Fifth Abel Conference : Celebrating the Mathematical Impact of John F. Nash Jr. and Louis Nirenberg

Transcription

1 Fifth Abel Conference : Celebrating the Mathematical Impact of John F. Nash Jr. and Louis Nirenberg Some analytic aspects of second order conformally invariant equations Yanyan Li Rutgers University November 4, 2015

2 Liouville u = 0, u > 0, R n implies u = Constant. Gidas, Ni, Nirenberg (1981) u = n(n 2)u n+2 n 2, u > 0, R n and some decay assumption implies u(x) = ( a ) n a 2 x x 2, a > 0, x R n. Method of Moving Planes

3 Liouville u = 0, u > 0, R n implies u = Constant. Gidas, Ni, Nirenberg (1981) u = n(n 2)u n+2 n 2, u > 0, R n and some decay assumption implies u(x) = ( a ) n a 2 x x 2, a > 0, x R n. Caffarelli, Gidas, Spruck (1989) Removed the decay assumption Method of Moving Planes

4 Both equations conformally invariant

5 Both equations conformally invariant Möbius transformation ψ : R n R n (a) x x (b) x ax, (a > 0) (c) x x x 2 u ψ := J ψ n 2 2n u ψ (a) u ψ (x) = u(x + x) (b) u ψ (x) = a n 2 2 u(ax) (c) u ψ (x) = 1 u( x ) x n 2 x 2

6 Conformally invariant operator: H(, u ψ, u ψ, 2 ψ) = H(, u, u, 2 u) ψ u > 0, ψ Möbius if and only if H(, u, u, 2 u) = f (λ(a u )), where A u := 2 n+2 u n 2 2 u + 2n 2n u n 2 n 2 (n 2) 2 u u 2 (n 2) 2 u 2n n 2 u 2 I, f is a symmetric function of λ = (λ 1,, λ n ).

7 Taking f (λ) = σ 1 (λ) := λ λ n. Then u = n(n 2)u n+2 n 2 is f (λ(a u )) = 2n u = 0 is f (λ(a u )) = 0.

8 Let Γ R n open,convex,symmetric cone,vertex at origin Γ n Γ Γ 1 n Γ n := {λ R n λ i > 0 i}, Γ 1 := {λ R n λ i > 0} i=1 f C 1 (Γ) C 0 (Γ) symmetric function f λi > 0 in Γ i, f > 0 in Γ, f = 0 on Γ Consider f (λ(a u )) = 1, λ(a u ) Γ and f (λ(a u )) = 0 or, the same as, λ(a u ) Γ

9 Let Γ R n open,convex,symmetric cone,vertex at origin Γ n Γ Γ 1 n Γ n := {λ R n λ i > 0 i}, Γ 1 := {λ R n λ i > 0} i=1 f C 1 (Γ) C 0 (Γ) symmetric function f λi > 0 in Γ i, f > 0 in Γ, f = 0 on Γ Consider and f (λ(a u )) = 1, λ(a u ) Γ f (λ(a u )) = 0 or, the same as, λ(a u ) Γ f λi > 0 in Γ means that the first equation is elliptic. The second equation is degenerate elliptic.

10 Motivation

11 Motivation Riemannian manifold (M, g) of dimension n 3, denote the Schouten tensor A g = (n 2) 1 (Ric g [2(n 1)] 1 R g g), where Ric g and R g denote the Ricci and scalar curvature. λ(a g ) = (λ 1,, λ n ) denote the eigenvalues of A g with respect to g. A fully nonlinear version of the Yamabe problem: Assume λ(a g ) Γ on M, does there exist ĝ = u 4 n 2 g such that f (λ(aĝ )) = 1, λ(aĝ ) Γ, on M? If (f, Γ) = (σ 1, Γ 1 ), the Yamabe problem.

12 Answer is Yes if: (i) (f, Γ), f concave, homogeneous of degree 1, (M, g) is locally conformally flat, (ii) (f, Γ) = (σ 1 k k, Γ k ), and k n 2, (iii) (f, Γ) = (σ 1 k k, Γ k ), k = 2.

