Some Conditions for Maximal Monotonicity of Bifunctions

We present necessary and sufficient conditions for a monotone bifunction to be maximally monotone, based on a recent characterization of maximally monotone operators. These conditions state the existence of solutions to equilibrium problems obtained by perturbing the defining bifunction in a suitable way.


Introduction
This paper deals with monotonicity in the context of a reflexive Banach space X. Monotonicity for set valued operators and bifunctions appear in the literature as an elegant and useful mathematical notion. Monotone bifunctions have been extensively studied since the following equilibrium problem appeared in the seminal work of Blum and Oettli [3]: given a bifunction F : X × X → R = R ∪ {+∞, −∞} and a set K ⊆ X such that F (x, x) = 0 for every x ∈ K, find x ∈ K such that F (x, y) 0 for all y ∈ K.
The aim of this paper is to present some necessary and sufficient conditions for a monotone bifunction to be maximal in the pointwise sense. The conditions we will present are equivalent to the existence of solutions to a class of equilibrium problems obtained by perturbing a given bifunction in a suitable way.
The paper is organized as follows. In the next section, we begin fixing notations and reviewing some important points of monotone operator theory. Notation not explicitly defined there is standard and as in [11]. Based on the definitions in [1] and [5], we introduce suitable notions of monotone and maximally monotone bifunctions and show that there is a bijection between special classes of monotone bifunctions and monotone operators. We finish the section by introducing pointwise maximal monotonicity. In Section 3, we use a recent characterization of maximally monotone operators due to the third author [9], in order to characterize maximally monotone bifunctions in terms of the solution sets of equilibrium problems. We also get a generalization of the following characterization of maximality: a monotone operator A is maximally monotone if and only if for each x ∈ X, there exists x ∈ X such that 0 ∈ J (x − x) + A(x ), with J denoting the duality mapping (see, for instance, [6, p. 324]

Notation and Preliminary Results
In the following, X is a reflexive Banach space, X * its dual and ·, · : X × X * → R is the duality pairing. The indicator function of a set C ⊆ X is the function δ C : Given a multivalued operator T : X ⇒ X * , its domain and graph are, respectivelly, the sets The operator is called monotone if for every x, y ∈ X and x * ∈ T (x), y * ∈ T (y), It is called maximally monotone if it is monotone and its graph is not properly included in the graph of any other monotone operator. If the inequality is strict whenever x = y, we say that T is strictly monotone. The inverse of T is the operator T −1 : X * ⇒ X defined by Given a proper convex function f : X → R∪{+∞}, its Fenchel conjugate is the function The subdifferential of a lower semicontinuous (l.s.c.) proper convex function f at a point x ∈ X is the set This defines an operator ∂f : X ⇒ X * which is maximally monotone [12]. If we take f to be the function 1 2 · 2 , then its subdifferential operator is a maximally monotone operator called the duality mapping of X; it is denoted by J , and satisfies (1) For any operator T : X ⇒ X * with D(T ) = ∅, the Fitzpatrick function of T is the function ϕ T : X × X * → R ∪ {+∞} defined by If T is maximally monotone, then ϕ T ≥ ·, · with equality exactly on GraphT ; moreover, it is the smallest convex function having this property. Given Monotonicity and maximal monotonicity is preserved by the translation. In addition, (see Section 3.1 in [10]). We will make use of the following result from [9]: Theorem 1 For every monotone operator T : X ⇒ X * , the following statements are equivalent: (a) T is maximally monotone; (b) Graph(T ) + Graph(−B) = X × X * for every maximally monotone operator B : X ⇒ X * such that ϕ B is finite-valued. (c) There exists a maximally monotone operator B : X ⇒ X * such that ϕ B is finite-valued and Graph(T ) + Graph(−B) = X × X * , and there exists (p, p * ) ∈ Graph(B) such that p − y, p * − y * > 0 for every (y, y * ) ∈ Graph(B)\{(p, p * )}.

Remark 1
As it is obvious from the proof of the above theorem in [9], the assumptions that B is monotone (and in particular maximally monotone) and ϕ B is finite-valued, are not used in the proof of the implication (c) ⇒ (a). Hence, if there exists an operator B : A bifunction is, by definition, any function F : The domain of a bifunction F is defined to be the set domF = {x ∈ X : ∀y ∈ X, F (x, y) > −∞}.
If F is monotone, then for every x, y ∈ domF one will have Given any bifunction F : X × X → R one defines the operator A F : X ⇒ X * by This happens in particular whenever x ∈ D(A F ) and F is monotone; indeed, in this case for any A F corresponds to one of the two "diagonal subdifferential operators" introduced in [7], see also [8].
Remark 2 It is easy to check that whenever a bifunction F is monotone, the operator A F is also monotone. The converse is not true, as can be seen by the following example. Define Then F is not monotone, and it is easy to show that A F = 0 so it is monotone. This leads to other examples, by taking the sum of the above bifunction with another monotone bifunction. For instance, if T : X → X * is any monotone, bounded linear operator, then the bifunction For every operator T : X ⇒ X * one defines the bifunction G T by The mappings T → G T and F → A F are not one-to-one [5]; however, if we restrict their domain and range, they become bijections as we shall see. We introduce first the following notation: For every C * ∈ C (X * ), recall that its support function is This means that This bijection satisfies ψ(−∞) = ∅.
The following simple proposition holds: is a bijection, whose inverse is the mapping

