An analytical proof of Hardy-like inequalities related to the Dirac operator

Transcription

1 An analytical poof of Hady-like inequalities elated to the Diac opeato Jean Dolbeault a,1 Maia J. Esteban a,1 Michael Loss b,2 Luis Vega c,1 a CEREMADE, Univesité Pais-Dauphine, Pais Cedex 16, Fance b School of Mathematics, Geogia Tech, Atlanta, GA 30332, USA c Univesidad del País Vasco, Depatamento de Matemáticas, Facultad de Ciencias, Apatado 644, Bilbao, Spain Abstact We pove some shap Hady type inequalities elated to the Diac opeato by elementay, diect methods. Some of these inequalities have been obtained peviously using spectal infomation about the Diac-Coulomb opeato. Ou esults ae stated unde optimal conditions on the asymptotics of the potentials nea zeo and nea infinity. Key wods: Hady inequality, Diac opeato, Optimal constants, Diac-Coulomb Hamiltonian, elativistic Hydogen atom 2000 MSC : Pimay : 35Q40, 35Q75, 46N50, 81Q10 ; Seconday : 34L40, 35P05, 47A05, 47F05, 47N50, 81V45, 81V55 1 Intoduction and main esults The uncetainty pinciple is without any doubt a fundamental attibute of quantum mechanics [17]. In the case of the Laplacian it states that fo all c 2003 by the authos. This pape may be epoduced, in its entiety, fo noncommecial puposes. 1 PICS CNRS, Pais-Bilbao and Euopean Pogams HPRN-CT # & Wok patially suppoted by U.S. National Science Foundation gant DMS Septembe 2003

2 functions f C 0 R 3, R 3 f 2 dx 1 4 R3 f 2 dx. 1 2 This inequality is also known as Hady s inequality. By scaling, the powe of the potential is seen to be optimal but also the constant 1 cannot be impoved. 4 Howeve, it is still possible to impove the inequality by adding lowe ode tems. In ecent yeas thee has been a geat effot to find optimal impoved Hadytype inequalities in the case of the Laplacian. The pioneeing wok in this diection is due to Bezis and Vázquez [7] in the case of Diichlet bounday conditions, and to Lieb and Yau [16] in the case without bounday conditions. Futhe impovements have been obtained in [1 6,11,18]. An analogue of this inequality fo a elativistic vesion of the Schödinge equation whee the Laplacian is eplaced by is an inequality due to Kato : 2 R3 f 2 f, f dx 2 π see [14,15]. In this inequality the powe and constant ae again optimal. An immediate consequence of this inequality is that the elativistic model of the hydogenic atom with kinetic enegy and with nuclea chage Z is stable if and only if ν := Zα π/2, whee α 1/ is the fine stuctue constant. Lieb and Yau [16] also discoveed some genealizations fo to balls. The Diac elativistic hydogenic atom is stable only if ν < 1. The Diac Hamiltonian is unbounded fom below and instability has to be intepeted in the diffeent, moe subtle, sense of a beakdown of selfadjointness of the Diac-Coulomb Hamiltonian. If the Coulomb singulaity is smeaed out, then the theshold fo stability is eached when the lowest eigenvalue in the gap eaches the uppe bound of the negative continuum. This happens in geneal fo lage values of ν. In the case of the Diac-Coulomb Hamiltonian, the stability is a consequence of the following Hady-type inequality. Theoem 1 [9] Let σ = σ i i=1,2,3 be the Pauli-matices : σ 1 = , σ 2 = 0 i i 0, σ 3 = Then fo evey ϕ H 1 R 3, C 2, R 3 σ ϕ ϕ 2 dx R3 ϕ 2 dx. 3 2

