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 sec- tions 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 IR^{n}. For a function f from IR to IR, one can define
a corresponding vector-valued function f^{L}^{θ} on IR^{n} by applying f to the spectral values
of the spectral decomposition of x ∈ IR^{n} 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 in-
volved circular cone constraints.

Keywords. Circular cone, vector-valued function, semismooth function, spectral de- composition, 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.

### 1 Introduction

The circular cone is a pointed closed convex cone having hyperspherical sections orthog-
onal 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 IR^{n}. Then, the n-dimensional circular cone L_{θ} is expressed as

L_{θ} := {x = (x_{1}, x_{2})^{T} ∈ IR × IR^{n−1}| cos θkxk ≤ x_{1}}.

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 = (z_{1}, z_{2}) ∈ IR × IR^{n−1}, one has

z = λ_{1}(z)u^{1}_{z} + λ_{2}(z)u^{2}_{z}, (1)
where

λ_{1}(z) = z_{1}− kz_{2}kctanθ
λ_{2}(z) = z_{1}+ kz_{2}k tan θ
and

u^{1}_{z} = 1
1 + ctan^{2}θ

1 0

0 ctanθ · I

1

−¯z2

u^{2}_{z} = 1
1 + tan^{2}θ

1 0

0 tan θ · I

1

¯
z_{2}

with ¯z_{2} := z_{2}/kz_{2}k if z_{2} 6= 0, and ¯z_{2} being any vector w ∈ IR^{n−1} satisfying kwk = 1
if z_{2} = 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} : IR^{n} → IR^{n} as

f^{L}^{θ}(z) = f (λ_{1}(z)) u^{1}_{z} + f (λ_{2}(z)) u^{2}_{z}. (2)
It is not hard to see that f^{L}^{θ} is well-defined for all z. In particular, if z_{2} = 0, then

f^{L}^{θ}(z) = f (z_{1})
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 (λ_{1}(x))u^{(1)}_{x} + f (λ_{2}(x))u^{(2)}_{x} ∀x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} (3)

where λi(x) = x1+ (−1)^{i}kx2k and u^{(i)}x = ^{1}_{2}(1, (−1)^{i}x¯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 ∈ IR^{n} if and only if f is continuous at λ_{i}(z) for i = 1, 2;

(b) f^{L}^{θ} is directionally differentiable at z ∈ IR^{n} if and only if f is directionally differ-
entiable at λ_{i}(z) for i = 1, 2;

(c) f^{L}^{θ} is (Fr´echet) differentiable at z ∈ IR^{n} 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 ∈ IR^{n} if and only if f is continuously contin-
uous at λ_{i}(z) for i = 1, 2;

(e) f^{L}^{θ} is strictly continuous at z ∈ IR^{n} 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 contin-
uous 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 IR^{n}the space of n-dimensional
real column vectors and let e = (1, 0, · · · , 0) ∈ IR^{n}. Given x, y ∈ IR^{n}, the Euclidean inner
product and norm are hx, yi = x^{T}y and kxk = √

x^{T}x. For a linear mapping H : IR^{n} →
IR^{m}, its operator norm is kHk := maxkxk=1kHxk. For α ∈ IR and s ∈ IR^{n}, 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 : IR^{n} → IR^{m}, we say

(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 ∈ IR^{n} in the Hadamard sense if the directional
derivative F^{0}(x; d) exists for all d ∈ IR^{n} and

F^{0}(x; d) = lim

d0→d t↓0

F (x + td^{0}) − 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 ∂_{B}F (x), where

∂_{B}F (x) :=n

z→xlim∇F (z)

z ∈ D_{F}o
,
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^{α}F (x), ∀x ∈ IR^{n} and k ≥ 0;

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

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

This section is devoted to study the properties of directional differentiability, strict conti-
nuity, 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 = (h_{1}, h_{2}) ∈ IR × IR^{n−1}, we have
f^{L}^{θ}0

