An alternative approach for a distance inequality associated with the second-order cone and the circular cone

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to appear in Journal of Inequalities and Applications, 2016

An alternative approach for a distance inequality associated with the second-order cone and the circular cone

Xin-He Miao¹

Department of Mathematics Tianjin University, China

Tianjin 300072, China

Yen-chi Roger Lin ² Department of Mathematics National Taiwan Normal University

Taipei 11677, Taiwan

Jein-Shan Chen ³ Department of Mathematics National Taiwan Normal University

Taipei 11677, Taiwan.

April 28, 2016

(revised on November 10, 2016)

Abstract. It is known that the second-order cone and the circular cone have many anal- ogous properties. In particular, there exists an important distance inequality associated with second-order cone and circular cone. The inequality indicates that the distances of arbitrary points to the second-order cone and the circular cone are equivalent, which is crucial in analyzing the tangent cone and normal cone for the circular cone. In this paper, we provide an alternative approach to achieve the aforementioned inequality. Al- though the proof is a bit longer than the existing one, the new approach offers a way to clarify when the equality holds. Such clarification is helpful for further studying in the relationship between the second-order cone programming problems and the circular cone programming problems.

Keywords. Second-order cone, circular cone, projection, distance.

1E-mail: xinhemiao@tju.edu.cn.

2E-mail: yclin@math.ntnu.edu.tw

3Corresponding author. E-mail:jschen@math.ntnu.edu.tw.

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1 Introduction

The circular cone [3, 14] is a pointed closed convex cone having hyperspherical sections orthogonal to its axis of revolution about which the cone is invariant to rotation. Let L_θ denote the circular cone in IRⁿ, which is defined by

Lθ := x = (x1, x2) ∈ IR × IRⁿ⁻¹ | kxk cos θ ≤ x1

= x = (x₁, x₂) ∈ IR × IRⁿ⁻¹ | kx₂k ≤ x₁tan θ , (1) with k · k denoting the Euclidean norm and θ ∈ (0,^π₂). When θ = ^π₄, the circular cone L_θ reduces to the well-known second-order cone (SOC) Kⁿ [2, 6] (also called the Lorentz cone), i.e.,

Kⁿ:=(x₁, x₂) ∈ IR × IRⁿ⁻¹ | kx₂k ≤ x₁ .

In particular, K¹ is the set of nonnegative reals IR+. It is well known that the second- order cone Kⁿ is a special kind of symmetric cones [4]. But when θ 6= ^π₄, the circular cone L_θ is a non-symmetric cone [3, 13, 15].

In [14], Zhou and Chen showed that there is a special relationship between the SOC and the circular cone as follows:

x ∈ L_θ ⇐⇒ Ax ∈ Kⁿ with A = tan θ 0 0 In−1

, (2)

where I_n−1is the (n − 1) × (n − 1) identity matrix. Based on the relationship (2) between the SOC and circular cone, Miao et al. [12] showed that circular cone complementarity problems can be transformed into the second-order cone complementarity problems.

Furthermore from the relationship (2), there hold

x ∈ int L_θ ⇐⇒ Ax ∈ int Kⁿ and x ∈ bd L_θ ⇐⇒ Ax ∈ bd Kⁿ.

Besides the relationship between second-order cone and circular cone, some topological structures play important roles in theoretical analysis for optimization problems. For example, the projection formula onto a cone facilitates designing algorithms for solving conic programming problems [1, 5, 7, 9, 16]; the distance formula from an element to a cone is an important factor in the approximation theory; the tangent cone and normal cone are crucial in analyzing the structure of the solution set for optimization problems [8, 10, 11]. From above illustrations, it raises an interesting question: What is the relationship between second-order cone and circular cone regarding the projection formula, the distance formula, the tangent cone and normal cone, and so on? The issue of the tangent cone and normal cone has been studied in [14, Theorem 2.3]. In this paper, we focus on the other two issues.

More specifically, we provide an alternative approach to achieve an inequality which was obtained in [14, Theorem 2.2]. Although the proof is a bit longer than the existing

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one, the new approach offers a way to clarify when the equality holds which is helpful for further studying in the relationship between the second-order cone programming problems and the circular cone programming problems.

