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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 27 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 6. Tuesday, Feb 24 By now the structure of the proofs should be getting somewhat routine – but I will go on to the point that I hope it all becomes clear! So, recall the definitions of a Lebesgue integrable function on the line (forming the linear space L 1 ( R )) and of a set of measure zero E ⊂ R . The first thing we want to show is that the putative norm on L 1 does make sense. Proposition 9. If f ∈ L 1 ( R ) then  f  ∈ L 1 ( R ) and if f n is an absolutely summable series of step functions converging to f almost everywhere then N (6.1) f = lim f k .   N →∞  k =1  So in some sense the definition of the Lebesgue integral ‘involves no cancellations’. There are extensions of the integral, we may even see one, which do exploit cancel lations. Proof. By definition if f ∈ L 1 ( R ) then it is the limit, on the set of absolute conver gence, of a summable series of step functions, { f n } . We need to make such a series for  f  . The idea in this case is the ‘obvious’ one. We know that n (6.2) f j ( x ) → f ( x ) if  f j ( x )  < ∞ . j =1 j So, set k k − 1 (6.3) g 1 ( x ) =  f 1 ( x )  , g k ( x ) =  f j ( x )  −  f j ( x )  ∀ x ∈ R . j =1 j =1 Then, for sure, N N (6.4) g k ( x ) =  f j ( x )  →  f ( x )  if  f j ( x )  < ∞ . k =1 j =1 j So, what we need to check, for a start, is that { g j } is an absolutely summable series of step functions. The triangle inequality in the form  v  −  w  ≤  v − w  shows that, for k > 1 , k k − 1 (6.5)  g k ( x )  =  f j ( x )  −  f j ( x )  ≤  f k ( x )  . j =1 j =1 Thus (6.6)  g k  ≤  f k  < ∞ k k so the g k ’s do indeed form an absolutely summable series. From its construction we know that N N (6.7) g k ( x ) =  f j ( x )  →  f ( x )  if  f n ( x )  < ∞ . k =1 j =1 n 28 LECTURE NOTES FOR 18.102, SPRING 2009 So, this is what we want except that the set on which  g k ( x )  < ∞ may be larger k than the set for which we have convergence here. Now, in the notes there is a result to handle this, but we can simply make the series converge less rapidly by adding a ‘pointless’ subseries. Namely replace g k by ⎧ ⎪ g k ( x ) if n = 3 k − 2 ⎨ (6.8) h n ( x ) = ⎪ f k ( x ) if n = 3 k − 1 ⎩ − f k ( x ) if n = 3 k. This series converges absolutely if and only if both the  g k ( x )  and  f k ( x )  series converge – the convergence of the latter implies the convergence of the former so (6.9)  h n ( x )  < ∞ ⇐⇒  f k ( x )  ....
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This note was uploaded on 11/08/2011 for the course PHY 18.102 taught by Professor Staff during the Spring '09 term at MIT.
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