Qualitative properties of positive solutions to mixed local and nonlocal critical problems in $\mathbb{R}^n$

TL;DR

This work addresses the qualitative behavior of positive solutions to the mixed local-nonlocal critical elliptic equation $-$ $\Delta u$ $+(-\Delta)^s u=\lambda h u^p+u^{2^*-1}$ in $\mathbb{R}^n$ with $n\ge4$, $s\in(0,1)$, $p\in(0,2^*-1)$ and weight $h$. It develops a bootstrapping framework to promote weak solutions to viscosity solutions and, for $p\in[1,2^*-1)$, to classical solutions, obtaining $u\in C^{\alpha}(\mathbb{R}^n)$ for $p\in(0,1)$ and $u\in C^{2,\alpha}(\mathbb{R}^n)$ with a priori bounds $\|u\|_{C^{2,\alpha}(\mathbb{R}^n)}\le C(\|h\|_{C^1}+\|u\|_{L^{\infty}})$. The paper proves power-type decay bounds $C_1/|x|^{n+2s}\le u(x)\le C_2/|x|^{n-2s}$ for large $|x|$ and establishes radial symmetry of all positive solutions via the method of moving planes. A key technical tool is a Fourier-based decomposition of the mixed operator through the fundamental solution $\mathcal{Z}$, obtained by integrating the heat kernel, and its split into $\mathcal{Z}_1+\mathcal{Z}_2$ to control regularity and asymptotics. These results extend the understanding of mixed-order critical problems and provide a foundation for further study of existence and qualitative properties of solutions.

Abstract

We consider the following mixed local and non-local critical elliptic equation: \begin{equation*}\label{0.1} \left\{ \begin{array}{lll} -Δu+(-Δ)^su=λh u^{p}+u^{2^*-1}, &\text{in}\,\, \mathbb{R}^n, u>0, &\text {in} \,\, \mathbb{R}^n, \lim\limits_{|x|\to\infty} u(x) = 0, \end{array} \right. \end{equation*} where $n\geqslant4, \,\, p\in (0,2^*-1),\,\, 2^*:=\frac{2n}{n-2}$ and $h$ is a positive function. We first show the existence and regularity results of viscosity solutions to the above critical elliptic equation. More precisely, from \cite{Su-Xu} weak solutions are obtained and we prove they are indeed viscosity solutions and their regularity is: \( u \in C^α(\mathbb{R}^n) \) for $p\in(0,1);$ \( u \in C^{2,β}(\mathbb{R}^n) \) for $p\in [1, 2^*-1).$ Moreover, for $p\in [1, 2^*-1)$, these viscosity solutions are indeed classical ones and we then prove the existence of positive solutions with the qualitative properties such as the decay estimates and the radial symmetry.

Theorems & Definitions (3)

• Theorem 1.1
• Remark 1.2
• Theorem 1.3: $C^{2,\alpha}$- regularity for $p\in[1,2^*-1)$] Let $u \in \mathcal{X}^{1,2}(\mathbb{R}^n)$ be a weak solution of \ref{['equ1']} with $p\in[1,2^*-1)$. Then $u \in C^{2,\alpha}(\mathbb{R}^n)$ for any $\alpha \in (0, 1)$ and $\Vert u\Vert _{C^{2,\alpha}(\mathbb{R}^n)}\leqslant C \left(\|h\|_{C^{1}(\mathbb{R}^n)}+ \|u\|_{L^\infty(\mathbb{R}^n)} \right)$ for some $C=C(n,s,p,\lambda, \alpha, 2^*).$ Moreover, $\lim_{|x|\to+\infty}u(x)=0.$ Unlike the subcritical problems consider e.g. in MR4808805, here the control constant $C$ in Theorem \ref{['thm:1.2']} is also influenced by the critical exponent. Regarding the regularity of solutions to either the classical Laplace or fractional Laplace equations on $\mathbb{R}^n$, there are already quite a number of results, see e.g. MR2270163CABRE201423MR2863859. Especially, in dipierro2024global, the Hölder regularity for the classical fractional Laplacian equation with perturbations has been established with $p\in(0,1)$. Here, we prove the corresponding Hölder or $C^{2,\alpha}-$regularity for mixed local and nonlocal equations driven by the sum of Laplacian and fractional Laplacian with $p\in(0,2^*-1]$. Finally, we obtain the power-type decay at infinity and the radial symmetry of positive classical solutions to \ref{['equ1']} by comparison arguments and the method of moving planes, respectively. Let $n \geqslant4,\, p\in[1,2^*-1)$ and $s \in (0,1)$. Let $u$ be a positive classical solution of \ref{['equ1']}. Then, there exist constants $C_1, C_2>0$ such that, for every $|x| \geqslant 1$, $\frac{C_1}{|x|^{n+2s}} \leqslant u(x) \leqslant \frac{C_2}{|x|^{n-2s}}.$ Moreover, all positive solutions of \ref{['equ1']} are radially symmetric about some point in $\mathbb{R}^n.$ Note that compared with the power-type decay in SerenaDipierro2025, we could only obtain the order of the upper bound $\frac{1}{ |x|^{n-2s}}$ here rather than $\frac{1}{ |x|^{n+2s}}$ for the Schrödinger equation in SerenaDipierro2025. The rest of this paper is organized as follows. In Section \ref{['sec2']}, we collect some elementary results of $\mathcal{X}^{1,2}(\mathbb{R}^n)$, introduce the functional setting, such as the notions of weak solutions, viscosity solutions and energy functional, and several fundamental properties of the heat kernel. Sections \ref{['sec3']}-\ref{['sec:existence of vs']} are devoted to the proofs of existence of viscosity solutions and their regularity (Theorems \ref{['thm:vis']} and \ref{['thm:1.2']}). In Sections \ref{['sec5']}-\ref{['sec6']}, we establish the power-type decay estimate at infinity and the radial symmetry of positive classical solutions (Theorem \ref{['thm4.8']}). In this section, we will introduce the functional setting for the mixed local and nonlocal critical elliptic problem \ref{['equ1']} and the properties of heat kernel associated to the mixed operator given in \ref{['heat kernel']}. Let $s\in(0,1)$. If $u:\mathbb{R}^n\to \mathbb{R}$ is a measurable function, we set $[u]_s:= \left(\int_{\mathbb{R}^n}|D^su(x)|^2\, dx\right)^{1/2} := \left(\iint_{\mathbb{R}^{2n}}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\, dxdy\right)^{1/2}$ and we refer to $[u]_s$ as the Gagliardo seminorm of $u$ (of order $s$). We define the function space $\mathcal{X}^{1,2}(\mathbb{R}^n) : = \left\{u\in L^{2^*}(\mathbb{R}^n):\triangledown u\in L^2(\mathbb{R}^n)\, \text{and}\, [u]_s< +\infty \right\}$ which is the completion of $C_0^{\infty}(\mathbb{R}^n)$ with respect to the norm $\Vert u\Vert:=\left(\Vert\triangledown u\Vert_{L^2(\mathbb{R}^n)}^2+[u]_s^2\right)^{1/2}, \quad u\in C_0^{\infty}(\mathbb{R}^n).$ In the literature, different notions of solutions are taken into account when dealing with elliptic equations, such as the weak (also called distributional, or variational, or energy) solutions (i.e. the solutions that belong to a suitable Sobolev space and satisfy the equation in a distributional sense, when integrated against a suitable set of test functions) and the viscosity solutions (i.e. all the smooth functions that touch either from above or below the continuous solution are required to be either viscosity subsolutions or viscosity supersolutions). We say that $u \in \mathcal{X}^{1,2}(\mathbb{R}^n)$ is a (weak) solution of \ref{['equ1']} if $\int_{\mathbb{R}^n}\triangledown u(x)\cdot\triangledown \varphi(x)\, dx+\iint_{\mathbb{R}^{2n}}\frac{\left(u(x)-u(y)\right)\left(\varphi(x)-\varphi(y)\right)}{|x-y|^{n+2s}}\, dxdy\,\,\,\,\,=\int_{\mathbb{R}^n}|u|^{2^*-2}u \varphi \, dx+\lambda\int_{\mathbb{R}^n}h|u|^{p-1}u \varphi \, dx$ for any $\varphi\in \mathcal{X}^{1,2}(\mathbb{R}^n).$ We recall that $u \in C(\mathbb{R}^n)$ is a viscosity subsolution of \ref{['equ1']} (more generally, as in MR3161511, one can define the notion of viscosity subsolution and supersolution for semicontinuous functions, so that this class is closed under $sup$ and $inf$). Let $f= \lambda h u^{p}+u^{2^*-1}$ and $x_0 \in \mathbb{R}^n$ and $\phi \in C^2(B_R(x_0))$ be such that $\phi(x_0) = u(x_0) \quad \text{and} \quad \phi \geqslant u \text{ in } B_R(x_0).$ If we define $v(x):= \phi(x)\text{if } x \in B_R(x_0),u(x)\text{otherwise},$ then we have $-\Delta v(x_0)+(-\Delta)^s v(x_0) \leqslant f(x_0, v(x_0))$. Similarly, we say that $u \in C(\mathbb{R}^n)$ is a viscosity supersolution of \ref{['equ1']} if for any $\psi \in C^2(B_R(x_0))$ such that $\psi(x_0) = u(x_0)$ and $\psi \leqslant u$ in $B_R(x_0)$, if we define $w(x) := \psi(x)\text{if } x \in B_R(x_0),u(x)\text{otherwise} ,$ then we have $-\Delta v(x_0)+(-\Delta)^s v(x_0) \geqslant f(x_0, v(x_0))$. A function $u$ is called a viscosity solution of problem \ref{['equ1']} if it is both a viscosity subsolution and a viscosity supersolution. First, we give several properties related to the heat kernels of the mixed local and nonlocal operators. Let $n \geqslant 1$ and $s \in (0, 1)$. Let $\mathcal{H}$ be as defined in \ref{['heat kernel']}. Then, $\mathcal{H}$ is nonnegative, radially symmetric, and nonincreasing with respect to $r = |x|$.There exist positive constants $C_1$ and $C_2$ such that $\mathcal{H}(x,t) \leqslant C_1 \left\{\frac{t}{|x|^{n+2s}} \vee \frac{t^{s}}{|x|^{n+2s}}\right\} \wedge \left\{t^{-\frac{n}{2s}} \wedge t^{-\frac{n}{2}}\right\},$ and $\mathcal{H}(x, t) \geqslant C_2 \frac{t}{|x|^{n+2s}}\text{if } 1 < t < |x|^{2s}e^{\frac{\pi |x|^2}{t}} t^{-\frac{n}{2}}\text{if } |x|^2 < t < |x|^{2s} < 1,$ where $a \land b := \min \{a, b\}$ and $a \lor b := \max \{a, b\}$. Then, define the auxiliary function $\mathcal{Z}(x)=\int_0^{+\infty}\mathcal{H}(x,t)dt.