Symmetry, quantitative Liouville theorems and analysis of large solutions of conformally invariant fully nonlinear elliptic equations

Abstract

We establish blow-up profiles for any blowing-up sequence of solutions of general conformally invariant fully nonlinear elliptic equations on Euclidean domains. We prove that (i) the distance between blow-up points is bounded from below by a universal positive number, (ii) the solutions are very close to a single standard bubble in a universal positive distance around each blow-up point, and (iii) the heights of these bubbles are comparable by a universal factor. As an application of this result, we establish a quantitative Liouville theorem.

1 Introduction

The main goal of this paper is to give a fine analysis of blow-up solutions of conformally invariant fully nonlinear second order elliptic equations.

Let $n \ge 3$ be an integer and

$$\begin{aligned} \Gamma \subset {\mathbb {R}}^n\ \hbox {be an open convex symmetric cone with vertex at the origin} \end{aligned}$$

(1)

satisfying

$$\begin{aligned} \Gamma _n\subset \Gamma \subset \Gamma _1, \end{aligned}$$

(2)

where

$$\begin{aligned} \Gamma _n :=\{\lambda \in {\mathbb {R}}^n\ |\ \lambda _i>0\ \forall \ i\} , \qquad \Gamma _1 :=\Big \{\lambda \in {\mathbb {R}}^n\ |\ \sum _{i=1}^n\lambda _i>0\Big \}. \end{aligned}$$

We assume that

$$\begin{aligned}&f\in C^1(\Gamma )\cap C^0(\overline{\Gamma })\ \hbox {is symmetric in } \ \lambda _i, \end{aligned}$$

(3)

$$\begin{aligned}&f>0, \ \ \frac{ \partial f}{ \partial \lambda _i}>0\ \hbox {in}\ \Gamma \ \forall \ i,\ \ f=0\ \hbox {on}\ \partial \Gamma , \end{aligned}$$

(4)

$$\begin{aligned}&f(\lambda )< 1\qquad \forall \ \lambda \in \Gamma \text { satisfying } \ \sum _{i=1}^n \lambda _i <\delta . \end{aligned}$$

(5)

In (1) and (3), the symmetric property of $\Gamma$ and f is understood in the sense that if $\lambda \in \Gamma$ and $\tilde{\lambda }$ is a permutation of $\lambda$, then $\tilde{\lambda } \in \Gamma$ and $f(\tilde{\lambda }) = f(\lambda )$. Also, throughout the paper,

$$\begin{aligned} \hbox {whenever we write }f(\lambda ),\hbox { we implicitly assume that }\lambda \in \overline{\Gamma }. \end{aligned}$$

When $\Gamma \ne \Gamma _1$, (5) is a consequence of (3) and (4) (cf. [13, Proposition B.1]). However, this does not have to be the case when $\Gamma = \Gamma _1$, for example when

$$\begin{aligned} f(\lambda ) = \Big (\sum _{i=1}^n \lambda _i\Big )^{\frac{n-1}{n+3}} \Big (\sum _{i=1}^n \lambda _i^2\Big )^{\frac{2}{n+3}}. \end{aligned}$$

Illuminating examples of $(f, \Gamma )$ are $(f, \Gamma )=(\sigma _k^{\frac{1}{k}}, \Gamma _k)$ where $\sigma _k(\lambda ) = \sum \lambda _{i_1} \ldots \lambda _{i_k}$ is the k-th elementary symmetric function and

$$\begin{aligned} \Gamma _k&= \text { the connected component of }\{\lambda \in {\mathbb {R}}^n: \sigma _k(\lambda )> 0\}\text { containing }\Gamma _n\\&= \{\lambda \in {\mathbb {R}}^n: \sigma _l(\lambda ) > 0 \text { for all } 1 \le l \le k\}. \end{aligned}$$

Besides (1)–(5), $(\sigma _k^{\frac{1}{k}}, \Gamma _k)$ enjoys other nice and helpful properties, such as concavity and homogeneity properties of $\sigma _k^{1/k}$, Newton’s inequalities, divergence and variational structures, etc., which we do not assume in this paper. In particular, we would like to note that no concavity or homogeneity assumption on f is being made in the present paper.

For a positive $C^2$ function u, let $A^u$ be the $n\times n$ matrix with entries

$$\begin{aligned} (A^u)_{ij} = - \frac{2}{n-2} u^{-\frac{n+2}{n-2}} \nabla _i \nabla _j u + \frac{2n}{(n-2)^2} u^{-\frac{2n}{n-2}}\,\nabla _i u \,\nabla _j u - \frac{2}{(n-2)^2} u^{-\frac{2n}{n-2}}\,|\nabla u|^2\,\delta _{ij}. \end{aligned}$$

This is sometimes referred to as the conformal Hessian of u.

The conformal Hessian $A^u$ arises naturally in conformal geometry as follows. Recall that the Riemann curvature $Riem_g$ of a Riemannian metric g can be decomposed into traced and traceless parts as

where $A_g = \frac{1}{n-2}(\mathrm{Ric}_g - \frac{1}{2(n-1)}R_g\,g)$, $\mathrm{Ric}_g$, $R_g$ and $W_g$ are the Schouten curvature, the Ricci curvature, the scalar curvature and the Weyl curvature of g and denotes the Kulkarni–Nomizu product. While the (1, 3)-valent Weyl curvature remains unchanged under a conformal change of the metric, the Schouten curvature is adjusted by a second order operator of the conformal factor. In particular, if we consider the metric $g_u := u^{\frac{4}{n-2}}g_{flat}$ conformal to the flat metric $g_{flat}$ on ${\mathbb {R}}^n$, then the Schouten curvature $A_{g_u}$ of $g_u$ is given by the conformal Hessian in the form

$$\begin{aligned} A_{g_u} = u^{\frac{4}{n-2}}\,(A^u)_{ij}\,dx^i\,dx^j. \end{aligned}$$

Consequently, we have

$$\begin{aligned} \lambda (A_{g_u}) = \lambda (A^u) \end{aligned}$$

where $\lambda (A_{g_u})$ denotes the eigenvalues of $A_{g_u}$ with respect to the metric $g_u$ and $\lambda (A^u)$ denotes those of the matrix $A^u$.

$A^u$ enjoys a conformal invariance property, inherited from the conformal structure of ${\mathbb {R}}^n$, which will be of special importance in our treatment. Recall that a map $\varphi : {\mathbb {R}}^n \cup \{\infty \} \rightarrow {\mathbb {R}}^n \cup \{\infty \}$ is called a Möbius transformation if it is the composition of finitely many of the following types of transformations:

• a translation: $x \mapsto x + \bar{x}$ where $\bar{x}$ is a given vector in ${\mathbb {R}}^n$,

• a dilation: $x \mapsto a\,x$ where a is a given positive scalar,

• a Kelvin transformation: $x \mapsto \frac{x}{|x|^2}$.

For a function u and a Möbius transformation $\varphi$, let

$$\begin{aligned} u_\varphi = |J_\varphi |^{\frac{n-2}{2n}}u \circ \varphi , \end{aligned}$$

(6)

where $J_\varphi$ is the Jacobian of $\varphi$. A calculation gives

$$\begin{aligned} A^{u_\varphi }(x) = O_\varphi (x)^t A^u(\varphi (x)) O_\varphi (x) \end{aligned}$$

for some orthogonal $n \times n$ matrix $O_\varphi (x)$. In particular,

$$\begin{aligned} \lambda (A^{u_\varphi }(x)) = \lambda (A^u( \varphi (x))). \end{aligned}$$

(7)

The main result of this paper concerns an analysis on the behavior of a sequence $\{u_k\}\in C^2(B_3(0))$ satisfying

$$\begin{aligned} f(\lambda (A^{u_k})) =1, \ u_k>0,\ \hbox {in}\ B_3(0), \end{aligned}$$

(8)

and

$$\begin{aligned} \sup _{B_1(0)}u_k\rightarrow \infty , \end{aligned}$$

(9)

where $(f, \Gamma )$ satisfies (1)–(5). Note that no other assumptions on $u_k$ is made.

As is known, Eq. (8) is a fully nonlinear elliptic equation. Fully nonlinear elliptic equations involving $f(\lambda (\nabla ^2 u))$ were investigated in the classic paper of Caffarelli et al. [2].

Our paper appears to be the first fine blow-up analysis in this fully nonlinear context. We expect this to serve as a crucial step in the study of the problem on Riemannian manifolds.

To obtain our result on fine analysis of blow-up solutions, we make use of the following Liouville theorems.

Theorem A

([8]) Let $(f, \Gamma )$ satisfy (1)–(4) and let $0<v\in C^2({\mathbb {R}}^n)$ satisfy

$$\begin{aligned} f(\lambda (A^v))=1\qquad \hbox {in}\ {\mathbb {R}}^n. \end{aligned}$$

(10)

Then

$$\begin{aligned} v(x)\equiv \left( \frac{a}{ 1+b^2|x-\bar{x}|^2} \right) ^{ \frac{n-2}{2} }, \qquad x\in {\mathbb {R}}^n \end{aligned}$$

(11)

for some $\bar{x}\in {\mathbb {R}}^n$ and some positive constants a and b satisfying

$$\begin{aligned} f(2b^2a^{-2}, \ldots , 2b^2a^{-2})=1. \end{aligned}$$

Theorem B

([12]) Let $\Gamma$ satisfy (1) and (2), and let $0< v\in C^{0,1}_{loc}({\mathbb {R}}^n {\setminus } \{0\})$ satisfy

$$\begin{aligned} \lambda (A^v) \in \partial \Gamma \hbox { on }\ {\mathbb {R}}^n {\setminus } \{0\} \end{aligned}$$

(12)

in the viscosity sense (see Definition 1.2 below). Then v is radially symmetric about the origin and v(r) is non-increasing in r.

For $(f, \Gamma )=(\frac{1}{2n}\sigma _1, \Gamma _1)$, Eq. (8) is the critical exponent equation $-\Delta u= n(n-2)u^{ (n+2) /(n-2) }$ and Theorem A was proved by Caffarelli et al. [1]. See also Gidas et al. [5] under some decay assumption of u at infinity. For $(f, \Gamma )=(\sigma _2^{1/2}, \Gamma _2)$ in ${\mathbb {R}}^4$ and $v \in C^{1,1}_{loc}({\mathbb {R}}^4)$, the result was proved by Chang et al. [3].

In fact we need a stronger version of Theorem A (see Theorem 1.1) and a variant of Theorem B (see Theorem 1.2). For simplicity, readers are advised that in the main body of the paper

$$\begin{aligned}&\hbox {all theorems, propositions and lemmas hold under }(1)\hbox {--}(5),\\&\qquad \hbox { instead of the stated weaker hypotheses on }(f,\Gamma ). \end{aligned}$$

Theorem 1.1

Let $(f, \Gamma )$ satisfy

$$\begin{aligned}&\Gamma \subset {\mathbb {R}}^n\ \hbox {be an open symmetric set}, \end{aligned}$$

(13)

$$\begin{aligned}&\Gamma \subset \Gamma _1 \text { and } \Gamma \cap \{\lambda + t\mu : t > 0\} \text { is convex for all }\lambda \in \Gamma , \mu \in \Gamma _n, \end{aligned}$$

(14)

$$\begin{aligned}&f \in C^1(\Gamma ) \text { is symmetric in } \lambda _i \text { and } \frac{ \partial f}{ \partial \lambda _i}>0\ \hbox {in}\ \Gamma \ \forall \ i. \end{aligned}$$

(15)

Assume that $0<v\in C^0({\mathbb {R}}^n)$, $0<v_k\in C^2(B_{R_k}(0))$, $R_k\rightarrow \infty$,

$$\begin{aligned} f(\lambda (A^{v_k}))=1 \quad \ \ \hbox {in}\ B_{R_k}(0), \end{aligned}$$

(16)

and

$$\begin{aligned} v_k\rightarrow v\ \hbox {in}\ C^0_{loc}({\mathbb {R}}^n). \end{aligned}$$

(17)

Then either v is constant or v is of the form (11) for some $\bar{x}\in {\mathbb {R}}^n$ and some positive constants a and b.

If it holds in addition that

$$\begin{aligned}&\text {there exists }t_0 > 0\text { such that }f(t_0, \ldots , t_0) < 1, \end{aligned}$$

(18)

then v cannot be constant. If it holds further that

$$\begin{aligned} \Gamma + \Gamma _n = \{\lambda + \mu : \lambda \in \Gamma , \mu \in \Gamma _n\} \subset \Gamma , \end{aligned}$$

(19)

then the constants a and b in (11) satisfy $(2b^2a^{-2}, \ldots , 2b^2a^{-2}) \in \Gamma$ and

$$\begin{aligned} f(2b^2a^{-2}, \ldots , 2b^2a^{-2})=1. \end{aligned}$$

Remark 1.1

In Theorem 1.1, if condition (18) is dropped, the case that v is constant can occur. See the counterexample in Remark 2.1.

Theorem 1.2

Let $(f, \Gamma )$ satisfy (13)–(15). Assume that $v_*\in C^0({\mathbb {R}}^n {\setminus } \{0\})$, $0<v_k\in C^2(B_{R_k}(0) {\setminus } \{0\})$, $R_k\rightarrow \infty$,

$$\begin{aligned} f(\lambda (A^{v_k}))=1 \quad \ \ \hbox {in}\ B_{R_k}(0) {\setminus } \{0\}, \end{aligned}$$

and, for some $M_k \rightarrow \infty$,

$$\begin{aligned} M_k\,v_k\rightarrow v_*\ \hbox {in}\ C^0_{loc}({\mathbb {R}}^n {\setminus } \{0\}). \end{aligned}$$

(20)

Then $v_*$ is radially symmetric about the origin, i.e. $v_*(x) = v_*(|x|)$. In particular, if $v_k \in C^2(B_{R_k}(0))$, $f(\lambda (A^{v_k}))=1$ in $B_{R_k}(0)$ and $M_kv_k$ converges to $v_*$ in $C^0_{loc}({\mathbb {R}}^n)$, then $v_*$ is constant.

Remark 1.2

When $(f,\Gamma )$ satisfies (1)–(4) and an additional hypothesis that f is homogeneous of positive degree, the function $v_*$ in Theorem 1.2 is a viscosity solution of (12) and the conclusion follows from Theorem B. However, when f is not homogeneous, $v_*$ is not necessarily a viscosity solution of (12).

It is not difficult to see that, under (1)–(4), the function v in Theorem 1.1 is a viscosity solution of (10) (see Remark B.2). We have the following conjecture.

Conjecture

Let $(f, \Gamma )$ satisfy (1)–(4), and let $0<v\in C^0_{loc}({\mathbb {R}}^n)$ be a viscosity solution of (10). Then v is of the form (11) for some $\bar{x}\in {\mathbb {R}}^n$ and some positive constants a and b.

The notion of viscosity solutions given below is consistent with that in [12].

Definition 1.1

A positive continuous function v in an open set $\Omega \subset {\mathbb {R}}^n$ is a viscosity supersolution (respectively, subsolution) of

$$\begin{aligned} f(\lambda (A^v))=1, \ \ \hbox {in}\ \Omega , \end{aligned}$$

when the following holds: if $x_0\in \Omega$, $\varphi \in C^2(\Omega )$, $(v-\varphi )(x_0)=0$, and $v-\varphi \ge 0$ near $x_0$, then

$$\begin{aligned} f(\lambda (A^\varphi (x_0))\ge 1 \end{aligned}$$

(respectively, if $(v-\varphi )(x_0)=0$, and $v-\varphi \le 0$ near $x_0$, then either $\lambda (A^\varphi (x_0)) \in {\mathbb {R}}^n{\setminus } \overline{\Gamma }$ or $f(\lambda (A^\varphi (x_0)))\le 1$). We say that v is a viscosity solution if it is both a viscosity supersolution and a viscosity subsolution.

Definition 1.2

A positive continuous function v in an open set $\Omega \subset {\mathbb {R}}^n$ is a viscosity supersolution (respectively, subsolution) of

$$\begin{aligned} \lambda (A^v) \in \partial \Gamma , \ \ \hbox {in}\ \Omega , \end{aligned}$$

when the following holds: if $x_0\in \Omega$, $\varphi \in C^2(\Omega )$, $(v-\varphi )(x_0)=0$, and $v-\varphi \ge 0$ near $x_0$, then

$$\begin{aligned} \lambda (A^\varphi (x_0) \in \bar{\Gamma } \end{aligned}$$

(respectively, if $(v-\varphi )(x_0)=0$, and $v-\varphi \le 0$ near $x_0$, then either $\lambda (A^\varphi (x_0)) \in {\mathbb {R}}^n{\setminus } \overline{\Gamma }$). We say that v is a viscosity solution if it is both a viscosity supersolution and a viscosity subsolution.