13 Answer is Yes if: (i) (f, Γ), f concave, homogeneous of degree 1, (M, g) is locally conformally flat, (ii) (f, Γ) = (σ 1 k k, Γ k ), and k n 2, (iii) (f, Γ) = (σ 1 k k, Γ k ), k = 2. Through work of many: [Alice Chang, Gursky, Paul Yang, 2002], [Pengfei Guan, Guofang Wang, 2003], [Aobing Li, L., 2003, 2005], [Gursky, Viaclovsky, 2004, 2007], [Yuxin Ge, Guofang Wang, 2006], [Weimin Sheng, Trudinger, Xujia Wang, 2007], [Luc Nguyen, L., 2014].

14 Answer is Yes if: (i) (f, Γ), f concave, homogeneous of degree 1, (M, g) is locally conformally flat, (ii) (f, Γ) = (σ 1 k k, Γ k ), and k n 2, (iii) (f, Γ) = (σ 1 k k, Γ k ), k = 2. Through work of many: [Alice Chang, Gursky, Paul Yang, 2002], [Pengfei Guan, Guofang Wang, 2003], [Aobing Li, L., 2003, 2005], [Gursky, Viaclovsky, 2004, 2007], [Yuxin Ge, Guofang Wang, 2006], [Weimin Sheng, Trudinger, Xujia Wang, 2007], [Luc Nguyen, L., 2014]. Open in particular if: (f, Γ) = (σ 1 k k, Γ k ), 3 k < n 2.

15 Answer is Yes if: (i) (f, Γ), f concave, homogeneous of degree 1, (M, g) is locally conformally flat, (ii) (f, Γ) = (σ 1 k k, Γ k ), and k n 2, (iii) (f, Γ) = (σ 1 k k, Γ k ), k = 2. Through work of many: [Alice Chang, Gursky, Paul Yang, 2002], [Pengfei Guan, Guofang Wang, 2003], [Aobing Li, L., 2003, 2005], [Gursky, Viaclovsky, 2004, 2007], [Yuxin Ge, Guofang Wang, 2006], [Weimin Sheng, Trudinger, Xujia Wang, 2007], [Luc Nguyen, L., 2014]. Open in particular if: (f, Γ) = (σ 1 k k, Γ k ), 3 k < n 2. Rescaling a blow up sequence of solutions of the geometric equation leads to f (λ(a u )) = 1, in R n, and f (λ(a u )) = 0, or, the same as, λ(a u ) Γ, in R n.

16 Caffarelli, Nirenberg, Spruck, 1985: Introduce (f, Γ) of such type, pioneering work on existence of smooth solutions for Dirichlet problem: { f (λ( 2 u)) = g(x), in Ω R n, u = h(x) on Ω.

17 Caffarelli, Nirenberg, Spruck, 1985: Introduce (f, Γ) of such type, pioneering work on existence of smooth solutions for Dirichlet problem: { f (λ( 2 u)) = g(x), in Ω R n, u = h(x) on Ω. Equation f (λ(a u )) = 1 resembles the above. Additional feature: conformal invariance of equation.

18 Illuminating examples: (f, Γ) = (σ 1 k k, Γ k ), 1 k n σ k (λ) := λ i1 < <λ ik λ i1 λ ik, the k-th elementary symmetric function Γ k : the connected component of {λ R n σ k (λ) > 0} containing the positive cone Γ n = {λ R n λ i > 0 i}

19 Recall assumption: Γ R n open,convex,symmetric cone,vertex at origin Γ n Γ Γ 1 f C 1 (Γ) C 0 (Γ) symmetric function f λi > 0 in Γ i, f > 0 in Γ, f = 0 on Γ With normalization: f (2, 2,, 2) = 1.

20 Recall assumption: Γ R n open,convex,symmetric cone,vertex at origin Γ n Γ Γ 1 f C 1 (Γ) C 0 (Γ) symmetric function f λi > 0 in Γ i, f > 0 in Γ, f = 0 on Γ With normalization: f (2, 2,, 2) = 1. Theorem 1 (Aobing Li and L., 2005) (Liouville type theorem) u C 2 (R n ), f (λ(a u )) = 1, u > 0, R n implies u(x) = ( a ) n a 2 x x 2, a > 0, x R n.

