is pseudomonotone. You can also define Â : V→V^{′} by
Aˆ(u)(t) ≡ A(u (t),t) a.e.
Then when can you obtain a useable limit condition for Â? I think the earliest solution to
this problem was given in [?]. These ideas were extended to set valued maps in [?] and to
another situation in [?].
Define V≡V_{p} by
V = Lp([0,T ];V ), p > 1,
where V is a separable Banach space and H is a Hilbert space such that
V ⊆ H = H ′ ⊆ V ′
with each space dense in the following one. The measure space is chosen to be
([0,T],ℬ([0,T ]),m )
where m is the Lebesgue measure and ℬ
([0,T ])
consists of all the
Borel sets, although one could use the σ algebra of Lebesgue measurable sets as well. We
denote by V_{p} or V the above space. If U is a Banach space, U_{r} will denote
L^{r}
([0,T],U)
.
We will assume the following measurability condition. For each u ∈V,
t → A (u (t),t) is a measurable multifunction (47.7.23)
(47.7.23)
In the case when A
(⋅,t)
is single-valued, bounded and pseudomonotone, this
measurability condition is satisfied and so it is measurable. Thus, this definition is a
generalization of what would be expected for single-valued operators. We use the
following lemma.
Lemma 47.7.1Let U be a separable reflexive Banach space.Suppose there is asequence
{uj (ω )}
_{j=1}^{∞}in U, where each ω → u_{j}
(ω)
is measurable and foreach ω, sup_{i}
∥ui(ω)∥
< ∞. Then, there exists u
(ω )
∈ U such that ω → u
(ω)
ismeasurable, and a subsequence n
(ω)
, that depends on ω, such that the weaklimit
lim u (ω) = u (ω )
n(ω)→∞ n(ω)
holds.
Proof. Let
{zi}
_{i=1}^{∞} be a countable dense subset of U^{′}. Let h : U →∏_{i=1}^{∞}ℝ be
defined by
∏∞
h(u) = 〈zi,u〉.
i=1
Let X = ∏_{i=1}^{∞}ℝ with the product topology. Then, this is a Polish space with the
metric defined as d
(x,y)
= ∑_{i=1}^{∞}
-|xi−yi|
1+|xi−yi|
2^{−i}. By compactness, for a fixed ω,the
h
(un(ω))
are contained in a compact subset of X. Next, define
Γ (ω) = ∪---h(u-(ω)),
n k≥n k
which is a nonempty compact subset of X.
Next, we claim that ω → Γ_{n}
(ω)
is a measurable multifunction.
The proof of the claim is as follows. It is necessary to show that Γ_{n}^{−}
(O )
defined as
{ω : Γ n(ω)∩ O ⁄= ∅}
is measurable whenever O is open. It suffices to verify this for O a
basic open set in the topology of X. Thus let O = ∏_{i=1}^{∞}O_{i} where each O_{i} is a proper
open subset of ℝ only for i ∈
which is a measurable set since u_{k} is measurable.
Then, it follows that ω → Γ_{n}
(ω )
is strongly measurable because it has compact values
in X, thanks to Tychonoff’s theorem. Thus Γ_{n}^{−}
(H)
=
{ω : H ∩ Γ n(ω) ⁄= ∅}
is
measurable whenever H is a closed set. Now, let Γ
(ω)
be defined as ∩_{n}Γ_{n}
(ω)
and then
for H closed,
Γ − (H ) = ∩nΓ −n (H )
and each set in the intersection is measurable, so this shows that ω → Γ
(ω )
is also
measurable. Therefore, it has a measurable selection g
(ω )
. It follows from the definition
of Γ
(ω)
that there exists a subsequence n
(ω)
such that
g(ω) = lim h(u (ω)) in X.
n(ω)→∞ n(ω)
In terms of components, we have
gi(ω) = lim 〈zi,un(ω)(ω )〉 .
n(ω)→ ∞
Furthermore, there is a further subsequence, still denoted with n
(ω)
, such that
u_{n(ω)
}
(ω)
→ u
(ω)
weakly. This means that for each i,
〈 〉
gi(ω) = n(liωm)→ ∞ zi,un(ω)(ω) = 〈zi,u(ω)〉.
