Proof. We assume ${\underline U}$ is a strict subsolution, the other case follows the same lines. Let $\Omega _n = \{x\in \Omega : \text {dist}(x,\,\partial \Omega )>1/n\}$. For $k,\, n \in \mathbb {N}$, let $w_{n, k}$ be a continuous function in $\bar \Omega \setminus \Omega _{n + k + 1}$ such that $w_{n,k} = \varphi$ on $\partial \Omega$, $w_{n,k} = {\underline U}$ on $\partial \Omega _{n + k + 1}$ (say, the harmonic function in $\Omega \setminus \bar \Omega _{n + k+ 1}$ satisfying the mentioned boundary conditions).

Now, let $U_{n,k} : \Omega _{n}^c \to \mathbb {R}$ given by

\[ U_{n,k}(x) = \left \{ \begin{array}{@{}ll} {\underline U}(x) \quad & \mbox{if} \ x \in \Omega_{n + k + 1} \setminus \Omega_n, \\ \min\{ w_{n,k}(x), {\underline U}(x) \} \quad & \mbox{if} \ x \in \Omega \setminus \Omega_{n + k + 1}, \\ \varphi(x) & \mbox{if} \ x \in \Omega^c. \end{array} \right . \]

Notice that $U_{n,k}$ is continuous, and $\min _{\partial \Omega } \{ \varphi \} \leq U_{n, k} \leq {\underline U}$ in $\Omega \setminus \Omega _{n + k + 1}$, from which, by dominated convergence theorem, we have that

\[ \int_{\Omega \setminus \Omega_{n + k + 1}} |U_{n,k}(y) - {\underline U}(y) |K(x - y) {\rm d}y \to 0, \]

as $k \to \infty$, uniformly in $x \in \Omega _n,\, K \in \mathcal {K}$, for $n$ fixed. Then, since ${\underline U}$ is a strict subsolution, the above estimate implies that for each $n$, there exists $k(n)$ such that, for each $k \geq k(n)$, the function $U_{n,k}$ is a viscosity subsolution to the Dirichlet problem

(2.4)\begin{align} \left \{ \begin{aligned} -\mathcal{I} u+|Du|^p+\lambda u & =f - \dfrac{1}{n} \quad & & \text{in }\Omega_n, \\ u & = U_{n,k} \quad & & \text{in }\Omega_n^c. \end{aligned} \right . \end{align}

By a similar argument, using that $\bar U$ is a viscosity supersolution for the problem in $\Omega$, we can construct $\bar U_{n,k} \in C(\mathbb {R}^N)$ a supersolution to (2.4) with $\bar U_{n,k} = \bar U$ in $\Omega _{n + k + 1}$, $U_{n,k} \leq \bar U_{n,k}$ in $\Omega$, and such that $\bar U_{n,k} = \varphi$ in $\Omega ^c$, for all $k \geq k(n)$ (relabelling $k(n)$ if necessary). Thus, by Theorem 1 in [Reference Barles, Chasseigne and Imbert3], there exists a unique viscosity solution $u_{n,k} \in C(\mathbb {R}^N)$ for (2.4). Moreover, it satisfies ${\underline U} \leq u_{n,k} \leq \bar U$ in $\Omega _{n + k + 1}$, for all $n$ and $k \geq k(n)$, and by construction we have $u_{n,k} \in L^1_\omega (\mathbb {R}^N)$ uniformly in $n$ and $k \geq k(n)$. Comparison principles are available by the assumption $\lambda > -\lambda _0 (\mathcal {I})$ for all $n,\, k$ large.

Thus, the family $\{ u_{n,k(n)} \}_{n \in \mathbb {N}}$ has uniform interior $C^{\alpha }$ estimates by the results of [Reference Guy, Emmanuel and Cyril4]. Using this and that the family is uniformly bounded in compact sets of $\Omega$, we can use stability results of viscosity solutions to conclude the result, taking $n \to +\infty$.

Proof of proposition 3.1. Proof of proposition 3.1

For an arbitrary linear operator $L = L_K$ in the family, and for $\eta$ as in lemma 3.2, we write

(3.4)\begin{equation} L d^\tau (x) = L[Q_\eta] d^\tau(x) + L[Q_\eta^c] d^\tau(x). \end{equation}

It is easy to see that if $\rho < \eta /4$, we have

\[ |L[Q_\eta^c] d^\tau(x)| \leq \int_{Q_\eta^c} |d^\tau(x + y) - \rho^\tau|K(y){\rm d}y \leq C\Lambda (c_\tau \eta^{-(N + 2s)} + \rho^\tau \eta^{{-}2s}), \]

for some $C > 0$ just depending on $N,\, s,\, \Omega$.

