Alternative proofs for some results of vector-valued functions associated with second-order cone

Journal of Nonlinear Analysis and Convex Analysis, vol. 6, pp. 297-325, 2005

Alternative proofs for some results of vector-valued functions associated with second-order cone

Jein-Shan Chen ¹ Department of Mathematics National Taiwan Normal University

Taipei, Taiwan 11677

September 14, 2004 (Revised, February 22, 2005)

Abstract. Let Kⁿ be the Lorentz/second-order cone in IRⁿ. For any function f from IR to IR, one can define a corresponding vector-valued function f^soc(x) on IRⁿ by applying f to the spectral values of the spectral decomposition of x ∈ IRⁿ with respect to Kⁿ. It was shown by J.-S. Chen, X. Chen and P. Tseng that this vector-valued function inher- its from f the properties of continuity, Lipschitz continuity, directional differentiability, Fr´echet differentiability, continuous differentiability, as well as semismoothness. It was also proved by D. Sun and J. Sun that the vector-valued Fischer-Burmeister function as- sociated with second-order cone is strongly semismooth. All proofs for the above results are based on a special relation between the vector-valued function and the matrix-valued function over symmetric matrices. In this paper, we provide a straightforward and in- tuitive way to prove all the above results by using the simple structure of second-order cone and spectral decomposition.

Key words. Second-order cone, vector-valued function, semismooth function, comple- mentarity, spectral decomposition.

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

1 Introduction

The second-order cone (SOC) in IRⁿ, also called the Lorentz cone, is defined to be Kⁿ:= {(x₁, x₂) ∈ IR × IRⁿ⁻¹| kx₂k ≤ x₁},

1E-mail: jschen@math.ntnu.edu.tw, TEL: 886-2-29320206, FAX: 886-2-29332342.

where k · k denotes the Euclidean norm. If n = 1, K¹ is the set of nonnegative reals IR+. Recently, there have been much study on second-order cone in optimization, partic- ularly in the context of applications and solution methods for second-order cone program (SOCP) [1, 2, 9, 12, 14, 18, 22]. For any x = (x₁, x₂) ∈ IR × IRⁿ⁻¹, we can decompose x as

x = λ1u⁽¹⁾+ λ2u⁽²⁾, (1)

where λ₁, λ₂ and u⁽¹⁾, u⁽²⁾ are the spectral values and the associated spectral vectors of x, with respect to Kⁿ, given by

λi = x1+ (−1)ⁱkx2k, (2)

u⁽ⁱ⁾ =







1 2

³1, (−1)ⁱ x₂ kx2k

´, if x₂ 6= 0,

1 2

³1, (−1)ⁱw^´, if x₂ = 0, (3)

for i = 1, 2, with w being any vector in IRⁿ⁻¹ satisfying kwk = 1. If x2 6= 0, the decomposition (1) is unique. With this spectral decomposition, for any function f : IR → IR, the following vector-valued function associated with Kⁿ (n ≥ 1) was considered (see [9]):

f^soc(x) = f (λ1)u⁽¹⁾+ f (λ2)u⁽²⁾ ∀x = (x1, x2) ∈ IR × IRⁿ⁻¹. (4) If f is defined only on a subset of IR, then f^soc is defined on the corresponding subset of IRⁿ. The definition (4) is unambiguous whether x₂ 6= 0 or x₂ = 0. The above definition (4) is analogous to one associated with the semidefinite cone Sⁿ, see [20, 21].

The study of this function is motivated by second-order cone complementarity prob- lem (SOCCP), see [3, 4] and references therein. In fact, in the paper [4], it studied the continuity and differentiability properties of the vector-valued function f^soc. In partic- ular, it showed that the properties of continuity, strict continuity, Lipschitz continuity, directional differentiability, differentiability, continuous differentiability, and (ρ-order) semismoothness are each inherited by f^soc from f . These results parallel those obtained recently in [5] for matrix-valued functions and are useful in the design and analysis of smoothing and nonsmooth methods for solving SOCP and SOCCP. The proofs are based on an elegant relation between the vector-valued function f^soc and its matrix-valued coun- terpart (see Lemma 4.1 of [4]). This relation enables applying the results from [5] for matrix-valued functions to the vector-valued function f^soc. In this paper, we study an intuitive way to prove all the aforementioned results without using the relation as will be seen in Sec. 3.

A popular approach to solving SOCCP is to reformulate it as an unconstrained min- imization problem. Specifically, it is to find a smooth (continuously differentiable) func- tion ψ : IRⁿ× IRⁿ→ IR₊ such that

ψ(x, y) = 0 ⇐⇒ hx, yi = 0, x ∈ Kⁿ, y ∈ Kⁿ, (5)

which yields that the SOCCP can be expressed as an unconstrained smooth (continuously differentiable) minimization problem: min

ζ∈IRⁿf (ζ) := ψ(F (ζ), G(ζ)), for some F and G. For detailed reformulation, please refer to [3]. Such a ψ is usually called a merit function. A popular choice of ψ is

ψ(x, y) = 1

2kφ(x, y)k² x, y ∈ IRⁿ, (6)

where

φ(x, y) := (x²+ y²)^1/2− x − y. (7) Here (·)² and (·)^1/2 are well-defined via the Jordan product as will be explained in Sec.

