The same growth of FB-SOC merit function and NR-SOC merit function 1

The same growth of FB-SOC merit function and NR-SOC merit function

Shaohua Pan

School of Mathematical Sciences South China University of Technology

Guangzhou 510640, China E-mail: shhpan@scut.edu.cn

Jein-Shan Chen Department of Mathematics National Taiwan Normal University

Taipei, Taiwan 11677 E-mail: jschen@math.ntnu.edu.tw

Jing-Fan Li

Department of Mathematics National Taiwan Normal University

Taipei, Taiwan 11677 E-mail: 697400011@ntnu.edu.tw

September 6, 2009

Abstract. We establish the same growth of the Fischer-Burmeister (FB) merit function and the natural residual (NR) merit function associated with second-order cones. This extends an important result proved by Tseng in [12, Lemma 3.1] to the setting of second- order cones. Particularly, using such relation, we obtain the global error bound property of the FB merit function for the second-order cone complementarity problem (SOCCP), which plays a key role in analyzing the convergence rates of those algorithms based on FB SOC complementarity function for the second-order cone program and the SOCCP.

Key Words. Fischer-Burmeister function, Natural-Residual function, merit function, second-order cone.

1A much simpler proof for this result can be seen in [1].

1 Introduction

A well-known approach for solving the nonlinear complementarity problem (NCP) is to reformulate it as a global minimization over IRⁿ via a certain merit function. For the approach to be effective, the choice of the merit function is crucial. A popular choice is the Fischer-Burmeister (FB) merit function:

ψ_FB(a, b) = 1

2|ϕFB(a, b)|², where ϕ_FB : IR× IR → IR is the FB NCP-function defined by

ϕ_FB(a, b) :=√

a² + b²− (a + b) ∀a, b ∈ IR.

It turns out that ϕ_FBand ψ_FB have many desirable properties; see [4, 5]. Another popular choice is the natural residual (NR) merit function:

ψ_NR(a, b) = 1

2|ϕNR(a, b)|², where ϕ_NR : IR× IR → IR is the NR NCP-function defined as

ϕ_NR(a, b) := a− (a − b)+ = min{a, b} ∀a, b ∈ IR.

The NR merit function ψ_NR is not differentiable, which is its main drawback compared with ψ_FB. For the two functions, Tseng [12] proved the following important inequality:

(2−√

2)|ϕNR(a, b)| ≤ |ϕFB(a, b)| ≤ (2 +√

2)|ϕNR(a, b)| ∀a, b ∈ IR (1) which says ψ_FB and ψ_NR have the same order of growth behavior. It was this relation that accounts for the global error bound property of the FB merit function, which plays a key role in analyzing the convergence rates of those algorithms based on ϕ_FB; see [9, 11, 13].

The second-order cone complementarity problem (SOCCP), as an extension of the NCP, is to find a vector x∈ IRⁿ such that

x∈ K, F (x) ∈ K, ⟨x, F (x)⟩ = 0, (2) where F : IRⁿ → IRⁿis a continuous mapping,⟨·, ·⟩ denotes the Euclidean inner product, and K is the Cartesian product of second-order cones (SOCs). In other words,

K = Kⁿ¹ × Kⁿ² × · · · × K^nm,

where n₁, . . . , n_m ≥ 1, n1+· · · + nm = n, and Kⁿⁱ is the SOC in IRⁿⁱ defined by Kⁿⁱ :=^{x = (x₁, x₂)∈ IR × IRⁿⁱ⁻¹ | x1 ≥ ∥x2∥^},

with∥·∥ denoting the Euclidean norm and K¹ being the set of nonnegative real numbers IR₊. Clearly, when m = n and n₁ =· · · = nm = 1, the SOCCP (2) becomes the NCP.