From now on, we concentrate on $L[Q_\eta ] d^\tau (x)$ in (3.4). For each $z = (z',\, z_N) \in \Omega \cap Q_\eta$, we have

\[ d(z) \leq z_N - \psi(z'), \]

from which we directly have

(3.5)\begin{equation} L d^\tau (x) \geq I, \end{equation}

with $I$ as in lemma 3.2.

On the other hand, by the smoothness of the domain, we use Lemma 3.1 in [Reference Chen, Felmer and Quaas9], from which we get the existence of a constant $C_\Omega > 0$ just depending on $\Omega$ and $N$ such that

\[ d(y) \geq (y_N - \psi(y'))(1 - C_\Omega|y'|^2), \]

for $y$ close to the boundary, near $x_0$. Thus, taking $\eta$ small enough in terms of $C_\Omega$, we also have

\[ L d^\tau (x) \leq \lim_{\epsilon \to 0^+} \int_{Q_\eta \setminus B_\epsilon} [(\rho + y_N - \psi(y'))_+^\tau (1 + C_\Omega |y'|^2) - \rho^\tau] K(y) {\rm d}y, \]

and from here, it is easy to see that

(3.6)\begin{equation} L d^\tau (x) \leq I + C_\Omega \int_{Q_\eta} (\rho + y_N - \psi(y'))^\tau_+ |y'|^2 K(y){\rm d}y. \end{equation}

Now, for the second term in the last expression, we can write

\begin{align*} & \int_{Q_\eta} (\rho + y_N - \psi(y'))^\tau_+ |y'|^2 K(y){\rm d}y \\ & \quad\leq \Lambda \int_{B_\eta'} |y'|^{2 - N - 2s} \int_{\psi(y') - \rho}^{\eta} (\rho + y_N - \psi(y'))^\tau {\rm d}y_N {\rm d}y' \\ & \quad\leq \Lambda \frac{1}{1 + \tau} (\rho + \eta/4)^{1 + \tau}\int_{B_\eta'} |y'|^{2 - N - 2s} {\rm d}y' \\ & \quad\leq C \frac{\eta^{1 + \tau}}{1 + \tau}, \end{align*}

where we have used the fact that $\psi (z') \leq \eta /4$ for $|z'| \leq \eta$. Hence, replacing in (3.6) and using (3.5), we conclude that

(3.7)\begin{equation} I \leq L d^\tau (x) \leq I + C C_\Omega \frac{1}{1 + \tau}, \end{equation}

where the constant $C > 0$ depends on $N,\, s,\, \Lambda$

Then, by lemma 3.2, we get

\[ L d^\tau(x) = \rho^{\tau - 2s} (c_K(\rho, \tau) + O(\rho^s) + O(\rho^{1 + \tau}) + O(\rho^{2\,s - \tau})). \]

We get from here that

\[ \mathcal{I} d^\tau(x) = \rho^{\tau - 2s} \Bigg{(} \inf_{i \in I} \sup_{j \in J} c_{K_{ij}}(\rho, \tau) + O(\rho^s) + O(\rho^{1 + \tau}) \Bigg{)}, \]

from which the first result follows.

For the last part of the proposition, since the operator is of form (1.11) we have $K_{ij}^\rho = K_{ij}$, then by Lemma 2.1 in [Reference Ros-Oton and Serra17] we arrive at

(3.8)\begin{equation} c(\tau):=\inf_{i \in I} \sup_{j \in J} c_{K_{ij}}(\tau) = \inf_i \sup_j \{ -\tilde a_{ij} \ (-\Delta)_{\mathbb{R}}^s w_\tau (1) \} \end{equation}

where $(\Delta _{\mathbb {R}})^s$ denotes the fractional Laplacian in dimension one, $w_\gamma (t) = t_+^{\tau }$ for $t \in \mathbb {R}$, and

\[ \tilde a_{ij} := \int_{S^{N - 1}} |\theta_N|^{2s} a_{ij}(\theta) d\sigma(\theta), \]

where $\sigma$ denotes the $N-1$ dimensional Hausdorff measure in the unit sphere. Now if we define $c_1(\tau )= (-\Delta _{\mathbb {R}})^s w_\tau (1)$, the qualitative properties follow since the function $c_1$ is strictly concave in (-1,2 s) by proposition 3.1 of [Reference Chen, Felmer and Quaas9] and $c_1(-1^+) =+\infty$, $c_1(2s^-) = +\infty$. Moreover, $c_1(s)= c_1(s-1) = 0$ by Lemma 6.2 of [Reference Ros-Oton and Serra17].

Proof of lemma 3.2. Proof of lemma 3.2

We concentrate on the case $\tau < 0$ since it is the most difficult due to the unboundedness of $d^\tau$.