2. The function φ is called Fischer-Burmeister function. It is the natural extension of Fischer-Burmeister function over IRⁿ to Kⁿ. D. Sun and J. Sun proved that φ is strongly semismooth in [19], while ψ was proved smooth (continuously differentiable) everywhere by J.-S. Chen and P. Tseng in [3]. In this paper, we also provide an alternative proof for property of strong semismoothness of φ in Sec. 4.

In what follows, for any differentiable (in the Fr´echet sense) mapping F : IRⁿ→ IR^m, we denote its Jacobian(not transposed) at x ∈ IRⁿ by ∇F (x) ∈ IR^m×n, i.e., (F (x + u) − F (x) − ∇F (x)u)/kuk → 0 as u → 0. “ := ” means “define”. We write z = O(α) (respectively, z = o(α)), with α ∈ IR and z ∈ IRⁿ, to mean kzk/|α| is uniformly bounded (respectively, tends to zero) as α → 0.

2 Basic Concepts and Known Results

In this section, we review some basic materials regarding vector-valued functions. These contain continuity, (local) Lipschitz continuity, directional differentiability, differentiabil- ity, continuous differentiability, as well as (ρ-order) semismoothness. We also recall some known results for vector-valued functions for which we will provide alternative proofs later.

Let the mapping F : IRⁿ → IR^m. Then F is continuous at x ∈ IRⁿ if F (y) → F (x) as y → x; and F is continuous if F is continuous at every x ∈ IRⁿ. We say F is strictly continuous (also called ‘locally Lipschitz continuous’) at x ∈ IRⁿ if there exist scalars κ > 0 and δ > 0 such that

kF (y) − F (z)k ≤ κky − zk ∀y, z ∈ IRⁿ with ky − xk ≤ δ, kz − xk ≤ δ;

and F is strictly continuous if F is strictly continuous at every x ∈ IRⁿ. We say F is directionally differentiable at x ∈ IRⁿ if

F⁰(x; h) := lim

t→0⁺

F (x + th) − F (x)

t exists ∀h ∈ IRⁿ;

and F is directionally differentiable if F is directionally differentiable at every x ∈ IRⁿ. F is differentiable (in the Fr´echet sense) at x ∈ IRⁿ if there exists a linear mapping

∇F (x) : IRⁿ → IR^m such that

F (x + h) − F (x) − ∇F (x)h = o(khk).

If F is differentiable at every x ∈ IRⁿ and ∇F is continuous, then F is continuously differentiable. We notice that, in the above expression about strict continuity of F , if δ can be taken to be ∞, then F is called Lipschitz continuous with Lipschitz constant κ.

It is well-known that if F is strictly continuous, then F is almost everywhere differ- entiable by Rademacher’s Theorem–see [6] and [17, Sec. 9J]. In this case, the generalized Jacobian ∂F (x) of F at x (in the Clarke sense) can be defined as the convex hull of the generalized Jacobian ∂_BF (x), where

∂BF (x) :=

½

xlim^j→x∇F (x^j)|F is differentiable at x^j ∈ IRⁿ

¾

.

The notation ∂_B is adopted from [15]. In [17, Chap. 9], the case of m = 1 is considered and the notations “ ¯∇” and “ ¯∂” are used instead of, respectively, “∂B” and “∂”. Assume F : IRⁿ → IR^m is strictly continuous, then F is said to be semismooth at x if F is directionally differentiable at x and, for any V ∈ ∂F (x + h), we have

F (x + h) − F (x) − V h = o(khk).

Moreover, F is called ρ-order semismooth at x (0 < ρ < ∞) if F is semismooth at x and, for any V ∈ ∂F (x + h), we have

F (x + h) − F (x) − V h = O(khk^1+ρ).

The following lemma, proven by D. Sun and J. Sun [20, Thm. 3.6] using the defini- tion of generalized Jacobian, enables one to study the semismooth property of f^soc by examining only those points x ∈ IRⁿwhere f^soc is differentiable and thus work only with the Jacobian of f^soc, rather than the generalized Jacobian. It is a very useful working lemma for verifying semismoothness property.

Lemma 2.1 Suppose F : IRⁿ→ IRⁿ is strictly continuous and directionally differentiable in a neighborhood of x ∈ IRⁿ. Then, for any 0 < ρ < ∞, the following two statements are equivalent:

(a) For any v ∈ ∂F (x + h) and h → 0,

F (x + h) − F (x) − vh = o(khk) (respectively, O(khk)^1+ρ).

