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Sobolev Spaces

# Sobolev Spaces - 4 36 BRUCE K DRIVER 23 Sobolev Spaces...

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436 BRUCE K. DRIVER 23. Sobolev Spaces De fi nition 23.1. For p [1 , ] , k N and an open subset of R d , let W k,p loc ( ) := { f L p ( ) : α f L p loc ( ) (weakly) for all | α | k } , W k,p ( ) := { f L p ( ) : α f L p ( ) (weakly) for all | α | k } , (23.1) k f k W k,p ( ) := X | α | k k α f k p L p ( ) 1 /p if p < and (23.2) k f k W k,p ( ) = X | α | k k α f k L ( ) if p = . In the special case of p = 2 , we write W k, 2 loc ( ) =: H k loc ( ) and W k, 2 ( ) =: H k ( ) in which case k·k W k, 2 ( ) = k·k H k ( ) is a Hilbertian norm associated to the inner product (23.3) ( f, g ) H k ( ) = X | α | k Z α f · α g dm. Theorem 23.2. The function, k·k W k,p ( ) , is a norm which makes W k,p ( ) into a Banach space. Proof. Let f, g W k,p ( ) , then the triangle inequality for the p — norms on L p ( ) and l p ( { α : | α | k } ) implies k f + g k W k,p ( ) = X | α | k k α f + α g k p L p ( ) 1 /p X | α | k h k α f k L p ( ) + k α g k L p ( ) i p 1 /p X | α | k k α f k p L p ( ) 1 /p + X | α | k k α g k p L p ( ) 1 /p = k f k W k,p ( ) + k g k W k,p ( ) . This shows k·k W k,p ( ) de fi ned in Eq. (23.1) is a norm. We now show completeness. If { f n } n =1 W k,p ( ) is a Cauchy sequence, then { α f n } n =1 is a Cauchy sequence in L p ( ) for all | α | k. By the completeness of L p ( ) , there exists g α L p ( ) such that g α = L p lim n →∞ α f n for all | α | k. Therefore, for all φ C c ( ) , h f, ∂ α φ i = lim n →∞ h f n , ∂ α φ i = ( 1) | α | lim n →∞ h α f n , φ i = ( 1) | α | lim n →∞ h g α , φ i . This shows α f exists weakly and g α = α f a.e. This shows f W k,p ( ) and that f n f W k,p ( ) as n → ∞ .

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ANALYSIS TOOLS WITH APPLICATIONS 437 Example 23.3. Let u ( x ) := | x | α for x R d and α R . Then Z B (0 ,R ) | u ( x ) | p dx = σ ¡ S d 1 ¢ Z R 0 1 r αp r d 1 dr = σ ¡ S d 1 ¢ Z R 0 r d αp 1 dr = σ ¡ S d 1 ¢ · ( R d αp d αp if d αp > 0 otherwise (23.4) and hence u L p loc ¡ R d ¢ i ff α < d/p. Now u ( x ) = α | x | α 1 ˆ x where ˆ x := x/ | x | . Hence if u ( x ) is to exist in L p loc ¡ R d ¢ it is given by α | x | α 1 ˆ x which is in L p loc ¡ R d ¢ i ff α + 1 < d/p, i.e. if α < d/p 1 = d p p . Let us not check that u W 1 ,p loc ¡ R d ¢ provided α < d/p 1 . To do this suppose φ C c ( R d ) and ± > 0 , then h u, ∂ i φ i = lim ± 0 Z | x | u ( x ) i φ ( x ) dx = lim ± 0 ( Z | x | i u ( x ) φ ( x ) dx + Z | x | = ± u ( x ) φ ( x ) x i ± ( x ) ) . Since ¯ ¯ ¯ ¯ ¯ Z | x | = ± u ( x ) φ ( x ) x i ± ( x ) ¯ ¯ ¯ ¯ ¯ k φ k σ ¡ S d 1 ¢ ± d 1 α 0 as ± 0 and i u ( x ) = α | x | α 1 ˆ x · e i is locally integrable we conclude that h u, ∂ i φ i = Z R d i u ( x ) φ ( x ) dx showing that the weak derivative i u exists and is given by the usual pointwise derivative.
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Sobolev Spaces - 4 36 BRUCE K DRIVER 23 Sobolev Spaces...

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