We write

\[ I = I_0 + \tilde I, \]

with

\begin{align*} I_0& = \mathrm{P.V.} \int_{Q_\eta} [(\rho + y_N)_+^\tau - \rho^{\tau}] K(y){\rm d}y, \\ \tilde I & = \int_{Q_\eta} [(\rho + y_N - \psi(y'))_+^\tau - (\rho + y_N)_+^\tau] K(y){\rm d}y, \end{align*}

where the last integral is well defined since, using (3.2), we have

\[ |(\rho + y_N - \psi(y'))_+^\tau - (\rho + y_N)^\tau| \leq C\rho^{\tau - 2} |y|^2, \]

for all $|y|$ small enough in terms of $\rho$. This is enough to compensate the singularity of the kernel $K$ and pass to the limit as $\epsilon \to 0$ using dominated convergence theorem.

1. Estimate for $\tilde I$. Rescaling, we have

\[ \tilde I = \rho^{\tau - 2s} \int_{Q_{\eta/\rho}} [(1 + z_N - \tilde \psi(z'))_+^\tau - (1 + z_N)_+^\tau] K^\rho(z){\rm d}z, \]

where $\tilde \psi (z') := \rho ^{-1} \psi (\rho z')$ and $K^\rho (z) = \rho ^{N + 2s} K(\rho z)$. Notice that by the ellipticity condition, we have

\[ \gamma |z|^{-(N + 2s)} \leq K^\rho(z) \leq \Lambda |z|^{-(N + 2s)}, \quad z \in \mathbb{R}^N \setminus \{ 0 \}, \ \rho > 0. \]

For this, we divide the integral in several parts. Namely, we consider the splitting

(3.9)\begin{equation} d^{2\,s - \tau} \tilde I = I_1 + I_2 + I_3, \end{equation}

where for $i = 1,\,2,\,3$, we denote

\begin{align*} & I_i = \int_{A_i} [(1 + z_N - \tilde \psi(z'))_+^\tau - (1 + z_N)_+^\tau] K^\rho(z){\rm d}z, \ \mbox{with} \\ & A_1 = B_1' \times \left(-\frac{\eta}{\rho}, \frac{\eta}{\rho}\right), \ A_2 = B_{\frac{\eta}{\sqrt{\rho}}}' \setminus B_1' \times \left(-\frac{\eta}{\rho}, \frac{\eta}{\rho}\right), A_3 = Q_{\frac{\eta}{\rho}} \setminus (A_1 \cup A_2). \end{align*}

For $I_1$, we notice that by the assumptions on the chart $\psi$ (c.f. (3.2)), we have $|\tilde \psi (z')| \leq \rho |z'|^2$ if $|z'| \leq 1$. Then, we perform the subdivision

\begin{align*} & I_1 = I_{11} + I_{12} + I_{13}, \\ & A_{11} = B_1' \times \left(-\frac{\eta}{\rho}, -\frac{1}{2}\right), \ A_{12} = B_1' \times \left(-\frac{1}{2}, \frac{1}{2}\right), \ A_{13} = B_1' \times \left(\frac{1}{2}, \frac{\eta}{\rho}\right), \end{align*}

where we have adopted a similar notation as in (3.9).

For $I_{11}$, we make a subdivision with the form

\begin{align*} I_{11} & \leq \int_{B_1'} \int_{\tilde \psi_-(z') - 1}^{{-}1} (1 + z_N - \tilde \psi(z'))^\tau K^\rho(z){\rm d}z \\ & \quad+ \int_{B_1'} \int_{\tilde \psi_+(z') - 1}^{{-}1} [(1 + z_N - \tilde \psi_+(z'))^\tau - (1 + z_N)^\tau]K^\rho(z){\rm d}z \\ & = : I_{111} + I_{112}, \end{align*}

where, for $a \in \mathbb {R}$, we have written $a = a_+ + a_-$.

