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Lectures on Maximal Monotone Operators

arXiv:math/9302209v1 [math.FA] 4 Feb 1993Lectures on Maximal Monotone OperatorsR. R. PhelpsDept.

Math. GN–50, Univ.

of Wash., Seattle WA 98195; phelps@math.washington.edu(Lectures given at Prague/Paseky Summer School, Czech Republic, August 15–28, 1993.)Introduction. These lectures will focus on those properties of maximal monotone op-erators which are valid in arbitrary real Banach spaces.

Most applications (to nonlinearpartial differential equations, optimization, calculus of variations, etc.) take place in re-flexive spaces, in part because several key properties have only been shown to hold in suchspaces.

(See, for instance, [De], [Pa–Sb] and [Ze].) We will generally isolate the reflexivityhypothesis, hoping that by doing so, it will eventually be possible to decide their validitywithout that hypothesis.

In Section 1 we define maximal monotone operators and provesome of their main elementary properties. Section 2 is devoted to the prototypical classof subdifferentials of convex functions.

Gossez’s subclass of monotone operators of type(D) is examined in Section 3 and Section 4 gives a brief treatment of another subclass,the locally maximal monotone operators. We have attempted to keep the exposition self–contained, using only standard tools of elementary functional analysis.

One exception isthe application of the Brouwer fixed–point theorem in the proof of the Debrunner–Flortheorem (Lemma 1.7).1. Monotone Operators.Definition 1.1.

A set-valued map T from a Banach space E into the subsets of its dualE∗is said to be a monotone operator provided⟨x∗−y∗, x −y⟩≥0∀x, y ∈E and x∗∈T(x), y∗∈T(y).We do not require that T(x) be nonempty. The domain (or effective domain) of T is theset D(T) = {x ∈E: T(x) ̸= ∅}.Examples 1.2.

(a) The simplest examples of such operators are linear and single–valued. For instance,if H is a real Hilbert space and T: H →H∗≡H is a linear map, then T is monotone ifand only if it is, in the usual sense, a positive operator: ⟨T(x), x⟩≥0 for all x.

(b) Let D be a nonempty subset of the real numbers R. A function ϕ: D →R∗≡Rdefines a monotone operator if and only if ϕ is monotone nondecreasing in the usual sense:That is,[ϕ(t2) −ϕ(t1)] · (t2 −t1) ≥0 ∀t1, t2 ∈Diffϕ(t1) ≤ϕ(t2) whenever t1 < t2. (c) Examples of set–valued monotone functions from R to subsets of R are easy toexhibit: For instance, let ϕ(x) = 0 if x < 0, ϕ(x) = 1 if x > 0 and let ϕ(0) be any subsetof [0, 1].1

(d) Here is an important single–valued but nonlinear example: Let f be a continuousreal–valued function on E which is Gateaux differentiable (that is, for each x ∈E the limitdf(x)(y) = limt→0f(x + ty) −f(x)t,y ∈Eexists and is a bounded linear functional of y). Such a function f is convex if and only ifthe mapping x →df(x) is monotone.

Indeed, suppose that f is convex; then for 0 < t < 1,convexity implies thatf(x + t(y −x)) −f(x)t≤(1 −t)f(x) + tf(y) −f(x)t= f(y) −f(x).It follows that df(x)(y −x) ≤f(y) −f(x), for any x, y ∈E.Thus, if x, y ∈E andx∗= df(x), y∗= df(y), then⟨x∗, y −x⟩≤f(y) −f(x) and −⟨y∗, y −x⟩= ⟨y∗, x −y⟩≤f(x) −f(y);now add these two inequalities. A proof of the converse may be found in [Ph, p.17].

(e) The next example arises in fixed-point theory. Let C be a bounded closed convexnonempty subset of Hilbert space H and let U be a (generally nonlinear) nonexpansivemap of C into itself: ∥U(x) −U(y)∥≤∥x −y∥for all x, y ∈C.

Let I denote the identitymap in H; then T = I −U is monotone, with D(T) = C. Indeed, for all x, y ∈C,⟨T(x) −T(y), x −y⟩= ⟨x −y −(U(x) −U(y)), x −y⟩== ∥x −y∥2 −⟨U(x) −U(y), x −y⟩≥∥x −y∥2 −∥U(x) −U(y)∥· ∥x −y∥≥0.Note that 0 is in the range of T if and only if U has a fixed point in C; this hints at theimportance for applications of studying the ranges of monotone operators. (f) Again, in Hilbert space, let C be a nonempty closed convex set and let P be themetric projection of H onto C; that is, P(x) is the unique element of C which satisfies∥x−P(x)∥= inf{∥x−y∥: y ∈C}.

We first prove the fundamental fact that the mapping Psatisfies (in fact, it is characterized by) the following variational inequality: For all x ∈H,⟨x −P(x), z −P(x)⟩≤0for all z ∈C. (1.1)Indeed, if z ∈C and 0 < t < 1, then zt ≡tz + (1 −t)P(x) ∈C and hence ∥x −P(x)∥≤∥x −zt∥= ∥(x −P(x)) −t(z −P(x))∥.

Squaring both sides of this inequality, expandingand then cancelling ∥x −P(x)∥2 on both sides yields0 ≤−2t⟨x −P(x), z −P(x)⟩+ t2∥z −P(x)∥2.If we then divide by t and take the limit as t →0 we obtain (1.1). Moreover, if y ∈H andwe write down (1.1) again, using y in place of x, then take z = P(y) in the first equation,z = P(x) in the second one and add the two, we obtain⟨x −y, P(x) −P(y)⟩≥∥P(x) −P(y)∥2for all x, y ∈H,(1.2)2

which shows that P is in monotone in a very strong sense. Note that P is an example of anonexpansive mapping in the sense of the previous example: We always have the inequality⟨x−y, P(x)−P(y)⟩≤∥x−y∥·∥P(x)−P(y)∥, so combined with the monotonicity inequalityabove, we have ∥P(x) −P(y)∥≤∥x −y∥for all x, y ∈H.

(g) Here is a fundamental example of a set–valued monotone mapping, the dualitymapping from E into 2E∗. For any x ∈E defineJ(x) = {x∗∈E∗: ⟨x∗, x⟩= ∥x∗∥· ∥x∥and ∥x∗∥= ∥x∥}.By the Hahn–Banach theorem, J(x) is nonempty for each x, so D(J) = E. Suppose thatx∗∈J(x) and y∗∈J(y).

Then⟨x∗−y∗, x−y⟩= ∥x∗∥2 −⟨x∗, y⟩−⟨y∗, x⟩+∥y∗∥2 ≥∥x∗∥2 −∥x∗∥·∥y∥−∥y∗∥·∥x∥+∥y∗∥2= ∥x∗∥2 −2∥x∗∥· ∥y∗∥+ ∥y∗∥2 = (∥x∗∥−∥y∗∥)2,so this is also monotone in a rather strong sense.In order to define maximal monotone operators we must consider their graphs.Definition 1.3. A subset G of E ×E∗is said to be monotone provided ⟨x∗−y∗, x−y⟩≥0whenever (x, x∗), (y, y∗) ∈G.

A set–valued mapping T: E →2E∗is a monotone operatorif and only if its graphG(T) = {(x, x∗) ∈E × E∗: x∗∈T(x)}is a monotone set. A monotone set is said to be maximal monotone if it is maximal in thefamily of monotone subsets of E × E∗, ordered by inclusion.

An element (x, x∗) ∈E × E∗is said to be monotonically related to the subset G provided⟨x∗−y∗, x −y⟩≥0for all (y, y∗) ∈G.We say that a monotone operator T is maximal monotone provided its graph is a maximalmonotone set.The most frequently used form of the definition of maximality of T is the following con-dition: Whenever (x, x∗) ∈E × E∗is monotonically related to G(T), then x ∈D(T) andx∗∈T(x).There is an obvious one-to-one correspondence between monotone sets and monotone op-erators. An easy application of Zorn’s lemma shows that every monotone operator T canbe extended to a maximal monotone operator T , in the sense that G(T) ⊂G(T).Definition 1.4.

If T: E →2E∗is a monotone operator, its inverse T −1 is the set valuedmapping from E∗to 2E defined by T −1(x∗) = {x ∈E: x∗∈T(x)}. Obviously,G(T −1) = {(x∗, x) ∈E∗× E: x∗∈T(x)},which is (within a permutation) the same as the monotone set G(T).

In particular,then,T −1 is maximal monotone if and only if T is maximal monotone.3

Examples 1.5. (a) The monotone mapping ϕ defined in Example 1.2 (c) is maximal if and only ifϕ(0) = [0, 1].

More generally, it is easily seen that a monotone nondecreasing function ϕon R is maximal monotone if and only if ϕ(x) = [ϕ(x−), ϕ(x+)] for each x ∈R (where,for instance, ϕ(x−) ≡limt→x−ϕ(t)). (b) Any positive linear operator T on Hilbert space is maximal monotone.

Indeed,suppose (x, x∗) ∈H ×H is monotonically related to G(T). Then, for any z ∈H and λ > 0we have0 ≤⟨T(x ± λz) −x∗, (x ± λz) −x⟩= ±λ⟨T(x) ± λT(z) −x∗, z⟩= ±λ⟨T(x) −x∗, z⟩+ λ2⟨T(z), z⟩.Dividing by λ and then letting λ →0 shows that ⟨T(x) −x∗, z⟩= 0 for all z ∈H, hencethat x∗= T(x).Exercise 1.6.

Prove that if T is maximal monotone, then T(x) is a convex set, for everyx ∈E.A major goal of these lectures is to examine the consequences of maximal monotonicitywith regards to questions of convexity of D(T) and R(T) (or convexity of their interiors orclosures). Other basic questions involve conditions under which D(T) = E or R(T) = E∗,or whether these sets might be dense in E or E∗, respectively.Perhaps the most fundamental result concerning monotone operators is the extensiontheorem of Debrunner–Flor [D–F].

An easy consequence of the following version of theirtheorem states that if T is maximal monotone and if its range is contained in a weak*compact convex set C, then D(T) = E. (Given x0 ∈E, let φ (below) be the constantmapping φ(x∗) = x0 for all x∗∈C, by the lemma, there exists x∗0 ∈C such that {(x0, x∗0)}∪G(T) is monotone; by maximality, x0 ∈D(T). )Lemma 1.7 (Debrunner–Flor).

