Analysis in weighted spaces : preliminary version Frank Pacard To cite this version: Frank Pacard. Analysis in weighted spaces : preliminary version. 3rd cycle. Téhéran (Iran, 2006, pp.75. <cel-00392164> HAL Id: cel-00392164 https://cel.archives-ouvertes.fr/cel-00392164 Submitted on 5 Jun 2009 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.
Analysis in weighted spaces : preliminary version May 29, 2006
2
Contents 1 Weighted L 2 analysis on a punctured ball 5 1.1 A simple model problem................................. 5 1.2 Analysis in weighted spaces in the punctured unit ball................ 9 1.3 The spectrum of the Laplacian on the unit sphere................... 10 1.4 Indicial roots....................................... 11 1.5 A crucial a priori estimate................................ 11 2 Weighted L 2 analysis on a punctured manifold 21 2.1 The Laplace-Beltrami operator in normal geodesic coordinates........... 21 2.2 Two global results.................................... 22 2.3 The kernel of the operator A.............................. 26 2.4 The range of the operator A.............................. 27 2.5 Fredholm properties for A............................... 29 2.6 The deficiency space................................... 30 2.6.1 The kernel of A revisited :........................... 37 2.6.2 The deficiency space :.............................. 37 3 Weighted C 2,α analysis on a punctured manifold 39 3.1 From weighted Lebesgue spaces to weighted Hölder spaces.............. 39 3.2 An example........................................ 48 4 Analysis on ALE spaces 51 4.1 Asymptotically Locally Euclidean spaces........................ 51 4.2 An example from conformal geometry......................... 53 5 Mean curvature of hypersurfaces 57 5.1 The mean curvature................................... 57 5.2 Jacobi operator and Jacobi fields............................ 59 3
4 CONTENTS 6 Minimal hypersurfaces with catenoidal ends 63 6.1 The n-catenoid...................................... 63 6.2 Unmarked space of minimal hypersurfaces with catenoidal ends........... 65 6.3 The marked space of minimal hypersurfaces with catenoidal ends.......... 67 7 Analysis on manifolds with cylindrical ends 69 7.1 Manifolds with cylindrical ends............................. 69 7.2 Manifolds with periodic-cylindrical ends........................ 71
Chapter 1 Weighted L 2 analysis on a punctured ball 1.1 A simple model problem Let B R (resp. BR denote the open (closed ball of radius R > 0 in R n and B R = B R {0} (resp. B R denote the corresponding punctured ball. Given ν R and a function satisfying f : B 1 R n R x ν f L (B 1 1 we would like to study the solvability of the equation { x 2 u = f in B1 (1.1 u = 0 on B 1 A solution of this equation is understood in the sense of distributions, namely u is a solution of (1.1 if u L 1 (B 1 B R, for all R (0, 1 and if u v dx = f v x 2 dx B 1 B 1 for all C functions v with compact support in B 1. We claim that : 5
6 CHAPTER 1. WEIGHTED L 2 ANALYSIS ON A PUNCTURED BALL Proposition 1.1.1. Assume that n 3 and ν (2 n, 0. Then there exists a constant c = c(n, ν > 0 and for all f L loc (B 1 there exists u a solution of (1.1 which satisfies x ν u L (B 1 1 The proof of this result is a simple consequence of the maximum principle. First, recall the expression of the Euclidean Laplacian in polar coordinates Using this expression we get at once = 2 r + n 1 r r + 1 r 2 S n 1 x 2 x ν = ν (2 n ν x ν away from the origin. Now, if ν (2 n, 0 (this is where we use the fact that n 3!, we observe that the constant c n,ν := γ (2 n ν > 0 The existence of a solution of (1.1 can then be obtained arguing as follows : Given R (0, 1/2, we first solve the problem { x 2 u R = f in B 1 B R u R = 0 on B 1 B R (1.2 Since f L (B 1 B R, the existence of a solution u R W 2,p (B 1 B R for any p (1, follows from the following classical result : Proposition 1.1.2 ([?], Theorem 9.15. Given p (1, and Ω a smooth bounded domain of R n, if g L p (Ω then there exists a unique solution of { v = g in Ω which belongs to W 2,p (Ω W 1,p 0 (Ω. In our case Ω = B 1 B R and for all p (1,, and hence v = 0 on Ω f L (B 1 B R L p (B 1 B R, u R W 2,p (B 1 B R W 1,p 0 (B 1 B R. One can use the Sobolev Imbedding Theorem to show that u R C 1,α ( B 1 B R for all α (0, 1.
1.1. A SIMPLE MODEL PROBLEM 7 Proposition 1.1.3 ([?], Theorem 7.26. If α = 1 n p, then provided Ω is a smooth bounded domain of R n. The maximum principle also implies that W 2,p (Ω C 1,α ( Ω for all x B 1 B R. Indeed, observe that the function is positive on B 1 B R. Moreover u R (x 1 c n,ν x ν (1.3 w(x = 1 c n,ν x ν u R (x w 0 in B 1 B R. Therefore one can apply the maximum principle Proposition 1.1.4 ([?], Theorem 8.1. Assume that v W 1,2 (Ω satisfies v 0 in some smooth bounded domain Ω R n. Then inf v inf (min(v, 0 Ω Ω This result applies to the function w in B 1 B R. We conclude that w 0 and hence u R 1 c n,ν x ν. Applying the same reasoning to u R we obtain the desired inequality. Observe that, in the case where u R is C 2, one can simply invoke the classical maximum principle ([?], Theorem 3.1. Now, we would like to pass to the limit, as R tends to 0. To this aim, we use the following estimates for solutions of (1.2 Proposition 1.1.5 ([?], Theorem 9.13. Given a smooth bounded domain Ω R n whose boundary has two disjoint components T 1 and T 2, Ω Ω T 1 and p (1,. There exists a constant c = c(n, p, Ω, Ω > 0 such that, if g L p (Ω and v W 2,p (Ω, satisfy { v = g in Ω v = 0 on T 1 then v W 2,p (Ω c ( v L p (Ω + g L p (Ω
8 CHAPTER 1. WEIGHTED L 2 ANALYSIS ON A PUNCTURED BALL Using this result with Ω = B 1 B R, T 1 = B 1 and Ω = B 1 B 2R, together with the a priori bound (1.3, we conclude that, for all R (0, 1/2 there exists a constant c = c(n, ν, R > 0 such that u R W 2,p (B 1 B 2R c for all R (0, R. It is now enough to apply the Sobolev Imbedding Theorem Proposition 1.1.6 ([?], Theorem 7.26. The imbedding is compact provided 0 < α < 1 n p W 1,p (Ω C 0,α ( Ω and Ω is a smooth bounded domain. It is now easy to use these two results together with a standard diagonal argument to show that there exists a sequence (R i i tending to 0 such that the sequence of functions u Ri converges to some continuous function u on compacts of B 1. Obviously u will be a solution of (1.1 and, passing to the limit in (1.3, will satisfy c n,ν x ν u L (B 1 1 (1.4 We have thus obtained a solution of (1.1 satisfying (1.4, provided ν (2 n, 0. completes the proof of Proposition 1.1.1. This Exercise 1.1.1. Given points x 1,..., x m R n, weights parameters µ, ν 1,..., ν m R, we define two positive smooth functions such that : g : R n {x 1,..., x n } R and h : R n {x 1,..., x n } R (i For each i = 1,..., m, g(x = x x i νi point x i. and h(x = x x i νi 2 in a neighborhood of the (ii g(x = x µ and h(x = x µ away from a compact subset of R n. Show that, provided n 3 and µ, ν 1,..., ν m (2 n, 0, given a function f : R n {x 1,..., x n } R satisfying f h it is possible to find a solution of the equation γ 2 u = f which satisfies u g
