Pointwise Bounds and Blow-up for Choquard-Pekar Inequalities at an Isolated Singularity
aa r X i v : . [ m a t h . A P ] D ec Pointwise Bounds and Blow-up for Choquard-Pekar Inequalities atan Isolated Singularity
Marius Ghergu ∗ and Steven D. Taliaferro †‡ Abstract
We study the behavior near the origin in R n , n ≥
3, of nonnegative functions u ∈ C ( R n \{ } ) ∩ L λ ( R n ) (0.1)satisfying the Choquard-Pekar type inequalities0 ≤ − ∆ u ≤ ( | x | − α ∗ u λ ) u σ in B (0) \{ } (0.2)where α ∈ (0 , n ) , λ > , and σ ≥ ∗ is the convolution operation in R n . Weprovide optimal conditions on α, λ , and σ such that nonnegative solutions u of (0.1,0.2) satisfypointwise bounds near the origin. Keywords : pointwise bound; blow-up; isolated singularity; Choquard-Pekar equation; Rieszpotential.
In this paper we study the behavior near the origin in R n , n ≥
3, of nonnegative functions u ∈ C ( R n \{ } ) ∩ L λ ( R n ) (1.1)satisfying the Choquard-Pekar type inequalities0 ≤ − ∆ u ≤ ( | x | − α ∗ u λ ) u σ in B (0) \{ } (1.2)where α ∈ (0 , n ) , λ >
0, and σ ≥ ∗ is the convolution operation in R n . Theregularity condition u ∈ L λ ( R n ) in (1.1) is required for the nonlocal convolution operation in (1.2)to make sense.A motivation for the study of (1.1,1.2) comes from the equation − ∆ u = ( | x | − α ∗ u λ ) | u | λ − u in R n , (1.3)where α ∈ (0 , n ) and λ >
1. For n = 3, α = 1, and λ = 2, equation (1.3) is known in the literatureas the Choquard-Pekar equation and was introduced in [16] as a model in quantum theory of aPolaron at rest (see also [2]). Later, the equation (1.3) appears as a model of an electron trappedin its own hole, in an approximation to Hartree-Fock theory of one-component plasma [7]. More ∗ School of Mathematics and Statistics, University College Dublin, Belfield, Dublin 4, Ireland; [email protected] † Mathematics Department, Texas A&M University, College Station, TX 77843-3368; [email protected] ‡ Corresponding author, Phone 001-979-845-3261, Fax 001-979-845-6028
Schr¨odinger-Newton equation .The Choquard-Pekar equation (1.3) has been investigated for a few decades by variationalmethods starting with the pioneering works of Lieb [7] and Lions [8, 9]. More recently, new andimproved techniques have been devised to deal with various forms of (1.3) (see, e.g., [10, 11, 13,14, 15, 19] and the references therein).Using nonvariational methods, the authors in [14] obtained sharp conditions for the nonexistenceof nonnegative solutions to − ∆ u ≥ ( | x | − α ∗ u λ ) u σ in an exterior domain of R n , n ≥ Question 1.
Suppose α ∈ (0 , n ) and λ > σ , ifany, does there exist a continuous function ϕ : (0 , → (0 , ∞ ) such that all nonnegative solutions u of (1.1,1.2) satisfy u ( x ) = O ( ϕ ( | x | )) as x → ϕ when it exists?We call the function ϕ in (1.4) a pointwise bound for u as x → Remark 1.
Let u λ ∈ C ( R n \{ } ) be a nonnegative function such that u λ = 0 in R n \ B (0) and u λ ( x ) = ( | x | − ( n − if 0 < λ < nn − λ ≥ nn − for 0 < | x | < . Then u λ ∈ L λ ( R n ) and − ∆ u λ = 0 in B (0) \{ } . Hence u λ is a solution of (1.1,1.2) for all α ∈ (0 , n ), λ >
0, and σ ≥
0. Thus any pointwise bound for nonnegative solutions u of (1.1,1.2) as x → u λ ( x ) and whenever u λ ( x ) is such a bound it is necessarily optimal. Inthis case we say u is harmonically bounded at 0.In order to state our results for Question 1, we define for each α ∈ (0 , n ) the continuous,piecewise linear function g α : (0 , ∞ ) → [0 , ∞ ) by g α ( λ ) = nn − if 0 < λ < n − αn − n − αn − − λ if n − αn − ≤ λ < nn − max { , − α − n λ } if λ ≥ nn − . (1.5)According to the following theorem, if the point ( λ, σ ) lies below the graph of σ = g α ( λ ) thenall nonnegative solutions u of (1.1,1.2) are harmonically bounded at 0. Theorem 1.1.
Suppose u is a nonnegative solution of (1.1,1.2) where α ∈ (0 , n ) , λ > , and ≤ σ < g α ( λ ) . Then u is harmonically bounded at , that is, as x → , u ( x ) = ( O ( | x | − ( n − ) if < λ < nn − O (1) if λ ≥ nn − . (1.6) Moreover, if λ ≥ nn − then u has a C extension to the origin, that is, u = w | R n \{ } for somefunction w ∈ C ( R n ) .