The merit function approach for the NCP can be extended to the SOCCP case; see [2]. This approach aims to find a function ψ : IRⁿ× IRⁿ→ IR+ satisfying

ψ(x, y) = 0 ⇐⇒ x ∈ Kⁿ, y ∈ Kⁿ, ⟨x, y⟩ = 0,

and then the SOCCP can be reformulated as an unconstrained minimization problem:

xmin∈IRⁿ Ψ(x) :=

∑m i=1

ψ(x_i, F_i(x)),

where x_i ∈ IRⁿⁱ and F_i : IRⁿ→ IRⁿⁱ. We call such ψ an SOC merit function. Analogous to the NCP case, a popular choice for ψ is the FB SOC merit function

ψ_FB(x, y) := 1

2∥ϕFB(x, y)∥², (3)

where ϕ_FB : IRⁿ× IRⁿ→ IRⁿ is the vector-valued FB function defined by

ϕ_FB(x, y) := (x²+ y²)^1/2− x − y (4) with x² = x◦x denoting the Jordan product of x and itself, and x^1/2 meaning the vector such that (x^1/2)² = x. Similarly, the NR merit function can be defined in the SOC case:

ψ_NR(x, y) := 1

2∥ϕNR(x, y)∥² (5)

where ϕ_NR : IRⁿ× IRⁿ → IRⁿ is the vector-valued natural residual function:

ϕ_NR(x, y) := x− (x − y)+ (6)

with (·)+ denoting the projection onto the coneKⁿ. The function ϕ_NR was employed in [6, 7] to develop the smoothing Newton methods for the SOCCP.

It has been an open question whether ψ_FB and ψ_NR have the same order of growth behavior in the SOC case. In other words, can the inequality (1) be established for the SOC case? We answer this question affirmatively in this paper. Particularly, using this result, we obtain that the square root of the following FB merit function

Ψ_FB(x) :=

∑m i=1

ψ_FB(x_i, F_i(x)) (7)

provides a global error bound for the solution of (2) under suitable conditions of F . This is a key to analyze the convergence rates of the algorithms based on ϕ_FB for the SOCCP.

Throughout this paper, we denote bd(Kⁿ) by the boundary ofKⁿ, and int(Kⁿ) by the interior ofKⁿ. The notation o(∥x^k∥²) means a function satisfying lim

k→∞

o(∥x^k∥²)

∥x^k∥² = 0, and O(∥x^k∥²) denotes a function satisfying |O(∥x^k∥²)| ≤ C∥x^k∥² for some constant C > 0.

2 Preliminaries

For any x = (x₁, x₂), y = (y₁, y₂)∈ IR × IRⁿ⁻¹, their Jordan product [3] is defined as x◦ y := (⟨x, y⟩, x1y₂+ y₁x₂),

which, unlike scalar or matrix multiplication, is not associative in general. The identity element under this product is e := (1, 0, . . . , 0)^T ∈ IRⁿ, i.e., e◦ x = x for any x ∈ IRⁿ. We recall from [6] that each x = (x1, x2) ∈ IR × IRⁿ⁻¹ admits a spectral factorization, associated with Kⁿ, of the form

x = λ₁(x)u⁽¹⁾_x + λ₂(x)u⁽²⁾_x , (8) where λ_i(x) and u⁽ⁱ⁾_x for i = 1, 2 are the spectral values and the associated spectral vectors of x, with respect to Kⁿ, given by

λ_i(x) := x₁+ (−1)ⁱ∥x2∥, u⁽ⁱ⁾_x := 1 2

(

1, (−1)ⁱx¯₂⁾, (9) with ¯x₂ = _∥x^x2

2∥ if x₂ ̸= 0 and otherwise being any vector in IRⁿ⁻¹ satisfying ∥¯x2∥ = 1.

The following lemmas are used in the subsequent analysis; see [2] for their proofs.

Lemma 2.1 For any x = (x1, x2), y = (y1, y2)∈ IR × IRⁿ⁻¹, if x²+ y² ∈ bd(Kⁿ), then x²₁ =∥x2∥², y₁² =∥y2∥², x₁y₁ = x^T₂y₂, x₁y₂ = y₁x₂.