1.2. ANALYSIS IN WEIGHTED SPACES IN THE PUNCTURED UNIT BALL 9 1.2 Analysis in weighted spaces in the punctured unit ball Given R we define the space L 2 (B 1 := x +1 L 2 (B 1 This space is endowed with the norm ( u L 2 (B1 := u 2 x 2 2 dx B 1 It is easy to check that 1/2 Lemma 1.2.1. The space (L 2 (B 1, L 2 (B 1 is a Banach space. Exercise 1.2.1. Provide a proof of Lemma 1.2.1. We define the unbounded operator A by A : L 2 (B 1 L 2 (B 1 u x 2 u The domain of this operator is the set of functions u L 2 (B 1 such that A u = f L 2 (B 1 in the sense of distributions : This means that u W 2,2 (B 1 B R, for all R (0, 1/2 and u v dx = f v x 2 dx B 1 B 1 for all C functions v with compact support in B 1. We start with some properties of A which are inherited from the corresponding classical properties for elliptic operators. Proposition 1.2.1. Assume that R is fixed. There exists a constant c = c(n, > 0 such that for all u, f L 2 (B 1 satisfying x 2 u = f in B 1 we have u L 2 1 (B 1/2 + 2 u L 2 2 (B 1/2 c ( f L 2 (B 1 + u L 2 (B 1 The proof of this result follows from the : Proposition 1.2.2 ([?], Theorem 9.11. Given a smooth bounded domain Ω R n, Ω Ω and p (1,. There exists a constant c = c(n, p, Ω, Ω > 0 such that, if g L p (Ω and v W 2,p (Ω, satisfy v = g in Ω then v W 2,p (Ω c ( v L p (Ω + g L p (Ω
10 CHAPTER 1. WEIGHTED L 2 ANALYSIS ON A PUNCTURED BALL The proof of the Proposition 1.2.1 goes as follows : Given R (0, 1/2 we define the functions Obviously, we have v(x := u(r x and g(x := f(r x x 2 v = g in B 2 B 1/2. We can then apply the result of Proposition 1.2.2 with Ω = B 2 B 1/2 and Ω = B 3/2 B 1, we conclude that v 2 W 2,2 (B 3/2 B 1 c ( v 2 L 2 (B 2 B 1/2 + g 2 L 2 (B 2 B 1/2 Performing the change of variables backward, we conclude that R 2 n u 2 L 2 (B 3R/2 B R + R4 n 2 u 2 L 2 (B 3R/2 B R ( c R n u 2 L 2 (B 2R B R/2 + R n f 2 L 2 (B 2R B R/2 It remains to multiply this inequality by R n 2 2, choose R = 1 3 ( 2 3 i, for i N and sum the result over i. We obtain u 2 L 2 1 (B 1/2 + 2 u 2 L 2 2 (B 1/2 c ( u 2 L 2 (B 1 + f 2 L 2 (B 1 This completes the proof of the result. 1.3 The spectrum of the Laplacian on the unit sphere We recall some well known facts about the spectrum of the Laplacian on the unit sphere. Proposition 1.3.1 ([?], Theorem??. The eigenvalues of S n 1 are given by λ j = j (n 2 + j where j N. The corresponding eigenspace will be denoted by E j and the corresponding eigenfunctions are the restrictions to S n 1 of the homogeneous harmonic polynomials on R n. One easy computation is the following : If P is a homogeneous harmonic polynomial of degree j, then P (x = x j P (x/ x and hence r r P = j P Using the expression of the Laplacian in polar coordinates, we find that r 2 P = j (n 2 + j P + S n 1P
1.4. INDICIAL ROOTS 11 Since P is assumed to be harmonic, when restricted to the unit sphere this equality leads to S n 1P = j (n 2 + j P This at least shows that the restrictions to S n 1 of the homogeneous harmonic polynomials of degree j on R n belong to E j. Exercise 1.3.1. What is the dimension of the j-th eigenspace E j? 1.4 Indicial roots We set j := n 2 2 Definition 1.4.1. The indicial roots of at the origin are the real numbers given by + j for j N. ν ± j := 2 n 2 ± j The indicial roots are related to the asymptotic behavior of the solutions of the homogeneous problem u = 0 in R n {0}. Indeed, a simple computation shows that if φ E j. ( x ν± j φ = 0 1.5 A crucial a priori estimate We now want to prove the key result which explains the importance of the parameters j in the study of the operator x 2 when defined between weighted L 2 -spaces. This is the purpose of the : Proposition 1.5.1. Assume that ± j for j N. Then there exists a constant c = c(n, > 0 such that, for all u, f L 2 (B 1 satisfying x 2 u = f in B 1, we have u L 2 (B 1 c ( f L 2 (B1 + u L2 (B 1 B 1/2
12 CHAPTER 1. WEIGHTED L 2 ANALYSIS ON A PUNCTURED BALL Observe that this result states that we can control the weighted L 2 -norm of u in terms of the weighted L 2 -norm of f and some information about the function u away from the origin. To prove the result let us perform the eigenfunction decomposition of both u and f. We write x = r θ where r = x and θ = x/ x S n 1 and we decompose u(r, θ = j 0 u j (r, θ and f(r, θ = j f j (r, θ where, for each j 0, the functions u j (r, and f j (r, belong to E j. In particular S n 1u j = λ j u j and S n 1f j = λ j f j, wherever this makes sense. Observe that B 1 u 2 x 2 2 dx = j 0 B 1 u j 2 x 2 2 dx = j 0 1 0 u j 2 L 2 (S n 1 rn 3 2 dr and f 2 x 2 2 dx = f j 2 x 2 2 dx = B 1 j 0 B 1 j 0 1 0 f j 2 L 2 (S n 1 rn 3 2 dr where L2 (S n 1 is the L 2 (S n 1 norm. In addition, the functions u j and f j satisfy in the sense of distributions in B1. Indeed, making use of u v dx = f v x 2 dx B 1 B 1 x 2 u j = f j (1.5 with test functions of the form v(r, θ = h(r φ(θ where φ E j and h is a smooth function with compact support in (0, 1, we find that u j is a E j -valued function solution of (1.5. Using the decomposition of the Laplacian in polar coordinates, we also find that in the sense of distribution. Moreover 1 0 u j 2 L 2 (S n 1 rn 3 2 dr < r 2 2 r u j + (n 1 r r u j λ j u j = f j (1.6 and 1 0 f j 2 L 2 (S n 1 rn 3 2 dr < The Sobolev Imbedding Theorem will help us justifying most of the forthcoming computation : Proposition 1.5.2 ([?], Theorem 7.26. If α = 1 n p then provided Ω is a smooth bounded domain. W 2,p (Ω C 1,α ( Ω
