Here is an assortment of nonlinear operators which are useful in applications
to nonlinear partial differential equations. Generalizations of the notion of a
pseudomonotone map will be presented later to include the case of set valued
pseudomonotone maps. This is on the single valued version of some of these and these
ideas originate with Brezis in the 1960’s. A good description is given in Lions
[?].
Definition 23.1.1For V a real Banach space, A : V → V^{′}is a pseudomonotonemap ifwhenever
un ⇀ u (23.1.1)
(23.1.1)
and
lim ns→up∞〈Aun,un − u〉 ≤ 0 (23.1.2)
(23.1.2)
it follows that for all v ∈ V,
lim inf 〈Aun,un − v〉 ≥ 〈Au,u− v〉. (23.1.3)
n→∞
(23.1.3)
The half arrows denote weak convergence.
If V is finite dimensional, then pseudomonotone maps are continuous. Also the
property of being pseudomonotone is preserved when restriction is made to finite
dimensional spaces. The notation is explained in the following diagram.
∗
W ′ i← V ′
W i→ V
The map i is just the inclusion map. iw ≡ w and i^{∗} is the usual adjoint map.
∗
〈i f,w〉
_{W′,W}≡
〈f,iw〉
_{V ′,V } =
〈f,w 〉
_{V ′,V }. Thus i^{∗}Ai
(w)
∈ W^{′} and it is defined
by
〈i∗Ai(w),z〉W ′,W ≡ 〈Aw,z〉V′,V
in other words, you restrict A to W and only consider what the resulting functional does
to things in W.
Proposition 23.1.2Let V be finite dimensional and let A : V → V^{′}bepseudomonotone and bounded (meaning A maps bounded sets to bounded sets). ThenA is continuous. Also, if A : V → V^{′}is pseudomonotone and bounded, and ifW ⊆ V is a finite dimensional subspace, then i^{∗}Ai is pseudomonotone as a mapfrom W to W^{′}.
Proof:Say u_{n}→ u. Does it follow that Au_{n}→ Au? If not, then there is a
subsequence such that Au_{n}→ ξ≠Au thanks to
{Aun}
being bounded. Then the limsup
condition holds obviously. In fact the limit of 〈Au_{n},u_{n}−u〉 exists and equals 0. Hence for
all v,
lim inf 〈Aun,un − v〉 ≥ 〈Au,u − v〉
n→∞
Therefore,
〈ξ,u− v〉 ≥ 〈Au, u− v〉
for all v and so in fact ξ = Au after all. Thus A must be continuous.
As to the second part of this proposition, if you have w_{n}⇀ w in W, then in fact
convergence takes place strongly because weak and strong convergence are the same in
finite dimensions. Hence the same argument given above holds to show that i^{∗}Ai is
continuous.
Definition 23.1.3A : V → V^{′}is monotoneif for all v,u ∈ V,
〈Au − Av,u− v〉 ≥ 0,
and A is Hemicontinuousif forall v,u ∈ V,
lim 〈A(u +t(v − u)),u − v〉 = 〈Au,u − v〉.
t→0+
Theorem 23.1.4Let V be a Banach space and let A : V → V^{′}be monotone andhemicontinuous. Then A is pseudomonotone.
Proof: Let A be monotone and Hemicontinuous. First here is a claim.
Claim:If 23.1.1 and 23.1.2 hold, then lim_{n→∞}〈Au_{n},u_{n}− u〉 = 0.
Next divide by t and use the Hemicontinuity of A to conclude that
lim inf〈Aun,u − v〉 ≥ 〈Au,u − v〉.
n→ ∞
From the claim,
liminf _{n→∞}〈Au_{n},u − v〉
= liminf _{n→∞}
(〈Aun, un − v〉+ 〈Aun, u− un〉)
= liminf _{n→∞}〈Au_{n},u_{n}− v〉≥〈Au,u − v〉.■
Monotonicity is very important in the above proof. The next example shows that even
if the operator is linear and bounded, it is not necessarily pseudomonotone.
