On the vector-valued functions associated with circular cones

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to appear in Abstract and Applied Analysis, 2014

On the vector-valued functions associated with circular cones

Jinchuan Zhou ¹ Department of Mathematics

School of Science

Shandong University of Technology Zibo 255049, Shandong, P.R.China

E-mail: jinchuanzhou@163.com

Jein-Shan Chen ² Department of Mathematics National Taiwan Normal University

Taipei 11677, Taiwan E-mail: jschen@math.ntnu.edu.tw

April 6, 2014

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 degree.

Let L_θ denote the circular cone in IRⁿ. For a function f from IR to IR, one can define a corresponding vector-valued function f^L^θ on IRⁿ by applying f to the spectral values of the spectral decomposition of x ∈ IRⁿ with respect to L_θ. In this paper, we study properties that this vector-valued function inherits from f , including H¨older continuity, B-subdifferentiability, ρ-order semismoothness, as well as positive homogeneity. These results will play crucial role in designing solution methods for optimization problem involved circular cone constraints.

Keywords. Circular cone, vector-valued function, semismooth function, spectral decomposition, positively homogeneous.

AMS subject classifications. 26A27, 26B05, 26B35, 49J52, 90C33, 65K05

1The author’s work is supported by National Natural Science Foundation of China (11101248, 11271233), Shandong Province Natural Science Foundation (ZR2010AQ026, ZR2012AM016), and Young Teacher Support Program of Shandong University of Technology.

2Corresponding author. Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The author’s work is supported by Ministry of Science and Technology, Taiwan.

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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 degree. Let L_θ denote the circular cone in IRⁿ. Then, the n-dimensional circular cone L_θ is expressed as

L_θ := {x = (x₁, x₂)^T ∈ IR × IRⁿ⁻¹| cos θkxk ≤ x₁}.

The application of L_θ lies in engineering field, for example, optimal grasping manipula- tion for multi-gingered robots, see [3].

In our previous work [21], we have explored some important features about circular cone, such as characterizing its tangent cone, normal cone, and second-order regularity, etc. In particular, the spectral decomposition associated with L_θ was discovered, i.e., for any z = (z₁, z₂) ∈ IR × IRⁿ⁻¹, one has

z = λ₁(z)u¹_z + λ₂(z)u²_z, (1) where

λ₁(z) = z₁− kz₂kctanθ λ₂(z) = z₁+ kz₂k tan θ and

u¹_z = 1 1 + ctan²θ

1 0

0 ctanθ · I

1

−¯z2

u²_z = 1 1 + tan²θ

1 0

0 tan θ · I

1

¯ z₂

with ¯z₂ := z₂/kz₂k if z₂ 6= 0, and ¯z₂ being any vector w ∈ IRⁿ⁻¹ satisfying kwk = 1 if z₂ = 0. With this spectral decomposition (1), analogous to so-called SOC-function f^soc (see [4, 5, 6]) and SDP-function f^mat (see [7, 16]), we define a vector-valued function associated with circular cone as below. More specifically, for f : IR → IR, we define f^L^θ : IRⁿ → IRⁿ as

f^L^θ(z) = f (λ₁(z)) u¹_z + f (λ₂(z)) u²_z. (2) It is not hard to see that f^L^θ is well-defined for all z. In particular, if z₂ = 0, then

f^L^θ(z) = f (z₁) 0

.

Note that when θ = 45^◦, L_θ reduces to the second-order cone (SOC) and the vector- valued function f^L^θ defined as in (2) corresponds to the SOC-function f^soc given by

f^soc(x) = f (λ₁(x))u⁽¹⁾_x + f (λ₂(x))u⁽²⁾_x ∀x = (x₁, x₂) ∈ IR × IRⁿ⁻¹ (3)

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where λi(x) = x1+ (−1)ⁱkx2k and u⁽ⁱ⁾x = ¹₂(1, (−1)ⁱx¯2)^T.

It is well known that the vector-valued function f^soc associated with second-order cone and matrix-valued function f^mat 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 f^soc and f^mat are inherited from f , such as continuity, strictly continuity, directional differentiability, Fr´echet 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 f^L^θ. In [3], the authors answer the questions from the following aspects:

(a) f^L^θ is continuous at z ∈ IRⁿ if and only if f is continuous at λ_i(z) for i = 1, 2;

(b) f^L^θ is directionally differentiable at z ∈ IRⁿ if and only if f is directionally differentiable at λ_i(z) for i = 1, 2;

(c) f^L^θ is (Fr´echet) differentiable at z ∈ IRⁿ if and only if f is (Fr´echet) differentiable at λ_i(z) for i = 1, 2;

(d) f^L^θ is continuously differentiable at z ∈ IRⁿ if and only if f is continuously continuous at λ_i(z) for i = 1, 2;

(e) f^L^θ is strictly continuous at z ∈ IRⁿ if and only if f is strictly continuous at λ_i(z) for i = 1, 2;

(f) f^L^θ is Lipschitz continuous with constant k > 0 if and only if f is Lipschitz continuous with constant k > 0;

(g) f^L^θ is semismooth at z if and only if f is semismooth at λ_i(z) for i = 1, 2.

