mu-differentiability of an internal function Ricardo Almeida and V - PowerPoint PPT Presentation

mu-differentiability of an internal function Ricardo Almeida and V ´ ıtor Neves University of Aveiro, Portugal May 2006

Definition [Reeken, 1992] Let E and F be normed spaces, U ⊆ E open set and f : ∗ U → ∗ F an internal function. f is m-differentiable if 1. for all a ∈ σ U there exist 0 ≈ δ a ∈ ∗ R + and a finite linear operator Df a ∈ ∗ L ( E, F ) such that, for all x ∈ ∗ U , there is some η ≈ 0 with δ a < | x − a | ≈ 0 ⇒ f ( x ) − f ( a ) = Df a ( x − a ) + | x − a | η 2. f ( ns ( ∗ U )) ⊆ ns ( ∗ F ). Theorem [Schlesinger, 1997] If E and F are finite dimensional and f : ∗ K → ∗ F an internal function, with K a compact set, then the following statements are equivalent: 1. f is S-continuous and m-differentiable; 2. There exists a differentiable standard function g : K → F with sup | f ( x ) − g ( x ) | ≈ 0 . x ∈ ∗ K 1

Definition Let E and F be normed spaces, U ⊆ E open set and f : ∗ U → ∗ F an internal function. f is mu-differentiable if 1. for each a ∈ σ U there exists a positive infinitesimal δ a such that, for all x ∈ µ ( a ), there exists a finite linear operator Df x ∈ ∗ L ( E, F ) for which holds ∀ y ∈ µ ( a ) | x − y | > δ a ⇒ f ( x ) − f ( y ) = Df x ( x − y ) + | x − y | η for some η ≈ 0. 2. f ( ns ( ∗ U )) ⊆ ns ( ∗ F ). Theorem Let f : ∗ U → ∗ F be a mu-differentiable function. Then, for all x, y ∈ ns ( ∗ U ) with x ≈ y , we have 1. f ( x ) ≈ f ( y ); 2. if d ∈ ∗ E with | d | = 1, Df x ( d ) ≈ Df y ( d ). 2

Theorem If E and F are finite dimensional and f : ∗ U → ∗ F an internal function, then: 1. If f is mu-differentiable then st ( f ) : U → F is a C 1 function, Dst ( f ) a = stDf a for a ∈ σ U and ∀ a ∈ σ U ∃ η 0 ≈ 0 ∀ x ≈ a | f ( x ) − st ( f )( x ) | ≤ η 0 . 2. If there exists a C 1 standard function g : U → F with ∀ a ∈ σ U ∃ η 0 ≈ 0 ∀ x ≈ a | f ( x ) − g ( x ) | ≤ η 0 , then f is mu-differentiable. The function g = st ( f ). Proof (1) (...) 3

(2) Fix a ∈ σ U and let δ a := √ η 0 . Fix x, y ∈ µ ( a ) with δ a < | x − y | . Since g is of class C 1 then g ( x ) − g ( y ) = Dg x ( x − y ) + | x − y | η for some η ≈ 0. Define ǫ 1 := g ( x ) − f ( x ) and ǫ 2 := g ( y ) − f ( y ). Then f ( x ) − f ( y ) = Dg x ( x − y ) + | x − y | η + ǫ 2 − ǫ 1 and | ǫ 1 − ǫ 2 | | x − y | ≤ | ǫ 1 | + | ǫ 2 | ≤ 2 η 0 ≈ 0 . √ η 0 | x − y | ✷ 4

Corollary If f : U → F is a standard function then f is of class C 1 ⇔ f is mu-differentiable Theorem If f : ∗ U → ∗ F is a mu-differentiable function, then • ∀ a ∈ σ U ∃ δ ≈ 0 ∀ d ∈ ∗ E ∃ L ∈ fin ( ∗ F ) ∀ x ∈ ∗ U � x ≈ a ⇒ f ( x + δd ) − f ( x ) � | d | = 1 ∧ ≈ L . δ • ∀ x ∈ ns ( ∗ U ) ∃ δ x ≈ 0 ∃ Df x ∈ ∗ L ( E, F ) ∀ y ∈ ∗ U ∃ η ≈ 0 | Df x | is finite ∧ [ δ x < | x − y | ≈ 0 ⇒ f ( x ) − f ( y ) = Df x ( x − y ) + | x − y | η ] . • ∀ a ∈ σ U ∃ δ a ≈ 0 ∃ Df a ∈ ∗ L ( E, F ) ∀ x, y ∈ µ ( a ) ∃ η ≈ 0 | Df a | is finite ∧ [ | x − y | > δ a ⇒ f ( x ) − f ( y ) = Df a ( x − y ) + | x − y | η ] . 5

