Estimating the number of negative eigenvalues of Schrödinger operators

Transcription

1 Estimating the number of negative eigenvalues of Schrödinger operators Alexander Grigor yan University of Bielefeld IMS, CUHK, Hong Kong, March-April 2012

2 1 Upper estimate in R n, n Introduction and statement Given a non-negative L 1 loc function V (x) on Rn, consider the Schrödinger type operator H V = Δ V where Δ = n k=1 2 x 2 k is the classical Laplace operator. More precisely, H V is defined as a form sum of Δ and V, so that, under certain assumptions about V, the operator H V is self-adjoint in L 2 (R n ). Denote by Neg (H V ) the number of negative eigenvalues of H V (counted with multiplicity), assuming that its spectrum in (, 0) is discrete. For example, the latter is the case when V (x) 0 as x. We are are interested in obtaining estimates of Neg (H V ) in terms of the potential V. Suppose that V is an attractive potential field in quantum mechanics. Then H V is the Hamiltonian of a particle that moves in this field, and the negative eigenvalues of H V correspond to so called bound states of the particle, that is, the negative energy levels E k that are inside a 1

3 potential well. -V(x) 0 E k x Hence, Neg (H V ) determines the number of bound states of the system. In particular, if V is the potential field of an electron in an atom, then Neg (H V ) is the maximal number of possible electron orbits in the atom. Estimates of Neg (H V ), especially upper bounds, are of paramount importance for quantum mechanics. 2

4 We start with a famous theorem of Cwikel-Lieb-Rozenblum. Theorem 1.1 Assume n 3 and V L n/2 (R n ). Then H V can be defined as a self-adjoint operator, its negative spectrum is discrete, and the following estimate is true Neg (H V ) C n R n V (x) n/2 dx. (1.1) This estimate was proved independently by the above named authors in Later Lieb used (1.1) to prove the stability of the matter in the framework of quantum mechanics. The estimate (1.1) implies that, for a large parameter α, Neg (αv ) = O ( α n/2) as α. (1.2) This is a so called semi-classical asymptotic (that corresponds to letting 0), and it is expected from another consideration that Neg (αv ) should behave as α n/2, at least for a reasonable class of potentials. 3

5 1.2 Counting function Before the proof of Theorem 1.1, let us give an exact definition of the operator H V and its counting function. Given a potential V in R n, that is, a non-negative function from L 1 loc (Rn ), define the bilinear energy form by E V (f, g) = f gdx V fgdx R n R n for all f, g D := C0 (R n ), and the corresponding quadratic form E V (f) := E V (f, f). For any open set R n, we consider a restriction of E V to D := (). The form (E V, D ) is called closable in L 2 () if C 0 1. it is semi-bounded below, that is, for some constant K 0, E V (f) K f 2 2 for all f D ; 2. and, for any sequence {f n } D, f n 2 0 and E V (f n f m ) 0 = E V (f n ) 0. 4

6 A closable form (E V, D ) has a unique extension to a subspace F V, of L 2 () so that F V, is a Hilbert space with respect to the inner product (f, g) E := E V (f, g) + (K + 1) (f, g), (1.3) (that is, (E V, F V, ) is closed) and D is dense in F V,. Being a closed form, (E V, F V, ) has the generator H V, that can be defined as an (unbounded) operator in L 2 () with a maximal possible domain dom (H V, ) F V, such that E V (f, g) = (H V, f, g) f dom(h V, ) and g F V,. (1.4) Then H V, is a self-adjoint operator in L 2 (). For example, for f, g D we have E V (f, g) = f gdx V fgdx = ( Δf V f) gdx so that H V, f = Δf V f. 5

7 Since the operator H V, is self-adjoint, the spectrum of H V, is real and semi-bounded below. The counting function N λ of H V, is defined by N λ (H V, ) = dim Im 1 (,λ) (H V, ), (1.5) where 1 (,λ) (H V, ) is the spectral projector of H V, of the interval (, λ). For example, if the spectrum of H V, is discrete and {ϕ k } is an orthonormal basis of eigenfunctions with eigenvalues {λ k } then 1 (,λ) (H V, ) is the projection on the subspace of L 2 () spanned by all ϕ k with λ k < λ. It follows that N λ (H V, ) is the number of eigenvalues λ k < λ counted with multiplicity. The definition (1.5) has advantage that it always makes sense. Lemma 1.2 The following identity is true for all real λ: N λ (H V, ) = sup { dim V : V D and E V (f) < λ f 2 2 f V \ {0} }, (1.6) where V D means that V is a subspace of D. In fact, it suffices to restrict sup to finite dimensional subspaces V. 6

8 For example, if the spectrum of H V, is discrete and {ϕ k } is an orthonormal basis of eigenfunctions with eigenvalues {λ k } then the condition E V (f) < λ f 2 2 is satisfied exactly for f = ϕ k provided λ k < λ, because E V (ϕ k ) = (H V, ϕ k, ϕ k ) = λ k (ϕ k, ϕ k ) < λ ϕ k 2 2. The optimal space V in (1.7) is spanned by all {ϕ k } with λ k < λ, and its dimension is equal to N λ (H V, ). There is also a version of counting function with non-strict inequality: Then the following identity is true: N λ (H V, ) = dim Im 1 (,λ] (H V, ). Nλ (H V, ) = sup {dim V : V F V, and E V [f] λμ [f] f V}. (1.7) 7

9 1.3 Reduction to operator 1 V Δ For the sake of proof of Theorem 1.1, we will assume that V > 0 and, moreover, 1 V L1 loc (Rn ). Then by approximation argument one can handle a general case. Set H V H V,R n. Our aim is to prove the upper bound N 0 (H V ) C n R n V n/2 dx for the number N 0 (H V ) of negative eigenvalues. By an approximation argument the same estimate will hold for the number N 0 (H V ) of nonpositive eigenvalues. For λ = 0 the identity (1.6) becomes N 0 (H V, ) = sup {dim V : V D and E V (f) < 0 f V \ {0}}. (1.8) The condition E V (f) < 0 here is equivalent to f 2 dx V f 2 dx < 0 (1.9) for all non-zero f V where V is a subspace of D. 8

10 We will interpret this inequality in terms of the counting function of another operator. Consider a new measure μ defined by and the energy form dμ = V (x) dx E (f) = f 2 dx R n for f D. Then (1.9) can be rewritten in the form E (f) < f 2 2,μ so that } N 0 (H V, ) = sup {dim V : V D and E (f) < f 22,μ f V \ {0}. (1.10) The right hand side here is the counting function of another operator. Indeed, denoted by L V, the generator of the energy form (E, D ) in L 2 (, μ). This form can be shown to be closable, so that its generator L V, is a self-adjoint operator in L 2 (, μ). Note also that this operator is positive definite because so is E. By definition, we have, for all f, g dom (L V, ), E (f, g) = (L V, f, g) μ. 9

11 In particular, for f, g D this implies (Δf) gdx = f g dx = (L V, f) gv dx, whence L V, f = 1 Δf that is, L V V, = 1 Δ. V The counting function N λ (L V, ) of the operator L V, is defined exactly as for H V,. Lemma 1.2 for this operator means that { } N λ (L V, ) = sup dim V : V D and E (f) < λ f 2 2,μ f V \ {0}. (1.11) For λ = 1 the right hand side of (1.11) coincides with that of (1.10), which implies N 0 (H V, ) = N 1 (L V, ). (1.12) In particular, for the case = R n, we have N 0 (H V ) = N 1 (L V ). The identity (1.12) is called Birman-Schwinger principle. Informally the identity (1.12) reflects the equivalence of the inequalities Δ V 0 and 1 Δ 1 that are understood in the sense of V quadratic forms. 10