21 Recall assumption: Γ R n open,convex,symmetric cone,vertex at origin Γ n Γ Γ 1 f C 1 (Γ) C 0 (Γ) symmetric function f λi > 0 in Γ i, f > 0 in Γ, f = 0 on Γ With normalization: f (2, 2,, 2) = 1. Theorem 1 (Aobing Li and L., 2005) (Liouville type theorem) u C 2 (R n ), f (λ(a u )) = 1, u > 0, R n implies u(x) = ( a ) n a 2 x x 2, a > 0, x R n. Alice Chang, Gursky, Paul Yang, 2002: (f, Γ) = (σ 1 2 2, Γ 2 ) in R 4.

22 Theorem 2 (L., 2009) (local gradient estimates) For constant 0 < a 1, u C 2 (B 2 ), f (λ(a u )) = a, 0 < u b, in B 2 implies log u C in B 1 C depends only on (f, Γ) and b, independent of a.

23 Theorem 2 (L., 2009) (local gradient estimates) For constant 0 < a 1, u C 2 (B 2 ), f (λ(a u )) = a, 0 < u b, in B 2 implies log u C in B 1 C depends only on (f, Γ) and b, independent of a. Pengfei Guan, Guofang Wang, 2003: (f, Γ) = (σ 1 k k, Γ k ), 1 k n.

24 Theorem 3 (L., 2009) u C 0,1 loc (Rn \ {0}), λ(a u ) Γ in R n \ {0}, viscosity sense implies u radially symmetric about the origin 0.

25 Theorem 4 (Luc Nguyen, L. ) Assume {u k } C 2 (B 2 ), f (λ(a u k )) = 1, u k > 0, in B 2, sup B 1 u k.

26 Theorem 4 (Luc Nguyen, L. ) Assume {u k } C 2 (B 2 ), f (λ(a u k )) = 1, u k > 0, in B 2, sup B 1 u k. Then ɛ > 0, after passing to a subsequence, {x 1 k,, x m k } B 2(0), xk i x j k δ > 0, k, i j, (1 ɛ)u xi k,u k(x i k ) (x) u k (x) (1+ɛ)U xi k,u k(x i k ) (x), x B δ/2 (x i k ). 0 < c u k(xk i ) u k (x j C <, i, j, k, k ) c u k (xk 1) u k(x) C u k (xk 1), in B 1(0) \ m i=1b δ/4 (xk i ), k. U x,µ (x) = µu(µ 2 n 2 (x x)), 1 U(x) = ( ) n 2 1+ x 2 2 satisfies f (λ(a U )) = 1, m, δ, c, C depend only on (f, Γ)..

27 Bôcher theorem for harmonic functions: u = 0, u > 0 in B 1 \ {0} R n, n 3, u(x) = implies a + smooth function, x n 2 a 0 constant.

28 Recall Γ R n open,convex,symmetric cone,vertex at origin Γ n Γ Γ 1

29 Recall Γ R n open,convex,symmetric cone,vertex at origin Γ n Γ Γ 1 Introduce µ Γ [0, n 1], unique number defined by ( µ Γ, 1,, 1) Γ

30 Recall Γ R n open,convex,symmetric cone,vertex at origin Γ n Γ Γ 1 Introduce µ Γ [0, n 1], unique number defined by ( µ Γ, 1,, 1) Γ Facts µ Γ = n 1 if and only if Γ = Γ 1 µ Γ = 0 if and only if Γ = Γ n µ Γk = n k k, 1 k n µ Γk > 1 for k < n 2 µ Γk = 1 for k = n 2 µ Γk < 1 for k > n 2

31 Theorem 5 (Luc Nguyen, L. ) (a Bôcher type theorem): Assume µ Γ > 1 and (1, 0,, 0) Γ. Let 0 < u C 0,1 loc (B 2 \ {0}), λ(a u ) Γ in B 2 \ {0}.

