Linear potentials and applications in conformal geometry

Shiguang Ma$\mbox{}^{\dagger}$ and Jie Qing$\mbox{}^{\ddagger}$

###### Abstract.

In this paper we derive estimates for linear potentials that hold away from
thin subsets. And, inspired by the celebrated work of Huber (cf. [16, 22]), we
verify that, for a subset that is thin at a point, there is always a geodesic
that reaches to the point and avoids the thin subset in general dimensions. As
applications of these estimates on linear potentials, we consider the scalar
curvature equations and slightly improve the results of Schoen-Yau [34, 35]
and Carron [11] on the Hausdorff dimensions of singular sets which represent
the ends of complete conformal metrics on domains in manifolds of dimensions
greater than 3. We also study $Q$-curvature equations in dimensions greater
than 4 and obtain stronger results on the Hausdorff dimensions of the singular
sets (cf [13]). More interestingly, our approach based on potential theory
yields a significantly stronger finiteness theorem on the singular sets for
$Q$-curvature equations in dimension 4 (cf. [15, 12]), which is a remarkable
analogue of Huber’s theorem [16, 22].

###### Key words and phrases:

Riesz potentials, Log potentials, outer capacities, $\alpha$-thinness, scalar
curvature equations, $Q$-curvature equations, Hausdorff diemsions

###### 1991 Mathematics Subject Classification:

53C21; 31B35; 53A30; 31B05

##  1\. Introduction

In this paper we employ the linear potential theory to study scalar curvature
equations and $Q$-curvature equations in conformal geometry. This is a
continuation of our recent work on $n$-superharmonic functions (cf, [8, 9, 24,
25]) inspired by the Huber’s theorem and related work in superharmonic
functions in dimension 2 (cf. [16, 22, 5, 21]).

Linear potential theory has always been a major subject in analysis and
partial differential equations. For this paper, we refer readers, for
instance, to [28, 2, 4], for good introductions on potential theory. For
clarity, the definitions of Riesz potentials and Log potentials are given in
Section 2. For our purpose, the kernel functions are not chosen for
discussions on the boundary behavior of potentials and we focus on the outer
capacity and thin subsets (please see Definition 2.2 and Definition 2.3 in
Section 2). We also push to make some of the potential theory work on
Riemannian manifolds. The interesting result on Riesz potentials we obtain is

######  Theorem 1.1.

Suppose that $(M^{n},g)$ is a complete Riemannian manifold and $\mu$ is a
finite nonnegative Radon measure on a bounded domain $G\subset M^{n}$. Let $S$
be a compact subset in $G$ such that its Hausdorff dimension is greater than
$d$, where $d<n-\alpha$ and $\alpha\in(1,n)$. Then there is a point $p\in S$
and a subset $E$ that is $\alpha$-thin at $p$ such that

(1.1)
$\int_{G}\frac{1}{d(x,y)^{n-\alpha}}d\mu\leq\frac{C}{d(x,p)^{n-\alpha-d}}$

for some constant $C$ and all $x\in B_{\delta}(p)\setminus E$ for some small
$\delta>0$.

The proof of Theorem 1.1 uses a general decomposition result [23, Proposition
1.4] and multi-scale analysis. We also give a proof of a slight extension of
[28, Theorem 6.3] for Log potentials on manifolds, which is closely related to
[16, 22, 5, 24, 25] for us. What makes these estimates useful is the following
key observation about thin subsets in general dimensions (cf. [16, 22, 5, 24,
25]).

######  Theorem 1.2.

Let $E$ be a subset in the Euclidean space $R^{n}$ and $p\in R^{n}$. Suppose
that $E$ is $\alpha$-thin at the point $p$ for $\alpha\in(1,n]$. Then there is
always a ray from $p$ that avoids $E$ at least within some small ball at $p$.

The proof of Theorem 1.2 uses only the scaling property (Lemma 2.2), the
contractive property (Lemma 2.3), and the calculation of
$C^{\alpha}(S^{n-1},B_{2}(0))$ (Lemma 2.4) for the outer capacity
$C^{\alpha}(E,\Omega)$ defined in Definition 2.2 and $\alpha$-thinness in
Definition 2.3.

In conformal geometry, the scalar curvature equation describes the conformal
transformation of the scalar curvature

(1.2)
$-\frac{4(n-1)}{n-2}\Delta[\bar{g}]u+R[\bar{g}]u=R[u^{\frac{4}{n-2}}\bar{g}]u^{\frac{n+2}{n-2}}.$

There have been many works on singular solutions after the seminal paper [34],
where the singularities represent the ends of complete conformal metrics on
domains in Riemannian manifolds (cf. for instance, [35, Chapter VI] [11] and
[33, 27, 26]).

######  Theorem 1.3.

Let $(M^{n},\bar{g})$ be a complete Riemannian manifold and $S$ be a compact
subset in $M^{n}$. And let $D$ be a bounded open neighborhood of $S$. Suppose
that $g=u^{\frac{4}{n-2}}\bar{g}$ is a conformal metric on $D\setminus S$ and
is geodesically complete near $S$. Then the Hausdorff dimension

(1.3) $\text{dim}_{\mathscr{H}}(S)\leq\frac{n-2}{2}$

provided $R^{-}[g]\in L^{\frac{2n}{n+2}}(D\setminus S,g)\bigcap
L^{p}(D\setminus S,g)$ for some $p>n/2$, where $R^{-}[g]$ is the negative part
of the scalar curvature of the metric $g$. Consequently, (1.3) holds when the
scalar curvature $R[g]$ of the conformal metric $g$ is nonnegative.

Theorem 1.3 is a slight improvement of [34, Theorem 2.7] and [11, Theorem C].
Our approach is based on Theorem 1.1 and Theorem 1.2. Particularly, Theorem
1.3 covers domains in general manifolds, while others (cf. [34, 11]) are
restricted to domains in round spheres. The use of auxiliary testing functions
built from the level sets is the key analytic technique (cf. [17, 7, 24, 25]).
We remark that, for our approach, the complement $M^{n}\setminus D$ is not
relevant (cf. Theorem 3.1 in Section 3).

In conformal geometry, one considers the Paneitz operator

$P_{4}=\Delta^{2}+\text{div}(4A\cdot\nabla-(n-2)J\nabla)+\frac{n-4}{2}Q_{4}$

and the associated $Q$-curvature,

$Q_{4}=-\Delta J+\frac{n}{2}J^{2}-2|A|^{2},$

where the Schouten curvature $A=\frac{1}{n-2}(Ric-Jg)$ and
$J=\frac{1}{2(n-1)}R$. The curvature $Q_{4}$, under a conformal change of the
metric, transforms by the $Q$-curvature equation:

(1.4)
$P_{4}[\bar{g}]u=\frac{n-4}{2}Q_{4}[u^{\frac{4}{n-4}}\bar{g}]u^{\frac{n+4}{n-4}}\text{
in dimensions $\geq 5$}$

and

(1.5) $P_{4}[\bar{g}]u+Q_{4}[\bar{g}]=Q_{4}[e^{2u}\bar{g}]e^{4u}\text{ in
dimension $4$}.$

On $Q$-curvature equations in dimensions greater than 4, we have

######  Theorem 1.4.

Let $(M^{n},\bar{g})$ be a complete Riemannian manifold for $n\geq 5$ and $S$
be a compact subset in $M^{n}$. And let $D$ be a bounded open neighborhood of
$S$. Suppose that $g=u^{\frac{4}{n-4}}\bar{g}$ is a conformal metric on
$D\setminus S$ with nonnegative scalar curvature $R[g]\geq 0$ and is
geodesically complete near $S$. And suppose also that

$Q_{4}^{-}[g]\in L^{\frac{2n}{n+4}}(D\setminus S,g),$

where $Q_{4}^{-}[g]$ is the negative part of $Q$-curvature of the metric $g$.
Then

(1.6) $\text{dim}_{\mathscr{H}}(S)\leq\frac{n-4}{2}.$

There have been a lot of works on the study of singular solutions to
$Q$-curvature equations on manifolds of dimensions greater than $4$, notably
[29, 30, 13, 20], for example. Theorem 1.4 is an improvement of [13, Theorem
1.2] in terms of curvature conditions and the coverage of domains in general
manifolds. The preliminary estimates in Lemma 4.1 serve to facilitate the
argument of treating the bi-Laplace as the iteration of the Laplace, which are
interesting alternatives to usual elliptic estimates of $Q$-curvature
equations. Again, the complement $M^{n}\setminus D$ is not relevant for our
approach (cf. Theorem 4.1 in Section 4).

On $Q$-curvature equations in dimension 4, there have been several attempts to
establish analogue results of Huber’s theorem on finiteness of singularities
(cf. [15, 12, 24, 25]). $Q$-curvature in dimension 4 indeed plays a similar
role as the Gaussian curvature does in dimension 2 (please see (1.5) for
instance). Our following result is a significant improvement of the finiteness
result of [15, Theorem 2] (cf. also [14]). It covers domains in general
manifolds and drops other additional curvature assumptions in [15, Theorem 2].
The potential theory approach here, particularly Theorem 1.1 and Theorem 1.2,
seems to be more effective. And the preliminary estimates in Lemma 4.3 are
interesting for $Q$-curvature equations in dimension 4 too. Once again, the
complement $M^{n}\setminus D$ is not relevant for our approach (cf. Theorem
4.2 in Section 4).

######  Theorem 1.5.

Let $(M^{4},\bar{g})$ be a complete Riemannian manifold and $S$ be a compact
subset in $M^{n}$. And let $D$ be a bounded open neighborhood of $S$. Suppose
that $g=e^{2u}\bar{g}$ is a conformal metric on $D\setminus S$ with
nonnegative scalar curvature $R[g]\geq 0$ and is geodesically complete near
$S$. And suppose that

$\int_{D}Q_{4}^{-}[g]dvol[g]<\infty,$

where $Q_{4}^{-}[g]$ is the negative part of $Q$-curvature of the metric $g$.
Then $S$ consists of only finitely many points.

The organization of this paper is as follows: In Section 2 we define linear
potentials and develop potential theory with the outer capacity and the notion
of $\alpha$-thinness. Then we prove Theorem 1.1 and Theorem 1.2. In Section 3
we build the framework to use potential theory developed in Section 2 to
estimate the Hausdorff dimension of singular sets which correspond to the ends
of complete conformal metrics on domains of manifolds. And we prove Theorem
1.3. In Section 4, based on the framework built in Section 3, we prepare some
preliminary estimates and prove Theorem 1.4 and Theorem 1.5 for $Q$-curvature
equations.

##  2\. On linear potentials

The study of linear potentials has been extensive and full of great
achievements. For this paper, readers are referred, for instance, to [28, 2,
4] for good introductions. In this section we will introduce the theory of
linear potential to facilitate the discussion of some estimates of linear
potential that is inspired by the one in [16, 22, 5, 24, 25]. The estimates
provide us some alternative tools to study the problems on the Hausdorff
dimensions of singularities of solutions to a class of geometric partial
differential equations in conformal geometry (cf. [34, 35, 13, 12] for
instance). We will introduce the potential theory in the way that is brief,
mostly self-contained, and suffices to serve our purpose.

###  2.1. Linear potential and the outer capacity in Euclidean spaces

For the purpose to relate potentials on Euclidean spaces to that on manifolds,
we want to introduce potentials that are possibly confined to an open subset
$\Omega\subseteq R^{n}$ in the Euclidean space. We will use the definition of
a Radon measure on locally compact Hausdorff spaces in [31, page 455].

######  Definition 2.1.

Let $\Omega\subseteq R^{n}$ be a bounded open subset in the Euclidean space
$R^{n}$. Then, for $x\in\Omega$, let

(2.1) $R^{\alpha,\Omega}_{\mu}(x)=\left\\{\begin{aligned}
\int_{\Omega}\frac{1}{|x-y|^{n-\alpha}}d\mu(y)\quad&\text{ when
$\alpha\in(1,n)$}\\\ \int_{\Omega}\log\frac{D}{|x-y|}d\mu(y)\quad&\text{ when
$\alpha=n$}\end{aligned}\right.$

for a Radon measure $\mu$ on $\Omega$, where $D$ is the diameter of $\Omega$.

For basic properties of the potential $R^{\alpha,\Omega}_{\mu}(x)$, readers
are referred to [28, Chapter 2]. Most facts, results, and arguments in [28,
Chapter 2] that are relevant for the discussions in this paper hold with
little or slight changes.

######  Definition 2.2.

Let $E$ be a subset in $\Omega$ and $\Omega$ be a bounded open subset in
$R^{n}$. For $\alpha\in(1,n]$, we define a capacity by

(2.2) $C^{\alpha}(E,\Omega)=\inf\\{\mu(\Omega):\mu\geq 0\text{ on $\Omega$ and
}R^{\alpha,\Omega}_{\mu}(x)\geq 1\text{ for all $x\in E$}\\}.$

Because of the choice of the kernel functions in Definition 2.1, the capacity
$C^{\alpha}(E,\Omega)$ in Definition 2.2 is not intended to be the same as
relative capacity where the kernel function is the Green function for a so-
called Greenian domain $\Omega$. Similar to [28, Theorem 4.1 in Chapter 2] and
[28, Section 2.6], we have

######  Lemma 2.1.

Let $C^{\alpha}$ be the capacity defined as in Definition 2.2 for
$\alpha\in(1,n]$.