(b) For any h → 0 such that F is differentiable at x + h,

F (x + h) − F (x) − ∇F (x + h)h = o(khk) (respectively, O(khk)^1+ρ).

We say F is semismooth (respectively, ρ-order semismooth) if F is semismooth (re- spectively, ρ-order semismooth) at every x ∈ IRⁿ. We say F is strongly semismooth if it is 1-order semismooth. Convex functions and piecewise continuously differentiable functions are examples of semismooth functions. The composition of two (respectively, ρ-order) semismooth functions is also a (respectively, ρ-order) semismooth function. The prop- erty of semismoothness, as introduced by Mifflin [13] for functionals and scalar-valued functions and further extended by L. Qi and J. Sun [16] for vector-valued functions, is of particular interest due to the key role it plays in the superlinear convergence analysis of certain generalized Newton methods [10, 11, 15, 16, 23]. For extensive discussions of semismooth functions, see [8, 13, 16].

For any x = (x₁, x₂) ∈ IR × IRⁿ⁻¹ and y = (y₁, y₂) ∈ IR × IRⁿ⁻¹, we define their Jordan product as

x ◦ y =^³x^Ty, y₁x₂+ x₁y₂^´. (8) We write x² to mean x ◦ x and write x + y to mean the usual componentwise addition of vectors. Then, ◦, +, together with e = (1, 0, . . . , 0) ∈ IRⁿ, give rise to a Jordan algebra associated with Kⁿ[7, Chap. II]. If x ∈ Kⁿ, then there exists a unique vector in Kⁿ, which we denote by x^1/2, such that (x^1/2)² = x^1/2◦ x^1/2 = x. For any x = (x₁, x₂) ∈ IR × IRⁿ⁻¹, we also define the symmetric matrix

L_x =

"

x₁ x^T₂ x₂ x₁I

#

, (9)

viewed as a linear mapping from IRⁿ to IR^n×n. The matrix L_x has various interesting properties that were studied in [9]. Especially, we have L_x· y = x ◦ y for any x, y ∈ IRⁿ. Now, we summarize the results shown in [4] for which we will provide alternative proofs that are straightforward and intuitive in the subsequent sections.

Proposition 2.2 For any f : IR → IR, the following results hold:

(a) f^soc is continuous at an x ∈ IRⁿ with eigenvalues λ₁, λ₂ if and only if f is continuous at λ₁, λ₂.

(b) f^soc is directionally differentiable at an x ∈ IRⁿ with eigenvalues λ₁, λ₂ if and only if f is directionally differentiable at λ₁, λ₂

(c) f^soc is differentiable at an x ∈ IRⁿ with eigenvalues λ₁, λ₂ if and only if f is differ- entiable at λ₁, λ₂.

(d) f^soc is continuously differentiable at an x ∈ IRⁿ with eigenvalues λ₁, λ₂ if and only if f is continuously differentiable at λ1, λ2.

(e) f^soc is strictly continuous at an x ∈ IRⁿ with eigenvalues λ₁, λ₂ if and only if f is strictly continuous at λ₁, λ₂.

(f) f^soc is Lipschitz continuous (with respect to k · k) with constant κ if and only if f is Lipschitz continuous with constant κ.

(g) f^soc is semismooth if and only if f is semismooth.

Proposition 2.3 The vector-valued Fischer-Burmeister function associated with second- order cone defined as (7) is strongly semismooth.

3 Alternative Proofs of Continuity and Differentia- bility

In this section, we present alternative proofs for Prop. 2.2 of Sec. 2 which is one of the main purposes of this paper. Unlike the existed proofs which employed an elegant lemma ([4, Lem. 4.1]), our arguments come from an intuitive way only using the simple structure of second-order cone and basic definitions. We need some technical lemmas before starting the alternative proofs.

Lemma 3.1 Let λ₁ ≤ λ₂ be the spectral values of x ∈ IRⁿ and m₁ ≤ m₂ be the spectral values of y ∈ IRⁿ. Then we have

|λ₁− m₁|²+ |λ₂− m₂|² ≤ 2 kx − yk² , (10) and hence, |λ_i− m_i| ≤√

2 kx − yk , ∀i = 1, 2.

Proof. The proof follows from a direct computation. 2

Lemma 3.2 Let x = (x₁, x₂) ∈ IR × IRⁿ⁻¹ and y = (y₁, y₂) ∈ IR × IRⁿ⁻¹. (a) If x₂ 6= 0, y₂ 6= 0, then we have

ku⁽ⁱ⁾− v⁽ⁱ⁾k ≤ 1

kx₂kkx − yk , ∀i = 1, 2 , (11) where u⁽ⁱ⁾, v⁽ⁱ⁾ are the unique spectral vectors of x and y, respectively.

(b) If either x₂ = 0 or y₂ = 0, then we can choose u⁽ⁱ⁾, v⁽ⁱ⁾ such that the left hand side of inequality (11) is zero.

