R E S E A R C H
Open Access
Circular cone convexity and some inequalities
associated with circular cones
Jinchuan Zhou
1, Jein-Shan Chen
2*and Hao-Feng Hung
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
The study of this paper consists of two aspects. One is characterizing the so-called circular cone convexity of f by exploiting the second-order differentiability of fLθ; the
other is introducing the concepts of determinant and trace associated with circular cone and establishing their basic inequalities. These results show the essential role played by the angle
θ
, which gives us a new insight when looking into properties about circular cone.MSC: 26A27; 26B05; 26B35; 49J52; 90C33; 65K05 Keywords: circular cone; convexity; determinant; trace
1 Introduction
Recently, much attention has been paid to the nonsymmetric cone optimization problems, see [–] and the references therein. Unlike symmetric cones [], there is no unified struc-ture for nonsymmetric cones. Hence, how to tackle nonsymmetric cone optimization is still an issue. For symmetric cone optimization, the algebraic structure associated with symmetric cones, including second-order cone and positive semi-definite matrix cones, allows us to study them via exploiting the unified Euclidean Jordan algebra []. In gen-eral, the way to deal with nonsymmetric cone optimization depends on the feature of the associated nonsymmetric cone. In this paper, we focus on a special nonsymmetric cone, circular coneLθ. The circular cone [–] is a pointed closed convex cone having
hyper-spherical sections orthogonal to its axis of revolution about which the cone is invariant to rotation. Let its half-aperture angle be θ with θ∈ (, ◦). Then, it is mathematically expressed as Lθ:= x= (x, x)T∈ R × Rn–| x≥ x cos θ =x= (x, x)T∈ R × Rn–| x≥ x cot θ .
Real applications of a circular cone lie in some engineering problems, for example, in the formulation for optimal grasping manipulation for multi-fingered robots, the grasping force of ith finger is subject to a circular cone constraint, see [, ] and references for more details.
AlthoughLθis a nonsymmetric cone, we can, due to its special structure, establish the
explicit form of orthogonal decomposition (or spectral decomposition) [] as
x= λ(x)· u()x + λ(x)· u()x , ()
©2013Zhou et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribu-tion License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribuAttribu-tion, and reproducAttribu-tion in any medium, provided the original work is properly cited.
where ⎧ ⎨ ⎩ λ(x) = x–x cot θ, λ(x) = x+x tan θ and ⎧ ⎪ ⎨ ⎪ ⎩ u()x =+cotθ cot θ In– –¯x = sinθ –(sin θ cos θ )¯x , u()x =+tanθ tan θ In– ¯x = cosθ (sin θ cos θ )¯x
with ¯x= x/x if x= , and ¯xbeing any vector w inRn–satisfyingw = if x= . Clearly, x∈Lθif and only if λ(x)≥ .
The formula () allows us to define the following vector-valued function: fLθ(x) := fλ (x) u()x + fλ(x) u()x , ()
where f is a real-valued function from J toR with J being a subset in R. Let S be the set of all x∈ Rnwhose spectral values λ
i(x) for i = , belong to J, i.e., S :={x ∈ Rn| λi(x)∈ J, i =
, }. According to [], we know that S is open if and only if J is open. In addition, as J is an interval, then S is convex because
minλ(x), λ(y) ≤ λ βx+ ( – β)y≤ λ βx+ ( – β)y ≤ maxλ(x), λ(y) , ∀β ∈ [, ].
Throughout this paper, we always assume that J is an interval inR. Clearly, as θ = ◦, L◦reduces to the second-order cone and the above expressions () and () correspond to the spectral decomposition and the SOC-function associated with the second-order cone, respectively (see [, ] for more information regarding fsoc).
It is well known that in dealing with symmetric cone optimization problems, such as second-order cone optimization problems and positive semi-definite optimization prob-lems, this type of vector-valued functions plays an essential role. Inspired by this, we study the properties of fLθ, which is crucial for circular cone optimization problems. In
our previous works, we have studied the smooth and nonsmooth analysis of fLθ [, ];
and the circular cone monotonicity and second-order differentiability of fLθ []. From
the aforementioned research, there is an interesting observation: some properties com-monly shared by fsocand fLθ are independent of the angle θ ; for example, fLθ is
direc-tionally differentiable, Fréchet differentiable, semi-smooth if and only if f is direcdirec-tionally differentiable, Fréchet differentiable, semi-smooth; while some properties are dependent on the angle θ ; for example, fLθ with f (t) = –/t for t > is circular cone monotone as
θ∈ [◦, ◦), but not circular cone monotone as θ∈ (, ◦).