  1. (1)

$C^{\alpha}$ is nondecreasing, that is,

$C^{\alpha}(E_{1},\Omega)\leq C^{\alpha}(E_{2},\Omega)$

when $E_{1}\subseteq E_{2}\subseteq\Omega\subseteq R^{n}$.

  2. (2)

$C^{\alpha}$ is countably subadditive, that is,

$C^{\alpha}(\bigcup_{i=1}^{\infty}E_{i},\Omega)\leq\sum_{i=1}^{\infty}C^{\alpha}(E_{i},\Omega)$

for subsets $E_{i}\subseteq\Omega$.

  3. (3)

$C^{\alpha}$ is an outer capacity, that is,

$C^{\alpha}(E,\Omega)=\inf\\{C^{\alpha}(U,\Omega):E\subseteq U\text{ and
$U\subseteq\Omega$ open}\\}.$

The immediate and important property of the outer capacity $C^{\alpha}$ in
Definition 2.2 is the scaling property (cf. [4, page 135]).

######  Lemma 2.2.

For a positive number $\lambda$, let

$A_{\lambda}=\\{\lambda x:x\in A\\}$

for any subset $A$ in $R^{n}$. Then, for $\alpha\in(1,n]$,

$C^{\alpha}(E_{\lambda},\Omega_{\lambda})=\lambda^{n-\alpha}C^{\alpha}(E,\Omega).$

###### Proof.

For a nonnegative Radon measure $\mu$ on $\Omega$, we associate it with a
nonnegative Radon measure

$\mu^{*}(A_{\lambda})=\mu(A)$

on $\Omega_{\lambda}$. Then

$R^{\alpha,\Omega_{\lambda}}_{\mu^{*}}(\lambda
x)=\lambda^{\alpha-n}R^{\alpha,\Omega}_{\mu}(x)$

for $x\in\Omega$. Therefore

$\displaystyle C^{\alpha}$
$\displaystyle(E_{\lambda},\Omega_{\lambda})=\inf\\{\mu^{*}(\Omega_{\lambda}):R^{\alpha,\Omega_{\lambda}}_{\mu^{*}}(\lambda
x)\geq 1\text{ for all $x\in E$}\\}$
$\displaystyle=\lambda^{n-\alpha}\inf\\{\lambda^{\alpha-n}\mu(E):R^{\alpha,\Omega}_{\lambda^{\alpha-n}\mu}(x)\geq
1\text{ for all $x\in E$}\\}$
$\displaystyle=\lambda^{n-\alpha}C^{\alpha}(E,\Omega).$

∎

Next important property of the outer capacity $C^{\alpha}$ in Definition 2.2
is the contractive property (cf. [28, 2, 4]).

######  Lemma 2.3.

Suppose that

$\Phi:\Omega\to\Omega$

is a contractive map, that is,

$|\Phi(x)-\Phi(y)|\leq|x-y|$

for all $x,y\in\Omega$. Then, for $\alpha\in(1,n]$,

$C^{\alpha}(\Phi(E),\Omega)\leq C^{\alpha}(E,\Omega)$

for any subset $E\subseteq\Omega$.

###### Proof.

Let $\mu$ be a nonnegative Radon measure on $\Omega$ such that
$R^{\alpha,\Omega}_{\mu}(x)\geq 1$ for all $x\in E$. Then let $\mu^{*}$ be a
nonnegative Radon measure on $\Omega$ such that $\mu^{*}(A)=\mu(\Phi^{-1}(A))$
for any $A\subseteq\Omega$ and therefore

$\int_{\Omega}f(\tilde{y})d\mu^{*}(\tilde{y})=\int_{\Omega}f\circ\Phi(y)d\mu(y).$

Notice that

$\displaystyle R^{\alpha,\Omega}_{\mu^{*}}(\Phi(x))$
$\displaystyle=\int_{\Omega}\frac{1}{|\Phi(x)-\tilde{y}|^{n-\alpha}}d\mu^{*}(\tilde{y})=\int_{\Omega}\frac{1}{|\Phi(x)-\Phi(y)|^{n-\alpha}}d\mu(y)$
$\displaystyle\geq\int_{\Omega}\frac{1}{|x-y|^{n-\alpha}}d\mu(y)=R^{\alpha,\Omega}_{\mu}(x)\geq
1.$

Thus

$\displaystyle C^{\alpha}(\Phi(E),\Omega)$
$\displaystyle=\inf\\{\nu(\Omega):\nu\geq 0\text{ on $\Omega$ and
}R^{\alpha,\Omega}_{\nu}(x)\geq 1\text{ for all $x\in\Phi(E)$}\\}$
$\displaystyle\leq\inf\\{\mu^{*}(\Omega):\mu^{*}\text{ induced from $\mu$ and
}R^{\alpha,\Omega}_{\mu^{*}}(\Phi(x))\geq 1\text{ for all $x\in E$}\\}$
$\displaystyle=\inf\\{\mu(\Omega):\mu\geq 0\text{ on $\Omega$ and
}R^{\alpha,\Omega}_{\mu}(x)\geq 1\text{ for all $x\in
E$}\\}=C^{\alpha}(E,\Omega).$

The argument for $\alpha=n$ is similar and the proof is complete. ∎

Before we introduce the notion of thinness by $C^{\alpha}$, for completeness,
let us calculate the outer capacity $C^{\alpha}(S^{n-1},B_{2})$, where

$B_{2}=\\{x\in R^{n}:|x|<2\\}\text{ and }S^{n-1}=\\{x\in R^{n}:|x|=1\\}.$

######  Lemma 2.4.

([28, Example 5.4.3]) For $\alpha\in(1,n]$,

$C^{\alpha}(S^{n-1},B_{2})=c(n,\alpha)$

for some positive constant $c(n,\alpha)$.

###### Proof.

It suffices to show that $C^{\alpha}(S^{n-1},B_{2})$ is finite and positive.
Let $\sigma$ be the volume measure for the unit sphere so that the total
measure of $S^{n-1}$ is $1$. First we realize that the potential, for
$\alpha\in(1,n]$ and $x\in S^{n-1}$,

$R^{\alpha,B_{2}}_{\sigma}(x)\geq m$

for some $m=m(n,\alpha)>0$. Therefore
$C^{\alpha}(S^{n-1},B_{2})\leq\frac{1}{m}<\infty$ by Definition 2.2. To see
that $C^{\alpha}(S^{n-1},B_{2})>0$, for any $\mu$ on $B_{2}$, we use Lemma 2.5
below to pick up a point $p\in S^{n-1}$ such that (2.3) holds and calculate,
for $\alpha\in(1,n)$,

$\displaystyle R^{\alpha,B_{2}}_{\mu}(p)$
$\displaystyle=(n-\alpha)\int_{0}^{\infty}\mu(\\{\frac{1}{r}-\frac{1}{3}>s\\}\bigcap
B_{2})\frac{1}{(s+\frac{1}{3})^{n-\alpha+1}}ds+\frac{1}{3^{n-\alpha}}\mu(B_{2})$
$\displaystyle=(n-\alpha)\int_{0}^{3}\mu(B_{r}(p)\bigcap
B_{2})r^{\alpha-n-1}dr+\frac{1}{3^{n-\alpha}}\mu(B_{2})$ $\displaystyle\leq
M(n,\alpha)\mu(B_{2})$

for some $M(n,\alpha)>0$ and $r=|x-p|$. For $\alpha=n$,

$\displaystyle R^{\alpha,B_{2}}_{\mu}(p)$
$\displaystyle=\int_{0}^{\infty}\mu(\\{\frac{3}{r}-1>s\\}\bigcap
B_{2})\frac{1}{1+s}ds+\log\frac{4}{3}\mu(B_{2})$
$\displaystyle=\int_{0}^{3}\mu(B_{r}(p)\bigcap
B_{2})\frac{1}{r}dr+\log\frac{4}{3}\,\mu(B_{2})$ $\displaystyle\leq
M(n,n)\mu(B_{2})$

for some $M(n,n)>0$. In the above we used [32, Theorem 8.16]. This implies
$C^{\alpha}(S^{n-1},B_{2})\geq\frac{1}{M(n,\alpha)}>0$ by Definition 2.2. Thus
the proof is complete. ∎

By the Vitali covering lemma, we prove the following fact used in the above.

######  Lemma 2.5.

Let $n\geq 2$ and $\mu$ be a finite nonnegative Radon measure on $B_{2}\subset
R^{n}$. Then there is a point $p\in S^{n-1}$ such that

(2.3) $\mu(B_{r}(p)\bigcap B_{2})\leq c(n)\mu(B_{2})r^{n-1}\text{ for all
$r>0$}.$

for some dimensional constant $c=c(n)$.

###### Proof.

For convenience, let $\mu(B_{2})=1$. Assume otherwise, for any $q\in S^{n-1}$,
there is $r_{q}>0$ such that

$\mu(B_{r_{q}}(q)\bigcap B_{2})\geq c(n)r_{q}^{n-1}.$

Using Vitali covering lemma, we have $\\{q_{1},q_{2},\cdots,q_{k}\\}\subset
S^{n-1}$ such that the collection of balls

$\\{B_{r_{q_{1}}}(q_{1}),B_{r_{q_{2}}}(q_{2}),\cdots B_{r_{q_{k}}}(q_{k})\\}$

are disjoint but the collection of balls

$\\{B_{3r_{q_{1}}}(q_{1}),B_{3r_{q_{2}}}(q_{2}),\cdots
B_{3r_{q_{k}}}(q_{k})\\}$

cover the sphere $S^{n-1}$. Therefore, on one hand,

$c(n)\sum_{i=1}^{k}r_{q_{i}}^{n-1}\leq\sum_{i=1}^{k}\mu(B_{r_{q_{i}}}(q_{i})\bigcap
B_{2})\leq\mu(B_{2})=1.$

On the other hand,

$|S^{n-1}|\leq\sum_{i=1}^{k}|B_{3r_{q_{i}}}(q_{i})\bigcap
S^{n-1}|<|S^{n-1}|c(n)\sum_{i=1}^{k}r_{q_{i}}^{n-1}$

when $c(n)$ is sufficiently large, where $|\cdot|$ stands for the Lebesgue
measure on $S^{n-1}$. Therefore the lemma is proven by contradiction. ∎

Now let us introduce the geometric definition of thinness. For notions of
thinness in terms of the fine topology and Wiener criterion, readers are
referred, for instance, to [28, 2, 4]. Let

$\omega^{\delta}_{i}(p)=\\{x\in
R^{n}:|x-p|\in[2^{-i}\delta,2^{-i+1}\delta]\\}$

and

$\Omega^{\delta}_{i}(p)=\\{x\in
R^{n}:|x-p|\in(2^{-i-1}\delta,2^{-i+2}\delta)\\}.$

######  Definition 2.3.

Let $E$ be a subset in the Euclidean space $R^{n}$ and $p\in R^{n}$ be a point
in $R^{n}$. The subset $E$ is said to be $\alpha$-thin at the point $p$ for
$\alpha\in(1,n)$ if

$\sum_{i\geq
1}\frac{C^{\alpha}(E\bigcap\omega^{\delta}_{i}(p),\Omega^{\delta}_{i}(p))}{C^{\alpha}(\partial
B_{2^{-i}\delta}(p),B_{2^{-i+1}\delta}(p))}<\infty$

for some small $\delta>0$. The subset $E$ is said to be $n$-thin at $p$ if

$\sum_{i\geq
1}iC^{n}(E\bigcap\omega^{\delta}_{i}(p),\Omega^{\delta}_{i}(p))<\infty$

for some small $\delta>0$.

Combining Lemma 2.1-2.4 with the above definition, we observe the following
important property of $\alpha$-thin sets, inspired by [5] (see also [24, 25]).
We recall Theorem 1.2 from the introduction for readers’ convenience.

######  Theorem 2.1.

Let $E$ be a subset in the Euclidean space $R^{n}$ and $p\in R^{n}$ be a
point. Suppose that $E$ is $\alpha$-thin at the point $p$ for
$\alpha\in(1,n]$. Then there is a ray from $p$ that avoids $E$ at least within
some small ball at $p$.

