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 Lθ denote the circular cone in IRn. For a function f from IR to IR, one can define a corresponding vector-valued function fLθ on IRn by applying f to the spectral values of the spectral decomposition of x ∈ IRn with respect to Lθ. 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 Lθ denote the circular cone in IRn. Then, the n-dimensional circular cone Lθ is expressed as
Lθ := {x = (x1, x2)T ∈ IR × IRn−1| cos θkxk ≤ x1}.
The application of Lθ 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 Lθ was discovered, i.e., for any z = (z1, z2) ∈ IR × IRn−1, one has
z = λ1(z)u1z + λ2(z)u2z, (1) where
λ1(z) = z1− kz2kctanθ λ2(z) = z1+ kz2k tan θ and
u1z = 1 1 + ctan2θ
1 0
0 ctanθ · I
1
−¯z2
u2z = 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 fsoc (see [4, 5, 6]) and SDP-function fmat (see [7, 16]), we define a vector-valued function associated with circular cone as below. More specifically, for f : IR → IR, we define fLθ : IRn → IRn as
fLθ(z) = f (λ1(z)) u1z + f (λ2(z)) u2z. (2) It is not hard to see that fLθ is well-defined for all z. In particular, if z2 = 0, then
fLθ(z) = f (z1) 0
.
Note that when θ = 45◦, Lθ reduces to the second-order cone (SOC) and the vector- valued function fLθ defined as in (2) corresponds to the SOC-function fsoc given by
fsoc(x) = f (λ1(x))u(1)x + f (λ2(x))u(2)x ∀x = (x1, x2) ∈ IR × IRn−1 (3)
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 fsoc associated with second-order cone and matrix-valued function fmat associated with positive semidefinite cone play crucial role in the theory and numerical algorithm for second-order cone programming and semidefinite programming, respectively. In particular, many properties of fsoc and fmat are inherited from f , such as continuity, strictly continuity, directional differentiability, 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 fLθ. In [3], the authors answer the questions from the following aspects:
(a) fLθ is continuous at z ∈ IRn if and only if f is continuous at λi(z) for i = 1, 2;
(b) fLθ is directionally differentiable at z ∈ IRn if and only if f is directionally differ- entiable at λi(z) for i = 1, 2;
(c) fLθ is (Fr´echet) differentiable at z ∈ IRn if and only if f is (Fr´echet) differentiable at λi(z) for i = 1, 2;
(d) fLθ is continuously differentiable at z ∈ IRn if and only if f is continuously contin- uous at λi(z) for i = 1, 2;
(e) fLθ is strictly continuous at z ∈ IRn if and only if f is strictly continuous at λi(z) for i = 1, 2;
(f) fLθ is Lipschitz continuous with constant k > 0 if and only if f is Lipschitz contin- uous with constant k > 0;
(g) fLθ is semismooth at z if and only if f is semismooth at λi(z) for i = 1, 2.
In this paper, we further study some other properties associated with fLθ, 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 fsoc and fLθ always share the same properties. In- deed, they do not. There exists some property that holds for fsoc and f , but fails for fLθ 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 = xTy and kxk = √
xTx. For a linear mapping H : IRn → IRm, 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 → IRm, 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 − ykα < +∞;
(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 ∈ DFo , where DF 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(khk1+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) = kαF (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.
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 fLθ 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 fLθ be defined as in (2). Then, fLθ is directionally differentiable at z if and only if f is directionally differentiable at λi(z) for i = 1, 2.
Moreover, for any h = (h1, h2) ∈ IR × IRn−1, we have fLθ0
(z; h) = f0(z1; h1) 0
= f0(z1; h1)e when z2 = 0 and h2 = 0;
fLθ0
(z; h) = 1
1 + ctan2θf0(z1; h1− kh2kctanθ) 1 0 0 ctanθ · I
"
1
−khh2
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
fLθ0
(z; h) = 1
1 + ctan2θf0
λ1(z); h1− zT2h2
kz2kctanθ 1 0 0 ctanθ · I
"
1
−kzz2
2k
#
− ctanθ 1 + ctan2θ
f (λ1(z)) kz2k Mz2h
+ 1
1 + tan2θf0
λ2(z); h1+z2Th2
kz2k tan θ 1 0 0 tan θ · I
"
1
z2
kz2k
#
+ tan θ 1 + tan2θ
f (λ2(z)) kz2k Mz2h, where
Mz2 :=
0 0
0 I − z2zT2 kz2k2
.
