42.5. MOREAU’S THEOREM 1397

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 (42.5.35)

It follows easily that a minimizing sequence is bounded and so from lower semicontinuitywhich implies weak lower semicontinuity, 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) =1

2λ

∣∣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 follows asin Lemma 42.3.23 that

1λ(x− yx) ∈ ∂φ (yx)

Thusx ∈ yx +λ∂φ (yx)

and soyx = Jλ x.

Thus

φ λ (x) =1

2λ|x− Jλ x|2 +φ (Jλ (x)) =

λ

2|Aλ x|2 +φ (Jλ x)

Note that Jλ x ∈ D(∂φ) and so it must also be in D(φ) . Now also

Aλ x≡ xλ− 1

λJλ x ∈ ∂φ (Jλ x)

This is so if and only if

x ∈ Jλ x+λ∂φ (Jλ x) = (I +λ∂φ)(Jλ x) = (I +λ∂φ)(I +λ∂φ)−1 x

which is clearly true by definition.Next consider the claim about differentiability.

φ λ (y)−φ λ (x) =λ

2|Aλ y|2 +φ (Jλ y)−

(λ

2|Aλ x|2 +φ (Jλ x)

)