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# Two classes of merit function for infinite-dimensional SOCCPs

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Numerical Functional Analysis and Optimization, vol. 31, pp. 387-413, 2010

### Two classes of merit function for infinite-dimensional SOCCPs

Juhe Sun 1 School of Science

Shenyang Aerospace University Shenyang 110136, China

E-mail: juhesun@abel.math.ntnu.edu.tw

Jein-Shan Chen 2 Department of Mathematics National Taiwan Normal University

Taipei 11677, Taiwan E-mail: jschen@math.ntnu.edu.tw

November 16, 2009

Abstract. In this paper, we extend two classes of merit functions for the second-order complementarity problem (SOCCP) to infinite-dimensional SOCCP. These two classes of merit functions include several popular merit functions, which are used in NCP (nonlinear complementarity problem), SDCP (semidefinite complementarity problem), and SOCCP, as special cases. We give conditions under which the infinite-dimensional SOCCP has a unique solution and show that all these merit functions provide an error bound for infinite-dimensional SOCCP and have bounded level sets. These results are very useful for designing solution methods for infinite-dimensional SOCCP.

Key words. Hilbert space, second-order cone, merit functions, fixed point, error bound, level set.

### 1 Introduction

Let H be a Hilbert space endowed with an inner product h·, ·i, and write the norm induced by h·, ·i as k · k. The conic complementarity problem CP (K, F, G) in H is, for any given

1also affiliated with Department of Mathematics of National Taiwan Normal University.

2Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The author’s work is supported by National Science Council of Taiwan.

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closed convex cone K ⊂ H and functions F, G : H → H, to find points x, y, ζ ∈ H such that

hx, yi = 0, x ∈ K, y ∈ K, x = F (ζ), y = G(ζ),

where K := {x ∈ H| hx, yi ≥ 0, ∀y ∈ K} is the dual cone of K. A closed convex cone K ⊂ H is called self-dual if K coincides with its dual cone K, for example, the nonnegative orthant cone Rn+ := {(x1, · · · , xn) ∈ Rn| xj ≥ 0, j = 1, 2, . . . , n} and the second-order cone (also called Lorentz cone) Kn := {(x1, x2) ∈ R × Rn−1| x1 ≥ kx2k}.

This paper focuses on the conic complementarity problem associated with the infinite- dimensional second-order cone K in H (will be defined as in (10)) which is closed, convex, and self-dual (see Section 2 for details). Since K is self-dual, the conic complementarity problem reduces to CP (K, F, G), which is to find x, y, ζ ∈ H such that

hx, yi = 0, x ∈ K, y ∈ K,

x = F (ζ), y = G(ζ). (1)

For finite-dimensional second-order cone optimization and complementarity problems, there have proposed various methods, including the interior point methods [1, 15, 18], the smoothing and semismooth Newton methods [3, 7, 10, 11, 13], and the merit function method [2, 4]. As far as we know, only very few of aforementioned methods are extended to infinite-dimensional SOCCP case. More precisely, for infinite-dimensional second- order cone optimization and complementarity problems, some particular interior point method was employed in , and a merit function method was considered in  where its merit function is ψFB : H × H → R+ given by

ψFB(x, y) := 1

2kφFB(x, y)k2, (2)

which is induced by the Fischer-Burmeister (FB) function φFB : H × H → H defined as φFB(x, y) := (x2+ y2)1/2− (x + y). (3) Here x2 means x • x, where • will be introduced in Section 2. In this paper, we also concern with the merit function method for (1). In other words, we aim to seek a smooth function Ψ : H × H → R+ such that, for any x, y ∈ H,

Ψ(x, y) = 0 ⇐⇒ x ∈ K, y ∈ K, hx, yi = 0, (4) and then the problem CP (K, F, G) can be transformed into a smooth minimization problem:

minζ∈Hf (ζ) := Ψ(F (ζ), G(ζ)).

Traditionally, such a f or Ψ is called a merit function associated with K.

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The two classes of merit functions that we will investigate come intuitively from the finite-dimensional case where H equals Rn and is associated with the Lorentz cone Kn, which was studied in . The first class is

fLT(ζ) := ψ0(hF (ζ), G(ζ)i) + ψ(F (ζ), G(ζ)), (5) where ψ0 : R → R+ is any smooth function satisfying

ψ0(t) = 0 ∀t ≤ 0 and ψ00(t) > 0 ∀t > 0, (6) and ψ : H × H → R+ satisfies

ψ(x, y) = 0, hx, yi ≤ 0 ⇐⇒ (x, y) ∈ K × K, hx, yi = 0. (7) The second class is

fcLT(ζ) := ψ0(F (ζ) • G(ζ)) + ψ(F (ζ), G(ζ)), (8) where ψ0 : H → R+ is given by

ψ0(w) = 1

2k(w)+k2 (9)

and ψ : H × H → R+ satisfies (7). The function fLT was originally proposed by Luo and Tseng for NCP case  and was extended to the SDCP case by Tseng , then to the SOCCP case by Chen . We explore the extension to the infinite-dimensional SOCCPs as will be seen in Sections 3. The second class of merit functions for SDCP case was recently studied by Goes and Oliveira  and a variant of cfLT was also studied by Chen  for SOCCP case.

As mentioned, we will define and study these merit functions associated with K in Hilbert space H. Three examples of ψ will be studied in Section 3. In Section 4, we will show that, under certain conditions, the infinite-dimensional SOCCP has a unique solution and both fLT and cfLT provide global error bound, which plays an important role in analyzing the convergence rate of some iterative methods for solving CP (K, F, G).

Besides, under the condition that F and G are jointly monotone and a strictly feasible solution exists, we will prove that both fLT and cfLT have bounded level sets which will ensure that the sequence generated by a decent algorithm has at least an accumulation point. All these properties will make it possible to construct a decent algorithm for solving the equivalent unconstrained reformulation of CP (K, F, G). Moreover, we will show that both fLT and cfLT are Fr´echet differentiable and their derivatives have computable formulas.

