MIT18_102s09_lec13

MIT18_102s09_lec13 - MIT OpenCourseWare http:/ocw.mit.edu...

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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 .
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82 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 13. Thursday, Mar 19: Baire’s theorem Note from Lecture 9, modiFed and considerably extended. Theorem 8 (Baire) . If M is a non-empty complete metric space and C n M, n N , are closed subsets such that [ (13.1) M = C n n then at least one of the C n ’s has an interior point. Proof. So, choose p 1 / C 1 , which must exist since otherwise C 1 contains an open ball. Since C 1 is closed there exists 1 > 0 such that B ( p 1 , 1 ) C 1 = . Next choose p 2 B ( p 1 , 1 / 3) which is not in C 2 , which is possible since otherwise B ( p 1 , 1 / 3) C 2 , and 2 > 0 , 2 < 1 / 3 such that B ( p 2 , 2 ) C 2 = . So we have used both the fact that C 2 has empty interior and the fact that it is closed. Now, proceed, inductively. Assume that there is a Fnite sequence p i , i = 1 ,...,k and positive numbers 0 < k < k 1 / 3 < k 2 / 3 2 < < 1 / 3 k 1 < 3 k such that p j B ( p j 1 , j 1 / 3) and ··· B ( p j , j ) C j = . Then we can add another p k +1 by using the properties of C k it has non-empty interior so there is some point in B ( p k , k / 3) which is not in C k +1 and then B ( p k +1 , k +1 ) C k +1 = where k +1 > 0 but k +1 < k / 3 . Thus, we have constructe and inFnite sequence { p k } in M. Since d ( p k +1 ,p k ) < k / 3 this is a Cauchy sequence. In fact (13.2) d ( p k ,p k + l ) < k / 3 + + k + l 1 / 3 < 3 k < 2 k / 3 ··· for all l > 0 , and this tends to zero as k → ∞ . Since M is complete this sequence converges. ±rom (13.2) the limit, q M must lie in the closure of B ( p k , 2 k / 3) for every k. Hence q / C k for any k which contradicts (13.1). Thus, at least one of the C n must have non-empty interior. One application of this is often called the uniform boundedness principle, I will just call it: Theorem 9 (Uniform boundedness) . Let B be a Banach space and suppose that T n is a sequence of bounded (i.e. continuous) linear operators T n : B −→ V where V is a normed space. Suppose that for each b B the set { T n ( b ) } ⊂ V is bounded (in norm of course) then sup n T n < . Proof.
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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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MIT18_102s09_lec13 - MIT OpenCourseWare http:/ocw.mit.edu...

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