YanYan Li. Department of Mathematics. Rutgers University. 110 Frelinghuysen Rd. Piscataway, NJ 08854. 1 Introduction. For n ⥠3, consider. ââu = n(n â 2)u.
Remark on some conformally invariant integral equations: the method of moving spheres YanYan Li Department of Mathematics Rutgers University 110 Frelinghuysen Rd. Piscataway, NJ 08854
1
Introduction
For n ≥ 3, consider
n+2
−∆u = n(n − 2)u n−2 ,
on Rn .
(1)
It was proved by Gidas, Ni and Nirenberg [21] that any positive C 2 solution of (1) satisfying (2) lim inf |x|n−2 u(x) < ∞, |x|→∞
must be of the form u(x) ≡
a 2 1 + a |x − x¯|2
! n−2 2
,
where a > 0 is some constant and x¯ ∈ Rn . Hypothesis (2) was removed by Caffarelli, Gidas and Spruck in [8]; this is important for applications. Such Liouville type theorems have been extended to general conformally invariant fully nonlinear equations by Li and Li ([24]-[27]); see also related works of Viaclovsky ([39]-[40]) and Chang, Gursky and Yang ([13]-[14]). The method used in [21], as well as in much of the above cited work, is the method of moving planes. The method of moving planes has become a very powerful tool in the study of nonlinear elliptic equations, see Alexandrov [1], Serrin [37], Gidas, Ni and Nirenberg [21]-[22], Berestycki and Nirenberg [2], and others. In [30], Li and Zhu gave a proof of the above mentioned theorem of Caffarelli, Gidas and Spruck using the method of moving spheres (i.e. the method of moving 1
2 planes together with the conformal invariance), which fully exploits the conformal invariance of the problem and, as a result, captures the solutions directly rather than going through the usual procedure of proving radial symmetry of solutions and then classifying radial solutions. Significant simplifications to the proof in [30] have been made in Li and Zhang [29]. The method of moving spheres has been used in [24]-[27]. Liouville type theorems for various conformally invariant equations have received much attention, see, in addition to the above cited papers, [23], [17], [15], [33], [41] and [42]. In this paper we study some conformally invariant integral equations. Lieb proved in [31], among other things, that there exist maximizing functions, f, for the Hardy-Littlewood-Sobolev inequality on Rn : k
Z Rn
f(y) dykLq (Rn) ≤ Np,λ,n kfkLp (Rn ) , | · −y|λ
with Np,λ,n being the sharp constant and 1p + λn = 1 + 1q , 1 < p, q, nλ < ∞, n ≥ 1. q When p = q 0 = q−1 or p = 2 or q = 2, Np,λ,n and the maximizing f 0 s are explicitly 2n evaluated. When p = q 0, i.e., p = 2n−λ and q = 2n , the Euler-Lagrange equation for λ a maximizing f is, modulo a positive constant multiple, Z
f(x)p−1 =
Rn
f(y) dy. |x − y|λ
(3)
Writing λ = n − α and u = f p−1 , then 0 < α < n, and equation (3) becomes Z
u(x) =
n+α
Rn
u(y) n−α dy, |x − y|n−α
∀ x ∈ Rn .
(4)
As mentioned above, maximizing solutions f of (3) are classified in [31] and they are, in terms of u, of the form n−α a u(x) ≡ ( ) 2 , (5) 2 d + |x − x¯| where a, d > 0 and some x¯ ∈ Rn . Of course, a is a fixed constant depending only on n and α, while d and x¯ are free. Equation (4), or (3), is conformally invariant in the following sense. Let v be a positive function on Rn , for x ∈ Rn and λ > 0, we define vx,λ(ξ) = (
λ )n−α v(ξ x,λ ), |ξ − x|
ξ ∈ Rn ,
(6)
3 where ξ x,λ = x +
λ2 (ξ − x) . |ξ − x|2
(7)
Then, if u is a solution of (4), so is ux,λ for any x ∈ Rn and λ > 0. The conformal invariance of (4) was used in [31]. More studies on issues concerning the HardyLittlewood-Sobolev inequality, among other things, were given by Carlen and Loss in [9]-[12], where the conformal invariance of the problem was further exploited. After classifying all maximizing solutions of (3), Lieb raised the beautiful question (page 361 of [31]) on the (essentially) uniqueness of solutions of (3), or, equivalently, of (4). He produced (page 363 of [31]) a nontrivial 2n parameter family of solutions of equation (3), or (4), which are not as regular as the maximizers. For α−n instance, modulo a positive constant, |x| 2 is a solution of of (4). In a recent paper, Chen, Li and Ou established the following result which answers n the question of Lieb in the class of L∞ loc (R ). n Theorem 1.1 ([18]) Let u ∈ L∞ loc (R ) be a positive function satisfying (4). Then u is given by (5) for some constants a, d > 0 and some x¯ ∈ Rn .
In an earlier version of the present paper ([28, version one]), we gave a simpler proof of Theorem 1.1. The proof, in the spirit of [30] and [29] and following Section 2 of [29], fully exploits the conformal invariance of the integral equation. It is different from the one in [18]. In particular, we do not follow the usual procedure of proving radial symmetry of solutions and then classifying radial solutions, and we do not need to distinguish n ≥ 2 and n = 1. For the method of moving spheres or moving planes, there are, roughly speaking, three steps, one is to get started with the procedure, second is to prove that the function and the reflected one coincide if the procedure stops, and the third is to handle the case when the procedure never stops. Our arguments are also different for handling these steps. This proof is presented in Section 2. Lieb pointed out to us that his question also concerns functions which are not n ∞ n in L∞ loc (R ). In particular, it is not known a priori that maximizers are in Lloc (R ). This has led us to study the question further and to establish 2n
n−α Theorem 1.2 For n ≥ 1, 0 < α < n, let u ∈ Lloc (Rn ) be a positive solution of (4). Then u ∈ C ∞ (Rn ). 2n n−α An answer to the question of Lieb is therefore known in the class of Lloc (Rn ).
The above mentioned solution |x|
α−n 2
2n
n−α does not belong to Lloc (Rn ), though it belongs
4 2n to Ltloc (Rn ) for any t < n−α . The question remains unanswered for the class of 2n t n Lloc (R ) for t < n−α . In the process of proving Theorem 1.2, we have established the following result which should be of independent interest. n For n ≥ 1 and 0 < α < n, let V ∈ L α (B3 ) be a non-negative function, set
δ(V ) := kV kL nα (B3) .