32 Theorem 5 (Luc Nguyen, L. ) (a Bôcher type theorem): Assume µ Γ > 1 and (1, 0,, 0) Γ. Let 0 < u C 0,1 loc (B 2 \ {0}), λ(a u ) Γ in B 2 \ {0}. Then u(x) = a ( 1 x µ Γ 1 + ẘ ) n 2 µ Γ 1 where a = inf B 1 \{0} x n 2 u(x) = lim x 0 x n 2 u(x) 0, ẘ nonnegenative function in C 0,1 loc (B 2 \ {0}) L (B 1 ), either inf B 1 \{0} ẘ > 0 or ẘ 0 in B 2 ẘ C α (B 1 ) 0 < α < 1, if a = 0 or 1 < µ Γ < 2.

33 Theorem 5 (Luc Nguyen, L. ) (a Bôcher type theorem): Assume µ Γ > 1 and (1, 0,, 0) Γ. Let 0 < u C 0,1 loc (B 2 \ {0}), λ(a u ) Γ in B 2 \ {0}. Then u(x) = a ( 1 x µ Γ 1 + ẘ ) n 2 µ Γ 1 where a = inf B 1 \{0} x n 2 u(x) = lim x 0 x n 2 u(x) 0, ẘ nonnegenative function in C 0,1 loc (B 2 \ {0}) L (B 1 ), either inf B 1 \{0} ẘ > 0 or ẘ 0 in B 2 ẘ C α (B 1 ) 0 < α < 1, if a = 0 or 1 < µ Γ < 2. Not hold for harmonic functions, i.e. not for Γ = Γ 1. Neither µ Γ > 1 nor (1, 0,, 0) Γ could be removed Γ = Γ k satisfies µ Γ > 1 and (1, 0,, 0) / Γ if 2 k < n 2 1 < µ Γk < 2 means n 3 < k < n 2

34 Proof of Theorem 4 (Luc Nguyen, L. ) Assume {u k } C 2 (B 2 ), f (λ(a u k )) = 1, u k > 0, in B 2, sup B 1 u k. Then ɛ > 0, after passing to subsequence, {x 1 k,, x m k } B 2(0), xk i x j k δ > 0, k, i j, (1 ɛ)u xi k,u k(x i k ) (x) u k (x) (1+ɛ)U xi k,u k(x i k ) (x), x B δ/2 (x i k ). 0 < c u k(xk i ) u k (x j C <, i, j, k, k ) c u k (xk 1) u k(x) C u k (xk 1), in B 1(0) \ m i=1b δ/4 (xk i ), k. U x,µ (x) = µu(µ 2 n 2 (x x)), 1 U(x) = ( ) n 2 1+ x 2 2 satisfies f (λ(a U )) = 1, m, δ, c, C depend only on (f, Γ).

35 Proposition 1. Assume 0 < v C 0 (R n ), 0 < v k C 2 (B Rk ), R k, f (λ(a v k )) = 1 in B Rk, v k v in Cloc 0 (Rn ). Then v(x) = ( a ) n a 2 x x 2, a > 0, x R n.

36 Proposition 1. Assume 0 < v C 0 (R n ), 0 < v k C 2 (B Rk ), R k, f (λ(a v k )) = 1 in B Rk, v k v in Cloc 0 (Rn ). Then v(x) = ( a ) n a 2 x x 2, a > 0, x R n. v satisfies f (λ(a v )) = 1, in R n in viscosity sense.

37 Proposition 1. Assume 0 < v C 0 (R n ), 0 < v k C 2 (B Rk ), R k, f (λ(a v k )) = 1 in B Rk, v k v in Cloc 0 (Rn ). Then v(x) = ( a ) n a 2 x x 2, a > 0, x R n. v satisfies f (λ(a v )) = 1, in R n in viscosity sense. Open Problem. Let v C 0 loc (Rn ) satisfy f (λ(a v )) = 1, in R n in viscosity sense. Is it true that v(x) = ( a ) n a 2 x x 2, a > 0, x R n?

