Negative Power Nonlinear Integral Equations on Bounded Domains
aa r X i v : . [ m a t h . A P ] A p r NEGATIVE POWER NONLINEAR INTEGRAL EQUATIONS ONBOUNDED DOMAINS
DEDICATE TO HA¨IM BREZIS ON OCCASION OF HIS 75TH BIRTHDAY
JINGBO DOU, QIANQIAO GUO AND MEIJUN ZHU
Abstract
This is the continuation of our previous work [5], where we introducedand studied some nonlinear integral equations on bounded domains that are re-lated to the sharp Hardy-Littlewood-Sobolev inequality. In this paper, we intro-duce some nonlinear integral equations on bounded domains that are related tothe sharp reversed Hardy-Littlewood-Sobolev inequality. These are integral equa-tions with nonlinear term involving negative exponents. Existence results as wellas nonexistence results are obtained.
Keywords
Reversed sharp Hardy-Littlewood-Sobolev inequality, Integral equa-tions, Existence, nonexistence
Mathematics Subject Classification(2000).
Introduction
In [5], motivated by the study of certain semi-linear equations and the sharpSobolev inequality, we introduced and studied the integral equations (with positivepower) related to the sharp Hardy-Littlewood-Sobolev (HLS for short) inequality.Let us briefly recall these as the follows.For 0 < α < n , on any bounded domain Ω ⊂ R n with smooth boundary, weconsideredˆ ξ α (Ω) = sup f ∈ L nn + α (Ω) ,f =0 R Ω R Ω f ( x ) | x − y | − ( n − α ) f ( y ) dxdy || f || L nn + α (Ω) . It was showed in [5] that ˆ ξ α (Ω) = N α , where N α is the best constant of the classicalsharp HLS inequality (due to Lieb [7]); And ˆ ξ α (Ω) is not attained by any functionsif Ω = R n . This indicates that there is not any energy maximizing solution to f n − αn + α ( x ) = Z Ω f ( y ) | x − y | n − α dy, f ≥ , x ∈ Ω . We then considered a general integral equation f q − ( x ) = Z Ω f ( y ) | x − y | n − α dy + λ Z Ω f ( y ) | x − y | n − α − dy, f ≥ , x ∈ Ω , (1.1)for α < n , and studied the existence and nonexistence of positive solutions fordifferent power q and parameter λ .In this paper we consider integral equation (1.1) for α > n . This case is relatedto so called sharp reversed HLS inequality, which was discovered by Dou and Zhu[4].Recall the sharp reversed HLS inequality from [4] (see also related work by Ngˆo,Nguyen [8] and by Beckner [1]): heorem A. For α > n , | Z R n Z R n f ( x ) | x − y | − ( n − α ) g ( y ) dxdy | ≥ N α || f || L nn + α ( R n ) || g || L nn + α ( R n ) (1.2) holds for all non-negative functions f, g ∈ L nn + α ( R n ) , where N α = N ( 2 nn + α , α, n ) = π ( n − α ) / Γ( α/ n/ α/ { Γ( n/ n ) } − α/n ; (1.3) And the equality holds if and only if f = c g = c (cid:0) c + | x − x | (cid:1) n + α , (1.4) where c , c , c are any positive constants, and x ∈ R n . Similar to what is done in [5]: for any smooth domain Ω ⊂ R n , we consider ξ α (Ω) = inf f ∈ L nn + α (Ω) ,f ≥ ,f =0 R Ω R Ω f ( x ) | x − y | − ( n − α ) f ( y ) dxdy || f || L nn + α (Ω) . We will show that ξ α (Ω) = N α , (1.5)and ξ α (Ω) is not attained by any functions if Ω = R n (see Proposition 2.1 below).Again, we notice that the corresponding Euler-Lagrange equation for the minimizer(if the minimum is attained) is integral equation (1.1) with λ = 0 and a negativepower ( n − αn + α < α > n there is no energy minimizingsolution to integral equation (1.1) for q = 2 n/ ( n + α ) and λ = 0.Let p α = 2 n/ ( n − α ), q α = 2 n/ ( n + α ) and d (Ω) = sup x,y ∈ Ω | x − y | be the diameterof the bounded domain Ω. In this paper, we consider integral equation (1.1) for α > n . We shall prove Theorem 1.1.
Assume α > n and Ω is a bounded domain with smooth boundary.(1) For < q < q α (subcritical case), and − d (Ω) < λ , there is a positive solution f ∈ C (Ω) to equation (1.1) .(2) For q = q α (critical case), and − d (Ω) < λ < , there is a positive solution f ∈ C (Ω) to equation (1.1) .(3) For q α ≤ q < (critical and supercritical cases), and λ ≥ , if Ω is a star-shapeddomain, then there is not any positive C (Ω) solution to (1.1) . We emphasis here that q α < α > n ). Thus equation (1.1) has a nonlin-ear term with a negative power. Our results indicate that even though the integralequation, which is related to the reversed HLS inequality, is of negative nonlinear-ity, similar phenomena to the integral equation with positive power can be seen.Contrary to integral equations with positive nonlinearity, it seems that no compactembedding can be used directly for the existence result to the integral equationwith subcritical negative power. Different techniques are needed for deriving theexistence as well as the nonexistence results. See more details in Section 3 below.We organize the paper as follows: In Section 2, we focus on the nonexistenceresult (part (3) of Theorem 1.1). In Section 3, we first obtain the existence result(part (1) of Theorem 1.1), and then we show the symmetric and monotonically in-creasing properties of solutions to the integral equation on a ball, even no pointwise oundary condition is given (Theorem 3.3 below). In Section 4, we come back tothe integral equations with critical exponent and a lower order term and prove theexistence result (part (2) of Theorem 1.1).Notation: for any function f ( x ) defined on Ω, we always use ˜ f ( x ) to representits trivial extension in R n , namely,˜ f ( x ) := ( f ( x ) x ∈ Ω , x ∈ R n \ Ω . We also denote I α f ( x ) := Z R n f ( y ) | x − y | n − α dy, I α, Ω f ( x ) := Z Ω f ( y ) | x − y | n − α dy and L q + (Ω) := { f ∈ L q (Ω) \ { } : f ≥ } . nonexistence for critical and supercritical cases In this section, we first derive energy estimate (1.5) for any domain Ω ⊂ R n , andthen show that the infimum ξ α (Ω) is not achieved by any function once Ω = R n . Proposition 2.1.
