# The Vector-Valued Functions Associated with Circular Cones

Authors: Jinchuan Zhou; Jein-Shan Chen
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

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, which includes second-order cone as a special case when the rotation angle is 45 degrees. LetLθ denote the circular cone in Rn. For a function f from R to R, one can define a corresponding vector-valued function fLθ on Rn by applying f to the spectral values of the spectral decomposition of x∈Rn with respect to Lθ. In this paper, we study properties that this vector-valued function inherits from f, including Hölder continuity, B-subdifferentiability, ρ-order semismoothness, and positive homogeneity. These results will play crucial role in designing solution methods for optimization problem involved in circular cone constraints.


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## Body

## 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, which includes second-order cone as a special case when the rotation angle is 45 degrees. LetLθ denote the circular cone in Rn. Then, the n-dimensional circular cone Lθ is expressed as
(1)Lθ∶={x=(x1,x2)T∈R×Rn-1∣cos⁡θ∥x∥≤x1}.
The application of Lθ lies in engineering field, for example, optimal grasping manipulation for multigingered robots; see [1].In our previous work [2], we have explored some important features about circular cone, such as characterizing its tangent cone, normal cone, and second-order regularity. In particular, the spectral decomposition associated with Lθ was discovered; that is, for any z=(z1,z2)∈R×Rn-1, one has
(2)z=λ1(z)uz1+λ2(z)uz2,
where
(3)λ1(z)=z1-∥z2∥ctanθ,λ2(z)=z1+∥z2∥tanθ,uz1=11+ctan2θ[100ctanθ·I][1-z¯2],uz2=11+tan2θ[100tanθ·I][1z¯2],
with z¯2∶=z2/∥z2∥ if z2≠0, and z¯2 being any vector w∈Rn-1 satisfying ∥w∥=1 if z2=0. With this spectral decomposition (2), analogous to the so-called SOC-function fsoc (see [3–5]) and SDP-function fmat (see [6, 7]), we define a vector-valued function associated with circular cone as below. More specifically, for f:R→R, we define fLθ:Rn→Rn as
(4)fLθ(z)=f(λ1(z))uz1+f(λ2(z))uz2.
It is not hard to see that fLθ is well-defined for all z. In particular, if z2=0, then
(5)fLθ(z)=[f(z1)0].Note that whenθ=45°, Lθ reduces to the second-order cone (SOC) and the vector-valued function fLθ defined as in (4) corresponds to the SOC-function fsoc given by
(6)fsoc(x)=f(λ1(x))ux(1)+f(λ2(x))ux(2),fsoc(x)=ff∀x=(x1,x2)∈R×Rn-1,
where λi(x)=x1+(-1)i∥x2∥ and ux(i)=(1/2)(1,(-1)ix¯2)T.It is well known that the vector-valued functionfsoc associated with second-order cone and matrix-valued function fmat associated with positive semidefinite cone play crucial role in the theory and numerical algorithm for second-order cone programming and semidefinite programming, respectively. In particular, many properties of fsoc and fmat are inherited from f, such as continuity, strictly continuity, directional differentiability, Fréchet differentiability, continuous differentiability, and semismoothness. It should be mentioned that, compared with second-order cone and positive semidefinite cone, Lθ is a nonsymmetric cone. Hence a natural question arises whether these properties are still true for fLθ. In [1], the authors answer the questions from the following aspects:(a)
fLθ is continuous at z∈Rn if and only if f is continuous at λi(z) for i=1,2;(b)
fLθ is directionally differentiable at z∈Rn if and only if f is directionally differentiable at λi(z) for i=1,2;(c)
fLθ is (Fréchet) differentiable at z∈Rn if and only if f is (Fréchet) differentiable at λi(z) for i=1,2;(d)
fLθ is continuously differentiable at z∈Rn if and only if f is continuously continuous at λi(z) for i=1,2;(e)
fLθ is strictly continuous at z∈Rn if and only if f is strictly continuous at λi(z) for i=1,2;(f)
fLθ is Lipschitz continuous with constant k>0 if and only if f is Lipschitz continuous with constant k>0;(g)
fLθ is semismooth at z if and only if f is semismooth at λi(z) for i=1,2.In this paper, we further study some other properties associated withfLθ, such as Hölder continuity, ρ-order semismoothness, directionally differentiability in the Hadamard sense, the characterization of B-subdifferential, positive homogeneity, and boundedness. Of course, one may wonder whether fsoc and fLθ always share the same properties. Indeed, they do not. There exists some property that holds for fsoc and f but fails for fLθ and f. A counterexample is presented in the final section.To end the third section, we briefly review our notations and some basic concepts which will be needed for subsequent analysis. First, we denote byRn the space of n-dimensional real column vectors and let e=(1,0,…,0)∈Rn. Given x,y∈Rn, the Euclidean inner product and norm are 〈x,y〉=xTy and ∥x∥=xTx. For a linear mapping H:Rn→Rm, its operator norm is ∥H∥∶=max⁡∥x∥=1∥Hx∥. For α∈R and s∈Rn, we write s=O(α) (resp., s=o(α)) to means ∥s∥/|α| is uniformly bounded (resp., tends to zero) as α→0. In addition, given a function F:Rn→Rm, we say the following:(a)
F is Hölder continuous with exponent α∈(0,1], if
(7)[F]α∶=sup⁡x≠y∥F(x)-F(y)∥∥x-y∥α<+∞;(b)
F is directionally differentiable at x∈Rn in the Hadamard sense, if the directional derivative F′(x;d) exists for all d∈Rn and
(8)F′(x;d)=lim⁡d′→dt↓0F(x+td′)-F(x)t;(c)
F is B-differentiable (Bouligand-differentiable) at x, if F is Lipschitz continuous near x and directionally differentiable at x;(d)
ifF is strictly continuous (locally Lipschitz continuous), the generalized Jacobian ∂F(x) is the convex hull of the ∂BF(x), where
(9)∂BF(x)∶={lim⁡z→x∇F(z)∣z∈DF},whereDF denotes the set of all differentiable points of F;(e)
F is semismooth at x, if F is strictly continuous near x, directionally differentiable at x, and for any V∈∂F(x+h),
(10)F(x+h)-F(x)-Vh=o(∥h∥);(f)
F is ρ-order semismooth at x (ρ>0) if F is semismooth at x and for any V∈∂F(x+h),
(11)F(x+h)-F(x)-Vh=O(∥h∥1+ρ);in particular, we sayF is strongly semismooth if it is 1-order semismooth;(g)
F is positively homogeneous with exponent α>0, if
(12)F(kx)=kαF(x),∀x∈Rn,k≥0;(h)
F is bounded if there exists a positive scalar M>0 such that
(13)∥F(x)∥≤M,∀x∈Rn.


## 2. Directional Differentiability, Strict Continuity, Hölder Continuity, andB-Differentiability

This section is devoted to study the properties of directional differentiability, strict continuity, and Hölder continuity. The relationship of directional differentiability betweenfLθ and f has been given in [1, Theorem 3.2] without giving the exact formula of directional differentiability. Nonetheless, such formulas can be easily obtained from its proof. Here we just list them as follows.Lemma 1.
Letf:R→R and fLθ be defined as in (4). Then, fLθ is directionally differentiable at z if and only if f is directionally differentiable at λi(z) for i=1,2. Moreover, for any h=(h1,h2)∈R×Rn-1, we have
(14)(fLθ)′(z;h)=[f′(z1;h1)0]=f′(z1;h1)e,
when z2=0 and h2=0. Consider
(15)(fLθ)′(z;h)=11+ctan2θf′(z1;h1-∥h2∥ctanθ)×[100ctanθ·I][1-h2∥h2∥]+11+tan2θf′(z1;h1+∥h2∥tanθ)×[100tanθ·I][1h2∥h2∥],
when z2=0 and h2≠0; otherwise,
(16)(fLθ)′(z;h)=11+ctan2θf′(λ1(z);h1-z2Th2∥z2∥ctanθ)×[100ctanθ·I][1-z2∥z2∥]-ctanθ1+ctan2θf(λ1(z))∥z2∥Mz2h+11+tan2θf′(λ2(z);h1+z2Th2∥z2∥tanθ)×[100tanθ·I][1z2∥z2∥]+tanθ1+tan2θf(λ2(z))∥z2∥Mz2h,
where
(17)Mz2∶=[000I-z2z2T∥z2∥2].Lemma 2.
Letf:R→R and fLθ be defined as in (4). Then, the following hold.(a)
fLθ is differentiable at z if and only if f is differentiable at λi(z) for i=1,2. Moreover, if z2=0, then
(18)∇fLθ(z)=f′(z1)I;otherwise,(19)∇fLθ(z)=[ξϱz2T∥z2∥ϱz2∥z2∥aI+(η-a)z2z2T∥z2∥2],where(20)a=tanθ1+tan2θf(λ2(z))∥z2∥-ctanθ1+ctan2θf(λ1(z))∥z2∥=f(λ2(z))-f(λ1(z))λ2(z)-λ1(z),ξ=f′(λ1(z))1+ctan2θ+f′(λ2(z))1+tan2θ,η=ξ-ϱ(ctanθ-tanθ),ϱ=-ctanθ1+ctan2θf′(λ1(z))+tanθ1+tan2θf′(λ2(z)).(b)
fLθ is continuously differentiable (smooth) at z if and only if f is continuously differentiable (smooth) at λi(z) for i=1,2.Note that the formula of gradient∇fLθ given in [1, Theorem 3.3] and Lemma 2 is the same by using the following facts:
(21)11+ctan2θ=sin2θ,11+tan2θ=cos⁡2θ,ctanθ1+ctan2θ=tanθ1+tan2θ=sinθcos⁡θ.The following result indicating thatλi is Lipschitz continuous on Rn for i=1,2 will be used in proving the Lipschitz continuity between fLθ and f.Lemma 3.