(8)
n Theorem 1.3 For n ≥ 1, 0 < α < n, ν > r > n−α , there exist positive constants ¯ δ < 1 and C ≥ 1, depending only on n, α, r and ν, such that for any 0 ≤ V ∈ n ¯ h ∈ Lν (B2 ) and 0 ≤ u ∈ Lr (B3 ) satisfying L α (B3 ), with δ(V ) ≤ δ,
Z
u(x) ≤
B3
V (y)u(y) dy + h(x), |x − y|n−α
we have
x ∈ B2 ,
kukLν (B 1 ) ≤ C kukLr (B3 ) + khkLν (B2) .
(9)
(10)
2
n Corollary 1.1 For n ≥ 1, 0 < α < n, ν > r > n−α , R2 > R1 > 0, let 0 ≤ V ∈ n ν r L α (BR2 ), h ∈ L (BR1 ) that 0 ≤ u ∈ L (BR2 ) satisfy
u(x) ≤
Z BR2
V (y)u(y) dy + h(x), |x − y|n−α
x ∈ BR1 .
Then, for some > 0, u ∈ Lν (B ). Remark 1.1 After we proved Theorem 1.2 and Theorem 1.3 in [28, version two], a revision of [18] was made which included another proof of Theorem 1.2. For α = 2 and n ≥ 3, Theorem 1.3 is essentially equivalent to a result of Brezis and Kato (Theorem 2.3 in [6]), so it can be viewed as an integral equation analogue of their theorem. After informing Brezis of Theorem 1.3, he kindly pointed out that it is similar to, though not the same as, Lemma A.1 in [7]. Indeed, our proof of the theorem makes use of special properties of the potential |x|α−n, and it is not clear to us at this point whether the conclusion of the theorem still holds when replacing n n n−α |x|α−n by any Y ∈ Lw , the weak L n−α space, as in Lemma A.1 of [7]. Theorem 1.2, Theorem 1.3 and Corollary 1.1 are established in Section 2.
5 We also study some equations similar to (4), though they do not have the same kind of conformal invariance property. For n ≥ 1, 0 < α < n and µ > 0, let u be positive Lebesgue measurable function in Rn satisfying Z
u(x) =
Rn
u(y)µ dy, |x − y|n−α
∀ x ∈ Rn .
(11)
Theorem 1.4 Let n ≥ 1 and 0 < α < n. Then n (i) For 0 < µ < n−α , equation (11) does not have any positive Lebesgue measurable solution u, unless u ≡ ∞. n ≤ µ < n+α , equation (11) does not have any positive solution (ii) For n−α n−α n(µ−1)
u ∈ Lloc α (Rn ). For µ >
n+α , n−α
we know from Lemma 4.2 that if u is a positive solution in
n(µ−1) α
Lloc (Rn ), u must be in C ∞ (Rn ). Theorem 1.4 is proved in Section 4. In [24]-[27], all conformally invariant second order fully nonlinear equations are classified and Liouville type theorems are established for the elliptic ones. It would be interesting to identify as many as possible conformally invariant integral equations for which (essentially) uniqueness of solutions can be obtained. One class of such equations, similar to (4), is Z
u(x) =
Rn
|x − y|pu(y)−
2n+p p
∀ x ∈ Rn ,
dy,
where n ≥ 1 and p > 0. We study more general equations, similar to (11), including those which are not conformally invariant. For n ≥ 1, p, q > 0, let u be a non-negative Lebesgue measurable function in Rn satisfying Z u(x) = |x − y|p u(y)−q dy, ∀ x ∈ Rn . (12) Rn
Theorem 1.5 For n ≥ 1, p > 0 and 0 < q ≤ 1 + 2n , let u be a non-negative p n Lebesgue measurable function in R satisfying (12). Then q = 1 + 2n and, for some p n constants a, d > 0 and some x¯ ∈ R , u(x) ≡
d + |x − x¯|2 a
!p 2
.
(13)
6 Remark 1.2 For some a = a(n, p) > 0, (13) indeed solves (12) with q = 1 + 2n . p This is proved in Appendix A. The argument also shows that, modulo a constant, (5) is a solution of (4), a known fact whose proofs can be found in [38] (page 131), [31], and, for n ≥ 2, in [35]. Our proof is different. The proof of Theorem 1.5, similar to our proof of Theorem 1.1, is given in Section 5. It turns out that for n = 3, p = 1 and q = 7, integral equation (12) is associated with some fourth order conformal covariant operator on 3−dimensional compact Riemannian manifolds, arising from the study of conformal geometry. See, e.g., Paneitz [36], Fefferman and Graham [19], Branson [3] and Chang and Yang [16]. , associated with Question 1 Is equation (12), in the case p > 0 and q = 1 + 2n p some kind of pseudo-differential conformal covariant operators on n−dimensional compact Riemannian manifolds, the same way the case n = 3, p = 1 and q = 7 is associated with the above mentioned fourth order conformal covariant operator? After posting [28, version one] on the Archive and essentially completing the proof of Theorem 1.5, we became aware of some recent work of Xu [43] where he proved Theorem 1.5 in the special case n = 3, p = 1 and u ∈ C 4(R3 ). He also proved in the same paper that for n = 3, p = 1 and q > 7(= 1 + 2n ), equation (12) does not p 4 3 admit any non-negative solution u in C (R ). Radial solutions of the biharmonic equations corresponding to (12) with n = 3 and p = 1 were studied by McKenna and Reichel in [34]. Question 2 Is it true that for all n ≥ 1, p > 0 and q > 1 + not admit any positive solutions?
2n p
equation (12) does
We point out that if we consider the integral equations of the following form Z
u(x) =
Rn
G(|x − y|, u(y))dy,
u > 0,
∀ x ∈ Rn ,
(14)
and consider the transformation of the form ux,λ(ξ) = h((
λ )2n)u(ξ x,λ ), |ξ − x|
where ξ x,λ is given by (7), and wish that Z Z λ 2n x,λ h(( ) ) G(|ξ − y|, u(y))dy ≡ G(|ξ − z|, ux,λ(z))dz |ξ − x| Rn Rn
(15)
7 for all x, ξ ∈ Rn , λ > 0 and all positive function u, then we are only led to equation together with the transformations we use in (4) and equation (12) with q = 1 + 2n p the paper. Note that condition (15) guarantees that whenever u is a solution of (14) λ so is ux,λ for all x ∈ Rn and λ > 0. The quantity ( |ξ−x| )2n is the Jacobian of the conformal transformation ξ to ξ x,λ . It looks worthwhile to study equation (12) on a bounded domain (existence of solutions, etc.). In this connection, we draw readers0 attention to some works of Brezis and Cabre [4] and Brezis, Dupaigne and Tesei [5].
