One of the nice properties of pseudomonotone maps is that when you add two of them,
you get another one. I will give a proof in the case that the two pseudomonotone maps
are both bounded. It is probably true in general, but as just noted, it is less trouble to
verify if you don’t have to worry about as many conditions. I will also assume the spaces
are all real so it will not be necessary to constantly write the real part. Actually, we do a
slightly more general version which says that a bounded pseudomonotone added to a
modified bounded pseudomonotone is a modified bounded pseudomonotone.
First is the theorem about the sum of two bounded set valued pseudomonotone
operators.
Theorem 23.5.1Say A,B are set valuedbounded pseudomonotone operators.Thentheir sum is also a set valued bounded pseudomonotone operator. Also, if u_{n}→ u weakly,z_{n}→ z weakly, z_{n}∈ A
(un)
, and w_{n}→ w weakly with w_{n}∈ B
(un)
, then if
limsup_{n→∞}
〈zn + wn,un − u 〉
≤ 0, it follows that
lim inf 〈zn + wn,un − v〉 ≥ 〈z(v)+ w(v),u − v〉, z(v) ∈ A (u),w(v) ∈ B (u),
n→∞
and z ∈ A
(u)
,w ∈ B
(u)
.
Proof:Say z_{n}∈ A
(un)
,w_{n}∈ B
(un)
,u_{n}→ u weakly and
lim ns→up∞ 〈zn + wn,un − u〉 ≤ 0.
Claim:Both of limsup_{n→∞}
〈z ,u − u〉
n n
,limsup_{n→∞}
〈w ,u − u〉
n n
are no larger
than 0.
Proof of the claim: Suppose limsup_{n→∞}
〈w ,u − u〉
n n
= δ > 0. Then take a
subsequence such that the limsup equals lim. The limsup only gets smaller when you go
to a subsequence. Thus, continuing to denote the subsequence with n we still have
limsup_{n→∞}
〈z + w ,u − u 〉
n n n
≤ 0. But from the fact that we just took a subsequence for
which the limsup = lim,
for all v which could be violated using
separation theorems if z is not in A
(u)
. Thus z ∈ A
(u)
. Similarly w ∈ B
(u)
.
■
The above is the main result but we can attempt to see what happens if one of the
operators is only modified pseudomonotone.
Note that if B is bounded pseudomonotone, then it is certainly modified bounded
pseudomonotone.
Theorem 23.5.2Suppose A,B : X → P
(X′)
are bothpseudomonotone andbounded. Then so is their sum.If A is bounded pseudomonotone and B is modifiedbounded pseudomonotone, then A + B is modified bounded pseudomonotone.
Proof: It is clear that Ax + Bx is closed and convex because this is true
of both of the sets in the sum. It is also bounded because both terms in the
sum are bounded. It only remains to verify the limit condition. Suppose then
that
un → u weakly
Will the limit condition hold for A + B when applied to this further subsequence?
Suppose z_{n}∈ Ax_{n},w_{n}∈ Bx_{n} and
lim sn→up∞ 〈zn + wn,un − u〉 ≤ 0 (23.5.26)
(23.5.26)
Is there a subsequence such that the liminf condition holds? From the above,
Then using that A is bounded pseudomonotone, lim_{n→∞}
〈zn,un − u〉
= 0 also. If follows
for any v,
lim inf 〈zn,un − v〉 ≥ 〈z(v),u − v〉
n→∞
Then from this it is routine to establish the modified pseudomonotone limit
condition for the sum A + B. For the subsequence just described, still denoted with
n,
lim sup 〈zn + wn,un − u〉 ≤ 0
n→ ∞
and ∗. In fact, you would have for any v,
lim inf 〈zn,un − v〉 ≥ 〈z(v),u− v〉,z(v) ∈ A (u )
n→∞
lim ni→nf∞〈wn,un − v〉 ≥ 〈w(v),u − v〉,w (v) ∈ A (u)
We continue to be in the situation of 23.5.26 and we are asking for a subsequence such
that the liminf condition will hold for the subsequence. Suppose this liminf condition is
not obtained for any subsequence. The desired liminf condition will hold for a
subsequence if either limsup_{n→∞}
〈z ,u − u〉
n n
or limsup_{n→∞}
〈w ,u − u〉
n n
is ≤ 0. This
was shown above. Therefore, if there is no subsequence yielding the liminf condition, you
must have both of these strictly positive. Say δ > 0 is smaller than both. Let n denote a
subsequence such that
then you could apply the above argument and obtain a further subsequence for which
the liminf condition would hold for the sum. Thus, we must have for this new
subsequence,
which is a contradiction. Thus the liminf condition must hold for some subsequence. In
case both are bounded and pseudomonotone, things are easier. You don’t have to take a
subsequence. ■
It is not entirely clear whether the sum of modified bounded pseudomonotone
operators is modified bounded pseudomonotone. This is because when you go to a
subsequence, the limsup gets smaller and so it is not entirely clear whether the
subsequence for A will continue to yield the limit condition if a further subsequence is
taken.
In fact, you can add a bounded pseudomonotone to a generalized bounded
pseudomonotone and get a generalized bounded pseudomonotone. The proof is just like
the above and is given next.
Theorem 23.5.3Suppose A,B : X →P
(X ′)
.If A is bounded pseudomonotoneand B is generalized bounded pseudomonotone, then A + B is generalized boundedpseudomonotone.
Proof: It is clear that Ax + Bx is closed and convex because this is true
of both of the sets in the sum. It is also bounded because both terms in the
sum are bounded. It only remains to verify the limit condition. Suppose then
that
u → u weakly
n
Will the limit condition hold for A + B when applied to this further subsequence?
Suppose z_{n}∈ Ax_{n},w_{n}∈ Bx_{n} and
lim sn→up∞ 〈zn + wn,un − u〉 ≤ 0 (23.5.28)
(23.5.28)
If v is given, is there a subsequence such that the liminf condition holds? From the
above,
and so, if the second term ≤ 0, since B is modified bounded pseudomonotone, there is a
subsequence, still denoted with n for which
lim inf 〈w ,u − v〉 ≥ 〈w (v),u− v〉, w (v) ∈ B (u) (*)
n→∞ n n
(*)
lim inf 〈wn,un − u〉 ≥ 〈w (u),u− u〉 = 0
n→∞
You just get a subsequence which works for v and note that the limsup condition is only
strengthened for the subsequence and then obtain a further subsequence which goes with
u to get the second condition along with the first. Note thatliminf gets bigger when you
go to a subsequence so if ∗ holds for the first subsequence, then it holds even better for
the second.
Hence you would have, for this subsequence depending on v