The Strichartz inequality for orthonormal functions Rupert L. Frank - PowerPoint PPT Presentation

The Strichartz inequality for orthonormal functions Rupert L. Frank Caltech Joint work with Mathieu Lewin, Elliott Lieb and Robert Seiringer Strichartz inequality for orthonormal functions J. Eur. Math. Soc., to appear. Preprint: arXiv:1306.1309 Joint work in progress with Julien Sabin TexAMP, Houston, October 26, 2013 R. Frank – Strichartz inequality for orthonormal functions – October 26, 2013 # 1

Introduction – The Schr¨ odinger equation By spectral theory the solution e − itH ψ of the time-dependent Schr¨ odinger equation i ∂ ∂t Ψ = H Ψ , Ψ | t =0 = ψ with H self-adjoint satisfies ∥ e − itH ψ ∥ = ∥ ψ ∥ for all t ∈ R . Here we are interested in the phenomenon of dispersion . Example: H = − ∆ in L 2 ( R d ) and ψ ( x ) = ( πσ 2 ) − d/ 4 e ip · x e − x 2 / 2 σ 2 . Then ) d/ 2 σ 2 ( � 2 = e − σ 2 ( x − 2 tp ) 2 / ( σ 4 +4 t 2 ) . e it ∆ ψ � � �( ) ( x ) π ( σ 4 + 4 t 2 ) Dispersion is quantified by Strichartz inequalities . Simplest form: ) ( d +2) /d (∫ ∫ ∫ � 2( d +2) /d dx dt ≤ C d R d | ψ ( x ) | 2 dx e it ∆ ψ �( � ) � ( x ) . R d R Due to Strichartz (1977); see also Lindblad–Sogge , Ginibre–Velo , Keel–Tao , Foschi , . . . R. Frank – Strichartz inequality for orthonormal functions – October 26, 2013 # 2

Goal – A Strichartz inequality for orthonormal functions Is there an inequality for � 2 ) ( d +2) /d ∫ ∫ (∑ � e it ∆ ψ j � �( ) ( x ) dx dt j R d R with ψ j orthonormal in L 2 ( R d ) ? Obvious answer: By triangle inequality (without using orthogonality!) � 2 ) ( d +2) /d (∑ N ∫ ∫ e it ∆ ψ j dx dt ≤ C d N ( d +2) /d � �( ) � ( x ) j =1 R d R Can we do better than that? Main result: Yes, we can! � 2 ) ( d +2) /d (∑ N ∫ ∫ e it ∆ ψ j dx dt ≤ C ′ d N ( d + 1 ) / d � � �( ) ( x ) j =1 R d R And this is best possible! R. Frank – Strichartz inequality for orthonormal functions – October 26, 2013 # 3

Compare with Lieb–Thirring inequalities The Sobolev interpolation inequality says that for γ ≥ 1 (and more) ) γ + d/ 2 − 1 ) − γ − 1 (∫ d/ 2 (∫ ∫ d/ 2 2( γ + d/ 2) R d |∇ ψ | 2 dx ≥ S d,γ R d | ψ | 2 dx γ + d/ 2 − 1 dx R d | ψ | . This was generalized by Lieb–Thirring (1976) to orthonormal functions ψ j γ + d/ 2 − 1   γ + d/ 2 d/ 2 (∑ N ) ∫ ∫ ∑ N γ + d/ 2 − 1 R d |∇ ψ j | 2 dx ≥ K d,γ N − γ − 1 j =1 | ψ j | 2 dx . d / 2   j =1 R d γ This is better than N − d/ 2 (from triangle inequality) and optimal in the semi-classical limit. Case γ = 1 is used in the Lieb–Thirring proof of stability of matter . Slightly more precise version: for any operator Γ ≥ 0 on L 2 ( R d ) , ) γ + d/ 2 − 1 ) − γ − 1 d/ 2 (∫ d/ 2 γ + d/ 2 ( γ γ + d/ 2 − 1 dx Tr( − ∆)Γ ≥ K d,γ Tr Γ R d Γ( x, x ) . γ − 1 R. Frank – Strichartz inequality for orthonormal functions – October 26, 2013 # 4

