MIT18_102s09_lec07

MIT18_102s09_lec07 - 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 . 41 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 7. Thursday, Feb 26 So, what was it with my little melt-down? I went too cheap on the monotonicity theorem and so was under-powered for Fatou’s Lemma. In my defense, I was trying to modify things on-the-fly to conform to how we are doing things here. I should also point out that at least one person in the audience made a comment which amounted to pointing out my error. So, here is something closer to what I should have said – it is not far from what I did say of course. Proposition 12. [Montonicity again] If f j ∈ L 1 ( R ) is a monotone sequence, either f j ( x ) ≥ f j +1 ( x ) for all x ∈ R and all j or f j ( x ) ≤ f j +1 ( x ) for all x ∈ R and all j, and f j is bounded then (7.1) { x ∈ R ; lim f j ( x ) is finite } = R \ E j →∞ where E has measure zero and f = lim f j ( x ) a.e. is an element of L 1 ( R ) (7.2) j →∞ with lim f − f j = 0 . j →∞ | | Moral of the story – drop the assumption of positivity and replace it with the bound on the integral. In the approach through measure theory this is not necessary because one has the concept of a measureable, non-negative, function for which the integral ‘exists but is infinite’ – we do not have this. Proof. Since we can change the sign of the f i (now) it suffices to assume that the f i are monotonically increasing. The sequence of integrals is therefore also montonic increasing and, being bounded, converges. Thus we can pass to a subsequence g i = f n i with the property that (7.3) | g j − g j − 1 | = g j − g j − 1 < 2 − j ∀ j > 1 . This means that the series h 1 = g 1 , h j = g j − g j − 1 , j > 1 , is absolutely summable. So we know for the result last time that it converges a.e., that the limit, f, is integrable and that j (7.4) f = lim h k = lim g j = lim f j . j →∞ k =1 j →∞ n →∞ In fact, everywhere that the series h j ( x ) , which is to say the sequence g k ( x ) , j converges so does f n ( x ) , since the former is a subsequence of the latter which is monotonic. So we have (7.1) and the first part of (7.2). The second part, corresponding to convergence...
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MIT18_102s09_lec07 - MIT OpenCourseWare http/ocw.mit.edu...

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