Throughout this paper, for any given Banach spaces X and Y, let L(X , Y) denote the Banach space of all continuous linear mappings from X into Y. We simply write L(X , Y) = L(X ) and denote GL(X ) the set of all invertible mappings in L(X ). The

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norm of any l ∈ L(X , Y) is defined by klk := sup{kl(x)k | x ∈ X and kxk = 1}. In addition, for any self-adjoint linear operator l from X → X , we write l  0 (respectively, l  0) to mean that l is positive definite (respectively, positive semidefinite). For any x ∈ H, (x)+ denotes the orthogonal projection of x onto K, whereas (x) means the orthogonal projection of x onto −K. A sequence of elements {xn} ⊂ H → x means limn→∞kxn− xk = 0. A sequence of operators {Tn} → T means limn→∞kTn− T k = 0.

### 2 Preliminaries

In this section, we recall some background materials and preliminary results that will be used later. We begin with introducing the infinite-dimensional second-order cone.

Recall that the finite-dimensional second-order cone (also called Lorentz cone) is defined as Kn := {(r, x0) ∈ R × Rn−1| r ≥ kx0k}. As discussed in , this Lorentz cone Kn can be rewritten as

Kn :=



x ∈ Rn

hx, ei ≥ 1

√2kxk



with e = (1, 0) ∈ R × Rn−1.

Motivated by this, the following closed convex cone in the Hilbert space H is considered:

K(e, r) := {x ∈ H | hx, ei ≥ rkxk},

where e ∈ H with kek = 1 and 0 < r < 1. Observe that K(e, r) is pointed, that is, K(e, r) ∩ (−K(e, r)) = {0}. Moreover, by denoting

hei := {x ∈ H | hx, ei = 0}, we may express the closed convex cone K(e, r) as

K(e, r) =



x0+ λe ∈ H

x0 ∈ hei and λ ≥ r

√1 − r2kx0k

 . When H = Rn and e = (1, 0) ∈ R × Rn−1, K(e,1

2) coincides with Kn. By this, we shall call K(e,12) the infinite-dimensional second-order cone (or infinite-dimensional Lorentz cone) in H determined by e. In the rest of this paper, we shall only consider any fixed unit vector e ∈ H, and denote

K := K

 e, 1

√2



(10) since two infinite-dimensional second-order cones K(e1,1

2) and K(e2,1

2) associated with different unit elements e1 and e2 in H are isometric.

Unless specifically stated otherwise, we shall alternatively write any point x ∈ H as x = x0+ λe with x0 ∈ hei and λ = hx, ei. In addition, for any x, y ∈ H, we shall write

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x K y (respectively, x K y) if x − y ∈ int(K) (respectively, x − y ∈ K). Now, we introduce the spectral decomposition for any element x ∈ H. For any x = x0+ λe ∈ H, we can decompose x as

x = α1(x)vx(1)+ α2(x)vx(2),

where α1(x), α2(x) and v(1)x , vx(2) are the spectral values and the associated spectral vectors of x with respect to K, given by

αi(x) = (−1)ikx0k + λ, v(i)x =

1 2



(−1)i x0 kx0k + e



, if x0 6= 0,

1

2((−1)iw + e), if x0 = 0,

for i = 1, 2 with w being any vector in hei satisfying kwk = 1. Its determinant and trace is defined as det(x) := α1(x)α2(x) and tr(x) := α1(x) + α2(x), respectively.

Next, we come to the Jordan product associated with the infinite-dimensional Lorentz cone K. For any x = x0+ λe ∈ H and y = y0+ µe ∈ H, we define the Jordan product of x and y by

x • y := (µx0+ λy0) + hx, yie. (11) Clearly, when H = Rn and e = (1, 0) ∈ R × Rn−1, this definition is the same as the one given by [6, Chapter II]. From the definition (11) and direct computation, it is easy to verify that the following properties hold.

Property 2.1 (a) x • y = y • x and x • e = x for all x, y ∈ H.

(b) (x + y) • z = x • z + y • z for all x, y, z ∈ H.

(c) hx, y • zi = hy, x • zi = hz, x • yi for all x, y, z ∈ H.

(d) For any x = x0+ λe ∈ H, x2 = x • x = 2λx0+ kxk2e ∈ K and hx2, ei = kxk2. (e) If x = x0+λe ∈ K, then there is a unique x1/2 ∈ K such that (x1/2)2 = (x1/2)•(x1/2) =

x, where

x1/2 =p

α1(x) vx(1)+p

α2(x) vx(2) =

0, if x = 0

x0

2τ + τ e, otherwise with τ =

qλ+

λ2−kx0k2

2 .

(f ) Every x = x0+ λe ∈ H with λ2 − kx0k2 6= 0 is invertible with respect to the Jordan product, i.e., there is a unique point x−1 ∈ H such that x • x−1 = e, where

x−1 = α1(x)−1vx(1)+ α2(x)−1vx(2) = −x0+ λe

det(x) = −x0 + λe λ2− kx0k2. Moreover, x ∈ int(K) if and only if x−1 ∈ int(K).

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Associated with every x ∈ H, we define a linear mapping Lx from H to H by

Lxy := x • y for any y ∈ H. (12)

It is clear that Lx ∈ L(H) and this mapping possesses the following favorable properties.

Property 2.2 [5, Lemma 2.2] For any x ∈ H, let Lx ∈ L(H) be defined as in (12).

Then, we have

(a) x K0 ⇐⇒ Lx  0 and x K0 ⇐⇒ Lx  0;

(b) if x = x0 + λe with λ 6= 0 and |λ| 6= kx0k, then Lx ∈ GL(H) with the inverse given by

L−1x y = 1

λ y0− hx−1, yix0 + hx−1, yie for any y = y0 + µe ∈ H.

