Existence of Solutions for Implicit Obstacle Problems of Fractional Laplacian Type Involving Set-Valued Operators

The paper is devoted to a new kind of implicit obstacle problem given by a fractional Laplacian-type operator and a set-valued term, which is described by a generalized gradient. An existence theorem for the considered implicit obstacle problem is established, using a surjectivity theorem for set-valued mappings, Kluge’s fixed point principle and nonsmooth analysis.

Recent works focus on systems governed by nonlocal operators and exhibiting setvalued terms in the form of generalized gradient of a locally Lipschitz function. Frassu et al. [18] proved the existence of three nontrivial solutions for a pseudo-differential inclusion driven by a nonlocal anisotropic operator and with generalized gradient of a locally Lipschitz potential. Teng [19] and Xi et al. [20] applied the nonsmooth critical point theory to obtain multiplicity results for nonlocal elliptic hemivariational inequalities. Liu and Tan [21] employed a surjectivity theorem for pseudomonotone and coercive operators to explore a nonlocal hemivariational inequality. For related works, we refer to [13,14].
In relevant situations encountered in engineering and economic models, such as Nash equilibrium with shared constraints and transport optimization feedback control, the constraint sets depend on the unknown state variable. For this reason, the theory of quasi variational/hemivariational inequalities was prompted to become one of the most promising research domains in applied mathematics. Yet, as far as we know, there is no publication considering differential inclusion problems with nonlocal operators and implicit obstacle effect (i.e., the constraints depend on the unknown function). It is the goal of the present work to study an implicit obstacle problem containing a generalized fractional Laplace operator and a generalized gradient term. Specifically, we establish a general existence theorem for this new type of problem.
The rest of the paper is organized as follows. In Sect. 2, we formulate the problem and survey some preliminary material needed in the study of Problem 2.1. Section 3 provides the variational formulation of Problem 2.1, whereas Sect. 4 comprises the study of variational selection associated to our problem. Section 5 presents our existence theorem with its proof employing Kluge's fixed point principle, a surjectivity theorem for multivalued operators and nonsmooth analysis. Section 6 sets forth our existence result without the relaxed monotonicity condition. Section 7 is devoted to conclusions.

Problem Formulation and Mathematical Background
Let Ω be a bounded domain in R N with Lipschitz boundary, let s ∈]0, 1[ with N > 2s and let f ∈ L 2 (Ω). Denoting Ω c := R N \Ω, we formulate the following implicit obstacle problem: where L K stands for the generalized nonlocal fractional Laplace operator given by R is a given function, the multivalued term ∂ j(x, ·) denotes the generalized gradient (see Definition 2.2 below) of a locally Lipschitz function r → j(x, r ), and ζ ∈ L 2 (Ω) belongs to the (exterior) normal cone . Throughout the paper, K is assumed to fulfill the conditions: (1), then Problem 2.1 reduces to the nonlocal inclusion problem which has been studied by Migórski et al. [22] for the particular case of kernel K (x) := |x| −(N +2s) for all x ∈ R N \{0}, i.e., L K is the fractional Laplace oper- dy for a.e. x ∈ R N . Then, we briefly review basic notation and results which are needed in the sequel. For more details, we refer to monographs [23][24][25][26].
Let us begin with some definitions and properties for set-valued mappings.
If this holds for every x ∈ X , then F is called upper semicontinuous. (ii) We say that F is sequentially closed at x 0 ∈ X , if for every sequence {(x n , y n )} ⊂ Gr(F) with (x n , y n ) → (x 0 , y 0 ) in X × Y , then it holds (x 0 , y 0 ) ∈ Gr(F), where Gr(F) is the graph of the set-valued mapping F defined by Gr(F) := {(x, y) ∈ X × Y : y ∈ F(x)}. We say that F is sequentially closed, if it is sequentially closed at every x 0 ∈ X .
The next proposition characterizes the upper semicontinuity of set-valued maps.
Let E be a Banach space with its dual E * . A function J : E → R is said to be locally Lipschitz at u ∈ E, if there exist a neighborhood N (u) of u and a constant The next proposition collects some basic results (see, e.g., [25,Proposition 3.23]).