2y Remark 1 the bound (1.6) for u is optimal.By the next theorem, if the point ( λ, σ ) lies above the graph of σ = g α ( λ ) then there does notexist a pointwise bound for nonnegative solutions of (1.1,1.2) as x → Theorem 1.2.
Suppose α, λ , and σ are constants satisfying α ∈ (0 , n ) , λ > , and σ > g α ( λ ) . Let ϕ : (0 , → (0 , ∞ ) be a continuous function satisfying lim t → + ϕ ( t ) = ∞ . Then there exists a nonnegative solution u of (1.1,1.2) such that u ( x ) = O ( ϕ ( | x | )) as x → . Theorems 1.1 and 1.2 completely answer Question 1 when the point ( λ, σ ) does not lie on thegraph of g α . Concerning the critical case that ( λ, σ ) lies on the graph of g α we have the followingresult. Theorem 1.3.
Suppose α ∈ (0 , n ) .(i) If < λ < n − αn − and σ = g α ( λ ) then all nonnegative solutions u of (1.1,1.2) are harmonicallybounded at .(ii) If λ = n − αn − and σ = g α ( λ ) then there does not exist a pointwise bound for nonnegativesolutions u of (1.1,1.2) as x → .(iii) If α ∈ (2 , n ) , λ > nα − , and σ = g α ( λ ) then there does not exist a pointwise bound fornonnegative solutions u of (1.1,1.2) as x → . If u is a nonnegative solution of (1.1,1.2) where ( λ, σ ) lies in the first quadrant of the λσ -planeand σ = g α ( λ ) then according to Theorems 1.1 and 1.2 either(i) u is bounded around the origin and can be extended to a C function in the whole R n ; or(ii) u can be unbounded around the origin but must satisfy u = O ( | x | − ( n − ) as x →
0; or(iii) no pointwise a priori bound exists for u as x →
0, that is solutions can be arbitrarily largearound the origin.The regions in which these three possibilities occur are depicted in Figs. 1–3 below.If α ∈ (0 , n ) and λ > < λ < n − αn − ;(ii) n − αn − ≤ λ < nn − ;(iii) nn − ≤ λ < ∞ .The proofs of Theorems 1.1–1.3 in case (i)(resp. (ii), (iii)) are given in Section 3 (resp. 4, 5).In Section 2 we provide some lemmas needed for these proofs. Our approach relies on an integralrepresentation formula for nonnegative superharmonic functions due to Brezis and Lions [1] (seeLemma 2.1 below) together with various integral estimates for Riesz potentials.Finally we mention that throughout this paper ω n denotes the volume of the unit ball in R n and by Riesz potential estimates we mean the estimates given in [5, Lemma 7.12] and [17, Chapter5, Theorem 1]. See also [4, Appendix C]. 3 − α n − 2 n − 2 n n − α n − 2 n − 2 n α 2 n − λ σ = − − 2α n u(x)=O ( | x | −(n−2) ) u(x)=O ( ) σ Arbitrarily large solutions λ Figure 1: Case α ∈ (2 , n ). n − 2 n n − 2 n u(x)=O ( ) σ = u(x)=O ( | x | −(n−2) ) σ λ Arbitrarily large solutions
Figure 2: Case α = 2. n − α n − 2 n − 2 n n − α n − 2 n − 2 n α 2 n − λ σ = − u(x)=O ( ) u(x)=O ( | x | −(n−2) ) σ λ Arbitrarily large solutions
Figure 3: Case α ∈ (0 , Preliminary lemmas
In this section we provide some lemmas needed for the proofs of our results in Sections 3–5.
Lemma 2.1.