$ We give some basic properties of $\mathcal{Z}(x)$, which are essential for proving Theorem \ref{['thm4.8']}, as summarized below. To start with, we show the existence of the upper bound of $\mathcal{Z}(x)$. Let $n \geqslant 2$ and $s \in (0, 1)$. Then, $\mathcal{Z}(x)\leqslant \frac{c_1}{|x|^{n-2s}}\quad \text{ for } |x| > 1;\mathcal{Z}(x)\leqslant c_2 \frac{1}{|x|^{n-2}}\text{if } n \geqslant 3,2|\ln |x||\text{if } n = 2, \quad \text{ for } |x| \leqslant 1$ for some positive constants $c_1$ and $c_2$ depending on $n$ and $s$. From Theorem \ref{['heat']}, we derive that $0 \leqslant \mathcal{H}(x, t) \leqslant {C_1} \left\{ t^{-\frac{n}{2s}} \wedge \frac{t}{|x|^{n+2s}} \right\} \quad\text{for all}\quad t > 1,$ $\text{and}\quad 0 \leqslant \mathcal{H}(x, t) \leqslant {C_1} \left\{ t^{-\frac{n}{2}} \wedge \frac{t^s}{|x|^{n+2s}} \right\}\quad\text{for all}\quad t \in (0, 1).$ From these formulas, we conclude that, if $|x| > 1$, $\mathcal{Z}(x)= \int_{1}^{+\infty} \mathcal{H}(x, t) dt + \int_{0}^{1} \mathcal{H}(x, t) dt\leqslant {C_1} \left( \int_{1}^{|x|^{2s}} \frac{t}{|x|^{n+2s}} dt + \int_{|x|^{2s}}^{+\infty} t^{-\frac{n}{2s}} dt + \int_{0}^{1} \frac{t^s}{|x|^{n+2s}} dt \right) \leqslant \frac{C_1}{ |x|^{n-2s}},$ and, if $|x| \leqslant 1$, $\mathcal{Z}(x)= \int_1^{+\infty} \mathcal{H}(x,t) dt + \int_0^1 \mathcal{H}(x,t) dt\leqslant {C_1} \left( \int_1^{+\infty} t^{-\frac{n}{2s}} dt + \int_0^{|x|^2} \frac{t^s}{|x|^{n+2s}} dt + \int_{|x|^2}^1 t^{-\frac{n}{2}} dt \right)\leqslant \frac{C_2}{|x|^{n-2}}\text{if } n \geqslant 32 \ln |x|\text{if } n = 2 .$ By combining \ref{['equ:4.1']} with \ref{['equ:4.2']}, we complete the proof. We shall make use of the nonnegativity of $\mathcal{H}$ and SerenaDipierro2025 to prove the strict positivity of $\mathcal{Z}$ and obtain its lower bound. Let $n \geqslant 2$ and $s \in (0, 1)$. Then, $\mathcal{Z}$ is positive and $\mathcal{Z}(x) \geqslant \frac{c_3}{|x|^{n+2s}} \quad \text{if } |x| > 1 \quad \text{and} \quad \mathcal{Z}(x) \geqslant \frac{c_4}{|x|^{n-2}} \quad \text{if } |x| \leqslant 1$ for some positive constants $c_3$ and $c_4$ depending only on $n$ and $s$. From Theorem \ref{['heat']}, we know that $\mathcal{H}(x, t) \geqslant C_2 \frac{t}{|x|^{n+2s}} \quad \text{if } 1 < t < |x|^{2s}.$ Recalling the fact that $\mathcal{H}$ is nonnegative, one deduces that, for any $|x| > 2$, $\mathcal{Z}(x) = \int_{0}^{+\infty} \mathcal{H}(x, t) dt \geqslant C_2 \int_{1}^{|x|^{2s}} \mathcal{H}(x, t) dt \geqslant C_2 \int_{1}^{2} \frac{t}{|x|^{n+2s}} dt \geqslant \frac{C_3}{|x|^{n+2s}}.$ Moreover, in light of Theorem \ref{['heat']}, for every $|x|^2 < t < |x|^{2s} < 1$, $\mathcal{H}(x, t) > C_2 e^{\pi \frac{|x|^2}{t}} t^{-\frac{n}{2}}.$ As a consequence, by the definition of $\mathcal{Z}$, one has that, for every $|x| < \frac{1}{2}$, $\mathcal{Z}(x)\geqslant \int_{|x|^2}^{|x|^{2s}} C_2 e^{\pi \frac{|x|^2}{t}} t^{-\frac{n}{2}} dt\geqslant C_2 \int_{|x|^{2-2s}}^{1} e^{\pi y |x|^{-n+2}} y^{\frac{n}{2}-2} dy\geqslant \frac{ C_2}{|x|^{n-2}} \int_{{\frac{1}{2}}^{2-2s}}^{1} e^{\pi y} y^{\frac{n}{2}-2} dy = \frac{c_{n,s}}{|x|^{n-2}}.$ In addition, we recall that $\mathcal{Z} \geqslant 0$, and $\mathcal{Z}$ is nonincreasing in $r = |x|$. According to \ref{['equ:4.3']} and \ref{['equ:4.4']} , one concludes that $\mathcal{Z}(x) > 0$ for every $x \in \mathbb{R}^n$. To sum up, we can obtain the following theorem. Let $n \geqslant 2$ and $s \in (0, 1)$. Then, $\mathcal{Z}$ is positive and $\frac{c_1}{|x|^{n+2s}}\leqslant\mathcal{Z}(x) \leqslant \frac{c_2}{|x|^{n-2s}} \quad \text{if } |x| > 1,$ and $\frac{c_3}{|x|^{n-2}}\leqslant\mathcal{Z}(x)\leqslant \frac{c_4}{|x|^{n-2}}\text{if } n \geqslant 32 \ln |x|\text{if } n = 2 . \quad \text{if } |x| \leqslant 1$ for some constants $c_1,$ $c_2$, $c_3$ and $c_4$ depending only on $n$ and $s$. Let $n\geqslant 2$ and $\Phi : \mathbb{R}^n \to \mathbb{C}$ be in the Schwartz space. Then, the function $\mathbb{R}^n \times (0, +\infty) \ni (\xi, t) \mapsto e^{-t(|\xi|^{2s} + |\xi|^2 )} \Phi(\xi)$ belongs to $L^1(\mathbb{R}^n \times (0, +\infty), \mathbb{C})$ and $\iint_{\mathbb{R}^n \times (0, +\infty)} e^{-t(|\xi|^{2s} + |\xi|^2 )} \Phi(\xi) d\xi dt = \int_{\mathbb{R}^n} \frac{\Phi(\xi)}{ |\xi|^{2s} + |\xi|^2} d\xi.$ We have that $\int_0^{+\infty} e^{-t(|\xi|^{2s} + |\xi|^2 )} dt = \frac{1}{ |\xi|^{2s} + |\xi|^2}.$ Therefore, by the Fubini-Tonelli's Theorem, \iint_{\mathbb{R}^n \times (0, +\infty)} |e^{-t(|\xi|^{2s}+|\xi|^2)} \Phi(\xi)| d\xi dt= \iint_{\mathbb{R}^n \times (0, +\infty)} e^{-t(|\xi|^{2s}+|\xi|^2)} |\Phi(\xi)| d\xi dt= \int_{\mathbb{R}^n} \frac{|\Phi(\xi)|}{|\xi|^{2s}+|\xi|^2} d\xi, which is finite. □ With this preparatory work, we can now check that $\mathcal{Z}$ is the fundamental solution of the mixed order operator $-\Delta +(-\Delta)^s$, as clarified by the following result. Let $n\geqslant2,$ the Fourier transform of $\mathcal{Z}$ equals $\frac{1}{|\xi|^{2s}+|\xi|^2}$ in the sense of distribution. More explicitly, for every $\phi : \mathbb{R}^n \to \mathbb{R}$ in the Schwartz space, we have that $\int_{\mathbb{R}^n} \mathcal{Z}(x) \phi(x) dx = \int_{\mathbb{R}^n} \frac{\hat{\phi}(\xi)}{|\xi|^{2s}+|\xi|^2} d\xi.$ By \ref{['Bessel']} and the Fubini-Tonelli's Theorem (whose validity is a consequence of Lemma \ref{['lem4.2']}), we have that \int_{\mathbb{R}^n} \mathcal{Z}(x) \phi(x) dx= \int_{\mathbb{R}^n} \left( \int_0^{+\infty} \mathcal{H}(x,t) dt \right) \phi(x) dx= \int_0^{+\infty} \left( \int_{\mathbb{R}^n} \mathcal{H}(x,t) \phi(x) dx \right) dt. Hence, by \ref{['heat kernel']}, $\int_{\mathbb{R}^n} \mathcal{Z}(x) \phi(x) dx = \int_0^{+\infty} \left( \int_{\mathbb{R}^n} \left( \int_{\mathbb{R}^n} e^{-t(|\xi|^{2s}+|\xi|^2)+2\pi i x \cdot \xi} \phi(x) d\xi \right) dx \right) dt.$ This and the Fubini-Tonelli's Theorem (whose validity is a consequence of SerenaDipierro2025) yield that \int_{\mathbb{R}^n} \mathcal{Z}(x) \phi(x) dx= \int_0^{+\infty} \left( \int_{\mathbb{R}^n} \left( \int_{\mathbb{R}^n} e^{-t(|\xi|^{2s}+|\xi|^2)+2\pi i x \cdot \xi} \phi(x) dx \right) d\xi \right) dt= \int_0^{+\infty} \left( \int_{\mathbb{R}^n} e^{-t(|\xi|^{2s}+|\xi|^2)} \hat{\phi}(\xi) d\xi \right) dt. From this and \ref{['equ:4.6']} (utilized here with $\Phi := \hat{\phi}$) we arrive at \ref{['lemm4.7']}, as desired. In this section, we mainly address the regularity of weak solutions of \ref{['equ1']} (also shown to be viscosity solutions). In Sections \ref{['sec3']} and \ref{['sec4']}, we first obtain the $C^{1,\alpha}$ and $C^{2,\alpha}$-regularity of weak solutions to equation \ref{['equ1']}. Based on this regularity result, in Section \ref{['sec:existence of vs']}, we prove that these weak solutions are also viscosity solutions. In this section we prove a Hölder estimate of solution to the equation \ref{['equ1']} when $p\in(0,1).$ Our proof is based on the arguments developed in MR2735074MR2863859, similar Hölder estimates with very different proofs can be found in MR2494809 for symmetric kernels and in MR2974279 for non-symmetric kernels. Let $s \in (0,1)$, $\alpha \in (0, \min\{1, 2s\})$, and define $\phi(x) = C_1 |x|^\alpha$. Let $Q= \int_{\mathcal{C}} \frac{\phi(a + 2z) + \phi(a - 2z) - 2\phi(a)}{|z|^{n+2s}} dz,$ and $\mathcal{C} = \left\{ z \in \mathbb{R}^n : |z| < \eta_1 |a|,\ |z \cdot a| \geqslant (1 - \eta_2)|a||z| \right\},$ with $\eta_1, \eta_2 \in (0, 1/2)$ being constants depending only on $\alpha$. Then there exists a constant $N = N(n, \alpha) > 0$ such that $Q \leqslant -N C_1 |a|^{\alpha - 2s}.$ Define $\varphi(t):= \phi(a + 2tz) = C_1 |a + 2tz|^\alpha.$ We compute the first and second derivatives of $\varphi$: First derivative: $\varphi'(t) = 2C_1 \alpha |a + 2tz|^{\alpha - 2} (a + 2tz) \cdot z,$Second derivative: $\varphi"(t) = 4C_1 \alpha \left[ (\alpha - 2) |a + 2tz|^{\alpha - 4} \left( (a + 2tz) \cdot z \right)^2 + |a + 2tz|^{\alpha - 2} |z|^2 \right].$ On the set $\mathcal{C}$, we have the estimates, for all $t\in[-1,1],$ |a + 2tz|\leqslant (1 + 2\eta_1)|a|,|(a + 2tz) \cdot z|\geqslant (1 - \eta_2)|a||z| - 2|z|^2 \geqslant (1 - 2\eta_1 - \eta_2)|a||z|. Since $\alpha - 2 < 0$, we obtain: $\varphi"(t) \leqslant 4C_1 \alpha |a + 2tz|^{\alpha - 4} \left[ (\alpha - 2)(1 - 2\eta_1 - \eta_2)^2 + (1 + 2\eta_1)^2 \right] |a|^2 |z|^2.