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 K** ^{n}* be the Lorentz/second-order cone in IR

^{n}*. For any function f from IR*

*to IR, one can define a corresponding vector-valued function f*

^{soc}

*(x) on IR*

*by applying*

^{n}*f to the spectral values of the spectral decomposition of x ∈ IR*

^{n}*with respect to K*

*. It was shown by J.-S. Chen, X. Chen and P. Tseng that this vector-valued function inher-*

^{n}*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 IR* ^{n}*, also called the Lorentz cone, is defined to be

*K*

^{n}*:= {(x*

_{1}

*, x*

_{2}

*) ∈ IR × IR*

^{n−1}*| kx*

_{2}

*k ≤ x*

_{1}

*},*

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, K*^{1} 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 = (x*_{1}*, x*_{2}*) ∈ IR × IR*^{n−1}*, we can decompose x*
as

*x = λ*1*u*^{(1)}*+ λ*2*u*^{(2)}*,* (1)

*where λ*_{1}*, λ*_{2} *and u*^{(1)}*, u*^{(2)} *are the spectral values and the associated spectral vectors of*
*x, with respect to K** ^{n}*, given by

*λ**i* *= x*1*+ (−1)*^{i}*kx*2*k,* (2)

*u** ^{(i)}* =

1 2

³*1, (−1)*^{i}*x*_{2}
*kx*2*k*

´*, if x*_{2} *6= 0,*

1 2

³*1, (−1)*^{i}*w*^{´}*,* *if x*_{2} *= 0,* (3)

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

*f*^{soc}*(x) = f (λ*1*)u*^{(1)}*+ f (λ*2*)u*^{(2)} *∀x = (x*1*, x*2*) ∈ IR × IR*^{n−1}*.* (4)
*If f is defined only on a subset of IR, then f*^{soc} is defined on the corresponding subset of
IR^{n}*. The definition (4) is unambiguous whether x*_{2} *6= 0 or x*_{2} = 0. The above definition
*(4) is analogous to one associated with the semidefinite cone S** ^{n}*, 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 : IR*^{n}*× IR*^{n}*→ IR*_{+} such that

*ψ(x, y) = 0* *⇐⇒* *hx, yi = 0,* *x ∈ K*^{n}*,* *y ∈ K*^{n}*,* (5)

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

*ζ∈IR*^{n}*f (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

2*kφ(x, y)k*^{2} *x, y ∈ IR*^{n}*,* (6)

where

*φ(x, y) := (x*^{2}*+ y*^{2})^{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 IR^{n}*to K*^{n}*. 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 : IR*^{n}*→ IR** ^{m}*,

*we denote its Jacobian(not transposed) at x ∈ IR*

^{n}*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 ∈ IR*

^{n}*, 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 : IR*^{n}*→ IR*^{m}*. Then F is continuous at x ∈ IR*^{n}*if F (y) → F (x)*
*as y → x; and F is continuous if F is continuous at every x ∈ IR*^{n}*. We say F is strictly*
*continuous (also called ‘locally Lipschitz continuous’) at x ∈ IR** ^{n}* if there exist scalars

*κ > 0 and δ > 0 such that*

*kF (y) − F (z)k ≤ κky − zk ∀y, z ∈ IR*^{n}*with ky − xk ≤ δ, kz − xk ≤ δ;*

*and F is strictly continuous if F is strictly continuous at every x ∈ IR*^{n}*. We say F is*
*directionally differentiable at x ∈ IR** ^{n}* if

*F*^{0}*(x; h) := lim*

*t→0*^{+}

*F (x + th) − F (x)*

*t* exists *∀h ∈ IR** ^{n}*;

*and F is directionally differentiable if F is directionally differentiable at every x ∈ IR** ^{n}*.

*F is differentiable (in the Fr´echet sense) at x ∈ IR*

*if there exists a linear mapping*

^{n}*∇F (x) : IR*^{n}*→ IR** ^{m}* such that

*F (x + h) − F (x) − ∇F (x)h = o(khk).*

*If F is differentiable at every x ∈ IR*^{n}*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 ∂*_{B}*F (x), where*

