Existence of Solutions to a Conformally Invariant Integral Equation Involving Poisson-Type Kernels

Abstract

In this paper, we study existence of solutions to a conformally invariant integral equation involving Poisson-type kernels. Such integral equation has a stronger non-local feature and is not the dual of any PDE. We obtain the existence of solutions in the antipodal symmetry class.

1 Introduction

In [9], Hang–Wang–Yan established the following sharp integral inequality:

$$\begin{aligned} \Vert v \Vert _{ L^\frac{2 n}{n - 2} (B_1) } \le n^{ - \frac{n - 2}{2 (n - 1)} } \omega _n^{ - \frac{n - 2}{2 n (n - 1)} } \Vert v \Vert _{ L^\frac{2 (n - 1)}{n - 2} (\partial B_1) } \end{aligned}$$

(1.1)

for every harmonic function v on the unit ball $B_1 \subset \mathbb R^n$ ($n \ge 3$), where $\omega _n$ is the Euclidean volume of $B_1$. They also classified all the maximizers by showing that the equality holds if and only if $v = \pm 1$ up to a conformal transform on the unit sphere $\partial B_1$. This is actually a higher dimensional generalization of Carleman’s inequality [2], which was used by Carleman to prove the classical isoperimetric inequality. Let $g_{\mathbb R^n}$ be the Euclidean metric on $\mathbb R^n$. Then for a positive harmonic function v on $B_1$, the scalar curvature of $g = v^\frac{4}{n - 2} g_{\mathbb R^n}$ on $B_1$ is identically zero. Moreover, under the metric g, the volume of $B_1$ and the area of $\partial B_1$ are equal to $\int _{B_1} v^\frac{2 n}{n - 2} d \xi$ and $\int _{\partial B_1} v^\frac{2 (n - 1)}{n - 2} d s$, respectively. Hence, the inequality (1.1) can be considered as an isoperimetric inequality in the conformal class of $g_{\mathbb R^n}$ for which the scalar curvature vanishes. In [10], Hang-Wang-Yan further obtained a generalization of (1.1) on a smooth compact Riemannian manifold of dimension $n \ge 3$ with non-empty boundary by introducing an isoperimetric ratio over the scalar-flat conformal class. It was conjectured there that unless the manifold is conformally diffeomorphic to the Euclidean ball, the supremum of the isoperimetric ratio over the scalar-flat conformal class is always strictly larger than that in the Euclidean ball, so that the maximizers would exist. This conjecture was confirmed in higher dimensions under certain geometric assumptions by Jin-Xiong [13] and Chen-Jin-Ruan [4], and also was confirmed for balls with a small hole by Gluck-Zhu [8].

Using the Möbius transformation in (1.5), the equivalent form of (1.1) in the upper half-space is given by

$$\begin{aligned} \Vert \mathcal {P} u \Vert _{ L^\frac{2 n}{n - 2} (\mathbb R^n_+) } \le n^{ - \frac{n - 2}{2 (n - 1)} } \omega _n^{ - \frac{n - 2}{2 n (n - 1)} } \Vert u \Vert _{ L^\frac{2 (n - 1)}{n - 2} (\mathbb R^{n - 1}) }, \end{aligned}$$

(1.2)

where $\mathbb R^{n - 1}$ is the boundary of $\mathbb R^n_+$ and $\mathcal {P} u$ is the Poisson integral of u in the upper half-space. The maximizers are $u(y') = c (\lambda ^2 + |y' - y'_0|^2)^{ - \frac{n - 2}{2} }$ for some constant c, positive constant $\lambda$, and $y'_0 \in \mathbb R^{n - 1}$. In [3], Chen proved an analogous inequality for a one-parameter family $\{ \mathcal {P}_a \}_{2 - n< a < 1}$ of Poisson-type kernels in $\mathbb R^n_+$. More specifically, let the parameter a satisfy $2 - n< a < 1$ with $n \ge 2$, and define the Poisson-type kernels

$$\begin{aligned} P_a (y', x) = c_{n, a} \frac{ x_n^{1 - a} }{ (|x' - y'|^2 + x_n^2)^\frac{n - a}{2} } ~~~~~~ for ~ y' \in \mathbb R^{n - 1}, ~ x \in \mathbb R_+^n, \end{aligned}$$

where $x = (x', x_n) \in \mathbb R_+^n = \mathbb R^{n - 1} \times (0, + \infty )$ and $c_{n, a}$ is the positive normalization constant such that $\int _{ \mathbb R^{n - 1} } P_a (y', x) d y' = 1$. Consider the following Poisson-type integral

$$\begin{aligned} (\mathcal {P}_a u) (x) = \int _{ \mathbb R^{n - 1} } P_a (y', x) u(y') d y' ~~~~~~ for ~ x \in \mathbb R_+^n. \end{aligned}$$

(1.3)

It becomes the Poisson integral when $a = 0$ (i.e., $\mathcal {P}_0 = \mathcal {P}$). Chen [3] proved the following sharp integral inequality

$$\begin{aligned} \Vert \mathcal {P}_a u \Vert _{ L^\frac{2 n}{n + a - 2} (\mathbb R^n_+) } \le \mathcal {S}_{n, a} \Vert u \Vert _{ L^\frac{2 (n - 1)}{n + a - 2} (\mathbb R^{n - 1}) }, \end{aligned}$$

(1.4)

where the sharp constant $\mathcal {S}_{n, a}$ depends only on n and a. This Poisson-type integral (1.3) was used earlier by Caffarelli-Silvestre [1] to localize the fractional Laplacian operator. Indeed, when $-1<a<1$, then it was shown in [1] that

$$\begin{aligned} \begin{aligned} \text{ div } [x_n^{a}\nabla (\mathcal {P}_a u)]&=0\quad \text{ in } \mathbb R^n_+,\\ -\lim _{x_n\rightarrow 0^+} x_n^{a}\partial _{x_n} (\mathcal {P}_a u)&= C_{n,a} (-\Delta )^{\frac{1-a}{2}} u\quad \text{ on } \mathbb R^{n-1}, \end{aligned} \end{aligned}$$

where $C_{n,a}$ is a positive constant and $(-\Delta )^{\frac{1-a}{2}}$ is the fractional Laplacian operator. See also Yang [19] for higher order extensions for the fractional Laplacian. We refer to Dou-Guo-Zhu [5], Gluck [7] and the references therein for other related integral inequalities.

One can define the Poisson-type integral $\widetilde{ \mathcal {P} }_a v$ on $B_1$ as the pull back operator of $\mathcal {P}_a$ via the Möbius transformation:

$$\begin{aligned} F: \ \overline{\mathbb R_+^n} \rightarrow \overline{B}_1, ~~~~~~ x \mapsto \frac{2 (x + e_n)}{|x + e_n|^2} - e_n, \end{aligned}$$

(1.5)

where $e_n = (0, \dots , 0, 1) \in \mathbb R^n$. Then for $y'\in \mathbb R^{n-1}$,

$$\begin{aligned} F(y', 0) = \bigg ( \frac{2 y'}{1 + |y'|^2}, \frac{1 - |y'|^2 }{1 + |y'|^2} \bigg ) \in \partial B_1 \end{aligned}$$

is the inverse of the stereographic projection. For $v \in L^\frac{2 (n - 1)}{n + a - 2} (\partial B_1)$, let

$$\begin{aligned} u(y') = \bigg ( \frac{ \sqrt{2} }{|(y',0) + e_n|} \bigg )^{n + a - 2} v( F(y', 0) ), \end{aligned}$$

and define

$$\begin{aligned} (\widetilde{ \mathcal {P} }_a v) ( F(x) )=\bigg ( \frac{|x + e_n|}{ \sqrt{2} } \bigg )^{n + a - 2} (\mathcal {P}_a u) (x). \end{aligned}$$

That is,

$$\begin{aligned} (\widetilde{ \mathcal {P} }_a v) \circ F(x) = |x + e_n|^{n + a - 2} \mathcal {P}_a \bigg ( \frac{ v \circ F(y', 0) }{ |(y',0) + e_n|^{n + a - 2} } \bigg ) ~~~~~~ for ~ v \in L^\frac{2 (n - 1)}{n - 2 + a} (\partial B_1). \end{aligned}$$