Theorem If E and F are finite dimensional and f : ∗ U → ∗ F an internal function, then: then st ( f ) : U → F is a C k function, 1. If f is k -times mu-differentiable D j st ( f ) a = stD j f a for j = 1 , 2 , . . . , k and a ∈ σ U . Furthermore ∀ j ∈ { 0 , 1 , . . . , k − 1 } ∀ a ∈ σ U ∃ η j ≈ 0 ∀ x ≈ a | D j f x − D j st ( f ) x | ≤ η j . 2. If there exists a C k standard function g : U → F with ∀ j ∈ { 0 , 1 , . . . , k − 1 } ∀ a ∈ σ U ∃ η j ≈ 0 ∀ x ≈ a | D j f x − D j g x | ≤ η j then f is k -times mu-differentiable and g = st ( f ). 6

Taylor’s Theorem. Let E and F be two standard finite dimensional normed ∗ U → ∗ F a function k -times mu- spaces, U a standard open set and f : differentiable, k ∈ σ N . Then, 1. for every x ∈ ns ( ∗ U ), there exists ǫ ≈ 0 such that, whenever y ∈ ∗ U with ǫ < | y − x | ≈ 0, there exists η ≈ 0 satisfying f ( y ) = f ( x )+ Df x ( y − x )+ 1 2! D 2 f x ( y − x ) (2) + ... + 1 k ! D k f x ( y − x ) ( k ) + | y − x | k η. 2. for every x ∈ ns ( ∗ U ), there exists ǫ ≈ 0 such that, whenever y ∈ ∗ U with ǫ < | y − x | ≈ 0, there exists η ≈ 0 satisfying f ( y ) = st ( f )( x ) + Dst ( f ) x ( y − x ) + 1 2! D 2 st ( f ) x ( y − x ) (2) + . . . + 1 k ! D k st ( f ) x ( y − x ) ( k ) + | y − x | k η. 7

Proof (1) Define the sequence ( ǫ i ) i = − 1 ,...,k − 1 by • f ( y ) = st ( f )( y ) + ǫ − 1 , ( ǫ − 1 ≤ η 0 ); • f ( x ) = st ( f )( x ) + ǫ 0 , ( ǫ 0 ≤ η 0 ); • Df x ( y − x ) = Dst ( f ) x ( y − x ) + | y − x | ǫ 1 , ( ǫ 1 ≤ η 1 ); • ... • D k − 1 f x ( y − x ) ( k − 1) = D k − 1 st ( f ) x ( y − x ) ( k − 1) + | y − x | k − 1 ǫ k − 1 , ( ǫ k − 1 ≤ η k − 1 ); Furthermore � y − x � y − x � y − x � y − x � ( k ) � ( k ) � ( k ) � ( k ) D k f x ≈ D k f a ≈ D k st ( f ) a ≈ D k st ( f ) x , | y − x | | y − x | | y − x | | y − x | so there exists ǫ k ≈ 0 with D k f x ( y − x ) ( k ) = D k st ( f ) x ( y − x ) ( k ) + | y − x | k ǫ k . 8

1 1 1 k − 1 } and take y ∈ ∗ U with ǫ < | y − x | ≈ 0. Define ǫ = max { η k +1 0 , η 1 , ..., η 2 k Since st ( f ) is of class C k , then st ( f )( y ) = st ( f )( x )+ Dst ( f ) x ( y − x )+ 1 2! D 2 st ( f ) x ( y − x ) (2) + ... + 1 k ! D k st ( f ) x ( y − x ) ( k ) + | y − x | k η Consequently f ( y ) = f ( x ) + Df x ( y − x ) + 1 2! D 2 f x ( y − x ) (2) + ... + 1 k ! D k f x ( y − x ) ( k ) + | y − x | k η + + ǫ − 1 − ǫ 0 − | y − x | ǫ 1 − | y − x | 2 ǫ 2 − ... − | y − x | k − 1 ǫ k − 1 − | y − x | k ǫ k and | ǫ − 1 | | ǫ 0 | | ǫ 1 | | y − x | k − 2 + . . . + | ǫ k − 1 | | ǫ 2 | | y − x | k + | y − x | k + | y − x | k − 1 + | y − x | ≤ η 0 + η 0 + η 1 + η 2 + ... + η k − 1 ≤ ≈ 0 . k k k − 1 k − 2 1 k +1 k +1 η k k − 1 η 2 η η η 1 k − 1 0 0 2 1 (2) Define ǫ := η k +1 , (...) 0 ✷ 9