Using the ellipticity condition and integrating by parts, we can write

\begin{align*} I_{112} &\le \Gamma \int_{B_{{1}}^{\prime}} {\int_{{\tilde{\psi }}_ + ({z}^{\prime})-1}^{-1/2} {\displaystyle{{{(1 + z_N-{\tilde{\psi }}_ + ({z}^{\prime}))^\tau} -{(1 + z_N)^\tau} } \over {|z|^{N + 2s}}}} } {\rm d}z_N{\rm d}{z}^{\prime} \\ & = \displaystyle{\Gamma \over {1 + \tau }}\int_{B_{{1}}^{\prime}} {\left\{ {\displaystyle{{{(1 + z_N-{\tilde{\psi }}_ + ({z}^{\prime}))}^{\tau + 1}-{(1 + z_N)}^{\tau + 1}} \over {|z|^{N + 2s}}}\Bigg{|}_{z_N = {\tilde{\psi }}_ + ({z}^{\prime})-1}^{z_N = -1/2} } \right.} {\rm } \\ & \quad + (N + 2{\mkern 1mu} s)\left. {\int_{{\tilde{\psi }}_ + -1}^{-1/2} {\displaystyle{{{(1 + z_N-{\tilde{\psi }}_ + ({z}^{\prime}))}^{\tau + 1}-{(1 + z_N)}^{\tau + 1}} \over {|z|^{N + 2{\kern 1pt} s + 1}}}} z_N{\rm d}z_N} \right\}{\rm d}{z}^{\prime} \\ & \le \displaystyle{{C\Gamma } \over {1 + \tau }}\int_{B_{{1}}^{\prime}} {\left\{ {\displaystyle{{(\tau + 1)\rho |{z}^{\prime}|^2 + \rho ^{\tau + 1}|{z}^{\prime}|^{2(\tau + 1)}} \over {{(1 + |{z}^{\prime}|^2)}^{(N + 2s)/2}}}} \right.} {\rm } \\ & \quad + \left. {\int_{{\tilde{\psi }}_ + ({z}^{\prime})-1}^{-1/2} {\displaystyle{{{(1 + z_N-{\tilde{\psi }}_ + ({z}^{\prime}))}^{\tau + 1}-{(1 + z_N)}^{\tau + 1}} \over {|z|^{N + 2{\kern 1pt} s + 1}}}} z_N{\rm d}z_N} \right\}{\rm d}{z}^{\prime}{\rm } \\ & \le \displaystyle{{C\Gamma } \over {1 + \tau }}\left\{ {\rho ^{\tau + 1} + \int_{B_{{1}}^{\prime}} {\int_{{\tilde{\psi }}_ + ({z}^{\prime})-1}^{-1/2} {} } } \right. \\ & \qquad \qquad \quad \left. { \times \displaystyle{{{(1 + z_N-{\tilde{\psi }}_ + ({z}^{\prime}))^{\tau + 1}}-{(1 + z_N)^{\tau + 1}}} \over {|z|^{N + 2{\kern 1pt} s + 1}}}z_N{\rm d}z_N{\rm d}{z}^{\prime}} \right\},\end{align*}

for some universal constant $C > 0$. Here, we have used that $|\tilde \psi (z')| \leq C \rho |z'|^2$ for $|z'| \leq 1$. Since $1 + \tau > 0$, we see that

\[I_{112} \le \displaystyle{{C\Gamma } \over {1 + \tau }}{\rm }\left\{ {\rho ^{\tau + 1} + \int_{B_{{1}}^{\prime}} {\displaystyle{{\rho |{z}^{\prime}|^2} \over {{(1 + |{z}^{\prime}|^2)^{{{N + 2{\kern 1pt} s + 1} \over 2}}}}}} \int_{{\tilde{\psi }}_ + -1}^{-1/2} {{(1 + z_N)}^\tau } {\rm d}z_N{\rm d}{z}^{\prime}} \right\}, \]

and from this, we conclude that

\[ I_{112} \leq \frac{C \Gamma}{1 + \tau} \rho^{\tau + 1}. \]

For $I_{111}$, by direct integration and the estimates for $\tilde \psi$, we see that

\[ I_{111} \leq C\Gamma \int_{B_1'} |\tilde \psi(z')|^{1 + \tau} {\rm d}z \leq C \Gamma \rho^{1 + \tau}. \]

For the lower bound, we proceed similarly, noticing that this time we can write

(3.10)\begin{equation} \begin{aligned} I_{11} & \geq{-}\int_{B_1'} \int_{{-}1}^{- 1 + \tilde \psi_+(z')} (1 + z_N)^\tau K^\rho(z){\rm d}z \\ & \quad+ \int_{B_1'} \int_{- 1 + \tilde \psi_+(z')}^{{-}1/2} [(1 + z_N - \tilde \psi_+(z'))^\tau - (1 + z_N)^\tau] K^\rho(z){\rm d}z. \end{aligned} \end{equation}

From here, by direct integration in the first term in the last inequality, and a similar procedure leading to the estimate concerning $I_{112}$ above for the second term, we conclude that

(3.11)\begin{equation} -\frac{C\Gamma}{1 + \tau} \rho^{\tau + 1} \leq I_{11} \leq \frac{C \Gamma}{1 + \tau} \rho^{\tau + 1}, \end{equation}

For $I_{12}$, we use that $|\tilde \psi (z')| \leq C \rho |z'|^2$ to perform a first-order Taylor expansion to get

(3.12)\begin{equation} -C \Gamma \rho \leq I_{12} \leq C \Gamma \int_{B_1'} \int_{{-}1/2}^{1/2} \frac{\rho |z'|^2}{|z|^{N + 2s}} {\rm d}z_N {\rm d}z' \leq C \Gamma \rho. \end{equation}

