• 沒有找到結果。

SOC-monotone and SOC-convex functions vs.

N/A
N/A
Protected

Share "SOC-monotone and SOC-convex functions vs."

Copied!
21
0
0

(1)

Contents lists available atSciVerse ScienceDirect

Linear Algebra and its Applications

journal homepage:w w w . e l s e v i e r . c o m / l o c a t e / l a a

matrix-monotone and matrix-convex functions<

a

b

, Jein-Shan Chen

c,,1,2

aDepartment of Mathematics, South China University of Technology, Wushan Road 381, Tianhe District, Guangzhou 510641, China bDepartment of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 80424, Taiwan

cDepartment of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan

A R T I C L E I N F O A B S T R A C T

Article history:

Received 30 September 2010 Accepted 9 April 2012 Available online 12 May 2012 Submitted by R.A. Brualdi AMS classification:

26B05 26B35 90C33 65K05

Keywords:

Hilbert space Second-order cone SOC-monotonicity SOC-convexity

The SOC-monotone function (respectively, SOC-convex function) is a scalar valued function that induces a map to preserve the monotone order (respectively, the convex order), when imposed on the spectral factorization of vectors associated with second-order cones (SOCs) in general Hilbert spaces. In this paper, we provide the sufficient and necessary characterizations for the two classes of functions, and particularly establish that the set of continuous SOC-monotone (re- spectively, SOC-convex) functions coincides with that of continuous matrix monotone (respectively, matrix convex) functions of order 2.

1. Introduction

Let

H

be a real Hilbert space of dimension dim

(H) ≥

3 endowed with an inner product

·, ·

and its induced norm

 · 

. Fix a unit vector e

and denote by

e



the orthogonal complementary

< This work was supported by National Young Natural Science Foundation (No. 10901058) and the Fundamental Research Funds for the Central Universities.

Corresponding author.

E-mail addresses:shhpan@scut.edu.cn(S. Pan),chiangyy@math.nsysu.edu.tw(Y. Chiang),jschen@math.ntnu.edu.tw(J.-S. Chen).

1 The author’s work is supported by National Science Council of Taiwan, Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan.

2 Also a Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office.

http://dx.doi.org/10.1016/j.laa.2012.04.030

(2)

space of e, i.e.,

e

x

x

e

0

} .

Then each x can be written as x

xe

x0e for some xe

e

and x0

∈ R.

The second-order cone (SOC) in

H

, also called the Lorentz cone, is a set defined by K



x

x

e

1 2

x



xe

x0e

x0

xe



.

From [7, Section 2], we know that K is a pointed closed convex self-dual cone. Hence,

H

becomes a partially ordered space via the relationK. In the sequel, for any x

y

∈ H

, we always write xK y (respectively, xK y) when x

y

∈

K (respectively, x

y

∈

intK); and denote xeby the vectorxxe

e if xe

=

0, and otherwise by any unit vector from

e



.

Associated with the second-order cone K, each x

xe

x0e

∈H

can be decomposed as

x

1

x

u1

x

2

x

u2

x

(1)

where

i

x

and ui

x

for i

1

,

2 are the spectral values and the associated spectral vectors of x, defined by

i

x

x0

1

i

xe

ui

x

1 2

e

1

ixe



(2)

Clearly, when xe

=

0, the spectral factorization of x is unique by definition.

Let f

J

⊆ R → R

be a scalar valued function, where J is an interval (finite or infinite, closed or open) in

R

. Let S be the set of all x

∈ H

whose spectral values

1

x

and

2

x

)

belong to J. Unless otherwise stated, in this paper S is always taken in this way. By the spectral factorization of x in (1) and (2), it is natural to define fsoc

S

by

fsoc

x

f

1

x

u1

x

f

2

x

u2

x

x

S

.

(3) It is easy to see that the function fsocis well defined whether xe

=

0 or not. For example, by taking f

t

) =

t2, we have that fsoc

x

x2

x

x, where “

◦

” means the Jordan product and the detailed definition is see in the next section. Note that

1

x

1

y

2

2

x

2

y

2

2

x

2

y

2

2x0y0

2

xe

ye

2

x

2

y

2

2

x

y

2

x

y

2

.

We may verify that the domain S of fsocis open in

H

if and only if J is open in

R

. Also, S is always convex since, for any x

xe

x0e

y

ye

y0e

S and

0

1

,

1[

x

1

y]



x0

1

y0



xe

1

ye

min

1

x

1

y

2[

x

1

y]



x0

1

y0



xe

1

ye

max

2

x

2

y

)},

which implies that

x

1

y

S. Thus, fsoc

x

1

y

)

is well defined.