Suppose that C is a weak* compact convex subset ofE∗, that φ: C →E is weak* to norm continuous and that M ⊂E × C is a monotone set.Then there exists x∗0 ∈C such that {(φ(x∗0), x∗0)} ∪M is a monotone set.Proof. For each element (y, y∗) ∈M letU(y, y∗) = {x∗∈C: ⟨x∗−y∗, φ(x∗) −y⟩< 0}.Since x∗→⟨x∗−y∗, φ(x∗) −y⟩is weak* continuous on the bounded set C, each ofthese sets is relatively weak* open.If the conclusion of the lemma fails, then C =S{U(y, y∗): (y, y∗) ∈M}.

By compactness, there must exist (y1, y∗1), (y2, y∗2), ..., (yn, y∗n)in M such that C = Sni=1{U(yi, y∗i )}. Let β1, β2, .

. ., βn be a partition of unity subordinateto this cover of C; that is, each βi is weak* continuous on C, 0 ≤βi ≤1, Σβi = 1 and{x∗∈C: βi(x∗) > 0} ⊂U(yi, y∗i ) for each i.

Let K = co{y∗i } ⊂C and define the weak*continuous map p of K into itself byp(x∗) = Σβi(x∗)y∗i ,x∗∈K.4

Note that K is a finite dimensional compact convex set which (since the weak* topologyis the same as the norm topology in finite dimensional spaces) is homeomorphic to a finitedimensional ball. Thus, the Brouwer fixed–point theorem is applicable.

(See [Fr] for severalproofs of the latter.) It follows that there exists z∗∈K such that p(z∗) = z∗.

We thereforehave0 = ⟨p(z∗) −z∗, Σβj(z∗)(yj −φ(z∗))⟩= ⟨Σβi(z∗)(y∗i −z∗), Σβj(z∗)(yj −φ(z∗))⟩== Σi,jβi(z∗)βj(z∗)⟨y∗i −z∗, yj −φ(z∗)⟩.Define αij = ⟨y∗i −z∗, yj −φ(z∗)⟩. It is straightforward to verify thatαij + αji = αii + αjj + ⟨y∗i −y∗j , yj −yi⟩≤αii + αjj,the inequality following from the monotonicity of M.Next, to simplify notation, letβi(z∗) = βi.

Note that for all i, j,βiβjαij + βjβiαji = βiβj(αij + αji2) + βjβi(αij + αji2).It follows that0 = Σβiβjαij = Σβiβj(αij + αji2) ≤Σβiβj(αii + αjj2).We claim that this inequality implies that βiβj = 0 for all i, j. Indeed, for every pair i, jsuch that βiβj > 0 we must have z∗∈U(yi, y∗i )∩U(yj, y∗j ), hence both αii < 0 and αjj < 0so that Σβiβj( αii+αjj2) < 0, a contradiction.

We conclude that βi ≡βi(z∗) = 0 for all i,an impossibility, since Σβi(z∗) = 1.Definition 1.8. A set–valued mapping T: E →2E∗is said to be locally bounded at thepoint x ∈E provided there exists a neighborhood U of x such that T(U) is a bounded set.Note that this does not require that the point x actually be in D(T).

Thus, it is true (butnot interesting) that T is locally bounded at each point of E \ D(T).There are at least four proofs (all based on the Baire category theorem) that a maximalmonotone operator T is locally bounded at the interior points of D(T). (See the discussionin [Ph, Sec.

2].) The first one was by Rockafellar [Ro2], who showed that much of it canbe carried out in locally convex spaces.

His proof is the longest one, but it has the greatadvantage of simultaneously proving that the interior of D(T) is convex. By specializinghis proof to Banach spaces (below), it is possible to shorten it considerably.

We first makesome simple general observations about an arbitrary subset D ⊂E, its convex hull coDand its interior int D.(i) Always, D ⊂coD, so int D ⊂int(coD). Suppose it is shown that int(coD) ⊂D.

Sinceint(coD) is open, it is therefore a subset of int D and hence int(coD) = int D, showingthat int D is convex. (None of these assertions assume that int D is nonempty!

)5

(ii) However, if int(coD) is nonempty and contained in D (so that int(coD) = int D), thenD = int(coD), hence is convex. [This follows from the fact that for any convex set C withinterior we have C ⊂int C, hence D ⊂coD ⊂int(coD) therefore D ⊂int(coD) = int D ⊂D.

](iii) Another useful elementary fact is the following: If {Cn} is an increasing sequenceof closed convex sets having nonempty interior, then int S Cn ⊂S int Cn (and, in fact,equality holds). [Here is a proof, the reader might have a simpler one: If x ∈int S Cn,then it is in some Cn, hence in the closure of S int Cn.

The latter is an open convex set, soif it doesn’t contain x, there must exist a closed half–space H supporting its closure at x.Thus, int Cn ⊂H for each n, therefore Cn = int Cn ⊂H. This leads to the contradictionthat x ∈int S Cn ⊂int H.]Notation.

We will denote the closed unit ball in E [resp. in E∗] by B [resp.

B∗]. Thus,if r > 0, say, thenrB∗= {x∗∈E∗: ∥x∗∥≤r}.Theorem 1.9 (Rockafellar).

Suppose that T is maximal monotone and that int coD(T)is nonempty. Then int D(T) = int coD(T) (so int D(T) is convex) and T is locally boundedat each point of int D(T).

Moreover, D(T) = int D(T), hence it is also convex.Proof. Let C = int coD(T).

For each n ≥1 letSn = {x ∈nB: T(x) ∩nB∗̸= ∅}.Then Sn ⊂Sn+1 and D(T) = S Sn ⊂S coSn. Since the Sn’s are increasing, this lastunion is convex, so it contains coD(T) and therefore contains C. As an open subset of aBanach space, C has the Baire property and therefore there exists an integer n0 such thatthe closure (relative to C) of C ∩coSn has nonempty interior for all n ≥n0.

In particular,the larger set int(coSn) is nonempty for each such n. We haveint coD(T) ⊂int[n≥n0coSn ⊂[n≥n0int(coSn),the last inclusion being a special case of the elementary fact (iii) described above. Wewill show two things; first, that T is locally bounded at any point in each set int(coSn)(n ≥n0) and second, that each such point is in D(T) (which, as noted in (i) above, willimply that int D(T) is convex).

For the first step, then, suppose that x0 ∈int(coSn)(for a fixed n ≥n0). Assume without loss of generality that n is sufficiently large thatR(T) ∩nB∗̸= ∅and for each m ≥n letMm = {(u, u∗) ∈E × E∗: u ∈D(T) and u∗∈T(u) ∩mB∗};this is a nonempty monotone subset of E × mB∗.

For each m ≥n and x ∈E defineAm(x) = {x∗∈E∗: ⟨x∗−u∗, x −u⟩≥0∀u ∈D(T) and u∗∈T(u) ∩mB∗}.6

Since Am(x) is the set of all x∗such that (x, x∗) is monotonically related to Mm, theDebrunner–Flor Lemma 1.7 guarantees that it is nonempty. For each such m and everyx ∈E we have Am+1(x) ⊂Am(x) and T(x) ⊂Am(x).

Moreover, as the intersectionof weak*–closed half–spaces, each Am(x) is weak* closed. Suppose that u∗∈nB∗; thenSn ⊂{x ∈E: |⟨u∗, x⟩| ≤n2} and therefore coSn is contained in the same set.Byhypothesis, there exists ǫ > 0 such that x0 + 2ǫB ⊂coSn.

Now, choose any x ∈x0 + ǫBand x∗∈An(x). For all u ∈Sn and u∗∈T(u) ∩nB∗we must have⟨x∗, u −x⟩≤⟨u∗, u −x⟩≤2n2,which shows that Sn ⊂{u ∈E: ⟨x∗, u −x⟩≤2n2} hence coSn is contained in this same(closed and convex) set.

Thus,x0 + 2ǫB ⊂coSn ⊂{u ∈E: ⟨x∗, u −x⟩≤2n2}.Suppose that ∥v∥≤ǫ, so that x + v ∈x0 + 2ǫB ⊂coSn.We then have ⟨x∗, v⟩=⟨x∗, (x + v) −x⟩≤2n2. Thus, ǫ∥x∗∥= sup{⟨x∗, v⟩: ∥v∥≤ǫ} ≤2n2, which implies that∥x∗∥≤2n2/ǫ.

We have shown, then, that if x ∈x0+ǫB and x∗∈An(x), then x∗∈2n2ǫ B∗.SinceT(x0 + ǫB) =[{T(x): x ∈x0 + ǫB} ⊂[{An(x): x ∈x0 + ǫB} ⊂(2n2/ǫ)B∗,we see that T is locally bounded at x0. (Note that this part of the proof did not requirethat T be maximal.

)To show that x0 ∈D(T), note that An(x0) ⊂(2n2/ǫ)B∗and is weak* closed, henceis weak* compact. For m ≥n we have Am(x0) ⊂An(x0) and therefore the sequence{Am(x0)} is a decreasing family of nonempty weak* compact sets.

Let x∗0 ∈Tm≥n Am(x0).If u ∈D(T) and u∗∈T(u), then ∥u∗∥≤m for some m ≥n. By definition of Am(x0) wehave ⟨x∗0 −u∗, x0 −u⟩≥0.

By maximality of T, this implies that x∗0 ∈T(x0) and thereforex0 ∈D(T).Corollary 1.10. Suppose that E is reflexive and that T: E →2E∗is maximal monotone.Then int R(T) is convex.

If int R(T) is nonempty, then R(T) is convex.Proof.Since T −1 is a maximal monotone operator from E∗to 2E, by the previoustheorem, int D(T −1) ≡int R(T) is convex.That the first part of this corollary can fail in a nonreflexive space will be shown byExample 2.21 (below).Definition 1.11. A subset A ⊂E (not necessarily convex) which contains the origin issaid to be absorbing if E = ∪{λA : λ > 0}.

Equivalently, A is absorbing if for each x ∈Ethere exists t > 0 such that tx ∈A. A point x ∈A is called an absorbing point of A if thetranslate A −x is absorbing.It is obvious that any interior point of a set is an absorbing point.

If A1 is the union ofthe unit sphere and {0}, then A1 is absorbing, even though it has empty interior. A proofof the following theorem of J. Borwein and S. Fitzpatrick [B–F] may be found in [Ph].7

Theorem 1.12. Suppose that T: E →2E∗is monotone and that x ∈D(T).

If x is anabsorbing point of D(T) (in particular, if x ∈int D(T)), then T is locally bounded at x.Note that the foregoing result does not require that D(T) be convex nor that T be maximal.There are trivial examples which show that 0 can be an absorbing point of D(T) but notan interior point (for instance, let T be the restriction of the duality mapping J to the setA1 defined above). Even if D(T) is convex and T is maximal monotone, D(T) can haveempty interior, as shown by the following example.

(In this example, T is an unboundedlinear operator, hence it is not locally bounded at any point and therefore D(T) has noabsorbing points. )Example 1.13.