Letz,y∈Rn with spectral values λi(z), λi(y), respectively. Then, we have
(22)|λi(z)-λi(y)|≤2max⁡{tanθ,ctanθ}∥z-y∥,|λi(z)-λi(y)|≤2max⁡{tanθ,ctt}fori=1,2.Proof.
First, we observe that(23)|λ1(z)-λ1(y)|=|z1-∥z2∥ctanθ-y1+∥y2∥ctanθ|≤|z1-y2|+∥z2-y2∥ctanθ≤max⁡{1,ctanθ}(|z1-y1|+∥z2-y2∥)≤max⁡{1,ctanθ}2|z1-y1|2+∥z2-y2∥2=max⁡{1,ctanθ}2∥z-y∥.
Applying the similar argument to λ2 yields
(24)|λ2(z)-λ2(y)|≤max⁡{1,tanθ}2∥z-y∥.
Then, the desired result follows from the fact that max⁡{1,ctanθ,tanθ}max⁡{ctanθ,tanθ}.Theorem 4.
Letf:R→R and fLθ be defined as in (4). Then, fLθ is strictly continuous (local Lipschitz continuity) at z if and only if f is strictly continuous (local Lipschitz continuity) at λi(z) for i=1,2.Proof.
“⇐” Suppose that f is strictly continuous at λi(z), for i=1,2; that is, there exist ki>0 and δi>0, for i=1,2 such that
(25)∀τζ∈[λi(z)]|f(τ)-f(ζ)|≤ki|τ-ζ|,∀τ,ζ∈[λi(z)-δi,λi(z)+δi],i=1,2.
Let δ¯∶=min⁡{δ1,δ2} and C∶=[λ1(z)-δ¯,λ1(z)+δ¯]∪[λ2(z)-δ¯,λ2(z)+δ¯]. Define
(26)f~(τ)∶={f(τ)ifτ∈C,(1-t)f(λ1(z)+δ¯)+tf(λ2(z)-δ¯)ifλ1(z)+δ¯<λ2(z)-δ¯,τ=(1-t)(λ1(z)+δ¯)+t(λ2(z)-δ¯)witht∈(0,1)f(λ1(z)-δ¯)ifτ<λ1(z)-δ¯f(λ2(z)+δ¯)ifτ>λ2(z)+δ¯.
Clearly, f~ is Lipschitz continuous on R; that is, there exists k>0 such that lipf~(τ)≤k, for all τ∈R. Since C~∶=conv(C) is compact, according to [6, Lemma 4.5] or [5, Lemma 3], there exist continuously differentiable functions fv:R→R for v=1,2,… converging uniformly to f~ on C~ such that
(27)|(fv)′(τ)|≤k∀τ∈C~,∀v.
Now, let :=δ¯/(2max⁡{tanθ,ctanθ}). Then, from Lemma 3, we know that C~ contains all spectral values of y∈B(z,δ). Therefore, for any w∈B(z,δ), we have λi(w)∈C~ for i=1,2 and
(28)∥(fv)Lθ(w)-fLθ(w)∥2=∥[fv(λ1(w))-f(λ1(w))]uw1kkkkk+[fv(λ2(w))-f(λ2(w))]uw2∥2=[fv(λ1(w))-f(λ1(w))]2∥uw1∥2+[fv(λ2(w))-f(λ2(w))]2∥uw2∥2=11+ctan2θ|fv(λ1(w))-f(λ1(w))|2+11+tan2θ|fv(λ2(w))-f(λ2(w))|2,
where we have used the facts that ∥uw1∥=1/1+ctan2θ, ∥uw2∥=1/1+tan2θ, and 〈uw1,uw2〉=0. Since {fv}v=1∞ converges uniformly to f on C~, the above equations show that {(fv)Lθ}v=1∞ converges uniformly to fLθ on B(z,δ). If w2=0, then it follows from Lemma 2 that ∇(fv)Lθ(w)=(fv)′(w1)I. Hence it follows from (27) that
(29)∥∇(fv)Lθ(w)∥=|(fv)′(w1)|≤k,
since in this case λi(w)=w1∈C~. If w2≠0, then
(30)∇(fv)Lθ(w)=[ξϱw2T∥w2∥ϱw2∥w2∥aI+(ξ-ϱ(ctanθ-tanθ)-a)w2w2T∥w2∥2]=[ξϱw2T∥w2∥ϱw2∥w2∥aI+(ξ-a)w2w2T∥w2∥2]+[000[-ϱ(ctanθ-tanθ)]w2w2T∥w2∥2]=[ξϱw2T∥w2∥ϱw2∥w2∥ξI]+(a-ξ)[000I-w2w2T∥w2∥2]-ϱ(ctanθ-tanθ)[000w2w2T∥w2∥2],
where a, ξ, ϱ are given as in (20) with λi(z) replaced by λi(w) for i=1,2 and f replaced by fv. For simplicity of notations, let us denote
(31)A∶=[ξϱw2T∥w2∥ϱw2∥w2∥ξI]+(a-ξ)[000I-w2w2T∥w2∥2],B∶=-ϱ(ctanθ-tanθ)[000w2w2T∥w2∥2].
Note that
(32)|a|=|fv(λ2(w))-fv(λ1(w))λ2(w)-λ1(w)|≤k,
where the inequality comes from the fact that fv is continuously differentiable on C~ and (27). Besides, we also note that
(33)|ξ|=|(fv)′(λ1(w))1+ctan2θ+(fv)′(λ2(w))1+tan2θ|≤11+ctan2θ|(fv)′(λ1(w))|+11+tan2θ|(fv)′(λ2(w))|≤[11+ctan2θ+11+tan2θ]k=k,(34)|ϱ|=|-ctanθ1+ctan2θ(fv)′(λ1(w))+tanθ1+tan2θ(fv)′(λ2(w))|≤[|-ctanθ1+ctan2θ|+|tanθ1+tan2θ|]k=[ctanθ1+ctan2θ+tanθ1+tan2θ]k=2tanθ1+tan2θk≤k.(i)
Forϱ=0, then ∇(fv)Lθ(w) takes the form of ξI+(a-ξ)Mw2 whose eigenvalues are ξ and a by [5, Lemma 1]. In other words, in this case, we get from (32) and (33) that
(35)∥∇(fv)Lθ(w)∥=max⁡{|a|,|ξ|}≤k.(ii)
Forϱ≠0, since B=-ϱ(ctanθ-tanθ)(0,w2/∥w2∥)T(0,w2/∥w2∥), the eigenvalues of B are -ϱ(ctanθ-tanθ) and 0 with multiplicity n-1. Note that
(36)|ϱ(ctanθ-tanθ)|=|1-ctan2θ1+ctan2θ(fv)′(λ1(w))+1-tan2θ1+tan2θ(fv)′(λ2(w))|≤[|1-ctan2θ1+ctan2θ|+|1-tan2θ1+tan2θ|]k=|ctan2θ-11+ctan2θ+1-tan2θ1+tan2θ|k=2|1-tan2θ1+tan2θ|k≤2k.Note that
(37)A=ϱ∥w2∥Lw~+(a-ξ)Mw2=ϱ∥w2∥[Lw~+(a-ξ)∥w2∥ϱMw~2],
where w~=(ξ∥w2∥/ϱ,w2) and
(38)Lw~∶=[w~1w~2Tw~2w~1I].
In this case the matrix A has eigenvalues of ξ±ϱ and a with multiplicity n-2. Hence,
(39)∥∇(fv)Lθ(w)∥≤max⁡{|ξ+ϱ|,|ξ-ϱ|,|a|}+|ϱ(ctanθ-tanθ)|≤max⁡{|ξ|+|ϱ|,|a|}+|ϱ(ctanθ-tanθ)|≤4k,
where the last step is due to (32), (33), (34), and (36).
Putting (29), (35), and (39) together, we know that
(40)∥∇(fv)Lθ(w)∥≤4k∀w∈B(z,δ),∀v.
Fix any x,y∈B(z,δ) with x≠y. Since {(fv)Lθ}v=1∞ converges uniformly to fLθ on B(z,δ), then for any ϵ>0 there exists v0 such that
(41)∥(fv)Lθ(w)-fLθ(w)∥≤ϵ,∀w∈B(z,δ),∀v≥v0.
Since fv is continuously differentiable, (fv)Lθ is continuously differentiable by Lemma 2. Thus,
(42)∥fLθ(x)-fLθ(y)∥=∥fLθ(x)-(fv)Lθ(x)+(fv)Lθ(x)-(fv)Lθ(y)kkkkk+(fv)Lθ(y)-fLθ(y)∥≤∥fLθ(x)-(fv)Lθ(x)∥+∥(fv)Lθ(x)-(fv)Lθ(y)∥+∥(fv)Lθ(y)-fLθ(y)∥≤2ϵ+∥∫01∇(fv)Lθ(y+t(x-y))(x-y)dt∥≤2ϵ+4k∥x-y∥.
Because ϵ>0 is arbitrary, this ensures that
(43)∥fLθ(x)-fLθ(y)∥≤4k∥x-y∥∀x,y∈B(z,δ),
which says fLθ is strictly continuous at z.
“⇒” Suppose that fLθ is strictly continuous at z, then there exist k>0 and δ>0 such that
(44)∥fLθ(x)-fLθ(y)∥≤k∥x-y∥∀x,y∈B(z,δ).Case 1. z2≠0. Take θ,μ∈[λ1(z)-δ1,λ1(z)+δ1] with δ1:=min⁡{δ,λ2(z)-λ1(z)}. Let
(45)x∶=θuz1+λ2(z)uz2,y∶=μuz1+λ2(z)uz2.
Then, ∥x-z∥≤δ and ∥y-z∥≤δ and it follows from (44) that
(46)|f(θ)-f(μ)|=1∥uz1∥∥fLθ(x)-fLθ(y)∥≤k∥uz1∥∥x-y∥=k∥uz1∥|θ-μ|∥uz1∥=k|θ-μ|,
which says f is strictly continuous at λ1(z). The similar argument shows the strict continuity of f at λ2(z).