3 As in 1 and 2, the powes of and the constants ae optimal. Inequality 3 has been established using a chaacteization of the eigenvalues of a self-adjoint opeato in a gap of its essential spectum by means of a paticula min-max. See below and efe to [9] fo moe details. Fo othe esults on minmax chaacteizations of the eigenvalues of Diac opeatos, see [10,13,12,8]. By scaling, if we eplace ϕ by ε 1 ϕε 1 and take the limit ε 0, 3 implies that fo all ϕ H 1 R 3, C 2, R 3 σ ϕ 2 dx R3 ϕ 2 dx. 4 This inequality is slightly genealized fom take ϕ = g, 0 and conside independently the cases whee g takes eithe eal o puely imaginay values of the following inequality : fo all g H 1 R 3, C, R3 g 2 g 2 dx dx, 5 R 3 which is itself equivalent to 1 : take f = g. Note that the lagest space in which Inequality 3 holds is lage than H 1 R 3, C 2 and contained in H 1/2 R 3, C 2. Fo moe details, see [9]. In [9], the poof of 3 has been caied out by using explicit knowledge on the point-spectum of the Diac-Coulomb opeato H ν := i α + β ν, whee the matices β, α k M 4 4 C, k = 1, 2, 3, ae defined by α k = 0 σ k σ k 0, β = Id 0 0 Id. Id is the identity matix in C 2 and ν is a eal paamete taking its values in 0, 1. It is well-known [19] that fo any ν 0, 1 H ν can be defined as a selfadjoint opeato with domain D ν satisfying : H 1 R 3, C 4 D ν H 1/2 R 3, C 4 and spectum σh ν = σ ess H 0 { } λ ν 1, λ ν 2,, σ ess H 0 =, 1] [1, +, whee {λ ν k} k 1 is the nondeceasing sequence of eigenvalues of H ν, all contained in the inteval 0, 1 and such that : λ ν 1 = 1 ν 2, lim k + λν k = 1 fo evey ν 0, 1. Fo a lage set of potentials V with singulaities not stonge than Coulombic ones, moe pecisely, fo all those satisfying : ν lim V x = 0 and + c 1 V c 2 = supv, 6 3

4 with ν 0, 1, c 1, c 2 R, c 1, c 2 0, c 1 + c 2 1 < 1 ν 2, the following esult was poved in [9] : Theoem 2 [9] Let V be a adially symmetic function satisfying 6 and define λ 1 V as the smallest eigenvalue of H 0 + V in the inteval 1, 1. Then fo all ϕ H 1 R 3, C 2, R 3 σ ϕ λ 1 V V + 1 λ 1 V ϕ 2 dx V ϕ 2 dx. 7 R 3 Inequality 7 is achieved by the lage component, i.e., the two-spino made of the fist two complex valued components of the fou-spino, of any eigenfunction associated with λ 1 V. In paticula if V = ν, ν 0, 1, we get Coollay 3 [9] Fo any ν 0, 1, fo all ϕ H 1 R 3, C 2, R 3 σ ϕ ν 2 + ν ν 2 ϕ 2 dx ν R3 ϕ 2 dx. 8 This inequality is achieved in L 2 R 3, 1 dx 4. Inequality 3 is obtained fom 8 by taking the limit ν 1. Theoem 1 is theefoe a staightfowad consequence of Coollay 3. Note that 3 is not achieved in L 2 R 3, 1 dx 4. The aim of this pape is twofold. On the one hand, we give a diect analytical poof of Theoem 1 which does not use any a pioi spectal knowledge on the opeato H ν. On the othe hand, we pove moe geneal inequalities by showing that fo some continuous functions W > 1 a.e. and constants R > 0 and CR 0, the inequality R 3 σ ϕ W + ϕ 2 dx R 3 W ϕ 2 dx + CR ϕ 2 dµ R S R 9 holds fo all ϕ H 1 R 3, C 2. Hee µ R is the suface measue induced by Lebesgue s measue on the sphee S R := {x R 3 : = R}. Note that this inequality is elevant fo the Diac opeato with potential V = W. Impoved inequalities like 3 o 9 with the opeato σ eplaced by can easily be obtained by consideing sepaately the eal and the imaginay pats of the components of the two-spinos. We ae actually inteested in undestanding fo which functions W Inequality 9 holds and what is the optimal behavio of the function W nea 0 o nea +. By optimal at s = 0 o s = +, we mean optimal at each ode in the sense that we look fo a expansion of the fom W = k=1 c k W k such 4