In this paper, we further study the circular cone convexity of f . More precisely, a real-valued function f : J→ R is said to beLθ-convex of order n on S if for any x, y∈ S,
fLθβx+ ( – β)y
The characterization ofLθ-convexity is based on the observation that f isLθ-convex if and
only if (fLθ)(x)(h, h)∈Lθfor all h∈ Rn. Our result shows that the circular cone convexity
requires that the angle θ belongs in [◦, ◦). In particular, we show that f isLθ-convex
of order if and only if θ∈ [◦, ◦) and f is convex.
On the other hand, using the spectral decomposition (), we define the determinant and traceof x in the framework of circular cone as
det(x) := λ(x)λ(x) and tr(x) := λ(x) + λ(x),
respectively. In the symmetric cone setting, the concepts of determinant and trace are the key ingredients of barrier and penalty functions which are used in barrier and penalty methods (including interior point methods) for symmetric cone optimization, see [– ]. Here we further study some basic inequalities of det(x) and tr(x) in the framework of circular cone. As seen in Section , the obtained inequalities are classified into three categories: (i) the first class is independent of the angle (i.e., still holds in the framework of circular cone); (ii) the second class is dependent on the angle, for example, for x, y∈Lθ,
the inequality
det(e + x + y)≤ det(e + x) det(e + y),
where e = (, , . . . , )∈ Rn, fails as θ ∈ (, ◦) but holds as θ ∈ [◦, ◦); (iii) the third class always fails no matter what value of θ is chosen. These results give us a new insight into a circular cone and make us focus more on the role played by the angle θ .
The notation used in this paper is standard. For example, denote by Rn the
n-dimensional Euclidean space and by R+ the set of all nonnegative real scalars, i.e., R+={t ∈ R | t ≥ }. For x, y ∈ Rn, the inner product is denoted by xTy. LetSn mean the spaces of all real symmetric matrices inRn×n, and letSn
+denote the cone of positive semi-definite matrices. We write xLθ yto stand for x – y∈Lθ. Finally, we define :=
for convenience.
2 Circular cone convexity
The main purpose of this section is to provide characterizations ofLθ-convex functions.
First, we need the following technical lemma.
Lemma . Given αi∈ R for i = , . . . , and βi∈ R for i = , , , we define
F(β, β, β) := αβ+ αβ+ αββ+ αββ+ αββ+ αβββ. () IfF(β, β, β)≥ for all (β, β, β)∈ R, then
α≥ , α≥ , α≥ , α≥ , α≥ –√αα. Furthermore, if α≤ ⎧ ⎨ ⎩ αα for α≥ , [α– (α/α)]α for α∈ [–√αα, ), ()
Proof If β = , thenF(β, β, β) = β[αβ+ αβ]. From F(β, β, β)≥ , we have αβ+ αβ≥ . Thus, α≥ by letting β→ and α≥ by letting β→ .
If β= , thenF(β, β, β) = β[αβ+ αβ]. FromF(β, β, β)≥ , we obtain α≥ and α≥ . If β= , then F(β, β, β) = αβ+ αβ+ αββ= ββ α β β + α+ α β β () whenever β= and β= . Let t = β/β. FromF(β, β, β)≥ , equation () implies
α≥ –αt– α /t, ∀t = , i.e., α≥ max t= –αt– α /t= – min t= αt+ α /t= –√αα. Furthermore, if α≥ , then F(β, β, β)≥ αβ+ αββ+ αβββ= β ββ α α/ α/ α β ββ ≥ , where the last step is due to
α α/ α/ α S +O,
which is ensured by condition (). Similarly, if α∈ [–√αα, ) (implying α= in this case), then F(β, β, β) = √ αβ+ α √α β + α– α α β+ αββ+ αββ+ αβββ ≥ α– α α β+ αββ+ αβββ =β ββ α– (α/α) α/ α/ α β ββ ≥ , where the last step is due to
α– (α/α) α/ α/ α S +O,
which is ensured by condition () and the fact α– (α/α)≥ since –√αα≤ α< .
This completes the proof.