Thus, for each z_{i} in a dense set, ω →
〈zi,u(ω)〉
is measurable. Since the z_{i} are dense,
this implies ω →
〈z,u (ω )〉
is measurable for every z ∈ U^{′} and so by the Pettis theorem,
ω → u
(ω)
is measurable. _
Also is a definition.
Definition 47.7.2Let A
(⋅,t)
: V →P
(V′)
. Then,the Nemytskii operator associatedwith A,
p ( p′ ′)
Aˆ: L ([0,T];V) → P L ([0,T];V ) ,
is given by
z ∈ ˆA(u) if and only if z ∈ Lp′ ([0,T];V′)and z(t) ∈ A (u(t),t) a.e. t.
Growth and coercivity
The next three conditions on the operator A are similar to the conditions proposed by
Bian and Webb, [?] See also Berkovitz and Mustonen [?] which seems to be the paper
where these ideas originated. These specific and reasonable conditions, together with
a fourth one we add below, allow us to prove an appropriate limit condition
that is based on the assumption that u → A
(u,t)
is a set-valued, bounded
and pseudomonotone map from V to P
(V ′)
and t → A
(u,t)
has a measurable
selection.
Our aim is to provide reasonable conditions under which an assumption of
pseudomonotonicity on u → A
(u,t)
transfers to a useable limit condition for the operator
Â defined on V = L^{p}
([0,T];V)
.
It is obvious that Âu is convex because this is true of A
(u,t)
. It is also closed. To see
this, suppose z_{n}∈Â
(u)
. Then z_{n}
(t)
∈ A
(u (t),t)
for a.e.t. Taking the union of the
exceptional sets, we can assume this inclusion holds off a single set of measure zero for all
n. If you have z_{n}→ w strongly in V^{′}, then a subsequence converges pointwise
a.e. Therefore, by upper semicontinuity of the pointwise operator u → A
(u,t)
,
it follows that w
(t)
∈ A
(u(t),t)
for a.e. t. Thus Âu is convex and strongly
closed.
We assume the following conditions on A.
A
(⋅,t)
: V →P
′
(V )
is pseudomonotone and bounded: A
(u,t)
is a closed convex
set for each t, u → A
(u,t)
is bounded, and if
lim sup 〈A (un,t),un − u〉 ≤ 0
n→∞
then for any v ∈ V,
lim inf 〈A(un,t),un − v〉 ≥ 〈z(v),u− v〉 some z(v) ∈ A (u,t)
n→ ∞
A
(⋅,t)
satisfies the estimates: There exists b_{1}≥ 0 and b_{2}≥ 0, such that
||z||V ′ ≤ b1||u||pV−1+ b2(t), (47.7.24)
(47.7.24)
for all z ∈ A
(u,t)
, b_{2}
(⋅)
∈ L^{p′
}
([0,T])
.
There exist a positive constant b_{3} and a nonnegative function b_{4} that is ℬ
is a measurable multifunction with respect to ℱ where ℱ will be the σ algebra of
Lebesgue measurable sets whenever t → u
(t)
is in V_{p}.
For u ∈V_{p}, we define Â
(u)
∈P
( ′)
Vp
as follows: z ∈Â
(u)
means that
z
(t)
∈ A
(u (t),t)
a.e. t. Thus this is the Nemytskii operator for A
(⋅,t)
.
In the following theorem and in arguments which take place below, U will be a
Hilbert space dense in V with the inclusion map compact. Such a Hilbert space always
exists and is important in probability theory where
(i,U,V )
is an abstract Wiener space.
However, in most applications from partial differential equations, it suffices to take U as a
suitable Sobolev space.