5 that at each ode k 0, 0 W k0 +1 = ow 1 k 0 k=1 c k W k and c k0 +1 is the lagest possible constant. What we ae going to pove is a little bit involved: it is not clea that c k is a constant see Appendix C and we ae only going to pove that the maximum of its lim inf is achieved. As in Hady-like inequalities fo the Laplacian opeato on balls centeed at the oigin, we will see that the optimal behavio of the function W nea 0 is a logaithmic petubation of the constant 1. Moe pecisely the optimal behavio nea the oigin is given by functions of the fom W x = k=1 X 1 2 X k 2, whee X 1 s := a logs 1 fo some a > 1, X k s := X 1 X k 1. The functions X k and W ae well defined fo = s < e a 1 see Appendix A fo basic popeties of the functions X k. These asymptotics ae optimal in the above sense, with W k = X 1 2 X k 2 and lim inf c k = 1. On the othe 8 hand, as +, the optimal gowth fo W is given by, i.e., the fist tem in the l.h.s. of 9 does not help. Theoem 4 Assume that fo some R > 0 Inequality 9 holds fo evey spino ϕ C 0 R 3, C 2, whee W is a adially symmetic continuous function fom R + to R +. Assume moeove that W 0 > 0 and W is nondeceasing in a neighbouhood of 0 +. Then W 0 1, and fo all k 1, lim inf s 0 + W s k lim sup W s/s 1 s + X1s 2 Xj 2 s X1 2 s Xk+1 2 s As soon as W 1, CR must be negative. Moeove, thee ae continuous functions W 1 fo which 9 holds with some R > 0, CR < 0, such that lim s + W s/s = 1 and 10 holds with equality fo all k 1. Note that this esult is independent of the paticula value of a > 1 which appeas in the definition of the functions X k. The fact that we have to intoduce a discontinuity at some s = R is not contadictoy with the known esults fo the usual Hady inequality, fo which bounded domains ae consideed and q is taken lage enough. In Section 2 we give a diect analytical poof of Coollay 3 togethe with seveal auxiliay esults. We ecall that Theoem 1 is a staightfowad consequence of Coollay 3. Section 3 is devoted to the poof of Theoem 4. Popeties of the functions X k, an existence esult fo a singula ODE needed 5

6 fo Theoem 4 and an example illustating why we have to conside a lim inf in this theoem ae given in thee appendices. 2 Poof of Theoem 1 In this Section, we actually pove Coollay 3, which is slightly moe than Theoem 1. Fist we fix some notations. The spino ϕ = ϕ 1 ϕ 2 takes its values in C 2 and by ϕ 2, ϕ 2 and σ ϕ 2 we denote, espectively, the quantities ϕ ϕ 2 2, Σ 3 k=1 k ϕ k ϕ 2 2 and 3 ϕ ϕ 2 i 2 ϕ ϕ 1 + i 2 ϕ 1 3 ϕ 2 2. Futhe, we notice that the Pauli matices ae Hemitian and satisfy the following popeties : σ j σ k + σ k σ j = 2 δ jk Id, j, k = 1, 2, 3. With a standad abuse of notations, each time a scala δ appeas in an identity involving opeatos acting on two-spinos, it has to be undestood as δ Id, whee Id is the identity opeato. On the othe hand, fo all vectos a, b C 3, we have σ aσ b = a b + i σ a b. Applying this fomula to a = x and b = i, we obtain the following expession fo the commutato of σ and σ x : [σ, σ x] = x x + 2 σ L = σ L, whee L = i x is the obital angula momentum opeato. The main point to note hee is that L acts only on the angula vaiables. Fo simplicity, fo any function h : R + R, we denote the functions x h and x h by h and h espectively. Now if such a function h is diffeentiable a.e. in R + and continuous in [0, R R, +, we have [σ, σ x h] = h σ L h + R [h] R δ R = σ L h + h + h + R [h] R δ R, whee by [h] R := hr + hr we denote the possible jump of h at R and δ R is the Diac delta function at = R, in spheical coodinates. The spectum of the opeato 1 + σ L is the discete set {±1, ±2, } 6

7 see [19]. This can be seen by noticing that 1 + σ L = J 2 L , J = L + σ 2. Then, the fact that the spectum of J 2 esp. L 2 is the set {jj + 1 ; j = 1, 3,... } esp. {ll + 1 ; l = j ± 1, j = 1, 3,... } poves the above esult The main point hee is that 0 is not in the spectum of 1 + σ L. If we denote by X + esp. X the positive esp. negative spectal space of 1 + σ L, and by P ± = ± 1+σ L 1+σ L the coesponding pojectos on H 1 R 3, C 2, fo all ϕ H 1 R 3, C 2, fo all h as above, ϕ +, [σ, σ x h] ϕ + 3h + h ϕ + 2 dx + R [h] R R 3 ϕ + 2 dµ R, S R ϕ, [σ, σ x h] ϕ h + h ϕ 2 dx + R [h] R R 3 ϕ 2 dµ R, S R whee ϕ ± := P ± ϕ. By Cauchy-Schwatz inequality, fo any measuable function g : R + R +, R 3 3h + h ϕ + 2 dx g σ ϕ + 2 dx R 3 11 R3 2 h 2 + ϕ + 2 dx R [h] R ϕ + g S 2 dµ R, R whee again we abbeviate g by g. Define now W and m : R + R by gs = s W s + s and ms = s hs, and assume that W is positive on R +. With the same notation as above, we can ewite 11 as R 3 1 s 2m + s m s + W m 2 s= ϕ + 2 dx 12 R3 σ ϕ W dx R [h] R ϕ + S 2 dµ R. R Similaly, fo ϕ := P ϕ, we find 7