Lemma 2.1
Let Ω be a bounded domain in R N with Lipschitz boundary and let s ∈ ]0, 1[ with N > 2s. Then we have: From Lemma 2.1 (ii) it is seen that the norm · X 0 on X 0 defined by 1 2 is equivalent to the norm induced by · X . We end the section by recalling the fixed point theorem of Kluge [29] and the surjectivity theorem for set-valued mappings of Le [30, Theorem 2.2], which will be used in the proof of our existence result.

Remark 2.1
In the statement of Theorem 2.1, we took advantage of Referee's comment pointing out that the usual formulation "If either C is bounded or Ψ (C) is bounded" of hypothesis in Kluge's fixed point theorem is equivalent to "If Ψ (C) is bounded". This is true because Ψ (C) ⊂ C.

Hypotheses and Variational Formulation
In this section, we set forth our assumptions and give the variational formulation. In what follows, we assume that the functions j : Ω × R → R and U : L 2 (Ω) → R verify the following conditions: is measurable on Ω for all r ∈ R and there exists l ∈ L 2 (Ω) such that j(·, l(·)) belongs to L 1 (Ω); (ii) j(x, ·) is locally Lipschitz for a.e. x ∈ Ω; (iii) There exists a constant m j ≥ 0 such that for all r 1 , r 2 ∈ R and a.e.
where ∂ j(x, r ) stands for the generalized gradient of j with respect to the variable r ; (iv) There exist constants α j ≥ 0 and β j ∈]0, 1/c(2) 2 [ such that for all r ∈ R and a.e.
is usually called relaxed monotone condition (see, e.g., [25]) for the locally Lipschitz function j(x, ·). It is equivalent to the inequality

Remark 3.2 The relaxed monotone condition H ( j)(iii) reads as
we infer that the function r → g(x, r ) := j(x, r ) + m j r 2 2 is convex (see [23, Proposition 2.2.9]) with the subdifferential ∂ C g(x, r ) = ∂ j(x, r ) + m j r . Consequently, the variational-hemivariational inequality in Problem 2.1 can be equivalently rewritten as a linearly perturbed variational inequality by replacing ∂ j(x, u) with ∂ C g(x, u)−m j u.
Define the set-valued mapping C : X 0 ⇒ X 0 by (i.e., the set C(u) introduced in the statement of Problem 2.1) for all u ∈ X 0 . We note that the set C(u) in (2) is closed and convex in X 0 , and 0 X 0 ∈ C(u).

Remark 3.3
The constraint in Problem 2.1 can be expressed as u X 0 ≤ U (u). This is an implicit nonlocal formulation substantially different with respect to classical statements as the pointwise constraints u(x) ≤ f (x) in obstacle problem. The motivation is to locate the solution in a nonlocal way relying on the continuous embedding To obtain the variational formulation of Problem 2.1, let u : R N → R be a smooth function such that (1) holds. For any v ∈ C(u), we act on the inclusion ( By virtue of the definitions of X 0 and generalized gradient we find Taking into account the preceding discussion, the variational formulation of Problem 2.1 reads as follows. (2)).
Further, let us introduce the function J : for all u ∈ L 2 (Ω). On account of hypothesis H ( j) and the definition of J in (3), the next lemma is a direct consequence of [25, Theorem 3.47].

Lemma 3.1 If H ( j) holds , then J defined in (3) has the properties:
with some c J > 0.

Variational Selection
The section is concerned with the existence of solutions to Problem 3.1. To this end, we pass through a related inequality problem.
Lemma 3.1(ii) ensures that if u ∈ X 0 is a solution to Problem 4.1, then u solves Problem 3.1 as well. Consequently, it is enough to establish the solvability of Problem 4.1. This will be achieved by means of the auxiliary problem: We are going to find a fixed point of the set-valued mapping S : X 0 ⇒ X 0 that we call variational selection, which is defined by

Proof Let
dx for all u, v ∈ X 0 , and consider the indicator function of C(w), that is the function I C(w) : X 0 → R := R {+∞} given by From the respective definitions and the fact that f ∈ L 2 (Ω) ⊂ X * 0 (see Lemma 2.1), Problem 4.2 can be expressed as the variational-hemivariational inequality: find u ∈ X 0 such that for all v ∈ X 0 .

