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

^{1}

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.

1**A 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 IR* ^{n}* 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*|ϕ*FB*(a, b)|*^{2}*,*
*where ϕ*_{FB} : IR*× IR → IR is the FB NCP-function deﬁned by*

*ϕ*_{FB}*(a, b) :=√*

*a*^{2} *+ b*^{2}*− (a + b)* *∀a, b ∈ IR.*

*It turns out that ϕ*_{FB}*and 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*|ϕ*NR*(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)*|ϕ*NR*(a, b)| ≤ |ϕ*FB*(a, b)| ≤ (2 +√*

2)*|ϕ*NR*(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∈ IR** ^{n}* such that

*x∈ K, F (x) ∈ K, ⟨x, F (x)⟩ = 0,* (2)
*where F : IR*^{n}*→ IR** ^{n}*is a continuous mapping,

*⟨·, ·⟩ denotes the Euclidean inner product,*and

*K is the Cartesian product of second-order cones (SOCs). In other words,*

*K = K*^{n}^{1} *× K*^{n}^{2} *× · · · × K*^{n}^{m}*,*

*where n*_{1}*, . . . , n*_{m}*≥ 1, n*1+*· · · + n**m* *= n, and* *K*^{n}* ^{i}* is the SOC in IR

^{n}*deﬁned by*

^{i}*K*

^{n}*:=*

^{i}^{{}

*x = (x*

_{1}

*, x*

_{2})

*∈ IR × IR*

^{n}

^{i}

^{−1}*| x*1

*≥ ∥x*2

*∥*

^{}}

*,*

with*∥·∥ denoting the Euclidean norm and K*^{1} being the set of nonnegative real numbers
IR_{+}*. Clearly, when m = n and n*_{1} =*· · · = n**m* = 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 : IR*^{n}*× IR*^{n}*→ IR*+ satisfying

*ψ(x, y) = 0* *⇐⇒ x ∈ K*^{n}*, y* *∈ K*^{n}*,* *⟨x, y⟩ = 0,*

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

*x*min*∈IR*^{n}*Ψ(x) :=*

∑*m*
*i=1*

*ψ(x*_{i}*, F*_{i}*(x)),*

*where x*_{i}*∈ IR*^{n}^{i}*and F** _{i}* : IR

^{n}*→ IR*

^{n}

^{i}*. 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*∥ϕ*FB*(x, y)∥*^{2}*,* (3)

*where ϕ*_{FB} : IR^{n}*× IR*^{n}*→ IR** ^{n}* is the vector-valued FB function deﬁned by

*ϕ*_{FB}*(x, y) := (x*^{2}*+ y*^{2})^{1/2}*− x − y* (4)
*with x*^{2} *= 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*∥ϕ*NR*(x, y)∥*^{2} (5)

*where ϕ*_{NR} : IR^{n}*× IR*^{n}*→ IR** ^{n}* is the vector-valued natural residual function:

*ϕ*_{NR}*(x, y) := x− (x − y)*+ (6)

with (*·)*+ denoting the projection onto the cone*K*^{n}*. 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(*K** ^{n}*) by the boundary of

*K*

*, and int(*

^{n}*K*

*) by the interior of*

^{n}*K*

^{n}*. 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 = (x*_{1}*, x*_{2}*), y = (y*_{1}*, y*_{2})*∈ IR × IR*^{n}^{−1}*, their Jordan product [3] is deﬁned as*
*x◦ y := (⟨x, y⟩, x*1*y*_{2}*+ y*_{1}*x*_{2}*),*

which, unlike scalar or matrix multiplication, is not associative in general. The identity
*element under this product is e := (1, 0, . . . , 0)*^{T}*∈ IR*^{n}*, i.e., e◦ x = x for any x ∈ IR** ^{n}*.