Lemma 2.2. Let f : IR → IR and fLθ be defined as in (2). Then, the following hold.
(a) fLθ is differentiable at z if and only if f is differentiable at λi(z) for i = 1, 2.
Moreover, if z2 = 0, then
∇fLθ(z) = f0(z1)I;
otherwise
∇fLθ(z) =
ξ %z2T
kz2k
%z2
kz2k aI + (η − a)z2zT2 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) fLθ is continuously differentiable (smooth) at z if and only if f is continuously differentiable (smooth) at λi(z) for i = 1, 2.
Note that the formula of gradient ∇fLθ given in [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 λi is Lipschitz continuous on IRn for i = 1, 2 will be used in proving the Lipschitz continuity between fLθ and f .
Lemma 2.3. Let z, y ∈ IRn with spectral values λi(z), λi(y), respectively. Then, we have
|λi(z) − λi(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.
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 fLθ be defined as in (2). Then, fLθ is strictly continuous (local Lipschitz continuity) at z if and only if f is strictly continuous (local Lipschitz continuity) at λi(z) for i = 1, 2.
Proof. “⇐” Suppose f is strictly continuous at λi(z) for i = 1, 2, i.e., there exists ki > 0 and δi > 0 for i = 1, 2 such that
|f (τ ) − f (ζ)| ≤ ki|τ − ζ| ∀τ, ζ ∈ [λi(z) − δi, λi(z) + δi], i = 1, 2.
Let ¯δ := min{δ1, δ2} and C := [λ1(z) − ¯δ, λ1(z) + ¯δ] ∪ [λ2(z) − ¯δ, λ2(z) + ¯δ]. Define
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 fv : IR → IR for v = 1, 2, · · · converging uniformly to ˜f on eC such that
|(fv)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
(fv)Lθ(w) − fLθ(w)
2
=
[fv(λ1(w)) − f (λ1(w))] u1w+ [fv(λ2(w)) − f (λ2(w))] u2w
2
= fv(λ1(w)) − f (λ1(w))2
ku1wk2+fv(λ2(w)) − f (λ2(w))2
ku2wk2
= 1
1 + ctan2θ|fv(λ1(w)) − f (λ1(w))|2+ 1
1 + tan2θ |fv(λ2(w)) − f (λ2(w))|2, where we have used the facts that ku1wk = 1/√
1 + ctan2θ, ku2wk = 1/√
1 + tan2θ, and hu1w, u2wi = 0. Since {fv}∞v=1 converges uniformly to f on eC, the above equations show that {(fv)Lθ}∞v=1 converges uniformly to fLθ on B(z, δ). If w2 = 0, then it follows from Lemma 2.2 that ∇(fv)Lθ(w) = (fv)0(w1)I. Hence it follows from (6) that
k∇(fv)Lθ(w)k = |(fv)0(w1)| ≤ k, (7)
since in this case λi(w) = w1 ∈ eC. If w2 6= 0, then
∇(fv)Lθ(w) =
ξ % w2T/kw2k
% w2/kw2k aI + (ξ − %(ctanθ − tan θ) − a)w2wT2 kw2k2
=
ξ % wT2/kw2k
% w2/kw2k aI + (ξ − a)w2w2T kw2k2
+
0 0
0 [−%(ctanθ − tan θ)] w2w2T kw2k2
=
ξ % w2T/kw2k
% w2/kw2k ξI
+ (a − ξ)
0 0
0 I − w2wT2 kw2k2
−%(ctanθ − tan θ)
0 0
0 w2w2T kw2k2
where a, ξ, % are given as in (5) with λi(z) replaced by λi(w) for i = 1, 2 and f replaced by fv. For simplicity of notations, let us denote
A :=
ξ % wT2/kw2k
% w2/kw2k ξI
+ (a − ξ)
0 0
0 I − w2wT2 kw2k2
and
