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There is a close resemblance between SOFT-IMPUTEand Nesterov’s gradient method (Nesterov, 2007, Section 3). However, as mentioned earlier the original motivation of our algorithm is very different.

The techniques used in this proof are adapted from Nesterov (2007). Proof Plugging Zλk=Z in (11), we have˜

Qλ(Z|Zλk) = fλ(Zk

λ) +12kPΩ⊥(ZλkZ)k2F (55)

fλ(Zkλ).

Let Zλk(θ)denote a convex combination of the optimal solution (Zλ) and the kthiterate (Zλk):

Using the convexity of fλ(·)we get:

fλ(Zk

λ(θ))≤(1−θ)fλ(Zkλ) +θfλ(Zλ∞). (57)

Expanding Zλk(θ)using (56), and simplifying P⊥(Zk

λ−Zλk(θ))we have:

kP⊥(Zk

λ−Zλk(θ))k2F = θ2kPΩ⊥(ZλkZλ∞)k2F

≤ θ2kZλkZλ∞k2F (58)

≤ θ2kZλ0−Zλ∞k2F. (59) Line 59 follows from (58) by observing thatkZλmZλ∞k2

F ≤ kZλm−1−Zλ∞k2F,∀m—a consequence of

the inequalities (19) and (18), established in Theorem 1.

Using (55), the value of fλ(Z)at the(k+1)thiterate satisfies the following chain of inequalities: fλ(Zkλ+1) ≤ min Z fλ(Z) +1 2kP ⊥ Ω(Zkλ−Z)k2F ≤ min θ∈[0,1] fλ(Zλk(θ)) +1 2kP ⊥ Ω(ZλkZkλ(θ))k2F (60) ≤ min θ∈[0,1] fλ(Zλk) +θ(fλ(Zλ∞)−fλ(Zλk)) +1 2θ 2kZ0 λ−Z∞λk2F . (61) Line 61 follows from Line 60, by using (57) and (59).

The r.h.s. expression in (61), is minimized atbθ(k+1)given by

b θ(k+1) = min{1,θk} ∈[0,1], where, θk = fλ(Zk λ)−fλ(Zλ∞) kZλ0−Zλk2 F . IfkZλ0−Zλ∞k2 F =0,then we takeθk=∞.

Note thatθk is a decreasing sequence. This implies that ifθk0 ≤1 thenθm1 for all mk0. Suppose,θ0>1.Thenbθ(1) =1. Hence using (61) we have:

fλ(Zλ1)−fλ(Zλ∞)≤1 2kZ 0 λ−Z∞λk2F =⇒ θ1≤ 1 2. Thus we get back to the former case.

Henceθk1 for all k≥1.

In addition, observe the previous deductions show that, ifθ0>1 then (20) holds true for k=1.

Combining the above observations, plugging in the value ofbθin (61) and simplifying, we get:

fλ(Zk+1 λ )−fλ(Zλk)≤ − (fλ(Zk λ)−fλ(Zλ∞))2 2kZλ0−Zλ∞k2 F . (62)

For the sake of notational convenience, we define the sequenceαk = fλ(Zkλ)−fλ(Zλ∞).It is easily

Using this notation in (62) we get: αk ≥ α2 k 2kZλ0−Zλ∞k2 Fk+1 (Sinceαk↓) ≥ αkαk+1 2kZλ0−Zλ∞k2 Fk+1. (63)

Dividing both sides of the inequality in (63), byαkαk+1we have:

α−1 k+1≥ 1 2kZλ0−Zλ∞k2 F +α−k1. (64)

Summing both sides of (64) over 1≤k≤(k−1)we get: α−1 kk−1 2kZλ0−Zλ∞k2 F +α−11. (65)

Sinceθ1≤1,we observeα1/(2kZλ0−Zλ∞k2F)≤1/2—using this in (65) and rearranging terms we

get, the desired inequality (20)—completing the proof of the Theorem.

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