Property 2.3 [5, Lemma 5.1] Let K be the infinite-dimensional Lorentz cone in H given as in (10). For any x, y ∈ H and z K0, the following implications hold:

z2 K x2 + y2 =⇒ L2z− L2y− L2x  0, z2 Kx2 =⇒ z Kx.

Moreover, the above implications remain true when “  ” is replaced by “  ”.

The following describes some important relations when x2+ y2 lies on the boundary of K.

Property 2.4 [5, Lemma 2.3] For any x = x0+λe, y = y0+µe ∈ H with x2+y2 ∈ int(K),/ we have

λ2 = kx0k2, µ2 = ky0k2, λµ = hx0, y0i, λy0 = µx0.

Property 2.5 Let K be any closed convex cone in H. For each x ∈ H, let x+K and xK denote the minimum distance projection of x onto K and −K, respectively. The following results hold.

(a) For any x ∈ H, we have x = x+K+ xK and kxk2 = kx+Kk2+ kxKk2. (b) For any x ∈ H and y ∈ K, we have hx, yi ≤ hx+K, yi.

(c) If K is self-dual, then for any x ∈ H and y ∈ K, we have k(x + y)+Kk ≥ kx+Kk.

(d) For any x ∈ K and y ∈ H with x2− y2 ∈ K, we have x − y ∈ K.

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Proof. These results are true for general closed convex cone whose proofs are the same as in [4, Lemma 5.1]. 2

To close this section, we review some definitions that will be used in subsequent analysis.

Definition 2.1 Let F, G : H → H be single-valued mappings.

(a) F is said to be η-strongly monotone if there exists a constant η > 0 satisfying hF (x) − F (y), x − yi ≥ ηkx − yk2, ∀x, y ∈ H.

(b) F is said to be Lipschitz continuous with constant γ if

kF (x) − F (y)k ≤ γkx − yk, ∀x, y ∈ H.

(c) F and G are said to be ρ-jointly strongly monotone if there exists a constant ρ > 0 satisfying

hF (x) − F (y), G(x) − G(y)i ≥ ρkx − yk2, ∀x, y ∈ H.

We also recall the concept of Fr´echet differentiability. For given Banach spaces X and Y, a mapping f from a nonempty open subset X of X into Y is said to be Fr´echet differentiable at x ∈ X if there exists lx ∈ L(X , Y) such that

h→0lim

f (x + h) − f (x) − lxh

khk = 0,

and lx is called the Fr´echet differential of f at x, denoted by Df (x). When f is Fr´echet differentiable at every point of X, we say that f is Fr´echet differentiable on X. If f is Fr´echet differentiable on a neighborhood U ∈ X of a point x0 ∈ X, and if, as a mapping from U into the Banach space L(X , Y), the mapping x → Df (x) is continuous at x0, then f is said to be continuously Fr´echet differentiable at x0. The mapping f is called continuously Fr´echet differentiable on X if it is continuously Fr´echet differentiable at every point of X.

### 3 Two classes of merit functions

In this section, we elaborate more about the two classes of merit functions for (1). We are motivated by a class of merit functions proposed by Luo and Tseng  for the NCP case originally which was already extended to the SDP and SOCCP by Tseng  and Chen , respectively. We introduce them as below. Let fLT be given as (5), i.e.,

fLT(ζ) := ψ0(hF (ζ), G(ζ)i) + ψ(F (ζ), G(ζ)),

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where ψ0 : R → R+ satisfies (6) and ψ : H × H → R+ satisfies (7). We notice that ψ0 is differentiable and strictly increasing on [0, ∞). Let Ψ+ denote the collection of ψ : H × H → R+ satisfying (7) that are Fr´echet differentiable and their derivatives satisfy the following conditions:

 hDxψ(x, y), Dyψ(x, y)i ≥ 0, ∀(x, y) ∈ H × H.

hx, Dxψ(x, y)i + hy, Dyψ(x, y)i ≥ 0, ∀(x, y) ∈ H × H. (13)

We will give an example of ψ belonging to Ψ+ in Proposition 3.1. Before that, we need the following three lemmas which will be used for proving Proposition 3.1 and Proposition 3.2.

Lemma 3.1 (a) For any x ∈ H, hx, (x)i = k(x)k2 and hx, (x)+i = k(x)+k2 (b) For any x ∈ H and y ∈ H, we have

x ∈ K ⇐⇒ hx, yi ≥ 0, ∀y ∈ K.

Proof. The results follow by Property 2.5 and self-duality of K. 2

Lemma 3.2  For x 6= 0 ∈ H, the following hold.

(a) If g(x) = kxk, we have Dg(x)h = hx, hi kxk . (b) If g(x) = kxk2, we have Dg(x)h = 2hx, hi.

(c) If g(x) = x

kxk, we have Dg(x)h = h

kxk − hx, hi kxk3 x.

Proof. The results can be verified by direct computation, also see . 2

Lemma 3.3 Let φFB and ψFB be given as in (3) and (2), respectively. Then, (a) φFB(x, y) = 0 ⇐⇒ x ∈ K, y ∈ K, x • y = 0 ⇐⇒ x ∈ K, y ∈ K, hx, yi = 0.

(b) for any x, y ∈ H, there holds

FB(x, y) ≥ 2kφFB(x, y)+k2 ≥ k(−x)+k2+ k(−y)+k2.

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Proof. (a) This is shown in [5, Lemma 3.1].

(b) The first inequality follows from Property 2.5(a). Since (x2 + y2)1/2 − x ∈ K, by Property 2.5(c), we can deduce that k((x2 + y2)1/2− x − y)+k2 ≥ k(−y)+k2. Similarly, we can get k((x2+ y2)1/2− x − y)+k2 ≥ k(−x)+k2. Adding the above two inequalities yields the desired second inequality. This completes the proof. 2

Proposition 3.1 Let ψ1 : H × H → R+ be given by ψ1(x, y) := 1

2 k(−x)+k2+ k(−y)+k2

(14) Then, the following results hold.

