Kenneth Kuttler

25.7. MAXIMAL MONOTONE OPERATORS 885

and so ⟨x∗nk

,xnk − y0⟩≥⟨y∗0,xnk − y0

⟩and the right side is bounded below because it converges to

⟨y∗0,x− y0

⟩and this is a con-

tradiction.Does the same proof work if x is a limit point of D(A)? No. Suppose x is a limit point

of D(A) . If A is not locally bounded, then there exists xn→ x,xn ∈D(A) and x∗n ∈ Axn and∥x∗n∥ → ∞. Then there is y0 close to x such that

⟨x∗nk

,xnk − y0⟩→−∞ but now everything

crashes in flames because it is not known that y0 ∈ D(A).It follows from the above theorem that if A is defined on all of X and is maximal

monotone, then it is locally bounded everywhere. Now here is a very interesting resultwhich is like the one which involves monotone and hemicontinuous conditions. It is in[55].

Theorem 25.7.21 Let A : X→P (X ′) be monotone and satisfies the following conditions:

1. If λ n → λ ,λ n ∈ [0,1] and zn ∈ A(u+λ n (v−u)) , then if B is any weakly open setcontaining 0, zn ∈ A(u)+B for all n large enough. (Upper semicontinuous into weaktopology along a line segment)

2. A(x) is closed and convex.

Then one can conclude that A is maximal monotone.

Proof: Let Â be a monotone extension of A. Let [û, ŵ] be such that ŵ ∈ Â(û). Now alsoby assumption, A(x) is not just convex but also closed.

If [û, ŵ] is not in the graph of A, then by separation theorems, there is u such that

⟨x∗,u⟩< ⟨ŵ,u⟩ for all x∗ ∈ A(û)

Then for λ > 0, let xλ ≡ û+λu, x∗λ∈ A(xλ ) . Then from monotonicity of Â,

0≤⟨x∗

λ− ŵ,xλ − û

⟩= λ

⟨x∗

λ− ŵ,u

⟩Thus ⟨

x∗λ− ŵ,u

⟩≥ 0

By Theorem 25.7.20, the monotonicity of A on X implies A is locally bounded also. Thusin particular, Axλ for small λ is contained in a bounded set. Now by that hemicontinuityassumption, you can get a subsequence λ n→ 0 for which x∗

λ nconverges weakly to x∗ ∈ Aû.

Therefore, passing to the limit in the above, we get

⟨x∗− ŵ,u⟩ ≥ 0

⟨x∗,u⟩ ≥ ⟨ŵ,u⟩> ⟨x∗,u⟩

a contradiction. Thus there is no proper extension and this shows that A is maximal mono-tone.

Recall the definition of a pseudomonotone operator.

25.7. MAXIMAL MONOTONE OPERATORS 885and so(xg Xm —yo) = (yo.Xn, —yo)and the right side is bounded below because it converges to (yp.x — yo) and this is a con-tradiction. JjDoes the same proof work if x is a limit point of D(A)? No. Suppose x is a limit pointof D(A). If A is not locally bounded, then there exists x, — x,x, € D(A) and x* € Ax, and\|xj |] + ce. Then there is yo close to x such that (x), ,%n, —yo) —> —ce but now everythingcrashes in flames because it is not known that yo € D(A).It follows from the above theorem that if A is defined on all of X and is maximalmonotone, then it is locally bounded everywhere. Now here is a very interesting resultwhich is like the one which involves monotone and hemicontinuous conditions. It is in[SS].Theorem 25.7.21 Let A: X — # (X') be monotone and satisfies the following conditions:1. If An 4 A,An € [0,1] and z, € A(u+Ay(v—u)), then if B is any weakly open setcontaining 0, Z, € A(u) +B for all n large enough. (Upper semicontinuous into weaktopology along a line segment)2. A(x) is closed and convex.Then one can conclude that A is maximal monotone.Proof: Let A be a monotone extension of A. Let [i, #] be such that v € A (@). Now alsoby assumption, A (x) is not just convex but also closed.If [a, 0] is not in the graph of A, then by separation theorems, there is u such that(x*,u) < (Wu) for all x* € A (i)Then for A > 0, let x, =@+Au, x3 €A(x,). Then from monotonicity of A,0< (xy —W, xy — it) =A (xy —vW,u)Thus(xy — W,u) >0By Theorem 25.7.20, the monotonicity of A on X implies A is locally bounded also. Thusin particular, Ax, for small A is contained in a bounded set. Now by that hemicontinuityassumption, you can get a subsequence A, — 0 for which x2, converges weakly to x* € Ai.Therefore, passing to the limit in the above, we get(x*,u) > (W,u) > (x*,u)a contradiction. Thus there is no proper extension and this shows that A is maximal mono-tone. JfRecall the definition of a pseudomonotone operator.