PMATH450_S2015.pdf

# Note note that f being measurable implies f p is

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Note. Note that f being measurable implies | f | p is measurable. We will look at the special case p = 2, E = [0 , 1]: k f k L 2 ([0 , 1]) = ✓Z 1 0 | f | 2 1 / 2 ) k f k 2 2 = Z 1 0 | f | 2 = Z 1 0 f f = h f, f i using the inner product h f, g i = R f g . Thus k · k 2 is a norm that comes from an inner product. Lecture 12: June 1 Recall (Norm Properties) . A norm should be real-valued and defined on a vector space, and have the following properties: k f k p 0 with k f k p = 0 i f = 0. k cf k p = | c | · k f k p .

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2 INTRODUCTION TO LEBESGUE INTEGRATION 18 k f + g k p  k f k p + k g k p . Note that if f = 0 a.e. then k f k p = 0, so k f k p is not a norm. Moreover if f = g a.e. , then k f k p = k g k p . Proof: Suppose f = g except on some A with m ( A ) = 0. Then k f k p = ( R A | f | p + R A C | g | p ) 1 /p = ( R A | g | p + R A C | g | p ) 1 /p = k g k p . Definition. First define the equivalence relation f g if f = g a.e. . Define L p ( E ) to be the set of equivalence classes under where k f k p < 1 . Example. Suppose E = [ a, b ]. Then C [ a, b ] L p [ a, b ]. Indeed suppose f 2 C [ a, b ], so 9 M such that | f | M on [ a, b ], so R [ a,b ] | f | p M p ( b - a ) < 1 . Moreover, all bounded measurable functions are in L P [ a, b ]. Example. C ( R ) 6✓ L p ( R ) since, for example, f ( x ) = 1 / 2 L p ( R ). Proposition. L p ( E ) is a vector space. Proof. Let f, g 2 L p ( E ) and c be a scalar. Then cf is measurable and k cf k p = ✓Z E | cf | p 1 /p = ✓Z E | c | p | f | p 1 /p = | c | ✓Z E | f | p 1 /p = | c | k f k p < 1 . To check f + g 2 L p ( E ), we use the inequality: for a, b 0, a p + b p 2 ( a + b 2 ) p . Then Z E k f + g k p 2 p Z E | f | + | g | 2 p 2 p Z E | f | p + | g | p 2 2 p - 1 ( k f k p p + k g k p p ) < 1 . Thus f + g 2 L p ( E ), so L p is a vector space. Proposition. k f k p = 0 only if f = 0 a.e. , i.e. f is the zero element in L p ( E ). Proof. Let E n = { x : | f ( x ) | 1 n } . Then R E n | f | p 1 n m ( E n ). But R E n | f | p R E | f | p = 0, so m ( E n ) = 0. Let F = S 1 n =1 E n = { x : f ( x ) 6 = 0 } , so m ( F ) P 1 n =1 E n = 0, so f = 0 a.e. . Thus k [ f ] k p = 0 () [ f ] = [0]. Definition. Let f be a measurable function defined on E . Define the 1 -norm (or essential supremum norm ) by k f k L 1 ( E ) = k f k 1 = inf A meas m ( E \ A )=0 { sup | f ( x ) | : x 2 A } . In particular, if E = R , then k f k L 1 ( R ) = inf m ( A C )=0 { sup | f ( x ) | : x 2 A } . Note. It is always the case that k f k L 1 ( E ) sup {| f ( x ) | : x 2 E } , since E is one of the sets A . It can be less. Example. 1. If f = 0 a.e. take A = { x : f ( x ) = 0 } . Then m ( A C ) = 0, and sup {| f ( x ) | : x 2 A } = 0, so inf m ( A C )=0 { sup | f ( x ) | : x 2 A } = 0. 2. Suppose f is continuous on R . We claim that k f k R = sup {| f ( x ) | : x 2 R } . Proof. Suppose M < sup | f | . Then there is some x 0 where | f ( x 0 ) | M 1 > M . By continuity of f , there exists δ > 0 such that if x 2 ( x 0 - δ , x 0 + δ ), then | f ( x ) | > M . Suppose k f k L 1 ( R ) < M and m ( A C ) = 0. Then A C 6◆ ( x 0 - δ , x 0 + δ ) because m ( x 0 - δ , x 0 + δ ) = 2 δ > 0. Thus A \ ( x 0 - δ , x 0 + δ ) 6 = ; . Pick y 2 ( x 0 - δ , x 0 + δ ). But then | f ( y ) | > M since y 2 ( x 0 - δ , x 0 + δ ). Thus sup {| f ( x ) | : x 2 A } > M . Since A was an arbitrary set with m ( A C ) = 0, R m ( A C )=0 { sup | f ( x ) | : x 2 A } M , a contradiction.
2 INTRODUCTION TO LEBESGUE INTEGRATION 19 Lecture 13: June 3 Definition. L 1 ( E ) is the set of equivalence classes of measurable f where k f k L 1 ( E ) < 1 .

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

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