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

Info icon This preview shows pages 42–45. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 42

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 43
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 .
Image of page 44

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 45
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern