Operator equations and domain dependence Hermann Knig Kiel, Germany - PowerPoint PPT Presentation

Operator equations and domain dependence Hermann König Kiel, Germany Bedlewo, July 2014 Hermann König (Kiel) Operator equations Bedlewo, July 2014 1 / 37

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Operator equations Basic questions Olek’s interest included: Isomorphic classification of C k (Ω) -spaces and of Sobolev spaces. These are natural domains for differential operators. We consider operator equations in C k -spaces modeling derivatives. Joint work with Vitali Milman. Hermann König (Kiel) Operator equations Bedlewo, July 2014 8 / 37

Operator equations Basic questions Olek’s interest included: Isomorphic classification of C k (Ω) -spaces and of Sobolev spaces. These are natural domains for differential operators. We consider operator equations in C k -spaces modeling derivatives. Joint work with Vitali Milman. Aim: Characterize derivatives by simple properties, like the Chain rule: f , g ∈ C 1 ( R ) , D ( f ◦ g ) = ( Df ) ◦ g · Dg ; D 2 ( f ◦ g ) = ( D 2 f ) ◦ g · ( Dg ) 2 + ( Df ) ◦ g · D 2 g f , g ∈ C 2 ( R ) . ; Hermann König (Kiel) Operator equations Bedlewo, July 2014 8 / 37

Operator equations Basic questions Olek’s interest included: Isomorphic classification of C k (Ω) -spaces and of Sobolev spaces. These are natural domains for differential operators. We consider operator equations in C k -spaces modeling derivatives. Joint work with Vitali Milman. Aim: Characterize derivatives by simple properties, like the Chain rule: f , g ∈ C 1 ( R ) , D ( f ◦ g ) = ( Df ) ◦ g · Dg ; D 2 ( f ◦ g ) = ( D 2 f ) ◦ g · ( Dg ) 2 + ( Df ) ◦ g · D 2 g f , g ∈ C 2 ( R ) . ; Replace D or D 2 by a general operator T : C k ( R ) → C ( R ) and study f , g ∈ C k ( R ) , T ( f ◦ g ) = ( Tf ) ◦ g · Tg ; (1) f , g ∈ C k ( R ) , T ( f ◦ g ) = ( Tf ) ◦ g · A 1 g + ( A 2 f ) ◦ g · Tg ; (2) with A 1 , A 2 : C l ( R ) → C ( R ) , l < k . Similar equations for the Leibniz rule. Hermann König (Kiel) Operator equations Bedlewo, July 2014 8 / 37

Operator equations Basic questions Olek’s interest included: Isomorphic classification of C k (Ω) -spaces and of Sobolev spaces. These are natural domains for differential operators. We consider operator equations in C k -spaces modeling derivatives. Joint work with Vitali Milman. Aim: Characterize derivatives by simple properties, like the Chain rule: f , g ∈ C 1 ( R ) , D ( f ◦ g ) = ( Df ) ◦ g · Dg ; D 2 ( f ◦ g ) = ( D 2 f ) ◦ g · ( Dg ) 2 + ( Df ) ◦ g · D 2 g f , g ∈ C 2 ( R ) . ; Replace D or D 2 by a general operator T : C k ( R ) → C ( R ) and study f , g ∈ C k ( R ) , T ( f ◦ g ) = ( Tf ) ◦ g · Tg ; (1) f , g ∈ C k ( R ) , T ( f ◦ g ) = ( Tf ) ◦ g · A 1 g + ( A 2 f ) ◦ g · Tg ; (2) with A 1 , A 2 : C l ( R ) → C ( R ) , l < k . Similar equations for the Leibniz rule. - General solutions of (1) and (2) ? - Dependence of the solutions on the domain and range spaces? Hermann König (Kiel) Operator equations Bedlewo, July 2014 8 / 37

Chain rule operator equation The chain rule functional equation The chain rule functional equation Chain rule for f , g ∈ C 1 ( R ) : D ( f ◦ g ) = ( Df ) ◦ g · Dg . Let T : C 1 ( R ) → C ( R ) be an operator satisfying the functional equation f , g ∈ C 1 ( R ) , x ∈ R . T ( f ◦ g )( x ) = ( Tf )( g ( x )) · ( Tg )( x ); (1) Which operators T satisfy (1)? Hermann König (Kiel) Operator equations Bedlewo, July 2014 9 / 37

Chain rule operator equation The chain rule functional equation The chain rule functional equation Chain rule for f , g ∈ C 1 ( R ) : D ( f ◦ g ) = ( Df ) ◦ g · Dg . Let T : C 1 ( R ) → C ( R ) be an operator satisfying the functional equation f , g ∈ C 1 ( R ) , x ∈ R . T ( f ◦ g )( x ) = ( Tf )( g ( x )) · ( Tg )( x ); (1) Which operators T satisfy (1)? Examples: ( Tf )( x ) = | f ′ ( x ) | p ( Tf )( x ) = sgn f ′ ( x ) | f ′ ( x ) | p a) p > 0 , and both satisfy (1). b) Let H ∈ C ( R ) , H > 0. Define ( Tf )( x ) = H ( f ( x )) / H ( x ) . Then T satisfies (1). c) Consider f ′ ( x ) � f ∈ C 1 ( R ) bijective � T : C 1 ( R ) → C ( R ) , ( Tf )( x ) = . 0 else Then T satisfies (1). Hermann König (Kiel) Operator equations Bedlewo, July 2014 9 / 37