*We recall from [6] that each x = (x*1

*, x*2)

*∈ IR × IR*

^{n}*admits a spectral factorization, associated with*

^{−1}*K*

*, of the form*

^{n}*x = λ*_{1}*(x)u*^{(1)}_{x}*+ λ*_{2}*(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* *K** ^{n}*, given by

*λ*_{i}*(x) := x*_{1}+ (*−1)*^{i}*∥x*2*∥, u*^{(i)}* _{x}* := 1
2

(

*1, (−1)*^{i}*x*¯_{2}^{)}*,* (9)
with ¯*x*_{2} = _{∥x}^{x}^{2}

2*∥* *if x*_{2} *̸= 0 and otherwise being any vector in IR*^{n}* ^{−1}* satisfying

*∥¯x*2

*∥ = 1.*

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

* Lemma 2.1 For any x = (x*1

*, x*2

*), y = (y*1

*, y*2)

*∈ IR × IR*

^{n}

^{−1}*, if x*

^{2}

*+ y*

^{2}

*∈ bd(K*

^{n}*), then*

*x*

^{2}

_{1}=

*∥x*2

*∥*

^{2}

*, y*

_{1}

^{2}=

*∥y*2

*∥*

^{2}

*, x*

_{1}

*y*

_{1}

*= x*

^{T}_{2}

*y*

_{2}

*, x*

_{1}

*y*

_{2}

*= y*

_{1}

*x*

_{2}

*.*

**Lemma 2.2 Let ϕ**_{FB} *and ϕ*_{NR} *be deﬁned by (4) and (6), respectively. Then,*
*ϕ*_{FB}*(x, y) = 0* *⇐⇒ ϕ*NR*(x, y) = 0* *⇐⇒ x ∈ K*^{n}*, y∈ K*^{n}*,* *⟨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 = (x*_{1}*, x*_{2}*), y = (y*_{1}*, y*_{2})*∈ IR × IR*^{n}^{−1}*with x*^{2}*+ y*^{2} *∈ bd(K*^{n}*), we have*

(2*−√*

2)*∥ϕ*NR*(x, y)∥ ≤ ∥ϕ*FB*(x, y)∥ ≤ (2 +√*

2)*∥ϕ*NR*(x, y)∥.*

**Proof. Fix any x = (x**_{1}*, x*_{2}*), y = (y*_{1}*, y*_{2})*∈ IR × IR*^{n}^{−1}*. If (x, y) = (0, 0), then the result*
*is direct by Lemma 2.2. We next suppose (x, y)* *̸= (0, 0). Let w = (w*1*, w*_{2}*) := x*^{2}*+ y*^{2}.
*From (x, y)̸= (0, 0) and x*^{2} *+ y*^{2} *∈ bd(K** ^{n}*), it follows that

*w*_{1} =*∥w*2*∥ = 2∥x*1*x*_{2}*+ y*_{1}*y*_{2}*∥ ̸= 0, λ*1*(w) = 0 and λ*_{2}*(w) = 4(x*^{2}_{1} *+ y*^{2}_{1}*),*

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) =*

√

*x*^{2}_{1}*+ y*^{2}_{1}
*x*_{1}*x*_{2}*+ y*_{1}*y*_{2}

√

*x*^{2}_{1}*+ y*^{2}_{1}

*−*

( *x*_{1} *+ y*_{1}
*x*_{2} *+ y*_{2}

)

*.*

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

*x*_{1}*x*_{2}*+ y*_{1}*y*_{2}

√

*x*^{2}_{1}*+ y*_{1}^{2} *− (x*2*+ y*2)

2

=

(√

*x*^{2}_{1}*+ y*^{2}_{1}*− (x*1*+ y*1)

)_{2}

*.*
From the last two equalities, we immediately obtain

*∥ϕ*FB*(x, y)∥*^{2} = 2

(√

*x*^{2}_{1}*+ y*_{1}^{2}*− (x*1*+ y*_{1})

)_{2}

*,*
which together with the inequality (1) yields

2(2*−√*

2)^{2}(min*{x*1*, y*_{1}*})*^{2} *≤ ∥ϕ*FB*(x, y)∥*^{2} *≤ 2(2 +√*

2)^{2}(min*{x*1*, y*_{1}*})*^{2}*.* (10)
*Now we consider (x− y) ∈ K*^{n}*, (x− y) ∈ −K*^{n}*and (x− y) ̸∈ −K*^{n}*∪ K** ^{n}*, respectively.

**Case 1: (x**− y) ∈ K^{n}*. Under this case, ϕ*_{NR}*(x, y) = y, and applying Lemma 2.1,*

*∥ϕ*NR*(x, y)∥*^{2} *= y*_{1}^{2}+*∥y*2*∥*^{2} *= 2y*_{1}^{2}*.*

*In addition, since (x− y) ∈ K*^{n}*implies x*_{1} *≥ y*1, the inequality (10) is equivalent to
2(2*−√*

2)^{2}*y*^{2}_{1} *≤ ∥ϕ*FB*(x, y)∥*^{2} *≤ 2(2 +√*
2)^{2}*y*^{2}_{1}*.*
Combining the last two equations, we readily obtain the desired result.