(z; h) = f^{0}(z_{1}; h_{1})
0

= f^{0}(z_{1}; h_{1})e
when z_{2} = 0 and h_{2} = 0;

f^{L}^{θ}0

(z; h) = 1

1 + ctan^{2}θf^{0}(z_{1}; h_{1}− kh_{2}kctanθ) 1 0
0 ctanθ · I

"

1

−_{kh}^{h}^{2}

2k

#

+ 1

1 + tan^{2}θf^{0}(z_{1}; h_{1}+ kh_{2}k tan θ) 1 0
0 tan θ · I

"

1

h2

kh2k

#

when z_{2} = 0 and h_{2} 6= 0; otherwise

f^{L}^{θ}0

(z; h) = 1

1 + ctan^{2}θf^{0}

λ_{1}(z); h_{1}− z^{T}_{2}h_{2}

kz_{2}kctanθ 1 0
0 ctanθ · I

"

1

−_{kz}^{z}^{2}

2k

#

− ctanθ
1 + ctan^{2}θ

f (λ_{1}(z))
kz_{2}k M_{z}_{2}h

+ 1

1 + tan^{2}θf^{0}

λ_{2}(z); h_{1}+z_{2}^{T}h_{2}

kz2k tan θ 1 0 0 tan θ · I

"

1

z2

kz_{2}k

#

+ tan θ
1 + tan^{2}θ

f (λ_{2}(z))
kz2k M_{z}_{2}h,
where

M_{z}_{2} :=

0 0

0 I − z_{2}z^{T}_{2}
kz_{2}k^{2}

.

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

(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 z_{2} = 0, then

∇f^{L}^{θ}(z) = f^{0}(z_{1})I;

otherwise

∇f^{L}^{θ}(z) =

ξ %z_{2}^{T}

kz2k

%z_{2}

kz_{2}k aI + (η − a)z_{2}z^{T}_{2}
kz_{2}k^{2}

,

where

a = tan θ
1 + tan^{2}θ

f (λ2(z))

kz_{2}k − ctanθ
1 + ctan^{2}θ

f (λ1(z))

kz_{2}k = f (λ2(z)) − f (λ1(z))
λ_{2}(z) − λ_{1}(z) ,
ξ = f^{0}(λ_{1}(z))

1 + ctan^{2}θ + f^{0}(λ_{2}(z))

1 + tan^{2}θ, η = ξ − %(ctanθ − tan θ), (5)

% = − ctanθ

1 + ctan^{2}θf^{0}(λ_{1}(z)) + tan θ

1 + tan^{2}θf^{0}(λ_{2}(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 + ctan^{2}θ = sin^{2}θ, 1

1 + tan^{2}θ = cos^{2}θ, ctanθ

1 + ctan^{2}θ = tan θ

1 + tan^{2}θ = sin θ cos θ.

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

Lemma 2.3. Let z, y ∈ IR^{n} 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

|λ_{1}(z) − λ_{1}(y)| = |z_{1}− kz_{2}kctanθ − y_{1}+ ky_{2}kctanθ|

≤ |z_{1}− y_{2}| + kz_{2}− y_{2}kctanθ

≤ max{1, ctanθ}(|z_{1} − y_{1}| + kz_{2}− y_{2}k)

≤ max{1, ctanθ}√

2p|z_{1}− y_{1}|^{2}+ kz_{2}− y_{2}k^{2}

= max{1, ctanθ}√

2kz − yk.

Applying the similar argument to λ_{2} yields

|λ_{2}(z) − λ_{2}(y)| ≤ max{1, tan θ}√

2kz − yk.

