Energy identity and removable singularities of maps from a Riemann surface with tension field unbounded in L 2
aa r X i v : . [ m a t h . A P ] M a y Energy identity and removable singularities ofmaps from a Riemann surface with tension fieldunbounded in L ∗ Abstract
We prove the removal singularity results for maps with bounded energy from theunit disk B of R centered at the origin to a closed Riemannian manifold whose tensionfield is unbounded in L ( B ) but satisfies the following condition:( Z B t \ B t | τ ( u ) | ) ≤ C ( 1 t ) a , for some 0 < a < < t <
1, where C is a constant independent of t .We will also prove that if a sequence { u n } has uniformly bounded energy and satisfies( Z B t \ B t | τ ( u n ) | ) ≤ C ( 1 t ) a , for some 0 < a < < t <
1, where C is a constant independent of n and t , then the energy identity holds for this sequence and there will be no neck formationduring the blow up process. Let (
M, g ) be a Riemannian manifold and (
N, h ) be a Riemannian manifold without bound-ary. For a W , ( M, N ) map u , the energy density of u is defined by e ( u ) = 12 |∇ u | = T r g ( u ∗ h ) . where u ∗ h is the pullback of the metric tensor h .The energy functional of the mapping u is defined as E ( u ) = Z M e ( u ) dV. A map u ∈ C ( M, N ) is called a harmonic map if it is a critical point of the energy.By Nash embedding theorem N can be isometrically embedded into an Euclidean space R K for some positive integer K . Then ( N, h ) can be viewed as a submanifold of R K and amap u ∈ W , ( M, N ) is a map in W , ( M, R K ) whose image lies on N . The space C ( M, N ) ∗ The author is supported by the DFG Collaborative Research Center SFB/Transregio 71. △ u = A ( u )( ∇ u, ∇ u ) . The tension field of a map u , τ ( u ), is defined by τ ( u ) = △ u − A ( u )( ∇ u, ∇ u ) , where A is the second fundamental form of N in R K . So u is a harmonic map if and onlyif τ ( u ) = 0.Notice that when M is a Riemann surface the functional E ( u ) is conformal invariantand harmonic maps are of special interest in this case. Consider a harmonic map u from aRiemann surface M to N , recall that in the fundamental paper [S-U] of Sacks and Uhlenbeckthe well-known removable singularity theorem has long been established by using a class ofpiecewise smooth harmonic functions to approximate the weak harmonic map. In the paper[Li-W1], the authors gave a slightly different proof of the following removable singularitytheorem: Theorem 1.1. [Li-W1] Let B be the unit disk in R centered at the origin. If u : B \ { } → N is a W , loc ( B \ { } , N ) ∩ W , ( B, N ) map and u satisfies τ ( u ) = g ∈ L ( B, R K ) , then u may be extended to a map belonging to W , ( B, N ) . In this direction we will prove the following result:
Proposition 1.2.
Let B be the unit disk in R centered at the origin. If u : B \ { } → N is a W , loc ( B \ { } , N ) ∩ W , ( B, N ) map and u satisfies the following condition: ( Z B t \ B t | τ ( u ) | ) ≤ C ( 1 t ) a , for some < a < and for any < t < , where C is a constant independent of t , thenthere exists some s > such that ∇ u ∈ L s ( B ) . A direct corollary of this result is the following removable singularity theorem:
Theorem 1.3.
Assume that u ∈ W , loc ( B \{ } , N ) ∩ W , ( B, N ) and u satisfies the followingcondition: ( Z B t \ B t | τ ( u ) | ) ≤ C ( 1 t ) a , for some < a < and for any < t < , where C is a constant independent of t .Then we have u ∈ \
1. In fact, they proved this resultunder a scaling invariant condition which can be deduced from the uniform boundness ofthe tension field in L p .By virtue of Fanghua Lin and Changyou Wang’s result, it is natural to ask the followingquestion: Qestion:
Let { u n } be a sequence from a closed Riemann surface to a closed Riemannianmanifold with tension field uniformly bounded in L p for some p >
1. Do the energy identityand bubble tree convergence results hold true during blowing up for such a sequence?
Remark 1.4.
Parker ([P] constructed a sequence from a Riemann surface whose tensionfield is uniformly bounded in L , in which the energy identity fails. In [Li-Z] the authors proved the following theorem:
Theorem 1.5. [Li-Z] Let { u n } be a sequence of maps from B to N in W , ( B, N ) withtension field τ ( u n ) , where B is the unit disk of R centered at the origin. If(I) k u n k W , ( B ) + k τ ( u n ) k W ,p ( B ) ≤ Λ for some p ≥ ,(II) u n → u strongly in W , loc ( B \ { } , N ) as n → ∞ ,then there exists a subsequence of { u n } (still denoted by { u n } ) and some nonnegative integer k , such that for any i = 1 , ..., k , there are some points x in , positive numbers r in and anonconstant harmonic sphere ω i (which is viewed as a map from R ∪ {∞} → N ) such that(1) x in → , r in → as n → ∞ ;(2) lim n →∞ ( r in r jn + r jn r in + | x in − x jn | r in + r jn ) = ∞ for any i = j ;(3) ω i is the weak limit or strong limit of u n ( x in + r in x ) in W , loc ( R , N ) ;(4) Energy identity: lim n →∞ E ( u n , B ) = E ( u, B ) + k X i =1 E ( ω i , R ) . (5) Neckless: the image u ( B ) S ki =1 ω i ( R ) is a connected set. emma 1.6. If τ ( u ) satisfies ( Z B t \ B t | τ ( u ) | ) ≤ C ( 1 t ) a , for some < a < and for any < t < , where C is a constant independent of t , then τ ( u ) is bounded in L p ( B ) for some p ≥ . Proof. Z B − k +1 \ B − k | τ ( u ) | p ≤ C (2 − k ) − p k τ ( u ) k pL ( B − k +1 \ B − k ) ≤ C (2 − k ) − p − ap , hence Z B | τ ( u ) | p ≤ C ∞ X k =1 (2 − k ) − p − ap . When 0 < a < , we can choose some p ≥ such that 2 − p − ap > P ∞ k =1 (2 − k ) − p − ap ≤ C , which implies that τ ( u ) is bounded in L p ( B ) for some p ≥ .Thus theorem 1.5 holds true for sequences { u n } satisfying the following conditions(I) k u n k W , ( B ) ≤ Λ and ( R B t \ B t | τ ( u n ) | ) ≤ C ( t ) a , for some 0 < a < and for any0 < t <
1, where C is independent of n , t ;(II) u n → u strongly in W , loc ( B \ { } , N ) as n → ∞ .With the help of this observation, we find the following theorem: Theorem 1.7.
