Complete Metrics with Constant Fractional Higher Order Q-Curvature on the Punctured Sphere

Abstract

This manuscript is devoted to constructing complete metrics with constant higher fractional curvature on punctured spheres with finitely many isolated singularities. Analytically, this problem is reduced to constructing singular solutions for a conformally invariant integro-differential equation that generalizes the critical GJMS problem. Our proof follows the earlier construction in Ao et al. (Math Ann 369:109–151, 2017), based on a gluing method, which we briefly describe. Our main contribution is to provide a unified approach for fractional and higher order cases. This method relies on proving Fredholm properties for the linearized operator around a suitably chosen approximate solution. The main challenge in our approach is that the solutions to the related blow-up limit problem near isolated singularities need to be fully classified; hence we are not allowed to use a simplified ODE method. To overcome this issue, we approximate solutions near each isolated singularity by a family of half-bubble tower solutions. Then, we reduce our problem to solving an (infinite-dimensional) Toda-type system arising from the interaction between the bubble towers at each isolated singularity. Finally, we prove that this system’s solvability is equivalent to the existence of a balanced configuration.

Appendices

Appendix A: Estimates on the Bubble-Towers Interactions

In this appendix, we quote some important integrals in our proof. The following expressions may be found in [8, Appendix 7] for $\sigma \in {\mathbb {R}}_+$. Let $\lambda _1,\lambda _2\lambda _3>0$ and $x_1,x_2\in {\mathbb {R}}^n$ with $x\ne 0$, we define

$$\begin{aligned} U_1:=U_{0,\lambda _1}, \quad U_2:=U_{0,\lambda _2}, \quad {\textrm{and}} \quad U_3:=U_{x,\lambda _3}, \end{aligned}$$

where $w_{x_0,\lambda }$ is given by (4.4). We also recall

$$\begin{aligned} \gamma _{\sigma }:=\frac{n-2\sigma }{2} \quad {\textrm{and}} \quad \gamma _{\sigma }^{\prime }:=\frac{n+2\sigma }{2} \end{aligned}$$

to be the Fowler rescaling exponent and its Lebesgue conjugate, respectively.

In what follows, we use the constants below:

$$\begin{aligned} A_1= & {} \frac{(n+2 \sigma )(n-2 \sigma )}{n} \int _{{\mathbb {R}}^n}\left( {|x|^{2\gamma _{\sigma }}\left( 1+|x|^2\right) ^{\gamma _{\sigma }^{\prime }}}+1\right) ^{-1} \textrm{d}x>0, \end{aligned}$$

(A.1)

$$\begin{aligned} A_2= & {} \frac{n+2 \sigma }{2} \int _{{\mathbb {R}}^n} {(|x|^2-1)}\left( 1+|x|^2\right) ^{-\gamma _{\sigma }-1}\textrm{d}x>0, \end{aligned}$$

(A.2)

and

$$\begin{aligned} A_3=-\frac{(n-2 \sigma )^2}{n} \int _{{\mathbb {R}}^n} {|x|^2}\left( 1+|x|^2\right) ^{-\gamma _{\sigma }-1} \textrm{d}x<0 . \end{aligned}$$

(A.3)

Lemma A.1

For any $\lambda _1,\lambda _2>0$. It holds

$$\begin{aligned} \int _{{\mathbb {R}}^n} f_{\sigma }^{\prime }(U_1) U_2 {\partial _{\lambda _1} U_1} \textrm{d}x=\frac{1}{\lambda _1} \Psi \left( \left| \log \frac{\lambda _2}{\lambda _1}\right| \right) \frac{\log \frac{\lambda _2}{\lambda _1}}{\left| \log \frac{\lambda _2}{\lambda _1}\right| }, \end{aligned}$$

where

$$\begin{aligned} \Psi (\ell )=e^{-\gamma _{\sigma } \ell }(1+\textrm{o}(1)) \quad {\textrm{as}} \quad \ell \rightarrow +\infty \end{aligned}$$

with

$$\begin{aligned} \Psi (\ell ):=\int _{{\mathbb {R}}} f_{\sigma }^{\prime }(v_{\textrm{sph}}(t)) v_{\textrm{sph}}(t+\ell ) v^{\prime }(t) \textrm{d}t. \end{aligned}$$

