1372 CHAPTER 42. MAXIMAL MONOTONE OPERATORS, HILBERT SPACE

Theorem 42.1.7 Let A and B be maximal monotone, let

y ∈ xλ +Bλ xλ +Axλ ,

and suppose Bλ xλ is bounded independent of λ . Then there exists x ∈ D(A)∩D(B) suchthat y = x+Ax+Bx.

Proof: First of all, it follows from Proposition 42.1.6 that there exists a unique xλ . Note

y− xλ −Bλ xλ ∈ Axλ

y− xµ −Bµ xµ ∈ Axµ

and so by monotonicity of A,(xµ − xλ +Bµ xµ −Bλ xλ ,xλ − xµ

)≥ 0

and so ∣∣xλ − xµ

∣∣2 ≤(Bµ xµ −Bλ xλ ,xλ − xµ

)= −

(Bλ xλ −Bµ xµ ,xλ − xµ

)(42.1.3)

I want to write as many things as possible in terms of the Bλ and Bµ . Denote as Jλ (B) theoperator (I +λB)−1 . Then

Bλ xλ =1λ(xλ − Jλ (B)xλ )

and soxλ = λBλ xλ + Jλ (B)xλ

Thus 42.1.3 becomes ∣∣xλ − xµ

∣∣2 =−(Bλ xλ −Bµ xµ ,λBλ xλ + Jλ (B)xλ −

(µBµ xµ + Jµ (B)xµ

))= −

(Bλ xλ −Bµ xµ ,λBλ xλ −µBµ xµ

)+(Bµ xµ −Bλ xλ ,Jλ (B)xλ − Jµ (B)xµ

)= −

(Bλ xλ −Bµ xµ ,λBλ xλ −λBµ xµ

)−(Bλ xλ −Bµ xµ ,(λ −µ)Bµ xµ

)−(Bλ xλ −Bµ xµ ,Jλ (B)xλ − Jµ (B)xµ

)Now recall Bµ x ∈ BJµ (B)x. Then by monotonicity the first and last terms to the right ofthe equal sign in the above are negative. Therefore,∣∣xλ − xµ

∣∣2 ≤ ∣∣(Bλ xλ −Bµ xµ ,(λ −µ)Bµ xµ

)∣∣≤C |λ −µ|

where C is some constant which comes from the assumption the Bλ xλ are bounded.