad \text{and } \quad \mathop{\rm{Tr}}(V_1-V_2)=0 \end{equation*}

thus $V_1=V_2$. Finally V solves the minimising problem (19) in view of (17) and a density argument.

Proof. Let us consider the map $\Phi:\mathbb{R}\times H^1_{0,S_r^+}(B_r^+,y^{1-2s}) \to (H^1_{0,S_r^+}(B_r^+,y^{1-2s}))^*$ defined as

\begin{align*} \Phi(\lambda,V)(W)&:=\int_{B_{r_0}^+} y^{1-2s} \nabla V\cdot\nabla W \, dz-\int_{B_{r_0}^+} y^{1-2s}\frac{\alpha}{|x|_k^2+\lambda^2}V W \, dz\\ &\quad-c_{N,s}\int_{B_{r_0}'} g \mathop{\rm{Tr}}(V)\mathop{\rm{Tr}}(W) \, dx+\int_{B_{r_0}^+} y^{1-2s}\left(\frac{\alpha}{|x|_k^2+\lambda^2}-\frac{\alpha}{|x|^2_k}\right) U W \, dz, \end{align*}

for any $W \in H^1_{0,S_{r_0}^+}(B_{r_0}^+,y^{1-2s})$. It is clear that Φ is well defined and that Φ is continuous in $(0,0)$ in view of Hölder’s inequality, Proposition 3.7, (3), and (40). Furthermore $\Phi(0,0)=0$.

Let us prove that $\Phi_V(0,0) \in \mathcal{L}(H^1_{0,S_{r_0}^+}(B_{r_0}^+,y^{1-2s}),(H^1_{0,S_{r_0}^+}(B_{r_0}^+,y^{1-2s})^*)$ is an isomorphism, where $\Phi_V$ is the partial derivative with respect to V of Φ. For any $W_1,W_2 \in H^1_{0,S_{r_0}^+}(B_{r_0}^+,y^{1-2s})$:

\begin{equation*} \sideset{_{(H^1_{0,S_{r_0}^+}(B_{r_0}^+,y^{1-2s}))^*}}{_{H^1_{0,S_{r_0}^+}(B_{r_0}^+,y^{1-2s})}}{\mathop{\left\langle \Phi_V(0,0)(W_1), W_2 \right\rangle}} = \left( W_1, W_2 \right)_{g,\alpha,k,0}. \end{equation*}

Hence, by Remark 4.1, $\Phi_V(0,0)$ is the Rietz isomorphism associated to the norm $\left\lVert\cdot\right\rVert_{g,\alpha,k,0}$.

We are now in position to apply the Implicit Function Theorem to Φ in the point $(0,0)$ and conclude that there exist $\tilde{\lambda} \gt 0$, ρ > 0 and a function:

(47)\begin{equation} f:(-\tilde\lambda,\tilde\lambda) \to B_\rho(0), \end{equation}

continuous in 0, such that $\Phi(\lambda,V)=0$ if and only if $V=f(\lambda)$ for any $\lambda \in (-\tilde\lambda,\tilde{\lambda})$ and $V \in B_\rho(0)$. The set $B_\rho(0)$ in (47) is defined as $B_\rho(0)=\{V \in H^1_{0,S_{r_0}^+}(B_{r_0}^+,y^{1-2s}):\left\lVert V\right\rVert_{H^1(B_{r_0}^+,y^{1-2s})} \lt \rho\}$.

It follows that $U_\lambda:=U-f(\lambda)$ solves (46) for any $\lambda \in (0,\tilde \lambda)$ since U is a solution of (33). Furthermore, $U_\lambda \to U$ strongly in $H^1(B_{r_0}^+,y^{1-2s})$ as $\lambda \to 0^+$ since f is continuous in 0 and $f(0)=0$.

Proof. We proceed in the spirit of [Reference Felli and Siclari19, Proposition 2.3], since $(|x|^2_k+\lambda^2)^{-1}U_\lambda \in L^2(B_r^+,y^{1-2s})$ and $g \in W^{1,\infty}_{loc}(\Omega \setminus\{0\})$. Then by [Reference Felli and Siclari19, Theorem 2.1, Proposition 3.6] and the proof of [Reference Felli and Siclari19, Proposition 2.2], for any $r\in (0,r_0)$ and $\rho \in (0,r)$,

(50)\begin{align} \nabla_x U_\lambda \in H^1(B_r^+\setminus B_{\rho}^+,y^{1-2s}), \quad \text{and } \quad y^{1-2s}\frac{\partial U_\lambda}{\partial y} \in H^1(B_r^+\setminus B_{\rho}^+,y^{2s-1}), \end{align}

(51)\begin{align} &\mathop{\rm{Tr}}(U_\lambda) \in H^{1+s}(B_r'\setminus B_{\rho}'), \quad \text{and } \quad \mathop{\rm{Tr}}(\nabla_x U_\lambda)=\nabla \mathop{\rm{Tr}}(U_\lambda),\nonumber\\ & \nabla U_\lambda \cdot z \in H^1(B_r^+\setminus B_{\rho}^+,y^{1-2s}), \quad \text{and } \quad \mathop{\rm{Tr}}(\nabla U_\lambda \cdot z)=\mathop{\rm{Tr}}(\nabla U_\lambda) \cdot x, \end{align}

where $H^{1+s}(B_r'\setminus B_{\rho}'):=\{w \in H^1(B_r'\setminus B_{\rho}'): \frac{\partial w}{\partial x_i} \in W^{s,2}(B_r'\setminus B_{\rho}') \,\text{for any } i=1,\dots, N\}$. We also have, in view of (46), the following identity:

(52)\begin{equation} \mathop{\rm{div}}(y^{1-2s}|\nabla U_\lambda|^2 z-2y^{1-2s}\nabla U_\lambda \cdot z \nabla U_\lambda)=(N-2s)|\nabla U_\lambda|^2 +2\frac{\alpha}{|x|_k^2+\lambda^2}U_\lambda \nabla U_\lambda \cdot z, \end{equation}

in a distributional sense in $B_r^+\setminus B_{\rho}^+$. Furthermore, thanks to (50),

(53)\begin{equation} \mathop{\rm{div}}(y^{1-2s}\nabla U_\lambda \cdot z \nabla U_\lambda)= -y^{1-2s}\frac{\alpha}{|x|_k^2+\lambda^2}U_\lambda \nabla U_\lambda \cdot z+ y^{1-2s}\nabla U_\lambda \cdot \nabla(\nabla U_\lambda\cdot z) \in L^1(B_r^+\setminus B_{\rho}^+), \end{equation}

and so by (52):

\begin{equation*} \mathop{\rm{div}}(y^{1-2s}|\nabla U_\lambda|^2 z) \in L^1(B_r^+\setminus B_{\rho}^+). \end{equation*}

Let, for any $\delta \in (0,r)$,

(54)\begin{equation} B^+_{r,\delta}:=\{(x,y)\ \in B_r^+:y \gt \delta\} \quad \text {and } \quad S^+_{r,\delta}:=\{(x,y)\ \in S_r^+:y \gt \delta\}. \end{equation}

Integrating by part on $B_r^+\setminus B_{\rho}^+$ we obtain, for any $\delta\in (0,\rho)$,

(55)\begin{align} \int_{B_{r,\delta}^+\setminus B_{\rho,\delta}^+}\mathop{\rm{div}}(y^{1-2s}|\nabla U_\lambda|^2 z) \, dz &= r \int_{S_{r,\delta}^+}y^{1-2s}|\nabla U_\lambda|^2 \, dS -\rho \int_{S_{\rho,\delta}^+}y^{1-2s}|\nabla U_\lambda|^2 \, dS \nonumber\\ &\quad-\delta^{2-2s}\int_{B'_{\sqrt{r^2-\delta^2}}\setminus B'_{\sqrt{\rho^2-\delta^2}}}|\nabla U_\lambda|^2(x,\delta)\, dx. \end{align}

We claim that there exists a sequence $\delta_n \to 0^+$ such that:

(56)\begin{equation} \lim_{n \to \infty} \delta^{2-2s}\int_{B'_{\sqrt{r^2-\delta_n^2}}\setminus B'_{\sqrt{\rho^2-\delta_n^2}}}|\nabla U_\lambda|^2(x,\delta)\, dx=0, \end{equation}

arguing by contradiction. If the claim does not hold than there exist a constant C > 0 and $\delta_0 \in (0,\rho)$ such that $B_r'\times(0,\delta_0)\subseteq B_{r_0}^+$ and:

(57)\begin{equation} \delta^{1-2s}\int_{B'_{\sqrt{r^2-\delta^2}}\setminus B'_{\sqrt{\rho^2-\delta^2}}}|\nabla U_\lambda|^2(x,\delta)\, dx \ge \frac{C}{\delta}, \quad \text{for any } \delta \in (0,\delta_0). \end{equation}

Then integrating (57) over $(0,\delta_0)$ we obtain:

\begin{equation*} \int_{0}^{\delta_0}\left(\delta^{1-2s}\int_{B'_r}|\nabla U_\lambda|^2(x,\delta)\, dx\right) \, d \delta \ge \int_{0}^{\delta_0} \frac{C}{\delta} d\delta = +\infty, \end{equation*}

which is a contradiction in view of the Fubini-Tonelli Theorem. Then we can pass to the limit as $\delta=\delta_n$ in (55) and conclude that, thanks to the Dominate Convergence Theorem and the Monotone Convergence Theorem,

(58)\begin{equation} \int_{B_{r}^+\setminus B_{\rho}^+}\mathop{\rm{div}}(y^{1-2s}|\nabla U_\lambda|^2 z) \, dz = r \int_{S_{r}^+}y^{1-2s}|\nabla U_\lambda|^2 \, dS -\rho \int_{S_{\rho}^+}y^{1-2s}|\nabla U_\lambda|^2 \, dS, \end{equation}

for a.e $r\in (0,r_0)$ and a.e. $\rho \in (0,r)$. Testing (48) with $\nabla U \cdot z$ we obtain, in view of (53) and Remark 4.3,

(59)\begin{align} &\int_{B^+_r\setminus B_\rho^+}\mathop{\rm{div}}(y^{1-2s}\nabla U_\lambda \cdot z \nabla U_\lambda) \, dz\nonumber\\ &\quad=\int_{B_r^+\setminus B_\rho^+} y^{1-2s}\nabla U_\lambda \cdot \nabla(\nabla U_\lambda\cdot z)\, dz-\int_{B_r^+\setminus B_\rho^+} y^{1-2s} \frac{\alpha}{|x|_k^2+\lambda^2}U_\lambda \nabla U_\lambda \cdot z\, dz \nonumber\\ &\quad=\frac{1}{r}\int_{S_r^+} y^{1-2s}|\nabla U_\lambda \cdot z|^2\, dS-\frac{1}{\rho}\int_{S_\rho^+} y^{1-2s}|\nabla U_\lambda \cdot z|^2\, dS\nonumber \\ & \qquad + c_{N,s}\int_{B_r'\setminus B_\rho'}g \mathop{\rm{Tr}}(U_\lambda)\, \nabla_x\mathop{\rm{Tr}}(U_\lambda)\cdot x \, dx. \end{align}

We note that $g \mathop{\rm{Tr}}(U_\lambda)^2 x \in W^{1,1}(B_r'\setminus B_\rho',\mathbb{R}^N)$ by (3) and (51) hence integrating by part we obtain

(60)\begin{align} \int_{B_r'\setminus B_\rho'}g\, \mathop{\rm{Tr}}(U_\lambda) \nabla_x\mathop{\rm{Tr}}(U_\lambda)\cdot x \, dx &=-\frac{1}{2}\int_{B_r'\setminus B_\rho'}(Ng+x \cdot \nabla g) \mathop{\rm{Tr}}(U_\lambda)^2 \, dx\nonumber\\ &\quad+\frac{r}{2}\int_{S_r'}g|\mathop{\rm{Tr}}(U_\lambda)|^2 dS'-\frac{\rho}{2}\int_{S_\rho'}g|\mathop{\rm{Tr}}(U_\lambda)|^2 dS'. \end{align}

Arguing as in the proof of (56), we see that there exists a sequence $\rho_n \to 0^+$ such that:

\begin{align*} \lim_{n \to \infty}\rho_n \int_{S_{\rho_n}^+}y^{1-2s}|\nabla U_\lambda|^2 \, dS=\lim_{n \to \infty}\rho_n \int_{S_{\rho_n}^+} y^{1-2s}&\left|\nabla U_\lambda \cdot \frac{z}{|z|}\right|^2\, dS\\ &=\lim_{n \to \infty}\rho_n\int_{S_{\rho_n}'}g|\mathop{\rm{Tr}}(U_\lambda)|^2 dS'=0. \end{align*}

Then by the Dominated Convergence Theorem, we can pass to the limit as $\rho=\rho_n$ and $n \to \infty$ in (58), (59), (60) and conclude that (49) holds in view of (52).

Proof. Let $r\in (0,r_0)$ and $B^+_{r,\delta}$, $S^+_{r,\delta}$ be as in (54) for any $\delta \in (0,r)$. Then, by (1),

(62)\begin{align} &\mathop{\rm{div}}\left(y^{1-2s} \frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda^2 \, z\right)\nonumber\\ &\quad= y^{1-2s}\left(2\frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda \nabla U_\lambda \cdot z+(N+2-2s) \frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda^2-2\frac{\alpha |x|^2_k}{(|x|^2_k+\lambda^2)^2}U_\lambda^2\right) \end{align}

and $y^{1-2s} \frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda^2 z \in W^{1,1}(B^+_{r,\delta},\mathbb{R}^{N+1})$. Integrating (62) by part in $B_{r,\delta}^+$ we obtain

(63)\begin{align} & r\int_{S_{r,\delta}^+}y^{1-2s}\frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda^2 \, dS- \delta^{2-2s} \int_{B'_{\sqrt{r^2-\delta^2}}} \frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda^2(x,\delta) \, dx\nonumber\\ &\quad= \int_{B_{r,\delta}^+}y^{1-2s}\left(2\frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda \nabla U_\lambda \cdot z+(N+2-2s) \frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda^2\right.\\ \nonumber & \qquad\qquad \left.-2\frac{\alpha |x|^2_k}{(|x|^2_k+\lambda^2)^2}U_\lambda^2\right) \, dz. \end{align}

We claim that there exists a sequence $\delta_n \to 0^+$ as $n \to \infty$ such that:

(64)\begin{equation} \lim_{n \to \infty}\delta_n^{2-2s} \int_{B'_{\sqrt{r^2-\delta_n^2}}} \frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda^2(x,\delta_n) \, dx=0, \end{equation}

arguing by contradiction. If (64) does not hold, then there exists a constant C > 0 and $\delta_0 \in (0,r)$ such that $(0,\delta_0) \times B_r'\subseteq B_{r_0}^+$, and

\begin{equation*} \delta^{1-2s} \int_{B'_{\sqrt{r^2-\delta^2}}} \frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda^2(x,\delta) \, dx \ge \frac{C}{\delta} \end{equation*}

for any $\delta \in (0,\delta_0)$. Integrating over $(0,\delta_0)$ we obtain:

\begin{equation*} +\infty \gt \int_{0}^{\delta_0}\delta^{1-2s} \left(\int_{B'_r} \frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda^2(x,\delta) \, dx\right) d\delta \ge \int_{0}^{\delta_0}\frac{C}{\delta} \, d \delta, \end{equation*}

a contradiction in view of the Fubini-Tonelli Theorem. Passing to the limit for $\delta =\delta_n$ as $n \to \infty$ in (63) we conclude that

(65)\begin{align} \int_{B_r^+}y^{1-2s}\frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda \nabla U_\lambda& \cdot z \, dz =\frac{r}{2}\int_{S_r^+}y^{1-2s}\frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda^2 \, dS\nonumber\\ &\quad- \frac{1}{2}\int_{B_r^+}y^{1-2s}\left((N+2-2s)\frac{\alpha}{|x|^2_k+\lambda^2} U_\lambda^2\right.\nonumber\\ & \qquad\qquad \left.-2\frac{\alpha |x|^2_k}{(|x|^2_k+\lambda^2)^2}U_\lambda^2\right) \, dz. \end{align}

Now we pass to the limit as $\lambda \to 0^+$, eventually along a suitable sequence $\lambda_n \to 0^+$, in each term of (49) taking into account (65). We recall that, by Theorem 4.2, $U_\lambda \to U$ strongly in $H^1(B_r^+,y^{1-2s})$ for any $r \in (0,r_0]$. It is clear that for any $ r \in (0,r_0)$:

\begin{equation*} \lim_{\lambda \to 0^+}\int_{B_r^+}y^{1-2s} |\nabla U_\lambda |^2 \, dz=\int_{B_r^+}y^{1-2s} |\nabla U |^2 \, dz. \end{equation*}

Furthermore, there exists a sequence $\lambda_n \to 0$ as $n \to \infty$ and $G \in L^2(B_{r_0}^+,y^{1-2s}|x|_k^{-2})$ such that:

(66)\begin{align} &(N+2-2s)\frac{\alpha}{|x|^2_k+\lambda_n^2} U_{\lambda_n}^2-2\frac{\alpha |x|^2_k}{(|x|^2_k+\lambda_n^2)^2}U_{\lambda_n}^2 \to (N-2s)\frac{\alpha}{|x|^2_k} U^2, \quad \text{for a.e. } z \in B_{r_0}^+,\nonumber\\ &\frac{\alpha}{|x|^2_k+\lambda_n^2}U_{\lambda_n}-\frac{\alpha}{|x|^2_k}U \to 0, \quad \text{for a.e. } z \in B_{r_0}^+,\\ & |U_{\lambda_n}|\le |G|, \quad \text{for a.e. } z \in B_{r_0}^+ \text{and any } n \in \mathbb{N}.\nonumber \end{align}

Then by the Dominated Convergence Theorem, we conclude that for any $r \in (0,r_0)$

\begin{align*} \lim_{n \to \infty}\int_{B_r^+}y^{1-2s}&\left((N+2-2s)\frac{\alpha}{|x|^2_k+\lambda_n^2} U_{\lambda_n}^2-2\frac{\alpha |x|^2_k}{(|x|^2_k+\lambda_n^2)^2}U_{\lambda_n}^2\right) \, dz\\ &\quad=(N-2s)\int_{B_r^+}y^{1-2s}\frac{\alpha}{|x|^2_k} U^2\, dz, \end{align*}

and

(67)\begin{equation} \lim_{n \to \infty}\int_{B_r^+}y^{1-2s}\left|\frac{\alpha}{|x|^2_k+\lambda_n^2}U_{\lambda_n}^2-\frac{\alpha}{|x|^2_k}U^2\right| \, dz=0. \end{equation}

By (3), (42), (40) and Proposition 3.1:

(68)\begin{equation} \lim_{\lambda \to 0^+}\int_{B_r'} |Ng +\nabla g\cdot x|\, |\mathop{\rm{Tr}}(U_\lambda)-\mathop{\rm{Tr}}(U)|^2 \, dx=0, \end{equation}

hence, for any $r \in (0,r_0)$,

\begin{equation*} \lim_{\lambda \to 0^+}\int_{B_r'} (Ng +x\cdot \nabla g) |\mathop{\rm{Tr}}(U_\lambda)|^2 \, dx= \int_{B_r'} (Ng +\nabla g\cdot x)|\mathop{\rm{Tr}}(U)|^2\, dx. \end{equation*}

