1226 CHAPTER 34. GELFAND TRIPLES AND RELATED STUFF

=∫ T

0e−(1/λ )t

φ (u0)dt +∫ T

0φ (u(s))

∫ T

se−(1/λ )(t−s) 1

λdtds

Now∫ T

s e−(1/λ )(t−s) 1λ

dt = 1− e1λ

s− 1λ

T < 1 and since φ ≥ 0, this shows that

Φ(uλ )≤ φ (u0)∫

∞

0e−(1/λ )tdt +

∫ T

0φ (u(s))dt

soΦ(uλ )≤ φ (u0)λ +Φ(u)

It follows that the conditions of Theorem 25.7.55 on Page 919 are satisfied and so L+∂Φ

is maximal monotone. Thus if f ∈H , there exists u ∈ D(L)∩D(∂Φ) such that for eachv ∈H there exists a solution uv to

u′v + zv +uv = f + v, uv (0) = u0 ∈ D(φ) .

where zv ∈ ∂Φ(uv). I will show that v→ uv has a fixed point. Note that if zi ∈ ∂Φ(ui) ,then ∫ t

t−h(z1− z2,u1−u2)ds≥ 0, any h≤ t

To see this, you could simply let u2 = u1 off [t−h, t] and pick z1 = z2 also on this set.Also, you can conclude that u ∈ ∂Φ implies u(t) ∈ ∂φ for a.e. t. I show this now. Let[a,b] ∈ G (∂φ) . Then as just noted, for [u,z] ∈ G (∂Φ) ,∫ t

t−h(z−b,u−a)ds≥ 0

Then by the fundamental theorem of calculus, for a.e. t,(z(t)−b,u(t)−a) ≥ 0 a.e. Let-ting {[ai,bi]}∞

i=1 be a dense subset of G (∂φ) , one can take the union of countably manysets of measure zero, one for each [ai,bi] and conclude that off this set of measure zero,(z(t)−bi,u(t)−ai) ≥ 0 for all i. Hence this is also true for all [a,b] ∈ G (∂φ) and soz(t) ∈ ∂φ (u(t)) for a.e. t.

Then if you have vi, i = 1,2

Luv1 −Luv2 + zv1 − zv2 +uv1 −uv2 = v1− v2

Then taking inner products with uv1 −uv2 and integrating up to t,

12|(uv1 −uv2)(t)|

2H +

∫ t

0(zv1 − zv2 ,uv1 −uv2)dt +

∫ t

0|uv1 −uv2 |

2 ds

≤ 12

∫ t

0|v1− v2|2 ds+

12

∫ t

0|uv1 −uv2 |

2 ds

Now by monotonicity of φ and the above,

|(uv1 −uv2)(t)|2H ≤

∫ t

0|v1− v2|2 ds