In the Hilbert space ℓ2 let D = {x = (xn) ∈ℓ2: (2nxn) ∈ℓ2} and defineTx = (2nxn), x ∈D. Then D(T) = D is a proper dense linear subspace of ℓ2 and T is apositive operator, hence – by Example 1.5 (b) – it is maximal monotone.It is conceivable that for a maximal monotone T, any absorbing point of D(T) isactually an interior point.

That this is true if D(T) is assumed to be convex is shown bycombining Theorem 1.12 with the following result.Theorem 1.14 (Libor Vesel´y). Suppose that T is maximal monotone and that D(T)is convex.

If x ∈D(T) and T is locally bounded at x, then x ∈int D(T).Proof. The first step doesn’t use the convexity hypothesis: Suppose that T is a maximalmonotone operator which is locally bounded at the point x ∈D(T); then x ∈D(T).

Indeed,by hypothesis, there exists a neighborhood U of x such that T(U) is a bounded set.Choose a sequence {xn} ⊂D(T) ∩U such that xn →x and choose x∗n ∈T(xn). Byweak* compactness of bounded subsets of E∗there exists a subnet of {(xn, x∗n)} – call it{(xα, x∗α)} – and x∗∈E∗such that x∗α →x∗(weak*).

It follows that for all (y, y∗) ∈G(T),⟨x∗−y∗, x −y⟩= limα ⟨x∗α −y∗, xα −y⟩≥0;by maximal monotonicity, x∗∈T(x), so x ∈D(T). Next, if x is in the boundary ofthe closed convex set D(T), then T is not locally bounded at x: Suppose there were aneighborhood U of x such that T(U) were bounded.

By the Bishop–Phelps theorem therewould exist a point z ∈U ∩D(T) and a nonzero element w∗∈E∗which supported D(T)at z; that is, ⟨w∗, z⟩= sup ⟨z∗, D(T)⟩. Now, T would also be locally bounded at z ∈U,so by the first step, z ∈D(T) and we could choose z∗∈T(z).

For any (y, y∗) ∈G(T) andany λ ≥0 we would have⟨z∗+ λw∗−y∗, z −y⟩= ⟨z∗−y∗, z −y⟩+ λ⟨w∗, z −y⟩≥0.By the maximality of T this would imply that z∗+ λw∗∈T(z) for each λ ≥0, whichshows that T(z) is not bounded, a contradiction. Since x /∈bdry D(T), it must be inint D(T).

By local boundedness, we can choose an open set U such that x ∈U ⊂int D(T)and T(U) is bounded. Thus, T is locally bounded at every point of U, which – by the firststep proved above – implies that U ⊂D(T) and therefore x ∈int D(T).Note that this result, combined with Theorem 1.9, implies that for any maximal monotoneT, if int D(T) is nonempty, then it is precisely the set of points in D(T) where T is locallybounded.

It remains open as to what happens if int D(T) is empty and D(T) is not convex.8

Exercise 1.15. Prove that if T is maximal monotone, then for all x ∈int D(T) the setT(x) is weak* compact and convex.Definition 1.16.

Let X and Y be Hausdorffspaces and suppose that T: X →2Y is aset–valued mapping. We say that T is upper semicontinuous at the point x ∈X if thefollowing holds: For every open set U ⊂Y such that T(x) ⊂U there exists an open subsetV of X such that x ∈V and T(V ) ⊂U.

Upper semicontinuity on a set is defined in theobvious way.Exercise 1.17. Prove that if T: E →2E∗is maximal monotone, then it is norm–to–weak*upper semicontinuous on int D(T).The following “fixed–point” result, which will be useful to us in Section 3, illus-trates both the utility of the upper semicontinuity property as well as the strength of theDebrunner–Flor Lemma 1.7.

It is a special case of Theorem 4 of [Bro].Lemma 1.18. Suppose that E is a reflexive Banach space and that K is a nonemptycompact convex subset of E. Let R: K →2K be an upper semicontinuous mapping suchthat R(u) is nonempty, closed and convex, for each u ∈K.

Then there exists u0 ∈K suchthat u0 ∈R(u0).Proof. Suppose there were no such point; then 0 /∈u −R(u) for each u ∈K.

By theseparation theorem applied to the compact convex set u −R(u) and 0, for each u therewould exist x∗∈E∗, ∥x∗∥= 1, and δ > 0 such that ⟨x∗, v⟩> δ for each v ∈u −R(u). Foreach x∗∈E∗, defineW(x∗) = {u ∈K: ⟨x∗, v⟩> 0∀v ∈u −R(u)}.For each ∥x∗∥= 1 let U(x∗) = {v ∈E: ⟨x∗, v⟩> 0}, so u ∈W(x∗) if and only if u ∈Kand u −R(u) ⊂U(x∗).

Now, if u ∈K, then our supposition implies that there exists∥x∗∥= 1 and δ > 0 such that u −R(u) + δB ⊂U(x∗). Upper semicontinuity of R (henceof −R) at u implies that for some 0 < ǫ < δ/2 we will have −R(y) ⊂−R(u)+ δ2B whenevery ∈(u + ǫB) ∩K.

It follows that for all such y, we havey −R(y) ⊂u + δ2B −R(u) + δ2B ⊂U(x∗),that is, (u + ǫB) ∩K ⊂W(x∗). This shows that every point u ∈K is in the interior ofsome W(x∗), so that the sets {int W(x∗)} form an open cover of K. As in the proof ofLemma 1.7, there is a finite subcover {int W(x∗j)}nj=1 of K and a continuous partition ofunity {β1, β2, .

. ., βn} subordinate to this cover.

Definer(x) =Xβj(x)x∗j,x ∈K.This is a continuous map from K into E∗and for all u ∈K and v ∈u −R(u) wehave ⟨r(u), v⟩= P βj(u)⟨x∗j, v⟩> 0 (since βj(u) > 0 implies that u ∈W(x∗j) hence that⟨x∗j, v⟩> 0). In Lemma 1.7 let φ = −r, reverse the roles of E and E∗(reflexivity permits9

this) and let M be the monotone set K×{0} to obtain u0 ∈K such that −⟨r(u0), u0−u⟩≥0for all u ∈K. In particular, this is true if u = v0 where v0 is any element of R(u0); thatis, the element v ≡u0 −v0 ∈u0 −R(u0) satisfies ⟨r(u0), v⟩≤0, in contradiction to thefact that ⟨r(u0), v⟩> 0, completing the proof.2.

Subdifferentials of Convex FunctionsThis section introduces what is perhaps the most basic class of maximal monotoneoperators. We assume that the reader has some familiarity with real–valued convex func-tions.

(A good reference for their elementary properties is [R–V].) Their use in optimizationand convex analysis is most simply handled by introducing the seeming complication ofadmitting extended real-valued functions, that is, functions with values in R ∪{∞}.Definition 2.1.

Let X be a Hausdorffspace and let f: X →R ∪{∞}. The effectivedomain of f is the set dom(f) = {x ∈X: f(x) < ∞}.

Recall that f is lower semicontinuousprovided {x ∈X: f(x) ≤r} is closed in X for every r ∈R. This is equivalent to sayingthat the epigraph of fepi(f) = {(x, r) ∈X × R: r ≥f(x)}is closed in X × R. Equivalently, f is lower semicontinuous providedf(x) ≤lim inf f(xα)whenever x ∈X and (xα) is a net in X converging to x.We say that f is proper ifdom(f) ̸= ∅.Note that if f is defined on a Banach space E and is convex, then so is dom(f).

Also,a function f is convex if and only if epi(f) is convex.This last fact is important; itimplies that certain properties of lower semicontinuous convex functions can be deducedfrom properties of these (rather special) closed convex subsets of E × R. One can viewthis as saying that the study of lower semicontinuous convex functions is a special case ofthe study of closed convex sets.Examples 2.2. (a) Let C be a nonempty convex subset of E; then the indicator functionδC, defined by δC(x) = 0 if x ∈C,= ∞otherwise, is a proper convex function which islower semicontinuous if and only if C is closed.This example is one reason for introducing extended real-valued functions, since it makesit possible to deduce certain properties of a closed convex set from properties of its lowersemicontinuous convex indicator function.Thus, one can cite this example to supportthe view that the study of closed convex sets is a special case of the study of lower semi-continuous convex functions.

It’s all a matter of which approach is more convenient, thegeometrical or the analytical. It is useful to be able to switch easily from one to the other.

(b) Let A be any nonempty subset of E∗such that the weak* closed convex hull of Ais not all of E∗(or, more simply, let A be a weak* closed convex proper subset of E∗) anddefine the support function σA of A byσA(x) = sup{⟨x∗, x⟩: x∗∈A},x ∈E.10

This is easily seen to be a proper lower semicontinuous convex function. (c) If f is a continuous convex function defined on a nonempty closed convex setC, extend f to be ∞at the points of E \ C; the resulting function is a proper lowersemicontinuous convex function.The next proposition uses completeness of E to describe a set where a lower semicon-tinuous convex function is necessarily continuous.Proposition 2.3.

Suppose that f is a proper lower semicontinuous convex function on aBanach space E and that D = int dom(f) is nonempty; then f is continuous on D.Proof. We need only show that f is locally bounded in D, since this implies that it islocally Lipschitzian in D (see [Ph, Prop.

1.6]). First, note that if f is bounded above (byM, say) in B(x; δ) ⊂D for some δ > 0, then it is bounded below in B(x; δ).

Indeed, if yis in B(x; δ), then so is 2x −y andf(x) ≤12[f(y) + f(2x −y)] ≤12[f(y) + M]so f(y) ≥2f(x) −M for all y ∈B(x; δ).Thus, to show that f is locally boundedin D, it suffices to show that it is locally bounded above in D.For each n ≥1, letDn = {x ∈D : f(x) ≤n}. The sets Dn are closed and D = ∪Dn; since D is a Baire space,for some n we must have U ≡int Dn nonempty.

We know that f is bounded above byn in U; without loss of generality, we can assume that B(0; δ) ⊂U for some δ > 0. If yis in D, with y ̸= 0, then there exists µ > 1 such that z = µy ∈D and hence (letting0 < λ = µ−1 < 1), the setV = λz + (1 −λ)B(0; δ) = y + (1 −λ)B(0; δ)is a neighborhood of y in D. For any point v = (1 −λ)x + λz ∈V(wherex ∈B(0; δ))we havef(v) ≤(1 −λ)n + λf(z),so f is bounded above in V and the proof is complete.Examples 2.4.

(a) The function f defined by f(x) = 1/x on (0, ∞), f(x) = ∞on (−∞, 0] shows thatf can be continuous at a boundary point x of dom(f) where f(x) = ∞. (Recall that theneighborhoods of ∞in (−∞, ∞] are all the sets (a, ∞], a ∈R.