Suppose u is a nonnegative solution of (1.1,1.2) for some constants α ∈ (0 , n ) , λ > ,and σ ≥ . Let v = u + 1 . Then v ∈ C ( R n \{ } ) ∩ L λ ( B (0)) (2.1) and, for some positive constant C , v satisfies ≤ − ∆ v ≤ C [ I n − α ( v λ )] v σ v ≥ ) in B (0) \{ } , (2.2) where ( I β f )( x ) := Z | y | < f ( y ) dy | x − y | n − β for β ∈ (0 , n ) . (2.3) Also − ∆ v, v µ ∈ L ( B (0)) for all µ ∈ [1 , nn − and v ( x ) = m | x | n − + h ( x ) + C Z | y | < − ∆ v ( y ) dy | x − y | n − for < | x | < where m ≥ and C > are constants and h is harmonic and bounded in B (0) .Proof. (2.1) follows from (1.1) and the definition of v .For 0 < | x | < Z | y | > u ( y ) λ dy | x − y | α ≤ (cid:18) max ≤| y |≤ u ( y ) λ (cid:19) Z < | y | < dy | x − y | α + Z | y | > u ( y ) λ dy ≤ C ≤ C min | z |≤ Z | y | < dy | z − y | α , where, as usual, C is a positive constant whose value may change from line to line. Thus for0 < | x | < Z R n u ( y ) λ dy | x − y | α ≤ Z | y | < u ( y ) λ dy | x − y | α + C Z | y | < λ | x − y | α dy ≤ C Z | y | < ( u ( y ) + 1) λ | x − y | α dy = C [ I n − α ( v λ )]( x ) . Hence, since ∆ u = ∆ v and u < v we see that (2.2) follows from (1.2). Also (2.1), (2.2), and [1]imply (2.4) with µ = 1 and (2.5), which together with Riesz potential estimates, yield the completestatement (2.4).The following lemma will be needed for the proof of Theorem 1.2 when 0 < λ ≤ nn − .5 emma 2.2. Suppose α ∈ (0 , n ) and λ ∈ (0 , nn − ] . Let { x j } ⊂ R n , n ≥ , and { r j } , { ε j } ⊂ (0 , be sequences satisfying < | x j +1 | < | x j | < / , (2.6)0 < r j < | x j | / and ∞ X j =1 ( ε λj + ε j ) < ∞ . (2.7) Then there exists a nonnegative function u ∈ C ∞ ( R n \{ } ) ∩ L λ ( R n ) (2.8) such that ≤ − ∆ u ≤ ε j r nj if < λ < nn − ε j r nj (log rj ) n − n if λ = nn − in B r j ( x j ) , (2.9) − ∆ u = 0 in B (0) \ ( { } ∪ ∞ [ j =1 B r j ( x j )) , (2.10) u ≥ A ε j r n − j if < λ < nn − ε j r n − j (log rj ) n − n if λ = nn − in B r j ( x j ) , (2.11) and for x ∈ B r j ( x j ) Z | y | < u ( y ) λ dy | x − y | α ≥ B ε λj if < λ < n − αn − ε λj log r j if λ = n − αn − ε λj r n − α − ( n − λj if n − αn − < λ < nn − ε λj r − αj (log r j ) − if λ = nn − (2.12) where A = A ( n ) and B = B ( n, λ, α ) are positive constants.Proof. Let ψ : R n → [0 ,
1] be a C ∞ function whose support is B (0). Define ψ j , f j : R n → [0 , ∞ )by ψ j ( y ) = ψ ( η ) where y = x j + r j η and f j = M j ψ j where M j = ε j r nj δ j and δ j = ( < λ < nn − (log r j ) n − n if λ = nn − . Since Z R n f j ( y ) dy = M j Z R n ψ ( η ) r nj dη ≤ ε j Z R n ψ ( η ) dη and by (2.6) and (2.7) the supports B r j ( x j ) of the functions f j are disjoint and contained in B / (0) we see by (2.7) that f := ∞ X j =1 f j ∈ C ∞ ( R n \{ } ) ∩ L ( R n ) and supp( f ) ⊂ B (0) . (2.13)Defining v j ( y ) = Z R n f j ( z ) dz | y − z | n − x = x j + r j ξ, y = x j + r j η, and z = x j + r j ζ, we find for β ∈ [0 , n ) and R ∈ [ ,
2] that Z | y − x j | 1) then using (2.16) we see that n ( n − ω n k v k L λ ( B (0) = k ∞ X j =1 v j k L λ ( B (0)) = Z B (0) ∞ X j =1 v j ( y ) λ dy /λ ≤ Z B (0) ⊂ B ( x j ) ∞ X j =1 v j ( y ) λ dy /λ ≤ ∞ X j =1 Z B ( x j ) v j ( y ) λ dy /λ ≤ ∞ X j =1 Cε λj /λ < ∞ by (2.7). Thus by (2.13) v ∈ C ∞ ( R n \{ } ) ∩ L λ ( B (0)) (2.17)7nd − ∆ v = f in R n \{ } . (2.18)Taking β = α ∈ (0 , n ) and R = in (2.14) and using (2.15) we find for | x − x j | < r j (i.e. | ξ | < n ( n − ω n ) λ Z | y | < v ( y ) λ dy | x − y | α ≥ Z | y − x j | < / v j ( y ) λ dy | x − y | α ≥ Cε λj δ − λj r n − α − ( n − λj I j ( ξ )where I j ( ξ ) := Z | η | < rj (cid:16) | η | n − +1 (cid:17) λ | ξ − η | α dη ≥ Z | η | < (cid:16) n − +1 (cid:17) λ | ξ − η | α dη + Z < | η | < rj (cid:16) | η | n − (cid:17) λ | ξ − η | α dη ≥ C ( n, λ, α ) + (cid:18) (cid:19) α λ Z < | η | < rj | η | ( n − λ + α dη ≥ C ( n, λ, α ) r n − α − ( n − λj if 0 < λ < n − αn − log r j if λ = n − αn − n − αn − < λ < nn − λ = nn − . Thus for | x − x j | < r j we have Z | y | < v ( y ) λ dy | x − y | α ≥ C ( n, λ, α ) ε λj if 0 < λ < n − αn − ε λj log r j if λ = n − αn − ε λj r n − α − ( n − λj if n − αn − < λ < nn − ε λj r − αj (log r j ) − if λ = nn − . (2.19)Also, for | x − x j | < r j we have v ( x ) ≥ n ( n − ω n Z B rj ( x j ) f ( y ) dy | x − y | n − = C ( n ) Z B rj ( x j ) M j ψ j ( y ) | x − y | n − dy = C ( n ) Z | η | < M j ψ ( η ) r nj r n − j | ξ − η | n − dη = C ( n ) ε j r n − j δ j Z | η | < ψ ( η ) dη | ξ − η | n − ≥ A ε j r n − j δ j (2.20)where A = C ( n ) min | ξ |≤ Z | η | < ψ ( η ) dη | ξ − η | n − > . Finally, letting u = χv where χ ∈ C ∞ ( R n → [0 , χ = 1 in B (0) and χ = 0 in R n \ B (0), it follows from (2.17)–(2.20) that u satisfies (2.8)–(2.12).8he following lemma will be needed for the proof of Theorem 1.2 when λ > nn − . Lemma 2.3. Suppose α ∈ (0 , n ) and λ > nn − . Let { x j } ⊂ R n , n ≥ , and { r j } , { ε j } ⊂ (0 , besequences satisfying < | x j +1 | < | x j | < / , (2.21)0 < r j < | x j | / and ∞ X j =1 ε j < ∞ . (2.22) Then there exists a positive function u ∈ C ∞ ( R n \{ } ) ∩ L λ ( R n ) (2.23) such that ≤ − ∆ u ≤ ε j r n/λj in B r j ( x j ) (2.24) − ∆ u = 0 in R n \ ( { } ∪ ∞ [ j =1 B r j ( x j )) (2.25) u ≥ Aε j r n/λj in B r j ( x j ) (2.26) and Z | y − x j | 1] be a C ∞ function whose support is B (0). Define ψ j , f j : R n → [0 , ∞ )by ψ j ( y ) = ψ ( η ) where y = x j + r j η and f j = M j ψ j where M j = ε j r n/λj . Since Z R n f j ( y ) dy = M j Z R n ψ ( η ) r nj dη = ε j r ( n − λ − nλ j Z R n ψ ( η ) dη ≤ ε j Z R n ψ ( η ) dη and by (2.21) and (2.22) the supports B r j ( x j ) of the functions f j are disjoint and contained in B / (0) we see by (2.22) that f := ∞ X j =1 f j ∈ C ∞ ( R n \{ } ) ∩ L ( R n ) and supp( f ) ⊂ B (0) . (2.28)Defining u j ( y ) = Z R n f j ( z ) dz | y − z | n − and making the change of variables x = x j + r j ξ, y = x j + r j η, and z = x j + r j ζ, 9e find for β ∈ [0 , n ) that Z R n or B rj ( x j ) u j ( y ) λ dy | x − y | β = Z R n or | η | < (cid:18)R R n M j ψ ( ζ ) r nj dζr n − j | η − ζ | n − (cid:19) λ r βj | ξ − η | β r nj dη = ε λj r − βj Z R n or | η | < (cid:16)R R n ψ ( ζ ) dζ | η − ζ | n − (cid:17) λ | ξ − η | β dη. (2.29)Also 0 < C ( n ) < R R n ψ ( ζ ) dζ | η − ζ | n − | η | n − +1 < C ( n ) < ∞ for η ∈ R n . (2.30)Taking β = 0 in (2.29) and using (2.30) we get Z R n u j ( y ) λ dy ≤ C ( n, λ ) ε λj Z R n (cid:18) | η | n − + 1 (cid:19) λ dη ≤ C ( n, λ ) ε λj (2.31)because λ > n/ ( n − u ( x ) := 1 n ( n − ω n Z R n f ( y ) dy | x − y | n − = 1 n ( n − ω n ∞ X j =1 u j ( x ) for x ∈ R n (2.32)and using (2.31) we get n ( n − ω n k u k L λ ( R n ) ≤ ∞ X j =1 k u j k L λ ( R n ) ≤ C ∞ X j =1 ε j < ∞ by (2.22). Thus (2.28) and (2.32) imply (2.23) and − ∆ u = f in R n \{ } . Hence (2.24) and (2.25)hold.Taking β = α ∈ (0 , n ) in (2.29) and using (2.30) we getmin x ∈ B rj ( x j ) Z | y − x j | Lemma 2.4. Suppose for some constants α ∈ (0 , n ) , λ > , and σ ≥ that u is a nonnegativesolution of (1.1,1.2) and u ( x ) = O (1) as x → . Then u has a C extension to the origin, that is, u = w | R n \{ } for some function w ∈ C ( R n ) .Proof. Let v = u + 1. Then by Lemma 2.1, v satisfies (2.5). Since u , and hence v , is bounded in B (0) \ { } , the constant m in (2.5) is