By Fatou’s Lemma and the Coarea Formula,

\begin{equation*} \int_{0}^{r_0}\left(\liminf_{\lambda \to 0^+}\int_{S_r^+}y^{1-2s}|\nabla U_\lambda -\nabla U|^2 \, dS \right) dr \le \liminf_{\lambda \to 0^+}\int_{B_{r_0}^+}y^{1-2s}|\nabla U_\lambda -\nabla U|^2 \, dS=0, \end{equation*}

and so

\begin{equation*} \liminf_{\lambda \to 0^+}\int_{S_{r}^+}y^{1-2s}|\nabla U_\lambda|^2 \, dS= \int_{S_{r}^+}y^{1-2s}|\nabla U|^2 \, dS, \end{equation*}

for a.e. $r \in (0,r_0)$. Similarly, for a.e. $r \in (0,r_0)$:

\begin{equation*} \liminf_{\lambda \to 0^+}\int_{S_{r}^+}y^{1-2s}|\nabla U_\lambda\cdot \nu|^2 \, dS= \int_{S_{r}^+}y^{1-2s}|\nabla U\cdot \nu|^2 \, dS, \end{equation*}

and, by (68) and Fatou’s Lemma,

\begin{equation*} \liminf_{\lambda \to 0^+}\int_{S'_r}g |\mathop{\rm{Tr}}(U_\lambda)|^2 \, d'S= \int_{S'_r}g |\mathop{\rm{Tr}}(U)|^2\, dS'. \end{equation*}

Furthermore passing to the limit for $\lambda=\lambda_n$ as $n \to \infty$ and λ_n is as in (66), we obtain:

\begin{equation*} \lim_{n \to \infty}\int_{S_{r}^+}y^{1-2s}\frac{\alpha}{|x|^2_k+\lambda_n^2}U_{\lambda_n}^2 \,dS=\int_{S_{r}^+}y^{1-2s}\frac{\alpha}{|x|^2_k}U^2 \, dS, \end{equation*}

for a.e. $r \in (0,r_0)$, thanks to Fatou’s Lemma and (67). In conclusion (61) holds.

Proof. Assume by contradiction that there exists $r \in (0,r_0]$ such that $H(r)=0$. From (33) and Remark 4.1 we deduce that $U\equiv 0$ on $B_r^+$. Let $\Sigma_k$ be as in (39). The function U is a solution of an elliptic equation with bounded coefficients away from $\Sigma_k$ and $\mathbb{R}^N \times\{0\}$. Then the claim follows from classical unique continuation principles, see for example [Reference Wolff34].

Proof. By the Coarea Formula

\begin{align*} D'(r)=(2s-N)&r^{-N+2s-1}\left(\int_{B_r^+} y^{1-2s}\left(|\nabla U|^2-\frac{\alpha}{|x|^2_k}U^2\right) \, dz-c_{N,s}\int_{B_r'}g|\mathop{\rm{Tr}}(U)|^2 \, dx\right)\\ +&r^{-N+2s}\left(\int_{S_r^+} y^{1-2s}\left(|\nabla U|^2-\frac{\alpha}{|x|^2_k}U^2\right) \, dS-c_{N,s}\int_{S_r'}g|\mathop{\rm{Tr}}(U)|^2 \, dS'\right), \end{align*}

and so (72) follows from (61). Furthermore $D \in W^{1,1}_{loc}((0,r_0])$ by (72), (70) and the Coarea Formula.

Proof. Since $1/H,D \in W^{1,1}_{loc}((0,r_0])$ it follows that $\mathcal{N} \in W^{1,1}_{loc}((0,r_0])$. We can deduce (74) directly from (44) and (73).

Furthermore by (71):

\begin{equation*} \frac{d}{dr}\mathcal{N}'(r)=\frac{D'(r)H(r)-D(r)H'(r)}{H^2(r)}=\frac{D'(r)H(r)-\frac{r}{2}(H'(r))^2}{H^2(r)}, \end{equation*}

and so (75) follows from (69), (70) and (72). Finally (77) is a consequence of the Cauchy-Schwartz inequality in $L^2(S_r^+,y^{1-2s})$ between the vectors U and $\frac{\partial U}{\partial\nu}$.

Proof. The claim follows from (3), (42), (44) and (76).

Proof. Thanks to Proposition 5.4, for a.e. $r \in (0,r_0)$:

\begin{equation*} \left(\mathcal{N}+\frac{N-2s}{2}\right)'(r) \ge v_2(r)\ge -Cr^{-1+\varepsilon}\left(\mathcal{N}(r)+\frac{N-2s}{2}\right). \end{equation*}

Hence an integration over $(r,r_0)$ yields:

\begin{equation*} \mathcal{N}(r) \le -\frac{N-2s}{2}+\left(\mathcal{N}(r_0)+\frac{N-2s}{2}\right)e^{\frac{C}{\varepsilon}r_0^\varepsilon} \end{equation*}

for any $r \in (0,r_0)$.

Proof. Since $\mathcal{N} \in W^{1,1}_{loc}((0,r_0])$ by Proposition 5.4, for any $r \in (0,r_0)$:

(81)\begin{equation} \mathcal{N}(r)= \mathcal{N}(r_0) -\int_{r}^{r_0} \mathcal{N}'(r) \, dr =\mathcal{N}(r_0) -\int_{r}^{r_0}v_1(r) \, dr-\int_{r}^{r_0}v_2(r) \, dr. \end{equation}

Since $v_1 \ge 0$ by (77) and $v_2 \in L^1(0,r_0)$ by (78) and (79), we can pass to the limit as $r \to 0^+$ in (81) and conclude that the limit (80) exists. From (74) and (79) it is finite.

Proof. For any $\lambda \in (0,r_0)$, thanks to (44), (85) and a change of variables,

\begin{align*} \mathcal{N}(\lambda)&=\frac{{\lambda^{2s-N}}}{H(\lambda)}\left(\int_{B_\lambda^+} y^{1-2s}\left(|\nabla U|^2-\frac{\alpha}{|x|^2_k}U^2\right) \, dz-c_{N,s}\int_{B_\lambda'}g|\mathop{\rm{Tr}}(U)|^2 \, dx\right)\\ &\ge \left(1-\alpha\left(\frac{2}{k-2}\right)^2+ c_{N,s}k_{N,s,g}r_0^\varepsilon\right)\frac{{\lambda^{2s-N}}}{H(\lambda)}\left(\int_{B_{\lambda}^+} y^{1-2s} |\nabla U|^2 \, dz\right)-\frac{N-2s}{2}\\ &=\left(1-\alpha\left(\frac{2}{k-2}\right)^2+ c_{N,s}k_{N,s,g}r_0^\varepsilon\right)\left(\int_{B_{1}^+} y^{1-2s} |\nabla V^\lambda|^2 \, dz\right)-\frac{N-2s}{2}. \end{align*}

Hence the claim follows from (79), (86) and (40).

Proof. By (71), (74), and (79):

\begin{equation*} -\frac{N-2s}{r} \le \frac{H'(r)}{H(r)}=\frac{2\mathcal{N}(r)}{r} \le \frac{2C_1}{r}\quad \text{for a.e. } r \in (0,r_0). \end{equation*}

An integration over $(\lambda,R \lambda )$ with $R \in (1,2]$ yields:

\begin{equation*} R^{2s-N}\le \frac{H(R\lambda)}{H(\lambda)} \le R^{2C_1}, \end{equation*}

thus (87) holds for $R \in (1,2]$ while if R = 1 it is obvious.

Furthermore for any $\lambda \in (0,r_0)$, by (87) and a change of variables,

\begin{align*} \int_{B_R^+}y^{1-2s} |V^\lambda|^2 \, dz & =\frac{\lambda^{-N-2+2s}}{H(\lambda)}\int_{B_{R\lambda}^+} y^{1-2s}|U|^2 \, dz\le C_3\frac{\lambda^{-N-2+2s}}{H(\lambda R)}\int_{B_{R\lambda}^+} y^{1-2s}|U|^2 \, dz\\ &= C_3R^{N+2-2s}\int_{B_1^+} y^{1-2s}|V^{\lambda R}|^2 \, dz \le C_32^{N+2-2s}\int_{B_1^+} y^{1-2s}|V^{\lambda R}|^2 \, dz, \end{align*}

for any $R \in [1,2]$. Hence, we have proved (88) and (89) follows from (87) in the same way.

