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

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.


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 ≥ 3 be an integer and Γ ⊂ R n be an open convex symmetric cone with vertex at the origin (1) where Γ n := {λ ∈ R n | λ i > 0 ∀ i}, We assume that In (1) and (3), the symmetric property of Γ and f is understood in the sense that if λ ∈ Γ andλ is a permutation of λ, thenλ ∈ Γ and f (λ) = f (λ). Also, throughout the paper, whenever we write f (λ), we implicitly assume that λ ∈ Γ.
Besides (1)-(5), (σ 1 k k , Γ k ) enjoys other nice and helpful properties, such as concavity and homogeneity properties of σ 1/k 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 × n matrix with entries 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 = 1 n−2 (Ric g − 1 2(n−1) R g g), 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 4 n−2 g f lat conformal to the flat metric g f lat on R n , then the Schouten curvature A gu of g u is given by the conformal Hessian in the form Consequently, we have where λ(A gu ) denotes the eigenvalues of A gu with respect to the metric g u and λ(A u ) denotes those of the matrix A u .
A u enjoys a conformal invariance property, inherited from the conformal structure of R n , which will be of special importance in our treatment. Recall that a map ϕ : R n ∪ {∞} → R n ∪ {∞} is called a Möbius transformation if it is the composition of finitely many of the following types of transformations: • a translation: x → x +x wherex is a given vector in R n , • a dilation: x → a x where a is a given positive scalar, • a Kelvin transformation: x → x |x| 2 . For a function u and a Möbius transformation ϕ, let where J ϕ is the Jacobian of ϕ. A calculation gives for some orthogonal n × n matrix O ϕ (x). In particular, λ(A uϕ (x)) = λ(A u (ϕ(x))).
The main result of this paper concerns an analysis on the behavior of a sequence {u k } ∈ C 2 (B 3 (0)) satisfying f (λ(A u k )) = 1, u k > 0, in B 3 (0), (8) and sup where (f, Γ) satisfies (1)- (5). Note that no other assumptions on u k is made. As is known, equation (8) is a fully nonlinear elliptic equation. Fully nonlinear elliptic equations involving f (λ(∇ 2 u)) were investigated in the classic paper of Caffarelli, Nirenberg and Spruck [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.
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 all theorems, propositions and lemmas hold under (1)-(5), instead of the stated weaker hypotheses on (f, Γ). and Then either v is constant or v is of the form (11) for somex ∈ R n and some positive constants a and b.
and, for some M k → ∞, Then v * is radially symmetric about the origin, i.e. v * (x) = v * (|x|). In particular, (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, Γ) satisfy (1)- (4), and let 0 < v ∈ C 0 loc (R n ) be a viscosity solution of (10). Then v is of the form (11) for somex ∈ R n and some positive constants a and b.
The notion of viscosity solutions given below is consistent with that in [12].
. We say that v is a viscosity solution if it is both a viscosity supersolution and a viscosity subsolution.
when the following holds: if . 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 notion of viscosity solutions and classical solutions coincide. Also, viscosity super-and sub-solutions are stable under uniform convergence, see Appendix B.
The following is a quantitative version of Theorem A, and is related to Theorem 1.1 and Theorem 1.3. and then, for somex ∈ R n satisfying v(x) = max there holds 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, Γ) satisfy (13)-(15) and let v ∈ C 2 (B 2 (0)) satisfy, for some constant b > 0, and Then, for some constant C depending only on (f, Γ) and b, For (f, Γ) = (σ 1/k k , Γ k ), the result was proved by Guan and Wang [6]. When (f, Γ) 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 Section 2 with the proof of Theorems 1.1 and 1.2. We then prove Theorem 1.6 in Section 3. In Section 4, we first establish an intermediate quantitative Liouville result and then use it to prove Theorem 1.3. In Section 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.