By a direct calculation, for $v \in L^\frac{2 (n - 1)}{n + a - 2} (\partial B_1)$, the Poisson-type integral $\widetilde{\mathcal {P}}_a v$ on the unit ball has the following explicit form:

$$\begin{aligned} (\widetilde{ \mathcal {P} }_a v) (\xi ) = \int _{\partial B_1} \widetilde{P}_a (\eta , \xi ) v(\eta ) d s_\eta ~~~~~~ for ~ \xi \in B_1, \end{aligned}$$

(1.6)

where

$$\begin{aligned} \widetilde{P}_a (\eta , \xi ) = 2^{a - 1} c_{n, a} \frac{ (1 - |\xi |^2)^{1 - a} }{ |\xi - \eta |^{n - a} }. \end{aligned}$$

Then, it follows from (1.4) that we have the following sharp inequality

$$\begin{aligned} \Vert \widetilde{ \mathcal {P} }_a v \Vert _{ L^\frac{2 n}{n + a - 2} (B_1) } \le \mathcal {S}_{n, a} \Vert v \Vert _{ L^\frac{2 (n - 1)}{n + a - 2} (\partial B_1) }. \end{aligned}$$

(1.7)

From now on, for simplicity, we will use the unified notation $\mathcal {P}_a v$ to denote either the Poisson-type integral (1.3) of v on the upper half space or the Poisson-type integral (1.6) of v on the unit ball, whenever there is no confusion. Inspired by Hang-Wang-Yan [10] on the proof of inequality (1.1), for a positive function $K \in C^1 (\partial B_1)$ we consider the weighted isoperimetric ratio

$$\begin{aligned} I(v, K) = \frac{ \int _{B_1} |\mathcal {P}_a v|^\frac{2 n}{n + a - 2} d \xi }{ \Big ( \int _{\partial B_1} K |v|^\frac{2 (n - 1)}{n + a - 2} d s \Big )^\frac{n}{n - 1} } ~~~~~~ for ~ v \in L^\frac{2 (n - 1)}{n + a - 2} (\partial B_1). \end{aligned}$$

In this paper, motivated by the classical Nirenberg problem we would like to study existence of positive solutions to the Euler-Lagrange equation of the functional I(v, K) for a given function $K > 0$. The Euler-Lagrange equation can be written as the following integral equation

$$\begin{aligned} K(\eta ) v(\eta )^\frac{n - a}{n + a - 2} = \int _{B_1} P_a (\eta , \xi ) \left[ (\mathcal {P}_a v) (\xi )\right] ^\frac{n - a + 2}{n + a - 2} d \xi , ~~~~~~ v > 0 ~~~~~~ on ~ \partial B_1. \end{aligned}$$

(1.8)

This equation is critical and conformally invariant. Moreover, it is not always solvable by a Kazdan-Warner type obstruction (see Lemma 3.1 of Hang-Wang-Yan [10] for $a=0$). In this paper, we show the following existence result.

Theorem 1.1

Suppose that $n \ge 2$ and $2 - n< a < 1$. Let $K \in C^1 (\partial B_1)$ be a positive function satisfying $K(\xi ) = K(- \xi )$ for every $\xi \in \partial B_1$. If

$$\begin{aligned} \frac{ \max _{\partial B_1} K }{ \min _{\partial B_1} K } < 2^\frac{1}{n}, \end{aligned}$$

(1.9)

then equation (1.8) has at least one positive Hölder continuous solution.

The existence of solutions to the Nirenberg problem for prescribed antipodal symmetric functions was established by Moser [14] in dimension two, and by Escobar-Schoen [6] in higher dimensions under a flatness assumption near the prescribed function’s maximum point. For the generalized Nirenberg problem for Q-curvature and fractional Q-curvatures, similar results have been obtained by Robert [15] and Jin-Li-Xiong [11, 12], respectively. In the case $a = 0$, the existence of solutions to (1.8) with antipodal symmetric functions K has been proved by Xiong [18] under a global flatness condition at K’s minimum point. Our condition is slightly weaker, although it is still a (not arbitrarily small, though) perturbation result. We do not know whether a local flatness condition would be sufficient. The difficulty is that the antipodal symmetry does not provide a desirable positive mass in our setting, which is different from the Nirenberg problem or the Yamabe problem. Note that equation (1.8) has a stronger non-local feature and is not the dual of any PDE. This, as already shown in [18], will lead to some differences from the classical Nirenberg problem [12].

This paper is organized as follows. In Sect. 2, we collect some elementary properties of the Poisson extension as a preparation. In Sect. 3, we show the blow up procedure for the non-linear integral equation (1.8). In Sect. 4, we use a variational method to prove Theorem 1.1.

2 Preliminaries

From now on, we denote $x = (x', x_n) \in \mathbb R^{n - 1} \times \mathbb R$ as the point in $\mathbb R^n$, $B_R (x)$ as the open ball of $\mathbb R^n$ with radius R and center x, $B_R^+(x)$ as $B_R(x) \cap \mathbb {R}_+^n$, and $B'_R (x')$ as the open ball in $\mathbb R^{n - 1}$ with radius R and center $x'$. For simplicity, we also write $B_R(0)$, $B_R^+(0)$ and $B'_R (0)$ as $B_R$, $B_R^+$ and $B'_R$, respectively.

Here we list several properties of the Poisson-type extension operator $\mathcal {P}_a$.

Proposition 2.1

Suppose that $n \ge 2$ and $2 - n< a < 1$. If $1 \le p < \infty$ and $1 \le q < \frac{n p}{n - 1}$, then the operator

$$\begin{aligned} \mathcal {P}_a : L^p (\mathbb R^{n - 1}) \rightarrow L_{loc}^q (\overline{\mathbb R_+^n}) \end{aligned}$$

is compact.

Proof

The proof is the same as that of [9, Corollary 2.2]. $\square$

Corollary 2.2

Suppose that $n \ge 2$ and $2 - n< a < 1$. If $1 \le p < \infty$ and $1 \le q < \frac{n p}{n - 1}$, then the operator

$$\begin{aligned} \mathcal {P}_a : L^p (\partial B_1) \rightarrow L^q (B_1) \end{aligned}$$

is compact.

Proof

The proof is the same as that of [10, Corollary 2.1]. $\square$

In order to establish regularity, we need the following simple fact

$$\begin{aligned} |\nabla _{x'}^k P_a (y', x)| = |\nabla _{y'}^k P_a (y', x)| \le C(n, a, k) x_n^{1 - a} (|x' - y'|^2 + x_n^2)^{ - \frac{n - a + k}{2} } \end{aligned}$$

(2.1)

for $x' \ne y'$ and $k \ge 1$.

Theorem 2.3

Suppose that $n \ge 2$, $2 - n< a < 1$ and $\frac{2 (n - 1)}{n + a - 2} \le p < \infty$. Let $K \in C^1 (\mathbb R^{n - 1})$ be a positive function. If $u \in L_{loc}^p (\mathbb R^{n - 1})$ is non-negative, not identically zero and satisfies

$$\begin{aligned} K (y') u(y')^{p - 1} = \int _{\mathbb R_+^n} P_a (y', x) \left[ (\mathcal {P}_a u) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x, \end{aligned}$$

(2.2)

then $u \in C_{loc}^\beta (\mathbb R^{n - 1})$ for any $\beta \in (0, 1)$.

The proof of this Hölder regularity is given in the appendix.

Finally, we also need the following Liouville theorem proved by Wang-Zhu [17].

Theorem 2.4

(Wang-Zhu [17]) Suppose that $n \ge 2$ and $a < 1$. If $\Phi \in C^2 (\mathbb R_+^n) \cap C^0 (\overline{\mathbb R_+^n})$ is a solution of

$$\begin{aligned} \left\{ \begin{aligned} - \, \textrm{div} (x_n^{a} \nabla \Phi )&= 0 ~~~~~~ in ~ \mathbb R_+^n, \\ \Phi&= 0 ~~~~~~ on ~ \mathbb R^{n - 1} \end{aligned} \right. \end{aligned}$$

and is bounded from below in $\mathbb R_+^n$. Then

$$\begin{aligned} \Phi (x) = C x_n^{1 - a} \end{aligned}$$

for some constant $C \ge 0$.

3 A Blow-Up Analysis

The local blow up analysis for the non-local integral equation (2.2) is as follows.