**Case 2: (x**− y) ∈ −K^{n}*. Now ϕ*_{NR}*(x, y) = x and x*_{1} *≤ y*1. Along with Lemma 2.1,

*∥ϕ*NR*(x, y)∥*^{2} =*∥x∥*^{2} *= 2x*^{2}_{1} = 2 min*{x*1*, y*_{1}*}*^{2}*,*
and the desired result follows from (10) directly.

**Case 3: (x**− y) ̸∈ −K^{n}*∪ K** ^{n}*. In this case, we must have

*∥x*2

*− y*2

*∥ > |x*1

*− y*1

*|. On the*other hand, by Lemma 2.1,

*∥x*2

*− y*2

*∥ = |x*1

*− y*1

*|. 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)∈ IR*^{n}*× IR*^{n}*| x ̸= 0, y ̸= 0, ∥ϕ*FB*(x, y)∥ ̸= 0, x*^{2}*+ y*^{2} *∈ int(K** ^{n}*)

^{}}

*.*

*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 ¯c*_{1} *and ¯c*_{2} *such that, for all (x, y)∈ S,*

¯

*c*_{1}*∥ϕ*NR*(x, y)∥ ≤ ∥ϕ*FB*(x, y)∥ ≤ ¯c*2*∥ϕ*NR*(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*

*)*

^{′}*∈ IR*

^{n}*× IR*

^{n}*| x*

^{′}*̸= 0, y*

^{′}*̸= 0, (x*

*)*

^{′}^{2}

*+ (y*

*)*

^{′}^{2}

*∈ int(K*

^{n}*),∥x*

^{′}*∥*

^{2}+

*∥y*

^{′}*∥*

^{2}

*≤ 2*

^{}}

*.*

*Clearly, S*

^{′}*is bounded, and therefore, for all (x*

^{′}*, y*

*)*

^{′}*∈ S*

*,*

^{′}¯

*c*1*∥ϕ*NR*(x*^{′}*, y** ^{′}*)

*∥ ≤ ∥ϕ*FB

*(x*

^{′}*, y*

*)*

^{′}*∥ ≤ ¯c*2

*∥ϕ*NR

*(x*

^{′}*, y*

*)*

^{′}*∥ .*

If *∥x∥ ≥ ∥y∥, then (*_{∥x∥}^{x}*,*_{∥x∥}* ^{y}* )

*∈ S*

*. From the last inequality, it follows that*

^{′}¯
*c*_{1}

*ϕ*_{NR}

( *x*

*∥x∥,* *y*

*∥x∥*

)
*≤*^{
}
*ϕ*_{FB}

( *x*

*∥x∥,* *y*

*∥x∥*

)
*≤ ¯c*2

*ϕ*_{NR}

( *x*

*∥x∥,* *y*

*∥x∥*

)
*.*

Multiplying the two sides by *∥x∥ and using the expressions of ϕ*FB *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 ¯***c*_{2}does not exist. Then there must have a sequence*{(x*^{k}*, y** ^{k}*)

*} ⊆ S*such that

*∥ϕ*NR*(x*^{k}*, y** ^{k}*)

*∥ <*1

*k∥ϕ*FB*(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*

^{∗}*{ϕ*FB

*(x*

^{k}*, y*

*)*

^{k}*} is bounded and ∥ϕ*NR

*(x, y)∥ is continuous, taking*the limit to the both sides of (11) yields

*∥ϕ*NR

*(x*

^{∗}*, y*

*)*

^{∗}*∥ = 0. By Lemma 2.2,*

*x*^{∗}*∈ K*^{n}*, y*^{∗}*∈ K*^{n}*,* *⟨x*^{∗}*, y*^{∗}*⟩ = 0.* (12)
*We proceed the arguments by two cases (x** ^{∗}*)

^{2}

*+(y*

*)*

^{∗}^{2}

*∈int(K*

^{n}*) and (x*

*)*

^{∗}^{2}

*+(y*

*)*

^{∗}^{2}

*∈bd(K*

*).*

^{n}*Unless otherwise stated, the index k appearing in the sequel are all suﬃciently large.*

