In the abstract theory of partial differential equations and variational inequalities, it is
important to consider set-valued maps from a Banach space to the power set of its dual.
In this section we give an introduction to this theory by proving a general result on
surjectivity for a class of such operators.
To begin with, if A : X →P
(Y )
is a set-valued map, define the graph of A
by
G (A) ≡ {(x,y) : y ∈ Ax} .
First consider a map A which maps ℂ^{n} to P
n
(ℂ )
which satisfies
Ax is compact and convex. (23.4.12)
(23.4.12)
and also the condition that if O is open and O ⊇ Ax, then there exists δ > 0 such that
if
y ∈ B(x,δ), then Ay ⊆O. (23.4.13)
(23.4.13)
This last condition is sometimes referred to as upper semicontinuity. In words,
A is upper semicontinuous and has values which are compact and convex. As
to the last condition of upper semi continuity, here is the formal definition.
Definition 23.4.1Let F : X → P
(Y)
be a set valued function. Then F isupper semicontinuous at x if for every open V ⊇ F
(x)
there exists an open set Ucontaining x such that whenever
ˆx
∈ U, it follows that F
(xˆ)
⊆ V .
Lemma 23.4.2Let A satisfy 23.4.13. Then AK is a subset of a compact setwhenever K is compact. Also the graph of A is closed if Ax is closed.
Proof:Let x ∈ K. Then Ax is compact and contained in some open set whose
closure is compact, U_{x}. By assumption 23.4.13 there exists an open set V_{x} containing x
such that if y ∈V_{x}, then Ay ⊆U_{x}. Let V_{x1},
⋅⋅⋅
,V_{xm} cover K. Then AK ⊆∪_{k=1}^{m}U_{xk}, a
compact set.
To see the graph of A is closed when Ax is closed, let x_{k}→x,y_{k}→ y where
y_{k}∈ Ax_{k}. Then letting O = Ax+B
(0,r)
it follows from 23.4.13 that y_{k}∈ Ax_{k}⊆ O for
all k large enough. Therefore, y ∈ Ax+B
(0,2r)
and since r > 0 is arbitrary and Ax is
closed it follows y ∈ Ax.■
Also, there is a general consideration relative to upper semicontinuous functions.
Lemma 23.4.3If f is upper semicontinuous on some set K and g is continuousand defined on f
(K )
, theng ∘ fis also upper semicontinuous.
Proof:Let x_{n}→ x in K. Let U ⊇g ∘ f
(x)
. Is g ∘ f
(xn)
∈ U for all n large enough?
We have f
(x)
∈ g^{−1}
(U )
, an open set. Therefore, if n is large enough, f
(xn)
∈ g^{−1}
(U )
. It
follows that for large enough n,g ∘ f
(xn)
∈ U and so g ∘ f is upper semicontinuous on K.
■
The next theorem is an application of the Brouwer fixed point theorem. First define
an n simplex, denoted by
_{i=1}^{n} is independent, the t_{i} are uniquely determined. If two of them
are
∑n ∑n
tixi = sixi
i=0 i=0
Then
∑n ∑n
ti(xi − x0) = si(xi − x0)
i=0 i=0
so t_{i} = s_{i} for i ≥ 1. Since the s_{i} and t_{i} sum to 1, it follows that also s_{0} = t_{0}. If n ≤ 2, the
simplex is a triangle, line segment, or point. If n ≤ 3, it is a tetrahedron, triangle, line
segment or point. To say that
{xi − x0}
_{i=1}^{n} are independent is to say that
{xi − xr}
_{i≠r}
are independent for each fixed r. Indeed, if x_{i}−x_{r} = ∑_{j≠i,r}c_{j}
(xj − xr)
, then you would
have
( )
x − x + x − x = ∑ c(x − x )+ ( ∑ c ) x
i 0 0 r j⁄=i,r j j 0 j⁄=i,r j 0
and it follows that x_{i}− x_{0} is a linear combination of the x_{j}− x_{0} for j≠i, contrary to
assumption. A collection of simplices is a tiling of ℝ^{n} if ℝ^{n} is contained in their union
and if S_{1},S_{2} are two simplices in the tiling, with
or else the two simplices do not intersect. The collection of simplices is said to be locally
finite if, for every point, there exists a ball containing that point which also intersects
only finitely many of the simplices in the collection. It is left to the reader to verify that
for each ε > 0, there exists a locally finite tiling of ℝ^{n} which is composed of simplices
which have diameters less than ε. The local finiteness ensures that for each ε
the vertices have no limit point. To see how to do this, consider the case of
ℝ^{2}. Tile the plane with identical small squares and then form the triangles
indicated in the following picture. It is clear something similar can be done in any
dimension. Making the squares identical ensures that the little triangles are locally
finite.