###### Proof.

First of all, due to the translation invariance, we may simply assume $p$ is
the origin of the Euclidean space. Then, by the scaling property of the outer
capacity $C^{\alpha}$ in Lemma 2.2, one notices that

$\frac{C^{\alpha}(E\bigcap\omega^{\delta}_{i},\Omega^{\delta}_{i})}{C^{\alpha}(\partial
B_{2^{-i}\delta},B_{2^{-i+1}\delta})}=\frac{C^{\alpha}(S_{i}(E)\bigcap\omega^{1}_{0},\Omega^{1}_{0})}{C^{\alpha}(\partial
B_{1},B_{2})}$

where the scaling map: $S_{i}(v)=\frac{2^{i}}{\delta}v$. Then we consider the
projection

$P(v)=\left\\{\begin{aligned} \frac{v}{|v|}\quad&\text{ when $v\in R^{n}$ and
$|v|\geq 1$}\\\ v\quad&\text{ when $v\in R^{n}$ and
$|v|<1$}\end{aligned}\right.,$

which is contractive. Therefore, in the light of Lemma 2.3, we have

$C^{\alpha}(P(S_{i}(E)\cap\omega^{1}_{0}),\Omega^{1}_{0})\leq
C^{\alpha}(S_{i}(E)\cap\omega^{1}_{0},\Omega^{1}_{0}).$

Next, using the countable sub-additivity in Lemma 2.1, we have

$C^{\alpha}(\bigcup_{i\geq
k}P(S_{i}(E)\cap\omega^{1}_{0}),\Omega^{1}_{0})\leq\sum_{i\geq
k}C^{\alpha}(P(S_{i}(E)\cap\omega^{1}_{0}),\Omega^{1}_{0}).$

Thus,

$\displaystyle C^{\alpha}$ $\displaystyle(\bigcup_{i\geq
k}P(S_{i}(E)\cap\omega^{1}_{0}),\Omega^{1}_{0})\leq\sum_{i\geq
k}C^{\alpha}(S_{i}(E)\cap\omega^{1}_{0},\Omega^{1}_{0})$ $\displaystyle\leq
C^{\alpha}(\partial B_{1},B_{2})\sum_{i\geq
k}\frac{C^{\alpha}(S_{i}(E)\bigcap\omega^{1}_{0},\Omega^{1}_{0})}{C^{\alpha}(\partial
B_{1},B_{2})}$ $\displaystyle\leq C^{\alpha}(\partial B_{1},B_{2})\sum_{i\geq
k}\frac{C^{\alpha}(E\bigcap\omega^{\delta}_{i},\Omega^{\delta}_{i})}{C^{\alpha}(\partial
B_{2^{-i}\delta},B_{2^{-i+1}\delta})}$

which is arbitrarily small when $k$ is appropriately large using Lemma 2.4 for
$C^{\alpha}(\partial B_{1},B_{2})$. And then this implies that

$\partial B_{1}\setminus\bigcup_{i\geq
k}P(S_{i}(E)\cap\omega^{1}_{0})\neq\emptyset.$

The argument for $\alpha=n$ is similar and easier. And the proof is complete.
∎

###  2.2. Linear potential on manifolds

On a given complete Riemannian manifold $(M^{n},g)$, let $d(\cdot,\cdot)$ be
the distance function associated with the given Riemannian metric $g$.

######  Definition 2.4.

Suppose that $(M^{n},g)$ is a complete Riemannian manifold and $U\subseteq
M^{n}$ is a bounded open subset. For $\alpha\in(1,n]$, the linear potential on
the Riemannian manifold $(M^{n},g)$ of order $\alpha$ for a Radon measure
$\mu$ on $U$ is given as

$\mathscr{R}^{\alpha,U}_{\mu}(x)=\left\\{\begin{aligned}
\int_{U}\frac{1}{d(x,y)^{n-\alpha}}d\mu(y)\quad&\text{ when
$\alpha\in(1,n)$}\\\ \int_{\Omega}\log\frac{D}{d(x,y)}d\mu(y)\quad&\text{ when
$\alpha=n$}\end{aligned}\right.,$

where $D$ is the diameter of $U$.

From the discussion in the previous subsection, it is easily seen that one may
generate an outer capacity $\mathscr{C}^{\alpha}(E,U)$ for any subset
$E\subseteq U\subseteq M^{n}$ that behaves like the counter part in Euclidean
spaces. To use $R^{\alpha,\Omega}_{\mu}(x)$ and $C^{\alpha}(E,\Omega)$ on
Euclidean spaces in the previous subsection to study
$\mathscr{R}^{\alpha,U}_{\mu}(p)$ and $\mathscr{C}^{\alpha}(A,U)$ on
manifolds, we first introduce the correspondence between Radon measures on the
tangent space $T_{p}M^{n}$ at each point $p\in M^{n}$ and those on
$(M^{n},g)$. Suppose that $(M^{n},g)$ is a complete Riemannian manifold. Let
$p\in M^{n}$ and $U$ be a convex normal coordinate neighborhood at $p$ on
$(M^{n},g)$, where the exponential map serves as the convex normal coordinate

$\text{exp}|_{p}:\Omega\to U.$

The domain $U$ is said to be convex if the unique geodesic joining any two
points in $U$ stays in $U$. Moreover, we may assume in the coordinate chart
$U$ the exponential map be uniformly bi-Lipschitz throughout this paper.

Then, for a Radon measure $\mu$ on $U\subseteq M^{n}$, one may introduce the
Radon measure $\mu^{*}$ on $\Omega\subset T_{p}M^{n}$ such that, for a subset
$E\subseteq\Omega$,

$\mu^{*}(E)=\mu(\text{exp}|_{p}E)\text{ and
}\int_{\Omega}f\circ\text{exp}|_{p}d\mu^{*}=\int_{U}fd\mu.$

It is then easily seen that the following equivalence between the linear
potential $R^{\alpha,\Omega}_{\mu^{*}}$, the outer capacities
$C^{\alpha}(\cdot,\Omega)$ and the corresponding
$\mathscr{R}^{\alpha,U}_{\mu}$, $\mathscr{C}^{\alpha}(\cdot,U)$ holds. Namely,

######  Lemma 2.6.

Suppose that $(M^{n},g)$ be a complete Riemannian manifold and $p\in M^{n}$.
Let

$\text{exp}|_{p}:\Omega\to U$

be the convex normal coordinate chart, where the exponential map is uniformly
bi-Lipschitz. And let $\alpha\in(1,n]$. Then, for $A\subset U$ and
$E=(\text{exp}|_{p})^{-1}A\subset\Omega$,

$\displaystyle C^{-1}R^{\alpha,\Omega}_{\mu^{*}}\leq$
$\displaystyle\mathscr{R}^{\alpha,U}_{\mu}\leq CR^{\alpha,\Omega}_{\mu^{*}}$
$\displaystyle C^{-1}C^{\alpha}(E,\Omega)\leq$
$\displaystyle\mathscr{C}^{\alpha}(A,U)\leq CC^{\alpha}(E,\Omega)$

for some constant $C=C(M^{n},g,U,p)$. Consequently, a subset $A\subset U$ is
$\alpha$-thin at $p$ if and only if $E=(\text{exp}|_{p})^{-1}(A)\subset\Omega$
is $\alpha$-thin at the origin of $T_{p}M^{n}$.

###### Proof.

The proof is straightforward based on the properties of the convex normal
coordinate chart at a point in a complete Riemannain manifold, where the
exponential map is bi-Lipschtiz. ∎

###  2.3. Estimates of Riesz potential

In this subsection, we introduce our estimates of Riesz potentials on
manifolds. We will recall some well known estimates for Riesz potential in
Euclidean spaces [28, Chapter 2].

Our estimates on Riesz potentials are designed to help understand the
Hausdorff dimensions of singularities of solutions of partial differential
equations on manifolds. Let us start with a general decomposition theorem for
nonnegative Radon measures on a complete Riemannian manifold based on [23,
Proposition 1.4], which is related to Lemma 2.5 and a broad generalization of
the Lebesgue Differentiation Theorem in some way.

######  Lemma 2.7.

( [23, Proposition 1.4]) Let $\mu$ be a nonnegative Radon measure on a
complete Riemannian manifold $(M^{n},g)$ and let

$G_{d}^{\infty}=\\{x\in M^{n}:\limsup_{r\to 0}r^{-d}\mu(B_{r}(x))=+\infty\\}$

for any $d\in[0,n]$. Then

$\mathscr{H}_{d}(G^{\infty}_{d})=0$

where $\mathscr{H}_{d}$ is the Hausdorff measure of dimension $d$.

###### Proof.

Based on the general decomposition theorem [23, Proposition 1.4] on the
Euclidean space and the correspondence of Radon measures in Lemma 2.6, this
lemma is easily seen. Specifically, we first prove the statement for Radon
measures supported in a convex normal coordinate chart used in Lemma 2.6. Then
the lemma follows by using a countable covering for $(M,g)$ by convex normal
coordinate charts. ∎

Now we are ready to state and prove one crucial analytic result in this paper
on the behavior of the Riesz potentials. For readers’ convenience, we recall
Theorem 1.1 from the introduction.

######  Theorem 2.2.

Suppose that $(M^{n},g)$ is a complete Riemannian manifold and $\mu$ is a
finite Radon measure on a bounded domain $G\subset M^{n}$. Let $S$ be a
compact subset in $G$ such that its Hausdorff dimension is greater than $d$.
And let $\alpha\in(1,n)$ and $d<n-\alpha$. Then there is a point $p\in S$ and
a subset $E$ that is $\alpha$-thin at $p$ such that

$\int_{G}\frac{1}{d(x,y)^{n-\alpha}}d\mu\leq\frac{C}{d(x,p)^{n-\alpha-d}}$

for some constant $C$ and all $x\in B_{\delta}(p)\setminus E$ for some
$\delta>0$.

###### Proof.

First, due to the assumption that the Hausdorff dimension of $S$ is greater
than $d$,

$\mathscr{H}_{d+\epsilon}(S)=\infty$

for some small $\epsilon>0$. Then, in the light of Lemma 2.7, there is a point
$p\in S$ such that

$\limsup_{r\to 0}r^{-(d+\epsilon)}\mu(B_{r}(p))\leq C<\infty.$

That is to say

(2.4) $\mu(B_{r}(p))\leq Cr^{d+\epsilon}$

when $r$ is appropriately small. Secondly, we may confine ourselves to a
convex normal coordinate neighborhood $U$ of $p$ and we may work on the
Euclidean space without the loss of generality in the light of the discussion
in the previous subsection, particularly, Lemma 2.6, where
$\text{exp}|_{p}:\Omega\to U$ and $\text{exp}|_{p}(0)=p$. For convenience, we
will not differentiate $\mu$ and $\mu^{*}$ if no confusion rises. Therefore,
for $x\in\omega_{i}^{\delta}\subset\Omega$ when $\delta$ is sufficiently small
and $i$ is appropriately large,

(2.5) $\displaystyle
R^{\alpha,\Omega}_{\mu}(x)=\int_{\Omega}\frac{1}{|x-y|^{n-\alpha}}d\mu$
$\displaystyle=\int_{\Omega\setminus
B_{2^{-i_{0}+2}\delta}}\frac{1}{|x-y|^{n-\alpha}}d\mu+\int_{B_{2^{-i_{0}+2}\delta}\setminus\Omega^{\delta}_{i}}\frac{1}{|x-y|^{n-\alpha}}d\mu+\int_{\Omega^{\delta}_{i}}\frac{1}{|x-y|^{n-\alpha}}d\mu,$

where $i_{0}\leq i$ to be fixed. For the first term in the right hand side of
(2.5),

$I=\int_{\Omega\setminus
B_{2^{-i_{0}+2}\delta}}\frac{1}{|x-y|^{n-\alpha}}d\mu\leq(\frac{1}{(2^{-i_{0}+2}\delta-2^{-i+1}\delta})^{n-\alpha}\mu(\Omega)\leq(\frac{1}{2^{-i_{0}+1}\delta})^{n-\alpha}\mu(\Omega).$

Recall that $2^{-i}\delta\leq|x|\leq 2^{-i+1}\delta$ for $x\in\omega_{i}$, we
have

(2.6)
$I\leq\mu(\Omega)\frac{(2^{-i+1}\delta)^{n-\alpha-d}}{(2^{-i_{0}+1}\delta)^{n-\alpha}}\frac{1}{|x|^{n-\alpha-d}}\leq
C\frac{1}{|x|^{n-\alpha-d}}$

where $C=C(n,\alpha,d,\delta,i_{0})$. For the second term in the right hand
side of (2.5),

$\displaystyle\int_{B_{2^{-i_{0}+2}\delta}\setminus\Omega_{i}^{\delta}}\frac{1}{|x-y|^{n-\alpha}}d\mu=\int_{B_{2^{-i_{0}+2}\delta}\setminus
B_{2^{-i+2}\delta}}\frac{1}{|x-y|^{n-\alpha}}d\mu+\int_{B_{2^{-i-1}\delta}}\frac{1}{|x-y|^{n-\alpha}}d\mu$
$\displaystyle\leq\int_{B_{2^{-i_{0}+2}\delta}\setminus
B_{2^{-i+2}\delta}}\frac{1}{|x-y|^{n-\alpha}}d\mu+(\frac{1}{2^{-i-1}\delta})^{n-\alpha}\mu(B_{2^{-i-1}\delta})$
$\displaystyle\leq\sum_{k=i_{0}}^{i-1}\int_{B_{2^{-k+2}}\delta\setminus
B_{2^{-k+1}\delta}}\frac{1}{|x-y|^{n-\alpha}}d\mu+(\frac{1}{2^{-i-1}\delta})^{n-\alpha}\mu(B_{2^{-i-1}\delta})$
$\displaystyle\leq\sum_{k=i_{0}}^{i-1}(\frac{1}{2^{-k}\delta})^{n-\alpha}\mu(B_{2^{-k+2}\delta})+(\frac{1}{2^{-i-1}\delta})^{n-\alpha}\mu(B_{2^{-i-1}\delta}).$