)(b) Suppose that C is nonempty closed and convex; then the lower semicontinuousconvex indicator function δC is continuous at x ∈C if and only if x ∈int C. Thus, ifint C = ∅, then δC is not continuous at any point of C = dom(δC).Definition 2.5. Recall that if E is a Banach space, then so is E×R, under any norm whichrestricts to give the original topology on the subspace E, for instance, ∥(x, r)∥= ∥x∥+|r|.Recall, also, that (E × R)∗can be identified with E∗× R, using the pairing⟨(x∗, r∗), (x, r)⟩= ⟨x∗, x⟩+ r∗· r.11

Remark 2.6. If a proper lower semicontinuous convex function f is continuous at somepoint x0 ∈dom(f), then dom(f) has nonempty interior and epi(f) has nonempty interiorin E × R.(Indeed, f(x) = ∞outside of dom(f), so x0 cannot be a boundary pointof the latter.

Moreover, there exists an open neighborhood U of x0 in dom(f) in whichf(x) < f(x0) + 1, so the open product set U × {r : r > f(x0) + 1} is contained in epi(f). )Definition 2.7.

If x ∈dom(f), define the subdifferential mapping ∂f by∂f(x) = {x∗∈E∗: ⟨x∗, y −x⟩≤f(y) −f(x) for all y ∈E}= {x∗∈E∗: ⟨x∗, y⟩≤f(x + y) −f(x) for all y ∈E},while ∂f(x) = ∅if x ∈E \ dom(f). It may also be empty at points of dom(f), as shownin the first example below.It is easy to see that ∂f is a monotone operator: If x∗∈∂f(x) and y∗∈∂f(y), then⟨x∗, y −x⟩≤f(y) −f(x) and −⟨y∗, y −x⟩= ⟨y∗, x −y⟩≤f(x) −f(y);now add these two inequalities.It is not obvious that ∂f is maximal monotone; in fact, it is not even obvious that it isnontrivial (i.e., that D(∂f) ̸= ∅.) Much of the rest of this section is devoted to establishingthese properties.Definition 2.8.

If x ∈dom(f) we define its right–hand directional derivative d+f(x) byd+f(x)(y) = limt→0+ t−1[f(x + ty) −f(x)],y ∈E.It follows from the convexity of f that this limit always exists (see [Ph]).Note thatd+f(x)(y) = ∞if x+ty ∈E \dom(f) for all t > 0. (It is also possible to have d+f(x)(y) =−∞; consider, for instance, d+f(0)(1) when f(x) = −x1/2 for x ≥0, = ∞elsewhere.) Wehave the following important relationship: For any point x ∈dom(f),x∗∈∂f(x) if and only if ⟨x∗, y⟩≤d+f(x)(y) for all y ∈E.It follows from this that for the example given above (f(x) = −√x for x ≥0), it mustbe true that ∂f(0) = ∅.

In the first example below, one sees that it is possible to have∂f(x) = ∅for a dense set of points x ∈dom(f).Examples 2.9. (a) Let C be the closed (in fact, compact) convex subset of ℓ2 defined byC = {x ∈ℓ2 : |xn| ≤2−n, n = 1, 2, 3, .

. .

}and define f on C by f(x) = Σ[−(2−n + xn)1/2]. Since each of the functions x →−(2−n + xn)1/2 is continuous, convex and bounded in absolute value by 2(−n+1)/2, the12

series converges uniformly, so f is continuous and convex. We claim that ∂f(x) = ∅forany x ∈C such that xn > −2−n for infinitely many n. Indeed, let en denote the n-th unitvector in ℓ2.

If x∗∈∂f(x) (so that, as noted above, x∗≤d+f(x)), then for all n suchthat xn > −2−n, we have−∥x∗∥≤⟨x∗, en⟩≤d+f(x)(en) = −(1/2)(2−n + xn)−1/2,an impossibility which implies that ∂f(x) = ∅. Note that if we make the usual extension(setting f(x) = ∞for x ∈ℓ2 \ C), then f is lower semicontinuous, but not continuous atany point of C (= bdry C).

(b) Let C be a nonempty closed convex subset of E; then for any x ∈C, the sub-differential ∂δC(x) of the indicator function δC is the cone with vertex 0 of all x∗∈E∗which “support” C at x, that is, which satisfy⟨x∗, x⟩= sup{⟨x∗, y⟩: y ∈C} ≡σC(x∗). (Indeed, x∗∈∂δC(x) if and only if ⟨x∗, y −x⟩≤δC(y) −δC(x), while x∗attains itssupremum on C at x if and only if the left hand side of this latter inequality is at most 0,while the right side is always greater or equal to 0.

)The following notion of an approximate subdifferential is useful in many parts of convexanalysis.Definition 2.10. Let f be a proper convex lower semicontinuous function and supposex ∈dom(f).

For any ǫ > 0 define the ǫ–subdifferential ∂ǫf(x) by∂ǫf(x) = {x∗: ⟨x∗, y⟩≤f(x + y) −f(x) + ǫ for all y ∈E}.It follows easily from the definition that ∂ǫf(x) convex and weak* closed. That fact that itis nonempty for every x ∈dom(f) follows from the convexity of epi(f) and the separationtheorem (in E × R) (see [Ph; Prop.

3.14]).The basic maximality technique which was used in proving the Bishop–Phelps theorem[Ph] was applied in E×R by Brøndsted and Rockafellar to prove the following fundamentallemma, which shows, among other things, that ∂f is nontrivial.Lemma 2.11. Suppose that f is a convex proper lower semicontinuous function on theBanach space E. Then given any point x0 ∈dom(f), ǫ > 0, λ > 0 and any functionalx∗0 ∈∂ǫf(x0), there exist x ∈dom(f) and x∗∈E∗such thatx∗∈∂f(x),∥x −x0∥≤ǫ/λand∥x∗−x∗0∥≤λ.In particular, the domain of ∂f is dense in dom(f), so D(∂f) = dom(f) is convex.The next result can be looked at in two ways.On the one hand, it is simply averification that (under a certain hypothesis), the subdifferential operation is additive.

(Weuse the usual vector sum of sets.) On the other hand, it will be seen later as a verificationin a special case that the sum of maximal monotone operators is again maximal monotone.13

Theorem 2.12. Suppose that f and g are convex proper lower semicontinuous functionson the Banach space E and that there is a point in dom(f) ∩dom(g) where one of them,say f, is continuous.

Then∂(f + g)(x) = ∂f(x) + ∂g(x),x ∈D(∂f) ∩D(∂g).Remark. It is immediate from the definitions that for x ∈dom(f + g) (which isidentical to dom(f) ∩dom(g)), one must have∂f(x) + ∂g(x) ⊂∂(f + g)(x).This inclusion can be proper.

To see this, let E = R2, let f denote the indicator functionδC and let g = δL, where C is the epigraph of the quadratic function y = x2 and L isthe x-axis. Obviously, C and L intersect only at the origin 0 and it is easily verified that∂f(0) = R−e, where e is the vector (0, 1), and ∂g(0) = Re, while∂(f + g)(0) = R2 ̸= ∂f(0) + ∂g(0).Proof.

Suppose that x∗0 ∈∂(f +g)(x0). In order to simplify the argument, we can replacef and g by the functionsf1(x) = f(x + x0) −f(x0) −⟨x∗0, x⟩and g1(x) = g(x + x0) −g(x0),x ∈E;it is readily verified from the definitions that if x∗0 ∈∂(f + g)(x0), then 0 ∈∂(f1 + g1)(0)and if 0 ∈∂f1(0) + ∂g1(0), then x∗0 ∈∂f(x0) + ∂g(x0).

Without loss of generality, then,we assume that x0 = 0, x∗0 = 0, f(0) = 0 and g(0) = 0. We want to conclude that 0 is inthe sum ∂f(0) + ∂g(0), under the hypothesis that 0 ∈∂(f + g)(0).

This last means that(f + g)(x) ≥(f + g)(0) = 0 for all x ∈E. (2.1)We now apply the separation theorem in E × R to the two closed convex sets C1 = epi(f)and C2 = {(x, r): r ≤−g(x)}; this is possible because f has a point of continuity indom(f) ∩dom(g) and hence – recall Remark 2.6 – C1 has nonempty interior.

Moreover,it follows from (2.1) that C2 misses the interior of C1 = {(x, r): r > f(x)}. Since (0, 0) iscommon to both sets, it is contained in any separating hyperplane.

Thus, there exists afunctional (x∗, r∗) ∈E∗× R, (0, 0) ̸= (x∗, r∗), such that⟨x∗, x⟩+ r∗· r ≥0 if r ≥f(x) and ⟨x∗, x⟩+ r∗· r ≤0 if r ≤−g(x).Since 1 > f(0) = 0 we see immediately that r∗≥0. To see that r∗̸= 0, (that is, thatthe separating hyperplane is not “vertical”), we argue by contradiction: If r∗= 0, then wemust have x∗̸= 0; also ⟨x∗, x⟩≥0 for all x ∈dom(f) and ⟨x∗, x⟩≤0 for all x ∈dom(g).This says that x∗separates these two sets.

This is impossible; by the continuity hypothesis,their intersection contains an interior point of dom(f). Without loss of generality, then,we can assume that r∗= 1 and hence, for any x ∈E,⟨−x∗, x −0⟩≤f(x) −f(0) and ⟨x∗, x −0⟩≤g(x) −g(0),that is, 0 = −x∗+ x∗∈∂f(0) + ∂g(0), which completes the proof.The next two lemmas lead easily to S. Simons’ recent proof [Si1,3] of Rockafellar’s maximalmonotonicity theorem for subdifferentials.14

Lemma 2.13. Suppose that f is a lower semicontinuous proper convex function on E. Ifα, β > 0, x0 ∈E and f(x0) < infEf + αβ, then there exist x ∈E and x∗∈∂f(x) suchthat ∥x −x0∥< β and ∥x∗∥< α.Proof.

Choose ǫ > 0 such that f(x0) −infEf < ǫ < αβ and then choose λ such thatǫ/β < λ < α. It follows that 0 ∈∂ǫf(x0) so by the Brøndsted–Rockafellar Lemma 2.11,there exist x ∈dom(f) and x∗∈∂f(x) such that ∥x∗∥≤λ < α and ∥x −x0∥≤ǫ/λ < β.Lemma 2.14.

With f as in the previous lemma, suppose that x ∈E (not necessarily indom(f)) and that infEf < f(x). Then there exist z ∈dom(f) and z∗∈∂f(z) such thatf(z) < f(x) and ⟨z∗, x −z⟩> 0.Proof.