Lemma .[, Theorem .] Let f : J→ R and fLθbe defined as in(). Then fLθis
i= , . Moreover, for u, v∈ Rn, if x = , then fLθ(x)(u, v) = ⎧ ⎪ ⎨ ⎪ ⎩ f(x) uTv uv+vu , either u= or v= , f(x)uTv
f(x)(vu+uv)+f(x)(tan θ –cot θ )(uv+¯uT¯vvu)
, otherwise. If x= , then fLθ(x)(u, v) = I I , where I:= vu˜ξ + ˜ u¯xTv+ v¯xTu +˜avTu+ (˜η – ˜a)¯xTv¯xTu, I:= (˜η – ˜a)u¯xTv+ ( – ˜d)¯xTv¯xTu+˜vu+ (˜η – ˜a)v¯xTu ¯x + ˜d¯xTuv+ vTu¯x+¯xTvu +˜a(uv+ vu) with ˜a =f(λ(x)) – f(λ(x)) λ(x) – λ(x) , ˜ξ = f(λ(x)) + cotθ + f(λ(x)) + tanθ, ˜ = – cot θ + cotθf λ (x) + tan θ + tanθf λ (x) , ˜η = cotθ + cotθf λ (x) + tan θ + tanθf λ (x) , ˜d = x cotθ + cotθf λ (x) + tan θ + tanθf λ (x) –f(λ(x)) – f (λ(x)) λ(x) – λ(x) , = – cot θ + cotθf λ (x) + tan θ + tanθf λ (x) .
The characterization ofLθ-convexity is established below, which can be regarded as the
extension of some results given in [, –] from the second-order cone setting to the circular cone setting.
Theorem . Suppose that f : J→ R is second-order continuously differentiable. If f is Lθ-convex of order n on S, then tan θ≥ , f is convex on J, and for all τ, τ∈ J with τ≤ τ,
f(τ)δ(τ, τ)≥ (τ– τ) δ(τ, τ) () and tanθ δ(τ, τ) + tanθ– δ(τ, τ) f(τ) – (τ– τ) δ(τ, τ) ≥ –f(τ ) tanθ– δ(τ , τ) tanθ δ(τ , τ) + tanθ– δ(τ , τ) . ()
Furthermore, if tanθ δ(τ, τ) + tanθ– δ(τ, τ) f(τ)≥ (τ– τ) δ(τ, τ) () and δ(τ, τ)δ(τ, τ)≤ tanθ δ(τ, τ) + tanθ– δ(τ, τ) f(τ)f(τ)(τ– τ), () or if tanθ δ(τ, τ) + tanθ– δ(τ, τ) f(τ) < (τ– τ) δ(τ, τ) and δ(τ, τ)δ(τ, τ) tanθ– f(τ) ≤tanθ– f(τ) tanθ δ(τ , τ) + tanθ– δ(τ, τ) δ(τ, τ) –tanθ δ(τ, τ) + tanθ– δ(τ, τ) f(τ) – (τ– τ) δ(τ, τ) × f(τ )(τ– τ), ()
then f isLθ-convex. Here δ(τ , τ) := f (τ ) – f (τ) – f(τ)(τ – τ) for τ , τ∈ J.
Proof According to [, Theorem .], f isLθ-convex if and only if (fLθ)(x)(h, h)∈Lθfor
all x∈ S and h ∈ Rn. We proceed the proof by considering the following three cases.
Case . For x= and h= , it follows from Lemma . that fLθ(x)(h, h) = f(x ) h .
Hence, (fLθ)(x)(h, h)∈Lθif and only if f(x
)≥ .
Case . For x= and h= , it follows from Lemma . that fLθ(x)(h, h) = f(x)h f(x)hh+ f(x)(tan θ – cot θ )hh . Hence, (fLθ)(x)(h, h)∈L
θif and only if f(x)≥ and
tan θh≥h+ (tan θ – cot θ )hh,
i.e., – tan θh+h ≤h+ (tan θ – cot θ )h h ≤ tan θ h+h .
Dividing byhand letting t = h/h yields – tan θt+ ≤ t + tan θ – cot θ ≤ tan θt+
⇐⇒ max
t∈R – tan θ
t+ – t≤ tan θ – cot θ ≤ min
t∈Rtan θ
t+ – t ⇐⇒ cot θ – tan θ ≤ tan θ – cot θ ≤ tan θ – cot θ
⇐⇒ tan θ ≥ .