1.5. A CRUCIAL A PRIORI ESTIMATE 13 Observe that u j W 2,2 ((r, 1 for all r (0, 1 and hence we find that u j C 1,1/2 loc ((0, 1]. Also, using the result of Proposition 1.2.1, we conclude that r u j x 2 dx < and r 2 u j x 2 2 dx < (1.7 B 1 B 1 Let j 0 denote the least index in N such that The proof of Proposition 1.5.1 is now decomposed into two parts. < j0 (1.8 Part 1 : The case where < j. Let χ be a cutoff function equal to 1 in B 1/2 and equal to 0 outside B 1, let us further assume that χ is radial. We multiply the equation (1.6 by χ 2 r 2 2 u j and integrate over B 1. We obtain using polar coordinates χ 2 r 2 u j r (r n 1 r u j dr dθ λ j χ 2 u 2 j r n 3 2 dr dθ = χ 2 u j f j r n 3 2 dr dθ B 1 B 1 B 1 where dθ denotes the volume form on S n 1 and hence the Euclidean volume form is given by dx = r n 1 dr dθ. We integrate the first integral by parts to get χ 2 r (χ u j 2 r 2 dx + (λ j + (n 2 2 χ 2 u 2 j r 2 2 dx B 1 B 1 = ( r (χ 2 r r χ 2 r 1 2 u 2 j dx χ 2 u j f j r 2 2 dx B 1 B 1 Even is formally, this computation is correct, some care is needed to justify the integration by parts at 0. Let us explain how the integration by parts is performed : We write χ 2 r 2 u j r (r n 1 r u j = r n 1 2 r (χ u j 2 + (n 2 2 χ 2 r n 3 2 u 2 j + ( r (χ 2 r χ 2 r n 2 2 u 2 j ( r χ 2 r n 1 2 u j r u j + χ 2 r n 2 2 u 2 j For all R (0, 1, we integrate this equality over [R, 1] S n 1 with respect to the measure dr dθ to get χ 2 r 2+1 n u j r (r n 1 r u j dx = r 2 r (χ u j 2 dx B 1 B R B 1 B R + (n 2 2 χ 2 u 2 B 1 B j r 2 2 dx R + ( r (χ 2 r χ 2 r 1 2 u 2 B 1 B j dx R + χ 2 r 2 ( r u j r u j + r 1 u 2 j r n 1 dθ BR (1.9 (1.10
14 CHAPTER 1. WEIGHTED L 2 ANALYSIS ON A PUNCTURED BALL Now, use the fact that, thanks to (1.7 ( 1 0 B r r 2 ( u j r u j + r 1 u 2 j r n 1 dθ r 1 dr < to show that, for a sequence of R i tending to 0 we have lim χ 2 r 2 ( u j r u j + r 1 u 2 j r n 1 dθ = 0 R i 0 BRi We now use this sequence of radii and pass to the limit in (1.10 to get χ 2 r 2+1 n u j r (r n 1 r u j dx = r 2 r (χ u j 2 dx B 1 B 1 + (n 2 2 χ 2 u 2 j r 2 2 dx B 1 + ( r (χ 2 r χ 2 r 1 2 u 2 j dx B 1 All subsequent integrations by parts can be justified using similar arguments, we shall leave the details to the reader. We shall now make use of the following Hardy type inequality Lemma 1.5.1. The following inequality holds (n 2 2 R 2 r 2 2 u 2 dx 4 r 2 r u 2 dx n R n provided the integral on the left hand side is finite. Using this Lemma together with (1.9 we conclude that ( 2 j 2 χ 2 u 2 j x 2 2 dx χ 2 f j u j x 2 2 dx + c u 2 j dx B1 B 1 B 1 B 1/2 where the constant c = c(n, > 0 does not depend on j. Now, we set η := 2 j 0 2 > 0 This is where it is important that ± j. Using Cauchy-Schwarz inequality together with the inequality 2 a b η a 2 + η 1 b 2 we get ( 2( 2 j 2 η χ 2 u 2 j x 2 2 dx η 1 χ B1 B 2 fj 2 x 2 2 dx + 2 c u 2 j dx. (1.11 1 B 1 B 1/2
1.5. A CRUCIAL A PRIORI ESTIMATE 15 Observe that, for j j 0, 2(j 2 2 η 2(j 2 0 2 η = η. We can sum all the inequalities (1.11 over j j 0 to conclude that η χ 2 ũ 2 x 2 2 dx η 1 χ 2 f 2 x 2 2 dx + 2 c ũ 2 dx. (1.12 B 1 B 1 B 1 B 12 where we have set ũ := j j 0 u j and f := j j 0 f j Proof of Lemma 1.5.1: We now provide a proof of the Hardy type inequality we have used. Assume that n 2 2 and also that r u 2 x 2 dx < R n since otherwise there is nothing to prove. Then, start with the identity (n 2 2 0 v 2 r n 3 2 dr = 0 v 2 r (r n 2 2 dr = 2 0 v r v r n 2 2 dr where the last equality follows from an integration by parts. Use Cauchy-Schwarz inequality to conclude that (n 2 2 2 v 2 r n 3 2 dr 4 r v 2 r n 2 2 dr 0 The inequality in Lemma 1.5.1 follows from the integration of this inequality over S n 1. Observe that, in order to justify the integration by parts, it is enough to assume that v 2 x 2 2 dx 0 converges. Part 2 : The case where > j and j 0. It remains to estimate u j, for j = 0,..., j 0 1. Here we simply use the fact that we have an explicit expression for u j in terms of f j. In order to simplify the discussion, we first assume that j 0. Then, we define ũ j by ũ j (r, = 1 2 j ( r 2 n 2 +j r t n 4 2 j f j (t, dt r 2 n 0 2 j r t n 4 2 +j f j (t, dt where has to be chosen according to the position of with respect to ± j. In fact (see below we will choose = 0 when > j and = 1 when < j. It is easy to check that in B 1. x 2 ũ j = f j
16 CHAPTER 1. WEIGHTED L 2 ANALYSIS ON A PUNCTURED BALL Basic strategy : Consider a quantity of the form u(r = r 2 n 2 ±j r t n 4 2 j f(t dt where we assume that 1 f 2 (r r n 3 2 dr < It is a simple exercise to compute, using an integration by parts that 1 ( u 2 (r r n 3 2 1 dr = u(1 2 R n 2 2 u(r 2 2 2(± j R This is where, once again, it is important that ± j. A simple application of Cauchy-Schwarz inequality, yields ( 1 u(r 2 r2+2 n f 2 (t t n 3 2 dt, 2 ± j 0 0 1 R f(r u(r r n 3 2 dr (1.13 provided we choose = 0 when > j and = 1 when < j. This is where the choice of is crucial. Plugging this information in (1.13 and using Cauchy-Schwarz inequality, immediately implies that 1 ( u 2 r n 3 2 1 1 dr R 2 ± j 2 f 2 (r r n 3 2 dr 0 ( 1 1 1/2 ( 1 1/2 + u 2 (r r n 3 2 dr f 2 (r r n 3 2 dr ± j It is a simple exercise to check that this implies that 1 for some constant c = c(, n, j > 0. R u 2 r n 3 2 dr c 0 1 0 0 f 2 (r r n 3 2 dr Using this result, and passing to the limit as R tends to 0, we conclude that ũ 2 j x 2 2 dx B 1 c fj 2 x 2 2 dx B 1 (1.14 for some constant c = c(, n, j > 0. It remains to evaluate the difference between the the functions u j and ũ j. Since x 2 (u j ũ j = 0
1.5. A CRUCIAL A PRIORI ESTIMATE 17 we find that u j ũ j = r 2 n 2 +j φ + r 2 n 2 j ψ where φ, ψ E j. Remembering that u j ũ j L 2 (B 1 we find, in the case where > j, that the only possibility is φ = ψ = 0. Therefore, in this case the proof is already complete since (1.14 provides the desired estimate. When < j, it is very likely that φ and ψ are not equal to 0. In this case, we evaluate φ L2 (S n 1 + ψ L2 (S n 1 c ũ j u j L2 (B 1 B 1/2 for some constant c = c(, j, n > 0. To obtain this estimate without much work observe that the space of functions {r 2 n 2 +j φ + r 2 n 2 j ψ : φ, ψ E j } is finite dimensional and that we have two (equivalent norms on it. Namely N 1 (r 2 n 2 +j φ + r 2 n 2 j ψ := φ L 2 (S n 1 + ψ L 2 (S n 1 and N 2 (r 2 n 2 +j φ + r 2 n 2 j ψ := r 2 n 2 +j φ + r 2 n 2 j ψ L 2 (B 1 B 1/2. Observe that we have implicitly used the fact that j 0 and hence the functions r r 2 n 2 +j and r r 2 n 2 j are linearly independent. Granted this estimate, we conclude that ( u j L 2 (B1 c f j L 2 (B1 + u j L 2 (B 1 B 1/2 This completes the proof of the result when all j 0. Collecting this estimates together with (1.12 this completes the proof of the Proposition 1.5.1 when j 0, for all j N. Part 3 : The case where > j = 0. We now turn to the case where j = 0. This case happens when n = 2 and j = 0. The equation satisfied by u 0 reads This time, the explicit formula we will use is ũ 0 (r := r 2 2 r u 0 + r r u 0 = f 0 r s 1 ( s t 1 f 0 (t dt ds where will be chosen appropriately, namely = 0 when > 0 and = 1 when < 0. Again, one can check directly that x 2 ũ 0 = f 0. To start with use the strategy developed above to prove that r ũ 0 L 2 1 (B 1 c f 0 L 2 (B 1
