Δ 2 n p δ b a 2 lecture 16 june 10 now suppose f 2

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δ (2 N ) p + δ ( b - a ) < 2 " Lecture 16: June 10 Now suppose f 2 L p [ a, b ] is arbitrary. Define f n ( x ) = ( f ( x ) | f ( x ) | N 0 otherwise . k f N ( x ) k  N is bounded in L p [ a, b ]. Notice that f N ! f a.e. pointwise, so | f ( x ) - f N ( x ) | p ! 0 a.e. pointwise and | f ( x ) - F N ( x ) | p | f | p 2 L 1 [ a, b ]. By DCT, k f N - f k p p = Z b a | f - f n | p ! Z b a 0 = 0 Pick N such that k f N - f k p < " 2 , and get h continuous such that k f N - h k p < " 2 (by the first part of the proof). Then k f - h k p  k f - f N k p + k f N - h k p < " . Finally we show that C [ a, b ] is not dense in L 1 [ a, b ]. Find c 2 ( a, b ). Let f [ a,c ) = 0 and f [ c,b ] = 1. We claim there is not a continuous g such that f = g a.e. . Suppose there is. For any δ > 0, ( c, c + δ ) has positive measure, so g = f ( x ) at some x 2 ( c, c + δ ). Hence g ( x ) = 1 at some x 2 ( c, c + δ ), so by continuity, g (0) = 1. Similarly, approaching from the other side gives g (0) = 0, a contradiction. So f 2 L 1 [ a, b ] and the equivalence class containing f contains no continuous function. Now suppose there is a sequence f n 2 C [ a, b ] such that f n ! f in L 1 -norm. ( f n ) is a Cauchy sequence in L 1 -norm, but the L 1 norm is the sup norm on the continuous functions, so really ( f n ) is Cauchy in C ([ a, b ]) with the usual sup norm. C [ a, b ] is complete with its usual sup norm, so f n ! g in sup norm where g is continuous. So k f n - g k sup ! 0, but f n ! f in L 1 , so f = g a.e. , a contradiction. Thus C [ a, b ] is a closed subspace of L 1 [ a, b ], whereas cl C [ a, b ] = L p [ a, b ] for 1 p < 1 . Corollary. Polynomials in L p [0 , 1] for p < 1 . Proof. Note that k f k p p = R 1 0 | f | p 1 /p  k f k p 1 . Let " > 0. Take f 2 L p and get g continuous such that k g - f k p < " 2 by Lusin’s theorem. Now by the Stone-Weierstrass, there exists a polynomial Q ( x ) such that k Q ( x ) - g ( x ) k sup < " 2 . Then k Q - g k p  k Q - g k 1 = k Q - g k sup < " 2 so k f - Q k p  k f - g k p + k g - Q k p < " 2 + " 2 = " . 2.7 Interchanging the Order of Integration Fubini’s Theorem. Suppose f : R 2 ! R where f - 1 ( O ) is Borel in R 2 for any open set O R . If Z R ✓Z R | f ( x, y ) | dm ( x ) dm ( y ) < 1
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3 FOURIER ANALYSIS 25 then Z R ✓Z R f ( x, y ) dm ( x ) dm ( y ) = Z R ✓Z R f ( x, y ) dm ( y ) dm ( x ) . Proof. To be completed in PMATH 451. End of midterm material Midterm: Definitions, state theorems, proofs done in class except for Lebesgue’s characterization of Riemann inte- grability. 3 Fourier Analysis 3.1 Hilbert Spaces Definition. An inner product space is a vector space V over C , with a map h · , · i : V V ! C satisfying: 1. h f, f i ≥ 0 and h f, f i = 0 () f = 0. 2. h f + g, h i = h f, h i + h g, h i . 3. h f, g i = h g, f i Note. h f, β g i = h β g, f i = β h g, f i = β h f, g i . h f, g + h i = h g + h, f i = h g, f i + h h, f i = h f, g i + h f, h i . Example. (a) C n with h x, y i = P n i =1 x i y i . (b) ` 2 = n ( x n ) 1 n =1 : k ( x n ) k ` 2 = (P 1 i =1 | x i | 2 ) 1 / 2 < 1 o with h x, y i = P 1 i =1 x i y i . (c) C [0 , 1] with h f, g i = R 1 0 f g . (d) L 2 [0 , 1] with h f, g i = R 1 0 f g . Notice that if f, g 2 L 2 , then by Holder, R | f g |  k f k 2 k g k 2 < 1 , so f g is integrable and R f g is well-defined.
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