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{δ_{1}, δ_{2}} and C := [λ_{1}(z) − ¯δ, λ_{1}(z) + ¯δ] ∪ [λ_{2}(z) − ¯δ, λ_{2}(z) + ¯δ]. Define

f (τ ) :=˜

f (τ ) if τ ∈ C,

(1 − t)f (λ_{1}(z) + ¯δ) + tf (λ_{2}(z) − ¯δ) if λ_{1}(z) + ¯δ < λ_{2}(z) − ¯δ and

τ = (1 − t)(λ_{1}(z) + ¯δ) + t(λ_{2}(z) − ¯δ)
with t ∈ (0, 1)

f (λ_{1}(z) − ¯δ) if τ < λ_{1}(z) − ¯δ
f (λ_{2}(z) + ¯δ) if τ > λ_{2}(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}(λ_{1}(w)) − f (λ_{1}(w))] u^{1}_{w}+ [f^{v}(λ_{2}(w)) − f (λ_{2}(w))] u^{2}_{w}

2

= f^{v}(λ_{1}(w)) − f (λ_{1}(w))2

ku^{1}_{w}k^{2}+f^{v}(λ_{2}(w)) − f (λ_{2}(w))2

ku^{2}_{w}k^{2}

= 1

1 + ctan^{2}θ|f^{v}(λ_{1}(w)) − f (λ_{1}(w))|^{2}+ 1

1 + tan^{2}θ |f^{v}(λ_{2}(w)) − f (λ_{2}(w))|^{2},
where we have used the facts that ku^{1}_{w}k = 1/√

1 + ctan^{2}θ, ku^{2}_{w}k = 1/√

1 + tan^{2}θ, and
hu^{1}_{w}, u^{2}_{w}i = 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}(w_{1})I. Hence it follows from (6) that

k∇(f^{v})^{L}^{θ}(w)k = |(f^{v})^{0}(w_{1})| ≤ k, (7)

since in this case λ_{i}(w) = w_{1} ∈ eC. If w_{2} 6= 0, then

∇(f^{v})^{L}^{θ}(w) =

ξ % w_{2}^{T}/kw_{2}k

% w_{2}/kw_{2}k aI + (ξ − %(ctanθ − tan θ) − a)w_{2}w^{T}_{2}
kw_{2}k^{2}

=

ξ % w^{T}_{2}/kw_{2}k

% w_{2}/kw_{2}k aI + (ξ − a)w2w_{2}^{T}
kw_{2}k^{2}

+

0 0

0 [−%(ctanθ − tan θ)] w2w_{2}^{T}
kw_{2}k^{2}

=

ξ % w_{2}^{T}/kw_{2}k

% w2/kw2k ξI

+ (a − ξ)

0 0

0 I − w_{2}w^{T}_{2}
kw_{2}k^{2}

−%(ctanθ − tan θ)

0 0

0 w_{2}w_{2}^{T}
kw_{2}k^{2}

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^{T}_{2}/kw_{2}k

% w_{2}/kw_{2}k ξI

+ (a − ξ)

0 0

0 I − w_{2}w^{T}_{2}
kw_{2}k^{2}

and

B := −%(ctanθ − tan θ)

0 0

0 w_{2}w^{T}_{2}
kw_{2}k^{2}

. Note that

|a| =

f^{v} λ_{2}(w) − f^{v} λ_{1}(w)
λ_{2}(w) − λ_{1}(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} λ_{1}(w)

1 + ctan^{2}θ +(f^{v})^{0} λ_{2}(w)
1 + tan^{2}θ

≤ 1

1 + ctan^{2}θ|(f^{v})^{0} λ_{1}(w)| + 1

1 + tan^{2}θ|(f^{v})^{0} λ_{2}(w)|

≤

1

1 + ctan^{2}θ + 1
1 + tan^{2}θ

k

= k (9)

and

|%| = | − ctanθ

1 + ctan^{2}θ(f^{v})^{0}(λ_{1}(w)) + tan θ

1 + tan^{2}θ(f^{v})^{0}(λ_{2}(w))|

≤

− ctanθ
1 + ctan^{2}θ

+

tan θ
1 + tan^{2}θ

k

=

ctanθ

1 + ctan^{2}θ + tan θ
1 + tan^{2}θ

k

= 2 tan θ
1 + tan^{2}θk

≤ k. (10)