Let { u n } be a sequence of maps from B to N in W , ( B, N ) with tensionfield τ ( u n ) , where B is the unit disk of R centered at the origin. If(I) k u n k W , ( B ) ≤ Λ and ( Z B t \ B t | τ ( u n ) | ) ≤ C ( 1 t ) a , for some < a < and for any < t < , where C is independent of n , t ,(II) u n → u strongly in W , loc ( B \ { } , N ) as n → ∞ ,then there exists a subsequence of { u n } (still denote it by { u n } ) and some nonnegative integer k , such that for any i = 1 , ..., k , there are some points x in , positive numbers r in and anonconstant harmonic sphere ω i (which is viewed as a map from R ∪ {∞} → N ) such that(1) x in → , r in → as n → ∞ ;(2) lim n →∞ ( r in r jn + r jn r in + | x in − x jn | r in + r jn ) = ∞ for any i = j ;(3) ω i is the weak limit or strong limit of u n ( x in + r in x ) in W , loc ( R , N ) ;(4) Energy identity: lim n →∞ E ( u n , B ) = E ( u, B ) + k X i =1 E ( ω i , R ) . (5) Neckless: the image u ( B ) S ki =1 ω i ( R ) is a connected set. emark 1.8. When ( Z B t \ B t | τ ( u n ) | ) ≤ C ( 1 t ) a , for some < a < and for any < t < , where C is independent of n , t , we can deducethat τ ( u n ) is uniformly bounded in L p ( B ) for any p < a and when a → , p → . Henceour condition is stronger than the condition that the tension field is bounded in L p , forsome p > , and this result suggests that we probably have a positive answer to the questionon page 3. This paper is organized as follows: In section 2 we will quote and prove results which willbe important in the following sections. In section 3 we will prove the removable singularityresult and the last theorem, theorem 1.7, will be proved in section 4.Throughout this paper, without illustration the letter C will denote positive constantsvarying from line to line, and we do not always distinguish sequences and its subsequences. Acknowledgement
I would like to thank Professor Guofang Wang for pointing outmany typing errors and I also appreciate Professor Youde Wang for bringing my attentionto the paper [Li-W1]. My interest in this kind of problem began at a class given by ProfessorYuxiang Li at Tinghua university, and I had many useful discussions with him . ǫ -regularity lemma and the Pohozeav inequality In this section we give the well known small energy regularity lemma for approximated har-monic maps and a version of Pohozeav inequality, which will be important in the followingsections. We assume that the disk B ⊆ R is the unit disk centered at the origin, whichhas the standard flat metric. Lemma 2.1.
Suppose that u ∈ W , ( B, N ) and τ ( u ) = g ∈ L ( B, R K ) , then there existsan ε > such that if R B |∇ u | ≤ ε , we have k u − ¯ u k W , ( B ) ≤ C ( k∇ u k L ( B ) + k g k L ( B ) ) . (2.1) Here ¯ u is the mean value of u over B .Proof. We can find a complete proof of this lemma in [D-T]. ✷ Using the standard elliptic estimates and the embedding theorems we can derive fromthe above lemma that
Corollary 2.2.
Under the assumptions of proposition 1.2, we have
Osc B r \ B r u ≤ C ( k∇ u k L ( B r \ B r ) + r k g k L ( B r \ B r ) ) ≤ C ( k∇ u k L ( B r \ B r ) + r − a ) . (2.2)In order to prove proposition 1.2, we need the following Pohozeav inequality Lemma 2.3.
Under the assumptions of proposition 1.2, for < t < t < , there holdstrue that Z ∂ ( B t \ B t ) r ( | ∂u∂r | − |∇ u | ) ≤ t k∇ u k L ( B t \ B t ) k g k L ( B t \ B t ) . (2.3)5 roof. Multiplying the both sides of the equation τ ( u ) = g by r ∂u∂r , we get Z B t \ B t r ∂u∂r △ u = Z B t \ B t r ∂u∂r g. By integral by parts we get Z B t \ B t r ∂u∂r △ udx = Z ∂ ( B t \ B t ) r | ∂u∂r | − Z B t \ B t ∇ ( r ∂u∂r ) ∇ udx, and Z B t \ B t ∇ ( r ∂u∂r ) ∇ udx = Z B t \ B t ∇ ( x k ∂u∂x k ) ∇ udx = Z B t \ B t |∇ u | + Z t t Z π r ∂∂r |∇ u | rdθdr = Z B t \ B t |∇ u | + 12 Z ∂ ( B t \ B t ) |∇ u | r − Z B t \ B t |∇ u | = 12 Z ∂ ( B t \ B t ) |∇ u | r. Clearly it implies the conclusion of the lemma. ✷ By the above lemma, we can deduce that
Corollary 2.4.
Under the assumptions of proposition 1.2, we have ∂∂t Z B t \ B t | ∂u∂r | − |∇ u | ≤ C k∇ u k L ( B t \ B t ) t − a . Proof.
