MOSER-TRUDINGER INEQUALITIES OF VECTOR BUNDLE OVER A COMPACT RIEMANNIAN MANIFOLD OF DIMENSION Introduction and Main results

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OSER-TRUDINGER INEQUALITIES OF VECTOR BUNDLE OVER A COPACT RIEANNIAN ANIFOLD OF DIENSION 2 YUXIANG LI, PAN LIU, YUNYAN YANG Abstract. Let (, g) be a 2-dimensional compact Riemannian manifold. In this paper, we use the method of blowing up analysis to prove several oser-trudinger type inequalities for vector bundle over (, g). We also derive an upper bound of such inequalities under the assumption that blowing up occurs. 1. Introduction and ain results Let (, g) be a 2-dimensional compact Riemannian manifold. One of Fontana s results (see [F]) says { < + if α 4π sup e αu2 = = + if α > 4π, R u 2 dv g=1, R udvg=0 which extends Trudinger and oser s inequalities (see [T], []). A wea form of the above inequality is (1.1) log e u dv g 1 u 2 dv g + udv g + C 16π for all u H 1,2 (), where the constant C depends only on the geometry of (see [], [A]). The inequality (1.1) has been extensively used in many mathematical and physical problems, for instance in the problem of prescribing Gaussian curvature ([Ch], [C-Y], [D-J-L-W]), the mean field equation and the abelian Chern-Simons model ([D-J-L-W2], [D-J-L-W3], [J-W]), ect. In this note we want to derive some new oser-trudinger type inequalities. We will consider a smooth vector bundle E with metric h and connection over. Here, we do not assume h = 0. We write and H 1 = H0, i.e. H 1 = {σ H 1,2 (, E) : Our main results are the follows: H 0 = {σ H 1,2 (, E) : σ = 0}, σ, ζ dv g = 0, for all ζ H 0 }. The research of the second author was partially supported by NSFC grant and the Foundation of Shanghai for Priority Academic Discipline. 1

2 YUXIANG LI, PAN LIU, YUNYAN YANG Theorem 1.1 Let (, g) be a 2-dimensional compact Riemannian manifold, (E, h) be a smooth vector bundle over (, g), and H 1 be defined as above. Denote H 1 = {σ H 1 : σ, σ dv g = 1}. Then we have sup e 4π σ 2 dv g <, σ H 1 where σ 2 = σ, σ h, and the constant 4π is sharp, which means that for any α > 4π, e α σ 2 dv g =. sup σ H 1 As a consequence of Theorem 1.1, we have Corollary 1.2 With the assumption in Theorem 1.1., there exists a constant C such that e σ dv g Ce 1 R 16π σ, σ dv g holds for all σ H 1. We remar that: If E is a trivial bundle, e i is global basis of E, with e i, e j h = h ij and e i = 0, then { } H 1 = U H 1,2 (, R n ) : UH U T dv g = 1, UdV g = 0, and we have the following: Corollary 1.3 Let (, g) be a 2-dimensional compact Riemannian manifold. C (, GL n ), s.t. H(x) are symmetric for any x. Then we have e 4πUHU T dv g < +, where 4π is sharp. sup U H 1 Let H If E is a trivial line bundle with e, e h = f(x), Corollary 1.3 is exactly Yang s result[y]. An analogue of Theorem 1.1 is the following: Theorem 1.4. With the assumption in Theorem 1.1, we denote H 2 = {σ H 1 : σ, σ + σ 2 )dv g = 1}. Then we have where the constant 4π is sharp. sup e 4π σ 2 dv g <, σ H 2

OSER-TRUDINGER INEQUALITIES OF VECTOR BUNDLES 3 The proof of Theorem 1.1 is outlined as follows. Let us define J α (σ) = eα σ 2 dv g. We first show that the sup of J α can be attained in H 1 if α < 4π. So we can choose α converging to 4π increasingly, and σ H 1 satisfying J α (σ ) = sup σ H 1 J α (σ). Denote c = σ (x ) = max x σ (x). By passing to a subsequence, we assume p = x. Without loss of generality, we may assume blow-up occur, that is, c +. Tae a local coordinate system (Ω, x) around p. Using the idea in [S] and [A-S], we define a sequence of functions, for some r > 0, where x Ω with We then prove that, for suitable r, ϕ (x) = 2α σ (x ), σ (x + r x) σ (x ) Ω = {x R 2 x + r x Ω}. ϕ 2 log(1 + π x 2 ) in C 2 loc (R2 ). Then we prove that c σ G in C 2 (Ω), Ω \ {p}, where G is certain Green section. Finally, with the asymptotic behavior of σ described above, we can establish the desired inequality and thus Theorem 1.1. In fact, we can give an upper bound of the functional J 4π in case that the blow-up happens. The proof of Theorem 1.4 proceeds similarly. Though we mainly follow the ideas in [L] (also see [A-D]) and [L-L], we should point out that, in this paper, the convergence of c σ is derived differently from that of [L], [A-D] and [L-L]: The ey gradient in [L] and [A-D] is the energy estimate (1.2) u 2 dv g = 1 {u c A } A. In this paper, though similar identity is used, the calculations are based on the local Pohozaev identities. We than Professor Weiyue Ding who notice us possible application of the Pohozaev identity when we study the extremal function for Fontana s inequality on 4-dimensional manifold (see [L-Y]). oreover, the method we get the upper bound of J 4π is also new: instead of capacity technique in [L], we use a result of Carleson and Chang ([C-C]) as follows: Theorem A Let B be unit ball in R 2. Given any sequence u H 1,2 0 (B), if u 0, and B u 2 dx 1, then we have sup (e 4πu2 1)dx πe. B

