# Existence Results for Constrained Quasivariational Inequalities

Authors: V. V. Motreanu
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

We deal with a constrained quasivariational inequality under a general form. We study existence of solutions in two situations depending on whether the set of constraints is bounded or possibly unbounded.


---

## Body

## 1. Introduction and Statement of Main Results

LetX be a real reflexive and separable Banach space assumed to be compactly embedded in a Banach space Y. We denote by X* the dual space of X, by Y* the dual space of Y, by 〈·,·〉X the duality brackets between X* and X, by 〈·,·〉Y the duality brackets between Y* and Y, by ∥·∥X the norm of X, and by ∥·∥Y the norm of Y. Given a function ψ:X→ℝ∪{+∞}, we denote by D(ψ):={x∈X:ψ(x)<+∞} the effective domain of ψ.In this paper we deal with the following problem(1)Findu∈Ksuchthat(u,u)∈D(Φ),〈Au,v-u〉X+Φ(u,v)-Φ(u,u)+J0(u;v-u)≥〈f,v-u〉X,∀v∈K.We describe the data entering problem (1): (i)
K⊂X is a nonempty, convex, closed subset;(ii)
A:X→X* is a (possibly nonlinear) operator;(iii)
Φ:X×X→ℝ∪{+∞} is such that, for all η∈K, the function Φ(η,·):X→ℝ∪{+∞} is convex with K∩D(Φ(η,·))≠∅; moreover, we will denote by ∂Φ(η,·) the convex subdifferential of Φ(η,·); that is,
(2)∂Φ(η,u)={w∈X*:Φ(η,v)-Φ(η,u)mmmmmmm,,mmm≥〈w,v-u〉X,∀v∈X};(iv)
J:Y→ℝ is a locally Lipschitz function, and the notation J0 stands for its generalized directional derivative in the sense of Clarke [1]; that is,
(3)J0(u;v)=limsupw→uλ→0+J(w+λv)-J(w)λ,∀u,v∈Y.In addition, we will denote by∂J the generalized gradient of J; that is,
(4)∂J(u)={w∈Y*:J0(u;v)≥〈w,v〉Y,∀v∈Y},∀u∈Y;(v)
f∈X*.Problem (1) is called a constrained quasivariational problem. Typically, we can choose X to be the Sobolev space (H01(Ω),∥∇·∥L2(Ω)) defined as the closure of Cc∞(Ω) in H1(Ω) for a bounded domain Ω⊂ℝN (N≥1), Y to be the Lebesgue space Lp(Ω) for 1≤p<2* (where 2*=+∞ if N∈{1,2} and 2*=2N/(N-2) if N≥3), K={u∈H01(Ω):u≥0a.e.inΩ}, A=-Δ (the negative Laplacian operator), Φ(u,v)=∫Ω‍g(u,v)dx where g:ℝ2→ℝ+ is convex in the second variable (then D(Φ)={(u,v)∈H01(Ω)×H01(Ω):g(u,v)∈L1(Ω)}), and J(u)=∫Ω‍j(x,u(x))dx where j:Ω×ℝ→ℝ is locally Lipschitz in the second variable. Constrained quasivariational problems were extensively studied; we refer, for example, to [2–5] and to the references therein. We point out three aspects which make our approach natural and general. First, we deal with the general setting of a pair of Banach spaces (X,Y) instead of focusing on spaces of functions; in particular, our results can be applied to problems with different boundary conditions. Second, the set of constraints K may be unbounded. Third, the form of the studied problem allows both variational and hemivariational constraints as it involves both a convex term Φ(u,·) and a generalized directional derivative J0; this type of problems models important processes in mechanics and engineering (see [6, 7]).In this paper, we consider the following hypotheses on the data described above:(H1)
for every sequence{un}n≥1⊂K with un⇀u in X, for some u∈K, one has
(5)〈Au,u-v〉X≤limsupn→∞〈Aun,un-v〉X,∀v∈K;(H2)
whenever{(ηn,un)}n≥1⊂(K×K)∩D(Φ), ηn⇀η in X, un⇀u in X, one has (η,u)∈(K×K)∩D(Φ) and
(6)limsupn→∞(Φ(ηn,v)-Φ(ηn,un))≤Φ(η,v)-Φ(η,u),∀v∈K;(H3)
givenη∈K, if u1,u2∈K satisfy (η,u1)∈D(Φ), (η,u2)∈D(Φ) and
(7)J0(η;u2-u1)+J0(η;u1-u2)≥〈Au2-Au1,u2-u1〉X,thenu1=u2.Remark 1.
We emphasize certain situations when hypotheses(H1)–(H3) are satisfied.