38 Proposition 1. Assume 0 < v C 0 (R n ), 0 < v k C 2 (B Rk ), R k, f (λ(a v k )) = 1 in B Rk, v k v in Cloc 0 (Rn ). Then v(x) = ( a ) n a 2 x x 2, a > 0, x R n. v satisfies f (λ(a v )) = 1, in R n in viscosity sense. Open Problem. Let v C 0 loc (Rn ) satisfy f (λ(a v )) = 1, in R n in viscosity sense. Is it true that v(x) = ( a ) n a 2 x x 2, a > 0, x R n? Proof of Proposition 1. 0 < v superharmonic, so y n 2 v(y) 2c 0 > 0, y 1. Passing to subsequence, shrinking R k, shrinking c 0, may assume v k (y) v(y) (R k ) n, v k (y) c 0 (R k ) 2 n, y R k.

39 Define for x R n, x + 1 R k /4, λ k (x) = sup{0 < µ R k 4 (v k) x,λ v k in B Rk (0)\B λ (x), 0 < λ < µ}, λ where (v k ) x,λ (y) := ( y x )n 2 v k (x + λ2 (y x) ), the Kelvin y x 2 transformation.

40 Define for x R n, x + 1 R k /4, λ k (x) = sup{0 < µ R k 4 (v k) x,λ v k in B Rk (0)\B λ (x), 0 < λ < µ}, λ where (v k ) x,λ (y) := ( y x )n 2 v k (x + λ2 (y x) ), the Kelvin y x 2 transformation. λ k (x) well defined and C(x) > 0 such that 0 < 1 C(x) λ k (x) R k 4, k.

41 Define for x R n, x + 1 R k /4, λ k (x) = sup{0 < µ R k 4 (v k) x,λ v k in B Rk (0)\B λ (x), 0 < λ < µ}, λ where (v k ) x,λ (y) := ( y x )n 2 v k (x + λ2 (y x) ), the Kelvin y x 2 transformation. λ k (x) well defined and C(x) > 0 such that 0 < 1 C(x) λ k (x) R k 4, k. Reasons: v k (y) c 0 (R k ) 2 n, y = R k, v k, (v k ) 1 and v k bounded above in a fixed radius ball centered at x. To obtain the above, used the gradient estimates: Theorem 2 (L., 2009) For constant 0 < a 1, u C 2 (B 2 ), f (λ(a u )) = 1, 0 < u b, in B 2 implies log u C in B 1 C depends only on (f, Γ) and b, independent of a.

42 Set λ(x) =lim inf λ(x) (0, ]. k

43 Set λ(x) =lim inf λ(x) (0, ]. k Claim. For every x R n, λ(x) < if and only if lim inf y y n 2 v(y) <.

44 Set λ(x) =lim inf λ(x) (0, ]. k Claim. For every x R n, λ(x) < if and only if lim inf y y n 2 v(y) <. Proof of Claim: Used strong maximum principle, Hopf Lemma, the elliptic equation satisfied by v k, the C 0 convergence of v k to v in a fixed ball centered at x, v k v L ( B Rk (0)) = ((R k ) 2 n ).

45 Set λ(x) =lim inf λ(x) (0, ]. k Claim. For every x R n, λ(x) < if and only if lim inf y y n 2 v(y) <. Proof of Claim: Used strong maximum principle, Hopf Lemma, the elliptic equation satisfied by v k, the C 0 convergence of v k to v in a fixed ball centered at x, v k v L ( B Rk (0)) = ((R k ) 2 n ). Consequently, either λ(x) or λ(x) < x. λ(x) leads to: v Constant, which can be ruled out. λ(x) < x leads to: lim y y n 2 v x, λ(x) (y) = α, x. where α := lim inf y y n 2 v(y) <.

46 We have arrived at: 0 < v C 0,1 loc (Rn ), v 0 in R n, for every x R n, there exists 0 < λ(x) < such that v x, λ(x) (y) v(y), y x λ(x), lim y y n 2 v x, λ(x) (y) = α := lim inf y y n 2 v(y), x.