2
Proof of Theorem 1.3, Corollary 1.1 and Theorem 1.2
In this section we prove Theorem 1.3. Let Z
ξ(x) := B3
V (y)u(y) dy + h(x) − u(x) ≥ 0, |x − y|n−α
x ∈ B2 .
Then u(x) = (Lu)(x) + f(x) + h(x) − ξ(x), where
Z
V (y)u(y) dy, |x − y|n−α
(Lu)(x) = B2
and
Z
f(x) = 2 0 such that
∇y |y|
n−α 2
u(y) · y > 0,
∀ 0 < |y| < r0 .
Consequently uλ(y) < u(y),
∀ 0 < λ < |y| < r0.
(36)
By (29) and the positivity and continuity of u, u(z) ≥
1 C(r0)|z|n−α
∀ |z| ≥ r0 .
(37)
For small λ0 ∈ (0, r0 ) and for 0 < λ < λ0 , uλ(y) = (
λ n−α λ2 y λ0 ) u( 2 ) ≤ ( )n−α sup u ≤ u(y), |y| |y| |y| Br0
∀ |y| ≥ r0 .
Estimate (35), with x = 0 and λ0 (x) = λ0 , follows from (36) and the above.
2
Define, for x ∈ R , n
¯ λ(x) = sup{µ > 0 | ux,λ(y) ≤ u(y) ∀ 0 < λ < µ, |y − x| ≥ λ}. ¯ x) < ∞ for some x¯ ∈ Rn , then Lemma 3.2 If λ(¯ ux¯,λ(¯ ¯ x) ≡ u
on Rn .
(38)
15 Proof of Lemma 3.2. Without loss of generality, we may assume x¯ = 0, and we ¯ ¯ = λ(0), ¯ use notations λ uλ = u0,λ. By the definition of λ, uλ¯ (y) ≤ u(y)
¯ ∀ |y| ≥ λ.
(39)
¯ and the positivity of the kernel, either uλ¯ (y) = u(y) By (34), with x = 0 and λ = λ, ¯ ¯ which we for all |y| ≥ λ—-then we are done—–or uλ¯ (y) < u(y) for all |y| > λ, assume below. By the Fatou lemma, lim inf |y|n−α (u − uλ¯ )(y) |y|→∞
Z
= lim inf |y|→∞
≥
Z
¯ |z|≥λ
¯ |z|≥λ
n+α n+α ¯ y, z)[u(z) n−α |y|n−α K(0, λ; − uλ¯ (z) n−α ]dz
!
¯ n+α n+α λ 1 − ( )n−α [u(z) n−α − uλ¯ (z) n−α ]dz > 0. |z|
Consequently, there exists 1 ∈ (0, 1) such that (u − uλ¯ )(y) ≥
1 |y|n−α
¯ + 1. ∀ |y| ≥ λ
By the above and the explicit formula of uλ, there exists 0 < 2 < 1 such that (u − uλ)(y) ≥
1 1 ¯ + 1, λ ¯≤λ≤λ ¯ + 2. (40) + (uλ¯ − uλ)(y) ≥ ∀ |y| ≥ λ n−α |y| 2|y|n−α
¯≤λ≤λ ¯ + and for Now, for ∈ (0, 2) which we choose below, we have, for λ ¯ + 1, λ ≤ |y| ≤ λ (u − uλ)(y) = ≥
Z
n+α
|z|≥λ
Z
n+α
¯ λ≤|z|≤λ+1
n+α
¯ ¯ λ+2≤|z|≤ λ+3
Z
n+α
K(0, λ; y, z)[u(z) n−α − uλ(z) n−α ]dz n+α
¯ λ≤|z|≤λ+1
n+α
K(0, λ; y, z)[uλ¯ (z) n−α − uλ(z) n−α ]dz
Z
+
n+α
K(0, λ; y, z)[u(z) n−α − uλ (z) n−α ]dz
Z
+ ≥
n+α
K(0, λ; y, z)[u(z) n−α − uλ (z) n−α ]dz
n+α
¯ ¯ λ+2≤|z|≤ λ+3
Because of (40), there exists δ1 > 0 such that n+α
n+α
n+α
K(0, λ; y, z)[u(z) n−α − uλ(z) n−α ]dz.
u(z) n−α − uλ (z) n−α ≥ δ1 ,
¯ + 2 ≤ |z| ≤ λ ¯ + 3. λ
16 Since ∇y K(0, λ; y, z) · y
K(0, λ; y, z) = 0,
|y|=λ
∀ |y| = λ,
= (n − α)|y − z|α−n−2 (|z|2 − |y|2) > 0,
¯ + 2 ≤ |z| ≤ λ ¯ + 3, ∀λ
and the function is smooth in the relevant region, we have, using also the positivity of the kernel, ¯ ≤ λ ≤ |y| ≤ λ ¯ + 1, λ ¯ + 2 ≤ |z| ≤ λ ¯ + 3, K(0, λ; y, z) ≥ δ2(|y| − λ), ∀ λ where δ2 > 0 is some constant independent of . It is easy to see that for some ¯≤λ≤λ ¯ + , constant C > 0 independent of , we have, for λ n+α n+α ¯ ≤ C, |uλ¯ (z) n−α − uλ(z) n−α | ≤ C(λ − λ)
¯ ≤ λ ≤ |z| ≤ λ ¯ + 1, ∀λ
¯ + 1) and (recall that λ ≤ |y| ≤ λ Z
!
Z
1 1 K(0, λ; y, z)dz ≤ | − λ dz| n−α ¯ ¯ |y − z| |y − z|n−α λ≤|z|≤λ+1 λ≤|z|≤λ+1 Z λ n−α 1 ( ) − 1 dz + λ ¯ |y| |y − z|n−α λ≤|z|≤λ+1 ≤ C|y λ − y| + C(|y| − λ) ≤ C(|y| − λ).
¯≤λ≤λ ¯ + and It follows from the above that for small > 0 we have, for λ ¯ + 1, λ ≤ |y| ≤ λ (u − uλ )(y) ≥ −C ≥ (δ1δ2
Z ¯ λ≤|z|≤λ+1
K(0, λ; y, z)dz + δ1 δ2(|y| − λ)
Z
¯ ¯ λ+2≤|z|≤ λ+3
Z ¯ ¯ λ+2≤|z|≤ λ+3
dz
dz − C)(|y| − λ) ≥ 0.