Proof. (a) From the spectral factorization (1)-(3), we know that u⁽ⁱ⁾ = 1

2

µ

1 , (−1)ⁱ x₂ kx₂k

¶

, v⁽ⁱ⁾ = 1 2

µ

1 , (−1)ⁱ y₂ ky₂k

¶

,

where uⁱ⁾, v⁽ⁱ⁾ are unique. Thus, we have u⁽ⁱ⁾− v⁽ⁱ⁾ = ¹₂

µ

0 , (−1)ⁱ(_kx^x2

2k − _ky^y2

2k)

¶

. Then

ku⁽ⁱ⁾− v⁽ⁱ⁾k = 1 2

°°

°°

°

x2

kx₂k − y2

ky₂k

°°

°°

°= 1 2

°°

°°

°

x2− y2

kx₂k +(ky2k − kx2k)y2

kx₂k · ky₂k

°°

°°

°

≤ 1 2

µ 1

kx₂kkx₂− y₂k + 1 kx₂k

¯¯

¯¯ky₂k − kx₂k

¯¯

¯¯

¶

≤ 1 2

µ 1

kx₂kkx₂− y₂k + 1

kx₂kkx₂− y₂k

¶

≤ 1

kx₂kkx − yk, where the first inequality follows from the triangle inequality.

(b) We can choose the same spectral vectors for x and y by the spectral factorization (1)-(3) since either x₂ = 0 or y₂ = 0. Then, it is obvious. 2

Lemma 3.3 For any w 6= 0 ∈ IRⁿ, we have ∇_w

µ w kwk

¶

= 1

kwk

µ

I − ww^T kwk²

¶

.

Proof. The verification is routine, so we omit it. 2

Now, we are ready to present our alternative proofs for Prop. 2.2. As mentioned, all of our proofs are from intuitive definitions as well as the structure of second-order cone.

Some portion of the proofs are similar to the original ones, we omit them when there is the case.

Proof. (a) “⇐” Suppose f is continuous at λ1, λ2. For any fixed x ∈ IRⁿand y → x, let the spectral factorizations of x, y be x = λ₁u⁽¹⁾+ λ₂u⁽²⁾ and y = m₁v⁽¹⁾+ m₂v⁽²⁾. Then, we discuss two cases.

Case (i): If x₂ 6= 0, then we have

f^soc(y) − f^soc(x) (12)

= f (m₁)

µ

v⁽¹⁾− u⁽¹⁾

¶

+

µ

f (m₁) − f (λ₁)

¶

u⁽¹⁾+ f (m₂)

µ

v⁽²⁾− u⁽²⁾

¶

+

µ

f (m₂) − f (λ₂)

¶

u⁽²⁾. Since f is continuous at λ₁, λ₂, and from Lemma 3.1, |m_i− λ_i| ≤√

2 ky − xk, we obtain f (m_i) −→ f (λ_i) as y → x. Also by Lemma 3.2, we know that kv⁽ⁱ⁾− u⁽ⁱ⁾k −→ 0 as y →

x. Thus, equation (12) yields f^soc(y) −→ f^soc(x) as y → x, since both f (m_i) and ku⁽ⁱ⁾k are bounded. Hence, f^soc is continuous at x ∈ IRⁿ .

Case (ii): If x₂ = 0, no matter y₂ is zero or not, we can arrange that x, y have the same spectral vectors. Thus, f^soc(y) − f^soc(x) =

µ

f (m₁) − f (λ₁)

¶

u⁽¹⁾+

µ

f (m₂) − f (λ₂)

¶

u⁽²⁾ . Then, f^soc is continuous at x ∈ IRⁿ by similar arguments.

“⇒” The proof for this direction is straightforward and similar to the arguments in [4, Prop. 5.2]. 2

Proof. (b) “⇐” Suppose f is directionally differentiable at λ₁, λ₂. Fix any x = (x₁, x₂) ∈ IR × IRⁿ⁻¹, then we discuss two cases as below.

Case (i): If x₂ 6= 0, then we have f^soc(x) = f (λ₁)u⁽¹⁾+f (λ₂)u⁽²⁾where λ_i = x₁+(−1)ⁱkx₂k and u⁽ⁱ⁾ = 1

2

µ

1 , (−1)ⁱ x₂ kx₂k

¶

for all i = 1, 2. From Lemma 3.3, we know that u⁽ⁱ⁾ is Fr´echet-differentiable with respect to x, with

∇_xu⁽ⁱ⁾ = (−1)ⁱ 2kx₂k





0 0

0 I − x₂x^T₂ kx₂k²



 , ∀i = 1, 2. (13)

Also by the expression of λi, we know that λi is Fr´echet-differentiable with respect to x, with