In order to study the relationship between second-order cone and circular cone, we need to recall some background materials. For any vector x = (x1, x2) ∈ IR × IRⁿ⁻¹, the spectral decomposition of x with respect to second-order cone is given by

x = λ₁(x)u⁽¹⁾_x + λ₂(x)u⁽²⁾_x , (3) where λ₁(x), λ₂(x), u⁽¹⁾x , and u⁽²⁾x are expressed as

λ_i(x) = x₁+ (−1)ⁱkx₂k and u⁽ⁱ⁾_x = 1 2

1

(−1)ⁱw

i = 1, 2, (4) with w = _kx^x²

2k if x₂ 6= 0, or any vector in IRⁿ⁻¹ satisfying kwk = 1 if x₂ = 0. In the setting of circular cone L_θ, Zhou and Chen [14] gave the following spectral decomposition of x ∈ IRⁿ with respect to L_θ:

x = µ₁(x)v_x⁽¹⁾+ µ₂(x)v_x⁽²⁾, (5) where µ₁(x), µ₂(x), vx⁽¹⁾, and vx⁽²⁾ are expressed as

µ₁(x) = x₁− kx₂k cot θ,

µ₂(x) = x₁+ kx₂k tan θ, and











v⁽¹⁾x = _1+cot¹ 2θ

1

− cot θ · w

, v⁽²⁾x = _1+tan¹ 2θ

1

tan θ · w

,

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with w = _kx^x²

2k if x2 6= 0, or any vector in IRⁿ⁻¹ satisfying kwk = 1 if x2 = 0. Moreover, λ₁(x), λ₂(x) and u⁽¹⁾x , u⁽²⁾x are called the spectral values and the spectral vectors of x associated with Kⁿ, whereas µ₁(x), µ₂(x) and vx⁽¹⁾, vx⁽²⁾ are called the spectral values and the spectral vectors of x associated with L_θ, respectively.

To proceed, we denote x₊ (resp. x^θ₊) the projection of x onto Kⁿ (resp. L_θ); also we set a₊ = max{a, 0} for any real number a ∈ IR. According to the spectral decompositions (3) and (5) of x, the expressions of x₊ and x^θ₊ can be obtained explicitly, as stated in the following lemma.

Lemma 1.1 ([6, 14]). Let x = (x₁, x₂) ∈ IR×IRⁿ⁻¹have the spectral decompositions given as (3) and (5) with respect to SOC and circular cone, respectively. Then the following hold:

(a)

x₊ = (x₁− kx₂k)₊u⁽¹⁾_x + (x₁+ kx₂k)₊u⁽²⁾_x

=







x, if x ∈ Kⁿ,

0, if x ∈ −(Kⁿ)^∗ = −Kⁿ, u, otherwise,

where u =

" _x

1+kx2k x1+kx22k

2 x2

kx2k

#

;

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(b)

x^θ₊ = (x₁− kx₂k cot θ)₊u⁽¹⁾_x + (x₁+ kx₂k tan θ)₊u⁽²⁾_x

=







x, if x ∈ L_θ,

0, if x ∈ −(L_θ)^∗ = −L^π

2−θ, v, otherwise,

where v =

" _x₁_+kx₂k tan θ 1+tan²θ

x1+kx2k tan θ

1+tan²θ tan θ

x2

kx₂k

# .

Based on the expression of the projection x^θ₊onto Lθ in Lemma 1.1, it is easy to obtain that for any x = (x₁, x₂) ∈ IR × IRⁿ⁻¹, the explicit formula of projection of x ∈ IRⁿ onto the dual cone L^∗_θ (denoted by (x^θ)^∗₊) is as below:

(x^θ)^∗₊ = (x₁− kx₂k tan θ)₊u⁽¹⁾_x + (x₁+ kx₂k cot θ)₊u⁽²⁾_x

=







x, if x ∈ L^∗_θ = L^π

2−θ, 0, if x ∈ −(L^∗_θ)^∗ = −L_θ, ω, otherwise,

where ω =

" _x₁_+kx₂_{k cot θ}

1+cot²θ

_x

1+kx2k cot θ

1+cot²θ cot θ

x2

kx₂k

# .

2 Main results

In this section, we give the main results of this paper.

Theorem 2.1. For any x ∈ IRⁿ, let x^θ₊ and (x^θ)^∗₊ be the projections of x onto the circular cone Lθ and its dual cone L^∗_θ, respectively. Let A be the matrix defined as in (2). Then the following hold:

(a) If Ax ∈ Kⁿ, then (Ax)₊= Ax^θ₊.

