Lecture 32 july 20 hardy tauberian theorem let f 2 l

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Lecture 32: July 20 Hardy Tauberian Theorem. Let f 2 L 1 ( T ) and assume there exists c such that | ˆ f ( n ) | c | n | for all n 6 = 0, then ( S n f ( t )) converges if and only if ( σ n f ( t )) converges, and the converge to the same limit. Moreover, if ( σ n f ( t )) converges uniformly on an interval I , then ( S n f ( t )) also converges uniformly on I .
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3 FOURIER ANALYSIS 50 Corollary. If f is continuous and | ˆ f ( n ) | c | n | 8 n 6 = 0 for some c , then ( σ n f ( t )) converges to f ( t ) uniformly on T , and hence so does ( S n f ( t )). Proof of Theorem. If ( S n f ( t )) converges then so does ( σ n f ( t )), and to the same limit. [See assignment.] Conversely, suppose ( σ n f ( t )) converges. Let [ r ] denote the integer part of r 2 R . Fix λ > 1 and look at X n< | j | [ λ n ] | ˆ f ( j ) | 2 X n<j [ λ n ] c | j | 2 c Z [ λ n ] n 1 x dx = 2 c log [ λ n ] n 2 c log λ n n = 2 c log λ . Now fix " > 0 and choose λ > 1 so that 2 c log λ < " . Then P n< | j | [ λ n ] | ˆ f ( j ) | < " for every n . Temporarily fix n large enough that n 6 = [ λ n ]. It is true that S n f ( t ) = [ λ n ] + 1 [ λ n ] - n σ [ λ n ] f ( t ) - n + 1 [ λ n ] - n σ n f ( t ) | {z } (1) - [ λ n ] + 1 [ λ n ] - n X n< | j | [ λ n ] 1 - | j | [ λ n ] + 1 ˆ f ( j ) e ijt | {z } (2) . Proof of this is left as an exercise. | (2) | [ λ n ] + 1 [ λ n ] - n X n< | j | [ λ n ] 1 - | j | [ λ n ] + 1 ˆ f ( j ) [ λ n ] + 1 [ λ n ] - n 1 - n + 1 [ λ n ] + 1 X n< | j | [ λ n ] ˆ f ( j ) = [ λ n ] + 1 [ λ n ] - n [ λ n ] - n [ λ n ] + 1 X n< | j | [ λ n ] ˆ f ( j ) = X n< | j | [ λ n ] ˆ f ( j ) < " Now assume σ n f ( t ) ! F ( t ). Then | (1) - F ( t ) | = [ λ n ] + 1 [ λ n ] - n σ [ λ n ] f - n + 1 [ λ n ] - n σ n f - [ λ n ] + 1 - ( n + 1) [ λ n ] - n F ( t ) = [ λ n ] + 1 [ λ n ] - n ( σ [ λ n ] f - f ) + n + 1 [ λ n ] - n ( F - σ n f ) [ λ n ] + 1 [ λ n ] - n σ [ λ n ] f - f | {z } ( A ) + n + 1 [ λ n ] - n | F - σ n f | | {z } ( B ) Now choose N 0 such that if n N 0 , then | σ m f - F | < " 1 . If n N 0 , then [ λ n ] N 0 , so ( A ) [ λ n ] + 1 [ λ n ] - n " 1 , ( B ) n + 1 [ λ n ] - n " 1 so | (1) - F ( t ) | [ λ n ] + 1 + n + 1 [ λ n ] - n " 1 λ n + n + 1 λ n - 1 - n " 1 = λ + 1 + 2 n λ - 1 - 1 n ! " 1 ! λ + 1 λ - 1 " 1 Pick N 1 such that if n N 1 , then λ +1 - 2 n λ - 1 - 1 n λ +2 λ - 1 , and pick " 1 so " 1 λ +2 λ - 1 < " . Then if n N 1 , | (1) - F | < " . Now pick N = max( N 0 , N 1 ), so if n N , then | S n f ( t ) - F ( t ) | | (1) + (2) - F | | (1) - F | + | (2) | < 2 " . Notice that if σ n f ( t ) ! F ( t ) uniformly over I , then N 0 can be chosen independently of t , and N 1 was already chosen independently of t , so N can be chosen independently of t . Thus S n f ( t ) ! F ( t ) uniformly on I . Proposition. If f is di erentiable and f 0 2 L 1 ( T ), then | n ˆ f ( n ) | ! 0 as n ! ± 1 . Thus | n ˆ f ( n ) | 1 for large n , so | ˆ f ( n ) | 1 | n | for large n , and thus | ˆ f ( n ) | c | n | for all n .
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3 FOURIER ANALYSIS 51 Corollary. If f 2 C 1 ( T ) (i.e. continuous derivative), then | ˆ f ( n ) | c | n | . Proof of Proposition. ˆ f ( n ) = 1 2 Z 2 0 f ( x ) e - inx dx Use integration by parts with u = f ( x ), dv = e - inx dx to get ˆ f ( n ) = 1 2 f ( x ) e - inx - in 2 0 | {z } 0 - 1 2 Z 2 0 f 0 ( x ) e - inx - in dx = 1 in 1 2 Z 2 0 f 0 ( x ) e - inx = 1 in b f 0 ( n ) Thus in ˆ f ( n ) = b f 0 ( n ), so n ˆ f ( n ) = b f 0 ( n ) ! 0 as n ! ± 1 by the Riemann-Lebesgue lemma. Corollary. If f 2 C k ( T ) (i.e. first k derivatives exist and are continuous), then n k ˆ f ( n ) ! 0, so ˆ f ( n ) c | n | k .
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