Basic Analysis – Gently Done Topological Vector Spaces

8: Banach Spaces 87
Proposition 8.4 The normedspace X iscomplete ifand onlyiftheseries � ∞
n=1 x n
converges, where (x n ) is any sequence in X satisfying � ∞
n=1 �x n
� < ∞. In other
words, a normed space is complete if and only if every absolutely convergent series
is convergent.
Proof Suppose that X is complete, and let (xn ) be any sequence in X such that
�∞ n=1�xn� < ∞. Let ε > 0 be given and put yn = �n k=1xk . Then, for n > m,
�
�
n�
�
�
�ym −yn� = �
� x �
k�
≤
< ε
k=m+1
n�
k=m+1
�x k �
for all sufficiently large m and n, since �∞ n=1�xn� < ∞. Hence (yn ) is a Cauchy
sequence and so converges since X is complete, by hypothesis.
Conversely, suppose �∞ n=1xn converges in X whenever �∞ n=1�xn� < ∞. Let
(yn ) be any Cauchy sequence in X. We must show that (yn ) converges. Now,
since (yn ) is Cauchy, there is n1 ∈ N such that �yn1 −y 1
m� < 2 whenever m > n1 .
Furthermore, there is n2 > n1 such that �yn2 − y 1
m� < 4 whenever m > n2 .
Continuing in this way, we see that there is n1 < n2 < n3 < ... such that
�y nk −y m
� < 1
2 k whenever m > n k . In particular, we have
for k ∈ N. Set xk = ynk+1 −y . Then nk
�ynk+1 −y 1
� < nk 2k n�
�xk� =
k=1
<
n�
k=1
n�
k=1
�y nk+1 −y nk �
It follows that �∞ k=1�xk� < ∞. By hypothesis, there is x ∈ X such that
�m k=1xk → x as m → ∞, that is,
m� m�
xk = (ynk+1 −ynk )
k=1
k=1
1
2 k.
= y nm+1 −y n1
→ x.
Hence y nm → x + y n1 in X as m → ∞. Thus the Cauchy sequence (y n ) has a
convergent subsequence and so must itself converge.