‘Semi-classical’ intuition behind Strichartz � 2 ) (2+ d ) /d ∫∫ (∑ N d N ( d + 1 ) / d best possible? e it ∆ ψ j dx dt ≤ C ′ � � �( ) ( x ) Why is j =1 Heuristics: At t = 0 consider N electrons in a box of size L with const. density ρ = L − d N . For | t | ≥ T the electrons have (approximately) disjoint supports and therefore � 2 ) (2+ d ) /d (∑ N ∫∫ dx dt ≈ N ≪ N ( d +1) /d . e it ∆ ψ j � � �( ) ( x ) j =1 | t |≥ T We think of T as the typical time it takes an electron to move a distance comparable with the size of the system. By Thomas–Fermi theory the expected momentum per particle is ≈ ρ 1 /d and therefore, if the electrons move ballistically T ≈ Lρ − 1 /d . Thus, � 2 ) (2+ d ) /d (∑ N ∫∫ dx dt ≈ TL d ρ (2+ d ) /d ≈ N ( d +1) /d . e it ∆ ψ j � �( ) � ( x ) j =1 | t |≤ T R. Frank – Strichartz inequality for orthonormal functions – October 26, 2013 # 5

The main result Theorem 1. Let d ≥ 1 and assume that 1 < p, q < ∞ satisfy 1 < q ≤ 1 + 2 2 p + d q = d . and d Then, for any orthonormal ψ j and any n j ∈ C ) p ) p ( q +1) q 2 q � � (∫ (∑ ∫ q 2 q � 2 ∑ dt ≤ C p e it ∆ ψ j � � � � �( ) j n j ( x ) dx j | n j | q +1 . (1) � � d,q R d � � R that is, with the notations γ ( t ) = e it ∆ γe − it ∆ and ρ γ ( x ) = γ ( x, x ) , � � � ρ γ ( t ) x ( R d )) ≤ C d,q ∥ γ ∥ q +1 . 2 q � L p t ( R ,L q S This is best possible in the sense that � � � ρ γ ( t ) 2 q � L p t ( R ,L q x ( R d )) sup = ∞ if r > q + 1 . ∥ γ ∥ S r γ R. Frank – Strichartz inequality for orthonormal functions – October 26, 2013 # 6

Remarks Recall if 1 < q ≤ 1 + 2 � � � ρ γ ( t ) x ( R d )) ≤ C d,q ∥ γ ∥ d . 2 q � L p t ( R ,L q q +1 S Remarks. (1) The inequality with the trace norm ∥ γ ∥ S 1 on the right side is known, even for the full range 1 ≤ p, q ≤ ∞ with ( p, q, d ) ̸ = (1 , ∞ , 2) (plus scaling condition). (2) This implies an inhomogeneous Strichartz inequality : if i ˙ γ ( t ) = [ − ∆ , γ ( t )] + iR ( t ) , γ ( t 0 ) = 0 , with R ( t ) self-adjoint, then for q as in our theorem � � ∫ e − is ∆ | R ( s ) | e is ∆ ds � � � � � ρ γ ( t ) x ( R d )) ≤ C . � L p t ( R ,L q � � 2 q � � R q +1 S (3) We prove that the inequality fails for q ≥ ( d + 1) / ( d − 1) . How about the range 1 + 2 /d < q < ( d + 1) / ( d − 1) ? R. Frank – Strichartz inequality for orthonormal functions – October 26, 2013 # 7

A new result The following solves the endpoint case . This is joint work with J. Sabin. Theorem 2. Let d ≥ 1 , q = ( d + 1) / ( d − 1) and p = ( d + 1) /d . Then, with the notations γ ( t ) = e it ∆ γe − it ∆ and ρ γ ( x ) = γ ( x, x ) , x ( R d )) ≤ C ′ � � � ρ γ ( t ) d ∥ γ ∥ q +1 , 1 . 2 q � L p t ( R ,L q S Note the Lorentz- 1 norm (dual of weak norm) on the right side! Via real interpolation, we get the full result. Corollary 3. Let d ≥ 1 and assume that 1 < p, q < ∞ satisfy 1 < q < d + 1 2 p + d and q = d . d − 1 Then, � � � ρ γ ( t ) x ( R d )) ≤ C d,q ∥ γ ∥ q +1 . 2 q � L p t ( R ,L q S R. Frank – Strichartz inequality for orthonormal functions – October 26, 2013 # 8