zero and by (1.1,1.2) − ∆ u , and hence − ∆ v , is bounded in B (0) \{ } . It therefore follows from (2.5) that v , and hence u , has a C extension to the origin. < λ < n − αn − In this section we prove Theorem 1.1–1.3 when 0 < λ < n − αn − . For these values of λ , the followingtheorem implies Theorems 1.1 and 1.3. Theorem 3.1. Suppose u is a nonnegative solution of (1.1,1.2) for some constants α ∈ (0 , n ) , < λ < n − αn − and ≤ σ ≤ nn − . (3.1) Then u ( x ) = O ( | x | − n ) as x → . (3.2) Proof. Let v = u + 1. Then by Lemma 2.1 we have that (2.1)–(2.5) hold. To prove (3.2), it clearlysuffices to prove v ( x ) = O ( | x | − n ) as x → . (3.3)Choose ε ∈ (0 , 1) such that λ < n − αn − ε . (3.4)By (2.4) we have v ∈ L nn − ε ( B (0)) which implies v λ ∈ L n ( n − ε ) λ ( B (0)) . Thus, since (3.4) implies λ ( n − ε ) n < n − αn , we have by Riesz potential estimates that I n − α ( v λ ) ∈ L ∞ ( B (0)) . Hence by (2.1) and (2.2), v is a C positive solution of0 ≤ − ∆ v ≤ Cv σ in B (0) \{ } . Thus by (3.1) and [18, Theorem 2.1], v satisfies (3.3).Our next result implies Theorem 1.2 when 0 < λ < n − αn − .11 heorem 3.2. Suppose α, λ , and σ are constants satisfying α ∈ (0 , n )0 < λ < n − αn − and σ > nn − . Let ϕ : (0 , → (0 , ∞ ) be a continuous function satisfying lim t → + ϕ ( t ) = ∞ . Then there exists a nonnegative solution u of (1.1,1.2) such that u ( x ) = O ( ϕ ( | x | )) as x → . (3.5) Proof. Let { x j } ⊂ R n and { r j } , { ε j } ⊂ (0 , 1) be sequences satisfying (2.6) and (2.7). Holding x j and ε j fixed and decreasing r j to a sufficiently small positive number we can assume Aε j r n − j > jϕ ( | x j | ) for j = 1 , , ... (3.6)and r ( n − σ − nj < A σ Bε λ + σ − j for j = 1 , , ... (3.7)where A and B are as in Lemma 2.2.Let u be as in Lemma 2.2. By (2.10), u satisfies (1.2) in B (0) \ ( { } ∪ ∪ ∞ j =1 B r j ( x j )). Also, for x ∈ B r j ( x j ), it follows from (2.9), (3.7), (2.12), and (2.11) that0 ≤ − ∆ u ≤ ε j r nj = r ( n − σ − nj A σ Bε λ + σ − j ( Bε λj ) Aε j r n − j ! σ ≤ ( | x | − α ∗ u λ ) u σ . Thus u satisfies (1.2) in B (0) \{ } . Finally by (2.11) and (3.6) we have u ( x j ) ≥ Aε j r n − j > jϕ ( | x j | )and thus (3.5) holds. n − αn − ≤ λ < nn − In this section we prove Theorems 1.1–1.3 when n − αn − ≤ λ < nn − . For these values of λ , theresult below implies Theorem 1.1. Theorem 4.1. Suppose u is a nonnegative solution of (1.1,1.2) for some constants α ∈ (0 , n ) , n − αn − ≤ λ < nn − and ≤ σ < n − αn − − λ. (4.1) Then u ( x ) = O ( | x | − n ) as x → . (4.2)12 roof. Let v = u + 1. Then by Lemma 2.1 we have that (2.1)–(2.5) hold. To prove (4.2), it clearlysuffices to prove v ( x ) = O ( | x | − ( n − ) as x → . (4.3)Since increasing λ or σ increases the right side of the second inequality in (2.2) , we can assumeinstead of (4.1) that n − αn − < λ < nn − , σ > , and 1 < λ + σ < n − αn − . (4.4)Since the increased value of λ is less than nn − , it follows from (2.4) that (2.1) still holds.By (4.4) there exists ε = ε ( n, λ, σ, α ) ∈ (0 , 1) such that (cid:18) n + 2 − αn + 2 − α − ε (cid:19) n − αn − < λ < nn − ε and λ + σ < n − αn − ε (4.5)which implies σ < n − αn − ε − λ < n − αn − ε − n − αn − < nn − ε . (4.6)Suppose for contradiction that (4.3) is false. Then there is a sequence { x j } ⊂ B / (0) \{ } suchthat x j → j → ∞ and lim j →∞ | x j | n − v ( x j ) = ∞ . (4.7)Since for | x − x j | < | x j | / Z | y − x j | > | x j | / | y | < − ∆ v ( y ) | x − y | n − dy ≤ (cid:18) | x j | (cid:19) n − Z | y | < − ∆ v ( y ) dy, it follows from (2.4) and (2.5) that v ( x ) ≤ C " | x j | n − + Z | y − x j | < | x