The restriction of this mapping to O m (X) is a bijection between O m (X) and B m (X).
Proof For every T ∈ O(X) and x ∈ X we have, using successively (5), (4) and (3): Similarly, for every F ∈ B s (X) and x ∈ X we obtain: . Thus, G A F = F . The remaining assertions follow immediately.
A usual definition of maximal monotonicity for bifunctions is the following: Definition 2 A monotone bifunction F : X × X → R is said to be a maximally monotone bifunction (MMB) if the operator A F is maximally monotone.
We also introduce another definition, on a more restricted class of bifunctions: The following proposition shows the relation between the two notions:

Proposition 2 Let F ∈ B m (X). Then F is an MMB if and only if F is a PMMB.
Proof Let F ∈ B m (X) be a PMMB, and let T be a maximally monotone extension of A F . Then A F (x) ⊆ T (x) for all x ∈ X, hence G A F ≤ G T . By Proposition 1, F = G A F , thus F ≤ G T . Since F is a PMMB and G T ∈ B m (X), we obtain F = G T . Hence, A F = A G T = T by Proposition 1. Hence A F is maximally monotone and F is an MMB.

Conversely, let F ∈ B m (X) be an MMB. Assume that H ∈ B m (X) satisfies F ≤ H . Then A F (x) ⊆ A H (x) for all x ∈ X. Since A F is a maximally monotone operator, A F = A H . In view of Proposition 1, this implies that
Hence, F is a PMMB.
Given any bifunction F , its Fitzpatrick transform [1,2] is defined as the function F : X × X * → R given by If T is any operator, then G T = ϕ T [1].
Now we show the following useful proposition:

Proposition 3 Let
A : X ⇒ X * be monotone and B : X ⇒ X * be maximally monotone and such that ϕ B is finite-valued. Consider the following statements: Then the following implications hold true: 1) ⇒ 2) ⇒ 3). If, moreover, B is single-valued and strictly monotone, then these statements are equivalent.
Then T is a monotone operator and by relation (2), ϕ T is finite-valued. By using the first part of the Corollary 2.7 of [9] we obtain that for each x ∈ X, it holds R(A + B(· − x)) = X * .
2) ⇒ 3) is obvious. Now assume that B is also single-valued and strictly monotone. Define B 1 by B 1 (x) = −B(−x). It is easy to see that B 1 is maximally monotone, single-valued, strictly monotone, and ϕ B 1 (x, x * ) = ϕ B (−x, −x * ) < +∞. Also, B −1 1 is maximally monotone, single valued, strictly monotone and such that ϕ B −1 By using Corollary 2.7 of [9] we deduce that A −1 is maximally monotone, hence A is maximally monotone.

Remark 3 The implication 3) ⇒ 1) does not hold in general if the operator B is not singlevalued. For example, consider the operators A and B defined on R by
Then A is monotone but not maximally monotone, B is maximally monotone and strictly monotone, with ϕ B finite-valued. To see this, it is enough to show that for given (x, x * ) ∈ R 2 it holds inf (y,y * )∈Graph(B) (x − y)(x * − y * ) > −∞. For this to hold, it is enough to have inf y∈R\{0} (x − y)(x * + s − y) > −∞, for s ∈ {−1, 1} . The latter is obviously true since we have the infimum of a quadratic function in y with positive leading coefficient. Hence ϕ B is finite-valued. For every x ∈ R we set x = x. We see that statement 3) holds since B(0) = [−1, 1] while A(x) is one of the sets {−1}, {0} and {1}. However, statement 1) does not hold.
The equivalence 1) ⇐⇒ 3) in Proposition 3 generalizes the following characterization of maximality: a monotone operator A is maximally monotone if and only if for each x ∈ X, there exists x ∈ X such that 0 ∈ J (x − x) + A(x ). See for instance [6, p. 324].

Main Results
Our first main result gives conditions for a bifunction to be maximally monotone.

