has a fixed point. This is
the main result and comes from the Brouwer fixed point theorem in ℝ^{n}. It is an
approximate fixed point.
Lemma 47.2.8Let f
(K (ω),ω)
be compact. For each r > 0, there exists x_{r}
(ω )
∈convex hull of f
(K (ω ),ω)
⊆ K
(ω)
such that
fr(xr(ω),ω) = xr (ω ), ∥fr(x,ω )− f (x,ω)∥ < r for all x ∈ K (ω)
and ω → x_{r}
(ω)
is measurable.
Proof:The upper limit in the sum of the above lemma n
(ω)
is a measurable
function. One can partition the measure space according to the value of n
(ω)
. This
gives a countable set of disjoint measurable subsets
{Ωn}
_{n=1}^{∞} in the partition
such that on the measurable set Ω_{n}, n
(ω)
= n. Specializing to the measurable
space consisting of Ω_{n}, we will assume here that n
(ω)
= n and show that there
exists a measurable fixed point x_{r}
(ω)
∈ K
(ω)
for ω ∈ Ω_{n}. Then the result
follows by letting x_{r}
(ω )
be that which has been obtained on Ω_{n}. Thus, from
now on, simply denote as n the upper limit and let ω ∈ Ω_{n}. If f_{r}
(xr,ω)
= x_{r}
and
∑n
xr = aiyi(ω)
i=1
for ∑_{i=1}^{n}a_{i} = 1 and the y_{i} described in the above lemma, we need
n
f (x,ω ) ≡ ∑ y (ω )ψ (f (x ,ω),ω)
r r i=1 i i r
n ( ( n ) ) n
= ∑ yj(ω )ψj f ∑ aiyi,ω ,ω = ∑ ajyj(ω) = xr.
j=1 i=1 j=1
Also, if this is satisfied, then we have the desired fixed point.
This will be satisfied if for each j = 1,
⋅⋅⋅
,n,
( ( ) )
∑n
aj = ψj f aiyi,ω ,ω (47.2.7)
i=1
(47.2.7)
so, let
{ ∑n }
Σn −1 ≡ a ∈ ℝn : ai = 1,ai ≥ 0
i=1
and let h
(⋅,ω)
: Σ_{n−1}→ Σ_{n−1} be given by
( ( ) )
∑n
h (a,ω)j ≡ ψj f aiyi,ω ,ω
i=1
Can we obtain a fixed point a
(ω)
such that ω → a
(ω )
is measurable? Since h
(⋅,ω )
is a
continuous function of a and ω → h
(x,ω)
is measurable, such a measurable fixed point
exists thanks to Theorem 47.2.3 or the much easier Theorem 47.2.6 above. Then
x_{r}
(ω)
= ∑_{i=1}^{n}a_{i}
(ω )
y_{i}
(ω )
so x_{r} is measurable. ■
The following is the Schauder fixed point theorem for measurable fixed points.
Theorem 47.2.9Let ω → K
(ω )
be a measurable multifunction which has convexand closed values in a separable Banach space. Let f
(⋅,ω)
: K
(ω)
→ K
(ω)
becontinuous and ω → f
(x,ω)
is measurable and f
(K (ω),ω)
is compact. Thenf
(⋅,ω)
has a fixed point x
(ω)
such that ω → x
(ω)
is measurable.
Proof: Recall that f
(xr (ω ),ω )
−f_{r}
(xr(ω),ω)
∈ B
(0,r)
and f_{r}
(xr (ω ),ω)
= x_{r}
(ω)
with x_{r}
(ω)
∈ convex hull of f
(K (ω),ω )
⊆ K
(ω)
. Here x_{r} is measurable. By Lemma
47.2.2 there is a measurable function x
(ω)
which equals the weak lim_{r(ω)
→0}x_{r(ω)
}
(ω)
.