$ By choosing $\eta_1, \eta_2$ sufficiently small (depending only on $\alpha\in(0,1)$) such that $(\alpha - 2)(1 - 2\eta_1 - \eta_2)^2 + (1 + 2\eta_1)^2 \leqslant \frac{\alpha - 1}{2} < 0,$ we get: $\varphi"(t) \leqslant -2C_1 \alpha (1 - \alpha) |a + 2tz|^{\alpha - 4} |a|^2 |z|^2.$ Using the bound $|a + 2tz|^{\alpha - 4} \geqslant 2^{\alpha - 4} |a|^{\alpha - 4}$, we further obtain: $\varphi"(t) \leqslant -2^{\alpha - 3} C_1 \alpha (1 - \alpha) |a|^{\alpha - 2} |z|^2\quad t\in[-1,1],\quad z\in\mathcal{C}.$ By the mean value theorem for second differences, there exists $t_0 \in (-1, 1)$ such that $\phi(a + 2z) + \phi(a - 2z) - 2\phi(a) = \varphi(1) + \varphi(-1) - 2\varphi(0) = \varphi"(t_0).$ Therefore, $Q\leqslant -\int_{\mathcal{C}}\frac{ 2^{\alpha - 3} C_1 \alpha (1 - \alpha) |a|^{\alpha - 2} |z|^2}{|z|^{n+2s}} dz=-2^{\alpha-3}C_1\alpha(1-\alpha)|a|^{\alpha-2}\int_{\mathcal{C}}\frac{|z|^2}{|z|^{n+2s}}dz=-NC_1|a|^{\alpha-2s},$ where $N=N(\alpha,n)>0.$ This completes the proof. Let $0 < s < 1$, $1/2 \leqslant r < R < 1$, and $f \in L^\infty(B_1)$. Let $u \in C_{loc}^2(B_1) \cap L_1(\mathbb{R}^n, \omega)$ with $\omega(x) = \frac{1}{1 + |x|^{n+2s}}$ such that $\mathcal{L}u:= f \quad \text{in } B_R.$ Then for any $\alpha \in (0, \min\{1, 2s\})$, we have [u]_{C^\alpha(B_r)}\leqslant N \left( (R - r)^{-\alpha} \sup_{B_R} |u|+ (R - r)^{2s-\alpha-2} \sup_{B_R} |u| + (R - r)^{-n-\alpha} \|u\|_{L_1(\mathbb{R}^n, \omega)} + (R - r)^{2s - \alpha} \text{osc}_{B_R} f \right), where $N = N(n, s, \alpha)$. 1. Localization problem: Denote $r_1 = (R - r)/2$, and $\bar{r} = (R + r)/2$. Set $w(x) = \mathds{1}_{B_R}(x) u(x)$. For $x \in B_{\bar{r}}$, we have $\Delta u(x) = \Delta w(x)$ and (-\Delta)^s u(x)= C_{n,s} P.V.\int_{\mathbb{R}^n}\frac{u(x)-u(x+z)}{|z|^{n+2s}} dz=C_{n,s} P.V. \int_{|z|<r_1}\frac{u(x)-u(x+z)}{|z|^{n+2s}} dz + \int_{|z|\geqslant r_1}\frac{u(x)-u(x+z)}{|z|^{n+2s}} dz= C_{n,s} P.V. \int_{|z|<r_1}\frac{w(x)-w(x+z)}{|z|^{n+2s}} dz + \int_{|z|\geqslant r_1}\frac{w(x)-u(x+z)}{|z|^{n+2s}} dz=C_{n,s} P.V. \int_{|z|<r_1}\frac{w(x)-w(x+z)}{|z|^{n+2s}} dz+\int_{|z|\geqslant r_1}\frac{w(x)-w(x+z)}{|z|^{n+2s}} dz\quad+ \int_{|z|\geqslant r_1}\frac{w(x+z)-u(x+z)}{|z|^{n+2s}} dz=(-\Delta)^sw-\int_{|z|\geqslant r_1}\frac{u(x+z)-w(x+z)}{|z|^{n+2s}} dz. Hence in $B_{\bar{r}}$, $- \mathcal{L}w(x) = g(x) - f(x),$ where $g(x) = -\int_{|z| \geqslant r_1} \frac{u(x+z)-w(x+z)}{|z|^{n+2s}} dz.$ Note that $\|g\|_{L^\infty(B_R)} \leqslant N r_1^{-n-2s} \|u\|_{L_1(\mathbb{R}^n, \omega)},$ where $N = N(n,s)$. 2. Construct comparison functions: For $x_0 \in B_r$, set $M(x, y) := w(x) - w(y) - \phi(x - y) - \Gamma(x),$ where $\phi(z) = C_1 |z|^{\alpha}$, $\alpha \in (0, \min\{1, 2s\})$, and $\Gamma(x) = C_2 |x - x_0|^2$. We will find $C_1, C_2 \in (0, \infty)$ depending only on $n, s, r_1, \|u\|_{L^\infty(B_R)}, \|u\|_{L_1(\mathbb{R}^n, \omega)}, \text{osc}_{B_R} f,$ but independent of the choice of $x_0 \in B_r$, such that $\sup_{x, y \in \mathbb{R}^n} M(x, y) \leqslant 0.$ To prove \ref{['eq:2.1']}, we first take $C_2 := 8 r_1^{-2} \|u\|_{L^\infty(B_R)}.$ Then, for $x \in \mathbb{R}^n \setminus B_{r_1/2}(x_0)$, $w(x) - w(y) \leqslant 2 \|u\|_{L^\infty(B_R)} \leqslant C_2 |x - x_0|^2.$ This shows that $M(x, y) \leqslant 0, \quad x \in \mathbb{R}^n \setminus B_{r_1/2}(x_0).$ To get a contradiction, assume there exist $x, y \in \mathbb{R}^n$ such that $M(x, y) > 0$. By \ref{['eq:2.2']} we know that $x \in B_{r_1/2}(x_0) \subset B_{(\bar{r}+r)/2}$. Moreover, if $M(x, y) > 0$, then $w(x) - w(y) > C_1 |x - y|^{\alpha}, \quad \text{i.e.,} \quad |x - y|^{\alpha} < \frac{2 \|u\|_{L^\infty(B_R)}}{C_1}.$ If we take a sufficiently large $C_1$ so that $C_1 \geqslant 2^{1+\alpha} r_1^{-\alpha} \|u\|_{L^\infty(B_R)}$, the above inequalities show that $y \in B_{\bar{r}}$. Therefore, the assumption that $M(x, y) > 0$ for some $x, y \in \mathbb{R}^n$ (and the continuity of $u$ on $B_R$) enables us to assume that there exist $\bar{x}, \bar{y} \in B_{\bar{r}}$ satisfying $\sup_{x,y \in \mathbb{R}^n} M(x, y) = M(\bar{x}, \bar{y}) > 0$. 3. Construct a contradiction: Note that at $\bar{x}, \bar{y} \in B_{\bar{r}}$ we have g(\bar{y}) - f(\bar{y})= - \mathcal{L}w(\bar{y}),-g(\bar{x}) + f(\bar{x})= \mathcal{L}w(\bar{x}). Thus, it follows that $-2 \|g\|_{\mathcal{L}^\infty(B_R)} - \text{osc}_{B_R} f \leqslant \mathcal{L}w(\bar{x}) - \mathcal{L}w(\bar{y}) := I.$ Set $J(\bar{x}, \bar{y}, z) = w(\bar{x} + z) + w(\bar{x} - z) - 2w(\bar{x}) - w(\bar{y} + z) - w(\bar{y} - z) + 2w(\bar{y}).$ Set $a=\bar{x}-\bar{y},$ since $M(x, y)$ attains its maximum at $\bar{x}, \bar{y}$, we have $\Delta w(\overline{x})\leqslant\Delta\phi(\bar{x} - \bar{y})+\Delta\Gamma(\bar{x})=N(\alpha,n)C_1|a|^{\alpha-2}+N(n)C_2,\Delta w(\bar{y})\geqslant-\Delta\phi(\bar{x} - \bar{y})=-N(\alpha,n)C_1|a|^{\alpha-2},$ and w(\bar{x} + z) - w(\bar{y} + z) - \phi(\bar{x} - \bar{y}) - \Gamma(\bar{x} + z)\leqslant w(\bar{x}) - w(\bar{y}) - \phi(\bar{x} - \bar{y}) - \Gamma(\bar{x}),w(\bar{x} - z) - w(\bar{y} - z) - \phi(\bar{x} - \bar{y}) - \Gamma(\bar{x} - z)\leqslant w(\bar{x}) - w(\bar{y}) - \phi(\bar{x} - \bar{y}) - \Gamma(\bar{x}) for all $z \in \mathbb{R}^n$. These two inequalities lead us to $J(\bar{x}, \bar{y}, z) \leqslant \Gamma(\bar{x} + z) + \Gamma(\bar{x} - z) - 2\Gamma(\bar{x}), \quad z \in \mathbb{R}^n.$ By again the assumption that $M(x, y)$ has the maximum at $\bar{x}, \bar{y}$, we have w(\bar{x} + z) - w(\bar{y} - z) - \phi(\bar{x} - \bar{y} + 2z) - \Gamma(\bar{x} + z)\leqslant w(\bar{x}) - w(\bar{y}) - \phi(\bar{x} - \bar{y}) - \Gamma(\bar{x}),w(\bar{x} - z) - w(\bar{y} + z) - \phi(\bar{x} - \bar{y} - 2z) - \Gamma(\bar{x} - z)\leqslant w(\bar{x}) - w(\bar{y}) - \phi(\bar{x} - \bar{y}) - \Gamma(\bar{x}) for all $z \in \mathbb{R}^n$. Hence it follows that, for any $z \in \mathbb{R}^n$, $J(\bar{x}, \bar{y}, z) \leqslant \phi(\bar{x} - \bar{y} + 2z) + \phi(\bar{x} - \bar{y} - 2z) - 2\phi(\bar{x} - \bar{y}) + \Gamma(\bar{x} + z) + \Gamma(\bar{x} - z) - 2\Gamma(\bar{x}).$ Since $\bar{x}, \bar{y}$ satisfy \ref{['eq:2.3']}, we have $|a| < r_1/2$. Also set, for some $\eta_1, \eta_2 \in (0, 1/2)$, $\mathcal{C} = \{ |z| < \eta_1 |a| : |z \cdot a| \geqslant (1 - \eta_2)|a||z| \}.$ Then $\mathcal{C} \subset B_{r_1}$ and $|2I|\leqslant \int_{|z| \geqslant r_1} \frac{J(\bar{x}, \bar{y}, z)}{|z|^{n+2s}} dz + \int_{B_{r_1} \setminus \mathcal{C}} \frac{J(\bar{x}, \bar{y}, z)}{|z|^{n+2s}} dz + \int_{\mathcal{C}} \frac{J(\bar{x}, \bar{y}, z)}{|z|^{n+2s}} dz := J_1 + J_2 + J_3.$ Note that $J_1 \leqslant N(n,s) r_1^{-2s} \|u\|_{L^\infty(B_R)}.$ By \ref{['eq:2.4']} it follows $J_2 \leqslant \int_{B_{r_1} \setminus \mathcal{C}} \frac{\Gamma(\bar{x} + z) + \Gamma(\bar{x} - z) - 2\Gamma(\bar{x})}{|z|^{n+2s}} dz \leqslant N r_1^{2-2s} C_2,$ where $N=N(n),$ which is independent of $\eta_1, \eta_2$ in the definition of $\mathcal{C}$. Now using \ref{['eq:2.5']} we obtain $J_3\leqslant \int_{\mathcal{C}} \frac{\phi(\bar{x} - \bar{y} + 2z) + \phi(\bar{x} - \bar{y} - 2z) - 2\phi(\bar{x} - \bar{y})}{|z|^{n+2s}} dz + \int_{\mathcal{C}} \frac{\Gamma(\bar{x} + z) + \Gamma(\bar{x} - z) - 2\Gamma(\bar{x})}{|z|^{n+2s}}:= J_{3,1} + J_{3,2}.$ The term $J_{3,2}$ is again bounded by $N r_1^{2-2s} C_2$, where $N = N(n)$. Finally, by Lemma \ref{['thm:4.1']}, $J_{3,1} \leqslant -N(n, \alpha) C_1 |a|^{\alpha-2s}.$ Thus, we get from \ref{['eq:2.6']} and the choice of $C_2$ that $I \leqslant N(n, s) r_1^{-2s} \|u\|_{L^\infty(B_R)} - N(n, \alpha) C_1 |a|^{\alpha-2s}.$ Combining \ref{['eq:2.1.']}, \ref{['eq:2.5']}, \ref{['eq:2.8']}, \ref{['eq:2.6']} and \ref{['eq:2.10']} we finally have $0 \leqslantN(s,n,\alpha) (\text{osc}_{B_R} f + r_1^{-2s} \|u\|_{L^\infty(B_R)} +r_1^{-2} \|u\|_{L^\infty(B_R)}+ r_1^{-n-2s} \|u\|_{L_1(\mathbb{R}^n,\omega)})\quad- N(n, \alpha) C_1 (|a|^{\alpha -2s} -|a|^{\alpha -2}):= G.$ Choose $C_1$ so that $C_1 \geqslant 2^{1+\alpha} r_1^{-\alpha} \|u\|_{L^\infty(B_R)}$ as well as $C_1 \geqslant N(n, s,\alpha) r_1^{2s - \alpha} (\text{osc}_{B_R} f + r_1^{-2s} \|u\|_{L^\infty(B_R)}+r_1^{-2} \|u\|_{L^\infty(B_R)} + r_1^{-n-2s} \|u\|_{L_1(\mathbb{R}^n,\omega)}) / N(n, \alpha).$ Then, for $\alpha \in (0, \min\{1, 2s\})$, by \ref{['eq:2.3']} $|a|^{\alpha - 2s} r_1^{2s - \alpha} > 1$ and $G\leqslant N(n,s) (\text{osc}_{B_R} f + r_1^{-2s} \|u\|_{L^\infty(B_R)}+r_1^{-2} \|u\|_{L^\infty(B_R)} + r_1^{-n-2s} \|u\|_{L_1(\mathbb{R}^n,\omega)})\quad (1 - |a|^{\alpha - 2s} r_1^{2s - \alpha}-r_1^{2s-\alpha}|a|^{\alpha-2}) < 0.