4 YUXIANG LI, PAN LIU, YUNYAN YANG The paper is organized as follows: In section 2, we settle some notations for use later. In section 3 we prove that 4π is the best constant. Section 4 is blowing up analysis. We will prove the convergence of c σ in section 5. Then we finish the proof of the Theorem 1.1 in section 6. We hope our results can be a useful tool in studying some problems arising from geometry and mathematical physics. In a forthcoming paper, we shall extend our results to high dimensional case and find some geometrical and physical applications. 2. Preinaries In this section, we clarify some notations. Tae finite coordinate domains {Ω } which cover. Let σ be a smooth section of E, on each Ω, we can set σ = n i=1 ui e i. We define σ 2 = ( ui Ω x j 2 + u i 2 )dx i,j i and σ H 1,2 (,E) = σ. Clearly, this norm is equivalent to ( ) 1 = ( 2 + 2 )dv g Let σ be a parallel section, i.e. σ H 0. Then we have σ H 1,2 C σ L 2. By the compactness of Sobolev embedding from H 1,2 into L 2, we have 1. H 0 is a finite dimensional vector space. We will use ζ 1,, ζ m to denote an orthogonal basis of H 0. 2. Poincaré inequality holds on H 1, i.e. for any σ H 1, we have σ 2 dv g C σ, σ dv g, so, on H 1, we now the following norm is equivalent to the H 1,2 norm. H 1,2 1 ( ) 1 = ( 2 2 )dv g 2. Throughout this paper, we will use (Ω p ; x 1, x 2 ) to denote an isothermal coordinate system around p with p = (0, 0). We can write g = e f p (d(x 1 ) 2 + d(x 2 ) 2 ) with f p (0) = 0. oreover, we always assume e 1, e 2,, e n to be an orthogonal basis of E in Ω p. We should explain some notations involving Φ. Locally, if Φ = u e is a section of E, then denotes the connection of E, and Φ, Φ = e f p u e, x i u e = x i u e, x i u e 0. x i i=1,2 i=1,2

OSER-TRUDINGER INEQUALITIES OF VECTOR BUNDLES 5 If Φ is a function, then Φ is just the tangent vector Φ x 1 + Φ x 1 x 2 x 2, and 0 Φ 2 = Φ x 1 2 + Φ x 2 2, Φ 2 = e f p 0 Φ 2. If Φ = (u 1,, u n ) H 1,2 (Ω) H 1,2 (Ω), we use 0 Φ 2 to denote n 0 u 2. and Finally, we set =1 B r = {(x 1, x 2 ) : x 1 2 + x 2 2 < r 2 } Ω p, 0 = 2 x 1 2 + 2 x 2 2. 3. The best constants In this section, we will prove the following statement: sup σ H1 eα σ 2 dv g < + for any α < 4π, and sup σ H1 eα σ 2 dv g = + for any α > 4π. Firstly, we need a result in [Ch] (cf. Theorem 2.50 in [Au]): Lemma 3.1. Let B r be a ball in R 2, then for any β < 2π, we have sup dx < C(r). Then we have the following: R Br ( u 2 +u 2 )dx=1 B r e βu2 Corollary 3.2. Let B r be a ball in R 2, and u 1,, u n H 1,2 (B r ). Then we have np u i 2 sup i=1 dx < C(r). for any β < 2π. np i=1 R Br ( ui 2 + u i 2 )dx=1 B r e β Proof. We set λ i = B r ( u i 2 + u i 2 )dx, then B r e β u i 2 λ i dx < C. Hence e β np i=1 B r u i 2 dx n ( i=1 B r e β u i 2 λ i np dx) λ λ i i C i=1 = C. Remar 3.1. Let (Ω p ; x 1, x 2 ) be a coordinate system around p, and B r Ω p. Given a section σ = u e, we set U = (u 1,, u n ). Clearly, 0 U 2 + U 2 C( σ, σ + σ 2 ). Hence, for any α < 2π C, we have sup dv g < +. R Br ( σ, σ + σ 2 )dv g =1 As an consequence, we have the following: B r e α σ 2