(i) For % = 0, then ∇(f^{v})^{L}^{θ}(w) takes the form of ξI + (a − ξ)M_{w}_{2} 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, w_{2}

kw_{2}k)^{T}(0, w_{2}

kw_{2}k), the eigenvalues of B are

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

|%(ctanθ − tan θ)| =

1 − ctan^{2}θ

1 + ctan^{2}θ(f^{v})^{0} λ_{1}(w) + 1 − tan^{2}θ

1 + tan^{2}θ(f^{v})^{0} λ_{2}(w)

≤

1 − ctan^{2}θ
1 + ctan^{2}θ

+

1 − tan^{2}θ
1 + tan^{2}θ

k

=

ctan^{2}θ − 1

1 + ctan^{2}θ +1 − tan^{2}θ
1 + tan^{2}θ
k

= 2

1 − tan^{2}θ
1 + tan^{2}θ
k

≤ 2k. (12)

Note that

A = %

kw_{2}kL_{w}_{˜} + (a − ξ)M_{w}_{2} = %
kw_{2}k

L_{w}_{˜} + (a − ξ)kw_{2}k

% M_{w}_{˜}_{2}

,

where ˜w = ξkw2k/%, w2 and

L_{w}_{˜} := ˜w_{1} 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).

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: z_{2} 6= 0. Take θ, µ ∈ [λ_{1}(z) − δ_{1}, λ_{1}(z) + δ_{1}] with δ_{1} := min{δ, λ_{2}(z) − λ_{1}(z)}. Let

x := θu^{1}_{z}+ λ_{2}(z)u^{2}_{z} and y := µu^{1}_{z}+ λ_{2}(z)u^{2}_{z}.
Then, kx − zk ≤ δ and ky − zk ≤ δ and it follows from (14) that

|f (θ) − f (µ)| = 1

ku^{1}_{z}kkf^{L}^{θ}(x) − f^{L}^{θ}(y)k ≤ k

ku^{1}_{z}kkx − yk = k

ku^{1}_{z}k|θ − µ|ku^{1}_{z}k = k|θ − µ|,
which says f is strictly continuous at λ_{1}(z). The similar argument shows the strict
continuity of f at λ_{2}(z).

Case 2: z_{2} = 0. For any θ, µ ∈ [z_{1} − δ, z_{1} + δ], we have kθe − zk = |θ − z_{1}| ≤ 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) = z_{1} for i = 1, 2. 2

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}^{θ}k 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: z_{2} 6= 0 and y_{2} 6= 0. We assume without loss of generality that kz_{2}k ≥ ky_{2}k.

Thus,

kf^{L}^{θ}(z) − f^{L}^{θ}(y)k

= kf (λ_{1}(z))u^{1}_{z}+ f (λ_{2}(z))u^{2}_{z}− f (λ_{1}(y))u^{1}_{y} − f (λ_{2}(y))u^{2}_{y}k

=

f (λ_{1}(z))u^{1}_{z}− u^{1}_{y} + f (λ2(z))u^{2}_{z} − u^{2}_{y}

+f (λ_{1}(z)) − f (λ_{1}(y))u^{1}_{y}+f (λ_{2}(z)) − f (λ_{2}(y))u^{2}_{y}

≤

f (λ1(z))u^{1}_{z}− u^{1}_{y} + f (λ2(z))u^{2}_{z} − u^{2}_{y}
+ |f (λ_{1}(z)) − f (λ_{1}(y))| ·