Using (2.4) for $\epsilon=0$, we continue from the above,

(2.7) $\displaystyle II$ $\displaystyle\leq
C(4^{d}\sum_{k=i_{0}}^{i-1}(\frac{1}{2^{-k}\delta})^{n-\alpha-d}+(\frac{1}{2^{-i-1}\delta})^{n-\alpha-d})$
$\displaystyle\leq
C(\frac{4^{d}}{1-2^{-(n-\alpha-d)}}(\frac{1}{2^{-i+1}\delta})^{n-\alpha-d}+(\frac{1}{2^{-i-1}\delta})^{n-\alpha-d})$
$\displaystyle\leq C\frac{1}{|x|^{n-\alpha-d}},$

where $C=C(n,\alpha,d,\delta,i_{0})$. To handle the third term in the right
hand side of (2.5), we let

$E_{i}^{\lambda}=\\{x\in\omega^{\delta}_{i}:\int_{\Omega^{\delta}_{i}}\frac{1}{|x-y|^{n-\alpha}}d\mu\geq\lambda
2^{i(n-\alpha-d)}\\},$

where $\lambda>0$ to be fixed. By Definition 2.2, we know

$C^{\alpha}(E^{\lambda}_{i},\Omega^{\delta}_{i})\leq\frac{\mu(\Omega^{\delta}_{i})}{\lambda
2^{i(n-\alpha-d)}}\leq\frac{C}{\lambda}\frac{(2^{-i+2}\delta)^{d+\epsilon}}{2^{i(n-\alpha-d)}}=\frac{C4^{d+\epsilon}}{\lambda}2^{-i\epsilon}(2^{-i})^{n-\alpha},$

where (2.4) for some $\epsilon>0$ is used and $\Omega^{\delta}_{i}\subset
B_{2^{-i+2}\delta}$. Now, from Lemma 2.4 and the scaling property, we know

$C^{\alpha}(\partial
B_{2^{-i}\delta},B_{2^{-i+1}\delta})=C(n,\alpha)(2^{-i}\delta)^{n-\alpha}$

and

$\sum_{i\geq
i_{0}}\frac{C^{\alpha}(E^{\lambda}_{i},\Omega^{\delta}_{i})}{C^{\alpha}(\partial
B_{2^{-i}\delta},B_{2^{-i+1}\delta})}\leq\frac{C}{\lambda}\sum_{i\geq
i_{0}}2^{-\epsilon i}<\infty.$

Thus, by Definition 2.3, the proof is completed. ∎

As a consequence of Theorem 2.1 and 2.2, we have

######  Corollary 2.1.

Suppose that $(M^{n},g)$ is a complete Riemannian manifold and $\mu$ is a
finite Radon measure on a bounded domain $G\subset M^{n}$. Let $S$ be a
compact subset in $G$ such that its Hausdorff dimension is greater than $d$.
And let $\alpha\in(1,n)$ and $d<n-\alpha$. Then there is a point $p\in S$ and
a subset $E$ such that

$\int_{G}\frac{1}{d(x,y)^{n-\alpha}}d\mu\leq\frac{C}{d(x,p)^{n-\alpha-d}}$

for some constant $C$ and for all $x$ along a geodesic ray from $p$ at least
within a small geodesic ball.

###  2.4. Estimates of Log potential

First, as stated in [28, Theorem 6.3], for the Log potential $U_{n}\mu(x)$ on
Euclidean spaces defined in [28, page 82],

$\lim_{x\to p\text{ and }x\in\Omega\setminus
E}\frac{U_{n}\mu(x)}{\log\frac{1}{|x-p|}}=\mu(\\{p\\}).$

The following is our version of [28, Theorem 6.3] on manifolds. For us it is a
generalization of [5, Theorem 1.3] in higher dimensions and linear version of
such behaviors for $n$-superharmonic functions (cf. [22, 8, 9, 24, 25]). For
convenience, we present a brief but full proof based on the potential theory
developed in previous subsections in this paper.

######  Theorem 2.3.

Suppose $(M^{n},g)$ is a complete Riemannian manifold. Let $\mu$ be a finite
Radon measure on a bounded domain $G\subset M^{n}$. Then, for $p\in G$, there
is a subset $A$ that is n-thin at $p$ and

$\lim_{x\to p\text{ and }x\in M^{n}\setminus
A}\frac{\int_{G}\log\frac{1}{d(x,p)}d\mu(x)}{\log\frac{1}{d(x,p)}}=\mu(\\{p\\}).$

###### Proof.

Let

$\text{exp}|_{p}:\Omega\to U$

be a convex normal coordinate at $p\in M^{n}$. Clearly, it suffices to show
that, there is a subset $A$ in $U$, which is $n$-thin at $p$, such that

(2.8) $\lim_{x\to p\text{ and }x\in U\setminus
A}\frac{\mathscr{R}^{n,U}_{\mu}(x)}{\log\frac{1}{d(x,p)}}=\mu(\\{p\\}).$

Therefore, for $x\in\omega^{\delta}_{i}(p)$, we write

(2.9)
$\displaystyle\mathscr{R}^{n,U}_{\mu}(x)=\int_{U}\log\frac{D}{d(x,y)}d\mu(y)$
$\displaystyle=\int_{U\setminus
B_{2^{-i_{0}+2}\delta}}\log\frac{D}{d(x,y)}d\mu+\int_{B_{2^{-i_{0}+2}\delta}\setminus\Omega^{\delta}_{i}}\log\frac{D}{d(x,y)}d\mu+\int_{\Omega^{\delta}_{i}}\log\frac{D}{d(x,y)}d\mu.$

Here we omit the center $p$ for each ball or annulus for simplicity. For the
first term in the right hand side of (2.9),

(2.10) $I=\int_{U\setminus
B_{2^{-i_{0}+2}\delta}}\log\frac{D}{d(x,y)}d\mu\leq\mu(U)\log\frac{D}{2^{-i_{0}+1}\delta}=o(1)\log\frac{1}{d(x,p)}\text{
as $x\to p$}.$

For the second term in the right hand side of (2.9),

$\displaystyle\int_{B_{2^{-i_{0}+2}\delta}\setminus\Omega_{i}^{\delta}}\log\frac{D}{d(x,y)}d\mu(y)=\int_{B_{2^{-i_{0}+2}\delta}\setminus
B_{2^{-i+1}\delta}}\log\frac{D}{d(x,y)}d\mu+\int_{B_{2^{-i-2}\delta}}\log\frac{D}{d(x,y)}d\mu$
$\displaystyle\leq C[\sum_{k=i_{0}}^{i}k\mu(B_{2^{-k+2}\delta}\setminus
B_{2^{-k+1}\delta})]+\mu(B_{2^{-i-2}\delta})\log\frac{D}{2^{-i-2}\delta}.$

Due to the regularity of Radon measures and
$d(x,p)\in[2^{-i-1}\delta,2^{-i}\delta]$, we know

(2.11)
$\mu(B_{2^{-i-2}\delta})\log\frac{D}{2^{-i-2}\delta}=\mu(\\{p\\})\log\frac{1}{d(x,p)}+o(\log\frac{1}{d(x,p)})\text{
as $x\to p$}$

and

(2.12) $\sum_{k=i_{0}}^{i}k\mu(B_{2^{-k+2}\delta}\setminus
B_{2^{-k+1}\delta})=o(1)i=o(1)\log\frac{1}{d(x,p)}$

as $i\to\infty$ or equivalently $x\to p$. To see (2.12), for any $\epsilon>0$,
we first find $k_{0}$ such that

$\mu(B_{2^{-l+2}\delta}\setminus B_{2^{-m+1}\delta})\leq\frac{1}{2}\epsilon$

for all $m\geq l\geq k_{0}$ due to the regularity of $\mu$. Next, we find $N$
such that

$\frac{\sum_{k=i_{0}}^{k_{0}}k\mu(B_{2^{-k+2}\delta}\setminus
B_{2^{-k+1}\delta})}{i}\leq\frac{1}{2}\epsilon$

for all $i\geq N$. Together, this gives

$\displaystyle\frac{\sum_{k=i_{0}}^{i}k\mu(B_{2^{-k+2}\delta}\setminus
B_{2^{-k+1}\delta})}{i}$
$\displaystyle=\frac{\sum_{k=i_{0}}^{k_{0}}k\mu(B_{2^{-k+2}\delta}\setminus
B_{2^{-k+1}\delta})}{i}+\frac{\sum_{k=k_{0}+1}^{i}k\mu(B_{2^{-k+2}\delta}\setminus
B_{2^{-k+1}\delta})}{i}$
$\displaystyle=\frac{\sum_{k=i_{0}}^{k_{0}}k\mu(B_{2^{-k+2}\delta}\setminus
B_{2^{-k+1}\delta})}{i}+\sum_{k=k_{0}+1}^{i}\mu(B_{2^{-k+2}\delta}\setminus
B_{2^{-k+1}\delta})$ $\displaystyle\leq\epsilon$

for all $i\geq N$. Thus we conclude that

(2.13) $II=(\mu(\\{p\\})+o(1))\log\frac{1}{d(x,p)}\text{as $x\to p$}.$

To handle the third term in the right side of (2.9), for $\lambda_{i}>0$ to be
determined, we consider

$A^{\lambda_{i}}=\\{x\in\omega^{\delta}_{i}:\int_{\Omega^{\delta}_{i}}\log\frac{D_{i}}{d(x,y)}d\mu\geq
i\lambda_{i}\\},$

where $D_{i}$ is the dimeter of $\Omega_{\delta}^{i}$. By Definition 2.2,

$\mathscr{C}^{n}(A^{\lambda_{i}},\Omega^{\delta}_{i})\leq\frac{\mu(\Omega^{\delta}_{i})}{i\lambda_{i}}.$

In the light of Definition 2.3, we consider

$\sum_{i\geq
i_{0}}i\mathscr{C}^{n}(A^{\lambda_{i}},\Omega^{\delta}_{i})\leq\sum_{i\geq
i_{0}}\frac{\mu(\Omega^{\delta}_{i})}{\lambda_{i}}$

and pick up $\lambda_{i}\to 0$ as $i\to\infty$ by the classic Paul du Bois-
Reymond Theorem (cf. [10, (5) Page 40] and [18]) for infinite series such that
$\sum_{i\geq i_{0}}\frac{\mu(\Omega^{\delta}_{i})}{\lambda_{i}}$ converges
when $\sum_{i\geq i_{0}}\mu(\Omega^{\delta}_{i})$ converges. This is to say
that the third term in the right side of (2.9)

(2.14) $\displaystyle III$
$\displaystyle=\int_{\Omega^{\delta}_{i}}\log\frac{D}{d(x,y)}d\mu(y)=\int_{\Omega^{\delta}_{i}}\log\frac{D_{i}}{d(x,y)}d\mu(y)+\log\frac{D}{D_{i}}\mu(\Omega^{i}_{\delta})$
$\displaystyle\leq(\lambda_{i}+(1+\frac{1}{i}\log\frac{1}{\delta})\mu(\Omega^{i}_{\delta}))\log\frac{D}{d(x,p)}$
$\displaystyle=o(1)\log\frac{1}{d(x,p)}\text{ as
$x\in\omega^{\delta}_{i}\setminus E^{\lambda_{i}}$ and $x\to p$}.$

Finally, if let $A=\bigcup_{i}A^{\lambda_{i}}$, we have

$\lim_{x\to p\text{ and }x\in U\setminus
A}\frac{\mathscr{R}^{n,U}_{\mu}(x)}{\log\frac{1}{d(x,p)}}=\mu(\\{p\\}),$

where $A$ is $n$-thin at $p$. The proof is complete. ∎

##  3\. On scalar curvature equations

In this section we focus on the scalar curvature equations for conformal
deformation of metrics. Let $(M^{n},\bar{g})$ be a compact Riemannian manifold
for $n\geq 3$. Let $R_{ijkl}[\bar{g}]$ be the Riemann curvature tensor,
$R_{ij}[\bar{g}]=R_{ijkl}\bar{g}^{kl}$ be the Ricci curvature tensor, and
$R[\bar{g}]=R_{ij}\bar{g}^{ij}$ be the scalar curvature. The scalar curvature
equation in conformal geometry is

(3.1)
$-\frac{4(n-1)}{n-2}\Delta[\bar{g}]u+R[\bar{g}]u=R[u^{\frac{4}{n-2}}\bar{g}]u^{\frac{n+2}{n-2}}$

for a positive function $u$. The scalar curvature equation describes how the
scalar curvature transforms under conformal change of metrics. In this section
we want to use the estimates for the Newton potential in the previous section
to study the Hausdorff dimensions of the singularities of solutions $u$ to the
scalar equations which represent the ends of a complete conformal metric
$u^{\frac{4}{n-2}}\bar{g}$.