Chain rule operator equation Solutions of the chain rule operator equation Solutions of the chain rule operator equation T : C k ( R ) → C ( R ) is C k -non-degenerate if T | C k b ( R ) � = 0 where C k b ( R ) are the (half-) bounded functions in C k ( R ) . Here k ∈ N 0 . Multiplying two solutions of the chain rule yields again a solution. Hermann König (Kiel) Operator equations Bedlewo, July 2014 10 / 37

Chain rule operator equation Solutions of the chain rule operator equation Solutions of the chain rule operator equation T : C k ( R ) → C ( R ) is C k -non-degenerate if T | C k b ( R ) � = 0 where C k b ( R ) are the (half-) bounded functions in C k ( R ) . Here k ∈ N 0 . Multiplying two solutions of the chain rule yields again a solution. Theorem Assume T : C k ( R ) → C ( R ) satisfies the chain rule f , g ∈ C k ( R ) T ( f ◦ g ) = ( Tf ) ◦ g · Tg ; for k ∈ N 0 and that T is C k -non-degenerate. Then there is p ≥ 0 and H ∈ C > 0 ( R ) such that for any f ∈ C k ( R ) ( Tf )( x ) = H ( f ( x )) | f ′ ( x ) | p { sgn f ′ ( x ) } , H ( x ) and this also holds for k = ∞ , i.e. f ∈ C ∞ ( R ) . Hermann König (Kiel) Operator equations Bedlewo, July 2014 10 / 37

Chain rule operator equation Chain rule operator equation Steps in the Proof a) Show localization: There is F : R k + 2 → R such that Tf ( x ) = F ( x , f ( x ) , ..., f ( k ) ( x )) for all f ∈ C k ( R ) and x ∈ R . b) Analyze the structure of the representing function F : F ( x , α 0 , ..., α k ) = H ( α 0 ) H ( x ) K ( α 1 ) , K multiplicative, F independent of α 2 , ..., α k if k ≥ 2. c) Show the measurability and then the continuity of the coefficient functions occurring in F . Hermann König (Kiel) Operator equations Bedlewo, July 2014 11 / 37

Chain rule operator equation Localization Localization in the case k = 1 Proposition If T satisfies the chain rule (1) in C 1 ( R ) and is C 1 -non-degenerate, there is F : R 3 → R such that ( Tf )( x ) = F ( x , f ( x ) , f ′ ( x )); x ∈ R , f , g ∈ C 1 ( R ) . Hermann König (Kiel) Operator equations Bedlewo, July 2014 12 / 37

Chain rule operator equation Localization Localization in the case k = 1 Proposition If T satisfies the chain rule (1) in C 1 ( R ) and is C 1 -non-degenerate, there is F : R 3 → R such that ( Tf )( x ) = F ( x , f ( x ) , f ′ ( x )); x ∈ R , f , g ∈ C 1 ( R ) . Step 1. First show localization on open intervals J ⊂ R : a) If g | J = Id , Tg | J = 1: For x ∈ J , there are h ∈ C 1 ( R ) , y ∈ R with Im h ⊂ J , h ( y ) = x and Th ( y ) � = 0. Then g ◦ h = h and 0 � = Th ( y ) = Tg ( x ) Th ( y ) , i.e. Tg ( x ) = 1, Tg | J = 1. Hermann König (Kiel) Operator equations Bedlewo, July 2014 12 / 37

Chain rule operator equation Localization Localization in the case k = 1 Proposition If T satisfies the chain rule (1) in C 1 ( R ) and is C 1 -non-degenerate, there is F : R 3 → R such that ( Tf )( x ) = F ( x , f ( x ) , f ′ ( x )); x ∈ R , f , g ∈ C 1 ( R ) . Step 1. First show localization on open intervals J ⊂ R : a) If g | J = Id , Tg | J = 1: For x ∈ J , there are h ∈ C 1 ( R ) , y ∈ R with Im h ⊂ J , h ( y ) = x and Th ( y ) � = 0. Then g ◦ h = h and 0 � = Th ( y ) = Tg ( x ) Th ( y ) , i.e. Tg ( x ) = 1, Tg | J = 1. b) If f 1 | J = f 2 | J for f 1 , f 2 ∈ C 1 ( R ) , then ( Tf 1 ) | J = ( Tf 2 ) | J : For x ∈ J , there are J 1 ⊂ J , g ∈ C 1 ( R ) with x ∈ J 1 , g | J 1 = Id , Im g ⊂ J . Hence Tg ( x ) = 1, f 1 ◦ g = f 2 ◦ g . Thus Tf 1 ( x ) = Tf 1 ( g ( x ) Tg ( x ) = T ( f 1 ◦ g )( x ) = T ( f 2 ◦ g )( x ) = Tf 2 ( x ) , i.e. Tf 1 | J = Tf 2 | J . Hermann König (Kiel) Operator equations Bedlewo, July 2014 12 / 37

Chain rule operator equation Localization Step 2. For f ∈ C 1 ( R ) and x 0 ∈ R , define g to be the tangent line at x 0 , g ( x ) := f ( x 0 ) + f ′ ( x 0 )( x − x 0 ) ; x ∈ R . � f ( x ) � x < x 0 . By construction h ∈ C 1 ( R ) . Let h ( x ) := g ( x ) x ≥ x 0 Let I 1 = ( −∞ , x 0 ) , I 2 = ( x 0 , ∞ ) . Then f | I 1 = h | I 1 , h | I 2 = g | I 2 . By Step 1, ( Tf ) | I 1 = ( Th ) | I 1 , ( Th ) | I 2 = ( Tg ) | I 2 . Hermann König (Kiel) Operator equations Bedlewo, July 2014 13 / 37