FUNCTIONAL ANALYSIS LECTURE NOTES CHAPTER 3. BANACH ...

6 CHRISTOPHER HEIL
Proposition 1.22. Let X be a normed space. Prove that X is a Banach space if and only
if every absolutely convergent series in X converges in X.
Proof. ⇒. This is Exercise 1.21.
⇐. Suppose that every absolutely convergent series is convergent. Let {fn}n∈N be a
Cauchy sequence in X. By Exercise 1.18, there exists a subsequence {fnk }k∈N such that
fnk+1 − fnk < 2−k for every k. The series
(fnk+1 k − fnk ) is absolutely convergent,
because
∞
− fnk ≤
∞
2 −k = 1 < ∞.
fnk+1
k=1
k=1
By hypothesis, we conclude that
(fnk+1 k − fnk ) converges in X, say to f. In terms of the
partial sums, this says that
fnk+1 − fn1 =
k
j=1
(fnj+1
− fnj ) → f as k → ∞,
or fnk → g = f +fn1 as k → ∞. Thus {fn}n∈N is a Cauchy sequence which has a subsequence
that converges to g. It therefore follows from Exercise 1.17 that fn → g. Therefore X is
complete.
Example 1.23 (The Harmonic Series). To illustrate the difference between convergence and
absolute convergence, consider the one-dimensional case, i.e., X = F, the field of scalars. Let
xn = 1
. Then n n xn = 1
n n is the harmonic series, and the partial sums sN = N n=1
this series are unbounded. Thus the harmonic series does not converge. Since sN → ∞, we
usually say that
1
1
n diverges to infinity, and write n n = ∞. n
On the other hand, consider the alternating series 1
n (−1)n . Since the terms alternate
n
signs and since 1 → 0, it follows that this series does converge to a finite scalar (in fact, it
n
converges to − ln 2). However, it does not converge absolutely.
Definition 1.24 (Unconditionally Convergent Series). Let X be a Banach space and let
{fn}n∈N be a sequence of elements of X. The series f = ∞ n=1 fn is said to converge unconditionally
if every rearrangement of the series converges. That is, f = ∞ n=1 fn converges
unconditionally if for each bijection σ : N → N the series
∞
converges.
n=1
fσ(n)
Remark 1.25. It is not obvious, but it can be shown that if ∞ n=1 fσ(n) is unconditionally
convergent, then every rearrangement of the series must converge to the same sum, i.e., there
is a single f such that f = ∞ n=1 fσ(n) for every permutation σ.
1
n of