8 R 3 1 s 2m s m s + W m 2 s= ϕ 2 dx 13 R3 σ ϕ W dx + R [h] R ϕ S 2 dµ R. R Note that fo any measuable adial function b, the spaces X + and X ae also othogonal in L 2 R 3 ; b dx. Moeove, we have Lemma 5 P σ 2P+ P + σ 2P 0 in H 1 R 3, C 2. Poof. A diect computation shows that the anti-commutato {σ, 1 + σ L} = 0, i.e., σ anticommutes with 1 + σ L. Hence σ 2 commutes with 1 + σ L. Now, let Φ ± X ± be two eigenfunctions of 1 + σ L with eigenvalues λ ±, λ < 0 < λ +. Then, σ Φ, σ Φ + = 1 λ + Φ, σ σ LΦ + = 1 Φ, 1 + σ Lσ 2 Φ + λ + = σ LΦ, σ 2 Φ + λ + = λ λ + Φ, σ 2 Φ +, which is impossible except if σ Φ, σ Φ + = 0. Adding 12 and 13, we get the following esult. Poposition 6 Let W be a positive measuable function on R + and conside two functions m ± : R + R such that the maps s m ± s/s ae continuous on [0, R R, + and diffeentiable a.e. on R +. Then fo any ϕ H 1 R 3, C 2, R3 σ ϕ W dx [m ± ] R P ± ϕ ± S 2 dµ R R 1 2m± ± s m ± R 3 ± s + W m± 2 s P ±ϕ 2 dx. s= 8

9 In ode to pove Inequality 9, we have to find two functions, m + and m, and a continuous function W 1 such that fo s R + a.e., This means 2m ± ± s m ± W + s m 2 ± W s. 1 W 2m ± ± s m ± s m 2 ± + s m 2 ± Moeove W has to be as lage as possible nea the oigin and nea infinity, i.e. optimal in the sense of Section 1. Then 9 follows with CR = max[m ] R, [m + ] R. In the sequel, fo evey function m as in Poposition 6, we will use the notation W ±,m := 2m ± s m s m 2 + s 1 + m Poof of Theoem 1 and Coollay 3. This is simply done by choosing m + m 1 o m + m 1 1 ν 2 in Poposition 6. Since in both cases the ν functions s m ± /s ae continous in R +, thee is no suface integal tem in those inequalities : CR = 0. Remak The above aguments leading to the poof of Theoem 1 can also be viewed as a completing the squae stategy : fo all ϕ H 1 R 3, h, g CR +, R +, h diffeentiable a.e. in R +, it is clea that R 3 g σ P± ϕ ± σ x h g P ± ϕ 2 dx 0. Expanding the squaes in the above expessions, integating by pats the coss tems and adding the two inequalities, we find 9 if W satisfies Poof of Theoem Diect estimates We stat this section by finding the optimal behavio nea the oigin and nea infinity fo continuous functions W fo which 9 holds, whee W is given o not by some function m as defined in 15. This is done in a seies of intemediate esults. 9

10 Lemma 7 Let W be any function satisfying W 1 on R + and fo which Inequality 9 holds. Then, necessaily, lim W s = 1 and lim sup s 0 + s + W s s 1. Poof. By assumption, W 1. If we had W 0 > 1, it would be easy to contadict the fact that 1 is the best constant in 4 and 5. As s +, the esult follows fom the simple obsevation that by scaling we can easily constuct functions ϕ n := n 3/2 ϕ /n such that R 3 ϕ n 2 dx = 1, R 3 σ ϕ n 2 dx 0 as n +, so that the gadient tem does not play any ole. Poposition 8 Let m ± C[0, δ fo some δ > 0. Conside W ±,m ± defined accoding to 15 and let W := minw +,m +, W,m be a function fo which W 1. Then 9 holds, m ± 0 = 1, m ± 1 in a neighbouhood of s = 0 + and lim sup m ± s 1 logs < s 0 + Poof. We pove this fo m +, the poof fo m being identical. Let us wite m + = n + 1. Then 14 is equivalent to 0 W 1 = s n n 2 2 s n s n n + n Fom this inequality, we infe that n nn n 2 /s, so that thee ae two possibilities fo the behavio of n nea 0 : eithe n is monotone and lim s 0 + ns = a, + ], o n oscillates nea 0 in the inteval 2, 0. The latte case is impossible because on the sequence of local minima appoaching 0, the.h.s. of 17 would eventually be negative. So lim s 0 + ns = a, + ] and if a 0, fo s > 0 small, s n n 2, which by integation implies that nea 0, ns 1 log s + C, C 0 18 fo some constant C 0, a contadiction. Hence, necessaily, a = 0 and the esult follows fom 18, which still holds tue when a = 0. Next we pove the following asymptotic esult : 10