¯ Lemma 3.2 is established. This and (40) violate the definition of λ.
2 ¯ By the definition of λ(x), ux,λ(y) ≤ u(y),
¯ ∀ 0 < λ < λ(x), |y − x| ≥ λ.
Multiplying the above by |y|n−α and sending |y| to infinity yields β = lim inf |y|n−α u(y) ≥ λn−α u(x), |y|→∞
¯ ∀ 0 < λ < λ(x).
(41)
17 ¯ x) < ∞, we use Lemma 3.2 and multiply (38) by |y|n−α and On the other hand, if λ(¯ then send |y| to infinity to obtain ¯ x)n−α u(¯ β = lim |y|n−α u(y) = λ(¯ x) < ∞. |y|→∞
(42)
¯ x) < ∞, then, Proof of Theorem 1.1. (i) If there exists some x¯ ∈ Rn such that λ(¯ n ¯ by (42) and (41), λ(x) < ∞ for all x ∈ R . Applying Lemma 3.2, we have ≡u ux,λ(x) ¯
on Rn , ∀ x ∈ Rn .
By a calculus lemma (lemma 11.1 in [29], see also lemma 2.5 in [30] for α = 2), any C 1 positive function u satisfying the above must be of the form (5). (ii) If λ(x) = ∞ for all x ∈ Rn , then ux,λ(y) ≤ u(y)
∀ |y − x| ≥ λ > 0, x ∈ Rn .
By another calculus lemma (lemma 11.2 in [29], see also lemma 2.2 in [30] for α = 2), u ≡ constant, violating (4). Theorem 1.1 is established.
2
4
Proof of Theorem 1.4
In this section we establish Theorem 1.4. Lemma 4.1 For n ≥ 1, 0 < α < n and µ > 0, let u be a Lebesgue measurable n positive solution of (11) which is not identically equal to ∞. Then, for any t < n−α , µ n t n u ∈ Lloc (R ) ∩ Lloc (R ), β := lim inf (|x|n−α u(x)) ≥ |x|→∞
and
Z |y|>2
Z
Rn
u(y)µdy > 0,
u(y)µ dy < ∞ |y|n−α
(43)
(44)
Proof of Lemma 4.1. Multiplying (11) by |x|n−α , we obtain (43) by applying the Fatou lemma. Since u is not identically equal to ∞, we see from (11) that u is finite almost everywhere. So, for some x1, x2 ∈ B1 , x1 6= x2, we have 2 Z X n i=1 R
u(y)µ dy ≤ u(x1 ) + u(x2) < ∞. |xi − y|n−α
18 It follows that u ∈ Lµloc (Rn ) and (44) holds. For R > 0, we write Z
u(x) = IR(x) + IIR(x) :=
|y|2R |x − y|n−α
(45)
Since u ∈ Lµloc (Rn ) and (44) holds, IIR ∈ L∞ (BR ). On the other hand, for any n 1 < t < n−α , we have, by Cauchy-Schwartz inequality, kIRkLt (BR ) ≤
Z |y| 0, write u as in (45). As usual, IIR ∈ C (BR ). For any 1 < p < µ(n−α) , n . By the property of the Riesz potential, let 1q = 1p − αn . Then q > n−α kIR kLq (BR) ≤ Ckuµ kLp (B2R ) = CkukµLpµ(B2R) < ∞. 0
So u ∈ Lqloc (Rn ). Let µ0 = max(1, µ), since uµ ≤ C + Cuµ , we have u(x) ≤ C
Z |y|2R n
x ∈ BR , u(y)µ dy. |x − y|n−α
n α By (44), h ∈ L∞ (BR ). Since n(µα−1) < n−α , V ∈ Lloc (Rn ). Since u ∈ Lqloc (Rn ) with n q > n−α , we have, by applying Corollary 1.1, u ∈ Lν (B(ν) ) for any ν > 0, where
19 (ν) > 0. Now, back to (45), IR is C ∞ near the origin by bootstrapping. By the translation invariance of the problem, u ∈ C ∞(Rn ). n (ii) For µ ≥ n−α , let V (y) = u(y)µ−1 . We know from Lemma 4.1 that u ∈ Ltloc (Rn ) for all t < n
n . n−α
n(µ−1)
Since u ∈ Lloc α (Rn ) by the assumption, we also have
α (Rn ). Now, for any R2 > R1 > 0, let V ∈ Lloc
Z
h(y) = Then u ∈ Lr (BR2 ) with r = ν > r, and
n(µ−1) , α
Z
u(x) =
|y|>R2
|y|>R2
u(y)µ dy. |x − y|n−µ
V ∈ L α (BR2 ) h ∈ L∞ (BR1 ) ⊂ Lν (BR1 ) for any n
V (y)u(y) dy + h(x), |x − y|n−α
x ∈ BR1 .
By Corollary 1.1, u ∈ Lr (BR1 ). Since R1 > 0 is arbitrary, u ∈ Lrloc (Rn ) for all r > 1. Bootstrap as usual, u ∈ C ∞ (Rn ).
2
For x ∈ Rn , λ > 0 and a positive function v on Rn , let vx,λ be as in (6). Lemma 4.3 For n ≥ 1, 0 < α < n and µ > 0, let u be a positive solution of (11). Then !n+α−µ(n−α) Z ux,λ(z)µ λ dz, ∀ ξ ∈ Rn , (46) ux,λ(ξ) = n−α n |z − x| R |ξ − z| and Z
λ u(ξ) − ux,λ(ξ) = K(x, λ; ξ, z)[u(z) − |z − x| |z−x|≥λ
!n+α−µ(n−α)
µ
where K(x, λ; ξ, z) =
ux,λ (z)µ ]dz, (47)
1 λ 1 −( )n−α x,λ . n−α |ξ − z| |ξ − x| |ξ − z|n−α
Moreover, K(x, λ; ξ, z) > 0,
∀ |ξ − x|, |z − x| > λ > 0.
Proof of Lemma 4.3. The lemma for µ = n+α is established in Section 3. The n−α proof works for all µ > 0 with minor modification.
2
20 Lemma 4.4 For n ≥ 1, 0 < α < n and µ > 0, let u ∈ C 1(Rn ) be a positive solution of (11). Then for any x ∈ Rn , there exists λ0 (x) > 0 such that ux,λ(y) ≤ u(y),
∀ 0 < λ < λ0 (x), |y − x| ≥ λ.
