where H is a Hilbert space and V is a Banach space contained in H and each of the
above inclusions is continuous and each space is dense in the next one. The standard
example of a Gelfand triple is H_{0}^{1}
(D )
⊆ L^{2}
(D )
⊆
( )
H10 (D)
^{′} with the convention that
L^{2}
(D )
is identified with its dual space. Thus for f ∈ L^{2}
(D )
, f is considered as
something in
( )
H10 (D )
^{′} according to the rule
〈f,ϕ〉 ≡ (f,ϕ)L2(D)
This is a very pleasant thing to contemplate and it is natural and transparent. However,
there are other ways to come up with a Gelfand triple which are much more perverse.
The following is an example of such a thing along with an application. See [?] and
references given there. I think this idea is due to Lions.
First consider the following situation.
X →θ Y
where θ is continuous, linear and one to one and X is a Banach space. Then θ
(X )
⊆ Y
and you could define
||θx||θ(X) ≡ ||x||X .
Then θ
(X )
can be considered the same thing as X because θ preserves distances and all
algebraic properties. Thus people write X ⊆ Y to save space. In the above simple
example, it is obvious what θ is. This is because the things in H_{0}^{1} and things in L^{2} are
both functions defined on D and we can simply take θ to be the identity map. However,
you might have H be the dual space of something. Thus it consists of bounded
linear transformations defined on some Banach space. Then it becomes necessary
to specify the manner in which vectors in V can be considered as vectors of
H.
Let ∞ > p ≥ 2. Then letting D be a bounded open set, H_{0}^{1}
(D )
embedds
continuously into L^{p′
}
(D )
. That is
||ϕ||p′ ≤ C ||ϕ|| 1. (32.1.1)
L H 0
(32.1.1)
Here
1′
p
+
1
p
= 1. Also note that an equivalent inner product on H_{0}^{1}
(D )
is
∫
(f,g)H10 ≡ D ∇f ⋅∇gdx
Then with respect to this inner product, the Riesz map is given by −Δ.
^{′}? If
it is not dense, then there exists g^{∗}∈
(Lp(D ))
^{′} which is not the limit of vectors of H.
Then since L^{p}
(D )
is reflexive, an application of the Hahn Banach theorem shows there
exists f ∈ L^{p}
(D )
such that
〈g∗,f〉 p ′ p ⁄= 0, 〈− Δϕ,f 〉 p ′ p = 0 (32.1.2)
(L(D)),L(D) (L (D )),L (D)
(32.1.2)
for all −Δϕ ∈ H. However, it was just shown H could be considered a subset of
(Lp (D))
^{′} in the manner described above. Therefore, the last equation in the above is of
the form
∫
0 = (− Δ ϕ,f) = 〈f,ϕ〉 = f ϕdx
H D
and since this holds for all ϕ ∈ H_{0}^{1}
(D)
, it follows by density of H_{0}^{1}
(D )
in L^{p′
}
(D )
, that
f = 0 and now this contradicts the inequality in 32.1.2.
Now Δ is defined on H_{0}^{1}
(D )
and it delivers something in
(H10)
^{′}≡ H. Of course
H_{0}^{1}
(D)
is dense in L^{p′
}
(D)
. Can Δ be extended to all of L^{p′
}
(D)
? The answer is yes and
it is more of the same given above. For ϕ ∈ H_{0}^{1}
(D )
,−Δϕ ∈ H ⊆
(Lp (D))
^{′}. Then by the
above, for ϕ ∈ H_{0}^{1}
(D )
and f ∈ L^{p}
(D )
,
∫
〈− Δ ϕ,f〉 ≡ 〈f,ϕ〉 ≡ fϕdx
D
|∫ |
|〈− Δ ϕ,f〉| ≡ |〈f,ϕ〉| ≡ || f ϕds||≤ ||ϕ|| ′ ||f ||
| D | Lp(D) Lp(D)
and so −Δ is a continuous linear mapping defined on a dense subspace H_{0}^{1}
(D)
of
L^{p′
}
(D )
and so this does indeed extend to a continuous linear map defined on all of
L^{p′
}
(D )
given by the formula
∫
〈− Δg,f〉 ≡ D fgdx
This proves the lemma.
Thus letting V ≡ L^{p}
(D )
, and H ≡
(H1 (D ))
0
^{′}, it follows V ⊆ H ⊆ V^{′} is a Gelfand
triple with the understanding of what it means for one space to be included in another
described above. To emphasize the above, for −Δϕ ∈ H,f ∈ L^{p},
∫
〈− Δ ϕ,f〉 ≡ (− Δϕ,f) ≡ 〈f,ϕ〉 ≡ fϕdx
H D
More generally, for g ∈ L^{p′
}
(D)
,−Δg ∈
(Lp(D))
^{′} according to the rule
∫
〈− Δg,f〉 ≡ Dfgdx.
With this example of a Gelfand triple, one can define a “porous medium
operator” A : V → V^{′}. Let Ψ be a real valued function defined on ℝ which
satisfies