(a) Hypothesis(H1) is satisfied, for instance, if A is weakly strongly continuous, that is, A is continuous from X endowed with the weak topology to X* endowed with the norm topology.
(b) Note that(H1) is satisfied, for instance, for X=H01(Ω), any closed, convex subset K⊂X, and A:H01(Ω)→H01(Ω)* defined by A=-Δ, where Δ:H01(Ω)→H01(Ω)* is the Laplacian operator, with Ω⊂ℝN (N≥1) a bounded domain. Indeed, let a sequence {un}n≥1⊂K with un⇀u in H01(Ω), for some u∈K. Using the weak lower semicontinuity of the norm, we can write
(8)limsupn→∞〈-Δun,un-v〉=limsupn→∞(∥un∥H01(Ω)2-(un,v)H01(Ω))limsupn→∞〈-Δun,un-v〉≥liminfn→∞∥un∥H01(Ω)2-(u,v)H01(Ω)limsupn→∞〈-Δun,un-v〉≥∥u∥H01(Ω)2-(u,v)H01(Ω)limsupn→∞〈-Δun,un-v〉=〈-Δu,u-v〉
for all v∈H01(Ω). Here, 〈·,·〉 are the duality brackets for the pair (H01(Ω)*,H01(Ω)) and (u,v)H01(Ω)=∫Ω‍∇u·∇vdx denotes the scalar product on H01(Ω). Whence (H1) holds in this case.
(c) Hypothesis(H2) is fulfilled in the case where Φ is sequentially weakly lower semicontinuous, D(Φ) is weakly closed, and Φ(·,u) is weakly strongly continuous on its effective domain for all u∈X.
(d) IfA is strongly monotone, that is, there exists a constant m>0 such that
(9)〈Au2-Au1,u2-u1〉X≥m∥u1-u2∥X2,∀u1,u2∈K,
and ∂J is bounded on K in the sense that
(10)∥ζ∥Y*≤c∥u∥Y,∀ζ∈∂J(u),∀u∈K,
with a positive constant c<m/(2c-), where c->0 is the best constant satisfying ∥u∥Y≤c-∥u∥X, for all u∈X (which exists by the continuity of the embedding of X in Y), then condition (H3) is satisfied.
(e) IfA is strictly monotone and J is Gâteaux differentiable and regular (see [1, Definition 2.3.4]), then condition (H3) is satisfied. In particular, if A is strictly monotone and J is continuously differentiable, then (H3) is satisfied.In this paper, we distinguish two cases depending on whether the setK is bounded or not necessarily bounded. The following result concerns the former situation.Theorem 2.
Assume that conditions(H1)–(H3) are satisfied and that the closed, convex set K is bounded in X. Then problem (1) has at least one solution.Remark 3.
Note that the existence of a solution of problem (1), which is the conclusion of Theorem 2, forces the intersection diag(K)∩D(Φ) to be nonempty, where the notation diag(K) stands for the diagonal of the set K; that is, diag(K)={(v,v):v∈K}. The nonemptiness of this intersection is not directly implied by the hypotheses (H1)–(H3), nor by the assumption made that K∩D(Φ(η,·))≠∅ for all η∈K. However, Theorem 4 below incorporates hypothesis (H4) which assumes in particular that diag(K)∩D(Φ)≠∅.Now, we deal with the case whereK is not assumed to be bounded. In this case, we additionally suppose the following: (H4)
there exist an elementv0∈K with (η,v0)∈D(Φ) for all η∈K and a real p≥1 such that
(11)limsup∥w∥X→∞〈Aw,w-v0〉X∥w∥Xp=+∞;(H5)
there exists a constantc0>0 such that we have
(12)〈z,v0-u〉X≤c0(1+∥u∥Xp),∀z∈∂Φ(u,·)(v0),∥z∥Y*≤c0(1+∥u∥Yp-1),∀z∈∂J(u),for allu∈K with (u,u)∈D(Φ), where v0 and p≥1 are as in (H4).We state now our main result for problem (1) dealing with the case where the set K is possibly unbounded.Theorem 4.
Assume that conditions(H1)–(H5) are satisfied. Then problem (1) has at least a solution.The rest of the paper is organized as follows. In Section2, we present the proof of Theorem 2, where we apply a version of the Schauder fixed point theorem. In Section 3, we give the proof of Theorem 4, which is actually based on Theorem 2.


