Endpoint resolvent estimates for compact Riemannian manifolds joint - PowerPoint PPT Presentation

Endpoint resolvent estimates for compact Riemannian manifolds joint work with R. L. Frank to appear in J. Funct. Anal. (arXiv:1611.00462) Lukas Schimmer California Institute of Technology 13 February 2017 Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 1 / 15

Resolvent estimates on R n Investigate L p ( R n ) → L q ( R n ) mapping properties of the resolvent ( − ∆ − z ) − 1 on R n . Theorem (Kenig, Ruiz and Sogge, 1987) For n ≥ 3 and z ∈ C \ (0 , ∞ ) 2( n + 1) ≤ 1 n + 3 p ≤ n + 2 � ( − ∆ − z ) − 1 � � L p → L p ′ ≤ C p , n | z | − n / 2+ n / p − 1 , . � � 2 n � Im z 1 / q 1 1 Re z 2 1 / p 1 1 / 2 Figure 1: Admissible values of p Figure 2: Admissible values of z Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 2 / 15

Applications These inequalities and their extensions have found many applications in analysis and PDE, including: Unique continuation problems and absence of positive eigenvalues (Koch and Tataru 2005; 2006) Limiting absorption principles (Goldberg and Schlag 2004; Ionescu and Schlag 2006) Absolute continuity of the spectrum of periodic Schr¨ odinger operators (Shen 2001) Eigenvalue bounds for Schr¨ odinger operators with complex potentials (Frank 2011) Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 3 / 15

Resolvent estimates on compact Riemannian manifolds M Let M be a compact Riemannian manifold without boundary of dimension n ≥ 3. Investigate L p ( M ) → L q ( M ) mapping properties of the resolvent ( − ∆ − z ) − 1 on M . Theorem (Dos Santos Ferreira, Kenig and Salo, 2014) For Im √ z ≥ δ with some arbitrary, but fixed δ > 0 2( n + 1) < 1 n + 3 p ≤ n + 2 L p → L p ′ ≤ C p , n ,δ | z | − n / 2+ n / p − 1 , � � ( − ∆ − z ) − 1 � . � � � 2 n Im z 1 / q 1 1 2 Re z 1 / p 1 1 / 2 Figure 3: Admissible values of p Figure 4: Admissible values of z Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 4 / 15

Optimality of the parabolic area of exclusion Theorem (Bourgain, Shao, Sogge and Yao 2015) Let M be a compact Riemannian manifold without boundary of dimension n ≥ 3 which is Zoll. Then there is a constant C > 0 such that for any function δ : R → (0 , ∞ ) with lim | κ |→∞ δ ( κ ) = 0 and lim inf | κ |→∞ | κ | δ ( κ ) ≥ C, n / 2 − n / p +1 � − ∆ − ( κ + i δ ( κ )) 2 � − 1 � � ( κ + i δ ( κ )) 2 � � � � � lim sup L p → L p ′ = ∞ . � � � � � � � | κ |→∞ If M = S n (with the standard metric), this holds also with C = 0 . Im z Note that Im √ z = δ can be written as ( Im z ) 2 = 4 δ 2 ( Re z + δ 2 ) Re z and that z = ( κ + i δ ( κ )) 2 is on the curve ( Im z ) 2 = 4 δ ( κ ) 2 ( Re z + δ ( κ ) 2 ) Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 5 / 15

Resolvent estimates on compact Riemannian manifolds M Let M be a compact Riemannian manifold without boundary of dimension n ≥ 2. Investigate L p ( M ) → L q ( M ) mapping properties of the resolvent ( − ∆ − z ) − 1 on M . Theorem (Frank and S., 2016; Burq, Dos Santos Ferreira, Krupchyk 2016) For Im √ z ≥ δ with some arbitrary, but fixed δ > 0 2( n + 1) = 1 n + 3 L p → L p ′ ≤ C p , n ,δ | z | − n / 2+ n / p − 1 = C p , n ,δ | z | − � � ( − ∆ − z ) − 1 � 1 n +1 , . � � � p Im z 1 / q 1 1 2 Re z 1 / p 1 1 / 2 Figure 6: Admissible values of p Figure 7: Admissible values of z Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 6 / 15