Fix λ ∈R such that infEf < λ < f(x) and letK = supy∈E,y̸=xλ −f(y)∥y −x∥.We first show that 0 < K < ∞. To that end, let F = {y ∈E: f(y) ≤λ}, so F is closed,nonempty and x /∈F.

Since dom(f) ̸= ∅, one can apply the separation theorem in E × Rto find u∗∈E∗and r ∈R such that f ≥u∗+ r. Suppose that y ∈E and that y ̸= x. Ify ∈F, thenλ −f(y) ≤λ −⟨u∗, y⟩−r ≤|λ −⟨u∗, x⟩−r| + ⟨u∗, x −y⟩henceλ −f(y)∥y −x∥≤|λ −⟨u∗, x⟩−r|dist(x, F)+ ∥u∗∥.If y /∈F, then λ−f(y)∥y−x∥< 0. In either case, there is an upper bound for λ−f(y)∥y−x∥, so K < ∞.To see that K > 0, pick any y ∈E such that f(y) < λ.

Since λ < f(x), we have y ̸= xand K ≥λ−f(y)∥y−x∥> 0.Suppose, now, that 0 < ǫ < 1, so that (1 −ǫ)K < K and hence, by definition of K,there exists x0 ∈E such that x0 ̸= x andλ −f(x0)∥x0 −x∥> (1 −ǫ)K.For z ∈E, let N(z) = K∥z −x∥; we have shown that (1 −ǫ)N(x0) + f(x0) < λ, thatis, (N + f)(x0) < λ + ǫN(y). We claim that λ ≤infE(N + f) Indeed, if z = x, then wehave λ < f(x) = (N + f)(z), while if z ̸= x, then λ−f(z)∥z−x∥≤K, from which it follows thatλ ≤(N + f)(z).

Thus, we have shown that there is a point x0 ∈E, x0 ̸= x, such that(N + f)(x0) < infE(N + f) + ǫK∥x0 −x∥.We now apply Lemma 2.13 to N + f, with β = ∥x0 −x∥and α = ǫK. Thus, there existsz ∈dom(N + f) ≡dom(f) and w∗∈∂(N + f)(z) such that ∥z −x0∥< ∥x −x0∥and∥w∗∥< ǫK.

It follows that ∥z −x∥> 0. From the sum formula (Theorem 2.12),∂(N + f)(z) = ∂N(z) + ∂f(z),15

so there exist y∗∈∂N(z) and z∗∈∂f(z) such that w∗= y∗+ z∗. Since y∗∈∂N(z), wemust have ⟨y∗, z −x⟩≥N(z) −N(x) = K∥z −x∥.

Thus⟨z∗, x −z⟩= ⟨y∗, z −x⟩+ ⟨w∗, x −z⟩≥K∥z −x∥−∥w∗∥· ∥x −z∥>> (1 −ǫ)K∥z −x∥> 0.Since z∗∈∂f(z), we have f(x) ≥f(z) + ⟨z∗, x −z⟩> f(z), which completes the proof.Theorem 2.15 (Rockafellar). If f is a proper lower semicontinuous convex function ona Banach space E, then its subdifferential ∂f is a maximal monotone operator.Proof.

Suppose that x ∈E, that x∗∈E∗and that x∗/∈∂f(x). Thus, 0 /∈∂(f −x∗)(x),which implies that infE(f −x∗) < (f −x∗)(x).

By Lemma 2.14 there exists z ∈dom(f −x∗) ≡dom(f) and z∗∈∂(f −x∗)(z) such that ⟨z∗, z−x⟩< 0. Thus, there exists y∗∈∂f(z)such that z∗= y∗−x∗, so that ⟨y∗−x∗, z −x⟩< 0.Example 2.16.

In a Banach space E define j(x) = (1/2)∥x∥2; this is clearly continuousand convex. The monotone duality mapping J is actually the subdifferential of j, andhence is maximal monotone.Proof.

It is readily computed that d+j(x)(y) = ∥x∥·d+∥x∥(y). If x = 0, then d+j(0)(y) =0 for all y, hence is linear and therefore ∂j(0) = {0}.

Suppose, then, that x ̸= 0. We know(from the remark following Definition 2.8) that x∗∈∂j(x) if and only if x∗≤d+j(x),that is, if and only if ∥x∥−1x∗≤d+∥x∥, which is equivalent to y∗≡∥x∥−1x∗∈∂∥x∥, thatis, if and only if ⟨y∗, y −x⟩≤∥y∥−∥x∥for all y ∈E.

If, in this last inequality, we takey = x+z, ∥z∥≤1 and apply the triangle inequality, we conclude that ∥y∗∥≤1. If we takey = 0, we conclude that ∥x∥≤⟨y∗, x⟩≤∥y∗∥·∥x∥, so ∥y∗∥= 1 and ⟨y∗, x⟩= ∥x∥, which isequivalent to what we want to prove.

The converse is easy: If ∥y∗∥= 1 and ⟨y∗, x⟩= ∥x∥,then for all y in E, we necessarily have ⟨y∗, y −x⟩≤∥y∥−∥x∥, so y∗∈∂∥x∥.There are other interesting and useful properties of the duality mapping J.Forinstance, it is immediate from the original definition that it satisfies J(−x) = −J(x)and J(λx) = λJ(x) for λ > 0. Since it is the subdifferential of the function 12∥x∥2, it isnot surprising that it reflects properties of the norm.Proposition 2.17.

(a) The norm in E is Gateaux differentiable (at nonzero points) if and only if J issingle valued. (b) The mapping J is “one–to–one” (that is, J(x) ∩J(y) = ∅whenever x ̸= y) if andonly if the norm in E is strictly convex; that is, ∥x + y∥< 2 whenever ∥x∥= 1, ∥y∥= 1and x ̸= y.

(c) Surjectivity of J is equivalent to reflexivity of E.Proof. Parts (a) and (b) are straightforward exercises.

Part (c) is an easy consequence ofR. C. James’ deep result that a Banach space E is reflexive if (and only if) each functionalin E∗attains its supremum on the unit ball of E at some point.

Indeed, if J is surjective16

and x∗∈E∗, then there exists x ∈E such that x∗∈J(x), hence ⟨x∗,x∥x∥⟩= ∥x∗∥≡sup{⟨x∗, y⟩: ∥y∥≤1}, showing x∗attains its supremum on the unit ball atx∥x∥.Definition 2.18. If S and T are monotone operators and x ∈D(S) ∩D(T), we define(S + T)(x) = S(x) + T(x) ≡{x∗+ y∗: x∗∈S(x), y∗∈T(x)},while (S + T)(x) = ∅otherwise.It is immediate that S+T is also a monotone operator and that – by definition – D(S+T) =D(S) ∩D(T).The following theorem of Rockafellar [Ro1] is basic to many applications of maximalmonotone operators.Theorem 2.19 (Rockafellar).

Suppose that E is reflexive, that S and T are maximalmonotone operators on E and that D(T) ∩int D(S) ̸= ∅. Then S + T is maximal.We refer to [Ro1] for the proof of this theorem (see, also, [B-C-P]), which relies partly onthe fact that any reflexive space can be renormed so that both it and its dual norm areGateaux differentiable (at nonzero points) [D–G–Z].

Since maximal monotonicity does notdepend on which norm defines the topology of E, this guarantees that the duality mappingJ can be assumed to be single–valued, one–to–one and onto, a useful step in the proof.The situation is wide open in arbitrary Banach spaces.Problem 2.20. Suppose that E is a nonreflexive Banach space and that S and T aremaximal monotone operators such that D(T)∩int D(S) ̸= ∅; is S+T necessarily maximal?What about the special case when S is the subdifferential of the indicator function δC ofa closed convex set C for which int C ∩D(T) ̸= ∅?Remark.

The answer to the first question is affirmative when D(S) = E = D(T); thiswas pointed out to us by Martin Heisler. One can use the local boundedness of S and Tand weak* compactness to prove that the graph of G(S +T) is closed in the norm × weak*topology in E × E∗; Lemma 2.2 of [F-P1] then applies to show that S + T is maximal.Rockafellar [Ro1, Theorem 3] has shown that the answer to the second question is affirma-tive for certain single–valued monotone operators T.The answer to the first question is also affirmative whenever S = ∂f and T = ∂g, wheref and g are proper lower semicontinuous convex functions; this is a consequence of The-orem 2.12 (that the subdifferential of the sum of two convex functions is the sum oftheir subdifferentials).Indeed, if intD(T) ̸= ∅, then [since int dom(∂g) ⊂int dom(g)and g is continuous on int dom(g)] we conclude that g is continous at some point ofD(S) ∩D(T) ⊂dom(f) ∩dom(g), so Theorem 2.12 implies that S + T = ∂(f + g) andTheorem 2.15 implies that the latter is maximal monotone.

The remark following thestatement of Theorem 2.12 shows that, even in a two dimensional Banach space, maximal-ity of a sum can fail if D(S) ∩intD(T) is empty; in that example, the graph of ∂f + ∂g isa proper subset of the graph of the maximal monotone operator ∂(f + g).17

The foregoing discussion is an example of how subdifferentials fulfill their role as proto-types when considering general questions about maximal monotone operators. Whenevera property is valid for subdifferentials in arbitrary Banach spaces there is some hope that italso holds for all maximal monotone operators.

On the other hand, of course, if it fails forsubdifferentials on nonreflexive spaces, the situation is obviously hopeless. The followingexample illustrates this with respect to the first assertion in Corollary 1.10 (that int R(T)is convex when E is reflexive).Example 2.21 (Simon Fitzpatrick).

There exists a continuous convex function f onthe Banach space c0 such that the interior of R(∂f) is not convex.Proof. With the usual supremum norm on c0, define g(x) = ∥x∥and h(x) = ∥x −e1∥,where e1 = (1, 0, 0, .

. .

), and let f = g + h. Since g and h are continous and convex, wehave ∂f(x) = ∂g(x) + ∂h(x) for each x ∈c0. It is straightforward to compute that ∂g(x)is either B∗if x = 0, or is contained in the set F of all finitely nonzero sequences in ℓ1otherwise.

It is also easy to see that ∂h(x) = ∂g(x −e1) for each x. It follows (letting e∗1denote the corresponding element of ℓ1) that ∂f(0) = −e∗1 + B∗and ∂f(e1) = e∗1 + B∗,while ∂f(x) is contained in F if x ̸= 0, e1.

Since int R(∂f) ⊃intB∗± e∗1, if it were convex,it would contain 0 = 12(e∗1 −e∗1) and hence a neighborhood U of 0. But, for any λ > 0, theelement (0, λ22 , λ23 , λ24 , .