In this paper, we further study some other properties associated with f^L^θ, such as H¨older 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 f^soc and f^L^θ always share the same properties. In- deed, they do not. There exists some property that holds for f^soc and f , but fails for f^L^θ and f . An counter-example is presented in the final section.

To end thid section, we briefly review our notations and some basic concepts which will be needed for subsequent analysis. First, we denote by IRⁿthe space of n-dimensional real column vectors and let e = (1, 0, · · · , 0) ∈ IRⁿ. Given x, y ∈ IRⁿ, the Euclidean inner product and norm are hx, yi = x^Ty and kxk = √

x^Tx. For a linear mapping H : IRⁿ → IR^m, its operator norm is kHk := maxkxk=1kHxk. For α ∈ IR and s ∈ IRⁿ, write s = O(α) (respectively s = o(α)) to means ksk/|α| is uniformly bounded (respectively, tends to zero) as α → 0. In addition, given a function F : IRⁿ → IR^m, we say

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(a) F is H¨older continuous with exponent α ∈ (0, 1] if

[F ]_α := sup

x6=y

kF (x) − F (y)k

kx − yk^α < +∞;

(b) F is directionally differentiable at x ∈ IRⁿ in the Hadamard sense if the directional derivative F⁰(x; d) exists for all d ∈ IRⁿ and

F⁰(x; d) = lim

d0→d t↓0

F (x + td⁰) − 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) If F is strictly continuous (locally Lipschitz continuous), the generalized Jacobian

∂F (x) is the convex hull of the ∂_BF (x), where

∂_BF (x) :=n

z→xlim∇F (z)

z ∈ D_Fo , where D_F 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),

F (x + h) − F (x) − V h = o(khk);

(f ) F is ρ-order semismooth at x (ρ > 0) if F is semismooth at x and for any V ∈

∂F (x + h),

F (x + h) − F (x) − V h = O(khk^1+ρ); (4) In particular, we say F is strongly semismooth if it is 1-order semismooth.

(g) F is positively homogeneous with exponent α > 0 if

F (kx) = k^αF (x), ∀x ∈ IRⁿ and k ≥ 0;

(h) F is bounded if there exists a positive scalar M > 0 such that kF (x)k ≤ M, ∀x ∈ IRⁿ.

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2 Directional differentiability, strict continuity, H¨ older continuity, and B-differentiability

This section is devoted to study the properties of directional differentiability, strict continuity, and H¨older continuity. The relationship of directional differentiability between f^L^θ and f has been given in [3, 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 below.

Lemma 2.1. Let f : IR → IR and f^L^θ be defined as in (2). Then, f^L^θ is directionally differentiable at z if and only if f is directionally differentiable at λ_i(z) for i = 1, 2.

Moreover, for any h = (h₁, h₂) ∈ IR × IRⁿ⁻¹, we have f^L^θ0

(z; h) = f⁰(z₁; h₁) 0

= f⁰(z₁; h₁)e when z₂ = 0 and h₂ = 0;

f^L^θ0

(z; h) = 1

1 + ctan²θf⁰(z₁; h₁− kh₂kctanθ) 1 0 0 ctanθ · I

"

1

−_kh^h²

2k

#

+ 1

1 + tan²θf⁰(z₁; h₁+ kh₂k tan θ) 1 0 0 tan θ · I

"

1

h2

kh2k

#

when z₂ = 0 and h₂ 6= 0; otherwise

f^L^θ0

(z; h) = 1

1 + ctan²θf⁰

λ₁(z); h₁− z^T₂h₂

kz₂kctanθ 1 0 0 ctanθ · I

"

1

−_kz^z²

2k

#

− ctanθ 1 + ctan²θ

f (λ₁(z)) kz₂k M_z₂h

+ 1

1 + tan²θf⁰

λ₂(z); h₁+z₂^Th₂

kz2k tan θ 1 0 0 tan θ · I

"

1

z2

kz₂k

#

+ tan θ 1 + tan²θ

f (λ₂(z)) kz2k M_z₂h, where

M_z₂ :=





0 0

0 I − z₂z^T₂ kz₂k²



.

Lemma 2.2. Let f : IR → IR and f^L^θ be defined as in (2). Then, the following hold.

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(a) f^L^θ is differentiable at z if and only if f is differentiable at λ_i(z) for i = 1, 2.

Moreover, if z₂ = 0, then

∇f^L^θ(z) = f⁰(z₁)I;

otherwise

∇f^L^θ(z) =







ξ %z₂^T

kz2k

%z₂

kz₂k aI + (η − a)z₂z^T₂ kz₂k²





 ,

where

a = tan θ 1 + tan²θ

f (λ2(z))

kz₂k − ctanθ 1 + ctan²θ

f (λ1(z))

kz₂k = f (λ2(z)) − f (λ1(z)) λ₂(z) − λ₁(z) , ξ = f⁰(λ₁(z))

1 + ctan²θ + f⁰(λ₂(z))

1 + tan²θ, η = ξ − %(ctanθ − tan θ), (5)

% = − ctanθ

1 + ctan²θf⁰(λ₁(z)) + tan θ

1 + tan²θf⁰(λ₂(z)).