Necessary parametrix and remainder bounds The proof relies on the construction of a parmetrix T ( z ) ( − ∆ − z ) T ( z ) = I + S ( z ) . we obtain ( − ∆ − z ) − 1 = T ( z ) − T ( z ) ∗ S ( z ) + S ( z ) ∗ ( − ∆ − z ) − 1 S ( z ) . and thus � � ( − ∆ − z ) − 1 � L p → L p ′ ≤ �T ( z ) � L p → L p ′ + �T ( z ) ∗ � L 2 → L p ′ �S ( z ) � L p → L 2 � � � � � ( − ∆ − z ) − 1 � + �S ( z ) ∗ � L 2 → L p ′ L 2 → L 2 �S ( z ) � L p → L 2 � � � We need the following mapping properties: �T ( z ) � L p → L p ′ � | z | − 1 / ( n +1) (Frank and S. 2016) �T ( z ) � L p → L 2 � | z | − ( n +3) / (4( n +1)) ( n ≥ 3: Dos Santos Ferreira, Kenig and Salo 2014) �S ( z ) � L p → L 2 � | z | ( n − 1) / (4( n +1)) ( n ≥ 3: Dos Santos Ferreira, Kenig and Salo 2014) 2 ( Im √ z ) − 1 ≤ | z | − 1 / 2 δ − 1 L 2 → L 2 ≤ | z | − 1 � ( − ∆ − z ) − 1 � � � Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 7 / 15

Necessary parametrix and remainder bounds The proof relies on the construction of a parmetrix T ( z ) ( − ∆ − z ) T ( z ) = I + S ( z ) . we obtain ( − ∆ − z ) − 1 = T ( z ) − T ( z ) ∗ S ( z ) + S ( z ) ∗ ( − ∆ − z ) − 1 S ( z ) . and thus � � ( − ∆ − z ) − 1 � L p → L p ′ ≤ �T ( z ) � L p → L p ′ + �T ( z ) � L p → L 2 �S ( z ) � L p → L 2 � � � � � ( − ∆ − z ) − 1 � + �S ( z ) � L p → L 2 L 2 → L 2 �S ( z ) � L p → L 2 � � � We need the following mapping properties: �T ( z ) � L p → L p ′ � | z | − 1 / ( n +1) (Frank and S. 2016) �T ( z ) � L p → L 2 � | z | − ( n +3) / (4( n +1)) ( n ≥ 3: Dos Santos Ferreira, Kenig and Salo 2014) �S ( z ) � L p → L 2 � | z | ( n − 1) / (4( n +1)) ( n ≥ 3: Dos Santos Ferreira, Kenig and Salo 2014) 2 ( Im √ z ) − 1 ≤ | z | − 1 / 2 δ − 1 L 2 → L 2 ≤ | z | − 1 � ( − ∆ − z ) − 1 � � � Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 7 / 15

The parametrix construction We will use the Hadamard parametrix T ( z ). The construction is local � ( T ( z ) u )( x ) = χ ( x ) F ( x , y , z ) χ ( y ) u ( y ) d µ g ( y ) ˜ M with d µ g denoting the volume form on M and, N � F ( x , y , z ) = α j ( x , y ) F j ( d g ( x , y ) , z ) j =0 with smooth coefficients α j and the Bessel potentials e i x ξ � j ! F j ( r , z ) = ( | ξ | 2 − z ) 1+ j d ξ . (2 π ) n R n Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 8 / 15