6 YUXIANG LI, PAN LIU, YUNYAN YANG Lemma 3.3. There exists a positive number α such that sup σ H1 eα σ 2 dv g < +. Denote α = sup{α : sup σ H1 eα σ 2 dv g < + }. We shall prove that α = 4π. Lemma 3.4. α 4π. Proof. Let Ω p be a coordinate domain. By oser s result [], we can find a sequence {u } H 1,2 0 (B 1 ) such that B 1 0 u 2 dx = B 1 u 2 dv g = 1, and B 1 e (4π+ 1 )u2 dvg > as +. We set σ = u e 1 Since u L 2 0, we get σ, σ dv g 1, m u e 1, ζ i L 2ζ i H 1. i=1 and e (4π+ 1 ) σ 2 dv g. Next, we prove an energy concentration result which is a generation of Lions wor: Lemma 3.5. Let σ H 1. If (3.1) e α σ 2 dv g = + for all α > α, then by passing to a subsequence,σ 0 in H 1,2 (, E), and for some p. σ, σ dv g δ p Proof. Without loss of generality, we assume σ σ 0 wealy in H 1,2 (, E) σ σ 0 strongly in L 2 (, E). We first claim that σ 0 = 0. Otherwise, we have (σ σ 0 ), (σ σ 0 ) dv g 1 σ 0, σ 0 dv g < 1. Hence we can find a α 1 > α such that eα 1 σ 2 dv g is bounded, which contradicts (3.1). Secondly we suppose σ, σ dv g µ δ p, p. Taing η C0 (B δ(p)), η 1 on B δ/2 (p), one can easily see that sup (ησ ), (ησ ) dv g < 1. Hence e ησ 2 is bounded in L α (, E) for some α > α. A covering argument implies that e σ 2 is bounded in L α (, E) for some α > α, which contradicts (3.1).

OSER-TRUDINGER INEQUALITIES OF VECTOR BUNDLES 7 Corollary 3.6. α = 4π Proof. By the definition of α, we can find a sequence σ H 1 such that e ( α+ 1 ) σ 2 dv g = +. Then, applying Lemma 3.5, we get σ, σ dv g δ p, and σ 2 dv g 0. Hence, for any η which is 0 near p, we have ( (ησ ) 2 + ησ 2 )dv g 0. Applying Lemma 3.3, one can find a subsequence (still denoted by σ ) such that e q ησ 2 dv g = V ol(ω) for any Ω \ {p} and q > 0. In coordinate around p, we set σ = u i e i. It is easy to chec that (η u i ) x j 2 dx = 1, Ω B 2r i,j where η is a cut-off function which is 1 on B r. Then, similar to the proof of Corollary 3.2, we can deduce from oser s result [] that q σ 2 dx < + for any q < 4π. Hence α 4π, which together with Lemma 3.4 implies α = 4π. B r e In a similar way, we can prove the following Proposition 3.7 For any α < 4π, and for any α > 4π, sup σ H 2 e α σ 2 dv g < +, sup e α σ 2 dv g = +. σ H 2 4. Blowing up analysis Let α be an increasing sequence which converges to 4π. First of all, we prove the following Lemma 4.1. The functional J α (σ) defined on the space H 1 admits a smooth maximizer σ H 1. oreover, we have (4.1) e α σ 2 dv g = sup e 4π σ 2 dv g. σ H 1

8 YUXIANG LI, PAN LIU, YUNYAN YANG Proof. It is easy to find σ H 1 such that J α (σ ) = sup σ H 1 J α (σ). One can chec that σ satisfies the following Euler-Lagrange equation: (4.2) σ = σ λ e α σ 2 m γ i ζ i, where = is defined as follows: for any τ H 1,2 (, E) σ, τ dv g = σ, τ dv g. It is easy to chec that (4.3) λ = i=1 σ 2 e α σ 2 dv g, and γ i = Let Ω p be an coordinate system around p. We set σ = u i e i, and U = (u 1, u2, un ), σ, ζ i λ e α σ 2 dv g. and e j = Γ l jidx i e l. Then, we get a local version of (4.2) (4.4) 0 u j = T s ji u s x i + Rj i ui + ef p(x +r x) ( uj e α σ 2 λ where T ji s m γ i ζ i, e p ), and R j i are smooth on Ω p. For simplicity, we write (4.4) as follows i=1 0 U = e f p(x +r x) U λ e α U 2 + T ( U ) + R(U ) i γ i U, ζ i. The standard elliptic estimates implie that σ C (). Since for any fixed σ H, e α σ 2 dv g e α σ 2 dv g, (4.1) follows immediately. We assume that (4.5) e α σ 2 dv g = + for all α > 4π (Otherwise, by the wealy compactness of L p, passing to a subsequence, we have e α σ 2 dv g = e α σ 0 2 dv g,