$ This contradicts the fact that $G \geqslant 0$. This proves the assertion in the theorem. More specifically, using the fact that $C_1$ and $C_2$ are independent of the choice of $x_0 \in B_r$, we obtain $|u(x) - u(y)| \leqslant C_1 |x - y|^{\alpha}, \quad x, y \in B_r,$ where $C_1$ is the right-hand side of the Hölder estimate in Lemma \ref{['thm:4.1']}. Let $0 < s < 1$, and $f \in L^\infty(B_1)$. Let $u \in C^2_{\mathrm{loc}}(B_1) \cap L_1(\mathbb{R}^n, \omega)$ with $\omega(x) = 1/(1 + |x|^{n+2s})$ such that $\mathcal{L}u = f$ in $B_1$. Then for any $\alpha \in (0, \min\{1, 2s\})$, we have $[u]_{C^\alpha(B_{1/2})} \leqslant N\|u\|_{L_1(\mathbb{R}^n, \omega)} + N \text{osc}_{B_1} f,$ where $N = N(n,s, \alpha)$. Set $r_k = 1 - 2^{-k-1}, \quad B_{(k)} = B_{r_k}, \quad k = 0, 1, 2, \ldots.$ Theorem \ref{['thm:4.1']} gives, for $k = 0, 1, 2, \ldots$, $[u]_{C^\alpha(B_{(k)})} \leqslant N_1 \left( 2^{2k} \sup_{B_{(k+1)}} |u| + 2^{(n+\alpha)k} \|u\|_{L_1(\mathbb{R}^n, \omega)} + \text{osc}_{B_{(k+1)}} f \right),$ where $N_1 = N_1(n,s, \alpha)$ is a constant independent of $n$. To estimate the first term on the right-hand side of \ref{['eq:2.11']}, by the well-known interpolation inequality, we have $\sup_{B_{(k+1)}} |u| \leqslant \varepsilon [u]_{C^\alpha(B_{(k+1)})} + N \varepsilon^{-n/\alpha} \|u\|_{L_1(B_{(k+1)})}, \quad \forall \epsilon \in (0, 1).$ Multiplying both sides of Equation \ref{['eq:2.12']} by $2^{2k}$, and upon taking $\epsilon = (N_1 2^{2k+3n/\alpha})^{-1},$ we have $2^{2k}\sup_{B_{(k+1)}} |u|\leqslant 2^{2k}\varepsilon [u]_{C^\alpha(B_{(k+1)})} + N2^{2k} \varepsilon^{-n/\alpha} \|u\|_{L_1(B_{(k+1)})}\leqslant \frac{1}{N_12^{\frac{3n}{\alpha}}}[u]_{C^\alpha(B_{(k+1)})}+C_12^{\frac{2kn}{\alpha}}\|u\|_{L_1(B_{(k+1)})}\leqslant \frac{1}{N_12^{\frac{3n}{\alpha}}}[u]_{C^\alpha(B_{(k+1)})}+C_12^{\frac{2kn}{\alpha}}\|u\|_{L_1(B_1)}.$ We multiply both sides of the above equation by $2^{-3kn/\alpha}$ and and substitute into \ref{['eq:2.11']} gives $2^{-\frac{3kn}{\alpha}}[u]_{C^\alpha(B_{(k)})} \leqslant 2^{-\frac{3n(k+1)}{\alpha}}[u]_{C^\alpha(B_{(k+1)})}+N 2^{-\frac{kn}{\alpha}} \|u\|_{L_1(B_1)} +N 2^{(n+\alpha)k-\frac{3kn}{\alpha}} \|u\|_{L_1(\mathbb{R}^n, \omega)} + N 2^{-\frac{3kn}{\alpha}}\text{osc}_{B_{1}} f,$ let $A_k=2^{-\frac{3kn}{\alpha}}[u]_{C^\alpha(B_{(k)})},$ then sum over $k$ to obtain \sum_{k=0}^{\infty} A_k\leqslant \sum_{k=0}^{\infty} A_{k+1} + N \sum_{k=0}^{\infty} 2^{-nk/\alpha} \|u\|_{L_1(B_1)}\quad + N \sum_{k=0}^{\infty} 2^{-3nk/\alpha + (n+\alpha)k} \|u\|_{L_1(\mathbb{R}^n,\omega)} + N \sum_{k=0}^{\infty} 2^{-3nk/\alpha} \text{osc}_{B_1}f, which immediately yields \ref{['equ:2.11']}. The corollary is proved. □ Let $f \in L^\infty(B_1)$ and let $u \in \mathcal{X}^{1,2}(B_3) \cap L^\infty(\mathbb{R}^n)$ be a weak solution of $\mathcal{L} u = f \quad \text{in} \quad B_1.$ Then, $u \in C^\alpha(B_{1/4})$ for any $0 < \alpha < \min\{2s, 1\}$ and $\|u\|_{C^\alpha(B_{1/4})} \leqslant C \left( \|u\|_{L^\infty(\mathbb{R}^n)} + \|f\|_{L^\infty(B_1)}\right),$ for a suitable $C > 0$ depending on $n$, $s$ and $\alpha$. We take $\rho \in C_0^\infty(B_1;[0,1])$ and we consider the mollifier $\rho_\varepsilon(x) := \varepsilon^{-n} \rho(x/\varepsilon)$, $\varepsilon > 0$. We define $u_\varepsilon := u * \rho_\varepsilon$ and $f_\varepsilon := f * \rho_\varepsilon$. Of course, $u_\varepsilon$ is a smooth function, by construction. Let us show that $u_\varepsilon$ solves the following equation $-\Delta u_\varepsilon+(-\Delta)^s u_\varepsilon = f_\varepsilon \quad \text{in} \quad B_{1/2}.$ For this, first we observe that for any test function $\phi\in C_0^{\infty}(B_{1/2})$, we have $\int_{\mathbb{R}^n}-\Delta u_{\epsilon}\phi dx=\int_{\mathbb{R}^n}\triangledown u_{\epsilon}\cdot\triangledown\phi dx=\int_{\mathbb{R}^n}[(\triangledown u)*\rho_{\epsilon}]\cdot\triangledown\phi dx=\int_{\mathbb{R}^n}\left[\int_{\mathbb{R}^n}\triangledown u(x+z)\rho_{\epsilon}(z)\triangledown\phi(x)dx\right]dz=\int_{B_{\epsilon}}\left[\int_{\mathbb{R}^n}\triangledown u(\tilde{x})\rho_{\epsilon}(z)\triangledown\phi(\tilde{x}-z)d\tilde{x}\right]dz,$ and since $u \in \mathcal{X}^{1,2}(B_3) \cap L^\infty(\mathbb{R}^n)$, \int_{\mathbb{R}^n} \left[ \int_{\mathbb{R}^{2n}} \frac{|u(x+z) - u(y+z)| |\phi(x) - \phi(y)| \rho_\varepsilon(z)}{|x-y|^{n+2s}} dx dy \right] dz= \int_{B_\varepsilon} \left[ \int_{Q(B_1)} \frac{|u(x+z) - u(y+z)| |\phi(x) - \phi(y)| \rho_\varepsilon(z)}{|x-y|^{n+2s}} dx dy \right] dz\leqslant \frac{\varepsilon^{-n}}{2} \int_{B_\varepsilon} \left[ \int_{Q(B_2)} \frac{|u(x) - u(y)|^2}{|x-y|^{n+2s}} dx dy + \int_{Q(B_1)} \frac{|\phi(x) - \phi(y)|^2}{|x-y|^{n+2s}} dx dy \right] dz< +\infty. Here we used the notation $Q(B_r) := \mathbb{R}^{2n} \setminus \left( (\mathbb{R}^n \setminus B_r) \times (\mathbb{R}^n \setminus B_r)\right), \quad r > 0.$ Hence, by Tonelli's Theorem the function $\mathbb{R}^{2n} \times \mathbb{R}^n \ni (x, y, z) \mapsto \frac{(u(x + z) - u(y + z)) (\phi(x) - \phi(y)) \rho_\varepsilon(z)}{|x - y|^{n+2s}}$ belongs to $L^1(\mathbb{R}^{2n} \times \mathbb{R}^n)$. Then, by Fubini's Theorem and the definition of $u_{\varepsilon},$ \int_{\mathbb{R}^n} \left[ \int_{\mathbb{R}^{2n}} \frac{(u(x+z)-u(y+z)) (\phi(x)-\phi(y)) \rho_\varepsilon(z)}{|x-y|^{n+2s}} dx dy \right] dz= \int_{\mathbb{R}^{2n}} \left[ \int_{\mathbb{R}^n} \frac{(u(x+z)-u(y+z)) (\phi(x)-\phi(y)) \rho_\varepsilon(z)}{|x-y|^{n+2s}} dx dy \right]= \int_{\mathbb{R}^{2n}} \frac{(u_\varepsilon(x)-u_\varepsilon(y)) (\phi(x)-\phi(y))}{|x-y|^{n+2s}} dx dy. Therefore, we may use Fubini's Theorem and obtain that for any $\phi \in C_0^\infty(B_{1/2}),$ \int_{\mathbb{R}^n} f_\varepsilon(x) \phi(x) dx= \int_{\mathbb{R}^n} \left[ \int_{\mathbb{R}^n} f(x + z) \phi(x) \rho_\varepsilon(z) dx \right] dz= \int_{B_\varepsilon} \left[ \int_{\mathbb{R}^n} f(\tilde{x}) \phi(\tilde{x} - z) \rho_\varepsilon(z) d\tilde{x} \right] dz= \int_{B_\varepsilon} \left[ \int_{\mathbb{R}^{2n}} \frac{(u(\tilde{x}) - u(\tilde{y})) (\phi(\tilde{x} - z) - \phi(\tilde{y} - z)) \rho_\varepsilon(z)}{|\tilde{x} - \tilde{y}|^{n+2s}} d\tilde{x} d\tilde{y} \right] dz +\int_{B_{\epsilon}}\left[\int_{\mathbb{R}^n}\triangledown u(\tilde{x})\rho_{\epsilon}(z)\triangledown\phi(\tilde{x}-z)d\tilde{x}\right]dz= \int_{\mathbb{R}^n} \left[ \int_{\mathbb{R}^{2n}} \frac{(u(x + z) - u(y + z)) (\phi(x) - \phi(y)) \rho_\varepsilon(z)}{|x - y|^{n+2s}} dx dy \right] dz+\int_{\mathbb{R}^n}\left[\int_{\mathbb{R}^n}\triangledown u(x+z)\rho_{\epsilon}(z)\triangledown\phi(x)d\tilde{x}\right]dz= \int_{\mathbb{R}^{2n}} \frac{(u_\varepsilon(x) - u_\varepsilon(y)) (\phi(x) - \phi(y))}{|x-y|^{n+2s}} dx dy+\int_{\mathbb{R}^n}\triangledown u_{\epsilon}\triangledown\phi dx, thanks to the fact that $u$ is a weak solution of \ref{['equ:2.1']}. This means that $u_\varepsilon$ is a smooth solution of \ref{['eq:3.2']}. As a consequence of this, we may apply Corollary \ref{['cor:4.3']} and obtain that $u_\varepsilon \in C^\alpha(B_{1/4})$ for any $0 < \alpha < \min\{2s, 1\}$ and $\|u_\varepsilon\|_{C^\alpha(B_{1/4})} \leqslant C\left(\|u_\varepsilon\|_{L^\infty(\mathbb{R}^n)} + \|f_\varepsilon\|_{L^\infty(B_1)}\right),$ for a suitable $C > 0$ depending on $n$, $s$ and $\alpha$. Hence, the desired result follows by sending $\varepsilon \to 0$. Let $u \in \mathcal{X}^{1,2}(\mathbb{R}^n)$ be a weak solution of \ref{['equ1']} with $p\in(0,1)$. Then $u \in C^\alpha(\mathbb{R}^n)$, for any $\alpha \in (0, \min\{2s, 1\})$, and $\Vert u\Vert _{C^{\alpha}(\mathbb{R}^n)}\leqslant C,$ for some $C>0$, depending only on $n,s,h,\alpha$ and $p.$ The Hölder regularity of $u$ follows from Proposition \ref{['pro:2.4']} and the $L^{\infty}(\mathbb{R}^n)$-regularity in Su-Xu, being $u$ a solution to \ref{['equ1']} with $p\in(0,1).$ The goal of this section is to establish the $C^{2,\alpha}$-regularity result for solutions of problem \ref{['equ1']} when $1\leqslant p < 2^*-1,$ as stated in Theorem \ref{['thm:1.2']}. Using techniques of smoothing and truncation MR4808805, we can obtain the $C^{2,\alpha}$ regularity of the solution of \ref{['equ1']}. However, here we adopt a different approach from MR4808805 by utilizing the properties of the Fourier transform and the Riesz kernel. Throughout this section $\mathcal{S}$ stands for the Schwartz space of rapidly decreasing $C^{\infty}$ functions in $\mathbb{R}^n$. We define the Riesz potentials by $(I_\alpha f)(x) = \frac{1}{\gamma(\alpha)} \int_{\mathbb{R}^n} |x - y|^{-n+\alpha} f(y) dy, \quad 0<\alpha<n$ with $\gamma(\alpha) = \pi^{n/2} 2^\alpha \frac{\Gamma(\alpha/2)}{\Gamma \left( \frac{n}{2} - \frac{\alpha}{2} \right)}.$ In Fourier space, we have $\widehat{\mathcal{L}u}(\xi) = (|\xi|^2 + |\xi|^{2s}) \hat{u}(\xi).$ Define $m(\xi) = |\xi|^2 + |\xi|^{2s},$ and $M(\xi) = \frac{1}{m(\xi)}.