Linear Program Relaxation of Sparse Nonnegative Recovery in Compressive Sensing Microarrays

Abstract

Compressive sensing microarrays (CSM) are DNA-based sensors that operate using group testing and compressive sensing principles. Mathematically, one can cast the CSM as sparse nonnegative recovery (SNR) which is to find the sparsest solutions subjected to an underdetermined system of linear equations and nonnegative restriction. In this paper, we discuss the 𝑙₁ relaxation of the SNR. By defining nonnegative restricted isometry/orthogonality constants, we give a nonnegative restricted property condition which guarantees that the SNR and the 𝑙₁ relaxation share the common unique solution. Besides, we show that any solution to the SNR must be one of the extreme points of the underlying feasible set.

1. Introduction

Nowadays, with the rapid development of molecular biology techniques, scientists use compressive sensing microarrays to collect the gene expression changes of patients suffer from specific diseases and test a lot of different drugs on cells genetically to look for medicine being able to change the abnormal gene expression [1, 2]. A DNA microarray is a collection of microscopic DNA spots attached to a solid surface. Each DNA spot contains a string of specific DNA sequences, known as probes. These can be a short section of a gene or other DNA element that are used to hybridize an organism's genetic sample under high-stringency conditions. Probe-target hybridization is usually detected and quantified by detection of chemiluminescence-labeled targets to infer the genetic makeup in the test sample.

Although the number of DNA sequences is extremely large, not all agents are expected to be present in a significant concentration at a given time and location. In traditional microarrays, this results in many inactive probes during sensing. On the other hand, we are often interested in only a small quantity of certain harmful biological agents. Therefore, it is important to not just detect the presence of agents in a sample but also estimate the concentrations with which they are present.

Assume that there are 𝑚 spots and 𝑛 labeled targets, and we have far fewer spots than target agents such that 𝑚≪𝑛. Mathematically, one can represent the DNA concentration of each organism as an element in a vector 𝑥∈ℝ^𝑛 and the measurements as 𝑏∈ℝ^𝑚. For 1≤𝑖≤𝑚 and 1≤𝑗≤𝑛, the probe at spot 𝑖 hybridizes to target 𝑗 with probability 𝑎_𝑖𝑗. The target 𝑗 occurs in the tested DNA sample with concentration 𝑥_𝑗, which is clearly nonnegative. Denoting by 𝐴∶=(𝑎_𝑖𝑗)_𝑚×𝑛, the process of DNA microarrays leads to the sparse nonnegative recovery (SNR) which is to find the sparsest solutions subjected to an underdetermined system of linear equations and nonnegative constraints, with the mathematical model as follows: min‖𝑥‖₀,s.t.𝐴𝑥=𝑏,𝑥≥0, 𝑃₀ where the variable vector 𝑥∈ℝ^𝑛, ‖𝑥‖₀ denotes the number of the nonzero entries of 𝑥, 𝐴∈ℝ^𝑚×𝑛 is the measurement matrix with full row rank, 𝑚≪𝑛, and 𝑏∈ℝ^𝑚.

SNR can be regarded as a special case of the sparse recovery, which is related to program min{‖𝑥‖₀∣𝐴𝑥=𝑏}. This program has sparked the significant concern and rapid development in recent years [3–5] owing to its wide applications. However, with the nonnegativity prior information about the object to be recovered in various applications such as CSM, solutions on 𝑃₀ tend to be closer to the actual situations and lead to substantial improvements in the image reconstruction. Moreover, with the nonnegative constraints, the feasible set becomes a polyhedral set instead of an affine subspace. This will bring us essential hardness in projecting on the feasible set. Thus, 𝑃₀ is more likely difficult to solve. Therefore, SNR deserves specific study.

Problem 𝑃₀ has been shown to be NP-hard [6, 7] in general from the perspective of computational complexity. One popular approach is to reconstruct the vector via the 𝑙₁ relaxation, which refers to min‖𝑥‖₁,s.t.𝐴𝑥=𝑏,𝑥≥0. 𝑃₁ Since 𝑃₁ is a standard linear program, it is easy to solve. An important issue is how to guarantee the equivalence of 𝑃₀ and 𝑃₁ in the sense that they have the same unique 𝑘-sparse solution under some conditions. Here, we call a vector 𝑥  𝑘-sparse if the number of its nonzero entries is no more than 𝑘. There has been some increasing interest and activity in this area; see, for example, [8–14]. Donoho and Tanner [9] firstly proposed that 𝑃₀ and 𝑃₁ share the common 𝑘-sparse unique solution if the polytope 𝐴𝑇 is outwardly 𝑘-neighborly, where 𝑇 is the standard simplex in ℝ^𝑛. Zhang [13] proved that 𝑃₀ and 𝑃₁ share the common 𝑘-sparse unique solution if the null space of 𝐴 is strictly half 𝑘-balance. Juditsky et al. [11] developed several different necessary and sufficient conditions for the 𝑃₀-𝑃₁ equivalence in the case of general type sign restrictions, including the nonnegative constraints as its special case. When the feasible set of 𝑃₀ is a singleton, the unknown can be recovered by optimizing any objective function over this constraint set, and 𝑃₀ and 𝑃₁ definitely get the same unique solution. In this case, Bruckstein et al. [8] got the uniqueness of the feasible solution under a sufficient condition that 𝐴 has a rowspan intersecting the positive orthant. Furthermore, Wang et al. [14] proved that the above sufficient condition is also necessary to the uniqueness of the feasible solution. Donoho and Tanner [10] proved that the underlying feasible set is a singleton if and only if the polytope 𝐴ℝ^𝑛₊ and ℝ^𝑛₊ have the same number of 𝑘-faces. Khajehnejad et al. [12] gave another equivalent condition of the uniqueness property by characterizing the support size of vectors in the null space of 𝐴.