Case 2. z2=0. For any θ,μ∈[z1-δ,z1+δ], we have ∥θe-z∥=|θ-z1|≤δ and ∥μe-z∥≤δ as well; that is, θe,μe∈B(z,δ). It then follows from (44) that
(47)|f(θ)-f(μ)|=∥(f(θ)-f(μ)0)∥=∥fLθ(θe)-fLθ(μe)∥≤k∥θe-μe∥=k|θ-μ|.
This means f is strictly continuous at λi(z)=z1 for i=1,2.Remark 5.
As mentioned in Section1, the strict continuity between fLθ and f has been given in [1, Theorem 3.5]. Here we provide an alternative proof, since our analysis technique is different from that in [1, Theorem 3.5]. In particular, we achieve an estimate regarding ∥∇(fv)Lθ∥ via its eigenvalues, which may have other applications.According to Lemma1 and Theorem 4, we obtain the following result immediately.Theorem 6.
Letf:R→R and fLθ be defined as in (4). Then, fLθ is B-differentiable at z if and only if f is B-differentiable at λi(z), for i=1,2.Next, inspired by [8, 9], we further study the Hölder continuity relation between f and fLθ.Theorem 7.
Letf:R→R and fLθ be defined as in (4). Then, fLθ is Hölder continuous with exponent α∈(0,1] if and only if f is Hölder continuous with exponent α∈(0,1].Proof.
“⇐” Suppose that f is Hölder continuous with exponent α∈(0,1]. To proceed the proof, we consider the following two cases.
Case 1. z2≠0 and y2≠0. We assume without loss of generality that ∥z2∥≥∥y2∥. Thus,
(48)∥fLθ(z)-fLθ(y)∥=∥f(λ1(z))uz1+f(λ2(z))uz2-f(λ1(y))uy1-f(λ2(y))uy2∥=∥f(λ1(z))[uz1-uy1]+f(λ2(z))[uz2-uy2]+[f(λ1(z))-f(λ1(y))]uy1+[f(λ2(z))-f(λ2(y))]uy2∥≤∥f(λ1(z))[uz1-uy1]+f(λ2(z))[uz2-uy2]∥+|f(λ1(z))-f(λ1(y))|·∥uy1∥+|f(λ2(z))-f(λ2(y))|·∥uy2∥.
Let us analyze each term in the above inequality. First, we look into the first term:
(49)∥f(λ1(z))[uz1-uy1]+f(λ2(z))[uz2-uy2]∥=tanθ1+tan2θ|f(λ1(z))-f(λ2(z))|·∥z2∥z2∥-y2∥y2∥∥≤tanθ1+tan2θ[f]α|λ1(z)-λ2(z)|α∥z2∥z2∥-y2∥y2∥∥=tanθ1+tan2θ[f]α(tanθ+ctanθ)α∥z2∥α·∥z2∥z2∥-y2∥y2∥∥≤tanθ1+tan2θ[f]α(tanθ+ctanθ)α∥z2∥α2∥z2∥∥z2-y2∥=2tanθ1+tan2θ[f]α(tanθ+ctanθ)α∥z2-y2∥z2∥∥1-α∥z2-y2∥α≤tanθ1+tan2θ[f]α(tanθ+ctanθ)α22-α∥z2-y2∥α≤tanθ1+tan2θ[f]α(tanθ+ctanθ)α22-α∥z-y∥α,
where the first inequality is due to the Hölder continuity of f, the second inequality comes from the fact that ∥(z2/∥z2∥)-(y2/∥y2∥)∥≤(2/∥z2∥)∥z2-y2∥ (cf. [8, Lemma 2.2]), and the third inequality follows from the fact that ∥z2-y2∥≤∥z2∥+∥y2∥≤2∥z2∥ (since ∥y2∥≤∥z2∥). Next, we look into the second term:
(50)|f(λ1(z))-f(λ1(y))|∥uy1∥≤[f]α|λ1(z)-λ1(y)|α11+ctan2θ≤[f]α(2max⁡{tanθ,ctanθ})α∥z-y∥α.
Similarly, the third term also satisfies
(51)|f(λ2(z))-f(λ2(y))|∥uy2∥≤[f]α|λ2(z)-λ2(y)|α11+tan2θ≤[f]α(2max⁡{tanθ,ctanθ})α∥z-y∥α.
Combining (49)–(51) proves that fLθ is Hölder continuous with exponent α∈(0,1].
Case 2. Either z2=0 or y2=0. In this case, we take uzi=uyi, for i=1,2 according to the spectral decomposition. Therefore, we obtain
(52)∥fLθ(z)-fLθ(y)∥=∥f(λ1(z))uz1+f(λ2(z))uz2-f(λ1(y))uy1-f(λ2(y))uy2∥=∥[f(λ1(z))-f(λ1(y))]uz1+[f(λ2(z))-f(λ2(y))]uz2∥≤|f(λ1(z))-f(λ1(y))|·∥uz1∥+|f(λ2(z))-f(λ2(y))|·∥uz2∥≤[f]α|λ1(z)-λ1(y)|α11+ctan2θ+[f]α|λ2(y)-λ2(z)|α11+tan2θ≤2[f]α(2max⁡{tanθ,ctanθ})α∥z-y∥α,
which says fLθ is Hölder continuous.
“⇒” Recall that fLθ(τe)=(f(τ),0)T. Hence, for any τ,ζ∈R,
(53)|f(τ)-f(ζ)|=∥fLθ(τe)-fLθ(ζe)∥≤[fLθ]α·∥τe-ζe∥α=[fLθ]α·|τ-ζ|α,
which says f is Hölder continuous.


## 3.ρ-Order Semismoothness and B-Subdifferential Formula

The property of semismoothness plays an important role in nonsmooth Newton methods [10, 11]. For more information on semismooth functions, see [12–15]. The relationship of semismooth between fLθ and f has been given in [1, Theorem 4.1]. But the exact formula of the B-subdifferential ∂B(fLθ) is not presented. Hence the main aim of this section is twofold: one is establishing the exact formula of B-subdifferential; another is studing the ρ-order semismoothness for ρ>0.Lemma 8.
Defineψ(z)=∥z∥ and Φ(z)=z/∥z∥ for z≠0. Then, ψ and Φ are strongly semismooth at z≠0.Proof.
Sincez≠0, it is clear that ψ and Φ are twice continuously differentiable and hence the gradient is Lipschitz continuous near z. Therefore, ψ and Φ are strongly semismooth at z, see [16, Proposition 7.4.5].The relationship ofρ-order semismoothness between fLθ and f is given below. Recall from [7] that in the definition of ρ-order semismooth, we can restrict x+h in (11) belonging to differentiable points.Theorem 9.
Letf:R→R and fLθ be defined as in (4). Given ρ>0, then the following statements hold. (a)
Iff is ρ-order semismooth at λi(z) for i=1,2, then fLθ is min⁡{1,ρ}-order semismooth at z.(b)
IffLθ is ρ-order semismooth at z, then f is ρ-semismooth at λi(z) for i=1,2.(c)
Forz2=0, fLθ is ρ-semismooth at z if and only if f is ρ-order semismooth at λi(z)=z1 for i=1,2.Proof.
(a) Takeh∈Rn satisfying z+h∈DfLθ. We consider the following two cases to complete the proof.
Case 1. For z2≠0, z2+h2≠0 as h is sufficiently close to 0. Since z+h∈DfLθ, we know that λi(z+h)∈Df for i=1,2 by Lemma 2. Then, according to Lemma 1, the first component of
(54)fLθ(z+h)-fLθ(z)-(fLθ)′(z+h;h)
is expressed as
(55)f(λ1(z+h))1+ctan2θ-f(λ1(z))1+ctan2θ-11+ctan2θ×f′(λ1(z+h);h1-(z2+h2)Th2∥z2+h2∥ctanθ)+f(λ2(z+h))1+tan2θ-f(λ2(z))1+tan2θ-11+tan2θ×f′(λ2(z+h);h1+(z2+h2)Th2∥z2+h2∥tanθ).
Because ∥·∥ is continuously differentiable over z2≠0, it is strongly semismooth at z2 by Lemma 8. Therefore,
(56)∥z2+h2∥=∥z2∥+(z2+h2)Th2∥z2+h2∥+O(∥h2∥2)=∥z2∥+(z2+h2)Th2∥z2+h2∥+O(∥h∥2).
Combining this and the ρ-semismoothness of f at λ1(z), we have
(57)f(λ1(z+h))=f(λ1(z))+f′(λ1(z+h))(λ1(z+h)-λ1(z))+O(|λ1(z+h)-λ1(z)|1+ρ)=f(λ1(z))+f′(λ1(z+h))(λ1(z+h)-λ1(z))+O(∥h∥1+ρ)=f(λ1(z))+f′(λ1(z+h))×(h1-(∥z2+h2∥-∥z2∥)ctanθ)+O(∥h∥1+ρ)=f(λ1(z))+f′(λ1(z+h))(h1-(z2+h2)Th2∥z2+h2∥ctanθ)+O(∥h∥2)+O(∥h∥1+ρ)=f(λ1(z))+f′(λ1(z+h))(h1-(z2+h2)Th2∥z2+h2∥ctanθ)+O(∥h∥1+min⁡{1,ρ}),
where the second equation is due to Lemma 3 and the last equality comes from the boundedness of f′, since f is strictly continuous at λ1(z). Similar argument holds for f(λ2(z+h)). Hence the first component of (54) is O(∥h∥1+min⁡{1,ρ}).
Next, let us look into the second component of (54), which involved λ1(z). By Lemma 1 again, it can be expressed as
(58)-ctanθ1+ctan2θf(λ1(z+h))z2+h2∥z2+h2∥+ctanθ1+ctan2θf′(λ1(z+h);h1-(z2+h2)Th2∥z2+h2∥ctanθ)×z2+h2∥z2+h2∥+ctanθ1+ctan2θf(λ1(z))z2∥z2∥+ctanθ1+ctan2θf(λ1(z+h))∥z2+h2∥M(z2+h2)h.