Claim 1
For each w ∈ X 0 , I C(w) : X 0 → R is a proper, convex, and lower semicontinuous function with 0 X 0 ∈ D(I C(w) ) (the effective domain).
This follows readily from the expression of the set C(w) (see (2)). Claim 1 confirms that the inclusion problem: find u ∈ X 0 such that is meaningful, where the notation ∂ C (I C(w) ) stands for the subdifferential of I C(w) in the sense of convex analysis. Furthermore, Lemma 3.1 shows that every solution to inclusion (5) is a solution to problem (4) too.
Next we aim to apply Theorem 2.2 to show that problem (5) is solvable.

Claim 2
A : X 0 → X * 0 is a linear, continuous and strongly monotone operator.
Notice that for all u, v ∈ X 0 , where the change of variable and symmetry requirement H ( j)(iii) have been used. Thus we infer that Au X * 0 = u X 0 and Au, u X 0 = u 2 X 0 for all u ∈ X 0 .
From (6), it follows that the linear operator A is bounded and strongly elliptic.
is a bounded and pseudomonotone set-valued operator such that for each u ∈ X 0 , the set A(u) + ∂ J (u) is closed and convex in X * 0 .
Indeed, Proposition 2.2 and Lemma 3.1 ensure that for each u ∈ X 0 , the set Au+∂ J (u) is nonempty, closed and convex in X * 0 . But, Lemma 3.1(iii) and equality (6) imply A(u) + ξ X * 0 ≤ u X 0 + c J 1 + c(2) u X 0 ) for all u ∈ X 0 and ξ ∈ ∂ J (u). We infer that A + ∂ J is a bounded map.
Next we apply Proposition 2.1 to prove that the set-valued mapping A + ∂ J is upper semicontinuous from X 0 to X * 0 with weak topology. It is sufficient to check that for each weakly closed subset D in X * 0 , the set (A + ∂ J ) − (D) is closed in X 0 . Let a sequence {u n } ⊂ (A + ∂ J ) − (D) be such that u n → u in X 0 as n → ∞, for some u ∈ X 0 .
Consequently, A + ∂ J is upper semicontinuous from X 0 to X * 0 with weak topology. To prove that A + ∂ J is pseudomonotone, let sequences {u n } and {u * n } be such that Our goal is to produce for each v ∈ X 0 an element u Indeed, by (9) we can find a sequence {ξ n } ⊂ X * 0 such that for each n ∈ N, ξ n ∈ ∂ J (u n ) and u * n = A(u n ) + ξ n . From (9) and the above equality it follows Using (8) and the compact embedding of X 0 into L 2 (Ω) (see Lemma 2.1), it holds u n → u in L 2 (Ω) as n → ∞. Moreover, in view of [32,Thoerem 2.2], it turns out ∂(J | X 0 )(u) ⊂ ∂(J | L 2 (Ω) )(u) for all u ∈ X 0 , which leads to Lemma 3.1 and the boundedness of {u n } in X 0 entail that the sequence {ξ n } is bounded both in L 2 (Ω) and X * 0 . Then, through (12), we pass to the limit as n → ∞ to get lim n→∞ ξ n , u n − u X 0 = lim n→∞ ξ n , u n − u L 2 (Ω) = 0. The latter, in conjunction with (11) and (6), yields This means that u n → u in X 0 . On the other hand, the reflexivity of X * 0 and boundedness of {ξ n } ⊂ X * 0 permit us to suppose that ξ n → ξ weakly in X * 0 for some ξ ∈ X * 0 . Now we can assert that ξ ∈ ∂ J (u) (see, e.g., [31,Theorem 1.1.4]). Since it is clear that (10) is verified with u * = A(u) + ξ ∈ A(u) + ∂ J (u). We conclude that A + ∂ J is pseudomonotone.