We remark here that all of the results in this section hold if we assume $S$
is compact, $D\subset M^{n}$ is a bounded domain that contains $S$, and
$(M^{n},\bar{g})$ is just complete, because the possible noncompact part
$M^{n}\setminus\bar{D}$ is not relevant for the purpose here.

###  3.1. Preliminaries

Let us start with [25, Lemma 3.1], which is a slight improvement of [13,
Proposition 8.1].

######  Lemma 3.1.

( [25, Lemma 3.1]) Let $(M^{n},\bar{g})$ be a compact Riemannian manifold and
$S$ be a closed subset in $M^{n}$. And let $D$ be an open neighborhood of $S$.
Suppose that $g=u^{\frac{4}{n-2}}\bar{g}$ is a conformal metric on $D\setminus
S$ and is geodesically complete near $S$. Then

$u(x)\to+\infty\text{ as $x\to S$}$

if $R^{-}[g]\in L^{p}(D\setminus S,g)$ for some $p>n/2$, where
$R^{-}[g]=\max\\{-R[g],0\\}$ stands for the negative part of the scalar
curvature $R[g]$ and $L^{p}(D\setminus S,g)$ is the $L^{p}$ space with respect
to the metric $g$.

For a preliminary estimate on the Hausdorff dimension of $S$, we follow the
proof of [25, Theorem 3.1] and get

######  Proposition 3.1.

Let $(M^{n},\bar{g})$ be a compact Riemannian manifold and $S$ be a closed
subset in $M^{n}$. And let $D$ be an open neighborhood of $S$ where the scalar
curvature $R[\bar{g}]$ is nonpositive. Suppose that
$g=u^{\frac{4}{n-2}}\bar{g}$ is a conformal metric on $D\setminus S$ and is
geodesically complete near $S$. Then the Newton capacity of $S$ is zero and
therefore the Hausdorff dimension

$\text{dim}_{\mathscr{H}}(S)\leq n-2,$

provided that

$R^{-}[g]\in L^{\frac{2n}{n+2}}(D\setminus S,g)\bigcap L^{p}(D\setminus S,g)$

for some $p>n/2$.

###### Proof.

Recall the scalar curvature equation

(3.2) $-\frac{4(n-1)}{n-2}\Delta
u=-Ru+R^{+}[g]u^{\frac{n+2}{n-2}}-R^{-}[g]u^{\frac{n+2}{n-2}}\text{ in
$D\setminus S$},$

where

(3.3) $\displaystyle\int_{D\setminus S}R^{-}[g]u^{\frac{n+2}{n-2}}dvol$
$\displaystyle\leq(\int_{D\setminus
S}(R^{-}[g])^{\frac{2n}{n+2}}u^{\frac{2n}{n-2}}dvol)^{\frac{n+2}{2n}}\text{vol}(D)^{\frac{n-2}{2n}}$
$\displaystyle\leq(\int_{D\setminus
S}(R^{-}[g])^{\frac{2n}{n+2}}dvol[g])^{\frac{n+2}{2n}}\text{vol}(D)^{\frac{n-2}{2n}}$
$\displaystyle<\infty.$

Here, and from now on in the following, all geometric quantities are under the
background metric $\bar{g}$ unless indicated otherwise. And, in the light of
Lemma 3.1, we know

$u(x)\to+\infty\text{ as $x\to S$}.$

As in the proof of [25, Theorem 3.1] (adopted from [7, Lemma 1.2]), we use the
following test functions. First we let

$u_{\alpha,\beta}=\left\\{\begin{aligned} \beta&\quad u\geq\alpha+\beta;\\\
u-\alpha&\quad u<\alpha+\beta.\end{aligned}\right.\text{ and
}\phi_{\alpha,\beta}=u_{\alpha,\beta}-\beta+\beta(1-\eta),$

where $\eta\in C^{\infty}_{c}(\Sigma_{\alpha})$ is a fixed cut-off function
that is equal to one in a neighborhood of $S$ and $\Sigma_{\alpha}=\\{x\in
D:u(x)>\alpha\\}$. Notice that, for $\beta$ sufficiently large,

$u_{\alpha,\beta}\in(0,\beta]\text{ in $\Sigma_{\alpha}$ and
}\phi_{\alpha,\beta}=0\text{ on $\\{x\in D:u(x)=\alpha\\}\bigcup\\{x\in
D:u\geq\alpha+\beta$}\\}$

and

$\nabla\phi_{\alpha,\beta}=\nabla u_{\alpha,\beta}+\beta\nabla\eta\text{ and
}\nabla u=\nabla u_{\alpha,\beta}\text{ when }\nabla u_{\alpha,\beta}\neq 0.$

We then multiply $\phi_{\alpha,\beta}$ to the equation (3.2) and get

$\frac{4(n-1)}{n-2}\int_{\Sigma_{\alpha}}\nabla
u\cdot\nabla\phi_{\alpha,\beta}dvol=\int_{\Sigma_{\alpha}}(-Ru+R[g]u^{\frac{n+2}{n-2}})\phi_{\alpha,\beta}dvol.$

Therefore

(3.4) $\displaystyle\frac{4(n-1)}{n-2}$
$\displaystyle\int_{\Sigma_{\alpha}}|\nabla u_{\alpha,\beta}|^{2}dvol$
$\displaystyle=\beta\int_{\Sigma_{\alpha}}(\frac{n-2}{4(n-1)}\nabla
u\cdot\nabla\eta+(-Ru+R[g]u^{\frac{n+2}{n-2}})(1-\eta))dvol$
$\displaystyle-\int_{\Sigma_{\alpha}}(-Ru+R^{+}[g]u^{\frac{n+2}{n-2}})(\beta-
u_{\alpha,\beta})dvol$
$\displaystyle+\int_{\Sigma_{\alpha}}R^{-}[g]u^{\frac{n+2}{n-2}}(\beta-
u_{\alpha,\beta})dvol$ $\displaystyle\leq C\beta,$

where $C$ depends on $\alpha$ and $\eta$ but does not depend on $\beta$, due
the support of $1-\eta$ and (3.3). That is

$\int_{\Sigma_{\alpha}}|\nabla\frac{u_{\alpha,\beta}}{\beta}|^{2}dvol\leq\frac{C}{\beta}\to
0$

as $\beta\to\infty$, where $\frac{u_{\alpha,\beta}}{\beta}$ is a function that
is identically one in a neighborhood of $S$. This implies the Newton capacity
$\text{Cap}_{2}(S,D)$ of $S$ is zero. Consequently, we know $S$ is of the
Hausdorff dimensions not greater than $n-2$ (cf. [3] and [35, Theorem 2.10 in
Chapter VI]). So the proof is complete. ∎

###  3.2. $-\Delta u$ is a Radon measure on $D$

In order to use the estimates of potentials in the previous section, we need
the following lemma (cf. [25, Lemma 3.2 - 3.4]).

######  Lemma 3.2.

Let $(M^{n},\bar{g})$ be a compact Riemannian manifold and $S$ be a closed
subset in $M^{n}$. And let $D$ be an open neighborhood of $S$ where the scalar
curvature $R[\bar{g}]$ is nonpositive. Suppose that
$g=u^{\frac{4}{n-2}}\bar{g}$ is a conformal metric on $D\setminus S$ and is
geodesically complete near $S$. Then $-\Delta u$ is a Radon measure on $D$ and
$-\Delta u|_{S}\geq 0$, provided that

$R^{-}[g]\in L^{\frac{2n}{n+2}}(D\setminus S,g)\bigcap L^{p}(D\setminus S,g)$

for some $p>n/2$.

###### Proof.

Again, recall the scalar curvature equation

(3.5) $-\frac{4(n-1)}{n-2}\Delta
u=-Ru+R^{+}[g]u^{\frac{n+2}{n-2}}-R^{-}[g]u^{\frac{n+2}{n-2}}=f\text{ in
$D\setminus S$},$

where

$\int_{D}R^{-}[g]u^{\frac{n+2}{n-2}}dvol<\infty.$

And, in the light of Lemma 3.1, we know

$u(x)\to\infty\text{ as $x\to S$}.$

Then we claim the right hand side $f$ of the equation (3.5) is in $L^{1}(D)$.
To prove this claim, we follow the argument in the proof of [25, Theorem 3.2]
(stated as [25, Lemma 3.2]) . Let

$\alpha_{s}(t)=\left\\{\begin{aligned} t&\quad t\leq s;\\\
\text{increasing}&\quad t\in[s,10s];\\\ 2s&\quad t\geq
10s\end{aligned}\right.$

(this functions was used in [17]). Notice that one may require
$\alpha_{s}^{\prime}\in[0,1]$ and $\alpha_{s}^{\prime\prime}\leq 0$. We
calculate

$-\Delta\alpha_{s}(u)=-\alpha^{\prime\prime}(u)|\nabla
u|^{2}+\alpha_{s}^{\prime}(u)(-\Delta u)$

and, for $s>\max\\{u(x):x\in\partial D\\}$,

$\int_{\partial D}\frac{\partial
u}{\partial\nu}d\sigma=\int_{D}\Delta\alpha_{s}(u)dvol=\int_{D}(-\alpha^{\prime\prime}(u)|\nabla
u|^{2}+\alpha^{\prime}(u)\frac{n-2}{4(n-1)}f)dvol.$

Hence

$\int_{D}(-\alpha_{s}^{\prime\prime}(u)|\nabla
u|^{2}+\alpha_{s}^{\prime}(u)\frac{n-2}{4(n-1)}f^{+})\,dvol\leq\int_{\partial
D}\frac{\partial
u}{\partial\nu}d\sigma+\frac{n-2}{4(n-1)}\int_{D}R^{-}[g]u^{\frac{n+2}{n-2}}dvol$

and

$\int_{D}|\Delta\alpha_{s}(u)|dvol=\int_{D}(-\alpha^{\prime\prime}(u)|\nabla
u|^{2}+\alpha^{\prime}(u)\frac{n-2}{4(n-1)}(f^{+}+f^{-}))dvol.$

By Fatou’s lemma, as $s\to\infty$, we have

$\int_{D}f^{+}dvol\leq\frac{4(n-1)}{n-2}\int_{\partial D}\frac{\partial
u}{\partial\nu}d\sigma+\int_{D}R^{-}[g]u^{\frac{n+2}{n-2}}dvol.$

So the claim is proven. Moreover,

$\int_{D}|\Delta\alpha_{s}(u)|dvol\leq\int_{\partial D}\frac{\partial
u}{\partial\nu}d\sigma+\frac{n-2}{2(n-1)}\int_{D}R^{-}[g]u^{\frac{n+2}{n-2}}dvol.$

Consequently, for $\phi\in C^{\infty}_{c}(D)$,

$\displaystyle|-\Delta$
$\displaystyle\alpha_{s}(u)(\phi)|=|\int_{D}(-\Delta\alpha_{s}(u))\phi dvol|$
$\displaystyle\leq\int_{D}|\Delta\alpha_{s}(u)|dvol\|\phi\|_{C^{0}(D)}$
$\displaystyle\leq(\int_{\partial D}\frac{\partial
u}{\partial\nu}d\sigma+\frac{n-2}{2(n-1)}\int_{D\setminus
S}R^{-}[g]u^{\frac{n+2}{n-2}}dvol)\|\phi\|_{C^{0}(D)},$

for any $s$ larger. Before we show $-\Delta u$ is a Radon measure, let us
state and prove a lemma which is useful for the proof now and later in the
following sections.

######  Lemma 3.3.

Let $(M^{n},\bar{g})$ be a compact Riemannian manifold and $S$ be a closed
subset in $M^{n}$. And let $D$ be an open neighborhood of $S$ where the scalar
curvature $R[\bar{g}]$ is nonpositive. Suppose that
$g=u^{\frac{4}{n-2}}\bar{g}$ is a conformal metric on $D\setminus S$ and is
geodesically complete near $S$. Then

(3.6) $\nabla u\in L^{p}(D)\text{ and }u\in L^{q}(D)$

for $p\in[1,\frac{n}{n-1})$ and $q\in[1,\frac{n}{n-2})$, provided that

$R^{-}[g]\in L^{\frac{2n}{n+2}}(D\setminus S,g)\bigcap L^{p}(D\setminus S,g)$

for some $p>n/2$.