47 We have arrived at: 0 < v C 0,1 loc (Rn ), v 0 in R n, for every x R n, there exists 0 < λ(x) < such that v x, λ(x) (y) v(y), y x λ(x), lim y y n 2 v x, λ(x) (y) = α := lim inf y y n 2 v(y), x. Claim. We can deduce from the above that v(x) = b ( a ) n a 2 x x 2, a, b > 0, x R n. Since v k v in C 0 loc (Rn ) and f (λ(a v k )) = 1, we see easily that b = 1.

48 Proof of Claim. Let ψ(y) = y y 2, and w (x) := (v x, λ(x) ) ψ, x R n.

49 Proof of Claim. Let ψ(y) = y y 2, and w (x) := (v x, λ(x) ) ψ, x R n. Then, for some δ(x) > 0, v ψ w (x) in B δ(x) \ {0}, w (x) (0) = α = lim inf y 0 v ψ(y), v ψ 0, in B δ(x) \ {0},

50 Proof of Claim. Let ψ(y) = y y 2, and w (x) := (v x, λ(x) ) ψ, x R n. Then, for some δ(x) > 0, v ψ w (x) in B δ(x) \ {0}, w (x) (0) = α = lim inf y 0 v ψ(y), v ψ 0, in B δ(x) \ {0}, Let D = {x R n v is differentiable at x}. Since v C 0,1 loc (Rn ), Lebesgue measure R n \ D = 0.

51 Proof of Claim. Let ψ(y) = y y 2, and w (x) := (v x, λ(x) ) ψ, x R n. Then, for some δ(x) > 0, v ψ w (x) in B δ(x) \ {0}, w (x) (0) = α = lim inf y 0 v ψ(y), v ψ 0, in B δ(x) \ {0}, Let D = {x R n v is differentiable at x}. Since v C 0,1 loc (Rn ), Lebesgue measure R n \ D = 0.

52 Proof of Claim. Let ψ(y) = y y 2, and w (x) := (v x, λ(x) ) ψ, x R n. Then, for some δ(x) > 0, v ψ w (x) in B δ(x) \ {0}, w (x) (0) = α = lim inf y 0 v ψ(y), v ψ 0, in B δ(x) \ {0}, Let D = {x R n v is differentiable at x}. Since v C 0,1 loc (Rn ), Lebesgue measure R n \ D = 0. By a Lemma, Namely, for some V R n, w (x) (0) = w ( x) (0), x, x D. w (x) (0) = V, x D.

53 A calculation yields w (x) (0) = (n 2)αx + α n n 2 v(x) n n 2 v(x).

54 A calculation yields Thus w (x) (0) = (n 2)αx + α n n 2 v(x) n n 2 v(x). x (n 2 2 α n n 2 v(x) 2 n 2 n 2 2 α x 2 + V x ) = 0, x D.

55 A calculation yields Thus w (x) (0) = (n 2)αx + α n n 2 v(x) n n 2 v(x). x (n 2 2 α n n 2 v(x) 2 n 2 n 2 2 α x 2 + V x ) = 0, x D. Consequently, for some x R n and d R, v(x) 2 n 2 α 2 n 2 x x 2 + dα 2 n 2.

56 A calculation yields Thus w (x) (0) = (n 2)αx + α n n 2 v(x) n n 2 v(x). x (n 2 2 α n n 2 v(x) 2 n 2 n 2 2 α x 2 + V x ) = 0, x D. Consequently, for some x R n and d R, v(x) 2 n 2 α 2 n 2 x x 2 + dα 2 n 2. Since v > 0, we must have d > 0, so Claim proved. v(x) ( α 2 n 2 d + x x 2 ) n 2 2.

57 The Lemma. For n 2, B 1 R n, w 1, w 2 C 0 (B 1 ), w 1, w 2 differentiable at 0, u L 1 loc (B 1 \ {0}), u 0 in B 1 \ {0}, u(y) max{w 1 (y), w 2 (y)}, y B 1 \ {0}, Then w 1 (0) = w 2 (0) = lim inf y 0 w 1 (0) = w 2 (0). u(y).

