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# 1.Introduction JinchuanZhou andJein-ShanChen TheVector-ValuedFunctionsAssociatedwithCircularCones ResearchArticle

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## The Vector-Valued Functions Associated with Circular Cones

1

### and Jein-Shan Chen

2,3

1Department of Mathematics, School of Science, Shandong University of Technology, Zibo, Shandong 255049, China

2Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan

3Mathematics Division, National Center for Theoretical Sciences, Taipei, Taiwan

Correspondence should be addressed to Jein-Shan Chen; jschen@math.ntnu.edu.tw Received 6 April 2014; Accepted 15 May 2014; Published 22 June 2014

Copyright © 2014 J. Zhou and J.-S. Chen. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The circular cone is a pointed closed convex cone having hyperspherical sections orthogonal to its axis of revolution about which the cone is invariant to rotation, which includes second-order cone as a special case when the rotation angle is 45 degrees. LetL𝜃 denote the circular cone inR𝑛. For a function𝑓 from R to R, one can define a corresponding vector-valued function 𝑓L𝜃onR𝑛 by applying𝑓 to the spectral values of the spectral decomposition of 𝑥 ∈ R𝑛with respect toL𝜃. In this paper, we study properties that this vector-valued function inherits from𝑓, including H¨older continuity, 𝐵-subdifferentiability, 𝜌-order semismoothness, and positive homogeneity. These results will play crucial role in designing solution methods for optimization problem involved in circular cone constraints.

### 1. Introduction

The circular cone is a pointed closed convex cone having hyperspherical sections orthogonal to its axis of revolution about which the cone is invariant to rotation, which includes second-order cone as a special case when the rotation angle is 45 degrees. LetL𝜃denote the circular cone inR𝑛. Then, the 𝑛-dimensional circular cone L𝜃is expressed as

L𝜃:= {𝑥 = (𝑥1, 𝑥2)𝑇∈ R × R𝑛−1| cos 𝜃 ‖𝑥‖ ≤ 𝑥1} . (1)

The application ofL𝜃lies in engineering field, for example, optimal grasping manipulation for multigingered robots;

see [1].

In our previous work [2], we have explored some impor- tant features about circular cone, such as characterizing its tangent cone, normal cone, and second-order regularity. In particular, the spectral decomposition associated with L𝜃 was discovered; that is, for any𝑧 = (𝑧1, 𝑧2) ∈ R × R𝑛−1, one has

𝑧 = 𝜆1(𝑧) 𝑢1𝑧+ 𝜆2(𝑧) 𝑢2𝑧, (2)

where

𝜆1(𝑧) = 𝑧1− 󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩ctan𝜃, 𝜆2(𝑧) = 𝑧1+ 󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩tan𝜃, 𝑢1𝑧= 1

1 + ctan2𝜃[1 0

0 ctan 𝜃 ⋅ 𝐼] [ 1

−𝑧2] , 𝑢2𝑧= 1

1 + tan2𝜃[1 0 0 tan 𝜃 ⋅ 𝐼] [1

𝑧2] ,

(3)

with𝑧2:= 𝑧2/‖𝑧2‖ if 𝑧2 ̸= 0, and 𝑧2being any vector𝑤 ∈ R𝑛−1 satisfying‖𝑤‖ = 1 if 𝑧2= 0. With this spectral decomposition (2), analogous to the so-called SOC-function𝑓soc (see [3–

5]) and SDP-function𝑓mat (see [6, 7]), we define a vector- valued function associated with circular cone as below. More specifically, for𝑓 : R → R, we define 𝑓L𝜃: R𝑛 → R𝑛as

𝑓L𝜃(𝑧) = 𝑓 (𝜆1(𝑧)) 𝑢1𝑧+ 𝑓 (𝜆2(𝑧)) 𝑢𝑧2. (4) It is not hard to see that 𝑓L𝜃 is well-defined for all𝑧. In particular, if𝑧2= 0, then

𝑓L𝜃(𝑧) = [𝑓 (𝑧1)

0 ] . (5)

Volume 2014, Article ID 603542, 21 pages http://dx.doi.org/10.1155/2014/603542

(2)

Note that when𝜃 = 45,L𝜃reduces to the second-order cone (SOC) and the vector-valued function𝑓L𝜃defined as in (4) corresponds to the SOC-function𝑓socgiven by

𝑓soc(𝑥) = 𝑓 (𝜆1(𝑥)) 𝑢(1)𝑥 + 𝑓 (𝜆2(𝑥)) 𝑢(2)𝑥 ,

∀𝑥 = (𝑥1, 𝑥2) ∈ R × R𝑛−1, (6) where𝜆𝑖(𝑥) = 𝑥1+ (−1)𝑖‖𝑥2‖ and 𝑢(𝑖)𝑥 = (1/2)(1, (−1)𝑖𝑥2)𝑇.

It is well known that the vector-valued function𝑓socasso- ciated 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 of𝑓socand𝑓mat are inherited from𝑓, such as continuity, strictly continuity, directional differentiability, Fr´echet differentiability, continu- ous differentiability, and semismoothness. It should be men- tioned 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 for𝑓L𝜃. In [1], the authors answer the questions from the following aspects:

(a)𝑓L𝜃is continuous at𝑧 ∈ R𝑛if and only if𝑓 is contin- uous at𝜆𝑖(𝑧) for 𝑖 = 1, 2;

(b)𝑓L𝜃 is directionally differentiable at𝑧 ∈ R𝑛 if and only if 𝑓 is directionally differentiable at 𝜆𝑖(𝑧) for 𝑖 = 1, 2;

(c)𝑓L𝜃is (Fr´echet) differentiable at𝑧 ∈ R𝑛if and only if 𝑓 is (Fr´echet) differentiable at 𝜆𝑖(𝑧) for 𝑖 = 1, 2;

(d)𝑓L𝜃 is continuously differentiable at 𝑧 ∈ R𝑛 if and only if𝑓 is continuously continuous at 𝜆𝑖(𝑧) for 𝑖 = 1, 2;

(e)𝑓L𝜃is strictly continuous at𝑧 ∈ R𝑛if and only if𝑓 is strictly continuous at𝜆𝑖(𝑧) for 𝑖 = 1, 2;

(f)𝑓L𝜃is Lipschitz continuous with constant𝑘 > 0 if and only if𝑓 is Lipschitz continuous with constant 𝑘 > 0;

(g)𝑓L𝜃is semismooth at𝑧 if and only if 𝑓 is semismooth at𝜆𝑖(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 Hada- mard sense, the characterization of B-subdifferential, positive homogeneity, and boundedness. Of course, one may wonder whether 𝑓soc and 𝑓L𝜃 always share the same properties.

Indeed, they do not. There exists some property that holds for𝑓socand𝑓 but fails for 𝑓L𝜃 and𝑓. A counterexample is presented in the final section.