OSER-TRUDINGER INEQUALITIES OF VECTOR BUNDLES 9 where σ 0 is the wea it of σ. Hence, Theorem 1.1 holds). It follows from Lemma 3.5 that σ, σ dv g δ p. Given any Ω \ {p}, we tae a cut-off function η which is 0 at p, and 1 on Ω, then ησ, ησ dv g = η 2 σ, σ + 2η σ dη, σ dv g + σ dη, σ dη dv g 0, hence Ω ( ησ 2 + ησ 2 )dv g 0. By Lemma 3.7, e α σ 2 is bounded in L p (Ω) for any p > 0. Standard elliptic estimates imply that σ 0 in C (Ω). Lemma 4.2. Let 1 β = σ λ e α σ 2 dv g. Then, we have inf β λ = +, inf = +, and sup γ < +. β Proof. For any fixed N > 0, we have Then If e α σ 2 V ol() + e α σ 2 dv g = σ N 1 N 2 λ < +, letting N +, we get e α σ 2 dv g V ol(). However, it follows from (4.1) that Therefore, inf λ = +. In a similar way, we have λ β e α σ 2 dv g = sup σ H 1 e 0 = V ol(). σ 2 e α σ 2 λ dv g V ol() + σ >N N 2. +. Then, we have γ i c 1 + c 2 { σ 2 1} e 4π σ 2 dv g > V ol(). σ 2 λ e α σ 2 dv g C for some constants c 1, c 2 and C depending only on. oreover, we have σ e α σ 2 dv g ( e4πn 2 + 1 λ λ λ N σ 2 e α σ 2 dv g ). Letting +, and then N +, we get 1 β 0. Let c = σ (x ) = max x σ (x). Then it follows from (4.4) and (4.5) that c +. By passing to a subsequence, we can assume x p. Set r 2 = λ c 2 e α c 2. Since σ 2 1 = e α σ 2 dv g c2 e α 2 c 2 e α 2 σ 2 dv g C c2 e α 2 c 2, λ λ λ we get r 2 e α 2 c 2 0, as +.

10 YUXIANG LI, PAN LIU, YUNYAN YANG Let Ω p be a local coordinate system around p, σ = u i e i and U = (u 1,, un ). Denote V (x) = U (x + r x), D = U (x ) and C = (c 1, c2,, cn ) = 1 V dx. B 1 B 1 A direct calculation shows that (4.6) 0 (V C ) = c 2 V e α ( σ 2 c 2 )+f(x +r x) + F. where F L 2 loc 0. Hence 0 (V C ) 0 in L 2 loc (R2, R n ), and B L (V C ) 2 dx 1. Then, by Poincare inequality, one has V C L 2 (B L ) C(L) for any fixed L > 0. By the standard elliptic estimates, V C V in C 0,α loc (R2, R n ) for some V H 1,2 (R 2, R n ). Notice that R 2 0 V 0 V T dx 1, Liouville s theorem gives V 0. Furthermore we obtain (4.7) V D = (V C ) + (C V (0)) 0 in C 0,α loc (R2, R n ). Let Then we have w (x) = 2α D (U (x + r x) D ) T. (4.8) 0 w = 2α e f(x +r x) (1 + D (V D ) T )e α ( σ 2 (x +r x) c 2 ) + F, where F L 2 loc 0. It is easy to see that on B L Hence Then the identity (D U T )(x + r x) D U T (x + r x) c 2. c 2 w (x) w (0) = 0. α ( σ 2 c 2 ) = w + α (V D )(V D ) T, together with Harnac inequality and standard elliptic estimates, gives w w in C 0,α loc (R2 ), where w satisfies 0 w = 8πe w in R 2 w(0) = sup R 2 w = 0 R 2 e w dx 1. By a result of [C-L], we have w(x) = 2 log(1 + π x 2 ) in R 2 and R 2 e w = 1. In the rest of this section, we will discuss the convergence of β σ.

