42.4. SOME EXAMPLES 1395

Using convexity of φ ,

≤ t2

2|z− x0|2 + tαφ (z)− tαφ (x0)

Divide by t and let t→ 0 to obtain that

(y− x0,z− x0)≤ αφ (z)−αφ (x0)

and soy− x0 ∈ ∂ (αφ (x0))

Thus y = x0 +α∂φ (x0) because ∂ (αφ) = α∂φ . Thus ∂φ is maximal monotone.

42.4 Some ExamplesHere are a few examples of how some of the above theory of maximal monotone operatorsin Hilbert space can be used.

Theorem 42.4.1 Let Ω be a bounded open set with Lipschitz boundary and let

φ : H ≡ L2 (Ω)→ R, V a closed subset of H1 (Ω) containing C∞c (Ω)

be given by φ (u)≡{ 1

2∫

Ωa(x)∇u(x) ·∇u(x)dx+

∫∂Ω

b(x)γu(x)dσ if u ∈V+∞ if u /∈V

Then φ is

convex proper and lower semicontinuous on H. Here 0< δ < a(x)≤∆<∞ for all x,a mea-surable. The subgradient ∂φ (u) will be−∇·a∇u at u∈D(∂φ). If V =

{u ∈ H1 : u = 0 on ΓD

}then u will satisfy the boundary conditions u = 0 on ΓD and ∇u(x) · n + b(x) = 0 onΓN ≡ ∂Ω\ΓD.

Proof: First of all, it is obviously proper and convex. Why would it be lower semicon-tinuous? Say un→ u in H. Is it the case that liminfn→∞ φ (un)≥ φ (u)?

Case 1: φ (u) = ∞. In this case, if liminfn→∞ φ (un) ̸= ∞ there would be a subsequencestill called n such that ∥un∥V is bounded and un → w weakly in V. However, φ is convexand strongly lower semicontinuous on V and so it is weakly lower semicontinuous on V .It follows that φ (w)≤ liminfn→∞ φ (un)< ∞. However, w = u because un→ u in H henceun→ u weakly in H. This is a contradiction to φ (u) = ∞.

Case 2: φ (u) < ∞. In this case, u ∈ V . If liminfn→∞ φ (un) = ∞, we are done. Ifliminfn→∞ φ (un)< ∞, there is a subsequence uk→ w weakly in V,

lim infn→∞

φ (un) = limn→∞

φ (uk) .

As above, this will imply w = u and since φ is strongly lower semicontinuous on V andalso convex, it is weakly l.s.c. and so φ (u)≤ liminfn→∞ φ (un) = liminfk→∞ φ (uk) .

Now say u ∈D(∂φ) . z ∈ ∂φ (u) , means (z,y−u)≤ φ (y)−φ (u) for all y. In particularthis would need to hold for y = u+ tv where v ∈V . Thus for t > 0,

(z, tv)≤ φ (u+ tv)−φ (u) =