PMATH450_S2015.pdf

# Recall that on monday we proved d n f 6 f even

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Recall that on Monday we proved D N f 6! f even pointwise, so this is a powerful theorem since ( D n ) is almost a summability kernel (it only fails (ii)). Proof of Theorem. We must show sup x | ( K n f )( x ) - f ( x ) | ! 0. Notice that | ( K n f )( x ) - f ( x ) | = Z T K n ( t ) f ( x - t ) dm ( t ) - f ( x ) Z T K n ( t ) dm ( t ) | {z } 1 = Z T K n ( t )( f ( x - t ) - f ( x )) dm ( t ) Z T | K n ( t ) || f ( x - t ) - f ( x ) | dm ( t ) Let " > 0. Choose δ > 0 by the uniform continuity of f so | f ( x - t ) - f ( x ) | < " M for all x when | t | < δ . Pick N such

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3 FOURIER ANALYSIS 43 that R 2 - δ δ | K n ( t ) | dm ( t ) < " 2 k f k 1 8 n N , by (iii). Then | ( K n f )( x ) - f ( x ) | Z T | K n ( t ) || f ( x - t ) - f ( x ) | dm ( t ) = Z δ - δ | K n ( t ) || f ( x - t ) - f ( x ) | dm ( t ) + Z δ < | t | | K n ( t ) || f ( x - t ) - f ( x ) | dm ( t ) " M Z δ - δ | K n ( t ) | dm (1) + 2 k f k 1 Z | t | > δ | K n ( t ) | dm ( t ) " M M + 2 k f k 1 " 2 k f k 1 = " + " = 2 " for n N . Lecture 28: July 10 Fejer Kernel. Let F N ( t ) = P N j = - N 1 - | j | N +1 e ijt . ( F n ) is called the Fejer kernel . Note. Note that ˆ F N (0) = 1 for all N , so the Fejer kernel satisfies the first property of being a summability kernel. Also F n ( t ) = 1 n +1 ( D 0 + D 1 + · · · + D N ). Lemma. F n ( t ) = 1 n +1 sin( n +1 2 ) t sin 2 t 2 0. Note. By the lemma, the Fejer kernel satisfies property 2 since k F n k 1 = ˆ F n (0) = R F n = 1. The third property follows, because if t 2 [ δ , 2 - δ ], then | sin t 2 | sin δ 2 . Thus Z 2 - δ δ | F n | 1 n + 1 Z (sin n +1 2 t ) 2 sin 2 δ 2 dm ( t ) 1 n + 1 1 sin 2 δ 2 Z T 1 dm ( t ) = 1 n + 1 1 sin 2 δ 2 ! 0 . Proof of Lemma. Notice that sin 2 t 2 = e it/ 2 - e - it/ 2 2 i ! 2 = - 1 4 ( e it - 2 + e - it ) sin 2 t 2 F n ( t ) = 1 2 - e it + e - it 4 ◆◆ n X j = - n 1 - | j | n + 1 e ijt Left as an exercise to work out = 1 n + 1 - 1 2 e - i ( n +1) t + 1 2 - 1 4 e i ( n +1) t = 1 n + 1 sin 2 n + 1 2 t Note. F n (0) = n + 1. Corollary. For f 2 C ( T ), F n f ! f uniformly.
3 FOURIER ANALYSIS 44 Note. It is common to denote F n f = σ n ( f ). In this case, F n f = P n j = - n \ F n f ( j ) e ijt = P n j = - n 1 - | j | n +1 ˆ f ( j ) e ijt . Also, notice that this gives a constructive proof that the trig polynomials are dense in C ( T ). Theorem. If f 2 L 1 ( T ) and ( K n ) is a summability kernel, then K n f ! f in the L 1 norm. Proof. If f is continuous, we already know that K n f ! f uniformly. But then k K n f - f k 1 = Z T | K n f - f |  k K n f - f k 1 Z T 1 = k K n f - f k 1 ! 0 so K n f ! f in the L 1 norm. Now take f 2 L 1 ( T ) and let " > 0. Get g 2 C ( T ) such that k g - f k 1 < " by Lusin’s theorem. k K n f - f k 1 = k K n f - K n g + K n g - g + g - f k 1  k K n ( f - g ) k 1 | {z } ? + k K n g - g k 1 + k g - f k 1  k K n k 1 k f - g k 1 + k K n g - g k 1 + " M " + " + " = " ( M + 2) for n N by the first part of the proof. Note that ? uses the fact that is a linear operator, which is left as an exercise. Corollary. For all f 2 L 1 ( T ), σ n ( f ) ! f in L 1 norm and there is a subsequence ( σ n j ( f )) such that σ n j ( f ) ! f pointwise a.e.. Uniqueness Theorem. If f 2 L 1 and ˆ f ( n ) = 0 8 n , then f = 0 in L 1 (a.e.). Proof. ˆ f ( n ) = 0 8 n ) σ n ( f ) = 0 8 n . But σ n ( f ) ! f in L 1 , so f = 0 in L 1 .

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

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