(48)
Proof of Lemma 4.4. This has been proved in Section 3 for µ = proof applies for all µ > 0.
n+α . n−α
The same
2 Define, for x ∈ Rn , ¯ λ(x) = sup{µ0 > 0 | ux,λ(y) ≤ u(y) ∀ 0 < λ < µ0 , |y − x| ≥ λ}. Lemma 4.5 For n ≥ 1, 0 < α < n and 0 < µ < ¯ solution of (11). Then λ(x) = ∞ for all x ∈ Rn .
let u ∈ C 1 (Rn ) be a positive
n+α , n−α
Proof of Lemma 4.5. We prove it by contradiction argument. Suppose that ¯ x) < ∞ for some x¯ ∈ Rn . Without loss of generality, we may assume x¯ = 0, and λ(¯ ¯ ¯ = λ(0), ¯ we use notations λ uλ = u0,λ. By the definition of λ, uλ¯ (y) ≤ u(y)
¯ ∀ |y| ≥ λ.
(49)
¯ n+α−µ(n−α) λ ¯ So, by (49) and (47) ) < 1 for |z| > λ. Since n + α − µ(n − α) > 0, ( |z| ¯ and the positivity of the kernel, we have, for |y| > λ, ¯ with x = 0 and λ = λ,
(u − uλ¯ )(y) = ≥
Z
¯ y, z)[u(z) − λ K(0, λ; ¯ |z| |z|≥λ
!n+α−µ(n−α)
µ
Z
¯ y, z)[1 − λ K(0, λ; ¯ |z| |z|≥λ
uλ¯ (z)µ ]dz
!n+α−µ(n−α)
]uλ¯ (z)µ dz > 0.
Thus, by the Fatou lemma and the above, lim inf |y|n−α (u − uλ¯ )(y) |y|→∞
≥ lim inf |y|→∞
≥
Z
¯ |z|≥λ
Z ¯ |z|≥λ
¯ y, z)[u(z)µ − uλ¯ (z)µ ]dz |y|n−α K(0, λ; !
¯ λ 1 − ( )n−α [u(z)µ − uλ¯ (z)µ ]dz > 0. |z|
21 Consequently, there exists 1 ∈ (0, 1) such that (u − uλ¯ )(y) ≥
1 |y|n−α
¯ + 1. ∀ |y| ≥ λ
By the above and the explicit formula of uλ, there exists 0 < 2 < 1 such that (u − uλ)(y) ≥
1 1 ¯ + 1, λ ¯≤λ≤λ ¯ + 2. (50) + (uλ¯ − uλ)(y) ≥ ∀ |y| ≥ λ n−α |y| 2|y|n−α
Now, using (49) and (50) as in Section 3, for ∈ (0, 2 ) which we choose below, we ¯≤λ≤λ ¯ + and for λ ≤ |y| ≤ λ ¯ + 1, have, for λ Z
(u − uλ)(y) ≥
¯ λ≤|z|≤λ+1
K(0, λ; y, z)[uλ¯ (z)µ − uλ(z)µ ]dz
Z
+
¯ ¯ λ+2≤|z|≤ λ+3
K(0, λ; y, z)[u(z)µ − uλ(z)µ ]dz.
Because of (50), there exists δ1 > 0 such that u(z)µ − uλ (z)µ ≥ δ1 ,
¯ + 2 ≤ |z| ≤ λ ¯ + 3. λ
It was shown in Section 3 that ¯ ≤ λ ≤ |y| ≤ λ ¯ + 1, λ ¯ + 2 ≤ |z| ≤ λ ¯ + 3, K(0, λ; y, z) ≥ δ2(|y| − λ), ∀ λ where δ2 > 0 is some constant independent of . It is easy to see that for some ¯≤λ≤λ ¯ + , constant C > 0 independent of , we have, for λ ¯ ≤ C, |uλ¯ (z)µ − uλ(z)µ | ≤ C(λ − λ)
¯ ≤ λ ≤ |z| ≤ λ ¯ + 1, ∀λ
¯ + 1), as in Section 3, and (recall that λ ≤ |y| ≤ λ Z ¯ λ≤|z|≤λ+1
K(0, λ; y, z)dz ≤ C(|y| − λ).
¯ ≤ λ ≤ λ ¯ + and It follows from the above that for small > 0 we have, for λ ¯ + 1, λ ≤ |y| ≤ λ (u − uλ )(y) ≥ −C ≥ (δ1δ2
Z ¯ λ≤|z|≤λ+1
K(0, λ; y, z)dz + δ1 δ2(|y| − λ)
Z
¯ ¯ λ+2≤|z|≤ λ+3
dz − C)(|y| − λ) ≥ 0.
Z ¯ ¯ λ+2≤|z|≤ λ+3
dz
22 ¯ Lemma 4.5 is established. This and (64) violate the definition of λ.
2 ¯ Proof of Theorem 1.4. According to Lemma 4.5, λ(x) = ∞ for all x ∈ Rn , i.e., ux,λ(y) ≤ u(y)
∀ |y − x| ≥ λ > 0, x ∈ Rn .
By a calculus lemma (lemma 11.2 in [29], see also lemma 2.2 in [30] for α = 2), u ≡ constant, violating (11). Theorem 1.4 is established.
2
5
Proof of Theorem 1.5
In this section we establish Theorem 1.5. Lemma 5.1 For n ≥ 1, p, q > 0, let u be a non-negative Lebesgue measurable function in Rn satisfying (12). Then Z Rn
γ := lim |x|−p u(x) = lim |x|→∞
(1 + |y|p)u(y)−q dy < ∞,
Z
|x|→∞ Rn
(51)
Z |x − y|p −q u(y) dy = u(y)−q dy ∈ (0, ∞), (52) n |x|p R
and, for some constant C ≥ 1, 1 + |x|p ≤ u(x) ≤ C(1 + |x|p), C
∀ x ∈ Rn .
(53)
Proof of Lemma 5.1. We see from (12) that u must be positive everywhere and |{y ∈ Rn | u(y) < ∞}| > 0, where | · | denotes the Lebesgue measure of the set. So there exist R > 1 and some measurable set E such that E ⊂ {y | u(y) < R} ∩ BR, and |E| ≥
1 . R
23 By (12), Z
u(x) =
−q
|x − y| u(y) dy ≥ p
Rn
−q
≥ (R)
Z
E
|x − y|p dy,
Z E
|x − y|p u(y)−q dy
∀ x ∈ Rn .
The first inequality in (53) follows from the above. For some 1 ≤ |¯ x| ≤ 2, Z Rn
|¯ x − y|pu(y)−q dy = u(¯ x) < ∞.
