The reasons behind some classical constructions in analysis V. - PowerPoint PPT Presentation

The reasons behind some classical constructions in analysis V. Milman Tel-Aviv University In the memory of a great scientist and a good friend, Aleksander (Olek) Pełczyński June 2014, Będlewo, Poland 2/25

Instead of an Introduction: some geometric results What should we call “duality”? Consider the class Cvx ( R n ) of all lower-semi-continuous convex functions f : R n → R ∪ { + ∞ } . The Legendre transform is the map � � x , y � − ϕ ( y ) L ϕ ( x ) = sup � . y ∈ R n Of course, there are many “Legendre transforms”: We may select 0 of the space, a scalar product and a shift for a function. Theorem (Artstein–Milman) 1. Assume T : Cvx ( R n ) → Cvx ( R n ) satisfies: (a) T · T ϕ = ϕ (for any ϕ ∈ Cvx ( R n )) ; (b) ϕ ≤ ψ implies T ϕ ≥ T ψ . Then T is a Legendre transform. It means that Cvx ( R n ) has a unique duality structure! 3/25

Let us embed K ( R n ) = { K ⊆ R n | closed convex } into Cvx ( R n ) by “convex characteristic” functions: � 0 x ∈ K , → 1 ∞ K − K = + ∞ x �∈ K . For K ∈ K 0 ( R n ) = { K ∈ K ( R n ) | 0 ∈ K } , define its gauge function (or Minkowski functional M ( 1 ∞ K ) ) – 1-homogeneous convex function � x � K , “generalized” norm, s.t. K = { x ∈ R n | � x � K ≤ 1 } . Let H 0 = {� x � K | K ∈ K 0 ( R n ) } . Define the Minkowski map M ( 1 ∞ K ) = � x � K ∈ H 0 and the support map S ( 1 ∞ K ) = sup {� x , y � | y ∈ K } ∈ H 0 . Obviously, M : K 0 ( R n ) → H 0 is an order preserving map (1-1 and onto) and S : K 0 ( R n ) → H 0 is an order reversing map. 4/25

We continue the theorem: Theorem (Artstein–Milman) 2. There is a unique order reversing extension of the support map S to Cvx ( R n ) which is the Legendre transform. And what is the polarity map on K 0 ( R n ) K → K ◦ = { x ∈ R n | � x , y � ≤ 1 ∀ y ∈ K } ? K ◦ is defined if 0 ∈ K . So, let Cvx 0 ( R n ) = { f ∈ Cvx ( R n ) | f ( x ) ≥ 0 and f ( 0 ) = 0 } be the class of geometric convex functions. 3. There is a unique order reversing extension of the polarity map { 1 ∞ K → 1 ∞ K ◦ | K ∈ K 0 } to Cvx 0 ( R n ) \ 0 defined by A f = sup � x , y � − 1 , f ( y ) and A ( 0 ) : = 1 ∞ { 0 } . (By extension we mean that A 1 ∞ K = 1 ∞ K ◦ ) 5/25

There are ONLY two dualities on Cvx 0 ( R n ) – L and A : Theorem (Artstein–Milman) Let n ≥ 2 . The maps L and A are (essentially) the only order reversing involutions on Cvx 0 ( R n ) . Precisely: if T : Cvx 0 → Cvx 0 is 1. involution T · T = Id. 2. order reversing: ∀ f , g ∈ Cvx 0 we have f ≤ g ⇒ Tf ≥ Tg , then ∃ C > 0 and B ∈ GL n , symmetric, s.t. either ∀ f ∈ Cvx 0 , Tf = L ( f ( Bx )) or ∀ f ∈ Cvx 0 , Tf = C A ( f ( Bx )) . (when n = 1 there are 8 such different dualities) 6/25

Consider the order preserving map (involution) J = LA = AL which connects two dualities (supporting map – Legendre transform L , and geometric duality A ). J is a very interesting map Cvx 0 → Cvx 0 , order preserving. It is the gauge map: [Fact]Artstein-Milman. J is the only order preserving extension of the Minkowski map M onto Cvx 0 ( R n ) , i.e. J ( 1 ∞ K ) ≡ M ( 1 ∞ K ) = � x � K . So, on the class of convex functions we have the notion of support function, Minkowski functional and polarity! 7/25