(A.4)

Proof

See [8, Lemma 7.1]. $\square$

Lemma A.2

If $\lambda _3={\mathcal {O}}(\lambda _1)$, then the following estimates hold

$$\begin{aligned} \int _{{\mathbb {R}}^n} f_{\sigma }^{\prime }(U_1) U_3 \partial _{\lambda _1} U_1 \textrm{d}x=A_2 |x_2|^{2 \sigma -n}\frac{\left( \lambda _1 \lambda _3\right) ^{\gamma _{\sigma }}}{\lambda _1}\left( 1+{\mathcal {O}}\left( \lambda _1\right) ^2\right) \end{aligned}$$

and

$$\begin{aligned} \int _{{\mathbb {R}}^n} f_{\sigma }^{\prime }(U_1) U_3 \partial _{x_\ell } U_1 \textrm{d}x=A_3 {x_\ell }{|x|^{2 \sigma -n-2}}\left( \lambda _1 \lambda _3\right) ^{\gamma _{\sigma }}\left( 1+{\mathcal {O}}\left( \lambda _1^2\right) \right) \quad {\textrm{for}} \quad \ell \in \{0,\dots ,n\}. \end{aligned}$$

Proof

See [8, Lemma 7.2]. $\square$

Lemma A.3

Let $\lambda _1,\lambda _2>0$ and $a\in {\mathbb {R}}^n$. If $|a| \leqslant \max \left\{ \lambda _1^2, \lambda _2^2\right\} \ll 1$ and $\min \left\{ \frac{\lambda _1}{\lambda _2}, \frac{\lambda _2}{\lambda _1}\right\} \ll 1$, then the following estimates holds

$$\begin{aligned}&\int _{{\mathbb {R}}^n} (\partial _a U_{a,\lambda _1}^{\gamma _{\sigma ^{\prime }}} )U_{0,\lambda _2}^{\gamma _{\sigma }} \textrm{d}x=-A_0 c_{\lambda _1,\lambda _2}C_{\lambda _1,\lambda _2}+c_{\lambda _1,\lambda _2}{\mathcal {O}}\left( C_{\lambda _1,\lambda _2}^2+c_{\lambda _1,\lambda _2}^2C_{\lambda _1,\lambda _2}^2\right) , \end{aligned}$$

(A.5)

where

$$\begin{aligned} c_{\lambda _1,\lambda _2}=\min \left\{ \left( \frac{\lambda _1}{\lambda _2}\right) ^{\gamma _{\sigma }},\left( \frac{\lambda _2}{\lambda _1}\right) ^{\gamma _{\sigma }}\right\} \quad {\textrm{and}} \quad C_{\lambda _1,\lambda _2}=\frac{a}{\max \left\{ \lambda _1^2, \lambda _2^2\right\} }. \end{aligned}$$

Proof

See [8, Lemma 7.3]. $\square$

Appendix B: Nondegeneracy of the Bubble Solution

In this section, we add the proof of the nondegeneracy of the spherical solution.

Proof of Lemma 4.6

Let us start with $\phi \in H^{\sigma }({\mathbb {R}}^n)$. Using the statement in [42, Lemma 5.1], it suffices to know that $\phi \in L^{\infty }\left( {\mathbb {R}}^n\right)$. We will divide the proof of this fact into three cases, which we describe as follows:

Case 1: $n>6\sigma$.

Indeed, notice that, from (4.14) since $f^{\prime }_\sigma (u_{\textrm{sph}})\in L^{\infty }({\mathbb {R}}^n)$, one can find a large constant $C\gg 1$ depending only on $n, \sigma$ such that

$$\begin{aligned} |\phi (x)| \leqslant \int _{{\mathbb {R}}^n} {C}{|x-y|^{2\sigma -n}}\left( \frac{|\phi (y)|}{1+|y|^{4 \sigma }}+\frac{1}{1+|y|^{n-2\sigma }}\right) \textrm{d}y \quad \text { for } \quad x \in {\mathbb {R}}^n. \end{aligned}$$

(B.1)

