PMATH450_S2015.pdf

# 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

• Notes
• 58

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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,

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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
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 ).

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• Spring '12
• N.Spronk
• FN, Lebesgue measure, Lebesgue integration, Eastern Orthodox liturgical days, Lebesgue

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