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.) (15) and

A gradient estimate
and Then, for some constant C depending only on n and θ, 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, Γ). 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 R n , we define wherever the expression makes sense. We will use w λ to denote w 0,λ . We start with a simple result. .
Proof. Write w in polar coordinates w(r, θ). It is easy to see that (30) is equivalent to r n−2 Estimate (31) is readily seen from the estimates It is easy to see that, for some We then define, for x ∈ B 4/3 (0), By the conformal invariance (7), v x,λ(x) satisfies Using the above two displayed equations, the definition ofλ(x), and using the ellipticity of the equation satisfied by v and v x,λ(x) , we can apply the strong maximum principle and Hopf Lemma to infer that eitherλ(x) = 5/3 − |x| or there exists some In the latter case, (29) implies that In either case, we obtain that It is readily seen that f (t, . . . , t) > 1 for all (t, . . . , t) ∈ Γ, v k satisfies (16) (cf. [13, Theorem 1.6]) and v k → 1 in C 0 loc (R n ).
Proof of Theorem 1.1. We may assume that R k ≥ 5 for all k.
Clearly, for every β > 1, there exists some positive constant C(β), independent of k, such that . It follows, after passing to a subsequence, that for and v is super-harmonic on R n . Using the positivity, the superharmonic of v, and the maximum principle, we can find Passing to a subsequence and shrinking R k and c 0 > 0, if necessary, we may assume that and Proof. For |x| ≤ R k 5 , we have, by (33) and (34), for all k that 1 where r 1 (x) = 1 4 + |x|, and c 1 (x) = max 1 + sup By (35) and Theorem 2.1, there exists c 2 (x) > 0, independent of k, such that Thus, by Lemma 2.1, we can find For 0 < λ < λ 1 (x), we have, using (35), that 1 2 (1 + |y|) < |y − x| and we obtain, using (37) and (34), that When and we obtain, using (37) and (34), that Letting we derive from (38) and (39) that Lemma 2.2 follows from (36) and (40).
Step 1. Ifλ(x) < ∞ for some x ∈ R n , then Sinceλ(x) < ∞, we have, along a subsequence,λ k (x) →λ(x) -but for simplicity, we still use {λ k (x)}, {v k }, etc to denote the subsequence. By the definition By the conformal invariance (7), Using (16), (41), (42), the definition ofλ k (x), and using the ellipticity of the equation satisfied by v k and (v k ) x,λ k (x) , we can apply the strong maximum principle and Hopf Lemma to infer the existence of some y k ∈ ∂B R k (0) such that It follows that This implies, in view of (33), that On the other hand, ifŷ i is such that |ŷ i | → ∞ and Step 1 is established.
Step 2. It remains to show that either v is constant or, for every x ∈ R n ,λ(x) < ∞.
To this end, we show that ifλ(x) = ∞ for some x ∈ R n , then v is constant. Indeed, assume thatλ k (x) → ∞ as k → ∞. We easily derive from this and the The above is equivalent to the property that for every fixed unit vector e, r n−2 In particular, α = lim inf |y|→∞ |y| n−2 v(y) = ∞. This implies, by Step 1, thatλ(x) = ∞ for every x ∈ R n , and therefore (43) holds for every x ∈ R n . This implies that v is a constant, see Corollary C.1. (15) and (18). Then the function v in Theorem 1.1 cannot be constant.
Proof. Fix some t > 0 for the moment. Set ϕ(x) = v(0) − t |x| 2 and fix some r > 0 such that ϕ > 0 in B r (0) and ϕ < v k on ∂B r (0) for all sufficiently large k. Let and lim k→∞ γ k = sup Noting that there is some C > 0 independent of δ and k such that, for large k, Thus, we can select t and δ such that where t 0 is the constant in (18). Since f (λ(A v k (x k ))) = 1, this contradicts (14), (15) and (18).
Recall that 0 < v ∈ C 0,1 loc (R n ), ∆v ≤ 0 in R n , and it remains to consider the case that, for every x ∈ R n , there exists 0 If v is in C 2 (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 ∈ C 0,1 loc (R n ). For readers' convenience, we outline the proof below.
We know that λ(A v ψ (y)) = λ(A v (ψ(y)). Namely, A v ψ (y) and A v (ψ(y)) differ only by an orthogonal conjugation. Introduce We deduce from the above properties of v that for every x ∈ R n , there exists some By [8,Lemma 4.1], Namely, for some V ∈ R n , A calculation yields Consequently, for somex ∈ R n and d ∈ R, Since We have proved that v is of the form (11) for somex ∈ R n and some positive constants a and b.
Theorem 1.1 is established.

