Completion
Theorem 1 (Completion)
If
(
V
,
a·
,
·A
V
)
is any inner product space, then there exists a Hilbert
space
(
H
,
, ,
H
)
and a map
U
:
V → H
such that
(i)
U
is 1–1
(ii)
U
is linear
(iii)
a
U
x
,U
y
A
H
=
a
x
,
y
A
V
for all
x
,
y
∈ V
(iv)
U
(
V
) =
b
U
x
v
v
x
∈ V
B
is dense in
H
. If
V
is complete, then
U
(
V
) =
H
.
Remark 2
(a)
H
is called the completion of
V
.
(b)
U
gives a oneto–one correspondance between elements of
V
and elements of
U
(
V
). So we can
think of
U
as giving the new name
U
x
to each
x
∈ V
and we can think of
U
(
V
) as being the same
as
V
but with the names of the elements changed. Thus, we can think of
V
as being
U
(
V
)
⊂ H
.
Using this point of view, the above theorem says that any inner product space can be extended to
a complete inner product space. I.e. can have its holes Flled in.
Motivation.
The hard part of the proof is to make a guess as to what
H
should be. That’s what we’ll do
now. A good strategy is to work backwards. Suppose that, somehow, we have found a suitable
H
with
V ⊂ H
. If we can Fnd a way to describe each element of
H
purely in terms of elements of
V
,
then we can turn around and take that as the deFnition of
H
.
Because
V
is dense in
H
, each element of
H
may be written as the limit of a sequence in
V
,
and each such sequence is Cauchy. Thus specifying an element of
H
is equivalent to specifying
a Cauchy sequence in
V
. But there is not a one–to–one correspondance between elements of
H
and Cauchy sequences in
V
, because many di±erent Cauchy sequences in
V
converge to the same
element of
H
. ²or example, if
b
x
n
B
n
∈
IN
⊂ V
converges to
x
∈ H
, then
b
x
n
+
1
2
n
x
1
B
n
∈
⊂ V
and
b
e
1
/n
x
n
+
(
−
1)
n
n
x
2
B
n
∈
⊂ V
also converge to
x
. To get a one–to–one correspondance, we can
identify each
x
∈ H
with the
set of all
Cauchy sequences in
V
that converge to
x
.
Outline of Proof.
²irst deFne
V
′
=
±
{
x
n
}
n
∈
v
v
v
{
x
n
}
n
∈
is a Cauchy sequence in
V
²
That is,
V
′
is the set of all Cauchy sequences in
V
. Next we deFne two Cauchy sequences
{
x
n
}
n
∈
and
{
y
n
}
n
∈
in
V
to be “equivalent”, written
{
x
n
}
n
∈
∼ {
y
n
}
n
∈
, if and only if
lim
n
→∞
b
x
n
−
y
n
b
V
= 0
This deFnition is rigged so that any two convergent sequences have the same limit if and only if
they are equivalent. Next, if
{
x
n
}
n
∈
∈ V
′
, we deFne the “equivalence class of
{
x
n
}
n
∈
” to be
the set
³
{
x
n
}
n
∈
´
=
±
{
y
n
}
n
∈
∈ V
′
v
v
v
{
y
n
}
n
∈
∼ {
x
n
}
n
∈
²
of all Cauchy sequences that are equivalent to
{
x
n
}
n
∈
. We shall shortly prove
September 7, 2011
Completion
1
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View Full DocumentLemma 3
∼
is an equivalence relation
(1)
. In particular, if
{
x
n
}
n
∈
IN
,
{
y
n
}
n
∈
∈ V
′
then
either
b
{
x
n
}
n
∈
B
=
b
{
y
n
}
n
∈
B
or
b
{
x
n
}
n
∈
B
∩
b
{
y
n
}
n
∈
B
=
∅
If you think of a Cauchy sequence as one person and an equivalence class of Cauchy sequences as
a “family” of related people, then the above Lemma says, that the whole world is divided into a
collection of nonoverlapping families. Next, we deFne
H
=
±
b
{
x
n
}
n
∈
B
v
v
v
{
x
n
}
n
∈
∈ V
′
²
as the set of all “families” and prove
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 Spring '13
 JoelFeldman
 Metric space, Limit of a sequence, Hilbert space, Cauchy sequence, Cauchy

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