Note that Φ is continuous differentiable (and hence is semismooth) with ∇Φ(z2)=(1/∥z2∥)(I-(z2z2T/∥z2∥2)) and M(z2+h2)h=∥z2+h2∥∇Φ(z2+h2)h2. Thus, expression (58) can be rewritten as
(59)-ctanθ1+ctan2θf(λ1(z+h))Φ(z2+h2)+ctanθ1+ctan2θf′(λ1(z+h);h1-(z2+h2)Th2∥z2+h2∥ctanθ)×Φ(z2+h2)+ctanθ1+ctan2θf(λ1(z))Φ(z2)+ctanθ1+ctan2θf(λ1(z+h))∇Φ(z2+h2)h2=ctanθ1+ctan2θ×[f′(λ1(z+h);h1-(z2+h2)Th2∥z2+h2∥ctanθ)-f(λ1(z+h))+f(λ1(z))+f′(λ1(z+h);h1-(z2+h2)Th2∥z2+h2∥ctanθ)]×Φ(z2+h2)+f(λ1(z))ctanθ1+ctan2θ×[-Φ(z2+h2)+Φ(z2)+∇Φ(z2+h2)h2]+ctanθ1+ctan2θ∇Φ(z2+h2)h2×[f(λ1(z+h))-f(λ1(z))]=O(∥h∥1+min⁡{1,ρ})+O(∥h∥2)+O(∥h∥2)=O(∥h∥1+min⁡{1,ρ}).
The second equation comes from (57), strongly semismoothness of Φ at z2, and
(60)∥∇Φ(z2+h2)h2[f(λ1(z+h))-f(λ1(z))]∥=O(∥h∥2),
since f is Lipschitz at λ1(z) (which is ensured by the ρ-order semismoothness of f). Analogous arguments apply for the second component of (54) involving λ2(z). From all the above, we may conclude that
(61)fLθ(z+h)-fLθ(z)-(fLθ)′(z+h;h)=O(∥h∥1+min⁡{1,ρ}),
which says fLθ is min⁡{1,ρ}-order semismooth at z under this case.
Case 2. For z2=0, if h2=0, then the proof is trivial. If h2≠0, then the first component of (54) satisfies
(62)11+ctan2θ×[f(λ1(z+h))-f(z1)f′(λ1(z+h);h1-∥h2∥ctanθ),k-f(λ1(z+h))-f(z1)f′(λ1(z+h);h1-∥h2∥ctanθ)]+11+tan2θ×[-f′(λ2(z+h);h1+∥h2∥tanθ)f(λ2(z+h))-f(z1)k-f′(λ2(z+h);h1+∥h2∥tanθ)]=O(|h1-∥h2∥ctanθ|1+ρ)+O(|h1+∥h2∥tanθ|1+ρ)=O(∥h∥1+ρ),
because f is ρ-order semismooth at z1. The second component of (54), by letting z¯2=h¯2 and Mh2h=0, takes the form
(63)-ctanθ1+ctan2θf(λ1(z+h))h¯2+ctanθ1+ctan2θf′(λ1(z+h);h1-∥h2∥ctanθ)h¯2+tanθ1+tan2θf(λ2(z+h))h¯2-tanθ1+tan2θf′(λ2(z+h);h1+∥h2∥tanθ)h¯2=-1tanθ+ctanθ[-f′(λ1(z+h);h1-∥h2∥ctanθ)f(λ1(z+h))-f(z1)-f′(λ1(z+h);h1-∥h2∥ctanθ)]h¯2+1tanθ+ctanθ[-f′(λ1(z+h);h1-∥h2∥ctanθ)f(λ2(z+h))-f(z1)-f′(λ2(z+h);h1+∥h2∥tanθ)]h¯2=O(∥h∥1+ρ),
where the last step is due to the ρ-order semismoothness of f.
(b) Suppose thatfLθ is ρ-order semismooth at z. Let t∈R such that f is differentiable at λ1(z)+t. We discuss the following two cases.
Case 1. For z2≠0, from f being Lipschitz at λ2(z) (and hence the differentiable points are dense near λ2(z)), there exists β(t)∈R such that β(t)=O(|t|1+ρ) and f is differentiable at λ2(z)+β(t) and λ2(z)+β(t)>λ1(z)+t as t is sufficiently small (since λ2(z)>λ1(z) by z2≠0). Denote h∶=tuz1+β(t)uz2. Then, z+h=[λ1(z)+t]uz1+[λ2(z)+β(t)]uz2 which implies λ1(z+h)=λ1(z)+t and λ2(z+h)=λ2(z)+β(t). Since f is differentiable at λ1(z)+t and λ2(z)+β(t), fLθ is also differentiable at z+h by Lemma 2. Notice that
(64)h=[h1h2]=[11+ctan2θt+11+tan2θβ(t)(-ctanθ1+ctan2θt+tanθ1+tan2θβ(t))z2∥z2∥],(65)z2+h2=(∥z2∥-ctanθ1+ctan2θt+tanθ1+tan2θβ(t))z2∥z2∥.
Hence,
(66)(z2+h2)Th2∥z2+h2∥=-ctanθ1+ctan2θt+tanθ1+tan2θβ(t),
which follows from the fact that ∥z2∥≠0 and t can be arbitrarily small (hence ∥z2∥-(ctanθ/(1+ctan2θ))t+(tanθ/(1+tan2θ))β(t)>0). Thus, it is clear that
(67)h1-(z2+h2)Th2∥z2+h2∥ctanθ=(11+ctan2θt+11+tan2θβ(t))-(-ctanθ1+ctan2θt+tanθ1+tan2θβ(t))ctanθ=t,h1+(z2+h2)Th2∥z2+h2∥tanθ=(11+ctan2θt+11+tan2θβ(t))+(-ctanθ1+ctan2θt+tanθ1+tan2θβ(t))tanθ=β(t).
In addition, it can be verified that
(68)|h1|=|11+ctan2θt+11+tan2θβ(t)|≤11+ctan2θ|t|+11+tan2θ|t|=|t|,
since β(t)=O(|t|1+ρ)≤|t| as t is sufficiently small. Similarly,
(69)∥h2∥=∥[-ctanθ1+ctan2θt+tanθ1+tan2θβ(t)]z2∥z2∥∥=|-ctanθ1+ctan2θt+tanθ1+tan2θβ(t)|≤ctanθ1+ctan2θ|t|+tanθ1+tan2θ|t|≤|t|.
Therefore, we obtain ∥h∥=O(t) which further implies O(∥h∥1+ρ)=O(|t|1+ρ). Then, by the hypothesis fLθ being ρ-order semismooth at z, that is,
(70)fLθ(z+h)-fLθ(z)-(fLθ)′(z+h;h)=O(∥h∥1+ρ),
we have
(71)〈fLθ(z+h)-fLθ(z)-(fLθ)′(z+h;h),e〉=O(∥h∥1+ρ)=O(|t|1+ρ).
In fact, the left-hand side of (71) takes the form of
(72)f(λ1(z+h))1+ctan2θ+f(λ2(z+h))1+tan2θ-f(λ1(z))1+ctan2θ-f(λ2(z))1+tan2θ-f′(λ1(z+h))1+ctan2θ(h1-(z2+h2)Th2∥z2+h2∥ctanθ)-f′(λ2(z+h))1+tan2θ(h1+(z2+h2)Th2∥z2+h2∥tanθ)=f(λ1(z)+t)1+ctan2θ+f(λ2(z)+β(t))1+tan2θ-f(λ1(z))1+ctan2θ-f(λ2(z))1+tan2θ-f′(λ1(z)+t)1+ctan2θt-f′(λ2(z)+β(t))1+tan2θβ(t)=11+ctan2θ×[f(λ1(z)+t)-f(λ1(z))-f′(λ1(z)+t)t]+11+tan2θ[f′(λ2(z)+β(t))β(t)f(λ2(z)+β(t))-f(λ2(z))-f′(λ2(z)+β(t))β(t)]=11+ctan2θ×[f(λ1(z)+t)-f(λ1(z))-f′(λ1(z)+t)t]+O(|t|1+ρ),
where the last step is due to the fact that f′ is bounded and
(73)f(λ2(z)+β(t))-f(λ2(z))=O(|t|1+ρ),
since f is Lipschitz at λ2(z). Hence (71) means
(74)f(λ1(z)+t)-f(λ1(z))-f′(λ1(z)+t)t=O(|t|1+ρ),
which says f is ρ-order semismooth at λ1(z). Applying similar arguments show that f is ρ-order semismooth at λ2(z).
Case 2. For z2=0, letting h=te. Since f is differentiable at λ1(z)+t=z1+t and λi(z+h)=z1+t, for i=1,2, fLθ is differentiable at z+h by Lemma 2; that is, z+h∈DfLθ. Because fLθ is ρ-order semismooth at z, we have
(75)fLθ(z+h)-fLθ(z)-(fLθ)′(z+h;h)=O(∥h∥1+ρ),
which, together with the fact ∥h∥=|t|, is equivalent to
(76)f(z1+t)-f(z1)-f′(z1+t)t=O(|t|1+ρ).
This clearly proves that f is ρ-order semismooth at λi(z)=z1, for i=1,2.
(c) The necessity comes from part (b), and the sufficiency follows from (62) and (63).Corollary 10.
Letf:R→R and fLθ be defined as in (4). For ρ∈(0,1], fLθ is ρ-order semismooth at z if and only if f is ρ-order semismooth at λi(z), for i=1,2.Remark 11.
In the framework of second-order cone and positive semidefinite cone, the corresponding result to part (a) has been established; see [5, Proposition 7] and [6, Proposition 4.10]. In [17], the author study the ρ-order semismoothness of the spectral operator for ρ∈(0,1]. Here we further show that if fLθ is ρ-semismooth, then f is ρ-semismooth for all ρ>0. In addition, if z2=0, then the ρ-order semismoothness of fLθ and f coincide with each other for all ρ>0.Inspired by [5, Lemma 4], we also obtain the following result.Theorem 12.
