4Hilbert spaces
II Linear Analysis
4.2 Riesz representation theorem
Our next theorem allows us to completely understand the duals of Hilbert spaces.
Consider the following map. Given a Hilbert space
H
and
v ∈ H
, consider
φ
v
∈ H
∗
defined by
φ
v
(w) = hw, vi.
Note that this construction requires the existence of a inner product.
Notice that this is indeed a bounded linear map, where boundedness comes
from the Cauchy-Schwarz inequality
|hw, vi| ≤ kvk
H
kwk
H
.
Therefore, φ taking v 7→ φ
v
is a map φ : H → H
∗
.
Using this simple construction, we have managed to produce a lot of members
of the dual. Are there any more things in the dual? The answer is no, and this
is given by the Riesz representation theorem.
Proposition
(Riesz representation theorem)
.
Let
H
be a Hilbert space. Then
φ
:
H → H
∗
defined by
v 7→ h·, vi
is an isometric anti-isomorphism, i.e. it is
isometric, bijective and
φ(λv + µw) =
¯
λφ(v) + ¯µφ(v).
Proof. We first prove all the easy bits, namely everything but surjectivity.
–
To show injectivity, if
φ
v
=
φ
u
, then
hw, vi
=
hw, ui
for all
w
by definition.
So
hw, v−ui
= 0 for all
w
. In particular,
hv−w, v −wi
= 0. So
v−w
= 0.
–
To show that it is an anti-homomorphism, let
v, w, y ∈ H
and
λ, µ ∈ F
.
Then
φ
λv+µw
(y) = hy, λv + µwi =
¯
λhy, vi + ¯µhy, wi =
¯
λφ
v
(y) + ¯µφ
w
(y).
– To show it is isometric, let v, w ∈ H and kwk
H
= 1. Then
|φ
v
(w)| = |hw, vi| ≤ kwk
H
kvk
H
= kvk
H
.
Hence, for all
v
,
kφ
v
k
H
∗
≤ kvk
H
for all
v ∈ H
. To show
kφ
v
k
H
∗
is exactly
kvk
H
, it suffices to note that
|φ
v
(v)| = hv, vi = kvk
2
H
.
So kφ
v
k
H
∗
≥ kvk
2
H
/kvk
H
= kvk
H
.
Finally, we show surjectivity. Let ξ ∈ H
∗
. If ξ = 0, then ξ = φ
0
.
Otherwise, suppose
ξ 6
= 0. The idea is that (
ker ξ
)
⊥
is one-dimensional, and
then the
v
we are looking for will be an element in this complement. So we
arbitrarily pick one, and then scale it appropriately.
We now write out the argument carefully. First, we note that since
ξ
is
continuous,
ker ξ
is closed, since it is the inverse image of the closed set
{
0
}
. So
ker ξ is complete, and thus we have
H = ker ξ ⊕ (ker ξ)
⊥
.
The next claim is that
dim
(
ker ξ
) = 1. This is an immediate consequence of the
first isomorphism theorem, whose proof is the usual one, but since we didn’t
prove that, we will run the argument manually.
We pick any two elements
v
1
, v
2
∈
(
ker ξ
)
⊥
. Then we can always find some
λ, µ not both zero such that
λξ(v
1
) + µξ(v
2
) = 0.
So
λv
1
+
µv
2
∈ ker ξ
. But they are also in (
ker ξ
)
⊥
by linearity. Since
ker ξ
and
(
ker ξ
)
⊥
have trivial intersection, we deduce that
λv
1
+
µv
2
= 0. Thus, any two
vectors in (
ker ξ
)
⊥
are dependent. Since
ξ 6
= 0, we know that
ker ξ
has dimension
1.
Now pick any
v ∈
(
ker ξ
)
⊥
such that
ξ
(
v
)
6
= 0. By scaling it appropriately,
we can obtain a v such that
ξ(v) = hv, vi.
Finally, we show that
ξ
=
φ
v
. To prove this, let
w ∈ H
. We decompose
w
using
the previous theorem to get
w = αv + w
0
for some
w
0
∈ ker ξ
and
α ∈ F
. Note that by definition of (
ker ξ
)
⊥
, we know
that hw
0
, vi = 0. Hence we know that
ξ(w) = ξ(αv + w
0
) = ξ(αv) = αξ(v)
= αhv, vi = hαv, vi = hαv + w
0
, vi = hw, vi.
Since w was arbitrary, we are done.
Using this proposition twice, we know that all Hilbert spaces are reflexive,
i.e. H
∼
=
H
∗∗
.
We now return to the proof of the proposition we claimed at the beginning.
Proposition. For f ∈ C(S
1
), defined, for each k ∈ Z,
ˆ
f(k) =
1
2π
Z
π
−π
e
ikx
f(x) dx.
The partial sums are then defined as
S
N
(f)(x) =
N
X
n=−N
e
inx
ˆ
f(k).
Then we have
lim
N→∞
1
2π
Z
π
−π
|f(x) − S
N
(f)(x)|
2
dx = 0.
Proof.
Consider the following Hilbert space
L
2
(
S
1
) defined as the completion of
C
C
(S
1
) under the inner product
hf, gi =
1
2π
Z
π
−π
f(x)¯g(x) dx,
Consider the closed subspace
U
N
= span{e
inx
: |n| ≤ N}.
Then in fact S
N
defined above by
S
N
(f)(x) =
N
X
n=−N
e
−inx
ˆ
f(k)
is the projection operator onto
U
N
. This is since we have the orthonormal
condition
he
inx
, e
−imx
i =
1
2π
Z
π
−π
e
inx
e
−imx
dx =
(
1 n = m
0 n 6= m
Hence it is easy to check that if
f ∈ U
N
, say
f
=
P
N
n=−N
a
n
e
inx
, then
S
N
f
=
f
since
S
N
(f) =
N
X
n−−N
ˆ
f(k)e
−inx
=
N
X
n=−N
hf, e
inx
ie
−inx
=
N
X
n=−N
a
n
e
−inx
= f
using the orthogonality relation. But if f ∈ U
⊥
N
, then
1
2π
Z
π
−π
e
−inx
f(x) dx = 0
for all |n| < N. So S
N
(f) = 0. So this is indeed a projection map.
In particular, we will use the fact that projection maps have norms
≤
1.
Hence for any P (x), we have
1
2π
Z
π
−π
|S
N
(f)(x) − S
N
(P )(x)|
2
dx ≤
1
2π
Z
π
−π
|f(x) − P (x)|
2
dx
Now consider the algebra
A
generated
{e
inx
:
n ∈ Z}
. Notice that
A
separates
points and is closed under complex conjugation. Also, for every
x ∈ S
1
, there
exists
f ∈ A
such that
f
(
x
)
6
= 0 (using, say
f
(
x
) =
e
ix
). Hence, by Stone-
Weierstrass theorem,
¯
A
=
C
C
(
S
1
), i.e. for every
f ∈ C
C
(
S
1
) and
ε >
0, there
exists a polynomial P of e
ix
and e
−ix
such that
kP − fk < ε.
We are almost done. We now let
N > deg P
be a large number. Then in
particular, we have S
N
(P ) = P . Then
1
2π
Z
π
−π
|S
N
(f) − f|
2
dx
1
2
≤
1
2π
Z
π
−π
|S
N
(f) − S
N
(P )|
2
dx
1
2
+
1
2π
Z
π
−π
|S
N
(P ) −P |
2
dx
1
2
+
1
2π
Z
π
−π
|P − f|
2
dx
1
2
≤ ε + 0 + ε
= 2ε.
So done.