12 1.4 Case of small V Here we illustrate the usage of (1.12) by proving a particular case of Theorem 1.1 as follows. Proposition 1.3 If n 3 then there is a constant c n > 0 such that R n V n/2 dx < c n N 0 (H V ) = 0. Proof. By (1.12) we need to prove that the spectrum of L V below 1 is empty, that is, inf spec L V 1. This is equivalent to the claim that the operator L V invertible and L 1 V 1. The inverse operator is defined by L 1 V f = u L V u = f, in L 2 (R n, μ) is 11

13 where f L 2 (R n, μ) and u dom (L V ). Hence, it suffices to prove that L V u = f u 2,μ f 2,μ. Multiplying L V u = f by u and integrating against μ, we obtain E (u) = (L V u, u) μ = (f, u) μ that is, u 2 dx = ufdμ. R n R n By Sobolev inequality, we have R n u 2 dx c n ( n. R n u 2n n 2 dx ) n 2 Note that this is the only place where n > 2 is used. 12

14 Using the Hölder inequality and the above lines, we obtain whence R n u 2 V dx ( c 1 n = c 1 n c 1 n u 2 R ( n ) n 2 ( n n n 2 dx Rn ) ( V n 2 dx ) 2 n ) 2 u 2 dx V n n 2 dx R n R ( ) ( n ) 2 ufdμ V n n 2 dx R n R ( ) n 1/2 ( ) 1/2 ( f 2 dμ u 2 dμ R n R n u 2,μ c 1 n ( R n V n 2 dx ) 2 n f 2,μ. ) 2 V n n 2 dx R n (1.13) Clearly, if R n V n 2 dx small enough then u 2,μ f 2,μ, which was to be proved. The argument in the proof of Proposition 1.3 allows to prove another part of Theorem

15 Proposition 1.4 If V L n/2 (R n ) then the form (E V, D) is closable. Consequently, the operator H V is defined as a self-adjoint operator in L 2 (R n ). Proof. It follows from the hypothesis that, for any ε > 0, V can be split to a sum of two potentials V = V 1 + V 2 where It follows from (1.13) that E (u) c n ( V 1 n/2 ε and V 2 L. ) 2/n V n/2 1 dx u 2 V 1 dx c n ε 1 R n R n Choosing ε sufficiently small, we obtain c n ε 1 2 whence u 2 V dx = u 2 V 1 dx + u 2 V 2 dx R n R n R n R n u 2 V 1 dx. 1 2 E (u) + K u 2 2, (1.14) 14

16 where K = V 2 L. In particular, we see that E V (u) = E (u) u 2 V dx K u 2 2 R n so that the form E V is semi-bounded below. By a standard result from the theory of quadratic forms, (1.14) implies that the form E V is closed in the domain W 1,2 (R n ), which finishes the proof. 15

17 1.5 Proof of Theorem 1.1 in general case The proof below is due to Li and Yau 83 but it is presented here from somewhat different angle. In a precompact domain the operator L V, has discrete positive spectrum. Denote its eigenvalues by λ k (), where k = 1, 2,..., so that the sequence {λ k ()} is increasing, and each eigenvalue is counted with multiplicity. The main part of the proof of Theorem 1.1 is contained in the following statement. Theorem 1.5 (AG, Yau 2003) Assume that there is a Radon measure ν in R n and α > 0 such that, for all precompact open sets, λ 1 () ν () α. (1.15) Then, for any positive integer k and any precompact open set, ( ) α k λ k () c, (1.16) ν () where c = c (α) > 0. 16

18 For example, if V = 1 then L V, is the Laplace operator Δ with the Dirichlet boundary condition on. The hypothesis (1.15) is satisfies if ν is a multiple of the Lebesgue measure as by the Faber-Krahn inequality Then (1.16) becomes λ 1 () c n (vol ) 2/n. λ k () c n ( ) 2/n k, vol that is also known to be true. Moreover, it matches the Weyl s asymptotic ( formula λ k () c k 2/n n vol ) as k. The point of Theorem 1.5 is that V in the definition of L V, can be arbitrary and measure ν can be arbitrary. By the way, there is no restriction of the dimension n in Theorem 1.5. Moreover, exactly in this form it is true on any Riemannian manifold instead of R n. Let us show how Theorem 1.5 implies Theorem 1.1. Let us use the variational principle: λ 1 () = inf u D (L V, u, u) μ (u, u) μ 17 = inf u D E (u) (u, u) μ.

19 Using again the Sobolev inequality and the Hölder inequality we obtain (u, u) μ = u 2 dx c n ( ( u 2 V dx ) n 2 u 2n n n 2 dx ) n 2 ( ) 2 u 2n n n 2 dx V n/2 n dx, ( ) 2 E (u) c n V n/2 n dx. (u, u) μ Hence, setting dν = c n/2 n V n/2 dx and minimizing in u, we obtain By Theorem 1.5, we conclude that λ 1 () ν () 2/n. λ k () c ( ) 2/n k. (1.17) ν () 18

20 We need to estimate the counting function N 1 (L V, ) = # {k : λ k () < 1}. By (1.17), λ k () < 1 implies k Cν () whence also N 1 (L V, ) Cν () = C V n/2 dx. It follows by (1.12) that also N 0 (H V, ) C We are left to pass from H V, to H V,R n. Recall that V n/2 dx C V n/2 dx. (1.18) R n N 0 (H V,R n) = sup {dim V : V D R n, E V (f) < 0 f V \ {0}}, where V is a finite-dimensional subspace of D R n. For any such V there exists a precompact open set containing supp f for all f V (for it suffices to have supp f V for the elements of a basis of V). Hence, V D and by (1.18) dim V C R n V n/2 dx, whence the same estimate for N 0 (H V,R n) follows. 19

21 Brief summary We prove the following theorem. Theorem 1.1. If V is a non-negative potential in R n with n 3 then for the operator H V = Δ V, N 0 (H V ) C n V (x) R n/2 dx. (1.1) n This was reduced to the following theorem. Theorem 1.5. For any bounded domain R n, denote by λ k () the k-th eigenvalue of the operator L V, = 1 Δ (with the Dirichlet boundary V condition on ). Assume that there is a Radon measure ν in R n and α > 0 such that, for all bounded domains, λ 1 () ν () α. (1.15) Then, for any positive integer k and any precompact open set, ( ) α k λ k () c, (1.16) ν () where c = c (α) > 0. 20

22 1.6 Nash inequality For the proof of Theorem 1.5 we need a Nash type inequality. Lemma 1.6 Assume that (1.15) holds, that is, for all precompact open sets, λ 1 () ν() α. Then, for all such and non-negative f D, where c = 2 2α 1. ( ) 1+α ( α E (f) c f 2 dμ fdμ fdν), (1.19) Remark. If V 1 then both μ and ν are Lebesgue measures, α = 2/n, and (1.19) becomes which is a classical Nash inequality. E (f) f 2+4/n 2 f 4/n 1, 21