**Case 1: (x*** ^{∗}*)

^{2}

*+ (y*

*)*

^{∗}^{2}

*∈ int(K*

*). In this case, it suﬃces to consider the following two*

^{n}*subcases: (1.1) x*

^{∗}*∈ int(K*

^{n}*) or y*

^{∗}*∈ int(K*

^{n}*); (1.2) x*

^{∗}*∈ bd(K*

^{n}*) and y*

^{∗}*∈ bd(K*

*).*

^{n}**Subcase 1.1: x**^{∗}*∈ int(K*^{n}*) or y*^{∗}*∈ int(K** ^{n}*). Without loss of generality, we assume that

*y*

^{∗}*∈ int(K*

*). Along with*

^{n}*⟨x*

^{∗}*, y*

^{∗}*⟩ = 0 in (12), we have x*

^{∗}*= 0, and (x*

^{∗}*−y*

*)*

^{∗}*∈ −int(K*

*).*

^{n}*This means that (x*^{k}*− y** ^{k}*)

*∈ −int(K*

*), and consequently*

^{n}*∥ϕ*NR

*(x*

^{k}*, y*

*)*

^{k}*∥ = ∥x*

^{k}*∥*

^{2}

*.*

*Let w*^{k}*= (w*^{k}_{1}*, w*_{2}^{k}*) = (x** ^{k}*)

^{2}

*+ (y*

*)*

^{k}^{2}

*, and λ*

^{k}_{1}

*and λ*

^{k}_{2}

*be the spectral values of w*

*. Then,*

^{k}*λ*

^{k}

_{i}*= w*

_{1}

*+ (*

^{k}*−1)*

^{i}*∥w*2

^{k}*∥ = ∥x*

^{k}*∥*

^{2}+

*∥y*

^{k}*∥*

^{2}+ (

*−1)*

^{i}*∥2(x*

*1*

^{k}*x*

^{k}_{2}

*+ y*

_{1}

^{k}*y*

_{2}

*)*

^{k}*∥, i = 1, 2.*

**(a) Suppose that w**^{k}_{2} *̸= 0. By formulas (8)–(9), it follows that*

*(w** ^{k}*)

*=*

^{1/2}

*√**λ*^{k}_{2}+*√*

*λ*^{k}_{1}

*√* 2
*λ*^{k}_{2}*−**√*

*λ*^{k}_{1}
2

*w*_{2}^{k}

*∥w*2^{k}*∥*

=

*√**λ*^{k}_{2}+*√*

*λ*^{k}_{1}
2
*w*_{2}^{k}

*√**λ*^{k}_{2}+*√*

*λ*^{k}_{1}

*.* (13)

*Together with ϕ*_{FB}*(x*^{k}*, y*^{k}*) = (w** ^{k}*)

^{1/2}*− (x*

^{k}*+ y*

*), we have that 1*

^{k}2*∥ϕ*FB*(x*^{k}*, y** ^{k}*)

*∥*

^{2}=

*∥x*

^{k}*∥*

^{2}+

*∥y*

^{k}*∥*

^{2}

*+ (x*

*)*

^{k}

^{T}*y*

^{k}*−* *x*^{k}_{1}*+ y*_{1}^{k}

√

*λ*^{k}_{2} +

√

*λ*^{k}_{1}

(

*∥x*^{k}*∥*^{2}+*∥y*^{k}*∥*^{2}+

√

*λ*^{k}_{2}

√

*λ*^{k}_{1}

)

*−2(x*^{k}_{1}*∥x** ^{k}*2

*∥*

^{2}

*+ y*

_{1}

^{k}*∥y*2

^{k}*∥*

^{2}

*) + 2(x*

^{k}_{1}

*+ y*

_{1}

^{k}*)(x*

^{k}_{2})

^{T}*(y*

_{2}

*)*

^{k}√

*λ*^{k}_{2} +

√

*λ*^{k}_{1}

*= T*_{1}^{k}*+ T*_{2}^{k}*+ T*_{3}^{k}*,* (14)

where

*T*_{1}* ^{k}* :=

*∥x*

^{k}*∥*

^{2}

*−*

*(x*

_{√}

^{k}_{1}

*+ y*

_{1}

*)*

^{k}*∥x*

^{k}*∥*

^{2}

*λ*

^{k}_{2}+

√

*λ*^{k}_{1} *−* _{√}*2x*^{k}_{1}*∥x** ^{k}*2

*∥*

^{2}

*λ*

^{k}_{2}+

√

*λ*^{k}_{1}
*,*

*T*_{2}* ^{k}* :=

*∥y*

^{k}*∥*

^{2}

*−*

*y*

_{1}

^{k}(

*∥y*^{k}*∥*^{2} +

√

*λ*^{k}_{2}

√

*λ*^{k}_{1}

)

