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R E S E A R C H

Open Access

An alternative approach for a distance

inequality associated with the second-order

cone and the circular cone

Xin-He Miao

1

, Yen-chi Roger Lin

2

and Jein-Shan Chen

2*

*Correspondence: [email protected] 2Department of Mathematics, National Taiwan Normal University, Taipei, 11677, Taiwan

Full list of author information is available at the end of the article

Abstract

It is well known that the second-order cone and the circular cone have many analogous properties. In particular, there exists an important distance inequality associated with the second-order cone and the 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. Although the proof is a bit longer than the existing one, the new approach offers a way to clarify when the equality holds. Such a clarification is helpful for further study of the relationship between the second-order cone programming problems and the circular cone programming problems.

Keywords: second-order cone; circular cone; projection; distance

1 Introduction

The circular cone [, ] is a pointed closed convex cone having hyperspherical sections orthogonal to its axis of revolution about which the cone is invariant to rotation. Let

denote the circular cone inRn, which is defined by

:=  x= (x, x)∈ R × Rn–| x cos θ ≤ x  =x= (x, x)∈ R × Rn–| x ≤ xtan θ  , ()

with ·  denoting the Euclidean norm and θ ∈ (,π). When θ =π, the circular cone

reduces to the well-known second-order cone (SOC)Kn[, ] (also called the Lorentz

cone), i.e.,

Kn:=(x

, x)∈ R × Rn–| x ≤ x

 . In particular,Kis the set of nonnegative realsR

+. It is well known that the second-order

coneKnis a special kind of symmetric cones []. But when θ=π

, the circular coneis

a non-symmetric cone [, , ].

©The Author(s) 2016. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro-vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

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

x ⇐⇒ Ax ∈Kn with A =  tan θ   In–  , ()

where In–is the (n – )× (n – ) identity matrix. Based on the relationship () between the

SOC and circular cone, Miao et al. [] showed that circular cone complementarity prob-lems can be transformed into the second-order cone complementarity probprob-lems. Further-more from the relationship (), we have

x∈ int ⇐⇒ Ax ∈ intKn and x∈ bd ⇐⇒ Ax ∈ bdKn.

Besides the relationship between second-order cone and circular cone, some topologi-cal structures play important roles in theoretitopologi-cal analysis for optimization problems. For example, the projection formula onto a cone facilitates designing algorithms for solving conic programming problems [–]; 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 [–]. From the above illustrations, an interesting question arises: What is the relationship be-tween 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 [], Theorem .. 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 [], Theorem .. Although the proof is a bit longer than the existing 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 = (x, x)∈ R × Rn–, the spectral

decomposition of x with respect to second-order cone is given by

x= λ(x)u()x + λ(x)u()x , ()

where λ(x), λ(x), u()x , and u()x are expressed as

λi(x) = x+ (–)ix and u(i)x =     (–)iw  , i= , , () with w = x

xif x= , or any vector in Rn–satisfyingw =  if x= . In the setting of the

circular coneLθ, Zhou and Chen [] gave the following spectral decomposition of x∈ Rn

with respect to:

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where μ(x), μ(x), v()x , and v()x are expressed as  μ(x) = x–x cot θ, μ(x) = x+x tan θ, and  v()x =+cotθ  – cot θ·w  , v()x =+tanθ  tan θ·w  , () with w = x

x if x= , or any vector in Rn–satisfyingw =  if x= . Moreover, λ(x), λ(x) and u()x , u()x are called the spectral values and the spectral vectors of x associated

withKn, whereas μ

(x), μ(x) and v()x , v()x are called the spectral values and the spectral

vectors of x associated withLθ, respectively.

To proceed, we denote x+(resp. xθ+) the projection of x ontoKn(resp.); also we set

a+= max{a, } for any real number a ∈ R. According to the spectral decompositions ()

and () of x, the expressions of x+and xθ+can be obtained explicitly, as stated in the

fol-lowing lemma.

Lemma .([, ]) Let x = (x, x)∈ R × Rn–have the spectral decompositions given as

() and () with respect to SOC and circular cone, respectively. Then the following hold: (a) x+= x–x +u () x + x+x +u () x = ⎧ ⎪ ⎨ ⎪ ⎩ x, if xKn, , if x∈ –(Kn)= –Kn, u, otherwise, where u =  x+x  x+x  xx  ; (b) += x–x cot θ +u () x + x+x tan θ +u () x = ⎧ ⎪ ⎨ ⎪ ⎩ x, if x, , if x∈ –()∗= ––θ, v, otherwise, where v =  x+x tan θ +tanθ

(x+x+tan tan θθ tan θ)

xx

 .

