42.6. A PERTURBATION THEOREM 1399

and so, |x− Jλ x| → 0 as λ → 0. If x is only in D(A), it also works because for y ∈ D(A)

|x− Jλ x| ≤ |x− y|+ |y− Jλ y|+ |Jλ y− Jλ x|≤ 2 |x− y|+ |y− Jλ y|

If ε is given, simply pick |y− x|< ε/2 and then

|x− Jλ x| ≤ ε + |y− Jλ y|

and the last converges to 0. Therefore, Jλ x→ x on D(A).Returning to the proof of the theorem, if x ∈ D(∂φ) then recall that

φ λ (x) =1

2λ|x− Jλ x|2 +φ (Jλ x)

and so,lim inf

λ→0φ λ (x)≥ lim inf

λ→0φ (Jλ x)≥ φ (x)≥ lim sup

λ→0φ λ (x)

which shows the desired result in case x ∈ D(∂φ) . Now consider the case where x /∈D(∂φ). In this case, there is a positive lower bound δ to |x− Jλ x| because each Jλ x ∈D(∂φ). Then from the definition and what was shown above,

φ λ (x) =λ

2|Aλ x|2 +φ (Jλ x)≥ λ

2|Aλ x|2 +(z∗,Jλ x)+ c

≥ λ

2|Aλ x|2 +(z∗,Jλ x− x)+(z∗,x)+ c

≥ 12|Aλ x| |x− Jλ x|− |z∗| |Jλ x− x|− |z∗| |x|+ c

≥ 12(|Aλ x|− |z∗|)δ −|z∗| |x|+ c

≥ 12

(δ

λ−|z∗|

)δ −|z∗| |x|+ c

Hence φ λ (x)→ ∞ and since φ (x)≥ φ λ (x) by construction, it follows that φ (x) = ∞. Theconstruction of φ λ also shows that as λ decreases, φ λ (x) increases.

Note that the last part of the argument shows that if x /∈ D(∂φ), then x /∈ D(φ) . Hencethis shows that

D(∂φ)⊆ D(φ)⊆ D(∂φ)

42.6 A Perturbation TheoremIn this section is a simple perturbation theorem found in [24] and [116].

Recall that for B a maximal monotone operator, Bλ ,the Yosida approximation, is de-fined by

Bλ x≡ 1λ(x− Jλ x) , Jλ x≡ (I +λB)−1 x.

This follows from Theorem 42.1.7 on Page 1372