For any domain Ω ⊂ R n , ξ α (Ω) = N α . Further, if Ω = R n ,then the infimum ξ α (Ω) is not achieved by any function in L q α (Ω) . Proof. If f ∈ L q α + (Ω), then ˜ f ∈ L q α + ( R n ) . It follows that ξ α (Ω) = inf f ∈ L qα + (Ω) \{ } R R n R R n ˜ f ( x ) | x − y | − ( n − α ) ˜ f ( y ) dxdy || ˜ f || L qα ( R n ) ≥ inf g ∈ L qα + ( R n ) \{ } R R n R R n g ( x ) | x − y | − ( n − α ) g ( y ) dxdy || g || L qα ( R n ) = N α . On the other hand, recall that f ( x ) = (cid:0) | x | (cid:1) n + α is an extremal function to thesharp reversed HLS inequality in Theorem A, as well as its conformal equivalentclass: f ǫ ( x ) = ǫ − n + α f ( | x − x ∗ | ǫ ) = (cid:0) ǫǫ + | x − x ∗ | (cid:1) n + α , (2.1)where x ∗ ∈ R n , ǫ >
0. Thus k I α f k L pα ( R n ) = k I α f ǫ k L pα ( R n ) , k f k L qα ( R n ) = k f ǫ k L qα ( R n ) . Choose x = x ∗ for some point x ∈ Ω and R small enough so that B R ( x ) ⊂ Ω.Then we define test function g ( x ) as g ( x ) = ( f ǫ ( x ) x ∈ B R ( x ) ⊂ Ω , x ∈ R n \ B R ( x ) . bviously, g ∈ L q α + ( R n ) . Thus, Z Ω Z Ω g ( x ) g ( y ) | x − y | n − α dxdy = Z R n Z R n f ǫ ( x ) f ǫ ( y ) | x − y | n − α dxdy − Z R n Z R n \ B R ( x ) f ǫ ( x ) f ǫ ( y ) | x − y | n − α dxdy + Z R n \ B R ( x ) Z R n \ B R ( x ) f ǫ ( x ) f ǫ ( y ) | x − y | n − α dxdy ≤ N α k f ǫ k L qα ( R n ) − I , where I := Z R n Z R n \ B R ( x ) f ǫ ( x ) f ǫ ( y ) | x − y | n − α dxdy. Notice that f ǫ ( x ) is an extremal function for the sharp reversed HLS inequality.Thus it satisfies integral equation: f n − αn + α ( x ) = B Z R n f ( y ) | x − y | n − α dy, where B is a suitable positive constant. We thus can estimate I in the following: I = C Z R n \ B R ( x ) f nn + α ǫ ( x ) dx = O ( Rǫ ) − n , as ǫ → . And we also have Z B R ( x ) f nn + α ǫ ( x ) dx = Z R n f nn + α ǫ ( x ) dx − Z R n \ B R ( x ) f nn + α ǫ ( x ) dx = Z R n f nn + α ǫ ( x ) dx − O ( Rǫ ) − n , as ǫ → . Hence, for small enough ǫ >
0, we have ξ α (Ω) ≤ R Ω R Ω g ( x ) g ( y ) | x − y | n − α dxdy k g k L qα (Ω) ≤ N α k f ǫ k L qα ( R n ) − I k f ǫ k L qα ( B R ( x )) = N α k f ǫ k L qα ( R n ) − O ( Rǫ ) − n k f ǫ k L qα ( R n ) − O ( Rǫ ) − n , which yields ξ α (Ω) ≤ N α as ǫ → ξ α (Ω) is not achieved if Ω = R n . In fact, if ξ α (Ω) wereattained by some function u ∈ L q α + (Ω), then ˜ u ∈ L q α + ( R n ) would be an extremalfunction to the sharp reversed HLS inequality on R n , which is impossible due toTheorem A. (cid:3) Proposition 2.1 indicates that for α > n there is not any minimizing energysolution to (1.1) for q = 2 n/ ( n + α ) and λ = 0. In fact, we will show that there isnot any positive C solution to (1.1) for α > n , q = 2 n/ ( n + α ) and λ = 0 on anystar-shaped domain. roof of part (3) in Theorem 1.1 (nonexistence part) . Without loss ofgenerality, here we assume that the origin is in Ω and the domain is star-shapedwith respect to the origin.Recall the following Pohozaev identity from [5]. Lemma 2.2.
Assume that the origin is in Ω and the domain is star-shaped withrespect to the origin. If u ∈ C (Ω) is a non-negative solution to u ( x ) = Z Ω u p − ( y ) | x − y | n − α dy + λ Z Ω u p − ( y ) | x − y | n − α − dy, x ∈ Ω , (2.2) where p = 0 , λ ∈ R , then ( np + α − n Z Ω u p ( x ) dx = − λ Z Ω Z Ω u p − ( x ) u p − ( y ) | x − y | n − α dydx + 1 p Z ∂ Ω ( x · ν ) u p ( x ) dσ, (2.3) where ν is the outward unit normal vector to ∂ Ω . Applying Lemma 2.2 to (1.1) for λ ≥ u ( x ) = f q − ( x ). Thus p = qq − = q ′ .Noticing that 1 > q ≥ q α is equivalent to p ≤ p α <
0, we have − λ Z Ω Z Ω u p − ( x ) u p − ( y ) | x − y | n − α − dydx + 1 p Z ∂ Ω ( x · ν ) u p − ( x ) dσ ≥ . Since Ω is star-shaped domain about the origin, we have x · ν > ∂ Ω. If λ > u ( x ) ≡ ∞ on Ω. If λ = 0, then u ≡ ∞ on ∂ Ω. Therefore we obtain acontradiction to that f ( x ) is a positive C (Ω) solution. Remark . Condition q < q >
1, one may call ita supercritical exponent (since it is bigger than q α ). However, in this case p = q/ ( q − > p α . Our nonexistence result may not betrue any more in this case.Note that the unit ball is conformally equivalent to the upper half space. Wehave Corollary 2.4.
There is no positive solution u ∈ L p α ( R n + ) ∩ C ( R n + ) to u ( x ) = Z R n + u p α − ( y ) | x − y | n − α dy, x ∈ R n + . Existence result for subcritical case
For subcritical exponents we have the following inequality:
Lemma 3.1.
Let q ∈ (0 , q α ) . There exists a positive constant C ( n, q, α, Ω) > such that Z Ω Z Ω f ( x ) | x − y | − ( n − α ) f ( y ) dxdy ≥ C ( n, q, α, Ω) k f k L q (Ω) (3.1) holds for any non-negative function f ∈ L q (Ω) . Proof.