## 2. Proof of Theorem2

For eachη∈K, we consider the auxiliary problem
(13)Findu∈Ksuchthat(η,u)∈D(Φ),〈Au,v-u〉X+Φ(η,v)-Φ(η,u)+J0(η;v-u)≥〈f,v-u〉X,∀v∈K.
Our first purpose, accomplished in Lemma 6 below, is to show that problem (13) has a unique solution. To do this, we need Fan’s lemma (see [8, page 208]) which we recall in the following statement.Theorem 5.
LetW be a Hausdorff topological vector space, let Z be a nonempty subset of W, and let F:Z→2W be such that (i)
F(x) is a nonempty, closed subset of W, for all x∈Z;(ii)
conv{x1,…,xn}⊂⋃i=1nF(xi) for all {x1,…,xn}⊂Z;(iii)
there isx-∈Z for which F(x-) is compact.Then ⋂x∈ZF(x)≠∅.Lemma 6.
Assume that hypotheses(H1)–(H3) are fulfilled and that the closed, convex set K is bounded in X. Then, for every η∈K, problem (13) has a unique solution.Proof.
Fixη∈K. Consider the set-valued mapping G:K∩D(Φ(η,·))→2X defined by
(14)G(v)={u∈K∩D(Φ(η,·)):〈Au-f,u-v〉Xmmmmmmmmmmmmmm,-J0(η;v-u)mmmmmmmmmmmm,mm+Φ(η,u)-Φ(η,v)≤0}
for all v∈K∩D(Φ(η,·)). We show that the assumptions of Theorem 5 are satisfied for W=X endowed with the weak topology, Z=K∩D(Φ(η,·)), and F=G.
For everyv∈K∩D(Φ(η,·)), we clearly have v∈G(v); hence G(v) is nonempty.
We check thatG(v) is weakly compact for every v∈K∩D(Φ(η,·)). To this end, we first prove that G(v) is sequentially weakly closed in X. Let a sequence {un}n≥1⊂G(v) with un⇀u in X, for some u∈X. Taking into account that X is compactly embedded in Y it follows that un→u in Y. Using the first part of assumption (H2), we have that u∈K∩D(Φ(η,·)). As un∈G(v), we know that
(15)〈Aun,un-v〉X≤〈f,un-v〉X+J0(η;v-un)+Φ(η,v)-Φ(η,un).
Passing to the limsup as n→∞, we find
(16)limsupn→∞〈Aun,un-v〉X≤〈f,u-v〉X+J0(η;v-u)+Φ(η,v)-Φ(η,u).
Here we made use of the weak convergence un⇀u in X, the continuity of J0(η;·) on Y, and the second part of (H2). Combining with (H1), we obtain that u∈G(v), thereby G(v) is sequentially weakly closed in X.
Using thatX is reflexive and separable and K is bounded, convex, and closed, we deduce that K is metrizable and weakly compact (see, e.g., [9, pages 44–50]). Since G(v)⊂K and using that G(v) is sequentially weakly closed, we derive that G(v) is weakly compact whenever v∈K∩D(Φ(η,·)). Therefore conditions (i) and (iii) in Theorem 5 are fulfilled.
We focus now on the verification of condition (ii) in Theorem5. Arguing by contradiction, we suppose that there exist v1,…,vn∈K∩D(Φ(η,·)) and u0∈conv{v1,…,vn} such that u0∉⋃i=1nG(vi). The convexity of the set K and of the function Φ(η,·) ensures that u0∈K∩D(Φ(η,·)). Then the assertion that u0∉⋃i=1nG(vi) reads as
(17)〈Au0-f,u0-vi〉X-J0(η;vi-u0)+Φ(η,u0)-Φ(η,vi)>0,∀i∈{1,…,n}.
Let
(18)Λ∶={v∈D(Φ(η,·)):〈Au0-f,u0-v〉Xmmmmmmmmmmm,-J0(η;v-u0),mmmmmmmmmmm+Φ(η,u0)-Φ(η,v)>0}.
It is clear that vi∈Λ for all i∈{1,…,n}. The convexity of the functions Φ(η,·) and J0(η;·) implies that Λ is a convex subset in X. We infer that conv{v1,…,vn}⊂Λ, so u0∈Λ, which is obviously impossible. This contradiction justifies condition (ii) in Theorem 5. Thus all the assumptions of Theorem 5 are satisfied.
Applying Theorem5, we obtain
(19)⋂v∈K∩D(Φ(η,·))‍G(v)≠∅.