23 Proof. Fix s > 0 and observe that E ( ) (f s) + E (f). (1.20) Set s := {x : f(x) > s} and note that supp (f s) + s. f s (f-s) + 0 s 22

24 that Since It follows from the variational property of λ 1 ( s ) and from (1.20), (f s) 2 + dμ = (f s) 2 + dμ E ( ) (f s)+ s λ 1 ( s ) we obtain by hypothesis ν ( s ) 1 s 1 λ 1 ( s ) ν ( s) α s α Substituting into (1.21) and using fdν ( f 2 2sf (f s) 2 +, fdν) a. E (f) λ 1 ( s ). (1.21) we obtain ( α f 2 dμ 2s fdμ s fdν) α E (f). (1.22) 23

25 Let us choose s from the condition 2s fdμ = 1 2 With this s we obtain whence ( 1 f 2 dμ 2 and (1.19) follows. ( 1 4 f 2 dμ fdμ f 2 dμ. ) α ( fdν) α E (f) ) 1+α ( ) α ( α f 2 dμ 2 2α+1 fdμ fdν) E (f), 24

26 1.7 Proof of Theorem 1.5 In the proof we work with the heat semigroup {P t } t 0 of the operator L V,, that is defined by Pt = e tl V,. Since L V, is a self-adjoint non-negative definite operator in L 2 (, μ), the operator Pt is bounded self-adjoint operator in L 2 (, μ) for any t 0. In fact, it is an integral operator: Pt f (x) = p t (x, y) f (y) dμ (y) where p t (x, y) is the heat kernel of L V,. We will use the following general properties of the heat kernel: 1. positivity: p t (x, y) 0; 2. the symmetry: p t (x, y) = p t (y, x) ; 3. the semigroup identity p t (x, z) p s (z, y) dμ (z) = p t+s (x, y) ; 25

27 4. the total mass inequality: p t (x, y) dμ (y) 1. The last step before the proof of Theorem 1.5 is the following lemma. Lemma 1.7 If (1.15) holds, that is, λ 1 () ν () α, then, for any precompact open set, p Cν () t (x, x) dμ (x). (1.23) t1/α where C = C (α). Proof. Fix s > 0, x and consider a function f = p s (x, ) and set u t = Pt f, that is, u t (y) = p t (y, z) f (z) dμ (z) = p t+s (x, y). 26

28 Then we have u 2 t dμ = p t+s(x, y)p t+s (y, x) dμ(y) = p 2(t+s) (x, x). On the other hand, by the Nash inequality we have Using and ( u 2 t dμ u t dμ = ) α α+1 u t dμ u t dν [CE (ut )] 1 α+1. p t+s (x, y) dμ (y) 1, (1.24) ( ) d E (u t ) = (L V, u t, u t ) μ = dt u t, u t = 1 d μ 2 dt (u t, u t ) μ we obtain u 2 t dμ ( ) α [ α+1 u t dν C 2 27 ] 1 d u 2 α+1 t dμ. dt

29 Recall that u t depends in fact on x. Setting v t (x) := u 2 t dμ = p 2(t+s) (x, x), rewrite the previous inequality in the form ( ) α [ α+1 v t (x) u t dν C 2 ] 1 v α+1 t. (1.25) t Integrating (1.25) against dμ(x) and using the Hölder inequality F α α+1 G 1 α+1 dμ [ ] α [ ] 1 α+1 α+1 F dμ Gdμ, we obtain v t (x) dμ (x) [ ] α [ α+1 u t dν C ] 1 v α+1 t dμ(x) 2 t }{{}}{{} [ F u t dνdμ(x) 28 ] α G α+1 [ C 2 ] 1 vt t dμ(x) α+1.

30 Observe that (1.24) implies ( u t (x, )dνdμ(x) = Denoting w (t) := we obtain from above v t (x) dμ (x) = w(t) ν() α α+1 ) u t (x, )dμ(x) dν dν = ν(). (1.26) ( C 2 p 2(t+s) (x, x) μ (x), ) 1 dw α+1. (1.27) dt Solving this differential inequality by separation of variables, we obtain w(t) C ν (). t 1/α Finally, choosing s = t we obtain p 4t (x, x) μ (x) C ν (), which was t 1/α to be proved. 29

31 Proof of Theorem 1.5. We need to show that ( ) α k λ k () c. ν () Note that p t (x, x) dμ (x) = trace P t. On the other hand, all the eigenvalues of Pt are equal to e tλk(), whence trace Pt = e tλk(). Hence, applying (1.23), we obtain k=1 e tλk() k=1 Cν () t 1/α. Restricting the summation to the first k terms, we obtain ke tλ k() 30 Cν () t 1/α

32 whence Choosing t from the condition λ k () 1 t ln kt1/α Cν (). that is, we obtain t = kt 1/α Cν () = e, ( Ce ν () ) α, k λ k () 1 ( ) α 1 t = k, Ce ν () which finishes the proof of Theorem

33 1.8 Minimal surfaces Let M be a two-dimensional manifold immersed in R 3 as an oriented minimal surface. The Riemannian metric on M is induced by the Euclidean structure of R 3. Denote by α the Riemannian area on M. For any function f C0 (M) and a real parameter ε, consider a deformation of M given by the mapping x x+εf(x)ν(x) where ν (x) is the unit normal vector field on M compatible with the orientation. Since M is a minimal surface, the first variation δα(f) of the area functional vanishes. For the second variation, the following formula is known: δ 2 α(f) = ( f 2 + 2Kf 2 )dα, (1.28) M where K = K(x) is the Gauss curvature of M at the point x M (since M is minimal, K(x) 0). If δ 2 α(f) 0 for all f then the minimal surface M is called stable. In particular, all area minimizers are stable. However, in general a minimal surface is not necessarily stable. By definition, the stability index ind(m) is the maximal dimension of a linear subspace V of C 0 (M) such that δ 2 α(f) < 0 for any f V \ {0}. 32

34 In other words, ind(m) = N 0 (H V ) where H V = Δ + 2K and Δ is the Laplace-Beltrami operator on M. It turns out that for this specific potential V = 2K the upper bound of Theorem 1.1 is satisfied. Theorem 1.8 (AG, Yau 2003) For any immersed oriented minimal surface M in R 3, we have ind(m) C K dα, (1.29) where C is an absolute constant. The proof goes in the same way as the one of Theorem 1.1 using Theorem 1.5. Using specific properties of Gauss curvature, we prove for the operator L V, = 1 Δ in M the eigenvalue estimate V M λ 1 () cμ () 1, 33

35 where dμ = K dα. By Theorem 1.5 this implies λ k () c k μ () and then as in the proof of Theorem 1.1, N 0 (H V ) Cμ (M) that is (1.29). 34

36 2 Lower estimates in R 2 Here we estimate N 0 (H V ) in R A counterexample to the upper bound In the case n = 2, the estimate (1.1) of Theorem 1.1 becomes N 0 (H V ) C V (x) dx, R 2 which however is wrong. To see that, consider in R 2 the potential V (x) = 1 x 2 ln 2 x if x > e and V (x) = 0 if x e. For this V we have V (x) dx <, R 2 35