In this paper we are interested in two classes of special scalar valued functions that induce the maps via (3) to preserve the monotone order and the convex order, respectively.

Definition 1.1. A function f

J

→ R

is said to be SOC-monotone if for any x

y

S,

x

K y

fsoc

x

Kfsoc

y

);

(4)

and f is said to be SOC-convex if, for any x

y

S and any

0

1

,

fsoc

x

1

y

K

fsoc

x

1

fsoc

y

).

(5) From Definition1.1and Eq. (3), it is easy to see that the set of SOC-monotone and SOC-convex functions are closed under positive linear combinations and pointwise limits.

(3)

The concept of SOC-monotone (respectively, SOC-convex) functions above is a direct extension of those given by [5,6] to general Hilbert spaces, and is analogous to that of matrix monotone (respec- tively, matrix convex) functions and more general operator monotone (respectively, operator convex) functions; see, e.g., [17,15,14,2,11,23]. Just as the importance of matrix monotone (respectively, matrix convex) functions to the solution of convex semidefinite programming [19,4], SOC-monotone (respec- tively, SOC-convex) functions also play a crucial role in the design and analysis of algorithms for convex second-order cone programming [3,22]. For matrix monotone and matrix convex functions, after the seminal work of Löwner [17] and Kraus [15], there have been systematic studies and perfect char- acterizations for them; see [8,16,4,13,12,21,20] and the references therein. However, the study on SOC-monotone and SOC-convex functions just begins with [5], and the characterizations for them are still imperfect. Particularly, it is not clear what is the relation between the SOC-monotone (respectively, SOC-convex) functions and the matrix monotone (respectively, matrix convex) functions.

In this work, we provide the sufficient and necessary characterizations for SOC-monotone and SOC- convex functions in the setting of Hilbert spaces, and show that the set of continuous SOC-monotone (SOC-convex) functions coincides with that of continuous matrix monotone (matrix convex) functions of order 2. Some of these results generalize those of [5,6] (see Propositions3.2and4.2), and some are new, which are difficult to achieve by using the techniques of [5,6] (see, for example, Proposition 4.4). In addition, we also discuss the relations between SOC-monotone functions and SOC-convex functions, verify Conjecture 4.2 in [5] under a little stronger condition (see Proposition6.2), and present a counterexample to show that Conjecture 4.1 in [5] generally does not hold. It is worthwhile to point out that the analysis in this paper depends only on the inner product of Hilbert spaces, whereas most of the results in [5,6] are obtained with the help of matrix operations.

Throughout this paper, all differentiability means Fréchet differentiability. If F

: H → H

is (twice) differentiable at x

∈ H

, we denote by F

x

(F

x

)

) the first-order F-derivative (the second-order F-derivative) of F at x. In addition, we use Cn

J

and C

J

)

to denote the set of n times and infinite times continuously differentiable real functions on J, respectively. When f

C1

J

)

, we denote by f[1] the function on J

J defined by

f[1]

⎧⎨

f(λ)−f(μ)

λ−μ if

f

if

and when f

C2

J

)

, denote by f[2]the function on J

J

×

J defined by f[2]

1

2

3

f[1]

1

2

f[1]

1

3

2

3

if

1

2

, τ

3are distinct, and for other values of

1

2

, τ

3, f[2]is defined by continuity; e.g., f[2]

1

1

3

f

3

f

1

f

1

3

1

3

1

2

f[2]

1

1

1

1 2f

1

).

For a linear operatorL from

into

, we writeL

≥

0 (respectively,L

>

0) to mean thatL is positive semidefinite (respectively, positive definite), i.e.,

h

Lh

0 for any h

(respectively,

h

Lh

0 for any 0

h

∈ H

).

2. Preliminaries

This section recalls some background material and gives several lemmas that will be used in the subsequent sections. We start with the definition of Jordan product [9]. For any x

xe

x0e

y

ye

y0e

∈ H

, the Jordan product of x and y is defined as

x

y

x0ye

y0xe

x

y

e

.

(4)

A simple computation can verify that for any x

y

z

∈ H

and the unit vector e, (i) e

e

e and e

x

x;

(ii) x

y

y

x; (iii) x

x2

y

x2

x

y

, where x2

x

x; (iv)

x

y

z

x

z

y

z. For any x

∈ H

, define its determinant by

det

x

1

x

2

x

x02

xe

2

Then each x

xe

x0e with det

x

) =

0 is invertible with respect to the Jordan product, i.e., there is a unique x1

xe

x0e

det

x

such that x

x1

=

e.