Classical constructions in analysis which appear (uniquely) from elementary (simplest) properties Now let us look into the continuation of the previous geometric ideas applied to the problems of Analysis . We start with a characterization of the classical Fourier transform F on R n : F f = � e − 2 π i � x , y � f ( y ) dy . Let S be the Schwartz class of “rapidly” decreasing (infinitely smooth) functions on R n . Theorem (Artstein, Faifman, Milman) Assume we are given a bijective transform F : S → S , s.t. ∀ f , g ∈ S we have F ( f · g ) = F f ∗ F g . Then ∃ diffeomorphism ω : R n → R n s.t. either ∀ f ∈ S, F f = F ( f ◦ ω ) or ∀ f ∈ S , F f = F ( f ◦ ω ) . 8/25

Real linearity and continuity of F is the automatic consequence. Previous versions contained more conditions and were proved jointly with S. Alesker. Joining these results with the previous theorem we may state that if F : S → S s.t. ∀ f , g ∈ S F ( f · g ) = F f ∗ F g , F ( f ∗ g ) = F f · F g , then ∃ linear A ∈ GL n , | det ( A ) | = 1, s.t. either ∀ f ∈ S , F f = F ( f ◦ A ) or ∀ f ∈ S , F f = F ( f ◦ A ) . 9/25

Derivative; Approach through the chain rule What algebraic property characterizes derivative on R ? Let f bounded from above or from � � C 1 f ∈ C 1 ( R ) | b ( R ) = . below (or both) write ( f ◦ g )( x ) : = f ( g ( x )) (composition). We say, T : Dom ( T ) = C 1 ( R ) → Im ( T ) ⊂ C ( R ) is non-degenerate if ∃ x 0 ∈ R , h 0 ∈ C 1 ( R ) , ( Th 0 )( x 0 ) = 0 , (1) ∃ x 2 ∈ R , h 2 ∈ C 1 b ( R ) , ( Th 2 )( x 2 ) < 0 . (2) (a very weak surjectivity type condition). 10/25

Theorem (Artstein, König, Milman) Let T : D om ( T ) : = C 1 ( R ) → Im ( T ) ⊂ C ( R ) be an operation satisfying the chain rule (3) T ( f ◦ g ) = ( Tf ) ◦ g · Tg ; f , g ∈ D ( T ) . Assume that T is non-degenerate in the sense of (1)+(2). (Normalizing conditions) T ( 2 Id ) = 2 (is a constant function), then Tf = f ′ is the only solution. No linearity or any kind of continuity of T is assumed, but is the consequence (under the above conditions) of the chain rule only. 11/25

Without the normalization condition the result is: ∃ strictly monotone continuously differentiable function G ( x ) and p > 0, s.t. p � d ( G ◦ f ) � · sgn ( f ′ ) . � � Tf = � � dG � � ( T : C k → C ) (and not Note: If T ( f ◦ g ) = Tf · Tg degenerate at any x ∈ R ) then Tf = 1 , ∀ f ∈ C k . So, we need ( Tf ) ◦ g · Tg to build a non-degenerate operation. The same answer may be written differently: ∃ H ∈ C ( R ) , H > 0 and p > 0 s.t. Tf = H ( f ( x )) � p sgn ( f ′ ) . � f ′ � � · H ( x ) 12/25

The chain rule (3) has a natural (and unique) domain on which it acts (with the image inside C ( R ) ), and it is C 1 ( R ) . Facts from (A-K-M). Let T : L → C ( R ) satisfy (3) for f , g , f ◦ g ∈ L . If: ◮ L = C ( R ) and ∃ g 0 ∈ C and x 0 ∈ R s.t. ( Tg 0 )( x 0 ) = 0 . Then T | C b ( R ) ≡ 0 . ◮ C ∞ ( R ) ⊂ L ⊂ C 1 ( R ) = ⇒ T maybe extended to C 1 ( R ) ; So the natural Dom ( T ) is C 1 ! 13/25