To end the third section, we briefly review our notations and some basic concepts which will be needed for subsequent analysis. First, we denote byR𝑛the space of𝑛-dimensional real column vectors and let𝑒 = (1, 0, . . . , 0) ∈ R𝑛. Given 𝑥, 𝑦 ∈ R𝑛, the Euclidean inner product and norm are⟨𝑥, 𝑦⟩=

𝑥𝑇𝑦 and ‖𝑥‖ = √𝑥𝑇𝑥. For a linear mapping 𝐻 : R𝑛 → R𝑚,

its operator norm is‖𝐻‖ := max‖𝑥‖=1‖𝐻𝑥‖. For 𝛼 ∈ R and 𝑠 ∈ R𝑛, we write 𝑠 = 𝑂(𝛼) (resp., 𝑠 = 𝑜(𝛼)) to means

‖𝑠‖/|𝛼| is uniformly bounded (resp., tends to zero) as 𝛼 → 0.

In addition, given a function𝐹 : R𝑛 → R𝑚, we say the following:

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

[𝐹]𝛼:= sup

𝑥 ̸= 𝑦

󵄩󵄩󵄩󵄩𝐹(𝑥) − 𝐹(𝑦)󵄩󵄩󵄩󵄩

󵄩󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩󵄩𝛼 < +∞; (7) (b)𝐹 is directionally differentiable at 𝑥 ∈ R𝑛in the Hada- mard sense, if the directional derivative𝐹󸀠(𝑥; 𝑑) exists for all𝑑 ∈ R𝑛and

𝐹󸀠(𝑥; 𝑑) = lim

𝑑󸀠→ 𝑑 𝑡↓0

𝐹 (𝑥 + 𝑡𝑑󸀠) − 𝐹 (𝑥)

𝑡 ; (8)

(c)𝐹 is 𝐵-differentiable (Bouligand-differentiable) at 𝑥, if𝐹 is Lipschitz continuous near 𝑥 and directionally differentiable at𝑥;

(d) if𝐹 is strictly continuous (locally Lipschitz continu- ous), the generalized Jacobian𝜕𝐹(𝑥) is the convex hull of the𝜕𝐵𝐹(𝑥), where

𝜕𝐵𝐹 (𝑥) := { lim𝑧 → 𝑥∇𝐹 (𝑧) | 𝑧 ∈ 𝐷𝐹} , (9)

where𝐷𝐹denotes 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) − 𝑉ℎ = 𝑜 (‖ℎ‖) ; (10)

(f)𝐹 is 𝜌-order semismooth at 𝑥 (𝜌 > 0) if 𝐹 is semi- smooth at𝑥 and for any 𝑉 ∈ 𝜕𝐹(𝑥 + ℎ),

𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) − 𝑉ℎ = 𝑂 (‖ℎ‖1+𝜌) ; (11)

in particular, we say𝐹 is strongly semismooth if it is 1-order semismooth;

(g)𝐹 is positively homogeneous with exponent 𝛼 > 0, if 𝐹 (𝑘𝑥) = 𝑘𝛼𝐹 (𝑥) , ∀𝑥 ∈ R𝑛, 𝑘 ≥ 0; (12)

(h)𝐹 is bounded if there exists a positive scalar 𝑀 > 0 such that

‖𝐹 (𝑥)‖ ≤ 𝑀, ∀𝑥 ∈ R𝑛. (13)

(3)

### Strict Continuity, Hölder Continuity,and𝐵-Differentiability

This section is devoted to study the properties of directional differentiability, strict continuity, and H¨older continuity. The relationship of directional differentiability between𝑓L𝜃 and 𝑓 has been given in [1, 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 follows.

Lemma 1. Let 𝑓 : R → R and 𝑓L𝜃 be defined as in (4).

Then,𝑓L𝜃is directionally differentiable at𝑧 if and only if 𝑓 is directionally differentiable at𝜆𝑖(𝑧) for 𝑖 = 1, 2. Moreover, for anyℎ = (ℎ1, ℎ2) ∈ R × R𝑛−1, we have

(𝑓L𝜃)󸀠(𝑧; ℎ) = [𝑓󸀠(𝑧1; ℎ1)

0 ] = 𝑓󸀠(𝑧1; ℎ1) 𝑒, (14) when𝑧2= 0 and ℎ2= 0. Consider

(𝑓L𝜃)󸀠(𝑧; ℎ) = 1

1 + ctan2𝜃𝑓󸀠(𝑧1; ℎ1− 󵄩󵄩󵄩󵄩ℎ2󵄩󵄩󵄩󵄩 ctan 𝜃)

× [1 0

0 ctan 𝜃 ⋅ 𝐼][[ [

1

− ℎ2

󵄩󵄩󵄩󵄩ℎ2󵄩󵄩󵄩󵄩

]] ]

+ 1

1 + tan2𝜃𝑓󸀠(𝑧1; ℎ1+ 󵄩󵄩󵄩󵄩ℎ2󵄩󵄩󵄩󵄩tan𝜃)

× [1 0

0 tan 𝜃 ⋅ 𝐼][[ [

1 ℎ2

󵄩󵄩󵄩󵄩ℎ2󵄩󵄩󵄩󵄩

]] ] ,

(15) when𝑧2= 0 and ℎ2 ̸= 0; otherwise,

(𝑓L𝜃)󸀠(𝑧; ℎ) = 1

1 + ctan2𝜃𝑓󸀠(𝜆1(𝑧) ; ℎ1−𝑧2𝑇2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 ctan 𝜃)

× [1 0

0 ctan 𝜃 ⋅ 𝐼][ [

1

− 𝑧2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩] ]

− ctan𝜃 1 + ctan2𝜃

𝑓 (𝜆1(𝑧))

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 𝑀𝑧2

+ 1

1 + tan2𝜃𝑓󸀠(𝜆2(𝑧) ; ℎ1+𝑧𝑇22

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 tan𝜃)

× [1 0

0 tan 𝜃 ⋅ 𝐼][ [

𝑧12

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩] ] + tan𝜃

1 + tan2𝜃

𝑓 (𝜆2(𝑧))

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 𝑀𝑧2ℎ,

(16)

where

𝑀𝑧2 := [[ [

0 0

0 𝐼 − 𝑧2𝑧𝑇2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩2 ]] ]

. (17)

Lemma 2. Let 𝑓 : R → R and 𝑓L𝜃be defined as in (4). Then, the following hold.

(a)𝑓L𝜃is differentiable at𝑧 if and only if 𝑓 is differentiable at𝜆𝑖(𝑧) for 𝑖 = 1, 2. Moreover, if 𝑧2= 0, then

∇𝑓L𝜃(𝑧) = 𝑓󸀠(𝑧1) 𝐼; (18) otherwise,

∇𝑓L𝜃(𝑧) = [[ [[ [ [

𝜉 󰜚𝑧𝑇2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩

󰜚𝑧2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 𝑎𝐼 + (𝜂 − 𝑎) 𝑧2𝑧𝑇2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩2 ]] ]] ] ]

, (19)

where 𝑎 = tan𝜃

1 + tan2𝜃

𝑓 (𝜆2(𝑧))

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 − ctan𝜃 1 + ctan2𝜃

𝑓 (𝜆1(𝑧))

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩

= 𝑓 (𝜆2(𝑧)) − 𝑓 (𝜆1(𝑧)) 𝜆2(𝑧) − 𝜆1(𝑧) ,

𝜉 = 𝑓󸀠(𝜆1(𝑧))

1 + ctan2𝜃+𝑓󸀠(𝜆2(𝑧)) 1 + tan2𝜃 , 𝜂 = 𝜉 − 󰜚 ( ctan 𝜃 − tan 𝜃) ,

󰜚 = − ctan𝜃

1 + ctan2𝜃𝑓󸀠(𝜆1(𝑧)) + tan𝜃

1 + tan2𝜃𝑓󸀠(𝜆2(𝑧)) . (20) (b)𝑓L𝜃is continuously differentiable (smooth) at𝑧 if and only if𝑓 is continuously differentiable (smooth) at 𝜆𝑖(𝑧) for𝑖 = 1, 2.