It is clear that for $C^2$ functions the notions of viscosity solutions and classical solutions coincide. Also, viscosity super- and sub-solutions are stable under uniform convergence, see “Appendix B”.

Note that for any $\lambda = (\lambda _1, \ldots , \lambda _n) \in \Gamma$, $t_1 = \max \lambda _i + 1 > 0$ and $(t_1, \ldots , t_1) \in \lambda + \Gamma _n$. In other words, the sets $\lambda + \Gamma _n$ have non-empty intersection with the ray $\{(t, \ldots , t): t > 0\}$. Thus, if $f^{-1}(1) \ne \emptyset$, then, in view of (4), there exists some $c>0$ such that $f(c,\ldots , c)=1$. In such situation, working with $\tilde{f}(\lambda ):=f(\frac{c}{2} \lambda )$ instead of f, we may assume without loss of generality the following normalization condition

$$\begin{aligned} f(2,\ldots , 2)=1. \end{aligned}$$

(21)

Let

$$\begin{aligned} U(x):= \left( \frac{1}{1+|x|^2}\right) ^{ \frac{n-2}{2} },\qquad x\in {\mathbb {R}}^n. \end{aligned}$$

A calculation gives

$$\begin{aligned} A^U\equiv 2I. \end{aligned}$$

With the normalization (21), U satisfies

$$\begin{aligned} f(\lambda (A^U))=1 \quad \hbox {on}\ {\mathbb {R}}^n. \end{aligned}$$

For $\bar{x}\in {\mathbb {R}}^n$ and $\mu >0$, let

$$\begin{aligned} U^{\bar{x}, \mu }(x)= \mu U(\mu ^{\frac{2}{n-2}}(x-\bar{x})) = \Big (\frac{\mu ^{\frac{2}{n-2}}}{1 + \mu ^{\frac{4}{n-2}} |x- \bar{x}|^2}\Big )^{\frac{n-2}{2}}. \end{aligned}$$

Note that, in the sense of (6), $U^{\bar{x}, \mu } = U_\varphi$ with $\varphi (x) = \mu ^{\frac{2}{n-2}}\,(x - \bar{x})$. Hence, by the conformal invariance (7), for any $\bar{x}\in {\mathbb {R}}^n$ and $\mu >0$,

$$\begin{aligned} f(\lambda (A^{U_{\bar{x}, \mu }})) = 1 \text { on } {\mathbb {R}}^n. \end{aligned}$$

Theorem 1.3

Let $(f, \Gamma )$ satisfy (13)–(15), (18)–(19), (5) and the normalization condition (21). Let $\epsilon \in (0,1/2]$. There exist constants $\bar{m} = \bar{m}(f,\Gamma ) \ge 1, {K}= {K}(f,\Gamma ) > 1$, ${\delta _*}= {\delta _*}(\epsilon , f,\Gamma )> 0, {C_*}= {C_*}(\epsilon , f,\Gamma ) > 1$ such that for any positive $u \in C^2(B_3(0))$ satisfying¹

$$\begin{aligned} f(\lambda (A^u)) = 1 \text { in } B_{3}(0) \text { and } \sup _{B_1(0)} u \ge {C_*}, \end{aligned}$$

there exists $\{x^1, \ldots , x^m\} \subset B_{2}(0)$ with $1 \le m \le \bar{m}$ satisfying

• (i) $u(x^1) \ge \sup _{B_1(0)} u$,

• (ii) $|x^i-x^j|\ge \frac{1}{{K}}$ for all $1 \le i\ne j \le m$,

• (iii) $\frac{1}{{K}} \le \frac{u(x^i)}{ u(x^j) }\le {K}$ for all $1 \le i, j \le m$,

• (iv) $|u(x) - U^{x^i, u(x^i)}(x)| \le \epsilon U^{x^i, u(x^i)}(x)$ for all $1 \le i \le m$, $x\in B_{{\delta _*}}(x^i)$,

• (v) $\frac{1}{{K}\delta _*^{n-2} u(x^1)}\le u(x) \le \frac{{K}}{ \delta _*^{n-2}u(x^1)}$ for all $x \in B_{\frac{3}{2}}(0){\setminus } \cup _{ i=1}^m B_{{\delta _*}}(x^i)$,

• (vi) $u(x^i) = \sup _{B_{{\delta _*}}(x^i)} u$.

Remark 1.3

If it holds further that $\sup _{B_1(0)} u > \frac{{K}^{1/2}}{{\delta _*}^{\frac{n-2}{2}}}$, then

$$\begin{aligned} \min _{1 \le i \le m} dist(x^i, B_1(0)) \le \Big [\sup _{B_1(0)} u\Big ]^{-\frac{2}{n-2}}. \end{aligned}$$

To see this, let $x_*$ be a point in $\bar{B}_1(0)$ such that $u(x_*) = \sup _{B_1(0)} u$. In view of (v) and the stated condition on $\sup _{B_1(0)} u$, $x_*$ belongs to some ball $B_{{\delta _*}}(x^{i_0})$. By (iv), we then have

$$\begin{aligned} u(x_*) \le 2U^{x^{i_0},u(x^{i_0})}(x_*) \le \frac{1}{|x_* - x^{i_0}|^{\frac{n-2}{2}}}, \end{aligned}$$

which implies the assertion.

Theorem 1.3 can be stated equivalently as follows.

Theorem 1.4

Let $(f, \Gamma )$ satisfy (13)–(15), (18)–(19), (5) and the normalization condition (21). Assume that $0 < u_k\in C^2(B_3(0))$ satisfy (8) and (9). Let $\epsilon \in (0,1/2]$. Then there exist $\bar{m} = \bar{m}(f,\Gamma ) \ge 1, {K}(f,\Gamma ) > 1$ and ${\delta _*}= {\delta _*}(\epsilon ,f,\Gamma ) > 0$ such that, after passing to a subsequence, still denoted by $u_k$, there exists $\{x_k^1, \ldots , x_k^m\}\subset B_{2}(0)$ ($1 \le m \le \bar{m}$) satisfying²

• (i) $u_k(x_k^1) \ge \sup _{B_1(0)} u_k$,

• (ii) $|x_k^i-x_k^j|\ge \frac{1}{{K}}$ for all $k \ge 1$, $1 \le i\ne j \le m$,

• (iii) $\frac{1}{{K}} \le \frac{ u_k(x_k^i) }{ u_k(x_k^j) }\le {K}$ for all $k \ge 1$, $1 \le i, j \le m$,

• (iv) $|u_k(x) - U^{x_k^i, u_k(x_k^i)}(x)| \le \epsilon U^{x_k^i, u_k(x_k^i)}(x)$ for all $k \ge 1$, $1 \le i \le m$, $x\in B_{{\delta _*}}(x_k^i)$,

• (v) $\frac{1}{{K}\delta _*^{n-2}u_k(x_k^1)}\le u_k(x) \le \frac{{K}}{ \delta _*^{n-2} u_k(x_k^1)}$ for all $k \ge 1$, $x \in B_{\frac{3}{2}}(0){\setminus } \cup _{ i=1}^m B_{{\delta _*}}(x_k^i)$,

• (vi) $u_k(x_k^i) = \sup _{B_{{\delta _*}}(x_k^i)} u_k$.

Remark 1.4

By Remark 1.3, we have

$$\begin{aligned} dist(\{x_k^1, \ldots , x_k^m\},B_1(0)) \rightarrow 0 \text { as } k \rightarrow \infty . \end{aligned}$$

When $(f,\Gamma )=(\frac{1}{2n}\sigma _1, \Gamma _1)$, Eq. (8) is $-\Delta u_k = n(n-2)\,u_k^{\frac{n+2}{n-2}}$ and Theorem 1.4 in this case was proved by Schoen [16].

See Li [9] and Chen and Lin [4] for analogous results for the equation $-\Delta u_k = K(x)u_k^{\frac{n+2}{n-2}}$.

In Theorems 1.3 and 1.4, $B_1(0)$, $B_2(0)$ and $B_3(0)$ can be replaced respectively by $B_{r_1}(0)$, $B_{r_2}(0)$ and $B_{r_3}(0)$, $0< r_1< r_2 < r_3$, and in this case the constants $\bar{m}$, ${K}$, ${\delta _*}$ and ${C_*}$ depend also on $r_1$, $r_2$ and $r_3$.

The following is a quantitative version of Theorem A, and is related to Theorems 1.1 and 1.3.

Theorem 1.5

(Quantitative Liouville Theorem) Let $(f, \Gamma )$ satisfy (13)–(15), (18)–(19), (5) and the normalization condition (21), and let $\gamma , r_1 > 0$ be constants. Then, for every $\epsilon \in (0,1/2]$, there exist some constants ${\delta _*}> 0, R^* > 0$, depending only on $(f, \Gamma )$, $\gamma , r_1$ and $\epsilon$, such that if $0< v\in C^2(B_R(0))$ for some $R\ge R^*$,

$$\begin{aligned} f(\lambda (A^v)=1\quad \hbox {in}\ B_R(0), \end{aligned}$$

(22)

and

$$\begin{aligned} v\ge \gamma \quad \hbox {in}\ \ B_{r_1}(0), \end{aligned}$$

(23)

then, for some $\bar{x}\in {\mathbb {R}}^n$ satisfying

$$\begin{aligned} v(\bar{x}) = \max _{B_{{\delta _*}R}(0)} v \le \frac{2^{n-1}}{\gamma \,r_1^{n-2}}, \qquad |\bar{x}| \le 2^{\frac{1}{n-2}} \gamma ^{-\frac{2}{n-2}}, \end{aligned}$$

(24)

there holds

$$\begin{aligned} |v(y)- U^{\bar{x}, v(\bar{x})}(y)| \le \epsilon \,U^{\bar{x}, v(\bar{x})}(y),\qquad \forall \ |y - \bar{x}|\le {\delta _*}R. \end{aligned}$$

(25)

Remark 1.5

The constant ${\delta _*}$ in Theorems 1.3 and 1.5 can be chosen the same.

Remark 1.6

An analogous result for the degenerate elliptic equation $\lambda (A^v) \in \partial \Gamma$ is a consequence of the local gradient estimate [13, Theorem 1.5].

An ingredient in our proof of Theorems 1.3 and 1.4 is the following local gradient estimate, which follows from Theorem 1.2 and the proof of [12, Theorem 1.10].

Theorem 1.6

Let $(f, \Gamma )$ satisfy (13)–(15) and let $v\in C^2(B_2(0))$ satisfy, for some constant $b>0$,

$$\begin{aligned} 0<v\le b \qquad \hbox {in}\ B_2(0), \end{aligned}$$

(26)

and

$$\begin{aligned} f(\lambda (A^v))=1 \qquad \hbox {in}\ B_2(0). \end{aligned}$$

(27)

Then, for some constant C depending only on $(f, \Gamma )$ and b,

$$\begin{aligned} |\nabla \ln v|\le C\quad \hbox {in}\ B_1(0). \end{aligned}$$

(28)

For $(f,\Gamma )=(\sigma _k^{1/k}, \Gamma _k)$, the result was proved by Guan and Wang [6].

When $(f,\Gamma )$ satisfies (1)–(4) and is homogeneous of positive degree, Theorem 1.6 was proved in [12].

The rest of the paper is organized as follows. We start in Sect. 2 with the proof of Theorems 1.1 and 1.2. We then prove Theorem 1.6 in Sect. 3. In Sect. 4, we first establish an intermediate quantitative Liouville result and then use it to prove Theorem 1.3. In Sect. 5, we prove Theorem 1.5 as an application of Theorem 1.3. In “Appendix A”, we present a lemma about super-harmonic functions which is used in the body of the paper. In “Appendix B”, we include a relevant remark on the limit of viscosity solutions of elliptic PDE. Finally we collect in “Appendix C” some relevant calculus lemmas.

2 Non-quantitative Liouville theorems

In this section, we prove Theorems 1.1 and 1.2. We use the method of moving spheres and establish along the way, as a tool, a gradient estimate which is in a sense weaker than that in Theorem 1.6 but suffices for the moment. (Note that the proof of Theorem 1.6 relies on Theorem 1.2.)

2.1 A gradient estimate

Theorem 2.1

Let $(f,\Gamma )$ satisfy (13), (15) and

$$\begin{aligned} \Gamma \cap \{\lambda + t\mu : t > 0\} \text { is convex for all } \lambda \in \Gamma , \mu \in \Gamma _n. \end{aligned}$$

Let $0 < v\in C^2(B_2(0))$ satisfy, for some constant $\theta > 1$,

$$\begin{aligned} \sup _{B_2(0)} v \le \theta \inf _{B_2(0)} v, \end{aligned}$$

(29)

and

$$\begin{aligned} f(\lambda (A^v))=1 \qquad \hbox {in}\ B_2(0). \end{aligned}$$

Then, for some constant C depending only on n and $\theta$,

$$\begin{aligned} |\nabla \ln v|\le C\quad \hbox {in}\ B_1(0). \end{aligned}$$

This type of gradient estimate was established and used in various work of the first named author and his collaborators under less general hypothesis on $(f,\Gamma )$. It turns out that the same proof works in the current situation. We give a detailed sketch here for completeness.

We use the method of moving spheres as in [7, 8, 14, 15]. For a function w defined on a subset of ${\mathbb {R}}^n$, we define

$$\begin{aligned} w_{x,\lambda }(y) = \frac{\lambda ^{n-2}}{|y - x|^{n-2}}w\Big (x + \frac{\lambda ^2(y -x )}{|y -x|^2}\Big ) \end{aligned}$$

wherever the expression makes sense. We will use $w_\lambda$ to denote $w_{0,\lambda }$. We start with a simple result.

Lemma 2.1

Let $R > 0$ and w be a positive Lipschitz function in $\bar{B}_R(0)$ such that, for some $L > 0$,

$$\begin{aligned} |\ln w(y) - \ln w(z)| \le L|y - z| \text { for all } y, z \in \bar{B}_R(0). \end{aligned}$$

Then for $\underline{\lambda } = \min (\frac{n-2}{2L},\frac{R}{2})$ we have

$$\begin{aligned} w_\lambda \le w\ \hbox {in}\ B_{\underline{\lambda }}(0){\setminus } B_\lambda , \forall 0<\lambda < \underline{\lambda }. \end{aligned}$$

(30)

Proof

Write w in polar coordinates $w(r, \theta )$. It is easy to see that (30) is equivalent to

$$\begin{aligned} r^{\frac{n-2}{2}} w(r, \theta ))\le \ s^{\frac{n-2}{2}} w(s, \theta ), \ \ \forall \ 0<r<s<\underline{\lambda },\ \forall \ \theta . \end{aligned}$$

(31)

Estimate (31) is readily seen from the estimates

$$\begin{aligned} \ln w(s, \theta ) -\ln w(r, \theta )&\ge - L|s - r| \ge - \frac{n-2}{2\underline{\lambda }} (s - r)\\&\ge - \frac{n-2}{2}[\ln s - \ln r]. \end{aligned}$$

Lemma 2.1 is established. $\square$

Proof of Theorem 2.1

By Lemma 2.1, there exists some $r_0 \in (0,1/3)$ such that

$$\begin{aligned} v_{x,\lambda } \le v \text { in } B_{r_0}(x) {\setminus } B_{\lambda }(x) \text { for all } \lambda \in (0,r_0) \text { and } x \in B_{4/3}(0). \end{aligned}$$

It is easy to see that, for some $r_1 \in (0,r_0)$,

$$\begin{aligned} v_{x,\lambda } \le v \text { in } B_{5/3}(0) {\setminus } B_{r_0}(x) \text { for all } \lambda \in (0,r_1) \text { and } x \in B_{4/3}(0). \end{aligned}$$

We then define, for $x \in B_{4/3}(0)$,

$$\begin{aligned} \bar{\lambda }(x) = \sup \Big \{\lambda \in (0, 5/3 - |x|): v_{x,\lambda } \le v \text { in } B_{5/3}(0) {\setminus } B_{\lambda }(x)\Big \}. \end{aligned}$$

We have

$$\begin{aligned} v_{x,\bar{\lambda }(x)} \le v\ \hbox {in}\ B_{5/3}(0){\setminus } B_{\bar{\lambda }(x)}(x), \end{aligned}$$

By the conformal invariance (7), $v_{x,\bar{\lambda }(x)}$ satisfies

$$\begin{aligned} f(\lambda (A^{v_{x,\bar{\lambda }(x)}}))=1, \quad \ \ \hbox {in}\ B_{5/3}(0){\setminus } \overline{ B_{\bar{\lambda }(x)}(x) }. \end{aligned}$$

Using the above two displayed equations, the definition of $\bar{\lambda }(x)$, and using the ellipticity of the equation satisfied by v and $v_{x,\bar{\lambda }(x)}$, we can apply the strong maximum principle and Hopf Lemma to infer that either $\bar{\lambda }(x) = 5/3 - |x|$ or there exists some $y \in \partial B_{5/3}(0)$ such that

$$\begin{aligned} v_{x,\bar{\lambda }(x)}(y)=v(y) \end{aligned}$$

—see the proof of [8, Lemma 4.5].

In the latter case, (29) implies that

$$\begin{aligned} \theta \ge \frac{v\big (x + \frac{\bar{\lambda }(x)^2(y-x)}{|y - x|^2}\big )}{v(y)} = \frac{|y - x|^{n-2}}{\bar{\lambda }(x)^{n-2}} \ge \frac{(5/3 - |x|)^{n-2}}{\bar{\lambda }(x)^{n-2}}. \end{aligned}$$

In either case, we obtain that

$$\begin{aligned} \bar{\lambda }(x) \ge c(n,\theta ) > 0 \text { for all } x \in B_{4/3}(0). \end{aligned}$$

The conclusion then follows from [8, Lemma A.2]. $\square$

2.2 Proof of Theorem 1.1

Remark 2.1

If we drop condition (18), the case that v is constant in Theorem 1.1 can occur. For example, consider $n \ge 3$ and

$$\begin{aligned} f(\lambda )&= \sigma _2(\lambda ) + 1, \qquad \Gamma = \Big \{\lambda \in {\mathbb {R}}^n\Big | \sum _{j=1}^n \lambda _j - \lambda _i > 0 \text { for all } i = 1, \ldots , n\Big \},\\ v_k(x)&= \Big (\frac{1}{R_k^{n+4}}|x - x_k|^{-\frac{n-4}{2}} + 1\Big )^{\frac{2(n-2)}{n-4}} \text { for } |x| < \frac{|x_k|}{2} = R_k \rightarrow \infty . \end{aligned}$$

It is readily seen that $f(t, \ldots , t) > 1$ for all $(t, \ldots , t) \in \Gamma$, $v_k$ satisfies (16) (cf. [13, Theorem 1.6]) and $v_k \rightarrow 1$ in $C^{0}_{loc}({\mathbb {R}}^n)$.

Proof of Theorem 1.1

We may assume that $R_k\ge 5$ for all k.

Clearly, for every $\beta >1$, there exists some positive constant $C(\beta )$, independent of k, such that $1/C(\beta )\le v_k\le C(\beta )$ in $B_\beta (0)$. It follows from Theorem 2.1 that $|\nabla \ln v_k| \le C'(\beta )$ in $B_{\beta /2}(0)$. It follows, after passing to a subsequence, that for every $0<\alpha <1$, $v_k\rightarrow v$ in $C^{\alpha }_{loc}({\mathbb {R}}^n)$, $v\in C^{0,1}_{loc}({\mathbb {R}}^n)$ and v is super-harmonic on ${\mathbb {R}}^n$.

Using the positivity, the superharmonic of v, and the maximum principle, we can find $c_0 > 0$ such that

$$\begin{aligned} v(y)\ge 2c_0 |y|^{2-n} ,\ \ \forall \ |y|\ge 1. \end{aligned}$$

(32)

Passing to a subsequence and shrinking $R_k$ and $c_0 > 0$, if necessary, we may assume that

$$\begin{aligned} |v_k(y)-v(y)|\le R_k^{-n},\quad \ \ \forall \ |y|\le R_k, \end{aligned}$$

(33)

and

$$\begin{aligned} v_k(y) \ge c_0(1+|y|)^{2-n}, \ \ \forall \ |y| < R_k. \end{aligned}$$

(34)

Lemma 2.2

Under the hypotheses of Theorem 1.1, there exists a function $\lambda ^{(0)}: {\mathbb {R}}^n \rightarrow (0,\infty )$ such that, for all k,

$$\begin{aligned} (v_k)_{x,\lambda } \le v_k\ \hbox {in}\ B_{R_k}(0){\setminus } B_\lambda (x), \forall \ 0<\lambda < \lambda ^{(0)}(x)\ \hbox {and}\ |x|\le \frac{R_k}{5}. \end{aligned}$$

Proof

For $|x| \le \frac{R_k}{5}$, we have, by (33) and (34), for all k that

$$\begin{aligned} \frac{1}{c_1(x)}\le v_k\le c_1(x)\ \hbox {in}\ B_{4r_1(x)}(x) \subset B_{R_k}(0), \end{aligned}$$

(35)

where

$$\begin{aligned} r_1(x) = \frac{1}{4} + |x|, \text { and } c_1(x) = \max \Big \{1+ \sup _{B_{4r_1(x)}(x)} v, \frac{1}{c_0} (1 + |x| + 4r_1(x))^{n-2}\Big \}. \end{aligned}$$

By (35) and Theorem 2.1, there exists $c_2(x)>0$, independent of k, such that

$$\begin{aligned} |\nabla \ln v_k|\le c_2(x)\ \hbox {in}\ B_{2r_1(x)}(x). \end{aligned}$$

Thus, by Lemma 2.1, we can find $0< \lambda _1(x) < r_1(x)$ independent of k such that

$$\begin{aligned} (v_k)_{x,\lambda } \le v_k\ \hbox {in}\ B_{\lambda _1(x)}(x){\setminus } B_\lambda (x) \ \ \forall \ 0< \lambda < \lambda _1(x). \end{aligned}$$

(36)

For $0< \lambda < \lambda _1(x)$, we have, using (35), that

$$\begin{aligned} (v_k)_{x,\lambda }(y) \le \frac{\lambda ^{n-2}c_1(x)}{|y - x|^{n-2}} , \quad \forall \ y \in B_{R_k}(0){\setminus } B_{\lambda }(x). \end{aligned}$$

(37)

When $y \in B_{R_k}(0) {\setminus } B_{4r_1(x)}(x)$, $\frac{1}{2}(1 + |y|) < |y - x|$ and we obtain, using (37) and (34), that

$$\begin{aligned} (v_k)_{x,\lambda }(y) \le \frac{(2\lambda )^{n-2}c_1(x)}{(1 + |y|)^{n-2}} \le \frac{(2\lambda )^{n-2}c_1(x)}{c_0}\,v_k(y). \end{aligned}$$

(38)

When $y \in B_{4r_1(x)}(x) {\setminus } B_{\lambda _1(x)}(x)$, $1 + |y| \le 2(1 + 3|x|)$, $|y - x| \ge \lambda _1(x)$ and we obtain, using (37) and (34), that

$$\begin{aligned} (v_k)_{x,\lambda }(y) \le \frac{\lambda ^{n-2} c_1(x)}{\lambda _1(x)^{n-2}} \le \frac{(2\lambda )^{n-2} c_1(x) (1 + 3|x|)^{n-2}}{\lambda _1(x)^{n-2} c_0} v_k(y). \end{aligned}$$

(39)

Letting

$$\begin{aligned} \lambda ^{(0)}(x) = \min \Big \{ \lambda _1(x), \frac{\lambda _1(x)}{2(1 + 3|x|)}\Big [\frac{c_1(x)}{c_0}\Big ]^{\frac{1}{n-2}}\Big \} \le \lambda _1(x), \end{aligned}$$

we derive from (38) and (39) that

$$\begin{aligned} (v_k)_{x,\lambda } \le v_k \ \ \ \hbox {in}\ B_{R_k}(0) {\setminus } B_{\lambda _1(x)}(x) \text { and } 0< \lambda < \lambda ^{(0)}(x). \end{aligned}$$

(40)

Lemma 2.2 follows from (36) and (40). $\square$

Define, for $x\in {\mathbb {R}}^n$ and $|x|\le R_k/5$, that

$$\begin{aligned} \bar{\lambda }_k(x) =\sup \Big \{0<\mu \le \frac{R_k}{5}\ |\ (v_k)_{x,\lambda } \le v_k\ \hbox {in}\ B_{R_k}(0){\setminus } B_\lambda (x), \forall 0<\lambda <\mu \Big \}. \end{aligned}$$

By Lemma 2.2,

$$\begin{aligned} \bar{\lambda }(x):= \displaystyle { \liminf _{k\rightarrow \infty }\bar{\lambda }_k(x) } \in [\lambda ^{(0)}(x), \infty ], \qquad x\in {\mathbb {R}}^n. \end{aligned}$$

By (32),

$$\begin{aligned} \alpha := \liminf _{|y|\rightarrow \infty } |y|^{n-2}v(y)\in (0, \infty ]. \end{aligned}$$

Lemma 2.3

Assume (13)–(15). Then either v is constant or

$$\begin{aligned} \alpha =\lim _{|y|\rightarrow \infty } |y|^{n-2}v_{x,\bar{\lambda }(x)}(y)= \bar{\lambda }(x)^{n-2}v(x)<\infty ,\qquad \forall \ x\in {\mathbb {R}}^n. \end{aligned}$$

Proof

Step 1. If $\bar{\lambda }(x)<\infty$ for some $x\in {\mathbb {R}}^n$, then

$$\begin{aligned} \alpha = \lim _{|y|\rightarrow \infty } |y|^{n-2}v_{x,\bar{\lambda }(x)}(y) =\bar{\lambda }(x)^{n-2}v(x) <\infty . \end{aligned}$$

Since $\bar{\lambda }(x)<\infty$, we have, along a subsequence, $\bar{\lambda }_k(x)\rightarrow \bar{\lambda }(x)$—but for simplicity, we still use $\{\bar{\lambda }_k(x)\}$, $\{v_k\}$, etc to denote the subsequence. By the definition of $\bar{\lambda }_k(x)$, we have

$$\begin{aligned} (v_k)_{x,\bar{\lambda }_k(x)} \le v_k\ \hbox {in}\ B_{R_k}(0){\setminus } B_{\bar{\lambda }_k(x)}(x), \end{aligned}$$

(41)

By the conformal invariance (7), $(v_k)_{x,\bar{\lambda }_k(x)}$ satisfies

$$\begin{aligned} f(\lambda (A^{(v_k)_{x,\bar{\lambda }_k(x)}}))=1, \quad \ \ \hbox {in}\ B_{R_k}(0){\setminus } \overline{ B_{\bar{\lambda }_k(x)}(x) }. \end{aligned}$$

(42)

Using (16), (41), (42), the definition of $\bar{\lambda }_k(x)$, and using the ellipticity of the equation satisfied by $v_k$ and $(v_k)_{x,\bar{\lambda }_k(x)}$, we can apply the strong maximum principle and Hopf Lemma to infer the existence of some $y_k\in \partial B_{R_k}(0)$ such that

$$\begin{aligned} (v_k)_{x,\bar{\lambda }_k(x)}(y_k)=v_k(y_k) \end{aligned}$$

—see the proof of [8, Lemma 4.5].

It follows that

$$\begin{aligned} \lim _{k\rightarrow \infty } |y_k|^{n-2} v_k(y_k) = \lim _{k\rightarrow \infty } |y_k|^{n-2} (v_k)_{x,\bar{\lambda }_k(x)}(y_k) =(\bar{\lambda }(x))^{n-2}v(x). \end{aligned}$$

This implies, in view of (33), that

$$\begin{aligned} \alpha \le \lim _{k\rightarrow \infty } |y_k|^{n-2} v(y_k)=\bar{\lambda }(x)^{n-2}v(x) =\lim _{|y|\rightarrow \infty } |y|^{n-2}v_{x,\bar{\lambda }(x)}(y)<\infty . \end{aligned}$$

On the other hand, if $\hat{y}_i$ is such that $|\hat{y}_i|\rightarrow \infty$ and

$$\begin{aligned} \alpha = \lim _{i\rightarrow \infty } |y_i|^{n-2}v(y_i), \end{aligned}$$

then, since $v_{ x, \bar{\lambda }(x) }\le v$ in ${\mathbb {R}}^n{\setminus } B_{\bar{\lambda }(x) }(x)$, we have

$$\begin{aligned} v(y_i)\ge \frac{ \bar{\lambda }(x) ^{n-2} }{ |y_i-x|^{n-2} } v\left( x+\frac{\bar{\lambda }(x)^2(y_i-x) }{ |y_i-x|^2 }\right) . \end{aligned}$$

This gives

$$\begin{aligned} \alpha = \lim _{i\rightarrow \infty } |y_i|^{n-2}v(y_i) \ge \bar{\lambda }(x)^{n-2}v(x). \end{aligned}$$

Step 1 is established.

Step 2. It remains to show that either v is constant or, for every $x\in {\mathbb {R}}^n$, $\bar{\lambda }(x)<\infty$.

To this end, we show that if $\bar{\lambda }(x)=\infty$ for some $x \in {\mathbb {R}}^n$, then v is constant. Indeed, assume that $\bar{\lambda }_k(x) \rightarrow \infty$ as $k\rightarrow \infty$. We easily derive from this and the convergence of $v_k$ to v that

$$\begin{aligned} v_{x,\lambda }\le v\quad \hbox {in}\ {\mathbb {R}}^n{\setminus } B_\lambda (x)\ \ \forall \ \lambda >0. \end{aligned}$$

(43)

The above is equivalent to the property that for every fixed unit vector e, $r^{ \frac{n-2}{2} }v(x+ re)$ is non-decreasing in r. Thus

$$\begin{aligned} r^{n-2} \min _{ \partial B_r(x) }v\ge r^{ \frac{n-2}{2} } \min _{ \partial B_1(x) }v\quad \forall \ r\ge 1. \end{aligned}$$

In particular, $\alpha =\liminf _{ |y|\rightarrow \infty } |y|^{n-2}v(y)=\infty$. This implies, by Step 1, that $\bar{\lambda }(x)=\infty$ for every $x\in {\mathbb {R}}^n$, and therefore (43) holds for every $x\in {\mathbb {R}}^n$. This implies that v is a constant, see Corollary C.1. $\square$

Lemma 2.4

Assume (13)–(15) and (18). Then the function v in Theorem 1.1 cannot be constant.

Proof

Fix some $t > 0$ for the moment. Set $\varphi (x) = v(0) - t\,|x|^2$ and fix some $r > 0$ such that $\varphi > 0$ in $B_{r}(0)$ and $\varphi < v_k$ on $\partial B_r(0)$ for all sufficiently large k. Let

$$\begin{aligned} \gamma _k = \sup _{B_r(0)} (\varphi - v_k) \text { and } \varphi _k = \varphi - \gamma _k. \end{aligned}$$

Then $\varphi _k \le v_k$ in $B_r(0)$ and $\varphi _k(x_k) = v_k(x_k)$ for some $x_k \in \bar{B}_r(0)$. Noting that

$$\begin{aligned} \gamma _k = (\varphi - v)(x_k) - (v_k - v)(x_k) = -t|x_k|^2 - (v_k - v)(x_k) \end{aligned}$$

and

$$\begin{aligned} \lim _{k \rightarrow \infty } \gamma _k = \sup _{B_r(0)} (\varphi - v) = 0 \end{aligned}$$

we deduce that $x_k \rightarrow 0$. This leads to

$$\begin{aligned} \varphi _k(x_k) = v_k(x_k), \nabla \varphi _k(x_k) = \nabla v_k(x_k), \nabla ^2\varphi _k(x_k) \le \nabla ^2 v_k(x_k) \end{aligned}$$

and

$$\begin{aligned} A^{\varphi _k}(x_k) \ge A^{v_k}(x_k). \end{aligned}$$

Noting that there is some $C > 0$ independent of $\delta$ and k such that, for large k,

$$\begin{aligned} A^{\varphi _k}(x_k) \le \Big (\frac{4}{n-2}v(0)^{-\frac{n+2}{n-2}}t + C\delta \Big )I. \end{aligned}$$

Thus, we can select t and $\delta$ such that

$$\begin{aligned} A^{v_k}(x_k) \le A^{\varphi _k}(x_k) \le t_0\,I. \end{aligned}$$

where $t_0$ is the constant in (18). Since $f(\lambda (A^{v_k}(x_k))) = 1$, this contradicts (14), (15) and (18). $\square$

Recall that $0<v\in C^{0,1}_{loc}(\mathbb R^n)$, $\Delta v\le 0\ \hbox {in}\ \mathbb R^n$, and it remains to consider the case that, for every $x\in \mathbb R^n$, there exists $0<\bar{\lambda }(x)<\infty$ such that

$$\begin{aligned} v_{x,\bar{\lambda }(x)}(y)\le v(y),\ \forall \ |y-x|\ge \bar{\lambda }(x), \end{aligned}$$

and

$$\begin{aligned} \lim _{|y|\rightarrow \infty } |y|^{n-2}v_{x,\bar{\lambda }(x)}(y)= \alpha :=\liminf _{|y|\rightarrow \infty }|y|^{n-2}v(y)<\infty . \end{aligned}$$

If v is in $C^2(\mathbb R^n)$, the conclusion of Theorem 1.1 follows from the proof of Theorem 1.3 in [8]. An observation made in [11] easily allows the proof to hold for $v\in C^{0,1}_{loc}(\mathbb R^n)$. For readers’ convenience, we outline the proof below.