(b) f^L^θ is continuously differentiable (smooth) at z if and only if f is continuously differentiable (smooth) at λ_i(z) for i = 1, 2.

Note that the formula of gradient ∇f^L^θ given in [3, Theorem 3.3] and Lemma 2.2 are the same by using the following facts

1

1 + ctan²θ = sin²θ, 1

1 + tan²θ = cos²θ, ctanθ

1 + ctan²θ = tan θ

1 + tan²θ = sin θ cos θ.

The following result indicating that λ_i is Lipschitz continuous on IRⁿ for i = 1, 2 will be used in proving the Lipschitz continuity between f^L^θ and f .

Lemma 2.3. Let z, y ∈ IRⁿ with spectral values λ_i(z), λ_i(y), respectively. Then, we have

|λ_i(z) − λ_i(y)| ≤√

2 max{tan θ, ctanθ}kz − yk for i = 1, 2.

Proof. First, we observe that

|λ₁(z) − λ₁(y)| = |z₁− kz₂kctanθ − y₁+ ky₂kctanθ|

≤ |z₁− y₂| + kz₂− y₂kctanθ

≤ max{1, ctanθ}(|z₁ − y₁| + kz₂− y₂k)

≤ max{1, ctanθ}√

2p|z₁− y₁|²+ kz₂− y₂k²

= max{1, ctanθ}√

2kz − yk.

Applying the similar argument to λ₂ yields

|λ₂(z) − λ₂(y)| ≤ max{1, tan θ}√

2kz − yk.

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Then, the desired result follows from the fact that max{1, ctanθ, tan θ} = max{ctanθ, tan θ}.

2

Theorem 2.1. Let f : IR → IR and f^L^θ be defined as in (2). Then, f^L^θ is strictly continuous (local Lipschitz continuity) at z if and only if f is strictly continuous (local Lipschitz continuity) at λ_i(z) for i = 1, 2.

Proof. “⇐” Suppose f is strictly continuous at λ_i(z) for i = 1, 2, i.e., there exists k_i > 0 and δi > 0 for i = 1, 2 such that

|f (τ ) − f (ζ)| ≤ k_i|τ − ζ| ∀τ, ζ ∈ [λ_i(z) − δ_i, λ_i(z) + δ_i], i = 1, 2.

Let ¯δ := min{δ₁, δ₂} and C := [λ₁(z) − ¯δ, λ₁(z) + ¯δ] ∪ [λ₂(z) − ¯δ, λ₂(z) + ¯δ]. Define

f (τ ) :=˜











f (τ ) if τ ∈ C,

(1 − t)f (λ₁(z) + ¯δ) + tf (λ₂(z) − ¯δ) if λ₁(z) + ¯δ < λ₂(z) − ¯δ and

τ = (1 − t)(λ₁(z) + ¯δ) + t(λ₂(z) − ¯δ) with t ∈ (0, 1)

f (λ₁(z) − ¯δ) if τ < λ₁(z) − ¯δ f (λ₂(z) + ¯δ) if τ > λ₂(z) + ¯δ.

Clearly, ˜f is Lipschitz continuous on IR, i.e., there exists k > 0 such that lip ˜f (τ ) ≤ k for all τ ∈ IR. Since eC := conv(C) is compact, according to Lemma [7, Lemma 4.5] or [6, Lemma], there exist continuously differentiable functions f^v : IR → IR for v = 1, 2, · · · converging uniformly to ˜f on eC such that

|(f^v)⁰(τ )| ≤ k ∀τ ∈ eC and ∀v. (6) Now, let δ := ¯δ/(√

2 max{tan θ, ctanθ}). Then, from Lemma 2.3, we know eC contains all spectral values of y ∈ B(z, δ). Therefore, for any w ∈ B(z, δ), we have λi(w) ∈ eC for i = 1, 2 and

(f^v)^L^θ(w) − f^L^θ(w)

2

=

[f^v(λ₁(w)) − f (λ₁(w))] u¹_w+ [f^v(λ₂(w)) − f (λ₂(w))] u²_w

2

= f^v(λ₁(w)) − f (λ₁(w))2

ku¹_wk²+f^v(λ₂(w)) − f (λ₂(w))2

ku²_wk²

= 1

1 + ctan²θ|f^v(λ₁(w)) − f (λ₁(w))|²+ 1

1 + tan²θ |f^v(λ₂(w)) − f (λ₂(w))|², where we have used the facts that ku¹_wk = 1/√

1 + ctan²θ, ku²_wk = 1/√

1 + tan²θ, and hu¹_w, u²_wi = 0. Since {f^v}^∞_v=1 converges uniformly to f on eC, the above equations show that {(f^v)^L^θ}^∞_v=1 converges uniformly to f^L^θ on B(z, δ). If w2 = 0, then it follows from Lemma 2.2 that ∇(f^v)^L^θ(w) = (f^v)⁰(w₁)I. Hence it follows from (6) that

k∇(f^v)^L^θ(w)k = |(f^v)⁰(w₁)| ≤ k, (7)

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since in this case λ_i(w) = w₁ ∈ eC. If w₂ 6= 0, then