*∂**B**F (x) :=*

½

*x*lim^{j}*→x**∇F (x*^{j}*)|F is differentiable at x*^{j}*∈ IR*^{n}

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 : IR*^{n}*→ 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 ∈ IR*^{n}*where 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 : IR*^{n}*→ IR*^{n}*is strictly continuous and directionally differentiable*
*in a neighborhood of x ∈ IR*^{n}*. 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 ∈ IR*^{n}*. 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 = (x*_{1}*, x*_{2}*) ∈ IR × IR*^{n−1}*and y = (y*_{1}*, y*_{2}*) ∈ IR × IR*^{n−1}*, we define their Jordan*
*product as*

*x ◦ y =*^{³}*x*^{T}*y, y*_{1}*x*_{2}*+ x*_{1}*y*_{2}^{´}*.* (8)
*We write x*^{2} *to mean x ◦ x and write x + y to mean the usual componentwise addition of*
*vectors. Then, ◦, +, together with e = (1, 0, . . . , 0) ∈ IR*^{n}*, give rise to a Jordan algebra*
*associated with K*^{n}*[7, Chap. II]. If x ∈ K*^{n}*, then there exists a unique vector in K** ^{n}*, which

*we denote by x*

^{1/2}*, such that (x*

*)*

^{1/2}^{2}

*= x*

^{1/2}*◦ x*

^{1/2}*= x. For any x = (x*

_{1}

*, x*

_{2}

*) ∈ IR × IR*

*, we also define the symmetric matrix*

^{n−1}*L** _{x}* =

"

*x*_{1} *x*^{T}_{2}
*x*_{2} *x*_{1}*I*

#

*,* (9)

viewed as a linear mapping from IR* ^{n}* to IR

^{n×n}*. The matrix L*

*has various interesting*

_{x}*properties that were studied in [9]. Especially, we have L*

_{x}*· y = x ◦ y for any x, y ∈ IR*

*. Now, we summarize the results shown in [4] for which we will provide alternative proofs that are straightforward and intuitive in the subsequent sections.*

^{n}*Proposition 2.2 For any f : IR → IR, the following results hold:*

*(a) f*^{soc} *is continuous at an x ∈ IR*^{n}*with eigenvalues λ*_{1}*, λ*_{2} *if and only if f is continuous*
*at λ*_{1}*, λ*_{2}*.*

*(b) f*^{soc} *is directionally differentiable at an x ∈ IR*^{n}*with eigenvalues λ*_{1}*, λ*_{2} *if and only if*
*f is directionally differentiable at λ*_{1}*, λ*_{2}

*(c) f*^{soc} *is differentiable at an x ∈ IR*^{n}*with eigenvalues λ*_{1}*, λ*_{2} *if and only if f is differ-*
*entiable at λ*_{1}*, λ*_{2}*.*

*(d) f*^{soc} *is continuously differentiable at an x ∈ IR*^{n}*with eigenvalues λ*_{1}*, λ*_{2} *if and only*
*if f is continuously differentiable at λ*1*, λ*2*.*

*(e) f*^{soc} *is strictly continuous at an x ∈ IR*^{n}*with eigenvalues λ*_{1}*, λ*_{2} *if and only if f is*
*strictly continuous at λ*_{1}*, λ*_{2}*.*

*(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 λ*_{1} *≤ λ*_{2} *be the spectral values of x ∈ IR*^{n}*and m*_{1} *≤ m*_{2} *be the spectral*
*values of y ∈ IR*^{n}*. Then we have*

*|λ*_{1}*− m*_{1}*|*^{2}*+ |λ*_{2}*− m*_{2}*|*^{2} *≤ 2 kx − yk*^{2} *,* (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 = (x*_{1}*, x*_{2}*) ∈ IR × IR*^{n−1}*and y = (y*_{1}*, y*_{2}*) ∈ IR × IR*^{n−1}*.*
*(a) If x*_{2} *6= 0, y*_{2} *6= 0, then we have*

*ku*^{(i)}*− v*^{(i)}*k ≤* 1

*kx*_{2}*kkx − yk , ∀i = 1, 2 ,* (11)
*where u*^{(i)}*, v*^{(i)}*are the unique spectral vectors of x and y, respectively.*

*(b) If either x*_{2} *= 0 or y*_{2} *= 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}*x*_{2}
*kx*_{2}*k*

¶

*, v** ^{(i)}* = 1
2

µ

*1 , (−1)*^{i}*y*_{2}
*ky*_{2}*k*

¶

*,*

*where u*^{i)}*, v*^{(i)}*are unique. Thus, we have u*^{(i)}*− v** ^{(i)}* =

^{1}

_{2}

µ

*0 , (−1)** ^{i}*(

_{kx}

^{x}^{2}

2*k* *−* _{ky}^{y}^{2}

2*k*)

¶

. Then

*ku*^{(i)}*− v*^{(i)}*k =* 1
2

°°

°°

°

*x*2

*kx*_{2}*k* *−* *y*2

*ky*_{2}*k*

°°

°°

°= 1 2

°°

°°

°

*x*2*− y*2

*kx*_{2}*k* +*(ky*2*k − kx*2*k)y*2

*kx*_{2}*k · ky*_{2}*k*

°°

°°

°

*≤* 1
2

µ 1

*kx*_{2}*kkx*_{2}*− y*_{2}*k +* 1
*kx*_{2}*k*

¯¯

¯¯*ky*_{2}*k − kx*_{2}*k*

¯¯

¯¯

¶

*≤* 1
2

µ 1

*kx*_{2}*kkx*_{2}*− y*_{2}*k +* 1

*kx*_{2}*kkx*_{2}*− y*_{2}*k*

¶

*≤* 1

*kx*_{2}*kkx − 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 x*_{2} *= 0 or y*_{2} = 0. Then, it is obvious. *2*

*Lemma 3.3 For any w 6= 0 ∈ IR*^{n}*, we have ∇*_{w}

µ *w*
*kwk*

¶

= 1

*kwk*

µ

*I −* *ww*^{T}*kwk*^{2}

¶

*.*

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 λ*1*, λ*2*. For any fixed x ∈ IR*^{n}*and y → x, let*
*the spectral factorizations of x, y be x = λ*_{1}*u*^{(1)}*+ λ*_{2}*u*^{(2)} *and y = m*_{1}*v*^{(1)}*+ m*_{2}*v*^{(2)}. Then,
we discuss two cases.

*Case (i): If x*_{2} *6= 0, then we have*

*f*^{soc}*(y) − f*^{soc}*(x)* (12)

*= f (m*_{1})

µ

*v*^{(1)}*− u*^{(1)}

¶

+

µ

*f (m*_{1}*) − f (λ*_{1})

¶

*u*^{(1)}*+ f (m*_{2})

µ

*v*^{(2)}*− u*^{(2)}

¶

+

µ

*f (m*_{2}*) − f (λ*_{2})

¶

*u*^{(2)}*.*
*Since f is continuous at λ*_{1}*, λ*_{2}*, 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 ∈ IR** ^{n}* .

*Case (ii): If x*_{2} *= 0, no matter y*_{2} *is zero or not, we can arrange that x, y have the same*
*spectral vectors. Thus, f*^{soc}*(y) − f*^{soc}*(x) =*

µ

*f (m*_{1}*) − f (λ*_{1})

¶

*u*^{(1)}+

µ

*f (m*_{2}*) − f (λ*_{2})

¶

*u*^{(2)} *.*
*Then, f*^{soc} *is continuous at x ∈ IR** ^{n}* 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 λ*_{1}*, λ*_{2}*. Fix any x = (x*_{1}*, x*_{2}*) ∈*
*IR × IR** ^{n−1}*, then we discuss two cases as below.

*Case (i): If x*_{2} *6= 0, then we have f*^{soc}*(x) = f (λ*_{1}*)u*^{(1)}*+f (λ*_{2}*)u*^{(2)}*where λ*_{i}*= x*_{1}*+(−1)*^{i}*kx*_{2}*k*
*and u** ^{(i)}* = 1

2

µ

*1 , (−1)*^{i}*x*_{2}
*kx*_{2}*k*

¶

*for all i = 1, 2. From Lemma 3.3, we know that u** ^{(i)}* is

*Fr´echet-differentiable with respect to x, with*

*∇*_{x}*u** ^{(i)}* =

*(−1)*

^{i}*2kx*

_{2}

*k*

0 0

*0 I −* *x*_{2}*x*^{T}_{2}
*kx*_{2}*k*^{2}

*,* *∀i = 1, 2.* (13)

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

*∇*_{x}*λ** _{i}* =

µ

*1 , (−1)*^{i}*x*_{2}
*kx*_{2}*k*

¶

*= 2u*^{(i)}*,* *∀i = 1, 2.* (14)
*Since f is directionally differentiable at λ*_{1}*, λ*_{2}, then the chain rule and product rule for
directional differentiation give