√

*λ*^{k}_{2} +

√

*λ*^{k}_{1} *−*_{√}*2y*^{k}_{1}*∥y** ^{k}*2

*∥*

^{2}

*λ*

^{k}_{2}+

√

*λ*^{k}_{1} *−* *2x*^{k}_{1}*(x*^{k}_{2})^{T}*y*_{2}^{k}

√

*λ*^{k}_{2} +

√

*λ*^{k}_{1}

*,* (15)

*T*_{3}^{k}*:= (x** ^{k}*)

^{T}*y*

^{k}*−*

*x*

^{k}_{1}

(

*∥y*^{k}*∥*^{2} +

√

*λ*^{k}_{2}

√

*λ*^{k}_{1}

)

√

*λ*^{k}_{2} +

√

*λ*^{k}_{1} *−* *2y*^{k}_{1}*(x*^{k}_{2})^{T}*y*^{k}_{2}

√

*λ*^{k}_{2}+

√

*λ*^{k}_{1}
*.*

*Since x*^{k}*→ x*^{∗}*= 0 and y*^{k}*→ y*^{∗}*∈ int(K** ^{n}*), it is not hard to verify that

*k*lim*→∞*

*T*_{1}^{k}

*∥x*^{k}*∥*^{2} = 1*−* *y*^{∗}_{1}

*y*^{∗}_{1} +*∥y** ^{∗}*2

*∥ + y*

*1*

^{∗}*− ∥y*

*2*

^{∗}*∥*= 1

2*.* (16)

*From the expression of T*_{2}* ^{k}*, it follows that

(√

*λ*^{k}_{2} +

√

*λ*^{k}_{1}

)

*T*_{2}^{k}*+ 2x*^{k}_{1}*(x*^{k}_{2})^{T}*y*_{2}* ^{k}* equals

(√

*λ*^{k}_{2}*− y** ^{k}*1

*− ∥y*

*2*

^{k}*∥*

^{) (}

*∥y*2

^{k}*∥*

^{2}

*+ y*

_{1}

^{k}*∥y*

*2*

^{k}*∥*

^{)}+

(√

*λ*^{k}_{1}*− y*1* ^{k}*+

*∥y*2

^{k}*∥*

^{) (}

*∥y*

^{k}*∥*

^{2}

*− y*1

^{k}√

*λ*^{k}_{2}

)

*.*
In addition, an elementary calculation yields that

√

*λ*^{k}_{2}*− y*1^{k}*− ∥y** ^{k}*2

*∥ =*

*4x*

^{k}_{1}

*y*

_{1}

^{k}*(x*

^{k}_{2})

^{T}*y*

_{2}

^{k}(√

*λ*^{k}_{2}*+ y*_{1}* ^{k}*+

*∥y*2

^{k}*∥*

^{) (}

*∥x*

*1*

^{k}*x*

^{k}_{2}

*+ y*

^{k}_{1}

*y*

_{2}

^{k}*∥ + y*1

^{k}*∥y*

*2*

^{k}*∥*

^{)}+

_{√}

*∥x*

^{k}*∥*

^{2}

*λ*^{k}_{2} *+ y*_{1}* ^{k}*+

*∥y*

*2*

^{k}*∥*

*+ o(∥x*

^{k}*∥*

^{2}

*),*(17)

√

*λ*^{k}_{1} *− y*1* ^{k}*+

*∥y*

*2*

^{k}*∥ = −*

*4x*

^{k}_{1}

*y*

_{1}

^{k}*(x*

^{k}_{2})

^{T}*y*

_{2}

^{k}(√

*λ*^{k}_{1} *+ y*_{1}^{k}*− ∥y*2^{k}*∥*^{) (}*∥x** ^{k}*1

*x*

^{k}_{2}

*+ y*

^{k}_{1}

*y*

^{k}_{2}

*∥ + y*1

^{k}*∥y*2

^{k}*∥*

^{)}+

_{√}

*∥x*

^{k}*∥*

^{2}

*λ*^{k}_{1} *+ y*_{1}^{k}*− ∥y** ^{k}*2

*∥*

*+ o(∥x*

^{k}*∥*

^{2}

*).*

The last two sides imply that

(√

*λ*^{k}_{2} +

√

*λ*^{k}_{1}

)