OSER-TRUDINGER INEQUALITIES OF VECTOR BUNDLES 11 Proposition 4.3. We have β σ G wealy in H 1,q (, E) for any 1 < q < 2, and β σ G in C 2 (Ω) for any Ω \ {p}, where G C 2 ( \ {p}) satisfies m (4.9) G = θδ p θ, ζ i ζ i for some θ E p. Proof. Since σ 0 in C ( \ ), and β λ 0, then for any ϕ C (, E), we have β ϕ, σ e α σ 2 dv g 0. λ On Ω p, we set \ θ i = and θ = θ i e i. It is easy to see that i=1 Ω p β u i λ e α σ 2 dv g, ϕ, σ e α σ 2 dv g ϕ, θ. λ oreover, we have β ϕ, σ e α σ 2 dv g = λ for any ϕ C (, E). Then, we get β γ i ϕ, ζ i dv g β γ i ϕ, ζ i dv g ϕ C 0, therefore, sup β γ < +. Let m 2 = β σ 2 dv g. Firstly, we need to prove sup m < +. If m +, then β u m L 1 0. The Proposition 7.1 in the Appendix yields that β σ 0 m in H 1,p for any p < 2. Hence it follows from Poincaré inequality and compact embedding of Sobolev space that β σ m 0 in L 2, which contradicts β σ m 2 dv g = 1. Since sup m < +, applying Proposition 7.1 again, we get β σ converges wealy in H 1,p for any p < 2. Therefore β σ L q C(q) for any q > 0. Assume β σ G wealy in H 1,q (, E). Then by Lemma 4.3 and standard elliptic estimate, we have β σ G in C 2 (Ω), Ω \ {p}. Testing the equation satisfied by β σ with φ C (, E), one has Hence φ, β σ dv g = φ, G dv g = θ, φ (p) φ,β σ λ e α σ 2 dv g m φ, β γ i ζ i dv g i=1 θ, φ(p) m i=1 θ, ζ i(p) φ, ζ i dv g. m θ, ζ i (p) i=1 φ, ζ i dv g,

12 YUXIANG LI, PAN LIU, YUNYAN YANG and whence (4.9) holds. This completes the proof of the lemma. 5. Applications of Pohozaev identity In this section, we will calculate eα σ 2 dv g. The ey gradient of the proof is to use the Pohozaev identity. Let Ω p. Testing equation (4.2) with η(r)r σ, we have η(r)r σ, σ dv g = η(r)r σ, σ eα σ 2 λ dx+ γ i η(r)r σ, ζ i dv g, where η is a cut-off function which is 0 outside B δ and 1 on B δ, r = (x 1 ) 2 + (x 2 ) 2. 2 η(r)r σ, σ 0 = j η(r)x p p σ, j σ 0 j = η(r) σ, σ 0 + η (r) xi x j r i σ, j σ 0 + η(r) x i j i σ, j σ 0 = η(r) σ, σ 0 + η (r) xi x j r i σ, j σ 0 + η(r) x i i j σ, j σ 0 +η(r) i,j x i R ij σ, j σ 0 = η(r) σ, σ 0 + η (r) xi x j r i σ, j σ 0 + 1 2 η(r)r σ, σ 0 +η(r) i,j x i R ij σ, j σ 0 + η(r)s 0 ( σ, σ ), where S 0 = e f S, and S(σ 1, σ 2 ) = 1 2 (r σ 1, σ 2 r σ 1, σ 2 σ 1, r σ 2 ). Since we do not assume h = 0, S 0 generally. We set 1 r δ η(r) = r δ a a δ r δ + a 0 r δ + a. Letting a 0, we get r σ, σ 0 dx = ( σ, σ 0 + 1 Ω p 2 r σ, σ 0 + x i R ij σ, j σ 0 )dx i,j x i x j r iσ, j σ 0 ds + S( σ, σ )dv g. Clearly, 1 2 r σ, σ 0 dx = = 1 2 δ πr 2 0 ( σ, σ 0 dθ)dr S 1 r σ, σ j 0 ds σ, σ 0 dx.

Then r σ, σ 0 dx = OSER-TRUDINGER INEQUALITIES OF VECTOR BUNDLES 13 + 1 2 On the other hand, we have e f e α σ 2 r σ, σ 0 dx = B δ λ and e f e α σ 2 r σ, σ 0 dx = B δ λ Therefore, we get (1 + O(r)) eα σ 2 dv g = α λ Ω p = + + i,j x i R ij σ, j σ 0 dx r σ, σ j 0 ds + e f r B δ σ e α σ 2, σ 0 dx + λ δ + 2πr 2 ( e α σ 2 0 S 1 r eα σ 2 ds g α λ e α σ 2 r f B δ α λ dv g. r eα σ 2 α λ x i x j x i x j r iσ, j σ 0 ds S( σ, σ )dv g. e f dθ)dr + 2α λ ds g S(σ, σ) e α σ 2 dv g B δ λ e α σ 2 dv g α λ e α σ 2 r f B δ α λ dv g. x i R ij σ, j σ 0 dx i,j r iσ, j σ 0 ds 1 2 γ i η(r)r σ, ζ i dv g B δ B δ (S( σ, σ ) S(σ, σ ) λ e α σ 2 )dv g. Let +, we have γ i η(r)r β σ, ζ i dv g C β σ H 1,q C G H 1,q (B δ ) B B δ δ and r σ, σ 0 ds ( x i R ij β σ, j β σ 0 +S)dx C β σ L q (B δ ) β σ L q (B δ ) C G L q (B δ ) G L q (B δ ), i,j where 1 < q < 2. By Lemma 7.2 in the Appendix, we get x i x j r iβ σ, j β σ 0 ds 0 + 1 r β σ, β σ j 0 ds 2 x i x j r ig, j G 0 ds 1 r G, G 0 ds 2 = 1 4π + O(δγ ). Since σ 0 in B δ, we get e α σ 2 α B δ 4π.

