lecture. - CASTLE Lab - Princeton University

Growth Conditions
Example(strongly convex function)
Let f(x) be a twice differentiable, strongly convex function over a convex set C
comprising its minimum: there exists a constant m > 0 such that ∇ 2 f(x) mI
for all x ∈ C. Since for any x,y ∈ C, there exists a point z = tx +(1−t)y with
t ∈ [0,1] such that
f(y)−f(x) = ∇f(x) ⊤ (y −x)+ 1
2 (y −x)⊤ ∇ 2 f(z)(y −x) ,
we have, by the strong convexity assumption,
f(y)−f(x) ≥ ∇f(x) ⊤ (y −x)+ m
2 ||y −x||2 .
Consider S = argmin Cf, s ∈ S, and write fS = minC f. We have
f(y)−fS ≥ ∇f(s) ⊤ (y −s)+ m
2 ||y −s||2 ≥ m
2 ||y −s||2 ,
using the fact that ∇f(s) ⊤ (y −s) ≥ 0 for all y ∈ C and s ∈ S, by optimality of
S. This proves that S is in fact reduced to {s}, and that the second-order
growth condition holds with modulus c = m/2:
f(x)−fS ≥ c[x −dist(x,S)] 2
for all x ∈ C.
Boris Defourny (ORFE) Lecture 6: Perturbation Analysis in Optimization March 15,2012 15 / 26