u^{1}_{y}

+ |f (λ_{2}(z)) − f (λ_{2}(y))| ·
u^{2}_{y}

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

f (λ_{1}(z))u^{1}_{z}− u^{1}_{y} + f (λ_{2}(z))u^{2}_{z}− u^{2}_{y}

= tan θ

1 + tan^{2}θ|f (λ_{1}(z)) − f (λ_{2}(z))| ·

z_{2}

kz_{2}k− y_{2}
ky_{2}k

≤ tan θ

1 + tan^{2}θ[f ]_{α}|λ_{1}(z) − λ_{2}(z)|^{α}

z_{2}

kz_{2}k − y_{2}
ky_{2}k

= tan θ

1 + tan^{2}θ[f ]_{α}(tan θ + ctanθ)^{α}kz_{2}k^{α}·

z_{2}

kz_{2}k− y_{2}
ky_{2}k

≤ tan θ

1 + tan^{2}θ[f ]α(tan θ + ctanθ)^{α}kz2k^{α} 2

kz_{2}kkz2− y2k

= 2 tan θ

1 + tan^{2}θ[f ]_{α}(tan θ + ctanθ)^{α}

z_{2}− y_{2}
kz_{2}k

1−α

kz_{2}− y_{2}k^{α}

≤ tan θ

1 + tan^{2}θ[f ]_{α}(tan θ + ctanθ)^{α}2^{2−α}kz_{2}− y_{2}k^{α}

≤ tan θ

1 + tan^{2}θ[f ]_{α}(tan θ + ctanθ)^{α}2^{2−α}kz − 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(z_{2}/kz_{2}k) − (y_{2}/ky_{2}k)k ≤ (2/kz_{2}k)kz_{2}− y_{2}k (cf. [2, Lemma
2.2]), and the third inequality follows from the fact that kz2− y_{2}k ≤ kz_{2}k + ky_{2}k ≤ 2kz_{2}k
(since ky_{2}k ≤ kz_{2}k). Next, we look into the second term:

|f (λ_{1}(z)) − f (λ_{1}(y))|ku^{1}_{y}k ≤ [f ]_{α}|λ_{1}(z) − λ_{1}(y)|^{α} 1

√

1 + ctan^{2}θ

≤ [f ]_{α} √

2 max{tan θ, ctanθ}α

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

|f (λ2(z)) − f (λ2(y))|ku^{2}_{y}k ≤ [f ]α|λ2(z) − λ2(y)|^{α} 1

√1 + tan^{2}θ

≤ [f ]_{α} √

2 max{tan θ, ctanθ}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 z_{2} = 0 or y_{2} = 0. In this case, we take u^{i}_{z} = u^{i}_{y} for i = 1, 2 according to
the spectral decomposition. Therefore, we obtain

kf^{L}^{θ}(z) − f^{L}^{θ}(y)k

= kf (λ_{1}(z))u^{1}_{z}+ f (λ_{2}(z))u^{2}_{z}− f (λ_{1}(y))u^{1}_{y}− f (λ_{2}(y))u^{2}_{y}k

=

f (λ_{1}(z)) − f (λ_{1}(y))u^{1}_{z} +f (λ_{2}(z)) − f (λ_{2}(y))u^{2}_{z}

≤ |f (λ_{1}(z)) − f (λ_{1}(y))| · ku^{1}_{z}k + |f (λ_{2}(z)) − f (λ_{2}(y))| · ku^{2}_{z}k

≤ [f ]_{α}|λ_{1}(z) − λ_{1}(y)|^{α} 1

√1 + ctan^{2}θ + [f ]_{α}|λ_{2}(y) − λ_{2}(z)|^{α} 1

√1 + tan^{2}θ

≤ 2[f ]_{α} √

2 max{tan θ, ctanθ}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

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

The property of semismoothness plays an important role in nonsmooth Newton meth-
ods [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 semis-
mooth 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 z_{2} = 0, f^{L}^{θ} is ρ-semismooth at z if and only if f is ρ-order semismooth at
λ_{i}(z) = z_{1} for i = 1, 2.

Proof. (a) Take h ∈ IR^{n} satisfying z + h ∈ D_{f}_{Lθ}. We consider the following two cases
to complete the proof.

Case 1: For z_{2} 6= 0, z_{2} + h_{2} 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 (λ_{1}(z + h))

1 + ctan^{2}θ − f (λ_{1}(z))

1 + ctan^{2}θ − 1

1 + ctan^{2}θf^{0}(λ_{1}(z + h); h_{1} −(z_{2}+ h_{2})^{T}h_{2}

kz_{2}+ h_{2}k ctanθ)
+ f (λ_{2}(z + h))

1 + tan^{2}θ − f (λ_{2}(z))

1 + tan^{2}θ − 1

1 + tan^{2}θf^{0}(λ2(z + h); h1+(z_{2}+ h_{2})^{T}h_{2}

kz_{2}+ h_{2}k tan θ).