∇(f^v)^L^θ(w) =





ξ % w₂^T/kw₂k

% w₂/kw₂k aI + (ξ − %(ctanθ − tan θ) − a)w₂w^T₂ kw₂k²





=





ξ % w^T₂/kw₂k

% w₂/kw₂k aI + (ξ − a)w2w₂^T kw₂k²



+





0 0

0 [−%(ctanθ − tan θ)] w2w₂^T kw₂k²





=

ξ % w₂^T/kw₂k

% w2/kw2k ξI

+ (a − ξ)





0 0

0 I − w₂w^T₂ kw₂k²





−%(ctanθ − tan θ)





0 0

0 w₂w₂^T kw₂k²





where a, ξ, % are given as in (5) with λ_i(z) replaced by λ_i(w) for i = 1, 2 and f replaced by f^v. For simplicity of notations, let us denote

A :=

ξ % w^T₂/kw₂k

% w₂/kw₂k ξI

+ (a − ξ)





0 0

0 I − w₂w^T₂ kw₂k²



 and

B := −%(ctanθ − tan θ)





0 0

0 w₂w^T₂ kw₂k²



. Note that

|a| =

f^v λ₂(w) − f^v λ₁(w) λ₂(w) − λ₁(w)

≤ k, (8)

where the inequality comes from the fact that f^v is continuously differentiable on eC and (6). Besides, we also note that

|ξ| =

(f^v)⁰ λ₁(w)

1 + ctan²θ +(f^v)⁰ λ₂(w) 1 + tan²θ

≤ 1

1 + ctan²θ|(f^v)⁰ λ₁(w)| + 1

1 + tan²θ|(f^v)⁰ λ₂(w)|

≤

1

1 + ctan²θ + 1 1 + tan²θ

k

= k (9)

and

|%| = | − ctanθ

1 + ctan²θ(f^v)⁰(λ₁(w)) + tan θ

1 + tan²θ(f^v)⁰(λ₂(w))|

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≤

− ctanθ 1 + ctan²θ

+

tan θ 1 + tan²θ

k

=

ctanθ

1 + ctan²θ + tan θ 1 + tan²θ

k

= 2 tan θ 1 + tan²θk

≤ k. (10)

(i) For % = 0, then ∇(f^v)^L^θ(w) takes the form of ξI + (a − ξ)M_w₂ whose eigenvalues are ξ and a by [6, Lemma 1]. In other words, in this case, we get from (8) and (9) that

k∇(f^v)^L^θ(w)k = max{|a|, |ξ|} ≤ k. (11)

(ii) For % 6= 0, since B = −%(ctanθ − tan θ)(0, w₂

kw₂k)^T(0, w₂

kw₂k), the eigenvalues of B are

−%(ctanθ − tan θ) and 0 with multiplicity n − 1. Note that

|%(ctanθ − tan θ)| =

1 − ctan²θ

1 + ctan²θ(f^v)⁰ λ₁(w) + 1 − tan²θ

1 + tan²θ(f^v)⁰ λ₂(w)

≤

1 − ctan²θ 1 + ctan²θ

+

1 − tan²θ 1 + tan²θ

k

=

ctan²θ − 1

1 + ctan²θ +1 − tan²θ 1 + tan²θ k

= 2

1 − tan²θ 1 + tan²θ k

≤ 2k. (12)

Note that

A = %

kw₂kL_w_˜ + (a − ξ)M_w₂ = % kw₂k

L_w_˜ + (a − ξ)kw₂k

% M_w_˜₂

,

where ˜w = ξkw2k/%, w2 and

L_w_˜ := ˜w₁ w˜₂^T

˜ w2 w˜1I

.

In this case the matrix A has eigenvalues of ξ ± % and a with multiplicity n − 2. Hence, k∇(f^v)^L^θ(w)k ≤ max{|ξ + %|, |ξ − %|, |a|} + |%(ctanθ − tan θ)|

≤ max{|ξ| + |%|, |a|} + |%(ctanθ − tan θ)|

≤ 4k, (13)

where the last step is due to (8), (9), (10), and (12).

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Putting (7), (11), and (13) together, we know

k∇(f^v)^L^θ(w)k ≤ 4k ∀w ∈ B(z, δ) and ∀v.

Fix any x, y ∈ B(z, δ) with x 6= y. Since {(f^v)^L^θ}^∞_v=1 converges uniformly to f^L^θ on B(z, δ), then for any > 0 there exists v0 such that

k(f^v)^L^θ(w) − f^L^θ(w)k ≤ , ∀w ∈ B(z, δ) and ∀v ≥ v0.

Since f^v is continuously differentiable, (f^v)^L^θ is continuously differentiable by Lemma 2.2. Thus,

kf^L^θ(x) − f^L^θ(y)k

= kf^L^θ(x) − (f^v)^L^θ(x) + (f^v)^L^θ(x) − (f^v)^L^θ(y) + (f^v)^L^θ(y) − f^L^θ(y)k

≤ kf^L^θ(x) − (f^v)^L^θ(x)k + k(f^v)^L^θ(x) − (f^v)^L^θ(y)k + k(f^v)^L^θ(y) − f^L^θ(y)k

≤ 2 +

Z 1 0

∇(f^v)^L^θ(y + t(x − y))(x − y)dt

≤ 2 + 4kkx − yk.