Claim 4 There exists R > 0 such that Au
with u X 0 = R, all ξ ∈ ∂ J (u) and all η ∈ ∂ C (I C(w) ).
By Claims 1-4 and Theorem 2.2, there exists u w ∈ X 0 resolving inclusion (5). Thus S(w) = ∅ holds true for each w ∈ X 0 . Now we claim that the solution set S(w) of Problem 4.2 is closed. Let {u n } ⊂ S(w) be such that u n → u in X 0 . For each n ∈ N, by (4), there holds for all v ∈ X 0 , or Au n , v − u n X 0 + J 0 (u n ; v − u n ) ≥ f , v − u n X 0 for all v ∈ C(w) because u n ∈ C(w). Passing to the upper limit as n → ∞ yields u ∈ S(w), hence the claim that S(w) is closed in X 0 is proved.
Moreover, we show that for each w ∈ X 0 , the set S(w) is convex. Toward this, we note that assumption m j c(2) 2 ≤ 1 turns A + ∂ J : X 0 → 2 X * 0 be monotone because for all u, v ∈ X 0 , ξ u ∈ ∂ J (u) and ξ v ∈ ∂ J (v) we obtain and v ∈ X 0 . The latter in conjunction with Proposition 2.2 (d) results in for all v ∈ C(w). Let z ∈ X 0 and λ ∈]0, 1[. Inserting v = z λ := λz + (1 − λ)u t in (14) gives Az λ , z − u t X 0 + J 0 (z λ ; z − u t ) ≥ f , z − u t X 0 . We pass to the upper limit as λ → 0 + finding Recall that z ∈ C(w) was arbitrary, so this renders u t ∈ S(w). Therefore the set S(w) is convex. Next we demonstrate that the set S(w) is bounded in X 0 for each w ∈ X 0 . Fix w ∈ X 0 . Arguing by contradiction, suppose that S(w) is unbounded, whence there exists a sequence {u n } ⊂ S(w) such that Since 0 X 0 ∈ C(w), we have Au n , u n X 0 − J 0 (u n ; −u n ) ≤ f , u n X 0 . Reasoning as in (13) enables us to find f X * 0 u n X 0 ≥ u n As it is known from assumption H ( j)(iv) that β j ∈]0, 1/c(2) 2 [, then (15) and the above estimate generate a contradiction, which completes the proof.

Existence Result
We are in a position to state our main result.

Claim 5 The graph of S is sequentially weakly closed.
Let {w n } ⊂ X 0 and {u n } ⊂ X 0 be sequences such that u n ∈ S(w n ), w n → w weakly in X 0 and u n → u weakly in X 0 . Hence, for each n ∈ N, it holds u n ∈ C(w n ), i.e., u n X 0 ≤ U (w n ). The compactness of the embedding of X 0 in L 2 (Ω) and the continuity of U postulated in condition H (U ) provide The fact that u n ∈ S(w n ) reads as for all v ∈ C(w n ), whereas the monotonicity of u → Au + ∂ J (u) reveals For each v ∈ C(w), consider the sequence {v n } defined by v n = U (w n ) We deduce that {v n } converges strongly to v in X 0 and v n ∈ C(w n ) for every n ∈ N. It is thus permitted to insert v = v n in (17). Passing to the upper limit as n → ∞ produces where we have used (6), the compact embedding of X 0 in L 2 (Ω) and that L 2 (Ω) × L 2 (Ω) (v, u) → J 0 (u; v) ∈ R is upper semicontinuous (see Lemma 3.1 and Proposition 2.2). The arbitrariness of v ∈ C(w) and Minty approach guarantee that u ∈ S(w). Therefore, Claim 5 is proved.
If the claim were not true, then there would exist sequences {u n } and {w n } with u n ∈ S(w n ) such that u n X 0 → ∞ as n → ∞.
For every n ∈ N, one has (16) for all v ∈ C(w n ). Bearing in mind that 0 X 0 ∈ C(w) for each w ∈ X 0 , we take v = 0 X 0 as test function in (16) obtaining Au n , u n X 0 − J 0 (u n ; −u n ) ≤ f X * 0 u n X 0 . The latter combined with (6) and Lemma 3.1(ii) shows This triggers a contradiction with (18) owing to β j ∈]0, 1/c(2) 2 [. We conclude that Claim 6 holds true.
On account of Claims 5 and 6, the required conditions to apply Theorem 2.1 are verified for the set-valued mapping S. Hence it has a fixed point in X 0 , which from Lemma 3.1 is a solution to Problem 3.1.