B := −%(ctanθ − tan θ)
0 0
0 w2wT2 kw2k2
. Note that
|a| =
fv λ2(w) − fv λ1(w) λ2(w) − λ1(w)
≤ k, (8)
where the inequality comes from the fact that fv is continuously differentiable on eC and (6). Besides, we also note that
|ξ| =
(fv)0 λ1(w)
1 + ctan2θ +(fv)0 λ2(w) 1 + tan2θ
≤ 1
1 + ctan2θ|(fv)0 λ1(w)| + 1
1 + tan2θ|(fv)0 λ2(w)|
≤
1
1 + ctan2θ + 1 1 + tan2θ
k
= k (9)
and
|%| = | − ctanθ
1 + ctan2θ(fv)0(λ1(w)) + tan θ
1 + tan2θ(fv)0(λ2(w))|
≤
− 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 ∇(fv)Lθ(w) takes the form of ξI + (a − ξ)Mw2 whose eigenvalues are ξ and a by [6, Lemma 1]. In other words, in this case, we get from (8) and (9) that
k∇(fv)Lθ(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(fv)0 λ1(w) + 1 − tan2θ
1 + tan2θ(fv)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 = %
kw2kLw˜ + (a − ξ)Mw2 = % kw2k
Lw˜ + (a − ξ)kw2k
% Mw˜2
,
where ˜w = ξkw2k/%, w2 and
Lw˜ := ˜w1 w˜2T
˜ w2 w˜1I
.
In this case the matrix A has eigenvalues of ξ ± % and a with multiplicity n − 2. Hence, k∇(fv)Lθ(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∇(fv)Lθ(w)k ≤ 4k ∀w ∈ B(z, δ) and ∀v.
Fix any x, y ∈ B(z, δ) with x 6= y. Since {(fv)Lθ}∞v=1 converges uniformly to fLθ on B(z, δ), then for any > 0 there exists v0 such that
k(fv)Lθ(w) − fLθ(w)k ≤ , ∀w ∈ B(z, δ) and ∀v ≥ v0.
Since fv is continuously differentiable, (fv)Lθ is continuously differentiable by Lemma 2.2. Thus,
kfLθ(x) − fLθ(y)k
= kfLθ(x) − (fv)Lθ(x) + (fv)Lθ(x) − (fv)Lθ(y) + (fv)Lθ(y) − fLθ(y)k
≤ kfLθ(x) − (fv)Lθ(x)k + k(fv)Lθ(x) − (fv)Lθ(y)k + k(fv)Lθ(y) − fLθ(y)k
≤ 2 +
Z 1 0
∇(fv)Lθ(y + t(x − y))(x − y)dt
≤ 2 + 4kkx − yk.
Because > 0 is arbitrary, this ensures fLθ(x) − fLθ(y)
≤ 4kkx − yk ∀x, y ∈ B(z, δ), which says fLθ is strictly continuous at z.
“⇒” Suppose fLθ is strictly continuous at z, then there exist k > 0 and δ > 0 such that kfLθ(x) − fLθ(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 := θu1z+ λ2(z)u2z and y := µu1z+ λ2(z)u2z. Then, kx − zk ≤ δ and ky − zk ≤ δ and it follows from (14) that
|f (θ) − f (µ)| = 1
ku1zkkfLθ(x) − fLθ(y)k ≤ k
ku1zkkx − yk = k
ku1zk|θ − µ|ku1zk = 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
= kfLθ(θe) − fLθ(µe)k ≤ kkθe − µek = k|θ − µ|.
This means f is strictly continuous at λi(z) = z1 for i = 1, 2. 2
Remark 2.1. As mentioned in the Section of Introduction, the strict continuity between fLθ 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∇(fv)Lθk 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 fLθ be defined as in (2). Then, fLθ is B-differentiable at z if and only if f is B-differentiable at λi(z) for i = 1, 2.
Next, inspired by [2, 19], we further study the H¨older continuity relation between f and fLθ.