Lemma 2.2 Let ϕ_FB and ϕ_NR be defined by (4) and (6), respectively. Then, ϕ_FB(x, y) = 0 ⇐⇒ ϕNR(x, y) = 0 ⇐⇒ x ∈ Kⁿ, y∈ Kⁿ, ⟨x, y⟩ = 0.

3 Main result

In this section, we concentrate on the main result of this paper which is stated as in Theorem 3.1. To prove the theorem, we need the following two crucial lemmas.

Lemma 3.1 Let ϕ_FB and ϕ_NR be defined by (4) and (6), respectively. Then, for any x = (x₁, x₂), y = (y₁, y₂)∈ IR × IRⁿ⁻¹ with x²+ y² ∈ bd(Kⁿ), we have

(2−√

2)∥ϕNR(x, y)∥ ≤ ∥ϕFB(x, y)∥ ≤ (2 +√

2)∥ϕNR(x, y)∥.

Proof. Fix any x = (x₁, x₂), y = (y₁, y₂)∈ IR × IRⁿ⁻¹. If (x, y) = (0, 0), then the result is direct by Lemma 2.2. We next suppose (x, y) ̸= (0, 0). Let w = (w1, w₂) := x²+ y². From (x, y)̸= (0, 0) and x² + y² ∈ bd(Kⁿ), it follows that

w₁ =∥w2∥ = 2∥x1x₂+ y₁y₂∥ ̸= 0, λ1(w) = 0 and λ₂(w) = 4(x²₁ + y²₁),

where the last equality is using Lemma 2.1. Together with the formulas (8)–(9), we get

ϕ_FB(x, y) = w^1/2− (x + y) =







√

x²₁+ y²₁ x₁x₂+ y₁y₂

√

x²₁+ y²₁





−

( x₁ + y₁ x₂ + y₂

)

.

In addition, using Lemma 2.1, it is not hard to calculate that

x₁x₂+ y₁y₂

√

x²₁+ y₁² − (x2+ y2)

2

=

(√

x²₁+ y²₁− (x1+ y1)

)₂

. From the last two equalities, we immediately obtain

∥ϕFB(x, y)∥² = 2

(√

x²₁+ y₁²− (x1+ y₁)

)₂

, which together with the inequality (1) yields

2(2−√

2)²(min{x1, y₁})² ≤ ∥ϕFB(x, y)∥² ≤ 2(2 +√

2)²(min{x1, y₁})². (10) Now we consider (x− y) ∈ Kⁿ, (x− y) ∈ −Kⁿ and (x− y) ̸∈ −Kⁿ∪ Kⁿ, respectively.

Case 1: (x− y) ∈ Kⁿ. Under this case, ϕ_NR(x, y) = y, and applying Lemma 2.1,

∥ϕNR(x, y)∥² = y₁²+∥y2∥² = 2y₁².

In addition, since (x− y) ∈ Kⁿ implies x₁ ≥ y1, the inequality (10) is equivalent to 2(2−√

2)²y²₁ ≤ ∥ϕFB(x, y)∥² ≤ 2(2 +√ 2)²y²₁. Combining the last two equations, we readily obtain the desired result.

Case 2: (x− y) ∈ −Kⁿ. Now ϕ_NR(x, y) = x and x₁ ≤ y1. Along with Lemma 2.1,

∥ϕNR(x, y)∥² =∥x∥² = 2x²₁ = 2 min{x1, y₁}², and the desired result follows from (10) directly.

Case 3: (x− y) ̸∈ −Kⁿ∪ Kⁿ. In this case, we must have∥x2− y2∥ > |x1− y1|. On the other hand, by Lemma 2.1, ∥x2 − y2∥ = |x1 − y1|. This leads to a contradiction. This shows that this subcase actually does not occur. The proof is completed. 2

For convenience, in the rest of this paper, the notation S always represents the set S :=^{(x, y)∈ IRⁿ× IRⁿ | x ̸= 0, y ̸= 0, ∥ϕFB(x, y)∥ ̸= 0, x²+ y² ∈ int(Kⁿ)^}. The following lemma shows that ϕ_FB and ϕ_NR have the same order of growth in S.