$ Formally, $u = \mathcal{F}^{-1}[M \hat{f}] = \mathcal{Z} * f,$ where $\mathcal{Z} = \mathcal{F}^{-1}[M]$. Take $\phi \in C_c^\infty(\mathbb{R}^n)$ with: $\phi(\xi) = 1$ for $|\xi| \leqslant 1$, $\phi(\xi) = 0$ for $|\xi| \geqslant 2$, $0 \leqslant \phi \leqslant 1$. Let $\psi(\xi) = 1 - \phi(\xi)$. Define: $M_1(\xi) = \frac{\phi(\xi)}{m(\xi)}, \quad M_2(\xi) = \frac{\psi(\xi)}{m(\xi)}.$ Clearly $M = M_1 + M_2$. Define the kernels: $\mathcal{Z}_1 = \mathcal{F}^{-1}[M_1], \quad \mathcal{Z}_2 = \mathcal{F}^{-1}[M_2].$ Then $\mathcal{Z} = \mathcal{Z}_1 + \mathcal{Z}_2$ and $u = \mathcal{Z}_1 * f + \mathcal{Z}_2 * f$. For $|\xi| \leqslant 2$: $M_1(\xi) = \frac{\phi(\xi)}{|\xi|^{2s}(1 + |\xi|^{2-2s})} = \frac{\phi(\xi)}{1 + |\xi|^{2-2s}} \cdot |\xi|^{-2s}:=A(\xi)\cdot|\xi|^{-2s}.$ Since $A(\xi) \in C_c^\infty$, we have $z:=\mathcal{F}^{-1}[A(\xi)]$. Thus, $\mathcal{Z}_1 = z * R_{2s},$ where $R_{2s}(x) = c_{n,s} |x|^{-(n-2s)}$ is the kernel of the Riesz potential $I_{2s} = (-\Delta)^{-s}$. On $\mathop{\mathrm{supp}}\nolimits \psi$, $m(\xi) \geqslant |\xi|^2$, so $|M_2(\xi)| \leqslant |\psi(\xi)|/|\xi|^2$. In fact, $M_2 \in C^\infty$ and all its derivatives decay at least as $|\xi|^{-2}$ at infinity, hence $M_2 \in \mathcal{S}$, and so $\mathcal{Z}_2 \in \mathcal{S}$. To obtain the $C^{2,\alpha}$-regularity of the solution of \ref{['equ1']}, we require the following two evident conclusions of Schwartz space, for which we do not provide detailed proofs here. Let $f \in \mathcal{S}(\mathbb{R}^n)$. Then for any multi-index $\beta\in Z_{+}^n$, we have $D^\beta f \in L^1(\mathbb{R}^n)$. In particular, $f \in L^1(\mathbb{R}^n)$. Let $Z \in \mathcal{S}(\mathbb{R}^n)$ and $f \in L^\infty(\mathbb{R}^n)$. Then for any $k \in \mathbb{N}$ and $\theta \in (0,1]$, the convolution $u = Z * f$ satisfies $u \in C^{k,\theta}(\mathbb{R}^n)$, and there exists a constant $C > 0$ (depending on $Z$, $k$, $\theta$) such that $\|u\|_{C^{k,\theta}} \leqslant C \|f\|_{L^\infty(\mathbb{R}^n)}.$ The proof proceeds in two steps: first we prove $u \in C^k$ and control the $C^k$-norm, then we prove that the $k$-th order derivatives are $\theta$-Hölder continuous. Step 1: Proof that $u \in C^k$ and $C^k$-norm estimate Since $Z \in \mathcal{S}$, classical convolution theory allows us to interchange the differentiation operator with convolution: $D^\beta u(x) = D^\beta (Z * f)(x) = (D^\beta Z) * f (x), \quad \forall \, |\beta| \leqslant k.$ For any $|\beta| \leqslant k$, apply Young's convolution inequality. Since $D^\beta Z \in L^1(\mathbb{R}^n)$ (Lemma \ref{['schwa']}) and $f \in L^\infty(\mathbb{R}^n)$, we have: $\|D^\beta u\|_{L^\infty(\mathbb{R}^n)} = \|(D^\beta Z) * f\|_{L^\infty(\mathbb{R}^n)} \leqslant \|D^\beta Z\|_{L^1(\mathbb{R}^n)} \|f\|_{L^\infty(\mathbb{R}^n)}.$ Define the constant $C_k^{(1)} = \max_{|\beta| \leqslant k} \|D^\beta Z\|_{L^1(\mathbb{R}^n)}.$ By Lemma \ref{['schwa']}, $C_k^{(1)} < \infty$. Therefore, $\|u\|_{C^k} = \max_{|\beta| \leqslant k} \|D^\beta u\|_{L^\infty(\mathbb{R}^n)} \leqslant C_k^{(1)} \|f\|_{L^\infty(\mathbb{R}^n)}.$ This proves $u \in C^k(\mathbb{R}^n)$. Step 2: Proof of $\theta$-Hölder Continuity of the $k$-th Order Derivatives Let $|\beta| = k$, and denote $g = D^\beta u = (D^\beta Z) * f$. We need to prove that $g \in C^{0,\theta}$, i.e., there exists a constant $C_{Z, \beta, \theta}$ such that $|g(x) - g(y)| \leqslant C_{Z,\beta, \theta} \|f\|_{L^\infty(\mathbb{R}^n)} |x - y|^\theta, \quad \forall x, y \in \mathbb{R}^n.$ From the definition of $g$, we have $|g(x) - g(y)|= \left| \int_{\mathbb{R}^n} [D^\beta Z(x - z) - D^\beta Z(y - z)] f(z) dz \right|\leqslant \|f\|_{L^\infty(\mathbb{R}^n)} \int_{\mathbb{R}^n} |D^\beta Z(x - z) - D^\beta Z(y - z)| dz.$ Let $h = x - y$, and make the change of variables $w = z - y$ to obtain \int_{\mathbb{R}^n} |D^\beta Z(h - w) - D^\beta Z(-w)| dw\leqslant |h| \int_0^1 \|\nabla(D^\beta Z)\|_{L^1(\mathbb{R}^n)} dt= |h| \cdot \|\nabla(D^\beta Z)\|_{L^1(\mathbb{R}^n)}. Therefore $|g(x) - g(y)| \leqslant \|f\|_{L^\infty(\mathbb{R}^n)} \cdot \|\nabla(D^\beta Z)\|_{L^1(\mathbb{R}^n)} \cdot |x - y|.$ This shows that $g$ is Lipschitz continuous. From Step 1, we know $g \in L^\infty(\mathbb{R}^n)$. Combining this with the above Lipschitz estimate, we can use interpolation to obtain a Hölder estimate of $g$ for any $\theta \in (0,1$: $|g(x) - g(y)| \leqslant 2 \|g\|_{L^\infty(\mathbb{R}^n)}^{1-\theta} \cdot \left( \|f\|_{L^\infty(\mathbb{R}^n)} \|\nabla(D^\beta Z)\|_{L^1(\mathbb{R}^n)} \right)^\theta |x - y|^\theta.$ Recalling the $C^k$-norm estimate $\|g\|_{L^\infty} \leqslant C_k^{(1)} \|f\|_{L^\infty}$ from Step 1, we finally conclude: $[g]_{C^{0,\theta}} \leqslant C \|f\|_{L^\infty(\mathbb{R}^n)},$ where $C$ is a positive constant depending on $Z, \beta$ and $\theta$. Employing the above two lemmas, we get the following regularity result of weak solutions of \ref{['equ1']}: Let $0 < s < 1$. If $u \in L^\infty(\mathbb{R}^n)$ and $\mathcal{L}u = f \in L^\infty(\mathbb{R}^n)$, then: If $2s \leqslant 1$, then $u \in C^{0,\alpha}(\mathbb{R}^n)$ for all $\alpha < 2s$, and $\|u\|_{C^{0,\alpha}} \leqslant C (\|f\|_{L^\infty}+\|u\|_{L^\infty})$ for a positive constant $C$ depending only on $n,\alpha, s.$If $2s > 1$, then $u \in C^{1,\alpha}(\mathbb{R}^n)$ for all $\alpha < 2s - 1$, and $\|u\|_{C^{1,\alpha}} \leqslant C (\|f\|_{L^\infty}+\|u\|_{L^\infty})$ for a positive constant $C$ depending only on $n,\alpha, s.$ Since $\mathcal{Z}_2 \in \mathcal{S}$, for any $k \in \mathbb{N}$, $\theta \in (0,1]$, by Lemma \ref{['con']}, we have $\|\mathcal{Z}_2 * f\|_{C^{ k,\theta}} \leqslant C_{k,\theta} \|f\|_{L^\infty}.$ For $\mathcal{Z}_1$, we have $\mathcal{Z}_1 = z* R_{2s}$. By Riesz potential theory (MR290095) and MR2270163, we first claim that : If $0 < \alpha < 1$ and $f \in L^\infty(\mathbb{R}^n)$, then $I_\alpha f(x)$ satisfies $|I_\alpha f(x) - I_\alpha f(y)| \leqslant C \|f\|_{L^\infty} |x-y|^\alpha,$ that is, $I_\alpha f \in C^{0,\alpha}(\mathbb{R}^n)$ . For $x, y \in \mathbb{R}^n$, let $\delta = |x - y|$ and write \quad I_\alpha f(x) - I_\alpha f(y)= c_{n,\alpha} \int_{\mathbb{R}^n} \left[ \frac{1}{|x - z|^{n-\alpha}} - \frac{1}{|y - z|^{n-\alpha}} \right] f(z) dz=\int_{B_{2\delta}(x)} \left[ \frac{1}{|x - z|^{n-\alpha}} - \frac{1}{|y - z|^{n-\alpha}} \right] f(z) dz+\int_{\mathbb{R}^n \setminus B_{2\delta}(x)} \left[ \frac{1}{|x - z|^{n-\alpha}} - \frac{1}{|y - z|^{n-\alpha}} \right] f(z) dz. For $z \in B_{2\delta}(x)$, we have $\left| \int_{B_{2\delta}(x)} \frac{f(z)}{|x-z|^{n-\alpha}} dz \right| \leqslant\|f\|_{L^\infty} \int_{B_{2\delta}} \frac{1}{|z|^{n-\alpha}} dz = C \|f\|_{L^\infty} \delta^\alpha.$ The same estimate holds for $y$. For $z \in \mathbb{R}^n \setminus B_{2\delta}(x)$, by the mean value theorem: $\left| \frac{1}{|x-z|^{n-\alpha}} - \frac{1}{|y-z|^{n-\alpha}} \right| \leqslant C \frac{|x-y|}{|\xi-z|^{n-\alpha+1}},$ where $\xi$ lies on the line segment joining $x$ and $y$. Since $|x-z| \geqslant 2\delta$ and $|x-y| = \delta$, we have $|\xi-z| \geqslant |x-z| - \delta \geqslant \frac{1}{2}|x-z|$. Therefore, $\left| \frac{1}{|x-z|^{n-\alpha}} - \frac{1}{|y-z|^{n-\alpha}} \right| \leqslant C \frac{\delta}{|x-z|^{n-\alpha+1}}.$ Integrating: $\int_{\mathbb{R}^n \setminus B_{2\delta}(x)} \frac{\delta}{|x-z|^{n-\alpha+1}} dz = C \delta \int_{2\delta}^{\infty} \frac{r^{n-1}}{r^{n-\alpha+1}} dr = C \delta \int_{2\delta}^{\infty} r^{\alpha-2} dr = C \delta^\alpha.$ So, $|I_\alpha f(x) - I_\alpha f(y)| \leqslant C\|f\|_{L^\infty} |x - y|^\alpha.$ This completes the proof of the Claim. Next, we have the other fact that : If $1 < \alpha < 2$ and $f \in L^\infty(\mathbb{R}^n)$, then $I_\alpha f \in C^{1,\alpha-1}(\mathbb{R}^n)$, and $\| \triangledown( I_\alpha f) \|_{C^{0,\alpha-1}} \leqslant C \|f\|_{L^\infty}.$ We first compute the gradient of $I_\alpha f$. For $1 < \alpha < 2$, we have, $\triangledown( I_\alpha f)(x) = -c_{n,\alpha}(n-\alpha) \int_{\mathbb{R}^n} \frac{x-z}{|x-z|^{n-\alpha+2}} f(z) dz.$ We need to show that $\triangledown( I_\alpha f) \in C^{0,\alpha-1}(\mathbb{R}^n)$. Take $x, y \in \mathbb{R}^n$ and let $\delta = |x-y|$. Consider $\triangledown( I_\alpha f)(x) - \triangledown( I_\alpha f)(y) = -c_{n,\alpha}(n-\alpha) \int_{\mathbb{R}^n} \left[ \frac{x-z}{|x-z|^{n-\alpha+2}} - \frac{y-z}{|y-z|^{n-\alpha+2}} \right] f(z) dz.$ As in the above computation, we have $\left| \int_{B_{2\delta}(x)} \frac{x-z}{|x-z|^{n-\alpha+2}} f(z) dz \right| \leqslant\|f\|_{L^\infty} \int_{B_{2\delta}} \frac{1}{|z|^{n-\alpha+1}} dz = C \|f\|_{L^\infty} \delta^{\alpha-1}.$ The same bound holds for the above term with $y$. For $z \in \mathbb{R}^n \setminus B_{2\delta}(x)$, define the kernel $K(x,z) = \frac{x-z}{|x-z|^{n-\alpha+2}}.