[Help] Proximal mapping

1. Properties of Proximal mapping

For a convex function h(x) , its proximal mapping is defined as:

Proxh(x)=argminu{h(u)+12∥u−x∥22}.

From the fact that the objective function is strictly convex, we know that Proxh(x) exists and is unique for all x. If u=Proxh(x) , we have

x−u∈∂h(u).

1.1 L1 Lipschitz and monotone

Theorem: Proximal mapping for any convex function h(x) is L1 Lipschitz and monotone, that is to say

• Lipschitz: ∥Proxh(x)−Proxh(y)∥2≤∥x−y∥2.
• Monotone: (Proxh(x)−Proxh(y))T(x−y)≥0.

proof: From the definition of proximal mapping, if u=Proxh(x) and u^=Proxh(x^) , we have

x−u∈∂h(u)x^−u^∈∂h(u^)

Then from the convexity,

h(u)h(u^)≥h(u^)+∂h(u^)T(u−u^)≥h(u)+∂h(u)T(u^−u)

we have

(∂h(u)−∂h(u^))T(u−u^)≥0,

which means that

(x−u−(x^−u^))T(u−u^)≥0,

Then,

0≤∥u−u^∥22≤(x−x^)T(u−u^)monotone≤∥x−x^∥2∥u−u^∥2Lipschitz.

1.2 Projection Property

Theorem: Proximal mapping for any convex function h(x) acts just like a Projection function, and its orthogonal projection is the proximal function corresponding to its conjugate function, that is to say

x=Proxh(x)+Proxh∗(x).

proof: If u=Proxh(x) , we have v=x−u∈∂h(u) . From the definition of conjugate function:

h∗(v)=maxz{vTz−h(z)}=vTu−h(u)

from which, we have u=x−v∈∂h∗(v), then v=Proxh∗(x).
So

x=u+v=Proxh(x)+Proxh∗(x).

1.3 Scaling and translation argument

Theorem: Let h(x)=f(tx+a) , then

Proxh(x)=1t(Proxt2f(tx+a)−a)

proof: Assume u=Proxh(x) , then we have

x−u∈∂h(u)=t∂f(tu+a)

Then

(tx+a)−(tu+a)∈∂t2f(tu+a)

so we have

tu+a=Proxt2f(tx+a)

so

u=Proxh(x)=1t(Proxt2f(tx+a)−a).