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# chap22 - A NALYSIS TOOLS W ITH APPLICATIONS 423 22 Banach...

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ANALYSIS TOOLS WITH APPLICATIONS 423 22. Banach Spaces III: Calculus In this section, X and Y will be Banach space and U will be an open subset of X. Notation 22.1 ( ±, O, and o notation) . Let 0 U o X, and f : U −→ Y be a function. We will write: (1) f ( x )= ± ( x ) if lim x 0 k f ( x ) k =0 . (2) f ( x )= O ( x ) if there are constants C< and r> 0 such that k f ( x ) k C k x k for all x B (0 ,r ) . This is equivalent to the condition that lim sup x 0 k f ( x ) k k x k < , where lim sup x 0 k f ( x ) k k x k lim r 0 sup {k f ( x ) k :0 < k x k r } . (3) f ( x )= o ( x ) if f ( x )= ± ( x ) O ( x ) , i.e. lim x 0 k f ( x ) k / k x k =0 . Example 22.2. Here are some examples of properties of these symbols. (1) A function f : U o X Y is continuous at x 0 U if f ( x 0 + h )= f ( x 0 )+ ± ( h ) . (2) If f ( x )= ± ( x ) and g ( x )= ± ( x ) then f ( x )+ g ( x )= ± ( x ) . Now let g : Y Z be another function where Z is another Banach space. (3) If f ( x )= O ( x ) and g ( y )= o ( y ) then g f ( x )= o ( x ) . (4) If f ( x )= ± ( x ) and g ( y )= ± ( y ) then g f ( x )= ± ( x ) . 22.1. The Di f erential. De f nition 22.3. Afunct ion f : U o X Y is di f erentiable at x 0 + h 0 U if there exists a linear transformation Λ L ( X,Y ) such that (22.1) f ( x 0 + h ) f ( x 0 + h 0 ) Λ h = o ( h ) . We denote Λ by f 0 ( x 0 ) or Df ( x 0 ) if it exists. As with continuity, f is di f erentiable on U if f is di f erentiable at all points in U. Remark 22.4 . The linear transformation Λ in De f nition 22.3 is necessarily unique. Indeed if Λ 1 is another linear transformation such that Eq. (22.1) holds with Λ replaced by Λ 1 , then ( Λ Λ 1 ) h = o ( h ) , i.e. lim sup h 0 k ( Λ Λ 1 ) h k k h k =0 . On the other hand, by de f nition of the operator norm, lim sup h 0 k ( Λ Λ 1 ) h k k h k = k Λ Λ 1 k . The last two equations show that Λ = Λ 1 . Exercise 22.1. Show that a function f :( a, b ) X is a di f erentiable at t ( a, b ) in the sense of De f nition 4.6 i f it is di f erentiable in the sense of De f nition 22.3. Also show Df ( t ) v = v ˙ f ( t ) for all v R

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424 BRUCE K. DRIVER Example 22.5. Assume that GL ( X,Y ) is non-empty. Then f : GL ( X,Y ) GL ( Y,X ) de f ned by f ( A ) A 1 is di f erentiable and f 0 ( A ) B = A 1 BA 1 for all B L ( X,Y ) . Indeed (by Eq. (3.13)), f ( A + H ) f ( A )=( A + H ) 1 A 1 =( A ¡ I + A 1 H ¢ ) 1 A 1 = ¡ I + A 1 H ¢ ) 1 A 1 A 1 = X n =0 ( A 1 H ) n · A 1 A 1 = A 1 HA 1 + X n =2 ( A 1 H ) n . Since
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chap22 - A NALYSIS TOOLS W ITH APPLICATIONS 423 22 Banach...

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