completeness-l_1-2010

completeness-l_1-2010 - Math 508 Jerry L. Kazdan...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 508 Jerry L. Kazdan Completeness of 1 Let 1 be the vector space of infinite sequences of real numbers X = (x1 , x2 , . . .) with finite norm X := ∑∞=1 |x j | . Here we show this space is complete. The proof is a bit fussy. j (n) (n) S TEP 1: F IND A CANDIDATE FOR THE LIMIT. Let Xn = (x1 , x2 , . . .) be a Cauchy sequence. Because (n) () |x1 − x1 | ≤ Xn − X , (n) the first coordinates x1 are a Cauchy sequence of real numbers and hence converge to some real (n) number z1 . Similarly, the other coordinates converge: lim x j → z j for some real numbers z j . We n→∞ thus suspect that Xn → Z := (z1 , z2 , . . .) . J S TEP 2: S HOW THAT THIS CANDIDATE Z IS IN Z = ∑|z j | = lim 1 . We show that j Now J J (n) ∑ |z j | = ∑ nlim |x j →∞ j=1 J | = lim j=1 (n) ∑ |x j n→∞ J →∞ ∑ |z j | < ∞ . j=1 |. j=1 Note that in the second equality there are no difficulties interchanging a limit with the sum of a finite number of real numbers. Since Cauchy sequences are bounded, there is an M such that Xn < M for all n . Thus, for any J J (n) ∑ |x j ∞ |≤ j=1 (n) ∑ |x j | = Xn < M . j=1 Letting n → ∞ we find that ∑J=1 |z j | ≤ Xn < M . Because J is arbitrary, we conclude that Z ≤ M j and hence Z ∈ 1 . S TEP 3: P ROVE THE CONVERGENCE. Let Finally we show that Xn − Z → 0. Given ε > 0 pick N so that if n and > N then Xn − X < ε . Consequently, for any fixed J and n, > N , we find J (n) ∑ |x j () −xj | ≤ j=1 ∞ (n) ∑ |x j () − x j | = Xn − X < ε. j=1 With n > N and J fixed, let → ∞ to find J (n) J ∑ |x j − z j | = lim j=1 (n) ∑ |x j →∞ j=1 () − x j | ≤ ε. But now this is true for all J so Xn − Z ≤ ε , as desired. Exercise Let norm X := 2 be the vector space of infinite sequences of real numbers X = (x1 , x2 , . . .) with finite ∑∞=1 |x j |2 . Show this space is complete. [This is Hilbert’s Hilbert Space] j ...
View Full Document

Ask a homework question - tutors are online