Let $\psi (y)=\frac{y}{|y|^2}$. We denote

$$\begin{aligned} v_\psi := |J_\psi |^{ \frac{n-2}{2n}} v\circ \psi . \end{aligned}$$

We know that $\lambda (A^{ v_\psi }(y))=\lambda (A^v (\psi (y))$. Namely, $A^{ v_\psi }(y)$ and $A^v (\psi (y))$ differ only by an orthogonal conjugation.

Introduce

$$\begin{aligned} w^{(x)}:=(v_{x,\bar{\lambda }(x)})_\psi ,\ x\in {\mathbb {R}}^n. \end{aligned}$$

We deduce from the above properties of v that for every $x\in \mathbb R^n$, there exists some $\delta (x)>0$ such that

$$\begin{aligned}&v_\psi \ge w^{(x)}\ \hbox {in}\ B_{\delta (x)}(0){\setminus }\{0\}, \\&w^{(x)}(0)=\alpha =\liminf _{y\rightarrow 0}v_\psi (y),\\&\Delta v_\psi \le 0,\ \hbox {in}\ B_{\delta (x)}(0){\setminus }\{0\}, \end{aligned}$$

Let $D=\{x\in {\mathbb {R}}^n\ |\ v\ \hbox {is differentiable at} \ x\}$. Since $v\in C^{0,1}_{loc}({\mathbb {R}}^n)$, the Lebesgue measure of $\mathbb R^n{\setminus } D$ is 0. It is clear that $w^{(x)}(y)$ is differentiable at $y=0$ if v is differentiable at x.

By [8, Lemma 4.1],

$$\begin{aligned} \nabla w^{(x)}(0)=\nabla w^{(\tilde{x})}(0),\ \ \forall \ x, \tilde{x}\in D. \end{aligned}$$

Namely, for some $V\in {\mathbb {R}}^n$,

$$\begin{aligned} \nabla w^{(x)}(0)=V,\ \forall \ x\in D. \end{aligned}$$

A calculation yields

$$\begin{aligned} \nabla w^{(x)}(0)=(n-2)\alpha x+\alpha ^{ \frac{n}{n-2} } v(x)^{ \frac{n}{n-2} } \nabla v(x). \end{aligned}$$

Thus

$$\begin{aligned} \nabla _x\left( \frac{n-2}{2} \alpha ^{ \frac{n}{n-2} } v(x) ^{ -\frac{2}{n-2} } - \frac{n-2}{2} \alpha |x|^2 +V\cdot x \right) =0,\ \forall \ x\in D. \end{aligned}$$

Consequently, for some $\bar{x}\in {\mathbb {R}}^n$ and $d\in {\mathbb {R}}$,

$$\begin{aligned} v(x) ^{ -\frac{2}{n-2} } \equiv \alpha ^{ -\frac{2}{n-2} } |x-\bar{x}|^2+d \alpha ^{ -\frac{2}{n-2} }. \end{aligned}$$

Since $v>0$, we must have $d>0$, so

$$\begin{aligned} v(x)\equiv \left( \frac{ \alpha ^{ \frac{2}{n-2} } }{d+|x-\bar{x}|^2}\right) ^{ \frac{n-2}{2} }. \end{aligned}$$

We have proved that v is of the form (11) for some $\bar{x}\in {\mathbb {R}}^n$ and some positive constants a and b.

To finish the proof, we show that $f(2b^2a^{-2}, \ldots , 2b^2a^{-2}) = 1$ when (18) and (19) are in effect. For $\delta > 0$, let

$$\begin{aligned} v_\delta (x) = v(x) - \delta |x|^2. \end{aligned}$$

Since $v_k \rightarrow v$ in $C^0(\bar{B}_\delta (0))$, there exists $\beta _k \rightarrow 0$ and $x_k \rightarrow 0$ such that $\hat{v}_k := v_\delta + \beta _k$ satisfies

$$\begin{aligned} (v_k - \hat{v}_k)(x_k) = 0 \text { and } v_k - \hat{v}_k \ge 0 \text { near } x_k. \end{aligned}$$

We have $A^{\hat{v}_k}(x_k) \ge A^{v_k}(x_k)$. Therefore, by (19), $\lambda (A^{\hat{v}_k}(x_k)) \in \Gamma$, and by (15),

$$\begin{aligned} f(\lambda (A^{\hat{v}_k}(x_k))) \ge f(\lambda (A^{v_k}(x_k))) = 1. \end{aligned}$$

(44)

Noting that $A^{\hat{v}_k}(x_k) \rightarrow 2b^2a^{-2}I$ as $\delta \rightarrow 0, k \rightarrow \infty$, we infer that $2b^2a^{-2} > t_0$. (Indeed, if $2b^2a^{-2} \le t_0$, then, for small $\rho > 0$, we have $A^{\hat{v}_k}(x_k) < (t_0 + \rho )I$ for small $\delta$ and large k, which implies, by (44), (14) and (15), that $1 \le f(\lambda (A^{\hat{v}_k}(x_k)) < f(t_0 + \rho , \ldots , t_0 + \rho )$, which contradicts (18).) In view of (19), this implies that $(2b^2a^{-2}, \ldots , 2b^2a^{-2}) \in \Gamma$. We can now send $k \rightarrow \infty$ and then $\delta \rightarrow 0$ in (44) to obtain

$$\begin{aligned} f(\lambda (A^v(0))) \ge 1, \text { i.e. } f(2b^2a^{-2}, \ldots , 2b^2a^{-2}) \ge 1. \end{aligned}$$

Using $v^\delta (x) = v(x) + \delta |x|^2$ instead of $v_\delta$ and the fact that $(2b^2a^{-2}, \ldots , 2b^2a^{-2}) \in \Gamma$, one can easily derive

$$\begin{aligned} f(2b^2a^{-2}, \ldots , 2b^2a^{-2}) \le 1. \end{aligned}$$

Theorem 1.1 is established. $\square$

2.3 Proof of Theorem 1.2

Proof of Theorem 1.2

We start with some preparation as in the proof of Theorem 1.1. We may assume that $R_k \ge 5$ for all k.

By hypotheses, $v_k$ is super-harmonic and positive on ${\mathbb {R}}^n {\setminus } \{0\}$. Therefore, $v_*$ is super-harmonic and non-negative on ${\mathbb {R}}^n {\setminus } \{0\}$. Hence either $v_* \equiv 0$ or $v_* > 0$ in ${\mathbb {R}}^n {\setminus } \{0\}$. In the former case we are done. We assume henceforth that the latter holds.

Now, for every $\beta >2$, there exists some positive constant $C(\beta )$, independent of k, such that $C(\beta )^{-1} \le M_k v_k\le C(\beta )$ in $B_\beta (0) {\setminus } B_{1/\beta }(0)$. It follows from Theorem 2.1 that $|\nabla \ln v_k| \le C'(\beta )$ in $B_{\beta /2}(0) {\setminus } B_{2/\beta }(0)$. It follows, after passing to a subsequence, that for every $0<\alpha <1$, $M_k v_k\rightarrow v_*$ in $C^{\alpha }_{loc}({\mathbb {R}}^n {\setminus } \{0\})$, $v_*\in C^{0,1}_{loc}({\mathbb {R}}^n {\setminus } \{0\})$ and

$$\begin{aligned} |\nabla \ln v_*| \le C'(\beta ) \text { in } B_{\beta /2}(0) {\setminus } B_{2/\beta }(0). \end{aligned}$$

(45)

By the super-harmonicity and the positivity of $v_*$, we can find $c_0 > 0$ such that

$$\begin{aligned} v_*(y)\ge 2c_0 |y|^{2-n} ,\ \ \forall \ |y|\ge 1. \end{aligned}$$

(46)

Hence, passing to a subsequence and shrinking $R_k$ and $c_0 > 0$ if necessary, we can assume without loss of generality that, for all k,

$$\begin{aligned} |M_kv_k(y)-v_*(y)|\le (R_k)^{-n},\quad \ \ \forall \ R_k^{-1}\le |y|\le R_k \end{aligned}$$

(47)

and

$$\begin{aligned} M_kv_k(y) \ge c_0(1+|y|)^{2-n} \ \ \forall 0< |y| < R_k. \end{aligned}$$

(48)

Denote

$$\begin{aligned} (v_k)_{x,\lambda }(y):=\left( \frac{\lambda }{|y-x|}\right) ^{n-2}v_k\left( x+ \frac{\lambda ^2(y-x)}{ |y-x|^2}\right) , \end{aligned}$$

the Kelvin transformation of $v_k$. We use $(v_k)_\lambda$ to denote $(v_k)_{0, \lambda }$.

Lemma 2.5

Under the hypotheses of Theorem 1.2, there exists a function $\lambda ^{(0)}: {\mathbb {R}}^n {\setminus } \{0\} \rightarrow (0,\infty )$ such that $\lambda ^{(0)}(x) \le |x|$ and, for all k,

$$\begin{aligned} (v_k)_{x,\lambda } \le v_k\ \hbox {in}\ B_{R_k}(0){\setminus } (B_\lambda (x) \cup \{0\}), \forall \ 0<\lambda < \lambda ^{(0)}(x)\ \hbox {and}\ |x|\le \frac{R_k}{5}. \end{aligned}$$

Proof

We adapt the proof of Lemma 2.2. For $0< |x| < \frac{R_k}{5}$, we have, by (47) and (48), for all k that

$$\begin{aligned} \frac{1}{c_1(x)}\le M_k v_k\le c_1(x)\ \hbox {in}\ B_{4r_1(x)}(x) \subset B_{R_k}(0), \end{aligned}$$

(49)

where

$$\begin{aligned} r_1(x) = \frac{1}{8}|x| \text { and } c_1(x) = \max \Big \{1 + \sup _{B_{4r_1(x)}(x)} v_*, \frac{1}{c_0}(1 + 2|x|)^{n-2}\Big \}. \end{aligned}$$

By Theorem 2.1 and (49), there exists $c_2(x)>0$, independent of k, such that

$$\begin{aligned} |\nabla \ln v_k|\le c_2(x)\ \hbox {in}\ B_{2r_1(x)}(x). \end{aligned}$$

Thus, by Lemma 2.1, we can find $0< \lambda _1(x) < r_1(x)$ independent of k such that

$$\begin{aligned} (v_k)_{x,\lambda } \le v_k \text { in } B_{\lambda _1(x)}(x) {\setminus } (B_\lambda (x) \cup \{0\}) \text { for all } 0< \lambda < \lambda _1(x). \end{aligned}$$

(50)

For $0< \lambda < \lambda _1(x)$, we have, using (49), that

$$\begin{aligned} (v_k)_{x,\lambda }(y) \le \frac{\lambda ^{n-2}c_1(x)}{|y - x|^{n-2}} \text { for } y \in B_{R_k}(0) {\setminus } (B_{\lambda }(x) \cup \{0\}). \end{aligned}$$

(51)

For $y \in B_{R_k}(0) {\setminus } (B_{1 + 4|x|}(x) \cup \{0\})$, we have $\frac{1}{2}(1 + |y|) \le |y - x|$ and we obtain, using (51) and (48), that

$$\begin{aligned} (v_k)_{x,\lambda }(y) \le \frac{(2\lambda )^{n-2} c_1(x)}{c_0} v_k(y). \end{aligned}$$

(52)

For $y \in B_{1 + 4|x|}(x) {\setminus } (B_{\lambda _1(x)}(x) \cup \{0\})$, we have $1 + |y| \le 2(1+3|x|)$, $|y - x| \ge \lambda _1(x)$ and we obtain, using (51) and (48), that

$$\begin{aligned} (v_k)_{x,\lambda }(y) \le \frac{(2\lambda )^{n-2} c_1(x) (1 + 3|x|)^{n-2}}{c_0 \lambda _1(x)^{n-2}} v_k(y). \end{aligned}$$

(53)

Letting

$$\begin{aligned} \lambda ^{(0)}(x) = \min \Big \{ \lambda _1(x), \frac{\lambda _1(x)}{2(1+3|x|)} \Big [\frac{c_1(x)}{c_0}\Big ]^{\frac{1}{n-2}}\Big \} \le \lambda _1(x), \end{aligned}$$

we see that the conclusion of Lemma 2.5 follows from (50), (52) and (53). $\square$

Define, for $0 < |x|\le R_k/5$, that

$$\begin{aligned} \bar{\lambda }_k(x) =\sup \Big \{0< \mu \le \min (|x|,\frac{R_k}{5})\ |\ (v_k)_{x,\lambda } \le v_k\ \hbox {in}\ B_{R_k}(0){\setminus } (B_\lambda (x) \cup \{0\}), \forall 0<\lambda <\mu \Big \}. \end{aligned}$$

By Lemma 2.5,

$$\begin{aligned} \bar{\lambda }(x):= \displaystyle { \liminf _{k\rightarrow \infty }\bar{\lambda }_k(x) } \in [\lambda ^{(0)}(x), |x|], \qquad x\in {\mathbb {R}}^n {\setminus } \{0\}. \end{aligned}$$

Clearly,

$$\begin{aligned} (v_*)_{x,\bar{\lambda }(x)} \le v_* \text { in } {\mathbb {R}}^n {\setminus } (B_{\bar{\lambda }(x)}(x) \cup \{0\}) \text { for all } x \in {\mathbb {R}}^n {\setminus } \{0\}. \end{aligned}$$

We have a dichotomy:

$$\begin{aligned} \text { either }&\bar{\lambda }(x) = |x| \text { for all }x \in {\mathbb {R}}^n {\setminus } \{0\}, \end{aligned}$$

(54)

$$\begin{aligned} \text { or }&\bar{\lambda }(x_0) < |x_0| \text { for some }x_0 \in {\mathbb {R}}^n {\setminus } \{0\}. \end{aligned}$$

(55)

In case (54), we obtain that $v_*$ is radially symmetric about the origin thanks to Lemma C.1. To finish the proof, we assume in the rest of the argument that (55) holds and derive a contradiction.

We first collect some properties of $\bar{\lambda }(x)$. We start with an analogue of Lemma 2.3. By (46), let

$$\begin{aligned} \alpha := \liminf _{|y|\rightarrow \infty } |y|^{n-2}v_*(y)\in (0, \infty ]. \end{aligned}$$

Lemma 2.6

Under the hypotheses of Theorem 1.2, if $\bar{\lambda }(x)< |x|$ for some $x\in {\mathbb {R}}^n {\setminus } \{0\}$, then

$$\begin{aligned} \alpha = \lim _{|y|\rightarrow \infty } |y|^{n-2}(v_*)_{x,\bar{\lambda }(x)}(y) =\bar{\lambda }(x)^{n-2}v_*(x) <\infty . \end{aligned}$$

Proof

We adapt Step 1 in the proof of Lemma 2.3. Assume that $\bar{\lambda }(x)<|x|$ and (without loss of generality) that $\bar{\lambda }_k(x)\rightarrow \bar{\lambda }(x)$. Arguing as before but using the strong maximum principle for solutions with isolated singularities [10, Theorem 1.6] instead of the standard strong maximum principle, this leads to the existence of some $y_k\in \partial B_{R_k}(0)$ such that

$$\begin{aligned} (v_k)_{x,\bar{\lambda }_k(x)}(y_k)=v_k(y_k). \end{aligned}$$

It follows that

$$\begin{aligned} \lim _{k\rightarrow \infty } |y_k|^{n-2} M_k v_k(y_k) = \lim _{k\rightarrow \infty } |y_k|^{n-2} (M_k v_k)_{x,\bar{\lambda }_k(x)}(y_k) =(\bar{\lambda }(x))^{n-2}v_*(x). \end{aligned}$$

This implies, in view of (47), that

$$\begin{aligned} \alpha \le \lim _{k\rightarrow \infty } |y_k|^{n-2} v_*(y_k)=\bar{\lambda }(x)^{n-2}v_*(x) =\lim _{|y|\rightarrow \infty } |y|^{n-2}(v_*)_{x,\bar{\lambda }(x)}(y)<\infty . \end{aligned}$$

On the other hand, as in the proof of Lemma 2.3, we can use $(v_*)_{ x, \bar{\lambda }(x) }\le v_*$ in ${\mathbb {R}}^n{\setminus } (B_{\bar{\lambda }(x) }(x) \cup \{0\})$ to show that

$$\begin{aligned} \alpha \ge \bar{\lambda }(x)^{n-2}v_*(x). \end{aligned}$$

The conclusion is readily seen. $\square$

Lemma 2.7

Under the hypotheses of Theorem 1.2, if $\bar{\lambda }(x_0) < |x_0|$ for some $x_0 \in {\mathbb {R}}^n {\setminus } \{0\}$, then

$$\begin{aligned} \limsup _{x \rightarrow x_0} \bar{\lambda }(x) \le \bar{\lambda }(x_0). \end{aligned}$$

Proof

Along a subsequence, we have $\bar{\lambda }_k(x_0) \rightarrow \bar{\lambda }(x_0)$.