11 Lemma 9 Let A denote the class of the functions n, continuous in the inteval [0, δ fo some δ > 0, and such that n0 = 0. Then, fo all k 1, sup lim inf n A s 0 + s n s n 2 s 1 k 1 4 X1s 2 Xj 2 s X 2 1 s Xk 2 s = 1 4. The fact that we ae dealing with a lim inf and not a lim sup may look supising at fist sight. Howeve, an uppe limit cannot be expected, as it is shown by the example given at the end of the pape. Poof. Fo 0 s < 1 and X 1 s := a log s 1, define implicitely n 1 by Then ns = 1 2 X 1s n 1 X 1 s s n s n 2 s = 1 4 t 2 + t n 1t n 2 1t t 2, t = 1 X 1 s. 1, +. Next, fo all k 1 and s > 1, let us define again n k+1 in tems of n k by Then n k s := 1 2 n k+1 t 1, t = 2 t 1 X 1 1/s 1, +. s n ks n 2 ks = 1 4 t + t n k+1t n 2 k+1t. 2 t 2 Hence, fo evey k 1 and evey 0 s < 1, with z = 1/X k s, we have s n s n 2 s = 1 k X 1 s 2 X j s 2 + X 1 s 2 X k s 2 z n 4 kz n 2 kz. 1 Choosing n k = 0 delives a function ns with s n s n 2 s = 1 k X 1 s 2 X j s 2. 4 Note that in this case, ns = k X 1 s X j s see Appendix A fo moe details. This shows that sup A lim inf s 0 + s n s n 2 s 1 k 1 4 X1s 2 Xj 2 s X 2 1 s Xk 2 s

12 2 Let now n be any function in A. Fo evey k, lim inf t + t n k n 2 k 0. If the above limit was to be lage than 0, say some constant b > 0, then, integating the inequality t n k n 2 k would show that n k tends to 0 at infinity, while on the othe hand, integating t n k b/2 would show that n k is unbounded nea infinity, which povides an obvious contadiction. Coollay 10 Let W be as in Poposition 8. Then 10 holds and the optimal asymptotic behavio nea the oigin is achieved. Poof. Close to s = 0 +, the fact that immediately povides lim inf s 0 + s n n 2 2 s n s n n + n s n n 2 W s k X1s 2 Xj 2 s X1 2 s Xk+1 2 s 1 8 and the optimal behaviou is achieved, fo instance, by W = W := minw +,1+ n, W,1 n, n := X 1 s X j s, so that W s = see Appendix A fo moe details. + X 2 1s X 2 j s + os 3.2 Estimates based on impoved Hady inequalities fo the Laplacian The above aguments show that the optimal gowth nea 0 and nea infinity fo any function W geneated as above by functions m ±, continuous nea the oigin and nea infinity, and fo which 9 holds, is given by 10 with equality fo each k 1, as in the statement of Theoem 4. On the othe hand, the optimality nea infinity was established in Lemma 7. Howeve, it emains to pove that thee is no function W not given by 15 with highe gowth at the oigin. This amounts to pove that thee is no adial function W with moe singula asymptotics nea the oigin and fo which the diffeential poblem s m = s m 2 s 2m m 2 W, m0 = 1, 12

