The Dual of a Hilbert Space
Let V denote a real Hilbert space with inner product, Ýu, vÞ V . Let V’ denote the space of
bounded linear functionals on V, equipped with the norm,
qFq V v = supá|FÝvÞ| : qvq V = 1â.
Then for any u 5 V, F u ÝvÞ := Ýu, vÞ V defines a unique bounded linear functional
Fu 5 Vv
with
|| F u || V v = || u|| V
Similarly, it follows from the Riesz theorem, that for any F 5 V v there is a unique u F 5 V
such that
FÝvÞ = Ýu F , vÞ V -v 5 V.
|| F|| V v = || u F || V
Moreover,
Evidently there exists an isometric isomorphism J from V onto V’
V:u $ JÝuÞ = F u 5 V v
In addition, J induces an inner product on Vv,
ÝF, GÞ V v = ÝJ ?1 F, J ?1 GÞ V = Ýu F , v G Þ V
One can define an abstract duality pairing on V v × V by
< F, v > V v ×V = FÝvÞ for F 5 V v , v 5 V
Then
ÝF, JvÞ V v = < F, v > V v ×V = ÝJ ?1 F, vÞ V
for F 5 V v , v 5 V
Now V v is an abstract dual space in the sense that we have no characterization for the
elements of V v other than the assertion that they are continuous linear functionals on V. In
order to have a concrete characterization for the elements in the dual space, we consider a
concrete Hilbert space, W. Then W is called a realization for the abstract space, Vv if there
exists an isometry j of W onto V v ; i.e.,
K = j ?1
J
V í Vv
EJ ¾ · j
W
1
Then for all f in W, Öf, v× W×V = jfÝvÞ for all v in V, and Öf, v× W×V is the realization of the
abstract duality pairing.
Examples
1. Take W = V. Then j = J and K = identity. In this case the duality pairing is just the V inner
product. This is what is meant when we say we ”identify a Hilbert space with its dual”.
Although this is frequently done, it is not the only realization for the dual of a Hilbert space
and in the treatment of problems in partial differential equations by variational methods, it is
not the best choice of realization for the dual.
2. Consider the Hilbert space
V = áu 5 L 2 : Ý1 + |z| 2 Þ s/2 ûÝzÞ 5 L 2 â = H s , s ³ 0,
with
Ýu, vÞ V = X n Ý1 + |z| 2 Þ s ûÝzÞv! ÝzÞdz,
R
Then V is dense in a larger Hilbert space H = H 0 and we can use the space H, called the
pivot space, to show that
W = H ?s = áf 5 S v : Ý1 + |z| 2 Þ ?s/2 !fÝzÞ 5 L 2 â
is a realization for the dual of V = H s .
For f 5 W and v 5 V define
jfÝvÞ = Öjf, v× W×V = X n fÝxÞvÝxÞdx = Ýf, vÞ 0
R
Then it follows that
|jfÝvÞ| = | XR n fÝxÞvÝxÞdx| = | XR n !fÝzÞv! ÝzÞdz|
= X n Ý1 + |z| 2 Þ ?s/2 !fÝzÞÝ1 + |z| 2 Þ s/2 v! ÝzÞdz
R
²
XR n Ý1 + |z| 2 Þ ?s |f!ÝzÞ| 2 dz
1/2
qvq V
This proves that jf 5 Vv for all f 5 W; i.e., j maps W into V v .
To show that j maps W onto V v , let G be an element of V v with G = Ju G for some u G 5 V.