Theorem 3.1

Suppose that $n \ge 2$ and $2 - n< a < 1$. Let $\frac{2 (n - 1)}{n + a - 2} \le p_i < \frac{2 n}{n + a - 2}$ be a sequence of numbers with $\lim _{i \rightarrow \infty } p_i = \frac{2 (n - 1)}{n + a - 2}$, and $K_i \in C^1 (B_1')$ be a sequence of positive functions satisfying

$$\begin{aligned} K_i \ge \frac{1}{c_0}, ~~~~~~ \Vert K_i \Vert _{C^1 (B_1')} \le c_0 \end{aligned}$$

for some constant $c_0 \ge 1$ independent of i. Suppose that $u_i \in C(\mathbb R^{n - 1})$ is a sequence of non-negative solutions of

$$\begin{aligned} K_i (y') u_i (y')^{p_i - 1} = \int _{\mathbb R_+^n} P_a (y', x) \left[ (\mathcal {P}_a u_i) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x ~~~~~~ for ~ y' \in B_1' \end{aligned}$$

(3.1)

and $u_i (0) \rightarrow + \infty$ as $i \rightarrow \infty$. Suppose that $R_i u_i (0)^{ p_i - \frac{2 n}{n + a - 2} } \rightarrow 0$ for some $R_i \rightarrow + \infty$ and

$$\begin{aligned} u_i (y') \le b u_i (0) ~~~~~~ for ~ |y'| < R_i u_i (0)^{ p_i - \frac{2 n}{n + a - 2} }, \end{aligned}$$

(3.2)

where $b > 0$ is independent of i. Then, after passing to a subsequence, we have

$$\begin{aligned} \phi _i (y') : = \frac{1}{u_i (0)} u_i \big ( u_i (0)^{ p_i - \frac{2 n}{n + a - 2} } y' \big ) \rightarrow \phi (y') ~~~~~~ in ~ C_{loc}^{1/2} (\mathbb R^{n - 1}), \end{aligned}$$

(3.3)

where $\phi > 0$ satisfies

$$\begin{aligned} K \phi (y')^\frac{n - a}{n + a - 2} = \int _{\mathbb R_+^n} P_a (y', x) \left[ (\mathcal {P}_a \phi ) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x ~~~~~~ for ~ y' \in \mathbb R^{n - 1} \end{aligned}$$

and $K : = \lim _{i \rightarrow \infty } K_i (0) > 0$ along the subsequence.

Proof

It follows from (3.1) and (3.3) that $\phi _i$ satisfies the equation

$$\begin{aligned} H_i (y') \phi _i (y')^{p_i - 1} = \int _{\mathbb R_+^n} P_a (y', x) \left[ (\mathcal {P}_a \phi _i) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x ~~~~~~ for ~ |y'| < R_i, \end{aligned}$$

(3.4)

where $H_i (y') : = K_i \big ( u_i (0)^{ p_i - \frac{2 n}{n + a - 2} } y' \big )$. Moreover, by (3.2), we have

$$\begin{aligned} 0 \le \phi _i (y') \le b ~~~~~~ for ~ |y'| < R_i. \end{aligned}$$

(3.5)

The proof consists of two steps.

Step 1. Estimate the locally uniform bound of $\{ \phi _i \}$ in some Hölder spaces.

Fixing $100< R < R_i/2$ for large i, we can define

$$\begin{aligned} \Phi _i' = \mathcal {P}_a (\chi _{B_R'} \phi _i) ~~~~~~ and ~~~~~~ \Phi _i'' = \mathcal {P}_a ( (1 - \chi _{B_R'}) \phi _i), \end{aligned}$$

where $\chi _{B_R'}$ is the characterization function of $B_R'$. Then

$$\begin{aligned} \mathcal {P}_a \phi _i = \Phi _i' + \Phi _i''. \end{aligned}$$

By (3.5) and the property of $\mathcal {P}_a$ we can get

$$\begin{aligned} 0 \le \Phi _i' \le b. \end{aligned}$$

(3.6)

Since $K_i \le c_0$ on $B'_1$, by (3.4) and (3.5), for any $|y'| < R - 2$ we have,

$$\begin{aligned} \begin{aligned} c_0 b^{p_i - 1}&\ge \int _{ B_{1/2} (y', 1) } P_a (y', x) \left[ (\mathcal {P}_a \phi _i) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x \\&\ge \frac{1}{C} \int _{ B_{1/2} (y', 1) } \left[ (\mathcal {P}_a \phi _i) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x \\&\ge \frac{1}{C} \left[ (\mathcal {P}_a \phi _i) (\bar{x})\right] ^\frac{n - a + 2}{n + a - 2} \end{aligned} \end{aligned}$$

for some $\bar{x} \in \overline{B}_{1/2} (y', 1)$, where we used the mean value theorem for integrals in the last inequality and $C > 0$ depends only on n and a. It follows that

$$\begin{aligned} \Phi _i'' (\bar{x}) \le (\mathcal {P}_a \phi _i) (\bar{x}) \le C b^\frac{(p_i - 1) (n + a - 2)}{n - a + 2}. \end{aligned}$$

Since $|\bar{x}'| \le R - 1$ and $\frac{1}{2} \le \bar{x}_n \le \frac{3}{2}$,

$$\begin{aligned} \begin{aligned} C b^\frac{(p_i - 1) (n + a - 2)}{n - a + 2} \ge \Phi _i'' (\bar{x})&= c_{n, a} \int _{ \mathbb R^{n - 1} \setminus B_R' } \frac{ \bar{x}_n^{1 - a} }{ (|\bar{x}' - z'|^2 + \bar{x}_n^2)^\frac{n - a}{2} } \phi _i (z') d z' \\&\ge \frac{1}{C} \int _{ \mathbb R^{n - 1} \setminus B_R' } \frac{\phi _i (z')}{ |\bar{x}' - z'|^{n - a} } d z'. \end{aligned} \end{aligned}$$

Therefore, for any $|y'| < R - 2$ and $x \in B_1' (y') \times (0, 1]$, we have

$$\begin{aligned} \begin{aligned} \frac{\Phi _i'' (x)}{x_n^{1 - a}}&\le C \int _{ \mathbb R^{n - 1} \setminus B_R' } \frac{\phi _i (z')}{ |x' - z'|^{n - a} } d z' \\&\le C \int _{ \mathbb R^{n - 1} \setminus B_R' } \frac{\phi _i (z')}{ |\bar{x}' - z'|^{n - a} } d z' \\&\le C b^\frac{(p_i - 1) (n + a - 2)}{n - a + 2}, \end{aligned} \end{aligned}$$

(3.7)

where the second inequality holds since

$$\begin{aligned} |\bar{x}' - z'| \le |\bar{x}' - x'| + |x' - z'| \le 2 + |x' - z'| \le 3 |x' - z'|. \end{aligned}$$

This together with (3.6) implies that

$$\begin{aligned} (\mathcal {P}_a \phi _i) (x) \le C(n, a, c_0, b) ~~~~~~ \forall ~ x \in B_{R - 2}' \times (0, 1]. \end{aligned}$$

Using the above estimate, we have by direct calculations that

$$\begin{aligned} \bigg \Vert \int _{B_{R - 2}' \times (0, 1]} P_a (y', x) \left[ (\mathcal {P}_a \phi _i) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x \bigg \Vert _{ C^\beta (B_{R - 3}') } \le C(n, a, b, c_0, R, \beta ) \end{aligned}$$

for any $\beta \in (0, 1)$. On the other hand, for $|y'| < R - 3$, by (2.1) we have

$$\begin{aligned} \begin{aligned}&\bigg | \nabla _{y'} \bigg ( \int _{ \mathbb R_+^n \setminus B_{R - 2}' \times (0, 1] } P_a (y', x) \left[ (\mathcal {P}_a \phi _i) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x \bigg ) \bigg | \\&\quad \le C \int _{ \mathbb R_+^n \setminus B_{R - 2}' \times (0, 1] } P_a (y', x) \left[ (\mathcal {P}_a \phi _i) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x \\&\quad \le C H_i (y') \phi _i (y')^{p_i - 1} \\&\quad \le C b^{p_i - 1} \\&\quad \le C(n, a, c_0, b). \end{aligned} \end{aligned}$$