For f ∈ L q (Ω), by using the reversed HLS inequality (1.2) we have h I α, Ω f, f i = Z R n Z R n ˜ f ( x ) | x − y | − ( n − α ) ˜ f ( y ) dxdy ≥ N α k ˜ f k L qα ( R n ) = N α k ˜ f k L qα (Ω) ≥ C ( n, q, α, Ω) k f k L q (Ω) . We would like to point out that one can also prove the above lemma directly viaa Young type inequality as that in Dou, Guo and Zhu [6].Based on the above lemma, we can obtain the existence result for subcriticalexponent (part (1) of Theorem 1.1). Notice that it is different from the case 0 <α < n (Lemma 3.2 in [5]) since no compact embedding can be used directly here.We follow a similar approach used in Dou, Guo and Zhu [6].
Lemma 3.2.
For < q < q α , λ > − d (Ω) , infimum ξ α,q (Ω) := inf f ∈ L q + (Ω) R Ω R Ω ( f ( x ) | x − y | − ( n − α ) f ( y ) + λf ( x ) | x − y | − ( n − α − f ( y )) dydx k f k L q (Ω) > , and it is attained by some nonnegative function in L q + (Ω) . Proof.
Notice: for x, y ∈ Ω and 0 > λ > − /d (Ω) , λ | x − y | ≥ λd (Ω) > ξ α,q (Ω) > { f j } ∞ j =1 in L q (Ω). Assume with-out loss of generality that f j ∈ L q α (Ω) (see, for example, Proposition 2.5 in [6]).Then we can normalize it such that k f j k L qα (Ω) = 1. It follows that there exists asubsequence such that f qj ⇀ f q ∗ weakly in L qαq (Ω) , as j → ∞ . Then Z Ω f qj → Z Ω f q ∗ , as j → ∞ . (3.2)Claim: k f j k L (Ω) ≤ C. We relegate the proof of this claim to the end.Once the claim is proved, we have R Ω f q ∗ > C > f qj ⇀ f q ∗ weakly in L q (Ω) . Then for any fixed x ∈ Ω, f − q ∗ ( y ) | x − y | α − n (1 + λ | x − y | ) ∈ L − q (Ω), thus, as j → ∞ , Z Ω f qj ( y ) f − q ∗ ( y ) | x − y | α − n (1 + λ | x − y | ) dy → Z Ω f ∗ ( y ) | x − y | α − n (1 + λ | x − y | ) dy. Further, we show that the above convergence is actually uniformly convergentfor all x ∈ Ω.By H¨older’s inequality we have Z Ω f qj ( y ) f − q ∗ ( y ) | x − y | α − n (1 + λ | x − y | ) dy ≤ ( Z Ω f q · q j ( y ) dy ) q ( Z Ω ( f − q ∗ ( y ) | x − y | α − n (1 + λ | x − y | )) − q dy ) − q ≤ C, that is, R Ω f qj ( y ) f − q ∗ ( y ) | x − y | α − n (1 + λ | x − y | ) dy is uniformly bounded for x ∈ Ω.Notice that for any x , x , y ∈ Ω , || x − y | α − n − | x − y | α − n | ≤ ( C | x − x | α − n , if 0 < α − n ≤ ,C | x − x | , if α − n > . hen for any x , x ∈ Ω , | Z Ω f qj ( y ) f − q ∗ ( y ) | x − y | α − n (1 + λ | x − y | ) dy − Z Ω f qj ( y ) f − q ∗ ( y ) | x − y | α − n (1 + λ | x − y | ) dy |≤ Z Ω f qj ( y ) f − q ∗ ( y ) || x − y | α − n (1 + λ | x − y | ) − | x − y | α − n (1 + λ | x − y | ) | dy ≤ C max( | x − x | α − n , | x − x | ) Z Ω f qj ( y ) f − q ∗ ( y ) dy ≤ C max( | x − x | α − n , | x − x | )( Z Ω f j ( y ) dy ) q ( Z Ω f ∗ ( y ) dy ) − q ≤ C max( | x − x | α − n , | x − x | ) . Thus R Ω f qj ( y ) f − q ∗ ( y ) | x − y | α − n (1 + λ | x − y | ) dy is equicontinuous in Ω. It followsthat, as j → ∞ , Z Ω f qj ( y ) f − q ∗ ( y ) | x − y | α − n (1 + λ | x − y | ) dy → Z Ω f ∗ ( y ) | x − y | α − n (1 + λ | x − y | ) dy uniformly for x ∈ Ω.Therefore for any ǫ > j ∈ N such that for any j > j , Z Ω Z Ω f qj ( x ) f − q ∗ ( x ) | x − y | α − n (1 + λ | x − y | ) f qj ( y ) f − q ∗ ( y ) dxdy ≥ Z Ω f qj ( x ) f − q ∗ ( x )[ Z Ω | x − y | α − n (1 + λ | x − y | ) f ∗ ( y ) dy − ǫ ] dx. By H¨older’s inequality we know R Ω f qj ( x ) f − q ∗ ( x ) dx ≤ C . Again, notice that Z Ω f qj ( x ) f − q ∗ ( x ) Z Ω | x − y | α − n (1 + λ | x − y | ) f ∗ ( y ) dydx → Z Ω Z Ω f ∗ ( x ) | x − y | α − n (1 + λ | x − y | ) f ∗ ( y ) dydx since f − q ∗ ( x ) R Ω | x − y | α − n (1 + λ | x − y | ) f ∗ ( y ) dy ∈ L − q (Ω). We have for j > j large enough, Z Ω Z Ω f qj ( x ) f − q ∗ ( x ) | x − y | α − n (1 + λ | x − y | ) f qj ( y ) f − q ∗ ( y ) dxdy ≥ Z Ω Z Ω f ∗ ( x ) | x − y | α − n (1 + λ | x − y | ) f ∗ ( y ) dydx − ǫ − Cǫ.
Similarly, we also have, for j > j large enough, Z Ω Z Ω f qj ( x ) f − q ∗ ( x ) | x − y | α − n (1 + λ | x − y | ) f qj ( y ) f − q ∗ ( y ) dxdy ≤ Z Ω Z Ω f ∗ ( x ) | x − y | α − n (1 + λ | x − y | ) f ∗ ( y ) dydx + ǫ + Cǫ. herefore, as j → ∞ , Z Ω Z Ω f qj ( x ) f − q ∗ ( x ) | x − y | α − n (1 + λ | x − y | ) f qj ( y ) f − q ∗ ( y ) dxdy → Z Ω Z Ω f ∗ ( x ) | x − y | α − n (1 + λ | x − y | ) f ∗ ( y ) dydx. It follows from the above and H¨older’s inequality thatlim inf j →∞ Z Ω Z Ω f j ( x ) | x − y | α − n (1 + λ | x − y | ) f j ( y ) dxdy ≥ Z Ω Z Ω f ∗ ( x ) | x − y | α − n (1 + λ | x − y | ) f ∗ ( y ) dxdy. Since || f j || L q (Ω) → || f ∗ || L q (Ω) >
0, the above inequality then implieslim inf j →∞ R Ω R Ω f j ( x ) | x − y | α − n (1 + λ | x − y | ) f j ( y ) dxdy k f j k L q (Ω) ≥ R Ω R Ω f ∗ ( x ) | x − y | α − n (1 + λ | x − y | ) f ∗ ( y ) dxdy k f ∗ k L q (Ω) . That is: f ∗ is a minimizer.Now we are left to prove the claim: k f j k L (Ω) ≤ C. From k f j k L qα (Ω) = 1, we have Z Ω Z Ω f j ( x ) | x − y | α − n (1 + λ | x − y | ) f j ( y ) dxdy ≥ (1 − | λ | d (Ω)) Z Ω Z Ω f j ( x ) | x − y | α − n f j ( y ) dxdy ≥ (1 − | λ | d (Ω)) N α k f j k L qα (Ω) ≥ C > . Since { f j } ∞ j =1 is a minimizing nonnegative sequence, we conclude that k f j k L q (Ω) ≥ C ( n, α, q ) >
0. By H¨older’s inequality, we have0 < C ( n, α, q ) ≤ k f j k L q (Ω) ≤ C ( n, α, q ) k f j k L qα (Ω) = C ( n, α, q ) . The upper bound on || f j k L q (Ω) indicates that Z Ω Z Ω f j ( x ) | x − y | α − n f j ( y ) dxdy < C ( n, α, q ) . It follows, via reversed H¨older’s inequality, that k I α, Ω f j k L q ′ (Ω) ≤ C ( n, α, q ) < ∞ . Further, H¨older’s inequality and reversed HLS inequality yield that ∞ > C ( n, α, q ) ≥ k I α, Ω f j k L q ′ (Ω) ≥ C ( n, α, q ) k I α, Ω f j k L pα (Ω) ≥ C ( n, α, q ) k f j k L qα (Ω) = C ( n, α, q ) > . Then we can show that for
M > M q ′ | Ω | < ( C ( n, α, q )) q ′ , thereexists 0 < δ < | Ω | such that m { x : I α, Ω f j ( x ) ≤ M } > δ, f or all j. (3.3) n fact, for Ω := { x : I α, Ω f j ( x ) ≤ M } ,( C ( n, α, q )) q ′ ≤ Z Ω ( I α, Ω f j ) q ′ dx = Z Ω \ Ω ( I α, Ω f j ) q ′ dx + Z Ω ( I α, Ω f j ) q ′ dx ≤ M q ′ | Ω | + ( Z Ω ( I α, Ω f i ) p α dx ) q ′ pα | Ω | − q ′ pα ≤ M q ′ | Ω | + ( Z Ω ( I α, Ω f i ) p α dx ) q ′ pα | Ω | − q ′ pα ≤ M q ′ | Ω | + ( C ( n, α, q ) C ( n, α, q ) ) q ′ | Ω | − q ′ pα <
12 ( C ( n, α, q )) q ′ + | Ω | − q ′ pα ( C ( n, α, q ) C ( n, α, q ) ) q ′ , which yields the existence of such δ > ǫ >
0, such that for any j , we can findtwo points x j , x j ∈ Ω with the properties that | x j − x j | ≥ ǫ and I α, Ω f j ( x j ) = Z Ω f j ( y ) | x j − y | α − n ≤ M, I α, Ω f j ( x j ) = Z Ω f j ( y ) | x j − y | α − n ≤ M. So Z Ω f j ( y ) dy ≤ Z Ω \ B ( x j , ǫ ) f j ( y ) dy + Z Ω \ B ( x j , ǫ ) f j ( y ) dy ≤ ( 4 ǫ ) α − n Z Ω \ B ( x j , ǫ ) f j ( y ) | x j − y | α − n dy +( 4 ǫ ) α − n Z Ω \ B ( x j , ǫ ) f j ( y ) | x j − y | α − n dy ≤ ( 4 ǫ ) α − n M, uniformly for all j . We thus verify the claim, and hereby, complete the proof ofLemma 3.2. (cid:3) It is standard to check that the minimizer f ( x ) for energy ξ α,q (Ω) is positive,and, up to a constant multiplier, satisfies the following equation: f q − ( x ) = Z Ω f ( y ) | x − y | n − α dy + λ Z Ω f ( y ) | x − y | n − α − dy, x ∈ Ω . (3.4)From the proof of Lemma 3.2, we also know that f ( x ) ∈ L (Ω), thus f q − ( x ) ∈ L ∞ (Ω) by equation (3.4).Writing u ( x ) = f q − ( x ) , p = q ′ , we thus find a weak positive solution u ( x ) ∈ L p (Ω) to u ( x ) = Z Ω u p − ( y ) | x − y | n − α dy + λ Z Ω u p − ( y ) | x − y | n − α − dy, x ∈ Ω (3.5)for 0 > p > nn − α = p α . To complete the proof of part (1) in Theorem 1.1, we needto show that u ∈ C (Ω). n fact, f ( x ) ∈ L (Ω) implies Z Ω u p − ( y ) dy < ∞ . It is easy to see that u ∈ C (Ω) from equation (3.5). To show u ∈ C (Ω), we candirectly compute, for i = 1 , ..., n , ∂ x i u ( x ) = − ( n − α ) Z Ω u p − ( y )( x i − y i ) | x − y | n − α +2 dy ∈ C (Ω) . Part (1) of Theorem 1.1 is hereby proved.It is interesting to study some properties about the positive solutions to the newintegral equation (3.5)(such as multiplicity of solutions, blowup behavior as q → q α in a star-shaped domain, etc.) In the rest of this section, as in [5], we will showthat even though the boundary condition is not given pointwise, the symmetricproperty for solutions to the integral equation (3.5) with λ = 0 on a unit ball stillholds. Contrary to the result in [5], here we will show that the solution is monotoneincreasing due to the monotone increasing property of the kernel.On B := B (0) = { x ∈ R n | | x | < , x ∈ R n } , for λ = 0 we rewrite the equation(3.5) as u ( x ) = Z B u p − ( y ) | x − y | n − α dy, x ∈ B (0) . (3.6)We have Theorem 3.3.