Theorem 2 Let F be a monotone bifunction such that F (x, ·) is convex and l.s.c., for each
x ∈ X. Then the following statements are equivalent.
There exist a finite-valued H ∈ B s (X) and p ∈ X such that (a) H (p, ·) is continuous affine, Proof 1) ⇒ 2) Let F be an MMB. If we set F 1 = G A F , then F 1 ∈ B m (X). By Proposition 1, A F 1 = A F , thus F 1 is an MMB. By Proposition 2, F 1 is a PMMB. 2) ⇒ 3) Let H be as in statement 3). Setting again F 1 = G A F , by Proposition 2 both F 1 and H are MMB. Since A F 1 = A F , the operators A F and A H are maximally monotone. In addition, ϕ A H = H is finite-valued.
Let B be the operator defined by B(x) = −A H (−x). It can be easily seen that B is maximally monotone, and ϕ B (x, Adding the two inequalities and setting x H = w we obtain F (x H , y) + H (x H , y) ≥ y − x H , w * + v * = 0, ∀y ∈ X.
3) ⇒ 4) We take p = 0 and H = G J , where J is the duality mapping. We will show that H satisfies all conditions of 4).

By Proposition 1, H ∈ B s (X).
Since y − x, x * and the set J (x) is nonempty and bounded, H is obviously finite-valued. Moreover, for all y ∈ X the definition of G J and (1) imply that G J (0, y) = 0 and G J (y, 0) = − y 2 . Hence, for y = 0, Given that G J (0, ·) = 0, both conditions 4(a) and 4(b) are satisfied. In order to obtain that H satisfies also condition 4(c), given (x 0 , x * 0 ) ∈ X × X * , let us consider T : X ⇒ X * defined by Then T is a maximally monotone operator, soH is a PMMB. We know that H = G T = ϕ T . Hence, by relation (2), H is finite-valued.
From 3) we obtain that there existsx ∈ X such that To conclude, let us calculateH (x, y) We set 4) ⇒ 1) Let H and p be as in 4). Since H (p, ·) : X → R is continuous affine, we have that A H (p) is singleton. By setting {p * } = A H (p), for every (y, y * ) ∈ Graph(A H )\{(p, p * )}, it holds that y = p so On the other hand, given (x 0 , x * 0 ) ∈ X × X * , by consideringx as in statement 4(c) we have Since H ∈ B s (X), g is convex and l.s.c.; it is also finite-valued by assumption 4). Obviously it satisfies F (x, y)+g(y) ≥ 0 for all y ∈ X, and F (x,x)+g(x) = 0. Hence, 0 ∈ ∂(F (x, ·)+g(·))(x). By the subdifferential sum rule (see for instance [6]), 0 ∈ ∂F (x, ·)(x) + ∂g(·)(x). Hence, there exists x * ∈ X * such that x * ∈ ∂F (x, ·)(x) = A F (x) and −x * ∈ ∂g(x). The last inclusion yields for every y ∈ X, Therefore, in view of Remark 1, we have obtained that A F and A H satisfy all assumptions (see (8) and (9) above) necessary to conclude that A F is a maximally monotone operator. This is equivalent to saying that F is an MMB, so 1) holds.
From the previous results we obtain the following.

Corollary 1
Let F : X × X → R be a monotone bifunction and let B : X → X * be a maximally monotone, single-valued and strictly monotone operator such that ϕ B is finitevalued. Then F is maximally monotone if, and only if, for every λ > 0 and for every x 0 ∈ X, there exists x λ ∈ X such that λF (x λ , y) + y − x λ , B(x λ − x 0 ) ≥ 0, ∀y ∈ X.  (2) we deduce that H is finite-valued. Set F 1 = G A F . Then F 1 ∈ B m (X) and A F 1 = A F , so F 1 is maximally monotone. For each λ > 0, from the equivalence (2) ⇔ (3) of the Theorem 2 applied to the bifunctions λF 1 and H , we deduce that there exists x λ ∈ X such that λF 1 (x λ , y) + H (x λ , y) ≥ 0, ∀ y ∈ X, that is, λF 1 (x λ , y) + y − x λ , B(x λ − x 0 ) ≥ 0, ∀ y ∈ X. On the other hand, it is easy to see that F 1 ≤ F . Therefore, for each λ > 0 we obtain that λF (x λ , y) + y − x λ , B(x λ − x 0 ) ≥ 0, ∀ y ∈ X.
(⇐) The inequality (10) is equivalent to −B(x λ − x) ∈ A λF (x λ ). Hence, given λ > 0, for each x ∈ X there exists x λ ∈ X such that 0 ∈ A λF (x λ ) + B(x λ − x). By the implication 3) =⇒ 1) of Proposition 3, we have that λF is an MMB, for each λ > 0 fixed. In particular, F is an MMB. Corollary 1 is a generalization of Proposition 2.6 obtained in [5]. We finish by observing that the statements 3) and 4) of the Theorem 2 and the Corollary 1 establish the existence of solutions to equilibrium problems obtained by perturbing the bifunction F according to a choice of a suitable bifunction H for each case.