For $I_{13}$, we perform a Taylor expansion again, from which we can write

\begin{align*} I_{13} & \leq C\Gamma \rho \int_{B_1'} |z'|^2 \int_{1/2}^{\eta/\rho} \frac{{\rm d}z_N}{(|z_N|^2 + |z'|^2)^{\frac{N + 2s} {2}}}{\rm d}z' \\ & \leq C\Gamma \rho \int_{B_1'} |z'|^2 |z'|^{-(N + 2\,s) + 1} \int_{0}^{+\infty} \frac{{\rm d}t}{(t^2 + 1)^{\frac{N + 2s} {2}}}{\rm d}z'. \end{align*}

A similar lower bound can be easily obtained, from which we conclude that $-C \Gamma \rho \leq I_{13} \leq C\Gamma \rho.$ Gathering this estimate together with (3.12) and (3.11) lead us to

(3.13)\begin{equation} I_1 = O(\rho^{\tau + 1}). \end{equation}

Now we proceed with $I_{2}$ in (3.9). We write

\begin{align*} & I_2 = I_{21} + I_{22}, \ \mbox{with} \\ & A_{21} = B_{\eta/\sqrt{\rho}}' \setminus B_1' \times \left(-\frac{\eta}{\rho}, -\frac{1}{2}\right), \ A_{22} = B_{\eta/\sqrt{\rho}}' \setminus B_1' \times \left(-\frac{1}{2}, \frac{\eta}{\rho}\right), \end{align*}

where we have adopted a similar notation as in (3.9).

We start by noticing that $|\tilde \psi (z')| \leq C \eta ^2$ when $|z'| \leq \eta /\sqrt {\rho }$, and from here, fixing $\eta > 0$ universally small, we can write

\begin{align*} I_{21} & \leq \int_{B_{\eta/\sqrt{\rho}}' \setminus B_1'} \int_{\tilde \psi_-(z') - 1}^{{-}1} (1 + z_N - \tilde \psi(z'))^\tau K^\rho(z){\rm d}z \\ & \qquad+ \int_{B_{\eta/\sqrt{\rho}}' \setminus B_1'} \int_{\tilde \psi_+(z') - 1}^{{-}1/2} [(1 + z_N - \tilde \psi_+(z'))^\tau - (1 + z_N)^\tau]K^\rho(z){\rm d}z \\ & \quad= : I_{211} + I_{212}. \end{align*}

For $I_{211}$, we have

\begin{align*} I_{211} \leq & C \Gamma \int_{B_{\eta/\sqrt{\rho}}' \setminus B_1'} |z'|^{-(N + 2s)} \int_{\tilde \psi_-(z') - 1}^{{-}1} (1 + z_N - \tilde \psi(z'))^\tau {\rm d}z_N {\rm d}z' \\ \leq & C \Gamma \rho^{\tau + 1} \int_{B_{\eta/\sqrt{\rho}}' \setminus B_1'} |z'|^{-(N + 2\,s) + 2(\tau + 1)} {\rm d}z' \\ \leq & \frac{C}{1 - 2s}\Gamma \rho^{1 + \tau}. \end{align*}

For $I_{212}$, by taking $\eta$ small enough but independent of $\rho$, we have

\begin{align*} I_{21} & \leq C \Lambda \int_{B_{\eta/\sqrt{\rho}}' \setminus B_1'} \frac{1}{(1 + |z'|)^{N+2s}} \int_{\tilde \psi_+(z') - 1}^{{-}1/2} [(1 + z_N - \tilde \psi_+(z'))^\tau\\ & \quad - (1 + z_N)^\tau] {\rm d}z_N {\rm d}z' \\ & \leq C \frac{\Lambda}{1 + \tau} \int_{B_{\eta/\sqrt{\rho}}' \setminus B_1'} \frac{1}{(1 + |z'|)^{N+2s}} \left(\left(\frac{1}{2} - \tilde \psi_+(z')\right)^{\tau + 1}\right.\\ & \quad \left.- \,\left(\frac{1}{2}\right)^{\tau + 1} + \tilde \psi_+(z')^{\tau + 1}\right) {\rm d}z' \\ & \leq C \frac{\Lambda}{1 + \tau} \Bigg{(} \rho \int_{B_{\eta/\sqrt{\rho}}' \setminus B_1'} \frac{|z'|^2}{(1 + |z'|)^{N+2s}} {\rm d}z'\\& \quad + \rho^{\tau + 1} \int_{B_{\eta/\sqrt{\rho}}' \setminus B_1'} \frac{|z'|^{2(\tau + 1)}}{(1 + |z'|)^{N+2s}} {\rm d}z' \Bigg{)} \\ & \leq C \rho^{1 + \tau}, \end{align*}

and from here, collecting the above estimates, we conclude that $I_{21} \leq C \Gamma \rho ^{1 + \tau }$. In a similar fashion as in (3.10) but applied to $I_{21}$, we arrive at the estimate