Letf:R→R be strictly continuous. Then, for any z∈Rn, the B-differential ∂B(fLθ)(z) is well defined and nonempty. Moreover,(i)
ifz2≠0, then
(77)∂B(fLθ)(z)={[ξϱz2T∥z2∥ϱz2∥z2∥aI+(η-a)z2z2T∥z2∥2]∣a=f(λ2(z))-f(λ1(z))λ2(z)-λ1(z),ξ-ϱctanθ∈∂Bf(λ1(z)),ξ+ϱtanθ∈∂Bf(λ2(z)),η=ξ-ϱ(ctanθ-tanθ)[ξϱz2T∥z2∥ϱz2∥z2∥aI+(η-a)z2z2T∥z2∥2]};(ii)
ifz2=0, then
(78)∂B(fLθ)(z)⊂{[ξϱwTϱwaI+(η-a)wwT]∣a∈∂f(λ1(z)),∥w∥=1,ξ-ϱctanθ∈∂Bf(λ1(z)),ξ+ϱtanθ∈∂Bf(λ1(z)),η=ξ-ϱ(ctanθ-tanθ)[ξϱwTϱwaI+(η-a)wwT]}.Proof.
Denote byΩ the set in the right side of (77).
Case 1. z2≠0. For any sequence {zk}k=1∞→z with zk∈D(fLθ). Then, we compute
(79)∇fLθ(zk)={[ξ(zk)ϱ(zk)(z2k∥z2k∥)Tϱ(zk)z2k∥z2k∥a(zk)I+(η(zk)-a(zk))z2k(zk)2T∥z2k∥2]∣η(zk)=ξ(zk)-ϱ(zk)(ctanθ-tanθ)[ξ(zk)ϱ(zk)(z2k∥z2k∥)Tϱ(zk)z2k∥z2k∥a(zk)I+(η(zk)-a(zk))z2k(zk)2T∥z2k∥2]}.
Since f is strictly continuous, we know that ξ(zk) and ϱ(zk) are bounded and hence have cluster points. We assume, without loss of generality, that ξ(zk)→ξ~ and ϱ(zk)→ϱ~. Note that f is differentiable at λi(zk) for i=1,2 by Lemma 2. Besides, from
(80)ξ(zk)-ϱ(zk)ctanθ=f′(λ1(zk)),ξ(zk)+ϱ(zk)tanθ=f′(λ2(zk)),
and the fact that any cluster point of {f′(λi(zk))}k=1∞ is in ∂Bf(λi(z)) by definition, we have
(81)ξ~-ϱ~ctanθ∈∂Bf(λ1(z)),ξ~+ϱ~tanθ∈∂Bf(λ2(z)).
This means that any cluster points of {∇(fLθ)(zk)}k=1∞ are element of Ω; that is, ∂B(fLθ)(z)⊂Ω.
Conversely, for anyξ and ϱ satisfying ξ-ϱctanθ∈∂Bf(λ1(z)) and ξ+ϱtanθ∈∂Bf(λ2(z)), there exist {λ1k}k=1∞→λ1(z) and {λ2k}k=1∞→λ2(z) with f being differentiable at λ1k and λ2k and {f′(λ1k)}k=1∞→ξ-ϱctanθ and {f′(λ2k)}k=1∞→ξ+ϱtanθ. Since λ2(z)=z1+∥z2∥tanθ>z1>z1-∥z2∥ctanθ=λ1(z), it implies that λ2k>λ1k is k large enough. Now, let
(82)zk∶=[z1kz2k]=[tan2θ1+tan2θλ1k+ctan2θ1+ctan2θλ2k(tanθ1+tan2θλ2k-ctanθ1+ctan2θλ1k)z2∥z2∥].
For points zk, it is easy to see that λi(zk)=λik for all k by using the following facts:
(83)tanθ1+tan2θλ2k=1ctanθ+tanθλ2k>1ctanθ+tanθλ1k=ctanθ1+ctan2θλ1k.
Hence, zk→z and fLθ are differentiable at zk by Lemma 2 (since f is differentiable at λi(zk)=λik for i=1,2). Then, we compute
(84)∇fLθ(zk)={[ξ(zk)ϱ(zk)(z2∥z2∥)Tϱ(zk)z2∥z2∥a(zk)I+(η(zk)-a(zk))z2z2T∥z2∥2]∣η(zk)=ξ(zk)-ϱ(zk)(ctanθ-tanθ)[ξ(zk)ϱ(zk)(z2∥z2∥)Tϱ(zk)z2∥z2∥a(zk)I+(η(zk)-a(zk))z2z2T∥z2∥2]|},
where
(85)a(zk)=f(λ2(zk))-f(λ1(zk))λ2(zk)-λ1(zk)⟶f(λ2(z))-f(λ1(z))λ2(z)-λ1(z)=a,ξ(zk)=f′(λ1k)1+ctan2θ+f′(λ2k)1+tan2θ⟶ξ-ϱctanθ1+ctan2θ+ξ+ϱtanθ1+tan2θ=ξ,ϱ(zk)=-ctanθ1+ctan2θf′(λ1k)+tanθ1+tan2θf′(λ2k)⟶-ctanθ1+ctan2θ(ξ-ϱctanθ)+tanθ1+tan2θ(ξ+ϱtanθ)=ϱ.
Since the limit of {∇fLθ(zk)}k=1∞ is an element of ∂BfLθ(z), we obtain Ω⊂∂BfLθ(z).
Case 2. z2=0. Consider any sequence {zk}k=1∞={(z1k,z2k)}k=1∞→z with fLθ being differentiable at zk for all k. By passing to a subsequence, we can assume that either z2k=0 for all k or z2k≠0 for all k. If z2k=0, then by Lemma 2 we know that f is differentiable at λi(zk)=z1k for i=1,2 and ∇fLθ(zk)=f′(z1k)I. Hence, the cluster point of {∇fLθ(zk)}k=1∞ is an element of (78) with ϱ=0 and ξ=a∈∂Bf(z1)⊂∂f(z1). If z2k≠0, by passing to a subsequence we can assume without loss of generality that {z2k/∥z2k∥}→w for some w with ∥w∥=1. Note that
(86)∇fLθ(zk)={[ξ(zk)ϱ(zk)(z2k∥z2k∥)Tϱ(zk)z2k∥z2k∥a(zk)I+(η(zk)-a(zk))z2z2T∥z2∥2]∣η(zk)=ξ(zk)-(zk)(ctanθ-tanθ)[ξ(zk)ϱ(zk)(z2k∥z2k∥)Tϱ(zk)z2k∥z2k∥a(zk)I+(η(zk)-a(zk))z2z2T∥z2∥2]}.
Moreover, from ξ(zk) and ϱ(zk) being bounded (due to the strictly continuous of f), we can assume that ξ(zk)→ξ~ and ϱ(zk)→ϱ~. Using (80) and any cluster point of {f′(λi(zk))}k=1∞ in ∂Bf(λi(z)), we have
(87)ξ~-ϱ~ctanθ∈∂Bf(λ1(z)),ξ~+ϱ~tanθ∈∂Bf(λ2(z)).
In addition,
(88)a(zk)=f(λ2(zk))-f(λ1(zk))λ2(zk)-λ1(zk)∈∂f(λ^k),
where λ^k∈[λ1(zk),λ2(zk)] and hence converges to λ1(z), since λi(zk)→λ1(z) for i=1,2, due to z2=0. Using the outer semicontinuity of ∂f we get that the cluster point of {a(zk)} belongs to ∂f(λ1(z)). Hence any cluster of {∇fLθ(zk)}k=1∞ belongs to an element of the set of the right side in (78).We point out one thing for Theorem12(ii). In the set of the right side in (78), a∈∂f(λ1(z)) cannot be replaced by a∈∂Bf(λ1(z)) because ∂Bf(λ1(z)) is usually a smaller subset of ∂f(λ1(z)). For example, letting f(t)=|t| and α=2, β=-1, we have
(89)f(α)-f(β)α-β=13∉∂Bf(t)∀t∈R,
while
(90)f(α)-f(β)α-β=13∈∂f(0)=[-1,1],
which is the main reason causing what we just pointed out.At present, we roughly describe∂B(fLθ) for z2=0. In other words, how to get the exact formula on ∂B(fLθ). Toward this end, we need to introduce the following definition. Given z∈Rn, define
(91)Γ(z)∶=lim⁡sup⁡tk→λ1(z),tk′→λ2(z)tk,tk′∈Df,tk<tk′(f′(tk),f′(tk′),f(tk′)-f(tk)tk′-tk),
where “limsup” is the outer limits in the sense of set-valued mapping; see [18, 19] for more details.First, forz2≠0, according to Lemma 2 let us write the gradient of ∇fLθ as
(92)∇fLθ(z)=1tanθ+ctanθ×[tanθ-(z2∥z2∥)-z2∥z2∥ctanθz2(z2)T∥z2∥2]f′(λ1(z))+1tanθ+ctanθ[ctanθ(z2∥z2∥)Tz2∥z2∥tanθz2(z2)T∥z2∥2]f′(λ2(z))+f(λ2(z))-f(λ1(z))λ2(z)-λ1(z)[000I-z2(z2)T∥z2∥2].The exact formula ofB-subdifferential ∂B(fLθ) is given below.Theorem 13.
Givenz∈Rn, the following statements hold. (a)
Ifz2≠0, then
(93)∂B(fLθ)(z)={1tanθ+ctanθ[tanθ-z¯2-z¯2ctanθz¯2z¯2T]β+1tanθ+ctanθ[ctanθz¯2Tz¯2tanθz¯2z¯2T]γ+[000I-z¯2z¯2T]ς∣(β,γ,ς)∈Γ(z)}1tanθ+ctanθ[tanθ-z¯2-z¯2ctanθz¯2z¯2T].(b)
Ifz2=0, then
(94)∂B(fLθ)(z)={[000I-wwT]ς|(β,γ,ς)∈Γ(z)∥w∥=11tanθ+ctanθ[tanθ-wT-wctanθwwT]β+1tanθ+ctanθ[ctanθwTwtanθwwT]γ+[000I-wwT]ς∣{1tanθ+ctanθ[tanθ-wT-wctanθwwT](β,γ,ς)∈Γ(z)∥w∥=1{[000I-wwT]ς|(β,γ,ς)∈Γ(z)∥w∥=11tanθ+ctanθ[tanθ-wT-wctanθwwT]}∪{τI∣τ∈∂Bf(z1)}.Proof.