37 whereas Neg (H V ) =. Indeed, consider the function f (x) = ( ) 1 ln x sin ln ln x 2 that satisfies in the region { x > e} the differential equation Δf + 1 V (x) f = 0. 2 For any positive integer k, function f has constant sign in the ring k := {x R 2 : πk < 12 } ln ln x < π (k + 1), and vanishes on k. For each function f k = f1 k we have E V (f k ) = f k 2 dx V fk 2 dx k k = f k Δf k dx V fk 2 dx k k = 1 V fk 2 dx < 0. 2 k 36

38 The same inequality holds for linear combination of functions f k since the intersection of their supports has measure 0. Hence, the space V = span {f k } has infinite dimension and E V (f) < 0 for all non-zero f V, which implies N 0 (H V ) =. In fact, one can show that no upper bound of the form N 0 (H V ) V (x) W (x) dx R 2 can be true, no matter how we choose a weight W (x). 37

39 2.2 Lower bound of N 0 (H V ) It turns out that in the case n = 2, instead of an upper bound, a lower bound in (1.1) is true. Theorem 2.1 (AG, Netrusov, Yau, 2004) For any non-negative potential V in R 2, N 0 (H V ) c V (x) dx R 2 (2.1) with some absolute constant c > 0. Let us describe an approach to the proof. Since N 0 (H V ) = sup {dim V : V D R 2 and E V (f) < 0 f V \ {0}}, it suffices to construct a subspace V of D R 2 such that E V is negative on V and dim V c V (x) dx. R 2 We will construct V as span {f k } where {f k } N k=1 is a sequence of functions with disjoint compact supports such that E V (f k ) < 0. Then E V (f) < 38

40 0 will be true for any non-zero function f span {f k }, and dim V = N. Hence, it suffices to construct a sequence {f k } N k=1 of functions with compact disjoint supports such that, for any k = 1,..., N, f k 2 dx < V fk 2 dx, R 2 R 2 and N c V (x) dx. R 2 Each function f k will be constructed as follows. Fix two reals 0 < r < R and consider the annulus A = { x R 2 : r < x < R } and denote by 2A the annulus 2A = {x R 2 : 12 } r < x < 2R. 39

41 Consider the following function 1, x A, 0, x / 2A, f (x) = 1 2 x ln, r x r, ln 2 r 2, R x 2R. 1 ln 2 ln 2R x x2 f=0 2A A f=1 f=0 r/2 r R 2R x1 40

42 This function f is harmonic in each of the four domains, whence we obtain f 2 dx = R 2 { r 2 x r } f 2 dx + f 2 dx {R x 2R} = { r 2 x r } f f ν dl + f f {R x 2R} ν dl = f (r) 2πr f (R) 2πR 1 = (ln 2) r 2πr + 1 (ln 2) R 2πR = 4π ln 2 < 20. Suppose that we have a sequence of annuli {A k } N k=1, with different centers and different radii, but such that the sequence {2A k } N k=1 is disjoint. Then, defining f k for each pair (A k, 2A k ) as above, we obtain a sequence of functions with disjoint supports and with f k 2 dx < 20. R 2 41

43 Note that V fk 2 dx V dx. R 2 A k Hence, the condition f R 2 k 2 dx < V f 2 R 2 k dx will be satisfied if V dx 20. A k Consider again measure μ given by dμ = V dx and restate our problem as follows: construct N annuli {A k } N k=1 such that (i) {2A k } N k=1 are disjoint, (ii) μ (A k ) 20 for each k, (iii) and N cμ (R 2 ). Of course, if μ (R 2 ) < 20 then such a sequence cannot be constructed. In this case we argue differently. Choose some 0 < r < R and consider 42

44 the function f (x) = 1, x r 0, x R, 1, r x R. ln R r ln R x x 2 f=0 f=1 r R x 1 43

45 while For this function R 2 f 2 dx = f (r) 2πr = 2π ln R r V f 2 dx R 2 { x r} V dx. Taking r and R r large enough, we obtain R 2 f 2 dx < R 2 V f 2 dx whence N 0 (H V ) 1. If μ (R 2 ) = R 2 V dx is bounded by some constant, say 20, then we obtain N 0 (H V ) cμ (R 2 ) just by taking c small enough. Hence, in the main part we can assume that μ (R 2 ) is large enough. In this case, the sequence of annuli satisfying (i)-(iii) can be always constructed. In fact, the positive answer is given by the following abstract theorem. 44

46 Theorem 2.2 Let (X, d) be a metric space and μ is a non-atomic Borel measure on X. Assume that the following properties are satisfied. 1. All metric balls B (x, r) = {y X : d (x, y) < r} are precompact. 2. There exists a constant M such that, for any ball B (x, r) there is a family of at most M balls of radii r/2 that cover B (x, r). Then there is a constant c = c (M) > 0 such that, for any 0 < v < μ (X) there exists at least c μ(x) annuli {A v k } such that (i) {2A k } are disjoint (ii) and μ (A k ) v for any k. Of course, R 2 satisfies all the hypotheses of Theorem 2.2. Taking v = 20 we obtain that if μ (R 2 ) > 20 then there exists at least c μ (R 2 ) annuli satisfying (i) and (ii), which finishes the proof of Theorem 2.1. We leave Theorem 2.2 without proof, only mentioning that it can be regarded as a sophisticated version of the ball covering argument. Note also that annuli in the statement cannot be replaced by balls. 45

47 2.3 Estimates of eigenvalues on S 2 Let us show one more application of Theorem 2.2. Theorem 2.3 Let λ k, k = 1, 2,..., be the k-th smallest eigenvalue of the Laplace-Beltrami operator Δ on (S 2, g), where g is an arbitrary Riemannian metric on S 2. Then, for any k, λ k C k 1 μ (S 2 ), (2.2) where C is a universal constant and μ is the Riemannian volume of the metric g. In fact, this theorem holds also for any closed Riemann surface, where the constant C depends also on the genus of the surface. However, the general case follows from the estimate for S 2. Note that λ 1 = 0 so that the case k = 1 is trivial. For k = 2 Theorem 2.3 was proved by Hersch in 1970 for the sphere and then for any Riemann surface by Yang and Yau in For a general k, Yau stated (2.2) as a conjecture, which was proved by Korevaar in

48 The main point of (2.2) that the constant C does not depend on the Riemannian metric g. The metric enters (2.2) only through the total area μ (S 2 ). This is essentially two-dimensional phenomenon as such estimates do not hold in higher dimensions. Let us show how Theorem 2.3 can be obtained from Theorem 2.2. Consider the counting function for Δ on (S 2, g): N λ = # {j 1 : λ j < λ}. Note that λ k < λ will follow from N λ k. We will prove that, for all λ > 0, N λ C 1 μ ( S 2) λ. (2.3) If (2.3) is already proved, then choosing here λ = C we obtain N λ k and, hence, which proves (2.2). λ k < λ = C k μ (S 2 ) 2C k 1 μ (S 2 ), k μ(s 2 ), where k 2, 47