*T*_{2}* ^{k}* is equal to

*∥x*^{k}_{√}*∥*^{2}(*∥y*^{k}_{2}*∥*^{2}*+ y*_{1}^{k}*∥y*_{2}^{k}*∥)*

*λ*^{k}_{2}*+ y*^{k}_{1} +*∥y*2^{k}*∥* +*∥x*^{k}*∥*^{2}(*∥y*^{k}*∥*^{2}*− y*1^{k}

√

*λ*^{k}_{2})

√

*λ*^{k}_{1}*+ y*_{1}^{k}*− ∥y*2^{k}*∥* *+ o(∥x*^{k}*∥*^{2})
+ *4x*^{k}_{1}*y*_{1}^{k}*(x*^{k}_{2})^{T}*y*^{k}_{2}(*∥y*2^{k}*∥*^{2}*+ y*^{k}_{1}*∥y** ^{k}*2

*∥)*

(√

*λ*^{k}_{2} *+ y*_{1}* ^{k}*+

*∥y*

*2*

^{k}*∥*

^{) (}

*∥x*

*1*

^{k}*x*

^{k}_{2}

*+ y*

_{1}

^{k}*y*

_{2}

^{k}*∥ + y*1

^{k}*∥y*

*2*

^{k}*∥*

^{)}

*−* *4x*^{k}_{1}*y*_{1}^{k}*(x*^{k}_{2})^{T}*y*_{2}^{k}

(

*∥y*^{k}*∥*^{2}*− y*1^{k}

√

*λ*^{k}_{2}

) (√

*λ*^{k}_{1}*+ y*^{k}_{1} *− ∥y*2^{k}*∥*^{) (}*∥x** ^{k}*1

*x*

^{k}_{2}

*+ y*

_{1}

^{k}*y*

_{2}

^{k}*∥ + y*1

^{k}*∥y*2

^{k}*∥*

^{)}

*− 2x*

*1*

^{k}*(x*

^{k}_{2})

^{T}*y*

_{2}

^{k}*.*

*Let T*_{21}^{k}*and T*_{22}* ^{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 T*

_{21}

*,*

^{k}*k*lim*→∞*

*T*_{21}^{k}

*∥x*^{k}*∥*^{2} = *∥y*_{2}^{∗}*∥*^{2}*+ y*_{1}^{∗}*∥y*^{∗}_{2}*∥*

*2(y*_{1}* ^{∗}*+

*∥y*2

^{∗}*∥)*+

*∥y*

^{∗}*∥*

^{2}

*− y*

_{1}

^{∗}*(y*

_{1}

*+*

^{∗}*∥y*

_{2}

^{∗}*∥)*

*2(y*

_{1}

^{∗}*− ∥y*2

^{∗}*∥)*

*= 0.*

*Whereas from the expression of T*_{22}* ^{k}*, it follows that

*T*

_{22}

^{k}*∥x*^{k}*∥*^{2} = *4x*^{k}_{1}*y*_{1}^{k}*(x*^{k}_{2})^{T}*y*_{2}^{k}

*∥x*^{k}*∥*^{2}

*∥y*^{k}_{2}*∥*^{2}*+ y*_{1}^{k}*∥y*_{2}^{k}*∥*

(√

*λ*^{k}_{2}*+ y*_{1}* ^{k}*+

*∥y*2

^{k}*∥*

^{) (}

*∥x*

*1*

^{k}*x*

^{k}_{2}

*+ y*

^{k}_{1}

*y*

_{2}

^{k}*∥ + y*1

^{k}*∥y*

*2*

^{k}*∥*

^{)}

*−*1

*2y*

^{k}_{1}

*−* *∥y*^{k}*∥*^{2}*− y*1^{k}

√

*λ*^{k}_{2}

(√

*λ*^{k}_{1} *+ y*_{1}^{k}*− ∥y*2^{k}*∥*^{) (}*∥x** ^{k}*1

*x*

^{k}_{2}

*+ y*

^{k}_{1}

*y*

^{k}_{2}

*∥ + y*1

^{k}*∥y*2

^{k}*∥*

^{)}

*.*

Since the sequence

{*4x*^{k}_{1}*y*_{1}^{k}*(x*^{k}_{2})^{T}*y*^{k}_{2}

*∥x*^{k}*∥*^{2}

}

is bounded, and the limit of the term in the bracket
*as k→ ∞ equals 0, we have lim**k**→∞**T*_{22}^{k}*/∥x*^{k}*∥*^{2} = 0. Thus, we get