Because k · k is continuously differentiable over z_{2} 6= 0, it is strongly semismooth at z_{2}
by Lemma 3.1. Therefore,

kz2+ h2k = kz2k + (z_{2}+ h_{2})^{T}h_{2}

kz_{2}+ h_{2}k + O(kh2k^{2}) = kz2k + (z_{2}+ h_{2})^{T}h_{2}

kz_{2}+ h_{2}k + O(khk^{2}).

Combining this and the ρ-semismoothness of f at λ_{1}(z), we have
f (λ_{1}(z + h)) = f (λ_{1}(z)) + f^{0} λ_{1}(z + h)

λ_{1}(z + h) − λ_{1}(z) + O(|λ_{1}(z + h) − λ_{1}(z)|^{1+ρ})

= f (λ1(z)) + f^{0} λ1(z + h)

λ1(z + h) − λ1(z) + O(khk^{1+ρ})

= f (λ_{1}(z)) + f^{0} λ_{1}(z + h)

h_{1} − (kz_{2} + h_{2}k − kz_{2}k)ctanθ + O(khk^{1+ρ})

= f (λ_{1}(z)) + f^{0} λ_{1}(z + h)

h_{1} −(z_{2}+ h_{2})^{T}h_{2}
kz_{2}+ h_{2}k ctanθ

+ O(khk^{2}) + O(khk^{1+ρ})

= f (λ1(z)) + f^{0} λ1(z + h)

h1 −(z_{2}+ h_{2})^{T}h_{2}
kz_{2}+ h_{2}k 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 f^{0} since f is strictly continuous at λ_{1}(z). Similar argument holds for
f (λ_{2}(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 + ctan^{2}θf (λ_{1}(z + h)) z_{2}+ h_{2}
kz_{2}+ h_{2}k
+ ctanθ

1 + ctan^{2}θf^{0}

λ_{1}(z + h); h_{1}− (z_{2}+ h_{2})^{T}h_{2}
kz_{2}+ h_{2}k ctanθ

z_{2}+ h_{2}
kz_{2}+ h_{2}k
+ ctanθ

1 + ctan^{2}θf (λ1(z)) z_{2}

kz_{2}k + ctanθ
1 + ctan^{2}θ

f (λ_{1}(z + h))

kz_{2}+ h_{2}k M_{(z}_{2}_{+h}_{2}_{)}h. (20)
Note that Φ is continuous differentiable (and hence is semismooth) with ∇Φ(z_{2}) =

1

kz_{2}k(I − z2z^{T}_{2}

kz_{2}k^{2}) and M_{(z}_{2}_{+h}_{2}_{)}h = kz_{2}+ h_{2}k∇Φ(z_{2}+ h_{2})h_{2}. Thus, expression (20) can be
rewritten as

− ctanθ

1 + ctan^{2}θf (λ1(z + h))Φ(z2+ h2)
+ ctanθ

1 + ctan^{2}θf^{0}

λ_{1}(z + h); h_{1}− (z_{2}+ h_{2})^{T}h_{2}
kz_{2}+ h_{2}k ctanθ

Φ(z_{2}+ h_{2})
+ ctanθ

1 + ctan^{2}θf (λ_{1}(z))Φ(z_{2}) + ctanθ

1 + ctan^{2}θf λ_{1}(z + h)∇Φ(z_{2}+ h_{2})h_{2}

= ctanθ

1 + ctan^{2}θ

−f (λ_{1}(z + h)) + f (λ_{1}(z)) + f^{0}

λ_{1}(z + h); h_{1}− (z_{2}+ h_{2})^{T}h_{2}
kz_{2}+ h_{2}k ctanθ

Φ(z_{2} + h_{2})
+f (λ1(z))ctanθ

1 + ctan^{2}θ [−Φ(z_{2}+ h_{2}) + Φ(z_{2}) + ∇Φ(z_{2}+ h_{2})h_{2}]

+ ctanθ

1 + ctan^{2}θ∇Φ(z_{2}+ h_{2})h_{2}f λ_{1}(z + h) − f (λ_{1}(z))

= O

khk^{1+min{1,ρ}}

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

= O

khk^{1+min{1,ρ}}
.