. .) is not in F, has distance from ±e∗1 equal to 1 + λ2 > 1 and (forsufficiently small λ) is in U.Another place one looks for prototypical properties is in Hilbert space (see, for in-stance, [Bre]), where things are made much easier by the fact that the duality mapping Jis replaced by the identity mapping.We conclude this section with a look at an additional property which characterizessubdifferentials within the class of maximal monotone operators.

Details may be found,for instance, in [Ph].Definition 2.22. A set-valued map T: E →2E∗is said to be n-cyclically monotoneprovidedX1≤k≤n⟨x∗k, xk −xk−1⟩≥0whenever n ≥2 and x0, x1, x2, .

. .xn ∈E, xn = x0, and x∗k ∈T(xk), k = 1, 2, 3, .

. ., n.We say that T is cyclically monotone if it is n-cyclically monotone for every n. Clearly, a2-cyclically monotone operator is monotone.Examples 2.23.

(a) The linear map in R2 defined by T(x1, x2) = (x2, −x1) is positive, hence maximalmonotone, but it is not 3-cyclically monotone: Look at the points (1, 1), (0, 1) and (1, 0). (b) Let f be a proper lower semicontinuous convex function; then ∂f is cyclicallymonotone.The final theorem of this section shows that this is the only such example [Ro3].18

Theorem 2.24 (Rockafellar). If T: E →2E∗is maximal monotone and cyclically mono-tone, with D(T) ̸= ∅, then there exists a proper convex lower semicontinuous function fon E such that T = ∂f.3.

Gossez’s monotone operators of type (D).In 1971, J.–P. Gossez [Go1] introduced the class of monotone operators of “dense type”in order to extend to nonreflexive spaces some of the basic known results about maximalmonotone operators on reflexive spaces.

He subsequently modified his definition [Go3,5] tothe one given below. One first identifies a Banach space E with its canonically embeddedimage bE in E∗∗.

Having done this, it is natural to consider the graph G(T) of a monotoneoperator T to be a subset of E∗∗× E∗.Definition 3.1. A monotone operator T: E →2E∗is said to of type (D) provided itsatisfies the following property: If (x∗∗, x∗) ∈E∗∗× E∗is monotonically related to G(T),then there exists a net (xα, x∗α) ∈G(T) such that xα →x∗∗in the σ(x∗∗, x∗) topology,(xα) is bounded and x∗α →x∗in norm.In using this definition it is convenient to extend T to a mapping T: E∗∗→2E∗as follows:We let T be the map whose graph G(T) ⊂E∗∗× E∗consists of all elements (x∗∗, x∗) ∈E∗∗× E∗which are monotonically related to G(T).The map T need not be monotone [Go4].

However, if T is monotone of type (D), then T ismaximal monotone. Indeed, suppose that (x∗∗, x∗) is monotonically related to G(T); it isthen obviously monotonically related to G(T), hence is in G(T).

It only remains to showthat T is monotone. Suppose that (x∗∗, x∗), (y∗∗, y∗) ∈G(T).

By hypothesis, there existsa net (xα, x∗α) ∈G(T) as above, hence0 ≤⟨y∗∗−ˆxα, y∗−x∗α⟩= ⟨y∗∗−x∗∗, y∗−x∗⟩+ ⟨x∗∗−ˆxα, y∗−x∗⟩+ ⟨y∗∗−ˆxα, x∗−x∗α⟩.Taking limits, we see that ⟨y∗∗−x∗∗, y∗−x∗⟩≥0.A word about terminology: When T is monotone of type (D), then any element of G(T)is the limit of a certain net in G(T), so we can consider the latter as being dense in theformer, hence Gossez’s earlier “dense type” and the later use of “type (D)”.Examples 3.2. (a) Define T: R\{0} →R by T(x) = 0 for all x ̸= 0.It is easily verified thatG(T) = R × {0} ⊂R2, hence is maximal monotone, while T is monotone of type (D) (butnot maximal).

(b) Suppose that E is reflexive. If T is maximal monotone, then T = T, hence T istrivially maximal monotone of type (D), that is, in reflexive spaces, the maximal monotoneoperators coincide with the maximal monotone operators of type (D).

(c) If f is a proper lower semicontinuous convex function on E, then ∂f is maximalmonotone of type (D) [Go1]. (This is not obvious.

In essence, it uses the first step ofRockafellar’s original proof [Ro3] of the maximal monotonicity of ∂f. )19

The fundamental fact about maximal monotone operators of type (D) is contained inTheorem 3.6 (below). In order to formulate it, we need to recall a few facts about convexfunctions and their Fenchel duals.Definition 3.3.

If f is a proper convex lower semicontinuous function on E, then thefunction f ∗defined on E∗byf ∗(x∗) = sup{⟨x∗, x⟩−f(x)},x∗∈E∗is proper, convex and weak* lower semicontinuous on E∗. One defines f ∗∗≡(f ∗)∗on E∗∗analogously.Exercises 3.4.

(a) If we let j(x) = 12∥x∥2, then j∗(x∗) = 12∥x∗∥2 and j∗∗(x∗∗) = 12∥x∗∗∥2. (b) If f is convex proper and lower semicontinuous, and ǫ ≥0, then ∂ǫf can becharacterized as follows: x∗∈∂ǫf(x) if and only if f(x) + f ∗(x∗) ≤⟨x∗, x⟩+ ǫ.

(Here, wetake ∂0f ≡∂f. )The following theorem, which is a very special case of a result of F. Browder [Bro,Theorem 10], provides a crucial step in proving Theorem 3.6 (below).Theorem 3.5.

Suppose that F is a finite dimensional Banach space with unit ball B andduality map J. Fix r > 0 and let A denote the restriction of −J to rB.

If T: rB →2F ∗is monotone, then there exists (x, x∗) ∈G(A) such that G(T) ∪{(x, x∗)} is monotone,that is, there exists x ∈rB and x∗∈−J(x) ⊂rB∗such that ⟨x∗−y∗, x −y⟩≥0 for all(y, y∗) ∈G(T).Proof. For each ǫ > 0 and finite subset G ⊂G(T), letHǫ,G = {(x, x∗) ∈G(A): ⟨x∗−y∗, x −y⟩≥−ǫ∀(y, y∗) ∈G}andH = {(x, x∗) ∈G(A): (x, x∗) is monotonically related to G(T)}.We will have proved the theorem if we show that the set H is nonempty.

It is straight-forward to verify that H = T{Hǫ,G: ǫ > 0,G finite, G ⊂G(T)}. Moreover, from thecompactness of rB × rB∗and the definition of J, it follows that each Hǫ,G is compact.The intersection of any finite number of these sets contains a set of the same form, so toshow that the family has the finite intersection property it suffices to show that each ofthem is nonempty; it will then follow by compactness that H is nonempty.

Fix ǫ and G.To show that Hǫ,G is nonempty, we define, for each x ∈rB,S(x) = {x∗∈rB∗: ⟨x∗−y∗, x −y⟩> −ǫ∀(y, y∗) ∈G}and for x∗∈rB∗,S−1(x∗) = {x ∈rB: ⟨x∗−y∗, x −y⟩> −ǫ∀(y, y∗) ∈G}.20

Each set S(x) is a finite intersection of relatively open subsets of rB∗, hence is relativelyopen and each set S−1(x∗) is convex. Moreover, each of the latter sets is nonempty: Thisis an application of Lemma 1.7, using the monotonicity of G, reversing the roles playedthere by E and E∗and letting φ: rB →rB∗be the constant map with value x∗.

Itfollows that the sets {S(x)} form an open cover of the compact set rB∗, hence thereexist {x1, x2, . .

., xn} ⊂rB such that rB∗= Snj=1 S(xj).As in the proof of Lemma1.7, there exists a partition of unity {β1, β2, . .

., βn} subordinate to this covering. Definethe continuous mapping p: rB∗→rB by p(x∗) = Pnj=1 βj(x∗)xj.

We claim that x∗∈S(p(x∗)) for all x∗∈rB∗. Indeed, for every j such that βj(x∗) > 0 we have x∗∈S(xj),that is, xj ∈S−1(x∗).

Since the latter is convex and since p(x∗) is a convex combinationof xj’s, it follows that p(x∗) ∈S−1(x∗), which is equivalent to x∗∈S(p(x∗)). Next, definethe set–valued mapping R: rB∗→2rB∗by R(x∗) = −J(p(x∗)).

Since p is continuous andJ is upper semicontinuous (Exercise 1.17 and Example 2.16), R is upper semicontinuous.Let K = rB∗; by Lemma 1.18 there exists x∗0 ∈rB∗such that x∗0 ∈R(x∗0), that is,x∗0 ∈−J(p(x∗0)) and, of course, x∗0 ∈S(p(x∗0)).Letting x0 = p(x∗0), this means thatx∗0 ∈(−J)(x0) ∩S(x0) ⊂Hǫ,G, which completes the proof.Note that since ∂j∗: E∗→2E∗∗, its inverse (∂j∗)−1 is a mapping from E∗∗to 2E∗.Theorem 3.6 (Gossez). If T is maximal monotone of type (D), then for all λ > 0,R(T + λ(∂j∗)−1) = E∗.Proof.

Since T is of type (D) if and only if the same is true of λ−1T −x∗for each λ > 0and x∗∈E∗, we need only show that 0 ∈R(T + (∂j∗)−1). Let F denote the directedfamily of all finite dimensional subspaces F ⊂E such that D(T)∩F ̸= ∅, partially orderedby inclusion.

For each such F, let iF : F →E denote the natural injection, with adjointi∗F : E∗→F ∗(the restriction mapping to F). Suppose, now that F ∈F and that r > 0.Apply Theorem 3.5 to F, F ∗, i∗F TiF and Kr = {x ∈F: ∥x∥≤r}, as follows: Let G be thegraph in Kr × F ∗of the restriction to Kr of the monotone operator i∗F TiF .

(We assumethat r is sufficiently large that Kr ∩D(T) ̸= ∅.) Let A: Kr →F ∗be −i∗F JiF .