We deduce (51) from the first inequality in (53) and the above. For |x| ≥ 1, |x − y|p | u(y)−q | ≤ (1 + |y|p)u(y)−q , |x|p so, in view of (51), (52) follows from the Lebesgue dominated convergence theorem. The second inequality in (53) follows from (12), (51) and (52).
2 Lemma 5.2 For n ≥ 1, p, q > 0, let u be a non-negative Lebesgue measurable function in Rn satisfying (12). Then u ∈ C ∞ (Rn ). Proof of Lemma 5.2. For R > 0, writing (12) as Z
u(x) = IR (x) + IIR(x) :=
|y|≤2R
|x − y|pu(y)−q dy +
Z |y|>2R
|x − y|pu(y)−q dy.
Because of (51), we can differentiate IIR(x) under the integral for |x| < R, and therefore IIR ∈ C ∞ (BR ). On the other hand, since u−q ∈ L∞ (B2R), clearly IR is at least H¨older continuous in BR . Since R > 0 is arbitrary, u is H¨older continuous in Rn . Now u−q is H¨older continuous in B2R, the regularity of IR further improves and, by bootstrap, we eventually have u ∈ C ∞ (Rn ). Lemma 5.2 is established.
2
Let v be a positive function on Rn . For x ∈ Rn and λ > 0, consider vx,λ(ξ) = (
|ξ − x| p x,λ ) v(ξ ), λ
ξ ∈ Rn ,
24 where ξ x,λ = x +
λ2 (ξ − x) . |ξ − x|2
Note that notation vx,λ in this section is different from that in Section 1-4. Making a change of variables y = z x,λ = x + we have dy = (
λ2 (z − x) , |z − x|2
λ )2n dz. |z − x|
Thus Z |y−x|≥λ
|ξ
x,λ
−q
− y| v(y) dy = p
Z |z−x|≤λ
Z
=
|z−x|≤λ
|ξ x,λ − z x,λ|p v(z x,λ)−q ( |ξ x,λ − z x,λ|p (
λ )2ndz |z − x|
λ )2n−pq vx,λ(z)−q dz. |z − x|
Using (30), we have (
Z
λ )−p |ξ − x|
|y−x|≥λ
Z
=
|z−x|≤λ
|ξ x,λ − y|pv(y)−q dy |ξ − z|p(
λ )2n−pq+p vx,λ(z)−q dz. |z − x|
(54)
Similarly, ( Z
=
λ )−p |ξ − x| |z−x|≥λ
Z |y−x|≤λ
|ξ − z|p (
|ξ x,λ − y|pv(y)−q dy
λ )2n−pq+p vx,λ(z)−q dz. |z − x|
(55)
Lemma 5.3 Let u be a positive solution of (12). Then Z
ux,λ (ξ) =
Rn
|ξ − z|p(
λ )2n−pq+p ux,λ(z)−q dz, |z − x|
∀ ξ ∈ Rn ,
(56)
and ux,λ(ξ) − u(ξ) =
Z |z−x|≥λ
k(x, λ; ξ, z)[u(z)−q − (
λ )2n−pq+p ux,λ(z)−q ]dz, |z − x|
(57)
25 where k(x, λ; ξ, z) = (
|ξ − x| p x,λ ) |ξ − z|p − |ξ − z|p . λ
Moreover ∀ |ξ − x|, |z − x| > λ > 0.
k(x, λ; ξ, z) > 0,
Proof of Lemma 5.3. Since (ξ x,λ )x,λ = ξ and (vx,λ)x,λ ≡ v, identity (56) follows from (12) and (54) and (55) with v = u. Similarly, using also (56), Z
u(ξ) =
|z−x|≥λ
Z
=
|z−x|≥λ
−q
|ξ − z| u(z) dz + p
Z |y−x| 0, let u be a solution of (12). Then for any x ∈ Rn , there exists λ0 (x) > 0 such that ux,λ(y) ≥ u(y),
∀ 0 < λ < λ0 (x), |y − x| ≥ λ.
(58)
Proof of Lemma 5.4. The proof is similar to that of lemma 2.1 in [29] and Lemma 3.1 in Section 3. Without loss of generality we may assume x = 0, and we use the notation uλ = u0,λ.
26 Since p > 0 and u is a positive C 1 function, there exists r0 > 0 such that
p
∇y |y|− 2 u(y) · y < 0,
∀ 0 < |y| < r0 .
Consequently ∀ 0 < λ < |y| < r0.
uλ(y) > u(y),
(59)
By (53), u(z) ≤ C(r0)|z|p
∀ |z| ≥ r0 .
(60)
For small λ0 ∈ (0, r0 ) and for 0 < λ < λ0 , we have, using (53) and (59), uλ(y) = (
|y| p λ2 y |y| ) u( 2 ) ≥ ( )p inf u ≥ u(y), λ |y| λ0 Br0
∀ |y| ≥ r0.
Estimate (58), with x = 0 and λ0 (x) = λ0 , follows from (59) and the above.
2
Define, for x ∈ R , n
¯ λ(x) = sup{µ > 0 | ux,λ(y) ≥ u(y) ∀ 0 < λ < µ, |y − x| ≥ λ}. Lemma 5.5 For n ≥ 1, p > 0 and 0 < q ≤ 1 + 2n , let u be a solution of (12). Then p ¯ λ(x) < ∞,
∀ x ∈ Rn ,
and ux,λ(x) ≡ u Consequently, q = 1 +
∀ x ∈ Rn .
on Rn ,
(61)
2n . p
¯ Proof of Lemma 5.5. By the definition of λ(x), ux,λ(y) ≥ u(y),
¯ ∀ 0 < λ < λ(x), |y − x| ≥ λ.
Multiplying the above by |y|−p and sending |y| to infinity yields, using (52), 0 < γ = lim |y|−pu(y) ≤ λ−p u(x), |y|→∞
¯ ∀ 0 < λ < λ(x).
(62)
¯ Thus λ(x) < ∞ for all x ∈ Rn . Now we prove (61). Without loss of generality, we may assume x = 0, and we ¯ = λ(0), ¯ ¯ use notations λ uλ = u0,λ and y λ = y 0,λ. By the definition of λ, uλ¯ (y) ≥ u(y)
¯ ∀ |y| ≥ λ.