Note that the formula of gradient ∇𝑓L𝜃 given in [1, Theorem 3.3] andLemma 2is the same by using the following facts:

1

1 + ctan2𝜃 = sin2𝜃, 1

1 + tan2𝜃 = cos2𝜃, ctan𝜃

1 + ctan2𝜃 = tan𝜃

1 + tan2𝜃 = sin 𝜃 cos 𝜃.

(21)

The following result indicating that𝜆𝑖is Lipschitz contin- uous onR𝑛for𝑖 = 1, 2 will be used in proving the Lipschitz continuity between𝑓L𝜃and𝑓.

Lemma 3. Let 𝑧, 𝑦 ∈ R𝑛 with spectral values 𝜆𝑖(𝑧), 𝜆𝑖(𝑦), respectively. Then, we have

󵄨󵄨󵄨󵄨𝜆𝑖(𝑧) − 𝜆𝑖(𝑦)󵄨󵄨󵄨󵄨 ≤ √2 max {tan 𝜃, ctan 𝜃}󵄩󵄩󵄩󵄩𝑧 − 𝑦󵄩󵄩󵄩󵄩, 𝑓𝑜𝑟 𝑖 = 1, 2. (22)

(4)

Proof. First, we observe that

󵄨󵄨󵄨󵄨𝜆1(𝑧) − 𝜆1(𝑦)󵄨󵄨󵄨󵄨

= 󵄨󵄨󵄨󵄨𝑧1− 󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩ctan𝜃 − 𝑦1+ 󵄩󵄩󵄩󵄩𝑦2󵄩󵄩󵄩󵄩ctan𝜃󵄨󵄨󵄨󵄨

≤ 󵄨󵄨󵄨󵄨𝑧1− 𝑦2󵄨󵄨󵄨󵄨 + 󵄩󵄩󵄩󵄩𝑧2− 𝑦2󵄩󵄩󵄩󵄩ctan𝜃

≤ max {1, ctan 𝜃} (󵄨󵄨󵄨󵄨𝑧1− 𝑦1󵄨󵄨󵄨󵄨 + 󵄩󵄩󵄩󵄩𝑧2− 𝑦2󵄩󵄩󵄩󵄩)

≤ max {1, ctan 𝜃} √2√󵄨󵄨󵄨󵄨𝑧1− 𝑦1󵄨󵄨󵄨󵄨2+ 󵄩󵄩󵄩󵄩𝑧2− 𝑦2󵄩󵄩󵄩󵄩2

= max {1, ctan 𝜃} √2 󵄩󵄩󵄩󵄩𝑧 − 𝑦󵄩󵄩󵄩󵄩.

(23)

Applying the similar argument to𝜆2yields

󵄨󵄨󵄨󵄨𝜆2(𝑧) − 𝜆2(𝑦)󵄨󵄨󵄨󵄨 ≤ max {1, tan 𝜃} √2󵄩󵄩󵄩󵄩𝑧 − 𝑦󵄩󵄩󵄩󵄩. (24) Then, the desired result follows from the fact that max{1, ctan𝜃, tan 𝜃} max{ctan 𝜃, tan 𝜃}.

Theorem 4. Let 𝑓 : R → R and 𝑓L𝜃 be defined as in (4).

Then,𝑓L𝜃is strictly continuous (local Lipschitz continuity) at𝑧 if and only if𝑓 is strictly continuous (local Lipschitz continuity) at𝜆𝑖(𝑧) for 𝑖 = 1, 2.

Proof. “⇐” Suppose that 𝑓 is strictly continuous at 𝜆𝑖(𝑧), for 𝑖 = 1, 2; that is, there exist 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. (25) Let𝛿 := min{𝛿1, 𝛿2} and 𝐶 := [𝜆1(𝑧) − 𝛿, 𝜆1(𝑧) + 𝛿] ∪ [𝜆2(𝑧) − 𝛿, 𝜆2(𝑧) + 𝛿]. Define

𝑓 (𝜏)̃

:=

{{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {

𝑓 (𝜏) if𝜏 ∈ 𝐶,

(1 − 𝑡) 𝑓 (𝜆1(𝑧) + 𝛿)

+ 𝑡𝑓 (𝜆2(𝑧) − 𝛿) if𝜆1(𝑧) + 𝛿 < 𝜆2(𝑧) − 𝛿, 𝜏 = (1 − 𝑡) (𝜆1(𝑧) + 𝛿)

+ 𝑡 (𝜆2(𝑧) − 𝛿) with𝑡 ∈ (0, 1) 𝑓 (𝜆1(𝑧) − 𝛿) if𝜏 < 𝜆1(𝑧) − 𝛿 𝑓 (𝜆2(𝑧) + 𝛿) if𝜏 > 𝜆2(𝑧) + 𝛿.

(26) Clearly, ̃𝑓 is Lipschitz continuous on R; that is, there exists 𝑘 > 0 such that lip ̃𝑓(𝜏) ≤ 𝑘, for all 𝜏 ∈ R. Since ̃𝐶 := conv(𝐶) is compact, according to [6, Lemma 4.5] or [5, Lemma 3], there exist continuously differentiable functions𝑓V: R → R forV = 1, 2, . . . converging uniformly to ̃𝑓 on ̃𝐶 such that

󵄨󵄨󵄨󵄨󵄨(𝑓V)󸀠(𝜏)󵄨󵄨󵄨󵄨󵄨 ≤ 𝑘 ∀𝜏 ∈𝐶, ∀V.̃ (27)

Now, let:= 𝛿/(√2 max{tan 𝜃, ctan 𝜃}). Then, fromLemma 3, we know that ̃𝐶 contains all spectral values of 𝑦 ∈ B(𝑧, 𝛿).