In the previous lemma, let t = t , t = t , then there holds true ∂∂t Z B t \ B t | ∂u∂r | − |∇ u | = Z ∂B t ( | ∂u∂r | − |∇ u | ) − Z ∂B t ( | ∂u∂r | − |∇ u | ) ≤ k g k L ( B t \ B t ) k∇ u k L ( B t \ B t ) ≤ C k∇ u k L ( B t \ B t ) t − a . ✷ A direct corollary of the above conclusion is the following
Corollary 2.5.
Under the assumptions of proposition 1.2, we can get that Z B t \ B t | ∂u∂r | − |∇ u | ≤ C k∇ u k L ( B t ) t − a . (2.4) Proof.
Integrating from 0 to t over the both sides of the inequality of the above corollary,and noting that k∇ u k L ( B s \ B s ) ≤ k∇ u k L ( B t ) , for any s ≤ t , we get this inequality directly. ✷ Removable of singularities
In this section we will discuss the issues on the removable of singularities of a class ofapproximated harmonic maps from the unit disk of R centered at the origin to a closedRiemannian manifold N . Lemma 3.1.
Assume that u satisfies the assumptions of proposition 1.2, then there areconstants λ > , C > , such that the following holds true Z B r |∇ u | ≤ Cr λ , (3.1) for r small enough.Proof. Because we only need to prove the lemma for r is small, we can assume that E ( u, B ) < ε . Let u ∗ ( r ) : (0 , → R K be a curve defined as follows u ∗ ( r ) = 12 π Z π u ( r, θ ) dθ. Then ∂u ∗ ∂r = 12 π Z π ∂u∂r dθ. On the one hand, we have Z B − kt \ B − k − t ∇ u ∇ ( u − u ∗ ) ≥ Z B − kt \ B − k − t ( |∇ u | − | ∂u∂r | ) ≥ Z B − kt \ B − k − t |∇ u | − C (2 − k t ) − a , where in the above second inequality we have used (2.4).Summing k from 0 to infinity we get Z B t ∇ u ∇ ( u − u ∗ ) ≥ Z B t |∇ u | − Ct − a . On the other hand, we have Z B − kt \ B − k − t ∇ u ∇ ( u − u ∗ ) = − Z B − kt \ B − k − t ( u − u ∗ ) △ u + Z ∂ ( B − kt \ B − k − t ) ∂u∂r ( u − u ∗ )= − Z B − kt \ B − k − t ( u − u ∗ )( τ ( u ) − A ( u )( ∇ u, ∇ u ))+ Z ∂ ( B − kt \ B − k − t ) ∂u∂r ( u − u ∗ ) . k from 0 to infinity we get Z B t ∇ u ∇ ( u − u ∗ ) ≤ ∞ X k =0 k u − u ∗ k L ∞ ( B − kt \ B − k − t ) [ k A k L ∞ Z B − kt \ B − k − t |∇ u | + C (2 − k t ) − a ] + Z ∂B t ∂u∂r ( u − u ∗ ) ≤ ǫ Z B t |∇ u | + Ct − a + Z ∂B t ∂u∂r ( u − u ∗ ) , note that here we have used corollary 2.2 and we let ǫ small by letting t small.Note that | Z ∂B t ∂u∂r ( u − u ∗ ) | ≤ ( Z ∂B t | ∂u∂r | ) ( Z ∂B t | u − u ∗ | ) ≤ ( Z π t | ∂u∂r | dθ ) ( Z π | ∂u∂θ | dθ ) ≤ Z π ( | ∂u∂θ | + t | ∂u∂r | ) dθ = t Z ∂B t |∇ u | . Combining the two sides of inequalities and let ǫ small(we can do this by letting t small),we conclude that there is a λ which is a positive constant smaller than one such that λ Z B t |∇ u | ≤ t Z ∂B t |∇ u | + Ct − a . Now denote f ( t ) = R B t |∇ u | , then we get the following ordinary differential inequality( f ( t ) t λ ) ′ ≥ − Ct − λ − a . Note that we can let λ small enough such that λ + a <