*(f*^{soc})^{0}*(x; h)*

*= f (λ*_{1}*)∇*_{x}*u*^{(1)}*h + u*^{(1)}*f*^{0}*(λ*_{1}*; h)(∇*_{x}*λ*_{1})^{T}*+ f (λ*_{2}*)∇*_{x}*u*^{(2)}*h + u*^{(2)}*f*^{0}*(λ*_{2}*; h)(∇*_{x}*λ*_{2})^{T}

= *f (λ*_{2}*) − f (λ*_{1})
*2kx*2*k*

0 0

*0 I −* *x*_{2}*x*^{T}_{2}
*kx*_{2}*k*^{2}

*h + 2f*^{0}*(λ*_{1}*; h)u*^{(1)}*(u*^{(1)})^{T}*+ 2f*^{0}*(λ*_{2}*; h)u*^{(2)}*(u*^{(2)})^{T}

= *f (λ*_{2}*) − f (λ*_{1})
*λ*_{2}*− λ*_{1}

0 0

*0 I −* *x*_{2}*x*^{T}_{2}
*kx*_{2}*k*^{2}

*h + 2f*^{0}*(λ*1*; h)u*^{(1)}*(u*^{(1)})^{T}*+ 2f*^{0}*(λ*2*; h)u*^{(2)}*(u*^{(2)})^{T}*,*

where the second equality uses equations (13) and (14), and the last equality employs
*the fact that λ*_{2}*− λ*_{1} *= 2kx*_{2}*k. Notice that we can obtain u*^{(i)}*(u** ^{(i)}*)

*as follows by direct computation :*

^{T}*u*^{(i)}*(u** ^{(i)}*)

*= 1 4*

^{T}

1 *(−1)*^{i}*x*^{T}_{2}
*kx*_{2}*k*
*(−1)*^{i}*x*_{2}

*kx*_{2}*k*

*x*_{2}*x*^{T}_{2}
*kx*_{2}*k*^{2}

*,* *∀i = 1, 2.*

Now let

*˜a =* *f (λ*_{2}*) − f (λ*_{1})

*λ*_{2}*− λ*_{1} *h ,* *˜b =* *f*^{0}*(λ*_{2}*; h) + f*^{0}*(λ*_{1}*; h)*

2 *,* *˜c =* *f*^{0}*(λ*_{2}*; h) − f*^{0}*(λ*_{1}*; 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 −* *x*_{2}*x*^{T}_{2}
*kx*2*k*^{2}

+

*˜b* *˜cx*^{T}_{2}
*kx*_{2}*k*

*˜cx*2

*kx*_{2}*k*

*˜bx*2*x*^{T}_{2}
*kx*_{2}*k*^{2}

=

*˜b* *˜cx*^{T}_{2}
*kx*_{2}*k*

*˜cx*_{2}

*kx*2*k* *˜aI + (˜b − ˜a)*_{kx}^{x}^{2}^{x}^{T}^{2}

2*k*^{2}

*.*

(16)
*This enables that f*^{soc} *is directionally differentiable at x when x*_{2} *6= 0 with (f*^{soc})^{0}*(x; h)*
being in form of (16).

*Case (ii): If x*2 *= 0, we compute the directional derivative (f*^{soc})^{0}*(x; h) at x for any*
*direction h by definition. Let h = (h*_{1}*, h*_{2}*) ∈ IR × IR** ^{n−1}*. We have two subcases. First,

*consider the subcase of h*

_{2}

*6= 0. From the spectral factorization, we can choose u*

*=*

^{(i)}1 2

µ

*1 , (−1)*^{i h}_{kh}^{2}_{2}_{k}

¶

*for all i = 1, 2, such that*

( *f*^{soc}*(x + th) = f (λ + 4λ*1*)u*^{(1)}*+ f (λ + 4λ*2*)u*^{(2)}
*f*^{soc}*(x)* *= f (λ)u*^{(1)}*+ f (λ)u*^{(2)}*,*

