47.3 A Set Valued Browder Lemma With Measurability
A simple application is a measurable version of the Browder lemma which is also valid for
upper semicontinuous set valued maps. In what follows, we do not assume that A
(⋅,ω )
is
a set valued measurable multifunction, only that it has a measurable selection which is a
weaker assumption. First is a general result on upper set valued maps
(u,ω)
→ A
(u,ω )
where u → A
(u,ω )
is upper semicontinuous and ω → A
(u,ω)
has a measurable
selection.
Theorem 47.3.1Let V be a reflexive separable Banach space. Suppose ω →A
(u,ω)
has a measurable selection in V^{′}, for each u ∈ V and ω ∈ Ω the set A
(u,ω)
is closed and convex in V^{′}and u → A
(u,ω)
is bounded. Also, suppose u → A
(u,ω)
is upper-semicontinuous from the strong topology of V to the weak topology of V^{′}.That is, if u_{n}→ u in V strongly, then if O is a weakly open set containing A
(u,ω)
,it follows that A
(u ,ω)
n
∈ O for all n large enough. Conclusion:Then, wheneverω → u
(ω)
is measurable into V there is a measurable selection for ω → A
(u(ω),ω)
into V^{′}.
Proof: Let ω → u
(ω )
be measurable into V, and let u_{n}
(ω )
→ u
(ω )
in V where u_{n} is
a simple function
m∑n n n n
un (ω) = ckXEnk (ω), the Ek disjoint, Ω = ∪kE k,
k=1
each c_{k}^{n} being in V . We can assume that
∥un(ω)∥
≤ 2
∥u(ω)∥
for all ω. Then, by
assumption, there is a measurable selection for ω → A
(cn,ω)
k
denoted as ω → y_{k}^{n}
(ω )
.
Thus, ω → y_{k}^{n}
(ω)
is measurable into V^{′} and y_{k}^{n}
(ω )
∈ A
(cn,ω)
k
for all ω ∈ Ω. Consider
now,
m∑n
yn (ω ) = ynk (ω) XEnk (ω).
k=1
It is measurable and for ω ∈ E_{k}^{n} it equals y_{k}^{n}
(ω)
∈ A
(cnk,ω)
= A
(un(ω),ω)
. Thus, y^{n}
is a measurable selection of ω → A
(un(ω),ω)
. By the assumption A
(⋅,ω)
is
bounded, for each ω these y^{n}
(ω)
lie in a bounded subset of V^{′}. The bound
might depend on ω of course. It follows now from Lemma 47.2.2 that there is a
subsequence
{ }
yn(ω)
that converges weakly to y
(ω)
, where ω → y
(ω)
is measurable.
But,
yn(ω)(ω) ∈ A (u (ω) ,ω )
n(ω)
is a convex closed set for which u → A
(u,ω )
is upper-semicontinuous and u_{n(ω)
}→ u,
hence, y
(ω)
∈ A
(u(ω),ω )
. This is the claimed measurable selection. ■
The next lemma is about the projection map onto a set valued map whose values are
closed convex sets.
Lemma 47.3.2Let ω →K
(ω )
be measurable into ℝ^{n}where K
(ω)
is closed andconvex. Then ω → P_{K(ω)
}u
(ω)
is also measurable into ℝ^{n}if ω → u
(ω)
is measurable.Here P_{K(ω)
}is the projection map giving the closest point.
Proof:It follows from standard results on measurable multi-functions [?] also in
Theorem 47.1.2 above that there is a countable collection
{wn (ω )}
, ω → w_{n}
(ω )
being
measurable and w_{n}
(ω )
∈K
(ω )
for each ω such that for each ω, K
(ω)
=∪_{n}w_{n}
(ω)
.
Let
dn(ω) ≡ min {∥u(ω)− wk (ω)∥,k ≤ n}
Let u_{1}
(ω)
≡ w_{1}
(ω)
. Let
u2(ω) = w1 (ω )
on the set
{ω : ∥u(ω)− w1 (ω )∥ < {∥u(ω)− w2 (ω )∥}}
and
u2(ω) ≡ w2(ω) off the above set.