For the 𝑙₁ relaxation of sparse recovery, one of the most significant conditions is the restricted isometry property (RIP), named by Candès and Tao [15] with the groundbreaking work of Donoho et al. [16, 17]. However, to the best of our knowledge, the nonnegative case of RIP has not been investigated. This paper will deal with this issue. We begin with investigating the solution property of SNR and show that any solution to the SNR must be one extreme point of its feasible set in Section 2. We prove in Section 3 the nonemptiness and boundedness of the solution set of 𝑃₁ and show that any solution of 𝑃₁ could be stated as the convex combination of its optimal extreme points. In Section 4, by defining the nonnegative restricted isometry/orthogonality constants, we derive a sufficient condition for exact recovery of the sparsest nonnegative image/signal via the linear program relaxation.

Now we give some notations used in the text. We use sol(⋅) and 𝑣(⋅) to denote the solution set and optimal value of problem (⋅). The 𝑒_𝑖∈ℝ^𝑛 would be the vector with only the 𝑖th entry 1 and the rest all 0. 𝑒^𝑙∈ℝ^𝑙 is the vector with each entry equal to 1; we also use 𝑒 to demonstrate that 𝑒^𝑛∈ℝ^𝑛 for short. The 𝑎_𝑖∈ℝ^𝑚 for 𝑖=1,…,𝑛 denote the column vectors of the matrix 𝐴 and 𝐴_𝐼=(𝑎_𝑖)_{𝑖∈𝐼⊂{1,…,𝑛}}. For any 𝑥∈ℝ^𝑛, 𝑥_𝑖 is the 𝑖th component and 𝐼(𝑥) is the support set of 𝑥; that is, 𝐼(𝑥)={𝑖∣𝑥_𝑖≠0,𝑖=1,…,𝑛}. For any subset 𝑇⊂{1,…,𝑛}, 𝑇^𝑐 denotes the complement set of 𝑇 out of {1,…,𝑛}.

2. Solution Property

Throughout the paper we assume that 𝒮∶=𝑥∈ℝ^𝑛∣𝐴𝑥=𝑏,𝑥≥0≠∅. (1) Apparently, 𝒮 is a polyhedral set in ℝ^𝑛. According to the representation theorem, any 𝑥∈𝒮 could be represented as follows: 𝑥=^𝑡_𝑖=1𝜆_𝑖𝑥^(𝑖)+^𝑞_𝑗=1𝜇_𝑗𝑑^(𝑗), (2) where ∑^𝑡_𝑖=1𝜆_𝑖=1, 𝜆_𝑖≥0, 𝑖=1,…,𝑡, and {𝑥^(𝑖)∈ℝ^𝑛∣𝑖=1,2,…,𝑡} are the extreme point set of 𝒮; 𝜇_𝑗≥0, 𝑗=1,…,𝑞, and {𝑑^(𝑗)∈ℝ^𝑛∣𝑗=1,2,…,𝑞} are the extreme direction set of 𝒮. Apparently, 𝐴𝑥^(𝑖)=𝑏, 𝑖=1,…,𝑡; 𝐴𝑑^(𝑗)=0,  𝑑^(𝑗)≥0, 𝑗=1,…,𝑞.

Define subsets of ℝ^𝑛 as follows: 𝑆₀{0},𝑆=sub₁=^𝑛_𝑖=1sub𝑒_𝑖⧵𝑆₀,𝑆₂=^𝑛_{𝑖1,𝑖2=1}sub𝑒_𝑖1,𝑒_𝑖2⧵¹_𝑗=0𝑆_𝑗,⋮𝑆_𝑟=^𝑛_{𝑖1,…,𝑖𝑟=1}sub𝑒_𝑖1,…,𝑒_𝑖𝑟⧵^𝑟−1_𝑗=0𝑆_𝑗,⋮𝑆_𝑛=sub𝑒₁,…,𝑒_𝑛⧵^𝑛−1_𝑗=0𝑆_𝑗, (3) where sub{𝑒_𝑖1,…,𝑒_𝑖𝑘} denotes the subspace spanned by the vectors 𝑒_𝑖1,…,𝑒_𝑖𝑘, 𝑘=1,…,𝑛. Clearly, {𝑆₁,𝑆₂,…,𝑆_𝑛} forms a partition of ℝ^𝑛; that is, ⋃^𝑛_𝑗=1𝑆_𝑗=ℝ^𝑛 and 𝑆_𝑖⋂𝑆_𝑗=∅, 𝑖≠𝑗. Moreover, ‖𝑥‖₀=𝑟 for any 𝑥∈𝑆_𝑟. Along with the nonemptiness of 𝒮, it is easy to see that 𝒮 must intersect one of these sets, hence 𝑣(𝑃₀)=min{𝑟∈ℝ∣𝒮⋂𝑆_𝑟≠∅} and sol(𝑃₀)≠∅. Furthermore, we have the following result for the optimal value of 𝑃₀.