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 ≥ 5 for all k. By hypotheses, v k is super-harmonic and positive on R n \ {0}. Therefore, v * is super-harmonic and non-negative on R n \ {0}. Hence either v * ≡ 0 or v * > 0 in R n \ {0}. In the former case we are done. We assume henceforth that the latter holds.
Now, for every β > 2, there exists some positive constant C(β), independent of k, such that . It follows, after passing to a subsequence, that for every 0 By the super-harmonicity and the positivity of v * , we can find c 0 > 0 such that v * (y) ≥ 2c 0 |y| 2−n , ∀ |y| ≥ 1.
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, and Denote the Kelvin transformation of v k . We use (v k ) λ to denote (v k ) 0,λ .
Define, for 0 < |x| ≤ R k /5, that By Lemma 2.5, Clearly, We have a dichotomy: 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 ofλ(x). We start with an analogue of Lemma 2.3. By (46), let α := lim inf |y|→∞ |y| n−2 v * (y) ∈ (0, ∞]. Lemma 2.6. Under the hypotheses of Theorem 1.2, ifλ(x) < |x| for some x ∈ R n \ {0}, then Proof. We adapt Step 1 in the proof of Lemma 2.3. Assume thatλ(x) < |x| and (without loss of generality) thatλ k (x) →λ(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 ∈ ∂B R k (0) such that It follows that This implies, in view of (47), that On the other hand, as in the proof of Lemma 2.3, we can use The conclusion is readily seen.
Proof. Along a subsequence, we haveλ k (x 0 ) →λ(x 0 ). As in the proof of Lemma 2.6, there exists y k ∈ ∂B R k (0) such that We know that Let m denote the modulus of continuity of v * in B |x 0 |/2 (x 0 ), i.e.
In the computation below, we use o(1) to denote quantities such that lim k→∞ o(1) = 0.

The conclusion follows.
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 thatλ(x) < |x| for all x ∈ 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 v * (x) = a 1 + b 2 |x −x| 2 n−2 2 x ∈ B r 0 (x 0 ).
for somex ∈ R n and some a, b > 0. For small δ > 0, let

Local gradient estimates
In this section, we adapt the argument in [12] to prove Theorem 1.6.
We can now apply Theorem 2.1 to obtain |∇ lnv i | ≤ C(β) in B β/2 (0) for all sufficiently large i.
On the other hand, if we definē then by the conformal invariance (7), we have This contradicts (59), in view of (62) and the convergence ofv i to v * . We have proved (57).
From (57), we can find some universal constant C > 1 such that Applying Theorem 2.1 again we obtain the required gradient estimate in B 1/4 (0).

Fine blow-up analysis 4.1 A quantitative centered Liouville-type result
In this subsection, we establish: Then for every ǫ > 0, there exists a constant δ 0 > 0, depending only on (f, Γ) and ǫ, such that, for all sufficiently large k, Recall that U = (1 + |x| 2 ) − n−2 2 , A U ≡ 2I and f (λ(A U )) = 1 on R n . Proposition 4.1 is equivalent to the following proposition.