PICT
In general, you could consider
[0,1]
^{n}. The point at the center is
(1∕2,⋅⋅⋅,1∕2)
. Then
there are 2n faces. Form the 2n pyramids having this point along with the 2^{n−1} vertices
of the face. Then use induction on each of these faces to form smaller dimensional
simplices tiling that face. Corresponding to each of these 2n pyramids, it is the union of
the simplices whose vertices consist of the center point along with those of these new
simplicies tiling the chosen face. In general, you can write any n dimensional cube as the
translate of a scaled
[0,1]
^{n}. Thus one can express each of identical cubes as a tiling of
m
(n)
simplices of the appropriate size and thereby obtain a tiling of ℝ^{n} with simplices.
A ball will intersect only finitely many of the cubes and hence finitely many of the
simplices. To get their diameters small as desired, just use
[0,r]
^{n} instead of
[0,1]
^{n}.
Thus one can give a function any value desired on these vertices and extend
appropriately to the rest of the simplex and obtain a continuous function.
The Kakutani fixed point theorem is a generalization of the Brouwer fixed point
theorem from continuous single valued maps to upper semicontinuous maps which have
closed convex values.
Theorem 23.4.4Let K be a compact convex subset of ℝ^{n}and let A : K →P
(K)
such that Ax is a closed convex subset of K and A is upper semicontinuous.Thenthere exists x such that x ∈ Ax. This is the “fixed point”.
Proof: Let there be a locally finite tiling of ℝ^{n} consisting of simplices having
diameter no more than ε. Let Px be the point in K which is closest to x. For each
vertex x_{k}, pick A_{ε}x_{k}∈ APx_{k} and define A_{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}∈
[0,1]
,∑_{i}t_{i} = 1,then
∑n
Aεx ≡ tkAεxk.
k=0
Now by construction A_{ε}x_{k}∈ APx_{k}∈ K and so A_{ε} is a continuous map defined on ℝ^{n}
with values in K thanks to the local finiteness of the collection of simplices. By the
Brouwer fixed point theorem A_{ε} has a fixed point x_{ε} in K, A_{ε}x_{ε} = x_{ε}.
∑n ε ε ε ε
xε = tkAεxk, Aεxk ∈ APx k ∈ K
k=0
where a simplex containing x_{ε} is
ε ε n∑ ε ε
[x0,⋅⋅⋅,xn], xε = tkxk
k=0
Also, x_{ε}∈ K and is closer than ε to each x_{k}^{ε} so each x_{k}^{ε} is within ε of K. It follows that
for each k,
ε ε
|Pxk − x k|
< ε and so
ε ε
lεim→0 |P xk − xk| = 0
By compactness of K, there exists a subsequence, still denoted with the subscript of ε
such that for each k, the following convergences hold as ε → 0
tε → t , A x ε→ y , P xε→ z, xε → z
k k ε k k k k k k
Any pair of the x_{k}^{ε} are within ε of each other. Hence, any pair of the Px_{k}^{ε} are within ε
of each other because P reduces distances. Therefore, in fact, z_{k} does not depend on
k.