Lemma 3.2 Let ϕ_FB and ϕ_NR be defined by (4) and (6), respectively. Then, there exist positive constants ¯c₁ and ¯c₂ such that, for all (x, y)∈ S,

¯

c₁∥ϕNR(x, y)∥ ≤ ∥ϕFB(x, y)∥ ≤ ¯c2∥ϕNR(x, y)∥.

Proof. We first argue the fact that, if the result holds when S is bounded, then it also holds when S is unbounded. Fix any (x, y)∈ S with S being unbounded. Let

S^′ :=^{(x^′, y^′)∈ IRⁿ× IRⁿ | x^′ ̸= 0, y^′ ̸= 0, (x^′)²+ (y^′)² ∈ int(Kⁿ),∥x^′∥²+∥y^′∥² ≤ 2^}. Clearly, S^′ is bounded, and therefore, for all (x^′, y^′)∈ S^′,

¯

c1∥ϕNR(x^′, y^′)∥ ≤ ∥ϕFB(x^′, y^′)∥ ≤ ¯c2∥ϕNR(x^′, y^′)∥ .

If ∥x∥ ≥ ∥y∥, then (_∥x∥^x ,_∥x∥^y )∈ S^′. From the last inequality, it follows that

¯ c₁

ϕ_NR

( x

∥x∥, y

∥x∥

) ≤ ϕ_FB

( x

∥x∥, y

∥x∥

) ≤ ¯c2

ϕ_NR

( x

∥x∥, y

∥x∥

) .

Multiplying the two sides by ∥x∥ and using the expressions of ϕFB and ϕ_NR yield the result. If ∥x∥ ≤ ∥y∥, then (_∥y∥^x ,_∥y∥^y )∈ S^′, and the result follows by similar arguments.

Next, we assume that the set S is bounded, and prove the result by two steps.

Step 1: Suppose such ¯c₂does not exist. Then there must have a sequence{(x^k, y^k)} ⊆ S such that

∥ϕNR(x^k, y^k)∥ < 1

k∥ϕFB(x^k, y^k)∥. (11) Without loss of generality, assume that the sequences {x^k} and {y^k} converge to x^∗ and y^∗, respectively. Since {ϕFB(x^k, y^k)} is bounded and ∥ϕNR(x, y)∥ is continuous, taking the limit to the both sides of (11) yields ∥ϕNR(x^∗, y^∗)∥ = 0. By Lemma 2.2,

x^∗ ∈ Kⁿ, y^∗ ∈ Kⁿ, ⟨x^∗, y^∗⟩ = 0. (12) We proceed the arguments by two cases (x^∗)²+(y^∗)² ∈int(Kⁿ) and (x^∗)²+(y^∗)² ∈bd(Kⁿ).

Unless otherwise stated, the index k appearing in the sequel are all sufficiently large.

Case 1: (x^∗)² + (y^∗)² ∈ int(Kⁿ). In this case, it suffices to consider the following two subcases: (1.1) x^∗ ∈ int(Kⁿ) or y^∗ ∈ int(Kⁿ); (1.2) x^∗ ∈ bd(Kⁿ) and y^∗ ∈ bd(Kⁿ).

Subcase 1.1: x^∗ ∈ int(Kⁿ) or y^∗ ∈ int(Kⁿ). Without loss of generality, we assume that y^∗ ∈ int(Kⁿ). Along with⟨x^∗, y^∗⟩ = 0 in (12), we have x^∗ = 0, and (x^∗−y^∗)∈ −int(Kⁿ).

This means that (x^k− y^k)∈ −int(Kⁿ), and consequently ∥ϕNR(x^k, y^k)∥ = ∥x^k∥².