(a) ψ1 satisfies (7).

(b) ψ1 is convex and Fr´echet differentiable at every (x, y) ∈ H × H with Dxψ1(x, y) = (x), Dyψ1(x, y) = (y).

(c) For every (x, y) ∈ H × H, we have

hDxψ1(x, y), Dyψ1(x, y)i ≥ 0.

(d) For every (x, y) ∈ H × H, we have

hx, Dxψ1(x, y)i + hy, Dyψ1(x, y)i = k(x)k2+ k(y)k2.

(e) ψ1 belongs to Ψ+.

Proof. The proofs are similar to those in [2, Proposition 3.1], so we omit them. 2

Next, we consider a further restriction on ψ. Let Ψ++ denote the collection of ψ ∈ Ψ+

satisfying the following conditions:

ψ(x, y) = 0, ∀(x, y) ∈ H × H whenever hDxψ(x, y), Dyψ(x, y)i = 0. (15) Two examples of such ψ are given in next two propositions.

Proposition 3.2 Let ψFB(x, y) be given by (2). Then, the following results hold.

(a) ψFB satisfies (7).

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(b) ψFB is Fr´echet differentiable at every (x, y) = (x0+ λe, y0+ µe) ∈ H × H. Moreover, DxψFB(0, 0) = DyψFB(0, 0) = 0. If (x, y) 6= (0, 0) and x2+ y2 ∈ int(K), then

DxψFB(x, y) = LxL−1(x2+y2)1/2φFB(x, y) − φFB(x, y), DyψFB(x, y) = LyL−1(x2+y2)1/2φFB(x, y) − φFB(x, y).

If (x, y) 6= (0, 0) and x2+ y2 ∈ int(K), then λ/ 2+ µ2 6= 0 and

DxψFB(x, y) = λ

2+ µ2 − 1

!

φFB(x, y),

DyψFB(x, y) = µ

2+ µ2 − 1

!

φFB(x, y)).

(c) For every (x, y) = (x0+ λe, y0+ µe) ∈ H × H, we have hDxψFB(x, y), DyψFB(x, y)i ≥ 0 and the equality holds whenever ψFB(x, y) = 0.

(d) For every (x, y) = (x0+ λe, y0+ µe) ∈ H × H, we have

hx, DxψFB(x, y)i + hy, DyψFB(x, y)i = kφFB(x, y)k2.

(e) ψFB belongs to Ψ++.

Proof. See [5, Theorem 4.1] and [5, Lemma 5.2]. 2

Proposition 3.2 tells us that ψFB defined as (2) belongs to Ψ++ which yields a merit function ψYF : H × H → R+ given as

ψYF(x, y) := ψ0(hx, yi) + ψFB(x, y), and studied by Yamashita and Fukushima .

Proposition 3.3 Let ψ2 : H × H → R+ be given by ψ2(x, y) = 1

2kφFB(x, y)+k2, (16)

where φFB is defined as (3). Then, the following results hold.

(a) ψ2 satisfies (7).

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(b) ψ2 is Fr´echet differentiable at every (x, y) = (x0+ λe, y0+ µe) ∈ H × H. Moreover, DxψFB(0, 0) = DyψFB(0, 0) = 0. If (x, y) 6= (0, 0) and x2+ y2 ∈ int(K), then

Dxψ2(x, y) = LxL−1(x2+y2)1/2φFB(x, y)+− φFB(x, y)+, Dyψ2(x, y) = LyL−1(x2+y2)1/2φFB(x, y)+− φFB(x, y)+. If (x, y) 6= (0, 0) and x2+ y2 ∈ int(K), then λ/ 2+ µ2 6= 0 and

Dxψ2(x, y) = λ

2+ µ2 − 1

!

φFB(x, y)+,

Dyψ2(x, y) = µ

2+ µ2 − 1

!

φFB(x, y)+.

(c) For every (x, y) = (x0+ λe, y0+ µe) ∈ H × H, we have hDxψ2(x, y), Dyψ2(x, y)i ≥ 0 and the equality holds whenever ψ2(x, y) = 0.

(d) For every (x, y) = (x0+ λe, y0+ µe) ∈ H × H, we have

hx, Dxψ2(x, y)i + hy, Dyψ2(x, y)i = kφFB(x, y)+k2.

(e) ψ2 belongs to Ψ++.

Proof. (a) Suppose ψ2(x, y) = 0 and hx, yi ≤ 0. Let z := −φFB(x, y). Then (−z)+ = φFB(x, y)+ = 0 which says z ∈ K. Since x + y = (x2+ y2)1/2+ z, squaring both sides and simplifying yield

2x • y = 2((x2 + y2)1/2• z) + z2.

Now, taking trace of both sides and using the fact tr(x • y) = 2hx, yi, we obtain

4hx, yi = 4h(x2+ y2)1/2, zi + 2kzk2. (17) Since (x2 + y2)1/2 ∈ K and z ∈ K, then we know h(x2 + y2)1/2, zi ≥ 0 by Lemma 3.1(b). Thus the right-hand of (17) is nonnegative, which together with hx, yi ≤ 0 implies hx, yi = 0. Therefore, with this, the equation (17) says z = 0 which is equivalent to φFB(x, y) = 0. Then by Lemma 3.3, we have x, y ∈ K. Conversely, if x, y ∈ K and hx, yi = 0, then again Lemma 3.3 yields φFB(x, y) = 0. Thus, ψ2(x, y) = 0 and hx, yi ≤ 0.

(b) For the proof of part (b), we need to discuss three cases.

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Case 1: If (x, y) = (0, 0), then for any h = h0+ ¯λe, k = k0 + ¯µe ∈ H, let µ1 ≤ µ2 be the spectral values and let v(1), v(2) be the corresponding spectral vectors of h2+ k2. Hence, by Property 2.1(e), we have

k(h2+ k2)1/2− h − kk = k√

µ1v(1)+√

µ2v(2)− h − kk

≤ √

µ1kv(1)k +√

µ2kv(2)k + khk + kkk

= (√

µ1+√ µ2)/√

2 + khk + kkk.