###### Proof.

In fact, we continue from the above, for $\phi\in C^{\infty}_{c}(D)$,

(3.7) $\displaystyle|\int_{D}$
$\displaystyle\nabla\alpha_{s}(u)\cdot\nabla\phi
dvol|=|\int_{D}(-\Delta[\bar{g}]\alpha_{s}(u)\phi)dvol|$
$\displaystyle\leq(\int_{\partial D}\frac{\partial
u}{\partial\nu}d\sigma+\frac{n-2}{2(n-1)}\int_{D\setminus
S}R^{-}[g]u^{\frac{n+2}{n-2}}dvol)\|\phi\|_{C^{0}(D)}$ $\displaystyle\leq
C(\int_{\partial D}\frac{\partial
u}{\partial\nu}d\sigma+\frac{n-2}{2(n-1)}\int_{D\setminus
S}R^{-}[g]u^{\frac{n+2}{n-2}}dvol)\|\nabla\phi\|_{L^{\lambda}(D)}$

for any $\lambda>n$ due to the Sobolev embedding theorem. Therefore, for any
$s$ appropriately large,

$\|\nabla\alpha_{s}(u)\|_{L^{p}(D)}\leq C\text{ and
}\|\alpha_{s}(u)\|_{L^{q}(D)}\leq C$

for some constant $C$ and $p=\lambda^{\prime}\in(1,\frac{n}{n-1})$ and
$q\in[1,\frac{n}{n-2})$, where $C$ is independent of $s$. Therefore we first
have, by Fatou’s lemma,

$\|u\|_{L^{q}(D)}\leq C$

for some $C$ and $q\in[1,\frac{n}{n-2})$. Moreover, we calculate

(3.8) $\displaystyle|\nabla u(\phi)|$ $\displaystyle=|\int_{D}u\nabla\phi
dvol|=|\lim_{s\to\infty}\int_{D}\alpha_{s}(u)\nabla\phi
dvol|=|\lim_{s\to\infty}\int_{D}\alpha_{s}^{\prime}(u)\nabla u\phi dvol|$
$\displaystyle\leq\limsup_{s\to\infty}\|\alpha^{\prime}_{s}(u)\nabla
u\|_{L^{p}(D)}\|\phi\|_{L^{\lambda}}\leq C\|\phi\|_{L^{\lambda}}.$

This implies

(3.9) $\nabla u\in L^{p}(D)\text{ and }u\in L^{q}(D)$

for $p\in[1,\frac{n}{n-1})$ and $q\in[1,\frac{n}{n-2})$. This lemma is proven.
∎

Back to the proof of Lemma 3.2,

(3.10) $\displaystyle(-\Delta u)(\phi)$ $\displaystyle=\int_{D}\nabla
u\cdot\nabla\phi\,dvol$
$\displaystyle=\lim_{s\to\infty}\int_{D}\alpha_{s}^{\prime}(u)\nabla
u\cdot\nabla\phi\,dvol$
$\displaystyle=\lim_{s\to\infty}(-\Delta\alpha_{s}(u))(\phi))$

where the dominated convergence theorem is applied due to $\nabla u\in
L^{1}(D)$. Thus, for $\phi\in C^{\infty}_{c}(D)$,

$|(-\Delta u)(\phi)|\leq\ (\int_{\partial D}\frac{\partial
u}{\partial\nu}d\sigma+\frac{n-2}{2(n-1)}\int_{D\setminus
S}R^{-}[g]u^{\frac{n+2}{n-2}}dvol)\|\phi\|_{C^{0}(D)},$

which implies that $-\Delta u$ is a Radon measure on $D$. To show that
$-\Delta u|_{S}\geq 0$, we calculate, for a nonnegative function $\phi\in
C^{\infty}_{c}(D)$,

$\displaystyle(-\Delta u)(\phi)$ $\displaystyle=\int_{D}\nabla
u\cdot\nabla\phi
dvol=\lim_{s\to\infty}\int_{D}\nabla\alpha_{s}(u)\cdot\nabla\phi dvol$
$\displaystyle=\lim_{s\to\infty}\int_{D}(-\Delta\alpha_{s}(u))\phi dvol$
$\displaystyle=\lim_{s\to\infty}\int_{D}[(\alpha_{s}^{\prime}(u)(\frac{n-2}{4(n-1)}(-Ru+R[g]u^{\frac{n+2}{n-2}})-\alpha_{s}^{\prime\prime}(u)|\nabla
u|^{2}]\phi dvol$
$\displaystyle\geq-\\{\frac{n-2}{4(n-1)}\int_{\text{supp}\phi\setminus
S}|-Ru+R[g]u^{\frac{n+2}{n-2}}|dvol\\}\|\phi\|_{C^{0}(D)}\to 0$

as $\int_{\text{supp}\phi\setminus S}dvol\to 0$ and $\|\phi\|_{C^{0}(D)}=1$,
which implies $-\Delta u|_{S}\geq 0$. ∎

###  3.3. Main result on the Hausdorff dimensions

Now we are ready to state and prove our result on the Hausdorff dimension of
the singular set $S$, which is a significant improvement of Proposition 3.1.
For convenience of readers, we recall Theorem 1.3 from the introduction.

######  Theorem 3.1.

Let $(M^{n},\bar{g})$ be a compact Riemannian manifold and $S$ be a closed
subset in $M^{n}$. And let $D$ be an open neighborhood of $S$. Suppose that
$g=u^{\frac{4}{n-2}}\bar{g}$ is a conformal metric on $D\setminus S$ and is
geodesically complete near $S$. Then the Hausdorff dimension

(3.11) $\text{dim}_{\mathscr{H}}(S)\leq\frac{n-2}{2}$

provided $R^{-}[g]\in L^{\frac{2n}{n+2}}(D\setminus S,g)\bigcap
L^{p}(D\setminus S,g)$ for some $p>n/2$. Consequently, (3.11) holds when the
scalar curvature $R[g]$ of the conformal metric $g$ is nonnegative.

###### Proof.

The outline of the proof is as follows: We first show that one may assume the
scalar curvature $R[\bar{g}]$ is nonpositive without loss of generality for
our purpose. Then we use the Green function to construct the integral
representation of the solution to the Laplace equation. Finally we apply Lemma
3.2, Theorem 2.2, and the geodesic completeness to complete the proof .

Step I In this step, we find a conformal change
$\bar{h}=v^{\frac{4}{n-2}}\bar{g}$ such that the scalar curvature $R[\bar{h}]$
is nonpositive (or even negative) in $D$, based on the similar idea used in
the proof of [25, Lemma 3.1]. This is trivial if the Yamabe constant of
$(M^{n},\bar{g})$ is nonpositive. Otherwise, take a point $p\in M^{n}\setminus
D$ and consider a connected sum of $M^{n}$ with another compact Riemannian
manifold $(M^{n}_{1},\bar{g}_{1})$ with very negative Yamabe constant in such
way that the conformal structure on the connected sum $M^{n}\sharp M^{n}_{1}$
is unchanged in $D\subset M^{n}\sharp M^{n}_{1}$. Then, by [19, Theorem 5],
the Yamabe constant of such connected sum is negative. Therefore one easily
finds a conformal metric $\bar{h}=v^{\frac{4}{n-2}}\bar{g}$ whose scalar
curvature is negative in $D$, where the function $v\in C^{\infty}(\bar{D})$
and

(3.12) $C^{-1}\leq v\leq C\text{ in $\bar{D}$}$

for some positive constant $C$. In any cases, we have
$g=u^{\frac{4}{n-2}}\bar{g}=(\frac{u}{v})^{\frac{4}{n-2}}\bar{h}$ and the
scalar curvature $R[\bar{h}]$ is nonpositive. In conclusion, due to (3.12), we
may simply assume $R[\bar{g}]$ is nonpositive (or even negative) in $D$
without loss of any generality for the purpose of obtaining the growth
estimate like the one given in Theorem 2.2.

Step II In this step, we use the Green function to construct the integral
representation of the solution $u$. In the light of Lemma 3.2, we may write

$-\Delta u=\mu\text{ in $D$}$

for a Radon measure $\mu$ on $D$. Let $G(x,y)$ be the Green function on $D$
given by [6, Theorem 4.17]. Then

$u=\int_{D}G(x,y)d\mu(y)+h$

for a smooth function $h$ that is harmonic in $D$. By [6, Theorem 4.17 (c)],
we have

$0<G(x,y)\leq\frac{C}{d(x,y)^{n-2}}$

for some constant $C$ and $x,y\in D$. We therefore arrive at, for $x\in D$,

(3.13) $u(x)\leq\int_{D}G(x,y)d\mu^{+}+h(x)\leq
C\mathscr{R}^{2,D}_{\mu^{+}}(x)+h(x).$

Step III Assume otherwise that $\text{dim}_{\mathscr{H}}(S)=d>\frac{n-2}{2}$.
From Theorem 2.2, there is a point $p\in S$ such that

$\mathscr{R}^{2,D}_{\mu^{+}}(x)\leq\frac{C}{d(x,p)^{n-2-d}}$

at least for $x$ along a short geodesic ray $\gamma$ from $p$, which implies

(3.14) $u(x)^{\frac{2}{n-2}}\leq\frac{C}{d(x,p)^{\frac{2(n-2-d)}{n-2}}}$

at least for $x$ along a short geodesic ray $\gamma$ from $p$, where

$\frac{2(n-2-d)}{n-2}=2-\frac{2d}{n-2}<1$

when $d>\frac{n-2}{2}$. Now the length of the curve $\gamma$ with respect to
the conformal metric $g=u^{\frac{4}{n-2}}\bar{g}$ is

$L(\gamma,g)\leq C\int_{0}^{l_{0}}\frac{1}{s^{\frac{2(n-2-d)}{n-2}}}ds<\infty$

when $d>\frac{n-2}{2}$, which contradicts with the geodesic completeness of
the conformal metric $g=u^{\frac{4}{n-2}}\bar{g}$. The proof is complete. ∎

The study of singular solutions to the scalar curvature equations started from
the seminal paper [34] (see also [35, Chapter VI] [11] and [33, 27, 26]) on
domains of the sphere. Theorem 3.1 here can be considered as a necessary
condition for the existences of singular solutions on domains in general
Riemannian manifolds and compared with [34, Theorem 2.7] and [11, Theorem C],
which stated the similar result for domains in the round sphere $S^{n}$ and
slightly stronger curvature assumptions. Clearly [34, Proposition 2.4] and the
quantity $d(M)$ there are not of local nature, while our approach here is very
much local in nature.

##  4\. On $Q$-curvature equations

In this section we will use linear potential theory developed in Section 2 to
study $Q$-curvature equations and prove our results on the Hausdorff
dimensions of the singular sets of positive solutions of $Q$-curvature
equations which correspond to ends of complete conformal metrics on domains of
a compact Riemannian manifold.

Again we remark here that all of the results in this section hold if we assume
$S$ is compact, $D\subset M^{n}$ is a bounded domain that contains $S$, and
$(M^{n},\bar{g})$ is just complete. Because the possible noncompact part
$M^{n}\setminus\bar{D}$ is not relevant for the purpose here.

###  4.1. $Q$-curvature equations in dimensions greater than $4$

In this subsection we focus on the equation (1.4) in dimensions greater than
$4$. We will always assume that the scalar curvature of the conformal metric
$g=u^{\frac{4}{n-4}}\bar{g}$ is nonnegative. We will first prove some
preliminary estimates based on discussions in the previous section. Our
strategy is to consider the bi-Laplace operator as the composition of the
Laplace operators. Let us write the scalar curvature equation and its
consequence:

(4.1) $-\Delta
u^{\frac{n-2}{n-4}}+\frac{n-2}{4(n-1)}Ru^{\frac{n-2}{n-4}}=\frac{n-2}{4(n-1)}R[g]u^{\frac{n+2}{n-4}}\text{
in $D\setminus S$}$

and

(4.2) $-\Delta u=\frac{2}{n-4}\frac{|\nabla
u|^{2}}{u}+\frac{n-4}{4(n-1)}(-Ru+R[g]u^{\frac{n}{n-4}})\quad\text{ in
$D\setminus S$}.$

Here, and from now on in the following, all geometric quantities are under the
background metric $\bar{g}$ unless indicated otherwise.