*where λ = x*_{1} *and 4λ*_{i}*= t*

µ

*h*_{1}*+ (−1)*^{i}*kh*_{2}*k*

¶

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

*f*^{soc}*(x + th) − f*^{soc}*(x) =*

µ

*f (λ + 4λ*_{1}*) − f (λ)*

¶

*u*^{(1)}+

µ

*f (λ + 4λ*_{2}*) − f (λ)*

¶

*u*^{(2)}*.*
The fact that

*t→0*lim^{+}

*f (λ + 4λ*_{1}*) − f (λ)*

*t* = lim

*t→0*^{+}

*f (λ + t(h*_{1}*− kh*_{2}*k)) − f (λ)*

*t* *= f*^{0}*(λ; h*_{1}*− kh*_{2}*k) ,*
and

*t→0*lim^{+}

*f (λ + 4λ*_{2}*) − f (λ)*

*t* = lim

*t→0*^{+}

*f (λ + t(h*_{1}*+ kh*_{2}*k)) − f (λ)*

*t* *= f*^{0}*(λ; h*_{1}*+ kh*_{2}*k),*
yields

*t→0*lim^{+}

*f*^{soc}*(x + th) − f*^{soc}*(x)*

*t* = lim

*t→0*^{+}

*f (λ + 4λ*_{1}*) − f (λ)*

*t* *u*^{(1)}+ lim

*t→0*^{+}

*f (λ + 4λ*_{2}*) − f (λ)*

*t* *u*^{(2)}

*= f*^{0}*(λ; h*1*− kh*2*k)u*^{(1)}*+ f*^{0}*(λ; h*1*+ kh*2*k)u*^{(2)} *,* (17)
*where u*^{(1)} = 1

2

µ

*1, −* *h*_{2}
*kh*_{2}*k*

¶

*, u*^{(2)} = 1
2

µ

*1,* *h*_{2}
*kh*_{2}*k*

¶

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

*Secondly, for the subcase of h*_{2} *= 0, the same argument applies except h*_{2}*/kh*_{2}*k is replaced*
*by any w ∈ IR*^{n−1}*with kwk = 1, i.e., choosing u** ^{(i)}* = 1

2*(1, (−1)*^{i}*w), for all i = 1, 2.*

Analogously,

*t→0*lim^{+}

*f*^{soc}*(x + th) − f*^{soc}*(x)*

*t* *= f*^{0}*(λ; h*1*)u*^{(1)}*+ f*^{0}*(λ; h*1*)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 x*2 *= 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 ∈ IR*^{n}*with spectral values λ*_{1}*, λ*_{2}, we
*will prove that f is directionally differentiable at λ*_{1}*, λ*_{2}*. For λ*_{1} *∈ IR and any direction*
*d*_{1} *∈ IR, let h := d*_{1}*u*^{(1)}*+ 0u*^{(2)} *where x = λ*_{1}*u*^{(1)}*+ λ*_{2}*u*^{(2)}*. Then, x + th = (λ*_{1}*+ td*_{1}*)u*^{(1)}+
*λ*2*u*^{(2)}, and

*f*^{soc}*(x + th) − f*^{soc}*(x)*

*t* = *f (λ*_{1}*+ td*_{1}*) − f (λ*_{1})
*t* *u*^{(1)}*.*

*Since f*^{soc} *is directionally differentiable at x, the above equation yields that*
*f*^{0}*(λ*1*; d*1) = lim

*t→0*^{+}

*f (λ*_{1}*+ td*_{1}*) − f (λ*_{1})

*t* *exists.*

*This means f is directionally differentiable at λ*_{1}*. Similarly, f is also directionally differ-*
*entiable at λ*2. *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 x*_{2} *6= 0, we have*

*∇f*^{soc}*(x) =*

*b* *cx*^{T}_{2}

*kx*_{2}*k*
*cx*2

*kx*_{2}*k* *aI + (b − a)x*2*x*^{T}_{2}
*kx*_{2}*k*^{2}

*,* (19)

where

*a =* *f (λ*_{2}*) − f (λ*_{1})

*λ*_{2}*− λ*_{1} *,* *b =* *f*^{0}*(λ*_{2}*) + f*^{0}*(λ*_{1})

2 *,* *c =* *f*^{0}*(λ*_{2}*) − f*^{0}*(λ*_{1})

2 *.* (20)

*If x*_{2} = 0, then

*∇f*^{soc}*(x) = f*^{0}*(λ)I.* (21)

*“⇒” Let f*^{soc} *be Fr´echet-differentiable at x ∈ IR*^{n}*with spectral eigenvalues λ*1*, λ*2, we
*will show that f is Fr´echet-differentiable at λ*_{1}*, λ*_{2}*. 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 = λ*_{1}*u*^{(1)} *+ λ*_{2}*u*^{(2)}*, let h := 1 · u*^{(1)} *+ 0 · u*^{(2)} *= u*^{(1)}. Then,
*x + th = (λ*_{1}*+ t)u*^{(1)}*+ λ*_{2}*u*^{(2)} *and f*^{soc}*(x + th) = f (λ*_{1}*+ t)u*^{(1)}*+ f (λ*_{2}*)u*^{(2)}. Thus,

*ν→∞*lim

*f*^{soc}*(x + t*^{ν}*h) − f*^{soc}*(x)*

*t** ^{ν}* = lim

*ν→∞*

*f*^{soc}*(λ*_{1} *+ t*^{ν}*) − f*^{soc}*(λ*_{1})

*t*^{ν}*u*^{(1)}

*ν→∞*lim

*f*^{soc}*(x + τ*^{ν}*h) − f*^{soc}*(x)*

*τ** ^{ν}* = lim

*ν→∞*

*f*^{soc}*(λ*_{1} *+ τ*^{ν}*) − f*^{soc}*(λ*_{1})

*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 x*_{2} *6= 0. It remains to show that ∇f*^{soc}
*is continuous at every x with x*_{2} *= 0. Fix any x = (x*_{1}*, 0) ∈ IR*^{n}*, so λ*_{1} *= λ*_{2} *= x*_{1}. Then,
from equation (20), we have

*y→x*lim*a = lim*

*y→x*

*f (λ*_{2}*) − f (λ*_{1})

*λ*_{2}*− λ*_{1} *= f*^{0}*(x*1)

*y→x*lim*b = lim*

*y→x*

1

2*(f*^{0}*(λ*_{2}*) + f*^{0}*(λ*_{1}*)) = f*^{0}*(x*_{1})

*y→x*lim*c = lim*

*y→x*

1

2*(f*^{0}*(λ*_{2}*) − f*^{0}*(λ*_{1}*)) = 0 .*
*Taking the limit in equation (19) as y → x yields lim*

*y→x**∇f*^{soc}*(y) = f*^{0}*(x*1*)I = ∇f*^{soc}*(x),*
*which says ∇f*^{soc} *is continuous at every x ∈ IR** ^{n}* .

*“⇒” 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 × IR** ^{n−1}* and

*h := (η, 0) ∈ IR × IR*^{n−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 = f*^{0}*(α+η)Ih =*

µ

*f*^{0}*(α+η)η , 0*

¶

*. Plugging this*
*into equation (22), it yields that f (α + η) − f (α) − f*^{0}*(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 ∈ IR*^{n}*and h ∈ IR*^{n}*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 x*_{2} *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 m*

*i*be the spectral

*values of z with the associated vectors v*

*. Hence, we have*

^{(i)}( *f*^{soc}*(x)* *= f (λ*_{1}*)u*^{(1)}*+ f (λ*_{2}*)u*^{(2)}*,*
*f*^{soc}*(x + h) = f (m*_{1}*)v*^{(1)}*+ f (m*_{2}*)v*^{(2)}*.*
In addition, from (19), we know

*∇f*^{soc}*(x + h) =*

*ˆb* *ˆcz*_{2}^{T}*kz*_{2}*k*
*ˆcz*_{2}

*kz*_{2}*k* *ˆaI − (ˆb − ˆa)z*_{2}*z*_{2}^{T}*kz*_{2}*k*^{2}

*,*

where

*ˆa =* *f (m*_{2}*) − f (m*_{1})

*m*_{2}*− m*_{1} *,* *ˆb =* *f*^{0}*(m*_{2}*) + f*^{0}*(m*_{1})

2 *,* *ˆc =* *f*^{0}*(m*_{2}*) − f*^{0}*(m*_{1})

2 *.*

*With this, we can write out f*^{soc}*(x+h)−f*^{soc}*(x)−∇f*^{soc}*(x+h)h := (Ξ*_{1}*, Ξ*_{2}) where Ξ_{1} *∈ IR*
and Ξ_{2} *∈ IR** ^{n−1}*. Since the expansion is very long, for simplicity, we denote Ξ

_{1}be the first component and Ξ2 be the second component of the expansion. We will show that Ξ

_{1}and Ξ

_{2}

*are both o(khk).*

First, we compute the first component Ξ_{1}:
Ξ_{1} = 1

2

(

*f (m*_{1}*) − f (λ*_{1}*) − f*^{0}*(m*_{1}*)(h*_{1}*−z*^{T}_{2}*h*_{2}
*kz*_{2}*k*)

)

+ 1 2

(

*f (m*_{2}*) − f (λ*_{2}*) − f*^{0}*(m*_{2}*)(h*_{1}+ *z*_{2}^{T}*h*_{2}
*kz*_{2}*k*)

)

= 1 2

½

*f (m*1*) − f (λ*1*) − f*^{0}*(m*1)

µ

*h*1*− (kz*2*k − kx*2*k)*

¶

*+ o(khk)*

¾

+1 2

½

*f (m*2*) − f (λ*2*) − f*^{0}*(m*2)

µ

*h*1*+ (kz*2*k − kx*2*k)*

¶

*+ o(khk)*

¾

*= o*

µ

*h*1*− (kz*2*k − kx*2*k)*

¶

*+ o(khk) + o*

µ

*h*1*+ (kz*2*k − kx*2*k)*

¶

*+ o(khk).*

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

*z*^{T}_{2}*h*_{2}

*kz*2*k* = *z*^{T}_{2}*(z*_{2}*− x*_{2})

*kz*2*k* *= kz*_{2}*k −* *kz*_{2}*kkx*_{2}*k*
*kz*2*k* *cos θ*

*= kz*_{2}*k − kx*_{2}*k*

µ

*1 + O(θ*^{2})

¶

*= kz*_{2}*k − kx*_{2}*k*

µ

*1 + O(khk*^{2})

¶

*= kz*_{2}*k − kx*_{2}*k*

µ

*1 + o(khk)*

¶

*,*

*where θ is the angle between x*_{2} *and z*_{2} *and note that z*_{2}*−x*_{2} *= h*_{2} *gives O(θ*^{2}*) = O(khk*^{2}).

Also the last equality in expression of Ξ_{1} *holds since f is semismooth and*
*m*_{i}*− λ*_{i}*= h*_{1}*+ (−1)*^{i}*(kz*_{2}*k − kx*_{2}*k).*

*On the other hand, due to h*_{1}*+(−1)*^{i}*(kz*_{2}*k−kx*_{2}*k) ≤ h*_{1}*+kz*_{2}*−x*_{2}*k = h*_{1}*+kh*_{2}*k , we observe*
*that when khk → 0 then h*1 *+ (−1)*^{i}*(kz*2*k − kx*2*k) → 0 and h*1*+ (−1)*^{i}*(kz*2*k − kx*2*k) =*
*O(khk). Thus, o*

µ

*h*_{1} *+ (−1)*^{i}*(kz*_{2}*k − kx*_{2}*k)*

¶

*= o(khk), which yields the first component*
Ξ_{1} *is o(khk).*

Now consider the second component Ξ_{2}:
Ξ2 *= −*1

2*f (m*1) *z*2

*kz*_{2}*k* +1

2*f (m*2) *z*2

*kz*_{2}*k*+ 1

2*f (λ*1) *x*2

*kx*_{2}*k* *−*1

2*f (λ*2) *x*2

*kx*_{2}*k*

*−*1
2

µ

*f*^{0}*(m*_{2}*) − f*^{0}*(m*_{1})

¶*z*_{2}*h*_{1}

*kz*_{2}*k* *−* *f (m*_{2}*) − f (m*_{1})
*m*_{2}*− m*_{1} *h*_{2}

*−*

(1 2

µ

*f*^{0}*(m*2*) − f*^{0}*(m*1)

¶

*−* *f (m*2*) − f (m*1)
*m*_{2}*− m*_{1}

)*z*2*z*_{2}^{T}*h*2

*kz*_{2}*k*^{2}

*= −*1
2

(

*f (m*_{1}) *z*_{2}

*kz*_{2}*k* *− f (λ*_{1}) *x*_{2}

*kx*_{2}*k* *− f*^{0}*(m*_{1})*z*_{2}*h*_{1}

*kz*_{2}*k* *−* *2f (m*_{1})
*m*_{2}*− m*_{1}*h*_{2}
+

µ

*f*^{0}*(m*_{1}) + *2f (m*_{1})
*m*2*− m*1

¶*z*_{2}*z*_{2}^{T}*h*_{2}
*kz*2*k*^{2}

)

+1 2

(

*f (m*_{2}) *z*_{2}

*kz*_{2}*k* *− f (λ*_{2}) *x*_{2}

*kx*_{2}*k* *− f*^{0}*(m*_{2})*z*_{2}*h*_{1}

*kz*_{2}*k−* *2f (m*_{2})
*m*_{2} *− m*_{1}*h*_{2}

*−*

µ

*f*^{0}*(m*_{2}) + *2f (m*_{2})
*m*_{2}*− m*_{1}

¶*z*_{2}*z*_{2}^{T}*h*_{2}
*kz*_{2}*k*^{2}

)