$ We estimate the difference $|K(x,z) - K(y,z)|$. By the mean value theorem, we have $|K(x,z) - K(y,z)| \leqslant|x-y| \sup_{\xi \in [x,y]} |D_\xi K(\xi,z)|,$ where $D_\xi K$ denotes the derivative matrix. Compute the derivative: $D_{x_j} K_i(x,z) = D_{x_j} \left( \frac{x_i - z_i}{|x-z|^{n-\alpha+2}} \right) = \frac{\delta_{ij}}{|x-z|^{n-\alpha+2}} - (n-\alpha+2) \frac{(x_i-z_i)(x_j-z_j)}{|x-z|^{n-\alpha+4}}.$ Thus, $|D_x K(x,z)| \leqslant\frac{C}{|x-z|^{n-\alpha+2}}.$ Since $|x-z| \geqslant 2\delta$ and $|\xi-z| \geqslant \frac{1}{2}|x-z|$ for $\xi \in [x,y]$, we have: $|K(x,z) - K(y,z)| \leqslant C \frac{\delta}{|x-z|^{n-\alpha+2}}.$ Therefore, $\int_{\mathbb{R}^n \setminus B_{2\delta}(x)} \frac{\delta}{|x-z|^{n-\alpha+2}} dz = C \delta \int_{2\delta}^{\infty} \frac{r^{n-1}}{r^{n-\alpha+2}} dr = C \delta \int_{2\delta}^{\infty} r^{\alpha-3} dr = C \delta^{\alpha-1}.$ Combining both regions, we obtain: $|\triangledown( I_\alpha f)(x) - \triangledown( I_\alpha f)(y)| \leqslant C \|f\|_{L^\infty} |x-y|^{\alpha-1}.$ This shows that $\triangledown( I_\alpha f) \in C^{0,\alpha-1}(\mathbb{R}^n)$. When $\alpha=1,$ we derive the result by a mollifier technique and a cutoff argument. Therefore, based on the above two conclusions, we have (1) If $2s \leqslant 1,$ we have $\|R_{2s} * f\|_{C^{0,\alpha}} \leqslant C (\|f\|_{L^\infty}+\|u\|_{L^\infty}).$ for $\alpha<2s.$ Since $\mathcal{Z}_1 * f = z * (R_{2s} * f)$ and $z\in L^1(\mathbb{R}^n),$ convolution preserves Hölder continuity: $\|\mathcal{Z}_1 * f\|_{C^{0,\alpha}} \leqslant C (\|f\|_{L^\infty}+\|u\|_{L^\infty}),$ (2) If $2s > 1$, we have $\|R_{2s} * f\|_{C^{1,\alpha}} \leqslant C (\|f\|_{L^\infty}+\|u\|_{L^\infty}),$ for $\alpha<2s-1.$ Since $D^\beta (z * (R_{2s} * f)) = (D^\beta z) * (R_{2s} * f)$ and $D^\beta z \in L^1$, we get $\|\mathcal{Z}_1 * f\|_{C^{1,\alpha}} \leqslant C (\|f\|_{L^\infty}+\|u\|_{L^\infty}).$ Combining the above estimates, we get $u = \mathcal{Z}_1 * f + \mathcal{Z}_2 * f.$ If $2s \leqslant 1$: $\mathcal{Z}_1 * f \in C^{0,\alpha}$ ($\alpha < 2s$), $\mathcal{Z}_2 * f \in C^\infty$, so $u \in C^{0,\alpha}$, and $\|u\|_{C^{0,\alpha}} \leqslant C (\|f\|_{L^\infty}+\|u\|_{L^\infty}).$If $2s > 1$: $\mathcal{Z}_1 * f \in C^{1,\alpha}$ ($\alpha < 2s - 1$), $\mathcal{Z}_2 * f \in C^\infty$, so $u \in C^{1,\alpha}$, and $\|u\|_{C^{1,\alpha}} \leqslant C (\|f\|_{L^\infty}+\|u\|_{L^\infty}).\qedhere$ Let $0 < s < 1$. If $u \in L^\infty(\mathbb{R}^n)$ and $\mathcal{L}u = f \in C^{0,\alpha}(\mathbb{R}^n)$ with $0 < \alpha \leqslant 1$, then: If $\alpha + 2s \leqslant 1$, then $u \in C^{0,\alpha+2s}(\mathbb{R}^n)$, and $\|u\|_{C^{0,\alpha+2s}} \leqslant C(\|u\|_{L^\infty} + \|f\|_{C^{0,\alpha}}).$If $\alpha + 2s > 1$, then $u \in C^{1,\alpha+2s-1}(\mathbb{R}^n)$, and $\|u\|_{C^{1,\alpha+2s-1}} \leqslant C(\|u\|_{L^\infty} + \|f\|_{C^{0,\alpha}}).$ By MR290095 and MR2270163, we have the following result: Let $0 < \alpha < 1$, $0 < \beta < 2$. If $f \in C^{0,\alpha}(\mathbb{R}^n)$, then $\text{if } \alpha + \beta < 1, I_{\beta}f \in C^{0,\alpha+\beta}(\mathbb{R}^n)$ and $\text{if } 1 < \alpha + \beta < 2, I_{\beta}f \in C^{1,\alpha+\beta-1}(\mathbb{R}^n) .$ Whence, analogue to the arguments in the proof of Proposition \ref{['pro 4.1']}, the validity of Proposition \ref{['pro 4.2']} is established. Let $u \in \mathcal{X}^{1,2}(\mathbb{R}^n)$ be a nontrivial solution of \ref{['equ1']} with $p\in[1,2^*-1)$. Then $u \in C^{1,\theta}(\mathbb{R}^n)$, for any $\theta \in (0, 1)$ and $\Vert u\Vert _{C^{1,\theta}(\mathbb{R}^n)}\leqslant C(n,s,\theta),$ for some $C=C(n,s,\theta).$ By the uniform $L^{\infty}$-estimate of $u$ (Su-Xu) and the assumption ($h_1$) that $h\in L^{\infty}(\mathbb{R}^n)$, we get $\lambda h u^p+u^{2^*-1}\leqslant C\lambda \|h\|_{L^{\infty}}\|u\|_{L^{\infty}}^p+\|u\|_{L^{\infty}}^{2^*-1}\leqslant C(h,p,\lambda,2^*).$ Thus, the right-hand side of \ref{['equ1']} is bounded in $\mathbb{R}^n.$ Applying Proposition \ref{['pro 4.1']}, we have if $s\leqslant 1/2$, then for any $\alpha<2s$, $u\in C^{0,\alpha}(\mathbb{R}^n).$if $s> 1/2$, then for any $\alpha<2s-1$, $u\in C^{1,\alpha}(\mathbb{R}^n).$ This implies in particular that $g(h,u):=\lambda h u^p+u^{2^*-1}$ is $C^{\alpha}(\mathbb{R}^n).$ Applying now Proposition \ref{['pro 4.2']}, we have if $\alpha+2s\leqslant 1$, then $u\in C^{0,\alpha+2s}(\mathbb{R}^n).$if $\alpha+2s> 1$, then $u\in C^{1,\alpha+2s-1}(\mathbb{R}^n).$ Indeed, if $\alpha+2s>1,$ then one can take $\theta=\alpha+2s-1.$ On the other hand if $\alpha+2s\leqslant1,$ we have that $g(h,u)$ is $C^{0,\alpha+2s}.$ As a sequence, one gets that $u$ is $C^{0,\alpha+4s}.$ Hence iterating a finite number of times, we will end up with $\alpha+2ks>1$ for some integer $k$. Therefore, iterating the procedure a finite number of times, one gets that $u\in C^{1,\theta}(\mathbb{R}^n)$ for some $\theta\in(0,1)$ depending only on $s.$ Next, we will give the proof of Theorem \ref{['thm:1.2']}. For this, we combine a suitable truncation argument for the solution $u$ with the $C^{1,\alpha}$-regularity argument. By analogous reasoning to the proof of SerenaDipierro2025, the $C^{2,\alpha}$- regularity of the solution follows. For the convenience of the reader, we sketch the proof in the following subsections. To begin with, we introduce some notations of semi-norms as follows. For $\alpha \in (0, 1)$, $k \in \mathbb{N}$, $x_0 \in B_{3/4}$, and $R \in (0, \frac{1}{20})$, we denote the interior norms as follows: $[u]_{\alpha;B_R(x_0)} := \sup_{x,y \in B_R(x_0)} \frac{|u(x) - u(y)|}{|x - y|^{\alpha}},$ $|u|'_{k;B_R(x_0)} := \sum_{j=0}^{k} R^j \|D^ju\|_{L^\infty(B_R(x_0))}$ and $|u|'_{k,\alpha;B_R(x_0)} := |u|'_{k;B_R(x_0)} + R^{k+\alpha}[D^ku]_{\alpha;B_R(x_0)}.$ Step 1. A mollifier technique and a truncation argument. Let $u \in \mathcal{X}^{1,2}(\mathbb{R}^n)$ solve \ref{['equ1']} and $\eta_\varepsilon$ be a standard mollifier. For every $x \in \mathbb{R}^n$, $R \in (0, \frac{1}{20})$, and $\varepsilon \in (0, R)$, we denote $u_\varepsilon(x) := (\eta_\varepsilon * u)(x) = \int_{|y| \leqslant \varepsilon} \eta_\varepsilon(y)u(x-y) dy.$ Also, we set $g_\varepsilon := \eta_\varepsilon * (\lambda hu^p + u^{2^*-1}).$ Then, we have that $-\Delta u_\varepsilon + (-\Delta)^s u_\varepsilon = g_\varepsilon \quad \text{in } B_1.$ Moreover, the following regularity estimates on $u_\varepsilon$ and $g_\varepsilon$ follow as a direct consequence of their definitions. Let $u \in L^\infty(\mathbb{R}^n)$. Then, $u_\varepsilon \in L^\infty(\mathbb{R}^n)$ and $\|u_\varepsilon\|_{L^\infty(\mathbb{R}^n)} \leqslant \|u\|_{L^\infty(\mathbb{R}^n)}.$ If in addition $u \in C^1(\mathbb{R}^n)$, then for every $y \in \mathbb{R}^n$, $\|g_\varepsilon\|_{C^\alpha(B_1(y))} \leqslant c_{p,2^*} \left(\lambda\|h\|_{C^1(\mathbb{R}^n)} \|u\|^p_{C^1(\mathbb{R}^n)} + \|u\|_{C^1(\mathbb{R}^n)}^{2^*-1} \right)$ where $c_{p,2^*}$ is some positive constant depending only on $p, 2^*$. We now use a cut off argument for $u_\varepsilon$ to get the $C^{2,\alpha}$-estimate for $u_\varepsilon$. Consider a cut-off function $\phi \in C_0^\infty (\mathbb{R}^n)$ satisfying $\phi \equiv 1 \text{ in } B_{3/2}, \quad \text{supp}(\phi) \subset B_2, \quad \text{and} \quad 0 \leqslant \phi \leqslant 1 \text{ in } \mathbb{R}^n$ and let $\phi^R(x) := \phi \left( \frac{x - x_0}{R} \right).$ We point out that $B_{4R}(x_0) \subset B_1 \quad \text{and} \quad \text{supp}(\phi^R) \subset B_{2R}(x_0).$ With this notation, one obtains the following result. Let $\alpha \in (0, 1)$ and $g \in C_{\text{loc}}^{\alpha}(B_1)$. Let $u \in C^{2,\alpha}(B_1) \cap L^\infty(\mathbb{R}^n)$ be a solution of $-\Delta u + (-\Delta)^s u = g \quad \text{in } B_1.$ Then, there exists $\psi \in C^{\alpha}(B_R(x_0))$ such that $v := \phi^R u$ satisfies $-\Delta v + (-\Delta)^s v = \psi \quad \text{in } B_1.$ In particular, $R^2 |\psi|'_{0,\alpha; B_R(x_0)} \leqslant C_{n,s} \left( R^2 |g|'_{0,\alpha; B_R(x_0)} + \|u\|_{L^\infty(\mathbb{R}^n)} \right),$ for some positive constant $C_{n,s}$. As a consequence of Lemma \ref{['dghs']}, setting $v_\varepsilon := \phi^R u_\varepsilon$, we have that there exists $\psi_\varepsilon \in C^{\alpha}(B_R(x_0))$ such that $v_\varepsilon$ satisfies $-\Delta v_\varepsilon + (-\Delta)^s v_\varepsilon = \psi_\varepsilon \quad \text{in } B_1.$ In particular, employing \ref{['equoo']}, one finds that $R^2 |\psi'_\varepsilon|_{0,\alpha; B_R(x_0)} \leqslant C_{n,s} \left( R^2 |g_\varepsilon|'_{0,\alpha; B_R(x_0)} + \|u_\varepsilon\|_{L^\infty(\mathbb{R}^n)} \right).$ We also observe that, since $v_\varepsilon \in C_0^\infty(B_{2R}(x_0))$, for all $\delta > 0$, there exists $C_{\delta} > 0$ such that R^2 |(-\Delta)^s v_\varepsilon|'_{0,\alpha; B_R(x_0)}= R^2 \|(-\Delta)^s v_\varepsilon\|_{L^\infty(B_R(x_0))} + R^{2+\alpha}[(-\Delta)^s v_\varepsilon]_{\alpha; B_R(x_0)}\leqslant C_{n,s} |u_\varepsilon|'_{2,\alpha_0; B_{2R}(x_0)} \leqslant \delta |u_\varepsilon|'_{2,\alpha; B_{2R}(x_0)} + C_{\delta} \|u_\varepsilon\|_{L^\infty(B_{2R}(x_0))}, where $\alpha_0 := 0\alpha < 2 - 2s,\alpha - (1-s)\alpha \geqslant 2 - 2s.$ Thus, combining this with MR1814364, one could deduce that for all $\delta > 0$ there exists $C_{\delta} > 0$ such that |v_\varepsilon|'_{2,\alpha; B_{R/2}(x_0)}\leqslant C \left( \|v_\varepsilon\|_{L^\infty(B_R(x_0))} + R^2 \left( |\psi_\varepsilon|'_{0,\alpha; B_R(x_0)} + |(-\Delta)^s v_\varepsilon|'_{0,\alpha; B_R(x_0)} \right) \right)\leqslant C \left( \|v_\varepsilon\|_{L^\infty(B_R(x_0))} + R^2 |\psi_\varepsilon|'_{0,\alpha; B_R(x_0)} + \delta |u_\varepsilon|'_{2,\alpha; B_{2R}(x_0)} + C_{\delta} \|u_\varepsilon\|_{L^\infty(B_{2R}(x_0))} \right). Therefore, for every $x_0 \in B_{3/4}$, recalling the definition of $v_\epsilon$ and exploiting \ref{['equhh']} and Lemma \ref{['ghgh']}, we conclude that, for all $\delta > 0$ above, there exists $C_\delta$ such that $|u_\epsilon|'_{2,\alpha;B_{R/2}(x_0)} = |v_\epsilon|'_{2,\alpha;B_{R/2}(x_0)}\leqslant C \left( R^2 |\psi_\epsilon|'_{0,\alpha;B_R(x_0)} + \delta |u_\epsilon|'_{2,\alpha;B_{2R}(x_0)} + C_\delta \|u\|_{L^\infty(B_{2R}(x_0))} \right)\leqslant C \left( R^2 |g_\epsilon|'_{0,\alpha;B_R(x_0)} + \|u_\epsilon\|_{L^\infty(\mathbb{R}^n)} + \delta |u_\epsilon|'_{2,\alpha;B_{2R}(x_0)} + C_\delta \|u\|_{L^\infty(B_{2R}(x_0))} \right)\leqslant C \left( \lambda\|h\|_{C^1(\mathbb{R}^n)} \|u\|^p_{C^1(\mathbb{R}^n)} + \|u\|_{C^1(\mathbb{R}^n)}^{2^*-1} + \|u\|_{L^\infty(\mathbb{R}^n)} + \delta |u_\epsilon|'_{2,\alpha;B_{2R}(x_0)} + C_\delta \|u\|_{L^\infty(B_{2R}(x_0))} \right)\leqslant C \left( \lambda\|h\|_{C^1(\mathbb{R}^n)} \|u\|^p_{C^1(\mathbb{R}^n)} + \|u\|_{C^1(\mathbb{R}^n)}^{2^*-1} + C_\delta \|u\|_{L^\infty(\mathbb{R}^n)} + \delta |u_\epsilon|'_{2,\alpha;B_{2R}(x_0)} \right),$ for some $C > 0$ depending on $n$, $s$, $p$, $2^*$ and $\alpha$. Step 2. Interior $C^{2,\alpha}$-regularity. The estimate in \ref{['equggh']}, coupled with the following statement, will allow us to obtain that the $C^{2,\alpha}$-norm of $u_\epsilon$ is bounded uniformly in some ball with respect to $\epsilon$, and thus use Arzelà-Ascoli theorem to complete the proof of Theorem \ref{['thm:1.2']}. The technical details go as follows. Let $y \in \mathbb{R}^n$, $d > 0$, and $u \in C^{2,\alpha}(B_d(y))$. Suppose that, for any $\delta > 0$, there exists $\Lambda_\delta > 0$ such that, for any $x \in B_d(y)$ and any $r \in (0,d - |x-y|]$, we have that $|u|'_{2,\alpha;B_{r/8}(x)} \leqslant \Lambda_\delta + \delta |u|'_{2,\alpha;B_{r/2}(x)}.$ Then, there exist constants $\delta_0, C > 0$, depending only on $n$, $\alpha$ and $d$, such that $\|u\|_{C^{2,\alpha}(B_{d/8}(y))} \leqslant C \Lambda_{\delta_0}.$ We will use Proposition \ref{['prohh']} with $d := 1/10$ so that, for every $y \in B_{1/2}$, $B_d(y) \subset B_{3/4} \quad \text{and} \quad d/4 < \frac{1}{20}.$ Moreover, we notice that the estimate in \ref{['equggh']} tells us that formula \ref{['hhhj']} is verified in our setting with $u$ replaced by $u_\epsilon$, $r := 4R$ and $\Lambda_\delta :=\lambda\|h\|_{C^1(\mathbb{R}^n)} \|u\|^p_{C^1(\mathbb{R}^n)} + \|u\|_{C^1(\mathbb{R}^n)}^{2^*-1} + C_\delta \|u\|_{L^\infty(\mathbb{R}^n)}.$ Therefore, we are in a position to exploit Proposition \ref{['prohh']}, thus obtaining that, for every $y \in B_{1/2}$, $\|u_\epsilon\|_{C^{2,\alpha}(\overline{B_{1/80}(y)})} \leqslant C \left( \lambda\|h\|_{C^1(\mathbb{R}^n)} \|u\|^p_{C^1(\mathbb{R}^n)} + \|u\|_{C^1(\mathbb{R}^n)}^{2^*-1} + C_\delta \|u\|_{L^\infty(\mathbb{R}^n)} \right).$ From the Arzelà-Ascoli Theorem, we obtain that $u \in C^{2,\alpha}(\overline{B_{1/80}(y)})$, for every $y \in B_{1/2}$, and $\|u\|_{C^{2,\alpha}(\overline{B_{1/80}(y)})} \leqslant C \left( \lambda\|h\|_{C^1(\mathbb{R}^n)} \|u\|^p_{C^1(\mathbb{R}^n)} + \|u\|_{C^1(\mathbb{R}^n)}^{2^*-1} + \|u\|_{L^\infty(\mathbb{R}^n)}\right).$ Hence, a covering argument and Theorem \ref{['regu']} give that \|u\|_{C^{2,\alpha}(\overline{B_{1/2}})}\leqslant C \left( \lambda\|h\|_{C^1(\mathbb{R}^n)} \|u\|^p_{C^1(\mathbb{R}^n)} + \|u\|_{C^1(\mathbb{R}^n)}^{2^*-1} + \|u\|_{L^\infty(\mathbb{R}^n)} \right)\leqslant C \left( \lambda\|h\|_{C^{1}(\mathbb{R}^n)}+ \|u\|_{L^\infty(\mathbb{R}^n)} \right), where the constant $C > 0$ depends on $n$, $s$, $\alpha$, $2^*$ and $p$. The ball $B_y$ is centered at the origin, but we may arbitrarily move it around $\mathbb{R}^n$. Covering $\mathbb{R}^n$ with these balls, we obtain the desired result. According to the above regularity theorem, we prove the weak solutions obtained in Theorem \ref{['thm:1.1']} are viscosity solutions. First, we present the relevant stability results for viscosity solutions that are required in the proof process. A sequence of lower-semicontinuous functions $u_{k}$ $\Gamma$-converges to $u$ in a set $\mathbb{R}^n$ if the two following conditions hold: for every sequence $x_{k} \to x$ in $\mathbb{R}^n$, $\liminf_{k \to \infty} u_{k}(x_{k}) \geqslant u(x)$.for every $x \in \mathbb{R}^n$, there is a sequence $x_{k} \to x$ in $\mathbb{R}^n$ such that $\limsup_{k \to \infty} u_{k}(x_{k}) = u(x).$ It is natural that a uniformly convergent sequence $u_{k}$ also converges in the $\Gamma$-convergence sense. A fundamental property of $\Gamma$-limits is that if $u_{k}$ $\Gamma$-converges to $u$ and $u$ has a strict local minimum at $x$, then there exists a sequence of points $x_{k} \to x$ such that $u_{k}$ has a local minimum at $x_{k}$. Let $u_{k}$ be a sequence of functions that are uniformly bounded in $\mathbb{R}^{n}$ and lower-semicontinuous in $\mathbb{R}^n$ such that $-\Delta u_k+(-\Delta)^s u_{k} \leqslant f_{k}$ in $\mathbb{R}^n$,$u_{k} \to u$ in the $\Gamma$ sense in $\mathbb{R}^n$,$u_{k} \to u$ a.e. in $\mathbb{R}^{n}$,$f_{k} \to f$ locally uniformly in $\mathbb{R}^n$ for some continuous function $f$. Then $-\Delta u+(-\Delta)^s u \leqslant f$ in $\mathbb{R}^n$. Let $\varphi$ be a test function from below for $u$ touching at a point $x$ in a neighborhood $N$. Since $u_{k}$ $\Gamma$-converges to $u$ in $\mathbb{R}^n$, for large $k$ we can find $x_{k}$ and $d_{k}$ such that $\varphi + d_{k}$ touches $u_{k}$ at $x_{k}$. Moreover, $x_{k} \to x$ and $d_{k} \to 0$ as $k \to +\infty$. Since $-\Delta u_k+(-\Delta)^s u_{k} \leqslant f_{k}$, if we let $v_{k} = \varphi + d_{k}\text{in } N,u_{k}\text{in } \mathbb{R}^{n} \setminus N,$ we have $-\Delta v_k(x_k)+(-\Delta)^s v_{k}(x_k) \leqslant f_{k}(x_{k}, v(x_k))$. Next, prove that $-\Delta v_k(z)+(-\Delta)^s v_k(z)\to -\Delta v(z)+(-\Delta)^s v(z)$ is locally uniform in $N.$ Since $v_k-v=d_k$ in $N$ and (iii), we have $(-\Delta+(-\Delta)^s )v_k\to (-\Delta+(-\Delta)^s )v$ locally uniformly in $N.$ Finally, we have that $-\Delta v+(-\Delta)^s v$ is continuous in $N$. We now compute \left|(-\Delta) v_k(x_k)+(-\Delta)^s v_k(x_k)-(-\Delta) v(x)+(-\Delta)^s v(x)\right|\leqslant \left|\left((-\Delta+(-\Delta)^s)\right)v_k(x_k)-\left((-\Delta+(-\Delta)^s)\right)v(x_k)\right|+\left|\left((-\Delta+(-\Delta)^s)\right)v(x_k)-\left((-\Delta+(-\Delta)^s)\right)v(x)\right|. So $-\Delta v_{k}(x_{k})+(-\Delta)^s v_k(x_k)$ converges to $-\Delta v(x_{k})+(-\Delta)^s v(x_k)$ as $k \to \infty$. Since $x_{k} \to x$ and $f_{k} \to f$ locally uniformly, we also have $f_{k}(x_{k}) \to f(x, v(x))$, which finally implies $-\Delta v(x)+(-\Delta)^s v(x) \leqslant f(x, v(x))$. By Lemma \ref{['regu']} and Theorem \ref{['thm:1.1']}, we have: $u \in C(\mathbb{R}^n),$ which satisfies the basic continuity requirement for the definition of viscosity solutions. Let $f(x,u(x)):=\lambda h u^p+u^{2^*-1}.$ Now, let us fix $x_0\in\mathbb{R}^n.$ We consider the mollifier $\rho_{\epsilon}(x):=\epsilon^{-n}\rho(x/\epsilon),\epsilon>0.$ We define $u_{\epsilon}:=u*\rho_{\epsilon}$ and $f_{\epsilon}:=f*\rho_{\epsilon}.$ Then we recalling the stability of viscosity solutions (see Lemma \ref{['stab']}) we may suppose that $u$ is also smooth in a neighborhood $\Omega$ of $x_0$ and, from \ref{['equu:2.5']}, for any $\varphi\in C_0^{\infty}(\Omega)$ $\int_{\mathbb{R}^n}-\Delta u \varphi dx+\int_{\mathbb{R}^{2n}}(-\Delta)^s u \varphi dx=\int_{\mathbb{R}^n}\triangledown u(x)\cdot\triangledown \varphi(x)\, dx+\iint_{\mathbb{R}^{2n}}\frac{\left(u(x)-u(y)\right)\left(\varphi(x)-\varphi(y)\right)}{|x-y|^{N+2s}}\, dxdy\,\,\,\,\,=\int_{\mathbb{R}^n}f(x,u(x)) \varphi \, dx.