Case . For x= , due to the simplification of notation, let us denote
μ:= h– cot θ¯xTh, μ:= h+ tan θ¯xTh, μ:= h– ¯xT h . () Then ¯xT h= μ– μ
tan θ+ cot θ and h=
tan θ μ+ cot θ μ
tan θ+ cot θ . ()
Note that μ, μ, and μcan take any value inR × R × R+by taking a suitable value of h (because the vector h has n variables). It follows from Lemma . that
fLθ(x)(h, h) = ˜ξh + ˜¯xThh+˜ah+ (˜η – ˜a)(¯xTh) [( – ˜d)(¯xT h)+ (˜η – ˜a)¯xThh]¯x+ [˜h+ ˜dh]¯x+ [˜ah+ ˜d¯xTh]h =: ¯x+ h , where = ˜ξ h+ ˜¯xThh+˜ah+ (˜η – ˜a) ¯xT h , = ( – ˜d) ¯xT h + (˜η – ˜a)¯xThh+˜h+ ˜dh, = ˜ah+ ˜d¯xTh . Hence, (fLθ)(x)(h, h)∈L θis equivalent to ≥ and tanθ≥ ¯x+ h. Note that = + cotθf λ (x) h– ¯xTh hcot θ+ ¯xT h cotθ + + tanθf λ (x) h+ ¯xTh htan θ+ ¯xT h tanθ +˜ah– ¯xT h = + cotθf λ (x) μ+ + tanθf λ (x) μ+˜aμ. ()
We now claim that ≥ for all h ∈ Rnif and only if fλ(x) ≥ , fλ(x) ≥ , and ˜a ≥ . ()
The sufficiency is clear. Let us show the necessity. In particular, choosing h = (– tan θ ,¯x) yields μ= and μ= . It then follows from ≥ that f(λ(x))≥ . If we choose h= (cot θ ,¯x), then we have f(λ(x))≥ . Finally, choosing h = (, kz) with k∈ R, z = and zT ¯x= gives = f(λ(x)) + cotθ + f(λ(x)) + tanθ +˜ak ≥ .
Dividing by k both sides and taking the limits as k→ ∞, we obtain ˜a ≥ . Since λ i(x)
can take an arbitrary value in J, it is clear that () is equivalent to saying that f(τ )≥ for all τ ∈ J, i.e., f is convex on J. Indeed, the condition ˜a ≥ is ensured by the fact that
˜a =f(λ(x))–f(λ(x)) λ(x)–λ(x) = f
(t
)≥ for some t∈ (λ(x), λ(x)). Now we calculate the values of and , respectively.
= – cot θ + cotθf λ (x) μ+ tan θ + tanθf λ (x) μ + ˜dμ– ˜d¯xTh+˜ah ¯xT h = – cot θ + cotθf λ (x) μ+ tan θ + tanθf λ (x) μ+ ˜dμ–¯xTh . ()
Meanwhile, it follows from () that
=
˜atan θ μ+ cot θ μ tan θ+ cot θ + ˜d μ– μ tan θ+ cot θ = tan θ+ cot θ
μ(˜a tan θ – ˜d) + μ(˜a cot θ + ˜d) . () Note that ¯x+ h= + ¯xTh+ h = + ¯xTh+ μ+¯xTh =+ ¯xTh + μ. ()
Putting () and ()-() together, the condition
tanθ ≥ ¯x + h can be rewritten equivalently as tanθ f(λ(x)) + cotθμ + f(λ(x)) + tanθμ +˜aμ ≥ – cot θ + cotθf λ (x) μ+ tan θ + tanθf λ (x) μ+ ˜dμ + (tan θ + cot θ )
μ(˜a tan θ – ˜d) + μ(˜a cot θ + ˜d)
i.e.,
tanθ– fλ(x)
μ+ (tan θ + cot θ )˜atanθ– ˜dμ + (tan θ + cot θ )˜a tanθ+ ˜dfλ(x)
– (˜a tan θ – ˜d)μμ + (tan θ + cot θ )(˜a tan θ – ˜d)fλ(x)
– (˜a cot θ + ˜d)μμ + tanθ+ fλ(x) fλ(x) μμ
– (˜a tan θ – ˜d)(˜a cot θ + ˜d)μμμ≥ . ()