Proof. By Proposition 6.1 $\{V^\lambda\}_{\lambda \in \left(0,\frac{r_0}{2}\right)}$ is bounded in $H^1(B_2^+,y^{1-2s})$. Hence

(91)\begin{equation} \limsup_{\lambda \to 0^+} \int_{B_2^+} y^{1-2s} (|\nabla V^\lambda|^2 +|V^\lambda|^2) \, dz \lt +\infty. \end{equation}

Let, for any $\lambda \in \left(0,\frac{r_0}{2}\right)$,

\begin{equation*} f_\lambda(R):=\int_{B_R^+} y^{1-2s} (|\nabla V^\lambda|^2 +|V^\lambda|^2) \, dz. \end{equation*}

The function f is absolutely continuous on $[1,2]$ and, thanks to the Coarea Formula, its distributional derivative is given by:

\begin{equation*} f'_\lambda(R)=\int_{S_R^+} y^{1-2s} (|\nabla V^\lambda|^2 +|V^\lambda|^2) \, dS \quad \text{for a.e. } R \in [1,2]. \end{equation*}

We argue by contradiction supposing that for any M > 0 there exists $\lambda_n \to 0^+$ such that:

\begin{equation*} \int_{S_R^+} y^{1-2s}(|\nabla V^{\lambda_n}|^2 +|V^{\lambda_n}|^2)\, dS \gt M \int_{B_R^+}y^{1-2s}(|\nabla V^{\lambda_n}|^2 +|V^{\lambda_n}|^2) \, dz, \end{equation*}

for any $n \in \mathbb{N}$ and any $R \in [1,2] \setminus {\frac{1}{\lambda_n}}\mathcal{M}$, hence for a.e. $R\in [1,2]$. Therefore,

\begin{equation*} f'_{\lambda_n}(R) \gt M f_{\lambda_n}(R), \quad \text{for a.e. } R\in [1,2] \,\text{and any } n \in \mathbb{N}. \end{equation*}

An integration over $[1,2]$ yields $f_{\lambda_n}(2) \gt e^Mf_{\lambda_n}(1)$ for any $n \in \mathbb{N}$. Hence

\begin{equation*} \liminf_{n \to \infty} f_{\lambda_n}(1)\le \limsup_{n \to \infty} f_{\lambda_n}(1)\le e^{-M}\limsup_{n \to \infty} f_{\lambda_n}(2), \end{equation*}

and so

\begin{equation*} \liminf_{\lambda \to 0^+} f_{\lambda}(1)\le e^{-M}\limsup_{\lambda \to 0^+} f_{\lambda}(2), \end{equation*}

for any M > 0. It follows that $\liminf_{\lambda \to 0^+} f_{\lambda}(1)=0$ by (91). We conclude that there exists a sequence $\lambda_n \to 0^+$ as $n\to \infty$ and $V \in H^1(B_1^+,y^{1-2s})$ such that:

\begin{equation*} \lim_{n \to \infty} \int_{B_1^+}y^{1-2s} (|\nabla V^{\lambda_n}|^2 +|V^{\lambda_n}|^2)\, dz=0, \end{equation*}

and $V_{\lambda_n} \rightharpoonup V$ weakly in $ H^1(B_1^+,y^{1-2s})$, taking into account Proposition 6.1. By Proposition 3.1, (86) and the lower semicontinuity of norms, we obtain,

\begin{equation*} \int_{B_1^+}y^{1-2s}(|\nabla V|^2 +|V|^2)\, dz=0 \quad \text{and } \quad \int_{\mathbb{S}^+}\theta_{N+1}^{1-2s} V^2 \, dS=1, \end{equation*}

which is a contradiction.

Proof. By a change of variables, the fact that $R_\lambda \in [1,2]$ and (85):

\begin{align*} \int_{\mathbb{S}^+}\theta_{N+1}^{1-2s} |\nabla V^{R_\lambda\lambda}|^2 dS &= R_\lambda^{-N+1+2s}\frac{H(\lambda)}{H(R_\lambda\lambda)}\int_{S_{R_\lambda}^+}y^{1-2s}|\nabla V^{\lambda}|^2 dS\\ &\le2 C_3 M \int_{B_{R_\lambda}^+}y^{1-2s}(|\nabla V^{\lambda}|^2+|V^{\lambda}|^2)\, dz \\ &\le 2^{N+3-2s}C_3^2 M\int_{B_1^+}y^{1-2s}(|\nabla V^{R_\lambda\lambda}|^2+|V^{R_\lambda\lambda}|^2)\, dz \le \overline{M} \lt +\infty, \end{align*}

for some $\overline M \gt 0$, in view of Proposition 6.1, (87), (88), (89), and (90).

Proof. Let V ^λ be as in (85) and R_λ as in Proposition 6.3. The family $\{V^{R_\lambda\lambda}\}_{\lambda \in \left(0,\min\left\{\lambda_0,\frac{r_0}{2}\right\}\right)}$ is bounded in $H^1(B_1^+,y^{1-2s})$, thanks to Proposition 6.1. Let $\lambda_p \to 0^+$ as $p\to \infty$. Then there exists a subsequence $\lambda_{p_q} \to 0^+$ as $q\to \infty$ and $V \in H^1(B_1^+,y^{1-2s})$ such that $V^{R_{\lambda_{p_q}}\lambda_{p_q}} \rightharpoonup V$ weakly in $H^1(B_1^+,y^{1-2s})$ as $q \to \infty$. By Proposition 3.1 the trace operator $\mathop{{\rm{Tr}}_{S^{+}_1}}$ is compact and so

(94)\begin{equation} \int_{\mathbb{S}^+} \theta_{N+1}^{1-2s} |V|^2 \, dS=1, \end{equation}

in view of (86). Hence V is non-trivial. We claim that:

(95)\begin{equation} V^{R_{\lambda_{p_q}}\lambda_{p_q}} \rightharpoonup V \quad \text{strongly in } H^1(B_1^+,y^{1-2s}) \text{as } q \to \infty. \end{equation}

For q sufficiently large $B_1^+\subseteq B^+_{r_0/(R_{\lambda_{p_q}}\lambda_{p_q})}$ and since $R_{\lambda_{p_q}}\lambda_{p_q} \not \in \mathcal{M}$, where $\mathcal{M}$ is as in Proposition 6.3, we have that

(96)\begin{align} &\int_{B_1^+} y^{1-2s} \left(\nabla V^{R_{\lambda_{p_q}}\lambda_{p_q}}\cdot\nabla W -\frac{\alpha}{|x|_k^2}V^{R_{\lambda_{p_q}}\lambda_{p_q}}W \right)\, dz\nonumber\\ &=\int_{\mathbb{S}^+} \theta_{N+1}^{1-2s} \frac{\partial V^{R_{\lambda_{p_q}}\lambda_{p_q}}}{\partial\nu} \,W \, dS+c_{N,s}(R_{\lambda_{p_q}}\lambda_{p_q})^{2s}\int_{B_1'} g(R_{\lambda_{p_q}}\lambda_{p_q}\cdot) \mathop{\rm{Tr}}(V^{R_{\lambda_{p_q}}\lambda_{p_q}})\mathop{\rm{Tr}}(W) \, dx, \end{align}

for any $W \in H^1(B_1^+,y^{1-2s})$, thanks to (33) and a change of variables. We will pass to the limit as $q \to \infty$ in (96). To this end we observe that, for any $W \in H^1(B_1^+,y^{1-2s})$,

\begin{align*} &\left|\lambda^{2s}\int_{B_1'} g(\lambda\cdot) \mathop{\rm{Tr}}(V^\lambda)\mathop{\rm{Tr}}(W) \, dx\right|=\left|\frac{\lambda^{2s-N}}{H(\lambda)}\int_{B_\lambda'} g(x) \mathop{\rm{Tr}}(U)(x)\mathop{\rm{Tr}}(W)(\lambda x) \, dx\right| \\ &\quad\le k_{N,s,g}\frac{\lambda^{2s+\varepsilon-N}}{H(\lambda)} \left|\int_{B_\lambda^+}y^{1-2s}|\nabla U|^2\, dz-\int_{B_\lambda^+}y^{1-2s}\frac{\alpha}{|x|_k^2}|U|^2\, dz\right.\\ & \qquad\qquad \left.+\frac{N-2s}{2\lambda}\int_{S^+_\lambda}y^{1-2s}|U|^2\, dS\right|^\frac{1}{2}\\ &\quad\times \left| \int_{B_\lambda^+}y^{1-2s}|\nabla W(\lambda \cdot)|^2\, dz-\int_{B_\lambda^+}y^{1-2s}\frac{\alpha}{|x|_k^2}W(\lambda \cdot)^2\, dz \right.\\ & \qquad\qquad \left. +\frac{N-2s}{2\lambda} \int_{S^+_\lambda}y^{1-2s}|W(\lambda\cdot)|^2\, dS \right|^{\frac {1}{2}}\\ &\quad=k_{N,s,g} \lambda^\varepsilon \left|\int_{B_1^+}y^{1-2s}|\nabla V^\lambda|^2\, dz-\int_{B_1^+}y^{1-2s}\frac{\alpha}{|x|_k^2}|V^\lambda|^2\, dz+\frac{N-2s}{2}\right|^\frac{1}{2}\\ &\quad\times \left|\int_{B_1^+}y^{1-2s}|\nabla W|^2\, dz-\int_{B_1^+}y^{1-2s}\frac{\alpha}{|x|_k^2}W^2\, dz+\frac{N-2s}{2}\int_{\mathbb{S}^+}\theta_{N+1}^{1-2s}|W|^2\, dS \right|^{\frac {1}{2}} \end{align*}

by a change of variables, the Hölder inequality, (3), (42), (85) and (86). We conclude that:

(97)\begin{equation} \lim_{\lambda \to 0^+}\left|\lambda^{2s}\int_{B_1'} g(\lambda\cdot) \mathop{\rm{Tr}}(V^\lambda)\mathop{\rm{Tr}}(W) \, dx\right|=0, \end{equation}

by Proposition 6.1 and (40). Thanks to (92), there exists a function $f \in L^2(\mathbb{S}^+,\theta_{N+1}^{1-2s})$ such that:

(98)\begin{equation} \frac{\partial V^{R_{\lambda_{p_q}}\lambda_{p_q}}}{\partial\nu} \rightharpoonup f \quad \text{weakly in } L^2(\mathbb{S}^+,\theta_{N+1}^{1-2s}) \text{as } q \to \infty, \end{equation}

up to a subsequence. Hence,

\begin{equation*} \lim_{q \to \infty}\int_{\mathbb{S}^+}\theta_{N+1}^{1-2s} \frac{\partial V^{R_{\lambda_{p_q}}\lambda_{p_q}}}{\partial\nu} \, W \, dS=\int_{\mathbb{S}^+} \theta_{N+1}^{1-2s} f W \, dS \end{equation*}

for any $W \in H^1(B_1^+,y^{1-2s})$. Furthermore,

\begin{align*} & \lim_{q \to \infty}\int_{B_1^+} y^{1-2s} \left(\nabla V^{R_{\lambda_{p_q}}\lambda_{p_q}}\cdot\nabla W -\frac{\alpha}{|x|_k^2}V^{R_{\lambda_{p_q}}\lambda_{p_q}}W \right)\, dz\\ & \qquad\qquad =\int_{B_1^+} y^{1-2s} \left(\nabla V\cdot\nabla W -\frac{\alpha}{|x|_k^2}VW \right)\, dz \end{align*}

by Remark 4.1. It follows that:

\begin{equation*} \int_{B_1^+} y^{1-2s} \left(\nabla V\cdot\nabla W -\frac{\alpha}{|x|_k^2}VW \right)\, dz=\int_{\mathbb{S}^+} \theta_{N+1}^{1-2s} f W \, dS, \end{equation*}

for any $W \in H^1(B_1^+,y^{1-2s})$, that is V is a weak solution of the problem:

(99)\begin{equation} \begin{cases} \mathop{\rm{div}}(y^{1-2s}\nabla V)=\frac{\alpha}{|x|_k^2}V, &\text{in } B_1^+,\\ -\lim_{y \to 0^+} y^{1-2s}\frac{\partial V}{\partial y}=0, &\text{on } B_1'. \end{cases} \end{equation}

Furthermore testing (96) with $ V^{R_{\lambda_{p_q}}\lambda_{p_q}}$,

\begin{align*} \lim_{q \to \infty}\int_{B_1^+}y^{1-2s}&\left(\left|\nabla V^{R_{\lambda_{p_q}}\lambda_{p_q}}\right|^2-\frac{\alpha}{|x|_k^2}\left|V^{R_{\lambda_{p_q}}\lambda_{p_q}}\right|^2 \right)\, dz\\ &\quad\quad=\lim_{q \to \infty}\int_{\mathbb{S}^+}\theta_{N+1}^{1-2s}\frac{\partial V^{R_{\lambda_{p_q}}\lambda_{p_q}}}{\partial\nu}V^{R_{\lambda_{p_q}}\lambda_{p_q}}\,dS=\int_{\mathbb{S}^+} \theta_{N+1}^{1-2s} f W \, dS, \end{align*}

thanks to (98) and the compactness of the trace operator $\mathop{{\rm{Tr}}_{S^+_1}}$, see Proposition 3.1. Hence from Remark 4.1 and (86) we deduce (95). Let for any $ r\in (0,1]$

\begin{align*} D_q(r)=\frac{1}{r^{N-2s}}\Big(\int_{B_r^+} y^{1-2s} &\left( |\nabla V^{R_{\lambda_{p_q}}\lambda_{p_q}}|^2 -\frac{\alpha}{|x|_k^2}|V^{R_{\lambda_{p_q}}\lambda_{p_q}}|^2 \right)\, dz\\ &-c_{N,s}(R_{\lambda_{p_q}}\lambda_{p_q})^{2s}\int_{B_r'} g(R_{\lambda_{p_q}}\lambda_{p_q}\cdot) |\mathop{\rm{Tr}}(V^{R_{\lambda_{p_q}}\lambda_{p_q}})|^2 \, dx\Big), \end{align*}

and

\begin{equation*} H_q(r)=\frac{1}{r^{N+1-2s}}\int_{S_r^+} y^{1-2s} |V^{R_{\lambda_{p_q}}\lambda_{p_q}}|^2\, dS. \end{equation*}

For any $r \in (0,1]$ we also define:

\begin{equation*} D_V(r)=\frac{1}{r^{N-2s}}\int_{B_r^+} y^{1-2s} \left( |\nabla V|^2 -\frac{\alpha}{|x|_k^2}|V|^2 \right)\, dz\\ \end{equation*}

and

(100)\begin{equation} H_V(r)=\frac{1}{r^{N+1-2s}}\int_{S_r^+} y^{1-2s} |V|^2\, dS. \end{equation}

Thanks to a scaling argument it is easy to see that:

\begin{equation*} \mathcal{N}_q(r):=\frac{D_q(r)}{H_q(r)}=\frac{D(R_{\lambda_{p_q}}\lambda_{p_q}r)}{H(R_{\lambda_{p_q}}\lambda_{p_q}r)}=\mathcal{N}(R_{\lambda_{p_q}}\lambda_{p_q}r), \quad \text{for any } r \in (0,1]. \end{equation*}

By (95), (97) and Remark 4.1, it follows that:

\begin{equation*} H_q(r) \to H_V(r) \quad \text{and } \quad D_q(r) \to D_V(r), \quad \text{as } q \to \infty, \text{for any } r \in (0,1]. \end{equation*}

Furthermore $H_V(r) \gt 0$ for any $r \in (0,1]$ by Proposition 5.2 in the case $g\equiv 0$ and $\Omega = B_2'$. In particular the function:

\begin{equation*} \mathcal{N}:(0,1] \to \mathbb{R}, \quad \mathcal{N}_V(r):=\frac{D_V(r)}{H_V(r)}, \end{equation*}

is well defined and $\mathcal{N}_V \in W^{1,1}_{loc}((0,1])$ by Proposition 5.4 in the case $g\equiv 0$ and $\Omega = B_2'$. In view of (100), (80):

(101)\begin{equation} \mathcal{N}_V(r)=\lim_{q \to \infty} \mathcal{N}(R_{\lambda_{p_q}}\lambda_{p_q}r)=\gamma \quad \text{for any } r \in (0,1]. \end{equation}

Hence $\mathcal{N}_V(r)$ is constant in $[0,1]$ and so:

\begin{equation*} \mathcal{N}'_V(r)\equiv 0, \text{for any } r \in (0,1]. \end{equation*}

By Proposition 5.4 it follows that:

\begin{equation*} \left(\int_{S_r^+}y^{1-2s}V^2\, dS\right)\left(\int_{S_r^+}y^{1-2s}\left|\frac{\partial V}{\partial\nu}\right|^2 \, dS\right)-\left(\int_{S_r^+}y^{1-2s}V \frac{\partial V}{\partial\nu} \, dS\right)^2=0, \end{equation*}

for a.e. $r \in (0,1)$, that is, equality holds in the Cauchy-Schwartz inequality for the vectors V and $\frac{\partial V}{\partial\nu}$ in $L^2(S_r^+,y^{1-2s})$ for a.e. $r \in (0,1)$.

Therefore, there exists a function $\eta(r)$ defined a.e. in $(0,1)$ such that:

\begin{equation*} \frac{\partial V}{\partial\nu}(r\theta)=\eta(r) V(r\theta), \quad \text{for a.e. r } \in (0,1) \,\text{and a.e. } \theta \in \mathbb{S}^+. \end{equation*}

Multiplying by $V(r\theta)$ and integrating over $\mathbb{S}^+$,

\begin{equation*} \int_{\mathbb{S}^+}\theta_{N+1}^{1-2s}\frac{\partial V}{\partial\nu}(r\theta)V(r\theta) \, dS=\eta(r)\int_{\mathbb{S}^+}\theta_{N+1}^{1-2s}|V(r\theta)|^2 \, dS, \text{for a.e. r } \in (0,1), \end{equation*}

and so $\eta(r)=\frac{H'_V(r)}{2H_V(r)}=\frac{\gamma}{r}$ for a.e. $r \in (0,1)$ by (71), (71) and (101). Since V is smooth away from $\Sigma_k$ by classical elliptic regularity theory (see (39)), an integration over $(r,1)$ yields:

(102)\begin{equation} V(r\theta)=r^\gamma V(1\theta)=r^\gamma Z(\theta), \quad \text{for any}\, r \in (0,1]\quad \text{and a.e. } \theta \in \mathbb{S}^+, \end{equation}

where $Z=V_{|\mathbb{S}^+}$ and $\left\lVert Z\right\rVert_{L^2(\mathbb{S}^+,\theta^{1-2s}_{N+1})}=1$ by (94). In view of [Reference Fall and Felli13, Lemma 1.1], (102) and (99) the function Z is an eigenfunction of problem (22) and the correspondent eigenvalue $\gamma_{\alpha,k,n}$ satisfies the relationship $\gamma(N-2s+\gamma)= \gamma_{\alpha,k,n}$, that is:

\begin{equation*} \gamma =-\frac{N-2s}{2}+\sqrt{\left(\frac{N-2s}{2}\right)^2 +\gamma_{\alpha,k,n}} \;\;\text{or } \;\;\gamma =-\frac{N-2s}{2} -\sqrt{\left(\frac{N-2s}{2}\right)^2 +\gamma_{\alpha,k,n}} \end{equation*}

Since $r^\gamma Z(\theta) \in H^1(B_1^+,y^{1-2s})$ by (102) then $r^{2\gamma-2} Z^2(\theta) \in L^1(B_1^+,y^{1-2s})$ by (40) and so we conclude that (93) must hold.