(3.14)\begin{equation} I_{21} = O(\rho^{1 + \tau}). \end{equation}

For $I_{22}$, taking $\eta$ universally small, there exist $0 < c,\, C < +\infty$ such that

\[ c (1 + z_N)^{\tau - 1} |\tilde \psi_+(z')| \leq (1 + z_N - \tilde \psi_+(z'))^\tau - (1 + z_N)^\tau = C (1 + z_N)^{\tau - 1} |\tilde \psi_+(z')|, \]

from which we can write

\[ I_{22}= O(1) \int_{B_{\eta/\sqrt{\rho}}' \setminus B_1'} \frac{\rho |z'|^2}{(1 + |z'|)^{N+2s}} \int_{{-}1/2}^{+\infty} (1 + z_N)^{\tau - 1} {\rm d}z_N {\rm d}z', \]

and from here we conclude that $I_{22} = O(\rho )$. This together with (3.14) lead us to

(3.15)\begin{equation} I_{2} = O(\rho^{1 + \tau}). \end{equation}

Now we deal with $I_3$. This time we consider the splitting

\begin{align*} & I_3 = I_{31} + I_{32}, \ \mbox{with} \\ & A_{31} = \{ (z', z_N) : z' \in B_{\eta/\rho}' \setminus B_{\eta/\sqrt{\rho}}', \ z_N \in (-\eta/\rho, \tilde \psi_+(z') + 1) \}, \\ & A_{32} = \{ (z', z_N) : z' \in B_{\eta/\rho}' \setminus B_{\eta/\sqrt{\rho}}', \ z_N \in (\tilde \psi_+(z') + 1, \eta/\rho) \}, \end{align*}

where we have adopted the notation in (3.9).

For $I_{31}$, we see that

\begin{align*} I_{31} & \leq \Lambda \int_{B_{\eta/\rho}' \setminus B_{\eta/\sqrt{\rho}}'} \int_{\tilde \psi_+(z') - 1}^{\tilde \psi_+(z') + 1} \frac{(1 + z_N - \tilde \psi_+(z'))^\tau}{|z'|^{N + 2s}}{\rm d}z_N {\rm d}z' \\ & \leq C\Lambda \int_{B_{\eta/\rho}' \setminus B_{\eta/\sqrt{\rho}}'} \frac{1}{|z'|^{N + 2s}} \int_{0}^{2} (1 + t)^\tau {\rm d}t {\rm d}z' \\ & \leq C\Lambda \int_{\eta/\sqrt{\rho}}^{\eta/\rho} r^{-(2 + 2s)}{\rm d}r, \end{align*}

from which we conclude that $I_{31} \leq C\Lambda \rho ^{s + 1/2}$.

On the other hand, since $\tau < 0$, for $I_{32}$, we can write

\begin{align*} I_{32} & \leq \Lambda \int_{B_{\eta/\rho}' \setminus B_{\eta/\sqrt{\rho}}'} \int_{\tilde \psi_+(z') + 1}^{+\infty} \frac{(1 + z_N - \tilde \psi_+(z'))^\tau}{(z_N^2 + |z'|^2)^{(N + 2s)/2}}{\rm d}z_N {\rm d}z' \\ & \leq \Lambda \int_{B_{\eta/\rho}' \setminus B_{\eta/\sqrt{\rho}}'} \int_{1}^{+\infty} \frac{1}{(z_N^2 + |z'|^2)^{(N + 2s)/2}}{\rm d}z_N {\rm d}z' \\ & \leq \Lambda \int_{B_{\eta/\rho}' \setminus B_{\eta/\sqrt{\rho}}'} \frac{1}{|z'|^{N + 2\,s - 1}} \int_{0}^{+\infty} \frac{1}{(t^2 + 1)^{(N + 2s)/2}}{\rm d}t {\rm d}z', \end{align*}

and from here, we conclude that

\[ I_{32} \leq C\Lambda \int_{\eta/\sqrt{\rho}}^{+\infty} r^{-(1 + 2s)} {\rm d}r \leq C \Lambda \rho^s. \]

Then, collecting the previous estimates, we conclude that

\[ I_3 \leq C \rho^s. \]

For the lower bound, we see that

\[ I_3 \geq{-} \int_{B_{\eta/\rho}' \setminus B_{\eta/\sqrt{\rho}}'} \int_{-\frac{\eta}{\rho}}^{{-}1} (1 + z_N)^\tau K^\rho(z){\rm d}z, \]

and arguing similarly as before, we arrive at $I_3 \geq -C \rho ^s$. Hence, we conclude that $I_3 = O(\rho ^s)$. Using this estimate, together with (3.13) and (3.15) and replacing them into (3.9), we conclude that

(3.16)\begin{equation} \tilde I = \rho^{\tau - 2s} (O(\rho^{1 + \tau}) + O(\rho^{s})), \end{equation}

where the $O$-term depends on $N,\, s,\, \Omega,\, \eta,\, \frac {1}{1 + \tau }$.