Lemma 1. Assume that 𝑣(𝑃₀)=𝑘, one must have 𝑘≤𝑚.

Proof. Suppose that the conclusion is not true; that is, there is 𝑥^∗∈sol(𝑃₀) with ‖𝑥^∗‖₀=𝑘>𝑚. Without loss of generality, let 𝑥^∗_𝑖>0, 𝑖=1,…,𝑘. We get 𝑥^∗₁𝑎₁+⋯+𝑥^∗_𝑘𝑎_𝑘=𝑏. (4) Meanwhile, rank(𝐴)=𝑚<𝑘. Thus, {𝑎_𝑖∈ℝ^𝑚∣𝑖=1,…,𝑘} must be linearly dependent; that is, there exist 𝑑₁,…,𝑑_𝑘, not all zero, such that 𝑑₁𝑎₁+⋯+𝑑_𝑘𝑎_𝑘=0. (5) Assume that 𝑑₁>0. By denoting 𝑑=(𝑑₁,…,𝑑_𝑘,0,…,0)^𝑇∈ℝ^𝑛, we get 𝐴𝑑=0. Taking 𝛿=min{𝑥^∗_𝑖/𝑑_𝑖∣𝑑_𝑖>0,𝑖=1,…,𝑘}, it holds that 𝐴𝑥^∗−𝛿𝑑=𝐴𝑥^∗𝑥−𝛿𝐴𝑑=𝑏,_∗𝑥−𝛿𝑑≥0,_∗−𝛿𝑑₀≤𝑘−1. (6) This is a contradiction with 𝑥^∗∈sol(𝑃₀). We complete the proof.

To characterize property of the solution set sol(𝑃₀), we need the next lemma. In particular, this brand new result will play a key role in proposing the sufficient condition of the uniqueness of sol(𝑃₀) in Section 4.

Lemma 2. Any two distinct solutions of 𝑃₀ must have different support sets.

Proof. Assume for contradiction that 𝑥^∗ and 𝑥 are two different solutions of 𝑃₀, 𝑥^∗≠𝑥. If 𝐼(𝑥^∗)=𝐼(𝑥), we have 𝑥^∗_𝑖>0, 𝑥_𝑖>0, for all 𝑖∈𝐼(𝑥^∗). Set 𝜆=min{min_{𝑖∈𝐼(𝑥∗)}{𝑥^∗_𝑖/𝑥_𝑖},min_{𝑖∈𝐼(𝑥∗)}{𝑥_𝑖/𝑥^∗_𝑖}}. Since 𝑥^∗≠𝑥, it must hold 𝜆<1. When 𝜆=𝑥^∗_𝑖0/𝑥_𝑖0, take 1𝑥=𝑥1−𝜆^∗−𝜆1−𝜆𝑥. (7) It is easy to see that 𝑥_𝑖=1𝑥1−𝜆^∗𝑖−𝜆𝑥1−𝜆^𝑖∈{0},𝑖=𝑖₀,ℝ₊,𝑖∈𝐼𝑥^∗⧵𝑖₀. (8) Thus, we have 𝐴𝑥=𝑏, 𝑥≥0 and ‖𝑥‖₀≤‖𝑥^∗‖₀−1. This is a contradiction with the optimality of 𝑥^∗. When 𝜆=𝑥_𝑗0/𝑥^∗_𝑗0, just taking 𝑥=(1/(1−𝜆))𝑥−(𝜆/(1−𝜆))𝑥^∗ instead, we get the contradiction by a similar way. The proof is completed.

Now we are in a position to give the main theorem in this section.

Theorem 3. Any solution of 𝑃₀ must be one of the extreme points of 𝒮.

