201A, Fall ’10, Thomases
Homework 3
Mihaela Ifrim
1. Let
X
be a normed linear space. A series in
∑
x
n
in
X
is
absolutely
convergent
if
∑
bardbl
x
n
bardbl
converges to a finite value in
R
. Prove
X
is a
Banach space if and only if every absolutely convergent series converges.
Proof.
•
Let
X
be a Banach space and suppose that
∑
(
x
n
) is absolutely
convergent; i.
e.,
∑
bardbl
x
n
bardbl
=
y
for some finite
y
.
For any
ε >
0, let
N
be sufficiently large so that
∑
∞
n
=
N
bardbl
x
n
bardbl
< ε
.
If
m, n
≥
N
, then
bardbl
∑
∞
i
=
m
x
i
−
∑
∞
i
=
m
(
x
i
)
bardbl
=
bardbl
∑
m
i
=
n
(
x
i
)
bardbl
<
∑
m
i
=
n
bardbl
x
i
bardbl
< ε
(applying the
triangle inequality). Therefore,
∑
n
x
n
is Cauchy and hence convergent.
We conclude that every absolutely convergent series in
X
converges.
•
Conversely, let
X
be a normed linear space in which every absolutely
convergent series converges, and suppose that
{
x
n
}
n
is a Cauchy se
quence. For each
k
∈
N
, choose
n
k
such that
bardbl
(
x
m
−
x
n
)
bardbl
<
2

k
for
m, n
≥
n
k
. In particular,
bardbl
x
n
k
+1
−
x
n
k
bardbl
<
2

k
. If we define
y
1
=
x
n
1
and
y
k
+1
=
x
n
k
+1
−
x
n
k
for
k
≥
1, it follows that
∑
(
bardbl
y
n
bardbl
)
≤ bardbl
x
n
1
bardbl
+ 1
:
i. e., (
y
n
) is absolutely convergent, and hence convergent.
More explicitly
∑
n
bardbl
y
n
bardbl
)
=
∑
k
bardbl
x
n
k
+1
−
x
n
k
bardbl
+
bardbl
x
n
1
bardbl ≤ bardbl
x
n
1
bardbl
+
∑
∞
k
1
2
k
=
bardbl
x
n
1
bardbl
+ 1. However, this implies that
∑
bardbl
y
n
bardbl ≤ ∞
. But we
considered that every absolute convergent series is convergent, which
means that
∑
y
n
≤ ∞
. But the sequence of partial sums of
y
n
is a sub
sequence of
x
n
. Since
x
n
is Cauchy and has a convergent subsequence,
then it will aslo converge to the same limit as the subsequence. Hence
we have started with a random Cauchy sequence in
X
and we got that
it is a convergent sequence, so we can conclude that
X
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 Spring '11
 Wiley
 Differential Equations, Calculus, Equations, Metric space, Compact space, Modes of convergence

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