Let α > n , p ∈ ( p α , . Then every positive solution u ∈ L p ( B ) to (3.6) is radially symmetric about the origin and strictly increasing in the radialdirection. Easy to see from the proof of Part (1) of Theorem 1.1 that u ∈ C ( B ). We willuse the method of moving planes to prove Theorem 3.3.Firstly, we recall the idea of the method of moving planes in B (see e.g. [3, 5]).For any real number λ ∈ ( − , T λ = { x ∈ R n | x = λ } , and x λ =(2 λ − x , x , · · · , x n ) as the reflection of point x = ( x , x , · · · , x n ) about plane T λ .Let Σ λ = { x = ( x , x , · · · , x n ) ∈ B | − < x < λ } , and Σ Cλ = B \ Σ λ be the complement of Σ λ in B . Set u λ ( x ) = u ( x λ ) . We shallcomplete the proof in two steps. In step 1, we show that for λ sufficiently close to − u ( x ) ≥ u λ ( x ) , ∀ x ∈ Σ λ . (3.7)Then we can start to move plane T λ along the x direction. In step 2, we move theplane to the right as long as inequality (3.7) holds. We show that the plane can bemoved to λ = 0. So u ( − x , x , · · · , x n ) ≥ u ( x , x , · · · , x n ) , ∀ x ∈ B , x ≥ . (3.8)Similarly, we can start to move plane T λ from a place close to λ = 1, and move itto the left limiting position T . Then u ( − x , x , · · · , x n ) ≤ u ( x , x , · · · , x n ) , ∀ x ∈ B , x ≥ . (3.9)By (3.8) and (3.9), we have that u ( x ) is symmetric about the plane x = 0. Sim-ilarly, we can show that u ( x ) is symmetric about any plane passing through the rigin, which then implies that u ( x ) is radially symmetric about the origin andstrictly increasing in the radial direction.First, we have following comparison inequality. Lemma 3.4.
For any x ∈ Σ λ with λ ∈ ( − , , it holds u ( x ) − u λ ( x ) ≥ Z Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) ( u p − ( y ) − u p − λ ( y )) dy. (3.10) Proof.
Let e Σ λ = { x λ | x ∈ Σ λ } be the reflection of Σ λ about plane T λ , then u ( x ) = Z Σ λ u p − ( y ) | x − y | n − α dy + Z e Σ λ u p − ( y ) | x − y | n − α dy + Z Σ Cλ \ e Σ λ u p − ( y ) | x − y | n − α dy. Noting that | x − y | > | x λ − y | in x ∈ Σ λ , y ∈ Σ Cλ \ e Σ λ , we have u ( x ) − u λ ( x ) = Z Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) u p − ( y ) dy + Z e Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) u p − ( y ) dy + Z Σ Cλ \ e Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) u p − ( y ) dy ≥ Z Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) u p − ( y ) dy + Z Σ λ (cid:2) | x − y λ | n − α − | x λ − y λ | n − α (cid:3) u p − λ ( y ) dy = Z Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) u p − ( y ) dy − Z Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) u p − λ ( y ) dy = Z Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) ( u p − ( y ) − u p − λ ( y )) dy. (cid:3) Proof of Theorem 3.3.Step 1.
Let u ∈ C ( B ) be a positive solution to equation (3.6). We show that for λ sufficiently close to −
1, inequality (3.7) holds.From (3.6) we have ∂u ( x ) ∂x | x = − = ( α − n ) Z B | x − y | α − n − ( − − y ) u p − ( y ) dy< . Therefore for λ sufficiently close to −
1, we have u ( x ) ≥ u λ ( x ) for x ∈ Σ λ . Step 2.
Plane T λ can be moved continuously towards right to its limiting positionas long as inequality (3.7) holds. efine λ = sup { λ ∈ [ − , | u ( y ) ≥ u µ ( y ) , ∀ y ∈ Σ µ , µ ≤ λ } . We claim that λ must be 0.We prove it by contradiction. Suppose not, that is, λ < u ( x ) > u λ ( x ) , in Σ λ . Hence, we have u ( x ) − u λ ( x ) > c > , in Σ λ − ǫ for ǫ > | x − y | < | x − y λ | for x, y ∈ Σ λ , we have, similar to the calculationin the proof of Lemma 3.4, that u ( x ) − u λ ( x ) = Z Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) ( u p − ( y ) − u p − λ ( y )) dy + Z Σ Cλ \ e Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) u p − ( y ) dy ≥ Z Σ Cλ \ e Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) u p − ( y ) dy. (3.11)If there exists some point x ∈ Σ λ such that u ( x ) = u λ ( x ), then since | x − y | > | x λ − y | for x ∈ Σ λ , y ∈ Σ Cλ , we deduce from (3.11) that u ( y ) ≡ ∞ , ∀ y ∈ Σ Cλ \ e Σ λ . This contradicts to the assumption that u ∈ C ( B ) is a positive solution.For some small δ >
0, we choose ε ∈ (0 , ǫ ) small enough such that for any λ ∈ [ λ , λ + ε ), there holds u ( x ) ≥ u λ ( x ) , ∀ x ∈ Σ λ − ε , and | | x − y | n − α − | x λ − y | n − α | ≤ δ for x ∈ Σ λ \ Σ λ − ε . Write Σ uλ = { x ∈ Σ λ | u λ ( x ) > u ( x ) } . It follows from (3.10) that for any x ∈ Σ uλ ,0 > u ( x ) − u λ ( x ) ≥ Z Σ λ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) ( u p − ( y ) − u p − λ ( y )) dy ≥ Z Σ uλ (cid:2) | x − y | n − α − | x λ − y | n − α (cid:3) ( u p − ( y ) − u p − λ ( y )) dy ≥ − δ Z Σ uλ ( u p − ( y ) − u p − λ ( y )) dy. ince u ∈ C ( B ), there exists a positive constant C such that C ≤ u ≤ C . Itfollows from the above Z Σ uλ ( u λ ( x ) − u ( x )) dx ≤ δ Z Σ uλ Z Σ uλ ( u p − ( y ) − u p − λ ( y )) dydx ≤ (1 − p ) δ Z Σ uλ Z Σ uλ u p − ( y )( u λ ( y ) − u ( y )) dydx ≤ Cδ ( ε + ε ) n Z Σ uλ ( u λ ( y ) − u ( y )) dy. It implies that k u λ − u k L (Σ uλ ) ≡ , for δ , ε, ε small enough, and hence Σ uλ must have measure zero.We thus have u ( x ) − u λ ( x ) ≥ , for any x ∈ Σ λ , ∀ λ ∈ [ λ , λ + ε )since u is continuous. This contradicts to the definition of λ . Hence, λ = 0. Wehereby complete the proof of Theorem 3.4. (cid:3) Existence result for critical case
In this section, we study the existence of positive solutions to the integral equa-tion with critical exponent.The non-existence of positive solution to (1.1) with critical exponent for λ ≥ λ <
0. To this end, we consider Q λ (Ω) := inf f ∈ L qα + (Ω) R Ω R Ω f ( x )( | x − y | − ( n − α ) + λ | x − y | − ( n − α − ) f ( y ) dydx k f k L qα (Ω) . Notice that the corresponding Euler-Lagrange equation for extremal functions, upto a constant multiplier, is integral equation (1.1) with q = q α .First, we show Lemma 4.1. Q λ (Ω) < N α for all λ < . Further, < Q λ (Ω) < N α for any λ ∈ ( − d (Ω) , . Proof.