58 The Lemma. For n 2, B 1 R n, w 1, w 2 C 0 (B 1 ), w 1, w 2 differentiable at 0, u L 1 loc (B 1 \ {0}), u 0 in B 1 \ {0}, u(y) max{w 1 (y), w 2 (y)}, y B 1 \ {0}, Then w 1 (0) = w 2 (0) = lim inf y 0 w 1 (0) = w 2 (0). u(y). Apply the lemma with w 1 = w (x), w 2 = w ( x), u = v ψ, x, x D.

59 Proposition 2. Assume {v k } C 2 (B Rk ), R k, f (λ(a v k ))(y) = 1, 0 < v k (y) v k (0) = 1, y R k. (1) Then ɛ > 0, k 0 = k 0 (ɛ) and δ = δ (ɛ) such that k > k 0, v k (y) U(y) 2ɛU(y), y δ R k. (2) Recall: ( ) n 2 U(x) := x 2 A U 2I, f (λ(a U )) 1

60 By gradient estimates, and by Proposition 1, after passing to subsequence, v k U, in C 0 loc (Rn ).

61 By gradient estimates, and by Proposition 1, after passing to subsequence, v k U, in C 0 loc (Rn ). Lemma 1. ɛ > 0, k 0, such that k k 0, min v k(y) (1 + ɛ)u(y), 0 < r < R k. y =r

62 By gradient estimates, and by Proposition 1, after passing to subsequence, v k U, in C 0 loc (Rn ). Lemma 1. ɛ > 0, k 0, such that k k 0, min v k(y) (1 + ɛ)u(y), 0 < r < R k. y =r Proof. Facts: U 0,λ (y) < U(y), 0 < λ < 1, y > λ, U 0,1 U. U 0,λ (y) > U(y), λ > 1, y > λ.

63 If for some ɛ > 0, r k Then r k, and Moving plane, min v k (y) > (1 + ɛ)u(y). y =r k (v k ) 0,λ (y) v k (y), 0 < λ < 1 + ɛ 2, y = r k. (v k ) 0,λ (y) v k (y), 0 < λ < 1 + ɛ 2, λ y r k. Sending k. U 0,λ (y) U(y), 0 < λ < 1 + ɛ 2, λ < y <. Contradiction. Lemma 1 proved.

64 Lemma 2. ɛ > 0, small δ 1 > 0, large r 1, k 1 > 0, such that k > k 1, v k (y) (1 ɛ)u(y), y δ 1 R k, r 1 y δ 1 R k v n+2 n 2 k ɛ.

65 Lemma 2. ɛ > 0, small δ 1 > 0, large r 1, k 1 > 0, such that k > k 1, v k (y) (1 ɛ)u(y), y δ 1 R k, r 1 y δ 1 R k v n+2 n 2 k ɛ. Proof. Since v k U in C 0 loc (Rn ), r 1, k 1 such that for all k k 1 v k (y) (1 ɛ 2 )U(y), y r 1, v k (y) (1 ɛ 2 )U(r 1 ) (1 2ɛ 2 )(r 1 ) 2 n, y = r 1,

66 Lemma 2. ɛ > 0, small δ 1 > 0, large r 1, k 1 > 0, such that k > k 1, v k (y) (1 ɛ)u(y), y δ 1 R k, r 1 y δ 1 R k v n+2 n 2 k ɛ. Proof. Since v k U in C 0 loc (Rn ), r 1, k 1 such that for all k k 1 v k (y) (1 ɛ 2 )U(y), y r 1, v k (y) (1 ɛ 2 )U(r 1 ) (1 2ɛ 2 )(r 1 ) 2 n, y = r 1, Since f (λ(a v k )) = 1, δ > 0, Trace (A v k ) δ, Thus v k (y) n 2 2 δv k(y) n+2 n 2 in r 1 y R k.

67 Superharmonicity of v k, maximum principle, we have v k (y) (1 ɛ 2 ) ( y 2 n (R k ) 2 n), r 1 y R k. Thus, δ 1 > 0 depending only on n and ɛ, v k (y) (1 2ɛ 2 ) y 2 n, r 1 y δ 1 R k. Lemma 2 follows, in view of Lemma 1.