As in the proof of Lemma 2.6, there exists $y_k\in \partial B_{R_k}(0)$ such that

$$\begin{aligned} (v_k)_{x_0,\bar{\lambda }_k(x_0)}(y_k)=v_k(y_k). \end{aligned}$$

(56)

We know that

$$\begin{aligned} \eta _k := \sup _{B_{|x_0|/2}(x_0)} |M_k v_k - v_*| \rightarrow 0 \text { as } k \rightarrow \infty . \end{aligned}$$

Let m denote the modulus of continuity of $v_*$ in $B_{|x_0|/2}(x_0)$, i.e.

$$\begin{aligned} m(r) = \sup \Big \{ |v_*(x) - v_*(y)|: x, y \in B_{|x_0|/2}(x_0), |x - y| < r\Big \}. \end{aligned}$$

In the computation below, we use o(1) to denote quantities such that

$$\begin{aligned} \lim _{k \rightarrow \infty } o(1) = 0. \end{aligned}$$

Fix some $\delta > 0$ and consider $|x - x_0| < |x_0|/2$. We note that

$$\begin{aligned} \Big |\Big (x + \frac{(\bar{\lambda }_k(x_0) + \delta )^2(y_k - x)}{|y_k - x|^2}\Big ) - \Big (x_0 + \frac{\bar{\lambda }_k(x_0)^2(y_k - x_0)}{|y_k - x_0|^2}\Big )\Big | = |x - x_0| + o(1). \end{aligned}$$

Thus,

$$\begin{aligned} \Big |v_*\Big (x + \frac{(\bar{\lambda }_k(x_0) + \delta )^2(y_k - x)}{|y_k - x|^2}\Big ) - v_*\Big (x_0 + \frac{\bar{\lambda }_k(x_0)^2(y_k - x_0)}{|y_k - x_0|^2}\Big )\Big | \le m(|x - x_0| + o(1)). \end{aligned}$$

It follows that

$$\begin{aligned} M_k (v_k)_{x, \bar{\lambda }_k(x_0) + \delta }(y_k)&= \Big (\frac{\bar{\lambda }_k(x_0) + \delta }{|y_k - x|}\Big )^{n-2} (M_kv_k)\Big (x + \frac{(\bar{\lambda }_k(x_0) + \delta )^2(y_k - x)}{|y_k - x|^2}\Big )\\&\ge \Big (\frac{\bar{\lambda }_k(x_0) + \delta }{|y_k - x|}\Big )^{n-2} \Big [(M_kv_k)\Big (x_0 + \frac{\bar{\lambda }_k(x_0) ^2(y_k - x_0)}{|y_k - x_0|^2}\Big )\\&\qquad \qquad \qquad -2\eta _k - m(|x - x_0| + o(1))\Big ]\\&= \Big (1 + \frac{\delta }{\bar{\lambda }_k(x_0)}\Big )^{n-2} M_k (v_k)_{x_0, \bar{\lambda }_k(x_0)}(y_k)\\&\qquad \qquad - \Big (\frac{\bar{\lambda }_k(x_0) + \delta }{|y_k - x|}\Big )^{n-2}[2\eta _k + m(|x - x_0| + o(1))]. \end{aligned}$$

Recalling (56), we arrive at

$$\begin{aligned} M_k (v_k)_{x, \bar{\lambda }_k(x_0) + \delta }(y_k)&\ge \Big (1 + \frac{\delta }{\bar{\lambda }_k(x_0)}\Big )^{n-2} M_k v_k(y_k)\\&\qquad \qquad - \Big (\frac{\bar{\lambda }_k(x_0) + \delta }{|y_k|}\Big )^{n-2}[o(1)+ m(|x - x_0| + o(1))]. \end{aligned}$$

Thus, in view of (48), we can find small $\bar{\epsilon } > 0$ depending only on $\delta$, c, $\bar{\lambda }(x_0)$ and the function $m(\cdot )$ such that, for all $|x - x_0| < \bar{\epsilon }$ and for large k,

$$\begin{aligned} M_k (v_k)_{x, \bar{\lambda }_k(x_0) + \delta }(y_k)&\ge \Big (1 + \frac{\delta }{4\bar{\lambda }_k(x_0)}\Big )^{n-2} M_k v_k(y_k). \end{aligned}$$

This implies that (cf. (56)), that

$$\begin{aligned} \bar{\lambda }_k(x) \le \bar{\lambda }_k(x_0) + \delta \text { for all } |x - x_0| < \bar{\epsilon } \text { and large } k. \end{aligned}$$

The conclusion follows. $\square$

We now return to drawing a contradiction from (55). By Lemma 2.7, we infer from (55) that there exists some $r_0 > 0$ such that $\bar{\lambda }(x) < |x|$ for all $x \in B_{r_0}(x_0)$. We can then argue as in the proof of Theorem 1.1, using Lemma 2.6 instead of Lemma 2.3 to obtain

$$\begin{aligned} v_*(x) = \Big (\frac{a}{1 + b^2|x - \bar{x}|^2}\Big )^{\frac{n-2}{2}} \qquad x \in B_{r_0}(x_0). \end{aligned}$$

for some $\bar{x} \in {\mathbb {R}}^n$ and some $a, b > 0$. For small $\delta > 0$, let

$$\begin{aligned} v_*^\delta (x) := v_*(x) + \delta |x - x_0|^2. \end{aligned}$$

Since $M_k v_* \rightarrow v_*$ in $C^{0}(\bar{B}_\delta (x_0))$, there exists $\beta _k \rightarrow 0$ and $x_k \rightarrow x_0$ such that the function $\xi _{k,\delta } := v_*^\delta + \beta _k$ satisfies

$$\begin{aligned} (M_k v_k - \xi _{k,\delta })(x_k) = 0 \text { and } M_k v_k - \xi _{k,\delta } \le 0 \text { near } x_k. \end{aligned}$$

It follows that

$$\begin{aligned} A^{v_k}(x_k) \ge A^{\frac{1}{M_k} \xi _{k,\delta }}(x_k) = M_k^{\frac{4}{n-2}} A^{\xi _{k,\delta }}(x_k) . \end{aligned}$$

On the other hand, by hypothesis, there is some $\lambda _* \in \Gamma$ such that $f(\lambda _*) = 1$ (e.g. $\lambda _* = \lambda (A^{v_1}(0))$). By (15), we can find $\hat{\lambda }_* \in \Gamma$ such that $f(\hat{\lambda }_*) > 1$. As $M_k \rightarrow \infty$ and $A^{\xi _{k,\delta }}(x_k) = 2b^{2}a^{-2}I + O(\delta )$, we can find k sufficiently large such that $M_k^{\frac{4}{n-2}} A^{\xi _{k,\delta }}(x_k) > \mathrm{diag}(\hat{\lambda }_*)$. We are thus led to

$$\begin{aligned} A^{v_k}(x_k) > \mathrm{diag}(\hat{\lambda }_*). \end{aligned}$$

As $f(\lambda (A^{v_k})) = 1$ and $f(\hat{\lambda }_*) > 1$, the above contradicts (14) and (15). $\square$

3 Local gradient estimates

In this section, we adapt the argument in [12] to prove Theorem 1.6.

For a locally Lipschitz function w in $B_2(0)$, $0< \alpha < 1$, $x \in B_2(0)$ and $0< \delta < 2 - |x|$, define

$$\begin{aligned}{}[w]_{\alpha ,\delta }(x)&= \sup _{0< |y - x|< \delta } \frac{|w(y) - w(x)|}{|y - x|^\alpha },\\ \delta (w,x,\alpha )&= \left\{ \begin{array}{ll} \infty &{} \text { if } (2 - |x|)^\alpha \,[w]_{\alpha ,2 - |x|}(x)< 1,\\ \mu &{} \text { where } 0 < \mu \le 2 - |x| \text { and } \mu ^\alpha \,[w]_{\alpha ,\mu }(x) = 1\\ &{} \text { if } (2 - |x|)^\alpha \,[w]_{\alpha ,2 - |x|}(x) \ge 1. \end{array}\right. \end{aligned}$$

Note that $\delta (w,x,\alpha )$ is well defined as $[w]_{\alpha ,\delta }(x)$ is continuous and non-decreasing in $\delta$. The object $\delta (w,x,\alpha )$ was introduced in [12]. Its reciprocal $\delta (w,x,\alpha )^{-1}$ plays a role similar to that of $|\nabla w(x)|$ in performing a rescaling argument for a sequence of functions blowing up in $C^\alpha$-norms. For example, when $\delta = \delta (w,x,\alpha ) < \infty$, the rescaled function $\hat{w}(y) := w(x + \delta y) - w(x)$ satisfies

$$\begin{aligned} \hat{w}(0) = 0 \text { and } [\hat{w}]_{\alpha ,1}(0) = \delta ^\alpha [\hat{w}]_{\alpha ,\delta }(x) = 1. \end{aligned}$$

Proof of Theorem 1.6

By the conformal invariance (7), it suffices to show bound $|\nabla \ln v|$ in $B_{1/4}(0)$.

We first claim that

$$\begin{aligned} \sup _{x \ne y \in B_{1/2}(0)} \frac{|\ln v(x) - \ln v(y)|}{|x - y|^\alpha } \le C(\Gamma , \alpha ) \text { for any } 0< \alpha < 1 . \end{aligned}$$

(57)

Assume otherwise that (57) fails for some $0< \alpha < 1$. Then there exist $0 < v_i \in C^2(B_2(0))$ such that $f(\lambda (A^{v_i})) = 1$ and $v_i \le b$ in $B_2(0)$ but

$$\begin{aligned} \sup _{x \ne y \in B_{1/2}(0)} \frac{|\ln v_i(x) - \ln v_i(y)|}{|x - y|^\alpha } \rightarrow \infty . \end{aligned}$$

This implies that, for any fixed $0< r < 1/2$,

$$\begin{aligned} \sup _{x \in B_{1/2}(0)} [\ln v_i]_{\alpha ,r}(x) \rightarrow \infty \text { and } \inf _{x \in B_{1/2}(0)} \delta (\ln v_i, x, \alpha ) \rightarrow 0. \end{aligned}$$

Therefore, there exists $x_i \in B_{1}(0)$,

$$\begin{aligned} \frac{1 - |x_i|}{\delta (\ln v_i, x_i, \alpha )} = \sup _{x \in B_{1}(0)} \frac{1 - |x|}{\delta (\ln v_i, x,\alpha )} \rightarrow \infty . \end{aligned}$$

Let $\sigma _i = \frac{1 - |x_i|}{2}$ and $\epsilon _i = \delta (\ln v_i, x_i, \alpha )$. Then

$$\begin{aligned} \frac{\sigma _i}{\epsilon _i} \rightarrow \infty , \epsilon _i \rightarrow 0, \text { and } \epsilon _i \le 2\,\delta (\ln v_i,z,\alpha ) \text { for any } |z - x_i| \le \sigma _i . \end{aligned}$$

(58)

We now define

$$\begin{aligned} \hat{v}_i(y) = \frac{1}{v_i(x_i)}\,v_i(x_i + \epsilon _i\,y) \text { for } |y| \le \frac{\sigma _i}{\epsilon _i}. \end{aligned}$$

Then

$$\begin{aligned}{}[\ln \hat{v}_i]_{\alpha ,1}(0) = \epsilon _i^\alpha \,[\ln v_i]_{\alpha ,\epsilon _i}(x_i) = 1 . \end{aligned}$$

(59)

Also, by (58), for any fixed $\beta > 1$ and $|y| < \beta$, there holds

$$\begin{aligned}{}[\ln \hat{v}_i]_{\alpha ,1}(y)&= \epsilon _i^\alpha \,[\ln v_i]_{\alpha ,\epsilon _i}(x_i + \epsilon _i\,y)\nonumber \\&\le 2^{-\alpha }\,\epsilon _i^\alpha \Big \{ \sup _{|z - (x_i + \epsilon _i y)| \le \epsilon _i} [\ln v_i]_{\alpha ,\epsilon _i/4}(z) + [\ln v_i]_{\alpha ,\epsilon _i/4}(x_i + \epsilon _i\,y)\Big \}\nonumber \\&\le \sup _{|z - (x_i + \epsilon _i y)| \le \epsilon _i} \delta (\ln v_i,z,\alpha )^\alpha \,[\ln v_i]_{\alpha ,\delta (\ln v_i,z,\alpha )}(z)\nonumber \\&\qquad \qquad + \delta (\ln v_i,x_i + \epsilon _i\,y,\alpha )^\alpha \,[\ln v_i]_{\alpha ,\delta (\ln v_i,x_i + \epsilon _i\,y,\alpha )}(x_i + \epsilon _i\,y)\nonumber \\&= 2 \end{aligned}$$

(60)

for all sufficiently large i. Since $\hat{v}_i(0) = 1$ by definition, we deduce from (59) and (60) that

$$\begin{aligned} \frac{1}{C(\beta )} \le \hat{v}_i(y) \le C(\beta ) \text { for } |y| \le \beta \text { and all sufficiently large }i . \end{aligned}$$

(61)

We can now apply Theorem 2.1 to obtain

$$\begin{aligned} |\nabla \ln \hat{v}_i| \le C(\beta ) \text { in } B_{\beta /2}(0) \text { for all sufficiently large }i. \end{aligned}$$

(62)

Passing to a subsequence and recalling (58) and (61), we see that $\hat{v}_i$ converges in $C^{0,\alpha '}$ ($\alpha< \alpha ' < 1$) on compact subsets of ${\mathbb {R}}^n$ to some positive, locally Lipschitz function $v_*$.

On the other hand, if we define

$$\begin{aligned} \bar{v}_i(y) = \epsilon _i^{\frac{n-2}{2}}\,v_i(x_i + \epsilon _i\,y) \text { for } |y| \le \frac{\sigma _i}{\epsilon _i}, \end{aligned}$$

then by the conformal invariance (7), we have

$$\begin{aligned} f(\lambda (A^{\bar{v}_i})) = 1 \text { in } B_{\sigma _i/\epsilon _i}(0). \end{aligned}$$

Since $\frac{\sigma _i}{\epsilon _i} \rightarrow \infty$, $\hat{v}_i = M_i\,\bar{v}_i$ where $M_i = v_i(x_i)^{-1} \epsilon _i^{-\frac{n-2}{2}} \rightarrow \infty$ (thanks to the bound $v_i \le b$), we then conclude from Theorem 1.2 that $v_*$ is constant, namely

$$\begin{aligned} v_* \equiv v_*(0) = \lim _{i \rightarrow \infty } \hat{v}_i(0) = 1. \end{aligned}$$

This contradicts (59), in view of (62) and the convergence of $\hat{v}_i$ to $v_*$. We have proved (57).

From (57), we can find some universal constant $C > 1$ such that

$$\begin{aligned} \frac{u(0)}{C} \le u \le C\,u(0) \text { in } B_{1/2}(0). \end{aligned}$$

Applying Theorem 2.1 again we obtain the required gradient estimate in $B_{1/4}(0)$. $\square$

4 Fine blow-up analysis

4.1 A quantitative centered Liouville-type result

In this subsection, we establish:

Proposition 4.1

Let $(f, \Gamma )$ satisfy (13)–(15), (18)–(19), (5) and the normalization condition (21). Assume that for a sequence $R_k\rightarrow \infty$, $0 < v_k \in C^2(B_{R_k})$ satisfy

$$\begin{aligned} f(\lambda (A^{v_k}))(y)=1,\ \ 0<v_k(y)\le v_k(0)=1, \quad |y|\le R_k. \end{aligned}$$

(63)

Then for every $\epsilon >0$, there exists a constant ${\delta _{0}}> 0$, depending only on $(f,\Gamma )$ and $\epsilon$, such that, for all sufficiently large k,

$$\begin{aligned} |v_k(y)-U(y)|\le 2 \epsilon U(y),\qquad \forall \ |y|\le {\delta _{0}}R_k. \end{aligned}$$

(64)

Recall that $U = (1 + |x|^2)^{-\frac{n-2}{2}}$, $A^U \equiv 2I$ and $f(\lambda (A^U)) = 1$ on ${\mathbb {R}}^n$.

