PMATH450_S2015.pdf

# N k 1 if 1 2 k d n k 1 c 2 log n step 2 use ubt with

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n k 1 if " < 1 2 k D n k 1 = c 2 log n . Step 2 Use UBT with X = C ( T ), Y = C , F = { T N : X ! Y : N 2 N } where T N ( f ) = S N ( f )(0). T N is linear (exercise). Notice that k T N k op = sup k f k 1 1 | T N ( f ) | = sup k f k 1 1 | S N ( f )(0) | and | S N ( f )(0) | = 1 2 Z 2 0 f ( t ) D N ( t ) dt 1 2 k f k 1 Z 2 0 | D N ( t ) | dt = k f k 1 k D N k 1 Thus k T N k op  k D N k 1 2 N + 1, so T N is bounded. By step 1, k T n k op | T n ( g n ) | = | S n ( g n )(0) | 1 2 k D n k 1 c 2 log n . Thus sup N k T N k op = 1 . By UBP, there is a continuous function f such that sup N | T N ( f ) | = 1 . Thus sup N | S N ( f )(0) | = sup N | T N ( f ) | = 1 so the Fourier series of f diverges at 0. Lecture 27: July 8 3.8 Convolution Definition. For f, g 2 L 1 the convolution of f and g is f g = Z T f ( x - t ) g ( t ) dm ( t )

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3 FOURIER ANALYSIS 41 Note. Notice that f g = Z T f ( x + t ) g ( - t ) dm ( t ) = Z T f ( u ) g ( x - u ) dm ( u ) = g f so convolution commutes. Facts. 1. f g 2 L 1 . Proof. Z T | ( f g )( x ) | dm ( x ) = Z T Z T f ( x - t ) g ( t ) dm ( t ) dm ( x ) Z T Z T | f ( x - t ) | g ( t ) | dm ( t ) dm ( x ) = Z T | g ( t ) | ✓Z T | f ( x - t ) | dm ( x ) dm ( t ) = Z T | g ( t ) | k f t k L 1 dm ( t ) = k f k L 1 Z T | g ( t ) | dm ( t ) = k f k L 1 k g k L 1 < 1 2. If f 2 L 1 and g 2 L 1 , then f g 2 L 1 . Proof. | ( f g )( x ) | = Z T f ( x - t ) g ( t ) dm ( t ) Z T | f ( x - t ) || g ( t ) | dm ( t )  k f k L 1 Z T | g ( t ) | dm ( t ) = k f k L 1 k g k L 1 Thus sup x | ( f g )( x ) |  k f k 1 k g k 1 < 1 . 3. S N ( f ) = f D N , so k S N ( f ) k 1  k f k 1 k D N k 1 . Proof. S N ( f )( x ) = Z T f ( x + t ) D N ( t ) dm ( t ) = Z T f ( x + t ) D N ( - t ) dm ( t ) = Z T f ( x - u ) D N ( u ) dm ( u ) = ( f D N )( x ) Example. Let g ( t ) = e int . Then ( f g )( x ) = ( g f )( x ) = Z T g ( x - t ) f ( t ) dm ( t ) = Z T e in ( x - t ) f ( t ) dm ( t ) = e inx Z T e - int f ( t ) dm ( t ) = e inx ˆ f ( n ) . Note. f N X k =1 c k g k = N X k =1 c k ( f g k ) so if P ( x ) = P N - N c n e inx then ( f P ( x ) = N X - N c n ( f e inx ) = N X - N c n ˆ f ( n ) e inx . Thus [ f P ( n ) = c n ˆ f ( n ) = ˆ P ( n ) ˆ f ( n ). More generally, [ f g ( n ) = ˆ f ( n g ( n ) for all f, g 2 L 1 .
3 FOURIER ANALYSIS 42 Proof. [ f g = Z ( f g )( x ) e - inx dm ( x ) = Z T ✓Z T f ( x - t ) g ( t ) dm ( t ) e - in ( x - t ) e - int dm ( x ) = Z T ✓Z T f ( x - t ) e - in ( x - t ) dm ( x ) g ( t ) e - int dm ( t ) = Z T ✓Z T f ( u ) e - inu dm ( u ) g ( t ) e - int dm ( t ) = Z T ˆ f ( n ) g ( t ) e - int dm ( t ) = ˆ f ( n g ( n ) Note. It would be nice if there were g 2 L 1 such that f g = f for all f 2 L 1 . It turns out that no such g exists. Proof. If f g = f then ˆ f ( n g ( n ) = [ f g ( n ) = ˆ f ( n ) for all n , so ˆ g ( n ) = 1 for all n , which is impossible by the Riemann Lebesgue lemma. 3.9 Bounded Approximate Identities Definition. A summability kernel is a sequence ( K n ) C ( T ) satisfying (i) ˆ K n (0) = R T K n ( x ) = 1. (ii) There is some M such that R T | K n ( x ) | dm ( x ) M for all n . (iii) For all 0 < δ < , lim n !1 R 2 - δ δ | K n ( x ) | dm ( x ) = 0. If K n 0 for all n , then (ii) follows from (i), and ( K n ) is called a positive summability kernel . Almost an Example. Let K n ( t ) = χ ( - 1 n , 1 n ) m ( - 1 n , 1 n ) . Notice that R T K n ( t ) = 1 for all n , and (ii) holds since K n 0. (iii) holds because if 1 N < δ , then K n ( t ) = 0 for all t 2 ( δ , 2 - δ ) for all n N . The only thing we are missing is continuity, which can be achieved with a small modification. (Exercise) Theorem. Let f 2 C ( T ) and ( K n ) be a summability kernel. Then K n f ! f uniformly.

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

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