We next give several lemmas where Lemma 2.1 is used in Section 3 to characterize SOC- monotonicity, and Lemmas2.2and2.3are used in Section4to characterize SOC-convexity.

Lemma 2.1. Let

z

e

z

1

}

. Then, for any given u

e

with

u

1 and

θ, λ ∈ R

, the following results hold.

(a)

u

z

0 for any z

if and only if

.

(b)

z

2

2

u

z

2for any z

if and only if

2

≥

0.

Proof. (a) Suppose that

u

z

0 for any z

. If

0, then

θ ≥ |λ|

clearly holds. If

0, take z

sign

u. Since

u

1, we have z

∈ B

, and consequently,

u

z

0 reduces to

θ − |λ| ≥

0. Conversely, if

θ ≥ |λ|

, then using the Cauchy–Schwartz inequality yields

u

z

0 for any z

.

(b) Suppose that

z

2

2

u

z

2for any z

∈ B

. Then we must have

2

≥

0. If not, for those z

with

z

1 but

u

z

u

z

, it holds that

2

u

z

2

2

u

2

z

2

z

2

,

which contradicts the given assumption. Conversely, if

2

≥

0, the Cauchy–Schwartz inequality implies that

2

u

z

2

z

2for any z

∈ B

. 

Lemma 2.2. For any given a

b

c

and x

xe

x0e with xe

=

0, the inequality a

he

2

he

xe

2

bh0

xe

he

2

ch0

xe

he

2

≥

0 (6) holds for all h

he

h0e

if and only if a

0

b

0 and c

≥

0.

Proof. Suppose that (6) holds for all h

he

h0e

. By letting he

xe

h0

1 and he

xe

h0

=

1, respectively, we get b

0 and c

0 from (6). If a

≥

0 does not hold, then by taking he

=

b+|ca+| 1zzee with

ze

xe

0 and h0

=

1, (6) gives a contradiction

1

≥

0. Conversely, if a

0

b

0 and c

≥

0, then (6) clearly holds for all h

. 

Lemma 2.3. Let f

C2

J

and ue

e

with

ue

1. For any h

he

h0e

, define

1

h

h0

ue

he

2

2

h

h0

ue

he

2

h



he

2

ue

he

2

.

Then, for any given a

d

and

1

2

∈

J, the following inequality 4f

1

f

2

1

h

2

2

h

2

2

a

d

f

2

2

h

2

h

2

2

a

d

f

1

1

h

2

h

2

a2

d2

h

4

2 [

a

d

1

h

a

d

2

h

]2

h

2

≥

0 (7) holds for all h

he

h0e

if and only if

a2

d2

0

f

2

a

d

a

d

2and f

1

a

d

a

d

2

.

(8)

(5)

Proof. Suppose that (7) holds for all h

he

h0e

. Taking h0

0 and he

0 with

he

ue

0, we have

1

h

0

2

h

0 and

h

he

 >

0, and then (7) gives a2

d2

0. Taking he

0 such that

ue

he

he

and h0

ue

he

0, we have

1

h

0

2

h

2h0and

h

) >

0, and then (7) reduces to the following inequality

4

a

d

f

2

a

d

2h20

a2

d2

he

2

h20

0

.

This implies that

a

d

f

2

a

d

2

≥

0. If not, by letting h0be sufficiently close to

he



, the last inequality yields a contradiction. Similarly, taking h with he

0 satisfying

ue

he

he

and h0

ue

he

, we get f

1

a

d

a

d

)

2from (7).

Next, suppose that (8) holds. Then, the inequalities f

2

a

d

a

d

2and f

1

a

d

a

d

)

2imply that the left-hand side of (7) is greater than

4f

1

f

2

1

h

2

2

h

2

4

a2

d2

1

h

2

h

h

2

a2

d2

h

4

,

which is obviously nonnegative if

1

h

2

h

) ≤

0. Now assume that

1

h

2

h

0. If a2

d2

=

0, then the last expression is clearly nonnegative, and if a2

d2

>

0, then the last two inequalities in (8) imply that f

1

f

2

a2

d2

0

and therefore,

4f

1

f

2

1

h

2

2

h

2

4

a2

d2

1

h

2

h

h

2

a2

d2

h

4

4

a2

d2

1

h

2

2

h

2

4

a2

d2

1

h

2

h

h

2

a2

d2

h

4

a2

d2

2

1

h

2

h

h

22

0

.

Thus, we prove that inequality (7) holds. The proof is complete. 

To close this section, we introduce the regularization of a locally integrable real function. Let

ϕ

be a real function of class Cwith the following properties:

0,

ϕ

is even, the support supp

1

1

, and

R

1. For each

0, let

ε

t

1ε

εt

. Then supp

ε

and

ϕ

εhas all the properties of

ϕ

listed above. If f is a locally integrable real function, we define its regularization of order

as the function

fε

s

 f

s

t

ε

t

dt

 f

s

t

t

dt

.

(9) Note that fεis a Cfunction for each

0, and limε→0fε

x

f

x

)

if f is continuous.

3. Characterizations of SOC-monotone functions

In this section we present some characterizations for SOC-monotone functions, by which the set of continuous SOC-monotone functions is shown to coincide with that of continuous matrix monotone functions of order 2. To this end, we need the following technical lemma.

Lemma 3.1. For any given f

J

→ R

with J open, let fsoc

S

→ H

be defined by (3).

(a) fsocis continuous on S if and only if f is continuous on J.

(b) fsocis (continuously) differentiable on S iff f is (continuously) differentiable on J. Also, when f is differentiable on J, for any x

xe

x0e

S and v

ve

v0e

,

fsoc



x

v

⎧⎪

⎪⎪

⎪⎪

⎪⎩

f

x0

v if xe

0

b1

x

a0

x

xe

ve

xe

c1

x

v0xe

a0

x

ve

b1

x

v0e

c1

x

xe

ve

e if xe

0

(10)

where a0

x

) =

fλ2(2x())−x)−λf1(1x()x)), b1

x

) =

f2(x))+2f1(x)), c1

x

) =

f2(x))−2f1(x)).

(6)

(c) If f is differentiable on J, then for any given x

S and all v

,

fsoc



x

e

f

soc

x

and

e

fsoc



x

v

v

f

soc

x



.

(d) If fis nonnegative (respectively, positive) on J, then for each x

S,

fsoc



x

0

respectively

fsoc



x

0

).

Proof. (a) Suppose that fsocis continuous. Letbe the set composed of those x

te with t

J.

Clearly,

⊆

S, and fsocis continuous on. Noting that fsoc

x

f

t

e for any x

∈

, it follows that f is continuous on J. Conversely, if f is continuous on J, then fsocis continuous at any x

xe

x0e

S with xe

0 since

i

x

and ui

x

for i

1

,

2 are continuous at such points. Next, let x

xe

+

x0e be an arbitrary element from S with xe

=

0, and we prove that fsocis continuous at x. Indeed, for any z

ze

z0e

∈

S sufficiently close to x, it is not hard to verify that

fsoc

z

fsoc

x

f

2

z

f

x0

2

f

1

z

f

x0

2

f

2

z

f

1

z

2

.

Since f is continuous on J, and

1

z

2

z

x0as z

→

x, it follows that f

1

z

f

x0

and f

2

z

f

x0

as z

x

.

The last two equations imply that fsocis continuous at x.

(b) When fsocis (continuously) differentiable, using the similar arguments as in part (a) can show that f is (continuously) differentiable. Next assume that f is differentiable. Fix any x

xe

x0e

∈

S. We first consider the case where xe

0. Since

i

x

for i

1

,

2 andxxeeare continuously differentiable at such x, it follows that f

i

x

and ui

x

)

are differentiable and continuously differentiable, respectively, at x. Then fsocis differentiable at such x by the definition of fsoc. Also, an elementary computation shows that

i

x

v

v

e

1

i

xe

v

v

e

e

xe

v0

1

i

xe

ve

xe

(11)

 xe

xe



v

v

v

e

e

xe

xe

v

v

e

e

xe

xe

3

ve

xe

xe

ve

xe

xe

3 (12)

for any v

ve

v0e

∈ H

, and consequently, [f

i

x

]v

f

i

x



v0

1

i

xe

ve

xe



[ui

x

]v

1

2

1

i

 ve

xe

xe

ve

xe

xe

3



.