Combing the above two estimates and using (3.4), we can obtain

$$\begin{aligned} \Vert \phi _i^{p_i - 1} \Vert _{ C^{3/4} (B_{R - 3}') } \le C(n, a, b, c_0, R). \end{aligned}$$

(3.8)

Since $\phi _i (0)^{p_i - 1} = 1$, by (3.8) there exists $\delta > 0$ depending only on n, a, b and $c_0$ such that $\phi _i (y')^{p_i - 1} \ge \frac{1}{2}$ for all $|y'| < \delta$. Hence,

$$\begin{aligned} (\mathcal {P}_a \phi _i) (x) \ge \frac{1}{C} \int _{B_\delta '} \frac{ x_n^{1 - a} }{ (|x' - y'|^2 + x_n^2)^\frac{n - a}{2} } 2^{ - \frac{1}{p_i - 1} } d y' \ge \frac{1}{C} \frac{ x_n^{1 - a} }{ (1 + |x|)^{n - a} }. \end{aligned}$$

Again, using (3.4) we can get for $|y'| < R - 3$,

$$\begin{aligned} \phi _i (y')^{p_i - 1} \ge \frac{1}{C(n, a, c_0, b, R)} > 0. \end{aligned}$$

This together with (3.8) implies that

$$\begin{aligned} \Vert \phi _i \Vert _{ C^{3/4} (B_{R - 3}') } \le C(n, a, c_0, b, R). \end{aligned}$$

(3.9)

Hence, (3.3) is proved.

Step 2. Show the convergence of $\mathcal {P}_a \phi _i$ and the equation of $\phi _i$.

Fixing $100< R < R_i/2$ for large i, we write (3.4) as

$$\begin{aligned} H_i (y') \phi _i (y')^{p_i - 1} = \int _{B_R^+} P_a (y', x) \left[ (\mathcal {P}_a \phi _i) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x + h_i (R, y'), \end{aligned}$$

(3.10)

where

$$\begin{aligned} h_i (R, y') = \int _{\mathbb R_+^n \setminus B_R^+} P_a (y', x) \left[ (\mathcal {P}_a \phi _i) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x \ge 0. \end{aligned}$$

By (2.1) and (3.4), for any $|y'| < R - 1$, we have

$$\begin{aligned} |\nabla h_i (R, y')| \le C h_i (R, y') \le C H_i (y') \phi _i (y')^{p_i - 1} \le C(n, a, c_0, b). \end{aligned}$$

Therefore, after passing to a subsequence,

$$\begin{aligned} h_i (R, y') \rightarrow h(R, y') \end{aligned}$$

for some non-negative function $h \in C^{3/4} (B_{R - 1})$.

Similar as in Step 1, we write $\mathcal {P}_a \phi _i$ into following two parts $\Phi _i'$ and $\Phi _i''$:

$$\begin{aligned} \Phi _i' = \mathcal {P}_a (\eta _R \phi _i) ~~~~~~ and ~~~~~~ \Phi _i'' = \mathcal {P}_a ( (1 - \eta _R) \phi _i), \end{aligned}$$

where $\eta _R$ is a smooth cut-off function satisfying $\eta _R \equiv 1$ in $B_{R-4}'$ and $\eta _R \equiv 0$ in $(B_{R-3}')^c$. By using (3.9) and noticing that

$$\begin{aligned} \Phi _i' (x) = c_{n,a}\int _{\mathbb {R}^{n-1}} \frac{1}{(|z'|^2 +1)^{\frac{n-a}{2}}} (\eta _R \phi _i)(x' - x_n z') dz', \end{aligned}$$

we can obtain $\Vert \Phi _i' \Vert _{ C^{\alpha } (B_{R/2}^+) } \le C(n, a, c_0, b, R)$ with $\alpha := \min \{3/4, 1-a\} > 0$. On the other hand, similar to (3.7) we have

$$\begin{aligned} \left\| \frac{\Phi _i'' }{x_n^{1 - a}} \right\| _{C^1(B_{R/2}^+)} \le C(n, a, c_0, b, R), \end{aligned}$$

and hence $\Vert \Phi _i'' \Vert _{ C^{\alpha } (B_{R/2}^+) } \le C(n, a, c_0, b, R)$. Therefore, after passing to a subsequence, we have

$$\begin{aligned} \mathcal {P}_a \phi _i \rightarrow \tilde{\Phi } ~~~~~~ in ~ C_{loc}^{\alpha /2} (\overline{\mathbb R_+^n}) \end{aligned}$$

for some $\tilde{\Phi } \ge 0$ satisfying

$$\begin{aligned} \left\{ \begin{aligned} -\,\textrm{div} (x_n^{a} \nabla \tilde{\Phi })&= 0 ~~~~~~ in ~ \mathbb R_+^n, \\ \tilde{\Phi }&= \phi ~~~~~~ on ~ \mathbb R^{n - 1}. \end{aligned} \right. \end{aligned}$$

From (3.5), we know that $0 \le \phi \le b$ in the whole $\mathbb R^{n - 1}$, and thus $\mathcal {P}_a \phi$ is bounded in $\mathbb R_+^n$. Hence, $\tilde{\Phi } - \mathcal {P}_a \phi \in C^2 (\mathbb R_+^n) \cap C^0 (\overline{\mathbb R_+^n})$ satisfies

$$\begin{aligned} \left\{ \begin{aligned} -\,\textrm{div} (x_n^{a} \nabla (\tilde{\Phi } - \mathcal {P}_a \phi ) )&= 0 ~~~~~~ in ~ \mathbb R_+^n, \\ \tilde{\Phi } - \mathcal {P}_a \phi&= 0 ~~~~~~ on ~ \mathbb R^{n - 1}. \end{aligned} \right. \end{aligned}$$

It follows from the Liouville-type result in Theorem 2.4 that

$$\begin{aligned} \tilde{\Phi } = \mathcal {P}_a \phi + c_1 x_n^{1 - a} \end{aligned}$$

(3.11)

for some constant $c_1 \ge 0$. Sending $i \rightarrow \infty$ in (3.10), we have

$$\begin{aligned} K \phi (y')^\frac{n - a}{n + a - 2} = \int _{B_R^+} P_a (y', x) \tilde{\Phi } (x)^\frac{n - a + 2}{n + a - 2} d x + h(R, y'). \end{aligned}$$

(3.12)

If $c_1 > 0$ in (3.11), taking $y' = 0$ and sending $R \rightarrow \infty$ we obtain that

$$\begin{aligned} K \phi (0)^\frac{n - a}{n + a - 2} \ge \int _{B_R^+} P_a (0, x) \tilde{\Phi } (x)^\frac{n - a + 2}{n + a - 2} d x \rightarrow \infty . \end{aligned}$$

This is a contradiction. Hence, $c_1 = 0$ and $\tilde{\Phi } = \mathcal {P}_a \phi$.

Now we adapt some arguments in [12, Proposition 2.9]. By (3.12), $h(R, y')$ is non-increasing with respect to R. Notice that for $R \gg |y'|$,

$$\begin{aligned} \begin{aligned} \frac{ R^{n - a} }{ (R + |y'|)^{n - a} } h_i (R, 0)&\le h_i (R, y') \\&= c_{n, a} \int _{\mathbb R_+^n \setminus B_R^+} \frac{ |x|^{n - a} }{ (|x' - y'|^2 + x_n^2)^\frac{n - a}{2} } \frac{ x_n^{1 - a} }{ |x|^{n - a} } \left[ (\mathcal {P}_a \phi _i) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x \\&\le \frac{ R^{n - a} }{ (R - |y'|)^{n - a} } h_i (R, 0). \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \lim _{R \rightarrow \infty } h(R, y') = \lim _{R \rightarrow \infty } h(R, 0) = : c_2 \ge 0. \end{aligned}$$

Sending R to $\infty$ in (3.12), by the Lebesgue’s monotone convergence theorem we have

$$\begin{aligned} K \phi (y')^\frac{n - a}{n + a - 2} = \int _{\mathbb R_+^n} P_a (y', x) \left[ (\mathcal {P}_a \phi ) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x + c_2. \end{aligned}$$

