Chapter 11- Convolutions and Approximations

Chapter 11- Convolutions and Approximations - A NALYSIS...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ANALYSIS TOOLS WITH APPLICATIONS 197 11. Approximation Theorems and Convolutions Let ( X, M ) be a measure space, A M an algebra. Notation 11.1. Let S f ( A ) denote those simple functions φ : X C such that φ 1 ( { λ } ) A for all λ C and µ ( φ 6 =0) < . For φ S f ( A ) and p [1 , ) , | φ | p = P z 6 =0 | z | p 1 { φ = z } and hence Z | φ | p = X z 6 =0 | z | p µ ( φ = z ) < so that S f ( A ) L p ( µ ) . Lemma 11.2 (Simple Functions are Dense) . The simple functions, S f ( M ) , form adensesubspaceo f L p ( µ ) for all 1 p< . Proof. Let { φ n } n =1 be the simple functions in the approximation Theorem 7.12. Since | φ n | | f | for all n, φ n S f ( M ) (verify!) and | f φ n | p ( | f | + | φ n | ) p 2 p | f | p L 1 . Therefore, by the dominated convergence theorem, lim n →∞ Z | f φ n | p = Z lim n →∞ | f φ n | p =0 . Theorem 11.3 (Separable Algebras implies Separability of L p —Spaces) . Suppose 1 and A M is an algebra such that σ ( A )= M and µ is σ - f nite on A . Then S f ( A ) is dense in L p ( µ ) . Moreover, if A is countable, then L p ( µ ) is separable and D = { X a j 1 A j : a j Q + i Q ,A j A with µ ( A j ) < } is a countable dense subset. Proof. First Proof. Let X k A be sets such that µ ( X k ) < and X k X as k →∞ . k N let H k denote those bounded M — measurable functions, f, on X such that 1 X k f S f ( A ) L p ( µ ) . It is easily seen that H k is a vector space closed under bounded convergence and this subspace contains 1 A for all A A . Therefore by Theorem 8.12, H k is the set of all bounded M — measurable functions on X. f L p ( µ ) , the dominated convergence theorem implies 1 X k {| f | k } f f in L p ( µ ) as k . We have just proved 1 X k {| f | k } f S f ( A ) L p ( µ ) for all k and hence it follows that f S f ( A ) L p ( µ ) . The last assertion of the theorem is a consequence of the easily veri f ed fact that D is dense in S f ( A ) relative to the L p ( µ ) —norm . Second Proof. Given ±> 0 , by Corollary 8.42, for all E M such that µ ( E ) < , there exists A A such that µ ( E 4 A ) <±. Therefore (11.1) Z | 1 E 1 A | p = µ ( E 4 A ) . This equation shows that any simple function in S f ( M ) may be approximated arbitrary well by an element from D and hence D is also dense in L p ( µ ) .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
198 BRUCE K. DRIVER Corollary 11.4 (Riemann Lebesgue Lemma) . Suppose that f L 1 ( R ,m ) , then lim λ ± Z R f ( x ) e iλx dm ( x )=0 . Proof. Let A denote the algebra on R generated by the half open intervals, i.e. A consists of sets of the form n a k =1 ( a k ,b k ] R where a k k ¯ R . By Theorem 11.3given ±> 0 there exists φ = P n k =1 c k 1 ( a k ,b k ] with a k k R such that Z R | f φ | dm < ±. Notice that Z R φ ( x ) e iλx dm ( x )= Z R n X k =1 c k 1 ( a k ,b k ] ( x ) e iλx dm ( x ) = n X k =1 c k Z b k a k e iλx dm ( x n X k =1 c k λ 1 e iλx | b k a k = λ 1 n X k =1 c k ¡ e iλb k e iλa k ¢ 0 as | λ | →∞ .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 25

Chapter 11- Convolutions and Approximations - A NALYSIS...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online