Therefore, for any𝑤 ∈ B(𝑧, 𝛿), we have 𝜆𝑖(𝑤) ∈ ̃𝐶 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󵄩󵄩󵄩󵄩󵄩𝑢1𝑤󵄩󵄩󵄩󵄩󵄩2 + [𝑓V(𝜆2(𝑤)) − 𝑓 (𝜆2(𝑤))]2󵄩󵄩󵄩󵄩󵄩𝑢2𝑤󵄩󵄩󵄩󵄩󵄩2

= 1

1 + ctan2𝜃󵄨󵄨󵄨󵄨𝑓V(𝜆1(𝑤)) − 𝑓 (𝜆1(𝑤))󵄨󵄨󵄨󵄨2

+ 1

1 + tan2𝜃󵄨󵄨󵄨󵄨𝑓V(𝜆2(𝑤)) − 𝑓(𝜆2(𝑤))󵄨󵄨󵄨󵄨2,

(28)

where we have used the facts that‖𝑢1𝑤‖ = 1/√1 + ctan2𝜃,

‖𝑢2𝑤‖ = 1/√1 + tan2𝜃, and ⟨𝑢1𝑤, 𝑢2𝑤⟩ = 0. Since {𝑓V}V=1con- verges uniformly to𝑓 on ̃𝐶, the above equations show that {(𝑓V)L𝜃}V=1converges uniformly to𝑓L𝜃onB(𝑧, 𝛿). If 𝑤2= 0, then it follows fromLemma 2that∇(𝑓V)L𝜃(𝑤) = (𝑓V)󸀠(𝑤1)𝐼.

Hence it follows from (27) that

󵄩󵄩󵄩󵄩󵄩󵄩∇(𝑓V)L𝜃(𝑤)󵄩󵄩󵄩󵄩󵄩󵄩 =󵄨󵄨󵄨󵄨󵄨(𝑓V)󸀠(𝑤1)󵄨󵄨󵄨󵄨󵄨 ≤ 𝑘, (29) since in this case𝜆𝑖(𝑤) = 𝑤1∈ ̃𝐶. If 𝑤2 ̸= 0, then

∇(𝑓V)L𝜃(𝑤)

=[[[[ [

𝜉 󰜚𝑤𝑇2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩

󰜚𝑤2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩 𝑎𝐼 + (𝜉 − 󰜚(ctan𝜃 − tan𝜃) − 𝑎)𝑤2𝑤𝑇2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩2 ]] ]] ]

=[[[[ [

𝜉 󰜚𝑤𝑇2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩

󰜚𝑤2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩 𝑎𝐼 + (𝜉 − 𝑎)𝑤2𝑤𝑇2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩2 ]] ]] ]

+ [[ [

0 0

0 [−󰜚 (ctan 𝜃 − tan 𝜃)]𝑤2𝑤𝑇2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩2 ]] ]

=[[[[ [

𝜉 󰜚𝑤𝑇2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩

󰜚𝑤2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩 𝜉𝐼 ]] ]] ]

+ (𝑎 − 𝜉) [[ [

0 0

0 𝐼 −𝑤2𝑤𝑇2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩2 ]] ]

− 󰜚 (ctan 𝜃 − tan 𝜃) [[ [

0 0

0 𝑤2𝑤2𝑇

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩2 ]] ] ,

(30)

(5)

where𝑎, 𝜉, 󰜚 are given as in (20) with𝜆𝑖(𝑧) replaced by 𝜆𝑖(𝑤) for𝑖 = 1, 2 and 𝑓 replaced by 𝑓V. For simplicity of notations, let us denote

𝐴 :=[[[ [

𝜉 󰜚𝑤𝑇2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩

󰜚𝑤2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩 𝜉𝐼 ]] ] ]

+ (𝑎 − 𝜉) [[ [

0 0

0 𝐼 −𝑤2𝑤𝑇2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩2 ]] ] ,

𝐵 := −󰜚 (ctan 𝜃 − tan 𝜃) [[ [

0 0

0 𝑤2𝑤𝑇2

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩2 ]] ] .

(31)

Note that

|𝑎| =󵄨󵄨󵄨󵄨

󵄨󵄨󵄨󵄨󵄨

𝑓V(𝜆2(𝑤)) − 𝑓V(𝜆1(𝑤)) 𝜆2(𝑤) − 𝜆1(𝑤) 󵄨󵄨󵄨󵄨

󵄨󵄨󵄨󵄨󵄨≤ 𝑘, (32) where the inequality comes from the fact that𝑓Vis continu- ously differentiable on ̃𝐶 and (27). Besides, we also note that

󵄨󵄨󵄨󵄨𝜉󵄨󵄨󵄨󵄨 =󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

(𝑓V)󸀠(𝜆1(𝑤))

1 + ctan2𝜃 +(𝑓V)󸀠(𝜆2(𝑤)) 1 + tan2𝜃 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

≤ 1

1 + ctan2𝜃󵄨󵄨󵄨󵄨󵄨(𝑓V)󸀠(𝜆1(𝑤))󵄨󵄨󵄨󵄨󵄨 + 1

1 + tan2𝜃󵄨󵄨󵄨󵄨󵄨(𝑓V)󸀠(𝜆2(𝑤))󵄨󵄨󵄨󵄨󵄨

≤ [ 1

1 + ctan2𝜃 + 1

1 + tan2𝜃] 𝑘 = 𝑘,

(33)

󵄨󵄨󵄨󵄨󰜚󵄨󵄨󵄨󵄨 =󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨− ctan𝜃

1 + ctan2𝜃(𝑓V)󸀠(𝜆1(𝑤)) + tan𝜃

1 + tan2𝜃(𝑓V)󸀠(𝜆2(𝑤))󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

≤ [󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨− ctan𝜃

1 + ctan2𝜃󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 +󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 tan𝜃 1 + tan2𝜃󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨]𝑘

= [ ctan𝜃

1 + ctan2𝜃+ tan𝜃 1 + tan2𝜃] 𝑘

= 2 tan 𝜃 1 + tan2𝜃𝑘 ≤ 𝑘.

(34)

(i) For󰜚 = 0, then ∇(𝑓V)L𝜃(𝑤) takes the form of 𝜉𝐼+(𝑎−

𝜉)𝑀𝑤2whose eigenvalues are𝜉 and 𝑎 by [5, Lemma 1].

In other words, in this case, we get from (32) and (33) that

󵄩󵄩󵄩󵄩󵄩󵄩∇(𝑓V)L𝜃(𝑤)󵄩󵄩󵄩󵄩󵄩󵄩 = max {|𝑎| ,󵄨󵄨󵄨󵄨𝜉󵄨󵄨󵄨󵄨} ≤ 𝑘. (35) (ii) For󰜚 ̸= 0, since 𝐵 = −󰜚(ctan 𝜃 − tan 𝜃)(0, 𝑤2/‖𝑤2‖)𝑇 (0, 𝑤2/‖𝑤2‖), the eigenvalues of 𝐵 are −󰜚(ctan 𝜃 − tan𝜃) and 0 with multiplicity 𝑛 − 1. Note that

󵄨󵄨󵄨󵄨󰜚(ctan𝜃 − tan𝜃)󵄨󵄨󵄨󵄨

=󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨

1 − ctan2𝜃

1 + ctan2𝜃(𝑓V)󸀠(𝜆1(𝑤)) +1 − tan2𝜃

1 + tan2𝜃(𝑓V)󸀠(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.

