MIT18_102s09_lec06

MIT18_102s09_lec06 - MIT OpenCourseWare http://ocw.mit.edu...

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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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MIT18_102s09_lec06 - MIT OpenCourseWare http://ocw.mit.edu...

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