Rigidity of the chain rule To what extent is the standard chain rule inferred by a much weaker version? Let V : C 1 ( R ) → C ( R ) be non-degenerate, i.e. 1. For any x ∈ R there is f ∈ C 1 ( R ) such that Vf ( x ) � = 0 , and 2. For any x ∈ R there are y ∈ R and f ∈ C 1 ( R ) such that f ( y ) = x and Vf ( y ) � = 0 . 14/25

Theorem (König-Milman) Assume that V , T 1 , T 2 : C 1 ( R ) → C ( R ) are operators such that the equation V ( f ◦ g ) = ( T 1 f ) ◦ g · ( T 2 g ) holds for all f , g ∈ C 1 ( R ) . Assume that V is non-degenerate. There is a solution T of the chain rule equation T ( f ◦ g ) = ( Tf ) ◦ g · Tg , s.t. ( Vf )( x ) = c 1 ( f ( x )) c 2 ( x ) · ( Tf )( x ) ( T 1 f )( x ) = c 1 ( f ( x )) · ( Tf )( x ) ( T 2 f )( x ) = c 2 ( x ) · ( Tf )( x ) , i.e. all three a priori different operators are essentially the same (a strong super rigidity). 15/25

Stability of the Chain Rule T : C 1 ( R ) → C ( R ) is locally non-degenerate if ∀ open interval J ⊂ R , ∀ x ∈ J , ∃ g ∈ C 1 ( R ) , y ∈ R , s.t. g ( y ) = x , Im ( g ) ⊂ J and Tg ( y ) � = 0. Theorem (König-Milman) Fix T : C 1 ( R ) → C ( R ) and B : R 3 → R such that ∀ f , g ∈ C 1 and ∀ x ∈ R T ( f ◦ g )( x ) = Tf ◦ g ( x ) · Tg ( x ) + B ( x , f ◦ g ( x ) , g ( x )) . Assume that T is locally non-degenerate and Tf depends non-trivially on f ′ . Then B = 0 (and T satisfies the chain rule). 16/25

Even more rigidity Consider the “chain rule inequality” ( ∗ ) T ( f ◦ g ) ≤ ( Tf ) ◦ g · Tg for T : C 1 ( R ) → C ( R ) , Dom ( T ) = C 1 ( R ) . Assume that T satisfies the following: ◮ non-degeneration: ∀ open interval I ⊂ R , ∀ x ∈ I , ∃ g ∈ C 1 ( R ) s.t. g ( x ) = x , Im ( g ) ⊂ I and Tg ( x ) > 1. ◮ T is pointwise continuous: ∀ f , f n ∈ C 1 ( R ) s.t f n → f , n → f ′ uniformly on compact subsets we have f ′ ( Tf n ) ( x ) → ( Tf ) ( x ) pointwise for all x ∈ R . 17/25

Theorem (König-Milman) For T as above assume also ∃ x ∈ R s.t T ( − Id )( x ) < 0 . Then ∃ H ∈ C ( R ) , H > 0 , ∃ p > 0 and A ≥ 1 s.t for f ′ ≥ 0 � H | f ′ | p H ◦ f Tf = − A H ◦ f H | f ′ | p for f ′ < 0 . Note: ◮ For A = 1, T satisfies the chain rule equation: we have equality in ( ∗ ) . ◮ For both f and g non-decreasing we automatically have equality in ( ∗ ) . ◮ Actually, the same is true if for some C > 0 T ( f ◦ g ) ≤ C · ( Tf ) ◦ g · Tg and even much more generally (the answer is slightly modified). 18/25

The following classically sound functional statement is used: We say that K : R → R is submultiplicative if K ( αβ ) ≤ K ( α ) K ( β ) , ∀ α , β ∈ R . Theorem (König-Milman) Let K be submultiplicative, measurable and continuous at 0 and at 1 . Assume K ( − 1 ) < 0 < K ( 1 ) . Then ∃ p > 0 s.t. � α p for α ≥ 0 K ( α ) = − A | α | p for α < 0 (and K ( − 1 ) = − A ≤ − 1 ). [every assumption in the theorem is needed] As a corollary, K must be multiplicative on R + . 19/25