Binary sparse signal recovery with binary least squares

Lemma 1 ([29])

Consider the ( k + 1 ) -th step in the BLS algorithm. Then BLS selects the index

Lemma 2 ([29])

Assuming that S 1 ∈ Ω , Φ is a column normalization matrix and satisfies RIP of order | S 1 | + 1 , then for any i ∈ Ω ∖ S 1 , we have that

Lemma 3 ([17])

Suppose that S 1 , S 2 satisfy | S 2 ∖ S 1 | > 1 , Φ is a column normalization matrix and satisfies RIP of order | S 1 ∪ S 2 | . Let w ∈ R n and supp ⁡ ( w ) = S 2 . One has

Consider y = Φ ⁢ x . For any constant α > 0 and k ∈ { 0 , 1 , … , K − 1 } , one has

for all j ∈ Ω ∖ T , provided that Φ satisfies δ K − k + 1 < 1 / α + 1 .

Proof

The proof of Lemma 4 is similar to [29] except for the form of the residual. Define

Let

where

Then one has

where (A.1) follows from

On the other hand, applying Lemma 3, we can have

where (a) is from

By combining (A.1) and (A.2), we can get

So it is easy to see that ∥ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 ∥ 2 2 > α ⁢ ∥ x T ∖ Λ k − 1 ∥ 2 ⁢ | ⟨ Φ T ∖ Λ k − 1 ⁢ x T ∖ Λ k − 1 , Φ j ⟩ | . ∎

Lemma 5 ([8, 25])

Let S 1 ⊂ T and w ∈ R | S 1 | . Suppose that Φ satisfies δ | S 1 | < 1 . One has δ | S 1 | ≤ ( | S 1 | − 1 ) ⁢ μ .

Proof

The proof of Theorem 1 is similar to [29] except for the form of the residual.

Our proof is based on mathematical induction. Assume that the BLS algorithm selects the correct index in the first k − 1 iterations, i.e., Λ k − 1 ⊂ T , so according to Algorithm 1, it is only necessary to prove s k ∈ T in the 𝑘-th iteration, i.e.,

According to Algorithm 1, we can get

Because Φ is the column normalization matrix, there is ∥ P Λ k − 1 ⟂ ⁢ Φ i ∥ 2 ≤ ∥ Φ i ∥ 2 = 1 . Then

where (a) follows from the Cauchy–Schwarz inequality.

On the right side of (B.1), one has

where

(a) follows from Lemma 2 and the monotonicity of RIP.

Next, to prove (B.1), we should prove

According to Lemma 4, letting

from (A.1) and (A.2), we can have

So

Note that

then we have that

and

So

Thus (B.4) can be changed into

In order to prove (B.3), by (B.8) and (B.7), we only need to show

Through (B.2)–(B.9), it can be easily obtained that, when

(B.3) holds. So, for 1 ≤ k ≤ | T | , we have s k ∈ T .

In the following, we will prove that the BLS algorithm iterates 𝐾 steps. Firstly, for 1 ≤ k ≤ K − 1 , one has

where (a) is from 1 − δ K + 1 2 ⁢ ( 1 − δ K + 1 ⁢ K + 1 ) ≤ 1 − K + 1 ⁢ δ K + 1 ≤ 1 − δ K + 1 ≤ 1 − δ K + 1 , and (b) is from (3.2). So the BLS algorithm iterates at least | T | steps. When k = K , ∥ r K ∥ 2 = ∥ Φ T ∖ T K ⁢ x T ∖ T K + ν ∥ 2 = ∥ ν ∥ 2 ≤ η . So, to sum up, the BLS algorithm iterates 𝐾 steps. ∎

Proof

We divide the proof into two steps. Firstly, it is proved that, for 1 ≤ k ≤ K , there is s k ∈ T ; then it is proved that the BLS algorithm iterates 𝐾 steps. Similarly, assuming that the correct index is selected in the first k − 1 steps, it is only necessary to show that (B.1) is valid in the 𝑘-th iteration.

For the left side of (B.1), we have

To the right of (B.1), we have

At this point, we need to prove

Let

The lower bound of 𝑝 is

where (a) is from Lemma 5. The upper bound of 𝑞 is

where (a) follows from Lemma 2, and (b) follows from Lemma 5.

Thus, according to (C.2) and (C.3), one has

where (a) is from ( | Λ k − 1 | ⁢ | T ∖ Λ k − 1 | − | Λ k − 1 | ) ⁢ μ 2 > 0 , (b) is from

and the value of

decreases as 𝑘 increases (i.e., the maximum value ( 2 ⁢ K − 1 ) ⁢ μ is obtained when k = 1 ).

On the other hand,

In order to prove (C.1), according to (3.3) and (C.4), we need to show

Inequality (3.4) can guarantee that (C.5) holds.

Now, we will prove that BLS iterates 𝐾 steps under ∥ r k ∥ 2 ≤ η . For 1 ≤ k ≤ K − 1 , one has

where (a) is from Lemmas 3 and 5, (b) is from (3.4). That is, the BLS algorithm iterates at least | T | steps. In particular,

so BLS iterates 𝐾 steps. ∎