49 Let us prove (2.3) for any λ > 0. variational characterization N λ = sup { dim V : V D S 2, E (f) < λ f 2 2 The counting function admits f V \ {0}} where E (f) = f 2 g dμ and f 2 2 = f 2 dμ. S 2 S 2 Hence, it suffices to construct at least N = C 1 μ (S 2 ) λ functions f with disjoint supports and with E (f) < λ f 2 2. If λ is small enough, namely, if C 1 μ (R 2 ) λ 1 then we need to construct only one function, and it always exists: f 1. Hence, we can assume that λ > C. μ(s 2 ) Any metric g on S 2 is conformally equivalent to the canonical metric g 0 on S 2. Denote by μ 0 the canonical Riemannian measure on S 2. Note that the energy is a conformal invariant: E (f) = f 2 g dμ = f 2 g 0 dμ 0. S 2 S 2 Let d be the geodesic distance on (S 2, g 0 ). As in R 2, one can show that, for any annulus A on S 2 (with respect to d) one can construct a test 48

50 function f supported in 2A and such that f A = 1 and E (f) < K where K is some constant. On the other hand, f 2 2 f 2 dμ = μ (A), A so that E (f) < λ f 2 2 will follow from K λμ (A). Hence, we need to construct at least N = C 1 μ (S 2 ) λ annuli A k on S 2 so that 2A k are disjoint and μ (A k ) K λ. Let us emphasize that measure μ is defined by the metric g, whereas the annuli are defined using the distance function of g 0. Let us apply Theorem 2.2 to the metric space (S 2, d) with measure μ. Set v := K < λ C 1 Kμ (S 2 ). Choosing C > K, we have v < μ (S 2 ) so that Theorem 2.2 can be applied. Hence, we obtain at least c μ (S 2 ) = c μ v K (S2 ) λ annuli A k with disjoint 2A k and with μ (A k ) v = K λ, which finishes the proof of (2.3) with C = K c. 49

51 3 Upper estimate in R Statement of the result Consider a tiling of R 2 into a sequence of annuli {U n } n Z defined by U n n<0 = {e 2 n < x < e 2 n 1 }, U 0 = {e 1 < x < e}, U n n>0 = {e 2n 1 < x < e 2n } x 2 e -1 e e 2n-1 e 2n x 1 U n n<0 U 0 U n n>0 50

52 Given a potential (=a non-negative L 1 loc -function) V (x) on R2 and p > 1, define for any n Z the following quantities: A n = V (x) (1 + ln x ) dx, B n = V p (x) x 2(p 1) dx U n {e n < x <e n+1 } The main result of this section is the following theorem. (3.1) Theorem 3.1 (AG, N.Nadirashvili, 2012) For any potential V in R 2 and for any p > 1, we have Neg (V ) 1 + C An + C B n, (3.2) {n Z:A n >c} {n Z:B n >c} where C, c are positive constants depending only on p. The additive term 1 in (3.2) reflects a special feature of R 2 : for any non-zero potential V, there is at least 1 negative eigenvalue of H V, no matter how small are the sums in (3.2), as it was shown in the course of the proof of Theorem /p

53 Let us compare (3.2) with previously known upper bounds. A simpler (and coarser) version of (3.2) is Neg (V ) 1 + C V (x) (1 + ln x ) dx + C B n. (3.3) R 2 n Z Indeed, if A n > c then A n c 1/2 A n so that the first sum in (3.2) can be replaced by n Z A n thus yielding (3.3). The estimate (3.3) was obtained by Solomyak in In fact, he proved a better version: Neg (V ) 1 + C A 1, + C n Z B n, (3.4) where A denotes the whole sequence {A n } n Z and A 1, is the weak l 1 -norm (the Lorentz norm) given by A 1, = sup s# {n : A n > s}. s>0 Clearly, A 1, A 1 so that (3.4) is better than (3.3). 52

54 However, (3.4) also follows from (3.2) using an observation that A 1, sup s 1/2 An 4 A 1,. s>0 {A n >s} In particular, we have An 4c 1/2 A 1,, {A n >c} so that (3.2) implies (3.4). As we will see below, our estimate (3.2) provides for certain potentials strictly better results than (3.4). In the case when V (x) is a radial function, that is, V (x) = V ( x ), the following estimate was proved by physicists Chadan, Khuri, Martin, Wu in 2003: Neg (V ) 1 + V (x) (1 + ln x ) dx. (3.5) R 2 Although this estimate is better than (3.3), we will see that our main estimate (3.2) gives for certain potentials strictly better estimates than (3.5). 53

55 Another upper estimate for a general potential in R 2 was obtained by Molchanov and Vainberg in 2010: Neg (V ) 1 + C V (x) ln x dx + C V (x) ln ( 2 + V (x) x 2) dx, R 2 R 2 (3.6) where x = e + x. However, due to the logarithmic term in the second integral, this estimate never implies the linear semi-classical asymptotic Neg (αv ) O (α) as α, (3.7) that is expected to be true for nice potentials. Observe that the Solomyak estimates (3.3) and (3.4) are linear in V so that they imply (3.7) whenever the right hand side is finite. Our estimate (3.2) gives both linear asymptotic (3.7) for nice potentials and non-linear asymptotics for some other potentials. Let us emphasize two main novelties in our estimate (3.2): using the square root of A n instead of linear expressions, and the restriction of the both sums in (3.2) to the values A n > c and B n > c, respectively, which allows to obtain significantly better results. 54

56 The reason for the terms A n in (3.2) can be explained as follows. Different parts of the potential V contributes differently to Neg (V ). The high values of V that are concentrated on relatively small areas, contribute to Neg (V ) via the terms B n, while the low values of V scattered over large areas, contribute via the terms A n. Since we integrate V over annuli, the long range effect of V becomes similar to that of an onedimensional potential. In R 1 one expects Neg (αv ) α as α which explains the appearance of the square root in (3.2). By the way, the following estimate of Neg (V ) in R 1 + was proved by Solomyak: Neg (V ) 1 + C an (3.8) where n=0 a n = V (x) (1 + x ) dx I n and I n = [2 n 1, 2 n ] if n > 0 and I 0 = [0, 1]. Clearly, the sum an here resembles A n in (3.2), which is not a coincidence. In fact, our method allows to improve (3.8) by restricting the sum to those n for which a n > c. 55

57 Returning to (3.3), one can apply a suitable Hölder inequality to combine the both terms of (3.3) in one as follows. Assume that W (r) is a positive monotone increasing function on (0, + ) that satisfies the following Dini type condition both at 0 and at : 0 r ln r p p 1 dr W (r) 1 p 1 <. (3.9) Then ( ) 1/p Neg (V ) 1 + C V p (x) W ( x ) dx, (3.10) R 2 where the constant C depends on p and W. Here is an example of a weight function W (r) that satisfies (3.9): W (r) = r 2(p 1) ln r 2p 1 ln p 1+ε ln r, (3.11) where ε > 0. In particular, for p = 2, we obtain the following estimate: ( ) 1/2 Neg (V ) 1 + C V 2 (x) x 2 ln x 3 ln 1+ε ln x dx. (3.12) R 2 56

58 3.2 Examples Example 1. Assume that, for all x R 2, V (x) α x 2 for a small enough positive constant α. Then, for all n Z, ( B n α {e n < x <e n+1 } ) 1/p 1 x 2 dx α so that B n < c and the last sum in (3.2) is void, whence we obtain Neg (V ) 1 + C V (x) (1 + ln x ) dx. (3.13) R 2 The estimate (3.13) in this case follows also from the estimate (3.6) of Molchanov and Vainberg. 57

