PMATH450_S2015.pdf

However this does not hold for r that is l p r 6 l q

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However, this does not hold for R . That is, L p ( R ) 6✓ L q ( R ) and L q ( R ) 6✓ L p ( R ) if p 6 = q . Proof. 1 p 1 < p 2 < 1 case: Suppose f 2 L p 2 [0 , 1], so k f k p 2 < 1 . Notice that p 2 p 1 > 1, so take q = 1 1 - p 1 /p 2 so ( p 2 p 1 , q ) is a conjugate pair. Then by Holder’s inequality: Z 1 0 | f | p 1 = Z 1 0 ( | f | p 1 ) (1)  k | f | p 1 k p 2 /p 1 k 1 k q = ✓Z 1 0 || f | p 1 | p 2 /p 1 p 1 /p 2 = ✓Z 1 0 | f | p 2 p 1 /p 2 = k f k p 1 p 2 < 1

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2 INTRODUCTION TO LEBESGUE INTEGRATION 21 so f 2 L p 1 [0 , 1]. To see that L p 1 [0 , 1] 6 = L p 2 [0 , 1], let a = 1 /p 2 , so p 2 a = 1 and since p 1 < p 2 , p 1 a < 1. Then note that n p 2 a n 2 = 1 n and n p 1 a n 2 = 1 n 1+ δ for some δ > 0. Now let f = P 1 n =2 n a χ ( 1 n , 1 n + 1 n 2 ) . Then Z 1 0 | f | p 1 = 1 X n =2 n p 1 a n 2 = 1 X n =2 1 n 1+ δ < 1 Z 1 0 | f | p 2 = 1 X n =2 n p 2 a n 2 = 1 X n =2 1 n = 1 so f 2 L p 2 [0 , 1] but f / 2 L p 1 [0 , 1]. Now consider the 1 p 1 < p 2 = 1 case. Suppose f 2 L 1 [0 , 1], so k f k 1 < 1 . Let A = { x 2 [0 , 1] : | f ( x ) | > k f k 1 } . We proved in class that m ( A ) = 0. Notice that on [0 , 1] \ A , f  k f k 1 . Thus Z 1 0 | f | p 1 = Z [0 , 1] \ A | f | p 1 + Z A | f | p 1 Z [0 , 1] \ A k f k p 1 1 + 0  k f k p 1 1 < 1 so f 2 L p 1 [0 , 1]. Finally, L 1 [0 , 1] 6 = L p 1 [0 , 1] since if f ( x ) = 1 x 2 , then f 2 L p 1 [0 , 1] but f / 2 L 1 [0 , 1]. Minkowski’s Inequality (Triangle Inequality) . If f, g are measurable, then k f + g k p  k f k p + k g k p for 1 p  1 . Remark. If f = g a.e. (say f = g except on E with m ( E ) = 0), then k f k 1 = k f χ E + f χ E C k 1  k f χ E k 1 + k f χ E C k 1 = 0+ k f χ E C k 1 = k g χ E C k 1 = k g - g χ E k 1  k g k 1 + k- g χ E } 1 = k g k 1 and similarly k g k 1  k f k 1 , so we have k f k 1 = k g k 1 . Proof of Minkowski’s Inequality. p = 1 case left as exercise. For p = 1, k f + g k 1 = Z | f + g | Z ( | f | + | g | ) = Z | f | + Z | g | = k f k 1 + k g k 1 . So suppose 1 < p < 1 . Assume f, g 2 L p (since if not, RHS = 1 , so we are done). Take q such that 1 p + 1 q = 1 ) 1 + p q = p ) q + p = pq ) p = q ( p - 1). Now by Holder’s inequality, Z | f + g | p - 1 | f |  k | f + g | p - 1 k q k f k p . Notice that k | f + g | p - 1 k q q = Z | f + g | ( p - 1) q = Z | f + g | p = k f + g k p p ) k | f + g | p - 1 k q = k f + g k p/q p Thus Z | f + g | p - 1 | f |  k | f + g | p - 1 k q k f k p = k f + g k p/q p k f k p . Similarly, Z | f + g | p - 1 | g |  k f + g k p/q p k g k p . Now k f + g k p p = Z | f + g | p = Z | f + g | p - 1 ( | f + g | ) Z | f + g | p - 1 ( | f | + | g | ) = Z | f + g | p - 1 | f | + Z | f + g | p - 1 | g |  k f + g k p/q p ( k f k p + k g k p )
2 INTRODUCTION TO LEBESGUE INTEGRATION 22 Note that k f + g k p < 1 since f, g 2 L p and L p is a vector space. Dividing both sides by k f + g k p/q p gives k f + g k p - p q p  k f k p + k g k p but p - p q = 1, so the above is actually k f + g k p  k f k p + k g k p . Corollary. k · k p is a norm and L p is a normed vector space, and hence a metric space with metric d ( f, g ) = k f - g k p . Definition. A Banach space is a complete normed space. Theorem (Riesz-Fisher) . L p is a Banach space for 1 p  1 . Proof. We first consider p < 1 . Let ( f n ) be a Cauchy sequence in L p . We must show that 9 f 2 L p such that f n ! f in L p -norm. It is enough to show that there exists a subsequence ( f n k ) converging to some f in L p -norm. Start by picking ( f n k ) such that k f n k +1 - f n k k p < 1 2 k . (First pick n 1 such that for all n, m n 1 , k f n - f m k p < 1 2 . Then pick n 2 > n 1 such that for all n, m n 2 , k f n - f m k p < 1 4 . Notice that k

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

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