If $c_2 > 0$, then $\phi \ge \big ( \frac{c_2}{c_0} \big )^\frac{n + a - 2}{n - a}$ and thus $\mathcal {P}_a \phi \ge \big ( \frac{c_2}{c_0} \big )^\frac{n + a - 2}{n - a}$. This is impossible, since otherwise the integral in the right-hand side is infinity. Hence $c_2 = 0$. The proof of Theorem 3.1 is completed. $\square$

4 A Variational Problem

Let $K \in C^1 (\partial B_1)$ be a positive function satisfying $K(\xi ) = K(- \xi )$, and $L_\textrm{as}^p (\partial B_1) \subset L^p (\partial B_1)$ ($p \ge 1$) be the set of antipodally symmetric functions. For $p \ge \frac{2 (n - 1)}{n + a - 2}$, define

$$\begin{aligned} \lambda _{ \textrm{as}, p } (K) = \sup \bigg \{ \int _{B_1} |\mathcal {P}_a v|^\frac{2 n}{n + a - 2} d \xi : v \in L_\textrm{as}^p (\partial B_1) ~ with ~ \int _{\partial B_1} K |v|^p d s = 1 \bigg \}. \end{aligned}$$

Denote

$$\begin{aligned} \lambda _{ \textrm{as}, \frac{2 (n - 1)}{n + a - 2} } (K) = \lambda _\textrm{as} (K). \end{aligned}$$

Proposition 4.1

If

$$\begin{aligned} \lambda _\textrm{as} (K) > \frac{ \mathcal {S}_{n, a}^\frac{2 n}{n + a - 2} }{ (\min _{\partial B_1} K)^\frac{n}{n - 1} 2^\frac{1}{n - 1} }, \end{aligned}$$

(4.1)

where $\mathcal {S}_{n, a}$ is the sharp constant in the inequality (1.4), then $\lambda _\textrm{as} (K)$ is achieved.

Proof

We claim that

$$\begin{aligned} \liminf _{ p \searrow \frac{2 (n - 1)}{n + a - 2} } \lambda _{ \textrm{as}, p } (K) \ge \lambda _\textrm{as} (K). \end{aligned}$$

For any $\varepsilon > 0$, by the definition of $\lambda _\textrm{as} (K)$, we can find a function $v \in L_\textrm{as}^\infty (\partial B_1)$ such that

$$\begin{aligned} \int _{B_1} |\mathcal {P}_a v|^\frac{2 n}{n + a - 2} d \xi > \lambda _\textrm{as} (K) - \varepsilon ~~~~~~ and ~~~~~ \int _{\partial B_1} K |v|^\frac{2 (n - 1)}{n + a - 2} d s = 1. \end{aligned}$$

Let $V_p : = \int _{\partial B_1} K |v|^p d s$. Since

$$\begin{aligned} \lim _{ p \rightarrow \frac{2 (n - 1)}{n + a - 2} } V_p = \int _{\partial B_1} K |v|^\frac{2 (n - 1)}{n + a - 2} d s = 1, \end{aligned}$$

we have, for p close to $\frac{2 (n - 1)}{n + a - 2}$ sufficiently, that

$$\begin{aligned} \lambda _{ \textrm{as}, p } (K) \ge \int _{B_1} \bigg | \mathcal {P}_a \bigg ( \frac{v}{ V_p^{1/p} } \bigg ) \bigg |^\frac{2 n}{n + a - 2} d \xi \ge \lambda _\textrm{as} (K) - 2 \varepsilon . \end{aligned}$$

Since $\varepsilon$ is arbitrary, the claim is proved.

By the above claim, we can find $p_i \searrow \frac{2 (n - 1)}{n + a - 2}$ as $i \rightarrow \infty$ such that $\lambda _{ \textrm{as}, p_i } (K) \rightarrow \lambda \ge \lambda _\textrm{as} (K)$. Since $K \in C^1 (\partial B_1)$ is positive, if follows from Corollary 2.2 that for $p_i > \frac{2 (n - 1)}{n + a - 2}$, $\lambda _{ \textrm{as}, p_i }$ is achieved, say, by $v_i$. Since $|\mathcal {P}_a v_i| \le \mathcal {P}_a |v_i|$, we can assume that $v_i$ is non-negative. Moreover,

$$\begin{aligned} \Vert v_i \Vert _{L^{p_i} (\partial B_1)}^{p_i} \le \frac{1}{\min _{\partial B_1} K}. \end{aligned}$$

Then, by (1.7) we have $\Vert \mathcal {P}_a v_i \Vert _{L^\frac{2 n}{n + a - 2} (B_1)} \le C$ for some $C > 0$ independent of i. It is easy to see that $v_i$ satisfies the Euler-Lagrange equation

$$\begin{aligned} \lambda _{ \textrm{as}, p_i } (K) K(\eta ) v_i (\eta )^{p_i - 1} = \int _{B_1} P_a (\eta , \xi ) \left[ (\mathcal {P}_a v_i) (\xi )\right] ^\frac{n - a + 2}{n + a - 2} d \xi ~~~~~ \forall ~ \eta \in \partial B_1. \end{aligned}$$

(4.2)

By the regularity result in Theorem 2.3, $v_i \in C^\beta (\partial B_1)$ for any $\beta \in (0, 1)$.

Next we will show that $v_i$ is uniformly bounded. Otherwise, we have

$$\begin{aligned} v_i (\eta _i) = \max _{\partial B_1} v_i \rightarrow \infty ~~~~~ as ~ i \rightarrow \infty . \end{aligned}$$

Let $\eta _i \rightarrow \bar{\eta }$ as $i \rightarrow \infty$. By the stereographic projection with $\eta _i$ as the south pole, equation (4.2) is transformed to

$$\begin{aligned} \lambda _{ \textrm{as}, p_i } (K) K_i (y') u_i (y')^{p_i - 1} = \int _{\mathbb R_+^n} P_a (y', x) \left[ (\mathcal {P}_a u_i) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x ~~~~~~ \forall ~ y' \in \mathbb R^{n - 1}, \end{aligned}$$

where

$$\begin{aligned} K_i (y') = \bigg ( \frac{ \sqrt{2} }{|y' + e_n|} \bigg )^{(n + a - 2) (p_i - 1) - n + a} K( F(y') ) \end{aligned}$$

and

$$\begin{aligned} u_i (y') = \bigg ( \frac{ \sqrt{2} }{|y' + e_n|} \bigg )^{n + a - 2} v_i ( F(y') ). \end{aligned}$$

Hence, $u_i (0) = \max _{ \mathbb R^{n - 1} } u_i \rightarrow \infty$ as $i \rightarrow \infty$. Taking $R_i = u_i(0)^{-\frac{1}{2} (p_i - \frac{2n}{n+a-2} ) } \rightarrow +\infty$ and using Theorem 3.1, we obtain that after passing to a subsequence,

$$\begin{aligned} \phi _i (y') : = \frac{1}{u_i (0)} u_i \big ( u_i (0)^{ p_i - \frac{2 n}{n + a - 2} } y' \big ) \rightarrow \phi (y') ~~~~~~ in ~ C_{loc}^{1/2} (\mathbb R^{n - 1}), \end{aligned}$$

where $\phi > 0$ satisfies

$$\begin{aligned} \lambda K(\bar{\eta }) \phi (y')^\frac{n - a}{n + a - 2} = \int _{\mathbb R_+^n} P_a (y', x) \left[ (\mathcal {P}_a \phi ) (x)\right] ^\frac{n - a + 2}{n + a - 2} d x ~~~~~~ for ~ y' \in \mathbb R^{n - 1}. \end{aligned}$$

By Tang-Dou [16], $\phi$ is classified.