(36) Note that

𝐴 = 󰜚

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩𝐿𝑤̃+ (𝑎 − 𝜉) 𝑀𝑤2

= 󰜚

󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩 [𝐿𝑤̃+ (𝑎 − 𝜉)󵄩󵄩󵄩󵄩𝑤2󵄩󵄩󵄩󵄩

󰜚 𝑀𝑤̃2] ,

(37)

where𝑤 = (𝜉‖𝑤̃ 2‖/󰜚, 𝑤2) and 𝐿𝑤̃:= [̃𝑤1 𝑤̃2𝑇

𝑤̃2 𝑤̃1𝐼] . (38) In this case the matrix𝐴 has eigenvalues of 𝜉 ± 󰜚 and 𝑎 with multiplicity𝑛 − 2. Hence,

󵄩󵄩󵄩󵄩󵄩󵄩∇(𝑓V)L𝜃(𝑤)󵄩󵄩󵄩󵄩󵄩󵄩

≤ max {󵄨󵄨󵄨󵄨𝜉 + 󰜚󵄨󵄨󵄨󵄨,󵄨󵄨󵄨󵄨𝜉 − 󰜚󵄨󵄨󵄨󵄨,|𝑎|} + 󵄨󵄨󵄨󵄨󰜚(ctan𝜃 − tan𝜃)󵄨󵄨󵄨󵄨

≤ max {󵄨󵄨󵄨󵄨𝜉󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨󰜚󵄨󵄨󵄨󵄨,|𝑎|} + 󵄨󵄨󵄨󵄨󰜚(ctan𝜃 − tan𝜃)󵄨󵄨󵄨󵄨 ≤ 4𝑘, (39)

where the last step is due to (32), (33), (34), and (36).

Putting (29), (35), and (39) together, we know that

󵄩󵄩󵄩󵄩󵄩󵄩∇(𝑓V)L𝜃(𝑤)󵄩󵄩󵄩󵄩󵄩󵄩 ≤ 4𝑘 ∀𝑤 ∈ B (𝑧, 𝛿) , ∀V. (40) Fix any𝑥, 𝑦 ∈ B(𝑧, 𝛿) with 𝑥 ̸= 𝑦. Since {(𝑓V)L𝜃}V=1converges uniformly to𝑓L𝜃onB(𝑧, 𝛿), then for any 𝜖 > 0 there exists V0such that

󵄩󵄩󵄩󵄩󵄩󵄩(𝑓V)L𝜃(𝑤) − 𝑓L𝜃(𝑤)󵄩󵄩󵄩󵄩󵄩󵄩 ≤ 𝜖, ∀𝑤 ∈ B (𝑧, 𝛿) , ∀V ≥ V0. (41) Since𝑓Vis continuously differentiable,(𝑓V)L𝜃is continuously differentiable byLemma 2. Thus,

󵄩󵄩󵄩󵄩󵄩𝑓L𝜃(𝑥) − 𝑓L𝜃(𝑦)󵄩󵄩󵄩󵄩󵄩

=󵄩󵄩󵄩󵄩󵄩󵄩𝑓L𝜃(𝑥) − (𝑓V)L𝜃(𝑥) + (𝑓V)L𝜃(𝑥) − (𝑓V)L𝜃(𝑦) + (𝑓V)L𝜃(𝑦) − 𝑓L𝜃(𝑦)󵄩󵄩󵄩󵄩󵄩󵄩

≤󵄩󵄩󵄩󵄩󵄩󵄩𝑓L𝜃(𝑥) − (𝑓V)L𝜃(𝑥)󵄩󵄩󵄩󵄩󵄩󵄩 +󵄩󵄩󵄩󵄩󵄩󵄩(𝑓V)L𝜃(𝑥) − (𝑓V)L𝜃(𝑦)󵄩󵄩󵄩󵄩󵄩󵄩

+󵄩󵄩󵄩󵄩󵄩󵄩(𝑓V)L𝜃(𝑦) − 𝑓L𝜃(𝑦)󵄩󵄩󵄩󵄩󵄩󵄩

≤ 2𝜖 +󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩∫1

0 ∇(𝑓V)L𝜃(𝑦 + 𝑡 (𝑥 − 𝑦)) (𝑥 − 𝑦) 𝑑𝑡󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩

≤ 2𝜖 + 4𝑘 󵄩󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩󵄩.

(42)

(6)

Because𝜖 > 0 is arbitrary, this ensures that

󵄩󵄩󵄩󵄩󵄩𝑓L𝜃(𝑥) − 𝑓L𝜃(𝑦)󵄩󵄩󵄩󵄩󵄩 ≤ 4𝑘󵄩󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩󵄩 ∀𝑥,𝑦 ∈ B(𝑧,𝛿), (43) which says𝑓L𝜃is strictly continuous at𝑧.

“⇒” Suppose that 𝑓L𝜃 is strictly continuous at 𝑧, then there exist𝑘 > 0 and 𝛿 > 0 such that

󵄩󵄩󵄩󵄩󵄩𝑓L𝜃(𝑥) − 𝑓L𝜃(𝑦)󵄩󵄩󵄩󵄩󵄩 ≤ 𝑘󵄩󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩󵄩 ∀𝑥,𝑦 ∈ B(𝑧,𝛿).

(44) Case 1.𝑧2 ̸= 0. Take 𝜃, 𝜇 ∈ [𝜆1(𝑧) − 𝛿1, 𝜆1(𝑧) + 𝛿1] with 𝛿1 :=

min{𝛿, 𝜆2(𝑧) − 𝜆1(𝑧)}. Let

𝑥 := 𝜃𝑢1𝑧+ 𝜆2(𝑧) 𝑢2𝑧, 𝑦 := 𝜇𝑢1𝑧+ 𝜆2(𝑧) 𝑢2𝑧. (45) Then,‖𝑥 − 𝑧‖ ≤ 𝛿 and ‖𝑦 − 𝑧‖ ≤ 𝛿 and it follows from (44) that

󵄨󵄨󵄨󵄨𝑓(𝜃) − 𝑓(𝜇)󵄨󵄨󵄨󵄨 = 1󵄩󵄩󵄩󵄩𝑢1𝑧󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩𝑓L𝜃(𝑥) − 𝑓L𝜃(𝑦)󵄩󵄩󵄩󵄩󵄩 ≤ 𝑘

󵄩󵄩󵄩󵄩𝑢1𝑧󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩󵄩

= 𝑘

󵄩󵄩󵄩󵄩𝑢1𝑧󵄩󵄩󵄩󵄩󵄨󵄨󵄨󵄨𝜃 − 𝜇󵄨󵄨󵄨󵄨󵄩󵄩󵄩󵄩󵄩𝑢1𝑧󵄩󵄩󵄩󵄩󵄩 = 𝑘󵄨󵄨󵄨󵄨𝜃 − 𝜇󵄨󵄨󵄨󵄨,

(46) which says 𝑓 is strictly continuous at 𝜆1(𝑧). The similar argument shows the strict continuity of𝑓 at 𝜆2(𝑧).

Case 2.𝑧2= 0. For any 𝜃, 𝜇 ∈ [𝑧1−𝛿, 𝑧1+𝛿], we have ‖𝜃𝑒−𝑧‖ =

|𝜃 − 𝑧1| ≤ 𝛿 and ‖𝜇𝑒 − 𝑧‖ ≤ 𝛿 as well; that is, 𝜃𝑒, 𝜇𝑒 ∈ B(𝑧, 𝛿).