(a) Denote byΞ the set in the right side of (93). Take V∈∂B(fLθ)(z). By definition, there exists a sequence {zk}k=1∞→z with zk∈D(fLθ) satisfying ∇fLθ(zk)→V. Since z2≠0, then z2k≠0 for k sufficiently large. Note that λi(zk)∈Df for i=1,2 by Lemma 2 and from (92)
(95)∇fLθ(zk)=1tanθ+ctanθ×[tanθ-(z2k∥z2k∥)-z2k∥z2k∥ctanθz2k(z2k)T∥z2k∥2]f′(λ1(zk))+1tanθ+ctanθ×[ctanθ(z2k∥z2k∥)Tz2k∥z2k∥tanθz2k(z2k)T∥z2k∥2]f′(λ2(zk))+f(λ2(zk))-f(λ1(zk))λ2(zk)-λ1(zk)×[000I-z2k(z2k)T∥z2k∥2].
Note also that z2k/∥z2k∥→z2/∥z2∥=z¯2, λi(zk)→λi(z), for i=1,2, and
(96)(f′(λ1(zk)),f′(λ2(zk)),f(λ2(zk))-f(λ1(zk))λ2(zk)-λ1(zk))⟶Γ(z).
Hence V=lim⁡k→∞∇fLθ(zk)∈Ξ. This establishes ∂B(fLθ)(z)⊂Ξ.
Conversely, takeV∈Ξ; that is, there exists (β,γ,ς)∈Γ(z) such that
(97)V=1tanθ+ctanθ[tanθ-z¯2-z¯2ctanθz¯2z¯2T]β+1tanθ+ctanθ[ctanθz¯2Tz¯2tanθz¯2z¯2T]γ+[000I-z¯2z¯2T]ς.
By definition of Γ(z), there exists tk,tk′∈Df such that tk→λ1(z), tk′→λ2(z), tk′>tk, and
(98)(f′(tk),f′(tk′),f(tk′)-f(tk)tk′-tk)⟶(β,γ,ς).
Let
(99)zk∶=[z1kz2k]=[tan2θ1+tan2θtk+ctan2θ1+ctan2θtk′(tanθ1+tan2θtk′-ctanθ1+ctan2θtk)z2∥z2∥].
Note that tk′>tk, it is easy to see the λ1(zk)=tk, λ2(zk)=tk′ and uzki=uzi for i=1,2. Hence, zk→z and fLθ are differentiable at zk by Lemma 2 (since f is differentiable at λi(zk) for i=1,2). Hence according to the formula of gradients ∇fLθ(zk), (95), (97), and (98), together with the fact z2k/∥z2k∥=z2/∥z2∥, we have V=lim⁡k→∞∇fLθ(zk), which in turn implies V∈∂B(fLθ)(z).
(b) TakeV∈∂B(fLθ)(z); then by definition there exists zk→z with zk∈DfLθ such that ∇fLθ(zk)→V. By passing to a subsequence, we can assume that either z2k=0 for all k or z2k≠0 for all k. If z2k=0, then by Lemma 2  ∇fLθ(zk)=f′(z1k)I. Hence the cluster point of {f′(z1k)I} is an element of{τI∣τ∈∂Bf(z1)}. If z2k≠0, by passing to a subsequence we can assume that {z2k/∥z2k∥}→w for some w with ∥w∥=1. Note that λi(zk)∈Df (since zk∈DfLθ by Lemma 2), λi(zk)→λi(z), for i=1,2, and λ1(zk)<λ2(zk); it follows from (95) that V belongs to an element of (94) with
(100)(β,γ,ς)∈lim⁡sup⁡k→∞(f(λ2(zk))-f(λ1(zk))λ2(zk)-λ1(zk)f′(λ1(zk)),f′(λ2(zk)),f(λ2(zk))-f(λ1(zk))λ2(zk)-λ1(zk)).
Conversely, take V belonging to the left side of (94); that is, there exists τ∈∂Bf(z1) such that V=τI or exist (β,γ,ς)∈Γ(z) and w∈Rn-1 with ∥w∥=1 such that
(101)V=1tanθ+ctanθ[tanθ-wT-wctanθwwT]β+1tanθ+ctanθ[ctanθwTwtanθwwT]γ+[000I-wwT]ς.
If V=τI for some τ∈∂Bf(z1), then there exists tk→z1 with tk∈Df such that f′(tk)→τ. Let zk=tke, then λi(zk)=tk, for i=1,2, zk→z, and zk∈DfLθ by Lemma 2. Thus, ∇f(zk)=f′(tk)I, which further implies V=τI=lim⁡k→∞f′(tk)I=lim⁡k→∞∇fLθ(zk); that is, V∈∂B(fLθ)(z). The remaining case can be proved by using the same argument following (97) by replacing z¯2 by w. The proof is complete.Due to the important role played byΓ(z), we present the estimate of Γ as below.Lemma 14.
Givenz∈Rn, the following statements hold. (a)
Ifz2≠0, then
(102)Γ(z)=(∂Bf(λ1(z)),∂Bf(λ2(z)),f(λ2(z))-f(λ1(z))λ2(z)-λ1(z)).(b)
Ifz2=0, then
(103)Γ(z)⊂(∂Bf(z1),∂Bf(z1),∂f(z1)).Proof.
The case ofz2≠0 is clear, while the case of z2=0 can be proved by using the similar argument following (88).The exact estimate ofΓ(z) at z2=0 can be obtained provided that additional assumption is imposed on f.Lemma 15.
Suppose thatf:R→R is strictly continuous and directionally differentiable function satisfying
(104)lim⁡t,v→τσt≠vf(t)-f(v)t-v=lim⁡t→τσt∈Dff′(t)=fσ′(τ),f(t)-f(v)t-vlim⁡t,v→τσt≠v∀τ∈R,σ∈{-,+}.
Then, for z=(z1,0)∈Rn, we have
(105)Γ(z)={(τ,τ,τ)∣τ∈∂Bf(z1)}∪{(f-′(z1),f+′(z1),∂f(z1))}.Proof.
It follows from the definition ofΓ(z) via (91) that
(106)Γ(z)=limsup⁡tk,tk′→z1tktk′∈Df,tk<tk′(f′(tk),f′(tk′),f(tk′)-f(tk)tk′-tk)=limsup⁡tk,tk′→z1tk,tk′∈Df,tk<tk′≤z1(f′(tk),f′(tk′),f(tk′)-f(tk)tk′-tk)+limsup⁡tk,tk′→z1tktk′∈Df,z1≤tk<tk′(f′(tk),f′(tk′),f(tk′)-f(tk)tk′-tk)+limsup⁡tk,tk′→z1tk,tk′∈Df,tk<z1<tk′(f′(tk),f′(tk′),f(tk′)-f(tk)tk′-tk).
According to (104), it is easy to see that
(107)∂Bf(z1)={f-′(z1),f+′(z1)},∂f(z1)=[f-′(z1),f+′(z1)],limsup⁡tk,tk′→z1tk,tk′∈Df,tk<tk′≤z1(f′(tk),f′(tk′),f(tk′)-f(tk)tk′-tk)=(f-′(z1),f-′(z1),f-′(z1)),limsup⁡tk,tk′→z1tk,tk′∈Df,z1≤tk<tk′(f′(tk),f′(tk′),f(tk′)-f(tk)tk′-tk)=(f+′(z1),f+′(z1),f+′(z1)).
These are the elements of (τ,τ,τ) with τ∈∂Bf(z1), since ∂Bf(z1)={f-′(z1),f+′(z1)}. Now we claim that
(108)limsup⁡tk,tk′→z1tk,tk′∈Df,tk<z1<tk′(f′(tk),f′(tk′),f(tk′)-f(tk)tk′-tk)=(f-′(z1),f+′(z1),∂f(z1)).
It only needs to show
(109)limsup⁡tk,tk′→z1tk,tk′∈Df,tk<z1<tk′f(tk′)-f(tk)tk′-tk=∂f(z1).
First, we observe that
(110)f(tk′)-f(tk)tk′-tk=tk′-z1tk′-tkf(tk′)-f(z1)tk′-z1+z1-tktk′-tkf(z1)-f(tk)z1-tk⟶(1-r)f+′(z1)+rf-′(z1)∈∂f(z1),
for some r∈[0,1]. Conversely, taking τ∈∂f(z1) and using ∂f(z1)=[f-′(z1),f+′(z1)] yield τ=(1-r)f+′(z1)+rf-′(z1) for some r∈[0,1]. Due to Df being dense in R, for any k≥1, we define
(111)T1∶=Df∩[z1-r1k-1k2,z1-r1k]≠∅T2∶=Df∩[z1+(1-r)1k,z1+(1-r)1k+1k2]≠∅.
Take tk∈T1 and tk′∈T2; then for tk′>z1>tk, we have
(112)(1-r)1k≤tk′-z1≤(1-r)1k+1k2,1k≤tk′-tk≤1k+2k2,
which imply
(113)(1-r)(1/k)(1/k)+(2/k2)≤tk′-z1tk′-tk≤(1-r)(1/k)+(1/k2)1/k.
Thus,
(114)tk′-z1tk′-tk⟶1-r.
Similarly, we have
(115)z1-tktk′-tk⟶r.
In summary,
(116)τ=(1-r)f+′(z1)+rf-′(z1)=lim⁡k→∞tk′-z1tk′-tkf(tk′)-f(z1)tk′-z1+z1-tktk′-tkf(z1)-f(tk)z1-tk=lim⁡k→∞f(tk′)-f(tk)tk′-tk.