######  Lemma 4.1.

Let $(M^{n},\bar{g})$ be a compact Riemannian manifold for $n\geq 5$ and $S$
be a closed subset in $M^{n}$. And let $D$ be an open neighborhood of $S$
where the scalar curvature $R\leq-c_{0}<0$. Suppose that
$g=u^{\frac{4}{n-4}}\bar{g}$ is a conformal metric on $D\setminus S$ with
nonnegative scalar curvature $R[g]\geq 0$ and is geodesically complete near
$S$. And suppose also that

$Q_{4}^{-}[g]\in L^{\frac{2n}{n+4}}(D\setminus S,g).$

Then

(4.3) $\displaystyle\text{as a function on $D\setminus S$, }-\Delta
u\to+\infty\text{ as $x\to S$};$ $\displaystyle\text{as a Radon measure on
$D$, }\Delta u|_{S}=0;$ $\displaystyle\text{in fact, }\Delta u\in
L^{p}(D)\text{ for any $p\in[1,\frac{n}{n-2})$}.$

###### Proof.

First, using Lemma 3.3 for $u^{\frac{n-2}{n-4}}$, we know that

(4.4) $u\in L^{p}(D)\text{ for $p\in[1,\frac{n}{n-4})$}.$

Also, from Lemma 3.1 for $u^{\frac{n-2}{n-4}}$,

(4.5) $u(x)\to+\infty\text{ as $x\to S$},$

which implies, by (4.2),

$-\Delta u\to+\infty\text{ as $x\to S$}.$

To prove $-\Delta u$ is an integrable function in distributional sense, we
first realize $-\Delta u$ is a Radon measure on $D$ following (4.2) and Lemma
3.2. And, as a side product, we also have

$\int_{D}[\frac{2}{n-4}\frac{|\nabla
u|^{2}}{u}+\frac{n-4}{4(n-1)}(-Ru+R[g]u^{\frac{n}{n-4}})]dvol<\infty.$

In fact, from (4.1) and Lemma 3.2, we also know $-\Delta u^{\frac{n-2}{n-4}}$
is a Radon measure on $D$. To use this fact we calculate

$-\Delta\alpha_{s}(u)=-\Delta(\alpha_{s}(u)^{\frac{n-2}{n-4}})^{\frac{n-4}{n-2}}=\frac{n-4}{n-2}\alpha_{s}(u)^{-\frac{2}{n-4}}(-\Delta\alpha_{s}(u)^{\frac{n-2}{n-4}})+\frac{2}{n-4}\frac{|\nabla\alpha_{s}(u)|^{2}}{\alpha_{s}(u)}.$

To prove $-\Delta u|_{S}=0$, we consider

$(-\Delta u)(\phi)=\int_{D}\nabla u\cdot\nabla\phi dvol$

where $\nabla u$ is integrable in distributional sense directly from (3.8) and
(3.9). Therefore

$\int_{D}\nabla u\cdot\nabla\phi
dvol=\lim_{s\to\infty}\int_{D}\alpha_{s}^{\prime}(u)\nabla u\cdot\nabla\phi
dvol=\lim_{s\to\infty}\int_{D}(-\Delta\alpha_{s}(u))\phi dvol$

and

$\displaystyle(-\Delta u)(\phi)$
$\displaystyle=\frac{n-4}{n-2}\lim_{s\to\infty}\int_{D}\alpha_{s}(u)^{-\frac{2}{n-4}}(-\Delta\alpha_{s}(u)^{\frac{n-2}{n-4}})\phi
dvol$
$\displaystyle\quad+\frac{2}{n-4}\lim_{s\to\infty}\int_{D}\frac{|\nabla\alpha_{s}(u)|^{2}}{\alpha_{s}(u)}\phi
dvol$ $\displaystyle=\frac{n-4}{n-2}u^{-\frac{2}{n-4}}(-\Delta
u^{\frac{n-2}{n-4}})(\phi)+\frac{2}{n-4}\int_{D}\frac{|\nabla u|^{2}}{u}\phi
dvol$ $\displaystyle\to 0$

as $\int_{\text{supp}\phi\setminus S}dvol[\bar{g}]\to 0$ and
$\|\phi\|_{C^{0}(D)}\leq 1$. The proof will be complete after the following
$L^{p}$ estimate. To get the $L^{p}$ estimate, we first calculate

(4.6) $\displaystyle\int_{D\setminus S}Q_{4}^{-}[g]u^{\frac{n+4}{n-4}}dvol$
$\displaystyle=(\int_{D\setminus
S}(Q_{4}^{-}[g])^{\frac{2n}{n+4}}u^{\frac{2n}{n-4}}dvol)^{\frac{n+4}{2n}}\text{vol}(D)^{\frac{2n}{n-4}}$
$\displaystyle=(\int_{D\setminus
S}(Q_{4}^{-}[g])^{\frac{2n}{n+4}}dvol[g])^{\frac{n+4}{2n}}\text{vol}(D)^{\frac{2n}{n-4}}<\infty.$

Then we continue to use notations in the proof of Proposition 3.1 and let

$\alpha=\max\\{u(x):x\in\partial D\\}$

and $\alpha<\beta$. And recall

$u_{\alpha,\beta}=\left\\{\begin{aligned} \beta&\quad
x\in\Sigma_{\alpha+\beta};\\\ u(x)-\alpha&\quad
x\in\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}\end{aligned}\right.$

and

$\phi_{\alpha,\beta}=\left\\{\begin{aligned}
u_{\alpha,\beta}-\beta\eta&=u-(\alpha+\beta)+\beta(1-\eta),\text{ in
$\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}$}\\\ &=0\text{ on
$\partial\Sigma_{\alpha}$}\\\ &=0\text{ on
$\partial\Sigma_{\alpha+\beta}$}\end{aligned}\right.,$

where $\eta$ is a fixed cut-off function in $C^{\infty}_{c}(\Sigma_{\alpha})$
and is identically one in a neighborhood of $S$, and $\beta$ is arbitrarily
large. We now first multiply $1-\eta$ to the $Q$-curvature equation (1.4),
integrate over $D$, apply integral by parts multiple times, and get

(4.7)
$\int_{D}(1-\eta)Q_{4}^{+}u^{\frac{n+4}{n-4}}dvol\leq\int_{D}Q_{4}^{-}u^{\frac{n+4}{n-4}}dvol+C$

for some constant $C$ depending on the cut-off function $\eta$, $u$ at
$\partial D$, and $\|u\|_{L^{1}(D)}$. We then multiply $\phi_{\alpha,\beta}$
to both sides of the $Q$-curvature equation (1.4), integrate over
$\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}$, and get

(4.8) $\displaystyle\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}$
$\displaystyle\Delta
u\Delta\phi_{\alpha,\beta}dvol-\int_{\partial\Sigma_{\alpha}}\Delta
u\frac{\partial
u}{\partial\nu}d\sigma-\int_{\partial\Sigma_{\alpha+\beta}}\frac{\partial
u}{\partial\nu}\Delta ud\sigma$
$\displaystyle-\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}(4A(\nabla
u,\nabla\phi_{\alpha,\beta})-(n-2)J\nabla
u\cdot\nabla\phi_{\alpha,\beta})dvol$
$\displaystyle+\frac{n-4}{2}\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}Qu\phi_{\alpha,\beta}dvol=\frac{n-4}{2}\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}Q_{4}[g]u^{\frac{n+4}{n-4}}\phi_{\alpha,\beta}dvol,$

where $\nu$ is the outward normal direction at the boundary and the boundary
term $\int_{\partial\Sigma_{\alpha+\beta}}\frac{\partial
u}{\partial\nu}(-\Delta u)d\sigma$ is nonnegative due to (4.2) and
$\frac{\partial u}{\partial\nu}|_{\partial\Sigma_{\alpha+\beta}}=|\nabla u|$.
Therefore,

(4.9) $\displaystyle\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}$
$\displaystyle(\Delta
u)^{2}dvol+\beta\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}(\Delta
u)(\Delta(1-\eta))dvol$
$\displaystyle\leq-\int_{\partial\Sigma_{\alpha}}(-\Delta u)\frac{\partial
u}{\partial\nu}d\sigma+C\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\nabla
u|^{2}dvol$ $\displaystyle\quad-\beta\int_{\Sigma_{\alpha}}(4A(\nabla
u,\nabla(1-\eta))-(n-2)J\nabla u\cdot\nabla(1-\eta))dvol$
$\displaystyle\quad+C\beta\int_{D}u\,dvol+C\beta\int_{D}Q_{4}^{-}[g]u^{\frac{n+4}{n-4}}dvol,$

where we use (4.7) and $|\phi|\leq\beta$ in
$\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}$. After applying integral by
parts, we get,

(4.10) $\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\Delta
u|^{2}dvol\leq C\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\nabla
u|^{2}dvol+C\beta$

for some constant $C$ depending on the cut-off function $\eta$, $u$ at
$\partial\Sigma_{\alpha}$, and $\|u\|_{L^{1}(D)}$, because

$\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}\Delta u\Delta\eta
dvol=\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}u\Delta^{2}\eta dvol$

and similarly we may unload all derivatives from $u$ by integral by parts for
the other terms in the above (4.9). Now, to get a priori estimate, we
calculate

$\displaystyle\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\nabla
u|^{2}dvol$
$\displaystyle\leq\frac{1}{(n-4)C}\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}\frac{|\nabla
u|^{4}}{u^{2}}dvol+\frac{(n-4)C}{4}\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}u^{2}dvol$
$\displaystyle\leq\frac{1}{2C}\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\Delta
u|^{2}dvol+\frac{(n-4)C}{4}(\alpha+\beta)\int_{D}u\,dvol,$

due to (4.2), which implies, from (4.10),

(4.11) $\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\Delta
u|^{2}dvol\leq C(\alpha+\beta).$

We claim that (4.11) implies

(4.12) $\Delta u\in L^{p}(D)$

for all $p\in[1,\frac{n}{n-2})$. To prove (4.12), we first derive from (4.11),

$2^{-i}\int_{\Sigma_{2^{i-1}}\setminus\Sigma_{2^{i}}}|\Delta u|^{2}dvol\leq
C,$

for $i\geq i_{0}$ large, which implies

$\int_{\Sigma_{2^{i-1}}\setminus\Sigma_{2^{i}}}\frac{|\Delta
u|^{2}}{u}dvol\leq 2C$

and, for $s>0$ appropriately small for any $p\in[1,\frac{n}{n-2})$,

$\displaystyle\int_{\Sigma_{2^{i_{0}-1}}\setminus S}$
$\displaystyle\frac{|\Delta
u|^{2}}{u^{1+s}}dvol=\sum_{i=i_{0}}^{\infty}\int_{\Sigma_{2^{i-1}}\setminus\Sigma_{2^{i}}}\frac{|\Delta
u|^{2}}{u^{1+s}}dvol$
$\displaystyle\leq\sum_{i=i_{0}}^{\infty}2^{s(-i+1)}\int_{\Sigma_{2^{i-1}}\setminus\Sigma_{2^{i}}}\frac{|\Delta
u|^{2}}{u}dvol<\infty.$

Thus

$\int_{D\setminus S}|\Delta u|^{p}dvol\leq(\int_{D\setminus S}\frac{|\Delta
u|^{2}}{u^{1+s}}dvol)^{\frac{p}{2}}(\int_{D\setminus
S}u^{\frac{(1+s)p}{2-p}}dvol)^{1-\frac{p}{2}}<\infty$

where

$\frac{(1+s)p}{2-p}<\frac{n}{n-4}.$

∎

######  Corollary 4.1.

Under the same assumptions as in Lemma 4.1 we have

$\text{dim}_{\mathscr{H}}(S)\leq n-4.$

###### Proof.

Consequently from (4.2) and (4.11), we have

$\displaystyle\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\nabla\frac{u_{\alpha,\beta}}{\beta}|^{4}dvol$
$\displaystyle\leq\frac{(\alpha+\beta)^{2}}{\beta^{4}}\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}\frac{|\nabla
u|^{4}}{u^{2}}dvol$
$\displaystyle\leq\frac{(\alpha+\beta)^{2}}{\beta^{4}}\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\Delta
u|^{2}dvol$ $\displaystyle\leq C\frac{(\alpha+\beta)^{3}}{\beta^{4}}$

for some $\alpha$ appropriately large and $\beta\to\infty$, which leads to
$\text{Cap}_{4}(S)=0$ and completes the proof similar to the proof of
Proposition 3.1 (cf. [3] and [35, Theorem 2.10 in Chapter VI]). ∎

######  Lemma 4.2.

Let $(M^{n},\bar{g})$ be a compact Riemannian manifold for $n\geq 5$ and $S$
be a closed subset in $M^{n}$. And let $D$ be an open neighborhood of $S$
where the scalar curvature $R[\bar{g}]\leq-c_{0}<0$. Suppose that
$g=u^{\frac{4}{n-4}}\bar{g}$ is a conformal metric on $D\setminus S$ with
nonnegative scalar curvature $R[g]\geq 0$ and is geodesically complete near
$S$. And suppose also that

$Q_{4}^{-}[g]\in L^{\frac{2n}{n+4}}(D\setminus S,g).$

Then $\Delta^{2}u$ is a Radon measure on $D$ and $\Delta^{2}u|_{S}\geq 0$.

###### Proof.

Let $v=-\Delta u$. We will follow the proof of Lemma 3.2 to show that $-\Delta
v$ is a Radon measure on $D$ using Lemma 4.1. We continue to use the notations
from the proof of Lemma 3.2. We calculate

$-\Delta\alpha_{s}(v)=\alpha^{\prime}(v)(-\Delta
v)-\alpha^{\prime\prime}(v)|\nabla v|^{2},$

where, by the $Q$-curvature equation (1.4), we have

$-\Delta v=-\text{div}(4A(\nabla u)-(n-2)J\nabla
u)-\frac{n-4}{2}Q_{4}u+Q_{4}[g]u^{\frac{n+4}{n-4}}\text{ in $D\setminus S$}$

and

$-\Delta\alpha_{s}(v)=-\alpha_{s}^{\prime\prime}(v)|\nabla
v|^{2}+\alpha^{\prime}(v)(-\text{div}(4A(\nabla u)-(n-2)J\nabla
u)-\frac{n-4}{2}Q_{4}u+Q_{4}[g]u^{\frac{n+4}{n-4}})$

in $D$. In the light of Lemma 4.1, terms in the right hand side of the above
equation are all integrable except $-\alpha^{\prime\prime}(v)|\nabla
v|^{2}+Q_{4}^{+}[g]u^{\frac{n+4}{n-4}}$. Therefore the argument in the proof
of Lemma 3.2 works from this point and completes the proof. ∎

We now are ready to state and prove our main results for $Q$-curvature
equations in dimensions greater than $4$. For this, we recall Theorem 1.4 from
the introduction.