59 Example 2. Assume that a potential V satisfies the following condition: for some constant K and all n Z, sup V K inf V. (3.14) {e n < x <e n+1 } {e n < x <e n+1 } For such potential we have B n {e n < x <e n+1 } V dx, (3.15) so that (3.3) implies Neg (V ) 1 + C V (x) (1 + ln x ) dx + C V (x) dx, R 2 R 2 where the constant C depends also on K. Of course, the second term here can be absorbed by the first one thus yielding (3.13) with C = C (K). The estimate (3.13) in this case can be obtained from the estimate (3.5) of Chadan, Khuri, Martin, Wu by comparing V with a radial potential. 58

60 Example 3. Let V (x) = α x 2 ( 1 + ln 2 x ), where α > 0 is small enough. Then as in the first example B n < c, while A n can be computed as follows: for n > 0 e 2 n e 2 n α A n e 2n 1 r 2 ln 2 (ln r) rdr = α d ln ln r α, (3.16) r e 2n 1 and the same for n 0, so that A n < c for all n. Hence, the both sums in (3.2) are void, and we obtain Neg (V ) = 1. This result cannot be obtained by any of the previously known estimates. Indeed, in the estimates of Chadan, Khuri, Martin, Wu and of Molchanov, Vainberg one has R 2 V (x) (1 + ln x ) dx =, and in the estimate (3.4) of Solomyak one has A 1, =. As will be shown below, if α > 1/4 then Neg (V ) can be. Hence, Neg (V ) exhibits a non-linear behavior with respect to the parameter α, which cannot be captured by linear estimates. 59

61 Example 4. Assume that V (x) is locally bounded and ( ) 1 V (x) = o x 2 as x. ln 2 x Similarly to the above computation we see that A n 0 and B n 0 as n, which implies that the both sums in (3.2) are finite and, hence, Neg (V ) <. This result is also new. Note that in this case the integral V (x) (1 + ln x ) dx R 2 may be divergent; moreover, the norm A 1, can also be as one can see in the next example. Example 5. Choose q > 0 and set V (x) = and V (x) = 0 for x e 2. For n 2 we have 1 x 2 ln 2 x (ln ln x ) q for x > e 2 (3.17) A n e 2 n 1 e 2n 1 r 2 ln 2 q (ln r) rdr = r (ln ln r) e 2 n e 2n 1 d ln ln r (ln ln r) q 1 n q, 60

62 and, by (3.15), B n e n+1 e n 1 r 2 ln 2 r (ln ln r) q rdr = Let α be a large real parameter. Then {A n (αv )>c} e n+1 e n d ln r ln 2 r (ln ln r) q 1 n 2 ln q n. A n (αv ) α n q, (3.18) and the condition A n (αv ) > c is satisfied for n Cα 1/q, whence we obtain Cα 1/q α An (αv ) C n C α ( α 1/q) 1 q/2 = Cα 1/q. q n=1 It is clear that n B n (αv ) α. Hence, we obtain from (3.2) Neg (αv ) C ( α 1/q + α ). (3.19) If q 1 then the leading term here is α. Combining this with (2.1), we obtain Neg (αv ) α as α. 61

63 If q > 1 then this follows also from (3.5) and (3.4); if q = 1 then only the estimate (3.4) of Solomyak gives the same result as in this case A n 1 n and A 1, <. If q < 1 then the leading term in (3.19) is α 1/q so that Neg (αv ) Cα 1/q. As was shown by Birman and Laptev, in this case, indeed, Neg (αv ) α 1/q as α. Observe that in this case A 1, =, and neither of the estimates previous estimates (3.3), (3.5), (3.4), (3.6) yields even the finiteness of Neg (αv ), leaving alone the correct rate of growth in α. Example 6. Let V be a potential in R 2 such that An (V ) + B n (V ) <. (3.20) n Z Applying (3.2) to αv, we obtain Neg (αv ) 1 + Cα 1/2 n Z n Z An (V ) + α B n (V ). n Z 62

64 Combining with the lower bound (2.1) and letting α, we see that cα V dx Neg (αv ) α B n (V ) + o (α), R 2 n Z in particular, Neg (αv ) α as α. Furthermore, if V satisfies the condition (3.14) then, using (3.15), we obtain a more precise estimate Neg (αv ) α V (x) dx as α. (3.21) R 2 For example, (3.20) is satisfied for the potential (3.17) of Example 5 with q > 2, as it follows from (3.18). By a more sophisticated argument, one can show that (3.21) holds also for q > 1. 63

65 Example 7. Set R = e 2m where m is a large integer, choose α > 1 4 and consider the following potential on R 2 V (x) = α x 2 ln 2 x if e < x < R and V (x) = 0 otherwise. Computing A n as in (3.16) we obtain A n α for any 1 n m, whence it follows that An = m An αm α ln ln R. n Z n=1 Also, we obtain by (3.15) B n a n 2, for 1 n < 2 m, whence B n (V ) n Z 2 m 1 n=1 α n 2 α. By (3.2) we obtain Neg (V ) C α ln ln R + Cα. (3.22) 64

66 Observe that both (3.4) and (3.5) give in this case a weaker estimate Neg (V ) Cα ln ln R. Let us estimate Neg (V ) from below. Considering the function ) ( α 14 ln ln x f (x) = ln x sin that satisfies in the region = {e < x < R} the differential equation Δf + V (x) f = 0, and counting the number N of rings { } α 14 ln ln x < π (k + 1) k := in, we obtain x R 2 : πk < Neg (V ) N α ln ln R (assuming that α >> 1 ). On the other hand, (2.1) yields Neg (V ) cα. 4 Combining these two estimates, we obtain the lower bound Neg (V ) c ( α ln ln R + α ), that matches the upper bound (3.22). 65

67 3.3 The energy form revisited We consider a somewhat different energy form than in R n, n 3. For any open set R 2, consider a function space { } ( ) F V, = f L 2 loc : f 2 dx <, V f 2 dx < and the quadratic form on F V, : E V, (f) = f 2 dx We will use the following quantity: V f 2 dx. (3.23) Neg (V, ) := sup {dim V : V F V, : E V, (f) 0 for all f V}. (3.24) Clearly, we have N 0 (H V ) Neg(V, R 2 ), but in R 2 we do not loose much when we estimate a larger quantity Neg instead of N 0. (Observe that F V,R 2 contains f = const and E (f) 0 so that Neg (V, R 2 ) 1, but as we know, N 0 (H V ) 1). Theorem 3.1 contains the estimate of Neg (V ) = Neg (V, R 2 ). 66

68 For bounded domains with smooth boundary, Neg (V, ) is equal to the number of non-positive eigenvalues of the Neumann problem in for Δ V. A useful feature of Neg (V, ) is subadditivity with respect to. We say that a sequence { k } of open sets k R 2 is a partition of if all the sets k are disjoint, k, and \ k k has measure 0. k 67

69 Lemma 3.2 If { k } is a partition of, then Neg (V, ) k Neg (V, k ). (3.25) The idea of the proof is the same as in the classical Weyl s argument: adding additional Neumann boundaries inside increases the space of test functions and, hence, the number of non-negative eigenvalues. 68