2. Estimate for $I_0$. Rescaling, we have

\begin{align*} I_0 & = \rho^{\tau - 2s} \mathrm{P.V.} \int_{Q_{\eta/\rho}} [(1 + z_N)_+^\tau - 1] K^\rho(z){\rm d}z \\ & = \rho^{\tau - 2s} \mathrm{P.V.} \int_{\mathbb{R}^N} [(1 + z_N)_+^\tau - 1] K^\rho (z){\rm d}z + \rho^{\tau - 2s} \int_{Q_{\eta/\rho}^c} [(1 + z_N)_+^\tau - 1] K^\rho(z){\rm d}z\\ & = : \rho^{\tau - 2s} c_K(\rho, \tau) + I_{01}. \end{align*}

Notice that

\[{-}C \Gamma \rho^{\tau - 2s} \rho^{2s} \leq I_{01}. \]

On the other hand

\begin{align*} I_{01} & \leq \Gamma \rho^{\tau - 2s} \int_{Q_{\eta/\rho}^c \cap \{ z_N >{-}1 \} } (1 + z_N)^{\tau} |z|^{-(N + 2s)} {\rm d}z \\ & \leq C\Gamma \rho^{- 2s} \int_{B_{\eta/\rho}'} \int_{\eta/\rho}^{+\infty} |z|^{-(N + 2s)}{\rm d}z + C\Gamma \rho^{\tau - 2s} \frac{1}{1 + \tau}\int_{B_{\eta/\rho}'^c} |z'|^{-(N + 2s)}{\rm d}z' \\ & \leq C \Gamma (1 + \rho^{1 + \tau}). \end{align*}

Summarizing, we have

\[ I_{01} = O(\rho^\tau). \]

Joining this together with (3.16), we conclude the result.

Proof of theorem 1.1. Proof of theorem 1.1

We prove each case separately. We give the general remark that both sub and supersolutions we construct here are strict.

Case 1: For $t > 0$ and $s-1 < \gamma < 2\,s - 1< s$ to be fixed, we consider

\[ U^-_t=td^{s-1} - C_1d^{\gamma}, \]

for some $C_1>0$ to be chosen.

Then, using proposition 3.1 together with assumption (1.11) , we have $c(s\!-\!1) = 0$ and $c^+(\gamma )<0$. Then using (4.1), we find

\begin{align*} -\mathcal{I} U^-_t +|DU^-_t|^p & \leq{-}\mathcal{I}(td^{s-1})+C_1 \mathcal{M}^+(d^\gamma)+|DU^-_t|^p \\ & \leq tO(d^{{-}1})+C_1c^+(\gamma) d^{\gamma-2s}+|(t (s-1) d^{s - 2} - C_1 \gamma d^{\gamma - 1}) Dd|^p \\ & = tO(d^{{-}1})+C_1c^+(\gamma) d^{\gamma-2s} \\ & \quad+|t (s-1)|^p d^{(s - 2)p}|1 - C_1 t^{{-}1} (s-1)^{{-}1}\gamma d^{\gamma + 1 - s}|^p. \end{align*}

Since $p < p_2$, we can take $\gamma > s - 1$ such that $(s - 2)p > \gamma -2\,s$ (notice that if $p < p_0$, then $\gamma$ can be taken positive).

For such a $\gamma$, we take

\[ \bar C_1 = |t (s-1)|^p/|c^+(\gamma)|, \]

to conclude that there exists $\bar c_1 > 0$ such that, for every $\epsilon > 0$, we take $C_1 = \bar C_1 - \epsilon$ in the expression above to obtain that

\[ -\mathcal{I} U^-_t +|DU^-_t|^p \leq tO(d^{{-}1}) - \bar c_1 \epsilon d^{\gamma - 2s}, \]

for each $d \leq d_\epsilon$ for some $d$ small enough in terms of $\epsilon,\, t,\, C_1,\, \gamma.$ Since $s > 1/2$, and $\gamma < 2s-1$, we can take $d_\epsilon$ smaller to conclude that

\[ -\mathcal{I} U^-_t +|DU^-_t|^p +{\leq}{-} \frac{\bar c_1 \epsilon}{2} d^{\gamma - 2s} \leq{-} \| f \|_\infty , \quad \mbox{for} \ d(x) \leq d_\epsilon, \]

and therefore we have constructed a subsolution near the boundary. A straightforward computation tells us that the function

\[ U_t^-{-} C\chi_\Omega \]

is a viscosity subsolution to the problem in $\Omega$, when $C = C_\epsilon$ is taken large . Thus, we have constructed the subsolution.

