Theorem if f is of bounded variation then ˆ f n c n

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Theorem. If f is of bounded variation, then | ˆ f ( n ) | c | n | for all n 6 = 0. Proof. First note that functions of bounded variation are Riemann integrable, being a di erence of increasing func- tions. This also follows from the fact that functions of bounded variation have only jump discontinuities, therefore
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3 FOURIER ANALYSIS 53 only countably many discontinuities, hence the Lebesgue measure of the set of discontinuities is of measure zero. Riemann integrability follows from Lebesgue’s characterization. It immediately follows that f ( x ) e - inx is Riemann integrable. Now ˆ f ( n ) = 1 2 Z 2 0 f ( x ) e - inx dx = lim P R ( fe - inx , P ) for whatever Riemann sum R we like. Now R ( fe - inx , P ) = X f ( r k ) e - inr k ( x k - x k - 1 ) where P : 0 = x 1 < · · · < x n = 2 and r k 2 [ x k - 1 , x k ]. Notice that e - inx k - e - inx k - 1 x k - x k - 1 = - ine - inr k by the MVT. Choose r k from the above for each interval [ x k - 1 , x k ]. Then R ( fe - inx , P ) = m X k =1 f ( r k ) 1 - in e - inx k - e - inx k - 1 x k - x k - 1 ( x k - x k - 1 ) = - 1 in m X k =1 f ( r k )( e - inx k - e - inx k - 1 ) = - 1 in - f ( r 0 ) e - inx 0 + m - 1 X k =1 e - inx k ( f ( r k ) - f ( r k +1 )) + f ( r m ) e - inx m ! 1 2 1 | n | | f ( r 0 ) | + | f ( r m ) | + m - 1 X k =1 | f ( r k ) - f ( r k - 1 ) | ! 1 2 1 | n | (2 k f k sup + V f (0 , 2 )) Corollary (Dirichlet-Jordan Theorems) . If f is of bounded variation, then S n f ( t 0 ) ! f ( t + 0 )+ f ( t - 0 ) 2 . Further, if f is continuous, then S n f ! f uniformly. Proof. Apply the Hardy Tauberian theorem. Lecture 34: July 24 3.13 Gibbs Phenomena Example. Suppose h ( x ) = ( 1 0 x - 1 - < x < 0 . This has a jump discontinuity at 0 (and another at 1). By A6Q5, S 0 h (0) ! h (0 - ) + h (0 + ) 2 = 0 . Also ˆ h ( n ) = ( 0 n even 2 in n odd .
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3 FOURIER ANALYSIS 54 Sh ( t ) = X n odd 2 in e int = 1 X n =1 n odd 4 in e int - e - int 2 i = X n =1 n odd 4 n sin nt = 1 X k =0 4 (2 k + 1) sin(2 k + 1) t Thus Sh (0) = 0. Now consider S 2 N - 1 h ( t ) = N - 1 X k =0 4 (2 k + 1) sin(2 k + 1) = S 2 N h ( t ) . To find the maximum of S 2 N - 1 h ( t ), we take the derivative: ( S 2 N - 1 h ) 0 ( t ) = N - 1 X k =0 4 (2 k + 1) (2 k + 1) cos(2 k + 1) t = Re N - 1 X k =0 4 e (2 k +1) it ! = 4 Re N - 1 X k =0 e it 2 k e it ! N - 1 X k =0 e it 2 k e it = e it N - 1 X k =0 ( e it 2 ) k = e it e it 2 N - 1 e 2 it - 1 = e it e itN ( e itN - e - itN ) e it ( e it - e - it ) = e itN sin Nt sin t = (cos Nt + i sin Nt ) sin Nt sin t ) Re N - 1 X k =0 e it 2 k e it ! = cos Nt sin Nt sin t = sin 2 Nt 2 sin t ) ( S 2 N - 1 h ) 0 ( t ) = 2 sin 2 Nt sin t Our critical points are points t with sin 2 Nt = 0, and the first (positive) critical point is t = 2 N . Notice that ( S 2 N - 1 h ) 0 ( t ) > 0 on (0 , 2 N ) and ( S 2 N - 1 h ) 0 ( t ) < 0 past 2 N so this is a local maximum. Now we have ( S 2 N - 1 h ) 2 N = 4 N - 1 X k =0 sin(2 k + 1) 2 N 2 n + 1 = 2 N - 1 X k =0 sin(2 k + 1) 2 N (2 n + 1) 2 N N We can think of this as a Riemann sum of f over [0 , 2 ] with intervals of length N , sampling each interval at its midpoint, x k = (2 k + 1) 2 N . Thus 2 N - 1 X k =0 sin(2 k + 1) 2 N (2 n + 1) 2 N N = 2 N - 1 X k =0 sin x k x k ` ( I k ) This is a Riemann sum for f ( x ) = sin x x for the uniform partition of [0 , ]. f is continuous and thus Riemann integrable, and hence 2 R ( f, P n ) !
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