1, thus we can get that f ( t ) = Z B t |∇ u | ≤ Ct λ , f or t small enough. ✷ Now we are in a position to give a complete proof of proposition 1.2. Let r k = 2 − k and v k ( x ) = u ( r k x ), then( Z B \ B |∇ v k | s ) s ≤ C k v k − ¯ v k k W , ( B \ B ) ≤ ( Z B \ B |∇ v k | ) + C ( Z B rk \ B rk r k | τ | ) . Therefore we deduce that Z B \ B |∇ v k | s ≤ C ( Z B \ B |∇ v k | ) s + C ( Z B rk \ B rk r k | τ | ) s ≤ C ( Z B \ B |∇ v k | ) s + Cr s (1 − a ) k . k is large enough, there holds true Z B rk \ B rk |∇ u | ≤ . Hence we have that r s − k Z B rk \ B rk |∇ u | s ≤ C ( Z B rk \ B rk |∇ u | ) s + Cr s (1 − a ) k ≤ C Z B rk \ B rk |∇ u | + Cr s (1 − a ) k . This implies that Z B rk \ B rk |∇ u | s ≤ Cr − sk r λk + Cr − sak . Now choose s > s − < λ and 2 − sa >
0. There exists a positive integer k such that when k ≥ k there holds true Z B − k +1 \ B − k |∇ u | s ≤ C (2 − λ k + 2 − k (2 − sa ) ) . Therefore Z B r |∇ u | s ≤ C ∞ X k = k (2 − λ k + 2 − k (2 − sa ) )) ≤ C, for any r ≤ − k +1 , and this completes the proof of proposition 1.2.Proof of theorem 1.3: Note that Z B − k \ B − k − | τ ( u ) | p ≤ C (2 − k ) − p ( Z B − k \ B − k − | τ ( u ) | ) p ≤ C (2 − k ) − p − pa . Hence we can deduce from this by summing k from 0 to infinity that Z B | τ ( u ) | p ≤ C f or p <
21 + a .
Recall that we have proved that ∇ u ∈ L s ( B ) for some s >
1, hence we can deduce that u ∈ \
Assume that { u n } is a sequence which is uniformly bounded in W , ( B, N ) and satisfies( Z B t \ B t | τ ( u n ) | ) ≤ C ( 1 t ) a , for some 0 < a < < t <
1, where C is a constant independent of n and t .In this section, we will prove the energy identity for this sequence and for conveniencewe will assume that there is only one bubble ω which is the strong limit of u n ( r n . ) in W , loc ( R , N ) . Becauselim n →∞ Z B |∇ u n | = lim n →∞ Z B \ B δ |∇ u n | + lim n →∞ Z B δ \ B Rrn |∇ u n | + lim n →∞ Z B Rrn |∇ u n | . and lim δ → lim n →∞ Z B \ B δ |∇ u n | = Z B |∇ u | , lim R →∞ lim n →∞ Z B Rrn |∇ u n | = Z R |∇ ω | , hence to prove the energy identity we only need to prove thatlim R →∞ lim δ → lim n →∞ Z B δ \ B Rrn |∇ u n | = 0 . (4.1)At first we note that under the assumption of only one bubble we can deduce by astandard blow up argument the following Lemma 4.1.