68 Superharmonicity of v k, maximum principle, we have v k (y) (1 ɛ 2 ) ( y 2 n (R k ) 2 n), r 1 y R k. Thus, δ 1 > 0 depending only on n and ɛ, v k (y) (1 2ɛ 2 ) y 2 n, r 1 y δ 1 R k. Lemma 2 follows, in view of Lemma 1. Small energy implies L bound consequence of Liouville, as showed before. Lemma 3. δ 0 > 0 and C 0 > 1 such that if 0 < u C 2 (B 2 ), f (λ(a u )) = 1, in B 2, u 2n n 2 δ0, B 2 then u C 0 in B 1.

69 Since v k 1, by Lemma 2, for any ɛ > 0, we have, for large k, 2n n 2 k ɛ. r 1 y δ 1 R k v Lemma 4. C, δ 4 > 0, independent of k, such that v k (y) CU(y), y δ 4 R k. Proof. 4r 1 < r < δ 1 R k /4, consider For large k, ṽ k (z) = r n 2 2 vk (rz), ṽ k (z) 2n n < z <4 where δ 0 > 0 is the number in Lemma 3. 1 < z < 4. 4 = v k (η) 2n n 2 ɛ := δ0, r 4 < η <4r

70 By Lemma 3, ṽ k (z) C, for some universal constant C. By local gradient estimates, 1 < z < 3, 3 log ṽ k (z) C, 1 < z < 2. 2 Thus i.e. max ṽk(z) min ṽk(z). z =1 z =1 max v k(x) C min v k(x) CU(r). x =r x =r - used Lemma 1 for last inequality. Lemma 4 follows immediately.

71 Proof of Proposition 2. Only need to prove that there exists δ and k 0 such that for any k k 0, v k (y) (1 + 2ɛ)U(y), y δ R k.

72 Proof of Proposition 2. Only need to prove that there exists δ and k 0 such that for any k k 0, v k (y) (1 + 2ɛ)U(y), y δ R k. Suppose the contrary, passing to subsequence, y k = δ k R k, δ k 0 +, but v k (y k ) = Since v k v in C 0 loc (Rn ), y k. Consider rescaling of v k : max v k (y) (1 + 2ɛ)U(y k ). y =δ k R k We have ˆv k (z) := y k n 2 v k ( y k z), z < δ 4R k y k. f k (λ(aˆv k ))(z) := y k 2 f (λ(a v k ))(z) = y k 2, z < δ 4R k y k. Since ˆv k C, we can apply gradient estimates to f k to obtain:

73 0 < α < β <, C(α, β) such that for large k, log ˆv k (z) C(α, β), α < z < β. We know from Lemma 1 and the above and min ˆv k(z) 1 + 5ɛ z =1 4, max ˆv k(z) 1 + 3ɛ z =1 2. Passing to subsequence, for some 0 < v C 0,1 loc (Rn \ {0}), ˆv k ˆv in C 1,α loc (Rn \ {0}), 0 < α < 1, and v satisfies in viscosity sense λ(aˆv ) Γ, R n \ {0}.

74 Recall Theorem 3 (L., 2009) u C 0,1 loc (Rn \ {0}), λ(a u ) Γ in R n \ {0}, viscosity sense So ˆv radially symmetric. Passing to subsequence, implies u radially symmetric about the origin 0. min ˆv (z) 1 + 5ɛ z =1 4, max ˆv (z) 1 + 3ɛ z =1 2. Contradiction. Proposition 2 proved.

75 Proof of Theorem 4. By an energy estimate of (Aobing Li, L., 2003), 2n n 2 k C. 1.8 < r 1 < r 2 < 1.9, B 1.9 u B r2 \B r1 u r 1 < r 3 < r 4 < r 2 such that 2n n 2 k δ 0. u k C, in B r4 \ B r3, Go to a maximum point of u k in B r4, and apply Proposition 2,..., then apply Proposition 2 again in the region... Since each time, it takes away a fixed amount of energy, it stops in finite times (the total energy is bounded by C). Theorem 4 is proved.

76 THANK YOU!