Proposition 4.1 is equivalent to the following proposition.

Proposition 4.2

Let $(f,\Gamma )$ satisfy (13)–(15), (18)–(19), (5) and the normalization condition (21). For any $\epsilon > 0$ there exist ${\delta _{0}}, {C_{0}}> 0$ depending only on $(f,\Gamma )$ and $\epsilon$ such that if $0 < u \in C^2(B_{R}(0))$, $R > 0$, satisfies

$$\begin{aligned} f(\lambda (A^u)) = 1 \text { in } B_{R}(0) \text { and } u(0) = \sup _{B_R(0)} u \ge {C_{0}}\, R^{-\frac{n-2}{2}}, \end{aligned}$$

then

$$\begin{aligned} |u(x) - U^{0,u(0)}(x)| \le 2\epsilon U^{0,u(0)}(x) \text { for all } x \in B_{{\delta _{0}}R}(0). \end{aligned}$$

Proof of the equivalence between Propositions 4.1 and 4.2

It is clear that Proposition 4.2 implies Proposition 4.1.

Consider the converse. Let ${\delta _{0}}= {\delta _{0}}(\epsilon )$ be as in Proposition 4.1. Arguing by contradiction, we assume that there are some $\epsilon > 0$ and a sequence of $R_k$ and $u_k \in C^2(B_{R_k}(0))$ such that

$$\begin{aligned} f(\lambda (A^{u_k})) = 1 \text { in } B_{R_k}(0) \text { and } u_k(0) = \sup _{B_{R_k}(0)} u_k \ge k\,R_k^{-\frac{n-2}{2}} \end{aligned}$$

but the last estimate in Proposition 4.2 fails for each k.

Define

$$\begin{aligned} \bar{u}_k(y) = \frac{1}{u_k(0)} u_k\Big (\frac{y}{u_k(0)^{\frac{2}{n-2}}}\Big ) \text { for } |y| \le R_k\,u_k(0)^{\frac{2}{n-2}} =: \bar{R}_k. \end{aligned}$$

Then $f(\lambda (A^{\bar{u}_k})) = 1$ in $B_{\bar{R}_k}(0)$, $\sup _{B_{\bar{R}_k}(0)} \bar{u}_k = \bar{u}_k(0) = 1$, and $\bar{R}_k \ge k^{\frac{2}{n-2}} \rightarrow \infty$. By Proposition 4.1,

$$\begin{aligned} |\bar{u}_k(y) - U(y)| \le 2\epsilon U(y) \text { in } B_{{\delta _{0}}\,\bar{R}_k}(0) \text { for all sufficiently large }k. \end{aligned}$$

Returning to the original sequence $u_k$ we arrive at a contradiction. $\square$

Lemma 4.1

Under the hypotheses of Proposition 4.1 except for (5), we have

$$\begin{aligned} v_k\rightarrow U,\qquad \hbox {in}\ C^\beta _{loc}({\mathbb {R}}^n),\ \ \forall \ 0<\beta <1. \end{aligned}$$

(65)

Moreover, for every $\epsilon >0$, there exists $k_0\ge 1$ such that

$$\begin{aligned} \min _{|y|=r}v_k(y)\le (1+\epsilon )U(r),\qquad \forall \ 0<r< R_k/5, \ k\ge k_0. \end{aligned}$$

(66)

Proof

We first prove (65). Since $v_k$ satisfies (63), we deduce from Theorem 1.6 that

$$\begin{aligned} |\nabla \ln v_k|\le C\quad \hbox {in}\ B_{ R_k-1},\ \ \forall \ k, \end{aligned}$$

where C is independent of k. This yields (65) in view of Theorem 1.1.

We now prove (66). Suppose the contrary, then there exists some $\epsilon >0$ and sequences of $k_i\rightarrow \infty$, $0<r_i<R_{k_i}/5$ such that

$$\begin{aligned} v_{k_i}> (1+\epsilon )U\ \ \ \hbox {on}\ \partial B_{ r_i}. \end{aligned}$$

(67)

Because of (65), $r_i\rightarrow \infty$.

As in the proof of Lemma 2.2, there exists $\lambda ^{(0)}_i>0$ such that

$$\begin{aligned} (v_{k_i})_\lambda \le v_{k_i}\ \hbox {in}\ B_{r_i}{\setminus } B_\lambda , \forall 0<\lambda < \lambda ^{(0)}_i\ \hbox {and}\ |x|\le r_i. \end{aligned}$$

(68)

By the explicit expression of U, there exists some small $\delta >0$ independent of i such that, for large i,

$$\begin{aligned} U_\lambda (y)\le (1+\frac{\epsilon }{4})U(y),\quad \forall \ y\in \partial B_{ r_i},\ \lambda ^{(0)}_i \le \lambda \le 1+\delta , \end{aligned}$$

By the uniform convergence of $v_{k_i}$ to U on compact subsets of ${\mathbb {R}}^n$, we have, for large i,

$$\begin{aligned} (v_{k_i})_\lambda \le (1+\frac{\epsilon }{2}) U(y),\quad \forall \ y\in \partial B_{ r_i},\ \lambda ^{(0)}_i \le \lambda \le 1+\delta , \end{aligned}$$

As in the proof of Lemma 2.2, the moving sphere procedure does not stop before reaching $\lambda =1+\delta$, namely we have, for large i,

$$\begin{aligned} (v_{k_i})_\lambda \le v_{k_i}\ \hbox {in}\ B_{r_i}{\setminus } B_\lambda , \forall 0<\lambda < 1+\delta \ \hbox {and}\ |x|\le r_i. \end{aligned}$$

Sending i to $\infty$ leads to

$$\begin{aligned} U_{1+\delta }(y)\le U(y),\ \ \forall \ 1+\delta \le |y|\le 2. \end{aligned}$$

A contradiction—since we see from the explicit expression of U that $U_{1+\delta }(y)>U(y)$ for all $1<1+\delta < |y|\le 2$. $\square$

Lemma 4.2

Under the hypotheses of Proposition 4.1, for any $\epsilon > 0$, there exist a small $\delta _1>0$ and a large $r_1 > 1$, depending only on $(f, \Gamma )$ and $\epsilon$, such that, for all sufficiently large k,

$$\begin{aligned}&v_k(y) \ge (1-\epsilon )U(y), \qquad \forall \ |y|\le \delta _1 R_k, \end{aligned}$$

(69)

$$\begin{aligned} \text { and }\qquad&\int _{ r_1\le |y|\le \delta _1 R_k } v_k^{\frac{n+2}{n-2}} \le \epsilon . \end{aligned}$$

(70)

Proof

Assume without loss of generality that $\epsilon \in (0,1/2)$. Since $v_k\rightarrow U$ in $C^0_{loc}({\mathbb {R}}^n)$, there exist $r_2>1$ and $k_1$, depending on $\epsilon$, such that for all $k\ge k_1$

$$\begin{aligned} v_k(y)&\ge (1-\epsilon ^2)U(y),\qquad \forall \ |y|\le r_2, \end{aligned}$$

(71)

$$\begin{aligned} v_k(y)&\ge (1-\epsilon ^2)U(r_2)\ge (1-2\epsilon ^2) r_2^{2-n}, \qquad \forall \ |y|=r_2. \end{aligned}$$

(72)

By (5),

$$\begin{aligned} Trace\ (A^{v_k})\ge \delta > 0, \end{aligned}$$

and therefore

$$\begin{aligned} -\Delta v_k(y)\ge \frac{n-2}{2} \delta v_k(y) ^{ \frac{n+2}{n-2} } \qquad \hbox {in}\ r_2\le |y| \le R_k. \end{aligned}$$

(73)

Using the superharmonicity of $v_k$ and the maximum principle, we obtain

$$\begin{aligned} v_k(y)\ge (1-\epsilon ^2) \left( |y|^{2-n}- R_k^{2-n}\right) ,\qquad r_2\le |y|\le R_k. \end{aligned}$$

Thus, for any $\delta _2 \in (0,\epsilon ^{\frac{2}{n-2}})$, we have for all sufficiently large k that

$$\begin{aligned} v_k(y)\ge (1-\epsilon ^2)(1 - \delta _2^{n-2}) |y|^{2-n} \ge (1-2\epsilon ^2) |y|^{2-n},\qquad r_2\le |y|\le \delta _2 R_k. \end{aligned}$$

(74)

Now if $\delta _1 < \delta _2$, (69) is readily seen from (71) and (74).

Let

$$\begin{aligned} \hat{v}_k(y):= v_k(y)- (1-2\epsilon ^2) |y|^{2-n}. \end{aligned}$$

Then

$$\begin{aligned} -\Delta \hat{v}_k(y)\ge \hat{f}(y):= \frac{n-2}{2} \delta v_k(y) ^{ \frac{n+2}{n-2} } \qquad \hbox {in}\ r_2\le |y| \le \delta _2 R_k, \end{aligned}$$

and

$$\begin{aligned} \hat{v}_k(y)\ge 0,\quad \hbox {for}\ y\in \partial (B_{\delta _2 R_k}{\setminus } B_{ r_2}). \end{aligned}$$

Let $R_k' = \frac{\delta _2R_k}{2}$. Enlarging $k_1$ if necessary, we can apply Corollary A.3 in “Appendix A” to get

$$\begin{aligned} \min _{ |x|= R_k'} \hat{v}_k(x) \ge C^{-1} (\delta _2 R_k)^{2-n} \int _{ 2r_2\le |y|\le \delta _2 R_k/8} \delta v_k(y) ^{ \frac{n+2}{n-2} } dy, \qquad \forall \ k \ge k_1, \end{aligned}$$

(75)

where here and below C is some positive constant depending only on n. On the other hand, by Lemma 4.1, we have (after enlarging $k_1$ if necessary)

$$\begin{aligned} \min _{ |x|= R_k'} v_k(x) \le (1+\epsilon ^2) U(R_k') \le (1+2\epsilon ^2)(R_k')^{2-n}, \qquad \forall \ k \ge k_1, \end{aligned}$$

which implies that

$$\begin{aligned} \min _{ |x|= R_k'} \hat{v}_k(x) \le C\,\epsilon ^2 (\delta _2 R_k)^{2-n}, \qquad \forall \ k \ge k_1. \end{aligned}$$

(76)

It now follows from (75) and (76) that

$$\begin{aligned} \int _{ 2r_2\le |y|\le \delta _2 R_k/8} \delta v_k(y) ^{ \frac{n+2}{n-2} } dy \le c_1\,\epsilon ^2 \end{aligned}$$

where $c_1$ depends only on n. (70) is then established for $\epsilon \le \frac{1}{c_1}$ with $r_1 = 2r_2$ and $\delta _1 = \delta _2/8$. The conclusion for $\epsilon > 1/c_1$ also follows. $\square$

Lemma 4.3

Let $(f, \Gamma )$ satisfy (13)–(15). Then there exist $\delta _3>0$ and $C_3>1$, depending only on $(f, \Gamma )$, such that if $u\in C^2(B_2(0))$ satisfies

$$\begin{aligned} f(\lambda (A^u))=1, u>0, \qquad \hbox {in}\ B_2(0), \end{aligned}$$

and

$$\begin{aligned} \int _{ B_2(0)} u^{ \frac{2n}{n-2} }\le \delta _3, \end{aligned}$$

then

$$\begin{aligned} u\le C_3\quad \hbox {in}\ B_1(0). \end{aligned}$$

If $(f,\Gamma )$ satisfies in addition the conditions (18), (19) and the normalization condition (21), then $\delta _3$ can be chosen to be any constant smaller than $\int _{{\mathbb {R}}^n} U^{\frac{2n}{n-2}}\,dx$.

Proof

We adapt the proof of [7, Lemma 6.4]. Arguing by contradiction, we can find a sequence of $0 < u_j \in C^2(B_2)$ such that $f(\lambda (A^{u_j})) = 1$ in $B_2(0)$,

$$\begin{aligned} \int _{B_2(0)} u_j ^{ \frac{2n}{n-2} } \rightarrow 0 \end{aligned}$$

but

$$\begin{aligned} d(y_j)^{\frac{n-2}{2}}u_j(y_j) = \max _{\bar{B}_{3/2}(0)} d(y)^{\frac{n-2}{2}}u_j(y) \rightarrow \infty , \end{aligned}$$

where $y_j \in B_{3/2}(0)$ and $d(y) = 3/2 - |y|$.

Let $\sigma _j = \frac{1}{2}d(y_j) > 0$,

$$\begin{aligned} v_j(z) = \frac{1}{u_j(y_j)}u_j\Big (y_j + \frac{1}{u_j(y_j)^{\frac{2}{n-2}}} z\Big ) \text { for } |z| < r_j := u_j(y_j)^{\frac{2}{n-2}} \sigma _j \rightarrow \infty . \end{aligned}$$

Then by the conformal invariance property (7), $f(\lambda (A^{v_j})) = 1$ in $B_{r_j}(0)$, $v_j(0) = 1$, $v_j \le 2^{\frac{n-2}{2}}$ in $B_{r_j}(0)$ and

$$\begin{aligned} \int _{B_{r_j}(0)} v_j^{\frac{2n}{n-2}} \rightarrow 0. \end{aligned}$$

(77)

By Theorem 1.6, there is a constant C independent of j such that

$$\begin{aligned} |\nabla \ln v_j| \le C \text { in } B_{r_j/2}(0). \end{aligned}$$

Thus, after passing to a subsequence, we can assume that $v_j$ converges in $C^0_{loc}({\mathbb {R}}^n)$ to some positive function v (as $v_j(0) = 1$). This contradicts (77).

The above argument can be adapted to prove the last assertion of the lemma: Eq. (77) is replaced by

$$\begin{aligned} \int _{B_{r_j}(0)} v_j^{\frac{2n}{n-2}} \le \delta _3 < \int _{{\mathbb {R}}^n} U^{\frac{2n}{n-2}}\,dx. \end{aligned}$$

On the other hand, by Theorem 1.1, we have $v_j \rightarrow U$ in $C^0_{loc}({\mathbb {R}}^n)$. This gives a contradiction.

$\square$

Lemma 4.4

Let $(f, \Gamma )$ satisfy (13)–(15) and let $\delta _3$, $C_3$ be as in Lemma 4.3. If $u\in C^2(B_{2R}(0))$ satisfies

$$\begin{aligned} f(\lambda (A^u))=1, u>0, \qquad \hbox {in}\ B_{2R}(0), \end{aligned}$$

and

$$\begin{aligned} \int _{ B_{2R(0)}} u^{ \frac{2n}{n-2} }\le \delta _3, \end{aligned}$$

then

$$\begin{aligned} u\le C_3\,R^{-\frac{n-2}{2}}\quad \hbox {in}\ B_{R}(0). \end{aligned}$$

Proof

This follows from Lemma 4.3 and a change of variables, $\tilde{u}(y) = R^{\frac{n-2}{2}}u(Ry)$ for $|y| \le 2$. $\square$

Lemma 4.5

Under the hypotheses of Proposition 4.1, there exist positive constants $\delta _4 > 0$ and $C_4 > 1$, depending only on $(f,\Gamma )$, such that, for all sufficiently large k,

$$\begin{aligned} v_k(y)\le C_4 U(y), \qquad \forall \ |y|\le \delta _4 R_k. \end{aligned}$$

(78)

Proof

Let $\delta _3$ be as in Lemma 4.3. Since $v_k \le 1$, we deduce from Lemma 4.2, there is $r_1 > 1$ and $\delta _1 > 0$ such that

$$\begin{aligned} \int _{ r_1\le |y|\le \delta _1 R_k} v_k^{ \frac{2n}{n-2} } \le \epsilon . \end{aligned}$$

(79)

For any $2r_1<r< \delta _1 R_k/2$, consider

$$\begin{aligned} \tilde{v}_k(z)= r^{ \frac{n-2}{2} } v_k(rz),\quad \frac{1}{2}<|z|<2. \end{aligned}$$

By (79), we have, for large k,

$$\begin{aligned} \int _{ \frac{1}{2}<|z|<2 } \tilde{v}_k(z)^{ \frac{2n}{n-2} } =\int _{ \frac{r}{2}<|\eta |< 2r} v_k(\eta )^{ \frac{2n}{n-2} } \le \delta _3. \end{aligned}$$

It follows from Lemma 4.4 that

$$\begin{aligned} \tilde{v}_k(z)\le C \qquad \forall \ \frac{2}{3}<|z|< \frac{5}{4}, \end{aligned}$$

for some universal constant C. Since $\tilde{v}_k$ also satisfies $f(\lambda (A^{\tilde{v}_k})) = 1$, we can apply Theorem 1.6 to obtain

$$\begin{aligned} |\nabla \ln \tilde{v}_k(z)| \le C \qquad \forall \ |z| = 1, \end{aligned}$$

which implies that $\max _{|z| = 1} \tilde{v}_k \le C\,\min _{\partial B_1}\tilde{v}_k$. Returning to $v_k$, we obtain

$$\begin{aligned} \max _{\partial B_r} v_k \le C\,\min _{\partial B_r} v_k \end{aligned}$$

where C is universal. The conclusion then follows from Lemma 4.1. $\square$

Proof of Proposition 4.1

Fix $\epsilon > 0$. In view of Lemma 4.2 (cf. (69)), we only need to prove that there exist ${\delta _{0}}> 0$ such that, for all sufficiently large k,

$$\begin{aligned} v_k(y)\le (1+2\epsilon )U(y), \qquad \forall \ |y|\le {\delta _{0}}R_k. \end{aligned}$$

(80)

Suppose the contrary of the above, then, after passing to a subsequence and renaming the subsequence still as $\{v_k\}$ and $\{R_k\}$, there exist $|y_k|=\delta _k R_k$, $\delta _k\rightarrow 0^+$, such that

$$\begin{aligned} v_k(y_k)=\max _{ |y|=\delta _k R_k} v_k(y)> (1+2\epsilon ) U(y_k). \end{aligned}$$

(81)

In view of the convergence of $v_k$ to U, $|y_k|\rightarrow \infty$ as $k\rightarrow \infty$.