Thus
∥u (ω )− u(ω)∥
2
= d_{2}. Let
{ ω : ∥u (ω )− w (ω)∥ }
u3 (ω ) = w1(ω) on < ∥u (ω )− w (ω)1∥,j = 2,3 ≡ S1
{ j }
u3 (ω ) = w2(ω) on S1 ∩ ω : ∥u (ω )− w1(ω)∥
< ∥u(ω) − wj (ω)∥,j = 3
u3 (ω ) = w3(ω) on the remainder of Ω
Thus
∥u3 (ω )− u(ω)∥
= d_{3}. Continue this way, obtaining u_{n}
(ω)
such that
∥u (ω)− u(ω)∥ = d (ω)
n n
and u_{n}
(ω)
∈K
(ω)
with u_{n} measurable. Thus, in effect one picks the closest of
all the w_{k}
(ω)
for k ≤ n as the value of u_{n}
(ω)
and u_{n} is measurable and by
density in K
(ω)
of
{wn(ω)}
for each ω,
{un (ω)}
must be a minimizing sequence
for
λ(ω) ≡ inf{∥u (ω )− z∥ : z ∈ K(ω)}
Then it follows that u_{n}
(ω)
→ P_{K(ω)
}u
(ω)
weakly in ℝ^{n}. Here is why: Suppose it fails to
converge to P_{K(ω)
}u
(ω)
. Since it is minimizing, it is a bounded sequence. Thus
there would be a subsequence, still denoted as u_{n}
(ω)
which converges to some
q
(ω)
≠P_{K(ω)
}u
(ω)
. Then
λ(ω) = lim ∥u(ω)− un (ω )∥ ≥ ∥u(ω)− q(ω)∥
n→ ∞
because convex and lower semicontinuous is weakly lower semicontinuous. But this
implies q
(ω)
= P_{K(ω)
}
(u (ω ))
because the projection map is well defined thanks to strict
convexity of the norm used. This is a contradiction. Hence P_{K(ω)
}u
(ω )
= lim_{n→∞}u_{n}
(ω )
and so is a measurable function. It follows that ω → P_{K(ω)
}
(u (ω ),ω )
is measurable into
ℝ^{n}. ■
One way to prove the following Theorem in simpler cases is to use a measurable
version of the Kakutani fixed point theorem. It is done this way in [?] without the
dependence on ω. See also [?] for a measurable version of this fixed point theorem.
However, one can also prove it by a generalization of the proof Browder gave for a single
valued case and this is summarized here. We want to include the case where A is a sum of
two set valued operators and this involves careful consideration of the details. Such a
situation occurs when one considers operators which are a sum, one dependent on the
boundary of a region, and the other from a partial differential inclusion. Also, we will
need to consider finite dimensional subspaces which depend on ω which further
complicates the considerations.
Theorem 47.3.3Assumtions:Let B
(⋅,ω )
: V →P
(V′)
,C
(⋅,ω)
: V →P
(V ′)
for Va separable Banach space. Suppose that ω → B
(x,ω)
,ω → C
(x,ω)
each has a measurableselection and x → B
(x,ω )
,x → C
(x,ω)
each is upper semicontinuous from strong toweak topologies. Also let E
(ω)
be an n dimensional subspace of V which has a basis
{b (ω ),⋅⋅⋅,b (ω)}
1 n
each of which is a measurable function into V, and thatK
(ω)
⊆ E
(ω )
where K
(ω )
is a measurable multifunction which has convexclosed bounded values. Also let y
(ω)
be given, a measurable function into V^{′}.Conclusion:There exist measurable functions w_{B}
(ω )
,w_{C}
(ω )
and x
(ω)
withw_{B}
(ω)
∈ B
(x(ω),ω )
,w_{C}
(ω)
∈ C
(x (ω ),ω )
, and x
(ω)
∈ K
(ω)
such that for allz ∈ K
(ω )
,
〈y (ω )− (wB (ω )+ wC (ω )),z − x (ω)〉 ≤ 0
Proof: The argument will refer to the following commutative diagram.
θ(ω)∗
E (ω )′ → ℝn
i(ω )∗ A(⋅,ω) ↑ ↑ θ(ω)∗i(ω)∗A (θ(ω)⋅,ω )
E (ω) θ(←ω) ℝn
where A
(⋅,ω)
will be either B
(⋅,ω)
or C
(⋅,ω)
. Here θ
(ω )
e_{i}≡ b_{i}
(ω )
and extended
linearly. Then it is clear that θ
(ω)
maps measurable functions to measurable
functions.
What of θ
(ω)
^{−1}? Is ω → θ
(ω)
^{−1}h
(ω)
measurable into ℝ^{n} whenever h is
measurable into V ? Let h
(ω)
have values in E
(ω )
and be measurable into V .
