PMATH450_S2015.pdf

# Example notice that f x x on 0 2 is not continuous

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Example. Notice that f ( x ) = x on [0 , 2 ) is not continuous when extended (through 2 -periodicity) to R . Thus σ n ( f )(0) ! w f (0) = 6 = f (0) because f is not continuous at 0. Proposition. If S n ( f )( t 0 ) converges, then it converges to the same limit as σ n ( f )( t 0 ). Note. This means that if f is continuous and S n ( f )( t 0 ) converges, then it must be converging to f ( t 0 ). Lecture 31: July 17 Theorem. Suppose f 2 L 1 ( T ) and f is di erentiable at t 0 2 T , then S N f ( t 0 ) ! f ( t 0 ). Proof. | S N f ( t 0 ) - f ( t 0 ) | = Z T f ( t 0 - t ) D N ( t ) dm ( t ) - f ( t 0 ) Z T D n ( t ) = Z T D N ( t )( f ( t 0 - t ) - f ( t 0 )) dm ( t ) = Z T ( f ( t 0 - t ) - f ( t 0 )) sin( N + 1 2 ) t sin 1 2 t dm ( t ) Now let g ( t ) = ( f ( t 0 - t ) - f ( t 0 )) t t sin 1 2 t . We claim that g 2 L 1 . lim t ! 0 g ( t ) = lim t ! 0 f ( t 0 - t ) - f ( t 0 ) t 2 = - 2 f 0 ( t 0 ) Pick δ > 0 such that | t | δ ) | g ( t ) | 2 | f 0 ( t 0 ) | + 1. We now check that g 2 L 1 . Z T | g | = Z δ - δ + Z 2 - δ δ (2 | f 0 ( t 0 ) | + 1) Z δ - δ 1 + 1 sin 1 2 δ Z | f ( t 0 - t ) | + | f ( t 0 ) | dm ( t ) c 1 + c 2 ( k f k 1 + | f ( t 0 ) | ) < 1 . Thus g 2 L 1 . Then | S N f ( t 0 ) - f ( t 0 ) | = Z g ( t ) sin N + 1 2 t dm ( t ) = Z g ( t ) sin Nt cos 1 2 t + cos Nt sin 1 2 t dm ( T ) Now g 2 L 1 ) g ( t ) cos 1 2 t 2 L 1 , since | g ( t ) cos 1 2 t | | g | and similarly g ( t ) sin 1 2 t 2 L 1 . Thus Z g ( t ) cos 1 2 t sin Nt ! 0 , Z g ( t ) sin 1 2 t cos Nt ! 0 by the Riemann-Lebesgue lemma. Thus | S N f ( t 0 ) - f ( t 0 ) | ! 0.

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3 FOURIER ANALYSIS 49 Corollary. If f 2 L 1 is constant c on an open interval, then S N f ! c on the interval. Proof. f is di erentiable on the open interval. Definition. We say that f satisfies a Lipschitz condition of order at t 0 if there exist c and δ > 0 such that if | t | δ then | f ( t 0 - t ) - f ( t 0 ) | c | t | . Remark. 1. Since | t | is smaller for big , this condition is harder to satisfy as gets bigger. 2. If = 0, then f is bounded near t 0 . 3. If f is Lip for some > 0, then f is continuous at t 0 . 4. If f is di erentiable at t 0 , then f ( t 0 - t ) - f ( t 0 ) t ! | f 0 ( t 0 ) | , so f 2 Lip1. 5. If f 2 Lip for > 1 on an open interval I , then f is constant on I . Proof. For all a 2 I , | f ( a - t ) - f ( a ) | | t | c | t | - 1 . But | f ( a - t ) - f ( a ) | | t | ! | f 0 ( a ) | and c | t | - 1 ! 0, so f 0 ( x ) = 0 on I and thus f is constant. Example. Let f ( x ) = x 1 / 3 . f is Lip 1 3 at t 0 = 0, but not di erentiable at 0. Note. Di erentiable ) Lip1 ) Lip 8 < 1 ) Lip0 ) Continuous Proposition. Assume f 2 L 1 and f is Lip at t 0 for some > 0. Then S N f ( t 0 ) ! f ( t 0 ). Proof. Let g ( t ) = f ( t 0 - t ) - f ( t 0 ) t t sin 1 2 t . Then for small t , | g ( t ) | = f ( t 0 - t ) - f ( t 0 ) t t sin 1 2 t c | t | | t | c 1 = c 2 | t | 1 - 2 L 1 ( T ) . So g 2 L 1 and the same argument from the previous theorem applies. Definition. We say f has a right hand derivative at t 0 if lim t ! 0 + f ( t 0 + t ) - f ( t + 0 ) t exists, where t + 0 = lim t ! t + 0 f ( t ). We similarly define left hand derivative . Corollary. If f 2 L 1 is continuous at t 0 has has right and left hand derivatives at t 0 , then S N f ( t 0 ) ! f ( t 0 ). Proof. Exercise - f is Lip1. Example. Let g ( x ) = x 2 - 2 x 2 C ( T ). Notice that g 0 (0) = - 2 , but g 0 (2 ) = 2 , so g is not di erentiable at 0. By the corollary, S N g (0) ! g (0). Notice that S ( g ) = - 2 2 3 + P n 6 =0 2 n 2 e inx = - 2 2 3 + P 1 n =1 4 n 2 cos nx . Now S N ( g 0 ) = - 2 2 3 + P N n =1 4 n 2 ! g (0) = 0. Thus P 1 n =1 4 n 2 = 2 2 3 ) P 1 n =1 1 n 2 = 2 6 .
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• Spring '12
• N.Spronk
• FN, Lebesgue measure, Lebesgue integration, Eastern Orthodox liturgical days, Lebesgue

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