18 CHAPTER 1. WEIGHTED L 2 ANALYSIS ON A PUNCTURED BALL We leave the details to the reader. Once this is done, use again the above strategy to show that 1 ( ũ 2 0 r 2 1 1 1 dr R 2 2 f0 2 (r r 2 1 dr 0 + 1 ( 1 1/2 ( 1 1/2 ũ 2 0(r r 2 1 dr ũ 0 2 (r r 2+1 dr 0 Collecting these two estimates, we conclude that ũ 0 2 L 2 (B 1 c ( f 0 2 L 2 (B 1 + f 0 L 2 (B 1 ũ 0 L 2 (B 1 from which it follows that ũ 0 L 2 (B 1 c f 0 L 2 (B 1 Once this estimate has been obtained, we observe that u 0 ũ 0 = α + β log r When > 0, α = β = 0 since u 0 ũ 0 L 2 (B 1 and when < 0 we can argue as what has been already done when j 0 to obtain α + β c ũ 0 u 0 L 2 (B 1 B 1/2 0 for some constant c = c(n, > 0. Collecting all the estimate, we conclude that ( u 0 L 2 (B1 c f 0 L 2 (B1 + u 0 L2 (B 1 B 1/2 (1.15 This completes the proof in all cases. Exercise 1.5.1. Observe that there is another formula we could have used for ũ 0, namely ũ 0 (r = log r r t 1 f 0 (t dt r Prove the estimate (1.15 starting from this formula. t 1 log t f 0 (t dt. Exercise 1.5.2. Show that, in the main estimate in the statement of Proposition 1.5.1, one can replace u L 2 (B 1 B 1/2 by u L 1 (B 1 B 1/2. Exercise 1.5.3. Let a : B 1 R be a function which satisfies the bound a(x c x 2+α in B 1, for some α > 0. Show that the result of Proposition 1.5.1 remains true if the operator x 2 is replaced by the operator x 2 ( + a.
1.5. A CRUCIAL A PRIORI ESTIMATE 19 Exercise 1.5.4. Show that the result of Proposition 1.5.1 remains true if the operator x 2 is replaced by the operator x 2 + d, where d R is fixed, provided we define ( (n 2 2 j = R + j + d 2 1/2
20 CHAPTER 1. WEIGHTED L 2 ANALYSIS ON A PUNCTURED BALL
Chapter 2 Weighted L 2 analysis on a punctured manifold 2.1 The Laplace-Beltrami operator in normal geodesic coordinates Given a Riemannian manifold (M, g, the Laplace-Beltrami operator is defined in local coordinates x 1,..., x n g = 1 ( x i det(g g ij x j detg where g ij are the coefficients of the inverse of the matrix (g ij i,j. i,j Recall that, in local coordinates, the volume form on M is given by dvol g = detg dx 1... dx n In particular, if u is a smooth function of M, we have u g u dvol g = g ij x iu x j u dvol g = M M M g ij u 2 g dvol g Using the exponential mapping, we can define normal geodesic coordinates in a neighborhood of a point p M as follows : first choose an orthonormal basis e 1,..., e m of T p M. Then define the mapping ( F (x 1,..., x m := Exp p x i e i i One can prove that F is a local diffeomorphism from a neighborhood of 0 in (M, g into a neighborhood of p in M. 21
22 CHAPTER 2. WEIGHTED L 2 ANALYSIS ON A PUNCTURED MANIFOLD Proposition 2.1.1. ([?], Theorem?? In normal geodesic coordinates, the coefficients of the metric g can be expanded as g ij = ij + O( x 2 The functions O( x 2 are smooth function which vanish quadratically at the origin. As a simple consequence of this result, we have the expansion of the Laplace-Beltrami operator in normal geodesic coordinates : g = eucl + O( x 2 x i x j + O( x x k (2.1 The operator O( x 2 x i x j is a second order differential operator whose coefficients are smooth and vanish quadratically at the origin and the operator O( x 2 x k is a first order differential operator whose coefficients are smooth and vanish at the origin. This last expansion follows from a direct computation using the formula of g in local coordinates and the result of Proposition 2.1.1. 2.2 Two global results Using the normal geodesic coordinates, we extend the results of Proposition 1.1.1 and Proposition 1.5.1 in a global setting. As in the previous section (M, g is a compact n-dimensional Riemannian manifold without boundary. We choose points p 1,..., p k M and denote by M := M {p 1,..., p k } Given R small enough, we define B R (p M (resp. BR (p M to be the open (resp. closed geodesic ball of radius R centered at p. The corresponding punctured balls are denoted by BR (p and B R (p. Finally, we set M R := M j BR (p j We fix a smooth function such that, for all j = 1,..., k in some neighborhood of p j. Given R we define the space This space is endowed with the norm ( u L 2 (M := Again, we have γ : M (0, γ(p = dist(p, p j L 2 (M := γ +1 L 2 (M M u 2 γ 2 2 dvol g 1/2
2.2. TWO GLOBAL RESULTS 23 Lemma 2.2.1. The space (L 2 (M, L 2 (M is a Banach space. We define the unbounded operator A by A : L 2 (M L 2 (M u γ 2 ( g u + a u where a is a smooth function on M. The domain D(A of this operator is the set of functions u L 2 (M such that A u = f L 2 (M in the sense of distributions : This means that u W 2,2 (M R, for all R > 0 small enough and u ( g v + a v dvol g = f v γ 2 dvol g M for all C functions v with compact support in M. It is easy to check that Lemma 2.2.2. The domain of the operator A is dense in L 2 (M and the graph of A is closed. M Exercise 2.2.1. Give a proof of Lemma 2.2.2. The result we have obtain in Proposition 1.2.1 translates immediately into : Proposition 2.2.1. Assume R is fixed. There exists a constant c = c(n, > 0 such that for all u, f L 2 (M satisfying γ 2 ( g u + a u = f in M we have u L 2 1 (M + 2 u L 2 2 (M c ( f L 2 (M + u L 2 (M The proof of the result goes as follows : First observe that the result of Proposition 1.1.1 remains true if one changes B1 with BR. In which case the estimate of Proposition 1.1.1 has to be replaced by u L 2 1 (BR + 2 u L 2 2 (BR c ( f L 2 (B R + u L 2 (B R (2.2 if u, f L 2 (B R satisfy x 2 u = f in BR. This can be seen easily by performing a simple change v(x = u(r x and g(x = f(r x so that v and g satisfy x 2 v = g in B1, then the estimate follows from the corresponding estimate in Proposition 1.1.1. Close to the puncture p j we use normal geodesic coordinates so that γ = x and write the equation γ 2 ( g u + a u = f as x 2 eucl u = f + x 2 ( eucl g u x 2 a u