14 YUXIANG LI, PAN LIU, YUNYAN YANG Finally, we set S(σ 1, σ 2 ) = Aσ 1, σ 2, where A C 0 = O(r). Tae a cut-off function η which is 0 outsider B 2δ and 1 in B δ with η < 1 δ. We have A ησ, σ = Aησ, σ dv g ( A)σ, σ dv g B 2δ (p) B 2δ (p) B 2δ (p) Aησ, σ = e α σ 2 dv g ( A)σ, σ dv g B 2δ (p) λ B 2δ (p) Hence, we get So, we get β 2 (S( σ, σ ) S(σ, σ ) eα σ 2 )dv g B δ λ O(1) ( A)ηG, G dv g + O(δ) B 2δ (p) where δ 0 h(δ) = 0. Clearly, (5.1) The following lemma is very important: Lemma 5.1. we have c β 1. Proof. Since η G 2 dv g. B 2δ (p)\ (1 + O(r))e α σ 2 dv g = B δ 4π + λ β 2 + h(δ), \ eα σ 2 dv g = \, we get c = β e α σ 2 dv g = V ol() + λ β 2 c σ e α σ 2 σ 2 e α σ 2 dv g, λ λ we have c β. Assume our statement is not true, then we can find A > 1 such that c A > Aβ for sufficiently large. Denote σ τ = = τ i e i. c Without loss of generality, we assume τ 1 > 0. By the equation (4.4), we have for any fixed L > 0, 0 η(u i u1 (x ) A )+ 2 dx = = = Ω p 0 u 1 0η(u 1 u1 (x ) A )+ dx + o (1) η(u 1 u1 (x ) Ω p A )+ u1 e α σ 2 dv g + o (1) λ (u 1 u1 (x ) A )u1 e α σ 2 dv g + o (1) B Lr λ (x ) ( u 1 ) (x )(A 1) + o (1) u 1 e ϕ (x)+o (1) dx + o (1). A B L (0) Letting +, then L +, we get inf η(u 1 u1 (x ) A )+ 2 dv g (1 1 A )τ 1 2.. c 2

Let σ A = ηua,i OSER-TRUDINGER INEQUALITIES OF VECTOR BUNDLES 15 e i = η(min{u 1 (x), u (x ) }e 1 + n u i e i), where η is a cut-off function which is 0 \ B 2δ (p), and 1 in. Since we get σ A 2 = η i=1 sup A i=2 0 u A,i 2 + O( σ A σa ) + O( σa 2 ), ( σ A 2 + σ A 2 )dv g 1 A 1 A τ 12 < 1. Then, by Proposition 3.7, e α σ A is bounded in L q for some q > 1. Since β σ Ω p {u 1 τ1 c e α σ 2 dv g β σ e α σa 2 dv g β σ A } λ λ Ω p λ L q () eα σ A 2 L q () 0. In the same way, for any τ i, we have Since τ = 1, we obtain 1 A β A = A. β { σ Aβ } This contradict to the choice of A. β σ Ω p { u i τi c e α σ 2 dv g 0 A } λ σ 2 e α σ 2 dv g λ σ e α σ 2 dv g λ σ e α σ 2 dv g A λ {x Ω p : u i Aτ i β, i} i { u i τi c β σ e α σ 2 dv g A } λ Corollary 5.2. We have τ = θ, and (5.2) (e α σ 2 1)dV g = L + e α σ 2 dv g. B Lr (x ) Proof. By a straightforward calculation, we have e α σ 2 dv g = λ B Lr (x ) c 2 e w dx(1 + o(1)). B L The above inequality together with (5.1) implies (5.2). It is not difficult to chec that β σ e α σ 2 dv g = 1, L + λ and L + B Lr (x ) B Lr (x ) β u i λ e α σ 2 dv g = τ i.