Based on the expression of the projection xθ

+ontoin Lemma ., it is easy to obtain,

for any x = (x, x)∈ R × Rn–, the explicit formula of projection of x∈ Rnonto the dual

coneLθ(denoted by (xθ)∗+): += x–x tan θ +u () x + x+x cot θ +u () x = ⎧ ⎪ ⎨ ⎪ ⎩ x, if xLθ=–θ, , if x∈ –(Lθ)∗= –, ω, otherwise, where ω =  x+x cot θ +cotθ

(x+x cot θ+cotθ cot θ)

xx

 .

2 Main results

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

Theorem . For any x∈ Rn, let xθ

+ and(xθ)∗+ be the projections of x onto the circular coneLθand its dual coneLθ, respectively. Let A be the matrix defined as in (). Then the

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(a) If AxKn, then (Ax)

+= Axθ+.

(b) If Ax∈ –Kn, then (Ax)

+= A(xθ)∗+= .

(c) If Ax /Kn∪ (–Kn), then (Ax)+= (+tan θ)A

–(xθ)

+, where (xθ)∗+retains its expression only in the case of x /Lθ∪ (–).

Theorem . Let x= (x, x)∈ R × Rn–have the spectral decompositions with respect to the SOC and the circular cone given as in() and (), respectively. Then the following hold:

(a) dist(Ax,Kn) =

(xtan θx)–+(xtan θ+x)–,

(b) dist(x,Lθ) =



cotθ

+cotθ(xtan θx)–+ tan θ

+tanθ(xcot θ+x)–, where(a)–= min{a, }.

Now, applying Theorem ., we can obtain the relation on the distance formulas asso-ciated with the second-order cone and the circular cone. Note that when θ =π, we know

=Kn and Ax = x. Thus, it is obvious that dist(Ax,Kn) = dist(x,Lθ). In the following

theorem, we only consider the case θ=π.

Theorem . For any x= (x, x)∈ R × Rn–, according to the expressions of the distance formulas dist(Ax,Kn) and dist(x,L

θ), the following hold.

(a) For θ∈ (,π

), we have

dist Ax,Kn≤ dist(x,Lθ)≤ cot θ · dist

Ax,Kn. (b) For θ∈ (π , π ), we have dist(x,Lθ)≤ dist Ax,Kn≤ tan θ · dist(x,Lθ).

3 Proofs of main results 3.1 Proof of Theorem 2.1

(a) If AxKn, by the relationship () between the SOC and the circular cone, we have

xLθ. Thus, it is easy to see that (Ax)+= Ax = Axθ+.

(b) If Ax∈ –Kn, we know that –AxKn, which implies (Ax)

+= . Besides, combining

with (), we have –xLθ, which leads to (xθ)+= . Hence, we have (Ax)+= A(xθ)∗+= .

(c) If Ax /Kn∪ (–Kn), from Lemma .(a), we have

(Ax)+=  xtan θ+x  xtan θ+x  xx  =tan θ   + cotθ  x+x cot θ +cotθ x+x cot θ +cotθ xx  = + tan θ· A –  x+x cot θ +cotθ x+x cot θ +cotθ · cot θ ·xx  = + tan θ· A – xθ∗ +.

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Remark . Here, we say a few more words as regards part (c) in Theorem .. Indeed, if Ax /Kn∪ (–Kn), there are two cases for the element x∈ Rn, i.e., x /L

θ∪ (–) or

xLθ. When x /Lθ∪ (–Lθ), the relationship between (Ax)+and (xθ)∗+is just as stated in

Theorem .(c), that is, (Ax)+= (+tan θ)A

–(xθ)

+. However, when xLθ, we have (xθ)∗+= x.