(b) If Ax ∈ −Kⁿ, then (Ax)₊ = A(x^θ)^∗₊ = 0.

(c) If Ax /∈ Kⁿ ∪ (−Kⁿ), then (Ax)₊ = ^1+tan₂ ²^θA⁻¹(x^θ)^∗₊, where (x^θ)^∗₊ retains its expression only in the case of x /∈ L^∗_θ∪ (−L_θ).

Theorem 2.2. Let x = (x₁, x₂) ∈ IR×IRⁿ⁻¹ have the spectral decompositions with respect to the SOC and the circular cone given as in (3) and (5), respectively. Then the following hold:

(a) dist(Ax, Kⁿ) = q1

2(x₁tan θ − kx₂k)²₋+ ¹₂(x₁tan θ + kx₂k)²₋. (b) dist(x, Lθ) =

q cot²θ

1+cot²θ(x1tan θ − kx2k)²₋+ _1+tan^tan²^θ2θ(x1cot θ + kx2k)²₋, where (a)− = min{a, 0}.

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Now, applying Theorem 2.2, we can obtain the relation on the distance formulas associated with the second-order cone and the circular cone. Note that when θ = ^π₄, we know L_θ = Kⁿ and Ax = x. Thus, it is obvious that dist(Ax, Kⁿ) = dist(x, L_θ). In the following theorem, we only consider the case θ 6= ^π₄.

Theorem 2.3. For any x = (x₁, x₂) ∈ IR × IRⁿ⁻¹, according to the expressions of the distance formulas dist(Ax, Kⁿ) and dist(x, L_θ), the following hold.

(a) For θ ∈ (0,^π₄), we have

dist(Ax, Kⁿ) ≤ dist(x, L_θ) ≤ cot θ · dist(Ax, Kⁿ).

(b) For θ ∈ (^π₄,^π₂), we have

dist(x, L_θ) ≤ dist(Ax, Kⁿ) ≤ tan θ · dist(x, L_θ).

3 Proofs of main results

3.1 Proof of Theorem 2.1

(a) If Ax ∈ Kⁿ, by the relationship (2) between the SOC and the circular cone, we have x ∈ L_θ. Thus, it is easy to see that (Ax)₊= Ax = Ax^θ₊.

(b) If Ax ∈ −Kⁿ, we know that −Ax ∈ Kⁿ, which implies (Ax)₊ = 0. Besides, combining with (2), we have −x ∈ L_θ, which leads to (x^θ)^∗₊= 0. Hence, we have (Ax)₊= A(x^θ)^∗₊= 0.

(c) If Ax /∈ Kⁿ∪ (−Kⁿ), from Lemma 1.1(a), we have

(Ax)+ =

" _x

1tan θ+kx2k x1tan θ+kx2 2k

2

x2

kx₂k

#

= tan θ

2 (1 + cot²θ)

" _x

1+kx2k cot θ 1+cot²θ x1+kx2k cot θ

1+cot²θ x2

kx₂k

#

= 1 + tan²θ 2 · A⁻¹

" _x

1+kx2k cot θ 1+cot²θ x1+kx2k cot θ

1+cot²θ · cot θ · _kx^x²

2k

#

= 1 + tan²θ

2 · A⁻¹(x^θ)^∗₊. The proof is complete. 2

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Remark 3.1. Here, we say a few more words for the part(c) in Theorem 2.1. Indeed, if Ax /∈ Kⁿ∪ (−Kⁿ), there are two cases for the element x ∈ IRⁿ, i.e., x /∈ L^∗_θ ∪ (−Lθ) or x ∈ L^∗_θ. When x /∈ L^∗_θ∪(−L_θ), the relationship between (Ax)₊and (x^θ)^∗₊ is just as stated in Theorem 2.1(c), that is, (Ax)₊= ^1+tan₂ ²^θA⁻¹(x^θ)^∗₊. However, when x ∈ L^∗_θ, we have (x^θ)^∗₊= x. This implies that the relationship between (Ax)₊ and (x^θ)^∗₊ is not very clear.

Hence, the relation between (Ax)+ and (x^θ)^∗₊ in Theorem 2.1(c) is a bit limited.