Because > 0 is arbitrary, this ensures f^L^θ(x) − f^L^θ(y)

≤ 4kkx − yk ∀x, y ∈ B(z, δ), which says f^L^θ is strictly continuous at z.

“⇒” Suppose f^L^θ is strictly continuous at z, then there exist k > 0 and δ > 0 such that kf^L^θ(x) − f^L^θ(y)k ≤ kkx − yk ∀x, y ∈ B(z, δ). (14) Case 1: z₂ 6= 0. Take θ, µ ∈ [λ₁(z) − δ₁, λ₁(z) + δ₁] with δ₁ := min{δ, λ₂(z) − λ₁(z)}. Let

x := θu¹_z+ λ₂(z)u²_z and y := µu¹_z+ λ₂(z)u²_z. Then, kx − zk ≤ δ and ky − zk ≤ δ and it follows from (14) that

|f (θ) − f (µ)| = 1

ku¹_zkkf^L^θ(x) − f^L^θ(y)k ≤ k

ku¹_zkkx − yk = k

ku¹_zk|θ − µ|ku¹_zk = 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 kθe − zk = |θ − z₁| ≤ δ and kµe − zk ≤ δ as well, i.e, θe, µe ∈ B(z, δ). It then follows from (14) that

|f (θ) − f (µ)| =

f (θ) − f (µ) 0

= kf^L^θ(θe) − f^L^θ(µe)k ≤ kkθe − µek = k|θ − µ|.

This means f is strictly continuous at λ_i(z) = z₁ for i = 1, 2. 2

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Remark 2.1. As mentioned in the Section of Introduction, the strict continuity between f^L^θ and f has been given in [3, Theorem 3.5]. Here we provide an alternative proof, since our analysis technique is different from that in [3, Theorem 3.5]. In particular, we achieve an estimate regarding k∇(f^v)^L^θk via its eigenvalues, which maybe have other applications.

According to Lemma 2.1 and Theorem 2.1, we obtain the following result immediately.

Theorem 2.2. Let f : IR → IR and f^L^θ be defined as in (2). Then, f^L^θ is B-differentiable at z if and only if f is B-differentiable at λi(z) for i = 1, 2.

Next, inspired by [2, 19], we further study the H¨older continuity relation between f and f^L^θ.

Theorem 2.3. Let f : IR → IR and f^L^θ be defined as in (2). Then, f^L^θ is H¨older continuous with exponent α ∈ (0, 1] if and only if f is H¨older continuous with exponent α ∈ (0, 1].

Proof. “⇐” Suppose f is H¨older continuous with exponent α ∈ (0, 1]. To proceed the proof, we consider the following two cases.

Case 1: z₂ 6= 0 and y₂ 6= 0. We assume without loss of generality that kz₂k ≥ ky₂k.

Thus,

kf^L^θ(z) − f^L^θ(y)k

= kf (λ₁(z))u¹_z+ f (λ₂(z))u²_z− f (λ₁(y))u¹_y − f (λ₂(y))u²_yk

=

f (λ₁(z))u¹_z− u¹_y + f (λ2(z))u²_z − u²_y

+f (λ₁(z)) − f (λ₁(y))u¹_y+f (λ₂(z)) − f (λ₂(y))u²_y

≤

f (λ1(z))u¹_z− u¹_y + f (λ2(z))u²_z − u²_y + |f (λ₁(z)) − f (λ₁(y))| ·

u¹_y

+ |f (λ₂(z)) − f (λ₂(y))| · u²_y

. Let us analyze each term in the above inequality. First, we look into the first term:

f (λ₁(z))u¹_z− u¹_y + f (λ₂(z))u²_z− u²_y

= tan θ

1 + tan²θ|f (λ₁(z)) − f (λ₂(z))| ·

z₂

kz₂k− y₂ ky₂k

≤ tan θ

1 + tan²θ[f ]_α|λ₁(z) − λ₂(z)|^α

z₂

kz₂k − y₂ ky₂k

= tan θ

1 + tan²θ[f ]_α(tan θ + ctanθ)^αkz₂k^α·

z₂

kz₂k− y₂ ky₂k

≤ tan θ

1 + tan²θ[f ]α(tan θ + ctanθ)^αkz2k^α 2

kz₂kkz2− y2k

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= 2 tan θ

1 + tan²θ[f ]_α(tan θ + ctanθ)^α

z₂− y₂ kz₂k

1−α

kz₂− y₂k^α

≤ tan θ

1 + tan²θ[f ]_α(tan θ + ctanθ)^α2^2−αkz₂− y₂k^α

≤ tan θ

1 + tan²θ[f ]_α(tan θ + ctanθ)^α2^2−αkz − yk^α, (15) where the first inequality is due to the H¨older continuity of f , the second inequality comes from the fact that k(z₂/kz₂k) − (y₂/ky₂k)k ≤ (2/kz₂k)kz₂− y₂k (cf. [2, Lemma 2.2]), and the third inequality follows from the fact that kz2− y₂k ≤ kz₂k + ky₂k ≤ 2kz₂k (since ky₂k ≤ kz₂k). Next, we look into the second term:

|f (λ₁(z)) − f (λ₁(y))|ku¹_yk ≤ [f ]_α|λ₁(z) − λ₁(y)|^α 1

√

1 + ctan²θ

≤ [f ]_α √

2 max{tan θ, ctanθ}α

kz − yk^α. (16) Similarly, the third term also satisfies

|f (λ2(z)) − f (λ2(y))|ku²_yk ≤ [f ]α|λ2(z) − λ2(y)|^α 1

√1 + tan²θ

≤ [f ]_α √

kz − yk^α. (17) Combining (15)-(17) proves that f^L^θ is H¨older continuous with exponent α ∈ (0, 1].

Case 2: either z₂ = 0 or y₂ = 0. In this case, we take uⁱ_z = uⁱ_y for i = 1, 2 according to the spectral decomposition. Therefore, we obtain

kf^L^θ(z) − f^L^θ(y)k

= kf (λ₁(z))u¹_z+ f (λ₂(z))u²_z− f (λ₁(y))u¹_y− f (λ₂(y))u²_yk

=

f (λ₁(z)) − f (λ₁(y))u¹_z +f (λ₂(z)) − f (λ₂(y))u²_z

≤ |f (λ₁(z)) − f (λ₁(y))| · ku¹_zk + |f (λ₂(z)) − f (λ₂(y))| · ku²_zk

≤ [f ]_α|λ₁(z) − λ₁(y)|^α 1

√1 + ctan²θ + [f ]_α|λ₂(y) − λ₂(z)|^α 1

√1 + tan²θ

≤ 2[f ]_α √

kz − yk^α which says f^L^θ is H¨older continuous.

“⇒” Recall that f^L^θ(τ e) = (f (τ ), 0)^T. Hence for any τ, ζ ∈ IR,

|f (τ ) − f (ζ)| =

f^L^θ(τ e) − f^L^θ(ζe)

≤ f^L^θ

α· kτ e − ζek^α

= f^L^θ

α· |τ − ζ|^α which says f is H¨older continuous. 2

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3 ρ-order semismoothness and B-subdifferential for- mula

The property of semismoothness plays an important role in nonsmooth Newton methods [11, 13]. For more information on semismooth functions, see [12, 17, 18, 20]. The relationship of semismooth between f^L^θ and f has been given in [3, Theorem 4.1]. But the exact formula of the B-subdifferential ∂_B(f^L^θ) does not presented. Hence the main aim of this section is twofold: one is establishing the exact formula of B-subdifferential;

another is study the ρ-order semismoothness for ρ > 0.

Lemma 3.1. Define ψ(z) = kzk and Φ(z) = z

kzk for z 6= 0. Then, ψ and Φ are strongly semismooth at z 6= 0.

Proof. Since z 6= 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 [10, Proposition 7.4.5]. 2

The relationship of ρ-order semismoothness between f^L^θ and f is given below. Recall from [16] that in the definition of ρ-order semismooth, we can restrict x + h in (4) belonging to differentiable points.

Theorem 3.1. Let f : IR → IR and f^L^θ be defined as in (2). Given ρ > 0, then the following statements hold.

(a) If f is ρ-order semismooth at λ_i(z) for i = 1, 2, then f^L^θ is min{1, ρ}-order semismooth at z;

(b) If f^L^θ is ρ-order semismooth at z, then f is ρ-semismooth at λ_i(z) for i = 1, 2;

(c) For z₂ = 0, f^L^θ is ρ-semismooth at z if and only if f is ρ-order semismooth at λ_i(z) = z₁ for i = 1, 2.

Proof. (a) Take h ∈ IRⁿ satisfying z + h ∈ D_f_Lθ. We consider the following two cases to complete the proof.

Case 1: For z₂ 6= 0, z₂ + h₂ 6= 0 as h sufficiently close to 0. Since z + h ∈ D_f_Lθ, we know λi(z + h) ∈ Df for i = 1, 2 by Lemma 2.2. Then, according to Lemma 2.1, the first component of

f^L^θ(z + h) − f^L^θ(z) − (f^L^θ)⁰(z + h; h) (18) is expressed as

f (λ₁(z + h))

1 + ctan²θ − f (λ₁(z))

1 + ctan²θ − 1

1 + ctan²θf⁰(λ₁(z + h); h₁ −(z₂+ h₂)^Th₂

kz₂+ h₂k ctanθ) + f (λ₂(z + h))

1 + tan²θ − f (λ₂(z))

1 + tan²θ − 1

1 + tan²θf⁰(λ2(z + h); h1+(z₂+ h₂)^Th₂

kz₂+ h₂k tan θ).

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Because k · k is continuously differentiable over z₂ 6= 0, it is strongly semismooth at z₂ by Lemma 3.1. Therefore,

kz2+ h2k = kz2k + (z₂+ h₂)^Th₂

kz₂+ h₂k + O(kh2k²) = kz2k + (z₂+ h₂)^Th₂

kz₂+ h₂k + O(khk²).

