25.7. MAXIMAL MONOTONE OPERATORS 917

because this was how Jλ x was defined. Therefore,

φ λ (x) =1

2λ∥x− Jλ x∥2 +φ (Jλ (x)) =

λ

2∥Aλ x∥2 +φ (Jλ x) , A = ∂φ

It follows from this equation that

φ (Jλ x)≤ φ λ (x)≤ φ (x) , (25.7.80)

the second inequality following from taking y = x in the definition of φ λ .Next consider the claim about φ λ (x) ↑ φ (x). First suppose that x ∈ D(φ) . Then from

Proposition 25.7.53, x ∈ D(∂φ) and so from the material on approximations, Theorem25.7.36, it follows that Jλ x→ x. Hence from 25.7.80 and lower semicontinuity of φ ,

φ (x)≤ lim infλ→0

φ (Jλ x)≤ lim infλ→0

φ λ (x)≤ lim supλ→0

φ λ (x)≤ φ (x)

showing that in this case, limλ→0 φ λ (x) = φ (x). Next suppose x /∈D(φ) so that φ (x) = ∞.Why does φ λ (x)→ ∞? Suppose not. Then from the description of φ λ given above andusing the fact that the epigraph is closed and convex, there would exist a subsequence, stilldenoted as λ such that

C ≥ φ λ (x) =1

2λ∥x− Jλ x∥2 +φ (Jλ (x))≥

12λ∥x− Jλ x∥2 + ⟨z∗,x− Jλ x⟩+b

Then multiplying by λ , it follows that for a suitable constant M,

∥x− Jλ x∥2 ≤Mλ +λM ∥x− Jλ x∥

and so a use of the quadratic formula implies

∥x− Jλ x∥ ≤ M2

(1+√

5)

λ

Hence Jλ x→ x and so in 25.7.80 it follows from lower semicontinuity again that

∞ = φ (x)≤ lim infλ→0

φ (Jλ x)≤ lim infλ→0

φ λ (x)≤ lim supλ→0

φ λ (x)≤ φ (x)

and so again, limλ→0 φ λ (x) = ∞. Also note that if λ > µ, then

miny∈X

(1

2λ∥x− y∥2 +φ (y)

)≤min

y∈X

(1

2µ∥x− y∥2 +φ (y)

)because for a given y, 1

2λ∥x− y∥2 +φ (y)≤ 1

2µ∥x− y∥2 +φ (y). Thus φ λ (x) ↑ φ (x).

Next consider the claim about the Gateaux differentiability. Using the description25.7.78

φ λ (y)−φ λ (x) =

12λ∥y− Jλ y∥2 +φ (Jλ (y))−

(1

2λ∥x− Jλ x∥2 +φ (Jλ (x))

)(25.7.81)