Let w^k = (w^k₁, w₂^k) = (x^k)²+ (y^k)², and λ^k₁ and λ^k₂ be the spectral values of w^k. Then, λ^k_i = w₁^k+ (−1)ⁱ∥w2^k∥ = ∥x^k∥² +∥y^k∥²+ (−1)ⁱ∥2(x^k1x^k₂ + y₁^ky₂^k)∥, i = 1, 2.

(a) Suppose that w^k₂ ̸= 0. By formulas (8)–(9), it follows that

(w^k)^1/2 =





√λ^k₂+√

λ^k₁

√ 2 λ^k₂−√

λ^k₁ 2

w₂^k

∥w2^k∥



=





√λ^k₂+√

λ^k₁ 2 w₂^k

√λ^k₂+√

λ^k₁



. (13)

Together with ϕ_FB(x^k, y^k) = (w^k)^1/2− (x^k+ y^k), we have that 1

2∥ϕFB(x^k, y^k)∥² = ∥x^k∥²+∥y^k∥²+ (x^k)^Ty^k

− x^k₁+ y₁^k

√

λ^k₂ +

√

λ^k₁

(

∥x^k∥²+∥y^k∥²+

√

λ^k₂

√

λ^k₁

)

−2(x^k₁∥x^k2∥²+ y₁^k∥y2^k∥²) + 2(x^k₁ + y₁^k)(x^k₂)^T(y₂^k)

√

λ^k₂ +

√

λ^k₁

= T₁^k+ T₂^k+ T₃^k, (14)

where

T₁^k := ∥x^k∥²− (x_√^k₁ + y₁^k)∥x^k∥² λ^k₂ +

√

λ^k₁ − _√2x^k₁∥x^k2∥² λ^k₂+

√

λ^k₁ ,

T₂^k := ∥y^k∥²− y₁^k

(

∥y^k∥² +

√

λ^k₂

√

λ^k₁

)

√

λ^k₂ +

√

λ^k₁ −_√2y^k₁∥y^k2∥² λ^k₂+

√

λ^k₁ − 2x^k₁(x^k₂)^Ty₂^k

√

λ^k₂ +

√

λ^k₁

, (15)

T₃^k := (x^k)^Ty^k− x^k₁

(

∥y^k∥² +

√

λ^k₂

√

λ^k₁

)

√

λ^k₂ +

√

λ^k₁ − 2y^k₁(x^k₂)^Ty^k₂

√

λ^k₂+

√

λ^k₁ .

Since x^k→ x^∗ = 0 and y^k → y^∗ ∈ int(Kⁿ), it is not hard to verify that

klim→∞

T₁^k

∥x^k∥² = 1− y^∗₁

y^∗₁ +∥y^∗2∥ + y^∗1− ∥y^∗2∥ = 1

2. (16)

From the expression of T₂^k, it follows that

(√

λ^k₂ +

√

λ^k₁

)

T₂^k+ 2x^k₁(x^k₂)^Ty₂^k equals

(√

λ^k₂− y^k1 − ∥y^k2∥^{) (}∥y2^k∥²+ y₁^k∥y^k2∥⁾+

(√

λ^k₁− y1^k+∥y2^k∥^{) (}∥y^k∥²− y1^k

√

λ^k₂

)

. In addition, an elementary calculation yields that

√

λ^k₂− y1^k− ∥y^k2∥ = 4x^k₁y₁^k(x^k₂)^Ty₂^k

(√

λ^k₂+ y₁^k+∥y2^k∥^{) (}∥x^k1x^k₂+ y^k₁y₂^k∥ + y1^k∥y^k2∥⁾ +_√ ∥x^k∥²

λ^k₂ + y₁^k+∥y^k2∥ + o(∥x^k∥²), (17)

√

λ^k₁ − y1^k+∥y^k2∥ = − 4x^k₁y₁^k(x^k₂)^Ty₂^k

(√

λ^k₁ + y₁^k− ∥y2^k∥^{) (}∥x^k1x^k₂ + y^k₁y^k₂∥ + y1^k∥y2^k∥⁾ +_√ ∥x^k∥²

λ^k₁ + y₁^k− ∥y^k2∥ + o(∥x^k∥²).