Let x ∗ ∈ Ω. For small positive ǫ and a fixed R > B R ( x ∗ ) ⊂ Ω,we define ˜ f ǫ ( x ) = ( f ǫ ( x ) x ∈ B R ( x ∗ ) ⊂ Ω , x ∈ R n \ B R ( x ∗ ) , here f ǫ is given by (2.1). Obviously, ˜ f ǫ ∈ L q α ( R n ) . Thus, similar to the proof ofProposition 2.1, we have Z Ω Z Ω (cid:0) | x − y | n − α + λ | x − y | n − α − (cid:1) ˜ f ǫ ( x ) ˜ f ǫ ( y ) dxdy = Z R n Z R n | x − y | n − α f ǫ ( x ) f ǫ ( y ) dxdy − Z R n Z R n \ B R ( x ∗ ) f ǫ ( x ) f ǫ ( y ) | x − y | n − α dxdy + Z R n \ B R ( x ∗ ) Z R n \ B R ( x ∗ ) f ǫ ( x ) f ǫ ( y ) | x − y | n − α dxdy + λ Z B R ( x ∗ ) Z B R ( x ∗ ) f ǫ ( x ) f ǫ ( y ) | x − y | n − α − dxdy ≤ N α k f ǫ k L qα ( R n ) − C ( ǫR ) n + λJ , where J := Z B R ( x ∗ ) Z B R ( x ∗ ) f ǫ ( x ) f ǫ ( y ) | x − y | n − α − dxdy = Z B R ( x ∗ ) Z B R ( x ∗ ) | x − y | − ( n − α − (cid:0) ǫǫ + | x − x ∗ | (cid:1) n + α (cid:0) ǫǫ + | y − x ∗ | (cid:1) n + α dxdy = ǫ − ( n − α − − ( n + α ) Z B R (0) Z B R (0) (cid:12)(cid:12) x − yǫ (cid:12)(cid:12) − ( n − α − (cid:0) (cid:12)(cid:12) xǫ (cid:12)(cid:12) (cid:1) − n + α (cid:0) (cid:12)(cid:12) yǫ (cid:12)(cid:12) (cid:1) − n + α dxdy = ǫ Z B Rǫ (0) Z B Rǫ (0) | ξ − η | − ( n − α − (cid:0) | ξ | (cid:1) − n + α (cid:0) | η | (cid:1) − n + α dξdη ≥ C ǫ. So, for λ < ǫ >
0, we have Z Ω Z Ω (cid:0) | x − y | n − α + λ | x − y | n − α − (cid:1) ˜ f ǫ ( x ) ˜ f ǫ ( y ) dxdy ≤ N α k f ǫ k L qα ( R n ) − C ( ǫR ) n + λC ǫ. This implies that Q λ (Ω) < N α for all λ < Q λ (Ω) > λ ∈ ( − d (Ω) , (cid:3) The existence of solutions to equation (1.1) will follow from the existence of aminimizer for energy Q λ (Ω). Proposition 4.2.
For any λ ∈ ( − d (Ω) , , infimum Q λ (Ω) is achieved by a positivefunction f ∗ ∈ L q α (Ω) . For q < q α , consider Q λ,q (Ω) = inf f ∈ L q + (Ω) R Ω R Ω f ( x ) | x − y | − ( n − α ) f ( y ) dxdy + λ R Ω R Ω f ( x ) | x − y | − ( n − α − f ( y ) dxdy k f k L q (Ω) . By Lemma 3.2, the infimum is attained by a positive function f q , which satisfiesthe subcritical equation Q λ,q (Ω) f q − ( x ) = Z Ω f ( y ) | x − y | n − α dy + λ Z Ω f ( y ) | x − y | n − α − dy, x ∈ Ω , (4.1)and k f q k L q (Ω) = 1 . Further, we can show easily that f q ∈ C (Ω) and Q λ,q → Q λ for q → ( q α ) − . emma 4.3. For λ ∈ ( − d (Ω) , and q ∈ (0 , q α ) , let f q > be a minimal energysolution to (4.1) with k f q k L q (Ω) = 1 . If < Q λ,q ≤ N α − ǫ for some ǫ > , thenthere exists C > such that C ≤ f q ( x ) ≤ C uniformly for all x ∈ Ω and q ∈ (0 , q α ) . Proof.