Theorem 2.3. Let f : IR → IR and fLθ be defined as in (2). Then, fLθ 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,
kfLθ(z) − fLθ(y)k
= kf (λ1(z))u1z+ f (λ2(z))u2z− f (λ1(y))u1y − f (λ2(y))u2yk
=
f (λ1(z))u1z− u1y + f (λ2(z))u2z − u2y
+f (λ1(z)) − f (λ1(y))u1y+f (λ2(z)) − f (λ2(y))u2y
≤
f (λ1(z))u1z− u1y + f (λ2(z))u2z − u2y + |f (λ1(z)) − f (λ1(y))| ·
u1y
+ |f (λ2(z)) − f (λ2(y))| · u2y
. Let us analyze each term in the above inequality. First, we look into the first term:
f (λ1(z))u1z− u1y + f (λ2(z))u2z− u2y
= 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)αkz2kα·
z2
kz2k− y2 ky2k
≤ tan θ
1 + tan2θ[f ]α(tan θ + ctanθ)αkz2kα 2
kz2kkz2− y2k
= 2 tan θ
1 + tan2θ[f ]α(tan θ + ctanθ)α
z2− y2 kz2k
1−α
kz2− y2kα
≤ tan θ
1 + tan2θ[f ]α(tan θ + ctanθ)α22−αkz2− y2kα
≤ tan θ
1 + tan2θ[f ]α(tan θ + ctanθ)α22−αkz − ykα, (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))|ku1yk ≤ [f ]α|λ1(z) − λ1(y)|α 1
√
1 + ctan2θ
≤ [f ]α √
2 max{tan θ, ctanθ}α
kz − ykα. (16) Similarly, the third term also satisfies
|f (λ2(z)) − f (λ2(y))|ku2yk ≤ [f ]α|λ2(z) − λ2(y)|α 1
√1 + tan2θ
≤ [f ]α √
2 max{tan θ, ctanθ}α
kz − ykα. (17) Combining (15)-(17) proves that fLθ is H¨older continuous with exponent α ∈ (0, 1].
Case 2: either z2 = 0 or y2 = 0. In this case, we take uiz = uiy for i = 1, 2 according to the spectral decomposition. Therefore, we obtain
kfLθ(z) − fLθ(y)k
= kf (λ1(z))u1z+ f (λ2(z))u2z− f (λ1(y))u1y− f (λ2(y))u2yk
=
f (λ1(z)) − f (λ1(y))u1z +f (λ2(z)) − f (λ2(y))u2z
≤ |f (λ1(z)) − f (λ1(y))| · ku1zk + |f (λ2(z)) − f (λ2(y))| · ku2zk
≤ [f ]α|λ1(z) − λ1(y)|α 1
√1 + ctan2θ + [f ]α|λ2(y) − λ2(z)|α 1
√1 + tan2θ
≤ 2[f ]α √
2 max{tan θ, ctanθ}α
kz − ykα which says fLθ is H¨older continuous.
“⇒” Recall that fLθ(τ e) = (f (τ ), 0)T. Hence for any τ, ζ ∈ IR,
|f (τ ) − f (ζ)| =
fLθ(τ e) − fLθ(ζe)
≤ fLθ
α· kτ e − ζekα
= fLθ
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 fLθ and f has been given in [3, Theorem 4.1]. But the exact formula of the B-subdifferential ∂B(fLθ) 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 fLθ 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 fLθ be defined as in (2). Given ρ > 0, then the following statements hold.
(a) If f is ρ-order semismooth at λi(z) for i = 1, 2, then fLθ is min{1, ρ}-order semis- mooth at z;
(b) If fLθ is ρ-order semismooth at z, then f is ρ-semismooth at λi(z) for i = 1, 2;
(c) For z2 = 0, fLθ is ρ-semismooth at z if and only if f is ρ-order semismooth at λi(z) = z1 for i = 1, 2.
Proof. (a) Take h ∈ IRn satisfying z + h ∈ DfLθ. 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 ∈ DfLθ, we know λi(z + h) ∈ Df for i = 1, 2 by Lemma 2.2. Then, according to Lemma 2.1, the first component of
fLθ(z + h) − fLθ(z) − (fLθ)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).