The second equation comes from (19), strongly semismoothness of Φ at z_{2}, and

∇Φ(z_{2}+ h_{2})h_{2}h

f λ_{1}(z + h) − f (λ_{1}(z))i

= O(khk^{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 (18) involved λ_{2}(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 + ctan^{2}θ

h

f (λ_{1}(z + h)) − f (z_{1}) − f^{0} λ_{1}(z + h); h_{1}− kh_{2}kctanθi

+ 1

1 + tan^{2}θ
h

f (λ_{2}(z + h)) − f (z_{1}) − f^{0} λ_{2}(z + h); h_{1}+ kh_{2}k tan θi

= O(|h_{1}− kh_{2}kctanθ|^{1+ρ}) + O(|h_{1} + kh_{2}k tan θ|^{1+ρ})

= O(khk^{1+ρ}) (21)

because f is ρ-order semismooth at z_{1}. The second component of (18), by letting ¯z_{2} = ¯h_{2}
and M_{h}_{2}h = 0, takes the form

− ctanθ

1 + ctan^{2}θf λ_{1}(z + h)¯h_{2}+ ctanθ

1 + ctan^{2}θf^{0} λ_{1}(z + h); h_{1}− kh_{2}kctanθ¯h_{2}
+ tan θ

1 + tan^{2}θf λ2(z + h)¯h2− tan θ

1 + tan^{2}θf^{0} λ2(z + h); h1+ kh2k tan θ¯h2

= − 1

tan θ + ctanθ h

f λ_{1}(z + h) − f (z_{1}) − f^{0} λ_{1}(z + h); h_{1}− kh_{2}kctanθi¯h_{2}

+ 1

tan θ + ctanθ h

f λ_{2}(z + h) − f (z_{1}) − f^{0} λ_{2}(z + h); h_{1} + kh_{2}k tan θi¯h_{2}

= 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
λ_{1}(z) + t. We discuss the following two cases.

Case 1: For z_{2} 6= 0, from f being Lipschitz at λ_{2}(z) (and hence the differentiable points
are dense near λ_{2}(z)), there exists β(t) ∈ IR such that β(t) = O(|t|^{1+ρ}) and f is differen-
tiable at λ_{2}(z)+β(t) and λ_{2}(z)+β(t) > λ_{1}(z)+t as t sufficiently small (since λ_{2}(z) > λ_{1}(z)
by z2 6= 0). Denote h := tu^{1}_{z}+ β(t)u^{2}_{z}. Then, z + h = [λ1(z) + t]u^{1}_{z}+ [λ2(z) + β(t)]u^{2}_{z} which
implies λ_{1}(z + h) = λ_{1}(z) + t and λ_{2}(z + h) = λ_{2}(z) + β(t). Since f is differentiable at
λ_{1}(z) + t and λ_{2}(z) + β(t), f^{L}^{θ} is also differentiable at z + h by Lemma 2.2. Notice that

h =h_{1}
h_{2}

=

1

1 + ctan^{2}θt + 1

1 + tan^{2}θβ(t)

− ctanθ

1 + ctan^{2}θt + tan θ

1 + tan^{2}θβ(t)

z_{2}
kz_{2}k

and

z_{2}+ h_{2} =

kz_{2}k − ctanθ

1 + ctan^{2}θt + tan θ

1 + tan^{2}θβ(t)

z_{2}
kz_{2}k.
Hence,

(z_{2}+ h_{2})^{T}h_{2}

kz_{2} + h_{2}k = − ctanθ

1 + ctan^{2}θt + tan θ

1 + tan^{2}θβ(t),

which follows from the fact that kz_{2}k 6= 0 and t can be arbitrarily small (hence kz_{2}k −

ctanθ

1+ctan^{2}θt + _{1+tan}^{tan θ}2θβ(t) > 0). Thus, it is clear that
h_{1}− (z_{2}+ h_{2})^{T}h_{2}

kz_{2}+ h_{2}k ctanθ

= 1

1 + ctan^{2}θt + 1

1 + tan^{2}θβ(t)