It is easy to see that max Ω f q ( x ) := f q ( x q ) ≤ C < ∞ uniformly for all x ∈ Ωand q ∈ (0 , q ) provided 0 < q < q α .We first prove by contradiction that max Ω f q ( x ) = f q ( x q ) ≤ C < ∞ uniformly forall x ∈ Ω and q ∈ (0 , q α ). Suppose not. Then f q ( x q ) → + ∞ for q → ( q α ) − . Let µ q = f − − qα q ( x q ) , and Ω µ = Ω − x q µ q := { z | z = x − x q µ q for x ∈ Ω } . Define g q ( z ) = µ α − q q f q ( µ q z + x q ) , for z ∈ Ω µ . (4.2)Then g q satisfies Q λ,q (Ω) g q − q ( z ) = Z Ω µ g q ( y ) | z − y | n − α dy + λµ q Z Ω µ g q ( y ) | z − y | n − α − dy, z ∈ Ω µ , (4.3)and g q (0) = 1, g q ( z ) ∈ (0 , h q ( z ) := g q − q ( z ). Then (4.3) is equivalent to Q λ,q (Ω) h q ( z ) = Z Ω µ h p − q ( y ) | z − y | n − α dy + λµ q Z Ω µ h p − q ( y ) | z − y | n − α − dy, z ∈ Ω µ , (4.4)where p + q = 1 , h q (0) = 1 , h q ( z ) ≥ C , C > z in a domain b Ω covered by Ω µ ,when q → ( q α ) − < C (1 + | z | α − n ) ≤ h q ( z ) ≤ C (1 + | z | α − n ) , uniformly. (4.5)We relegate the proof of this claim to the end.Once the claim is proved, we can prove that h q ( z ) is equicontinuous on anybounded domain b Ω ⊂ Ω µ when q → ( q α ) − . We write Q λ,q (Ω) h q ( z )= Z Ω µ \ B (0 ,R ) h p − q ( y ) | z − y | n − α dy + Z Ω µ ∩ B (0 ,R ) h p − q ( y ) | z − y | n − α dy + λµ q Z Ω µ \ B (0 ,R ) h p − q ( y ) | z − y | n − α − dy + λµ q Z Ω µ ∩ B (0 ,R ) h p − q ( y ) | z − y | n − α − dy. Notice that Z Ω µ \ B (0 ,R ) h p − q ( y ) | z − y | n − α (1+ λµ q | z − y | ) dy ≥ (1 −| λ | d (Ω)) Z Ω µ \ B (0 ,R ) h p − q ( y ) | z − y | n − α dy ≥ . hen for ǫ > ≤ Z Ω µ \ B (0 ,R ) h p − q ( y ) | z − y | n − α (1 + λµ q | z − y | ) dy ≤ C Z Ω µ \ B (0 ,R ) h p − q ( y ) | y | n − α dy ≤ C Z ∞ R r ( α − n )( p − α − dr = CR ( α − n )( p − α < ǫ (4.6)for any z ∈ b Ω, by taking
R > q close to q α . Similarly, we have | λµ q Z Ω µ ∩ B (0 ,R ) h p − q ( y ) | z − y | n − α − dy | < ǫ (4.7)by taking R > q close to q α . On the other hand, it is easy tosee that R Ω µ ∩ B (0 ,R ) h p − q ( y ) | z − y | n − α dy ∈ C ( b Ω). Hence for z , z ∈ b Ω, | Z Ω µ ∩ B (0 ,R ) h p − q ( y ) | z − y | n − α dy − Z Ω µ ∩ B (0 ,R ) h p − q ( y ) | z − y | n − α dy |≤ Z Ω µ ∩ B (0 ,R ) h p − q ( y ) 1 | ξ − y | n − α +1 dy | z − z |≤ Z B (0 ,R ) | ξ − y | n − α +1 dy | z − z | ≤ CR α − | z − z | , (4.8)where ξ = tz + (1 − t ) z for some t ∈ (0 , h q ( z ) is equicontinuous on bounded domain b Ω ∈ R n when q → ( q α ) − .As q → ( q α ) − , there are two cases:Case 1. Ω µ → R nT := { ( z , z , · · · , z n ) | z n > T ≥ } , and h q ( z ) → h ( z ) ∈ C ( R nT )uniformly in any compact set in R nT , where h ( z ) satisfies Q λ h ( z ) = Z R nT h p α − ( y ) | z − y | n − α dy, h (0) = 1 . (4.9)Also, direct computation yields1 = Z Ω f qq ( y ) dy = µ ( n + α − q ) · ( nn + α − q ) q · Z Ω µ g qq dz ≤ Z Ω µ g qq dz = Z Ω µ h pq dz. On the other hand, by (4.5) we have R Ω µ h pq dz ≤ C uniformly. Again by (4.5), R R nT h p α dz = lim q → ( q α ) − R Ω µ h pq dz ≥ . Denote g ( x ) = h p α − ( x ). Then R R nT h p α dz = R nT g q α dz . By (4.9), we have N α − ǫ ≥ Q λ = R R nT R R n g ( x ) g ( y ) | x − y | n − α dxdy k g k q α L qα ( R nT ) ≥ R R nT R R nT g ( x ) g ( y ) | x − y | n − α dxdy k g k L qα ( R nT ) = R R n R R n ˜ g ( x )˜ g ( y ) | x − y | n − α dxdy k ˜ g k L qα ( R n ) ≥ N α . Contradiction!Case 2. Ω µ → R n , and h q ( z ) → h ( z ) ∈ C ( R n ) uniformly in any compact set in R n , where h ( z ) satisfies Q λ h ( z ) = Z R n h p α − ( y ) | z − y | n − α dy, h (0) = 1 . (4.10)Similarly, C ≥ R R n h p α dz ≥ . Denote g ( x ) = h p α − ( x ). Then R R n h p α dz = R R n g q α dz . By (4.10) we have N α − ǫ ≥ Q λ = R R n R R n g ( x ) g ( y ) | x − y | n − α dxdy k g k q α L qα ( R n ) ≥ R R n R R n g ( x ) g ( y ) | x − y | n − α dxdy k g k L qα ( R n ) ≥ N α , which again implies a contradiction.Thus we conclude that there exists C > f q ( y ) ≤ C uniformly in y ∈ Ω and q ∈ (0 , q α ).On the other hand, if min Ω f q ( x ) := f q (˜ x q ) → q → ( q α ) − , by using f q ( y ) ≤ C uniformly in y ∈ Ω and q ∈ (0 , q α ) we have ∞ ← f q − q (˜ x q ) = Z Ω f q ( y ) | ˜ x q − y | n − α dy + λ Z Ω f q ( y ) | ˜ x q − y | n − α − dy ≤ C < ∞ as q → ( q α ) − , which gives a contradiction.Now we are left to prove claim (4.5).We first notice that Q λ,q (Ω) = Q λ,q (Ω) h q (0) = Z Ω µ h p − q ( y ) | y | n − α (1 + λµ q | y | ) dy. (4.11)Thus, Z Ω µ h p − q ( y ) | y | α − n dy = 1(1 − | λ | d (Ω)) Z Ω µ h p − q ( y ) | y | α − n (1 − | λ | d (Ω)) dy ≤ − | λ | d (Ω)) Z Ω µ h p − q ( y ) | y | α − n (1 + λµ q | y | ) dy ≤ C < ∞ , (4.12)uniformly as q → ( q α ) − . Since h q ≥ p <