To apply Lemma ., we need to compute each coefficient in (). By calculation, we have ˜a tan θ – ˜d =f (λ (x)) – f(λ(x)) λ(x) – λ(x) tan θ– x cotθ + cotθf λ (x) + tan θ + tanθf λ (x) + x f(λ(x)) – f (λ(x)) λ(x) – λ(x) =f (λ (x)) – f(λ(x)) λ(x) – λ(x) tan θ– x cot θ tan θ+ cot θf λ (x) + tan θ tan θ+ ctan θf λ (x) + x f(λ(x)) – f (λ(x)) λ(x) – λ(x) = –tan θ+ ctan θ λ(x) – λ(x) fλ(x) + tan θ+ ctan θ [λ(x) – λ(x)] fλ(x) – fλ(x) =(tan θ + cot θ )[f (λ(x)) – f (λ(x)) – f (λ (x))(λ(x) – λ(x))] [λ(x) – λ(x)] = tan θ+ cot θ [λ(x) – λ(x)] δλ(x), λ(x) ,
where the third equation follows from the fact λ(x) – λ(x) = (tan θ + ctan θ )x. Similarly, we have ˜a tan θ + ˜d =(tan θ + cot θ )[f (λ(x)) – f (λ(x)) + ( tan θ tan θ+cot θf(λ(x)) + cot θ–tan θ tan θ+cot θf(λ(x)))(λ(x) – λ(x))] [λ(x) – λ(x)] = tan θ+ cot θ [λ(x) – λ(x)] tanθ tan+δ λ(x), λ(x) +tan θ– tanθ+ δ λ(x), λ(x) , ˜a cot θ + ˜d =(tan θ + cot θ )[f (λ(x)) – f (λ(x)) – f(λ(x))(λ(x) – λ(x))]
[λ(x) – λ(x)] = tan θ+ cot θ [λ(x) – λ(x)] δλ(x), λ(x) , ˜a tanθ+ ˜d
=(tan θ + cot θ )[f (λ(x)) – f (λ(x)) – [tan
θf(λ (x)) + ( – tanθ)f(λ(x))](λ(x) – λ(x))] [λ(x) – λ(x)] = tan θ+ cot θ [λ(x) – λ(x)] tanθ δλ(x), λ(x) +tanθ– δλ(x), λ(x) .
Corresponding each coefficient in () to (), we know ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ α= (tanθ– )f(λ(x)), α= (tan θ +cot θ ) [λ(x)–λ(x)]δ(λ(x), λ(x))[ tanθ tanθ+δ(λ(x), λ(x)) +tan θ– tanθ+δ(λ(x), λ(x))], α= (tan θ +cot θ ) [λ(x)–λ(x)]{[tan θ δ(λ (x), λ(x)) + (tanθ– )δ(λ(x), λ(x))]f(λ(x)), – δ(λ(x),λ(x)) [λ(x)–λ(x)]}, α= (tan θ +cot θ ) [λ(x)–λ(x)][δ(λ(x), λ(x))f (λ (x)) – δ(λ(x),λ(x)) [λ(x)–λ(x)]], α= (tanθ+ )f(λ(x))f(λ(x)), α= –(tan θ +cot θ ) [λ(x)–λ(x)]δ(λ(x), λ(x))δ(λ(x), λ(x)).
In view of Lemma ., the condition α≥ means tan θ ≥ , α, α≥ is ensured by the convexity of f (see ()), α≥ corresponds to (), and α≥ –√ααcorresponds to (). In addition, condition () takes the special form () and (), respectively. Theorem . Suppose that f : J→ R is second-order continuously differentiable. Then f isLθ-convex of order on S if and only if tan θ≥ and f is convex on J.
Proof The necessity is clear from Theorem .. For sufficiency, note that in () μ= since¯x=± in this case. Hence, () takes the form of
tanθ– fλ(x) μ+ tanθ+ fλ(x) fλ(x) μμ≥ for all μand μ, which is equivalent to verifying
tan θ≥ and fλ(x)
fλ(x)
≥ .
This is ensued by the conditions that tan θ≥ and f is convex on J. Thus, the proof is
complete.
If, in particular, θ = ◦, then () and () reduce to [, () in Proposition .]; () re-duces to [, () in Proposition .]. In addition, due to (), () holds automatically in this case. The above results indicate that theLθ-convexity is dependent on the properties of f
and the angle θ together.
3 Inequalities associated with circular cone
In this section, we establish some inequalities associated with circular cone, which we be-lieve will be useful for further analyzing the properties of fLθand proving the convergence
of interior point methods for optimization problems involved in circular cones.
In [], the author establishes the following results in the framework of second-order cone. More specifically, for xL◦ and yL◦, then
(a) det(e + x)/≥ + det(x)/, (b) det(x + y)≥ det(x) + det(y),
(c) det(αx + ( – α)y)≥ αdet(x) + ( – α)det(y),∀α ∈ [, ], (d) det(e + x + y)≤ det(e + x) det(e + y),
(e) If xL◦yL◦, then det(x)≥ det(y), tr(x) ≥ tr(y), and λi(x)≥ λi(y)for i = , ,
(f ) tr(x + y) = tr(x) + tr(y) and det(γ x) = γdet(x)for all γ∈ R.
In the following, we show that, in the framework of circular cone, the above inequalities can be classified into three categories. The first class holds independent of the angle, e.g., (a); the second class holds dependent on the angle, e.g., (b)-(e); the third class fails no matter what value of the angle is chosen, e.g., (f ).
Theorem . Let x= (x, x)∈ R × Rn–possess spectral factorization associated with cir-cular cone given as in(). Then
(a) [det(e + x)]/≥ + det(x)/for all x∈L θ;
(b) If xLθ y, then λ(x)≥ λ(y).
Proof (a) Note that det(x)≥ and det(e + x) ≥ since x, x + e ∈Lθ. Therefore,
det(e + x)/≥ + det(x)/
⇐⇒ det(e + x) ≥ + det(x)/+ det(x) ⇐⇒ λ(e + x)λ(e + x)≥ + λ(x)λ(x) + λ(x)λ(x) ⇐⇒ x+ –x cot θ x+ +x tan θ ≥ + λ(x)λ(x) + λ(x)λ(x) ⇐⇒ λ(x) + λ(x) + ≥ + λ(x)λ(x) + λ(x)λ(x) ⇐⇒ λ(x)λ(x) + λ(x) + λ(x) + ≥ + λ(x)λ(x) + λ(x)λ(x) ⇐⇒ λ(x) + λ(x)≥ λ(x)λ(x) ⇐⇒ λ(x) + λ(x) ≥ λ(x)λ(x).
Hence, to prove the desired result, it suffices to show that λ(x) + λ(x)
≥
λ(x)λ(x),
which is clearly true by the arithmetic mean-geometric mean (AM-GM) inequality. (b) Since x – y∈Lθ, we know x– y≥ x– y cot θ ≥ x – y cot θ,
i.e., λ(x) = x–x cot θ ≥ y–y cot θ = λ(y). Theorem . Let x= (x, x)∈ R × Rn–possess spectral factorization associated with cir-cular cone given as in(). Then the following hold.
(a) For all x, y∈Lθ,
det(x + y)≥ det(x) + det(y) +x+y
cscθ–x+ ysecθ. In particular, when θ∈ (, ◦], we have
(b) For all x, y∈Lθand α∈ [, ], detαx+ ( – α)y ≥ αdet(x) + ( – α)det(y) +αx + ( – α)y cscθ –αx+ ( – α)ysecθ.
In particular, when θ∈ (, ◦], we have
detαx+ ( – α)y≥ αdet(x) + ( – α)det(y). (c) If x, y∈Lθand θ∈ [◦, ◦), then
det(e + x + y)≤ det(e + x) det(e + y). ()
(d) If xLθyLθ and θ∈ (, ◦], then
λ(x)≥ λ(y), det(x)≥ det(y), and tr(x) ≥ tr(y). () Proof (a) Notice that
det(x + y) = λ(x + y)· λ(x + y) =x+ y–x+ y cot θ x+ y+x+ y tan θ = (x+ y)+ (x+ y)x+ y tan θ – (x+ y)x+ y cot θ – x+ y and det(x) + det(y) = λ(x)λ(x) + λ(y)λ(y) =x–x cot θ x+x tan θ +y–y cot θ y+y tan θ
= x+ xx tan θ – xx cot θ – x+ y+ yy tan θ – yy cot θ – y = x+ y+ xx tan θ + yy tan θ – xx cot θ – yy cot θ – x–y. Then we have
det(x + y) – det(x) – det(y) = xy– xTy+ xx+ y + yx+ y – xx – yy tan θ –xx+ y + yx+ y – xx – yy cot θ. Using x, y∈Lθ(and hence x + y∈Lθ) gives
xtan θ≥ x, –xtan θ≤ –x, x≥ x cot θ, –x≤ –x cot θ, –(x+ y)≤ –x+ y cot θ.
