sparse image representation via combined transforms - Convex ...
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98 CHAPTER 4. COMBINED IMAGE REPRESENTATION<br />
and its corresponding version with a Lagrangian multiplier λ,<br />
(RIA λ )<br />
x (k+1) = argmin ‖y − Φx‖ 2 2 + λ<br />
x<br />
N∑<br />
i=1<br />
|x i |<br />
|x (k)<br />
i<br />
| + δ .<br />
We know the following results.<br />
Theorem 4.2 Suppose we add one more constraint on x: x i ≥ 0,i=1, 2,... ,N. The sequence<br />
generated by (RIA), {x (k)<br />
i<br />
,k =1, 2, 3,...}, converges in the sense that the difference<br />
of sequential elements goes to zero:<br />
|x (k+1)<br />
i<br />
− x (k)<br />
i<br />
|→0, as k → +∞.<br />
We learned this result from [95].<br />
Theorem 4.3 If the sequence {x (k)<br />
i<br />
,k = 1, 2, 3,...} generated by (RIA λ ) converges, it<br />
converges to a local minimum of (LO λ ).<br />
Some related works can be found in [36, 106]. There is also some ongoing research, for<br />
example, the work being carried out by Boyd, Lobo and Fazel in the Information Systems<br />
Laboratories, Stanford.<br />
4.12 Proofs<br />
4.12.1 Proof of Proposition 4.1<br />
When ̂x is the solution to the problem as in (4.2), the first order condition is<br />
0=λ∇ρ(x)+2Φ T Φx − 2Φ T y,<br />
where ∇ρ(x) is the gradient vector of ρ(x). Hence<br />
1<br />
2 λ∇ρ(x) =ΦT (y − Φx) . (4.15)<br />
Recall ρ(x) = ∑ N<br />
i=1 ¯ρ(x i). It’s easy to verify that ∀x i , |¯ρ ′ (x i )|≤1. Hence for the left-hand<br />
side of (4.15), we have ‖ 1 2 λ∇ρ(x)‖ ∞ ≤ λ 2<br />
. The bound (4.6) follows. ✷