This implies that the relationship between (Ax)+and (xθ)∗+ is not very clear. Hence, the

relation between (Ax)+and (xθ)∗+in Theorem .(c) is a bit limited. 3.2 Proof of Theorem 2.2

(a) For any x = (x, x)∈ R × Rn–, from the spectral decomposition () with respect to the

SOC, we have Ax = (xtan θ–x)u()x + (xtan θ+x)u()x , where u()x and u()x are given as

in (). It follows from Lemma .(a) that (Ax)+= (xtan θx)+ux()+ (xtan θ+x)+u()x .

Hence, we obtain the distance dist(Ax,Kn):

dist Ax,Kn=Ax– (Ax)+ = xtan θx –u () x + xtan θ+x –u () x  =    xtan θx  –+   xtan θ+x  –.

(b) For any x = (x, x)∈ R × Rn–, from the spectral decomposition () with respect to

circular cone and Lemma .(b), with the same argument, it is easy to see that

dist(x,Lθ) =x– xθ+ =  cotθ  + cotθ xtan θx  –+ tanθ  + tanθ xcot θ+x  –.  3.3 Proof of Theorem 2.3 (a) For θ∈ (,π

), we have  < tan θ <  < cot θ andLθK

nL

θ. We discuss three cases

according to xLθ, x∈ –Lθ, and x /∪ (–Lθ).

Case .If xLθ, then AxKn, which clearly yields dist(Ax,Kn) = dist(x,Lθ) = .

Case .If x∈ –Lθ, then xcot θ≤ –x and

dist(x,Lθ) =x =



x+x.

In this case, there are two subcases for the element Ax. If xcot θ≤ xtan θ ≤ –x, i.e., Ax∈ –Kn, it follows that dist Ax,Kn=Ax =  xtanθ+x ≤  x+x = dist(x,Lθ)≤  x +xcotθ = cot θ·  x tanθ+x = cot θ· dist Ax,Kn,

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where the first inequality holds since tan θ <  (it becomes an equality only in the case of

x= ), and the second inequality holds since cot θ >  (it becomes an equality only in the case of x= ). On the other hand, if xcot θ≤ –x < xtan θ≤ , we have

dist Ax,Kn=Ax– (Ax) + =    xtan θx  –+   xtan θ+x  – =    xtan θx  ≤xtanθ+x <  x+x = dist(x,Lθ)≤  x – xx cot θ <   x  +  x cotθ– xx cot θ = cot θ· dist Ax,Kn,

where the third inequality holds becausex ≤ –xcot θ, and the fourth inequality holds

sincex > –xtan θ≥ . Therefore, for the subcases of x ∈ –Lθ, 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 x= .

Case .If x /∪ (–Lθ), then –x tan θ < x<x cot θ, which yields xtan θ<x

and xcot θ> –x. Thus, we have

dist(x,Lθ) =x– xθ+ =  cotθ  + cotθ xtan θx  –+ tanθ  + tanθ xcot θ+x  – =  cotθ  + cotθ xtan θx  .

On the other hand, it follows from –x tan θ < x<x cot θ and θ ∈ (,π) that

–x < –x tanθ< xtan θ<x.

This implies that

dist Ax,Kn=Ax– (Ax)+ =    xtan θx  –+   xtan θ+x  – =    xtan θx  .

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From this and θ∈ (,π), we see that dist Ax,Kn=    xtan θx  <  cotθ  + cotθ xtan θx  = dist(x,Lθ) =    + tanθ    xtan θx  =    + tanθ dist Ax,Kn.

Therefore, in these cases of x /∪ (–Lθ), we can conclude

dist Ax,Kn< dist(x,Lθ) =    + tanθdist Ax,Kn.

To sum up, from all the above and the fact that max{cot θ, 

+tanθ} = cot θ for θ ∈ (,π),

we obtain

dist Ax,Kn≤ dist(x,Lθ)≤ cot θ · dist

Ax,Kn.

(b) For θ∈ (π,π), we have  < cot θ <  < tan θ andLθKn. Again we discuss the

following three cases.

Case .If xLθ, then AxKn, which implies that dist(Ax,Kn) = dist(x,Lθ) = .

Case .If x∈ –Lθ, then xcot θ≤ –x and

dist(x,Lθ) =x =



x+x.

It follows from xcot θ≤ –x and θ ∈ (π,π) that xtan θ≤ xcot θ≤ –x, which leads

to Ax∈ –Kn. Hence, we have

dist Ax,Kn=Ax = 

x

tanθ+x.