∇_xλ_i =

µ

1 , (−1)ⁱ x₂ kx₂k

¶

= 2u⁽ⁱ⁾ , ∀i = 1, 2. (14) Since f is directionally differentiable at λ₁, λ₂, then the chain rule and product rule for directional differentiation give

(f^soc)⁰(x; h)

= f (λ₁)∇_xu⁽¹⁾h + u⁽¹⁾f⁰(λ₁; h)(∇_xλ₁)^T + f (λ₂)∇_xu⁽²⁾h + u⁽²⁾f⁰(λ₂; h)(∇_xλ₂)^T

= f (λ₂) − f (λ₁) 2kx2k





0 0

0 I − x₂x^T₂ kx₂k²



h + 2f⁰(λ₁; h)u⁽¹⁾(u⁽¹⁾)^T + 2f⁰(λ₂; h)u⁽²⁾(u⁽²⁾)^T

= f (λ₂) − f (λ₁) λ₂− λ₁





0 0

0 I − x₂x^T₂ kx₂k²



h + 2f⁰(λ1; h)u⁽¹⁾(u⁽¹⁾)^T + 2f⁰(λ2; h)u⁽²⁾(u⁽²⁾)^T,

where the second equality uses equations (13) and (14), and the last equality employs the fact that λ₂− λ₁ = 2kx₂k. Notice that we can obtain u⁽ⁱ⁾(u⁽ⁱ⁾)^T as follows by direct computation :

u⁽ⁱ⁾(u⁽ⁱ⁾)^T = 1 4







1 (−1)ⁱ x^T₂ kx₂k (−1)ⁱ x₂

kx₂k

x₂x^T₂ kx₂k²





 , ∀i = 1, 2.

Now let

˜a = f (λ₂) − f (λ₁)

λ₂− λ₁ h , ˜b = f⁰(λ₂; h) + f⁰(λ₁; h)

2 , ˜c = f⁰(λ₂; h) − f⁰(λ₁; h)

2 . (15)

Then, we can rewrite the previous expression of (f^soc)⁰(x; h) as

(f^soc)⁰(x; h) = ˜a





0 0

0 I − x₂x^T₂ kx2k²



+









˜b ˜cx^T₂ kx₂k

˜cx2

kx₂k

˜bx2x^T₂ kx₂k²







=







˜b ˜cx^T₂ kx₂k

˜cx₂

kx2k ˜aI + (˜b − ˜a)_kx^x2xT2

2k²





 .

(16) This enables that f^soc is directionally differentiable at x when x₂ 6= 0 with (f^soc)⁰(x; h) being in form of (16).

Case (ii): If x2 = 0, we compute the directional derivative (f^soc)⁰(x; h) at x for any direction h by definition. Let h = (h₁, h₂) ∈ IR × IRⁿ⁻¹. We have two subcases. First, consider the subcase of h₂ 6= 0. From the spectral factorization, we can choose u⁽ⁱ⁾ =

1 2

µ

1 , (−1)^{i h}_kh²_2k

¶

for all i = 1, 2, such that

( f^soc(x + th) = f (λ + 4λ1)u⁽¹⁾+ f (λ + 4λ2)u⁽²⁾ f^soc(x) = f (λ)u⁽¹⁾+ f (λ)u⁽²⁾,

where λ = x₁ and 4λ_i = t

µ

h₁+ (−1)ⁱkh₂k

¶

for all i = 1, 2. Thus, we obtain

f^soc(x + th) − f^soc(x) =

µ

f (λ + 4λ₁) − f (λ)

¶

u⁽¹⁾+

µ

f (λ + 4λ₂) − f (λ)

¶

u⁽²⁾. The fact that

t→0lim⁺

f (λ + 4λ₁) − f (λ)

t = lim

t→0⁺

f (λ + t(h₁− kh₂k)) − f (λ)

t = f⁰(λ; h₁− kh₂k) , and

t→0lim⁺

f (λ + 4λ₂) − f (λ)

t = lim

t→0⁺

f (λ + t(h₁+ kh₂k)) − f (λ)

t = f⁰(λ; h₁+ kh₂k), yields

t→0lim⁺

f^soc(x + th) − f^soc(x)

t = lim

t→0⁺

f (λ + 4λ₁) − f (λ)

t u⁽¹⁾+ lim

t→0⁺

f (λ + 4λ₂) − f (λ)

t u⁽²⁾

= f⁰(λ; h1− kh2k)u⁽¹⁾+ f⁰(λ; h1+ kh2k)u⁽²⁾ , (17) where u⁽¹⁾ = 1

2

µ

1, − h₂ kh₂k

¶

, u⁽²⁾ = 1 2

µ

1, h₂ kh₂k

¶

. Hence, (f^soc)⁰(x; h) exists with form of (17).

Secondly, for the subcase of h₂ = 0, the same argument applies except h₂/kh₂k is replaced by any w ∈ IRⁿ⁻¹ with kwk = 1, i.e., choosing u⁽ⁱ⁾ = 1

2(1, (−1)ⁱw), for all i = 1, 2.

