PMATH450_S2015.pdf

# We will first look at soft methods of proving

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We will first look at “soft” methods of proving convergence of Fourier series. (Non-constructive methods.) 3.7 Functional Analysis Let X, Y be Banach spaces and F : X ! Y be a linear map. Definition. We say a linear map F : X ! Y is bounded if k f k op = sup k x k X 1 k F ( x ) k Y < 1 Note. Note that k f k op = sup x 6 =0 k F ( x ) k Y k x k X (exercise), so k F k op k x k X ≥ k F ( x ) k Y . Example. 1. Define F : L 1 ( T ) ! C by F ( f ) = ˆ f ( n ) for some fixed n . Then k F ( f ) k C = | F ( f ) | = | ˆ f ( n ) |  k f k L 1 ( T ) 1 if k f k L 1 ( T ) 1. Thus k F k op 1 < 1 so F is bounded. Indeed k F k op = 1: take f ( x ) = e inx so ˆ f ( n ) = 1 and k f k L 1 = R T | e inx | dm ( x ) = 1, so k F k op 1. 2. Define F : L 2 ( T ) ! ` 2 ( Z ) by F ( f ) = ( ˆ f ( n )) 1 n = -1 . Notice that k F ( f ) k ` 2 = k ( ˆ f ( n )) k ` 2 = k f k L 2 ( T ) 8 f so k F k op = 1. 3. Define S N : L 1 ! L 1 by S N ( f ) = P N - N ˆ f ( n ) e inx 2 Trig( T ) L 1 ( T ). Exercise: Check that S N is linear. Recall (from Monday) that S N ( f )( x ) = 1 2 R 2 0 f ( t + x ) D N ( t ) dt . Let f 2 L 1 with k f k 1 1. Then k S N ( f ) k 1 = Z T | S N ( f )( x ) | dm ( x ) = Z T Z T f ( t + x ) D N ( t ) dm ( t ) dm ( x ) Z T Z T | f ( t + x ) D N ( t ) dm ( t ) | dm ( x ) = Z T Z T | f ( t + x ) | dm ( x ) | D N ( t ) | dm ( x ) = Z T Z T | f - t ( x ) | dm ( x ) | D N ( t ) | dm ( x ) = Z T ( k f - t k L 1 ) | D N ( t ) | dm ( t ) = Z T ( k f k L 1 ) | D N ( t ) | dm ( t ) = k f k L 1 Z T | D N ( t ) | dm ( t ) = k f k L 1 k D N k L 1  k f k L 1 (2 N + 1) 2 N + 1 Since D N ( t ) = P N - N e int so | D N ( t ) | 2 N + 1. Thus k S N k op 2 N + 1 < 1 . Actually, k S N k op  k D N k L 1 . To show that k S N k op = k D N k L 1 , we’ll find a sequence ( f k ) in L 1 with k f k k 1 = 1 8 k and S N ( f k ) ---! k !1 D N in the L 1 norm. Assuming this, then k S N k op k S N ( f k ) k L 1 k f k k L 1 k !1 ---! k D N k 1 1 Hence k S N k op = k D N k 1 . To do this, take f k = χ [ - 1 k , 1 k ] m [ - 1 k , 1 k ]

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3 FOURIER ANALYSIS 38 so k f k k 1 = 1 m [ - 1 k , 1 k ] Z T χ [ - 1 k , 1 k ] ( t ) dm ( t ) = 1 m [ - 1 k , 1 k ] m - 1 k , 1 k = 1 . We now have S N ( f k )( x ) = Z T f k ( t + x ) D N ( t ) dm ( t ) = Z T f k ( t ) D N ( t - x ) dm ( t ) so k S N ( f k )( x ) - D N ( x ) k L 1 = Z T | S N ( f k )( x ) - D N ( x ) | dm ( x ) = Z T ✓Z T f k ( t ) D N ( t - x ) dm ( t ) - Z T f k ( t ) dt | {z } 1 D N ( x ) = Z T ✓Z T f k ( t ) D N ( t - x ) dm ( t ) - Z T f k ( t ) D N ( x ) dt = Z T Z T ( f k ( t ) D N ( t - x ) - f k ( t ) D N ( x )) dm ( t ) dm ( x ) = Z T Z T f k ( t ) ( D N ( t - x ) - D N ( x )) dm ( t ) dm ( x ) Z T Z T | f k ( t ) | | D N ( t - x ) - D N ( x ) | dm ( t ) dm ( x ) = Z T Z 1 k - 1 k | f k ( t ) | D N ( t - x ) | {z } = D N ( x - t ) - D N ( x ) dm ( t ) dm ( x ) Note that since D N is a trig polynomial, it is continuous, so when t is small, D N ( x - t ) is close to D N ( x ). In fact, since T is compact, D N is uniformly continuous. Let " > 0. Use the uniform continuity of D N to pick K such that | t | 1 K ) | D N ( x - t ) - D N ( x ) | < " 8 x 2 T . Let k K , so k S N ( f k ) - D N k 1 Z T Z 1 k - 1 k | f k ( t ) | " dm ( t ) ! dm ( x ) = " Z T ✓Z T | f k ( t ) | dm ( t ) dm ( x ) = Z T k f k k 1 dm ( x ) = " Z T 1 dm ( x ) = " . Hence S N ( f k ) ! D N in L 1 norm, so as argued earlier k S N k op = k D N k L 1 . Fact. A bounded linear map F : X ! Y is continuous. Proof. Let x n ! x where x n , x 2 X . Look at k F ( x n ) - F ( x ) k Y = k F ( x n - x ) k Y  k F k op k x n - x k X ! 0. Thus F is continuous.
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• Spring '12
• N.Spronk
• FN, Lebesgue measure, Lebesgue integration, Eastern Orthodox liturgical days, Lebesgue

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