Thus,
det(x + y) – det(x) – det(y)
≥ xy– xTy+xx+ y + yx+ y – xtanθ– ytanθ – x(x+ y) – y(x+ y) +xcotθ+ycotθ = xy– xTy+x+ y x + y –x+ ytanθ – (x+ y)+ x+y cotθ ≥ x+ y– x+ ytanθ– x– y– xTy+ x+y cotθ =x+y– x+ ytanθ– x– y+x+y cotθ =x+y + cotθ–x+ y + tanθ =x+y cscθ–x + y secθ,
which is the desired result.
When θ ∈ (, ◦], we know tan θ ≤ cot θ. Since x, y ∈Lθ, i.e., x ≥ x cot θ and y≥ y cot θ, there exist a, b ≥ such that x=x cot θ + a and y =y cot θ + b. Hence,
det(x + y) – det(x) – det(y) = xy– xTy+ xx+ y + yx+ y – xx – yy tan θ –xx+ y + yx+ y – xx – yy cot θ =x + y x + y – x+ y cotθ +x+ y x + y – x+ y + ab + a cot θy + x – x+ y + a tan θy cotθ+x+ y – x + b cot θy + x – x+ y + b tan θx cotθ+x+ y – y ≥ ,
where the last step is due to x + y ≥ x+ y, x cotθ +x + y – y ≥ x + x+ y – y ≥ , and y cotθ+x+ y – x ≥ y + x+ y – x ≥ since cot θ≥ , due to θ ∈ (, ◦].
(b) The result follows from the fact that det(γ x) = γdet(x) for all γ≥ .
(c) Since θ ∈ [◦, ◦), cot θ ≤ . For x, y ∈Lθ, there exist two nonnegative scalars a, b≥ such that x=x cot θ + a and y=y cot θ + b. This implies
det(e + x) =x+ –x cot θ x+ +x tan θ = (a + )(cot θ + tan θ )x + (a + ), det(e + y) =y+ –y cot θ y+ +y tan θ = (b + )(cot θ + tan θ )y + (b + ).
Thus, we obtain det(e + x) det(e + y)
= (a + )(b + )(cot θ + tan θ )xy + (a + )(b + )(cot θ + tan θ )x
+ (a + )(b + )(cot θ + tan θ )y + (a + )(b + ). () On the other hand,
det(e + x + y) =x+ y+ –x+ y cot θ x+ y+ +x+ y tan θ =x + y – x+ y cot θ+ (a + b + ) ×x + y cot θ+x+ y tan θ + (a + b + ) =x + y – x+ y x + y cotθ +x + y – x+ y x+ y + (a + b + ) x + y – x+ y cot θ + (a + b + )x + y cot θ+ (a + b + )x+ y tan θ + (a + b + ) = cotθxy + (a + b + ) cot θx + (a + b + ) cot θy + (a + b + )
+x+y cotθ+ – cotθx+ y x + y + (a + b + )(tan θ – cot θ )x+ y – x+ y ≤ cotθx
y + (a + b + ) cot θx + (a + b + ) cot θy + (a + b + ) +x+y cotθ+ – cotθx + y + (a + b + )(tan θ – cot θ )x + y –x+y + xy = cotθxy + (a + b + )(cot θ + tan θ)x + (a + b + )(cot θ + tan θ)y
+ (a + b + )+ – cotθx + y
–x+y
+ xy = xy + (a + b + )(cot θ + tan θ)x + (a + b + )(ctan θ + tan θ)y
+ (a + b + ). ()
Note that (a + )(b + )(cot θ + tan θ )≥ (cot θ + tan θ)≥ and (a + )(b + )≥ a + b + ,
(a + )(b + )≥ a + b + , (a + )(b + )≥ (a + b + ). Hence, comparing () and () yields
det(e + x + y)≤ det(e + x) det(e + y).
