1378 CHAPTER 42. MAXIMAL MONOTONE OPERATORS, HILBERT SPACE

Taking a further subsequence, you can assume

Aλ yλ ⇀ z weak ∗ in L∞ (0,T ;H) . (42.2.16)

Thus z ∈ L∞ (0,T ;H) . Recall Aλ yλ ∈ AJλ yλ .Now A can be considered a maximal monotone operator on L2 (0,T ;H) according to

the ruleAy(t)≡ A(y(t))

whereD(A)≡

{f ∈ L2 (0,T ;H) : f (t) ∈ D(A) a.e. t

}By Lemma 42.1.5 applied to A considered as a maximal monotone operator on L2 (0,T ;H)and using 42.2.15 and 42.2.16, it follows y(t) ∈ D(A) a.e. t and z(t) ∈ Ay(t) a.e. t. Thenpassing to the limit in 42.2.7 yields

y(t)− y0 +∫ t

0z(s)ds =

∫ t

0f (s)ds. (42.2.17)

Then by fundamental theorem of calculus, y′ (t) exists a.e. t and

y′+ z = f , y(0) = y0

where z(t)∈ Ay(t) a.e. Since z∈ L∞ (0,T ;H) , it follows from 42.2.17 and the fundamentaltheorem of calculus that y′ ∈ L∞ (0,T ;H) also.

It remains to verify uniqueness. Suppose [y1,z1] is another pair which works. Thenfrom 42.2.17,

y(t)− y1 (t)+∫ t

0(z(r)− z1 (r))dr = 0

y(s)− y1 (s)+∫ s

0(z(r)− z1 (r))dr = 0

Therefore for s < t,

y(t)− y1 (t)− (y(s)− y1 (s)) =∫ t

s(z(r)− z1 (r))dr

and so||y(t)− y1 (t)|− |y(s)− y1 (s)|| ≤ K |s− t|

for some K depending on ∥z∥L∞ ,∥z1∥L∞ . Since y,y1 are bounded, it follows that t →|y(t)− y1 (t)|2 is also Lipschitz. Therefore by Corollary 26.4.3, |y(t)− y1 (t)|2 is the in-tegral of its derivative which exists a.e. So what is this derivative? As before,

(Dh (y(t)− y1 (t)) ,y(t +h)− y1 (t +h))

+

(1h

∫ t+h

t(z(s)− z1 (s))ds,y(t +h)− y1 (t +h)

)= 0