Combining this and the ρ-semismoothness of f at λ₁(z), we have f (λ₁(z + h)) = f (λ₁(z)) + f⁰ λ₁(z + h)

λ₁(z + h) − λ₁(z) + O(|λ₁(z + h) − λ₁(z)|^1+ρ)

= f (λ1(z)) + f⁰ λ1(z + h)

λ1(z + h) − λ1(z) + O(khk^1+ρ)

= f (λ₁(z)) + f⁰ λ₁(z + h)

h₁ − (kz₂ + h₂k − kz₂k)ctanθ + O(khk^1+ρ)

= f (λ₁(z)) + f⁰ λ₁(z + h)

h₁ −(z₂+ h₂)^Th₂ kz₂+ h₂k ctanθ

+ O(khk²) + O(khk^1+ρ)

= f (λ1(z)) + f⁰ λ1(z + h)

h1 −(z₂+ h₂)^Th₂ kz₂+ h₂k ctanθ

+ O

khk^1+min{1,ρ} (19) where the second equation is due to Lemma 2.3 and the firth equation comes from the boundedness of f⁰ since f is strictly continuous at λ₁(z). Similar argument holds for f (λ₂(z + h)). Hence the first component of (18) is O(khk^1+min{1,ρ}).

Next, let us look into the second component of (18), which is involved λ1(z). By Lemma 2.1 again, it can be expressed as:

− ctanθ

1 + ctan²θf (λ₁(z + h)) z₂+ h₂ kz₂+ h₂k + ctanθ

1 + ctan²θf⁰

λ₁(z + h); h₁− (z₂+ h₂)^Th₂ kz₂+ h₂k ctanθ

z₂+ h₂ kz₂+ h₂k + ctanθ

1 + ctan²θf (λ1(z)) z₂

kz₂k + ctanθ 1 + ctan²θ

f (λ₁(z + h))

kz₂+ h₂k M_(z₂_+h₂₎h. (20) Note that Φ is continuous differentiable (and hence is semismooth) with ∇Φ(z₂) =

1

kz₂k(I − z2z^T₂

kz₂k²) and M_(z₂_+h₂₎h = kz₂+ h₂k∇Φ(z₂+ h₂)h₂. Thus, expression (20) can be rewritten as

− ctanθ

1 + ctan²θf (λ1(z + h))Φ(z2+ h2) + ctanθ

1 + ctan²θf⁰

Φ(z₂+ h₂) + ctanθ

1 + ctan²θf (λ₁(z))Φ(z₂) + ctanθ

1 + ctan²θf λ₁(z + h)∇Φ(z₂+ h₂)h₂

= ctanθ

1 + ctan²θ

−f (λ₁(z + h)) + f (λ₁(z)) + f⁰

Φ(z₂ + h₂) +f (λ1(z))ctanθ

1 + ctan²θ [−Φ(z₂+ h₂) + Φ(z₂) + ∇Φ(z₂+ h₂)h₂]

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+ ctanθ

1 + ctan²θ∇Φ(z₂+ h₂)h₂f λ₁(z + h) − f (λ₁(z))

= O

khk^1+min{1,ρ}

+ O(khk²) + O(khk²)

= O

khk^1+min{1,ρ} .

The second equation comes from (19), strongly semismoothness of Φ at z₂, and

∇Φ(z₂+ h₂)h₂h

f λ₁(z + h) − f (λ₁(z))i

= O(khk²),

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 (18) involved λ₂(z). From all the above, we may conclude that

f^L^θ(z + h) − f^L^θ(z) − (f^L^θ)⁰(z + h; h) = O(khk^1+min{1,ρ}), which says f^L^θ 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 6= 0, then the first component of (18) satisfies

1 1 + ctan²θ

h

f (λ₁(z + h)) − f (z₁) − f⁰ λ₁(z + h); h₁− kh₂kctanθi

+ 1

1 + tan²θ h

f (λ₂(z + h)) − f (z₁) − f⁰ λ₂(z + h); h₁+ kh₂k tan θi

= O(|h₁− kh₂kctanθ|^1+ρ) + O(|h₁ + kh₂k tan θ|^1+ρ)

= O(khk^1+ρ) (21)

because f is ρ-order semismooth at z₁. The second component of (18), by letting ¯z₂ = ¯h₂ and M_h₂h = 0, takes the form

− ctanθ

1 + ctan²θf λ₁(z + h)¯h₂+ ctanθ

1 + ctan²θf⁰ λ₁(z + h); h₁− kh₂kctanθ¯h₂ + tan θ

1 + tan²θf λ2(z + h)¯h2− tan θ

1 + tan²θf⁰ λ2(z + h); h1+ kh2k tan θ¯h2

= − 1

tan θ + ctanθ h

f λ₁(z + h) − f (z₁) − f⁰ λ₁(z + h); h₁− kh₂kctanθi¯h₂

+ 1

tan θ + ctanθ h

f λ₂(z + h) − f (z₁) − f⁰ λ₂(z + h); h₁ + kh₂k tan θi¯h₂

= O(khk^1+ρ), (22)

where the last step is due to the ρ-order semismoothness of f .