0, we have Z Ω µ h p − q ( y ) dy ≤ C < ∞ (4.13)uniformly as q → ( q α ) − . n the other hand, we also have Z Ω µ h p − q ( y ) dy ≥ c > , as q → ( q α ) − . (4.14)Otherwise, if there exists a sequence q n → ( q α ) − such that R Ω µ h p n − q n ( y ) dy → p n + q n = 1, then for given R > ǫ > R >> R large enough, such that for z ∈ Ω µ ∩ B (0 , R ), as q n close to ( q α ) − ,1 ≤ h q n ( z )= 1 Q λ,q n (Ω) (cid:0) Z Ω µ \ B (0 ,R ) h p n − q n ( y ) | z − y | n − α dy + Z Ω µ ∩ B (0 ,R ) h p n − q n ( y ) | z − y | n − α dy + λµ q n Z Ω µ \ B (0 ,R ) h p n − q n ( y ) | z − y | n − α − dy + λµ q n Z Ω µ ∩ B (0 ,R ) h p n − q n ( y ) | z − y | n − α − dy (cid:1) ≤ Q λ,q n (Ω) (cid:0) Z Ω µ \ B (0 ,R ) h p n − q n ( y ) | z − y | n − α dy + Z Ω µ ∩ B (0 ,R ) h p n − q n ( y ) | z − y | n − α dy + λµ q n Z Ω µ \ B (0 ,R ) h p n − q n ( y ) | z − y | n − α − dy (cid:1) ≤ Q λ,q n (Ω) (cid:0) (1 + R R ) α − n Z Ω µ \ B (0 ,R ) h p n − q n ( y ) | y | n − α dy + ( R + R ) α − n Z Ω µ ∩ B (0 ,R ) h p n − q n ( y ) dy + λµ q n (1 − R R ) α +1 − n Z Ω µ \ B (0 ,R ) h p n − q n ( y ) | y | n − α − dy (cid:1) = 1 Q λ,q n (Ω) (cid:0) (1 + R R ) α − n Z Ω µ \ B (0 ,R ) h p n − q n ( y ) | y | n − α dy + ( R + R ) α − n Z Ω µ ∩ B (0 ,R ) h p n − q n ( y ) dy + λµ q n (1 − R R ) α +1 − n Z Ω µ h p n − q n ( y ) | y | n − α − dy − λµ q n (1 − R R ) α +1 − n Z Ω µ ∩ B (0 ,R ) h p n − q n ( y ) | y | n − α − dy (cid:1) ≤ Q λ,q n (Ω) (cid:0) (1 + R R ) α − n Z Ω µ h p n − q n ( y ) | y | n − α dy + λµ q n (1 − R R ) α +1 − n Z Ω µ h p n − q n ( y ) | y | n − α − dy (cid:1) +( R + R ) α − n Q λ,q n (Ω) Z Ω µ h p n − q n ( y ) dy − λµ q n ( R − R ) α +1 − n Q λ,q n (Ω) Z Ω µ h p n − q n ( y ) dy ≤ ǫ. That is, h q n ( z ) → , z ∈ Ω µ ∩ B (0 , R ) uniformly as q n → ( q α ) − . Then for R > R Ω µ ∩ B (0 ,R ) h p n − q n ( y ) dy ≤ R Ω µ h p n − q n ( y ) dy and Ω µ goes to either R nT := { ( z , z , · · · , z n ) | z n > T ≥ } or R n , we obtain a contradiction to (4.13).By (4.4), we havelim | z |→∞ | z | α − n Z Ω µ h p − q ( y ) | z − y | n − α dy ≥ lim | z |→∞ Q λ,q (Ω) h q ( z ) | z | α − n = lim | z |→∞ R Ω µ h p − q ( y ) | z − y | n − α (1 + λµ q | z − y | ) dy | z | α − n ≥ (1 − | λ | d (Ω)) lim | z |→∞ R Ω µ h p − q ( y ) | z − y | n − α dy | z | α − n . (4.15) ince | z | α − n h p − q ( y ) | z − y | n − α ≤ α − n h p − q ( y )(1 + | y | α − n ) as | z | → ∞ , and R Ω µ h p − q ( y )(1 + | y | α − n ) dy ≤ C by (4.12) and (4.13), we then havelim | z |→∞ | z | α − n Z Ω µ h p − q ( y ) | z − y | n − α dy = Z Ω µ h p − q ( y ) dy. (4.16)Hence by (4.15), (4.16), (4.13) and (4.14) we obtain the claim (4.5). Hereby wecomplete the proof of Lemma 4.3. (cid:3) Proof of Proposition 4.2 . Let f q > q ∈ (0 , q α ),which are also the minimal energy functions to energy Q λ,q . Then by Lemma 4.3,we know that { f q } are uniformly bounded above and bounded below by a positiveconstant. Thus they are equicontinuous due to equation (4.1). It follows that f q → f ∗ as q → ( q α ) − in C (Ω) , and f ∗ is the energy minimizer for Q λ . (cid:3) Completion of the Proof of Theorem 1.1 . Lemma 4.1 and Proposition 4.2implies the existence of a positive solution f ∈ L p α (Ω) ∩ C (Ω) to the equation (1.1)for q = nn + α , λ ∈ ( − d (Ω) , f ∈ C (Ω). (cid:3) Acknowledgements
We dedicate this paper to Professor Ha¨ım Brezis to celebrate his seventy five birth-day. We thank him for his great influence on us in the study of elliptic equationsthrough his lectures and numerous papers, in particular, the paper with LouisNirenberg [2]. The project is partially supported by the National Natural ScienceFoundation of China (Grant No. 11571268) and the Fundamental Research Fundsfor the Central Universities (Grant No. GK201802015) and Simmons Collaboration(Grant No. 280487).
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E-mail address : [email protected] ianqiao Guo, Department of Applied Mathematics, Northwestern PolytechnicalUniversity, Xi’an, Shaanxi, 710129, China E-mail address : [email protected] Meijun Zhu, Department of Mathematics, The University of Oklahoma, Norman, OK73019, USA
E-mail address : [email protected]@math.ou.edu