By upper semicontinuity of A, for all ε small enough,
AP xε⊆ Az + B (0,r)
k
In particular, since A_{ε}x_{k}^{ε}∈ APx_{k}^{ε},
A εx εk ∈ Az + B (0,r) for ε small enough
Since r is arbitrary and Az is closed, it follows
y ∈ Az.
k
It follows that since K is closed,
∑n ∑n
xε → z = tkyk, tk ≥ 0, tk = 1
k=0 k=0
Now by convexity of Az and the fact just shown that y_{k}∈ Az,
∑n
z = tkyk ∈ Az
k=0
and so z ∈ Az. This is the fixed point. ■
One can replace ℝ^{n} with ℂ^{n} in the above theorem because it is essentially ℝ^{2n}. Also
the theorem holds with no change for any finite dimensional normed linear space since
these are homeomorpic to ℝ^{n} or ℂ^{n}.
Lemma 23.4.5Suppose A : ℂ^{n}→P
(ℂn)
satisfies Ax is compact and convex, and A isupper semicontinuous, 23.4.13and K is a nonempty compact convex set in ℂ^{n}. Then ify ∈ ℂ^{n}there exists
[x,w]
∈ G
(A)
such thatx ∈Kand
Re (y− w, z− x) ≤ 0
for allz ∈K.
Proof:Tile ℂ^{n} with 2n simplices such that the collection is locally finite and each
simplex has diameter less than ε < 1. This collection of simplices is determined by a
countable collection of vertices. For each vertex x, pick A_{ε}x ∈ Ax and define A_{ε} on all of
ℂ^{n} by the following rule. If
x ∈ [x0,⋅⋅⋅,x2n],
so x =∑_{i=0}^{2n}t_{i}x_{i},then
2n
Aεx ≡ ∑ tkAεxk.
k=0
Thus A_{ε} is a continuous map defined on ℂ^{n} thanks to the local finiteness of
the collection of simplices. Let P_{K} denote the projection on the convex set K.
By the Brouwer fixed point theorem, there exists a fixed point, x_{ε}∈ K such
that
so x_{ε} = ∑_{k=0}^{2n}t_{k}^{ε}x_{k}^{ε}. Then since x_{ε} is contained in K, a
compact set, and the diameter of each simplex is less than 1, it follows that A_{ε}x_{k}^{ε} is
contained in A(K + B
(0,1)
), which is contained in a compact set thanks to Lemma
23.4.2. The reason is that A is assumed to take bounded sets to bounded sets and
K + B
(0,1)
is a bounded set.
From the Heine Borel theorem, there exists a sequence ε → 0 such that
tεk → tk,xε → x ∈ K,A εxεk → yk
for k = 0,
⋅⋅⋅
,2n. Since the diameter of the simplex containing x_{ε} converges to 0, it
follows
xεk → x,Aεxεk → yk.
By upper semicontinuity, it follows that for all r > 0, Ax_{k}^{ε}⊆ Ax + B
(0,r)
for all ε small
enough. Since A_{ε}x_{k}^{ε}∈ Ax_{k}^{ε},and Ax is closed, this implies y_{k}∈ Ax. Since Ax is
convex,
∑2n
tkyk ∈ Ax.
k=1
Hence for all z ∈K,
( ) ( )
∑2n 2∑n ε ε
Re y − tkyk,z− x = εli→m0 Re y− tkAεxk,z− xε
k=1 k=1
= lim Re (y− A εxε,z − xε) ≤ 0.
ε→0
Let w =∑_{k=1}^{2n}t_{k}y_{k}. ■
You could replace A with A∘P_{K} in the above and assume only that A is only defined
on K. This is because x ∈ K.
Lemma 23.4.6Suppose in addition to 23.4.12and 23.4.13, (compact convex valuedand upper semicontinuous) A is coercive,
{ }
lim inf Re-(y,x): y ∈ Ax = ∞.
|x|→∞ |x|
Then A is onto.
Proof:Let y ∈ ℂ^{n} and let K_{r}≡B
(0,r)
. By Lemma 23.4.5 there exists x_{r}∈ K_{r} and
w_{r}∈ Ax_{r} such that
Re (y− wr, z− xr) ≤ 0 (23.4.14)
(23.4.14)
for all z ∈ K_{r}. Letting z = 0,
Re(w ,x ) ≤ Re(y,x ).
r r r
Therefore,
{ }
inf Re-(w,-xr): w ∈ Axr ≤ |y|.
|xr|
It follows from the assumption of coercivity that
|xr|
is bounded independent of r.