The last two sides imply that

(√

λ^k₂ +

√

λ^k₁

)

T₂^k is equal to

∥x^k_√∥²(∥y^k₂∥²+ y₁^k∥y₂^k∥)

λ^k₂+ y^k₁ +∥y2^k∥ +∥x^k∥²(∥y^k∥²− y1^k

√

λ^k₂)

√

λ^k₁+ y₁^k− ∥y2^k∥ + o(∥x^k∥²) + 4x^k₁y₁^k(x^k₂)^Ty^k₂(∥y2^k∥²+ y^k₁∥y^k2∥)

(√

λ^k₂ + y₁^k+∥y^k2∥^{) (}∥x^k1x^k₂+ y₁^ky₂^k∥ + y1^k∥y^k2∥⁾

− 4x^k₁y₁^k(x^k₂)^Ty₂^k

(

∥y^k∥²− y1^k

√

λ^k₂

) (√

λ^k₁+ y^k₁ − ∥y2^k∥^{) (}∥x^k1x^k₂ + y₁^ky₂^k∥ + y1^k∥y2^k∥⁾− 2x^k1(x^k₂)^Ty₂^k.

Let T₂₁^k and T₂₂^k denote the sums of the first three terms and the last three terms, respectively, on the right hand side of last equation. By the expression of T₂₁^k,

klim→∞

T₂₁^k

∥x^k∥² = ∥y₂^∗∥²+ y₁^∗∥y^∗₂∥

2(y₁^∗+∥y2^∗∥) +∥y^∗∥²− y₁^∗(y₁^∗+∥y₂^∗∥) 2(y₁^∗− ∥y2^∗∥) = 0.

Whereas from the expression of T₂₂^k, it follows that T₂₂^k

∥x^k∥² = 4x^k₁y₁^k(x^k₂)^Ty₂^k

∥x^k∥²





 ∥y^k₂∥²+ y₁^k∥y₂^k∥

(√

λ^k₂+ y₁^k+∥y2^k∥^{) (}∥x^k1x^k₂+ y^k₁y₂^k∥ + y1^k∥y^k2∥⁾ − 1 2y^k₁

− ∥y^k∥²− y1^k

√

λ^k₂

(√

λ^k₁ + y₁^k− ∥y2^k∥^{) (}∥x^k1x^k₂+ y^k₁y^k₂∥ + y1^k∥y2^k∥⁾





.

Since the sequence

{4x^k₁y₁^k(x^k₂)^Ty^k₂

∥x^k∥²

}

is bounded, and the limit of the term in the bracket as k→ ∞ equals 0, we have limk→∞T₂₂^k/∥x^k∥² = 0. Thus, we get

k→∞lim T₂^k

∥x^k∥² = lim

k→∞

T₂₁^k + T₂₂^k

∥x^k∥^2(√λ^k₂+

√

λ^k₁

) = 0. (18)

We next take a look at T₃^k. By the expression of T₃^k,

(√

λ^k₂+

√

λ^k₁

)

T₃^k equals (x^k)^Ty^k

(√

λ^k₂ +

√

λ^k₁

)

− x^k1

(

∥y^k∥²+

√

λ^k₂

√

λ^k₁

)

− 2y^k1(x^k₂)^Ty^k₂

=

(√

λ^k₂ − y1^k−∥y2^k∥^{) [}(x^k₂)^Ty₂^k+ x^k₁∥y2^k∥^]+

(√

λ^k₁ −y1^k+∥y2^k∥^{) [}(x^k)^Ty^k− x^k1

√

λ^k₂

]

which, together with (17), implies that

(√

λ^k₂ +

√

λ^k₁

)

T₃^k = o(∥x^k∥²). Therefore,

klim→∞

T₃^k

∥x^k∥² = 0.