Consider now the sequence $\{V^{\lambda_{p_q}}\}_{q \in \mathbb{N}}$. Up to a further subsequence, $V^{\lambda_{p_q}} \rightharpoonup \tilde{V}$ weakly in $H^1(B_1^+,y^{1-2s})$ as $q \to \infty$, for some $\tilde{V} \in H^1(B_1^+,y^{1-2s})$ and $R_{\lambda_{p_q}} \to \tilde{R}$, for some $\tilde{R} \in [1,2]$ as $q \to \infty$. The strong convergence of $\{V^{R_{\lambda_{p_q}}\lambda_{p_q}}\}_{q \in \mathbb{N}}$ to V in $H^1(B_1^+,y^{1-2s})$ implies that, up to a further subsequence, both $V^{R_{\lambda_{p_q}}\lambda_{p_q}}$ and $\left|\nabla V^{R_{\lambda_{p_q}}\lambda_{p_q}}\right|$ are dominated a.e. by a $L^2(B_1^+,y^{1-2s})$ function, uniformly with respect to $q \in \mathbb{N}$. Up to a further subsequence, we may also assume that the limit:

\begin{equation*} \ell =\lim_{q\to \infty}\frac{H(R_{\lambda_{p_q}}\lambda_{p_q})}{H(\lambda_{p_q})}, \end{equation*}

exists, it is finite and strictly positive, taking into account (87). Then from the Dominated Convergence Theorem and a change of variables we deduce that:

\begin{align*} &\lim_{q\to \infty}\int_{B_1^+} y^{1-2s} V^{\lambda_{p_q}}(z) \phi(z) \, dz= \lim_{q\to \infty}R_{\lambda_{p_q}}^{N+2-2s}\int_{B_{1/R_{\lambda_{p_q}}}^+} y^{1-2s}V^{\lambda_{p_q}}(R_{\lambda_{p_q}}z)\phi(R_{\lambda_{p_q}} z) \, dz\\ &\quad =\lim_{q\to \infty}R_{\lambda_{p_q}}^{N+2-2s}\sqrt{\frac{H(R_{\lambda_{p_q}}\lambda_{p_q})}{H(\lambda_{p_q})}} \int_{B_{1}^+} y^{1-2s}\chi_{B_{1/R_{\lambda_{p_q}}}^+}(z)V^{R_{\lambda_{p_q}}\lambda_{p_q}}(z) \phi(R_{\lambda_{p_q}} z) \, dz\\ &\quad =\tilde{R}^{N+2-2s} \sqrt{\ell} \int_{B_{1/\tilde{R}}^+} y^{1-2s}V(z) \phi(\tilde{R} z) \, dz= \sqrt{\ell} \int_{B_1^+} y^{1-2s}V(z/\tilde{R}) \phi( z) \, dz, \end{align*}

for any $\phi \in C^\infty(\overline{B_1^+})$. By density we conclude that $V^{\lambda_{p_q}} \rightharpoonup \sqrt{\ell}V(\cdot/\tilde{R})$ weakly in $L^2(B_1^+,y^{1-2s})$ as $q \to \infty$. Since $V^{\lambda_{p_q}} \rightharpoonup \tilde{V}$ weakly in $H^1(B_1^+,y^{1-2s})$ as $q \to \infty$ we conclude that $\tilde{V}=\sqrt{\ell}V(\cdot/\tilde{R})$ and so $V^{\lambda_{p_q}} \rightharpoonup \sqrt{\ell}V(\cdot/\tilde{R})$ weakly in $H^1(B_1^+,y^{1-2s})$ as $q \to \infty$. Furthermore

\begin{align*} &\lim_{q\to \infty} \int_{B_1^+}y^{1-2s} |\nabla V^{\lambda_{p_q}}(z)|^2 \, dz= \lim_{q\to \infty}R_{\lambda_{p_q}}^{N+2-2s}\int_{B_{1/R_{\lambda_{p_q}}}^+} y^{1-2s} |\nabla V^{\lambda_{p_q}}(R_{\lambda_{p_q}} z)|^2\, dz\\ &=\lim_{q\to \infty}R_{\lambda_{p_q}}^{N-2s}\frac{H(R_{\lambda_{p_q}}\lambda_{p_q})}{H(\lambda_{p_q})} \int_{B_{1}^+} y^{1-2s}\chi_{B_{1/R_{\lambda_{p_q}}}^+}(z)|\nabla V^{R_{\lambda_{p_q}}\lambda_{p_q}}(z)|^2 \, dz\\ &=\tilde{R}^{N-2s} \ell \int_{B_{1/\tilde{R}}^+} y^{1-2s}|\nabla V|^2 dz= \int_{B_1^+} y^{1-2s}|\sqrt{\ell}\nabla V(\cdot/\tilde{R})|^2\, dz, \end{align*}

by the Dominated Convergence Theorem and a change of variables. Hence $V^{\lambda_{p_q}} \to \sqrt{\ell}V(\cdot/\tilde{R})$ strongly in $H^1(B_1^+,y^{1-2s})$ as $q \to \infty$.

Thanks to (102), V is a homogeneous function of degree γ and so $\tilde{V}=\sqrt{\ell}\tilde{R}^{-\gamma} V$. Moreover, since $V^{\lambda_{p_q}} \to \tilde{V}$ strongly in $L^2(\mathbb{S}^+,\theta_{N+1}^{1-2s})$ as $q \to \infty$ by Proposition 3.1,

\begin{equation*} 1=\int_{\mathbb{S}^+} \theta_{N+1}^{1-2s} |\tilde{V}(\theta)|^2 dS=\sqrt{\ell}\tilde{R}^{-\gamma}\int_{\mathbb{S}^+} \theta_{N+1}^{1-2s} |V(\theta)|^2 dS=\sqrt{\ell}\tilde{R}^{-\gamma} \end{equation*}

in view of (86) and (94). We conclude that $\tilde{V} =V$ thus completing the proof.

Proof. Let $n \in \mathbb{N}$ and $i \in \{1,\dots,M_{\alpha,k,n}\}$. Let $f \in C_c^\infty(0,r_0)$. Then testing (21) with the function $|z|^{N+1-2s}f(|z|) Z_{\alpha,k,n,i}(z/|z|)$ and passing in polar coordinates, by (35), we obtain:

\begin{equation*} -\varphi_{n,i}''(\lambda)-\frac{N+1-2s}{\lambda }\varphi_{n,i}'(\lambda)+\frac{\gamma_{\alpha,k,n}}{\lambda^2}\varphi_{n,i}(\lambda)=\zeta_{n,i}(\lambda) \quad \text{in } (0,r_0), \end{equation*}

in a distributional sense, where the distribution $\zeta_{n,i} \in \mathcal{D}'(0,r_0)$ is define as:

(107)\begin{equation} \sideset{_{\mathcal{D}'(0,r_0)}}{_{\mathcal{D}(0,r_0)}}{\mathop{\left\langle \zeta_{n,i}, f \right\rangle}} = \int_{0}^{r_0} \frac{f(\lambda)}{\lambda^{2-2s}} \left( \int_{\mathbb{S}'} g(\lambda \cdot ) \mathop{\rm{Tr}}(U)(\lambda \cdot ) \mathop{\rm{Tr}}\left(Z_{\alpha,k,n,i}\left(\frac{\cdot}{|\cdot|}\right)\right)\, dS'\right) \, d \lambda, \end{equation}

for any $f \in C_c^\infty(0,r_0)$. In particular $\zeta_{n,i}$ belongs to $L^1_{loc}((0,r_0])$ by the Coarea Formula and a change of variables. If $\Upsilon_{n,i} $ is as in (104), a direct computation shows that:

\begin{equation*} \Upsilon'_{n,i}(\lambda)=\lambda^{N+1-2s}\zeta_{n,i}(\lambda) \quad \text{in } \mathcal{D}'(0,r_0), \end{equation*}

hence

(108)\begin{equation} -\left(\lambda^{N+1-2s+2\sigma_n}\left(\lambda^{-\sigma_n} \varphi_{n,i}(\lambda)\right)'\right)'=\lambda^{\sigma_n} \Upsilon'_{n,i}(\lambda) \quad \text{in } \mathcal{D}'(0,r_0), \end{equation}

where

(109)\begin{equation} \sigma_n:=-\frac{N-2s}{2}+\sqrt{\left(\frac{N-2s}{2}\right)^2+\gamma_{\alpha,k,n.}}. \end{equation}

From (108) and (107), we deduce that $\lambda \to \lambda^{N+1-2s+2\sigma_n}\left(\lambda^{-\sigma_n} \varphi_{n,i}'(\lambda)\right)$ belongs to $W^{1,1}_{loc}((0,r_0])$ hence an integration over $(\lambda, r)$ yields:

(110)\begin{align} &\left(\lambda^{-\sigma_n} \varphi_{n,i}(\lambda)\right)'=-\lambda^{-N-1+2s-\sigma_n}\Upsilon_{n,i}(\lambda)\nonumber\\ &\quad-\lambda^{-N-1+2s-2\sigma_n}\sigma_n\left(C(r)+\int_{\lambda}^r\rho^{\sigma_n-1}\Upsilon_{n,i}(\rho) \, d\rho\right), \end{align}

for any $ r\in (0,r_0]$, for some real number C(r) depending on $r, \alpha,k,n$ and i. Since in view of (110) $\lambda \to \lambda^{-\sigma_n} \varphi_{n,i}(\lambda)$ belongs to $W^{1,1}_{loc}((0,r_0])$, a further integration yields:

(111)\begin{align} &\varphi_{n,i}(\lambda)=\lambda^{\sigma_n}\Bigg(r^{-\sigma_n}\varphi_{n,i}(r)+ \int_{\lambda}^{r}\rho^{-N-1+2s-\sigma_n}\Upsilon_{n,i}(\rho )\, d \rho\nonumber\\ &\quad+\sigma_n\int_{\lambda}^r\rho^{-N-1+2s-2\sigma_n}\left(C(r)+\int_{\rho}^r t^{\sigma_n-1}\Upsilon_{n,i}(t) \, dt\right)\, d \rho\Bigg)\nonumber\\ &\quad=\lambda^{\sigma_n}\Bigg(r^{-\sigma_n}\varphi_{n,i}(r)+ \int_{\lambda}^{r}\rho^{-N-1+2s-\sigma_n}\Upsilon_{n,i}(\rho )\, d\rho +\frac{\sigma_nC(r)r^{-N+2s-2\sigma_n}}{-N+2s-2\sigma_n}\nonumber\\ &\quad-\frac{\sigma_nC(r)\lambda^{-N+2s-2\sigma_n}}{-N+2s-2\sigma_n}-\frac{\sigma_n \lambda^{-N+2s-2\sigma_n}}{-N+2s-2\sigma_n}\int_{\lambda}^r t^{\sigma_n-1}\Upsilon_{n,i}(t) \, dt\nonumber\\ &\quad+\frac{\sigma_n}{-N+2s-2\sigma_n}\int_{\lambda}^r\rho^{-N-1+2s-\sigma_n}\Upsilon_{n,i}(\rho )\, d\rho \Bigg)\nonumber\\ &\quad=\lambda^{\sigma_n}\Bigg(\frac{\varphi_{n,i}(r)}{r^{\sigma_n}}-\frac{\sigma_nC(r)r^{-N+2s-2\sigma_n}}{N-2s+2\sigma_n} +\frac{N-2s+\sigma_n}{N-2s+2 \sigma_n}\int_{\lambda}^{r}\rho^{-N-1+2s-\sigma_n}\Upsilon_{n,i}(\rho )\, d\rho\Bigg)\nonumber\\ &\quad+\frac{\sigma_n\lambda^{-N+2s-\sigma_n}}{N-2s+2\sigma_n}\left(C(r)+\int_{\lambda}^r t^{\sigma_n-1}\Upsilon_{n,i}(t) \, dt\right), \end{align}

for any $\lambda \in (0,r_0]$.

Let n ₀ be as in (105) and $i \in \{1,\dots,M_{\alpha,k,n_0}\}$. By (105) and (109), $\gamma=\sigma_{n_0}$ and

\begin{align*} &\lambda^{-N-1+2s-\gamma}\left|\Upsilon_{n_0,i}(\lambda)\right| \lt c_{N,s}\lambda^{-N-1+2s-\gamma} \int_{B'_\lambda} |g| |\mathop{\rm{Tr}}(U)\ \left|\mathop{\rm{Tr}}\left(Z_{\alpha,k,n,i}\left(\frac{\cdot}{|\cdot|}\right)\right)\right|\, dx \\ &\le \lambda^{-N-1+2s-\gamma}\left(\int_{B'_\lambda} |g| |\mathop{\rm{Tr}}(U)|^2 dx \right)^\frac12 \left(\int_{B'_\lambda} |g| \left|\mathop{\rm{Tr}}\left(Z_{\alpha,k,n,i}\left(\frac{\cdot}{|\cdot|}\right)\right)\right|^2 dx \right)^\frac12 \\ &\le k_{N,s,g}\lambda^{-N-1+2s-\gamma+\varepsilon}\left(\int_{B_\lambda^+}y^{1-2s}|\nabla U|^2\, dz-\int_{B_\lambda^+}y^{1-2s}\frac{\alpha}{|x|_k^2}U^2\, dz\right.\\ &\qquad\qquad \left.+\frac{N-2s}{2\lambda}\int_{S_\lambda^+}y^{1-2s}U^2\, dz\right)^\frac12\\ &\times \Bigg(\int_{B_\lambda^+}y^{1-2s}|\nabla Z_{\alpha,k,n,i}(z/|z|)|^2\, dz-\int_{B_\lambda^+}y^{1-2s}\frac{\alpha}{|x|_k^2} |Z_{\alpha,k,n,i}(z/|z|)|^2\, dz \\ &+\frac{N-2s}{2\lambda}\int_{S_\lambda^+}y^{1-2s} |Z_{\alpha,k,n,i}(z/|z|)|^2\, dz\Bigg)^\frac12 \\ &=k_{N,s,g}\lambda^{-1-\gamma+\varepsilon}\sqrt{H(\lambda)}\left(\int_{B_1^+}y^{1-2s}|\nabla V^\lambda|^2\, dz-\int_{B_1^+}y^{1-2s}\frac{\alpha}{|x|_k^2}|V^\lambda|^2\, dz+\frac{N-2s}{2}\right)^\frac12\\ &\times\left(\int_{B_1^+}y^{1-2s}|\nabla Z_{\alpha,k,n,i}(z/|z|)|^2\, dz-\int_{B_1^+}y^{1-2s}\frac{\alpha}{|x|_k^2}| Z_{\alpha,k,n,i}(z/|z|)|^2\, dz+\frac{N-2s}{2}\right)^\frac12\\ &\qquad\qquad \le {\rm{const}}\;\lambda^{-1+\varepsilon} \end{align*}

for any $\lambda \in (0,r_0]$, by Holder inequality, a change of variables, (3), (42), (82), (85), (86), (104). Hence

(112)\begin{equation} \left|\Upsilon_{n_0,i}(\lambda)\right|\le {\rm{const}}\ \lambda^{N-2s+\gamma+\varepsilon} \quad \text{for any } \lambda \in (0,r_0]. \end{equation}

Now we show that for any $r \in (0,r_0]$:

(113)\begin{equation} C(r)+\int_{0}^{r}\lambda^{-1+\gamma}\Upsilon_{n_0,i}(\lambda) d\lambda=0. \end{equation}

From (112) it is clear that $\int_{0}^{r_0}\lambda^{-1+\gamma}\Upsilon_{n_0,i}(\lambda) d\lambda \lt +\infty$. We argue by contradiction. Since $\sigma_{n_0}=\gamma \gt -\frac{N-2s}{2}$ by (105) and (109), then from (111) we deduce that:

\begin{equation*} \varphi_{n_0,i}(\lambda)\sim\frac{\gamma\lambda^{-N+2s-\gamma}}{N-2s+2\gamma}\left(C(r)+\int_{\lambda}^r t^{-1+\gamma}\Upsilon_{n_0,i}(t) \, dt\right) \quad \text{as } \lambda \to 0^+ \end{equation*}

and so by (105):

(114)\begin{equation} \int_{0}^{r_0}\lambda^{N-1-2s}|\varphi_{n_0,i}(\lambda)|^2 \, d \lambda =+\infty. \end{equation}

On the other hand by Hölder inequality, a change of variables, (103) and [Reference Fall and Felli13, Lemma 2.4]:

\begin{align*} & \int_{0}^{r_0}\lambda^{N-1-2s}|\varphi_{n_0,i}(\lambda)|^2 \, d \lambda \le\int_{0}^{r_0}\lambda^{N-1-2s}\left(\int_{\mathbb{S}^+}\theta_{N+1}^{1-2s} |U(\lambda\theta)|^2 \,dS \right) d \lambda \\ & = \int_{B^+_{r_0}} y^{1-2s} \frac{U^2}{|z|^2} \, dz \lt +\infty, \end{align*}

which contradicts (114). It follows that:

(115)\begin{align} & \lambda^{-N+2s-\gamma}\left|C(r)+\int_{\lambda}^{r}\lambda^{-1+\gamma}\Upsilon_{n_0,i}(\lambda) d\lambda\right|=\lambda^{-N+2s-\gamma}\left|\int_{0}^{\lambda}\lambda^{-1+\gamma}\Upsilon_{n_0,i}(\lambda) d\lambda\right|&\nonumber\\ &= O(\lambda^{\gamma+\varepsilon}), \end{align}

in view of (112). In conclusion (106) follows from (111), (113), and (115).

Proof. From (103), since $\{Z_{\alpha,k,n}\}_{n \in \mathbb{N}\setminus\{0\}}$ is a orthonormal basis of $L^2(\mathbb{S}^+,\theta_{N+1}^{1-2s})$, see $\S$ 3.1, we have that:

(116)\begin{equation} H(\lambda)=\int_{\mathbb{S}^+}\theta_{N+1}^{1-2s}|U(\lambda \theta)|^2 \, dS=\sum_{n=1}^\infty \sum_{i=1}^{M_{\alpha,k,n}} |\varphi_{n,i}(\lambda)|^2, \end{equation}

by (69) and a change of variables. We argue by contradiction supposing that:

\begin{equation*} \lim_{\lambda \to 0^+}\lambda^{-2\gamma}H(\lambda)=0. \end{equation*}

Let n ₀ be as in (105). By (116) for any $i\in\{1,\dots,M_{\alpha,k,n_0}\}$,

\begin{equation*} \lim_{\lambda \to 0^+}\lambda^{-2\gamma} |\varphi_{n_0,i}(\lambda)|^2=0. \end{equation*}

By (106), for any $i\in\{1,\dots,M_{\alpha,k,n_0}\}$ and any $r \in (0,r_0]$

(117)\begin{align} \frac{\varphi_{n,i}(r)}{r^{\gamma}}+\frac{\gamma r^{-N+2s-2\gamma}}{N-2s+2\gamma}\int_{0}^{r}\rho^{-1+\rho}\Upsilon_{n_0,i}(\rho)& d\rho\nonumber\\ &+\frac{N-2s+\gamma}{N-2s+2 \gamma}\int_{0}^{r}\rho^{-N-1+2s-\gamma}\Upsilon_{n,i}(\rho )\, d\rho=0. \end{align}

