76.7. OTHER EXAMPLES, INCLUSIONS 2585
Therefore,∥u∥2
L2(U)+∫
U∑i, j
ai j (x)∂iu∂ ju = ∥u∥2W =
= ( f ,u)≤ ∥ f∥L2(U) ∥u∥L2(U) ≤ ∥ f∥L2(U) ∥u∥Wwhich shows that the map B−1 : H = L2 (U)→W is continuous.
Next suppose that Φ∈ L∞([0,T ]×Ω;L2
(Q1/2U,H
)). Then by continuity of the map-
ping B−1, it follows that Ψ ≡ B−1Φ satisfies Ψ ∈ L∞([0,T ]×Ω;L2
(Q1/2U,W
)). Thus
Φ = BΨ. In addition to this, to simplify the presentation, assume in addition that
⟨A(t,u,ω)−A(t,v,ω) ,u− v⟩ ≥ δ2 ∥u− v∥p
V
⟨A(t,u,ω) ,u⟩ ≥ δ2 ∥u∥p
V
Also assume the uniqueness condition of Lemma 76.3.16 is satisfied. Consider the follow-ing graph.
−1/n
1/n
There is a monotone Lipschitz function Jnwhich is approximating a function with theindicated jump. For a convex function φ , we denote by ∂φ its subgradient. Thus fory ∈ ∂φ (x)
(y,u)≤ φ (x+u)−φ (x) .
Denote the Lipschitz function as Jn and the maximal monotone graph which it is approxi-mating as J. Thus J denotes the ordered pairs (x,y) which are of the form (0,y) for |y| ≤ 1along with ordered pairs (x,1) ,x > 0 and ordered pairs (x,−1) for x < 0. The graph of Jis illustrated in the above picture and is a maximal monotone graph. Thus J = ∂φ whereφ (r) = |r|. As illustrated in the graph, Jn is piecewise linear.
Let φ n (r)≡∫ r
0 Jn (s)ds. It follows easily that φ n (r)→ φ (r) uniformly on R. Also leth≥ 0 be progressively measurable and uniformly bounded by M and let u0 ∈ L2 (Ω,W ) ,u0F0 measurable. Then from the above theorems, there exists a unique solution to the integralequation
Bun (t)−Bu0 +∫ t
0A(s,un,ω)ds+
∫ t
0h(s,ω)Jn (un)ds = B
∫ t
0ΨdW,
the last term equaling∫ t
0 ΦdW. The integral equation holds off a set of measure zero and isprogressively measurable.
Then from the Ito formula, one obtains, using the monotonicity of Jn an estimate inwhich C does not depend on n
12
E ⟨Bun (t) ,un (t)⟩−12
E ⟨Bu0,u0⟩+E∫ t
0⟨Aun,un⟩V ds≤C