Since $v_i$ is non-negative and antipodally symmetric, for any small $\delta > 0$ we have

$$\begin{aligned} \begin{aligned} 1&= \int _{\partial B_1} K v_i^{p_i} d s \\&\ge 2 \int _{F(B_\delta ')} K v_i^{p_i} d s \\&= 2 \int _{B_\delta '} K_i u_i^{p_i} d z' \\&= 2 u_i (0)^{n \big (p_i - \frac{2 (n - 1)}{n + a - 2} \big )} \int _{ B'_{ \delta u_i (0)^{\frac{2 n}{n + a - 2} - p_i} } } K_i \big ( u_i (0)^{ p_i - \frac{2 n}{n + a - 2} } y' \big ) \phi _i (y')^{p_i} d y' \\&\ge 2 \int _{B_R'} K_i \big ( u_i (0)^{ p_i - \frac{2 n}{n + a - 2} } y' \big ) \phi _i (y')^{p_i} d y' \\&\rightarrow 2 K(\bar{\eta }) \int _{B_R'} \phi (y')^\frac{2 (n - 1)}{n + a - 2} d y' \end{aligned} \end{aligned}$$

as $i \rightarrow \infty$ for any fixed $R > 0$. It follows that

$$\begin{aligned} 1 \ge 2 K(\bar{\eta }) \int _{ \mathbb R^{n - 1} } \phi (y')^\frac{2 (n - 1)}{n + a - 2} d y'. \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned} \mathcal {S}_{n, a}^\frac{2 n}{n + a - 2}&\ge \frac{ \int _{\mathbb R_+^n} |\mathcal {P}_a \phi |^\frac{2 n}{n + a - 2} }{ \big ( \int _{ \mathbb R^{n - 1} } |\phi |^\frac{2 (n - 1)}{n + a - 2} \big )^\frac{n}{n - 1} } \\&= \lambda K(\bar{\eta }) \bigg ( \int _{ \mathbb R^{n - 1} } |\phi |^\frac{2 (n - 1)}{n + a - 2} \bigg )^{ - \frac{1}{n - 1} }\\&\ge \lambda K(\bar{\eta })^\frac{n}{n - 1} 2^\frac{1}{n - 1}. \end{aligned} \end{aligned}$$

It implies that

$$\begin{aligned} \lambda \le \frac{ \mathcal {S}_{n, a}^\frac{2 n}{n + a - 2} }{ (\min _{\partial B_1} K)^\frac{n}{n - 1} 2^\frac{1}{n - 1} }, \end{aligned}$$

which contradicts the assumption (4.1). Therefore, $\{ v_i \}$ is uniformly bounded on $\partial B_1$.

By Theorem 2.3, $\{ v_i \}$ is bounded in $C^{1/2} (\partial B_1)$. Thus, after passing to a subsequence, we have for some non-negative function $v \in C(\partial B_1)$,

$$\begin{aligned} \begin{aligned} v_i&\rightarrow v ~~~~~~&in ~&~ C(\partial B_1), \\ \end{aligned} \end{aligned}$$

and thus,

$$\begin{aligned} \begin{aligned} \mathcal {P}_a v_i&\rightarrow \mathcal {P}_a v&in ~&~ C(\overline{B}_1). \end{aligned} \end{aligned}$$

Letting $i \rightarrow \infty$ in (4.2), we obtain that v satisfies

$$\begin{aligned} \lambda K(\eta ) v(\eta )^\frac{n - a}{n + a - 2} = \int _{B_1} P_a (\eta , \xi ) \left[ (\mathcal {P}_a v) (\xi )\right] ^\frac{n - a + 2}{n + a - 2} d \xi . \end{aligned}$$

Moreover, since

$$\begin{aligned} 1 = \int _{\partial B_1} K(\eta ) v_i (\eta )^{p_i} d\eta \rightarrow \int _{\partial B_1} K(\eta ) v (\eta )^{\frac{2 (n - 1)}{n + a - 2}} d\eta , \end{aligned}$$

we have $v > 0$ on $\partial B_1$. These also imply that $\lambda = \lambda _\textrm{as}(K)$ and $\lambda _\textrm{as} (K)$ is achieved. The proof of Proposition 4.1 is completed. $\square$

Proof

Let $v = 1$, then

$$\begin{aligned} \lambda _\textrm{as} (K) \ge \frac{ \int _{B_1} |\mathcal {P}_a 1|^\frac{2 n}{n + a - 2} d \xi }{ \big ( \int _{\partial B_1} K d s \big )^\frac{n}{n - 1} } \ge \frac{ \mathcal {S}_{n, a}^\frac{2 n}{n + a - 2} }{ (\max _{\partial B_1} K)^\frac{n}{n - 1} } > \frac{ \mathcal {S}_{n, a}^\frac{2 n}{n + a - 2} }{ (\min _{\partial B_1} K)^\frac{n}{n - 1} 2^\frac{1}{n - 1} }, \end{aligned}$$

where we use (1.9) in the last inequality. By Proposition 4.1, we obtain the desired result. $\square$

Appendix A Hölder Regularity

This appendix is devoted to the proof of Theorem 2.3. We start with the improvement of integrability of the subsolutions to some nonlinear integral equations.

Proposition A.1

Suppose that $n \ge 2$ and $2 - n< a < 1$. Let $1 < r, s \le \infty$, $1 \le t < \infty$, $\frac{n}{n - 1}< p< q < \infty$ satisfy

$$\begin{aligned} \frac{1}{n}< \frac{t}{q} + \frac{1}{r} < \frac{t}{p} + \frac{1}{r} \le 1 \end{aligned}$$

and

$$\begin{aligned} \frac{n}{t r} + \frac{n - 1}{s} = \frac{1}{t}. \end{aligned}$$

Assume that $U, V \in L^p (B_R^+)$, $W \in L^r (B_R^+)$, $f \in L^s (B_R')$ are all non-negative functions, $V \in L^q (B_{R/2}^+)$,

$$\begin{aligned} \Vert W \Vert _{L^r (B_R^+)}^{1/t} \Vert f \Vert _{L^s (B_R')} \le \varepsilon (n, a, p, q, r, s, t) ~ \text{ small } \end{aligned}$$

and

$$\begin{aligned} U(x) \le \int _{B_R'} P_a (y', x) f(y') \bigg ( \int _{B_R^+} P_a (y', z) W(z) U(z)^t d z \bigg )^{1/t} d y' + V(x) \end{aligned}$$

for $x \in B_R^+$. Then $U \in L^q (B_{R/4}^+)$ and

$$\begin{aligned} \Vert U \Vert _{ L^q (B_{R/4}^+) } \le c(n, a, p, q, r, s, t)\Big ( R^{ \frac{n}{q} - \frac{n}{p} } \Vert U \Vert _{ L^p (B_R^+) } + \Vert V \Vert _{ L^q (B_{R/2}^+) } \Big ). \end{aligned}$$

The proof of Proposition A.1 is the same as that of [9, Proposition 5.2]. We also need the following two $L^p$-boundedness for the operator $\mathcal {P}_a$ and its adjoint operator.

Proposition A.2

(Chen [3]) Suppose that $n \ge 2$ and $2 - n< a < 1$. For $1 < p \le \infty$ we have

$$\begin{aligned} \Vert \mathcal {P}_a f \Vert _{ L^\frac{n p}{n - 1} (\mathbb R_+^n) } \le c(n, a, p) \Vert f \Vert _{ L^p (\mathbb R^{n - 1}) } \end{aligned}$$

for any $f \in L^p (\mathbb R^{n - 1})$.

For a function F on $\mathbb R_+^n$, define

$$\begin{aligned} (\mathcal {T}_a F) (y') = \int _{\mathbb R_+^n} P_a (y', x) F(x) d x. \end{aligned}$$

Then we have the following inequality by a duality argument. See also the similar proof in [9, Proposition 2.3].

Proposition A.3

Suppose that $n \ge 2$ and $2 - n< a < 1$. For $1 \le p < n$ we have

$$\begin{aligned} \Vert \mathcal {T}_a F \Vert _{ L^\frac{(n - 1) p}{n - p} (\mathbb R^{n - 1}) } \le C(n, a, p) \Vert F \Vert _{ L^p (\mathbb R_+^n) } \end{aligned}$$

for any $F \in L^p (\mathbb R_+^n)$.

Next we give the details of the proof of Theorem 2.3.

Proof of Theorem 2.3

Let $\tilde{u}_0 (y') = K(y') u(y')^{p - 1}$ and $U_0 (x) = (\mathcal {P}_a u) (x)$. Then

$$\begin{aligned} \tilde{u}_0 (y') = \int _{\mathbb R_+^n} P_a (y', x) U_0 (x)^\frac{n - a + 2}{n + a - 2} d x. \end{aligned}$$

Define

$$\begin{aligned} U_R (x)= & {} \int _{ \mathbb R^{n - 1} \setminus B_R' } P_a (y', x) u(y') d y',\\ \tilde{u}_R (y')= & {} \int _{\mathbb R_+^n \setminus B_R^+} P_a (y', x) U_0 (x)^\frac{n - a + 2}{n + a - 2} d x. \end{aligned}$$

Since $u \in L_{loc}^p (\mathbb R^{n - 1})$, by Proposition A.2 we get $\int _{B_R'} P_a (z', \cdot ) u(z') d z' \in L^\frac{n p}{n - 1} (\mathbb R_+^n)$. Notice that

$$\begin{aligned} \frac{n p}{n - 1} \ge \frac{2 n}{n + a - 2}. \end{aligned}$$

Step 1. We claim that $U_0 \in L_{loc}^\frac{n p}{n - 1} (\overline{\mathbb R_+^n})$ and $U_R \in L^\frac{n p}{n - 1} (B_R^+) \cap L_{loc}^\infty (B_R^+ \cup B_R')$.

Since $u \in L_{loc}^p (\mathbb R^{n - 1})$, we have $u < \infty$ a.e. on $\mathbb R^{n - 1}$. It implies that $U_0 < \infty$ a.e. on $\mathbb R_+^n$. Hence, there exists $x_0 \in B_R^+$ such that $U_0 (x_0) < \infty$. It follows that

$$\begin{aligned} \int _{ \mathbb R^{n - 1} \setminus B_R' } \frac{u(z')}{ (|x_0' - z'|^2 + x_{0, n}^2)^\frac{n - a}{2} } d z' < \infty . \end{aligned}$$

Thus,

$$\begin{aligned} \int _{ \mathbb R^{n - 1} \setminus B_R' } \frac{u(z')}{ |z'|^{n - a} } d z' < \infty . \end{aligned}$$

For $0< \theta < 1$ and $x \in B_{\theta R}^+$, we have

$$\begin{aligned} U_R (x) = \int _{ \mathbb R^{n - 1} \setminus B_R' } P_a (z', x) u(z') d z' \le \frac{ c_{n, a} R^{1 - a} }{ (1 - \theta )^{n - a} } \int _{ \mathbb R^{n - 1} \setminus B_R' } \frac{ u(z') }{ |z'|^{n - a} } d z'. \end{aligned}$$

It follows that $U_R \in L_{loc}^\infty (B_R^+ \cup B_R')$. Since $\int _{B_R'} P_a (z', \cdot ) u(z') d z' \in L^\frac{n p}{n - 1} (\mathbb R_+^n)$, we know that $U_0 \in L_{loc}^\frac{n p}{n - 1} (B_R^+ \cup B_R')$. Since $R>0$ is arbitrary, we deduce that $U_0 \in L_{loc}^\frac{n p}{n - 1} (\overline{\mathbb R_+^n})$ and hence $U_R \in L^\frac{n p}{n - 1} (B_R^+)$.

Step 2. We show that $\tilde{u}_R \in L^\frac{p}{p - 1} (B_R') \cap L_{loc}^\infty (B_R')$.

Since $\tilde{u}_0 \in L_{loc}^\frac{p}{p - 1} (\mathbb R^{n - 1})$, we obtain $\tilde{u}_0 \in L^\frac{p}{p - 1} (B_R')$ and thus $\tilde{u}_R \in L^\frac{p}{p - 1} (B_R')$. Hence, we can find $y_0' \in B_R'$ such that $\tilde{u}_R (y_0') < \infty$. That is,

$$\begin{aligned} \int _{\mathbb R_+^n \setminus B_R^+} \frac{ z_n^{1 - a} }{ (|z' - y_0'|^2 + z_n^2)^\frac{n - a}{2} } U_0 (z)^\frac{n - a + 2}{n + a - 2} d z < \infty . \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{\mathbb R_+^n \setminus B_R^+} \frac{ z_n^{1 - a} }{ |z|^{n - a} } U_0 (z)^\frac{n - a + 2}{n + a - 2} d z < \infty . \end{aligned}$$

For $0< \theta < 1$ and $y' \in B_{\theta R}'$, we have

$$\begin{aligned} \tilde{u}_R (y') = \int _{\mathbb R_+^n \setminus B_R^+} P_a (y', z) U_0 (z)^\frac{n - a + 2}{n + a - 2} d z \le \frac{ c_{n, a} }{ (1 - \theta )^{n - a} } \int _{\mathbb R_+^n \setminus B_R^+} \frac{ z_n^{1 - a} }{ |z|^{n - a} } U_0 (z)^\frac{n - a + 2}{n + a - 2} d z. \end{aligned}$$

This implies that $\tilde{u}_R \in L_{loc}^\infty (B_R')$.

Step 3. We prove that $\tilde{u}_0 \in L_{loc}^\infty (\mathbb R^{n - 1})$ and $U_0 \in L_{loc}^\infty (\overline{\mathbb R_+^n})$.

Case 1: $\frac{2 (n - 1)}{n + a - 2}< p < \infty$. This is the subcritical case, and we directly use the bootstrap method to prove the regularity.

From Proposition A.2 and Step 1, we know that $U_0^\frac{n - a + 2}{n + a - 2} \in L_{loc}^{q_0} (\overline{\mathbb R_+^n})$ with

$$\begin{aligned} q_0 := \frac{n p}{n - 1} \cdot \frac{n + a - 2}{n - a + 2}> \frac{2 n}{n + a - 2} \cdot \frac{n + a - 2}{n - a + 2} = \frac{2 n}{n - a + 2} > 1. \end{aligned}$$

If $q_0 \ge n$, by Proposition A.3 we know that $\int _{B_R^+} P_a (\cdot , z) U_0 (z)^\frac{n - a + 2}{n + a - 2} d z \in L^r (\mathbb R^{n - 1})$ for any $1 \le r < \infty$. This together with Step 2 implies that $\tilde{u}_0 \in L_{loc}^r (B_R')$ for any $1 \le r < \infty$. Since R is arbitrary, we obtain $\tilde{u}_0 \in L_{loc}^r (\mathbb R^{n - 1})$ for any $1 \le r < \infty$. Moreover, by Proposition A.2 we have $\int _{B_R'} P_a (z', \cdot ) u(z') d z' \in L^s (\mathbb R_+^n)$ for any $\frac{n}{n - 1}< s < \infty$. Combined with Step 1, we also have $U_0 \in L_{loc}^s (\overline{\mathbb R_+^n})$ for any $1 \le s < \infty$. If $q_0 < n$, then Proposition A.3 yields $\int _{B_R^+} P_a (\cdot , z) U_0 (z)^\frac{n - a + 2}{n + a - 2} d z \in L^\frac{(n - 1) q_0}{n - q_0} (\mathbb R^{n - 1})$. Combined with Step 2, we have $\tilde{u}_0 \in L_{loc}^\frac{(n - 1) q_0}{n - q_0} (B_R')$ for any $R>0$. Consequently, we deduce that $u \in L_{loc}^{p_1} (\mathbb R^{n - 1})$ with

$$\begin{aligned} p_1 := (p - 1) \cdot \frac{(n - 1) q_0}{n - q_0} = p \cdot \frac{ (p - 1) \frac{n + a - 2}{n - a + 2} }{ 1 - \frac{p (n + a - 2)}{(n - 1) (n - a + 2)} } > p, \end{aligned}$$

where the last inequality holds since $p > \frac{2 (n - 1)}{n + a - 2}$. From now on, we denote the constant

$$\begin{aligned} \gamma :=\frac{ (p - 1) \frac{n + a - 2}{n - a + 2} }{ 1 - \frac{p (n + a - 2)}{(n - 1) (n - a + 2)} }>1. \end{aligned}$$

We can see that the regularity of u is boosted to $L^{p_1}_{loc}(\mathbb R^{n - 1})$ with $p_1 =p \cdot \gamma$.

Using Proposition A.2 and Step 1 again, we obtain $U_0^\frac{n - a + 2}{n + a - 2} \in L_{loc}^{q_1} (\overline{\mathbb R_+^n})$ with

$$\begin{aligned} q_1 := \frac{n p_1}{n - 1} \cdot \frac{n + a - 2}{n - a + 2} = q_0 \cdot \gamma > q_0. \end{aligned}$$

If $q_1 \ge n$, then we easily obtain $U_0 \in L_{loc}^s (\overline{\mathbb R_+^n})$ for any $1 \le s < \infty$. If $q_1 < n$, by a similar argument as above we can obtain that $u \in L_{loc}^{p_2} (\mathbb R^{n - 1})$ with

$$\begin{aligned} p_2 := (p-1) \cdot \frac{(n-1)q_1}{n-q_1}= p_1 \cdot \frac{ (p - 1) \frac{n + a - 2}{n - a + 2} }{ 1 - \frac{p_1 (n + a - 2)}{(n - 1) (n - a + 2)} } > p_1 \cdot \gamma \end{aligned}$$

due to $p_1 > p$. Hence, the regularity of u is boosted to $L^{p_2}_{loc}(\mathbb R^{n - 1})$ with $p_2 > p_1 \cdot \gamma$. By Proposition A.2 and Step 1 again, we obtain $U_0^\frac{n - a + 2}{n + a - 2} \in L_{loc}^{q_2} (\overline{\mathbb R_+^n})$ with

$$\begin{aligned} q_2 := \frac{n p_2}{n - 1} \cdot \frac{n + a - 2}{n - a + 2} > q_1\cdot \gamma . \end{aligned}$$

Repeating this process with finite many steps, we can boost $U_0$ to $L_{loc}^q (\overline{\mathbb R_+^n})$ for any $1 \le q < \infty$. By Hölder inequality we get

$$\begin{aligned} \tilde{u}_0 (y') = \int _{ B_R^+} P_a (y', z) U_0 (z)^\frac{n - a + 2}{n + a - 2} d z + \tilde{u}_R (y') \le c(n, a, q) \Vert U_0 \Vert _{ L^q (B_R^+) }^\frac{n - a + 2}{n + a - 2} + \tilde{u}_R (y') \end{aligned}$$

for some $q > \frac{n(n - a + 2)}{n + a - 2}$. This together with Step 2 implies that $\tilde{u}_0 \in L_{loc}^\infty (B_R')$. Since R is arbitrary, we have $\tilde{u}_0 \in L_{loc}^\infty (\mathbb R^{n - 1})$ and hence $u \in L_{loc}^\infty (\mathbb R^{n - 1})$. Combined with Step 1, we see $U_0 \in L_{loc}^\infty (\overline{\mathbb R_+^n})$.

Case 2: $p = \frac{2 (n - 1)}{n + a - 2}$. For this critical case, the bootstrap method above does not work. We will use Proposition A.1 to establish the regularity.

In this case, we have $U_0 \in L_{loc}^\frac{2 n}{n + a - 2} (\overline{\mathbb R_+^n})$ and $U_R \in L^\frac{2 n}{n + a - 2} (B_R^+) \cap L_{loc}^\infty (B_R^+ \cup B_R')$. Since $a < 1$, we get $0< \frac{n + a - 2}{n - a} < 1$. Then,

$$\begin{aligned} \tilde{u}_0 (y')^\frac{n + a - 2}{n - a} \le \bigg ( \int _{B_R^+} P_a (y', z) U_0 (z)^\frac{n - a + 2}{n + a - 2} d z \bigg )^\frac{n + a - 2}{n - a} + \tilde{u}_R (y')^\frac{n + a - 2}{n - a}. \end{aligned}$$

Hence,

$$\begin{aligned} U_0 (x) =&\int _{B_R'} P_a (y', x) u (y') d y' + U_R (x) \\ =&\int _{B_R'} P_a (y', x) K(y')^{- \frac{n + a - 2}{n - a} } \tilde{u}_0 (y')^\frac{n + a - 2}{n - a} d y' + U_R (x) \\ \le&\int _{B_R'} P_a (y', x) K(y')^{- \frac{n + a - 2}{n - a} } \bigg ( \int _{B_R^+} P_a (y', z) U_0 (z)^\frac{2}{n + a - 2} U_0 (z)^\frac{n - a}{n + a - 2} d z \bigg )^\frac{n + a - 2}{n - a} d y'\\&+ V_R (x), \end{aligned}$$

where

$$\begin{aligned} V_R (x) = \int _{B_R'} P_a (y', x) K(y')^{ - \frac{n + a - 2}{n - a} } \tilde{u}_R (y')^\frac{n + a - 2}{n - a} d y' + U_R (x). \end{aligned}$$

Since $\tilde{u}_R \in L^\frac{2 (n - 1)}{n - a} (B_R')$, we have $V_R \in L^\frac{2 n}{n + a - 2} (B_R^+)$. On the other hand, for $0< \theta < 1$, $x \in B_{\theta R}^+$, we have

$$\begin{aligned} \begin{aligned}&\int _{B_R'} P_a (y', x) K(y')^{- \frac{n + a - 2}{n - a} } \tilde{u}_R (y')^\frac{n + a - 2}{n - a} d y' \\&\quad \le (\min \nolimits _{B_R'} K)^{- \frac{n + a - 2}{n - a} } \bigg [ \Vert \tilde{u}_R \Vert _{ L^\infty (B_{ \frac{1 + \theta }{2} R }) }^\frac{n + a - 2}{n - a} + \frac{c(n, a)}{ (1 - \theta )^{n - a} R^{n - 1} } \int _{ B_R' \setminus B_{ \frac{1 + \theta }{2} R } } \tilde{u}_R (y')^\frac{n + a - 2}{n - a} d y' \bigg ] \\&\quad \le (\min \nolimits _{B_R'} K)^{- \frac{n + a - 2}{n - a} } \bigg [ \Vert \tilde{u}_R \Vert _{ L^\infty (B_{ \frac{1 + \theta }{2} R }) }^\frac{n + a - 2}{n - a} + \frac{c(n, a)}{ (1 - \theta )^{n - a} R^\frac{n + a - 2}{2} } \Vert \tilde{u}_R \Vert _{ L^\frac{2 (n - 1)}{n - a} (B_R') }^\frac{n + a - 2}{n - a} \bigg ]. \end{aligned} \end{aligned}$$

Hence, $V_R \in L_{loc}^\infty (B_R^+ \cup B_R')$. It follows from Proposition A.1 that $U_0 \in L^q (B_{R/4}^+)$ for any $\frac{2 n}{n + a - 2}< q < \infty$ when R is sufficiently small. Therefore,

$$\begin{aligned} \tilde{u}_0 (y') = \int _{ B_{R/4}^+ } P_a (y', z) U_0 (z)^\frac{n - a + 2}{n + a - 2} d z + \tilde{u}_{R/4} (y') \le c(n, a, q) \Vert U_0 \Vert _{ L^q (B_{R/4}^+) }^\frac{n - a + 2}{n + a - 2} + \tilde{u}_{R/4} (y') \end{aligned}$$

for some $q > \frac{n(n - a + 2)}{n + a - 2}$. In particular, we see $\tilde{u}_0 \in L^\infty (B_{R/8}')$. Since every point can be viewed as a center, we get $\tilde{u}_0 \in L_{loc}^\infty (\mathbb R^{n - 1})$ and hence $U_0 \in L_{loc}^\infty (\overline{\mathbb R_+^n})$.

Step 4. We prove that $u \in C_{loc}^\beta (\mathbb R^{n - 1})$ for any $\beta \in (0, 1)$.

From Step 1 and 2, we know that for any $R > 0$,

$$\begin{aligned} \int _{ \mathbb R^{n - 1} \setminus B_R' } \frac{u(y')}{ |y'|^{n - a} } d y'< \infty ~~~~~~ and ~~~~~~ \int _{\mathbb R_+^n \setminus B_R^+} \frac{ x_n^{1 - a} }{ |x|^{n - a} } U_0 (x)^\frac{n - a + 2}{n + a - 2} d x < \infty . \end{aligned}$$

Therefore, $\tilde{u}_R \in C^\infty (B_R')$ and $U_R \in C^{1-a}(B_R^+ \cup B_R')$. It follows from Step 3 that $\tilde{u}_0 \in C_{loc}^\beta (\mathbb R^{n - 1})$ for any $0< \beta < 1$. By the continuity, $\tilde{u}_0 > 0$ in $\mathbb R^{n - 1}$. Consequently, $u \in C_{loc}^\beta (\mathbb R^{n - 1})$ for any $0< \beta < 1$ since K is a positive $C^1$ function in $\mathbb R^n$. $\square$