−

− ctanθ

1 + ctan^{2}θt + tan θ

1 + tan^{2}θβ(t)
ctanθ

= t and

h_{1}+ (z_{2}+ h_{2})^{T}h_{2}
kz_{2}+ h_{2}k tan θ

= 1

1 + ctan^{2}θt + 1

1 + tan^{2}θβ(t)
+

− ctanθ

1 + ctan^{2}θt + tan θ

1 + tan^{2}θβ(t)
tan θ

= β(t).

In addition, it can be verified that

|h_{1}| =

1

1 + ctan^{2}θt + 1

1 + tan^{2}θβ(t)

≤ 1

1 + ctan^{2}θ|t| + 1

1 + tan^{2}θ|t| = |t|,
since β(t) = O(|t|^{1+ρ}) ≤ |t| as t is sufficiently small. Similarly,

kh_{2}k =

− ctanθ

1 + ctan^{2}θt + tan θ

1 + tan^{2}θβ(t)

z_{2}
kz_{2}k

=

− ctanθ

1 + ctan^{2}θt + tan θ

1 + tan^{2}θβ(t)

≤ ctanθ

1 + ctan^{2}θ|t| + tan θ

1 + tan^{2}θ|t| ≤ |t|.

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 + ctan^{2}θ +f λ2(z + h)

1 + tan^{2}θ − f (λ_{1}(z))

1 + ctan^{2}θ − f λ2(z)
1 + tan^{2}θ

−f^{0} λ_{1}(z + h)
1 + ctan^{2}θ

h_{1}− (z_{2}+ h_{2})^{T}h_{2}
kz_{2}+ h_{2}k ctanθ

−f^{0} λ_{2}(z + h)
1 + tan^{2}θ

h_{1} +(z_{2}+ h_{2})^{T}h_{2}
kz_{2}+ h_{2}k tan θ

= f λ_{1}(z) + t

1 + ctan^{2}θ +f λ_{2}(z) + β(t)

1 + tan^{2}θ − f (λ_{1}(z))

1 + ctan^{2}θ − f λ_{2}(z)
1 + tan^{2}θ

−f^{0} λ_{1}(z) + t

1 + ctan^{2}θ t − f^{0} λ_{2}(z) + β(t)
1 + tan^{2}θ β(t)

= 1

1 + ctan^{2}θ
h

f λ_{1}(z) + t − f λ_{1}(z) − f^{0} λ_{1}(z) + tti

+ 1

1 + tan^{2}θ
h

f λ_{2}(z) + β(t) − f (λ2(z)) − f^{0} λ_{2}(z) + β(t)β(t)i

= 1

1 + ctan^{2}θ
h

f λ_{1}(z) + t − f λ_{1}(z) − f^{0} λ_{1}(z) + tti

+ O(|t|^{1+ρ}),
where the last step is due to the fact that f^{0} is bounded and

f (λ_{2}(z) + β(t)) − f (λ_{2}(z)) = O(|t|^{1+ρ}),
since f is Lipschitz at λ_{2}(z). Hence (23) means

f (λ_{1}(z) + t) − f (λ_{1}(z)) − f^{0}(λ_{1}(z) + t) t = O(|t|^{1+ρ}),

which says f is ρ-order semismooth at λ_{1}(z). Applying similar arguments show that f is
ρ-order semismooth at λ_{2}(z).

Case 2: For z_{2} = 0, letting h = te. Since f is differentiable at λ_{1}(z) + t = z_{1} + t and
λ_{i}(z +h) = z_{1}+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 (z_{1}+ t) − f (z_{1}) − f^{0}(z_{1}+ t)t = O(|t|^{1+ρ}).