Example 23.1.5Let H be any Hilbert space (complete inner product space, more onthese later) and let A : H → H^{′}be given by
〈Ax,y〉 ≡ (− x,y)H .
Then A fails to be pseudomonotone.
Proof:Let
{xn}
_{n=1}^{∞} be an orthonormal set of vectors in H. Then Parsevall’s
inequality implies
If A were pseudomonotone, we would need to be able to conclude that for all
y ∈ H,
lim inf 〈Axn,xn − y〉 ≥ 〈Ax,x − y〉 = 0.
n→ ∞
However,
lim inf 〈Axn,xn − 0〉 = − 1 < 0 = 〈A0,0− 0〉.
n→∞
The following proposition is useful.
Proposition 23.1.6Suppose A : V → V^{′}is pseudomonotone and bounded whereV is separable. Then it must be demicontinuous.This means that if u_{n}→ u, thenAu_{n}⇀ Au. In case that V is reflexive, you don’t need the assumption that V isseparable.
Proof:Since u_{n}→ u is strong convergence and since Au_{n} is bounded, it
follows
Suppose this is not so that Au_{n} converges weakly to Au. Since A is bounded, there exists
a subsequence, still denoted by n such that Au_{n}⇀ ξ weak ∗. I need to verify ξ = Au.
From the above, it follows that for all v ∈ V
An interesting result is the following which states that a monotone linear function
added to a type M is also type M.
Proposition 23.1.9Suppose A : V → V^{′}is type M and suppose L : V → V^{′}is monotone, bounded and linear.Then L + A is type M. Let V be separable orreflexive so that the weak convergences in the following argument are valid.
Proof:Suppose u_{n}⇀ u and Au_{n} + Lu_{n}⇀ ξ and also that
lim sup 〈Aun + Lun,un〉 ≤ 〈ξ,u〉
n→ ∞
Does it follow that ξ = Au + Lu? Suppose not. There exists a further subsequence, still
called n such that Lu_{n}⇀ Lu. This follows because L is linear and bounded. Then from
monotonicity,
〈Lun,un〉 ≥ 〈Lun,u 〉+ 〈L (u),un − u〉
Hence with this further subsequence, the limsup is no larger and so
It follows since A is type M that Au = ξ − Lu, which contradicts the assumption that
ξ≠Au + Lu. ■
There is also the following useful generalization of the above proposition.
Corollary 23.1.10Suppose A : V → V^{′}is type M and suppose L : W → W^{′}ismonotone, bounded and linear where V ⊆ W and V is dense in W so that W^{′}⊆ V^{′}.Then for u_{0}∈ W define M
(u)
≡ L
(u − u0)
. Then M + A is type M. Let V beseparable or reflexive so that the weak convergences in the following argument arevalid.
Proof: Suppose u_{n}⇀ u and Au_{n} + Mu_{n}⇀ ξ and also that
lim sup 〈Aun + M un,un〉 ≤ 〈ξ,u〉
n→∞
Does it follow that ξ = Au + Mu? Suppose not. By assumption, u_{n}⇀ u and
so,
un − u0 ⇀ u − u0 weak convergence in W
since L is bounded, there is a further subsequence, still called n such that
M un = L (un − u0) ⇀ L (u− u0) = M u.
Since M is monotone,
〈M un − M u,un − u〉 ≥ 0
Thus
〈M u ,u 〉− 〈M u ,u 〉− 〈M u,u 〉+ 〈M u,u〉 ≥ 0
n n n n
and so
〈M u ,u 〉 ≥ 〈M u ,u〉+ 〈M u,u − u〉
n n n n
Hence with this further subsequence, the limsup is no larger and so
It follows since A is type M that Au = ξ − Mu, which contradicts the assumption that
ξ≠Au + Mu. ■
The following is Browder’s lemma. It is a very interesting application of the Brouwer
fixed point theorem.