Together with equations (14), (16) and (18), we immediately obtain

klim→∞

∥ϕFB(x^k, y^k)∥²

∥ϕNR(x^k, y^k)∥² = lim

k→∞

2(T₁^k+ T₂^k+ T₃^k)

∥x^k∥² = 1, which contradicts the inequality (11).

(b) Suppose that w₂^k= 0. Under this subcase, since λ^k₁ = λ^k₂, we have that 1

2∥ϕFB(x^k, y^k)∥² =∥x^k∥²+∥y^k∥²+ (x^k)^Ty^k−(x^k₁ + y₁^k)(∥x^k∥²+∥y^k∥²)

√∥x^k∥²+∥y^k∥² .

Since y₁^k > 0 by y^∗ ∈ int(Kⁿ), from w₂^k= 0 we get y₂^k =−(x^k1/y₁^k)x^k₂, which means that

∥y^k2∥² = o(∥x^k∥²) and (x^k)^Ty^k= x^k₁y₁^k− (x^k1/y₁^k)∥x^k2∥² = x^k₁y₁^k+ o(∥x^k∥²).

Substituting the two equalities into the above ¹₂∥ϕFB(x^k, y^k)∥² yields that 1

2∥ϕFB(x^k, y^k)∥² =∥x^k∥²− _√(x^k₁ + y₁^k)∥x^k∥²

∥x^k∥²+∥y^k∥² + (y₁^k)²− (x^k₁+ y^k₁)(y₁^k)²

√∥x^k∥²+∥y^k∥² + x^k₁y^k₁ + o(∥x^k∥²).

In addition, using ∥y^k2∥² = o(∥x^k∥²) and an elementary calculation, we have that (y₁^k)²− (x^k₁ + y₁^k)(y₁^k)²

√∥x^k∥²+∥y^k∥² + x^k₁y₁^k= (y₁^k)²− (y^k₁)³

√∥x^k∥²+∥y^k∥² + x^k₁y₁^k− x^k₁(y₁^k)²

√∥x^k∥²+∥y^k∥²

= _√ (y₁^k)²∥x^k∥²

∥x^k∥²+∥y^k∥^2(√∥x^k∥²+∥y^k∥²+ y^k₁⁾

+ o(∥x^k∥²).

From the last two equations, x^k → x^∗ = 0 and y^k→ y^∗, it follows that

k→∞lim

∥ϕFB(x^k, y^k)∥²

∥ϕNR(x^k, y^k)∥² = lim

k→∞

∥ϕFB(x^k, y^k)∥²

∥x^k∥² = 2− 2y^∗₁

∥y^∗∥ + 2(y₁^∗)²

∥y^∗∥(∥y^∗∥ + y^∗1) = 1, which gives a contradiction to the inequality (11).

Subcase 1.2: x^∗ ∈ bd(Kⁿ) and y^∗ ∈ bd(Kⁿ). Since (x^∗)² + (y^∗)² ∈ int(Kⁿ), we must have x^∗₁ > 0 and y₁^∗ > 0, which implies x^k₁ > 0 and y^k₁ > 0. Also, x^k− y^k ̸∈ −Kⁿ∪ Kⁿ. If not, ϕ_NR(x^k, y^k) = x^k or y^k, which implies that ∥ϕNR(x^∗, y^∗)∥ ̸= 0. In addition, noting that 0 =⟨x^∗, y^∗⟩ = ∥x^∗2∥∥y^∗2∥ + (x^∗2)^Ty₂^∗ = limk→∞[(x^k₂)^Ty^k₂ +∥x^k2∥∥y2^k∥], we have

(x^k₂)^Ty₂^k =−∥x^k₂∥∥y₂^k∥ + αk with lim

k→∞α_k = 0. (19)