24 CHAPTER 2. WEIGHTED L 2 ANALYSIS ON A PUNCTURED MANIFOLD Using the result of Proposition 1.1.1 together with the result of Proposition 2.1.1, we evaluate ( x 2 ( eucl g u x 2 a u L 2 (BR c R 2 u L 2 (M + u L 2 1 (M + 2 u L 2 2 (M for some constant c = c(n, > 0 which does not depend on R > 0, small enough. Next we apply (2.2 to conclude that ( u L 2 1 (BR + 2 u L 2 2 (BR c f L 2 (BR + u L 2 (M + R 2 ( u L 2 1 (M + 2 u L 2 2 (M for some constant c = c(n, > 0 independent of R > 0 small enough. This can also be written as ( (1 c R 2 ( u L 2 1 (BR + 2 u L 2 2 (BR c f L 2 (BR + u L 2 (M If R > 0 is chosen so that c R 2 1/2 we conclude that ( u L 2 1 (BR + 2 u L 2 2 (BR 2 c f L 2 (BR + u L 2 (M We now use the elliptic estimates provided by Proposition 2.2.2. ([?], Theorem?? Assume we are given Ω M, Ω Ω and p (1,. Then there exists c = c(m, g, Ω, Ω > 0 such that, if v W 2,p and g L 2 (Ω satisfy g v = g in Ω, then v Lp (Ω + 2 v Lp (Ω c ( g L p (Ω + u L p (Ω with Ω = M R/2 and Ω = M R to show that u L 2 1 (M R + 2 u L 2 2 (M R c ( f L 2 (M R/2 + u L 2 (M R/2 for some constant c = c(n, R > 0. The estimate then follows from the sum of the two estimates we have obtained. The following result is a consequence of Proposition 1.5.1. Proposition 2.2.3. Assume that ± j for j N. Then there exists a constant c = c(n, and a compact K in M such that, for all u, f L 2 (M satisfying γ 2 ( g u + a u = f in M, we have u L 2 (M c ( f L 2 (M + u L2 (K
2.2. TWO GLOBAL RESULTS 25 Again this result states that we can control the weighted L 2 - norm of u in terms of the weighted L 2 -norm of f and some information about the function u away from the punctures. The proof of this second Proposition, also follows from a perturbation argument. First observe that the result of Proposition 1.5.1 remains true if one changes B1 with BR. In which case the estimate of Proposition 1.5.1 has to be replaced by ( u L 2 (BR c f L 2 (BR + R 1 u L 2 (B R B R/2 (2.3 if u, f L 2 (B R satisfy x 2 u = f in BR. This can be seen easily by performing a simple change v(x = u(r x and g(x = f(r x so that v and g satisfy x 2 v = g in B1, then the estimate follows from the corresponding estimate in Proposition 1.5.1. Close to the puncture p j we use normal geodesic coordinates so that γ = x and write the equation γ 2 ( g u + a u = f as x 2 eucl u = f + x 2 ( eucl g u x 2 a u Using the result of Proposition 2.2.1 together with the result of Proposition 2.1.1, we evaluate x 2 ( eucl g u x 2 a u L 2 (B R c R 2 u L 2 (M for some constant c = c(n, > 0 which does not depend on R > 0, small enough. Next we apply (2.3 to conclude that ( u L 2 (BR c f L 2 (BR + R 2 u L 2 (M + R 1 u L 2 (B R B R/2 for some constant c = c(n, > 0 independent of R > 0 small enough. Adding on both sides u L 2 (M R we conclude that ( u L 2 (M c f L 2 (BR + R 2 u L 2 (M + R 1 u L2 (M R/2 where c = c(n, > 0 does not depend on R > 0 small enough. In other words ( (1 c R 2 u L 2 (M c f L 2 (BR + R 1 u L 2 (M R/2 It remains to fix R > 0 such that c R 2 1/2 and let K = M R/2. this completes the proof of the result. Exercise 2.2.2. Show that the result of Proposition 2.2.3 remains true if the function a : M R only belongs to L loc (M and satisfies the bound in M, for some α > 0. a c γ 2+α
26 CHAPTER 2. WEIGHTED L 2 ANALYSIS ON A PUNCTURED MANIFOLD Exercise 2.2.3. Show that the result of Proposition 2.2.3 remains true if, near any of the p i, the function a can be decomposed as a = d γ 2 + ã i where d R is a constant and the function ã i satisfies the bound ã i c γ 2+αi in B Ri (p i, for some α i > 0 and provided we define ( (n 2 2 j = R + j + c 2 1/2 Exercise 2.2.4. Show that the result of Proposition 2.2.3 remains true if, near any of the p i there exists local coordinates x 1,..., x n in which the coefficients of the metric can be expanded as g ij = ij + O( x β and if in addition for some β > 0. g ij = O( x β 1 Exercise 2.2.5. Extend the result of Proposition 2.2.3 to handle the case where, near any of the p i, the function a can be decomposed as a = d i γ 2 + ã i where d i R are constants and the function ã i satisfies the bound in B Ri (p i, for some α i > 0. ã i c γ 2+αi 2.3 The kernel of the operator A The results of the previous sections will now be used to derive the functional analytic properties of the operator A. We start with the : Theorem 2.3.1. The kernel of A is finite dimensional. For the time being, let us assume that ± j. We argue by contradiction and assume that the result is not true. Then, there would exist a sequence (u m m of elements of L 2 (M which satisfy A u m = 0. Without loss of generality we can assume that the sequence is normalized so that u m 2 γ 2 2 dvol g = 1 (2.4 M
2.4. THE RANGE OF THE OPERATOR A 27 and also that for all m m. Using the result of Proposition 2.2.3 we obtain where c = c(n, > 0 does not depend on m. M u m u m γ 2 2 dvol g = 0. (2.5 u m u m L 2 (M c u m u m L 2 (K (2.6 Using (2.4 together with the result of Proposition 2.2.1 we conclude that u m is bounded in W 1,2 (K. Now, we apply Rellich s compactness result : Proposition 2.3.1. ([?], Theorem?? Given a smooth bounded domain Ω M, the imbedding is compact. W 1,2 (Ω L 2 (Ω This result allows us to extract some subsequence (which we will still denote by (u m m which converges in L 2 (K. In particular, the sequence (u m m is a Cauchy sequence in L 2 (K. In view of (2.6 we see that the sequence (u m m is a Cauchy sequence in L 2 (M. This space being a Banach space, we conclude that this sequence converges in L 2 (M to some function u. Clearly, passing to the limit in (2.4 we see that u 2 γ 2 2 dvol g = 1 While, passing to the limit m in (2.5, we get u m u r 2 2 dx = 0 and then passing to the limit as m tends to, we conclude that u 2 r 2 2 dx = 0 Ω Ω Ω Clearly a contradiction. This completes the proof when ± j, for all j N. In order to complete the proof is all cases it is enough to observe that if u KerA then u KerA for all. Therefore, one can always reduce to the case where ± j for all j N. 2.4 The range of the operator A We pursue our quest of the mapping properties of the operators A by studying the range of this operator. Thanks to the results of the previous sections, we are in a position to prove the :