Lemma 23.1.11(Browder) Let K be a convexclosed and bounded set in ℝ^{n}and letA : K → ℝ^{n}be continuous and f ∈ ℝ^{n}. Then there exists x ∈ K such that for ally ∈ K,
(f − Ax,y − x)ℝn ≤ 0
If K is convex, closed, bounded subset of V a finite dimensional vector space,then the same conclusion holds. If f ∈ V^{′}, there exists x ∈ K such that for ally ∈ K,
〈f − Ax,y− x〉V′,V ≤ 0
Proof: Let P_{K} denote the projection onto K. Thus P_{K} is Lipschitz continuous.
x → PK (f − Ax + x)
is a continuous map from K to K. By the Brouwer fixed point theorem, it has a fixed
point x ∈ K. Therefore, for all y ∈ K,
(f − Ax +x − x,y − x ) = (f − Ax,y − x) ≤ 0
As to the second claim. Consider the following diagram.
n θ∗ ′
ℝ ←θ V
ℝn → V
where
∑n
θ(x) = xivi
i=1
Thus θ and θ^{∗} are both continuous linear and one to one and onto. Hence there is
x ∈ θ^{−1}K a closed convex and bounded subset of ℝ^{n} such that x = θ^{−1}u,u ∈ K,
and
From this lemma, there is an interesting theorem on surjectivity.
Proposition 23.1.12Let A : V → V^{′}be continuous and coercive,
〈A(v +v0),v〉
∥lvim∥→ ∞ ----∥v∥----- = ∞
V
for some v_{0}. Then for all f ∈ V^{′}, there exists v ∈ V such that Av = f.
Proof:Define the closed convex sets B_{n}≡B
(v,n)
0
. By Browder’s lemma, there
exists x_{n} such that
(f − Avn, y− vn) ≤ 0
for all y ∈ B_{n}. Then taking y = v_{0},
〈Avn,vn − v0〉 ≤ 〈f,vn − v0〉
letting w_{n} = v_{n}− v_{0},
〈A (wn +v0),wn 〉 ≤ 〈f,wn 〉
and so
〈A(wn-+-v0),wn-〉≤ ∥f∥
∥wn∥
which implies that the
∥wn∥
and hence the
∥vn∥
are bounded. It follows that for large n,v_{n} is an interior point of B_{n}. Therefore,
〈f − Avn,z〉V′,V ≤ 0
for all z in some open ball centered at v_{0}. Hence f − Av_{n} = 0. ■
Lemma 23.1.13Let A : V → V^{′}be type M and bounded and suppose V isreflexive or V is separable. Then A is demicontinuous.
Proof:Suppose u_{n}→ u and Au_{n} fails to converge weakly to Au. Then there is a
further subsequence, still denoted as u_{n} such that Au_{n}⇀ ζ≠Au. Then thanks to the
strong convergence, you have
lim snu→p∞ 〈Aun,un〉 = 〈ζ,u〉
which implies ζ = Au after all. ■
With these lemmas and the above proposition, there is a very interesting surjectivity
result.
Theorem 23.1.14Let A : V → V^{′}be type M, bounded, and coercive
lim 〈A-(u+-u0),u〉= ∞, (23.1.4)
∥u∥→∞ ∥u∥
(23.1.4)
for some u_{0}, where V is a separable reflexive Banach space. Then A is surjective.
Proof:Since V is separable, there exists an increasing sequence of finite
dimensional subspaces
{Vn}
such that ∪_{n}V_{n} = V and each V_{n} contains u_{0}. Say
span
(v1,⋅⋅⋅,vn)
= V_{n}. Then consider the following diagram.
Vn′ ←i∗ V ′
V →i V
n
The map i is the inclusion map. Consider the map i^{∗}Ai. By Lemma 23.1.13 this map is
continuous.