This ensures the existence of an element u∈K∩D(Φ(η,·)) satisfying
(20)〈Au,v-u〉X+Φ(η,v)-Φ(η,u)+J0(η;v-u)≥〈f,v-u〉X
for all v∈K∩D(Φ(η,·)). The above inequality being also satisfied if v∉D(Φ(η,·)), we conclude that u is a solution of problem (13).
It remains to show that the solution of problem (13) is unique. If u1,u2∈K are solutions of (13), then we have that (η,u1)∈D(Φ), (η,u2)∈D(Φ), and
(21)〈Au1,v-u1〉X+Φ(η,v)-Φ(η,u1)+J0(η;v-u1)≥〈f,v-u1〉X,∀v∈K,〈Au2,v-u2〉X+Φ(η,v)-Φ(η,u2)+J0(η;v-u2)≥〈f,v-u2〉X,∀v∈K.
Letting v=u2 in the first inequality and v=u1 in the second one and then adding the obtained relations, we arrive at
(22)〈Au1-Au2,u2-u1〉X+J0(η;u2-u1)+J0(η;u1-u2)≥0.
By assumption (H3), we conclude that u1=u2. The proof is complete.Denote byuη∈K the unique solution of problem (13) corresponding to η∈K. Lemma 6 guarantees that uη exists and is unique. We define π:K→K by
(23)π(η)=uη,∀η∈K.Lemma 7.
Assume that hypotheses(H1)–(H3) are fulfilled and that the closed, convex set K is bounded in X. Then, the map π:K→K given in (23) is sequentially weakly continuous.Proof.
Let a sequence{ηn}n≥1⊂K such that ηn⇀η in X for some η∈K. We need to show that π(ηn)⇀π(η) as n→∞. To do this, it suffices to check that, for any relabeled subsequence {ηn}n≥1, there is a subsequence of {π(ηn)}n≥1 weakly converging to π(η).
By the compactness of the embedding ofX in Y, we have that ηn→η in Y. Denote, for simplicity, π(ηn)=un. The definition of π yields (ηn,un)∈D(Φ) and
(24)〈Aun,un-v〉X≤Φ(ηn,v)-Φ(ηn,un)+J0(ηn;v-un)+〈f,un-v〉X,∀v∈K.
SinceK is bounded, {un}n≥1⊂K and X is reflexive, we know that along a subsequence, denoted again by {un}n≥1, we have
(25)un⇀winXasn⟶∞,
for some w∈X. The first part of (H2) yields (η,w)∈(K×K)∩D(Φ). Moreover, the compactness of the embedding of X in Y implies that un→w in Y. Letting n→∞ in (24), by means of (H1), (H2), the convergences ηn→η and un→w in Y, and the upper semicontinuity of J0(·;·) on Y×Y, we get
(26)〈Aw,w-v〉X≤limsupn→∞〈Aun,un-v〉X≤limsupn→∞(Φ(ηn,v)-Φ(ηn,un))+limsupn→∞J0(ηn;v-un)+〈f,w-v〉X≤Φ(η,v)-Φ(η,w)+J0(η;v-w)+〈f,w-v〉X,∀v∈K.
This means that w∈K is a solution of problem (13). Lemma 6 ensures that w is the unique solution of (13). Thus, by (23), we have π(η)=w. Taking into account (25), it follows that π(ηn)⇀π(η) as n→∞ up to a subsequence. This completes the proof.Remark 8.
As noted in the proof of Lemma6, the closed, bounded, convex subset K⊂X is metrizable for the weak topology. Therefore, Lemma 7 implies that π is weakly continuous.We need the following version of the Schauder fixed point theorem (see [10, page 452]).Theorem 9.
Suppose that(i)
X is a reflexive, separable Banach space;(ii)
the mapT:M⊂X→M is sequentially weakly continuous;(iii)
the setM is nonempty, closed, bounded, and convex.Then T has a fixed point.We are now in position to prove Theorem2.Proof of Theorem2.
In view of Lemma7 and the assumptions on X and K, we may apply Theorem 9 which shows that the map π:K→K admits a fixed point u∈K; that is, π(u)=u. Using the definition of π (see (23)), we deduce that u∈K is a solution of problem (1).