j | / − ∆ v ( y ) | x − y | n − dy for | x − x j | < | x j | . (4.8)Substituting x = x j in (4.8) and using (4.7) we find that | x j | n − Z | y − x j | < | x j | / − ∆ v ( y ) | x j − y | n − dy → ∞ as j → ∞ . (4.9)Also by (2.4) we have Z | y − x j | < | x j | / − ∆ v ( y ) dy → j → ∞ . (4.10)Defining f j ( η ) = − r nj ∆ v ( x j + r j η ) where r j = | x j | / y = x j + r j η in (4.10) and (4.9) we get Z | η | < f j ( η ) dη → j → ∞ (4.11)and Z | η | < f j ( η ) dη | η | n − → ∞ as j → ∞ . (4.12)Let N ( y ) = Z | z | < − ∆ v ( z ) dz | y − z | n − for 0 < | y | < . 13y (2.4) and Riesz potential estimates, N ∈ L nn − ε ( B (0)). Thus N λ ∈ L nλ ( n − ε ) ( B (0)). Henceby H¨older’s inequality and (4.5) we have for R ∈ (0 , 1] and | x − x j | < R | x j | / Z | y | < N ( y ) λ dy | x − y | α − Z | y − x j | 12 = 12for | x − x j | < R | x j | / | y − x j | > R | x j | / N ( y ) ≤ C " | x j | n − + Z | z − x j | 1] we have − ∆ v ( x ) ≤ C | x j | ( n − ε ) λ − ( n − α ) + Z | y − x j | Suppose the sequence { f j } is bounded in L p ( B R (0)) (4.16) for some constants p ∈ [1 , n ] and R ∈ (0 , . Then there exists a positive constant C = C ( n, λ, σ, α ) such that the sequence { f j } is bounded in L q ( B R (0)) (4.17) for some q ∈ ( p, ∞ ) satisfying p − q ≥ C . (4.18) Proof. For R ∈ (0 , 1] we formally define operators N R and I R by( N R f )( ξ ) = Z | η | < R f ( η ) dη | ξ − η | n − and ( I R f )( ξ ) = Z | η | < R f ( η ) dη | ξ − η | α . Define p by 1 p − p = 2 − εn (4.19)where ε is as in (4.5). Then p ∈ ( p, ∞ ) and thus by Riesz potential estimates we have k ( N R f j ) λ k p /λ = k N R f j k λp ≤ C k f j k λp (4.20)and k ( N R f j ) σ k p /σ = k N R f j k σp ≤ C k f j k σp (4.21)where k · k p := k · k L p ( B R (0)) . Since1 p = 1 p − − εn ≤ − − εn = n − εn we see by (4.5) that p λ > . (4.22)Now there are two cases to consider. Case I. Suppose p λ < nn − α . (4.23)15efine p and q by λp − p = n − αn (4.24)and 1 q := 1 p + σp = λ + σp − n − αn . (4.25)It follows from (4.22)–(4.25), (4.19), and (4.4) that1 < p λ < p < ∞ , q > , (4.26)and 1 p − q = 1 p − (cid:18) ( λ + σ ) (cid:18) p − − εn (cid:19) − n − αn (cid:19) = (2 − ε )( λ + σ ) + ( n − α ) n − λ + σ − p ≥ (2 − ε )( λ + σ ) + ( n − α ) − n ( λ + σ − n = 2 n − α − ( n − ε )( λ + σ ) n . Thus (4.18) holds by (4.5).By (4.24), (4.26), (4.20), and Riesz potential estimates we find that k ( I R (( N R f j ) λ )) q k p /q = k I R (( N R f j ) λ ) k qp ≤ C k ( N R f j ) λ k qp /λ ≤ C k f j k λqp . Also by (4.21) we get k ( N R f j ) σq k p σq = k ( N R f j ) σ k qp /σ ≤ C k f j k σqp . It therefore follows from (4.15), (4.25), H¨older’s inequality, and (4.16) that (4.17) holds. Case II. Suppose p λ ≥ nn − α . (4.27)Then by Riesz potential estimates, (4.16), and (4.20) we find that the sequence { I R (( N R f j ) λ ) } is bounded in L γ ( B R (0)) for all γ ∈ (1 , ∞ ) . (4.28)Let ˆ q = p /σ . Then by (4.19), 1 p − q = 1 p − σp = 2 − εn + 1 − σp . Thus for σ ≤ p − q ≥ − εn > σ > p − q ≥ − εn − σ − nλn − α ≥ − εn − n − αn − − λ − nλn − α = n + 2 − α − εnλ (cid:18) λ − (cid:18) n + 2 − αn + 2 − α − ε (cid:19) n − αn − (cid:19) > . Thus defining q ∈ ( p, ˆ q ) by 1 q = p + q σ > p − q = 12 (cid:18) p − q (cid:19) ≥ C ( n, λ, σ, α ) > . That is (4.18) holds.Since qσ/p < ˆ qσ/p = 1 there exists γ ∈ ( q, ∞ ) such that qγ + qσp = 1 . (4.29)Also k ( I R (( N R f j ) λ )) q k γ/q = k I R (( N R f j ) λ ) k qγ and by (4.21) k ( N R f j ) σq k p σq = k ( N R f j ) σ k qp /σ ≤ C k f j k σqp . It therefore follows from (4.15), (4.29), H¨older’s inequality, (4.28), and (4.16) that (4.17) holds.We