Proof. Given any solution 𝑥^∗∈sol(𝑃₀) with representation 𝑥^∗=^𝑡_𝑖=1𝜆^∗_𝑖𝑥^(𝑖)+^𝑞_𝑗=1𝜇^∗_𝑗𝑑^(𝑗), (9) where ∑^𝑡_𝑖=1𝜆^∗_𝑖=1, 𝜆^∗_𝑖≥0, 𝑥^(𝑖)≥0, 𝐴𝑥^(𝑖)=𝑏, for all 𝑖=1,…,𝑡; 𝜇^∗_𝑗≥0, 𝑑^(𝑗)≥0, 𝐴𝑑^(𝑗)=0, for all 𝑗=1,…,𝑞. We only need to prove that ^𝑝_𝑖=1𝜆^∗_𝑖=1,𝜆^∗_𝑖∈{}𝜇0,1,𝑖=1,…,𝑡,_∗𝑗=0,𝑗=1,…,𝑞, (10) in (9). To this end, we have the following three steps.
Firstly, we claim that in (9), 𝐼𝑥^(𝑖)=𝐼𝑥^∗,𝑖∈𝑖∣𝜆^∗_𝑖,𝐼𝑑>0,𝑖=1,…,𝑡_(𝑗)⊂𝐼𝑥^∗,𝑗∈𝑗∣𝜇^∗_𝑗>0,𝑗=1,…,𝑞. (11) According to the fact that 𝑥^(𝑖)≥0, 𝑖=1,…,𝑡, and 𝜇^∗_𝑗≥0, 𝑑^(𝑗)≥0, 𝑗=1,…,𝑞, one has for any 𝑙∈𝐼(𝑥^(𝑖)), 𝑖∈{𝑖∣𝜆^∗_𝑖>0,𝑖=1,2,…,𝑡}, 𝑥^∗_𝑙=^𝑡_𝑖=1𝜆^∗_𝑖𝑥_𝑙^(𝑖)+^𝑞_𝑗=1𝜇^∗_𝑗𝑑_𝑙^(𝑗)>0, (12) which means that 𝑙∈𝐼(𝑥^∗). This implies 𝐼(𝑥^(𝑖))⊂𝐼(𝑥^∗). Similarly, we get 𝐼(𝑑^(𝑗))⊂𝐼(𝑥^∗). On the contrary, from the optimality of 𝑥^∗ and (9), we know that 𝐼(𝑥^∗)⊂𝐼(𝑥^(𝑖)).
Secondly, we will show that 𝜇^∗_𝑗=0, 𝑗=1,…,𝑞. If this is not true, there is an index 𝑗₀ such that 𝜇^∗_𝑗0>0, then 𝜇^∗_𝑗0𝑑^(𝑗0) has at least one positive component. Denote 𝑑=∑^𝑞_𝑗=1𝜇^∗_𝑗𝑑^(𝑗), so 𝐴𝑑=0 and {𝑑_𝑙>0∣𝑙=1,…,𝑛}≠∅. Noting that 𝑥^(𝑖)≥0, (11) implies the fact that 𝐼(∑^𝑡_𝑖=1𝜆^∗_𝑖𝑥^(𝑖))=𝐼(𝑥^∗),𝐼(𝑑)⊂𝐼(𝑥^∗). Take 𝛿=min{(∑^𝑡_𝑖=1𝜆^∗_𝑖𝑥^(𝑖))_𝑙/𝑑_𝑙∣𝑑_𝑙>0,𝑙=1,…,𝑛}, and 𝑥=^𝑡_𝑖=1𝜆^∗_𝑖𝑥^(𝑖)−𝛿𝑑. (13) Without loss of generality, set 𝛿=(∑^𝑡_𝑖=1𝜆^∗_𝑖𝑥^(𝑖))_𝑙0/𝑑_𝑙0. It is easy to verify that 𝑥_𝑗=^𝑡_𝑖=1𝜆^∗_𝑖𝑥^(𝑖)_𝑗−𝛿𝑑_𝑗∈{0},𝑗=𝑙₀ℝ₊,𝑖∈𝐼𝑥^∗⧵𝑙₀, (14) hence 𝐴‖𝑥‖𝑥=𝑏,𝑥≥0,₀≤𝑥^∗₀−1, (15) which is a contradiction with the optimality of 𝑥^∗.
Thirdly, we will prove that ∑^𝑡_𝑖=1𝜆^∗_𝑖=1, 𝜆^∗_𝑖∈{0,1}, 𝑖=1,…,𝑡. Suppose that there exist 𝜆^∗₁>0, 𝜆^∗₂>0, and 𝜆^∗₁+𝜆^∗₂=1 and two different extreme points of 𝒮, say 𝑥⁽¹⁾,𝑥⁽²⁾, such that 𝑥^∗=𝜆^∗₁𝑥⁽¹⁾+𝜆^∗₂𝑥⁽²⁾. (16) Based on (11), we have 𝐼(𝑥⁽¹⁾)=𝐼(𝑥⁽²⁾)=𝐼(𝑥^∗), hence 𝑥⁽¹⁾∈sol(𝑃₀), 𝑥⁽²⁾∈sol(𝑃₀). Nevertheless, by Lemma 2, this is impossible. Hence, we show that 𝑥^∗∈𝑥⁽¹⁾,𝑥⁽²⁾,…,𝑥^(𝑡). (17) We complete the argument.

Theorem 3 tells us that each solution of 𝑃₀ lies in the extreme point set of 𝒮. Here is a concrete example.

Example 4. Let 𝐴=¹¹⁻³₋₁₀₂,𝑏=³₋₂. Obviously, the solution set and optimal value of 𝑃₀ are sol𝑃₀={(2,1,0)},𝑣𝑃₀=2, (18) respectively. In this case, (2,1,0) is the only extreme point of 𝒮. While the solution set and optimal value of (𝑃)∶=min{‖𝑥‖₀∣𝐴𝑥=𝑏} are sol(𝑃)={(0,0,−1)},𝑣(𝑃)=1, (19) respectively.

At the end of this section, we consider the 𝑙_𝑝(0<𝑝<1) relaxation of 𝑃₀min‖𝑥‖^𝑝_𝑝=^𝑛_𝑖=1𝑥^𝑝_𝑖,s.t.𝐴𝑥=𝑏,𝑥≥0. 𝑃_𝑝 Clearly, 𝑃_𝑝 is a concave relaxation of 𝑃₀. For the program 𝑃_𝑝, Ge et al. [7] derived the useful result as the following.

Lemma 5. The set of all extreme points of 𝒮 is exactly the set of all local minimizers to 𝑃_𝑝.

This lemma implies that any global solution of 𝑙_𝑝 relaxation must be one of its extreme points. From Theorem 3 and Lemma 5, we immediately draw a new proposition.

Proposition 6. For any 𝑝∈(0,1), there exists an extreme point of 𝒮 that is both an exact solution of 𝑃₀ and a local minimizer of 𝑃_𝑝.