Also

µ1 ≤ µ2 = khk2 + kkk2+ 2k¯λh0+ ¯µk0k

≤ khk2 + kkk2+ 2|¯λ|kh0k + 2|¯µ|kk0k

≤ 2(khk2+ kkk2).

Combining the above two inequalities yields ψ2(h, k) − ψ2(0, 0) = 1

2kφFB(h, k)+k2

≤ kφFB(h, k)k2

= k(h2+ k2)1/2− h − kk2

≤ ((√

µ1+√ µ2)/√

2 + khk + kkk)2

≤ (2p

2khk2 + 2kkk2/√

2 + khk + kkk)2

= O(khk2+ kkk2),

where the first inequality is from Lemma 3.3. This shows that ψ2 is differentiable at (0, 0) with

Dxψ2(0, 0) = Dyψ2(0, 0) = 0.

Case 2: If (x, y) 6= (0, 0) and x2+y2 ∈ int(K), let z be factored as z = α1(z)u(1)z2(z)u(2)z

for any z ∈ H. Now, let g : H → H be defined as g(z) := 1

2((z)+)2 = ˆg(α1(z))u(1)z + ˆg(α2(z))u(2)z , where ˆg : R → R is given by ˆg(α) := 1

2(max(0, α))2. From the continuous differentiability of ˆg and , the vector-valued function g is continuously Fr´echet differentiable. Hence, the first component g1(z) = 1

2k(z)+k2 of g(z) is continuously Fr´echet differentiable as well. By an easy computation, we have Dg1(z) = (z)+. Since ψ2(x, y) = g1FB(x, y)) and φFB is Fr´echet differentiable at (x, y) 6= (0, 0) with x2+ y2 ∈ int(K) (see ). Hence, the chain rule yields

Dxψ2(x, y) = DxφFB(x, y)Dg1FB(x, y)) = LxL−1

(x2+y2)1/2φFB(x, y)+− φFB(x, y)+,

(13)

Dyψ2(x, y) = DyφFB(x, y)Dg1FB(x, y)) = LyL−1(x2+y2)1/2φFB(x, y)+− φFB(x, y)+.

Case 3: If (x, y) 6= (0, 0) and x2+ y2 ∈ int(K), by direct computation, we know kxk/ 2+ kyk2 = 2kλx0+ µy0k under this case. Since (x, y) 6= (0, 0), this also implies λx0+ µy0 6= 0.

We notice that we can not apply the chain rule as in Case 2 because φFB is no longer differentiable in this case. By the spectral factorization, we observe that

φFB(x, y)+ = φFB(x, y) ⇐⇒ φFB(x, y) ∈ K φFB(x, y)+= 0 ⇐⇒ φFB(x, y) ∈ −K φFB(x, y)+ = α2u(2) ⇐⇒ φFB(x, y) /∈ K ∪ −K,

(18)

where α2 is the bigger spectral value of φFB(x, y) and u(2) is the corresponding spectral vector. Indeed, by applying Property 2.4, we can simplify φFB as

φFB(x, y) = λx0+ µy0

2+ µ2 − (x0+ y0) +p

λ2+ µ2− (λ + µ)

e. (19)

Therefore, α2 and u(2) are given as below:

α2 = pλ2+ µ2− (λ + µ) + kw2k, u(2) = 1

2

 w2 kw2k + e



, (20)

where w2 = λx0+ µy0

2+ µ2 − (x0+ y0).

To prove the differentiability of ψ2 under this case, we shall discuss the following three subcases according to the above observation (18).

(i) If φFB(x, y) /∈ K ∪ −K then φFB(x, y)+ = α2u(2), where α2 and u(2) are given as in (20). From the fact that ku(2)k = 1/√

2, we obtain

ψ2(x, y) = 1

2kφFB(x, y)+k2 = 1 4α22

= 1

4 h

(p

λ2 + µ2− (λ + µ))2 +2p

λ2+ µ2− (λ + µ)

kw2k + kw2k2i .

Since (x, y) 6= (0, 0) in this case, ψ2 is Fr´echet differentiable clearly. For all h ∈ H, we

(14)

have

[Dxw2]h

= 1

2+ µ2 − λ2

2+ µ2)pλ2+ µ2

!

hh, eix0+ λ

2+ µ2 − 1

!

h − hh, eie

− λµy0

2+ µ2)pλ2+ µ2hh, ei

= 1

pλ2 + µ23



2+ µ2)x0− λ2x0− λµy0



hh, ei + λ pλ2+ µ2

!

h − hh, eie

= λ

2+ µ2 − 1

!

h − hh, eie,

where the last equality holds by Property 2.4. Using the product rule and chain rule for differentiation gives

[Dxψ2(x, y)]h = 1

2[Dxα2]h

= 1 2α2

"

λ

2+ µ2 − 1

!

hh, ei + hw2, [Dxw2]hi kw2k

#

= 1 2α2

"

λ

2+ µ2 − 1

!

hh, ei + λ

2+ µ2 − 1

!hw2, hi − hw2, eihh, ei kw2k

#

= 1

2 λ

2+ µ2 − 1

! w2

kw2k + e, h

. It then follows that

Dxψ2(x, y) = λ

2+ µ2 − 1

!

α2u(2) = λ

2+ µ2 − 1

!

φFB(x, y)+. (21) Similarly, we can obtain that

Dyψ2(x, y) = µ

2+ µ2 − 1

!

φFB(x, y)+.

(ii) If φFB(x, y) ∈ K then φFB(x, y)+ = φFB(x, y), and hence ψ2(x, y) = 12FB(x, y)+k2 =

1

2FB(x, y)k2. Thus, by Proposition 3.1, we know that the derivative of ψ2 under this subcase is as below:

Dxψ2(x, y) = λ

2+ µ2 − 1

!