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(khk1+ρ)
= f (λ1(z)) + f0 λ1(z + h)
h1 − (kz2 + h2k − kz2k)ctanθ + O(khk1+ρ)
= f (λ1(z)) + f0 λ1(z + h)
h1 −(z2+ h2)Th2 kz2+ h2k ctanθ
+ O(khk2) + O(khk1+ρ)
= f (λ1(z)) + f0 λ1(z + h)
h1 −(z2+ h2)Th2 kz2+ h2k ctanθ
+ O
khk1+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(khk1+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 − z2zT2
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
λ1(z + h); h1− (z2+ h2)Th2 kz2+ h2k ctanθ
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
λ1(z + h); h1− (z2+ h2)Th2 kz2+ h2k ctanθ
Φ(z2 + h2) +f (λ1(z))ctanθ
1 + ctan2θ [−Φ(z2+ h2) + Φ(z2) + ∇Φ(z2+ h2)h2]
+ ctanθ
1 + ctan2θ∇Φ(z2+ h2)h2f λ1(z + h) − f (λ1(z))
= O
khk1+min{1,ρ}
+ O(khk2) + O(khk2)
= O
khk1+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
fLθ(z + h) − fLθ(z) − (fLθ)0(z + h; h) = O(khk1+min{1,ρ}), which says fLθ is min{1, ρ}-order semismooth at z under this case.
Case 2: For z2 = 0, if h2 = 0, then the proof is trivial. If h2 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(khk1+ρ) (21)
because f is ρ-order semismooth at z1. The second component of (18), by letting ¯z2 = ¯h2 and Mh2h = 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(khk1+ρ), (22)
where the last step is due to the ρ-order semismoothness of f .
(b) Suppose fLθ is ρ-order semismooth at z. Let t ∈ IR such that f is differentiable at λ1(z) + t. We discuss the following two cases.
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 differen- tiable 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 := tu1z+ β(t)u2z. Then, z + h = [λ1(z) + t]u1z+ [λ2(z) + β(t)]u2z which implies λ1(z + h) = λ1(z) + t and λ2(z + h) = λ2(z) + β(t). Since f is differentiable at λ1(z) + t and λ2(z) + β(t), fLθ is also differentiable at z + h by Lemma 2.2. 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 + ctan2θt + tan θ
1 + tan2θβ(t)
z2 kz2k. Hence,
(z2+ h2)Th2
kz2 + h2k = − ctanθ
1 + ctan2θt + tan θ
1 + tan2θβ(t),
which follows from the fact that kz2k 6= 0 and t can be arbitrarily small (hence kz2k −
ctanθ
1+ctan2θt + 1+tantan θ2θβ(t) > 0). Thus, it is clear that h1− (z2+ h2)Th2
kz2+ h2k ctanθ
= 1
1 + ctan2θt + 1
1 + tan2θβ(t)
−
− ctanθ
1 + ctan2θt + tan θ
1 + tan2θβ(t) ctanθ
= t and
h1+ (z2+ h2)Th2 kz2+ h2k tan θ
= 1
1 + ctan2θt + 1
1 + tan2θβ(t) +
− ctanθ
1 + ctan2θt + tan θ
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 + ctan2θt + tan θ
1 + tan2θβ(t)
z2 kz2k
=
− ctanθ
1 + ctan2θt + tan θ
1 + tan2θβ(t)
≤ ctanθ
1 + ctan2θ|t| + tan θ
1 + tan2θ|t| ≤ |t|.
Therefore, we obtain khk = O(t) which further implies O(khk1+ρ) = O(|t|1+ρ). Then, by the hypothesis fLθ being ρ-order semismooth at z, i.e.,
fLθ(z + h) − fLθ(z) − (fLθ)0(z + h; h) = O(khk1+ρ), we have
fLθ(z + h) − fLθ(z) − (fLθ)0(z + h; h), e = O(khk1+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 λi(z +h) = z1+t for i = 1, 2, fLθ is differentiable at z +h by Lemma 2.2, i.e., z +h ∈ DfLθ. Because fLθ is ρ-order semismooth at z, we have
fLθ(z + h) − fLθ(z) − (fLθ)0(z + h; h) = O(khk1+ρ) which, together with the fact khk = |t|, is equivalent to
f (z1+ t) − f (z1) − f0(z1+ t)t = O(|t|1+ρ).