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.

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 $\mathbb R^n$. $\square$