*k→∞*lim
*T*_{2}^{k}

*∥x*^{k}*∥*^{2} = lim

*k→∞*

*T*_{21}^{k}*+ T*_{22}^{k}

*∥x*^{k}*∥*^{2}^{(√}*λ*^{k}_{2}+

√

*λ*^{k}_{1}

) *= 0.* (18)

*We next take a look at T*_{3}^{k}*. By the expression of T*_{3}* ^{k}*,

(√

*λ*^{k}_{2}+

√

*λ*^{k}_{1}

)

*T*_{3}* ^{k}* equals

*(x*

*)*

^{k}

^{T}*y*

^{k}(√

*λ*^{k}_{2} +

√

*λ*^{k}_{1}

)

*− x** ^{k}*1

(

*∥y*^{k}*∥*^{2}+

√

*λ*^{k}_{2}

√

*λ*^{k}_{1}

)

*− 2y** ^{k}*1

*(x*

^{k}_{2})

^{T}*y*

^{k}_{2}

=

(√

*λ*^{k}_{2} *− y*1^{k}*−∥y*2^{k}*∥*^{) [}*(x*^{k}_{2})^{T}*y*_{2}^{k}*+ x*^{k}_{1}*∥y*2^{k}*∥*^{]}+

(√

*λ*^{k}_{1} *−y*1* ^{k}*+

*∥y*2

^{k}*∥*

^{) [}

*(x*

*)*

^{k}

^{T}*y*

^{k}*− x*

*1*

^{k}√

*λ*^{k}_{2}

]

which, together with (17), implies that

(√

*λ*^{k}_{2} +

√

*λ*^{k}_{1}

)

*T*_{3}^{k}*= o(∥x*^{k}*∥*^{2}). Therefore,

*k*lim*→∞*

*T*_{3}^{k}

*∥x*^{k}*∥*^{2} *= 0.*

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

*k*lim*→∞*

*∥ϕ*FB*(x*^{k}*, y** ^{k}*)

*∥*

^{2}

*∥ϕ*NR*(x*^{k}*, y** ^{k}*)

*∥*

^{2}= lim

*k**→∞*

*2(T*_{1}^{k}*+ T*_{2}^{k}*+ T*_{3}* ^{k}*)

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

**(b) Suppose that w**_{2}^{k}*= 0. Under this subcase, since λ*^{k}_{1} *= λ*^{k}_{2}, we have that
1

2*∥ϕ*FB*(x*^{k}*, y** ^{k}*)

*∥*

^{2}=

*∥x*

^{k}*∥*

^{2}+

*∥y*

^{k}*∥*

^{2}

*+ (x*

*)*

^{k}

^{T}*y*

^{k}*−(x*

^{k}_{1}

*+ y*

_{1}

*)(*

^{k}*∥x*

^{k}*∥*

^{2}+

*∥y*

^{k}*∥*

^{2})

√*∥x*^{k}*∥*^{2}+*∥y*^{k}*∥*^{2} *.*

*Since y*_{1}^{k}*> 0 by y*^{∗}*∈ int(K*^{n}*), from w*_{2}^{k}*= 0 we get y*_{2}* ^{k}* =

*−(x*

*1*

^{k}*/y*

_{1}

^{k}*)x*

^{k}_{2}, which means that

*∥y** ^{k}*2

*∥*

^{2}

*= o(∥x*

^{k}*∥*

^{2}

*) and (x*

*)*

^{k}

^{T}*y*

^{k}*= x*

^{k}_{1}

*y*

_{1}

^{k}*− (x*

*1*

^{k}*/y*

_{1}

*)*

^{k}*∥x*

*2*

^{k}*∥*

^{2}

*= x*

^{k}_{1}

*y*

_{1}

^{k}*+ o(∥x*

^{k}*∥*

^{2}

*).*

Substituting the two equalities into the above ^{1}_{2}*∥ϕ*FB*(x*^{k}*, y** ^{k}*)

*∥*

^{2}yields that 1

2*∥ϕ*FB*(x*^{k}*, y** ^{k}*)