Consider the following two rescalings of $v_k$:

$$\begin{aligned} \hat{v}_k(z):= |y_k|^{n-2} v_k(|y_k|z) \text { and } \bar{v}_k(z) = |y_k|^{\frac{n-2}{2}}v_k(|y_k|z),\qquad |z|<\frac{ R_k}{ |y_k|}\rightarrow \infty . \end{aligned}$$

(82)

By Lemma 4.5, we have

$$\begin{aligned} \hat{v}_k(z)\le C|z|^{2-n} \text { and } \bar{v}_k(z) \le C\,|y_k|^{-\frac{n-2}{2}} |z|^{2-n} \end{aligned}$$

(83)

for some constant C independent of k.

In view of the conformal invariance (7) and (63),

$$\begin{aligned} f(\lambda (A^{\bar{v}_k}(z))) = 1 \text { for } |z|< \frac{ R_k}{ |y_k|}. \end{aligned}$$

Recalling (83), we can apply Theorem 1.6 to obtain that for all $0<\alpha<\beta <\infty$, there exists positive constant $C(\alpha , \beta )$ such that for large k,

$$\begin{aligned} |\nabla \ln \bar{v}_k(z)|\le C(\alpha , \beta ),\qquad \forall \ \alpha<|z|<\beta , \end{aligned}$$

(84)

which implies that

$$\begin{aligned} |\nabla \ln \hat{v}_k(z)|\le C(\alpha , \beta ),\qquad \forall \ \alpha<|z|<\beta . \end{aligned}$$

(85)

We know from (82), (81) and Lemma 4.1 that

$$\begin{aligned} \min _{ |z|=1} \hat{v}_k(z)\le (1+ \epsilon )\frac{|y_k|^{n-2}}{U(y_k)}, \end{aligned}$$

(86)

and

$$\begin{aligned} \max _{ |z|=1} \hat{v}_k(z)\ge (1+ 2\epsilon )\frac{|y_k|^{n-2}}{U(y_k)}. \end{aligned}$$

(87)

We deduce from (85), (86) and (87), after passing to a subsequence, that for some positive function $\hat{v}^*$ in $C^{0,1}_{loc}({\mathbb {R}}^n{\setminus }\{0\})$,

$$\begin{aligned} \hat{v}_k\rightarrow \hat{v}^*\qquad \hbox {in}\ C^{\alpha }_{loc}({\mathbb {R}}^n{\setminus } \{0\}),\ \forall \ 0<\alpha <1. \end{aligned}$$

(88)

By Theorem 1.2, $\hat{v}^*$ is radially symmetric. On the other hand, we deduce from (86) and (87) after passing to limit that

$$\begin{aligned} \min _{ |z|=1} \hat{v}^*(z)\le 1+\epsilon , \quad \hbox {and}\quad \max _{ |z|=1} \hat{v}^*(z)\ge 1+2\epsilon . \end{aligned}$$

(89)

The above violates the radial symmetry of $\hat{v}^*$. Proposition 4.1 is established. $\square$

4.2 Detailed blow-up landscape

The proof of Theorem 1.3 uses the following consequence of the Harnack-type inequality for conformally invariant equations, see [4, 7, 16].

Lemma 4.6

Let $(f,\Gamma )$ satisfy (13)–(15) and (5). There exists a constant $C_6$, depending only on $(f,\Gamma )$, such that if $u \in C^2(B_{3}(0))$ is a positive solution of

$$\begin{aligned} f(\lambda (A^u)) = 1 \text { in } B_{3}(0) \end{aligned}$$

then

$$\begin{aligned} \int _{B_1(0)} |u|^{\frac{2n}{n-2}}\,dx \le C_6. \end{aligned}$$

Proof

We give the proof here for completeness. By (5),

$$\begin{aligned} -\Delta u \ge \frac{n-2}{2}\delta \,u^{\frac{n+2}{n-2}} > 0 \text { in } B_2(0). \end{aligned}$$

Thus, by Corollary A.2 in “Appendix A” as well as the maximum principle,

$$\begin{aligned} \inf _{B_{2}(0)} u = \inf _{B_{2}(0) {\setminus } B_{3/2}(0)} u \ge \frac{1}{C} \int _{B_1(0)} u^{\frac{n+2}{n-2}}\,dx. \end{aligned}$$

It follows that

$$\begin{aligned} \int _{B_1(0)} u^{\frac{2n}{n-2}}\,dx \le C\sup _{B_1(0)} u \,\inf _{B_{2}(0)} u. \end{aligned}$$

The conclusion follows from the above estimate and the Harnack-type inequality [8, Theorem 1.2]. (Note that (5) is used again here.) $\square$

Proof of Theorem 1.3

In view of Proposition 4.2 and (vi), it suffices to establish the theorem for $\epsilon = \epsilon _0 := 1/2$.

By Lemma 4.6,

$$\begin{aligned} \int _{B_2(0)} u^{\frac{2n}{n-2}} \,dx \le C_6. \end{aligned}$$

(90)

The constant $\bar{m}$ in the result can be selected to be the least integer satisfying

$$\begin{aligned} \bar{m} \ge 2C_6 \Big (\int _{B_1} U^{\frac{2n}{n-2}}\,dx\Big )^{-1}. \end{aligned}$$

(Clearly, this is an obvious upper bound for m if the $x^i$’s satisfies (iii).)

Let $\delta _3$ and $C_3$ be the constants in Lemma 4.4. Fix some $N_0 > \frac{C_1}{\delta _3}$. Then there is some $r_0 \in (3/2, 2)$ such that

$$\begin{aligned} \int _{r_0< |x| < r_0 + \frac{1}{2N_0} } u^{\frac{2n}{n-2}}\,dx \le \delta _3. \end{aligned}$$

By Lemma 4.4, this implies that

$$\begin{aligned} u(x) \le C_3\,(8N_0)^{\frac{n-2}{2}} =: C_{7} \text { for all } r_0 + \frac{1}{8N_0}< |x| < r_0 + \frac{3}{8N_0}. \end{aligned}$$

(91)

Let ${C_{0}}$ and ${\delta _{0}}$ be as in Proposition 4.2 (corresponding to $\epsilon = \epsilon _0$). We can assume without loss of generality that

$$\begin{aligned} {C_{0}}> 2 \text { and }{\delta _{0}}< 1. \end{aligned}$$

(92)

We now declare

$$\begin{aligned} {C_*}= \max \Big (2C_7, {C_{0}}(2{\delta _{0}}^{-1})^{\frac{\bar{m}(n-2)}{2}}(4N_0)^{-\frac{n-2}{2}}\Big ). \end{aligned}$$

(93)

This choice of ${C_*}$ will become clear momentarily.

Let $U_1 = B_{r_0 + \frac{3}{8N_0}}(0)$ and $V_1 = B_{r_0 + \frac{1}{8N_0}}(0) \subset U_0$. By (93), ${C_*}\ge 2C_7$, and so, by (91), there is some $x^1 \in V_1 \subset B_2(0)$ such that

$$\begin{aligned} u(x^1) = \sup _{U_1} u \ge {C_*}. \end{aligned}$$

Let $R_1 = \frac{1}{4N_0}$, then (93) gives

$$\begin{aligned} {C_*}\ge {C_{0}}\,R_1^{-\frac{n-2}{2}}. \end{aligned}$$

Hence, an application of Proposition 4.2 to u on the ball $B_{R_1}(x^1)$ leads to

$$\begin{aligned} |u(x) - U^{x^1, u(x^1)}(x)| \le \epsilon _0\,U^{x^1, u(x^1)}(x), \ \ \forall \ x\in B_{{\delta _{0}}R_1}(x^1). \end{aligned}$$

In particular, for $\frac{{\delta _{0}}R_1}{2} \le |x - x^1| \le {\delta _{0}}\,R_1$,

$$\begin{aligned} u(x) \le 2U^{x^1, u(x^1)}(x) \le \frac{2}{u(x^1) |x - x^1|^{n-2}} \le \frac{2^{n-1}}{{C_*}({\delta _{0}}R_1)^{n-2}} \le \frac{{C_*}}{2}, \end{aligned}$$

(94)

where we have used (93) in the last estimate.

Let $U_2 = U_1 {\setminus } B_{{\delta _{0}}\,R_1/2}(x^1)$ and $V_2 = V_1 {\setminus } B_{{\delta _{0}}R_1}(x^1) \subset U_1$. If

$$\begin{aligned} \sup _{U_2} u \le {C_*}, \end{aligned}$$

we stop. Otherwise, in view of (94), there is some $x^2 \in V_2$ such that

$$\begin{aligned} u(x^2) = \sup _{U_2} u \ge {C_*}. \end{aligned}$$

We then let $R_2 = \frac{{\delta _{0}}R_1}{2}$ so that (93) implies

$$\begin{aligned} {C_*}\ge {C_{0}}\,R_2^{-\frac{n-2}{2}}. \end{aligned}$$

Hence, by Proposition 4.2,

$$\begin{aligned} |u(x) - U^{x^2, u(x^2)}(x)| \le \epsilon _0 U^{x^2, u(x^2)}(x), \ \ \forall \ x\in B_{{\delta _{0}}R_2}(x^2). \end{aligned}$$

We then repeat the above process to define $U_3$, $V_3$, and to decide if a local max $x^3$ can be selected in $V_3$, etc. As explain above, the number m of times this process can be repeated cannot exceed $\bar{m}$.

We have obtained the set of local maximum points $\{x^1, \ldots , x^m\}$ of u and have verified (i) and (iv) for

$$\begin{aligned} {\delta _*}= \Big (\frac{{\delta _{0}}}{2}\Big )^{\bar{m}} \frac{1}{2N_0} \le {\delta _{0}}\,R_{m}. \end{aligned}$$

(vi) is readily seen as

$$\begin{aligned} dist(x^i, \partial U_i) \ge R_i \ge {\delta _*}. \end{aligned}$$

(ii) is also clear for

$$\begin{aligned} {K}\ge \Big (\frac{2}{{\delta _{0}}}\Big )^{\bar{m}} 4N_0 \ge \frac{2}{{\delta _{0}}\,R_{m-1}}. \end{aligned}$$

From construction, we have

$$\begin{aligned} \sup _{U_{m+1}} u \le {C_*}. \end{aligned}$$

By Theorem 1.6, this implies that

$$\begin{aligned} |\nabla \ln u(x)| \le C_8 \text { for all } x \in V_m = V_0 {\setminus } \cup _{i = 1}^m B_{{\delta _{0}}R_i}(x^i). \end{aligned}$$

(95)

Also, note that, for ${\delta _*}< |x - x^i| < {\delta _{0}}R_i$, we have

$$\begin{aligned} \frac{1}{u(x^i)} \left( \frac{1}{({C_*})^{-\frac{4}{n-2}} + ({\delta _{0}}\,R_i)^2.}\right) ^{\frac{n-2}{2}} \le U^{x^i, u(x^i)}(x) \le \frac{1}{u(x^i)} ({\delta _*})^{-(n-2)} , \end{aligned}$$

and so

$$\begin{aligned} \frac{1}{C_9u(x^i)} \le u(x) \le \frac{C_9}{u(x^i)} ({C_*})^{-2}\,({\delta _*})^{-(n-2)} \end{aligned}$$

(96)

It is now clear that (iii) and (v) hold for ${K}$ sufficiently large. The proof is complete. $\square$

5 A quantitative Liouville theorem

Proof of Theorem 1.5

Assume by contradiction that, for some $\epsilon \in (0,1/2]$, there exist $v_k \in C^2(B_{3R_k}(0)$, $R_k \rightarrow \infty$, such that $f(\lambda (A^{v_k})) = 1$ in $B_{3R_k}(0)$ and $v_k \ge \gamma$ in $B_{r_1}(0)$ but, for each k,

$$\begin{aligned} (24)\text { and }(25)\text { cannot simultaneously hold for any }\bar{x}. \end{aligned}$$

(97)

Define

$$\begin{aligned} u_k(y) = R_k^{\frac{n-2}{2}}v_k(R_k\,y) \text { for } |y| \le 3. \end{aligned}$$

Then $f(\lambda (A^{u_k})) = 1$ in $B_3(0)$ and

$$\begin{aligned} u_k \ge R_k^{\frac{n-2}{2}} \gamma \text { in } B_{r_1/R_k}(0). \end{aligned}$$

(98)

Thus, by applying Theorem 1.4 and after passing to a subsequence, we can select sets of local maximum points $\{x_k^1, \ldots , x_k^m\}$ of $u_k$ such that assertions (i)–(vi) in Theorem 1.4 hold. We can also assume that $x_k^i \rightarrow x_*^i$, $1 \le i \le m$.

By assertions (i), (iv) and (v) of Theorem 1.4, $u_k$ converges locally uniformly to zero in $B_1 {\setminus } \{x_*^1, \ldots , x_*^m\}$. Thus, in view of (98), we must have $x_*^{i_0} = 0$ for some (unique) $1 \le i_0 \le m$. Clearly, $B_{r_1/R_k}(0) \subset B_{{\delta _*}}(x_k^{i_0})$ for large k.