return now to the proof of Theorem 4.1. By (4.11) the sequence { f j } is bounded in L ( B (0)) . (4.30)Starting with this fact and iterating Lemma 4.1 a finite number of times ( m times is enough if m > /C ) we see that there exists R ∈ (0 , 1) such that the sequence { f j } is bounded in L p ( B R (0))for some p > n/ 2. Hence by Riesz potential estimates the sequence { N R f j } is bounded in L ∞ ( B R (0)). Thus (4.15) implies the sequence { f j } is bounded in L ∞ ( B R (0)) . (4.31)Since Z | η | < f j ( η ) dη | η | n − ≤ Z | η | Suppose α, λ , and σ are constants satisfying α ∈ (0 , n ) , n − αn − ≤ λ ≤ nn − and σ ≥ nn − if λ = n − αn − σ > n − αn − − λ if n − αn − < λ < nn − σ > n − αn − if λ = nn − . Let ϕ : (0 , → (0 , ∞ ) be a continuous function satisfying lim t → + ϕ ( t ) = ∞ . Then there exists a nonnegative solution u of (1.1,1.2) such that u ( x ) = O ( ϕ ( | x | )) as x → . (4.32) Proof. Let { x j } ⊂ R n and { r j } , { ε j } ⊂ (0 , 1) be sequences satisfying (2.6) and (2.7). Holding x j and ε j fixed and decreasing r j to a sufficiently small positive number we can assume for j = 1 , , ... that jϕ ( | x j | ) ≤ Aε j r n − j if n − αn − ≤ λ < nn − Aε j r n − j (cid:18) log rj (cid:19) n − n if λ = nn − (4.33)and A σ Bε λ + σ − j ≥ r ( n − σ − nj (cid:16) log r j (cid:17) − if λ = n − αn − r ( n − σ − (2 n − α − ( n − λ ) j if n − αn − < λ < nn − r ( n − σ − ( n − α ) j (cid:16) log r j (cid:17) ( n − σ +2 n if λ = nn − (4.34)where A and B are as in Lemma 2.2. Let u be as in Lemma 2.2. By (2.10), u satisfies (1.2) in B (0) \ ( { } ∪ ∪ ∞ j =1 B r j ( x j )). Also, for x ∈ B r j ( x j ), it follows from (2.9), (4.34), (2.12), and (2.11)that for n − αn − ≤ λ < nn − we have0 ≤ − ∆ u ≤ ε j r nj = r ( n − σ − nj (cid:18) log rj (cid:19) − A σ Bε λ + σ − j ( Bε λj log r j ) (cid:18) A ε j r n − j (cid:19) σ if λ = n − αn − r ( n − σ − (2 n − α − ( n − λ ) j A σ Bε λ + σ − j (cid:16) Bε λj r ( n − α ) − ( n − λj (cid:17) (cid:18) A ε j r n − j (cid:19) σ if n − αn − < λ < nn − ≤ ( | x | α ∗ u λ ) u σ , λ = nn − , we have0 ≤ − ∆ u ≤ ε j r nj (log r j ) n − n = r ( n − σ − ( n − α ) j (log r j ) ( n − σ +2 n A σ Bε λ + σ − j Bε λj r − αj log r j ! Aε j r n − j (log r j ) n − n σ ≤ ( | x | − α ∗ u λ ) u σ . Thus u satisfies (1.2) in B (0) \{ } . Finally, by (2.11) and (4.33) we have u ( x j ) ≥ jϕ ( | x j | )and thus (4.32) holds. λ ≥ nn − In this section we prove Theorems 1.1–1.3 when λ ≥ nn − . For these values of λ , our next resultimplies Theorem 1.1. Theorem 5.1. Suppose u is a nonnegative solution of (1.1,1.2) for some constants α ∈ (0 , n ) , λ ≥ nn − and ≤ σ < − α − n λ. (5.1) Then u ( x ) = O (1) as x → and u has a C extension to the origin.Proof. Let v = u + 1. Then by Lemma 2.1 we have that (2.1)–(2.5) hold. To prove (5.2) it clearlysuffices to prove v ( x ) = O (1) as x → . (5.3)By (2.1) and (5.1), the constant m in (2.5) is zero and thus by (2.5) v ( x ) ≤ C " Z | y | < − ∆ v ( y ) | x − y | n − dy for 0 < | x | < C .Since increasing σ increases the right side of the second inequality in (2.2) , we can assumeinstead of (5.1) that λ ≥ nn − < σ < − α − n λ (5.5)which implies σλ < − αn + 1 λ ≤ − αn + n − n = n − αn . (5.6)By (5.5) there exists ε = ε ( n, λ, σ, α ) ∈ (0 , 1) such that α + ε < n and σ < − α + ε − n λ which implies σ − λ < − α − εn . (5.7)For the proof of Theorem 5.1 we will need the following lemma.19 emma 5.1. Suppose v ∈ L p ( B (0)) (5.8) for some constant p ∈ (cid:20) λ, nλn − α − ε (cid:19) . (5.9) Then either v ∈ L nλn − α − ε ( B (0)) (5.10) or there exists a positive constant C = C ( n, λ, σ, α ) such that v ∈ L q ( B (0)) (5.11) for some q ∈ ( p, ∞ ) satisfying p − q ≥ C . (5.12) Proof. Define p by λp − p = n − α − εn . (5.13)Then by (5.9) 1 ≤ pλ < p < ∞ and thus by Riesz potential estimates and (5.8) we have k I n − α v λ k p ≤ C k v λ k pλ = C k v k λp < ∞ (5.14)where I β is defined in (2.3).Define p > p = 1 p + σp . (5.15)Then by H¨older’s inequality k (( I n − α v λ ) v σ ) p k ≤ k ( I n − α v λ ) p k p p k v σp k pσp = k I n − α v λ k p p k v k σp p < ∞ by (5.8) and (5.14). Hence by (2.2) − ∆ v ∈ L p ( B (0)) . (5.16)Also by (5.15), (5.13), (5.9), and (5.7) we have1 p = λ + σp − n − α − εn ≤ λ + σλ − n − α − εn = σλ + α + εn < λ − α + ε − n + α + εn = 1 λ + 2 n . Thus by (5.5) we see that p > . (5.17)20 ase I. Suppose p ≥ n . Then by (5.16), (5.4), and Riesz potential estimates we have v ∈ L q ( B (0)) for all q > Case II. Suppose p < n . Define q by 1 p − q = 2 n . (5.18)Then by (5.17) 1 < p < q < ∞ . Hence by (5.16), (5.4) and Riesz potential estimates we have (5.11) holds.Also by (5.18), (5.15), (5.13). (5.9), and (5.7) we get1 p − q = 1 p + 2 n − p = 1 p + 2 n − σp − λp + 1 − α + εn = − λ + σ − p + 1 − α + ε − n ≥ − ( λ + σ ) λ + 1 + 2 − α − εn > . Thus (5.12) holds.We now return to the proof of Theorem 5.1. By (2.1), v ∈ L λ ( B (0)). Starting with this factand iterating Lemma 5.1 a finite number of times we see that (5.10) holds. In particular v ∈ L p ( B (0)) (5.19)for some p > nλn − α . (5.20)Hence v λ ∈ L pλ ( B (0)) and pλ > nn − α . Thus by Riesz potential estimates I n − α ( v λ ) ∈ L ∞ ( B (0)).So by (2.2) 0 ≤ − ∆ v < Cv σ in B (0) \{ } . (5.21)Hence by (5.19), − ∆ v ∈ L pσ ( B (0)) and by (5.20) and (5.6) pσ > nn − α λσ > (cid:18) nn − α (cid:19) > . Thus by (5.4) and Riesz potential estimates v ∈ L q ( B (0)) where q = ∞ if pσ ≥ n − ε σp − − εn if pσ < n − ε . (5.22)If q = ∞ then (5.3) holds. Hence we can assume pσ < n − ε . Then by (5.22)1 p − q = 1 − σp + 2 − εn . Thus, if σ ∈ (0 , 1] then 1 p − q > n . 21n the other hand, if σ > p − q = 2 − εn − σ − p> − εn − σ − λ n − αn> − εn − − α − εn = αn . Thus for σ > p − q > C ( n, α ) > . Hence, after a finite number of iterations of the procedure of going from (5.19) to (5.22) we get v ∈ L ∞ ( B (0)) and hence we see again that (5.3) holds.Finally by Lemma 2.4, u has a C extension to the origin.The result below implies Theorems 1.2 and 1.3 when λ ≥ nn − . Theorem 5.2. Suppose α, λ , and σ are constants satisfying α ∈ (0 , n ) , λ ≥ nn − , σ ≥ , and σ > − α − n λ. Let ϕ : (0 , → (0 , ∞ ) be a continuous function satisfying lim t → + ϕ ( t ) = ∞ . Then there exists a nonnegative solution u of (1.1,1.2) such that u ( x ) = O ( ϕ ( | x | )) as x → . (5.23) Proof. If λ = nn − then Theorem 5.2 follows from Theorem 4.2. Hence we can assume λ > nn − .Let { x j } ⊂ R n and { r j } , { e j } ⊂ (0 , 1) be sequences satisfying (2.21) and (2.22). Holding x j and ε j fixed and decreasing r j to a sufficiently small positive number we can assume Aε j r n/λj > jϕ ( | x j | ) for j = 1 , , ... (5.24)and r nλ ( σ − (1 − α − n λ )) j < A σ Bε λ + σ − j for j = 1 , , ... (5.25)where A and B are as in Lemma 2.3. Let u be as in Lemma 2.3. By (2.25) u satisfies (1.2) in B (0) \ ( { } ∪ ∪ ∞ j =1 B r j ( x j )). Also, for x ∈ B r j ( x j ), it follows from (2.24), (5.25), (2.27), and (2.26)that 0 ≤ − ∆ u ≤ ε j r n/λj = r nλ ( σ − (1 − α − n λ )) j A σ Bε λ + σ − j Bε λj r αj ! Aε j r n/λj ! σ ≤ ( | x | α ∗ u λ ) u σ . Thus u satisfies (1.2) in B (0) \{ } .Finally by (2.26) and (5.24) we have u ( x j ) ≥ Aε j r n/λj > jϕ ( | x j | )and thus (5.23) holds. 22 eferenceseferences