Chain Rule. Let g, f be two m-differentiable functions at a and g ( a ), respectively, where a and g ( a ) are two standards. If Dg a is invertible and | ( Dg a ) − 1 | is finite, then f ◦ g is m-differentiable at a and D ( f ◦ g ) a = Df g ( a ) Dg a Proof Take δ := max { δ a , 2 δ g ( a ) | ( Dg a ) − 1 |} and choose x with δ < | x − a | ≈ 0. Then � x − a � � � 0 ≈ | g ( x ) − g ( a ) | = | Dg a ( x − a )+ | x − a | η 1 | > 2 δ g ( a ) | ( Dg a ) − 1 | � � + η 1 � Dg a � > δ g ( a ) � � | x − a | and f ( g ( x )) − f ( g ( a )) = Df g ( a ) ( g ( x ) − g ( a ))+ | g ( x ) − g ( a ) | η 2 = Df g ( a ) Dg a ( x − a )+ | x − a | η for some η ≈ 0. ✷ 10

Let f : ∗ U → ∗ R be a mu-differentiable function Mean Value Theorem. with U open and convex. Then, for all x, y ∈ ns ( ∗ U ) with | x − y | > δ a , where a := st ( x ) ∃ c ∈ [ x, y ] f ( x ) − f ( y ) = Df c ( x − y ) + | x − y | η for some η ≈ 0 . Proof Define an hyper-finite sequence ( x n ) n ∈ I in the following way: Let x 1 = x and fix δ 1 ≈ 0 with, for all z ∈ ∗ U : δ 1 < | z − x 1 | ≈ 0 ⇒ f ( z ) − f ( x 1 ) = Df x 1 ( z − x 1 ) + | z − x 1 | η 1 . y − x | y − x | and fix 0 ≈ δ 2 > δ 1 with, for all z ∈ ∗ U : Let x 2 = x 1 + 2 δ 1 δ 2 < | z − x 2 | ≈ 0 ⇒ f ( z ) − f ( x 2 ) = Df x 2 ( z − x 2 ) + | z − x 2 | η 2 y − x and take x 3 = x 2 + 2 δ 2 | y − x | . 11

Repeating the process, we obtain a sequence { x n | 1 ≤ n ≤ N + 1 } which satisfies the conditions • x 1 = x ; • x n +1 = x n + 2 δ n y − x | y − x | , δ n ≈ 0 and δ n > δ 1 , n = 1 , . . . , N ; • f ( x n +1 ) − f ( x n ) = Df x n ( x n +1 − x n ) + | x n +1 − x n | η n , for some η n ≈ 0, n = 1 , . . . , N ; • x N +1 = y (if not, choose 0 ≈ δ > δ N with x N + 2 δ y − x | y − x | = y ). Then N N N � � � f ( x ) − f ( y ) = ( f ( x n ) − f ( x n +1 )) = Df x n ( x n − x n +1 ) + | x n − x n +1 | η n n =1 n =1 n =1 and � � �� N n =1 | x n − x n +1 | η n � � � ≈ 0 . | x − y | 12

We will prove now that there exists c ∈ [ x, y ] such that � x − y � N � n =1 Df x n ( x n − x n +1 ) Df c ≈ . | x − y | | x − y | x − y Letting d := | x − y | , it is true that � N � N � N n =1 Df x n ( x n − x n +1 ) n =1 Df x n ( x n − x n +1 ) n =1 2 δ n Df x n ( d ) = = . � N � N | x − y | n =1 | x n − x n +1 | n =1 2 δ n Choosing m, M ∈ { x 1 , ..., x N } with Df m ( d ) = min 1 ≤ n ≤ N Df x n ( d ) & Df M ( d ) = max 1 ≤ n ≤ N Df x n ( d ) , we get � N n =1 2 δ n Df x n ( d ) Df m ( d ) ≤ ≤ Df M ( d ) . � N n =1 2 δ n So, there exists c ∈ [ m, M ] ⊆ [ x, y ] with � N n =1 2 δ n Df x n ( d ) Df c ( d ) ≈ . ✷ � N n =1 2 δ n 13

Norm Mean Value Theorem Let f : ∗ U → ∗ F be a mu-differentiable function with U open and convex. Then, for all x, y ∈ ns ( ∗ U ) with | x − y | > δ a , where a := st ( x ) ∃ c ∈ [ x, y ] | f ( x ) − f ( y ) | ≤ | Df c ( x − y ) | + | x − y | η for some η ≈ 0. 14