Recalling assertions (iv), (vi) and returning to the original sequence $v_k$ we get, for $\bar{x}_k = R_k\,x_k^{i_0}$ and $\bar{\mu }_k = v_k(\bar{x}_k) = \sup _{B_{{\delta _*}R_k}(\bar{x}_k)} v_k$, that $B_{r_1}(0) \subset B_{{\delta _*}R_k}(\bar{x}_k)$, $\bar{\mu }_k \ge \gamma$ and

$$\begin{aligned} (1 - \epsilon ) U^{\bar{x}_k, \bar{\mu }_k} \le v_k \le (1 + \epsilon ) U^{\bar{x}_k, \bar{\mu }_k} \text { in } B_{{\delta _*}R_k}(\bar{x}_k). \end{aligned}$$

We then have

$$\begin{aligned} \gamma \le v_k(0) = 2\left( \frac{\bar{\mu }_k^{\frac{2}{n-2}}}{1 + \bar{\mu }_k^{\frac{4}{n-2}}|\bar{x}_k|^2}\right) ^{\frac{n-2}{2}} \le \frac{2}{\bar{\mu }_k\,|\bar{x}_k|^{n-2}} \le \frac{2}{\gamma \,|\bar{x}_k|^{n-2}} \end{aligned}$$

This implies

$$\begin{aligned} |\bar{x}_k| \le 2^{\frac{1}{n-2}}\,\gamma ^{-\frac{2}{n-2}}. \end{aligned}$$

On the other hand, since $B_{r_1}(0) {\setminus } B_{r_1/2}(\bar{x}_k) \ne \emptyset$, we can select some $y_k \in B_{r_1}(0)$ such that

$$\begin{aligned} |\bar{x}_k - y_k| \ge \frac{r_1}{2}. \end{aligned}$$

This implies that

$$\begin{aligned} \gamma \le v(y_k) \le 2U^{\bar{x}_k, \bar{\mu }_k}(y_k) \le \frac{2}{\bar{\mu }_k\,|y_k - \bar{x}_k|^{n-2}} \le \frac{2^{n-1}}{\bar{\mu }_k\,r_1^{n-2}}, \end{aligned}$$

and so

$$\begin{aligned} \bar{\mu }_k \le \frac{2^{n-1}}{\gamma \,r_1^{n-2}}. \end{aligned}$$

We have thus shown that $\bar{x} = \bar{x}_k$ satisfies both (24) and (25), which contradicts (97). $\square$

Appendices

Appendix A: A remark on positive superharmonic functions

Let G(x, y) be the Green’s function of $-\Delta$ in $B_1{\setminus } B_\rho \subset {\mathbb {R}}^n$, $0<\rho <1/2$:

$$\begin{aligned} G(x,y):= \frac{1}{ n(n-2) \alpha (n) } \left( \frac{1}{ |x-y|^{n-2} } -h(x,y)\right) , \qquad x,y\in B_1{\setminus } B_\rho ,, \end{aligned}$$

where $\alpha (n)$ denotes the volume of the unit ball in ${\mathbb {R}}^n$, and h(x, y) satisfies, for $x\in B_1{\setminus } B_\rho$,

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta _y h(x,y)= 0, &{} y\in B_1{\setminus } B_\rho ,\\ h(x,y)= \frac{1}{ |x-y|^{n-2} }, &{} y\in \partial (B_1{\setminus } B_\rho ). \end{array} \right. \end{aligned}$$

Lemma A.1

For any $0<2\rho< \rho _0<\rho _1<\rho _2<1$, there exists some constants $C, C'>1$, depending only on $n, \rho _0, \rho _1, \rho _2$, such that the Green’s function G for $B_1{\setminus } B_\rho$ satisfies

$$\begin{aligned} G(x, y)\ge \frac{1}{C} \left( 1-\frac{\rho ^{n-2}}{ |y|^{n-2} }\right) , \qquad \forall \ \rho \le |y|\le \rho _0, \quad \rho _1\le |x|\le \rho _2. \end{aligned}$$

Consequently,

$$\begin{aligned} G(x,y)\ge 1/C',\qquad \forall \ 2\rho \le |y|\le \rho _0, \rho _1\le |x|\le \rho _2. \end{aligned}$$

Proof

In the following we use $C_1, C_2,\ldots$ to denote positive constants depending only on $\rho _0, \rho _1, \rho _2$ and n. For a fixed x satisfying $\rho _1\le |x|\le \rho _2$, it follows from the maximum principle that for some positive constant $C_1$,

$$\begin{aligned} 0<h(x, y)\le C_1, \quad \forall y\in B_1{\setminus } B_\rho . \end{aligned}$$

It follows that for some positive constants $C_2$ and $C_3$,

$$\begin{aligned} G(x,y)\ge |x-y|^{2-n}-C_1\ge 1/C_2,\qquad \forall \ y\in B_{ 2/C_3}(x){\setminus } \{x\} \subset B_1{\setminus } B_{\rho _0}. \end{aligned}$$

Since G(x, y) is a positive harmonic function of y in $(B_1{\setminus } B_\rho )\setminus \{x\}$. We can apply the Harnack inequality to obtain, for some $C_4$,

$$\begin{aligned} G(x,y)\ge 1/C_4,\qquad \forall \ |y|=\rho _0. \end{aligned}$$

By the maximum principle,

$$\begin{aligned} G(x,y)\ge \frac{1}{C_4} \left( 1-\frac{\rho ^{n-2}}{ |y|^{n-2} }\right) , \qquad \rho <|y|\le \rho _0. \end{aligned}$$

It follows that for some ${K}$,

$$\begin{aligned} G(x,y)\ge 1/{K},\qquad 2\rho \le |y|\le \rho _0. \end{aligned}$$

Lemma A.1 is established. $\square$

Corollary A.1

For any $0<2\rho< \rho _0<\rho _1<\rho _2<1$, let

$$\begin{aligned} \left\{ \begin{array}{rl} -\Delta \tilde{v}= \tilde{f}\ge &{} 0,\qquad B_1{\setminus } B_\rho ,\\ \tilde{v}\ge &{}0, \qquad \partial (B_1{\setminus } B_\rho ). \end{array} \right. \end{aligned}$$

Then, for some constants $C, C'>1$ depending only on $n, \rho _0, \rho _1, \rho _2$,

$$\begin{aligned} \inf _{ \rho _1\le |x|\le \rho _2 } \tilde{v}(x)\ge \frac{1}{C} \int _{ \rho \le |y|\le \rho _0} \left( 1-\frac{\rho ^{n-2}}{ |y|^{n-2} }\right) \tilde{f}(y)dy \ge \frac{1}{C'} \int _{ 2\rho \le |y|\le \rho _0} \tilde{f}(y)dy. \end{aligned}$$

Proof

For $\rho _1\le |x|\le \rho _2$, we use the Green’s formula to obtain

$$\begin{aligned} \tilde{v} (x)=\int _{ B_1{\setminus } B_\rho } G(x,y)\tilde{f}(y). \end{aligned}$$

Corrollary A.1 follows from Lemma A.1. $\square$

Corollary A.2

For any $0< \rho _0<\rho _1<\rho _2<1$, let

$$\begin{aligned} \left\{ \begin{array}{rl} -\Delta \tilde{v}= \tilde{f}\ge &{} 0,\qquad B_1,\\ \tilde{v}\ge &{}0, \qquad \partial B_1. \end{array} \right. \end{aligned}$$

Then, for some constants C depending only on $n, \rho _0, \rho _1, \rho _2$,

$$\begin{aligned} \inf _{ \rho _1\le |x|\le \rho _2 } \tilde{v}(x) \ge \frac{1}{C} \int _{ |y|\le \rho _0} \tilde{f}(y)dy. \end{aligned}$$

Proof

This follows from Corollary A.1 by sending $\rho \rightarrow 0$. $\square$

Corollary A.3

For $0<r<R/2$, let

$$\begin{aligned} \left\{ \begin{array}{rl} -\Delta v= f\ge &{} 0,\qquad B_R{\setminus } B_r,\\ v\ge &{}0, \qquad \partial (B_R{\setminus } B_r). \end{array} \right. \end{aligned}$$

Then, for any $\frac{2r}{R}<\rho _0<\rho _2<\rho _2 < 1$, there exist some constants $C, C'>1$ depending only on $n, \rho _0, \rho _1, \rho _2$,

$$\begin{aligned} \inf _{ \rho _1 R\le |x|\le \rho _2 R } v(x)&\ge \frac{1}{CR^{n-2} } \int _{ r \le |y|\le \rho _0 R} \left( 1-\frac{r^{n-2}}{ |y|^{n-2} }\right) f(y)dy\\&\ge \frac{1}{C'R^{n-2} } \int _{ 2r\le |y|\le \rho _0 R} f(y)dy. \end{aligned}$$

Proof

Performing a change of variables

$$\begin{aligned} \tilde{v}(x):= R^{n-2} v(Rx), \qquad \tilde{f}(x):= R^n f(Rx), \rho = r/R, \end{aligned}$$

we obtain Corollary A.3 from Corollary A.1 $\square$

Appendix B: A remark on viscosity solutions

In this section we consider the convergence of viscosity solutions in a slightly more general context. Let ${\mathbb {R}}^{n \times n}$, $Sym^{n \times n}$, $Sym^{n \times n}_+$ denote the set of $n \times n$ matrices, symmetric matrices, and positive definite symmetric matrices, respectively. Let $\mathcal {M} = {\mathbb {R}}^{n \times n}$ or $\mathcal {M} = Sym^{n \times n}$. Let $\Omega \subset {\mathbb {R}}^n$, $U \subset \mathcal {M}$ be open, $F \in C(U)$, $A \in C(\Omega \times {\mathbb {R}}\times {\mathbb {R}}^n \times Sym^{n \times n}; \mathcal {M})$ and consider partial differential equations for the form

$$\begin{aligned} F(A(x,u,\nabla u, \nabla ^2 u)) = 0. \end{aligned}$$

To keep the notation simple, we will abbreviate $A[u] = A( \cdot ,u,\nabla u, \nabla ^2 u)$, and whenever we write F(M), we implicitly assume that $M \in U$.

In applications, it is frequently assumed that

$$\begin{aligned}&M + N \in U \text { for } M \in U \text { and } N \in Sym_+^{n\times n}, \end{aligned}$$

(99)

$$\begin{aligned}&F(M + N) \ge F(M) \text { for } M \in U \text { and } N \in Sym_+^{n\times n}, \end{aligned}$$

(100)

$$\begin{aligned}&A(x,z,p,M)- A(x,z,p,M + N) \text { is non-negative definite for } N \in Sym_+^{n \times n}. \end{aligned}$$

(101)

In the main body of the paper,

$$\begin{aligned} \mathcal {M}= & {} Sym^{n \times n}, U\hbox { is the set of symmetric matrices such that }\lambda (U) \in \Gamma , \nonumber \\ F(M)= & {} f(\lambda (M)) - 1,\hbox { and }A[v] = A^v. \end{aligned}$$

(102)

The following definition is “consistent” with the assumptions (99), (100) and (101) and with Definition 1.1.

Definition B.1

A continuous function v is an open set $\Omega \subset {\mathbb {R}}^n$ is a viscosity supersolution (respectively, subsolution) of

$$\begin{aligned} F(A[v])=0, \ \ \hbox {in}\ \Omega , \end{aligned}$$

when the folowing holds: if $x_0\in \Omega$, $\varphi \in C^2(\Omega )$, $(v-\varphi )(x_0)=0$, and $v-\varphi \ge 0$ near $x_0$, then

$$\begin{aligned} F(A[\varphi ](x_0))\ge 0. \end{aligned}$$

(respectively, if $(v-\varphi )(x_0)=0$, and $v-\varphi \le 0$ near $x_0$, then either $A[\varphi ](x_0) \notin U$ or $F(A[\varphi ](x_0))\le 0$). Equivalently, we write $F(A[v]) \ge 0$ (respectively, $F(A[v]) \le 0$) in $\Omega$ in the viscosity sense.

We say that v is a viscosity solution if it is both a viscosity supersolution and a viscosity subsolution.

Proposition B.1

Let $\Omega \subset {\mathbb {R}}^n$, $U \subset \mathcal {M}$ be open, $F \in C(U)$, $A \in C(\Omega \times {\mathbb {R}}\times {\mathbb {R}}^n \times Sym^{n \times n}; \mathcal {M})$ and assume that the structural conditions (99), (100) and (101) are in effect.

• (a) If $v_k$ satisfies $F(A[v_k]) \le 0$ in $\Omega$ in the viscosity sense and if $v_k$ converges in $C^0_{loc}(\Omega )$ to v, then v also satisfies $F(A[v]) \le 0$ in $\Omega$ in the viscosity sense.

• (b) Assume further that the closure of $F^{-1}([0,\infty ))$ in $\mathcal {M}$ is a subset of U, i.e.

  $$\begin{aligned} \text { if }A_k \in U, F(A_k) \ge 0\text { and }A_k \rightarrow A\text { as } k \rightarrow \infty ,\text { then }A \in U. \end{aligned}$$

  (103)

  If $v_k$ satisfies $F(A[v_k]) \ge 0$ in $\Omega$ in the viscosity sense and if $v_k$ converges in $C^0_{loc}(\Omega )$ to v, then v also satisfies $F(A[v]) \ge 0$ in $\Omega$ in the viscosity sense.

Remark B.1

In general, condition (103) cannot be dropped in (b).

As a first example, consider

$$\begin{aligned} F(M)&= tr(M), \qquad U = \{M \in Sym^{n\times n}: tr(M) > 0\},\\ A[v]&= - \nabla ^2 v. \end{aligned}$$

Clearly, (99), (100) and (101) are satisfied. Now, if $v_k = -\frac{1}{k}|x|^2$, then $A[v_k] \in U$ but $v = \lim _{k \rightarrow \infty } v_k = 0$ does not satisfy $A[v] \in U$.

The situation does not improve even if one imposes that $v_k$ is a solution and that U is a maximal set where the ellipticity condition (100) holds. See Remark 2.1.

Remark B.2

If $(f,\Gamma )$ satisfies (1)–(4) and if (F, U) are given by (102), then all hypotheses of Proposition B.1 are met. In particular if $f(\lambda (A^{v_k})) = 1$ in some open set $\Omega \subset {\mathbb {R}}^n$ in the viscosity sense and $v_k$ converges in $C^0_{loc}(\Omega )$ to v, then $f(\lambda (A^v)) = 1$ in $\Omega$ in the viscosity sense.

Proof

The proof is standard and we include it here for readers’ convenience. We will only show (b). The proof of (a) is similar. Fix $x_0 \in \Omega$ and assume that $\varphi \in C^2(\Omega )$ such that $(v - \varphi )(x_0) = 0$ and $v -\varphi \ge 0$ in some small ball $B_\rho (x_0)$. We need to show that $F(A[\varphi ](x_0)) \ge 0$.

For $\delta \in (0,\rho )$, let $\varphi _\delta (x) = \varphi (x) - \delta |x - x_0|^2$. Since $\varphi _\delta (x) = \varphi (x) -\delta ^3 \le v(x) - \delta ^3$ for $|x - x_0| = \delta$, the convergence of $v_k$ to v implies that

$$\begin{aligned} v_k(x) \ge \varphi _\delta (x) + \frac{1}{2}\delta ^3 \text { for all } |x - x_0| = \delta \text { and all sufficiently large } k. \end{aligned}$$

Thus, as $\min _{B_\delta (x_0)} (v - \varphi _\delta ) = 0$, there is some $x_k \in B_\delta (x_0)$ such that

$$\begin{aligned} \beta _k := (v_k - \varphi _\delta )(x_k) = \min _{\bar{B}_\delta (x_0)} (v_k - \varphi _\delta ) \rightarrow 0 \text { as } k \rightarrow \infty . \end{aligned}$$

Also, since

$$\begin{aligned} \beta _k = (v_k - \varphi )(x_k) + \delta |x_k - x_0|^2 \ge (v_k - v)(x_k) + \delta |x_k - x_0|^2, \end{aligned}$$

we have that $x_k \rightarrow x_0$ as $k \rightarrow \infty$.

Now since $\hat{\varphi }_k := \varphi _\delta + \beta _k$ satisfies $(v_k - \hat{\varphi }_k)(x_k) = 0$, $v_k - \hat{\varphi }_k \ge 0$ in $B_\delta (x_0)$, and since $v_k$ is a solution of $F(A[v_k]) = 0$, we infer that

$$\begin{aligned} F(A[\hat{\varphi }_k](x_k)) \ge 0. \end{aligned}$$

Sending $k \rightarrow \infty$ and then $\delta \rightarrow 0$ and using (103), we deduce that $A[\varphi ](x_0) \in U$ and

$$\begin{aligned} F(A[\varphi ](x_0)) \ge 0, \end{aligned}$$

as desired. (Note that (103) is not required to show that (a).) $\square$

Appendix C: A calculus lemma

We collect here a couple calculus statements which was used when we applied the method of moving spheres in the body of the paper.

Lemma C.1

Let $f\in C^0({\mathbb {R}}^n {\setminus } \{0\})$, $n\ge 1$, $\nu >0$. Assume that, for all $x \in {\mathbb {R}}^n$ and $0 < \lambda \le |x|$, there holds

$$\begin{aligned} \Big ( \frac{\lambda }{|y-x|}\Big )^\nu f\Big (x+ \frac{ \lambda ^2 (y-x) }{ |y-x|^2 }\Big ) \le f(y) \text { for } y \in {\mathbb {R}}^n {\setminus } (B_\lambda (x) \cup \{0\}). \end{aligned}$$

Then f is radially symmetric about the origin and $f: (0,\infty ) \rightarrow (0,\infty )$ is non-increasing.

Proof

This was established in [10]. See the argument following equation (110) therein. $\square$

Corollary C.1

Let $f\in C^0({\mathbb {R}}^n)$, $n\ge 1$, $\nu >0$. Assume that

$$\begin{aligned} \Big ( \frac{\lambda }{|y-x|}\Big )^\nu f\Big (x+ \frac{ \lambda ^2 (y-x) }{ |y-x|^2 }\Big ) \le f(y),\quad \forall \ \lambda >0, \ x\in {\mathbb {R}}^n,\ |y-x|\ge \lambda . \end{aligned}$$

Then $f\equiv$ constant.