φFB(x, y) = λ

2+ µ2 − 1

!

φFB(x, y)+,

Dyψ2(x, y) = µ

2+ µ2 − 1

!

φFB(x, y) = µ

2+ µ2 − 1

!

φFB(x, y)+.

(22)

(15)

If there is (x0, y0) such that φFB(x0, y0) /∈ K ∪ −K and φFB(x0, y0) → φFB(x, y) ∈ K (the neighborhood of point belonging to this subcase). From (21) and (22), it can be seen that

Dxψ2(x0, y0) → Dxψ2(x, y), Dyψ2(x0, y0) → Dyψ2(x, y).

Thus, ψ2 is differentiable under this subcase.

(iii) If φFB(x, y) ∈ −K then φFB(x, y)+ = 0. Thus, ψ2(x, y) = 12FB(x, y)+k2 = 0 and it is clear that its derivative under this subcase is

Dxψ2(x, y) = 0 = λ

2+ µ2 − 1

!

φFB(x, y)+,

Dyψ2(x, y) = 0 = µ

2+ µ2 − 1

!

φFB(x, y)+.

(23)

Again, if there is (x0, y0) such that φFB(x0, y0) /∈ K ∪ −K and φFB(x0, y0) → φFB(x, y) ∈ K (the neighborhood of point belonging to this subcase). From (21) and (23), it can be seen that

Dxψ2(x0, y0) → Dxψ2(x, y), Dyψ2(x0, y0) → Dyψ2(x, y).

Thus, ψ2 is differentiable under this subcase. From the above, we complete the proof of this case and therefore the argument for part (b) is done.

(c) We wish to show that hDxψ2(x, y), Dyψ2(x, y)i ≥ 0 and the equality holds if and only if ψ2(x, y) = 0. We follow the three cases as above.

Case 1: If (x, y) = (0, 0), by part (b), we know Dxψ2(x, y) = Dyψ2(x, y) = 0. Therefore, the desired equality holds.

Case 2: If (x, y) 6= (0, 0) and x2 + y2 ∈ intK, by part (b), we have hDxψ2(x, y), Dyψ2(x, y)i

= h(LxL−1z − I)(φFB)+, (LyL−1z − I)(φFB)+i

= h(Lx− Lz)L−1zFB)+, (Ly− Lz)L−1zFB)+i

= h(Ly− Lz)(Lx− Lz)L−1zFB)+, L−1zFB)+i,

(24)

where z =px2+ y2 and I ∈ L(H) is an identity mapping. From elementary calculation, we obtain that

(Lz− Lx)(Lz − Ly)+(Lz− Ly)(Lz− Lx) = (Lz− Lx− Ly)2+(L2z− L2x− L2y).

Since z ∈ K and z2 = x2+ y2, Property 2.3 implies L2z − L2x− L2y  0. Then (24) yields hDxψ2(x, y), Dyψ2(x, y)i ≥ 1

2k(Lz − Lx− Ly)L−1zFB)+k2

= 1 2kLφ

FBL−1zFB)+k2,

(16)

where the equality uses Lz− Lx− Ly = Lz−x−y = Lφ

FB. If the equality holds, then the above relation yields kLφ

FBL−1zFB)+k2 = 0 and, by Property 2.1(d), Lφ

FBL−1zFB)+= φFB • (L−1zFB)+) = (L−1zFB)+) • φFB = 0.

Since z = px2+ y2 ∈ int(K) so that L−1z  0 (see Property 2.1(d)), multiplying L−1z both side gives φFB• (φFB)+= 0. From definition of Jordan product and Lemma 3.1(a), it implies (φFB)+ = 0 and hence ψ2 = 0. Conversely, if (φFB)+ = 0, then it is clear that hDxψ2(x, y), Dyψ2(x, y)i = 0.

Case 3: If (x, y) 6= (0, 0) and x2 + y2 ∈ int(K), by part (b), we have/

hDxψ2(x, y), Dyψ2(x, y)i = λ

2+ µ2 − 1

! µ

2+ µ2 − 1

!

FB(x, y)+k2 ≥ 0.

If the equality holds, then either φFB(x, y)+ = 0 or √ λ

λ22 = 1 or √ µ

λ22 = 1. In the second case, we have µ = 0 and λ ≥ 0, so that Property 2.4 yields y0 = 0 and λ = kx0k. In the third case, we have λ = 0 and µ ≥ 0, so that Property 2.4 yields x0 = 0 and µ = ky0k.

Thus, in these cases, we have x • y = 0, x ∈ K, y ∈ K. Then, by (7), ψ2(x, y) = 0.

(d) Again, we need to discuss the three cases as below.

Case 1: If (x, y) = (0, 0), by part (b), we know Dxψ2(x, y) = Dxψ2(x, y) = 0. Therefore, the desired equality holds.

Case 2: If (x, y) 6= (0, 0) and x2 + y2 ∈ int(K), by part (b), we have Dxψ2(x, y) = (LxL−1z − I)φFB(x, y)+, Dyψ2(x, y) = (LyL−1z − I)φFB(x, y)+, where we let z =px2 + y2. Thus,

hx, Dxψ2(x, y)i + hy, Dyψ2(x, y)i

= hx, (LxL−1z − I)φFB(x, y)+i + hy, (LyL−1z − I)φFB(x, y)+i

= h(LxL−1z − I)x, φFB(x, y)+i + h(LyL−1z − I)y, φFB(x, y)+i

= hL−1z Lxx + L−1z Lyy − x − y, φFB(x, y)+i

= hL−1z (x2+ y2) − x − y, φFB(x, y)+i

= hL−1z z2− x − y, φFB(x, y)+i

= hz − x − y, φFB(x, y)+i

= kφFB(x, y)+k2,

where the next-to-last equality follows from Lzz = z2, so that L−1z z2 = z and the last equality is from Lemma 3.1(a).