This is different from the result of Fung and Mangasarian in [18], where they showed that for sufficiently small 𝑝∈(0,1), there exists an extreme point 𝑥 of the polyhedral set 𝒯, obtained by lifting the set 𝒮, such that 𝑥 is an exact solution of 𝑃₀ and a global solution of the 𝑙_𝑝 relaxation.

3. Linear Program Relaxation

Consider the linear program relaxation 𝑃₁. Since the linear objective function ⟨𝑒,𝑥⟩ is bounded below over the feasible set, based on the Frank-Wolfe theorem, the minimum of 𝑃₁ is attainable. Among all the extreme points of 𝒮, {𝑥^(𝑖)∈R^𝑛∣𝑖=1,…,𝑡}, we call 𝑥^(𝑖) an optimal extreme point if it also meets ⟨𝑒,𝑥^(𝑖)⟩=𝑣(𝑃₁).

Proposition 7. Any 𝑥^∗∈sol(𝑃₁) could be stated as the convex combination of optimal extreme points of 𝑃₁. Hence,sol(𝑃₁) is bounded.

Proof. Given any 𝑥^∗∈sol(𝑃₀) with representation (9). If there is 𝑖₀∈{1,…,𝑡} such that 𝑥^(𝑖0) is not an optimal extreme point of 𝑃₁ and 𝜆^∗_𝑖0>0, we have ⟨𝑒,𝑥^(𝑖0)⟩>𝑣(𝑃₁), hence 𝑣𝑃₁=𝑒,𝑥^∗=^𝑡_𝑖=1𝜆^∗_𝑖𝑒,𝑥^(𝑖)+^𝑞_𝑗=1𝜇^∗_𝑗𝑒,𝑑^(𝑗)≥^𝑡_𝑖=1𝜆^∗_𝑖𝑒,𝑥^(𝑖)>^𝑡_𝑖=1𝜆^∗_𝑖𝑣𝑃₁=𝑣𝑃₁, (20) which is a contradiction.
Similarly, if there is 𝑗₀∈{1,…,𝑞} such that 𝜇^∗_𝑗0>0, we have 𝜇^∗_𝑗0⟨𝑒,𝑑^(𝑗0)⟩>0, hence 𝑣𝑃₁=𝑒,𝑥^∗=^𝑡_𝑖=1𝜆^∗_𝑖𝑒,𝑥^(𝑖)+^𝑞_𝑗=1𝜇^∗_𝑗𝑒,𝑑^(𝑗)>^𝑡_𝑖=1𝜆^∗_𝑖𝑒,𝑥^(𝑖)≥^𝑡_𝑖=1𝜆^∗_𝑖𝑣𝑃₁=𝑣𝑃₁, (21) which is a contradiction. This completes the proof.

From the above proposition, we know that linear program 𝑃₁ has at least one optimal extreme point. Thus, we could use simplex method or interior point method to solve 𝑃₁.

4. Nonnegative Restricted Property

In the framework of 𝑙₁ relaxation, a significant problem is how to guarantee the exact recovery of sparse image/signal via the 𝑙₁ relaxation. One of the most important qualifications is the restricted isometry property; see [15]. Recall that the 𝑘-restricted isometry constants (RIC) 𝛿_𝑘 is the smallest scalar satisfying 1−𝛿_𝑘‖𝑥‖²₂≤‖𝐴𝑥‖²₂≤1+𝛿_𝑘‖𝑥‖²₂,∀‖𝑥‖₀≤𝑘. (22) Similarly, the 𝑘,𝑘^′-restricted orthogonality constants (ROC) 𝜃_𝑘,𝑘′ for 𝑘+𝑘^′≤𝑛 is defined as the smallest scalar satisfying 𝐴𝑥,𝐴𝑦≤𝜃_𝑘,𝑘′‖𝑥‖₂𝑦₂,∀‖𝑥‖₀≤𝑘;∀𝑦₀≤𝑘^′, (23) where 𝑥 and 𝑦 have disjoint support sets. The RIC 𝛿_𝑘 and ROC 𝜃_𝑘,𝑘′ measure how close each submatrix of 𝐴 with certain cardinality is behaving like an orthonormal system. Under some restricted isometry property, one can get the sparse recovery via its 𝑙₁ relaxation. Nevertheless, for the nonnegative case, the sparse recovery may maintain new characterizations. Above all, we define NRIC and NROC.

Definition 8. Let 𝐴∈ℝ^𝑚×𝑛. We define the nonnegative 𝑘-restricted isometry constants (NRIC) 𝛿⁺_𝑘 as the smallest number satisfying 1−𝛿⁺_𝑘‖𝑥‖²₂≤‖𝐴𝑥‖²₂≤1+𝛿⁺_𝑘‖𝑥‖²₂,∀𝑥≥0,‖𝑥‖₀≤𝑘. (24) Similarly, we define the nonnegative 𝑘,𝑘^′-restricted orthogonality constants (NROC) 𝜃⁺_𝑘,𝑘′ for 𝑘+𝑘^′≤𝑛 as the smallest number satisfying 𝐴𝑥,𝐴𝑦≤𝜃⁺_𝑘,𝑘′‖𝑥‖₂𝑦₂,∀𝑥≥0,‖𝑥‖₀≤𝑘,∀𝑦≥0,𝑦₀≤𝑘^′, (25) with 𝐼(𝑥) and 𝐼(𝑦) being disjoint sets.

