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]
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 Fall '10
 STAFF
 Real Numbers, Vector Space, Limit of a sequence, Hilbert space, Cauchy sequence, Xn

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