Thus, thereexists an element of G(A) – call it (xF,r, −x∗F,r) – which is montonely related to G; thatis, ∥xF,r∥≤r, x∗F,r ∈i∗F J(xF,r) (this uses JiF = J in F) and ⟨−x∗F,r −y∗, xF,r −y⟩≥0whenever ∥y∥≤r and y∗∈i∗F TiF (y).Suppose, now, that both r0 and F0 are sufficiently large so that there exists y0 ∈F0with ∥y0∥≤r0 and y∗0 ∈i∗F0TiF0(y0). Fix F ⊃F0; for each r ≥r0 we will show that thefollowing set Hr is bounded: Define Hr to be the set of all (xF,r, x∗F,r) ∈F × F ∗such thatx∗F,r ∈i∗F J(xF,r) and ⟨−x∗F,r −y∗, xF,r −y⟩≥0 whenever ∥y∥≤r and y∗∈i∗F TiF (y).For any such (xF,r, x∗F,r) we have12∥xF,r∥2 + 12∥x∗F,r∥2 = ⟨x∗F,r, xF,r⟩≤⟨x∗F,r, y0⟩−⟨y∗0, xF,r⟩+ ⟨y∗0, y0⟩≤≤∥x∗F,r∥· ∥y0∥+ ∥y∗0∥· ∥xF,r∥+ ⟨y∗0, y0⟩.The subset of the plane where a positive quadratic function is dominated by a linearfunction is necessarily bounded, so there exists an upper bound on each of the sets {∥xF,r∥}21

and {∥x∗F,r∥}. That is, each of the sets Hr is bounded.

Moreover, each of them is closed (inthe product of the norm topologies), since i∗F JiF is easily seen to have closed graph andthe function (xF,r, x∗F,r) →⟨−x∗F,r −y∗, xF,r −y⟩is continuous, for each (y, y∗) ∈F × F ∗.Clearly, Hr ⊃Hr′ whenever r′ > r > 0, so for increasing r, the Hr’s form a decreasingfamily of nonempty compact sets and they therefore have nonempty intersection. Thisshows that there exists xF ∈F and x∗F ∈i∗F JxF such that⟨−x∗F −y∗, xF −y⟩≥0 whenever y ∈F, and y∗∈i∗F TiF (y).

(1)Note that any Hahn-Banach extension of x∗F from F to all of E is in J(xF ), so we canassume that x∗F ∈J(xF ) ⊂E∗. Since the nets (xF ) and (x∗F ) are bounded, and since wecan regard the xF ’s as elements of E∗∗, we see that there exists a subnet (call it (xF , x∗F ))in E∗∗× E∗converging to an element (x∗∗, x∗) ∈E∗∗× E∗in the σ(E∗∗, E∗) × σ(E∗, E)topology.

We want to show that (x∗∗, −x∗) is monotonically related to G(T) ⊂E∗∗× E∗,that is,⟨x∗∗−ˆy, −x∗−y∗⟩≥0 whenever (y, y∗) ∈G(T). (2)To see this, note that (using the weak* lower semicontinuity of both j∗and j∗∗)⟨x∗∗, x∗⟩≤j∗∗(x∗∗) + j∗(x∗) ≤lim inf[j(xF ) + j∗(x∗F )] = lim inf⟨x∗F , xF ⟩,(3)while (1) implies that, for all (y, y∗) ∈G(T),lim sup⟨x∗F , xF ⟩≤⟨y∗, y⟩+ ⟨x∗, y⟩−⟨x∗∗, y∗⟩;(4)together, these yield (2).

Since T is assumed to be of type (D), there exists a net (yα, −y∗α)in G(T) such that (yα) is bounded, converges to x∗∗in the σ(E∗∗, E∗) topology and∥y∗α −x∗∥→0. This fact, applied to (4), shows thatlim sup⟨x∗F , xF ⟩≤⟨x∗∗, x∗⟩+ ⟨x∗∗, x∗⟩−⟨x∗∗, x∗⟩= ⟨x∗∗, x∗⟩.Now, from (3),⟨x∗∗, x∗⟩≤j∗∗(x∗∗) + j∗(x∗) ≤lim inf⟨x∗F , xF ⟩≤lim sup⟨x∗F , xF ⟩≤⟨x∗∗, x∗⟩,which shows that x∗∗∈∂j∗(x∗).Thus, −x∗∈T(x∗∗) and x∗∈∂(j∗)−1(x∗∗), whichcompletes the proof.Corollary 3.7.

If E is reflexive and T: E →2E∗is maximal monotone, then R(T +λJ) =E∗for every λ > 0.Proof.It follows directly from the definitions that for a reflexive space E, one has∂(j∗)−1 = J and, as has been noted earlier, T = T, so the corollary is immediate.Theorem 3.8. If T is of maximal monotone of type (D), then R(T) is convex.Before proving this, we need the fact that if T is maximal monotone of type (D) andx∗∈coR(T), then there exists x ∈E such thatsup(y∗∗,y∗)∈G(T)⟨y∗∗−ˆx, x∗−y∗⟩< ∞.This follows in a straightforward way from the following lemma and the definition of type(D).

Note that if T is of type (D), then R(T) ⊂R(T).22

Lemma 3.9. Suppose that E and F are linear spaces in duality and that T: E →2F ismonotone.

If x∗∈coR(T), then there exists x ∈coD(T) such thatsup(y,y∗)∈G(T )⟨y∗−x∗, x −y⟩< ∞.Proof. Suppose that x∗= P tix∗i where ti ≥0, P ti = 1 and x∗i ∈R(T), so there existxi ∈E such that x∗i ∈T(xi).

Take x = P tixi; then for any (y, y∗) ∈G(T),⟨y∗−x∗, x −y⟩= ⟨y∗−Xtix∗i ,Xtjxj −y⟩=Xi,jtitj⟨y∗−x∗i , xj −y⟩==Xi,jtitj⟨y∗−x∗j, xj −y⟩+Xi,jtitj⟨x∗j −x∗i , xj −y⟩≤Xi,jtitj⟨x∗j −x∗i , xj −y⟩==Xi

Suppose, then, that x∗∈coR(T). By Theorem3.6, for each λ > 0 there exists y∗λ ∈E∗, x∗∗λ∈j∗(y∗λ) and z∗λ ∈T(x∗∗λ ) such thatx∗= λy∗λ + z∗λ.

By the foregoing remark, there exists x ∈E such that⟨x∗∗λ −ˆx, x∗−z∗λ⟩≡λ⟨x∗∗λ −ˆx, y∗λ⟩is bounded above for all λ > 0. It follows that for some M > 0 (and all λ > 0),λ∥y∗α∥2 ≤λ∥x∗∗λ ∥2 + λ∥y∗λ∥2 = 2λ⟨x∗∗λ , y∗λ⟩≤M + 2λ⟨y∗λ, x⟩≤M + 2λ∥y∗λ∥· ∥x∥.¿From this we see that λy∗λ →0 as λ →0; indeed, if there were a sequence λn →0 suchthat ∥λny∗λn∥were bounded away from 0, then we would necessarily have ∥y∗λn∥→∞and dividing both sides of the inequality above by λn∥y∗λn∥would lead to a contradiction.Thus, x∗−z∗λ = λy∗λ →0; since z∗λ ∈R(T), this shows that x∗∈R(T) ⊂R(T).

(Since R(T) ⊂R(T), their closures are in fact equal. )Corollary 3.10.

If E is reflexive and T: E →2E∗is maximal monotone, then both R(T)and D(T) are convex.Proof. By Example 3.2(b), if T is maximal monotone, then it is maximal monotone oftype (D), so it follows from Theorem 3.8 that R(T) is convex.

By applying this result tothe maximal monotone operator T −1 we obtain convexity of R(T −1) ≡D(T).23

Definition 3.11. An operator T: E →2E∗is said to be coercive provided D(T) is boundedor there exists a function c: R+ →R such that c(r) →∞when r →∞and ⟨x∗, x⟩≥c(∥x∥) · ∥x∥for each (x, x∗) ∈G(T).Remark.

It is easily verified that if D(T) is unbounded, then T is coercive if and only iffor every M > 0 there exists r > 0 such that⟨x∗, x⟩∥x∥≥Mwhenever∥x∥≥r and x∗∈T(x).Indeed, if this holds, take c(r) = inf{ ⟨x∗,x⟩∥x∥:∥x∥≥r and x∗∈T(x)}.Examples 3.12. (a) The duality mapping J is an obvious example of a coercive operator,since ⟨x∗, x⟩= ∥x∥2 whenever x∗∈J(x).

(b) If T is a positive linear operator and λ > 0, then T + λJ is coercive: If x ∈Eand x∗∈(T + λJ)(x), then x∗= T(x) + λz∗for some z∗∈J(x) and hence ⟨x∗, x⟩=⟨T(x), x⟩+ λ⟨z∗, x⟩≥λ∥x∥2. From the Remark following Problem 2.20, it follows thatT + λJ is also maximal monotone.

(c) Recall that a function ϕ: R →R is a monotone operator if and only if it isnondecreasing. It is easily seen that ϕ is coercive if and only if ϕ(t) →±∞as t →±∞.Exercise 3.13.

Show that if T is coercive, then so is T: E∗∗→2E∗.Theorem 3.14 (Gossez). Suppose that T is a coercive maximal monotone operator oftype (D).

Then R(T) = E∗and hence R(T) = E∗.Proof. It is clear from the definition of type (D) that one always has R(T) = R(T).Suppose, then, that x∗∈E∗.By Theorem 3.6, for each λ > 0 there exist y∗λ ∈E∗,x∗∗λ ∈∂j∗(y∗λ) and z∗λ ∈T(x∗∗λ ) such that x∗= λy∗λ + z∗λ.

We have⟨x∗∗λ , x∗⟩= λ⟨x∗∗λ , y∗λ⟩+ ⟨x∗∗λ , z∗λ⟩= λ∥x∗∗λ ∥2 + ⟨x∗∗λ , z∗λ⟩.Since (Exercise 3.13) T is coercive, if {∥x∗∗λ ∥} were unbounded as λ →0, the right side of∥x∗∥≥⟨x∗∗λ , x∗⟩∥x∗∗λ ∥= λ∥x∗∗λ ∥+ ⟨x∗∗λ , z∗λ⟩∥x∗∗λ ∥would be unbounded, an impossibility. Thus, the bounded net {x∗∗λ } has a subnet (call it{x∗∗λ }) converging in the σ(E∗∗, E∗) topology to an element x∗∗∈E∗∗.

We will show thatx∗∈T(x∗∗) by showing that (x∗∗, x∗) is monotonically related to G(T). Suppose, then,that (u, u∗) ∈G(T).

Since x∗λ = x∗−λy∗λ, we have0 ≤⟨x∗∗λ −ˆu, x∗λ −u∗⟩= ⟨x∗∗λ −ˆu, x∗−u∗⟩−λ⟨x∗∗λ −ˆu, y∗λ⟩.Recall that x∗∗λ ∈∂j∗(y∗λ) implies boundedness of ∥y∗λ∥= ∥x∗∗λ ∥, so the second term on theright converges to 0 as λ →0, yielding 0 ≤⟨x∗∗−ˆu, x∗−u∗⟩.24

Corollary 3.15. If E is reflexive and T is a coercive maximal monotone operator on E,then R(T) = E∗.Proof.