Dropping Assumption H(j)(iii)
The aim of this section is to point out that assumption H ( j)(iii) can be dropped in the statement of Theorem 5.1. This important fact has been pointed out in one of Referee's reports, where it was also outlined the proof. We have preferred to keep the statement of Theorem 5.1 because our original approach was completely different relying on Theorem 2.2. Here is the improved statement. Proof Let us introduce the set-valued mapping T : X 0 × L 2 (Ω) ⇒ X 0 × L 2 (Ω) by T (v, w) = (u, F(u)), where u is the unique solution of the classical variational inequality in the Hilbert space X 0 : find u ∈ X 0 such that u ∈ C(v) and for all z ∈ C(v) and, for a constant R > 0, The set of constraints C(v) is given in (2), and it was already noted that C(v) is closed and convex in X 0 , and 0 X 0 ∈ C(v).
Setting z = 0 X 0 in (19) yields Combining with the continuous embedding X 0 ⊂ L 2 (Ω), there exists a constant R 0 > 0 (independent of R) for which u L 2 (Ω) ≤ R 0 . Now we fix R > R 0 . Then the definition of the set-valued mapping F entails F(u) = ∂ J (u), so by (19) with the fixed R, if (u, w) ∈ X 0 × L 2 (Ω) is a fixed point of T , i.e., (u, w) ∈ T (u, w), one has that u is a solution of Problem 4.1 and we are done. We prove that the set-valued mapping T possesses a fixed point by applying Theorem 2.1. To this end we check that the graph of T is sequentially weakly closed. Let the sequences {v n } ⊂ X 0 and {w n } ⊂ L 2 (Ω) satisfy v n → v weakly in X 0 , w n → w weakly in L 2 (Ω), and let (u n , σ n ) ∈ T (v n , w n ) be such that (u n , σ n ) → (u, σ ) weakly in X 0 × L 2 (Ω), with σ n ∈ ∂ J (u n ). By the definition of T and knowing that (u n , σ n ) ∈ T (v n , w n ), it turns out u n ∈ C(v n ) and for all z ∈ C(v n ). Due to u n → u weakly in X 0 and the compact embedding X 0 ⊂ L 2 (Ω) we have along a relabeled subsequence that u n → u strongly in L 2 (Ω), thereby u X 0 ≤ R 0 . Exploiting σ n → σ weakly in L 2 (Ω) in conjunction with σ n ∈ ∂ J (u n ) enables us to deduce that σ ∈ ∂ J (u), thus σ ∈ F(u). Since u n X 0 ≤ U (v n ), by the compact embedding of X 0 in L 2 (Ω) and the continuity of U on L 2 (Ω) (note hypotheesis H (U )), we infer that u X 0 ≤ U (v) or equivalently u ∈ C(v).
In order to conclude that the graph of T is sequentially weakly closed it remains to show that u ∈ X 0 is a solution of for all z ∈ C(v). We proceed on the pattern in the proof of Theorem 5.1. Let z ∈ C(v). For each n ∈ N, we construct z n = U (v n ) U (v) z. From the compact embedding X 0 ⊂ L 2 (Ω) and hypothesis H (U ), it follows that z n → z in X 0 in view of lim n→∞ z n − z X ensuring that z n ∈ C(v n ) for every n ∈ N, which allows us to insert z = z n in (21). We arrive at Through the weak lower semicontinuity of the norm, in the limit as n → ∞ we obtain (22).
Next we prove that the range T (X 0 × L 2 (Ω)) of the set-valued mapping T is bounded in X 0 × L 2 (Ω). From the definition of T , we see that it suffices to show that the first component u of T (v, w) is bounded. It is so since the function J in (3) is Lipschitz continuous on every bounded set in L 2 (Ω), which reflects in the fact that the generalized gradient ∂ J (u) is bounded whenever u is bounded. For z = 0 X 0 in (19) we are led to estimate (20). Hence, admitting that w is bounded in L 2 (Ω) we get that u is bounded in X 0 , whence T (X 0 × L 2 (Ω)) is bounded in X 0 × L 2 (Ω).
We have shown that the hypotheses of Theorem 2.1 are fulfilled in the case of the set-valued mapping T : X 0 × L 2 (Ω) → 2 X 0 ×L 2 (Ω) . Then Theorem 2.1 provides the existence of a fixed point of T . As remarked before, this completes the proof.

Conclusions
In this paper, we consider an implicit obstacle problem driven by a fractional Laplace operator and a set-valued mapping which is described by a generalized gradient. Under quite general assumptions on the data, and employing Kluge's fixed point principle for multivalued operators, and a surjectivity theorem, we prove that the set of weak solutions for the implicit obstacle problem is nonempty. Finally, implementing an idea suggested by one of the Referees, we improve our existence result. Problems of this type are encountered in transport optimization, Nash equilibrium theory and related fields. In the future we plan to apply the theoretical results established in the current paper to Nash equilibrium problems and population dynamics models.