3.2 Proof of Theorem 2.2

(a) For any x = (x₁, x₂) ∈ IR × IRⁿ⁻¹, from the spectral decomposition (3) with respect to the SOC, we have Ax = (x₁tan θ − kx₂k)u⁽¹⁾x + (x₁tan θ + kx₂k)u⁽²⁾x , where u⁽¹⁾x and u⁽²⁾x

are given as in (4). It follows from Lemma 1.1(a) that (Ax)₊ = (x₁tan θ − kx₂k)₊u⁽¹⁾x + (x₁tan θ + kx₂k)₊u⁽²⁾x . Hence, we obtain the distance dist(Ax, Kⁿ) as bellow:

dist(Ax, Kⁿ) = kAx − (Ax)₊k

=

(x₁tan θ − kx₂k)₋u⁽¹⁾_x + (x₁tan θ + kx₂k)₋u⁽²⁾_x

= r1

2(x₁tan θ − kx₂k)²₋+ 1

2(x₁tan θ + kx₂k)²₋.

(b) For any x = (x₁, x₂) ∈ IR × IRⁿ⁻¹, from the spectral decomposition (5) with respect to circular cone and Lemma 1.1(b), with the same argument, it is easy to get that

dist(x, L_θ) =

x − x^θ₊

= s

cot²θ

1 + cot²θ(x₁tan θ − kx₂k)²₋+ tan²θ

1 + tan²θ(x₁cot θ + kx₂k)²₋. 2

3.3 Proof of Theorem 2.3

(a) For θ ∈ (0,^π₄), we have 0 < tan θ < 1 < cot θ and L_θ ⊂ Kⁿ ⊂ L^∗_θ. We discuss three cases according to x ∈ L_θ, x ∈ −L^∗_θ and x /∈ L_θ∪ (−L^∗_θ).

Case 1. If x ∈ L_θ, then Ax ∈ Kⁿ which clearly yields dist(Ax, Kⁿ) = dist(x, L_θ) = 0.

Case 2. If x ∈ −L^∗_θ, then x₁cot θ ≤ −kx₂k and dist(x, L_θ) = kxk =

q

x²₁+ kx₂k².

Under this case, there are two subcases for the element Ax. If x₁cot θ ≤ x₁tan θ ≤

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−kx₂k, i.e., Ax ∈ −Kⁿ, it follows that dist(Ax, Kⁿ) = kAxk =

q

x²₁tan²θ + kx₂k² ≤ q

x²₁+ kx₂k²

= dist(x, L_θ) ≤ q

x²₁ + kx₂k²cot²θ

= cot θ · q

x²₁tan²θ + kx₂k²

= cot θ · dist(Ax, Kⁿ),

where the first inequality holds since tan θ < 1 (it becomes equality only in the case of x = 0), and the second inequality holds since cot θ > 1 (it becomes equality only in the case of x₂ = 0). On the other hand, if x₁cot θ ≤ −kx₂k < x₁tan θ ≤ 0, we have

= r1

2(x₁tan θ − kx₂k)²₋+ 1

2(x₁tan θ + kx₂k)²₋

= r1

2(x₁tan θ − kx₂k)² ≤ q

x²₁tan²θ + kx₂k² <

q

x²₁+ kx₂k²

= dist(x, L_θ)

≤ q

x²₁ − x₁kx₂k cot θ

<

r1

2x²₁+1

2kx₂k²cot²θ − x₁kx₂k cot θ

= cot θ · dist(Ax, Kⁿ),

where the third inequality holds because kx₂k ≤ −x₁cot θ, and the fourth inequality holds since kx₂k > −x₁tan θ ≥ 0. Therefore, for the subcases of x ∈ −L^∗_θ, we can conclude that

dist(Ax, Kⁿ) ≤ dist(x, Lθ) ≤ cot θ · dist(Ax, Kⁿ), and dist(x, Lθ) = cot θ · dist(Ax, Kⁿ) holds only in the case of x2 = 0.

Case 3. If x /∈ L_θ∪ (−L^∗_θ), then −kx₂k tan θ < x₁ < kx₂k cot θ, which yields x₁tan θ <

kx₂k and x₁cot θ > −kx₂k. Thus, we have dist(x, L_θ) =

x − x^θ₊

= s

cot²θ

1 + tan²θ(x₁cot θ + kx₂k)²₋

= s

cot²θ

1 + cot²θ(x₁tan θ − kx₂k)².