28 CHAPTER 2. WEIGHTED L 2 ANALYSIS ON A PUNCTURED MANIFOLD Theorem 2.4.1. Assume that ± j for j N. Then the range of A is closed. Let u m, f m L 2 (M be sequences such that f m := A u m converges to f in L 2 (M. Since we now know that Ker A is finite dimensional, it is closed and we can project every each u m onto { } u L 2 (M : u v r 2 2 dx = 0 v Ker A M the orthogonal complement of Ker A in L 2 (M, with respect to the scalar product associated to the weighted norm. Therefore, without loss of generality, we can assume that u m is L 2 -orthogonal to Ker A. Since f m converges in L 2 (M, there exists c > 0 such that f m L 2 (M c. (2.7 Now, we claim that the sequence (u m m is bounded in L 2 (M. To prove this claim, we argue by contradiction and assume that (at least for a subsequence still denoted (u m m We set u m v m := u m L 2 (M lim m + um L 2 (M = f m and g m := u m L 2 (M so that A v m = g m. Applying the result of Proposition 2.2.1, we conclude that the sequence (v m m is bounded in W 1,2 (K and hence, using Rellich s Theorem, we conclude that a subsequence (still denoted (v m m converges in L 2 (K. Now the result of Proposition 2.2.3 yields v m v m L 2 (M c ( g m g m L 2 (M + v m v m L2 (K. (2.8 On the right hand side, the sequence (g m m tends to 0 in L 2 (M and the sequence (v m m converges in L 2 (K. Therefore, we conclude that (v m m is a Cauchy sequence in L 2 (M and hence converges to v L 2 (M. To reach a contradiction, we first pass to the limit in the identity A v m = g m to get that the function v is a solution of A v = 0 and hence v Ker A. But by construction v L 2 (M = 1 and also v m v γ 2 2 dvol g = 0 M (since v Ker A and, passing to the limit in this last identity we find that v L 2 (M = 0. A contradiction. Now that the claim is proved, we use the result of Proposition 2.2.1 together with Rellich s Theorem to extract, form the sequence (u m m some subsequence which converges to u in L 2 (M. Once more, Proposition 2.2.3 implies that u m u m L 2 (M c ( f m f m L 2 (M + u m u m L2 (K. (2.9
2.5. FREDHOLM PROPERTIES FOR A 29 This time, on the right hand side, the sequence (f m m converges in L 2 (M and the sequence (u m m converges in L 2 (K. Therefore, we conclude that (u m m is a Cauchy sequence in L 2 (M and hence converges to u L 2 (M. Passing to the limit in the identity A u m = f m we conclude that A u = f and hence f belongs to the range of A. This completes the proof of the result. 2.5 Fredholm properties for A It will be convenient to identify the dual of L 2 (M with L 2 (M. This is done using the scalar product u, v := u v γ 2 dvol g (2.10 Clearly, given v L 2 (M, we can define T v ( L 2 (M by T v (u = u, v M Moreover, we have T v (L 2 (M = v L 2 (M Conversely, given T ( L 2 (M there exists a unique v L 2 (M such that u, v = T (u for all u L 2 (M. We define A, the adjoint of A A : ( L 2 (M ( L 2 (M is defined to be an unbounded operator. An element T ( L 2 (M domain of A, if and only if there exists S ( L 2 (M such that belongs to D(A, the for all v D(A. We will write A (T = S. T (A v = S(v Granted the above identification of ( L 2 (M with L 2 (M it is easy to check that we can identify A with A. Indeed, if we write T = T u and A (T = T f, for u, f L 2 (M, then, by definition T u (A v := u, A v and A (T (v := f, v for all v D(A. Hence, we have u( g + av dvol g = M M f v γ 2 dvol g
30 CHAPTER 2. WEIGHTED L 2 ANALYSIS ON A PUNCTURED MANIFOLD for all v D(A. This in particular implies that γ 2 ( g + a u = f in the sense of distributions. Since u, f L 2 (M, we conclude that u D(A and f = A u. Conversely, if u D(A, we can write for all v D(A u, A v = = M M = A u, v u ( g + av dvol g v ( g + au dvol g The integrations by parts can be justified since, according to the result of Proposition 2.2.1, we have v L 2 1 (M, u L 2 1 (M, 2 v L 2 2 (M and 2 u L 2 2 (M. Therefore T u D(A and A (T u = T A u. With these identifications in mind, we can state the Theorem 2.5.1. Assume that ± j for all j N. Then Ker A = (Im A and Im A = (Ker A The first part is a classical property for unbounded operators with closed graph and dense domain (see Corollary II.17 in [?]. The second result follows from classical results for unbounded operators with dense domains, closed graph and closed range (see Theorem II.18 in [?]. Observe that, because of our identifications, F is obtained from F using the scalar product defined in (2.10. Very useful for us will be the : Corollary 2.5.1. Assume that ± j for all j N. Then A is injective if and only if A is surjective. 2.6 The deficiency space Even though the previous results seem already a great achievement, since it will provide right inverses for some operators, we will need a more refined result. As usual, this result for operators defined on the punctured manifold M are obtained by perturbing the corresponding results in Euclidean space. To start with, let us prove the :
2.6. THE DEFICIENCY SPACE 31 Lemma 2.6.1. Assume that ± j, for j N. There exist an operator G : L 2 (B1 L 2 (B1 and c = c(n, > 0 such that for all f L 2 (B 1, the function u := G (f is a solution of x 2 u = f in B 1 and u L 2 (B 1 + u L 2 1 (B 1 + 2 u L 2 2 (B 1 c f L 2 (B 1 At first glance this result looks rather strange wince we are not imposing any boundary data. Nevertheless, some boundary data are hidden in the construction of the operator G. Observe that we state the existence of G and do not state any uniqueness of this operator! The proof of the existence of G relies on the eigenfunction decomposition of the function f. We decompose as usual f = j 0 f j where f(r, E j for all j N. Let j 0 N be the least index for which < j0 We set f = j j0 f j Clearly f L 2 (B 1 and, for all R (0, 1/2 one can solve x 2 ũ R = f in B 1 B R ũ R = 0 on B 1 B R The existence of ũ R follows from Proposition 1.1.2 and we have the estimate ũ R L2 (B 1 B R c f L 2 (B 1 B R for some constant c = c(n, R > 0. We claim that there exists a constant c = c(n, > 0 such that ũ R L 2 (B 1 B R c f L 2 (B 1 B R (2.11 Here the norm in L 2 (B 1 B R is nothing but the restriction of the restriction of the norm in L 2 (B 1 to functions which are defined in B 1 B R. The proof of the claim follows the Part 1 of the proof of Proposition 1.5.1. We omit the details.