$ Therefore, $-\Delta u+(-\Delta)^s u=f(x,u(x))$ for almost any $x\in\Omega$ and, in fact, for any $x\in\Omega.$ Then take a test function $\phi \in C^2(B_R(x_0))$ such that: $\phi(x_0) = u(x_0), \quad \phi(x) \geqslant u(x) \quad \forall x \in B_R(x_0).$ Define a function $v$ by $v(x) = \phi(x),x \in B_R(x_0),u(x),x \in \mathbb{R}^n \setminus B_R(x_0).$ We need to prove: $-\Delta v(x_0) + (-\Delta)^s v(x_0) \leqslant f(x_0,v(x_0)).$ At the minimum point, we have $-\Delta v(x_0) \leqslant -\Delta \phi(x_0),$ then $-\Delta v(x_0) \leqslant -\Delta u(x_0).$ Since $v\geqslant u$ in $\mathbb{R}^n$ and $v(x_0)=u(x_0)$, we obtain $(-\Delta)^s v(x_0)= c(n,s) \int_{\mathbb{R}^n} \frac{2v(x_0) - v(x_0+y) - v(x_0-y)}{|y|^{n+2s}} \, dy\leqslant c(n,s) \int_{\mathbb{R}^n} \frac{2u(x_0) - u(x_0+y) - u(x_0-y)}{|y|^{n+2s}} \, dy.$ Combining \ref{['equ4.2.']} and \ref{['equ4.3.']}, we get that $-\Delta v(x_0)+(-\Delta)^s v(x_0)\leqslant -\Delta u(x_0)+(-\Delta)^s u(x_0)=f(x_0, v(x_0)).$ This shows that $u$ is a viscosity subsolution, and one can prove similarly that $u$ is a viscosity supersolution. Since $u$ is both a viscosity subsolution and a viscosity supersolution of equation \ref{['equ1']}, we conclude that $u$ is a viscosity solution of \ref{['equ1']}. In this section, we will discuss the power-type decay estimates and radial symmetry of the positive classical solutions to equation \ref{['equ1']}. In Section \ref{['sec5']}, we apply the maximum principle and comparison arguments to obtain the decay at infinity of positive classical solutions. In Section \ref{['sec6']}, we employ the method of moving planes to show that these classical solutions are also radially symmetric. In this section, we shall apply the maximum principle and comparison arguments to obtain the decay at infinity of positive classical solutions. First, we devote to establishing the existence of positive classical solutions. Let $n\geqslant 4$ and $s\in(0,1)$. Then, problem \ref{['equ1']} has a classical solution, which satisfies $u>0$ in $\mathbb{R}^{n}$. Moreover, $\lim_{|x|\to+\infty}u(x)=0.$ From Theorem \ref{['thm:1.2']}, we have that there exists a nonnegative classical solution of \ref{['equ1']}. Now, we focus on proving that $u>0$ in $\mathbb{R}^{n}$. For this, we argue by contradiction, and we assume that there exists a global minimum point $x_{0}\in\mathbb{R}^{n}$ at which $u(x_{0})=0$. Accordingly, we have that $\Delta u(x_{0})\geqslant 0$ and $(-\Delta)^{s}u(x_{0})<0$. As a result, we deduce from \ref{['equ1']}, $h>0$ and $\lambda>0$ that $0=\lambda hu^{p}(x_{0})+u^{2^*-1}(x_0)=-\Delta u(x_{0})+(-\Delta)^{s}u(x_{0})+u(x_{0})<0,$ which is a contradiction. On the other hand, \ref{['5.10equ']} follows immediately from Theorem \ref{['regu']}. To prove the decay estimate of positive solutions, we start with the following two lemmas, which construct suitable subsolutions and supersolutions. Let $n>4s$, There exists a function $\omega \in C^{1,\alpha}(\mathbb{R}^n)$ satisfying $-\Delta \omega + (-\Delta)^s \omega = 0\text{in } \mathbb{R}^n \setminus B_1,\omega > 0\text{in } \mathbb{R}^n,\lim_{|x| \to +\infty} \omega(x) = 0$ in the distribution sense and for every $|x| > 1$, $\omega(x) \geqslant \frac{c_1}{|x|^{n+2s}},$ for some constant $c_1 > 0$. Define $\omega := \mathcal{Z} * \chi_{B_{1/2}}$, where $\chi_{B_{1/2}}$ is the characteristic function of the ball $B_{1/2}$. Since $\chi_{B_{1/2}}\in L^q(\mathbb{R}^n)$ for $q\geqslant1,$ we have that $\omega\in C^{1,\alpha}(\mathbb{R}^n)$ for any $\alpha\in(0,1).$ As a result, the limit in \ref{['equ4.10']} is also satisfied. By \ref{['sou']}, for any $f\in L^2(\mathbb{R}^n)$, $u=\mathcal{Z}*f$ satisfies: $-\Delta u+(-\Delta)^s u =f,$ Taking $f=\chi_{B_{1/2}}$ we have in $\mathbb{R}^n\setminus B_1$ that $f\equiv0$, hence $\omega > 0$ in $\mathbb{R}^n$ solves $-\Delta \omega + (-\Delta)^s \omega = 0 \quad \text{in } \mathbb{R}^n\setminus B_1,$ and therefore the equation in \ref{['equ4.10']} is satisfied. Using the lower bound on $\mathcal{Z}$, for any $|x| > 1$, one has that \omega(x)= \int_{B_{1/2}} \mathcal{Z}(x - y) dy \geqslant c \int_{B_{1/2}} \frac{dy}{|x - y|^{n+2s}}\geqslant c \int_{B_{1/2}} \frac{dy}{(|x| + |y|)^{n+2s}} \geqslant \frac{c}{|x|^{n+2s}}, up to renaming $c > 0$. As a consequence of this, we complete the proof of the lemma. Let $n>4,$ there exists $v \in C^{1,\alpha}(\mathbb{R}^n)$ satisfying $-\Delta v + (-\Delta)^s v = 0\text{in } \mathbb{R}^n \setminus B_1,v > 0\text{in } \mathbb{R}^n,\lim_{|x|\to+\infty} v(x) = 0.$ Also, when $|x| > 1$, we have that $0 < v(x) \leqslant \frac{c_2}{|x|^{n-2s}},$ for some constant $c_2 > 0$. The proof is similar to the lemma \ref{['lem4.6']}, we will now present the parts where it differs. We also consider the function $v := \mathcal{Z} * \chi_{B_{1/2}}$, and it satisfies $-\Delta v + (-\Delta)^s v = \chi_{B_{1/2}} \quad \text{in } \mathbb{R}^n.$ Moreover, from Lemma \ref{['lem4.1']}, when $|x| > 1$, we have that v(x)= \int_{B_{1/2}} \mathcal{Z}(x - y) dy \leqslant \int_{B_{1/2}} \frac{c dy}{|x - y|^{n-2s}}\leqslant \int_{B_{1/2}} \frac{c dy}{(|x| - 1/2)^{n-2s}} \leqslant \frac{c}{|x|^{n-2s}}, for some positive constant $c$ that may change from step to step, and also that $\lim_{|x| \to +\infty} v(x) = 0.$ We consider the function $\omega$ given by Lemma \ref{['lem4.6']}. Utilizing the positivity and continuity of $u$ and $\omega$ in $\mathbb{R}^n$, we can find a constant $\beta > 0$ such that $h := u - \beta \omega > 0$ in $B_1$. Furthermore, from \ref{['equ1']} and Lemma \ref{['lem4.6']}, we see that $-\Delta h + (-\Delta)^s h \geqslant 0\text{in } \mathbb{R}^n \setminus B_1,\lim_{|x| \to +\infty} h(x) = 0.$ We claim that $h(x) \geqslant 0 \quad \text{for all } |x| > 1.$ Indeed, let $h^- = \max(-h, 0)$ be the negative part of $u$. Since $h \in \mathcal{X}^{1,2}(\mathbb{R}^n)$, we have $h^- \in \mathcal{X}^{1,2}(\mathbb{R}^n)$. Using $h^-$ as a test function: \langle \mathcal{L} h, h^- \rangle= \int_{\mathbb{R}^n} \nabla h \cdot \nabla h^- \, dx + \iint_{\mathbb{R}^{2n}} \frac{(h(x)-h(y))(h^-(x)-h^-(y))}{|x-y|^{n+2s}} \, dx\, dy . Note that: $\nabla h \cdot \nabla h^- = -|\nabla h^-|^2$$(h(x)-h(y))(h^-(x)-h^-(y)) = -(h^-(x)-h^-(y))^2 - h^+(x)h^-(y) - h^-(x)h^+(y) \leqslant 0$ Hence $\langle \mathcal{L} h, h^- \rangle \leqslant 0$. But by assumption $\langle \mathcal{L} h, h^- \rangle \geqslant 0$, so we must have: $\langle \mathcal{L} h, h^- \rangle = 0$ This forces $h^- \equiv 0$, thus $h \geqslant 0$ in $|x|\geqslant 1$. Hence, for every $|x| > 1$, $u(x) \geqslant \beta \omega(x) \geqslant \frac{c_1}{|x|^{n+2s}}.$ This establishes the bound from below in Theorem \ref{['thm4.8']}. We now focus on the bound from above. Owing to $\lim\limits_{|x|\to+\infty}u(x)=0,$ we can find some $R>1$ such that $-\Delta u+(-\Delta)^{s}u= 0\qquad\text{in }\mathbb{R}^{n} \setminus B_{R}.$ In this case, we make use of the function $v$ given by Lemma \ref{['lem4.7']}. From the positivity and continuity of $u$ and $v$ in $\mathbb{R}^{n}$, there exists $\gamma>0$ such that $g:=v-\gamma u>0$ in $\overline{B_{R}}$. In view of Theorem \ref{['thm5.7']} and Lemma \ref{['lem4.7']}, one has that $-\Delta g+(-\Delta)^{s}g=0\text{in } \mathbb{R}^{n}\setminus B_{R},\lim\limits_{|x|\to+\infty}g(x)=0.$ We claim that $g(x)\geqslant 0\qquad\text{for all }|x|>R.$ Indeed, let $g^- = \max(-g, 0)$ be the negative part of $u$. Since $g \in \mathcal{X}^{1,2}(\mathbb{R}^n)$, we have $g^- \in \mathcal{X}^{1,2}(\mathbb{R}^n)$. Using $g^-$ as a test function: \langle \mathcal{L} g, g^- \rangle= \int_{\mathbb{R}^n} \nabla g \cdot \nabla g^- \, dx + \iint_{\mathbb{R}^{2n}} \frac{(g(x)-g(y))(g^-(x)-g^-(y))}{|x-y|^{n+2s}} \, dx\, dy . Hence $\langle \mathcal{L} g, g^- \rangle \leqslant 0$. But by assumption $\langle \mathcal{L} g, g^- \rangle \geqslant 0$, so we must have: $\langle \mathcal{L} g, g^- \rangle = 0.$ This forces $g^- \equiv 0$, thus $g \geqslant 0$ in $|x|\geqslant R$. In turn, \ref{['equ5.13']} implies that, for all $|x|>R$, $u(x)\leqslant\gamma v(x)\leqslant\frac{c_{2}}{|x|^{n-2s}}.$ Thus, the continuity of $u$ allows us to complete the proof of the bound from above in Theorem \ref{['thm4.8']}. Our aim is now to prove the radial symmetry of the positive solution (Theorem \ref{['thm4.8']}). The main statement of this section is the following. Let $n \geqslant 4$ and $s \in (0, 1)$. Then, all positive solutions of \ref{['equ1']} are radially symmetric about some point in $\mathbb{R}^n$ when $p\in[1, 2^*-1)$. To establish Theorem \ref{['thm5.1']}, we will exploit the moving planes method. To begin with, we recall some notations. 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