On the other hand, it follows from −kx₂k tan θ < x₁ < kx₂k cot θ and θ ∈ (0,^π₄) that

−kx₂k < −kx₂k tan²θ < x₁tan θ < kx₂k.

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This implies that

= r1

2(x₁tan θ − kx₂k)²₋+ 1

= r1

2(x₁tan θ − kx₂k)². From this and θ ∈ (0,^π₄), we see that

dist(Ax, Kⁿ) = r1

2(x₁tan θ − kx₂k)² <

s

cot²θ

1 + cot²θ(x₁tan θ − kx₂k)²

= dist(x, L_θ) =

r 2

1 + tan²θ r1

2(x₁tan θ − kx₂k)²

=

r 2

1 + tan²θdist(Ax, Kⁿ).

Therefore, in these cases of x /∈ L_θ∪ (−L^∗_θ), we can conclude dist(Ax, Kⁿ) < dist(x, L_θ) =

r 2

1 + tan²θdist(Ax, Kⁿ), To sum up, from all the above and the fact that max{cot θ,q

2

1+tan²θ} = cot θ for θ ∈ (0,^π₄), we obtain that

dist(Ax, Kⁿ) ≤ dist(x, Lθ) ≤ cot θ · dist(Ax, Kⁿ).

(b) For θ ∈ (^π₄,^π₂), we have 0 < cot θ < 1 < tan θ and L^∗_θ ⊂ Kⁿ ⊂ L_θ. Again we discuss the following three cases.

Case 1. If x ∈ L_θ, then Ax ∈ Kⁿ which implies that dist(Ax, Kⁿ) = dist(x, L_θ) = 0.

Case 2. If x ∈ −L^∗_θ, then x₁cot θ ≤ −kx₂k and dist(x, L_θ) = kxk =

q

x²₁+ kx₂k².

It follows from x₁cot θ ≤ −kx₂k and θ ∈ (^π₄,^π₂) that x₁tan θ ≤ x₁cot θ ≤ −kx₂k, which leads to Ax ∈ −Kⁿ. Hence, we have

dist(Ax, Kⁿ) = kAxk = q

x²₁tan²θ + kx₂k².

With this, it is easy to verify that dist(Ax, Kⁿ) ≥ dist(x, L_θ) for θ ∈ (^π₄,^π₂). Moreover, we note that

tan θ · dist(x, L_θ) = q

tan²θ(x²₁+ kx₂k²) ≥ q

x²₁tan²θ + kx₂k² = dist(Ax, Kⁿ).

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Thus, it follows that

dist(x, L_θ) ≤ dist(Ax, Kⁿ) ≤ tan θ · dist(x, L_θ), and dist(Ax, Kⁿ) = tan θ · dist(x, L_θ) holds only in the case of x₂ = 0.

Case 3. If x /∈ L_θ∪ (−L^∗_θ), then we have −kx₂k tan θ < x₁ < kx₂k cot θ and dist(x, L_θ) =

x − x^θ₊

= s

cot²θ

1 + tan²θ(x₁cot θ + kx₂k)²₋

= s

cot²θ

1 + cot²θ(x₁tan θ − kx₂k)².

Since −kx₂k tan θ < x₁ < kx₂k cot θ, it follows immediately that −kx₂k tan²θ < x₁tan θ <

kx2k. Again, there are two subcases for the element Ax. If −kx2k tan²θ < −kx2k <

x₁tan θ < kx₂k, then we have Ax /∈ Kⁿ∪ (−Kⁿ). Thus, it follows that dist(Ax, Kⁿ) = kAx − (Ax)₊k

= r1

2(x₁tan θ − kx₂k)²₋+ 1

= r1

2(x₁tan θ − kx₂k)². This together with θ ∈ (^π₄,^π₂) yields

dist(Ax, Kⁿ) > dist(x, Lθ).

Moreover, by the expressions of dist(Ax, Kⁿ) and dist(x, L_θ), it is easy to verify dist(Ax, Kⁿ) =

r 2

1 + tan²θdist(x, L_θ).

