Robust signal recovery via ℓ_1–2/ℓ_𝑝 minimization with partially known support

Proof of Lemma 2.3

Suppose ∥ y ∥ 0 = l ; then 𝒚 is an 𝑙-sparse vector.

Case 1: For l ≤ k 2 , by Lemma 2.2 and the fact that ∥ y ∥ 2 ≤ l ⁢ ∥ y ∥ ∞ ≤ k 2 ⁢ η , we have

Case 2: For l > k 2 , we will prove it by mathematical induction. The proof technique is similar to [2, Lemma 5.1]. Assume (2.1) holds for l - 1 . We can write 𝒚 as y = ∑ i = 1 l c i ⁢ z i , where c 1 ≥ c 2 ≥ ⋯ ≥ c l > 0 , { z i } i = 1 l are different unit vectors with one entry ± 1 and the other entries zero. As we know, ∥ y ∥ 1 = ∑ i = 1 l c i ≤ k 2 ⁢ η ≤ ( l - 1 ) ⁢ η . So

which implies Ω is not empty. We choose the largest element j ∈ Ω , which means

In order to decompose 𝒚 into the sum of a series of ( l - 1 ) -sparse signals, we define

Then, for j ≤ w ≤ l , we have

where the last inequality follows from (A.2). By simple computation, we also have

where we use (A.3) and the definition of ξ w in the second equality. On the other hand, from the definition of ζ w , we can get

where the second equality is from (A.3), and

where the second inequality is due to ∥ y ∥ ∞ ≤ η and (A.3), and the last inequality is because of (A.1). From the definition of ζ w , we know ζ w is ( l - 1 ) -sparse, so we can get the following inequality by the induction assumption:

Proof of Theorem 3.1

Let x ∗ = x + h , and let T 1 be the supports of the largest 𝑘 entries in absolute value of the h T c , where h T c is the restriction of 𝒉 to the index set T c . Denote T ¯ 0 = T ∪ T 0 , T ¯ 1 = T ∪ T 1 , T ¯ 0 c = ( T ∪ T 0 ) c , T ¯ 1 c = ( T ∪ T 1 ) c .

Since x ∗ is the solution of (1.4), we have

Due to the face that T c = T 0 ∪ T ¯ 0 c , it is easy to get

Combining (B.1), (B.2), (B.3), we can get

Then, based on the fact that

(B.4) becomes

We know that

Then, applying Lemma 2.3 with

we get

Therefore, we have

Additionally, by the Hölder inequality with a = p p - 1 and b = p ,

so by combing (B.6) and (B.7), we get

using condition (3.1) that guarantees 1 - δ k + s - C p ⁢ ( A , k + s , k ) > 0 , so that

From (B.5), we can get

and by applying Lemma 2.5 with d = k + s and γ = 2 ⁢ ∥ x T ¯ 0 c ∥ 1 + ∥ h ∥ 2 , we further get

Hence

Simply arranging the above inequality, then we have

where

Using condition (3.1), we can get B ⁢ ( k , s ) > 0 ; therefore,