It then follows from (44) that

󵄨󵄨󵄨󵄨𝑓(𝜃) − 𝑓(𝜇)󵄨󵄨󵄨󵄨 =󵄩󵄩󵄩󵄩

󵄩󵄩󵄩󵄩(𝑓 (𝜃) − 𝑓 (𝜇) 0 )󵄩󵄩󵄩󵄩

󵄩󵄩󵄩󵄩 =󵄩󵄩󵄩󵄩󵄩𝑓L𝜃(𝜃𝑒) − 𝑓L𝜃(𝜇𝑒)󵄩󵄩󵄩󵄩󵄩

≤ 𝑘 󵄩󵄩󵄩󵄩𝜃𝑒 − 𝜇𝑒󵄩󵄩󵄩󵄩 = 𝑘󵄨󵄨󵄨󵄨𝜃 − 𝜇󵄨󵄨󵄨󵄨.

(47) This means𝑓 is strictly continuous at 𝜆𝑖(𝑧) = 𝑧1for𝑖 = 1, 2.

Remark 5. As mentioned inSection 1, the strict continuity between𝑓L𝜃and𝑓 has been given in [1, Theorem 3.5]. Here we provide an alternative proof, since our analysis technique is different from that in [1, Theorem 3.5]. In particular, we achieve an estimate regarding‖∇(𝑓V)L𝜃‖ via its eigenvalues, which may have other applications.

According toLemma 1andTheorem 4, we obtain the fol- lowing result immediately.

Theorem 6. Let 𝑓 : R → R and 𝑓L𝜃 be defined as in (4).

Then,𝑓L𝜃is𝐵-differentiable at 𝑧 if and only if 𝑓 is 𝐵-differen- tiable at𝜆𝑖(𝑧), for 𝑖 = 1, 2.

Next, inspired by [8, 9], we further study the H¨older continuity relation between𝑓 and 𝑓L𝜃.

Theorem 7. Let 𝑓 : R → R and 𝑓L𝜃 be defined as in (4).

Then,𝑓L𝜃 is H¨older continuous with exponent𝛼 ∈ (0, 1] if and only if𝑓 is H¨older continuous with exponent 𝛼 ∈ (0, 1].

Proof. “⇐” Suppose that 𝑓 is H¨older continuous with expo- nent𝛼 ∈ (0, 1]. To proceed the proof, we consider the follow- ing two cases.

Case 1.𝑧2 ̸= 0 and 𝑦2 ̸= 0. We assume without loss of generality that‖𝑧2‖ ≥ ‖𝑦2‖. Thus,

󵄩󵄩󵄩󵄩󵄩𝑓L𝜃(𝑧) − 𝑓L𝜃(𝑦)󵄩󵄩󵄩󵄩󵄩

= 󵄩󵄩󵄩󵄩󵄩𝑓(𝜆1(𝑧)) 𝑢1𝑧+ 𝑓 (𝜆2(𝑧)) 𝑢2𝑧− 𝑓 (𝜆1(𝑦)) 𝑢𝑦1

−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󵄩󵄩󵄩󵄩󵄩

≤ 󵄩󵄩󵄩󵄩󵄩𝑓(𝜆1(𝑧)) [𝑢1𝑧− 𝑢1𝑦] + 𝑓 (𝜆2(𝑧)) [𝑢2𝑧− 𝑢2𝑦]󵄩󵄩󵄩󵄩󵄩

+ 󵄨󵄨󵄨󵄨𝑓 (𝜆1(𝑧)) − 𝑓 (𝜆1(𝑦))󵄨󵄨󵄨󵄨 ⋅󵄩󵄩󵄩󵄩󵄩𝑢1𝑦󵄩󵄩󵄩󵄩󵄩 + 󵄨󵄨󵄨󵄨𝑓 (𝜆2(𝑧)) − 𝑓 (𝜆2(𝑦))󵄨󵄨󵄨󵄨 ⋅󵄩󵄩󵄩󵄩󵄩𝑢2𝑦󵄩󵄩󵄩󵄩󵄩 .

(48)

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]󵄩󵄩󵄩󵄩󵄩

= tan𝜃

1 + tan2𝜃󵄨󵄨󵄨󵄨𝑓(𝜆1(𝑧)) − 𝑓 (𝜆2(𝑧))󵄨󵄨󵄨󵄨 ⋅󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩 𝑧2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 − 𝑦2

󵄩󵄩󵄩󵄩𝑦2󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩

≤ tan𝜃

1 + tan2𝜃[𝑓]𝛼󵄨󵄨󵄨󵄨𝜆1(𝑧) − 𝜆2(𝑧)󵄨󵄨󵄨󵄨𝛼󵄩󵄩󵄩󵄩

󵄩󵄩󵄩󵄩󵄩 𝑧2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 − 𝑦2

󵄩󵄩󵄩󵄩𝑦2󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩

󵄩󵄩󵄩󵄩󵄩

= tan𝜃

1 + tan2𝜃[𝑓]𝛼(tan 𝜃 + ctan 𝜃)𝛼󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩𝛼⋅󵄩󵄩󵄩󵄩

󵄩󵄩󵄩󵄩󵄩 𝑧2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 − 𝑦2

󵄩󵄩󵄩󵄩𝑦2󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩

󵄩󵄩󵄩󵄩󵄩

≤ tan𝜃

1 + tan2𝜃[𝑓]𝛼(tan 𝜃 + ctan 𝜃)𝛼󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩𝛼 2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩󵄩󵄩󵄩󵄩𝑧2− 𝑦2󵄩󵄩󵄩󵄩

= 2 tan𝜃

1 + tan2𝜃[𝑓]𝛼(tan 𝜃 + ctan 𝜃)𝛼󵄩󵄩󵄩󵄩

󵄩󵄩󵄩󵄩󵄩 𝑧2− 𝑦2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 󵄩󵄩󵄩󵄩

󵄩󵄩󵄩󵄩󵄩

1−𝛼󵄩󵄩󵄩󵄩𝑧2− 𝑦2󵄩󵄩󵄩󵄩𝛼

≤ tan𝜃

1 + tan2𝜃[𝑓]𝛼(tan 𝜃 + ctan 𝜃)𝛼22−𝛼󵄩󵄩󵄩󵄩𝑧2− 𝑦2󵄩󵄩󵄩󵄩𝛼

≤ tan𝜃

1 + tan2𝜃[𝑓]𝛼(tan 𝜃 + ctan 𝜃)𝛼22−𝛼󵄩󵄩󵄩󵄩𝑧 − 𝑦󵄩󵄩󵄩󵄩𝛼,

(49) where the first inequality is due to the H¨older continu- ity of 𝑓, the second inequality comes from the fact that

‖(𝑧2/‖𝑧2‖) − (𝑦2/‖𝑦2‖)‖ ≤ (2/‖𝑧2‖)‖𝑧2− 𝑦2‖ (cf. [8, Lemma 2.2]), and the third inequality follows from the fact that

(7)

‖𝑧2− 𝑦2‖ ≤ ‖𝑧2‖ + ‖𝑦2‖ ≤ 2‖𝑧2‖ (since ‖𝑦2‖ ≤ ‖𝑧2‖). Next, we look into the second term:

󵄨󵄨󵄨󵄨𝑓(𝜆1(𝑧)) − 𝑓 (𝜆1(𝑦))󵄨󵄨󵄨󵄨󵄩󵄩󵄩󵄩󵄩𝑢1𝑦󵄩󵄩󵄩󵄩󵄩

≤ [𝑓]𝛼󵄨󵄨󵄨󵄨𝜆1(𝑧) − 𝜆1(𝑦)󵄨󵄨󵄨󵄨𝛼 1

√1 + ctan2𝜃

≤ [𝑓]𝛼(√2 max {tan 𝜃, ctan 𝜃})𝛼󵄩󵄩󵄩󵄩𝑧 − 𝑦󵄩󵄩󵄩󵄩𝛼.

