PMATH450_S2015.pdf

# Note functions in l 1 e are often called the

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Note. Functions in L 1 ( E ) are often called the essentially bounded. Example. C [ a, b ] L 1 [ a, b ], since all continuous functions are measurable and all f 2 C [ a, b ] are bounded and thus k f k L 1 [ a,b ] sup x 2 [ a,b ] | f ( x ) | < 1 . Note that C ( R ) 6✓ L 1 ( R ). All truly bounded measurable functions are in L 1 . Exercise. Prove k f k 1 = inf { 2 R : m { x : | f ( x ) | > } = 0 } Note. This implies that if β > k f k 1 , then take k f k 1 < < β with m { x : | f ( x ) | > } , so m { x : | f ( x ) | > β } = 0. Theorem. If k f k 1 = 0 then f = 0 a.e. . Proof. By above, m { x : | f ( x ) | > 1 n } = 0 for all n 2 N . Then 1 [ n =1 x : | f ( x ) | > 1 n = { x : | f ( x ) | 6 = 0 } thus m { x : | f ( x ) | 6 = 0 } = 0 by σ -subadditivity, so f = 0 a.e. . Note. Similarly, m { x : | f ( x ) | > k f k 1 } = 0. Also R E | f |  k f k 1 m ( E ), since Z E | f | = Z E \ { x : | f ( x ) | k f k 1 } | f | + Z E \ { x : | f ( x ) | > k f k 1 } | f |  k f k 1 m ( E \ { x : | f ( x ) |  k f k 1 } ) + 0  k f k 1 m ( E ) Definition. Let 1 p  1 . Take q 2 [1 , 1 ] such that 1 p + 1 q = 1 where 1 1 = 0. q is called the conjugate index of p . Example. Examples of conjugate pairs: (1 , 1 ) , ( 1 , 1) , (2 , 2) , (4 , 4 3 ). Theorem (Holder’s Inequality) . Let 1 p, q  1 be conjugate indices, so 1 p + 1 q = 1. Then for all measurable functions f, g , Z | fg |  k f k p k g k q . In particular, if f 2 L p and g 2 L q , then fg 2 L 1 and k fg k 1  k f k p k g k q . Note. This generalizes Cauchy-Schwarz for L 2 : | h f, g i |  h f, f i 1 / 2 h g, g i 1 / 2 = k f kk g k . For L 2 , h f, g i = R f g , so h f, f i = R | f | 2 = k f k 2 L 2 , so CS is equivalently: Z f g  k f k L 2 k g k L 2 . Proof of Holder’s Inequality. We first consider the p = 1 , q = 1 case. If RHS = 1 or k f k 1 = 0 or k g k 1 = 0, we are done. In particular, we can assume k g k 1 < 1 . Let A = { x : | g ( x ) | > k g k 1 } , so m ( A ) = 0. Then Z | fg | = Z A | fg | + Z A C | fg | 0 + Z A C k g k 1 | f | = k g k 1 Z A C | f |  k g k 1 Z | f | = k g k 1 k f k 1 .

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2 INTRODUCTION TO LEBESGUE INTEGRATION 20 We now consider the 1 < p, q < 1 case. We use the fact that for all a, b > 0, ab 1 p a p + 1 q b q . To see that this is true, take s, t such that a = e s/p and b = e t/q . By convexity of y = e x , we have e 1 p s + 1 q t 1 p e s + 1 q e t , so ab = e s/p e t/q = e 1 p s + 1 q t 1 p e s + 1 q e t = 1 p a p + 1 q b q . Now assume (WLOG) that k f k p and k g k q are finite and positive. Apply the inequality with a = | f ( x ) | k f k p and b = | g ( x ) | k g k q for all x : | f ( x ) || g ( x ) | k f k p k g k q 1 p | f ( x ) | k f k p p + 1 q | g ( x ) | k g k q q . Now integrate both sides: Z | f ( x ) || g ( x ) | k f k p k g k q Z 1 p | f ( x ) | k f k p p + Z 1 q | g ( x ) | k g k q q Notice that Z 1 p | f ( x ) | k f k p p = 1 p 1 k f k p p Z | f ( x ) | p = 1 p 1 k f k p p k f k p p = 1 p and similarly with g . Thus we end up with Z | f ( x ) || g ( x ) | k f k p k g k p 1 p + 1 q = 1 ) Z | fg |  k f k p k g k q . Lecture 14: June 5 Interesting Fact. For any p 1, L p [0 , 1] L 1 [0 , 1]. Proof. p = 1 is an exercise. Consider 1 < p < 1 . Take f 2 L p [0 , 1] and look at k f k 1 = Z 1 0 | f | = Z 1 0 | f | 1  k f k p k 1 k q since k 1 k q = R 1 0 1 q 1 /q = 1 < 1 . Note. In fact, if 1 < p 1 < p 2 < 1 , then L 1 [0 , 1] ) L p 1 [0 , 1] ) L p 2 [0 , 1] ) L 1 [0 , 1] .
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