Also, by partitioning the Euclidean space as ${\mathbb {R}}^n=B_d(0)\cup B_d(x)\cup (B_d(0)\cup B_d(x))^c$ with $d:={|x|}/{2} \geqslant 1$, and integrating on each subpart, we obtain

$$\begin{aligned} \int _{{\mathbb {R}}^n} \frac{|x-y|^{2\sigma -n}}{1+|y|^{n-2\sigma }}{\textrm{d} y}&\lesssim \frac{1}{|x|^{n-4\sigma }} \quad \mathrm{for \ all} \quad x \in {\mathbb {R}}^n. \end{aligned}$$

(B.2)

Furthermore, by substituting the last inequality into (B.1), one has

$$\begin{aligned} |\phi (x)| \leqslant C\left[ \int _{{\mathbb {R}}^n} \frac{|x-y|^{2\sigma -n}}{1+|y|^{4\sigma }} {|\phi (x)|}\textrm{d}y+\frac{1}{1+|x|^{n-4\sigma }}\right] \quad \text { for } \quad x \in {\mathbb {R}}^n. \end{aligned}$$

(B.3)

Next, since $n>6\sigma$, one has that $[p_0,p_*)\ne \varnothing$, where $p_0=\frac{2 n}{n-2 \sigma }$ and $p_*=\frac{n}{2 \sigma }$, which allows us to use the Hardy–Littlewood–Sobolev inequality to get

$$\begin{aligned} \Vert \phi \Vert _{L^{p_1}({\mathbb {R}}^n)}&\lesssim \left\| \frac{|\phi (x)|}{1+|x|^{4 \sigma }} *{|x|^{2 \sigma -n}}\right\| _{L^{p_1}({\mathbb {R}}^n)} \nonumber \\&\lesssim \left\| \frac{|\phi (x)|}{1+|x|^{4 \sigma }}\right\| _{L^{q_1}({\mathbb {R}}^n)}\nonumber \\&\lesssim \Vert \phi \Vert _{L^{p_0}({\mathbb {R}}^n)}\left\| \frac{1}{1+|x|^{4\sigma }}\right\| _{L^{q_0}({\mathbb {R}}^n)}, \end{aligned}$$

(B.4)

for any $p \in [p_0, p_*)$ and $p_2=\frac{n p_0}{n-2\sigma p_0}$.

In what follows, we are based on the estimate (B.4) to run the bootstrap argument below and obtain the desired $L^\infty$-estimate. First, notice that from (B.4), we have $\phi \in L^{p_1}({\mathbb {R}}^n)$, and so $\phi \in L^{p_1}({\mathbb {R}}^n)$ for all $p\in [p_0,p_1]$. Second, we check whether $p_1\geqslant p_*$ or not. In the affirmative case, we apply (B.4) with $p=p_*-\varepsilon$ for $0<\varepsilon \ll 1$ small enough to obtain that $\phi \in L^{p_1}({\mathbb {R}}^n)$ for all $p\in [p_0,+\infty )$. In the negative case, we use (B.4) with $p=p_1$, which gives us that $\phi \in L^{p_2}({\mathbb {R}}^n)$ for all $p\in [p_0,p_2]$, where $p_2=\frac{np_1}{n-2\sigma p_1}$. Third, we repeat the same process for this new exponent.

More precisely, it is not hard to check that the bootstrap sequence $\{p_\ell \}_{\ell \in {\mathbb {N}}}\subset [p_0,+\infty )$ satisfies

$$\begin{aligned} p_{\ell +1}=\left( 1+\frac{4\sigma }{n-6\sigma }\right) p_\ell \quad \mathrm{for \ all} \quad \ell \in {\mathbb {N}}. \end{aligned}$$

Hence, $\lim _{\ell \rightarrow +\infty }p_{\ell }=+\infty$, which shows that the bootstrap technique terminates in a finite step.