However, since f
(K (ω),ω )
is compact, there is a subsequence still denoted with r
(ω)
such that f
( )
xr(ω),ω
converges strongly to some x ∈f
(K (ω ),ω )
. It follows that
f_{r(ω)
}
( )
xr(ω)(ω),ω
also converges to x strongly. But this equals x_{r(ω)
}
(ω)
which
shows that x_{r(ω)
}
(ω)
converges strongly to the measurable x
(ω)
. Therefore,
f (x(ω),ω) = lim f (xr(ω)(ω ),ω) = lim fr(ω)(xr(ω)(ω),ω )
r(ω)→0 r(ω)→0
= lim xr(ω)(ω) = x(ω). ■
r(ω)→0
As a special case of the above, here is a corollary which generalizes the earlier result
on the Brouwer fixed point theorem.
Corollary 47.2.10Let
(Ω,ℱ )
be a measurable space and let K
(ω)
be a convexand compact set in ℝ^{n}and ω → K
(ω)
is a measurable multifunction. Also letf
(⋅,ω)
: K
(ω)
→ K
(ω)
be continuous and for fixed x ∈ ℝ^{n},ω → f
(x,ω)
ismeasurable. Then there exists a fixed point x
(ω)
for f
(⋅,ω)
such that ω → x
(ω)
ismeasurable.
Note that in all of these considerations, there is no loss of generality in assuming
f
(⋅,ω)
is defined on the whole space X thanks to the theorem which says that a
continuous function defined on a convex closed set can be extended to a continuous
function defined on the whole space.
In the case of a single set, the following corollary is also obtained.
Corollary 47.2.11Let X be a Banach space and let K be a compact convex subset. Letf : K × Ω → K satisfy
x → f (x,ω) is continous
ω → f (x,ω) is measurable
Then f
(⋅,ω)
has a fixed point x
(ω)
such that ω → x
(ω)
is measurable.
Proof:The set K has a countable dense subset
{ki}
. You could consider Y as the
closure in X of the span of these k_{i}. Thus Y is a separable Banach space which contains
K. Now apply the above result. ■
If X is only a normed linear space, you could just consider its completion and apply
the above result. Since K is compact, it is automatically complete with respect to the
norm on X.
What of the Schaefer fixed point theorem? Is there a measurable version of it? A map
h : X → X for X a Banach space is a compact map if it is continuous and it
takes bounded sets to precompact sets. If you have such a compact map and
it maps a closed ball to a closed ball, then it must have a fixed point by the
Schauder theorem. If you have h = f
(⋅,ω)
where ω → f
(x,ω)
is measurable,
x → f
(x,ω )
compact, then if f
(⋅,ω)
maps a closed ball to a closed ball, it must have a
measurable fixed point x
(ω)
by the above. Now the following is a version of
the Schaefer fixed point theorem which can be used to get measurable fixed
points.
Theorem 47.2.12Let f
(⋅,ω)
: X → X be a compact map (takes bounded setsto precompact sets and continuous) where X is a Banach space. Also supposethat
sup{z ∈ f (B (0,r),ω)} ≤ C (r)
independent of ω. Here ω → f
(x,ω )
is measurable for each x ∈ X. Then either
There is a measurable fixed point x
(ω )
for tf
(⋅,ω)
for all t ∈
[0,1]
or
For every r > 0, there exists ω and t ∈
(0,1)
such that if x satisfiesx = tf
(x,ω )
, then
∥x∥
> r .
Proof:Suppose that alternative 2 does not hold and yet alternative 1 also fails to
hold. Since alternative 2 does not hold, there exists M_{0} such that for all ω, and for all
t ∈
(0,1)
, if x = tf
(x,ω)
, then
∥x(ω)∥
≤ M_{0}. If alternative 1 fails, then there is some t
with no measurable fixed point x
(ω)
for tf
(⋅,ω)
. So let M > M_{0}. By the measurable
Schauder fixed point theorem, Theorem 47.2.9, there is measurable x_{M}
(ω)
such
that
x (ω) = t(r f (x (ω),ω)), r y = y if ∥y∥ ≤ M, r y = M-y-if ∥y∥ > M
M M M M M ∥y∥