Proof of the equivalence between Proposition 4.1 and Proposition 4.2. It is clear that Proposition 4.2 implies Proposition 4.1.
Consider the converse. Let δ 0 = δ 0 (ǫ) be as in Proposition 4.1. Arguing by contradiction, we assume that there are some ǫ > 0 and a sequence of R k and u k ∈ C 2 (B R k (0)) such that Defineū Returning to the original sequence u k we arrive at a contradiction.
Moreover, for every ǫ > 0, there exists k 0 ≥ 1 such that Proof. We first prove (65). Since v k satisfies (63), we deduce from Theorem 1.6 that 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 ǫ > 0 and sequences of k i → ∞, 0 < r i < R k i /5 such that Because of (65), r i → ∞.
As in the proof of Lemma 2.2, there exists λ (0) i > 0 such that By the explicit expression of U, there exists some small δ > 0 independent of i such that, for large i, By the uniform convergence of v k i to U on compact subsets of R n , we have, for large i, i ≤ λ ≤ 1 + δ, As in the proof of Lemma 2.2, the moving sphere procedure does not stop before reaching λ = 1 + δ, namely we have, for large i, Sending i to ∞ leads to U 1+δ (y) ≤ U(y), ∀ 1 + δ ≤ |y| ≤ 2.
A contradiction -since we see from the explicit expression of U that U 1+δ (y) > U(y) for all 1 < 1 + δ < |y| ≤ 2. and Proof. Assume without loss of generality that ǫ ∈ (0, 1/2). Since v k → U in C 0 loc (R n ), there exist r 2 > 1 and k 1 , depending on ǫ, such that for all By (5), Using the superharmonicity of v k and the maximum principle, we obtain Thus, for any δ 2 ∈ (0, ǫ 2 n−2 ), we have for all sufficiently large k that Now if δ 1 < δ 2 , (69) is readily seen from (71) and (74). Letv Enlarging k 1 if necessary, we can apply Corollary A.3 in Appendix A to get 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) It now follows from (75) and (76) that where c 1 depends only on n. (70) is then established for ǫ ≤ 1 c 1 with r 1 = 2r 2 and δ 1 = δ 2 /8. The conclusion for ǫ > 1/c 1 also follows.  (15). Then there exist δ 3 > 0 and C 3 > 1, depending only on (f, Γ), such that if u ∈ C 2 (B 2 (0)) satisfies If (f, Γ) satisfies in addition the conditions (18), (19) and the normalization condition (21), then δ 3 can be chosen to be any constant smaller than R n U 2n n−2 dx.
Then by the conformal invariance property (7) in B r j (0) and By Theorem 1.6, there is a constant C independent of j such that Thus, after passing to a subsequence, we can assume that v j converges in C 0 loc (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: Equation (77) is replaced by On the other hand, by Theorem 1.1, we have v j → U in C 0 loc (R n ). This gives a contradiction.
Proof. Let δ 3 be as in Lemma 4.3. Since v k ≤ 1, we deduce from Lemma 4.2, there is r 1 > 1 and δ 1 > 0 such that For any 2r 1 < r < δ 1 R k /2, consider v k (z) = r n−2 By (79), we have, for large k, for some universal constant C. Sinceṽ k also satisfies f (λ(Aṽ k )) = 1, we can apply Theorem 1.6 to obtain where C is universal. The conclusion then follows from Lemma 4.1.
Proof of Proposition 4.1. Fix ǫ > 0. In view of Lemma 4.2 (cf. (69)), we only need to prove that there exist δ 0 > 0 such that, for all sufficiently large k, v k (y) ≤ (1 + 2ǫ)U(y), 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 | = δ k R k , δ k → 0 + , such that v k (y k ) = max In view of the convergence of v k to U, |y k | → ∞ as k → ∞. Consider the following two rescalings of v k : for some constant C independent of k.
The above violates the radial symmetry ofv * . Proposition 4.1 is established.

