{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MIT18_102s09_lec13

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

This preview shows pages 1–3. Sign up to view the full content.

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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
82 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 13. Thursday, Mar 19: Baire’s theorem Note from Lecture 9, modified 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 finite 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 infinite 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. From (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. This follows from a pretty direct application of Baire’s theorem to B. Con- sider the sets (13.3) S p = { b B, b � ≤ 1 , T n b V p n } , p N .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}