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Holder-spaces

# Holder-spaces - 4 56 BRUCE K DRIVER 24 Hlder Spaces...

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456 BRUCE K. DRIVER 24. Hölder Spaces Notation 24.1. Let be an open subset of R d ,BC ( ) and BC ( ¯ ) be the bounded continuous functions on and ¯ respectively. By identifying f BC ( ¯ ) with f | BC ( ) , we will consider BC ( ¯ ) as a subset of BC ( ) . For u BC ( ) and 0 1 let k u k u := sup x | u ( x ) | and [ u ] β := sup x,y x 6 = y ½ | u ( x ) u ( y ) | | x y | β ¾ . If [ u ] β < , then u is Hölder continuous with holder exponent 43 β. The collection of β — Hölder continuous function on will be denoted by C 0 ( ):= { u BC ( ):[ u ] β < } and for u C 0 ( ) let (24.1) k u k C 0 ( ) := k u k u +[ u ] β . Remark 24.2 . If u : C and [ u ] β < for some β> 1 , then u is constant on each connected component of . Indeed, if x and h R d then ¯ ¯ ¯ ¯ u ( x + th ) u ( x ) t ¯ ¯ ¯ ¯ [ u ] β t β /t 0 as t 0 which shows h u ( x )=0 for all x . If y is in the same connected component as x, then by Exercise 17.5 there exists a smooth curve σ :[0 , 1] such that σ (0) = x and σ (1) = y. So by the fundamental theorem of calculus and the chain rule, u ( y ) u ( x )= Z 1 0 d dt u ( σ ( t )) dt = Z 1 0 0 dt =0 . This is why we do not talk about Hölder spaces with Hölder exponents larger than 1 . Lemma 24.3. Suppose u C 1 ( ) BC ( ) and i u BC ( ) for i =1 , 2 ,...,d, then u C 0 , 1 ( ) , i.e. [ u ] 1 < . The proof of this lemma is left to the reader as Exercise 24.1. Theorem 24.4. Let be an open subset of R d . Then (1) Under the identi f cation of u BC ¡ ¯ ¢ with u | BC ( ) ,BC ( ¯ ) is a closed subspace of BC ( ) . (2) Every element u C 0 ( ) has a unique extension to a continuous func- tion (still denoted by u ) on ¯ . Therefore we may identify C 0 ( ) with C 0 ( ¯ ) BC ( ¯ ) . (In particular we may consider C 0 ( ) and C 0 ( ¯ ) to bethesamewhen β> 0 . ) (3) The function u C 0 ( ) k u k C 0 ( ) [0 , ) is a norm on C 0 ( ) which make C 0 ( ) into a Banach space. Proof. 1. The f rst item is trivial since for u BC ( ¯ ) , the sup-norm of u on ¯ agrees with the sup-norm on and BC ( ¯ ) is complete in this norm. 43 If β =1 ,u is is said to be Lipschitz continuous.

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ANALYSIS TOOLS WITH APPLICATIONS 457 2. Suppose that [ u ] β < and x 0 . Let { x n } n =1 be a sequence such that x 0 = lim n →∞ x n . Then | u ( x n ) u ( x m ) | [ u ] β | x n x m | β 0 as m, n →∞ showing { u ( x n ) } n =1 is Cauchy so that ¯ u ( x 0 ) := lim n →∞ u ( x n ) exists. If { y n } n =1 is another sequence converging to x 0 , then | u ( x n ) u ( y n ) | [ u ] β | x n y n |
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Holder-spaces - 4 56 BRUCE K DRIVER 24 Hlder Spaces...

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