This completes the proof.Corollary 16.
Supposef:R→R is strictly continuous and directionally differentiable function satisfying
(117)lim⁡t,v→τσt≠vf(t)-f(v)t-v=lim⁡t→τσt∈Dff′(t)=fσ′(τ),lim⁡t,v→τσt≠vf(t)-f(v)t-v∀τ∈R,σ∈{-,+}.
Then, for any z=(z1,0)∈Rn,
(118)∂B(fLθ)(z)={[ξϱwTϱwaI+(η-a)wwT]∣eithera=ξ∈∂Bf(z1),ϱ=0,ora∈∂f(z1),ξ-ϱctanθ=f-′(z1),ξ+ϱtanθ=f+′(z1),η=ξ-ϱ(ctanθ-tanθ),∥w∥=1[ξϱwTϱwaI+(η-a)wwT]}.Proof.
This result follows by combining Theorem13 and Lemma 15.We point out that ifθ=45∘, then Corollary 16 reduces to [5, Lemma 5].


## 4. Directional Differentiability in Hadamard Sense, Positive Homogeneity, and Boundedness

In this section, we study some other important properties betweenfLθ and f, such as directional differentiability in the Hadamard sense, positive homogeneity, and boundedness. These are new discoveries and are not studied in [1, 5, 6] or other settings. First, we note that if a function is directionally differentiable in the Hadamard senses, then it must be directionally differentiable. The converse statement holds true if this function is from R to R, since in this case the direction just has ±1. More precisely, for a real-valued function g:R→R, g is directionally differentiable if and only if g is directionally differentiable in the Hadamard sense. Indeed, in the proof of [5, Proposition 3], the authors already employ the property of directionally differentiable in the Hadamard sense and even the assumption is directionally differentiable. However, for general mappings g:Rn to R or g:Rn→Rm, these two concepts are not equivalent. In other words, if g is directionally differentiable in the Hadamard sense, then g is directionally differentiable. But, the converse is invalid in general, unless some additional assumption is imposed; for example, local Lipschitz continuity [20]. Nonetheless, we will show for the special function fLθ:Rn→Rn, these two concepts are still equivalent.Theorem 17.
Letf:R→R and fLθ be defined as in (4). Then, fLθ is directionally differentiable at z if and only if fLθ is directionally differentiable at z in the Hadamard sense.Proof.
“⇐” This direction is clear.
“⇒” Suppose that fLθ is directionally differentiable at z. Then, from Lemma 1, f is directionally differentiable, and hence f is directionally differentiable in the Hadamard sense, since f is a function from R to R. To proceed the proof, we consider the following three cases.
Case 1. z2=0 and h2=0. Let h′=(h1′,h2′)→h=(h1,h2). If h2′=0, then the proof is trivial. If h2′≠0, then(119)fLθ(z+th′)-fLθ(z)=[f(z1+th1′-t∥h2′∥ctanθ)1+ctan2θ+f(z1+th1′+t∥h2′∥tanθ)1+tan2θ(-f(z1+th1′-t∥h2′∥ctanθ)ctanθ1+ctan2θ+f(z1+th1′+t∥h2′∥tanθ)tanθ1+tan2θ)h2′∥h2′∥]-[f(z1)1+ctan2θ+f(z1)1+tan2θ(-f(z1)ctanθ1+ctan2θ+f(z1)tanθ1+tan2θ)w],where w is taken to be h2′/∥h2′∥. Since f is directionally differentiable in the Hadamard sense, then
(120)lim⁡h′→ht↓0f(z1+th1′-t∥h2′∥ctanθ)-f(z1)t=f′(z1;h1),lim⁡h′→ht↓0f(z1+th1′-t∥h2′∥tanθ)-f(z1)t=f′(z1;h1).
Therefore,
(121)1t[f(z1+th1′-t∥h2′∥ctanθ)1+ctan2θ+f(z1+th1′+t∥h2′∥tanθ)1+tan2θkkk-f(z1)1+ctan2θ-f(z1)1+tan2θf(z1+th1′-t∥h2′∥ctanθ)1+ctan2θ+f(z1+th1′+t∥h2′∥tanθ)1+tan2θ]⟶f′(z1;h1)1+ctan2θ+f′(z1;h1)1+tan2θ=f′(z1;h1),1t{[-f(z1+th1′-t∥h2′∥ctanθ)ctanθ1+ctan2θk+f(z1+th1′+t∥h2′∥tanθ)tanθ1+tan2θ]k-[-f(z1)ctanθ1+ctan2θ+f(z1)tanθ1+tan2θ][-f(z1+th′1-t∥h′2∥ctanθ)ctanθ1+ctan2θ}h2′∥h2′∥⟶0
because the term in big brace approaches to zero and h2′/∥h2′∥ is bounded. These two limits imply
(122)lim⁡h′→ht↓0fLθ(z+th′)-fLθ(z)t=[f′(z1;h1)0]=f′(z1;h1)e=(fLθ)′(z;h),
which says fLθ is directionally differentiable at z in the Hadamard sense.
Case 2. z2=0 and h2≠0. Note that(123)fLθ(z+th′)-fLθ(z)=[f(z1+th1′-t∥h2′∥ctanθ)1+ctan2θ+f(z1+th1′+t∥h2′∥tanθ)1+tan2θ(-f(z1+th1′-t∥h2′∥ctanθ)ctanθ1+ctan2θ+f(z1+th1′+t∥h2′∥tanθ)tanθ1+tan2θ)h2′∥h2′∥]-[f(z1)0]=[f(z1+th1′-t∥h2′∥ctanθ)1+ctan2θ+f(z1+th1′+t∥h2′∥tanθ)1+tan2θ-f(z1)(-f(z1+th1′-t∥h2′∥ctanθ)ctanθ1+ctan2θ+f(z1+th1′+t∥h2′∥tanθ)tanθ1+tan2θ)h2′∥h2′∥].For the first component, we compute that
(124)1t[f(z1+th1′-t∥h2′∥ctanθ)1+ctan2θ+f(z1+th1′+t∥h2′∥tanθ)1+tan2θ-f(z1)f(z1+th1′-t∥h2′∥ctanθ)1+ctan2θ]=1t[f(z1+t(h1′-∥h2′∥ctanθ))-f(z1)1+ctan2θk+f(z1+t(h1′+∥h2′∥tanθ))-f(z1)1+tan2θ]⟶f′(z1;h1-∥h2∥ctanθ)1+ctan2θ+f′(z1;h1+∥h2∥tanθ)1+tan2θkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkast↓0,h′→h,
where the last step is due to the fact that f is directionally differentiable in the Hadamard sense.
For the second component, we have(125)1t[-f(z1+th1′-t∥h2′∥ctanθ)ctanθ1+ctan2θk+f(z1+th1′+t∥h2′∥tanθ)tanθ1+tan2θ]h2′∥h2′∥⟶[-ctanθ1+ctan2θf′(z1;h1-∥h2∥ctanθ)+tanθ1+tan2θf′(z1;h1+∥h2∥tanθ)]h2∥h2∥.
The above two limits show that
(126)lim⁡h′→ht↓0fLθ(z+th′)-fLθ(z)t=11+ctan2θ×f′(z1;h1-∥h2∥ctanθ)[100ctanθ·I][1-h2∥h2∥]+11+tan2θf′(z1;h1+∥h2∥tanθ)[100tanθ·I][1h2∥h2∥]=(fLθ)′(z;h).
Case 3. z2≠0. Then(127)fLθ(z+th′)-fLθ(z)=[f(z1+th1′-∥z2+th2′∥ctanθ)1+ctan2θ+f(z1+th1′+∥z2+th2′∥tanθ)1+tan2θ(-f(z1+th1′-∥z2+th2′∥ctanθ)ctanθ1+ctan2θ+f(z1+th1′+∥z2+th2′∥tanθ)tanθ1+tan2θ)z2+th2′∥z2+th2′∥]-[f(z1-∥z2∥ctanθ)1+ctan2θ+f(z1+∥z2∥tanθ)1+tan2θ(-f(z1-∥z2∥ctanθ)ctanθ1+ctan2θ+f(z1+∥z2∥tanθ)tanθ1+tan2θ)z2∥z2∥].Then, the first component of (fLθ(z+th′)-fLθ(z))/t behaves as follows (when t↓0 and h′→h):
(128)1t[f(z1+th1′-∥z2+th2′∥ctanθ)1+ctan2θk+f(z1+th1′+∥z2+th2′∥tanθ)1+tan2θk-f(z1-∥z2∥ctanθ)1+ctan2θ-f(z1+∥z2∥tanθ)1+tan2θf(z1+th1′-∥z2+th2′∥ctanθ)1+ctan2θ]=1t[f(z1+th1′-∥z2+th2′∥ctanθ)-f(z1-∥z2∥ctanθ)1+ctan2θ+f(z1+th1′+∥z2+th2′∥tanθ)-f(z1+∥z2∥tanθ)1+tan2θ]=1t[(f(z1+th1′-∥z2∥ctanθ-tz2Th2′∥z2∥ctanθ+o(t))-f(z1-∥z2∥ctanθ)(z1+th′1-∥z2∥ctanθ-tz2Th2′∥z2∥ctanθ+o(t))f(z1+th′1+∥z2∥tanθ+tz2Th2′∥z2∥tanθ+o(t)))k×(1+ctan2θ)-1k+(f(z1+th1′+∥z2∥tanθ+tz2Th2′∥z2∥tanθ+o(t))k-f(z1+∥z2∥tanθ)f(z1+th′1+∥z2∥tanθ+tz2Th2′∥z2∥tanθ+o(t)))k×(1+tan2θ)-1(f(z1+th′1-∥z2∥ctanθ-tz2Th2′∥z2∥ctanθ+o(t))]⟶f′(z1-∥z2∥ctanθ;h1-(z2Th2/∥z2∥)ctanθ)1+ctan2θ+f′(z1+∥z2∥tanθ;h1+(z2Th2/∥z2∥)tanθ)1+tan2θ,
where in the last step we have used the fact that f is directionally differentiable in the Hadamard sense. Recall that Φ(z2)=z2/∥z2∥ is continuously differentiable at z2≠0. Then, the second component of (fLθ(z+th′)-fLθ(z))/t behaves as follows (when t↓0 and h′→h):
(129)1t{+f(z1+∥z2∥tanθ)tanθ1+tan2θ][-f(z1+th1′-∥z2+th2′∥ctanθ)ctanθ1+ctan2θ+f(z1+th1′+∥z2+th2′∥tanθ)tanθ1+tan2θ]z2+th2′∥z2+th2′∥-[-f(z1-∥z2∥ctanθ)ctanθ1+ctan2θk+f(z1+∥z2∥tanθ)tanθ1+tan2θ]z2∥z2∥[-f(z1+th1′-∥z2+th2′∥ctanθ)ctanθ1+ctan2θ}⟶[-ctanθ1+ctan2θf′(z1-∥z2∥ctanθ;h1-z2Th2∥z2∥ctanθ)+tanθ1+tan2θf′(z1+∥z2∥tanθ;h1+z2Th2∥z2∥tanθ)]×z2∥z2∥+[-f(z1-∥z2∥ctanθ)ctanθ1+ctan2θk+f(z1+∥z2∥tanθ)tanθ1+tan2θ]×[1∥z2∥(I-z2z2T∥z2∥2)]h2.
