On Birmans Sequence of Hardy-Rellich Type Inequalities Isaac B. - PowerPoint PPT Presentation

On Birman’s Sequence of Hardy-Rellich Type Inequalities Isaac B. Michael (joint with F. Gesztesy, L.L. Littlejohn and R. Wellman) IWOTA Conference - Chemnitz August 14-18, 2017 1 / 29

Introduction In 1961, M. ˇ S. Birman established the following sequence of integral inequalities: M. ˇ S Birman (1961) For n ∈ N and f ∈ C n 0 ((0 , ∞ )) , � � � ∞ � ∞ 2 � 2 dx ≥ ((2 n − 1)!!) 2 � � � � f ( x ) � f ( n ) ( x ) � � dx . ( I n ) � � 2 2 n x n 0 0 In particular, I 1 is the classical Hardy inequality � � � ∞ � ∞ 2 � � � � � 2 dx ≥ 1 f ( x ) � f ′ ( x ) � � dx , � � 4 x 0 0 and I 2 is the Rellich inequality � � � ∞ � ∞ 2 � � � � � 2 dx ≥ 9 f ( x ) � f ′′ ( x ) � � dx . � � x 2 16 0 0 2 / 29

Introduction Our joint paper shows: A new proof of Birman’s inequalities on a more general Hilbert space H n ([0 , ∞ )) of functions on [0 , ∞ ). For any 0 < b < ∞ , these inequalities hold on the standard Sobolev space H n 0 ((0 , b )). Birman’s constants ((2 n − 1)!!) 2 / 2 2 n in these inequalities are best possible and the only function that gives equality is the function identically zero in L 2 ((0 , ∞ )) . Birman’s inequalities are closely related to a sequence of generalized aro operators , { T n } , with interesting spectral continuous Ces´ properties. This generalized sequence of inequalities extends mutatis mutandis to H -valued functions, where H is a separable Hilbert space . 3 / 29

The Function Spaces H n ([0 , ∞ )) and H n ((0 , ∞ )) ′ Definition 1 (The Function Space H n ([0 , ∞ ))) Let n ∈ N . Define the function space H n ([0 , ∞ )) via � � � f ( j ) ∈ AC loc ([0 , ∞ )); f ( n ) ∈ L 2 ((0 , ∞ )); H n ([0 , ∞ )) := f : [0 , ∞ ) → C � f ( j ) (0) = 0 , j = 0 , 1 , . . . , n − 1 . Note: G. H. Hardy , J. E. Littlewood , and G. P´ olya proved the classical Hardy inequality I 1 on H 1 in 1934. The fact f ∈ H n ([0 , ∞ )) ⇒ f ′ ∈ H n − 1 ([0 , ∞ )) . is important in the new proof of Birman’s inequalities. When endowed with the inner product � ∞ f ( n ) ( x ) g ( n ) ( x ) dx , ( f , g ) H n ([0 , ∞ )) := 0 H n ([0 , ∞ )) is a Hilbert space . 4 / 29

The Function Spaces H n ([0 , ∞ )) and H n ((0 , ∞ )) ′ Proposition 1 The inner product space ( H n ([0 , ∞ )) , ( · , · ) H n ([0 , ∞ )) ) is actually a Hilbert space. In addition, C ∞ 0 ((0 , ∞ )) is dense in ( H n ([0 , ∞ )) , ( · , · ) H n ([0 , ∞ )) ) . Caution: We emphasize that H n ([0 , ∞ )) � = H n 0 ((0 , ∞ )) , n ∈ N , with H n 0 ((0 , ∞ )) the standard Sobolev space obtained upon completing C ∞ 0 ((0 , ∞ )) in the norm of H n ((0 , ∞ )). Indeed, define � f ∈ H n ([0 , ∞ )) via, � 0 , x near 0 , � f ( x ) = x (2 n − 1) / 2 / ln( x ) , x near ∞ , such that f ( j ) ∈ AC loc ([0 , ∞ )) , � j = 0 , 1 , . . . , n . f ( j ) / Calculations show � ∈ L 2 ((0 , ∞ )), 0 ≤ j ≤ n − 1 . 5 / 29