On the other hand, if −kx2k tan²θ < x1tan θ ≤ −kx2k < kx2k, then we have Ax ∈ −Kⁿ, which implies

dist(Ax, Kⁿ) = kAxk = q

x²₁tan²θ + kx2k². Therefore, it follows that

dist(x, L_θ) = s

cot²θ

1 + cot²θ(x₁tan θ − kx₂k)²

<

r1

2(x₁tan θ − kx₂k)² ≤ q

x²₁tan²θ + kx₂k²

= dist(Ax, Kⁿ).

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Since

tan²θ(x₁tan θ − kx₂k)²− (1 + tan²θ)(x²₁tan²θ + kx₂k²)

= −2x₁kx₂k tan³θ − x²₁tan²θ − kx₂k²

= −x₁tan θ · (kx₂k tan²θ + x₁tan θ) + kx₂k(−x₁tan³θ − kx₂k)

≥ kx₂k(−x₁tan³θ − kx₂k)

≥ 0,

where the first inequality holds due to −kx₂k tan²θ < x₁tan θ, and the second inequality holds due to x₁tan θ < −kx₂k and θ ∈ (^π₄,^π₂), we have

dist(Ax, Kⁿ) ≤ tan θ · dist(x, Lθ).

From all the above analysis and the fact that max{tan θ,q

2

1+tan²θ} = tan θ for θ ∈ (^π₄,^π₂), we can conclude that

dist(x, L_θ) ≤ dist(Ax, Kⁿ) ≤ tan θ · dist(x, L_θ).

Thus, the proof is complete. 2

Remark 3.2. We point out that Theorem 2.3 is equivalent to the results in [14, Theorem 2.2], that is, for any x, z ∈ IRⁿ, we have

kAk⁻¹ dist(Az, Kⁿ) ≤ dist(z, L_θ) ≤ kA⁻¹k dist(Az, Kⁿ) (7) and

kA⁻¹k⁻¹ dist(A⁻¹x, Lθ) ≤ dist(x, Kⁿ) ≤ kAk dist(A⁻¹x, Lθ). (8) However, the above inequalities depends on the factors kAk and kA⁻¹k. Here, we provide a more concrete and simple expression for the inequality. What is the benefit of such a new expression? Indeed, the new approach provides the situation where the equality holds, which is helpful for further studying in the relationship between the second-order cone programming problems and the circular cone programming problems. In particular, from the proof of Theorem 2.3, it is clear that dist(Ax, Kⁿ) = tan θ · dist(x, L_θ) holds only under the cases of x = (x1, x2) ∈ Lθ or x2 = 0; otherwise we would have the strict inequality dist(Ax, Kⁿ) < tan θ · dist(x, L_θ). To the contrast, it takes tedious algebraic manipulations to obtain such situations by using (7) and (8).

The following example elaborates more about why dist(Ax, Kⁿ) = tan θ · dist(x, L_θ) holds only under the cases of x = (x₁, x₂) ∈ L_θ or x₂ = 0.

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Example 3.1. Let x = (x₁, x₂) ∈ IR × IRⁿ⁻¹ and A = tan θ 0 0 In−1

. When x ∈ L_θ, we have Ax ∈ Kⁿ. It is clear to see that dist(Ax, Kⁿ) = tan θ · dist(x, L_θ) = 0. When x₂ = 0, i.e., x = (x₁, 0) ∈ IR × IRⁿ⁻¹, it follows that Ax = (x₁tan θ, 0). If x₁ ≥ 0, we have x ∈ L_θ and Ax ∈ Kⁿ, which implies that dist(Ax, Kⁿ) = tan θ · dist(x, L_θ) = 0. In other case, if x1 < 0, we get that x ∈ −L^∗_θ and Ax ∈ −Kⁿ. All the above gives that dist(Ax, Kⁿ) = kAxk = |x₁| tan θ = tan θ · dist(x, L_θ).

Competing interests The authors declare that none of the authors have any competing interests in the manuscript.

Authors contributions All authors participated in its design and coordination and helped to draft the manuscript. All authors read and approved the final manuscript.

Acknowledgments We would like to thank the Editor and the anonymous referees for their careful reading and constructive comments which have helped us to significantly improve the presentation of the paper. The first author’s work is supported by Na- tional Natural Science Foundation of China (No. 11471241). The third author’s work is supported by Ministry of Science and Technology, Taiwan.

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