874 CHAPTER 25. NONLINEAR OPERATORS

However, as noted above, it is always the case that

maxy∈B

minx∈A

H (x,y)≤minx∈A

maxy∈B

H (x,y)

Of course all of this works with no change if you have E,F reflexive Banach spacesand the sets A,B are just closed and bounded and convex. Then you just use the fact thatthe functional is weakly lower semicontinuous in the first variable and weakly upper semi-continuous in the second. Recall that lower semicontinuous and convex implies weaklylower semicontinuity. Then just use weak convergence instead of strong convergence inthe above argument. Recall that closed bounded and convex sets with the weak topologycan be considered metric spaces. I think the above is most interesting in finite dimensions.Of course in this case, you can simply assume the norm is the standard Euclidean norm andthere is then no need to assume one of the norms is strictly convex. It comes automatically.Just use an equivalent norm which is strictly convex.

25.7.2 Equivalent Conditions For Maximal MonotoneNext is the theorem about the graph being maximal being equivalent to the operator beingmaximal monotone. It is a very convenient result to have. The proof is a modified versionof one in Barbu [13]. It is based on the following lemma also in Barbu. This is a little likethe Browder lemma but is based on the min max theorem above. It is also a very interestingargument.

Lemma 25.7.7 Let E be a finite dimensional Banach space and let K be a convex andcompact subset of E. Let G (A) be a monotone subset of E×E ′ such that D(A)⊆ K and Bis a single valued monotone and continuous operator from E to E ′. Then there exists x ∈ Ksuch that

⟨Bx+ v,u− x⟩E ′,E ≥ 0 for all [u,v] ∈ G (A) .

If B is coercive

lim∥x∥→∞

⟨Bx,x⟩∥x∥

= ∞,

and 0 ∈ D(A), then one can assume only that K is convex and closed.

Proof: Let T : E→ K be the multivalued operator defined by

Ty≡{

x ∈ K : ⟨By+ v,u− x⟩E ′,E ≥ 0 for all [u,v] ∈ G (A)}

Here y ∈ E and it is desired to show that Ty ̸= /0 for all y ∈ K. For [u,v] ∈ G (A) , let

Ku,v ={

x ∈ K : ⟨By+ v,u− x⟩E ′,E ≥ 0}

Then Ku,v is a closed, hence compact subset of K. The thing to do is to show that

∩[u,v]∈G (A)Ku,v ≡ Ty ̸= /0