Clearly, 𝛿⁺_𝑘≤𝛿_𝑘,𝜃⁺_𝑘,𝑘′≤𝜃_𝑘,𝑘′. (26) Moreover, the numbers 𝛿⁺_𝑘 and 𝜃⁺_𝑘,𝑘′ are nondecreasing in 𝑘,𝑘^′.

By employing the projections of vectors in the null space of 𝐴 to ℝ^𝑛₊, we now provide a sufficient condition to determine a solution of 𝑃₀.

Theorem 9. Suppose that 𝑘≥1 is such that 𝛿⁺_𝑘+𝜃⁺_{𝑘,𝑘−1}<1 and 𝑥^∗∈𝒮 with ‖𝑥^∗‖₀=𝑘. Then, 𝑥^∗ is a solution of 𝑃₀.

Proof. We complete the proof by contradiction. If this is not true, there exists 𝑥 such that 𝐴𝑥=𝑏, 𝑥≥0,‖𝑥‖₀≤𝑘−1. Set ℎ=𝑥^∗−𝑥. Clearly, 𝐴ℎ=0. Take ℎ=ℎ⁺−ℎ⁻ with ℎ⁺ and ℎ⁻ being the projections of ℎ and −ℎ to ℝ^𝑛₊, respectively. We have ℎ⁺≥0, ℎ⁻≥0, ⟨ℎ⁺,ℎ⁻⟩=0. In particular, ℎ⁺_𝑖=𝑥^∗_𝑖−𝑥_𝑖+,𝑖∈𝐼𝑥^∗,(ℎ0,else,₋)_𝑖=𝑥^∗_𝑖−𝑥_𝑖−,𝑖∈𝐼(𝑥),0,else. (27) Therefore, 1≤‖ℎ‖₀≤𝑥^∗₀+‖𝑥‖₀ℎ≤2𝑘−1,1≤⁺₀‖ℎ≤𝑘,0≤⁻‖₀≤𝑘−1. (28) Thus, we get 0=‖𝐴ℎ‖²₂=𝐴ℎ⁺−𝐴ℎ⁻²₂=𝐴ℎ⁺²₂+‖𝐴ℎ⁻‖²₂−2𝐴ℎ⁺,𝐴ℎ⁻≥1−𝛿⁺_𝑘ℎ⁺²₂+1−𝛿⁺_𝑘−1‖ℎ⁻‖²₂−𝜃⁺_{𝑘,𝑘−1}ℎ⁺²₂+‖ℎ⁻‖²₂≥1−𝛿⁺_𝑘−𝜃⁺_{𝑘,𝑘−1}ℎ⁺²₂+1−𝛿⁺_𝑘−1−𝜃⁺_{𝑘,𝑘−1}‖ℎ⁻‖²₂>0, (29) in which the first inequality is due to (24), (25), and the fact that 2𝑎𝑏≤𝑎²+𝑏², and the last inequality is because of the assumption of 𝛿⁺_𝑘+𝜃⁺_{𝑘,𝑘−1}<1 and the monotonicity of 𝛿⁺_𝑘 in 𝑘. This is a contradiction. Therefore, 𝑥^∗∈sol(𝑃₀).

With the special result that any two solutions of 𝑃₀ have different support sets, we next derive a sufficient condition on the uniqueness of solution to 𝑃₀.

Theorem 10. Suppose that 𝑘≥1 is such that 𝛿⁺_𝑘+𝜃⁺_𝑘,𝑘<1 and 𝑥^∗∈𝒮 with ‖𝑥^∗‖₀=𝑘. Then, 𝑥^∗ is the unique solution of 𝑃₀.

Proof. Since 𝛿⁺_𝑘+𝜃⁺_𝑘,𝑘<1 implies 𝛿⁺_𝑘+𝜃⁺_{𝑘,𝑘−1}<1, we know that 𝑥^∗∈sol(𝑃₀). Now we just need to verify that 𝑥^∗ is the unique solution of 𝑃₀. Assume that this is not true; that is, there is another solution 𝑥≠𝑥^∗. According to Lemma 2, it must hold 𝐼(𝑥)≠𝐼(𝑥^∗). Take ℎ=𝑥^∗−𝑥. By the argument similar to that in the proof of Theorem 9, we get 2≤‖ℎ‖₀≤𝑥^∗₀+‖𝑥‖₀ℎ≤2𝑘,1≤⁺₀‖ℎ≤𝑘,1≤⁻‖₀≤𝑘, (30) and the contradiction. We conclude the proof.

Now we are ready to give the main result of this paper, which is called the nonnegative restricted property.

Theorem 11. Assume that 𝑘≥1 is such that 𝛿⁺_2𝑘+√2+1𝜃⁺_𝑘,2𝑘<1, (31) and 𝑥^∗∈𝒮 with ‖𝑥^∗‖₀=𝑘. Then 𝑥^∗ is exactly the common unique minimizer of 𝑃₀ and 𝑃₁.