######  Theorem 4.1.

Let $(M^{n},\bar{g})$ be a compact Riemannian manifold for $n\geq 5$ and $S$
be a closed subset in $M^{n}$. And let $D$ be an open neighborhood of $S$.
Suppose that $g=u^{\frac{4}{n-4}}\bar{g}$ is a conformal metric on $D\setminus
S$ with nonnegative scalar curvature $R[g]\geq 0$ and is geodesically complete
near $S$. And suppose also that

$Q_{4}^{-}[g]\in L^{\frac{2n}{n+4}}(D\setminus S,g).$

Then

$\text{dim}_{\mathscr{H}}(S)\leq\frac{n-4}{2}.$

###### Proof.

In the light of Step I in the proof of Theorem 3.1, we may assume the scalar
curvature $R\leq-c_{0}<0$ for some $c_{0}$ without loss of any generality.
Then we use Lemma 4.1 and 4.2 and conclude that

$\Delta^{2}u=\mu$

for a Radon measure $\mu$ on $D$. We use [6, Theorem 4.7] first to write

$-\Delta u=\int_{D}G(x,y)d\mu+h(x)$

for some harmonic function $h(x)$, where $G(x,y)$ is the Green function for
$-\Delta$. Then we have

$u(x)=\int_{D}G(x,z)\int_{D}G(z,y)d\mu(y)dvol(z)+b(x)$

where $b(x)$ is bi-harmonic, where

$\int_{D}G(x,z)\int_{D}G(z,y)d\mu(y)dvol(z)=\int_{D}(\int_{D}G(x,z)G(z,y)dvol(z))d\mu(y)$

and

$0<\int_{D}G(x,z)G(z,y)dvol(z)\leq\frac{C}{d(x,y)^{n-4}}$

for constant $C$ and $n\geq 5$ due to [6, Proposition 4.12], where [6,
Proposition 4.12] can be easily proven to be available for bounded domains in
Riemannian manifolds. Hence

$u(x)\leq\mathscr{R}^{4,D}_{\mu^{+}}(x)+b(x).$

From now on, using the same argument of the proof of Theorem 3.1, based on
Theorem 2.2 for $\alpha=4$ and $n\geq 5$, we conclude

$\text{dim}{{}_{\mathscr{H}}}(S)\leq\frac{n-4}{2}$

and finish the proof. ∎

There have been a lot of works on the study of singular solutions to
$Q$-curvature equations on manifolds of dimensions greater than $4$, notably
[29, 30, 13, 20], for example. Theorem 4.1, for instance, is an improvement of
[13, Theorem 1.2] in terms of curvature conditions. And the approach here is
different from [13].

###  4.2. $Q$-curvature equations in dimension $4$

In this subsection we will study the $Q$-curvature equation (1.5). Our
approach here in principle is similar to that in the previous subsection but
different in calculations and details. We will always assume that the scalar
curvature of the conformal metric $g=e^{2u}\bar{g}$ is nonnegative. We will
first derive some preliminary estimates from the scalar curvature equation for
$w=e^{u}$ and the $Q$-curvature equation (1.5) for $u$. Let us write the
scalar curvature equation for $e^{u}$

(4.13) $-\Delta e^{u}=\frac{1}{6}(-Re^{u}+R[g]e^{3u})\text{ in $D\setminus
S$}$

and consequently,

(4.14) $-\Delta u=|\nabla u|^{2}+\frac{1}{6}(-R+R[g]e^{2u})\text{ in
$D\setminus S$}.$

######  Lemma 4.3.

Let $(M^{4},\bar{g})$ be a compact Riemannian manifold and $S$ be a closed
subset in $M^{n}$. And let $D$ be an open neighborhood of $S$ where the scalar
curvature $R\leq 0$. Suppose that $g=e^{2u}\bar{g}$ is a conformal metric on
$D\setminus S$ with nonnegative scalar curvature $R[g]\geq 0$ and is
geodesically complete near $S$. And suppose also that

$Q_{4}^{-}[g]\in L^{1}(D\setminus S,g).$

Then

(4.15) $\displaystyle\text{as a Radon measure, }-\Delta u|_{S}=0;$
$\displaystyle\text{in fact, }\Delta u\in L^{p}(D)\text{ for any
$p\in[1,\frac{4}{3})$}.$

###### Proof.

First, by Lemma 3.1 for $e^{u}$, we have

$u(x)\to\infty\text{ as $x\to S$}.$

Then, by the proof of Lemma 3.2 and (4.14), we know that

  * •

$-\Delta u$ is a Radon measure on $D$;

  * •

$\nabla u\in L^{p}(D)$ for any $p\in[1,\frac{4}{3})$ and $u\in L^{p}(D)$ for
any $p\in[1,2)$;

  * •

$|\nabla u|^{2}+\frac{1}{6}(-R+R[g]e^{2u})\in L^{1}(D)$.

The same argument as in the proof of Lemma 4.1 we can prove that $-\Delta
u|_{S}=0$ as a Radon measure. Also, for the $L^{p}$ estimate, following the
proof of Lemma 4.1, we multiple $1-\eta$ to both sides of (1.5) and get

$\int_{D}(1-\eta)Q_{4}^{+}e^{4u}dvol\leq\int_{D}Q_{4}^{-}[g]dvol[g]+C.$

Next we multiple $\phi_{\alpha,\beta}$ to both sides of (1.5) and integrate

$\displaystyle\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}$
$\displaystyle\Delta
u\Delta\phi_{\alpha,\beta}dvol-\int_{\partial\Sigma_{\alpha}}\Delta
u\frac{\partial\phi_{\alpha,\beta}}{\partial\nu}d\sigma-\int_{\partial\Sigma_{\alpha+\beta}}\Delta
u\frac{\partial\phi_{\alpha,\beta}}{\partial\nu}d\sigma$
$\displaystyle-\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}(4A(\nabla
u,\nabla\phi_{\alpha})-2J\nabla u\cdot\nabla\phi_{\alpha,\beta}dvol$
$\displaystyle+\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}Q_{4}\phi_{\alpha,\beta}dvol=\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}Q_{4}[g]e^{4u}\phi_{\alpha,\beta}dvol$

and, again, the boundary term at $\partial\Sigma_{\alpha+\beta}$ is with the
sign in our favor, thanks to (4.14) and $\frac{\partial
u}{\partial\nu}|_{\Sigma_{\alpha+\beta}}=|\nabla u|$ for the outward normal
$\nu$ of $\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}$. Similar to the
estimates in the proof of Lemma 4.1, we get

$\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\Delta u|^{2}dvol\leq
C\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\nabla
u|^{2}dvol+C\beta.$

And we handle $\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\nabla
u|^{2}dvol$ similarly as before

$\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\nabla
u|^{2}dvol\leq\frac{1}{2C}\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\nabla
u|^{4}dvol+C\leq\frac{1}{2C}\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\Delta
u|^{2}dvol+C.$

due to (4.14). Therefore

(4.16) $\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\Delta
u|^{2}dvol\leq C\beta.$

Now, using the same idea as in the proof of Lemma 4.1, we rewrite (4.16) as

$\int_{\Sigma_{2^{i-1}}\setminus\Sigma_{2^{i}}}\frac{|\Delta
u|^{2}}{u}dvol\leq C$

and, for $s>0$ appropriately small for any $p\in[1,\frac{4}{3})$, we derive

$\int_{D\setminus S}\frac{|\Delta u|^{2}}{u^{1+s}}dvol\leq C,$

which implies

$\int_{D\setminus S}|\Delta u|^{p}dvol\leq(\int_{D\setminus S}\frac{|\Delta
u|^{2}}{u^{1+s}}dvol)^{\frac{p}{2}}(\int_{D\setminus
S}u^{\frac{(1+s)p}{2-p}}dvol)^{1-\frac{p}{2}}$

when

$\frac{(1+s)p}{2-p}<2.$

∎

######  Corollary 4.2.

Under the assumptions as in Lemma 4.3, we know the singular set $S$ is of zero
Hausdorff dimension.

###### Proof.

From (4.14) and (4.16) in the above we have

$\int_{\Sigma_{\alpha}\setminus\Sigma_{\alpha+\beta}}|\nabla\frac{u_{\alpha,\beta}}{\beta}|^{4}dvol\leq
C\beta^{-3}$

for some $\alpha$ appropriately large and $\beta\to\infty$, which leads to
$\text{Cap}_{4}(S,D)=0$ and completes the proof as in Proposition 3.1 (cf. [3]
and [35, Theorem 2.10 in Chapter VI]). ∎

What follows is to go beyond that $S$ is of zero Hausdorff dimension. We now
are ready to state and prove our main result on the finiteness of
singularities for the $Q$-curvature equation in dimension $4$. This is
inspired by [16, 22, 5, 24, 25]. We recall Theorem 1.5 from the introduction.

######  Theorem 4.2.

Let $(M^{4},\bar{g})$ be a compact Riemannian manifold and $S$ be a closed
subset in $M^{n}$. And let $D$ be an open neighborhood of $S$. Suppose that
$g=e^{2u}\bar{g}$ is a conformal metric on $D\setminus S$ with nonnegative
scalar curvature $R[g]\geq 0$ and is geodesically complete near $S$. And
suppose that

$\int_{D}Q_{4}^{-}[g]dvol[g]<\infty.$

Then $S$ consists of only finitely many points.

###### Proof.

As before, we use the argument in Step I on Theorem 3.1 to assume that the
scalar curvature of the background metric $\bar{g}$ is less than a negative
number, i.e. $R\leq-c_{0}<0$, without loss of any generality for our purpose.
Let

$v=-\Delta u+u$

and claim $-\Delta v$ is a Radon measure on $D$ with $-\Delta v|_{S}\geq 0$.
Let us start with

(4.17) $-\Delta v=\Delta^{2}u-\Delta u=-\text{div}(4A(\nabla u)-2J\nabla
u)-Q_{4}+Q_{4}[g]e^{4u}-\Delta u.$

By Lemma 4.3 and (4.14), we know

  * •

$v(x)\to\infty$ as $x\to S$;

  * •

All terms in the right side of (4.17) except $Q_{4}^{+}[g]e^{4u}$ is
integrable.

Therefore, folowing the same argument as in the proof of Lemma 4.2, the claim
is proven. Obviously, the same conclusion holds for $\Delta^{2}u=-\Delta
v+\Delta u$ from $-\Delta v$ and what we know about $\Delta u$ in Lemma 4.3.
Thus we let

$\Delta^{2}u=\mu$

for a Radon measure on $D$ with $\Delta^{2}u|_{S}\geq 0$. Like in the proof of
Theorem 4.1, we first write

$-\Delta u(x)=\int_{D}G(x,y)d\mu(y)+h(x)$

by [6, Theorem 4.17], where $h(x)$ is a harmonic function. Then we write

$u(x)=\int_{D}G(x,z)\int_{D}G(z,y)d\mu(y)dvol(z)+b(x)$

where $b(x)$ is a bi-harmonic function and, due to [6, Proposition 4.12],

$\int_{D}G(x,z)G(z,y)dvol(z)\leq C(1+\log\frac{1}{d(x,y)})$

for some constant $C$ in dimension $4$, where [6, Proposition 4.12] can be
easily made available on bounded domains in manifolds. Therefore

$u(x)\leq C\mathscr{R}^{4,D}_{\mu^{+}}(x)+b(x).$

Applying Theorem 2.3, we have

$\lim_{x\to p\text{ and }x\notin E}\frac{u(x)}{\log\frac{1}{d(x,p)}}\leq
C\mu^{+}(\\{p\\})=C\mu(\\{p\\})$

where $E$ is a subset that is $n$-thin at $p$. Next, in the light of Theorem
2.1, we conclude that $\mu(\\{p\\})\geq 1/C$ for each $p\in S$ by the
completeness of the metric $g$ near $S$, which indeed implies that $S$ can
only have finitely many points. So the proof is complete. ∎

Theorem 4.2 is a significant improvement of [15, Theorem 2] (please see also
[12, 14, 25]).

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$\mbox{}^{\dagger}$ Department of Mathematics, Nankai University, Tianjin,
China;
e-mail<EMAIL_ADDRESS>
$\mbox{}^{\ddagger}$ Department of Mathematics, University of California,
Santa Cruz, CA 95064;
e-mail<EMAIL_ADDRESS>