(b) Suppose f^L^θ is ρ-order semismooth at z. Let t ∈ IR such that f is differentiable at λ₁(z) + t. We discuss the following two cases.

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Case 1: For z₂ 6= 0, from f being Lipschitz at λ₂(z) (and hence the differentiable points are dense near λ₂(z)), there exists β(t) ∈ IR 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 sufficiently small (since λ₂(z) > λ₁(z) by z2 6= 0). Denote h := tu¹_z+ β(t)u²_z. Then, z + h = [λ1(z) + t]u¹_z+ [λ2(z) + β(t)]u²_z 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), f^L^θ is also differentiable at z + h by Lemma 2.2. Notice that

h =h₁ h₂

=







1

1 + ctan²θt + 1

1 + tan²θβ(t)

− ctanθ

1 + ctan²θt + tan θ

1 + tan²θβ(t)

z₂ kz₂k





 and

z₂+ h₂ =

kz₂k − ctanθ

1 + tan²θβ(t)

z₂ kz₂k. Hence,

(z₂+ h₂)^Th₂

kz₂ + h₂k = − ctanθ

1 + tan²θβ(t),

which follows from the fact that kz₂k 6= 0 and t can be arbitrarily small (hence kz₂k −

ctanθ

1+ctan²θt + _1+tan^{tan θ}2θβ(t) > 0). Thus, it is clear that h₁− (z₂+ h₂)^Th₂

kz₂+ h₂k ctanθ

= 1

1 + ctan²θt + 1

1 + tan²θβ(t)

−

− ctanθ

1 + tan²θβ(t) ctanθ

= t and

h₁+ (z₂+ h₂)^Th₂ kz₂+ h₂k tan θ

= 1

1 + ctan²θt + 1

1 + tan²θβ(t) +

− ctanθ

1 + tan²θβ(t) tan θ

= β(t).

In addition, it can be verified that

|h₁| =

1

1 + ctan²θt + 1

1 + tan²θβ(t)

≤ 1

1 + ctan²θ|t| + 1

1 + tan²θ|t| = |t|, since β(t) = O(|t|^1+ρ) ≤ |t| as t is sufficiently small. Similarly,

kh₂k =

− ctanθ

1 + tan²θβ(t)

z₂ kz₂k

=

− ctanθ

1 + tan²θβ(t)

≤ ctanθ

1 + ctan²θ|t| + tan θ

1 + tan²θ|t| ≤ |t|.

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Therefore, we obtain khk = O(t) which further implies O(khk^1+ρ) = O(|t|^1+ρ). Then, by the hypothesis f^L^θ being ρ-order semismooth at z, i.e.,

f^L^θ(z + h) − f^L^θ(z) − (f^L^θ)⁰(z + h; h) = O(khk^1+ρ), we have

f^L^θ(z + h) − f^L^θ(z) − (f^L^θ)⁰(z + h; h), e = O(khk^1+ρ) = O(|t|^1+ρ). (23) In fact, the left-hand side of (23) takes the form of

f λ1(z + h)

1 + ctan²θ +f λ2(z + h)

1 + tan²θ − f (λ₁(z))

1 + ctan²θ − f λ2(z) 1 + tan²θ

−f⁰ λ₁(z + h) 1 + ctan²θ

h₁− (z₂+ h₂)^Th₂ kz₂+ h₂k ctanθ

−f⁰ λ₂(z + h) 1 + tan²θ

h₁ +(z₂+ h₂)^Th₂ kz₂+ h₂k tan θ

= f λ₁(z) + t

1 + ctan²θ +f λ₂(z) + β(t)

1 + tan²θ − f (λ₁(z))

1 + ctan²θ − f λ₂(z) 1 + tan²θ

−f⁰ λ₁(z) + t

1 + ctan²θ t − f⁰ λ₂(z) + β(t) 1 + tan²θ β(t)

= 1

1 + ctan²θ h

f λ₁(z) + t − f λ₁(z) − f⁰ λ₁(z) + tti

+ 1

1 + tan²θ h

f λ₂(z) + β(t) − f (λ2(z)) − f⁰ λ₂(z) + β(t)β(t)i

= 1

1 + ctan²θ h

f λ₁(z) + t − f λ₁(z) − f⁰ λ₁(z) + tti

+ O(|t|^1+ρ), where the last step is due to the fact that f⁰ is bounded and

f (λ₂(z) + β(t)) − f (λ₂(z)) = O(|t|^1+ρ), since f is Lipschitz at λ₂(z). Hence (23) means

f (λ₁(z) + t) − f (λ₁(z)) − f⁰(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 λ_i(z +h) = z₁+t for i = 1, 2, f^L^θ is differentiable at z +h by Lemma 2.2, i.e., z +h ∈ D_f_Lθ. Because f^L^θ is ρ-order semismooth at z, we have

f^L^θ(z + h) − f^L^θ(z) − (f^L^θ)⁰(z + h; h) = O(khk^1+ρ) which, together with the fact khk = |t|, is equivalent to

f (z₁+ t) − f (z₁) − f⁰(z₁+ t)t = O(|t|^1+ρ).