Analogously,

t→0lim⁺

f^soc(x + th) − f^soc(x)

t = f⁰(λ; h1)u⁽¹⁾+ f⁰(λ; h1)u⁽²⁾. (18) Hence, (f^soc)⁰(x; h) exists with form of (18). From the above, it shows that f^soc is di- rectionally differentiable at x when x2 = 0 and its directional derivative (f^soc)⁰(x; h) is either in form of (17) or (18).

“⇒” Suppose f^soc is directionally differentiable at x ∈ IRⁿ with spectral values λ₁, λ₂, we will prove that f is directionally differentiable at λ₁, λ₂. For λ₁ ∈ IR and any direction d₁ ∈ IR, let h := d₁u⁽¹⁾+ 0u⁽²⁾ where x = λ₁u⁽¹⁾+ λ₂u⁽²⁾. Then, x + th = (λ₁+ td₁)u⁽¹⁾+ λ2u⁽²⁾, and

f^soc(x + th) − f^soc(x)

t = f (λ₁+ td₁) − f (λ₁) t u⁽¹⁾.

Since f^soc is directionally differentiable at x, the above equation yields that f⁰(λ1; d1) = lim

t→0⁺

f (λ₁+ td₁) − f (λ₁)

t exists.

This means f is directionally differentiable at λ₁. Similarly, f is also directionally differ- entiable at λ2. 2

Proof. (c) “⇐” The proof of this direction is identical to the proof shown as in (b), but with “directionally differentiable” replaced by “differentiable”. We omit the proof and only present the formula of f^soc(x) as below. For x₂ 6= 0, we have

∇f^soc(x) =







b cx^T₂

kx₂k cx2

kx₂k aI + (b − a)x2x^T₂ kx₂k²





, (19)

where

a = f (λ₂) − f (λ₁)

λ₂− λ₁ , b = f⁰(λ₂) + f⁰(λ₁)

2 , c = f⁰(λ₂) − f⁰(λ₁)

2 . (20)

If x₂ = 0, then

∇f^soc(x) = f⁰(λ)I. (21)

“⇒” Let f^soc be Fr´echet-differentiable at x ∈ IRⁿ with spectral eigenvalues λ1, λ2, we will show that f is Fr´echet-differentiable at λ₁, λ₂. Suppose not, then f is not Fr´echet- differentiable at λ_i for some i ∈ {1, 2}. Thus, either f is not directionally differentiable at λ_i or, if it is, the right- and left-directional derivatives of f at λ_i are unequal. In

either case, this implies that there exist two sequences of non-zero scalars t^ν and τ^ν, ν = 1, 2, . . . , converging to zero, such that the limits

ν→∞lim

f (λ_i+ t^ν) − f (λ_i)

t^ν , lim

ν→∞

f (λ_i+ τ^ν) − f (λ_i) τ^ν

exist (possible ∞ or −∞) and either are unequal or both equal to ∞ or are both equal to −∞. Now for any x = λ₁u⁽¹⁾ + λ₂u⁽²⁾, let h := 1 · u⁽¹⁾ + 0 · u⁽²⁾ = u⁽¹⁾. Then, x + th = (λ₁+ t)u⁽¹⁾+ λ₂u⁽²⁾ and f^soc(x + th) = f (λ₁+ t)u⁽¹⁾+ f (λ₂)u⁽²⁾. Thus,

ν→∞lim

f^soc(x + t^νh) − f^soc(x)

t^ν = lim

ν→∞

f^soc(λ₁ + t^ν) − f^soc(λ₁)

t^ν u⁽¹⁾

ν→∞lim

f^soc(x + τ^νh) − f^soc(x)

τ^ν = lim

ν→∞

f^soc(λ₁ + τ^ν) − f^soc(λ₁)

τ^ν u⁽¹⁾ .

It follows that these two limits either are unequal or are both non-finite. This implies that f^soc is not Fr´echet-differentiable at x where is a contradiction. 2

Proof. (d) “⇐” Suppose f is continuously differentiable. From equation (19), it can been seen that ∇f^soc is continuous at every x with x₂ 6= 0. It remains to show that ∇f^soc is continuous at every x with x₂ = 0. Fix any x = (x₁, 0) ∈ IRⁿ, so λ₁ = λ₂ = x₁. Then, from equation (20), we have

y→xlima = lim

y→x

f (λ₂) − f (λ₁)

λ₂− λ₁ = f⁰(x1)

y→xlimb = lim

y→x

1

2(f⁰(λ₂) + f⁰(λ₁)) = f⁰(x₁)

y→xlimc = lim

y→x

1

2(f⁰(λ₂) − f⁰(λ₁)) = 0 . Taking the limit in equation (19) as y → x yields lim

y→x∇f^soc(y) = f⁰(x1)I = ∇f^soc(x), which says ∇f^soc is continuous at every x ∈ IRⁿ .

