1 2 z 2 fg dx 1 2 z f p 1 p z g q 1 q 1 2 z f p 1 p 1

Info icon This preview shows pages 33–36. Sign up to view the full content.

1 2 Z 2 0 fg dx 1 2 ✓Z | f | p 1 /p ✓Z | g | q 1 /q = 1 2 Z | f | p 1 /p 1 2 Z | g | q 1 /q = k f k p k g k q Note. { e in } 1 n = -1 are orthonormal on T , since Z 2 0 | e inx | 2 dm ( x ) = m [0 , 2 ] = 1 , Z 2 0 e inx e - imx dx = 0 . Trigonometric Polynomials. The set of trigonometric polynomials is Trig( T ) = ( N X n = - M a n e inx : N, M 2 N , a n 2 C ) = sp { e inx : n 2 Z } If P 2 Trig( T ), then P (0) = P (2 ) and P is 2 -periodic, so Trig( T ) C ( T ). The values of n for which a n 6 = 0 are called the frequencies of P . For each n , a n = P, e inx = 1 2 Z 2 0 P ( x ) e inx dx. The degree of a trigonometric polynomial P is deg( P ) = max {| n | : a n 6 = 0 } . Theorem. Trig( T ) is dense in C ( T ) in the sup norm, and dense in L p ( T ) for p < 1 in the p -norm. Proof. To prove Trig( T ) is dense in C ( T ), we use Stone-Weierstrass theorem. We need to prove that Trig( T ) is closed under conjugation,
Image of page 33

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

3 FOURIER ANALYSIS 34 is an algebra (closed under addition and multiplication), contains constants, separates points (if x 6 = y then 9 f such that f ( x ) 6 = f ( y )). Lecture 22: June 24 – Midterm Lecture 23: June 26 Note. Bonus question on assignment 4: Does there exist a Lebesgue measurable set A such that m ( A \ U ) = M ( U ) 2 for every open set U ? Proof (Continued). Closed under conjugation: P a n e inx = P a n e i ( - n ) x 2 Trig( T ). Separates points: Suppose s 6 = t 2 T . For P ( x ) = e ix , we have P ( s ) 6 = P ( t ). By Stone-Weierstrass, Trig( T ) is dense in C ( T ) in the sup-norm, and therefore dense in the L 2 -norm. Since C ( T ) are dense in L 2 ( T ) in the L 2 -norm, Note. This means that sp { e inx : n 2 Z } = L 2 ( T ). Thus { e inx : n 2 Z } is a complete orthonormal set, hence a basis for L 2 ( T ). Consequences of Completeness of { e inx : n 2 Z } . 1. If f 2 L 2 ( T ), then there exists P n 2 Trig( T ) such that k P n - f k 2 ! 0. 2. f = P 1 n = -1 f, e inx e inx , where f, e inx = 1 2 R 2 0 f ( x ) e - inx ˆ f ( n ). Thus f = P 1 n = -1 ˆ f ( n ) e inx . If we let S N ( f )( x ) = P N n = - N ˆ f ( n ) e inx , then S N ( f ) ! f in the L 2 norm. 3. Parseval’s Theorem : 1 X n = -1 | ˆ f ( n ) | 2 = 1 X n = -1 | f, e inx | 2 = k f k 2 L 2 ( T ) . 4. ˆ f ( n ) = 0 8 n 2 Z ) f = 0. 5. F : L 2 ( T ) ! ` 2 ( Z ) : f 7! ( ˆ f ( n )) 1 n = -1 is an isometric isomorphism. (Onto because if ( a n ) 2 ` 2 ( Z ), then we know that f = P 1 n = -1 a n e inx 2 L 2 ( T ). Can we extend these ideas to L 1 ( T ) ?. Let f 2 L 1 ( T ). Then f ( x ) e - inx 2 L 1 ( T ). Thus ˆ f ( n ) = 1 2 R 2 0 f ( x ) e - inx dx is well defined. Another Form for Fourier Series. We can write e inx = cos nx + i sin nx . Then X ˆ f ( n ) e inx = X ˆ f ( n )(cos nx + i sin nx ) = ˆ f (0) + 1 X n =1 ˆ f ( n )(cos nx + i sin nx ) + 1 X n =1 ˆ ( f ( - n ) cos( - nx ) + i sin( - nx )) = ˆ f (0) + 1 X n =1 ( ˆ f ( n ) + ˆ f ( - n )) cos nx + i 1 X n =1 ( ˆ f ( n ) - ˆ f ( - n )) sin nx
Image of page 34
3 FOURIER ANALYSIS 35 This is often written as f ( x ) = ˆ f (0) + 1 X n =1 A n cos nx + B n sin nx where A n = ˆ f ( n ) + ˆ f ( - n ) = 1 Z 2 0 f ( x ) cos nx dx, B n = ˆ f ( n ) - ˆ f ( - n ) = 1 Z 2 0 f ( x ) sin nx dx. Properties. 1. ( [ f + g )( n ) = ˆ f ( n ) + ˆ g ( n ).
Image of page 35

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

Image of page 36
This is the end of the preview. Sign up to access the rest of the document.
  • Spring '12
  • N.Spronk
  • FN, Lebesgue measure, Lebesgue integration, Eastern Orthodox liturgical days, Lebesgue

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern