4Hilbert spaces
II Linear Analysis
4.4 The isomorphism with `
2
We ended the previous section with two examples. Both of them are Hilbert
spaces, and both have countable basis. Is there any way we can identify the
both? This is a reasonable thing to ask. If we are given a Hilbert space
H
of
finite dimension
dim H
=
n
, then we know that
H
is indeed isomorphic to
R
n
(or
C
n
) with the Euclidean norm. In some sense
`
2
is just an “infinite version”
of
R
n
. So we might expect all other Hilbert spaces with countable dimension to
be isomorphic to `
2
.
Recall that if we have a finite-dimensional Hilbert space
H
with
dim H
=
n
,
and an orthonormal basis {e
1
, ··· , e
n
}, then each x ∈ H can be written as
x =
n
X
i=1
hx, e
i
ie
i
,
and
kxk
2
=
n
X
i=1
|hx, e
i
i|
2
.
Thus
H
is isomorphic to
`
n
2
, the space
R
n
with the Euclidean norm, via the map
x 7→ (hx, e
1
i, ··· , hx, e
n
i).
Can we push this to the infinite dimensional case? Yes. We will have to replace
our finite sum
P
n
i=1
with an infinite sum. Of course, with an infinite sum, we
need to make sure things converge. This is guaranteed by Bessel’s inequality.
Lemma
(Bessel’s inequality)
.
Let
E
be Euclidean and
{e
i
}
N
i=1
with
N ∈ N∪{∞}
an orthonormal system. For any
x ∈ E
, define
x
i
=
hx, e
i
i
. Then for any
j ≤ N
,
we have
j
X
i=1
|x
i
|
2
≤ kxk
2
.
Proof. Consider the case where j is finite first. Define
F
j
= span{e
1
, ··· , e
j
}.
This is a finite dimensional subspace of
E
. Hence an orthogonal projection
P
F
j
exists. Moreover, we have an explicit formula for this:
P
F
j
=
j
X
i=1
hx, e
i
ie
i
.
Thus
j
X
i=1
|x
i
|
2
= kP
F
j
xk
2
≤ kxk
2
since we know that
kP
F
j
k ≤
1. Taking the limit as
j → ∞
proves the case for
infinite j.
The only thing we required in the proof is for the space to be Euclidean.
This is since we are talking about the sum
∞
X
i=1
|x
i
|
2
,
and this is a sum of numbers. However, if we want to investigate the sum
x =
∞
X
i=1
hx, e
i
ie
i
,
then we’d better require the space to be Hilbert, so that the sum has something
to converge to.
Proposition.
Let
H
be a separable Hilbert space, with a countable basis
{e
i
}
N
i=1
, where N ∈ N ∪ {∞}. Let x, y ∈ H and
x
i
= hx, e
i
i, y
i
= hy, e
i
i.
Then
x =
N
X
i=1
x
i
e
i
, y =
N
X
i=1
y
i
e
i
,
and
hx, yi =
N
X
i=1
x
i
¯y
i
.
Moreover, the sum converges absolutely.
Proof.
We only need to consider the case
N
=
∞
. Otherwise, it is just finite-
dimensional linear algebra.
First, note that our expression is written as an infinite sum. So we need to
make sure it converges. We define the partial sums to be
s
n
=
n
X
i=1
x
i
e
i
.
We want to show s
n
→ x. By Bessel’s inequality, we know that
∞
X
i=1
|x
i
|
2
≤ kxk
2
.
In particular, the sum is bounded, and hence converges.
For any m < n, we have
ks
n
− s
m
k =
n
X
i=m+1
|x
i
|
2
≤
∞
X
i=m+1
|x
i
|
2
.
As
m → ∞
, the series must go to 0. Thus
{s
n
}
is Cauchy. Since
H
is Hilbert,
s
n
converges, say
s
n
→ s =
∞
X
i=1
x
i
e
i
.
Now we want to prove that this sum is indeed
x
itself. Note that so far in the
proof, we have not used the fact that
{e
i
}
is a basis. We just used the fact that
it is orthogonal. Hence we should use this now. We notice that
hs, e
i
i = lim
n→∞
hs
n
, e
i
i = lim
n→∞
n
X
j=1
x
j
he
j
, e
i
i = x
i
.
Hence we know that
hx −s, e
i
i = 0.
for all
i
. So
x − s
is perpendicular to all
e
i
. Since
{e
i
}
is a basis, we must have
x − s = 0, i.e. x = s.
To show our formula for the inner product, we can compute
hx, yi = lim
n→∞
*
n
X
i=1
x
i
e
i
,
n
X
j=1
y
j
e
j
+
= lim
n→∞
n
X
i,j=1
x
i
¯y
j
he
i
, e
j
i
= lim
n→∞
n
X
i,j=1
x
i
¯y
j
δ
ij
= lim
n→∞
n
X
i=1
x
i
¯y
i
=
∞
X
i=1
x
i
¯y
i
.
Note that we know the limit exists, since the continuity of the inner product
ensures the first line is always valid.
Finally, to show absolute convergence, note that for all finite j, we have
j
X
i=1
|x
i
¯y
i
| ≤
v
u
u
t
n
X
i=1
|x
i
|
2
v
u
u
t
n
X
i=1
|y
i
|
2
≤ kxkkyk.
Since this is a uniform bound for any j, the sum converges absolutely.
Note that in the case of x = y, our formula for the inner product gives
kxk
2
=
N
X
i=1
|x
i
|
2
.
This is known as Parseval’s equality
What this proposition gives us is that given any separable Hilbert space,
we can find “coordinates” for it, and in terms of these coordinates, our inner
product and hence norm all act like `
2
. In particular, we have the map
x 7→ {hx, e
i
i}
N
i=1
that takes
H
into
`
2
. This is injective since by Parseval’s equality, if the image
of x is 0, then kxk
2
=
P
0 = 0. So x = 0.
This is good, but not good enough. We want the map to be an isomorphism.
Hence, we need to show it is surjective. In other words, every element in
`
2
is
obtained. This is a theorem by Riesz and Fisher, and is in fact easy to prove,
since there is an obvious candidate for the preimage of any {x
i
}
i∈N
.
Proposition.
Let
H
be a separable Hilbert space with orthonormal basis
{e
i
}
i∈N
. Let
{a
i
}
i∈N
∈ `
2
(
C
). Then there exists an
x ∈ H
with
hx, e
i
i
=
a
i
.
Moreover, this x is exactly
x =
∞
X
i=1
x
i
e
i
.
Proof.
The only thing we need to show is that this sum converges. For any
n ∈ N, define
s
n
=
n
X
i=1
a
i
e
i
∈ H.
For m < n, we have
ks
n
− s
m
k
2
=
n
X
m+1
|a
i
|
2
→ 0
as
m → ∞
because
{a
i
} ∈ `
2
. Hence
s
n
is Cauchy and as such converges to
x
.
Obviously, we have
hx, e
i
i = lim
n→∞
n
X
j=1
a
j
he
j
, e
i
i = a
i
.
So done.
This means we have a isomorphism between
`
2
and
H
. Moreover, this is
continuous and in fact isometric. So this is a very strong result. This says all
separable Hilbert spaces are `
2
.