70 3.4 One negative eigenvalue in a disc Denote by D r the open disk of radius r in R 2, that is, D r = {x R 2 : x < r}, and set D 1 D. Lemma 3.3 For any p > 1 there is ε > 0 such that, for any potential V in D, V L p (D) ε Neg (V, D) = 1. Sketch of proof. Since always Neg (V, D) 1, we need only to prove that Neg (V, D) 1. We will prove that if u F V,D then u 1 in L 2 (D) and E V,D (u) 0 u = 0, which will imply that Neg (V, D) 1. Extend u F V,D to R 2 using the inversion Φ (x) = x : for any x 2 x / D, set u (x) = u (Φ (x)). By conformal invariance of energy, we have u 2 dx = 2 u 2 dx 2 V u 2 dx. R 2 D D 69

71 Choose a cutoff function ϕ such that ϕ D2 1, ϕ R 2 \D 3 = 0 and set u = uϕ. Then it follows that u 2 dx C V u 2 dx, D 4 D with some absolute constant C. Since u 1, one uses in the proof the Poincaré inequality in D in the form u L 2 C u L 2. Next, we have by Hölder inequality D ( V u 2 dx D ) 1/p ( V p dx D u 2p p 1 dx ) 1 1/p, and by Sobolev inequality ( ) 1 1/p ( ) 1 1/p u 2p p 1 dx u 2p p 1 dx C u 2 dx. D D 4 D 4 Combining the above three lines, we obtain ( ) 1/p u 2 dx C V p dx u 2 dx. (3.26) D 4 D D 4 70

72 Assuming that V L p (D) is small enough, we see that (3.26) is only possible if u = const. Since u 1 in L 2 (D), it follows that u 0. Corollary 3.4 Let be a domain in R 2 that is bilipschitz equivalent to D r. Then V p dx cr 2 2p Neg (V, ) = 1. (3.27) where c > 0 depends on p and on the Lipschitz constant of the mapping between D r and. Proof. Indeed, if = D r then (3.27) follows from Lemma 3.3 by scaling transformation. For a general one shows that Neg (V, ) Neg (CV, D r ) where V is the pull-back of V under the bilipschitz mapping L : D r where the constant C depends on the Lipschitz constant. 71

73 3.5 Negative eigenvalues in a square Denote by Q the unit square in R 2. Lemma 3.5 For any p > 1 and for any potential V in Q, where C depends only on p. Neg (V, Q) 1 + C V L p (Q), (3.28) Proof. It suffices to construct a partition P of Q into a family of N disjoint subsets such that 1. Neg (V, ) = 1 for any P; 2. N 1 + C V L p (Q). Indeed, if such a partition exists then we obtain by Lemma 3.2 Neg (V, Q) P Neg (V, ) = N, (3.29) 72

74 and (3.28) follows from the above bound of N. The elements of a partition will be of two shapes: it is either a square of the side length 0 < l 1 or a step, that is, a set of the form = A \ B where A is a square of the side length l, and B is a square of the side length l/2 that is attached to one of corners of A. l/2 l l In the both cases we refer to l as the size of. By Corollary 3.4, the condition 1 for such a set will follow from V p dx cl 2 2p. (3.30) 73

75 Apart from the shape, we will distinguish also the type of a set P of size l as follows: we say that - is of a large type, if V p dx > cl 2 2p, - is of a medium type if c l 2 2p < V p dx cl 2 2p, (3.31) - and is of small type if V p dx c l 2 2p. (3.32) Here c is the constant from (3.30) and c (0, c) will be chosen below. In particular, if is of small or medium type then Neg (V, ) = 1. The construction of the partition P will be done by induction. At each step i 1 of induction we will have a partition P (i) of Q such that 74

76 1. each P (i) is either a square or a step; 2. If P (i) is a step then is of a medium type. At step 1 we have just one set: P (1) = {Q}. At any step i 1, partition P (i+1) is obtained from P (i) as follows. If P (i) is small or medium then becomes one of the elements of the partition P (i+1). If P (i) is large, then it is a square, and it will be further partitioned into a few sets that will become elements of P (i+1). Denoting by l the side length of the square, let us first split into four equal squares 1, 2, 3, 4 of side length l/2 and consider the following cases. small small 1 small medium Case 1. If among 1,..., 4 the number of small squares is at most 2, then all sets 1,..., 4 become elements of P (i+1). 75

77 Case 2. If among 1,..., 4 there are exactly 3 small squares, say, 2, 3, 4, then we have V p dx = V p dx 3c \ ( ) 2 2p l = 3c 2 2p 2 l 2 2p < cl 2 2p, 2 where we choose c to satisfy 3c 2 2p 2 < c. On the other hand, we have V p dx > cl 2 2p. Therefore, by reducing the size of 1 (but keeping 1 attached to the corner of ) one can achieve the equality \ 1 V p dx = cl 2 2p. Hence, we obtain a partition of into two sets 1 and \ 1, where the step \ 1 is of medium type, while the square 1 can be of any type. Both 1 and \ 1 become elements of P (i+1). 76

78 Case 3. Let us show that all 4 squares 1,..., 4 cannot be small. Indeed, in this case we would have V p dx = 4 k=1 k V p dx 4c ( ) 2 2p l = ( 4c 2 2p 2) l 2 2p. 2 Let us choose c so small that 4c 2 2p 2 < c. Then the above estimate contradicts the assumption that is of large type. As we see from construction, at each step i only large squares get partitioned further, and the size of the large type squares in P (i+1) reduces at least by a factor 2. If the size of a square is small enough then it is necessarily of small type, because the right hand side of (3.32) goes to as l 0. Hence, the process will stop after finitely many steps. After sufficiently many steps we obtain a partition P where all the elements are either of small or medium types. In particular, we have Neg (V, ) = 1 for any P. 77

79 Let N be a number of elements of P. We need to show that N 1 + C V L p (Q). (3.33) At each step of construction, denote by L the number of large elements, by M the number of medium elements, and by S the number of small elements. Let us show that the quantity 2L + 3M S is non-decreasing during the construction. Indeed, at each step we split one large square, so that by removing this square, L decreases by 1. However, we add 78

80 new elements of partitions, which contribute to the quantity 2L+3M S as follows. 1. If is split into s 2 small and 4 s medium/large squares as in Case 1, then the value of 2L + 3M S has the increment at least (4 s) s = 6 3s If is split into 1 square and 1 step as in Case 2, then one obtains at least 1 medium set and at most 1 small, so that 2L + 3M S has the increment at least = 0. (Luckily, Case 3 cannot occur. In that case, we would have 4 new small squares so that L and M would not have increased, whereas S would have increased at least by 3, so that no quantity of the type C 1 L+ C 2 M S would have been monotone increasing). Since for the partition P (1) we have 2L + 3M S 1, this inequality will remain true at all steps of construction and, in particular, it is 79

81 satisfied for the final partition P. For the final partition we have L = 0, whence it follows that S 1 + 3M and, hence, N = S + M 1 + 4M. (3.34) Let us estimate M. Let 1,..., M be the medium type elements of P and let l k be the size of k. Each k contains a square k k of the size l k /2, and all the squares { k }M k=1 are disjoint, which implies that M lk 2 4. (3.35) Using the Hölder inequality and (3.35), we obtain ( M 2 M ) 1/p ( M ) 1/p ( M p M = l 4 1/p k=1 2 k l p k k=1 l 2 k k=1 k=1 l 2p p k Since by (3.31) c l 2 2p k < k V p dx, it follows that ( M ) 1/p ( 1/p M C V p dx C V dx) p. k Q k=1 80 k=1 l 2 2p k ) 1/p.

