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FISTA energy recurrence with certified proximal errors

Statement

Let xk,yk,tk be the certified inexact FISTA iterates of Certified inexact proximal-gradient and FISTA updates, with errors ηk≥0, and let x∗ minimize F=f+g. Set t0=0, u0=x0−x∗, and uk=tkxk−(tk−1)xk−1−x∗. Then 2tk2L(F(xk)−F(x∗))+∥uk∥22≤2tk−12L(F(xk−1)−F(x∗))+∥uk−1∥22+2tk2Lηk, with the k=1 previous objective term interpreted as zero.

Facts & Assumptions

Given: The inexact FISTA recurrence, certified errors and a minimizer in the Statement.

[F1]

Certified inexact proximal-gradient and FISTA updates: t1=1, y1=x0, xk satisfies the stated certificate, tk+1=(1+1+4tk2)/2 and hence tk+1(tk+1−1)=tk2, and yk=xk−1+((tk−1−1)/tk)(xk−1−xk−2) for k≥2.

[F2]

One-step estimate for a certified inexact proximal call: F(z)≥F(xk)+(L/2)∥xk−yk∥22+L⟨yk−z,xk−yk⟩−ηk for every z.

Proof

1.1F1giveninductionalgebra

The recurrence [F1] gives tk≥1 for every k, so zk=(1−1/tk)xk−1+(1/tk)x∗ is a convex combination. It also gives tk(yk−zk)=uk−1 and uk=uk−1+tk(xk−yk), including k=1 with z1=x∗.

2.1F1F2step 1.1givenalgebra∎

For k≥2, convexity gives F(zk)≤(1−1/tk)F(xk−1)+(1/tk)F(x∗). Apply [F2] at yk,zk; the same multiplication and norm expansion as in step 1.1 yield 2tk2L(F(xk)−F(x∗))≤2tk(tk−1)L(F(xk−1)−F(x∗))+∥uk−1∥22−∥uk∥22+2tk2Lηk. Use tk(tk−1)=tk−12 to obtain the stated recurrence. For k=1, [F2] with z=x∗ gives the same base inequality directly, and t1=1 leaves the previous objective term zero.

Depends on

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