13 cannot be solved fo some function m, continuous at 0. The est of this section is devoted to this question. Step 1 : We fist emak that in this poblem the angula vaiables do not play any ole : only adially symmetic spinos of a paticula fom ae elevant to obtain the optimal asymptotics. Poposition 11 Let W : R + R + be a adially symmetic continuous and a.e. diffeentiable function. Assume that a.e. [0, R, Then, fo all ϕ H 1 0B R, C 2, R 3 W 3 W W σ ϕ 2 dx R 3 + W ϕ 2 dx, and the optimizes ae adially symmetic and of the fom ϕ = v 0. In paticula, if W 0 > 0 and W is nondeceasing nea 0, 20 holds tue fo R sufficiently small. Poof. Let =. By, we mean x. Fo all 2-spino ϕ with compact suppot in the ball B R, using σ x 2 = 1, we have R 3 + W σ ϕ 2 dx = R 3 + W σ x σ ϕ 2 dx = R 3 + W ϕ 1 2 σ Lϕ dx = ϕ R 3 + W σ 2 Lϕ 2 dx W < ϕ, σ Lϕ > d S 2 = ϕ R 3 + W σ 2 Lϕ 2 dx < ϕ, σ Lϕ > 0 + W S 2 Now, if we choose ϕ belonging to the class of spinos geneated by the eigenfunctions of σ L with eigenvalue n, we notice that d. 13

14 R 3 + W σ ϕ 2 dx = R 3 + W ϕ 2 n 2 dx + + W + n 2 ϕ 2 dx 2 + W n = R 3 + W ϕ n + W 1 + W n dx + ϕ 2 dx, + W 2 which implies that fo all ϕ suppoted in B R, R 3 + W σ ϕ 2 dx R 3 + W ϕ 2 dx, 21 and the optimizes fo this inequality ae adially symmetic. Indeed, emembe that the spectum of 1 + σ L, is the set {±1, ±2,... }. Hence, n {..., 3, 2, 0, 1, 2,... } But ou assumptions imply that the minimum of n 2 + 2n + W 1 + W n on B R is nonnegative fo n 0 and 0 fo n = 0. Hence, the optimizes fo 21 coespond to spinos which ae eigenfunctions of 1 + σ L with eigenvalue 1 n = 0. These spinos ae adially symmetic and thei second component is equal to 0 see [19]. The last assetion of the poposition tivially follows fom the fact that W having a finite limit at 0, lim 0 + W must be equal to 0. Step 2 : We pove a elation between Hady-like inequalities fo the Laplacian and fo the Diac opeato in the adially symmetic case. Conside a function W : R + R + such that W/ 3 is integable at infinity and define a new vaiable y := 1 s + W s 3 ds = t + W t t 3 dt Now, fo any u C 0 R +, R, we define qy := u, whee y and ae elated by the above change of vaiables. Then staightfowad computations show that the inequalities W u 2 d W u 2 d 23 0 and + y 2 q 2 dy 0 + ae equivalent, with V given in tems of = y by V y = W 4 y 2 W + = 0 V q 2 dy 24 W + 2 t + W t t dt W 1 14

15 Poposition 12 Let W : R + R + be such that W/ 3 is integable at infinity. Then Hady-like inequalities 23 and 24 ae equivalent, with W and V elated by 22 and 25. Remak Note that when dealing with functions which ae compactly suppoted in a fixed ball, the behavio of W nea infinity is ielevant, since W can be modified outside the ball, without changing the integals in the above inequalities. In paticula, this is the case when seaching fo the optimal asymptotics nea the oigin of the functions W fo which 9 holds. Step 3 : Let us focus now on impoved Hady inequalities fo the Laplacian. Compaed with 23, Inequality 24 is easie to deal with, because the potential appeas only in the.h.s. In [5,6,11] see also [1 4,18] fo elated esults we find the following optimality esult : Theoem 13 [11] The optimal asymptotical behavio nea the oigin fo potentials V fo which the Hady-like inequality 24 holds fo all q C 0 R 3, R is given at each ode by V s = X1s 2 Xj 2 s. 26 An elementay poof fo Theoem 13 in the adially symmetic case. Fo completion, let us give a simple poof of this esult. This can be done by using the same kind of changes of vaiables as those used in the poof of Lemma 9. Let a > 1 be the constant which appeas in the definition of X 1 and take R < e a. Fo all u H 1 0B R, fo evey k 1, define the functions g k by u = 1 1 g 1 X 1 1/ g k s := s g k+1 t, t = 1 X 1 1/s A simple computation shows that R 0 2 u 2 d = 1 4 R 0 u 2 d + + X 1 1 R g 1 2 dy. With the notation t = ts = 1/X 1 1/s = a + log s, it is clea that s dt ds = 1. Fom the definition of g k+1, we get, fo any k 1, g ks = 1 2 s g k+1t + s dt ds g k+1t. 15