## 3. Proof of Theorem4

It suffices to prove Theorem4 when the set K is unbounded because for a bounded set K the result is true according to Theorem 2. Let Km={x∈K:∥x∥X≤m}. Let m0≥1 be an integer such that ∥v0∥X≤m0, where v0 is the element entering (H4). We claim that Theorem 2 can be applied with K replaced by Km whenever m≥m0.Note thatv0∈Km0, so v0∈Km∩D(Φ(η,·)) for all η∈K, all m≥m0 (using the first part of (H4)). Thus, Km∩D(Φ(η,·))≠∅ for all η∈Km, all m≥m0. Since K is convex and closed in X, it turns out that Km is convex, closed, and bounded in X, for all m≥m0.We check that assumptions(H1)–(H3) of Theorem 2 remain valid when K is replaced by Km with m≥m0. Towards this, we fix some m≥m0. If {(ηn,un)}n≥1⊂(Km×Km)∩D(Φ) satisfies ηn⇀η in X and un⇀u in X, then assumption (H2) (for K) implies (η,u)∈(K×K)∩D(Φ). On the other hand, the weak convergences ensure that
(27)∥η∥X≤liminfn→∞∥ηn∥X≤m,∥u∥X≤liminfn→∞∥un∥X≤m.
Hence, (η,u)∈(Km×Km)∩D(Φ). The second part of (H2) for Km and conditions (H1) and (H3) for Km hold because (H1), (H2), and (H3) have been imposed for K, which contains Km. Thus it is permitted to apply Theorem 2 for Km in place of K, with any m≥m0.Applying Theorem2, we find a sequence {um}m≥m0 in X such that um∈Km, (um,um)∈D(Φ), and
(28)〈Aum,v-um〉X+Φ(um,v)-Φ(um,um)+J0(um;v-um)≥〈f,v-um〉X
for all v∈Km, all m≥m0. Letting v=v0 (see (H4)) in (28), we obtain
(29)〈Aum,um-v0〉X≤Φ(um,v0)-Φ(um,um)+J0(um;v0-um)+〈f,um-v0〉X
for all m≥m0. By the definition of the convex subdifferential ∂Φ(um,·), we have
(30)Φ(um,v0)-Φ(um,um)≤〈z,v0-um〉X,∀z∈∂Φ(um,·)(v0),∀m≥m0.
Then, invoking the growth condition for ∂Φ(um,·)(v0) in (H5), we see that
(31)Φ(um,v0)-Φ(um,um)≤c0(1+∥um∥Xp),∀m≥m0.
Recall that
(32)J0(u;v)=maxw∈∂J(u)〈w,v〉Y,∀u,v∈Y
(see [1, Proposition 2.1.2(b)]). This fact combined with the growth condition for the generalized gradient ∂J(um) as stated in (H5) enables us to write
(33)J0(um;v0-um)=maxw∈∂J(um)〈w,v0-um〉Y≤c0(1+∥um∥Yp-1)∥v0-um∥Y
for all m≥m0. By the continuity of the embedding X⊂Y, the inequality above leads to
(34)J0(um;v0-um)≤c1(1+∥um∥Xp-1)∥v0-um∥X,∀m≥m0,
where c1>0 is a constant. Combining (29), (31), and (34) yields
(35)〈Aum,um-v0〉X≤c0(1+∥um∥Xp)+[c1(1+∥um∥Xp-1)+∥f∥X*]∥v0-um∥X
for all m≥m0. Relation (35) ensures that the sequence {um}m≥m0 is bounded in X; indeed, if we suppose that we have ∥um∥X→+∞ along a (relabeled) subsequence, then it is seen from (35) that there is a constant c>0 such that
(36)limsupm→∞〈Aum,um-v0〉X∥um∥Xp≤c,
which contradicts hypothesis (H4).By the reflexivity ofX, there exists a subsequence of {um}m≥m0, denoted again by {um}m≥m0, such that
(37)um⇀uinXasm⟶∞,
for some u∈X. Using hypothesis (H2) with ηm=um, we derive that (u,u)∈(K×K)∩D(Φ).It remains to show thatu verifies the inequality in problem (1). Let an arbitrary element v∈K and let m1=m1(v)∈ℕ such that m1≥max{m0,∥v∥X}. Then v∈Km for each m≥m1 and so from (28), we have that
(38)〈Aum,um-v〉X≤Φ(um,v)-Φ(um,um)+J0(um;v-um)+〈f,um-v〉X.
The compactness of the embedding X⊂Y and (37) guarantee that um→u in Y as m→∞. Then the upper semicontinuity of J0(·;·) on Y×Y implies
(39)limsupm→∞J0(um;v-um)≤J0(u;v-u).
Assumptions (H1) and (H2) ensure that
(40)〈Au,u-v〉X≤limsupm→∞〈Aum,um-v〉X,limsupm→∞(Φ(um,v)-Φ(um,um))≤Φ(u,v)-Φ(u,u).
Passing to the limsup as m→∞ in (38) and using (39) and (40), we get that u∈K satisfies the inequality in (1). Since v was chosen arbitrarily in K, we conclude that u solves problem (1). The proof of Theorem 4 is complete.



---

*Source: 427908-2013-10-02.xml*