[Help] Proximal mapping
2016-02-18 03:02
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Properties of Proximal mapping
1 L1 Lipschitz and monotone
2 Projection Property
3 Scaling and translation argument
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).
Lipschitz:∥Proxh(x)−Proxh(y)∥2≤∥x−y∥2.
Monotone: (Proxh(x)−<
20000
span style="position: absolute; clip: rect(1.869em 1000em 2.883em -0.424em); top: -2.717em; left: 0.003em;">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^∥2Lipschitz.
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).
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).
1 L1 Lipschitz and monotone
2 Projection Property
3 Scaling and translation argument
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 sayLipschitz:∥Proxh(x)−Proxh(y)∥2≤∥x−y∥2.
Monotone: (Proxh(x)−<
20000
span style="position: absolute; clip: rect(1.869em 1000em 2.883em -0.424em); top: -2.717em; left: 0.003em;">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^∥2Lipschitz.
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 sayx=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), thenProxh(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).
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