For any ǫ > , there exist R and δ such that Z B λ \ B λ |∇ u n | ≤ ǫ , f or any λ ∈ ( Rr n , δ ) . (4.2)Now we have been prepared to prove the energy identity. The proof is a little similarwith the proof in the previous section. We assume that δ = 2 m n Rr n where m n is a positiveinteger.On the one hand, we have Z B kRrn \ B k − Rrn ∇ u n ∇ ( u n − u ∗ n ) ≥ Z B kRrn \ B k − Rrn ( |∇ u n | − | ∂u n ∂r | ) ≥ Z B kRrn \ B k − Rrn |∇ u n | − C (2 k Rr n ) − a . This implies that Z B δ \ B Rrn ∇ u n ∇ ( u n − u ∗ n ) ≥ Z B δ \ B Rrn |∇ u n | − Cδ − a .
10n the other hand, we have that Z B kRrn \ B k − Rrn ∇ u n ∇ ( u n − u ∗ n ) = − Z B kRrn \ B k − Rrn ( u n − u ∗ n ) △ u n + Z ∂ ( B kRrn \ B k − Rrn ) ∂u n ∂r ( u n − u ∗ n )= − Z B kRrn \ B k − Rrn ( u n − u ∗ n )( τ ( u n ) − A ( u n )( ∇ u n , ∇ u n ))+ Z ∂ ( B kRrn \ B k − Rrn ) ∂u n ∂r ( u n − u ∗ n ) . We deduce from the above by summing from 1 to m n that Z B δ \ B Rrn ∇ u n ∇ ( u n − u ∗ n ) ≤ m n X k =1 k u n − u ∗ n k L ∞ ( B kRrn \ B k − Rrn ) [ k A k L ∞ Z B kRrn \ B k − Rrn |∇ u n | + C (2 k Rr n ) − a ] + Z ∂ ( B δ \ B Rrn ) ∂u n ∂r ( u n − u ∗ n ) ≤ ǫ Z B δ \ B Rrn |∇ u n | + Cδ − a + Z ∂ ( B δ \ B Rrn ) ∂u n ∂r ( u n − u ∗ n ) . Comparing with the two sides we get that(1 − ǫ ) Z B δ \ B Rrn |∇ u n | ≤ Cδ − a + 2 Z ∂ ( B δ \ B Rrn ) ∂u n ∂r ( u n − u ∗ n ) . As for the boundary terms, we have Z ∂B δ ∂u n ∂r ( u n − u ∗ n ) ≤ ( Z ∂B δ | ∂u n ∂r | ) ( Z ∂B δ | u n − u ∗ n | ) ≤ ( Z π δ | ∂u n ∂r dθ | ) ( Z π | ∂u n ∂θ | dθ ) ≤ Z π δ | ∂u n ∂r dθ | + | ∂u n ∂θ | dθ = δ Z π |∇ u n | dθ. By trace embedding theorem, we have Z π |∇ u n ( ., δ ) | δdθ = Z ∂B δ |∇ u n ( ., δ ) | dS δ ≤ Cδ k∇ u n k W , ( B δ \ B δ ) ≤ Cδ k u n − ¯ u n k W , ( B δ \ B δ ) ≤ Cδ ( 1 δ k∇ u n k L ( B δ ) + k τ ( u n ) k L ( B δ \ B δ ) ) ≤ Cδ − a , δ small. From this we deduce that Z ∂B δ ∂u n ∂r ( u n − u ∗ n ) ≤ Cδ − a ) . Similarly we get Z ∂B Rrn ∂u n ∂r ( u n − u ∗ n ) ≤ C ( Rr n ) − a ) , for n big enough. Therefore(1 − ǫ ) Z B δ \ B Rrn |∇ u n | ≤ Cδ − a + Cδ − a ) + C ( Rr n ) − a ) . Clearly this implies that lim R →∞ lim δ → lim n →∞ Z B δ \ B Rrn |∇ u n | = 0 , (4.3)which completes the proof of the energy identity. ✷ In this part we will prove that there is no neck between the base map u and the bubble ω ,i.e. the C compactness of the sequence modulo bubbles.We only need to prove the followinglim R →∞ lim δ → lim n →∞ Osc