The above two limits show that
(130)lim⁡h′→ht↓0fLθ(z+th′)-f(z)t=11+ctan2θf′(λ1(z);h1-z2Th2∥z2∥ctanθ)×[100ctanθ·I][1-z2∥z2∥]-ctanθ1+ctan2θf(λ1(z))∥z2∥Mz2h+11+tan2θf′(λ2(z);h1+z2Th2∥z2∥tanθ)×[100tanθ·I][1z2∥z2∥]+tanθ1+tan2θf(λ2(z))∥z2∥Mz2h=(fLθ)′(z;h).
The proof is complete.Theorem 18.
Letf:R→R and fLθ be defined as in (4). Then, the following statements are equivalent.(a)
fLθ is directionally differentiable at z;(b)
fLθ is directionally differentiable at z in the Hadamard sense;(c)
f is directionally differentiable at λi(z), for i=1,2;(d)
f is directionally differentiable at λi(z), for i=1,2 in the Hadamard sense.Proof.
The equivalence(a)⇔(b) comes from Theorem 17; (c)⇔(d) is due to the fact that f:R→R; (a)⇔(c) follows from Lemma 1.Below we study the relationship of positive homogeneity and boundedness betweenfLθ and f.Theorem 19.
Letf:R→R and fLθ be defined as in (4). Then, fLθ is positively homogeneous at z with exponent α>0 if and only if f is positively homogeneous at λi(z) for i=1,2 with exponent α>0.Proof.
“⇐” Suppose that f is positively homogeneous at λi(z) for i=1,2 with exponent α>0. For any z=(z1,z2)∈R×Rn-1 with z=λ1(z)uz1+λ2(z)uz2, we observe that ukzi=uzi, i=1,2 for k≥0, whenever z2≠0 and λi(kz)=kλi(z), for i=1,2. Hence, when z2≠0, we have
(131)fLθ(kz)=f(λ1(kz))ukz1+f(λ2(kz))ukz2=f(kλ1(z))uz1+f(kλ2(z))uz2=kα[f(λ1(z))uz1+f(λ2(z))uz2]=kαf(z).
When z2=0, we know that λi(z)=z1 for i=1,2. Thus,
(132)fLθ(kz)=[f(kz1)0]=[kαf(z1)0]=kαfLθ(z).
All the above shows that fLθ is positively homogeneous at z with exponent α>0.
“⇒” Suppose that fLθ is positively homogeneous at z with exponent α>0; that is, fLθ(kz)=kαfLθ(z). Then, we have
(133)f(kλi(z))∥uzi∥2=〈fLθ(kz),uzi〉=kα〈fLθ(z),uzi〉=kαf(λi(z))∥uzi∥2,
which in turn implies f(kλi(z))=kαf(λi(z)), since ∥uzi∥≠0. Hence, f is positively homogenous at λi(z) for i=1,2 with exponent α>0.Theorem 20.
Letf:R→R and fLθ be defined as in (4). Then, fLθ is bounded if and only if f is bounded.Proof.
“⇐” Suppose that f is bounded by M. The proof for this direction follows from the following inequality:
(134)∥fLθ(z)∥=∥f(λ1(z))uz1+f(λ2(z))uz2∥≤|f(λ1(z))|·∥uz1∥+|f(λ2(z))|·∥uz2∥≤(11+ctan2θ+11+tan2θ)M.
“⇒” Suppose that fLθ is bounded by M. This direction is trivial because for any τ∈R, one has fLθ(τe)=(f(τ),0)T, and hence
(135)|f(τ)|=∥(f(τ)0)∥=∥fLθ(τe)∥≤M.


## 5. Final Remarks

So far, we have shown that many properties holding forfsoc can be extended to the setting for fLθ. One may wonder whether fsoc and fLθ always share the same properties. The answer is no! Here, we present a simple property that holds for fsoc but fails for fLθ. Some more different properties between fLθ and fsoc are discovered in [21].To see the counterexample, we recall that a functionF:Rn→Rn is said to be an odd (even, resp.) function if F(-x)=-F(x) (F(-x)=F(x), resp.) for all x∈Rn.Proposition 21.
Letf:R→R and fsoc be given as in (6). Then fsoc is an odd (even, resp.) function on Rn if and only if f is an odd (even, resp.) function on R.Proof.
“⇐” In the setting of second-order cone, we observe that
(136)u-z1=uz2,u-z2=uz1,λ1(-z)=-λ2(z),λ2(-z)=-λ1(z),
which implies(137)fsoc(-z)=f(λ1(-z))u-z1+f(λ2(-z))u-z2=f(-λ2(z))uz2+f(-λ1(z))uz1={-f(λ2(z))uz2-f(λ1(z))uz1iffisodd,f(λ2(z))uz2+f(λ1(z))uz1iffiseven,={-fsoc(z)iffisodd,fsoc(z)iffiseven.
“⇒” For τ∈R, we have
(138)f(-τ)=〈fLθ(-τe),e〉={〈-fLθ(τe),e〉=-f(τ)iffLθisodd,〈fLθ(τe),e〉=f(τ)iffLθiseven.The below example illustrates that the above relationship fails to hold forfLθ and f.Example 22.
LetfLθ be given as in (4) with tanθ=2 (θ=tan-12). Then,(a)
f(t)=t3 is an odd function, but fLθ is not an odd function at z=(1,1);(b)
f(t)=t2 is an even function, but fLθ is not an even function at z=(1,1).Forf(t)=t3, it is clear that f(t)=t3 is an odd function. Nonetheless, we verify that
(139)fLθ(z)=120(110,215),fLθ(-z)=120(-50,-35),
which says -fLθ(z)≠fLθ(-z). Thus, fLθ is not an odd function at z=(1,1).Similarly, forf(t)=t2 which is an even function, we compute
(140)fLθ(z)=110(20,35),fLθ(-z)=110(20,5).
Hence, fLθ(-z)≠fLθ(z) which says fLθ is not an even function at z=(1,1).Finally, let us discuss the relationship between circular cone and the (nonsymmetric) matrix cone introduced in [17, 22], where the authors study the epigraph of six different matrix norms, such as the Frobenius norm, the l∞ norm, l1 norm, the spectral or the operator norm, the nuclear norm, and the Ky Fan k-norm. If we regard a matrix as a high-dimensional vector, then the circular cone is equivalent to the matrix cone with Frobenius norm. More precisely, denote
(141)Km,nϵ∶={(t,X)∈R×Rm×n∣ϵ-1t≥∥X∥F},
where ∥X∥F denotes the Frobenius norm of X; that is, ∥X∥F∶=(∑i=1m∑j=1n|Xij|2)1/2. Notice that the circular cone Lθ can be equivalently written as
(142)Lθ={(x1,x2)∈R×Rn-1∣x1tanθ≥∥x2∥}.
Let t=x1, ϵ=ctanθ and X=diag⁡(x2) where diag denotes the diagonal matrix; then Kn-1,1ctanθ reduces to Lθ. Conversely, the matrix cone Km,nϵ can be also regarded as a circular cone with
(143)tanθ=ϵ-1,x1=t,x2=(X11,X12,…,X1n,…,Xi1,…,Xin,…,Xm1,k…,Xmn)T∈Rmn.
Therefore, Km,nϵ is a (mn+1)-dimensional circular cone.In addition, we know for a vectory∈Rn (regarding as a matrix in Rn×1) the singular value decomposition is
(144)y=UΣVT,
where V=1∈R, Σ=(∥y∥,0,…,0)T∈Rn×1, and U=[U1,…,Un]∈Rn×n with U1=y/∥y∥ and U2,…,Un are arbitrary orthonormal vectors orthogonal to y. It indicates that the singular value of y is ∥y∥ and 0 with multiplicity n-1. Hence the spectral/operator norm (largest singular value), the Ky Fan k-norm (the sum of k-largest singular value), or the nuclear norm (the sum of the singular values) are all ∥y∥. This means that Lθ is a special case of the matrix cone studied in [17, 22], where the properties of spectral operator are studied, such as well-definiteness, the directional differentiability, the Fréchet differentiability, the locally Lipschitz continuity, the ρ-order B-differentiability (0<ρ≤1), the ρ-order G-semismooth (0<ρ≤1), and the characterization of Clarke’s generalized Jacobian. In this paper, by using the special structure of circular cone, we mainly establish the B-subdifferential (the approach we considered here is more directly and depended on the special structure of circular cone), the directional differentiability in the Hadamard sense, and the ρ-order semismooth for ρ>1.



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*Source: 603542-2014-06-22.xml*