(50)

Similarly, the third term also satisfies

󵄨󵄨󵄨󵄨𝑓(𝜆2(𝑧)) − 𝑓 (𝜆2(𝑦))󵄨󵄨󵄨󵄨󵄩󵄩󵄩󵄩󵄩𝑢2𝑦󵄩󵄩󵄩󵄩󵄩

≤ [𝑓]𝛼󵄨󵄨󵄨󵄨𝜆2(𝑧) − 𝜆2(𝑦)󵄨󵄨󵄨󵄨𝛼 1

√1 + tan2𝜃

≤ [𝑓]𝛼(√2 max {tan 𝜃, ctan 𝜃})𝛼󵄩󵄩󵄩󵄩𝑧 − 𝑦󵄩󵄩󵄩󵄩𝛼. (51)

Combining (49)–(51) proves that𝑓L𝜃is H¨older continuous with exponent𝛼 ∈ (0, 1].

Case 2. Either𝑧2 = 0 or 𝑦2= 0. In this case, we take 𝑢𝑖𝑧 = 𝑢𝑖𝑦, for𝑖 = 1, 2 according to the spectral decomposition. There- fore, we obtain

󵄩󵄩󵄩󵄩󵄩𝑓L𝜃(𝑧) − 𝑓L𝜃(𝑦)󵄩󵄩󵄩󵄩󵄩

= 󵄩󵄩󵄩󵄩󵄩𝑓(𝜆1(𝑧)) 𝑢1𝑧+ 𝑓 (𝜆2(𝑧)) 𝑢2𝑧− 𝑓 (𝜆1(𝑦)) 𝑢1𝑦

−𝑓 (𝜆2(𝑦)) 𝑢2𝑦󵄩󵄩󵄩󵄩󵄩

= 󵄩󵄩󵄩󵄩󵄩[𝑓(𝜆1(𝑧)) − 𝑓 (𝜆1(𝑦))] 𝑢1𝑧 + [𝑓 (𝜆2(𝑧)) − 𝑓 (𝜆2(𝑦))] 𝑢2𝑧󵄩󵄩󵄩󵄩󵄩

≤ 󵄨󵄨󵄨󵄨𝑓 (𝜆1(𝑧)) − 𝑓 (𝜆1(𝑦))󵄨󵄨󵄨󵄨 ⋅󵄩󵄩󵄩󵄩󵄩𝑢𝑧1󵄩󵄩󵄩󵄩󵄩 + 󵄨󵄨󵄨󵄨𝑓 (𝜆2(𝑧)) − 𝑓 (𝜆2(𝑦))󵄨󵄨󵄨󵄨 ⋅󵄩󵄩󵄩󵄩󵄩𝑢2𝑧󵄩󵄩󵄩󵄩󵄩

≤ [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, ctan 𝜃})𝛼󵄩󵄩󵄩󵄩𝑧 − 𝑦󵄩󵄩󵄩󵄩𝛼,

(52)

which says𝑓L𝜃is H¨older continuous.

“⇒” Recall that 𝑓L𝜃(𝜏𝑒) = (𝑓(𝜏), 0)𝑇. Hence, for any 𝜏, 𝜁 ∈ R,

󵄨󵄨󵄨󵄨𝑓(𝜏) − 𝑓(𝜁)󵄨󵄨󵄨󵄨 = 󵄩󵄩󵄩󵄩󵄩𝑓L𝜃(𝜏𝑒) − 𝑓L𝜃(𝜁𝑒)󵄩󵄩󵄩󵄩󵄩

≤ [𝑓L𝜃]𝛼⋅ 󵄩󵄩󵄩󵄩𝜏𝑒 − 𝜁𝑒󵄩󵄩󵄩󵄩𝛼= [𝑓L𝜃]𝛼⋅ 󵄨󵄨󵄨󵄨𝜏 − 𝜁󵄨󵄨󵄨󵄨𝛼, (53) which says𝑓 is H¨older continuous.

### 3.𝜌-Order Semismoothnessand𝐵-Subdifferential Formula

The property of semismoothness plays an important role in nonsmooth Newton methods [10,11]. For more information on semismooth functions, see [12–15]. The relationship of semismooth between 𝑓L𝜃 and 𝑓 has been given in [1, Theorem 4.1]. But the exact formula of the𝐵-subdifferential

𝜕𝐵(𝑓L𝜃) is not presented. Hence the main aim of this section is twofold: one is establishing the exact formula of𝐵- subdifferential; another is studing the𝜌-order semismooth- ness for𝜌 > 0.

Lemma 8. Define 𝜓(𝑧) = ‖𝑧‖ and Φ(𝑧) = 𝑧/‖𝑧‖ for 𝑧 ̸= 0.

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

Proof. Since𝑧 ̸= 0, it is clear that 𝜓 and Φ are twice continu- ously differentiable and hence the gradient is Lipschitz con- tinuous near𝑧. Therefore, 𝜓 and Φ are strongly semismooth at𝑧, see [16, Proposition 7.4.5].

The relationship of 𝜌-order semismoothness between 𝑓L𝜃and𝑓 is given below. Recall from [7] that in the definition of𝜌-order semismooth, we can restrict 𝑥+ℎ in (11) belonging to differentiable points.

Theorem 9. Let 𝑓 : R → R and 𝑓L𝜃 be defined as in (4).

Given𝜌 > 0, then the following statements hold.

(a) If𝑓 is 𝜌-order semismooth at 𝜆𝑖(𝑧) for 𝑖 = 1, 2, then 𝑓L𝜃is min{1, 𝜌}-order semismooth at 𝑧.

(b) If𝑓L𝜃 is𝜌-order semismooth at 𝑧, then 𝑓 is 𝜌-semi- smooth at𝜆𝑖(𝑧) for 𝑖 = 1, 2.

(c) For𝑧2= 0, 𝑓L𝜃is𝜌-semismooth at 𝑧 if and only if 𝑓 is 𝜌-order semismooth at 𝜆𝑖(𝑧) = 𝑧1for𝑖 = 1, 2.

Proof. (a) Takeℎ ∈ R𝑛satisfying𝑧 + ℎ ∈ 𝐷𝑓L𝜃. We consider the following two cases to complete the proof.

Case 1. For𝑧2 ̸= 0, 𝑧2+ ℎ2 ̸= 0 as ℎ is sufficiently close to 0.