Together with the definition of fsoc, we calculate that

fsoc



x

v is equal to f

1

x

2



v0

xe

ve

xe

 

e

xxee



f

1

x

2

 ve

xe

xe

ve

xe

xe

3



f

2

x

2



v0

xe

ve

xe

 

e

xxee



f

2

x

2

 ve

xe

xe

ve

xe

xe

3



b1

x

v0e

c1

x

xe

ve

e

c1

x

v0xe

b1

x

xe

ve

xe

a0

x

ve

a0

x

xe

ve

xe

where

2

x

1

x

2

xe



is used for the last equality. Thus, we get (10) for xe

=

0. We next consider the case where xe

=

0. Under this case, for any v

ve

v0e

,

(7)

fsoc

x

v

fsoc

x

f

x0

v0

ve

2

e

ve

f

x0

v0

ve

2

e

ve

f

x0

e

f

x0

v0

ve

2 e

f

x0

v0

ve

2 e

f

x0

v0

ve

2 ve

f

x0

v0

ve

2 ve

o

v

f

x0

v0e

ve

ve

o

v

where ve

vveeif ve

=

0, and otherwise veis an arbitrary unit vector from

e

. Hence,

fsoc

x

v

fsoc

x

f

x0

v

o

v

).

This shows that fsocis differentiable at such x with

fsoc



x

v

f

x0

)

v.

Assume that f is continuously differentiable. From (10), it is easy to see that

fsoc



x

)

is continuous at every x with xe

=

0. We next argue that

fsoc



x

)

is continuous at every x with xe

0. Fix any x

x0e with x0

J. For any z

ze

z0e with ze

0, we have

fsoc



z

v

fsoc



x

v

b1

z

a0

z

ve

b1

z

f

x0

v0

a0

z

f

x0

ve

c1

z

v0

ve

).

(13) Since f is continuously differentiable on J and

2

z

x0

1

z

x0as z

x, we have

a0

z

f

x0

b1

z

f

x0

and c1

z

0

.

Together with Eq. (13), we obtain that

fsoc



z

fsoc



x

as z

→

x.

(c) The result is direct by the definition of

f

)

socand a simple computation from (10).

(d) Suppose that f

t

0 for all t

J. Fix any x

xe

x0e

S. If xe

=

0, the result is direct. It remains to consider the case xe

0. Since f

t

0 for all t

J, we have b1

x

0, b1

x

c1

x

f

1

x

0, b1

x

c1

x

f

2

x

0 and a0

x

) ≥

0. From part (b) and the definitions of b1

x

and c1

x

)

, it follows that for any h

he

h0e

,

h

fsoc



x

h

b1

x

a0

x

xe

he

2

2c1

x

h0

xe

he

b1

x

h20

a0

x

he

2

a0

x

he

2

xe

he

2

1

2

b1

x

c1

x

[h0

xe

he

]2

1

2

b1

x

c1

x

[h0

xe

he

]2

0

.

This implies that the operator

fsoc



x

)

is positive semidefinite. Particularly, if f

t

0 for all t

J, we have that

h

fsoc



x

h

0 for all h

=

0. The proof is complete. 

Lemma3.1(d) shows that the differential operator

fsoc



x

)

corresponding to a differentiable non- decreasing f is positive semidefinite. So, the differential operator

fsoc



x

)

associated with a differen- tiable SOC-monotone function is also positive semidefinite.

Proposition 3.1. Assume that f

C1

J

)

with J open. Then f is SOC-monotone if and only if

fsoc



x

h

K for any x

S and h

∈

K.

Proof. If f is SOC-monotone, then for any x

S, h

K and t

0, we have fsoc

x

th

fsoc

x

K 0

,

which, by the continuous differentiability of fsocand the closedness of K, implies that

fsoc



x

hK 0

.

Section 3 is devoted to developing proximal point method to solve the monotone second-order cone complementarity problem with a practical approximation criterion based on a new

These include so-called SOC means, SOC weighted means, and a few SOC trace versions of Young, H¨ older, Minkowski inequalities, and Powers-Størmer’s inequality.. All these materials

The purpose of this talk is to analyze new hybrid proximal point algorithms and solve the constrained minimization problem involving a convex functional in a uni- formly convex

From these characterizations, we particularly obtain that a continuously differentiable function defined in an open interval is SOC-monotone (SOC-convex) of order n ≥ 3 if and only

We point out that extending the concepts of r-convex and quasi-convex functions to the setting associated with second-order cone, which be- longs to symmetric cones, is not easy

strongly monotone or uniform P -function to obtain property of bounded level sets, see Proposition 3.5 of Chen and Pan (2006).. In this section, we establish that if F is either

Abstract Like the matrix-valued functions used in solutions methods for semidefinite pro- gram (SDP) and semidefinite complementarity problem (SDCP), the vector-valued func-

Chen, “Alternative proofs for some results of vector- valued functions associated with second-order cone,” Journal of Nonlinear and Convex Analysis, vol.. Chen, “The convex and