A Note on Approximating Curve with 1-Norm Regularization Method for the Split Feasibility Problem

Definition 2.1. Let φ : X → ℝ ∪ {+∞} be a convex functional, x₀ ∈ dom (φ) = {x ∈ X : φ(x)<+∞}. Set

If ∂φ(x₀) ≠ ∅, φ is said to be subdifferentiable at x₀ and ∂φ(x₀) is called the subdifferential of φ at x₀. For any ξ ∈ ∂φ(x₀), we say ξ is a subgradient of φ at x₀.

Lemma 2.2. There holds the following property:

where ∂(∥x∥) denotes the subdifferential of the functional ∥x∥ at x ∈ X.

Proof. The process of the proof will be divided into two parts.

Case  1. In the case of x = 0, for any x^* ∈ X^* such that ∥x^*∥≤1 and any y ∈ X, there holds the inequality

so we have x^* ∈ ∂(∥x∥), and thus, Conversely, for any x^* ∈ ∂(∥x∥), we have from the definition of subdifferential that Consequently, and this implies that ∥x^*∥≤1. Thus, we have verified that Combining (2.6) and (2.9), we immediately obtain

Case  2. If x ≠ 0, for any x^* ∈ {x^* ∈ X^* : ∥x^*∥ = 1, 〈x, x^*〉 = ∥x∥}, we obviously have

which means that x^* ∈ ∂(∥x∥), and thus, Conversely, if x^* ∈ ∂(∥x∥), we have hence, On the other hand, using (2.14), we get and consequently, that is, Equation (2.17) together with (2.15) implies that hence, ∥x^*∥≤1. Note that (2.14) implies that ∥x^*∥≥〈x, x^*〉/∥x∥ = 1; we assert that Thus we have from (2.14) and (2.19) that The proof is finished by combining (2.12) and (2.20).

Corollary 2.3. In l-dimensional Euclidean space ℝ^l, there holds the following result:

Let H be a Hilbert space and f : H → ℝ a functional. Recall that

• (i)

  fis convex if f(λx + (1 − λ)y) ≤ λf(x)+(1 − λ)f(y), for all 0 < λ < 1, for all x, y ∈ H;

• (ii)

  f is strictly convex if f(λx + (1 − λ)y) < λf(x)+(1 − λ)f(y), for all 0 < λ < 1, for all x, y ∈ H with x ≠ y;

• (iii)

  f is coercive if f(x) → ∞ whenever ∥x∥→∞. See [19] for more details about convex functions.

Lemma 2.4 (see [20].)Let H be a Hilbert space and C a nonempty closed convex subset of H. Let f : H → ℝ be a convex and subdifferentiable functional. Then x ∈ C is a solution of the problem

if and only if there exists some ξ ∈ ∂f(x) satisfying the following optimality condition:

Example 3.1. Let C = {(x, y) : x + y = 1}, Q = {(x, y) : x + y = 1/2} and

Then A : ℝ² → ℝ² is a bounded linear operator. Obviously, S_α = {(x, y) : x + y = 1, x ≥ 0, y ≥ 0} and it contains more than one element.

Proposition 3.2. For any α > 0, x_α ∈ S_α if and only if there exists some ξ ∈ ∂(∥x∥₁) satisfying the following inequality:

Proof. Let

then Since f is convex and differentiable with gradient f_α is convex, coercive, and subdifferentiable with the subdifferential that is, By Corollary 2.3 and Lemma 2.4, the proof is finished.

Theorem 3.3. Denote by x_α an arbitrary element of S_α, then the following assertions hold:

• (i)

  ∥x_α∥₁ is decreasing for α ∈ (0, ∞);

• (ii)

  ∥(I − P_Q)Ax_α∥ is increasing for α ∈ (0, ∞).

Proof. Let α > β > 0, for any x_α ∈ S_α, x_β ∈ S_β. We immediately obtain

Adding up (3.13) and (3.14) yields which implies ∥x_α∥₁ ≤ ∥x_β∥₁. Hence (i) holds.

Using (3.14) again, we have

which together with (i) implies and hence (ii) holds.

Definition 3.4 (see [14].)An element x^† ∈ ℱ is said to be the minimum-norm solution of SFP (1.1) in the sense of norm ∥·∥ if ∥x^†∥ = inf _x∈ℱ∥x∥. In other words, x^† is the projection of the origin onto the solution set ℱ of SFP (1.1). Thus there exists only one minimum-norm solution of SFP (1.1) in the sense of norm ∥·∥, which is always denoted by x^†.

We can also give the concept of minimum-norm solution of SFP (1.1) in other senses.

Definition 3.5. An element $$ is said to be a minimum-norm solution of SFP (1.1) in the sense of 1-norm if $$. We use ℱ₁ to stand for all minimum-norm solutions of SFP (1.1) in the sense of 1-norm and ℱ₁ is called the minimum-norm solution set of SFP (1.1) in the sense of 1-norm.

Obviously, ℱ₁ is a closed convex subset of ℱ. Moreover, it is easy to see that ℱ₁ ≠ ∅. Indeed, taking a sequence {x_n} ⊂ ℱ such that ∥x_n∥₁ → inf _x∈ℱ∥x∥₁ as n → ∞, then {x_n} is bounded. There exists a convergent subsequence $$ of {x_n}. Set $$, then $$ since ℱ is closed. On the other hand, using lower semicontinuity of the norm, we have

and this implies that $$.

However, ℱ₁ may contain more than one elements, in general (see Example 3.1, ℱ₁ = {(x, y) : x + y = 1, x, y ≥ 0}).

Theorem 3.6. Let α > 0 and x_α ∈ S_α. Then ω(x_α) ⊂ ℱ₁, where $$.

Proof. Taking $$ arbitrarily, for any α ∈ (0, ∞), we always have

Since $$ is a solution of SFP (1.1), $$. This implies that then, thus {x_α} is bounded.

Take ω ∈ ω(x_α) arbitrarily, then there exists a sequence {α_n} such that α_n → 0 and $$ as n → ∞. Put $$. By Proposition 3.2, we deduce that there exists some ξ_n ∈ ∂(∥x_n∥₁) such that

This implies that Since $$, the characterizing inequality (2.2) gives then, Combining (3.23) and (3.25), we have Consequently, we get Furthermore, noting the fact that x_n → ω and I − P_Q and A are all continuous operators, we have (I − P_Q)Aω = 0; that is, Aω ∈ Q; thus, ω ∈ ℱ. Since $$ is a minimum-norm solution of SFP (1.1) in the sense of 1-norm, using (3.21) again, we get Thus we can assert that ω ∈ ℱ₁ and this completes the proof.

Corollary 3.7. If ℱ₁ contains only one element $$, then $$, (α → 0).

Remark 3.8. It is worth noting that the minimum-norm solution of SFP (1.1) in the sense of norm ∥·∥ is very different from the minimum-norm solution of SFP (1.1) in the sense of 1-norm. In fact, x^† may not belong to ℱ₁! The following simple example shows this fact.

Example 3.9. Let C = {(x, y) : x + 2y ≥ 2, x ≥ 0, y ≥ 0},  Q = {(x, y) : x + y = 1, x ≥ 0, y ≥ 0}, and

It is not hard to see that A : ℝ² → ℝ² is a bounded linear operator and A(x, y) ^T = ((1/2)x, y) ^T, for  all (x, y) ∈ C. Obviously, ℱ = {(x, y) : x + 2y = 2, x ≥ 0, y ≥ 0}, x^† = (2/5, 4/5), but ℱ₁ = {(0,1)}. Hence, x^† ∈ ℱ∖ℱ₁.