Then for all v 5 V,
GÝvÞ = ÖG, v× V v ×V = Ýu G , vÞ V
2
= X n Ý1 + |z| 2 Þ s û G ÝzÞv! ÝzÞdz
R
Now let
2 s
g = T ?1
F Ý1 + |z| Þ û G ÝzÞ
= T ?1
F ßĝÝzÞà,
ÖG, v× V v ×V = XR n gÝxÞvÝxÞ dx = Öjg, v× V v ×V
so
Here G denotes the element of Vv and ĝ denotes the function Ý1 + |z| 2 Þ s û G ÝzÞ which is the
Fourier transform of g. Then the fact that Ý1 + |z| 2 Þ s û G ÝzÞ = ĝÝzÞ and u G 5 V = H s implies
Ý1 + |z| 2 Þ ?s/2 ĝÝzÞ = Ý1 + |z| 2 Þ s/2 û G ÝzÞ 5 L 2 ;
i.e., g belongs to W = H ?s . We have shown that GÝvÞ = Ýu G , vÞ V = jgÝvÞ, and
since jg = G, every G in V’ is the image under j of an element g in W which shows j is
onto. If we define
qgq W = supá|jgÝuÞ| : quq V = 1â,
then j is an isometry. Note that since jg = Ju G , where
2 s
g = j ?1 Ju G = T ?1
F Ý1 + |z| Þ û G ÝzÞ
then
2 s
Ku G = j ?1 Ju G = T ?1
F Ý1 + |z| Þ û G ÝzÞ
2 s
= T ?1
F Ý1 + |z| Þ T F Ýu G Þ
is the canonical isometry of V onto W. i.e.,
TF
Ps
Sv í Sv
·
¹ T ?1
F
Ví W
K
where P s denotes the multiplication by Ý1 + |z| 2 Þ s . Since
jfÝvÞ = X n fÝxÞvÝxÞdx = Öf, v× W×V
R
the realization of the duality pairing is just the H 0 inner product, extended to W × V. This
may be interpreted to mean that the space H = H 0 ÝR n Þ occupies a position precisely
midway between the space H s and its dual space, H ?s . We say that H 0 is the ”pivot space”
between H s and H ?s . Functions in H s may be viewed as being more regular than H 0
functions to precisely the same degree that H 0 functions are more regular than elements of
H ?s .
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3. Consider the Hilbert space V = H 10 ÝUÞ with
Ýu, vÞ V = X ßuÝxÞvÝxÞ + 4u 6 4vàdx.
U
Here we will use the pivot space H = H 0 ÝUÞ to show that a realization of the dual of V is
given by
n
W = áf 5 D v ÝUÞ : f = f 0 + > j=1 / j f j , f i 5 H 0 ÝUÞ -iâ
For f 5 W, and v 5 V define
jfÝvÞ = X fÝxÞvÝxÞdx = Ýf, vÞ 0 .
U
Then for any test function, d
jfÝdÞ = X Ýf 0 + > j=1 / j fÞdÝxÞdx. = X ßf 0 d ? > j=1 f j / j dÝxÞàdx.
n
n
U
U
and
n
|jfÝdÞ| ² qf 0 q 0 qdq 0 + > j=1 qf j q 0 q/ j dq 0 ²
n
qf j q 20
> j=0
1/2
qdq V -d 5 DÝUÞ
But DÝUÞ is dense in V, hence this estimate extends to all of V, showing that jf 5 Vv for all
f 5 W.
To show that j is onto V’, let F be an element of V’ with F = Ju F for some u F in V. Then for
any test function, d
FÝdÞ = Ýu F , dÞ V = X ßu F ÝxÞdÝxÞ + 4u F 6 4dàdx
U
= X ßÝu F ÝxÞ ? 4 2 u F ÞdÝxÞàdx
U
= X fÝxÞdÝxÞdx
U
fÝxÞ = Ýu F ÝxÞ ? 4 2 u F Þ.
for
Since u F 5 V, it follows that f 5 D v ÝUÞ. In addition,
n
fÝxÞ = u F ? 4 2 u F = u F ? > j=1 / j Ý/ j u F Þ
and
u F 5 V implies / j u F 5 H 0 ÝUÞ for each j = 1, ..., n. Thus we have shown,
n
f 5 D v ÝUÞ and fÝxÞ = u F ? > j=1 / j Ý/ j u F Þ 5 W
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Moreover, for all test functions d,
FÝdÞ = Ýu F , dÞ V = Ýf, dÞ 0 = jfÝdÞ
Once again, since DÝUÞ is dense in V, the result extends to all of V; i.e., F = jf for f in W
given by f = u F ? 4 2 u F ., u F 5 V .
Note that jf = Ju F implies
Ku F = j ?1 Ju F = Ý1 ? 4 2 Þu F = f.
i.e., K = 1 ? 4 2 is the canonical isometry of V = H 10 ÝUÞ onto its dual W which we denote by
H ?1 ÝUÞ. The realization of the duality pairing is
Öf, v× W×V = XU fÝxÞvÝxÞdx. = Ýf, vÞ 0 ;
showing that the duality pairing is just the extension of the H 0 ÝUÞ inner product to W × V.
That is to say that H 0 ÝUÞ acts as the pivot space between the spaces V = H 10 ÝUÞ and its
dual, H ?1 ÝUÞ,
H 10 ÝUÞ Ð H 0 ÝUÞ Ð H ?1 ÝUÞ.
As in the case of H s and its dual, H ?s , the pivot space H 0 ÝUÞoccupies a position that is
precisely midway (in terms of regularity) betweenH 10 ÝUÞ and its dual, H ?1 ÝUÞ. The functions
in H 0 (U) are first derivatives of functions from H 10 (U), and elements of
H ?1 ÝUÞ are first derivatives of functions from H 0 ÝUÞ.
We have considered here some examples of the situation where X and Y are linear spaces
where the notion of convergence is defined and
X Ð Y means that
and
a) X Ð Y
b) x n ¸ 0 in X implies iÝx n Þ ¸ 0 in Y.
In addition, F 5 X v means that x n ¸ 0 in X implies FÝx n Þ ¸ 0 in R, and it is then evident
that
X Ð Y implies Y v Ð X v ;
i.e., if
i:X¸Y
is continuous,
t
v
v
then
i:Y ¸X
is also continuous.
Note, however, that the transpose inclusion, t i, need not be injective. In particular, if X is a
closed subspace of Hilbert space, Y, consider
5
zF 5 Y
with
F = JÝz F Þ 5 Y v .
The projection theorem implies z F = z 1 + z 2 , with z 1 5 X and z 2 e X, so that t iÝFÞ = JÝz 1 Þ
which can be zero even if F is not zero.
t
However, if i : X ¸ Y has a dense image, then i : Y v ¸ X v is injective. To see this,
suppose that iÝXÞ is dense in Y, and t iÝFÞ = 0. Then,
t
x, iÝFÞ = 0 for all x 5 X.
But
t
x, iÝFÞ = ÖiÝxÞ, F×
and since the elements iÝxÞ are dense in Y, and F is continuous on Y, F must be zero
whenever t iÝFÞ is zero.
It follows from this discussion that the dual space any function space in which the test
functions are densely included (e.g. H m0 ÝUÞ, H m ÝR n Þ ) can be identified with a subspace of
the space of distributions. That is, if the inclusion, C K0 ÝUÞ Ð X has a dense image, then
X v Ð D v ÝUÞ is a continuous injection. For example, H ?1 ÝUÞ is a subspace of the
distributions, which means that equality in H ?1 ÝUÞ can be interpreted to mean equality in the
sense of distributions. On the other hand, the dual of H 1 ÝUÞ is not a subspace of the
distributions since the test functions are not dense in H 1 ÝUÞ. In fact, we have already seen
that the orthogonal complement of H 10 ÝUÞ in H 1 ÝUÞ is the subspace
N = áz 5 H 1 ÝUÞ : z ? 4 2 z 5 H 0 ÝUÞ, and z ? 4 2 z = 0 â
and it is clear that N is the null space of the duality mapping
J : H 1 ÝUÞ = H 10 ÝUÞ ã N ¸ H ?1 ÝUÞ .
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