In a similar way, we can construct a supersolution in $\Omega$ with the form

\[ U^+_t=td^{s-1} + (\bar C_1+ \epsilon) d^{\gamma} + \mathcal{C}_\epsilon \chi_\Omega. \]

We can apply the Perron method from § 2 to conclude the existence of a solution to problem (P) satisfying the desired rate near the boundary, where $\theta = \gamma + 1 - s$. If $p < p_0$, then we can take $\gamma > 0$ in the above analysis.

Similar arguments hold for $t<0$.

Case 2: Let $\beta$ as in the statement of the theorem. Hence, $-1< \beta < s-1$ and for some $s-1 < \gamma < 0$ and $T,\, C_1 > 0$ to be fixed, denote

\[ U^-{=} Td^\beta - d^\gamma. \]

We invoke proposition 3.1 again, noticing that if (1.11) holds, and by choice of $\beta$ and range of $\gamma$, we have $c(\beta )<0$ and $c^+(\gamma )<0$. Then using again (4.1), we find

\begin{align*} -\mathcal{I} U^-{+}|DU^-|^p & \leq{-}c(\beta) T d^{\beta - 2s} + O(d^{\beta - s}) + c^+(\gamma) d^{\gamma - 2s} \\ & \quad + (T |\beta|)^p d^{(\beta - 1)p}|1 - T^{{-}1} \beta^{{-}1} \gamma d^{\gamma - \beta}|^p \\ & \leq ({-}c(\beta) T + (T |\beta|)^p) d^{\beta - 2s} + O(d^{\beta - s}) + c^+(\gamma) d^{\gamma - 2s}. \end{align*}

Then, taking $T = \bar T$ such that

\[ c(\beta) \bar T = \bar T^p |\beta|^p \]

concludes that

\[ -\mathcal{I} U^-{+} |DU^-|^p \leq O(d^{\beta - s})+ c^+(\gamma) d^{\gamma - 2s}, \]

in a neighbourhood of $\partial \Omega$. Note that it is possible to fix $\gamma$ close enough to $s - 1$ in order to have $\gamma - 2\,s < \beta - s$, and then, for each point close to the boundary, we have

\[ -\mathcal{I} U^-{+} |DU^-|^p \leq \frac{c^+(\gamma)}{2} d^{\gamma - 2s}. \]

We conclude the existence of a subsolution in $\Omega$ in the same way as before.

For the supersolution, we consider

\[ U^+{=} \bar T d^{\beta} + d^\gamma, \]

and proceed as before.

Case 3: The proof is similar to the previous case, but we provide the details for completeness. Since $p>p_2$, we have $s-1<\beta <0$. For $T,\, C>0$ and $\gamma \in (0,\,2s-1)$ to be fixed, we define

\[ U ={-}Td^{\beta} - Cd^\gamma. \]

Notice that in this case $c(\gamma ) < 0$.

As before, writing $\tilde{\mathcal {I}}$ as the operator $- \mathcal {I} (- \cdot )$, using proposition 3.1 and (1.11), together with (4.1), we have that

\begin{align*} -\mathcal{I} U + |DU|^p & \leq T \tilde {\mathcal{I}} (d^\beta ) + C \mathcal{M}^+ (d^\gamma) + |T \beta d^{\beta - 1} + C\gamma d^{\gamma - 1}|^p \\ & \leq d^{\beta - 2s} (T \tilde c(\beta) + O(d^s)) + C c^+(\gamma) d^{\gamma - 2s} + |T \beta d^{\beta - 1} + C\gamma d^{\gamma - 1}|^p, \end{align*}

where $\tilde c(\beta ) < 0$ since $s-1<\beta <0$. At this point, we notice that since $\beta < 0$ and $C,\, \gamma > 0$, we have

\[ |T \beta d^{\beta - 1} + C\gamma d^{\gamma - 1}| \leq T |\beta| d^{\beta - 1}, \]

for all $d = d(x)$ small enough. Using this, we get that

(4.2)\begin{equation} -\mathcal{I} U + |DU|^p \leq d^{\beta - 2s} (T \tilde c(\beta) + O(d^s)) + C c^+(\gamma) d^{\gamma - 2s} + T^p |\beta|^p d^{(\beta - 1)p}, \end{equation}

and fixing $T = T^* := (-\tilde c(\beta ) |\beta |^p)^{\frac {1}{p - 1}} > 0$, since $\beta - 2\,s = (\beta - 1)p$, we conclude that

\[ -\mathcal{I} U + |DU|^p \leq O(d^{\beta - s}) + Cc^+(\gamma) d^{\gamma - 2s} . \]

Since we also have $\beta - s>-1>\gamma -2\,s$, by similar arguments used before, we can fix $C > 0$ large enough to find that $U$ is a subsolution.

Now, for $C > 0$ and $\gamma > 0$, consider the function $V = -T^* d^\beta + C d^\gamma$. Notice that $U \leq V$. By a similar computation as above, we get that $V$ is a supersolution for the problem.