MIT18_102s09_lec26

MIT18_102s09_lec26 - MIT OpenCourseWare http:/ocw.mit.edu...

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

View Full Document Right Arrow Icon
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 .
Background image of page 1

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

View Full DocumentRight Arrow Icon
147 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 26. Thursday, May 14:Review Now, there was one fnal request beFore I go through a quick review oF what we have done. Namely to state and prove the Hahn-Banach Theorem. This is about extension oF Functionals. Stately starkly, the basic question is: Does a normed space have any non-trivial continuous linear Functionals on it? That is, is the dual space always non-trivial (oF course there is always the zero linear Functional but that is not very amusing). We did not really encounter this problem since For a Hilbert space, or even a pre-Hilbert space, there is always the space itsefl, giving continuous linear Functionals through the pairing Riesz’ Theorem says that in the case oF a Hilbert space that is all there is. I could have used the Hahn-Banach Theorem to show that any normed space has a completion, but I gave a more direct argument For this, which was in any case much more relevant For the cases oF L 1 ( R ) and L 2 ( R ) For which we wanted concrete completions. Theorem 19 (Hahn-Banach) . If M V is a linear subspace of a normed space and u : M −→ C is a linear map such that (26.1) | u ( t ) | ≤ C t V t M then there exists a bounded linear functional U U : V −→ C with U C and = u. M ±irst, by computation, we show that we can extend any continuous linear Func- tional ‘a little bit’ without increasing the norm. Lemma 20. Suppose M V is a subspace of a normed linear space, x / M and u : M −→ C is a bounded linear functional as in (26.1) then there exists = { t V ; t = t + ax, a C such that u : M (26.2) u u ( t + ax ) C t + ax V , t M, a C . = u, | | M Proof. Note that the decompositon t = t + ax oF a point in M is unique, since t + ax
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 5

MIT18_102s09_lec26 - MIT OpenCourseWare http:/ocw.mit.edu...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online