Therefore, picking r strictly larger than this bound, 23.4.14 implies
Re (y− wr, v) ≤ 0
for all v in some open ball containing 0. Therefore, for all v in this ball
Re (y− wr, v) = 0
and hence this holds for all v ∈ ℂ^{n} and so y = w_{r}∈ Ax_{r}. This proves the
lemma.
Lemma 23.4.7Let F be a finite dimensional Banach space of dimension n, and let Tbe a mapping from F to P
′
(F )
such that 23.4.12and 23.4.13both hold for F^{′}in place ofℂ^{n}. Then if T is also coercive,
{ Re-y∗(u) ∗ }
||lu|im|→ ∞ inf ||u|| : y ∈ Tu = ∞, (23.4.15)
(23.4.15)
it follows T is onto.
Proof: Let
|⋅|
be an equivalent norm for F such that there is an isometry of ℂ^{n} and
F,θ. Now define A : ℂ^{n}→P
n
(ℂ )
by Ax ≡ θ^{∗}Tθx.
′ θ∗ n
P (F ) → ℂ
T ↑ ∘θ ↑ A
F ← ℂn
Thus y ∈ Ax means that there exists z^{∗}∈ Tθx such that
(w,y) n = z∗(θw)
ℂ
for all w ∈ ℂ^{n}. Then A satisfies the conditions of Lemma 23.4.6 and so A is onto.
Consequently T is also onto. ■
With these lemmas, it is possible to prove a very useful result about a class of
mappings which map a reflexive Banach space to the power set of its dual space. For
more theorems about these mappings and their applications, see [?]. In the discussion
below, we will use the symbol, ⇀, to denote weak convergence.
Definition 23.4.8Let V be a Reflexive Banach space. We say T : V →P
(V ′)
ispseudomonotoneif the following conditions hold.
Tu is closed, nonempty, convex. (23.4.16)
(23.4.16)
If F is a finite dimensional subspace of V , then if u ∈ F and W ⊇ Tu for W a weaklyopen set in V^{′}, then there exists δ > 0 such that
In the case that T takes bounded sets to bounded sets so it is a bounded set valued
operator, it turns out you don’t have to consider the second of the above conditions
about the upper semicontinuity. It follows from the other conditions. It is convenient to
use the notation
∗ ∗ ∗ ′
〈u ,v〉 ≡ u (v),u ∈ V ,v ∈ V.
and this will be used interchangeably with the earlier notation from now on.
The next lemma has to do with upper semicontinuity being obtained from simpler
conditions.
Lemma 23.4.9Let T : X →P
′
(X )
satisfy conditions 23.4.16and 23.4.18above andsuppose T is bounded (Tx for x in a bounded set is bounded). Then if x_{n}→ x in X, andif U is a weakly open set containing Tx, then Tx_{n}⊆ U for all n large enough. If fact thelimit condition 23.4.18can be weakened to the following more general condition: Ifu_{k}⇀ u, and
lim sup Reu∗ (u − u) ≤ 0, (**)
k→ ∞ k k
(**)
then there exists a subsequencestill denoted as
{uk}
, such that if u_{k}^{∗}∈ Tu_{k}, then forall v ∈ V , there exists u^{∗}
(v)
∈ Tu such that
lim inf Re u∗(uk − v) ≥ Reu ∗(v)(u − v). (23.4.20)
k→ ∞ k
(23.4.20)
(This weaker condition says that if the lim sup condition holds for the originalsequence, then there is a subsequence such that the lim inf condition holds forall v. In particular, for this subsequence, the lim sup condition continues tohold.)
Proof:If this is not true, there exists x_{n}→ x, also a weakly open set U, containing
Tx and z_{n}∈ Tx_{n}, but z_{n}
∕∈
U. Then, taking a further subsequence, we can assume
z_{n}→ z weakly and z
∈∕
U. Then the strong convergence implies
lim sup Re 〈zn,xn − x 〉 ≤ 0
n→ ∞
By assumption, there is a subsequence still denoted with n such that for any
y,
lim inf Re 〈zn,xn − y〉 ≥ Re 〈z(y),x− y〉, some z (y) ∈ T (x)
k→∞