16 Moeove, fo any A > 0, + A + A g k+1 ts ds = g s k+1 t 2 dt, X 1 1 A dt 2 + s ds g k+1ts ds = g k+1t 2 dt. X 1 Taking A > 0 small enough, this means + A g ks 2 ds = A X 1 1 A g k+1 t 2 dt + since g k+1 has a compact suppot in 0, +. Thus + X 1 k R g k 2 ds = + X 1 k+1 R g k+1 2 dt R 0 + X 1 1 A whee by X 1 k+1 we denote the invese function of X k+1, and R 0 2 u 2 d = 1 4 u = R 0 X 1 X k 1/2 gk+1 g k+1t 2 dt X 2 1 X 2 k u 2 d, 27 1 X k X X1X Xk 2 u 2 d+ g k+1 2 dt. X 1 k+1 R The asymptotical optimality shaed at evey ode by the functions defined in 26 follows fom the fact that fo evey A > 0, inf g DA,+, g 0 + A + g 2 dt A g 2 dt = 0. Hence, thee exists functions u such that the fist tem in the.h.s. of 27 is negligible w..t. the second one. Coollay 14 Let W : R + R + be a adially symmetic continuous and a.e. diffeentiable function satisfying 20. Then, the optimal asymptotic gowth at the oigin fo all functions W fo which 9 holds in H0B 1 R, C 2 is that of the function W s = X 2 8 1s Xj 2 s. 28 Poof. If W violates the asymptotics given by 28, a tedious calculation using 25 shows that the coesponding potential V violates the optimal asymptotics given by

17 3.3 Optimal functions The fist pat of Theoem 4 is poved by Lemma 7, Poposition 11 and Coollay 14. Fo the second pat, we have to match optimal functions nea the oigin and nea infinity. 1 Accoding to Coollaies 10 and 14, fo n := 1 2 W := minw +,1+ n, W,1 n = X 1 s X j s, X 2 1s X 2 j s + os as s 0 + is optimal nea the oigin see Appendix A fo moe details. A simple computation shows that W becomes smalle than 1 fo any s > R, fo some R 0, 1. A fist example of a function W 1 which has optimal behavio nea the oigin is theefoe given by W 1 := max W, 1 = minw +, m +, W, m, with 1 ± n if s < R ±, m ± = 1 if s R ±, whee [0, R ± ] is the suppot of W ±,1± n 1. 2 On the othe hand, if we compute W := minw +,1+ñ, W,1 ñ with ñ := 1 4 s 1, we notice that W 1 fo all s T fo T = 1 48 [ / /3 ]. Numeically, one finds T Hence, W 2 := max1, W 1 is an example of a function W 1 which has an optimal behavio at infinity : W s s as s +. Note that W 2 = minw +, m +, W, m, with 1 if s < T ±, m ± = 1 ± ñ if s T ±, whee [T ±, + is the suppot of W ±,1±ñ 1. The function W 2 has an additional nice popety : since fo s lage, W 2 s+ 1 8 s, if we scale Inequality 9 keeping the L 2 -nom constant, on one end of the scale we obtain Inequality 4, while on the othe end we find the uncetainty pinciple / classical Hady inequality 1. 3 Now we pove that one can optimize the behavio of W nea 0 and nea infinity simultaneously, with W > 1 on 0, +. 17

18 Case + : We take a lage enough so that the function W + := maxw, W +,1+ñ is well defined, continuous in 0, + and satisfies : W + W in [0, R], W + W +,1+ñ in [R, +, W + R > 1, fo some R > 0 numeically, a > 5 is enough. This amounts to define W + as W +,m +, with m if s R, m + s = W +,1+ñ if s R, whee m is the solution of the O.D.E. poblem s m = 2m + s m W 1 + m 2, m0 = 1. The existence of m is poved in Appendix B. Case : This is dealt with in the same manne, by patching this time W and W,1 ñ in an appopiate way. The function W 3 := minw, W + satisfies all the popeties stated in Theoem 4. In all the above examples whee W 1 the functions m ± have discontinuities and CR < 0. Indeed, this has to be the case wheneve W 1. Let m ± be defined by 15. Accoding to Poposition 8, m ± 1 in a neighbouhood of s = 0 +. Using W 1, we get s m ± m 1 2, and an easy O.D.E. agument shows that m ± cannot be globally defined, so it must have a discontinuity. The aguments used in the poof of Poposition 6 allow us to conclude. Appendix A : Popeties of the functions X k Let a > 1. Define X 1 s := 1 a log s fo any s 0, ea 1, and, by induction fo any k 1, X k+1 s := X 1 X k s. Note that 0 < s < e a 1 = 0 < X 1 s < 1 < e a 1, 18