“⇒” The proof for this direction is similar to the original proof of [4, Prop. 5.4], so we omit it. 2

Proof. (e) and (f) The original proofs in [4] use the working Lemma 2.1 directly which is the same idea as the one used for the whole paper, so the proofs for part(e) and (f) are identical to theirs. We therefore omit them. 2

Proof. (g) “⇒” Suppose f^soc is semismooth, then f^soc is strictly continuous and di- rectionally differentiable. By part (b) and (e), f is strictly continuous and directionally differentiable. Now, for any α ∈ IR and any η ∈ IR such that f is differentiable at α + η, part (c) yields that f^soc is differentiable at x + h, where x := (α, 0) ∈ IR × IRⁿ⁻¹ and

h := (η, 0) ∈ IR × IRⁿ⁻¹. Hence, we can choose the same spectral vectors for x + h and x

such that ₍

f^soc(x + h) = f (α + η)u⁽¹⁾+ f (α + η)u⁽²⁾, f^soc(x) = f (α)u⁽¹⁾+ f (α)u⁽²⁾.

Since f^soc is semismooth, by Lemma 2.1, we have

f^soc(x + h) − f^soc(x) − ∇f^soc(x + h)h = o(khk). (22) On the other hand, (21) says ∇f^soc(x+h)h = f⁰(α+η)Ih =

µ

f⁰(α+η)η , 0

¶

. Plugging this into equation (22), it yields that f (α + η) − f (α) − f⁰(α + η)η = o(|η|). Thus, by Lemma 2.1 again, it says that f is semismooth at α. Since α is arbitrary, f is semismooth.

“⇐” Suppose f is semismooth, then f is strictly continuous and directionally differen- tiable. By part (b) and (c), f^soc is strictly continuous and directionally differentiable.

For any x ∈ IRⁿ and h ∈ IRⁿ such that f^soc is differentiable at x + h, we will verify that f^soc(x + h) − f^soc(x) − ∇f^soc(x + h)h = o(khk).

Case (i): If x₂ 6= 0, let λ_i be the spectral eigenvalues of x and u⁽ⁱ⁾ be the associated vectors. We denote x + h by z for convenience, i.e., z := x + h and let mi be the spectral values of z with the associated vectors v⁽ⁱ⁾. Hence, we have

( f^soc(x) = f (λ₁)u⁽¹⁾+ f (λ₂)u⁽²⁾, f^soc(x + h) = f (m₁)v⁽¹⁾+ f (m₂)v⁽²⁾. In addition, from (19), we know

∇f^soc(x + h) =







ˆb ˆcz₂^T kz₂k ˆcz₂

kz₂k ˆaI − (ˆb − ˆa)z₂z₂^T kz₂k²





 ,

where

ˆa = f (m₂) − f (m₁)

m₂− m₁ , ˆb = f⁰(m₂) + f⁰(m₁)

2 , ˆc = f⁰(m₂) − f⁰(m₁)

2 .

With this, we can write out f^soc(x+h)−f^soc(x)−∇f^soc(x+h)h := (Ξ₁, Ξ₂) where Ξ₁ ∈ IR and Ξ₂ ∈ IRⁿ⁻¹. Since the expansion is very long, for simplicity, we denote Ξ₁ be the first component and Ξ2 be the second component of the expansion. We will show that Ξ₁ and Ξ₂ are both o(khk).

First, we compute the first component Ξ₁: Ξ₁ = 1

2

(

f (m₁) − f (λ₁) − f⁰(m₁)(h₁−z^T₂h₂ kz₂k)

)

+ 1 2

(

f (m₂) − f (λ₂) − f⁰(m₂)(h₁+ z₂^Th₂ kz₂k)

)

= 1 2

½

f (m1) − f (λ1) − f⁰(m1)

µ

h1− (kz2k − kx2k)

¶

+ o(khk)

¾

+1 2

½

f (m2) − f (λ2) − f⁰(m2)

µ

h1+ (kz2k − kx2k)

¶

+ o(khk)

¾

= o

µ

h1− (kz2k − kx2k)

¶

+ o(khk) + o

µ

h1+ (kz2k − kx2k)

¶

+ o(khk).

In the above expression of Ξ1, the third equality holds since the following:

z^T₂h₂

kz2k = z^T₂(z₂− x₂)

kz2k = kz₂k − kz₂kkx₂k kz2k cos θ

= kz₂k − kx₂k

µ

1 + O(θ²)

¶

= kz₂k − kx₂k

µ

1 + O(khk²)

¶

= kz₂k − kx₂k

µ

1 + o(khk)

¶

,

where θ is the angle between x₂ and z₂ and note that z₂−x₂ = h₂ gives O(θ²) = O(khk²).