16 YUXIANG LI, PAN LIU, YUNYAN YANG Hence θ i τ i = L + 1 L + Ω p\b Lr (x ) B Lr (x ) β u i e α σ 2 dv g λ β σ λ e α σ 2 dv g = 0. 6. The proof of theorem 1.1 On Ω p, we set Ũ = ( ũ 1,, ũn ). First, we have the following lemma: Lemma 6.1. We have 0 Ũ 2 dx where ρ(δ) is a continuous function of δ with ρ(0) = 0. σ, σ dv g + ρ(δ), c 2 Proof. It is well-nown that 0 Ũ 2 dx = We set 0 U 2 dx. σ = ( up x 1 + up x 2 )e p + A p i ui e p, where A p s are smooth functions. Hence, we get 0 U 2 = 0 U 2 2 A(σ ), σ + AU 2. Since c 2 A(σ ), σ dx C( A(c σ ) q dx) 1 q ( c σ p dx) 1 p C( c σ q dx) 1 q ( c σ p dx) 1 p we get C( G q dx) 1 q ( G p dx) 1 p, A(σ ), σ dv g ρ 1(δ). Similarly we have A(σ ) 2 ρ 2(δ) c 2. The lemma follows immediately from the above two inequalities. c 2 On Ω p, we can write G as follows : G = log r 2π τ + s p + θ.

OSER-TRUDINGER INEQUALITIES OF VECTOR BUNDLES 17 where s p is a constant section, and θ is continuous local section of E with θ(0) = 0. Then Theorem 1.1 follows from the following proposition: Proposition 6.2 If (4.5) holds, then we have e 4π σ 2 dv g V ol() + πe 1+4π τ,sp. sup σ H 1 Proof. Set Y = (Ũ L ) +, where Then Y H 1,2 0 (B δ, R n ), and 0 Y 2 dx 1 L = (l 1,, l n ) = sup Ũ = O( 1 ). c \ G 2 dv g + o δ (1) 1 c 2. By Lemma 6.1 and Proposition 7.2 in Appendix, we have Y 2 dx 1 1 2π log δ + τ, s p + ɛ(, δ), where δ +0 ϑ = ɛ(, δ) = 0. Let 1 1 1 2π log δ+ τ,s p +ɛ(,δ) c 2 c 2 = 1 + 1 2π log δ + τ, s p + ɛ(, δ) c 2 By Theorem A and the method we have used to prove Corollary 3.2, we have sup e α ϑ Y 2 dx δ 2 (πe + π). If ũ i < l, we have on B Lr (x ) τ i = 0, Hence, we have y i = ũi l + o δ(1) c on B Lr (x ), It is easy to chec that and on B Lr (x ), Recall that τ = 1, we get ũ i = o δ(1) c, l = o δ(1) c. + O( log2 δ c 4 ). for sufficiently large. A straightforward calculation shows ϑ Y 2 = Ũ 2 2ŨL + Ũ 2 ( 1 2π log δ + τ, s p ) + ɛ(δ, ). c L i i = τ 2π log δ + sign(τ i )s i p + o δ (1), Ũ c c 2 = ( τ 1,, τ n ) + o (1) c. Ũ L = 1 2π log δ + τ, s p + ɛ (δ, ),

18 YUXIANG LI, PAN LIU, YUNYAN YANG where δ 0 ɛ (δ, ) = 0. Since Ũ 2 /c 2 1 as +, we have Ũ 2 ϑ Y 2 1 2π log δ + τ, s p + ɛ (δ, ). Letting +, then L +, and δ 0, we get (6.1) e α σ 2 dv g πe 1+4π τ,sp L + B Lr (x ) Then by Corollary 5.2, we obtain Clearly, Hence L + L + L + B Lr (x ) e α σ 2 dv g = B ρ, ρ > 0. B ρ(p)\b Lr (x ) Y 2 Ũ 2 = σ 2, and α ϑ Y 2 α U 2 2 log δ + C. e α ϑ Y 2 dx O(δ 2 ) B ρ (p)\b Lr (x ) It is easy to see that for any fixed ρ > 0 e α ϑ Y 2 dv g B δ \ B ρ. \B ρ (p) Letting ρ 0, we get L + \B Lr (x ) e α ϑ Y 2 dx = πδ 2, L + Then the Proposition follows from (6.1) and Corollary 5.2. 7. Appendix e α ϑ Y 2 dx < +. e α σ 2 dv g = O(δ 2 ) B ρ (p). B ρ (p) e α ϑ Y 2 dx δ 2 πe. B Lr (x ) We will prove two propositions which have been used in section 5 and section 6. Since the proof is routine, we prove them in this appendix. The following result is an extension of Theorem 2.2 in [S2]: Proposition 7.1. Let σ be a smooth section of E satisfies the following equation σ = f. If σ L 2 γ and f L 1 1, then for any q < 2, there is a constant C which depends only on q, γ and, s.t. σ H 1,q C Proof. For any p, we tae a cut-off function which is 1 in B r and 0 outside B 2r. Write σ = u i e i. Given t > 0, we set u i,t = min{ηu i, t} and σ t = u i,t e i. Then f, η 2 σ t dv g = η 2 σ t, σ dv g. We get ηu i,t 2 dx C 1 t + C 2 C t when t is sufficiently large. B r (p) i