Simply use the fact that reflexivity implies that T = T.The fact (Proposition 2.17(c)) that E is reflexive if R(J) = E∗shows that one cannot omitreflexivity from this result.In order that R(T) be convex it is not necessary for T to be maximal monotone of type(D). This was shown by Gossez [Go2,3] with the help of the following example.Example 3.16.Let A be defined on the nonreflexive space ℓ1 as follows: For eachx = (xk) ∈ℓ1, let {A(x)n} be the ℓ∞sequence defined by(Ax)n = −Xknxk.It is not hard to verify that A is bounded, linear and antisymmetric (that is, ⟨Ax, y⟩=−⟨Ay, x⟩for all x, y ∈ℓ1) hence is monotone, satisfying ⟨Ax, x⟩= 0 for all x.

This lattermeans that, in particular, A is a positive operator, hence it is maximal monotone. Therange R(A) of A is a linear subspace (hence is convex and has convex closure) whichis properly contained in the proper closed subspace c of ℓ∞consisting of all convergentsequences.

(Indeed, limn→∞(Ax)n = −P∞k=1 xk. )Gossez [Go2] uses the operator A by showing that there exists λ > 0 such that R(A + λJ)is not dense in ℓ∞.

This shows that A+λJ is not of type (D), in view of Theorem 3.14 andthe fact that A + λJ is maximal monotone and coercive (Example 3.12(b)). Thus not allmaximal monotone operators (not even the coercive ones) are of type (D).

Subsequently,he showed [Go3] that the fact that R(A + λJ) is not dense in ℓ∞implies that its closure isnot convex, that is, there exists a coercive maximal monotone operator T on ℓ1 such thatR(T) is not convex.4. Locally maximal monotone operators.As we have seen, some of the nice properties of maximal monotone operators onreflexive spaces fail to hold in general, but are valid for the subclass of maximal monotoneoperators of type (D).

In this section we introduce another subclass which shares some ofthe same properties.Definition 4.1. A set-valued mapping T: E →2E∗is said to be locally maximal monotoneif, for each norm-open convex subset U ⊂E∗which intersects R(T), the restriction of theinverse operator T −1 to U is maximal monotone in U.

The latter means that the graphG((T −1)|U) ⊂U × E is a maximal monotone subset of U × E.The “working definition” of this property is the following: If U is an open convex subset ofE∗which intersects R(T) and if (x, x∗) ∈E × U is montonically related to each (y, y∗) ∈G(T) ∩(E × U), then (x, x∗) ∈G(T).25

It is clear (take U = E∗) that every locally maximal monotone operator is maximalmonotone.The locally maximal monotone operators were introduced in [F–P1] because theyare the precise class for which a certain approximation scheme is valid. While their exactposition within the class of all maximal monotone operators is still unclear, some importantproperties are known.Proposition 4.2.

(i) If T is locally maximal monotone, then R(T) is convex. (ii) If f is a proper lower semicontinuous convex function on E, then ∂f is locallymaximal monotone.The proof for (i) may be found in [F–P1].

Property (ii), which is a nontrivial extensionof Rockafellar’s maximality theorem (Theorem 2.15), was proved by S. Simons [Si2]; see,also, [Si3]. In order to see that maximal monotone operators in reflexive spaces are locallymaximal monotone, we first reformulate the definition.Proposition 4.3.

A monotone operator T on E is locally maximal monotone if and onlyif it satisfies the following condition: For any weak* closed convex and bounded subset Cof E∗such that R(T) ∩int C ̸= ∅and for each x ∈E and x∗∈int C with x∗/∈T(x), thereexists z ∈E and z∗∈T(z) ∩C such that ⟨x∗−z∗, x −z⟩< 0.Proof. In one direction, if T is locally maximal monotone and C is given, let U = int C. Inthe other direction, if U is open and convex in E∗, if u ∈E and x ∈E with u∗∈T(u) ∩Uand x∗∈U but x∗/∈T(x), then there exists ǫ > 0 such that u∗+ǫB∗⊂U and x∗+ǫB∗⊂U.

By convexity, C ≡[u∗, x∗] + ǫB∗is a weak* closed, convex and bounded subset of Uwhich can be used to verify that U has the required property.The only use of reflexivity in the next proposition is an application of Theorem 2.19(on the sum of two maximal monotone operators in a reflexive space).Proposition 4.4. If E is reflexive and T is maximal monotone on E, then it is locallymaximal monotone.Proof.

Suppose that C is weak* closed and convex and that int C ∩R(T) ̸= ∅. Supposealso that x∗∈int C but x∗/∈T(x).

Let T1 denote the inverse T −1 of T and let T2 = ∂δC.Since int D(T2) = int C, these are maximal monotone operators from E∗into E for whichD(T1) ∩int D(T2) ̸= ∅; by Theorem 2.19, their sum T1 + T2 is maximal monotone. Nowx∗/∈T(x) implies that x /∈T1(x∗), and since T2(x∗) = {0}, we see that x /∈T1(x∗)+T2(x∗).By maximality of T1+T2, there exists z∗∈D(T1)∩D(T2) ≡R(T)∩C and z ∈(T1+T2)(z∗)such that ⟨x∗−z∗, x−z⟩< 0.

We can write z = u+v, where u ∈T1(z∗) (that is, z∗∈T(u))and v ∈T2(z∗). The latter means that ⟨z∗−w∗, v⟩≥0 for all w∗∈C.

We have thusproduced z∗∈T(u) ∩C such that0 > ⟨x∗−z∗, x −u⟩−⟨x∗−z∗, v⟩≥⟨x∗−z∗, x −u⟩,26

showing that T satisfies the condition in Proposition 4.3 and is therefore locally maximalmonotone.Recall Gossez’s Example 3.15 of the linear maximal monotone operator A: ℓ1 →ℓ∞.Its interest in this context is the fact that, even though R(A) is linear (hence convex),A is not locally maximal monotone, so not every maximal monotone operator is locallymaximal monotone.Example 4.5. The operator A is not locally maximal monotone.Proof.

Let e = (1, 0, 0, . .

. ), considered as an element of either ℓ1 or ℓ∞, and letz = (−12, 123 , 124 , 125 , .

. .) ∈ℓ1.Some computations using the definition show that (Az)1 = 14 while for n ≥2, (Az)n =14 +12n +12n+1 .

Moreover, e −Ae = (1, 1, 1, . .

.) and ∥e −Az∥∞= 34, so if U is the openunit ball in ℓ∞, then x∗≡e −Az ∈U.

Let x = e −z. If u ∈ℓ1 and Au ∈U, thenlimn→∞|(Au)n| = | P∞k=1 uk| ≤1 and hence⟨x∗−Ax, u⟩= ⟨e −Ae, u⟩=∞Xk=1uk ≤1,while⟨x∗, x⟩= ⟨e, e⟩−⟨Az, e⟩−⟨e, z⟩= 1 −(Az)1 −z1 = 1 −14 + 12 > 1.Thus, x∗̸= Ax even though⟨x∗−Au, x −u⟩= ⟨x∗, x⟩−⟨Au, x⟩−⟨x∗, u⟩= ⟨x∗, x⟩+ ⟨Ax, u⟩−⟨x∗, u⟩≥0whenever Au ∈U, contradicting the definition of locally maximal monotone.We still do not know whether the class of maximal monotone operators of type (D)is actually different from the class of locally maximal monotone operators.

To see thatfor coercive operators, the former class is contained in the latter, we need two prelimi-nary results. The proof of the following identity consists of an elementary but tediouscomputation.Proposition 4.6.

If u, v, x ∈E, u∗, v∗, x∗∈E∗and λ ∈[0, 1], then⟨λu∗+ (1 −λ)v∗−x∗, λu + (1 −λ)v −x⟩== λ⟨u∗−x∗, u −x⟩+ (1 −λ)⟨v∗−x∗, v −x⟩−λ(1 −λ)⟨u∗−v∗, u −v⟩(4.1)Lemma 4.7. Suppose that T is a maximal monotone operator, that U is an open subsetof E∗and that z∗∈U \ T(z) is such that ⟨x∗−z∗, x −z⟩≥0 for all x∗∈T(x) ∩U.

Thenthere exist b ∈E, b∗∈U and r > 0 such that for all x∗∈T(x) ∩U,⟨x∗−b∗, x −b⟩≥r.27

Proof.Since T is maximal monotone there exists y∗∈T(y) such that ⟨y∗−z∗, y−z⟩< 0.Let 1 > λ > 0 be such that b∗:= λz∗+ (1 −λ)y∗∈U and let b = λz + (1 −λ)y. Then,using the identity (4.1), for all x∗∈T(x) ∩U we have⟨x∗−b∗, x −b⟩= λ⟨x∗−z∗, x −z⟩+ (1 −λ)⟨x∗−y∗, x −y⟩−λ(1 −λ)⟨z∗−y∗, z −y⟩≥−λ(1 −λ)⟨z∗−y∗, z −y⟩> 0so we may set r = −λ(1 −λ)⟨z∗−y∗, z −y⟩> 0.Theorem 4.8. Suppose that T is a maximal monotone operator such that either (i)R(T) = E∗or (ii) R(T) = E∗and T is coercive.

Then T is locally maximal monotone.Proof. Suppose, first, that R(T) = E∗, that U ⊂E∗is open and convex and that z ∈E,z∗∈U are such that ⟨z∗−x∗, z −x⟩≥0 for all x ∈E such that x∗∈T(x) ∩U.

Ifz∗/∈T(z), then there would exist b ∈E, b∗∈U and r > 0 as in Lemma 4.7. Sinceb∗∈R(T) by hypothesis, there exists x ∈E such that b∗∈T(x) ∩U and hence by Lemma4.7, ⟨b∗−b∗, x −b⟩≥r > 0, a contradiction.Suppose, next, that R(T) is dense in E∗and that T is coercive.

If T were not locallymaximal monotone, we could find an open convex subset U ⊂E∗with U ∩R(T) ̸= ∅andelements z ∈E and z∗∈U\T(z) such that ⟨z∗−x∗, z −x⟩≥0 whenever x ∈E andx∗∈T(x) ∩U. Choose b, b∗and r as in Lemma 4.7.

Since R(T) is dense in E∗, we canfind xn ∈E and x∗n ∈T(xn) such that ∥b∗−x∗n∥→0. But then for all sufficiently largen, we would have x∗n ∈U and hencer ≤⟨x∗n −b∗, xn −b⟩≤∥x∗n −b∗∥∥xn −b∥,which would imply that ∥xn −b∥→∞.

Coercivity would then imply that ∥x∗n∥→∞,again a contradiction.Corollary 4.9. If T is maximal monotone, coercive and of type (D), then it is locallymaximal monotone.Proof.

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