The Function Spaces H n ([0 , ∞ )) and H n ((0 , ∞ )) ′ Theorem 2 is used in the new proof of Birman’s inequalities. Theorem 2 Let f ∈ H n ([0 , ∞ )) . Then (i) f ( n − j ) / x j ∈ L 2 ((0 , ∞ )) for j = 0 , 1 , . . . n ; In particular, ⇒ f ′ ∈ H n − 1 ([0 , ∞ )) ; f ∈ H n ([0 , ∞ )) = � � 2 f ( j ) ( x ) j = 0 , 1 , . . . , n − 1; (ii) lim x →∞ x 2 n − 2 j − 1 = 0 , � � 2 f ( j ) ( x ) (iii) lim x ↓ 0 x 2 n − 2 j − 1 = 0 , j = 0 , 1 , . . . , n − 1 . The above is proved using an integral inequality independently due to G. Tomaselli (1969), G. Talenti (1969), R. S. Chisholm & W. N. Everitt (1971), and B. Muckenhoupt (1972). 6 / 29

The Function Spaces H n ([0 , ∞ )) and H n ((0 , ∞ )) ′ Consider the spaces H n ((0 , ∞ )) ′ and D n ([0 , ∞ )) given below. Definition 3 (The Function Space H n ((0 , ∞ )) ′ ) Let n ∈ N . Define the function space H n ((0 , ∞ )) ′ via � � H n ((0 , ∞ )) ′ := � f ( j ) ∈ AC loc ((0 , ∞ )) , f : (0 , ∞ ) → C � j = 0 , 1 , . . . , n − 1; f ( n ) , f / x n ∈ L 2 ((0 , ∞ )) . Definition 4 (The Function Space D n ([0 , ∞ ))) Let n ∈ N . Define the function space D n ([0 , ∞ )) via � �� x � t 1 � t n − 1 � � � � f ∈ L 2 ((0 , ∞ )) D n ([0 , ∞ )) := · · · f ( t ) dtdt n − 1 . . . dt 1 0 0 0 Surprisingly, H n ([0 , ∞ )) is equal to both spaces. Theorem 5 For each n ∈ N , H n ([0 , ∞ )) = H n ((0 , ∞ )) ′ = D n ([0 , ∞ )) . 7 / 29

A New Proof of Birman’s Hardy-Rellich Type Inequalities Theorem 6 (Birman’s Inequalities on H n ([0 , ∞ ))) Let n ∈ N and 0 � = f ∈ H n ([0 , ∞ )) . Then, � � � ∞ � ∞ � � dx > ((2 n − 1)!!) 2 2 � � 2 f ( x ) � � � f ( n ) ( x ) � � dx . � � � 2 2 n x n 0 0 Our new proof of Birman’s inequalities consists of iterating Hardy’s inequality, with repeated use of the elementary inequality 2 xy ≤ ε x 2 + ε − 1 y 2 , x , y ∈ R , ε > 0 . This, integration by parts, and the Cauchy–Schwarz inequality, results in � � ∞ � ∞ | f ( x ) | 2 ( − ε 2 + ε ) � � dx , n = 1 , � 2 dx ≥ � f ( n ) ( x ) x 2 0 � ∞ | f ( x ) | 2 2 2 n − 2 ( − ε 2 + (2 n − 1) ε ) (2 n − 3)!! x 2 n dx , n ≥ 2 . 0 0 Maximizing over ε ∈ (0 , ∞ ) proves the theorem. 8 / 29

Optimality of Birman’s Constant Theorem 7 (Optimality of Birman’s Constant) The constant ((2 n − 1)!!) 2 / 2 2 n in Birman’s inequalities is best possible on H n ([0 , ∞ )) for all n ∈ N . Recalling D n ([0 , ∞ )) = H n ([0 , ∞ )), where �� x � � � t 1 � t n − 1 � � � f ∈ L 2 ((0 , ∞ )) D n ([0 , ∞ )) := · · · f ( t ) dtdt n − 1 . . . dt 1 , 0 0 0 leads to the construction of an interesting linear operator T n . Definition 8 (The Linear Operator T n ) Let n ∈ N . Define the linear operator T n on L 2 ((0 , ∞ )) via � x � t 1 � t n − 1 ( T n f )( x ) := 1 f ∈ L 2 ((0 , ∞ )) . · · · f ( t ) dtdt n − 1 . . . dt 1 , x n 0 0 0 Note: T n f ∈ L 2 ((0 , ∞ )), f ∈ L 2 ((0 , ∞ )) by Thms. 5, 2 ( i ). 9 / 29

Generalized Continuous C´ esaro Operators T n The operator T n is a generalization of the continuous C´ esaro operator on L 2 ((0 , ∞ )), � x ( T 1 f )( x ) = 1 f ∈ L 2 ((0 , ∞ )) f ( x ) dx , x 0 also known as the classical Hardy (integral) operator . Birman’s inequalities are closely related to T n , which posseses several interesting properties of its own. Theorem 9 (Boundedness and Non-Compactness of T n ) Let n ∈ N and define T n as above. Then T n is bounded in L 2 ((0 , ∞ )) with operator norm 2 n � T n � = (2 n − 1)!! . T n is not compact ( it has purely a.c. spectrum ) . 10 / 29

Generalized Continuous C´ esaro Operators T n Theorem 10 (Invertibility of T n ) Define T n , n ∈ N , as above. Then T n is invertible and � � � � � � � f ∈ AC ( n − 1) T − 1 T − n f ∈ L 2 ((0 , ∞ )) dom = dom = ((0 , ∞ )); n 1 loc � x j f ( j ) ∈ L 2 ((0 , ∞ )) , j = 1 , . . . , n , � � ( x ) = d n � � T − 1 T − 1 dx n x n f ( x ) , f ∈ dom . n f n Note: T n is not boundedly invertible as 0 ∈ σ ( T n ), n ∈ N . n ) g , it is necessary that lim x → 0 + ( x n g ( x )) ( j ) = 0, For g = ( T n ◦ T − 1 0 ≤ j ≤ n − 1. Surprisingly, this is consequence of lying in the space above. Lemma 11 ((0 , ∞ )) and x k f ( k ) ∈ L 2 ((0 , ∞ )) for Let n ∈ N . Assume f ∈ AC ( n − 1) loc k = 0 , 1 , . . . , n. Then x ↓ 0 ( x n f ( x )) ( j ) = 0 , lim j = 0 , 1 , . . . , n − 1 . 11 / 29

Generalized Continuous C´ esaro Operators T n We introduce the unitary Mellin transform , M , given by  L 2 ((0 , ∞ ); dx ) → L 2 ( R ; d λ ) ,    � a f �→ ( M f )( λ ) ≡ f ∗ ( λ ) := (2 π ) − 1 / 2 s-lim a →∞ 1 / a f ( x ) x − (1 / 2)+ i λ dx M :   for a.e. λ ∈ R ,   L 2 ( R ; d λ ) → L 2 ((0 , ∞ ); dx ) ,   M − 1 :  � b f ∗ �→ ( M − 1 f ∗ )( x ) ≡ f ( x ) := (2 π ) − 1 / 2 s-lim b →∞ − b f ∗ ( λ ) x − (1 / 2) − i λ d λ   for a.e. x ∈ (0 , ∞ ).  The fact, � d � dx x − 1 x − (1 / 2) − i λ = λ x − (1 / 2) − i λ , i x ∈ (0 , ∞ ) , λ ∈ R , 2 leads to the following definition of the operator S 1 in L 2 ((0 , ∞ ); dx ), � � � � T − 1 − 2 − 1 I L 2 ((0 , ∞ )) T − 1 S 1 := i , dom( S 1 ) = dom , 1 1 and shows S 1 is unitarily equivalent to the operator of multiplication by the independent variable in L 2 ( R ; d λ ), � M S 1 M − 1 f ∗ � ( λ ) = λ f ∗ ( λ ) for a.e. λ ∈ R and for all f ∗ ∈ L 2 ( R ; d λ ) such that λ f ∗ ∈ L 2 ( R ; d λ ). 12 / 29

Generalized Continuous C´ esaro Operators T n Summarizing, the Mellin transform diagonalizes S 1 and hence T 1 . Thus, the spectrum, and normality, of T 1 can be determined through study of S 1 . Theorem 12 Define S 1 as above. Then S 1 is self-adjoint and hence T 1 is normal. Moreover, the spectra of S 1 and T 1 are simple and purely absolutely continuous . In particular, σ ( S 1 ) = σ ac ( S 1 ) = R , σ ( T 1 ) = σ ac ( T 1 ) = C (1; 1) . Here C ( z 0 ; r 0 ) ⊂ C denotes the circle of radius r 0 > 0 centered at z 0 ∈ C . Note: The spectrum of T 1 was originally computed by A. Brown , P. R. Halmos , and A. L. Shields in 1965. These preliminary results are important in determining the spectral properties of T n for all n ∈ N . 13 / 29