With this, it is easy to verify that dist(Ax,Kn)≥ dist(x,L

θ) for θ∈ (π,π). Moreover, we note that tan θ· dist(x,Lθ) =  tanθ x +x ≥xtanθ+x = dist Ax,Kn. Thus, it follows that

dist(x,Lθ)≤ dist

Ax,Kn≤ tan θ · dist(x,Lθ),

and dist(Ax,Kn) = tan θ· dist(x,L

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Case .If x /∪ (–∗), then we have –x tan θ < x<x cot θ and dist(x,Lθ) =x– xθ+ =  cotθ  + cotθ xtan θx  –+ tanθ  + tanθ xcot θ+x  – =  cotθ  + cotθ xtan θx  .

Since –x tan θ < x <x cot θ, it follows immediately that –x tanθ < xtan θ <

x. Again, there are two subcases for the element Ax. If –x tanθ< –x < xtan θ<

x, then we have Ax /∈Kn∪ (–Kn). Thus, it follows that

dist Ax,Kn=Ax– (Ax)+ =    xtan θx  –+   xtan θ+x  – =    xtan θx  . This together with θ∈ (π,π) yields

dist Ax,Kn> dist(x,Lθ).

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

dist Ax,Kn= 

 + tanθ dist(x,Lθ).

On the other hand, if –x tanθ< xtan θ≤ –x < x, then we have Ax ∈ –Kn, which

implies

dist Ax,Kn=Ax = 

x

tanθ+x.

Therefore, it follows that

dist(x,Lθ) =  cotθ  + cotθ xtan θx  <    xtan θx  ≤xtanθ+x = dist Ax,Kn. Since tanθ xtan θx  –  + tanθ xtanθ+x = –xx tanθ– xtanθx = –xtan θ· x tanθ+ xtan θ +x –xtanθx

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≥ x

–xtanθx

≥ ,

where the first inequality holds due to –x tanθ< xtan θ, and the second inequality

holds due to xtan θ< –x and θ ∈ (π,π), we have

dist Ax,Kn≤ tan θ · dist(x,Lθ).

From all the above analyses and the fact that max{tan θ, 

+tanθ} = tan θ for θ ∈ (

π

,

π

), we

can conclude that

dist(x,Lθ)≤ dist

Ax,Kn≤ tan θ · dist(x,L θ).

Thus, the proof is complete. 

Remark . We point out that Theorem . is equivalent to the results in [], Theo-rem ., that is, for any x, z∈ Rn, we have

A–dist Az,Kn≤ dist(z,L

θ)≤A–dist Az,Kn () and A––dist A–x,

≤ dist x,Kn≤ A dist A–x,

. ()

However, the above inequalities depends on the factorsA and A–. 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 study of the relationship between the second-order cone pro-gramming problems and the circular cone propro-gramming problems. In particular, from the proof of Theorem ., it is clear that dist(Ax,Kn) = tan θ· dist(x,L

θ) holds only

un-der the cases of x = (x, x)∈ or x= ; otherwise we would have the strict inequality

dist(Ax,Kn) < tan θ· dist(x,L

θ). In contrast, it takes tedious algebraic manipulations to

ob-tain such situations by using () and ().

The following example elaborates more why dist(Ax,Kn) = tan θ· dist(x,L

θ) holds only

in the cases of x = (x, x)∈Lθor x= .

Example . Let x = (x, x)∈ R × Rn–and

A=  tan θ   In–  .

When xLθ, we have AxKn. It is clear to see that dist(Ax,Kn) = tan θ· dist(x,Lθ) = .

When x= , i.e., x = (x, )∈ R × Rn–, it follows that Ax = (xtan θ, ). If x≥ , we have x and AxKn, which implies that dist(Ax,Kn) = tan θ· dist(x,Lθ) = . In the other

case, if x< , we see that x∈ –Lθand Ax∈ –Kn. All the above gives dist(Ax,Kn) =Ax =

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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.

Author details

1Department of Mathematics, Tianjin University, Tianjin, 300072, China.2Department of Mathematics, National Taiwan Normal University, Taipei, 11677, Taiwan.

Acknowledgements

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 National Natural Science Foundation of China (No. 11471241). The third author’s work is supported by Ministry of Science and Technology, Taiwan.

Received: 4 May 2016 Accepted: 14 November 2016

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