Now, let us fix some $p \gg 1$ large enough. using the same strategy as in (B.2), we find

$$\begin{aligned} \int _{{\mathbb {R}}^n}{|x-y|^{2\sigma -n}} \frac{|\phi (y)|}{1+|y|^{4\sigma }} \textrm{d}y&\lesssim \left( \int _{{\mathbb {R}}^n} \frac{|x-y|^{(2\sigma -n) p^{\prime }}}{1+|y|^{4\sigma p_0^{\prime }}} \textrm{d}y\right) ^{\frac{1}{p^{\prime }}}\Vert \phi \Vert _{L^p({\mathbb {R}}^n)} \nonumber \\&\lesssim \frac{1}{1+|x|^{\frac{n\left( p^{\prime }-1\right) }{p^{\prime }}+2\sigma }}\lesssim 1 \quad \mathrm{for \ all} \quad x\in {\mathbb {R}}^n, \end{aligned}$$

(B.5)

where $p^{\prime }=\frac{p-1}{p}$ is the conjugate Lebesgue exponent of p. Finally, from the last estimate combined with (B.3), we deduce that $\phi \in L^{\infty }({\mathbb {R}}^n)$; this finishes the first case.

Case 2: $n=6 \sigma$.

Here we observe that since for $n=6 \sigma$, it holds that $p_0=p_*=3$, one has $[3,3)=\varnothing$; thus (B.4) does not make sense for this case. However, we still have (B.3). In addition, since by Sobolev embedding, we know $\phi \in H^{\sigma }({\mathbb {R}}^n)\hookrightarrow L^3({\mathbb {R}}^n)$, which, as before, yields

$$\begin{aligned} \Vert \phi \Vert _{L^{p_1}({\mathbb {R}}^n)}&\lesssim \left\| \frac{|\phi (x)|}{1+|x|^{4\sigma }} *\frac{1}{|x|^{4\sigma }}\right\| _{L^{p_1}({\mathbb {R}}^n)}+\left\| \frac{1}{1+|x|^{2\sigma }}\right\| _{L^{p_1}({\mathbb {R}}^n)} \\&\lesssim \left\| \frac{|\phi (x)|}{1+|x|^{4 \sigma }}\right\| _{L^{q_1}({\mathbb {R}}^n)}+1\\&\lesssim \Vert \phi \Vert _{L^3({\mathbb {R}}^n)}\left\| \frac{1}{1+|x|^{4\sigma }}\right\| _{L^{q_0}({\mathbb {R}}^n)}+1, \end{aligned}$$

where $q_0 \in (3, +\infty ),\zeta _1=\frac{3 q_0}{q_0+3} \in (\frac{3}{2}, 3)$, and $p_1=\frac{3 q_1}{3-q_1} \in (3, +\infty )$.

This means that $\phi \in L^{p}({\mathbb {R}}^n)$ for all $p \geqslant 3$. More precisely, by taking $q_0\gg 1$, one can make $p\gg 1$ large enough. Finally, by the same argument in the last case, we have $\phi \in L^{\infty }({\mathbb {R}}^n)$, which concludes the argument for the second case.

Case 3: $2\sigma<n<6\sigma$.

In this case, using the Hardy–Littlewood–Sobolev inequality, it follows that

$$\begin{aligned} \Vert \phi \Vert _{L^{p_1}({\mathbb {R}}^n)}&\lesssim \left\| \frac{|\phi (x)|}{1+|x|^{4 \sigma }} *{|x|^{n-2 \sigma }}\right\| _{L^{p_1}({\mathbb {R}}^n)} \\&\lesssim \left\| \frac{|\phi (x)|}{1+|x|^{4 \sigma }}\right\| _{L^{q_1}({\mathbb {R}}^n)}\\&\lesssim \Vert \phi \Vert _{L^{p_0}({\mathbb {R}}^n)}\left\| \frac{1}{1+|x|^{4\sigma }}\right\| _{L^{q_0}({\mathbb {R}}^n)}, \end{aligned}$$

where $p_0=\frac{2 n}{n-2 \sigma }=2_\sigma ^*$, $q_0 \in (\frac{n}{2\sigma }, \frac{2 n}{6\sigma -n})$, $q_1=\frac{p_0 q_0}{q_0+p_0}$, and $p_1=\frac{n q_1}{n-2\sigma q_1} \in (p_0, +\infty )$. This means that $\phi \in L^{p}({\mathbb {R}}^n)$ for all $p \geqslant p_0$. From (B.5) we conclude that $\phi \in L^{\infty }({\mathbb {R}}^n)$, which finishes the proof of this case.

The lemma is proved. $\square$