Also the last equality in expression of Ξ₁ holds since f is semismooth and m_i− λ_i = h₁+ (−1)ⁱ(kz₂k − kx₂k).

On the other hand, due to h₁+(−1)ⁱ(kz₂k−kx₂k) ≤ h₁+kz₂−x₂k = h₁+kh₂k , we observe that when khk → 0 then h1 + (−1)ⁱ(kz2k − kx2k) → 0 and h1+ (−1)ⁱ(kz2k − kx2k) = O(khk). Thus, o

µ

h₁ + (−1)ⁱ(kz₂k − kx₂k)

¶

= o(khk), which yields the first component Ξ₁ is o(khk).

Now consider the second component Ξ₂: Ξ2 = −1

2f (m1) z2

kz₂k +1

2f (m2) z2

kz₂k+ 1

2f (λ1) x2

kx₂k −1

2f (λ2) x2

kx₂k

−1 2

µ

f⁰(m₂) − f⁰(m₁)

¶z₂h₁

kz₂k − f (m₂) − f (m₁) m₂− m₁ h₂

−

(1 2

µ

f⁰(m2) − f⁰(m1)

¶

− f (m2) − f (m1) m₂− m₁

)z2z₂^Th2

kz₂k²

= −1 2

(

f (m₁) z₂

kz₂k − f (λ₁) x₂

kx₂k − f⁰(m₁)z₂h₁

kz₂k − 2f (m₁) m₂− m₁h₂ +

µ

f⁰(m₁) + 2f (m₁) m2− m1

¶z₂z₂^Th₂ kz2k²

)

+1 2

(

f (m₂) z₂

kz₂k − f (λ₂) x₂

kx₂k − f⁰(m₂)z₂h₁

kz₂k− 2f (m₂) m₂ − m₁h₂

−

µ

f⁰(m₂) + 2f (m₂) m₂− m₁

¶z₂z₂^Th₂ kz₂k²

)

:= Ξ⁽¹⁾₂ + Ξ⁽²⁾₂ ,

where Ξ⁽¹⁾₂ denotes the first half part while Ξ⁽²⁾₂ denotes the second half part. We will show that both Ξ⁽¹⁾₂ and Ξ⁽²⁾₂ are o(khk). For symmetry, it is enough to show that Ξ⁽¹⁾₂ is o(khk). From the observations that







z2 = x2+ h2, m₂− m₁ = 2kz₂k,

m_i− λ_i = h₁ + (−1)ⁱ(kz₂k − kx₂k) = O(khk), we have the following:

Ξ⁽¹⁾₂ = −1 2

(

f (m₁) z2

kz₂k − f (λ₁) x2

kx₂k − f⁰(m₁)z2h1

kz₂k − 2f (m1) m₂− m₁h₂ +

µ

f⁰(m1) + 2f (m1) m₂− m₁

¶z2z₂^Th2

kz₂k²

)

= −1 2

z₂ kz₂k

(

f (m₁) − f (λ₁) − f⁰(m₁)

µ

h₁− z₂^Th₂ kz₂k

¶)

+1 2

µ

f (m₁) − f (λ₁)

¶µ−h₂

kz2k + z₂z₂^Th₂ kz2k³

¶

−1 2f (λ₁)

µ z2

kz₂k − x2

kx₂k− h2

kz₂k + z2z₂^Th2

kz₂k³

¶

.

Following the same arguments as in the first component Ξ1, it can be seen that f (m₁) − f (λ₁) − f⁰(m₁)

µ

h₁− z₂^Th₂ kz₂k

¶

= o(khk).

Since m₁− λ₁ = h₁− (kz₂k − kx₂k) = O(khk) and f is strictly continuous, then f (m₁) − f (λ₁) = O(khk). In addition, −h₂/kz₂k + z₂z₂^Th₂/kz₂k³ = O(khk). Hence,

µ

f (m₁) − f (λ₁)

¶µ−h₂

kz₂k +z₂z^T₂h₂ kz₂k³

¶

= O(khk²) = o(khk) .

Therefore, it remains to show that the last part of Ξ⁽¹⁾₂ is o(khk). Now, note that z₂

kz2k − x₂

kx2k − h₂

kz2k+ z₂z₂^Th₂ kz2k³ = x₂

µ 1

kz2k− 1

kx2k + z^T₂h₂ kz2k³

¶

+ O(khk²) .

Let θ(z₂) := −1/kz₂k, then ∇θ(z₂) = − −1 kz2k²

z₂

kz2k = z₂

kz2k³. This implies that 1

kz2k − 1

kx2k+ z₂^Th₂

kz2k³ = θ(x₂) − θ(z₂) − ∇θ(z₂)(x₂− z₂) = O(khk²), where the last equality is from first Taylor approximation. Thus, we obtain

f (λ₁)

µ z₂

kz₂k − x₂

kx₂k− h₂

kz₂k + z₂z₂^Th₂ kz₂k³

¶

= o(khk) .