completeness-l_1-2010

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

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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 ...
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