82 Combining this with N 1 + 4M, we obtain N 1 + C V L p (Q), thus finishing the proof. Corollary 3.6 Let be a domain in R 2 that is bilipschitz equivalent to D. Then ( 1/p Neg (V, ) 1 + C V dx) p, where C > 0 depends on p and on the Lipschitz constant of the mapping between D and. 81

83 3.6 One negative eigenvalue in R 2 Now we would like to obtain conditions for Neg (V, R 2 ) = 1 in terms of some weighted L 1 -norms. The method that we have used in the case n 3 (Proposition 1.3) was based on the operator L V = 1 Δ and V estimating of L 1 V in L 2 (R n, V dx). The hidden reason why it was possible is the existence of the positive c Green function g (x, y) = n of Δ. In fact, the operator L 1 x y n 2 V is given by L 1 V f = R n g (x, y) f (y) V (y) dy. The application of the Sobolev in the proof of Proposition 1.3 can be replaced by a direct estimate of the norm of this integral operator in L 2 (R n, V dx). In fact, the classical proof of the Sobolev inequality uses this approach. One of the difficulties in R 2 is the absence of a positive Green function of the Laplace operator. To overcome this difficulty, we introduce an auxiliary potential V 0 C0 (R 2 ), such that V 0 0 and V

84 Lemma 3.7 (AG, 2006) Operator H 0 = Δ + V 0 has a positive Green function g (x, y) that admits the following estimate g (x, y) ln x ln y + ln + 1 x y, (3.36) where x := e + x and means min. By Lemma 3.3 there exists V 0 such that Neg (V 0, R 2 ) = 1. Fix such V 0 and, hence, the Green function g (x, y) of H 0 for what follows. For a given potential V, define as measure ν by dν = V dx and consider the integral operator G V defined by G V f (x) = g (x, y) f (y) dν (y). R 2 Denote by G V the norm of G V in the space L 2 (R 2, ν). 83

85 Lemma 3.8 If G V 1 2 then Neg (V, R2 ) = 1. Sketch of the proof. The idea is that the operator G V is the inverse of the operator 1 H V 0 in L 2 (ν) so that G V 1 implies that the 2 spectrum of 1 H V 0 is confined in [2, ). This implies that H 0 2V in the sense of quadratic forms, that is, u 2 dx + V 0 u 2 dx 2 V u 2 dx R 2 R 2 R 2 for all u F V. If V is a subspace of F V where E V 0 then for any u V u 2 dx V u 2 dx. R 2 R 2 Combining the two lines, we obtain u 2 dx V 0 u 2 dx, R 2 R 2 that is, E V0 (u) 0. Taking sup dim V we obtain Neg ( V, R 2) Neg ( V 0, R 2) = 1. 84

86 The next step is estimating the norm G V in terms of V. g (x, y) is symmetric in x, y, we have a simple estimate G V sup g (x, y) dν (x), y R 2 Since which together with Lemma 3.7 leads to 1 G V C ln x dν (x) + C sup ln + dν (x). R 2 y R 2 R x y 2 However, G V admits a better estimate, as will be explained below. 85

87 3.7 Transformation to a strip It will be more convenient to estimate first Neg (V, S) where S is a strip in R 2 defined by S = { (x 1, x 2 ) R 2 : x 1 R, 0 < x 2 < π }. The strip S is the image of R 2 + under the conformal mapping z ln z: 2 + ln z π S 0 0 Let γ (x, y) be the push-forward of the Green function g (x, y) under this mapping, that is, γ (x, y) = g (e x, e y ). 86

88 Using the estimate (3.37) of g, it is possible to show that γ (x, y) C x 1 y 1 + C ln + 1 x y. (3.37) For example, x 1 arises from ln e x = ln e x 1+ix 2 = ln e x 1 = x 1. Consider also the corresponding integral operator Γ V f (x) = γ (x, y) f (y) dν (y), (3.38) S where measure ν is defined as above by dν = V (x) dx. Denote by Γ V the norm of Γ V in L 2 (S, ν). Lemma 3.8 implies the following. Lemma 3.9 Γ V 1 8 Neg (V, S) = 1. The main point in the proof is that the holomorphic mappings are conformal and, hence, preserve the Dirichlet integral. 87

89 3.8 Estimating Γ V For any n Z set Q n = S {n < x 1 < n + 1}, S n = S { 2 n < x 1 < 2 n 1} for n < 0, S 0 = S { 1 < x 1 < 1}, S n = S {2 n 1 < x 1 < 2 n } for n > 0, S -3 S -2 S -1 S 0 S 1 S 2 S 3 Q -8 Q -7 Q -6 Q -5 Q -4 Q -3 Q -2 Q -1 Q 0 Q 1 Q 2 Q 3 Q 4 Q 5 Q 6 Q 7 S x 1 a n (V ) = (1 + x 1 ) V (x) dx 2 n V (x) dx (3.39) S n S n ( ) 1/p b n (V ) = V p (x) dx. (3.40) Q n 88

90 Lemma 3.10 The operator Γ V admits the following norm estimate in L 2 (S, ν): Γ V C sup a n (V ) + C sup b n (V ). (3.41) n Z n Z Approach to the proof. Note that by (3.37) Γ V f (x) C (1 + x 1 y 1 ) f (y) V (y) dy S 1 +C ln + f (y) V (y) dy. (3.42) x y S The second integral operator can be estimated by the Hölder inequality: ( ( ) ) p 1/p 1 1 ln + V (y) dy ln + dy S x y B(x,1) x y ( 1/p V p (y) dy). B(x,1) S The first integral here is equal to a finite constant depending only on p, but independent of x. The second integral is bounded by C sup n b n (V ). 89

91 It is much more subtle to estimate the norm of the first integral operator in (3.42) via C sup n Z a n (V ). This problem is reduced to an one dimensional problem by integrating in the direction x 2. Then we apply a certain weighted Hardy inequality. We skip the details as the argument is quite lengthy. Corollary 3.11 There is a constant c > 0 such that sup a n (V ) c and n sup b n (V ) c Neg (V, S) = 1. n Proof. Assuming that the constant c here is small enough, we obtain from (3.41) that Γ V 1, whence by Lemma 3.9 Neg (V, S) =

92 3.9 Rectangles For all α [, + ), β (, + ] such that α < β, denote by P α,β the rectangle Note that P α,β S. P α,β = { (x 1, x 2 ) R 2 : α < x 1 < β, 0 < x 2 < π }. Lemma 3.12 For any potential V in a rectangle P α,β β α 1, we have with the length Neg (V, P α,β ) Neg (17V, S), where V is extended to S by setting V = 0 outside P α,β. Sketch of the proof. It suffices to show that any function u F V,P can be extended to F V,S so that u 2 dx 17 u 2 dx. (3.43) S P 91

93 Attach to P from each side one rectangle, say P from the left and P from the right, each having the length 4 (β α) (to ensure that the latter is > π). Extend function u to P by applying four times symmetries in the vertical sides, so that u 2 dx = 4 u 2 dx. P P x 2 x Φ'(x) P' P P'' Φ''(x) S x (α,0) (β,0) x 1 Then slightly reduce P by taking intersections with the circle of radii β α centered at (α, 0). Now we extend u from P to the left by using 92