B δ \ B Rrn u n = 0 . (4.4)Again we assume that δ = 2 m n Rr n and let Q ( t ) = B t + t Rr n \ B t − t Rr n , similarly to theproof of the previous part we can get(1 − ǫ ) Z Q ( k ) |∇ u n | ≤ k + t Rr n Z ∂B k + t Rrn |∇ u n | + 2 t − k Rr n Z ∂B t − kRrn |∇ u n | + C (2 k + t Rr n ) − a . Denote f ( t ) = R Q ( t ) |∇ u n | , then we have that(1 − ǫ ) f ( t ) ≤ (1 − ǫ ) f ( k + 1) ≤ f ′ ( k + 1) + C (2 k + t Rr n ) − a , for k ≤ t ≤ k + 1.Note that f ′ ( k + 1) − f ′ ( t ) = Z ∂ ( B k +1+ t Rrn \ B t + t Rrn ) ∂u n ∂r ( u n − u ∗ n ) + Z ∂ ( B t − tRrn \ B t − k − Rrn ) ∂u n ∂r ( u n − u ∗ n ) ≤ C (2 t + t Rr n ) − a ) . − ǫ ) f ( t ) ≤ f ′ ( t ) + C (2 t + t Rr n ) − a . (4.5)In virtue of the above inequality we get(2 − (1 − ǫ ) t f ( t )) ′ = 2 − (1 − ǫ ) t f ′ ( t ) − (1 − ǫ )2 − (1 − ǫ ) t f ( t ) log 2 ≥ − C (1 − a − (1 − ǫ )) t (2 t Rr n ) − a . Integrating this from 1 to L we get2 − (1 − ǫ ) L f ( L ) − − (1 − ǫ ) f (1) ≥ − C Z L (1 − a − (1 − ǫ )) t (2 t Rr n ) − a = − C (1 − a − (1 − ǫ )) t log 2(1 − a − (1 − ǫ )) | L (2 t Rr n ) − a ≥ − C (2 t Rr n ) − a . Therefore we have that f (1) ≤ f ( L )2 − (1 − ǫ )( L − + C (2 t Rr n ) − a . (4.6)Now let t = i and D i = B i +1 Rr n \ B i Rr n , then we have f (1) = Z D i S D i − |∇ u n | , and equality will hold true for L satisfying Q ( L ) ⊆ B δ \ B Rr n = B mn Rr n \ B Rr n . In other words L should satisfy(1) i − L ≥ i + L ≤ M n . If(I) i ≤ m n , let L = i and so f (1) = Z D i S D i − |∇ u n | ≤ CE ( u n , B δ \ B Rr n )2 − (1 − ǫ ) i + C (2 i Rr n ) − a ;(II) i > m n , let L = m n − i and so f (1) = Z D i S D i − |∇ u n | ≤ CE ( u n , B δ \ B Rr n )2 − (1 − ǫ )( m n − i ) + C (2 i Rr n ) − a . Hence we have m n X i =1 E ( u n , D i ) ≤ CE ( u n , B δ \ B Rr n )( X i ≤ m n − i − ǫ + X i> m n − ( m n − i ) − ǫ ) + C m n X i =1 (2 i Rr n ) − a ≤ CE ( u n , B δ \ B Rr n ) + Cδ − a . Osc B δ \ B Rrn u n ≤ C m n X i =1 ( E ( u n , D i ) + (2 i Rr n ) − a ) ≤ CE ( u n , B δ \ B Rr n ) + Cδ − a . Clearly the above inequality implies thatlim R →∞ lim δ → lim n →∞ Osc B δ \ B Rrn u n = 0 , which completes the proof of the Neckless. ✷ References [D] W. Ding: Lectures on the heat flow of harmonic maps.
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Houston J. Math. , 1996, 559-590. YONG LUOMathematisches Institut, Albert-Ludwigs-UniversitEckerstr. 1, 79104, Freiburg, Germany [email protected]@math.uni-freiburg.de