19 which implies that s a = lim k + X k s 0, 1 is independent of s the limit is unique since d 2 X 1 /ds 2 changes sign only once on 0, e a. Then s dx 1 ds = X2 1s and s Xk+1 1 dx k+1 ds = X k+1 s s Xk 1 dx k ds. Let π k s := k k X j s and σ k s := π j s. Since s X 1 dx k+1 k+1 ds = π k+1, it follows that s dπ k ds = π k σ k. By definition of X k, X k+1 s = X k t t=x1 s and σ k+1 s = t 1+σ k t and s dσ k+1 t=x 1 s ds = t dσ k dt t + σ kt + 1 t 2. t=x 1 s Using the two above identities, we can pove by induction the following fomula : Lemma 15 Fo any k 1, fo any s 0, e a 1, 2s dσ k ds s σ2 ks = k πj 2 s. We may now pass to the limit k +. Let σs := + π j s : 2s dσ ds s σ2 s = + π 2 j s. With the notations W := this means + π 2 j s and ns := 1 2 σ = π j s, Coollay 16 Fo any s 0, e a 1, s d n ds n2 = 2 W 1. Appendix B : Solving a singula O.D.E. Hee we solve the diffeential equation W 1 = s n n 2 2 s n s n n + n 2, n0 = 1, 29 in an inteval [0, δ], δ > 0, small, with W s = k=1 X 1 s 2 X k s 2. 19

20 Poposition 17 Thee exists δ > 0 such that 29 has a continuous solution in [0, δ. Note that this poblem is a limiting one in the sense that thee is no function W moe singula than W at the oigin, fo which the above poblem can be solved with continuity at the oigin. Poof. Let C, δ be two positive constants and define the set X C,δ := {u C[0, δ] : lim sup us logs C}. s 0 Let us wite n := n 1 + w = X 1 s X j s. Then n is a solution to 29 if and only if w is a solution to w = f 0 + f 1 w + f 2 w 2, 30 whee f 0, f 1, f 2 have the following behavio nea 0 : f 0 = s s log s 2 f 1 = s 0 1 s log s log log s f 2 s 0 1 s log s In ode to solve equation 30 togethe with the initial value w0 = 0, we intoduce the map T : X C,δ X C,δ defined by T ws := s 0 f 0 + f 1 w + f 2 w 2 dy, and look fo a fiwed point. By choosing C > 1/4 and δ < 1 small enough, T maps X C,δ into itself and it is a contaction. So, thee is a unique solution of 30 in X C,δ which means that 29 has a unique continuous solution n in the inteval [0, δ], with n/ n 1 in X C,δ, such that n0 = 0. 20

21 Appendix C : Why do we have a lim inf in Theoem 4? We ae going to give a qualitative example showing that only a lim inf can be achieved. Let W := W + s sn W whee ε n and s n ae such that n 0 ε n s n > 0, ε n > 0, lim s n = 0, n + ε n < +, n 0 ε n+1 < s n s n+1, and assume that W is a bounded function with compact suppot in 0, 1. Then lim sup s 0 + W s > 1 and the equations 2m ± ± s m ± s m 2 ± + s = W 1 + m 2 ± have no solution continuous up to s = 0. Acknowledgment : M.J.E. would like to thank the Geogia Tech School of Mathematics fo its hospitality and M.L. would like to thank CEREMADE whee some of this wok has been caied out. Refeences [1] Adimuthi. Hady-Sobolev inequality in H 1 Ω and its applications. To appea in Comm. Contemp. Math. [2] Adimuthi, M. J. Esteban. An impoved Hady-Sobolev inequality in W 1,p and its application to Schödinge opeato. To appea in NODEA. [3] Adimuthi, M. Ramaswamy, N. Chaudhui. An impoved Hady-Sobolev inequality and its applications. Poc. Am. Math. Soc. 130 no , P [4] Adimuthi, K. Sandeep. Existence and non-existence of the fist eigenvalue of the petubed Hady-Sobolev opeato. Poc. R. Soc. Edinb., Sect. A, Math. 132 no , p [5] G. Babatis, S. Filippas, A. Tetikas. Seies expansion fo L p -Hady inequalities. To appea in Indiana Univ. Math. J. 21

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