Parametrix bounds Lemma (Frank and S., 2016) Let δ > 0 and either ( p , q ) = ( 2 n ( n +1) n +1 , 2 n ( n +1) n − 1 ) or ( p , q ) = ( 2 n 2 n n 2 +4 n − 1 , n 2 − 2 n +1 ) . Then, if | z | ≥ δ , 1 � T ( z ) u � q , ∞ � | z | − n +1 � u � p , 1 . 1 / q 1 It is sufficient to consider characteristic functions u = I E . The statement is then equivalent to 1 1 1 1 q ≤ C | z | − n +1 µ g ( E ) sup λµ g ( A ) p 2 λ> 0 � > λ � � with A = � x ∈ M : �� T ( z ) I E � ( x ) � 1 / p 1 / 2 1 Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 9 / 15

Proof idea The kernel T ( z ) is decomposed dyadically | z | 1 / 2 d g ( x , y ) ∈ [ − 2 ν , − 2 ν − 1 ] ∪ [2 ν − 1 , 2 ν ] � T ( z ) = T ν ( z ) . ν ≥ 0 By the Carleson–Sj¨ olin theorem: Lemma ( n ≥ 3: Dos Santos Ferreira, Kenig, Salo 2014; n = 2: Frank and S. 2016) 2 ) � u � p 2 ≤ q ≤ ∞ , p ′ = n + 1 − ν ( n q ′ − n +1 2 p − n n 2 q − 1 2 � T ν ( z ) u � q ≤ C | z | if n − 1 q , ν ≥ 1 . Decompose 1 / q 1 � � T ν ( z ) = T (1) ( z )+ T (2) ( z ) T ( z ) − T 0 ( z ) = T ν ( z )+ 1 ≤ ν ≤ ρ ν>ρ and use H¨ older’s inequality to bound 1 2 1 1 λµ g ( A ) ≤ � T (1) ( z ) I E � q 1 µ g ( A ) 1 + � T (2) ( z ) I E � q 2 µ g ( A ) q ′ q ′ 2 Applying the above bounds and optimising over ρ 1 / p yields the desired bound. 1 1 / 2 Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 10 / 15

Parametrix bounds Lemma (Frank and S., 2016) Let δ > 0 and either ( p , q ) = ( 2 n ( n +1) n +1 , 2 n ( n +1) n − 1 ) or ( p , q ) = ( 2 n 2 n n 2 +4 n − 1 , n 2 − 2 n +1 ) . Then, if | z | ≥ δ , 1 � T ( z ) u � q , ∞ � | z | − n +1 � u � p , 1 . Corollary (Frank and S., 2016) Let δ > 0 and let 1 ≤ p ≤ 2 ≤ q with n 2 − 2 n + 1 p − 1 1 2 < 1 q < n − 1 q = n + 1 , . 2 n ( n + 1) 2 n Then, if | z | ≥ δ , 1 � T ( z ) u � q � | z | − n +1 � u � p . In fact, the interpolation yields the inequality 1 � T ( z ) u � q , s � | z | − n +1 � u � p , s for 1 ≤ s ≤ ∞ , which for p < s < q is stronger than the one stated above. Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 11 / 15

Optimality of the parabolic area of exclusion Theorem (Frank and S., 2016) Let M be a compact Riemannian manifold without boundary of dimension n ≥ 2 which is Zoll. Then there is a constant C > 0 such that for any function δ : R → (0 , ∞ ) with lim | κ |→∞ δ ( κ ) = 0 and lim inf | κ |→∞ | κ | δ ( κ ) ≥ C, 1 � − ∆ − ( κ + i δ ( κ )) 2 � − 1 � � ( κ + i δ ( κ )) 2 � � � n +1 � � lim sup L 2( n +1) / ( n +3) → L 2( n +1) / ( n − 1) = ∞ . � � � � � | κ |→∞ � � If M = S n (with the standard metric), this holds also with C = 0 . Im z Note that Im √ z = δ can be written as ( Im z ) 2 = 4 δ 2 ( Re z + δ 2 ) Re z and that z = ( κ + i δ ( κ )) 2 is on the curve ( Im z ) 2 = 4 δ ( κ ) 2 ( Re z + δ ( κ ) 2 ) Schimmer (Caltech) Endpoint resolvent estimates 02/13/20167 12 / 15