PMATH450_S2015.pdf

# 2 c f n ˆ f n 3 ˆ f n 1 2 r 2 f k f k l 1 t 8 n

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2. ( c f )( n ) = ˆ f ( n ). 3. | ˆ f ( n ) | 1 2 R 2 0 | f | = k f k L 1 ( T ) 8 n . Thus ( ˆ f ( n )) 1 n = -1 2 ` 1 ( Z ). Also k ( ˆ f ( n )) k ` 1 = sup n | ˆ f ( n ) |  k f k L 1 ( T ) . The map F : L 1 ( T ) ! ` 1 ( Z ) is linear and k F ( f ) k ` 1  k f k L 1 ( T ) . Lecture 24: June 29 4. ˆ f ( n ) = R T f ( x ) e - 2 nx dx = R f ( x ) e inx dx = ˆ f ( - n ). 5. Let f t ( x ) = f ( x - t ). Then k f t k L p ( T ) = k f k L p ( T ) (exercise). Also ˆ f t ( n ) = 1 2 Z 2 0 f t ( x ) e - inx dx = 1 2 Z 2 0 f ( x - t ) e - in ( x - t ) e int dx = 1 2 e - int Z 2 0 f ( x - t ) e in ( x - t ) dx = 1 2 e - int ˆ f ( n ) . 6. If f k ! f in the L 1 -norm, then ˆ f k ( n ) ! ˆ f ( n ) for all n , since | ˆ f k ( n ) - ˆ f ( n ) | = Z T ( f k ( x ) - f ( x )) e - inx dm ( x ) Z T | f k ( x ) - f ( x ) | dm ( x ) = k f k - f k 1 ! 0 . In fact, this convergence is uniform in n . Note. k ( ˆ f ( n )) k ` 1 = sup n | ˆ f ( n ) |  k f k 1 . Riemann-Lebesgue Lemma. If f 2 L 1 ( T ), then lim n ! ± 1 ˆ f ( n ) = 0. That is, ( ˆ f ( n )) 1 n = -1 2 c 0 . Note. This shows that not all ( a n ) 2 ` 1 are the Fourier coe ffi cients of some f 2 L 1 . Proof of Riemann-Lebesgue Lemma. Let " > 0. Fix f 2 L 1 and choose P 2 Trig( T ) such that k f - P k 1 < " . Let N = deg( P ). So ˆ P ( n ) = 0 if | n | > N . Then for | n | > N , | ˆ f ( n ) | = | ˆ f ( n ) - ˆ P ( n ) + ˆ P ( n ) | | ( \ f - P )( n ) | + | ˆ P ( n ) |  k f - p k 1 < " . Corollary. If f 2 L 1 , then R T f ( x ) sin nx dm ( x ) ! 0 and R T f ( x ) cos nx dm ( x ) ! 0 as n ! 1 .

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3 FOURIER ANALYSIS 36 3.6 Convergence of Fourier Series Dirichlet Kernel. Let D n ( x ) = P n k = - n e ikx 2 T rig ( T ). Note that sup x | D n ( x ) | = 2 n + 1. Also note that D n is an even function. Now ˆ D n ( k ) = ( 1 | k | n 0 otherwise . Also sin x 2 D n ( x ) = e i x 2 - e - i x 2 2 i ! n X k = - n e ikx = e ix ( n + 1 2 ) - e - ix ( n + 1 2 ) 2 i = sin ✓ n + 1 2 x . Thus D n ( x ) = sin [( n + 1 2 ) x ] sin ( x 2 ) , where we take the limit as x ! 0 for the value at 0. If δ > 0 and we restrict x 2 ( δ , 2 - δ ), then | D n ( x ) | 1 sin δ 2 . Now for arbitrary f , we have S N f ( x ) = N X n = - N ˆ f ( n ) e inx = N X n = - N 1 2 Z 2 0 f ( t ) e - int dt e inx = 1 2 Z 2 0 N X n = - N f ( t ) e in ( x - t ) dt = 1 2 Z 2 0 f ( t ) D N ( x - t ) dt = 1 2 Z 2 0 f ( t + x ) D N ( t ) dt Intuitive Idea for Why we Might Hope for S N f ( x ) ! f ( x ) for f Continuous. S N f ( x ) = 1 2 Z 2 0 f - x ( t ) sin ( N + 1 2 ) t sin t 2 ! dt = 1 2 Z 2 0 f - x ( t ) sin nt cos t 2 + cos nt sin t 2 sin t 2 dt = 1 2 Z 2 0 f - x ( t ) sin Nt cos t 2 sin t 2 + cos nT dt 1 2 Z 2 0 f - x ( t ) sin Nt cos t 2 sin t 2 dt = 1 2 Z δ 0 · · · + 1 2 Z 2 2 - δ · · · + 1 2 Z 2 - δ δ · · · | {z } (3) = ( ? ) Now (3) = 1 2 Z T f - x ( t ) cos t 2 sin t 2 χ [ δ , 2 - δ ] sin NT dt ! 0 by the Riemann-Lebesgue lemma since f - x ( t ) cos t 2 sin t 2 χ [ δ , 2 - δ ] | f - x | 1 sin δ 2 2 L 1 . Then ( ? ) 1 2 Z δ - δ f - x ( t ) cos t 2 sin t 2 sin NT dt = 1 2 Z δ - δ f ( t + x ) cos t 2 sin t 2 sin Nt dt 1 2 f ( x ) Z δ - δ cos t 2 sin t 2 sin Nt dt 1 2 f ( x ) Z T D N ( t ) dt = f ( x ) ˆ D N (0) = f ( x ) as f is continuous. However. This is not true. That is, the Fourier series of a continuous function does not necessarily converge. We will give two proofs of this fact: a constructive proof, and an abstract argument using functional analysis methods.
3 FOURIER ANALYSIS 37 Lecture 25: July 3 Extra o ffi ce hour: Tuesday 3-4.

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