(63)
27 ¯ 2n−pq+p λ ¯ So, by (63), (57), with x = 0 Since 2n − pq + p ≥ 0, ( |z| ) ≤ 1 for |z| ≥ λ. ¯ and the positivity of the kernel, either uλ¯ (y) = u(y) for all |y| ≥ λ—¯ and λ = λ, then we are done (using (57) to see that 2n − pq + p = 0)—–or uλ¯ (y) > u(y) for all ¯ which we assume below. |y| > λ, ¯ and the Fatou lemma By (57), with x = 0 and λ = λ,
lim inf |y|−p(uλ¯ − u)(y) |y|→∞
= lim inf |y|→∞
≥
Z
¯ |z|≥λ
Z
¯ ¯ y, z)[u(z)−q − ( λ )2n−pq+p uλ¯ (z)−q ]dz |y|−pk(0, λ; ¯ |z| |z|≥λ ! |z| ( ¯ )p − 1 [u(z)−q − uλ¯ (z)−q ]dz > 0. λ
Consequently, using also the positivity of (uλ¯ − u), there exists 1 ∈ (0, 1) such that (uλ¯ − u)(y) ≥ 1 |y|p
¯ + 1. ∀ |y| ≥ λ
By the above and the explicit formula of uλ, there exists 0 < 2 < 1 such that (uλ − u)(y) ≥ 1|y|p + (uλ − uλ¯ )(y) ≥
1 p ¯ + 1, λ ¯≤λ≤λ ¯ + 2. (64) |y| , ∀ |y| ≥ λ 2
λ 2n−pq+p Recall that 2n − pq + p ≥ 0 and therefore ( |z| ) ≤ 1 for |z| ≥ λ. For ¯ ≤ λ ≤ λ+ ¯ ¯ ∈ (0, 2) which we choose below, we have, for λ and for λ ≤ |y| ≤ λ+1,
(uλ − u)(y) ≥ ≥
Z |z|≥λ
k(0, λ; y, z)[u(z)−q − uλ (z)−q ]dz
Z
¯ λ≤|z|≤λ+1
k(0, λ; y, z)[u(z)−q − uλ (z)−q ]dz
Z
+ ≥
Z
¯ ¯ λ+2≤|z|≤ λ+3
¯ λ≤|z|≤λ+1
k(0, λ; y, z)[uλ¯ (z)−q − uλ(z)−q ]dz
Z
+
k(0, λ; y, z)[u(z)−q − uλ(z)−q ]dz
¯ ¯ λ+2≤|z|≤ λ+3
k(0, λ; y, z)[u(z)−q − uλ(z)−q ]dz.
Because of (64), there exists δ1 > 0 such that u(z)−q − uλ (z)−q ≥ δ1 ,
¯ + 2 ≤ |z| ≤ λ ¯ + 3. λ
Since k(0, λ; y, z) = 0,
∀ |y| = λ,
28 ∇y k(0, λ; y, z) · y
¯ + 2 ≤ |z| ≤ λ ¯ + 3, ∀λ
= p|y − z|p−2 (|z|2 − |y|2) > 0,
|y|=λ
and the function is smooth in the relevant region, we have, using also the positivity of the kernel, ¯ ≤ λ ≤ |y| ≤ λ ¯ + 1, λ ¯ + 2 ≤ |z| ≤ λ ¯ + 3, k(0, λ; y, z) ≥ δ2(|y| − λ), ∀ λ where δ2 > 0 is some constant independent of . It is easy to see that for some ¯≤λ≤λ ¯ + , constant C > 0 independent of , we have, for λ ¯ ≤ C, |uλ¯ (z)−q − uλ(z)−q | ≤ C(λ − λ) ¯ + 1) and (recall that λ ≤ |y| ≤ λ Z
¯ λ≤|z|≤λ+1
k(0, λ; y, z)dz ≤ C(|y| − λ) +
¯ ≤ λ ≤ |z| ≤ λ ¯ + 1, ∀λ
Z ¯ λ≤|z|≤λ+1 λ
|y λ − z|p − |y − z|p dz
≤ C(|y| − λ) + C|y − y| ≤ C(|y| − λ). ¯≤λ≤λ ¯ + and It follows from the above that for small > 0 we have, for λ ¯ + 1, λ ≤ |y| ≤ λ (uλ − u)(y) ≥ −C
Z
≥ (δ1δ2
¯ λ≤|z|≤λ+1
k(0, λ; y, z)dz + δ1 δ2(|y| − λ)
Z
¯ ¯ λ+2≤|z|≤ λ+3
Z
¯ ¯ λ+2≤|z|≤ λ+3
dz
dz − C)(|y| − λ) ≥ 0.
¯ Lemma 5.5 is established. This and (64) violate the definition of λ.
2 Proof of Theorem 1.5. According to Lemma 5.5, q = 1 + ux,λ(x) ≡u ¯
2n p
and
on Rn , ∀ x ∈ Rn .
By a calculus lemma (lemma 11.1 in [29], see also lemma 2.5 in [30] for α = 2), any C 1 positive function u satisfying the above must be of the form (13).
2 6
Appendix A
In this appendix, we show, as pointed out in Remark 1.2, that for some a = a(n, p) > 0, (13) solves (12) with q = 1 + 2n . Our proof works equally well for p equation (4).
29 For n ≥ 1 and p ∈ (−n, 0) ∪ (0, ∞), we consider integral equation Z
u(x) =
Rn
|x − y|pu(y)−
2n+p p
∀ x ∈ Rn .
dy,
(65)
Lemma 6.1 For n ≥ 1 and p ∈ (−n, 0) ∪ (0, ∞), there exists a unique a = a(n, p) > 0 such that for any x¯ ∈ Rn and d > 0, u(x) =
!p
d + |x − x¯|2 a
2
satisfies (65). . For a positive function v, and for x ∈ Rn and λ > 0, we use Proof. Let q = 1 + 2n p notation |ξ − x| p x,λ vx,λ(ξ) = ( ) v(ξ ), ξ ∈ Rn , λ where ξ x,λ is given by (7). By the conformal invariance, we only need to prove that modulo a positive constant multiple, p
u(x) := (1 + |x|2) 2 satisfies (65). Set
Z
u˜(x) =
Rn
|x − y|pu(y)−q dy,
x ∈ Rn .
We only need to show that u˜ is q a constant multiple of u. n For any x ∈ R , let λ(x) := 1 + |x|2. Observe that ux,λ(x) ≡ u,
on Rn ,
∀ x ∈ Rn .
(66)
Making a change of variables y = z x,λ(x) = x +
λ(x)2 (z − x) , |z − x|2
we have, using (66) and the conformal invariance of the equation ( (54), (55), (31) and (32)), Z |ξ − x| p Z x,λ(x) p −q u˜x,λ(x)(ξ) = ( ) |ξ − y| u(y) dy = |ξ − z|pux,λ(x)(z)−q dz = u˜(ξ). n n λ(x) R R
Multiplying this by |ξ|−p and sending |ξ| to ∞ leads to p
lim |ξ|−p u˜(ξ) = λ(x)−p u˜(x) = (1 + |x|2)− 2 u˜(x) = u(x)−1 u˜(x),
|ξ|→∞
∀ x ∈ Rn .
So u˜ is a constant multiple of u, and we are done.
2
30
7
Appendix B
In this Appendix, we present some calculus lemmas obtained with Nirenberg. These lemmas under the stronger assumption f ∈ C 1(Rn ) have been used repeatedly in some works on Liouville type theorems for conformally invariant equations, see, e.g., [29], [30], [24]-[27] and the present paper. Lemma 7.1 For n ≥ 1 and ν ∈ R, let f be a function defined on Rn and valued in [−∞, ∞] satisfying λ |y − x|
!ν
f(x +
λ2 (y − x) ) ≤ f(y), |y − x|2
∀ |x − y| > λ > 0.
(67)
Then f ≡ Constant or ±∞. Remark 7.1 If the first inequality in (67) is reversed, the conclusion still holds, since we can replace f by −f. Proof. For all b > 1, y, z ∈ Rn and y 6= z, let q
x = x(b) = y + b(z − y),
λ = λ(b) =
Then, z =x+ and, by (67), λ |y − x| Since
|z − x||y − x|.
λ2 (y − x) , |y − x|2
!ν
f(z) ≤ f(y). v u
u |z − x| λ lim = lim t = 1, b→∞ |y − x| b→∞ |y − x|
we have f(z) ≤ f(y). Lemma 7.1 follows since y 6= z are arbitrary.
2
Lemma 7.2 Let n ≥ 1, ν ∈ R and f ∈ C 0(Rn ). Suppose that for every x ∈ Rn , there exists λ(x) > 0 such that λ(x) |y − x|
!ν
f(x +
λ(x)2 (y − x) ) = f(y), |y − x|2
∀ y ∈ Rn \ {x}.
(68)
31 Then, for some a ≥ 0, d > 0 and x¯ ∈ Rn , 1 f(x) ≡ ±a d + |x − x¯|2
!ν
2
.
Proof of Lemma 7.2. By (68) and the continuity of f, α := lim |y|ν f(y) = λ(x)ν f(x), |y|→∞
∀ x ∈ Rn .
(69)
If ν = 0, then f ≡ α, and we are done. On the other hand, the case ν < 0 can 2(y−x) in (68). So we will easily be reduced to the case of ν > 0 if we let z = x + λ(x) |y−x|2 assume that ν > 0. If α = 0, then f ≡ 0, we are done. Otherwise, replacing f by a non-zero multiple of f, we may assume that α = 1. Since f(y) → 0 as |y| tends to ∞, and since f is continuous and positive, f has a maximum point, and we may assume that f has a maximum point at the origin. For any x ∈ Rn , we have, for large |y|, !ν
|y| λ(x)2 (y − x) |y| f(y) = λ(x) f(x + ) |y − x| |y − x|2 νx · y λ(x)2 (y − x) −2 + O(|y| )]f(x + ), = λ(x)ν [1 + |y|2 |y − x|2 ν
ν
and, by (69) and α = 1, λ(x)2 (y − x) ) − f(x)] |y − x|2 λ(x)2 (y − x) λ(x)ν νx · y +[ + O(|y|−1 )]f(x + ). |y| |y − x|2
|y|[|y|ν f(y) − 1] = |y|λ(x)ν [f(x +
(70)
Taking x = 0 in the above and using the fact that f has a maximum point at the origin, we obtain lim sup |y|[|y|ν f(y) − 1] ≤ 0. (71) |y|→∞
Claim. For any > 0, there exists M such that for any |y| ≥ M , there exists x˜ = x˜(y) satisfying x˜ +
λ(˜ x)2 (y − x˜) =0 |y − x˜|2
and
|˜ x| ≤ .
(72)
32 Indeed we know from (69) and α = 1 that λ(x) = f(x)−1/ν for all x ∈ Rn . For any ∈ (0, 1), pick M > 1 so that 2 (M − )−1 max f(x)− ν < , |x|≤ 2
then, for all |y| ≥ M , λ(x)2 (y max |x|≤
− x) −1 − ν2 −1 − ν2 . = max |y − x| f(x) ≤ (M − ) max f(x) < |x|≤ |x|≤ |y − x|2 2
Thus, by a degree argument using the continuity of f, there exists x˜ = x˜(y) satisfying (72). With x = x˜(y) in (70), we obtain lim inf |y|[|y|ν f(y) − 1] ≥ −νf(0) max f(z)− ν . 1
|y|→∞
|z|≤
Sending to 0, we have lim inf |y|[|y|ν f(y) − 1] ≥ 0. |y|→∞
Thus, in view of (71), lim |y|[|y|ν f(y) − 1] = 0.
|y|→∞
(73)
Let e1 = (1, 0, · · · , 0), · · · , en = (0, · · · , 0, 1). For any x ∈ Rn , 1 ≤ i ≤ n, and t ∈ R, let y = y(x, t, i) ∈ Rn be defined by tei =
λ(x)2 (y − x) . |y − x|2
Taking this y in (70) and sending t to 0, we obtain, in view of (73), 2 ∂f f(x + tei ) − f(x) νx · ei f(x) (x) = lim =− = −νxi f(x)1+ ν . 2 t→0 ∂xi t λ(x)
By the continuity of f, we know that f is in C 1 (Rn ), and we complete the proof 2 of Lemma 7.2 by writing the above system of PDEs as ∂x∂ i [f(x)− ν − |x|2] = 0 and solving it.
2 Acknowledgment. We thank Chen, Li and Ou for sending us their preprint [18] and thank Ou for an interesting talk on the work at a symposium. We are grateful to
33 Lieb for the encouragement and for pointing out the need to study the (essentially) n uniqueness of solutions of (4) beyond the L∞ loc (R ) class and bringing to our attention of solutions in [31] which are not maximizers. This has led us to study the regularity issue and to establish Theorem 1.2 and Theorem 1.3. We thank Brezis for pointing out the relation between Theorem 1.3 and Lemma A.1 in [7]. We thank Nirenberg for his interest in the work and for helping us to complete Appendix B (though not used in the present paper). The work was partially supported by NSF Grant DMS-0100819.
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