Since𝑧+ℎ ∈ 𝐷𝑓L𝜃, we know that𝜆𝑖(𝑧+ℎ) ∈ 𝐷𝑓for𝑖 = 1, 2 by Lemma 2. Then, according toLemma 1, the first component of

𝑓L𝜃(𝑧 + ℎ) − 𝑓L𝜃(𝑧) − (𝑓L𝜃)󸀠(𝑧 + ℎ; ℎ) (54) is expressed as

𝑓 (𝜆1(𝑧 + ℎ))

1 + ctan2𝜃 − 𝑓 (𝜆1(𝑧))

1 + ctan2𝜃 − 1 1 + ctan2𝜃

× 𝑓󸀠(𝜆1(𝑧 + ℎ) ; ℎ1−(𝑧2+ ℎ2)𝑇2

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 ctan𝜃) +𝑓 (𝜆2(𝑧 + ℎ))

1 + tan2𝜃 −𝑓 (𝜆2(𝑧)) 1 + tan2𝜃 − 1

1 + tan2𝜃

× 𝑓󸀠(𝜆2(𝑧 + ℎ) ; ℎ1+(𝑧2+ ℎ2)𝑇2

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 tan𝜃).

(55)

(8)

Because‖ ⋅ ‖ is continuously differentiable over 𝑧2 ̸= 0, it is strongly semismooth at𝑧2byLemma 8. Therefore,

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 = 󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 + (𝑧2+ ℎ2)𝑇2

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 + 𝑂(󵄩󵄩󵄩󵄩ℎ2󵄩󵄩󵄩󵄩2)

= 󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩 + (𝑧2+ ℎ2)𝑇2

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 + 𝑂(‖ℎ‖2) .

(56)

Combining this and the𝜌-semismoothness of 𝑓 at 𝜆1(𝑧), we have

𝑓 (𝜆1(𝑧 + ℎ))

= 𝑓 (𝜆1(𝑧)) + 𝑓󸀠(𝜆1(𝑧 + ℎ)) (𝜆1(𝑧 + ℎ) − 𝜆1(𝑧)) + 𝑂 (󵄨󵄨󵄨󵄨𝜆1(𝑧 + ℎ) − 𝜆1(𝑧)󵄨󵄨󵄨󵄨1+𝜌)

= 𝑓 (𝜆1(𝑧)) + 𝑓󸀠(𝜆1(𝑧 + ℎ)) (𝜆1(𝑧 + ℎ) − 𝜆1(𝑧)) + 𝑂 (‖ℎ‖1+𝜌)

= 𝑓 (𝜆1(𝑧)) + 𝑓󸀠(𝜆1(𝑧 + ℎ))

× (ℎ1− (󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 − 󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩)ctan𝜃) + 𝑂(‖ℎ‖1+𝜌)

= 𝑓 (𝜆1(𝑧)) + 𝑓󸀠(𝜆1(𝑧 + ℎ)) (ℎ1−(𝑧2+ ℎ2)𝑇2

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 ctan𝜃) + 𝑂 (‖ℎ‖2) + 𝑂 (‖ℎ‖1+𝜌)

= 𝑓 (𝜆1(𝑧)) + 𝑓󸀠(𝜆1(𝑧 + ℎ)) (ℎ1−(𝑧2+ ℎ2)𝑇2

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 ctan𝜃) + 𝑂 (‖ℎ‖1+min{1,𝜌}) ,

(57) where the second equation is due toLemma 3and the last equality comes from the boundedness of𝑓󸀠, since𝑓 is strictly continuous at𝜆1(𝑧). Similar argument holds for 𝑓(𝜆2(𝑧+ℎ)).

Hence the first component of (54) is𝑂(‖ℎ‖1+min{1,𝜌}).

Next, let us look into the second component of (54), which involved𝜆1(𝑧). ByLemma 1again, it can be expressed as

− ctan𝜃

1 + ctan2𝜃𝑓 (𝜆1(𝑧 + ℎ)) 𝑧2+ ℎ2

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩

+ ctan𝜃

1 + ctan2𝜃𝑓󸀠(𝜆1(𝑧 + ℎ) ; ℎ1−(𝑧2+ ℎ2)𝑇2

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 ctan𝜃)

× 𝑧2+ ℎ2

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩

+ ctan𝜃

1 + ctan2𝜃𝑓 (𝜆1(𝑧)) 𝑧2

󵄩󵄩󵄩󵄩𝑧2󵄩󵄩󵄩󵄩

+ ctan𝜃 1 + ctan2𝜃

𝑓 (𝜆1(𝑧 + ℎ))

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 𝑀(𝑧2+ℎ2)ℎ.

(58)

Note thatΦ is continuous differentiable (and hence is semi- smooth) with ∇Φ(𝑧2) = (1/‖𝑧2‖)(𝐼 − (𝑧2𝑧𝑇2/‖𝑧22)) and 𝑀(𝑧2+ℎ2)ℎ = ‖𝑧2 + ℎ2‖∇Φ(𝑧2+ ℎ2)ℎ2. Thus, expression (58) can be rewritten as

− ctan𝜃

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

1 + ctan2𝜃𝑓󸀠(𝜆1(𝑧 + ℎ) ; ℎ1−(𝑧2+ ℎ2)𝑇2

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 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))

+𝑓󸀠(𝜆1(𝑧 + ℎ) ; ℎ1−(𝑧2+ ℎ2)𝑇2

󵄩󵄩󵄩󵄩𝑧2+ ℎ2󵄩󵄩󵄩󵄩 ctan𝜃)]

× Φ (𝑧2+ ℎ2) +𝑓 (𝜆1(𝑧)) ctan 𝜃

1 + ctan2𝜃

× [−Φ (𝑧2+ ℎ2) + Φ (𝑧2) + ∇Φ (𝑧2+ ℎ2) ℎ2] + ctan𝜃

1 + ctan2𝜃∇Φ (𝑧2+ ℎ2) ℎ2

× [𝑓 (𝜆1(𝑧 + ℎ)) − 𝑓 (𝜆1(𝑧))]

= 𝑂 (‖ℎ‖1+min{1,𝜌}) + 𝑂 (‖ℎ‖2) + 𝑂 (‖ℎ‖2)

= 𝑂 (‖ℎ‖1+min{1,𝜌}) .

(59) The second equation comes from (57), strongly semismooth- ness ofΦ at 𝑧2, and

󵄩󵄩󵄩󵄩∇Φ(𝑧2+ ℎ2) ℎ2[𝑓 (𝜆1(𝑧 + ℎ)) − 𝑓 (𝜆1(𝑧))]󵄩󵄩󵄩󵄩 = 𝑂 (‖ℎ‖2) , (60) since𝑓 is Lipschitz at 𝜆1(𝑧) (which is ensured by the 𝜌-order semismoothness of𝑓). Analogous arguments apply for the second component of (54) involving 𝜆2(𝑧). From all the above, we may conclude that

𝑓L𝜃(𝑧 + ℎ) − 𝑓L𝜃(𝑧) − (𝑓L𝜃)󸀠(𝑧 + ℎ; ℎ)

= 𝑂 (‖ℎ‖1+min{1,𝜌}) , (61)

which says𝑓L𝜃is min{1, 𝜌}-order semismooth at 𝑧 under this case.

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