(17)

Case 3: If (x, y) 6= (0, 0) and x2 + y2 ∈ int(K), by part (b), we have/

Dxψ2(x, y) = λ

2+ µ2 − 1

!

φFB(x, y)+,

Dyψ2(x, y) = µ

2+ µ2 − 1

!

φFB(x, y)+.

Thus,

hx, Dxψ2(x, y)i + hy, Dyψ2(x, y)i

= λ

2+ µ2 − 1

!

hx, φFB(x, y)+i + µ

2+ µ2 − 1

!

hy, φFB(x, y)+i

=

* λx + µy

2+ µ2 − x − y, φFB(x, y)+ +

= hφFB(x, y), φFB(x, y)+i

= kφFB(x, y)+k2,

where the next-to-last equality uses (19) and the last equality is from Lemma 3.1(a) again.

(e) This is an immediate consequence of (a) through (d). 2

Proposition 3.4 Let fLT : H → R+ be given as (5) with ψ0 satisfying (6) and ψ satis- fying (7). Then, the following results hold.

(a) For all ζ ∈ H, we have fLT(ζ) ≥ 0 and fLT(ζ) = 0 if and only if ζ solves the infinite-dimensional SOCCP (1).

(b) Let Dψ0(hF (ζ), G(ζ)i) = Dtψ0(t) with t = hF (ζ), G(ζ)i. If ψ0, ψ and F, G are Fr´echet differentiable, then so is fLT and

DfLT(ζ) = Dψ0(hF (ζ), G(ζ)i)[(DF (ζ))TG(ζ) + (DG(ζ))TF (ζ)]

+(DF (ζ))TDxψ(F (ζ), G(ζ)) + (DG(ζ))TDyψ(F (ζ), G(ζ)).

(c) Assume F, G are are Fr´echet differentiable mappings on H and ψ belongs to Ψ+ (respectively, Ψ++). Then, for every ζ ∈ H where DF (ζ)[DG(ζ)]−1 is positive definite (respectively, positive semi-definite), either (i) fLT(ζ) = 0 or (ii) fLT(ζ) 6= 0 with hd(ζ), DfLT(ζ)i < 0, where

d(ζ) := −(DG(ζ)−1)[Dψ0(hF (ζ), G(ζ)i)G(ζ) + Dxψ(F (ζ), G(ζ))].

(18)

Proof. (a) This consequence follows from (5), (6) and (7).

(b) Fix any ζ ∈ H. From Theorem 4.2 in  and the Fr´echet differentiability of F and G, it follows that fLT : H → R+ is Fr´echet differentiable on H. By the chain rule of differential, we have, for any v ∈ H,

DfLT(ζ)v = Dψ0(hF (ζ), G(ζ)i)[hDF (ζ)v, G(ζ)i + hDG(ζ)v, F (ζ)i]

+hDxψ(F (ζ), G(ζ)), DF (ζ)vi + hDyψ(F (ζ), G(ζ)), DG(ζ)vi, which means

DfLT(ζ) = Dψ0(hF (ζ), G(ζ)i)[(DF (ζ))TG(ζ) + (DG(ζ))TF (ζ)]

+(DF (ζ))TDxψ(F (ζ), G(ζ)) + (DG(ζ))TDyψ(F (ζ), G(ζ)).

(c) First, we consider the case of ψ ∈ Ψ++ and fix ζ ∈ H, where DF (ζ)[DG(ζ)]−1 is positive semi-definite. Let α := Dψ0(hF (ζ), G(ζ)i) and drop the argument (ζ) for simplicity. Then

hd, DfLTi = h−(DG)−1(αG + Dxψ(F, G)), (DF )T(αG + Dxψ(F, G)) +(DG)T(αF + Dyψ(F, G))i

= −hαG + Dxψ(F, G), ((DG)−1)T(DF )T(αG + Dxψ(F, G))i

−hαG + Dxψ(F, G), αF + Dyψ(F, G)i

≤ −hαG + Dxψ(F, G), αF + Dyψ(F, G)i

= −α2hG, F i − α(hF, Dxψ(F, G)i) + hG, Dyψ(F, G)i)

−hDxψ(F, G), Dyψ(F, G)i

≤ −α2hG, F i − hDxψ(F, G), Dyψ(F, G)i,

where the first inequality holds since DF (DG)−1 is positive semi-definite and the in- equality follows from α ≥ 0 and equation (13). Now, we observe that tDψ0(t) > 0 if and only if t > 0 since ψ0 is strictly increasing on [0, ∞). Therefore, the first term on the right-hand side is non-positive and equals zero if hF, Gi ≤ 0. In addition, by equa- tions (13) and (15), the second term on the right-hand side is non-positive and equals zero if ψ(F, G) = 0. Thus, we have hd, DfLT(ζ)i ≤ 0 and the equality hold only when hF (ζ), G(ζ)i ≤ 0 and ψ(F (ζ), G(ζ)) = 0, in which equation (7) implies ζ satisfies (1), i.e., fLT(ζ) = 0.

Similar argument can be applied for the case of ψ ∈ Ψ+and DF (DG)−1being positive definite. 2

Next, we further consider another class of merit functions given as (8), i.e., fcLT(ζ) := ψ0(F (ζ) • G(ζ)) + ψ(F (ζ), G(ζ)),

(19)

where ψ0 : H → R+ is given as (9) and ψ : H × H → R+ satisfies (7). We notice that ψ0 possesses the following property:

ψ0(w) = 0 ⇐⇒ w K 0 which is a similar feature to (6) in some sense.

By imitating the steps for proving Proposition 3.4 and using the following Lemma 3.4 which has been proved in SOC case by Chen , we obtain Proposition 3.5 which is a result analogous to Proposition 3.4. We omit its proof.

Lemma 3.4 The function ψ0(x•y) := 1

2k(x•y)+k2is differentiable for all (x, y) ∈ H×H.