Proof. Since (31) implies 𝛿⁺_𝑘+𝜃⁺_𝑘,𝑘<1, sol(𝑃₀)={𝑥^∗} by Theorem 10.
Suppose that 𝑥 is a solution of 𝑃₁. Take ℎ=𝑥−𝑥^∗. To get sol(𝑃₁)={𝑥^∗}, it suffices to verify that ℎ=0. The proof includes three steps, the first two steps are parallel to that in [19], in the third step, we utilize the technique of projecting the null space of 𝐴 on ℝ^𝑛₊; for details, see (42) and the argument around it.
Firstly, we introduce a partition of {1,…,𝑛}. Let 𝑇₀ be the support set of 𝑥^∗, 𝑇₁ the index set including the first 𝑘 large components of 𝑥 in 𝑇^𝑐₀, 𝑇₂ the index set including the next 𝑘 large components of 𝑥 in 𝑇^𝑐₀⧵𝑇₁, and so on. Thus, 𝑇_𝑗𝑛=𝑘,𝑗=0,1,…,𝑘𝑇−1,_𝑗𝑛≤𝑘,𝑗=𝑘. (32) Moreover, for any 𝑗=0,1,…,[𝑛/𝑘], we define ℎ_𝑇𝑗𝑖=ℎ_𝑖,𝑖∈𝑇_𝑗,0,else, (33) which is exactly that for any 𝑗=1,…,[𝑛/𝑘]ℎ_𝑇𝑗𝑖=𝑥_𝑖−𝑥^∗_𝑖,𝑖∈𝑇₀,𝑥_𝑖,𝑖∈𝑇_𝑗,0,else. (34) Therefore, 𝐴ℎ=0,ℎ=∑^[𝑛/𝑘]_𝑗=0ℎ_𝑇𝑗, and ℎ_𝑇𝑗𝑛≥0,𝑗=1,…,𝑘,ℎ_𝑇𝑗0𝑛≤𝑘,𝑗=0,1,…,𝑘. (35)
Next, we show that ‖ℎ_{(𝑇0∪𝑇1)}^𝑐‖₂ is bounded by ‖ℎ_𝑇0‖₁. Note that for each 𝑗≥2, ℎ_𝑇𝑗2≤√𝑘ℎ_𝑇𝑗∞≤1√𝑘ℎ_{𝑇𝑗−11}, (36) where the second inequality is because of the monotonicity of ℎ_𝑖 on 𝑇^𝑐₀, and ^[𝑛/𝑘]_𝑗=2ℎ_𝑇𝑗2≤1√𝑘^{[𝑛/𝑘]−1}_𝑗=1ℎ_𝑇𝑗1≤1√𝑘ℎ_𝑇^𝑐₀₁. (37) This gives ℎ_{(𝑇0∪𝑇1)}^𝑐₂=^[𝑛/𝑘]_𝑗=2ℎ_𝑇𝑗2≤^[𝑛/𝑘]_𝑗=2ℎ_𝑇𝑗2≤1√𝑘ℎ_𝑇^𝑐₀₁. (38) In fact, 𝑥^∗_𝑇01=𝑥^∗₁≥‖𝑥‖₁=𝑥^∗+ℎ₁=𝑥^∗_𝑇0+ℎ_𝑇01+𝑥^∗_𝑇^𝑐₀+ℎ_𝑇^𝑐₀₁≥𝑥^∗_𝑇01−ℎ_𝑇01+ℎ_𝑇^𝑐₀₁−𝑥^∗_𝑇^𝑐₀₁=𝑥^∗_𝑇01−ℎ_𝑇01+ℎ_𝑇^𝑐₀₁, (39) which implies ℎ_𝑇^𝑐₀₁≤ℎ_𝑇01. (40) By applying (38) and (40), we have ℎ_{(𝑇0∪𝑇1)}^𝑐₂≤1√𝑘ℎ_𝑇^𝑐₀₁≤1√𝑘ℎ_𝑇01. (41)
Finally, we show that ‖ℎ_{𝑇0∪𝑇1}‖₂=0. By utilizing the projection ℎ=ℎ⁺−ℎ⁻, where ℎ⁺ and ℎ⁻ are projections of ℎ and −ℎ on ℝ^𝑛₊, we have ℎ_{(𝑇0∪𝑇1)}=ℎ⁺_{(𝑇0∪𝑇1)}−ℎ⁻_{(𝑇0∪𝑇1)}. Moreover, ℎ⁺_{𝑇0∪𝑇1𝑖}=𝑥_𝑖−𝑥^∗_𝑖+,𝑖∈𝑇₀,𝑥_𝑖,𝑖∈𝑇₁,ℎ0,else,_{−𝑇0∪𝑇1𝑖}=𝑥_𝑖−𝑥^∗_𝑖−,𝑖∈𝑇₀,0,𝑖∈𝑇₁,0,else. (42) It is easy to see ℎ⁺_{𝑇0∪𝑇10}ℎ≤2𝑘,_{−𝑇0∪𝑇10}≤𝑘. (43) From 𝐴ℎ=0, we compute 𝐴ℎ_{(𝑇0∪𝑇1)}²₂=𝐴ℎ_{(𝑇0∪𝑇1)},𝐴ℎ−𝐴ℎ_{(𝑇0∪𝑇1)}^𝑐=−^[𝑛/𝑘]_𝑗=2𝐴ℎ_{(𝑇0∪𝑇1)},𝐴ℎ_𝑇𝑗. (44) On one hand, based on the definition of NROC, (35), and (43), for 