*∥*

^{2}=

*∥x*

^{k}*∥*

^{2}

*−*

_{√}

*(x*

^{k}_{1}

*+ y*

_{1}

*)*

^{k}*∥x*

^{k}*∥*

^{2}

*∥x*^{k}*∥*^{2}+*∥y*^{k}*∥*^{2} *+ (y*_{1}* ^{k}*)

^{2}

*−*

*(x*

^{k}_{1}

*+ y*

^{k}_{1}

*)(y*

_{1}

*)*

^{k}^{2}

√*∥x*^{k}*∥*^{2}+*∥y*^{k}*∥*^{2} *+ x*^{k}_{1}*y*^{k}_{1} *+ o(∥x*^{k}*∥*^{2}*).*

In addition, using *∥y** ^{k}*2

*∥*

^{2}

*= o(∥x*

^{k}*∥*

^{2}) and an elementary calculation, we have that

*(y*

_{1}

*)*

^{k}^{2}

*−*

*(x*

^{k}_{1}

*+ y*

_{1}

^{k}*)(y*

_{1}

*)*

^{k}^{2}

√*∥x*^{k}*∥*^{2}+*∥y*^{k}*∥*^{2} *+ x*^{k}_{1}*y*_{1}^{k}*= (y*_{1}* ^{k}*)

^{2}

*−*

*(y*

^{k}_{1})

^{3}

√*∥x*^{k}*∥*^{2}+*∥y*^{k}*∥*^{2} *+ x*^{k}_{1}*y*_{1}^{k}*−* *x*^{k}_{1}*(y*_{1}* ^{k}*)

^{2}

√*∥x*^{k}*∥*^{2}+*∥y*^{k}*∥*^{2}

= _{√} *(y*_{1}* ^{k}*)

^{2}

*∥x*

^{k}*∥*

^{2}

*∥x*^{k}*∥*^{2}+*∥y*^{k}*∥*^{2}^{(√}*∥x*^{k}*∥*^{2}+*∥y*^{k}*∥*^{2}*+ y*^{k}_{1}^{)}

*+ o(∥x*^{k}*∥*^{2}*).*

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

*k→∞*lim

*∥ϕ*FB*(x*^{k}*, y** ^{k}*)

*∥*

^{2}

*∥ϕ*NR*(x*^{k}*, y** ^{k}*)

*∥*

^{2}= lim

*k→∞*

*∥ϕ*FB*(x*^{k}*, y** ^{k}*)

*∥*

^{2}

*∥x*^{k}*∥*^{2} = 2*−* *2y*^{∗}_{1}

*∥y*^{∗}*∥* + *2(y*_{1}* ^{∗}*)

^{2}

*∥y*^{∗}*∥(∥y*^{∗}*∥ + y** ^{∗}*1)

*= 1,*which gives a contradiction to the inequality (11).

**Subcase 1.2: x**^{∗}*∈ bd(K*^{n}*) and y*^{∗}*∈ bd(K*^{n}*). Since (x** ^{∗}*)

^{2}

*+ (y*

*)*

^{∗}^{2}

*∈ int(K*

*), we must*

^{n}*have x*

^{∗}_{1}

*> 0 and y*

_{1}

^{∗}*> 0, which implies x*

^{k}_{1}

*> 0 and y*

^{k}_{1}

*> 0. Also, x*

^{k}*− y*

^{k}*̸∈ −K*

^{n}*∪ K*

*.*

^{n}*If not, ϕ*

_{NR}

*(x*

^{k}*, y*

^{k}*) = x*

^{k}*or y*

*, which implies that*

^{k}*∥ϕ*NR

*(x*

^{∗}*, y*

*)*

^{∗}*∥ ̸= 0. In addition, noting*that 0 =

*⟨x*

^{∗}*, y*

^{∗}*⟩ = ∥x*

*2*

^{∗}*∥∥y*

*2*

^{∗}*∥ + (x*

*2)*

^{∗}

^{T}*y*

_{2}

*= lim*

^{∗}*k*

*→∞*

*[(x*

^{k}_{2})

^{T}*y*

^{k}_{2}+

*∥x*

*2*

^{k}*∥∥y*2

^{k}*∥], we have*

*(x*^{k}_{2})^{T}*y*_{2}* ^{k}* =

*−∥x*

^{k}_{2}

*∥∥y*

_{2}

^{k}*∥ + α*

*k*with lim

*k**→∞**α*_{k}*= 0.* (19)