Moreover, Dxψ0(x • y) = Ly(x • y)+, and Dyψ0(x • y) = Lx(x • y)+.

Proof. For any z ∈ H, we can factor z as z = α1(z)u(1)+ α2(z)u(2). Now, let g : H → H be defined as

g(z) := 1

2((z)+)2 = ˆg(α1(z))u(1)+ ˆg(α2(z))u(2), where ˆg : R → R is given by ˆg(α) := 1

2(max(0, α))2. From the continuous differentiability of ˆg and , the vector-valued function g is continuously Fr´echet differentiable. Hence, the first component g1(z) = 1

2k(z)+k2 of g(z) is continuously Fr´echet differentiable as well. By an easy computation, we have Dg1(z) = (z)+. Now, let

z(x, y) := x • y = (λy0+ µx0) + hx, yie,

then we have ψ0(x • y) = g1(z(x, y)). Fix x = x0 + λe ∈ H, y = y0 + µe ∈ H and h = h0+ le ∈ H, then we have

[Dxz(x, y)]h = hh, eiy0 + µ(h − hh, eie) + hh, yie

= ly0+ µh0 + hh, yie

= y • h

= Lyh.

Similarly, we can obtain [Dyz(x, y)]h = Lxh. Hence, applying the chain rule, the desired result follows. 2

Proposition 3.5 Let cfLT : H → R+ be given as (8) with ψ0 satisfying (9) and ψ satis- fying (7). Then, the following results hold.

(a) For all ζ ∈ H, we have cfLT(ζ) ≥ 0 and cfLT(ζ) = 0 if and only if ζ solves the infinite-dimensional SOCCP (1).

(20)

(b) If ψ0, ψ and F, G are Fr´echet differentiable, then so is cfLT and D cfLT(ζ) = [(DF (ζ))TLG(ζ)+ (DG(ζ))TLF (ζ)](F (ζ) • G(ζ))+

+(DF (ζ))TDxψ(F (ζ), G(ζ)) + (DG(ζ))TDyψ(F (ζ), G(ζ)).

### 4 Solution existence, error bound, and bounded level sets

In this section, using the above merit functions fLT and cfLT, we obtain error bounds for the solution of infinite-dimensional SOCCP (1). Meanwhile, we study the existence and uniqueness for the solution of CP (K, F, G). To reach our results, we need some lemmas as below.

Lemma 4.1 Let x = x0+ λe ∈ H and y = y0+ µe ∈ H. Then, we have hx, yi ≤√

2k(x • y)+k. (25)

Proof. First, we observe the fact that

x ∈ K ⇐⇒ (x)+ = x, x ∈ −K ⇐⇒ (x)+ = 0, x /∈ K ∪ −K ⇐⇒ (x)+ = α2u(2),

where α2 is the bigger spectral value of x with the corresponding spectral vector u(2) defined as in section 2. Hence, we have three cases.

Case 1: If x • y ∈ K, then (x • y)+ = x • y. By definition of Jordan product of x and y as (11), i.e., x • y := (µx0+ λy0) + hx, yie. It is clear that k(x • y)+k ≥ hx, yi and hence (25) holds.

Case 2: If x • y ∈ −K, then (x • y)+ = 0. Since x • y ∈ −K, by definition of Jordan product again, we have hx, yi ≤ 0. Hence, it is true that √

2k(x • y)+k ≥ hx, yi.

Case 3: If x • y /∈ K ∪ −K, then (x • y)+= α2u(2), where α2 = hx, yi + kµx0 + λy0k, u(2) = 1

2

 µx0+ λy0 kµx0+ λy0k+ e

 .

(21)

If hx, yi ≤ 0, then (25) is trivial. Thus, we can assume hx, yi > 0. In fact, the desired inequality (25) follows from the below.

k(x • y)+k2 = 1 2α22

= 1

2(hx, yi2+ 2hx, yikµx0+ λy0k + kµx0+ λy0k2)

≥ 1

2hx, yi2. Then, we complete the proof. 2

Lemma 4.2 Let ψFB, ψ1, ψ2 be given as (2), (14), (16), respectively. Then, ψFB, ψ1, and ψ2 satisfy the following inequality.

ψ(x, y) ≥ α(k(−x)+k2+ k(−x)+k2), ∀(x, y) ∈ H × H, for some positive constant α and  ∈ {FB, 1, 2}.

Proof. For ψ1, it is clear by definition (14) with α = 12. For ψFB and ψ2, the inequality is still true due to Lemma 3.3. 2

By the definition of ψ0 given as (9), we can obtain the following lemma easily.

Lemma 4.3 Let ψ0 be given as (9). Then, ψ0 satisfies ψ0(ω) ≥ βk(ω)+k2, ∀ω ∈ H, for some positive constant β.

Proposition 4.1 Let F and G are Lipschitz continuous with constants γ and δ, re- spectively. Suppose that F is η-strongly monotone and F and G are ρ-jointly strongly monotone mapping from H to H. Let fLT be given by (5) with ψ satisfying (7). If there exists a constant τ such that

τ − ρ

δ2

≤ pρ2− δ22− σ2)

δ2 , ρ > δp

γ2− σ2, (26)

where σ = 1 −p1 − 2η + γ2. Then, the infinite-dimensional conic complementarity prob- lem CP (K, F, G) given as in (1) has a unique solution ζ and there exists a scalar κ > 0 such that

κkζ − ζk2 ≤ max{1, hF (ζ), G(ζ)i} + k(−F (ζ))+k + k(−G(ζ))+k, ∀ζ ∈ H.

Moreover,

κkζ − ζk2 ≤ ψ0−1(fLT(ζ)) +

√2

√αfLT(ζ)1/2, ∀ζ ∈ H, where α is a positive constant.

Interior-point methods not amenable to warm start.. Chen,T

For example, Liu, Zhang and Wang  extended a class of merit functions proposed in  to the SCCP, Kong, Tuncel and Xiu  studied the extension of the implicit Lagrangian

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