32 CHAPTER 2. WEIGHTED L 2 ANALYSIS ON A PUNCTURED MANIFOLD Arguing as in the proof of Proposition 1.2.1 we conclude that there exists a constant c = c(n, > 0 such that ũ R L 2 1 (B 1 B R c f L 2 (B 1 B R In particular, given R (0, 1/2, there exists c = c(n,, R > 0 such that ũ R L 2 (B 1 B R + ũ R L 2 (B 1 B R c f L 2 (B 1 for all R (0, R. Then using Rellich s Theorem together with a simple diagonal argument, we conclude that there exists a sequence of radii R i tending to 0 such that the sequence (ũ Ri i converges in L 2 (B 1 B R, for all R (0, 1/2. Passing to the limit in the equation we obtain a solution ũ of Moreover, passing to the limit in (2.11, we have the estimate x 2 ũ = f in B1 (2.12 ũ = 0 on B 1 ũ L 2 (B 1 c f L 2 (B 1 To finish this study observe that the solution of (2.12 which belongs to L 2 (B 1 is unique. To see this, argue by contradiction. If the claim were not true there would exists two solutions and taking the difference we would obtain a function w L 2 (B 1 satisfying { x 2 w = 0 in B 1 Performing the eigenfunction decomposition of w as ũ = 0 on B 1 w = j j 0 w j we find that w j = r 2 n 2 +j φ j + r 2 n 2 j ψ j where φ j, ψ j E j. Using the fact that w j L 2 (B 1 we conclude that ψ j = 0. Next, using the fact that w j = 0 on B 1, we get φ j = 0 and hence w = 0. Therefore, we can define G ( f = ũ. It remains to understand the definition of G acting on f j, for j j 0 1. For the sake of simplicity, we assume that j = 0 (When j = 0, the formula has to be changed according to what we have already done in Part 3 of the proof of Proposition 1.5.1 and we use an explicit formula G (f j = 1 2 j ( r 2 n 2 +j r t n 4 2 j f j (t dt r 2 n 2 j r t n 4 2 +j f j (t dt
2.6. THE DEFICIENCY SPACE 33 where = 0 if > j and = 1 if < j. The estimate follows at once from the arguments developed in Part 2 of the proof of Proposition 1.5.1. We omit the details. Let us now provide a few applications of this result. Application # 1 : The first application is concerned with the extension of the previous result to the operator defined on the manifold in a neighborhood of one puncture. Lemma 2.6.2. Assume that ± j, for j N. Given p i M one of the punctures, there exists R i = R(p i, n, > 0, an operator G (i : L 2 (B R i (p i L 2 (B R i (p i and c = c(n,, p i > 0 such that for all f L 2 (B R i (p i, the function u := G (i (f is a solution of γ 2 ( g + a u = f in B R i (p i and u L 2 (B R i (p i + u L 2 1 (B R i (p i + 2 u L 2 2 (B R i (p i c f L 2 (B R i (p i This result follows from a simple perturbation argument. First observe that, a scaling argument shows that the result of Lemma 2.6.1 holds when the radius of the ball, which was chosen to be 1, is replaced by R. The corresponding operator will be denoted by G,R and the estimate holds with a constant which does not depend on R > 0. We leave this as an exercise. Thanks to the result of Proposition 2.1.1 we can write γ 2 ( g eucl + a u L 2 (BR (p i i c R ( u 2 L 2 (BR (p i i + u L 2 1 (BR (p i i + 2 u L 2 2 (BR (p i i provided R > 0 is small enough. This implies that f A G,R f L 2 (B R i (p i c R 2 f L 2 (B R i (p i for some constant c = c(n, > 0 which does not depend on R. This clearly implies that the operator A G,R is invertible provided R is fixed small enough, say R = R i. To obtain the result, it is enough to define G (i := G,Ri (A G,Ri. The relevant estimate then follows at once. Application # 2 : Recall that the functions x 2 n 2 ±j φ
34 CHAPTER 2. WEIGHTED L 2 ANALYSIS ON A PUNCTURED MANIFOLD are harmonic in B1 provided φ E j. Building on the result of the previous application, we now prove that one can perturb these functions to get, near any puncture p i a solution of the homogeneous problem associated with the operator γ 2 ( g + a. This is the content of the following : Lemma 2.6.3. For all puncture p i M, given j N and φ E j, there exists W ± (i j,φ defined in BR i (p i and which satisfies which is in B R i (p i. In addition, W ± (i j,φ for all < ± j + 2. Finally the mapping is linear. γ 2 ( g + a W ± (i j,φ = 0 x 2 n 2 ±j φ L 2 (B R i (p i φ E j W ± (i j,φ In this result, R i is the radius given in Lemma 2.6.2 and x are normal geodesic coordinates near p i. The proof of this Lemma uses the following computation which follows at once from Proposition 2.1.1 γ 2 ( g + a x 2 n 2 ±j φ = γ 2 ( g eucl + a x 2 n 2 ±j φ L 2 (B R i (p i for all < ± j + 2. The result then follows from Lemma 2.6.2. For each i = 1,..., k, we define χ (i to be a cutoff function which is identically equal to 1 in B Ri/2(p i and identically equal to 0 in M B 3Ri/4(p i. The main result of this section is : Proposition 2.6.1. Given <,, ± j, for all j N. Assume that u L 2 (M and f L 2 (M satisfy γ 2 ( g + a u = f in M. Then, there exists v L 2 (M such that In addition u v D, for some constant c = c(n,, > 0. := Span {χ (i W ±(i j,φ, : φ E j, < ± j < } v L 2 (M + u v D, c ( f L 2 (M + u L 2 (M
2.6. THE DEFICIENCY SPACE 35 The proof of this result relies on the corresponding result for the Laplacian in the punctured unit ball. Lemma 2.6.4. Given <,, ± j, for all j N. Assume that u L 2 (B 1 and f L 2 (B 1 satisfy x 2 u = f in B 1. Then, there exists v L 2 (B 1 such that In addition u v D, for some constant c = c(n,, > 0. := Span { x 2 n 2 ±j φ, : φ E j, < ± j < } v L 2 (B 1 + u v D, c ( f L 2 (B 1 + u L 2 (B 1 To prove the Lemma, we use the result of Lemma 2.6.1 and set v = G f L 2 (B 1. Therefore in B 1. We have x 2 (u v = 0 v L 2 (B 1 c f L 2 (B 1 for some constant c = c(n, > 0. We set w = u v which we decompose as usual w = j w j where w j (r, E j. We fix j 0 to be the least index for which < j0 and < j0 We define w = j j 0 w j We claim that w L 2 (B 1 and also that v L 2 (B 1 c w L 2 (B 1 B 1/2 for some constant c = c(n, > 0. The proof of the claim follows the arguments of Part 1 in the proof of Proposition 1.5.1. We omit the details. Next, observe that, for j = 0,..., j 0 1 the function w j is given by w j = x 2 n 2 +j φ j + x 2 n 2 j ψ j
36 CHAPTER 2. WEIGHTED L 2 ANALYSIS ON A PUNCTURED MANIFOLD for some φ j, ψ j E j. Observe that φ j = 0 if j < and ψ j = 0 if j < since w j L 2 (B 1. It is easy to see that φ j L 2 (S n 1 + φ j L 2 (S n 1 c w j L 2 (B 1 B 1/2 for some constant c = c(n, j > 0. We set so that v = v + w + u v = j=0,...,j 0 1, j=0,...,j 0 1, j> x 2 n 2 +j φ j + < j< x 2 n 2 +j φ j + j=0,...,j 0 1, j=0,...,j 0 1, j> x 2 n 2 j ψ j < j< x 2 n 2 j ψ j The estimate follows from collecting the above estimates. This completes the proof of Lemma 2.6.4. We proceed with the proof of Proposition 2.6.1. Choose inf(, + 1 such that ± j, for all j N. Using the result of Proposition 2.2.1 we have u L 2 1 (M + 2 u L 2 2 (M c ( f L 2 (M + u L 2 (M Using the decomposition given in Proposition 2.1.1, we conclude that, near any puncture p i, we have x 2 u = f x 2 ( g eucl + a L 2 (B R i We apply the previous result which yields the decomposition u = v + x 2 n 2 ±j φ <± j< where φ E j. Next use the result of Lemma 2.6.3 and replace all x 2 n 2 ±j φ by χ (i W ±(i j,φ get the decomposition u = v + ( x 2 n 2 ±j φ χ (i W ±(i j,φ + <± j< χ (i W ±(i j,φ <± j< Observe that the function ũ = v + ( x 2 n 2 ±j φ χ (i W ±(i j,φ L2 (M <± j< and also that γ 2 ( g + a ũ = f L 2 (M. If = then the roof is complete. If not, apply the same argument with u replaced by ũ, f replaced by f and replaced by and proceed until the gap between and is covered. We now give some important consequences of this result : to
2.6. THE DEFICIENCY SPACE 37 2.6.1 The kernel of A revisited : Thanks to the result of Proposition 2.6.1 we can state the : Lemma 2.6.5. Fix < such that, ± j, for j N. Assume that u L 2 (M satisfied γ 2 ( g u + a u = 0 in M. Then u L 2 (M provided the interval (, does not contain any ± j, for some j N. This Lemma is a direct consequence of the result of Proposition 2.6.1. It essentially states that the kernel of the operator A does not change as remains in some interval which does not contain any ± j, for j N. 2.6.2 The deficiency space : We now define Definition 2.6.1. Given > 0, j, for all j N, the deficiency space D is defined by D := Span {χ (i W ±(i j,φ, : φ E j, < ± j < } Observe that the dimension of D can be computed as follows As a first by product, we obtain dimd = 2 j, j< dim E j Proposition 2.6.2. Given > 0, j, for all j N. Assume that A is injective. Then the operator à : L 2 (M D L 2 (M is surjective and u γ 2 ( g u + a u Ker A = Ker à As a consequence of the previous Proposition, we have the Corollary 2.6.1. Given > 0, j, for all j N. Assume that A is injective. Then dim Ker A = codim Im A = 1 2 dim D
38 CHAPTER 2. WEIGHTED L 2 ANALYSIS ON A PUNCTURED MANIFOLD Under the assumptions of the Corollary, we have dim Ker A = dim Ker à and dim D = dim dim Ker A + codim Im A But, by duality, we have dim Ker A = codim Im A. The result then follows at once. Exercise 2.6.1. Extend the results of Corollary 2.6.1 to the case there A is not injective.
Chapter 3 Weighted C 2,α analysis on a punctured manifold 3.1 From weighted Lebesgue spaces to weighted Hölder spaces As far as linear analysis is concerned the results of the previous sections are sufficient. However, we would like to apply them to nonlinear problems for which is will be more convenient to work in the framework of Hölder spaces. The purpose of this section is to explain how the analysis of the previous section can be extended to weighted Hölder spaces. We begin with the definition of weighted Hölder spaces. Definition 3.1.1. Given l N, α (0, 1 and R, we define C l,α (M to be the space of functions u C l,α loc (M for which the following norm is finite. u C l,α k (M := u C l,α (M R + i=1 sup ρ n 2 2 u(exp pi (ρ C l,α ( B 2 B 1 T pi M ρ (0,R For example, the function γ 2 n 2 + C l,α (M if and only if. It also follows directly from this definition that C l,α (M L 2 (M for all >. Lemma 3.1.1. The space (C l,α (M, C l,α (M 39 is a Banach space.
40 CHAPTER 3. WEIGHTED C 2,α ANALYSIS ON A PUNCTURED MANIFOLD Exercise 3.1.1. Show that the embedding is compact provided l + α < l + α and <. C l,α (M C l,α (M The last easy observation is that the operator A : C l,α (M C l,α (M u γ 2 ( g u + au is well defined and bounded. The extension of our results to weighted Hölder spaces rely on the following regularity result. Proposition 3.1.1. Assume that, R are fixed with <. Further assume that the interval [, ] does not contain any ± j for j N. Then, there exists c = c(n,, > 0 such that for all u, f L 2 (B R (p i satisfying γ 2 ( g + a u = f in M, if f C 0,α (M then u C 2,α (M and ( u C 2,α (M c f C 0,α (M + u L 2 (M Before we proceed to the proof of this result, let us explain how it can be used. Application # 1: The first application of the result of Proposition 3.1.1 is concerned with the kernel of the operator A. Lemma 3.1.2. Assume that R is fixed with j, for j N. u L 2 (M is a solution of γ 2 ( g + a u = 0 in M. Then u C 2,α (M. Further assume that In other words, in order to check the injectivity of A, it is enough to check the injectivity of A, which in practical situation is easier to perform. Application # 2 : Observe that, if u L 2 (M is in the kernel of A then u is also in the kernel of A for all since L 2 (M L 2 (M. However, it follows from Proposition 3.1.1 the the following is also true : Lemma 3.1.3. Assume that R is fixed with j, for j N. Further assume that u L 2 (M is in the kernel of A. Then u is also in the kernel of A for all > for which [, ] does not contain any ± j, for j N
3.1. FROM WEIGHTED LEBESGUE SPACES TO WEIGHTED HÖLDER SPACES 41 Application # 3 : The third application of the result of Proposition 3.1.1 is concerned with the extension of the result of Proposition 2.6.1 to weighted Hölder spaces and this will be useful when dealing with nonlinear differential operators. We have the : Proposition 3.1.2. Given <,, ± j, for all j N. Assume that u L 2 (M and f C 0,α (M satisfy γ 2 ( g + a u = f in M. Then, there exists v C 2,α (M such that In addition u v D, for some constant c = c(n,, > 0. := Span {χ (i W ±(i j,φ, : φ E j, < ± j < } v C 2,α (M + u v D, c ( f C 0,α (M + u L 2 (M There are important by products of this result : Given, > 0, j for all j N. Assume that A is injective, then, according to the result of Corollary 2.5.1, the operator A is surjective and hence there exists G : L 2 (M L 2 (M. a right inverse for A (i.e. A G = I. In particular, given the function u := G f L 2 (M solves f C 0,α (M L 2 (M, A u = f in M. Applying the result of Proposition 3.1.2, we see that there exists v C 2,α (M such that and in addition u v D := Span {χ (i W ±(i j,φ, : φ E j, < ± j < } for some constant c = c(n, > 0. v C 2,α (M + u v D c ( f C 0,α (M + u L 2 (M If ( 0, 0 and if A is injective. Then, according to the result of Lemma 3.1.3 the operator A is also injective for all ( 0, 0. Therefore, according to the result of Corollary 2.5.1 the operator A is surjective. This implies that there exists G : L 2 (M L 2 (M.