The dual formulation Using H¨ older’s inequality (for operators and for functions) and the fact that ∫∫ ∫ e − it ∆ V ( t, · ) e it ∆ dt R × R d V ( t, x ) ρ γ ( t ) ( x ) dx dt = Tr γ R we see that Theorem 1 is equivalent to Theorem 4. Let d ≥ 1 and assume that 1 < p ′ , q ′ < ∞ satisfy 1 + d p ′ + d 2 2 ≤ q ′ < ∞ q ′ = 2 . and Then, with the same constant as in Theorem 1, � � ∫ e − it ∆ V ( t, · ) e it ∆ dt � � S 2 q ′ ≤ C d,q ∥ V ∥ L p ′ x ( R d )) . t ( R ,L q ′ � � � � R By interpolation it suffices to prove this for q ′ = p ′ = 1 + d/ 2 . R. Frank – Strichartz inequality for orthonormal functions – October 26, 2013 # 9

Proof of Theorem 2 For V ≥ 0 , d +2 ) d +2 � � (∫ ∫ e − it ∆ V ( t, · ) e it ∆ dt e − it ∆ V ( t, · ) e it ∆ dt � � S d +2 = Tr � � � � R R ∫ ∫ = · · · Tr V ( t 1 , x + 2 t 1 p ) · · · V ( t d +2 , x + 2 t d +2 p ) dt d +2 · · · dt 1 R R Here we use the notation f ( x + 2 tp ) = e − it ∆ f ( x ) e it ∆ . Lemma 5 ( Generalized Kato–Simon–Seiler ineq. ) . For α, β, γ, δ ∈ R and r ≥ 2 , ∥ f ∥ L r ( R d ) ∥ g ∥ L r ( R d ) ∥ f ( αx + βp ) g ( γx + δp ) ∥ S r ≤ . d d r | αδ − βγ | (2 π ) r Thus, ∥ V ( t 1 , · ) ∥ L 1+ d/ 2 · · · ∥ V ( t d +2 , · ) ∥ L 1+ d/ 2 � )� ( x x � Tr V ( t 1 , x + 2 t 1 p ) · · · V ( t d +2 , x + 2 t d +2 p ) � ≤ � � d d (4 π ) d | t 1 − t 2 | d +2 · · · | t d +2 − t 1 | d +2 R. Frank – Strichartz inequality for orthonormal functions – October 26, 2013 # 10

Proof of Theorem 2, cont’d We have shown that d +2 ∥ V ( t 1 , · ) ∥ L 1+ d/ 2 · · · ∥ V ( t d +2 , · ) ∥ L 1+ d/ 2 � � ∫ ∫ ∫ e − it ∆ V ( t, · ) e it ∆ dt � � S d +2 ≤ · · · x x d +2 dt d +2 · · · dt 1 � � d d (4 π ) d | t 1 − t 2 | d +2 · · · | t d +2 − t 1 | � � R R R Lemma 6 ( Multi-linear HLS inequality; Christ, Beckner ) . Assume that ( β ij ) 1 ≤ i,j ≤ N and ( r k ) 1 ≤ k ≤ N are real-numbers such that N N 1 β ik = 2( r k − 1) ∑ ∑ β ii = 0 , 0 ≤ β ij = β ji < 1 , r k > 1 , > 1 , . r k r k k =1 i =1 � � N Then ∫ ∫ f 1 ( t 1 ) · · · f N ( t N ) � � ∏ · · · i<j | t i − t j | β ij dt N · · · dt 1 � ≤ C ∥ f k ∥ L rk ( R ) . � � ∏ � � R R � k =1 For us, N = d + 2 , β ij = δ j,i +1 d/ ( d + 2) and r k = 1 + d/ 2 and thus d +2 � � ∫ e − it ∆ V ( t, · ) e it ∆ dt S d +2 ≤ C ∥ V ∥ d +2 � � . � � L 1+ d/ 2 � � t,x R R. Frank – Strichartz inequality for orthonormal functions – October 26, 2013 # 11