CLR implies LT H for > 0 and d 3 . Indeed, ( n ( V )) S ,d - PowerPoint PPT Presentation

R ECENT R ESULTS ON L IEB -T HIRRING I NEQUALITIES Ari Laptev and Timo Weidl Department of Mathematics Royal Institute of Technology SE-10044 Stockholm Sweden laptev@math.kth.se weidl@math.kth.se June 3, 2000 We give a survey of results on Lieb-Thirring inequalities for the eigenvalue moments of Schrödinger operators. In particular, we discuss the optimal values of the con- stants therein for higher dimensions. We elaborate on certain generalisations and some open problems as well.

0. I NTRODUCTION 1. Let H be the Schrödinger operator H ( V ; � ) = − � 2 ∆ − V ( x ) L 2 ( R d ) . on For suitable real-valued potential wells V the negative spectrum { λ n ( V ; � ) } of H is semi-bounded from below and discrete. 2. For σ ≥ 0 let � ( − λ n ( V ; � )) σ S σ,d ( V ; � ) = tr H σ − ( V ; � ) = n be the σ -Riesz mean of the negative spectrum. Moreover, let � � h< 0 ( − h ( ξ, x )) σ dxdξ S cl σ,d ( V ; � ) = (2 π � ) d be the σ -means of the symbol h = | ξ | 2 − V ( x ) .

For appropriate pairs of σ and d the Lieb-Thirring inequalities states that R ( σ, d ) S cl S σ,d ( V ; � ) ≤ σ,d ( V ; � ) 3. The Lieb-Thirring inequality captures the correct order of the semi-classical Weyl type asymptotics S σ,d ( V ; � ) = (1 + o (1)) S cl σ,d ( V ; � ) � → 0 . as The inequality holds for all positive values of � . It extracts hard information on the negative spectrum of Schrödinger operators from the classical systems in the non-asymptotical regime.

4. The ξ -integration in S cl σ,d evaluates to � σ + d S cl σ,d ( V ; � ) = L cl σ,d � − d 2 V dx, + where Γ( σ + 1) L cl σ,d = � . � σ + d 2 d π d/ 2 Γ 2 + 1 The Lieb-Thirring inequality turns into � σ + d ( − λ n ( V ; � )) σ ≤ L σ,d � − d � 2 V dx + n with the usual Lieb-Thirring constants L σ,d = R ( σ, d ) L cl σ,d .

In view of the Weyl asymptotics we have L σ,d ≥ L cl R ( σ, d ) ≥ 1 and σ,d . 5. One should ask the following questions: 1. For which σ and d does the inequality S σ,d ( V ; � ) ≤ R ( σ, d ) S cl σ,d ( V ; � ) actually hold? 2. What are the sharp values of R ( σ, d ) ? 3. For which σ and d is R ( σ, d ) = 1 ?

1. V ALIDITY OF L IEB -T HIRRING INEQUALITIES 1. Counterexamples: For d = 1 , 2 any arbitrary small attractive potential well will couple at least one � V d/ 2 dx can be bound state. Hence, we have S 0 ,d ( V ; � ) ≥ 1 while S cl 0 ,d ( V ; � ) ∼ arbitrary small. This contradicts to LT H for σ = 0 and d = 1 , 2 . For d = 1 the weakly coupled bound state satisfies � 2 �� λ 1 ( V ; � ) = − 1 + o ( � − 2 ) , V dx � → ∞ . 4 � 2 This implies S σ, 1 ( V ; � ) = O ( � − 2 σ ) while S cl σ, 1 ( V ; � ) = O ( � − 1 ) for � → ∞ . This excludes LT H for d = 1 and 0 < σ < 1 / 2 .

2. The LT H inequality holds true for σ ≥ 1 / 2 if d = 1 σ > 0 if d = 2 . σ ≥ 0 if d ≥ 3 σ > 1 / 2 , d = 1 σ > 0 , d ≥ 2 ; σ = 0 , d ≥ 3 ; [LTh] for and [C,L,R] for [W] for σ = 1 / 2 , d = 1 . The parameter � can be scaled out and we put � = 1 . 3. Borderline cases are the most complicated ones. In particular, for σ = 0 and d ≥ 3 LT H turns into the celebrated CLR estimate on the number of bound states � V d/ 2 rank H − ( V ) = S 0 ,d ( V ) ≤ L 0 ,d dx. +