(d) For θ∈ (, ◦], since cot θ≥ tan θ and x – y ∈Lθ, we know x– y≥ x– y cot θ ≥ x– y tan θ ≥
y – x
which means
λ(x) = x+x tan θ ≥ y+y tan θ = λ(y).
This together with the fact λ(x)≥ λ(y) by Part (b) in Theorem . and λi(x), λi(y)≥ for i= , (due to x, y∈Lθ) further yields
det(x) = λ(x)λ(x)≥ λ(y)λ(y) = det(y). Meanwhile, we obtain
tr(x) = λ(x) + λ(x)≥ λ(y) + λ(y) = tr(y).
Here are some remarks for Theorem ..
(i) Inequality () fails when θ∈ (◦, ◦). For example, let x = (, , ), y = (, –, –), and cot θ = .. Then det(x) = det(y) = / and det(x + y) = , which says
det(x + y) = < = det(x) + det(y).
(ii) Inequality () fails when θ∈ (, ◦). For example, let x = (/, /), y= (/, –/), and cot θ = . Then
det(e + x + y) = (.)> (.)= det(e + x) det(e + y).
(iii) Inequality () fails when θ∈ (◦, ◦). For example, for x = (., ), y = (, ), and cot θ= .. Then xLθy, λ(x) = . < = λ(y),
det(x) = (. – .)(. + ) = . < . = ( – .)( + ) = det(y), and tr(x) = . < . = tr(y).
Next, let us move from inequalities to equalities. In particular, we focus on two identities in the framework of second-order cone as below
tr(x + y) = tr(x) + tr(y) and det(γ x) = γdet(x), ∀γ ∈ R. () But these two identities fail to hold in the circular cone setting no matter what value of the angle is chosen. In fact, in the second-order cone case,
tr(x) = x and det(x) = x–x.
Hence, () holds trivially. For the circular cone setting, we have tr(x) = x+x(tan θ – cot θ) and det(x) =
x–x cot θ x+x tan θ . Thus, tr(x) is not linear any more, i.e., tr(x + y)= tr(x)+tr(y); e.g., for x = (, ) and y = (, –), and cot θ = / (or cot θ = ). Then
tr(x + y) = = = tr(x) + tr(y) or tr(x + y) = = – = tr(x) + tr(y).
In addition, det(γ x) = γdet(x) holds as γ ≥ but not true as γ < ; e.g., for x = (, ), γ = –, and cot θ = / (or cot θ = ), then
The precise relationship between tr(x + y) and tr(x) + tr(y) is provided as below. Theorem . tr(x + y) ⎧ ⎨ ⎩ ≥ tr(x) + tr(y) as θ ∈ (, ◦], ≤ tr(x) + tr(y) as θ ∈ [◦, ◦). Proof The result follows from the fact that
tr(x + y) – tr(x) – tr(y) =x + y – x+ y
(cot θ – tan θ ).
Note that tr(x) is positively homogeneous, i.e., tr(γ x) = γ tr(x) for all γ ≥ . This together with Theorem . yields the following result.
Corollary . The trace tr(x) is concave as θ∈ (, ◦] and is convex as θ∈ [◦, ◦). These results further indicate that the angle plays an essential role for a circular cone. As in symmetric cone optimization, we believe that these inequalities about det(x) and tr(x) are key ingredients in penalty and barrier functions which can be adapted in design-ing barrier and penalty algorithms (includdesign-ing interior point algorithm) for circular cone optimization. This merits our further research.
Competing interests
The authors declare that they have no competing interests. Authors’ contributions
All authors read and approved the final manuscript. Author details
1Department of Mathematics, School of Science, Shandong University of Technology, Zibo, 255049, P.R. China. 2Department of Mathematics, National Taiwan Normal University, Taipei, 11677, Taiwan.
Authors’ information
The second author is a member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. Acknowledgements
We are gratefully indebted to anonymous referees for their valuable suggestions that helped us to essentially improve the original presentation of the paper. The first author’s work is supported by the National Natural Science Foundation of China (11101248, 11271233) and Shandong Province Natural Science Foundation (ZR2010AQ026, ZR2012AM016). The second author’s work is supported by the National Science Council of Taiwan.
Received: 17 July 2013 Accepted: 13 November 2013 Published:04 Dec 2013 References
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10.1186/1029-242X-2013-571
Cite this article as: Zhou et al.: Circular cone convexity and some inequalities associated with circular cones. Journal