Thus
∑
h (ω) = ai(ω)bi(ω)
i
The question reduces to whether the a_{i} are measurable. To see that these are measurable,
consider first
∥h(ω)∥
< M for all ω. Let S_{r}≡
{ ∑ }
ω : inf|a|>r ∥ iaibi(ω)∥ > M
. Thus
this is a measurable set. Also every ω is in some S_{r} because if not, you could get a
sequence
r
|a |
→∞ and yet
∑ r
∥ iaibi(ω )∥
≤ M. But then, dividing by
r
|a |
and taking a
suitable subsequence, one can obtain ∑_{i}a_{i}b_{i}
(ω)
= 0 for some
|a|
= 1. Also the S_{r}
are increasing in r. Now for ω ∈ S_{r}, define Φ
(a,ω)
= −
∑
∥ iaibi(ω)− h(ω)∥
where we will let
|a|
≤ r + 1. Since
{bi(ω )}
is a basis, there exists a
(ω)
such
that Φ
(a(ω),ω)
= 0. This a must satisfy
|a|
≤ r + 1 because if not, then you
would have
∥∑ aibi(ω)∥
i
≥ M since ω ∈ S_{r}. But
∥∑ aibi(ω)∥
i
=
∥h (ω)∥
< M
. Thus the maximum of a → Φ
(a,ω )
occurs on the compact set
|a|
≤ r + 1
and is 0. By Kuratowski’s theorem, we have ω → a
(ω )
is measurable where
h
(ω )
= ∑_{i}a_{i}
(ω)
b_{i}
(ω )
on S_{r}. Thus, since every ω is in some S_{r}, we must have
ω → a_{i}
(ω)
is measurable in case
∥h(ω)∥
≤ M for all ω. In the general case, let a^{m}
(ω)
be
the measurable function which goes with h_{m}
(ω)
where h_{m}
(ω)
is given by a
truncation of h so that
∥hm (ω)∥
≤ m. For each ω,h_{m}
(ω )
is eventually smaller than
m, so h
(ω)
= h_{m}
(ω)
. Thus if a_{i}^{m}
(ω)
go with h_{m}
(ω)
, these are constant for
all m large enough. Thus letting a_{i}
(ω)
≡ lim_{m→∞}a_{i}^{m}
(ω)
,a_{i} is measurable
and
h(ω) = lim h (ω) = lim ∑ am (ω)b (ω) = ∑ a (ω)b (ω)
m→ ∞ m m→∞ i i i i i i
and so the a_{i} are indeed measurable. Thus the θ
(ω )
^{−1}h
(ω)
= ∑_{i}a_{i}
(ω)
e_{i} which shows
that θ
(ω )
^{−1} does map measurable functions to measurable functions. In particular,
θ
(ω )
^{−1}K
(ω )
is indeed a closed, bounded, convex, and measurable multifunction which
can be seen by considering a sequence
has a measurable selection and for fixed ω this is upper
semicontinuous in x. The second condition for fixed ω is obvious. Consider the first. It
was shown above that θ
(ω)
x is measurable into V . Thus, by Theorem 47.3.1, it follows
that ω → A
(θ(ω)x,ω )
has a measurable selection into V^{′}. Therefore, it suffices to show
that if z
(ω)
is measurable into V^{′} then θ
(ω)
^{∗}i
(ω )
^{∗}z
(ω)
is measurable into ℝ^{n}. Let
w ∈ ℝ^{n}. Then
(θ(ω)∗i(ω )∗ z(ω),w) n ≡ 〈i(ω )∗z (ω ),θ(ω)w 〉
ℝ 〈 〉
= i(ω)∗z(ω),∑ wibi(ω)
i
〈 ∑ 〉
= z(ω) , wibi(ω)
i V ′,V
which is measurable. By the Pettis theorem, ω → θ
(ω)
^{∗}i
(ω)
^{∗}z
(ω)
is measurable. Thus
Â
(⋅,ω)
has the properties claimed.
Now tile ℝ^{n} with n simplices, each having diameter less than ε < 1, the set of
simplices being locally finite. Define for A = B or C the single valued function Â_{ε} on all
of ℝ^{n} by the following rule. If
x ∈ [x0,⋅⋅⋅,xn ],
so x =∑_{i=0}^{n}t_{i}x_{i},t_{i}≥ 0,∑_{i}t_{i} = 1, then let Â_{ε}
(xk,ω )
be a measurable selection from
Â
(xk,ω )
for each x_{k} a vertex of the simplex. However, we chose Â_{ε}