CLR implies LT H for σ > 0 and d ≥ 3 . Indeed, � ( − λ n ( V )) σ S σ,d ( V ) = n � ∞ 1 dtt σ − 1 S 0 ,d ( V − t ) = σ 0 � �� � # { λ n < − t } � ∞ R (0 , d ) dtt σ − 1 S cl 0 ,d ( V − t ) ≤ σ 0 � �� � vol { ( ξ,x ): h< − t } (2 π ) d R (0 , d ) S cl σ,d ( V ) . ≤ In a similar way one shows that R ( σ ′ , d ) ≤ R ( σ, d ) for all σ ′ ≥ σ [Aizenman,Lieb]. 4. In the other borderline case σ = 1 / 2 and d = 1 for V ≥ 0 one finds in fact a two-sided estimate S cl 2 ( V ) ≤ 2 S cl ( V ) ≤ S 1 , 1 ( V ) 1 , 1 1 , 1 2 2 [GGM], [W], [HLT]. Note that σ = 1 / 2 and d = 1 is the only point in the Lieb- Thirring scale, where such a two-sided estimate is possible.

2. O N THE SHARP VALUES OF R ( σ, d ) . 1. The dimension d = 1 : Sharp constants appear already in [LTh], [AL] R ( σ, 1) = 1 for all σ ≥ 3 / 2 . It uses a trace identity for σ = 3 / 2 and the monotonicity argument [AL]. The only other case settled was σ = 1 / 2 with R (1 / 2 , 1) = 2 in [HLT]. This reflects the weak coupling behaviour. The optimal values of R ( σ, 1) for 1 / 2 < σ < 3 / 2 are unknown. An analysis of the lowest bound state shows that here σ − 1    σ − 1 ( − λ 1 ( V )) σ 2 2 R ( σ, 1) ≥ sup = 2 .  σ + 1 S cl σ, 1 ( V ) V ∈ L σ +1 2 2

2. Let { µ k } be the eigenvalues of the Dirichlet Laplacian H D Ω = − ∆ on an open domain Ω ⊂ R d . For any σ ≥ 1 , Λ ≥ 0 , d ∈ N , Ω ⊂ R d [Berezin ’72] 1 � � R d dξ ( | ξ | 2 − Λ) σ � ( µ k − Λ) σ ≤ Ω dx − − (2 π ) d k σ,d vol (Ω)Λ σ + d L cl ≤ 2 .

Proof. Let { φ k } be an o.n. eigenbase H D Ω . Put φ k ≡ 0 on R d \ Ω and � φ k ( ξ ) = (2 π ) − d/ 2 Ω φ k ( x ) e iξx dx. ˜ � φ k | 2 dξ = 1 ) gives R d | ˜ Jensen’s inequality ( σ ≥ 1 , � σ �� R d ( | ξ | 2 − Λ) | ˜ � � ( µ k − Λ) σ φ k ( ξ ) | 2 dξ = − − k k � R d ( | ξ | 2 − Λ) σ � φ k ( ξ ) | 2 dξ. | ˜ ≤ − k Parsevals inequality w.r.t. { φ k } in L 2 (Ω) implies � � dx dx φ k ( ξ ) | 2 = � Ω | e − ixξ | 2 | ˜ (2 π ) d = (2 π ) d . � Ω k

3. The Legendre transformed ˆ f ( p ) of a convex, non-negative function f ( t ) on R + is given by ˆ f ( p ) = sup ( pt − f ( t )) , p ≥ 0 . t ≥ 0 It reverses inequalities: f ( t ) ≤ g ( t ) for all t ≥ 0 implies ˆ f ( p ) ≥ ˆ g ( p ) for all p ≥ 0 . Note that [ p ] � � ( µ k − x ) − ) ∧ ( p ) ( = ( p − [ p ]) µ [ p ]+1 + µ k , k k =1 βp 1+ β − 1 cx 1+ β � ∧ ( p ) � = (1 + β ) 1+ β − 1 c β − 1 .

We put σ = 1 in Berezin‘s inequality 1 ,d vol (Ω)Λ 1+ d � ( µ k − Λ) − ≤ L cl 2 k and apply the Legendre transformation for x = Λ and p = n ∈ N � − 1 − 2 n � − 2 � d d 1 + d n 1+ 2 � d � L cl µ k ≥ 1 ,d vol (Ω) d 2 2 k =1 � − 2 d n 1+ 2 � L cl d 0 ,d vol (Ω) ≥ d 2 + d and recover a well-known result by Li and Yau.

4. The harmonic oscillator. Put m = ( m 1 , . . . , m d ) , d � m 2 k x 2 V ( x ) = Λ − Λ > 0 , m k > 0 . k , k =1 Then the operator H ( V ) has the eigenvalues d � λ τ ( V ) = − Λ + m k (1 + 2 τ k ) , k =1 with τ = ( τ 1 , . . . , τ d ) and τ k = 0 , 1 , 2 , . . . Λ 2 For σ, d = 1 it holds S cl 1 , 1 ( V ) = 4 m 1 and � S 1 , 1 ( V ) = ( m 1 (1 + 2 k ) − Λ) − k � � Λ 2 (2 m 1 ) 2 − t 2 = m 1 � Λ � 2 m 1 − 1 Λ where t = 1 + − 2 m 1 . 2

With the Lieb-Aizenman argument we get S σ, 1 ( V ) ≤ S cl σ, 1 ( V ) , σ ≥ 1 . A straightforward generalisation to higher dimensions is much more involved and gives [De la Bretèche] S σ,d ( V ) ≤ S cl σ,d ( V ) , σ ≥ 1 . The careful analysis of the same problem implies R ( σ, d ) > 1 for all σ < 1 [Helffer, Robert].

5. Alternatively, put V ( x ) = W ( x 1 , . . . , x d − 1 ) − m 2 d x 2 d . Integration in x d and ξ d gives � � � dxdξ � σ | ξ | 2 + m 2 S cl d x 2 d − W ( x ′ ) σ,d ( V ) = (2 π ) d − (2( σ + 1) m d ) − 1 S cl = σ +1 ,d − 1 ( W ) . Moreover, λ τ ′ ,τ d ( V ) = λ τ ′ ( W ( x ′ )) + m d (1 + 2 τ d ) . Evaluating the sum over τ d ≥ 0 first, it follows that � ( λ τ ′ ( W ) + m d (1 + 2 τ d )) σ S σ,d ( V ) = − τ ′ ,τ d � � � λ τ ′ ( W ) − m 2 d x 2 = S σ, 1 d τ ′ � � � λ τ ′ ( W ) − m 2 d x 2 S cl ≤ σ, 1 d τ ′ � � � dx d dξ d � σ | ξ d | 2 + m 2 � d x 2 d − λ τ ′ ( W ) ≤ − (2 π ) τ ′ (2( σ + 1) m d ) − 1 � ( − λ τ ′ ( W )) σ +1 ≤ τ ′

and for any σ ≥ 1 it holds S σ,d ( V ) /S cl σ,d ( V ) ≤ S σ +1 ,d − 1 ( W ) /S cl σ +1 ,d − 1 ( W ) . For V = Λ − � k m 2 k x 2 k iteration gives S 1 ,d ( V ) ≤ S cl 1 ,d ( V ) , σ ≥ 1 . In fact, this holds for all V ( x ) = W ( x ′ ) − m 2 d x 2 d !

6. We summarise R ( σ, 1) = 1 σ ≥ 3 / 2 , d = 1 , for R (1 / 2 , 1) = 2 σ = 1 / 2 , d = 1 , for σ − 1    σ − 1 2 1 2 < σ < 3 2 R ( σ, 1) ≥ 2 for 2 , d = 1 ,  σ + 1 2 R ( σ, d ) > 1 for σ < 1 , d ∈ N , R ( σ, 2) > 1 for σ < σ 0 , σ 0 ≃ 1 . 16 , d = 2 . The Dirichlet Laplacian and the harmonic oscillator permitt a LT H estimate with the classical constant if σ ≥ 1 . There exist certain explicite upper bounds on the constants R ( σ, d ) .

Lieb and Thirring posed the following Conjecture: In any dimension d there exists a finite critical value σ cr ( d ) , such that R ( σ, d ) = 1 for all σ ≥ σ cr ( d ) . In particular, one expects that σ cr ( d ) = 1 for d ≥ 3 , or L 1 ,d = L cl for d ≥ 3 . 1 ,d

3. L IEB -T HIRRING I NEQUALITIES FOR O PERATOR V ALUED P OTENTIALS 1. We consider a generalisation of LT H inequalities: G is a separable Hilbert space, 1 G is the identity on G . V : R d → S ∞ ( G ) is a compact s.-a. operator-valued fct. We study the negative spectrum { λ n ( V ) } of the operator L 2 ( R d ) ⊗ G. H ( V ) = − ∆ ⊗ 1 G − V ( x ) on