Hence by (106), (112) and (117):

\begin{equation*} \varphi_{n,i}(\lambda)=-\lambda^\gamma\frac{N-2s+\gamma}{N-2s+2 \gamma}\int_{0}^{\lambda}\rho^{-N-1+2s-\gamma}\Upsilon_{n,i}(\rho )\, d\rho+O(\lambda^{\gamma +\varepsilon})=O(\lambda^{\gamma +\varepsilon}), \end{equation*}

as $\lambda \to 0^+$ for any $i\in\{1,\dots,M_{\alpha,k,n_0}\}$. In view of (69) and (85), it follows that:

\begin{equation*} \sqrt{H(\lambda)}\int_{\mathbb{S}^+} \theta_{N+1}^{1-2s} V^\lambda Z \, dS=O(\lambda^{\gamma +\varepsilon}) \quad \text{as } \lambda \to 0^+, \end{equation*}

for any $Z \in V_{n_0}$, see (36). Then, in view of (83) with $\sigma=\frac{\varepsilon}{2}$,

(118)\begin{equation} \int_{\mathbb{S}^+} \theta_{N+1}^{1-2s} V^\lambda Z \, dS=O(\lambda^{\frac{\varepsilon}{2}}) \quad \text{as } \lambda \to 0^+, \end{equation}

for any $Z \in V_{n_0}$. On the other hand by Proposition 6.5 and Proposition 3.1, there exist $Z_0\in V_{n_0}$ with $\left\lVert Z_0\right\rVert_{L^2(\mathbb{S}^+,\theta_{N+1}^{1-2s})}=1$ and a sequence $\lambda_q \to 0^+$ as $q \to \infty$ such that:

(119)\begin{equation} \quad V^{\lambda_q} \to Z_0 \quad \text{strongly in } L^2(\mathbb{S}^+,\theta_{N+1}^{1-2s}) \,\text{as } q \to \infty. \end{equation}

Since $Z_0 \in V_{n_0}$, from the Parseval identity, (118), and (119) we deduce that $Z_0\equiv 0$ which contradicts the fact that $\left\lVert Z_0\right\rVert_{L^2(\mathbb{S}^+,\theta_{N+1}^{1-2s})}=1$.

Proof. In view of (80) and Proposition 6.5 we know that (120) holds for some $n \in \mathbb{N}\setminus\{0\}$. Furthermore for any sequence of strictly positive numbers $\lambda_p \to 0^+$ as $p \to \infty$ there exist a subsequence $\lambda_{p_q} \to 0^+$ as $q \to \infty$ and real numbers $\beta_1,\dots,\beta_{M_{\alpha,k,n}}$ such that:

(123)\begin{equation} \frac{U(\lambda z)}{\lambda^\gamma} \to |z|^\gamma\sum_{i=1}^{M_{\alpha,k,n}} \beta_i Z_{\alpha,k,n,i}(z/|z|) \quad \text{strongly in } H^1(B_1^+,y^{1-2s}) \,\text {as } q \to \infty^+, \end{equation}

taking into account Proposition 6.5 and (38). We claim that for any $i \in \{1,\dots,M_{\alpha,k,n}\}$ the number β_i does not depend neither on the sequence $\lambda_p \to 0^+$ nor on its subsequence $\lambda_{p_q}\to 0^+$. In view of (38), (103), (123) and Proposition 3.1

\begin{align*} &\lim_{q \to \infty}\lambda_{p_q}^{-\gamma}\varphi_{n,j}(\lambda_{p_q})=\lim_{q \to \infty}\int_{\mathbb{S}^+} \theta_{N+1}^{1-2s} \lambda_{p_q}^{-\gamma}U(\lambda_{p_q}\theta) Z_{\alpha,k,n,j}(\theta) \, dS\\ &\quad=\sum_{i=1}^{M_{\alpha,k,n}} \beta_i\int_{\mathbb{S}^+} \theta_{N+1}^{1-2s} Z_{\alpha,k,n,i} Z_{\alpha,k,n,j}\, dS =\beta_j, \end{align*}

for any $j \in \{1,\dots,M_{\alpha,k,n}\}$. On the other hand for any $r \in (0,r_0]$

\begin{align*} \lim_{q \to \infty}\lambda_{p_q}^{-\gamma}\varphi_{n,j}(\lambda_{p_q})=\frac{\varphi_{n,j}(r)}{r^{\gamma}}+\frac{\gamma r^{-N+2s-2\gamma}}{N-2s+2\gamma}\int_{0}^{r}\rho^{-1+\rho}\Upsilon_{n,j}(\rho) d\rho\\ +\frac{N-2s+\gamma}{N-2s+2 \gamma}\int_{0}^{r}\rho^{-N-1+2s-\gamma} \Upsilon_{n,j}(\rho )\, d\rho, \end{align*}

by (106). Hence

(124)\begin{align} \beta_j & =\frac{\varphi_{n,j}(r)}{r^{\gamma}}+\frac{\gamma r^{-N+2s-2\gamma}}{N-2s+2\gamma}\int_{0}^{r}\rho^{-1+\rho}\Upsilon_{n,j}(\rho) d\rho\nonumber\\ & +\frac{N-2s+\gamma}{N-2s+2 \gamma}\int_{0}^{r}\rho^{-N-1+2s-\gamma} \Upsilon_{n,j}(\rho )\, d\rho, \end{align}

for any $j \in \{1,\dots,M_{\alpha,k,n}\}$ and in particular β_j does not depend neither on the sequence $\lambda_p \to 0^+$ nor on its subsequence $\lambda_{p_q}\to 0^+$. Then by (124) and the Urysohn Subsequence Principle we conclude that (121) holds, thus completing the proof.

Proof of Corollary 2.12 and Corollary 2.13

We start by proving Corollary 2.12. Let U be a solution of (21) such that (26) holds and assume by contradiction that $U\not\equiv 0$ on $\Omega \times (0,\infty)$. Let γ be as in Theorem 6.8. Then there exists a sequence $\lambda_q\to 0^+$ such that:

\begin{equation*} \lim_{q \to \infty}\lambda_q^{-\gamma} U(\lambda_q z)=0, \quad \text{for a.e } z \in B_1^+. \end{equation*}

On the other hand by Theorem 2.9 there exists an eigenfunction Z of (22) such that:

\begin{equation*} \lim_{q \to \infty}\lambda_q^{-\gamma} U(\lambda_q z)=|z|^{\gamma}Z(z/|z|), \quad \text{for a.e. } z \in B_1^+, \end{equation*}

up to a further subsequence, which is a contradiction. Arguing in the same way, we can deduce Corollary 2.13 from Theorem 2.11, taking into account Remark 3.5.

Proof. Let $Y_{\alpha,k,1}$ be the first eigenfunction of (5) defined in $\S$ 2. In particular $Y_{\alpha,k,1}$ is positive. By [Reference Felli, Ferrero and Terracini17, Theorem 1.1] there exists an eigenfunction Ψ of problem (23), corresponding to the first eigenvalue $\eta_{\alpha,k,1}$, such that:

(125)\begin{equation} \lambda^{\frac{N-2}{2}-\sqrt{\left(\frac{N-2}{2}\right)^2+\eta_{\alpha,k,1}}}Y_{\alpha,k,1}(\lambda x) \to |x|^{-\frac{N-2}{2}+\sqrt{\left(\frac{N-2}{2}\right)^2+\eta_{\alpha,k,1}}} \Psi\left(\frac{x}{|x|}\right), \end{equation}

strongly in $H^1(B_1')$ as $\lambda \to 0^+$, since $Y_{\alpha,k,1}$ is positive. Furthermore for any $\phi \in C^\infty_c(\Omega)$:

\begin{equation*} \sideset{_{(\mathbb{H}_{\alpha,k}^s(\Omega))^*}}{_{\mathbb{H}_{\alpha,k}^s(\Omega)}}{\mathop{\left\langle L_{\alpha,k}^s Y_{\alpha,k,1}, \phi \right\rangle}} = (Y_{\alpha,k,1},\phi)_{\mathbb{H}_{\alpha,k}^s(\Omega)}= \mu_{\alpha,k,1}^s \int_{\Omega}Y_{\alpha,k,1} \phi \, dx, \end{equation*}

in view of (8), that is $Y_{\alpha,k,1}$ is weak solution of $L_{\alpha,k}^s Y_{\alpha,k,1}= \mu_{\alpha,k,1}^sY_{\alpha,k,1}$ in the sense given by (12). Let U be the extension of $Y_{\alpha,k,1}$ provided by Theorem 2.7. Since $Y_{\alpha,k,1}$ is positive then $|U|$ is the only solution to the minimisation problem (19) and so we conclude that U is positive. Then, in view of by Theorem 6.8 and Theorem 6.9,

(126)\begin{equation} \lambda^{\frac{N-2s}{2}-\sqrt{\left(\frac{N-2s}{2}\right)^2+\gamma_{\alpha,k,1}}}Y_{\alpha,k,1}(\lambda x)\to |x|^{-\frac{N-2s}{2}+\sqrt{\left(\frac{N-2s}{2}\right)^2+\gamma_{\alpha,k,1}}} \beta_1 \mathop{\rm{Tr}}(Z_{\alpha,k,1}((\cdot/|\cdot|))(x), \end{equation}

strongly in $H^s(B_1')$ as $\lambda \to 0^+$. Putting together (125) and (126) we obtain:

\begin{equation*} -\frac{N-2s}{2}+\sqrt{\left(\frac{N-2s}{2}\right)^2+\gamma_{\alpha,k,1}}=-\frac{N-2}{2}+\sqrt{\left(\frac{N-2}{2}\right)^2+\eta_{\alpha,k,1}} \end{equation*}

thus (24) follows from a direct computation. Finally, if k = N, problem (23) reduces to:

\begin{equation*} -\Delta_{\mathbb{S}'}\Psi-\alpha\Psi= \eta \Psi \quad \text{in } \mathbb{S}', \end{equation*}

which admits $-\alpha$ as first eigenvalue, hence we have proved (25) in view of (24).