916 CHAPTER 25. NONLINEAR OPERATORS

Then the function is well defined, convex, Gateaux differentiable,

Dzφ λ (x)≡ limt↓0

φ λ (x+ tz)−φ λ (x)t

= ⟨Aλ x,z⟩

so the Gateaux derivative is just Aλ x and for all x ∈ X ,

limλ→0

φ λ (x) = φ (x) ,

In addition,

φ λ (x) =1

2λ∥x− Jλ x∥2 +φ (Jλ (x)) (25.7.78)

where Jλ x is as before, the solution to

0 ∈ F (Jλ x− x)+λ∂φ (Jλ x)

Proof: First of all, why does the minimum take place? By the convexity, closed epi-graph, and assumption that φ is proper, separation theorems apply and one can say thatthere exists z∗ such that for all y ∈ H,

12λ∥x− y∥2 +φ (y)≥ 1

2λ∥x− y∥2 +(z∗,y)+ c (25.7.79)

It follows easily that a minimizing sequence is bounded and so from lower semicontinuitywhich implies weak lower semicontinuity due to convexity, there exists yx such that

miny∈H

(1

2λ∥x− y∥2 +φ (y)

)=

(1

2λ∥x− yx∥2 +φ (yx)

)Why is φ λ convex? For θ ∈ [0,1] ,

φ λ (θx+(1−θ)z)≡ 12λ

∥∥θx+(1−θ)z− y(θx+(1−θ)z)∥∥2

+φ(yθx+(1−θ)z

)≤ 1

2λ|θx+(1−θ)z− (θyx +(1−θ)yz)|2 +φ (θyx +(1−θ)yz)

≤ θ

2λ|x− yx|2 +

1−θ

2λ|z− yz|2 +θφ (yx)+(1−θ)φ (yz)

= θφ λ (x)+(1−θ)φ λ (z)

So is there a formula for yx? Since it involves minimization of the functional, it followsthat

0 ∈ − 1λ

F (x− yx)+∂φ (yx) =1λ

F (yx− x)+∂φ (yx)

Recall that if ψ (x) = 12 ∥x∥

2 , then ∂ψ (x) = F (x). Thus

yx = Jλ x