:= 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*

*z*2 *= x*2*+ h*2*,*
*m*_{2}*− m*_{1} *= 2kz*_{2}*k,*

*m*_{i}*− λ*_{i}*= h*_{1} *+ (−1)*^{i}*(kz*_{2}*k − kx*_{2}*k) = O(khk),*
we have the following:

Ξ^{(1)}_{2} *= −*1
2

(

*f (m*_{1}) *z*2

*kz*_{2}*k* *− f (λ*_{1}) *x*2

*kx*_{2}*k* *− f*^{0}*(m*_{1})*z*2*h*1

*kz*_{2}*k* *−* *2f (m*1)
*m*_{2}*− m*_{1}*h*_{2}
+

µ

*f*^{0}*(m*1) + *2f (m*1)
*m*_{2}*− m*_{1}

¶*z*2*z*_{2}^{T}*h*2

*kz*_{2}*k*^{2}

)

*= −*1
2

*z*_{2}
*kz*_{2}*k*

(

*f (m*_{1}*) − f (λ*_{1}*) − f*^{0}*(m*_{1})

µ

*h*_{1}*−* *z*_{2}^{T}*h*_{2}
*kz*_{2}*k*

¶)

+1 2

µ

*f (m*_{1}*) − f (λ*_{1})

¶µ*−h*_{2}

*kz*2*k* + *z*_{2}*z*_{2}^{T}*h*_{2}
*kz*2*k*^{3}

¶

*−*1
2*f (λ*_{1})

µ *z*2

*kz*_{2}*k* *−* *x*2

*kx*_{2}*k−* *h*2

*kz*_{2}*k* + *z*2*z*_{2}^{T}*h*2

*kz*_{2}*k*^{3}

¶

*.*

Following the same arguments as in the first component Ξ1, it can be seen that
*f (m*_{1}*) − f (λ*_{1}*) − f*^{0}*(m*_{1})

µ

*h*_{1}*−* *z*_{2}^{T}*h*_{2}
*kz*_{2}*k*

¶

*= o(khk).*

*Since m*_{1}*− λ*_{1} *= h*_{1}*− (kz*_{2}*k − kx*_{2}*k) = O(khk) and f is strictly continuous, then f (m*_{1}*) −*
*f (λ*_{1}*) = O(khk). In addition, −h*_{2}*/kz*_{2}*k + z*_{2}*z*_{2}^{T}*h*_{2}*/kz*_{2}*k*^{3} *= O(khk). Hence,*

µ

*f (m*_{1}*) − f (λ*_{1})

¶µ*−h*_{2}

*kz*_{2}*k* +*z*_{2}*z*^{T}_{2}*h*_{2}
*kz*_{2}*k*^{3}

¶

*= O(khk*^{2}*) = o(khk) .*

Therefore, it remains to show that the last part of Ξ^{(1)}_{2} *is o(khk). Now, note that*
*z*_{2}

*kz*2*k* *−* *x*_{2}

*kx*2*k* *−* *h*_{2}

*kz*2*k*+ *z*_{2}*z*_{2}^{T}*h*_{2}
*kz*2*k*^{3} *= x*_{2}

µ 1

*kz*2*k−* 1

*kx*2*k* + *z*^{T}_{2}*h*_{2}
*kz*2*k*^{3}

¶

*+ O(khk*^{2}*) .*

*Let θ(z*_{2}*) := −1/kz*_{2}*k, then ∇θ(z*_{2}*) = −* *−1*
*kz*2*k*^{2}

*z*_{2}

*kz*2*k* = *z*_{2}

*kz*2*k*^{3}. This implies that
1

*kz*2*k* *−* 1

*kx*2*k*+ *z*_{2}^{T}*h*_{2}

*kz*2*k*^{3} *= θ(x*_{2}*) − θ(z*_{2}*) − ∇θ(z*_{2}*)(x*_{2}*− z*_{2}*) = O(khk*^{2}*),*
where the last equality is from first Taylor approximation. Thus, we obtain

*f (λ*_{1})

µ *z*_{2}

*kz*_{2}*k* *−* *x*_{2}

*kx*_{2}*k−* *h*_{2}

*kz*_{2}*k* + *z*_{2}*z*_{2}^{T}*h*_{2}
*kz*_{2}*k*^{3}

¶

*= o(khk) .*