𝑗≥2, −𝐴ℎ_{(𝑇0∪𝑇1)},𝐴ℎ_𝑇𝑗=−𝐴ℎ⁺_{𝑇0∪𝑇1},𝐴ℎ_𝑇𝑗+𝐴ℎ⁻_{𝑇0∪𝑇1},𝐴ℎ_𝑇𝑗≤𝐴ℎ⁺_{𝑇0∪𝑇1},𝐴ℎ_𝑇𝑗+𝐴ℎ⁻_{𝑇0∪𝑇1},𝐴ℎ_𝑇𝑗≤𝜃⁺_𝑘,2𝑘ℎ⁺_{𝑇0∪𝑇12}⋅ℎ_𝑇𝑗2+𝜃⁺_𝑘,𝑘ℎ⁻_{𝑇0∪𝑇12}⋅ℎ_𝑇𝑗2≤𝜃⁺_𝑘,2𝑘ℎ_𝑇𝑗2⋅ℎ⁺_{𝑇0∪𝑇12}+ℎ⁻_{𝑇0∪𝑇12}≤√2𝜃⁺_𝑘,2𝑘ℎ_𝑇𝑗2⋅ℎ_{(𝑇0∪𝑇1)2}, (45) where the last inequality is by the fact that (𝑎+𝑏)²≤2(𝑎²+𝑏²). On the other hand, together with the definition of NRIC and (43), one has 𝐴ℎ_{(𝑇0∪𝑇1)}²₂=𝐴ℎ⁺_{𝑇0∪𝑇1}−𝐴ℎ⁻_{𝑇0∪𝑇1}²₂=𝐴ℎ⁺_{𝑇0∪𝑇1}²₂+𝐴ℎ⁻_{𝑇0∪𝑇1}²₂−2𝐴ℎ⁺_{𝑇0∪𝑇1},𝐴ℎ⁻_{𝑇0∪𝑇1}≥1−𝛿⁺_2𝑘ℎ⁺_{𝑇0∪𝑇1}²₂+1−𝛿⁺_𝑘ℎ⁻_{𝑇0∪𝑇1}²₂−2𝜃⁺_𝑘,2𝑘ℎ⁺_{𝑇0∪𝑇12}⋅ℎ⁻_{𝑇0∪𝑇12}≥1−𝛿⁺_2𝑘ℎ⁺_{𝑇0∪𝑇1}²₂+1−𝛿⁺_𝑘ℎ⁻_{𝑇0∪𝑇1}²₂−𝜃⁺_𝑘,2𝑘ℎ⁺_{𝑇0∪𝑇1}²₂+ℎ⁻_{𝑇0∪𝑇1}²₂=1−𝛿⁺_2𝑘−𝜃⁺_𝑘,2𝑘ℎ⁺_{𝑇0∪𝑇1}²₂+1−𝛿⁺_𝑘−𝜃⁺_𝑘,2𝑘ℎ⁻_{𝑇0∪𝑇1}²₂≥1−𝛿⁺_2𝑘−𝜃⁺_𝑘,2𝑘ℎ_{(𝑇0∪𝑇1)}²₂. (46) Therefore, we compute 1−𝛿⁺_2𝑘−𝜃⁺_𝑘,2𝑘ℎ_{(𝑇0∪𝑇1)}²₂≤√2𝜃⁺_𝑘,2𝑘ℎ_{(𝑇0∪𝑇1)2}⋅^[𝑛/𝑘]_𝑗=2ℎ_𝑇𝑗2(),()()≤√by4445,and462𝜃⁺_𝑘,2𝑘ℎ_{(𝑇0∪𝑇1)2}⋅1√𝑘ℎ_𝑇^𝑐₀₁()≤√by372𝜃⁺_𝑘,2𝑘ℎ_{(𝑇0∪𝑇1)2}⋅1√𝑘ℎ_𝑇01()≤√by402𝜃⁺_𝑘,2𝑘ℎ_{(𝑇0∪𝑇1)2}⋅ℎ_𝑇02≤√2𝜃⁺_𝑘,2𝑘ℎ_{(𝑇0∪𝑇1)}²₂, (47) where the forth inequality is from the fact that ‖ℎ_𝑇0‖²₁≤𝑘‖ℎ_𝑇0‖²₂. Then the assumption 𝛿⁺_2𝑘+(√2+1)𝜃⁺_𝑘,2𝑘<1 forces ℎ_{(𝑇0∪𝑇1)}²₂=0. (48) Thus, ℎ_{(𝑇0∪𝑇1)}=0. (49) Therefore, we get ℎ_{(𝑇0∪𝑇1)}^𝑐=0 by (41), hence ‖ℎ‖=0. This is exactly what we want. We complete the proof.

5. Conclusion

In this paper, we have derived a nonnegative restricted property condition, which ensures the exact recovery of sparse nonnegative image/signal via the linear program relaxation. Since the NRIC and NROC are defined in ℝ^𝑛₊, there may be more types of measurement matrices satisfying the nonnegative restricted property than that in the case of RIP, regardless of random matrices or deterministic matrices. As a byproduct of the main result, we have investigated the solution property of the sparse nonnegative recovery and shown that any solution of 𝑃₀ must be one of the extreme points of its feasible set. However, it is not clear whether a given extreme point of the feasible set is a solution to 𝑃₀. This can serve as a target for future work.

Acknowledgments

The work was supported in part by the National Basic Research Program of China (2010CB732501) and the National Natural Science Foundation of China (11171018).