OSER-TRUDINGER INEQUALITIES OF VECTOR BUNDLES 19 Let u i be the rearrangement of ηu i, and µ R 2(B ρ ) = µ R 2{ηu i,t t}. Then we have { } inf v 2 dx : v H 1,2 0 (B r) and v Bρ = t Ct. B r On the other hand, the inf is achieved by t log x r / log r ρ. By a direct computation, we have 2πt log r ρ C, µ ({x B r : ηu i t}) Cµ R n({x B r : ηu i t}) = Cµ R n(b ρ ) C(r, p)e A(r,p)t. Hence, we can find a constant A, s.t. In the same way, we get By the compactness of, we get {ηu i t} B r e At. {ηu i t} B r e A t. { σ t} e δ 0t for some δ 0 and sufficiently large t. Then, for any δ < δ 0, e δ σ dv g µ({m u m + 1})e δ(m+1) m=0 Locally, we denote U = (u 1,, u n ), then 0 u i = f i + T i 1( U) + T i 2(U), e (δ0 δ)m e δ C. where T 1 and T 2 are smooth and linear at each point. Testing the above equation with the function η 2 log 1+2ui+, where η is 1 on B 1+u i+ r and 0 outside B 2r. We obtain the following n 0 ηu i+ 2 dx log 2 + (T (1 + 2u i+ )(1 + u i+ 1 i ) ( ηu + ) + T2 i (U + ))dv g, B r B r i=1 where T 1 and T 2 are smooth and linear at each point. Since eδ σ dv g < C(δ), we have n u i+ 2 (1 + 2u i+ )(1 + u i+ ) C. B r i=1 Given q < 2, by Young s inequality, we have ( ) u i+ q 0 u i+ 2 dv g B r B r (1 + u i+ )(1 + 2u i+ ) + ((1 + ui+ )(1 + 2u i+ q )) 2 q dx ( ) 0 u i+ 2 (1 + u i+ )(1 + 2u i+ ) + i+ Ceδu dx. Similarly, we have and consequently, B r B r u i q dv g < C, σ H 1,q < C(q). m=0

20 YUXIANG LI, PAN LIU, YUNYAN YANG On Ω p, we can set we have the following Lemma: G = log r 2π τ + s p + θ, Lemma 7.2. There are constants γ (0, 1) and A > 0, s.t. θ (x) Ar γ 1, when x is near 0. Therefore, we have G 2 dv g = 1 2π log δ + τ, s p + o δ (1). \ Proof. Let θ = v i e i, and V = (v 1,, v n ), we have the equation of V 0 V = Q 1 V + 1 r Q Q 3 2 + F, where Q 1 Q 2 and Q 3 are smooth matrix, F is a smooth vector-valued function. Hence V W 2,p loc any p < 2. Clearly, we can write 1 Q Q x i Q = Q 1 x i where Λ are constant vector, and F 1 = o(1). Hence where F 2 L loc. Then for any q > 0. Hence, we get 0 x m V = Λ i x i x m r 2 r = Λ i 0 V = Λ i x i r 2 + F 2, x i r + F 1, + x m F 2 + V x m L q loc for x m V C 1 (B r ) V C 0 (B 2r )2r + C(1 + V L 2q)r 1 q for any q > 1. Notice that V (0) = 0, and then V = O(r γ ) for some γ > 0, we get x m V x i (x) Cr γ for some γ > 0, any i, m = 1, 2 and x 0. In the end, we get V (x) x γ 1. References [A] T. Aubin: Some Nonlinear Problems in Riemann Geometry. Spinger, 1998. [A-D] Adimurthi and O.Druet, Blow up analysis in dimension 2 and a sharp form of Trudinger oser inequality, Comm. PDE 29 No 1-2, (2004), 295-322. [A-S] Adimurthi and. Struwe: Global Compactness properties of semilinear elliptic equations with critical exponential growth. J. Funct. Anal., 175(1): (2000) 125 167. [C-C] L. Carleson and A. Chang: on the existence of an extremal function for an inequality of J.oser. Bull. Sc. ath., 110 (1986) 113-127.

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