Detailed blow-up landscape
The proof of Theorem 1.3 uses the following consequence of the Harnack-type inequality for conformally invariant equations, see [16,4,7].
Lemma 4. 6. Let (f, Γ) satisfy (13)- (15) and (5). There exists a constant C 6 , depending only on (f, Γ), such that if u ∈ C 2 (B 3 (0)) is a positive solution of Proof. We give the proof here for completeness. By (5), Thus, by Corollary A.2 in Appendix A as well as the maximum principle, It follows that The conclusion follows from the above estimate and the Harnack-type inequality [ The constantm in the result can be selected to be the least integer satisfyinḡ (Clearly, this is an obvious upper bound for m if the x i 's satisfies (iii).) Let δ 3 and C 3 be the constants in Lemma 4.4. Fix some N 0 > C 1 δ 3 . Then there is some r 0 ∈ (3/2, 2) such that

By Lemma 4.4, this implies that
Let C 0 and δ 0 be as in Proposition 4.2 (corresponding to ǫ = ǫ 0 ). We can assume without loss of generality that C 0 > 2 and δ 0 < 1. (92) We now declare This choice of C * will become clear momentarily. (93), C * ≥ 2C 7 , and so, by (93) gives Hence, an application of Proposition 4.2 to u on the ball B R 1 (x 1 ) leads to In particular, where we have used (93) in the last estimate.
we stop. Otherwise, in view of (94), there is some x 2 ∈ V 2 such that We then let R 2 = δ 0 R 1 2 so that (93) implies Hence, by Proposition 4.2, 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 exceedm.
We have obtained the set of local maximum points {x 1 , · · · , x m } of u and have verified (i) and (iv) for (ii) is also clear for From construction, we have sup By Theorem 1.6, this implies that Also, note that, for δ * < |x − x i | < δ 0 R i , we have and so 1 It is now clear that (iii) and (v) hold for K sufficiently large. The proof is complete. and soμ k ≤ 2 n−1 γ r n−2

A A remark on positive superharmonic functions
Let G(x, y) be the Green's function of −∆ in B 1 \ B ρ ⊂ R n , 0 < ρ < 1/2: where α(n) denotes the volume of the unit ball in R n , and h(x, y) satisfies, for Lemma A.1. For any 0 < 2ρ < ρ 0 < ρ 1 < ρ 2 < 1, there exists some constants C, C ′ > 1, depending only on n, ρ 0 , ρ 1 , ρ 2 , such that the Green's function G for B 1 \B ρ satisfies Consequently, Proof. In the following we use C 1 , C 2 , ... to denote positive constants depending only on ρ 0 , ρ 1 , ρ 2 and n. For a fixed x satisfying ρ 1 ≤ |x| ≤ ρ 2 , it follows from the maximum principle that for some positive constant C 1 , It follows that for some positive constants C 2 and C 3 , Since G(x, y) is a positive harmonic function of y in (B 1 \ B ρ ) \ {x}. We can apply the Harnack inequality to obtain, for some C 4 , By the maximum principle, It follows that for some K, Then, for some constants C, C ′ > 1 depending only on n, ρ 0 , ρ 1 , ρ 2 , (y)dy.

B A remark on viscosity solutions
In this section we consider the convergence of viscosity solutions in a slightly more general context. Let R n×n , Sym n×n , Sym n×n + denote the set of n × n matrices, symmetric matrices, and positive definite symmetric matrices, respectively. Let M = R n×n or M = Sym n×n . Let Ω ⊂ R n , U ⊂ M be open, F ∈ C(U), A ∈ C(Ω × R × R n × Sym n×n ; M) and consider partial differential equations for the form F (A(x, u, ∇u, ∇ 2 u)) = 0.
To keep the notation simple, we will abbreviate A[u] = A(·, u, ∇u, ∇ 2 u), and whenever we write F (M), we implicitly assume that M ∈ U.
In applications, it is frequently assumed that M + N ∈ U for M ∈ U and N ∈ Sym n×n + , The following definition is "consistent" with the assumptions (99), (100) and (101) and with Definition 1.1. in Ω in the viscosity sense.
We say that v is a viscosity solution if it is both a viscosity supersolution and a viscosity subsolution. 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. 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 ∈ Ω and assume that ϕ ∈ C 2 (Ω) such that (v − ϕ)(x 0 ) = 0 and v − ϕ ≥ 0 in some small ball B ρ (x 0 ). We need to show that F (A[ϕ](x 0 )) ≥ 0.

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