Two classes of merit function for infinite-dimensional SOCCPs

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.

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, · · · , x_n) ∈ Rⁿ| x_j ≥ 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 [8], and a merit function method was considered in [5] where its merit function is ψ_FB : H × H → R⁺ given by

ψ_FB(x, y) := 1

2kφ_FB(x, y)k², (2)

which is induced by the Fischer-Burmeister (FB) function φ_FB : H × H → H defined as φ_FB(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.

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 [2]. The first class is

f_LT(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

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

ψ^∗₀(w) = 1

2k(w)₊k² (9)

and ψ : H × H → R+ satisfies (7). The function f_LT was originally proposed by Luo and Tseng for NCP case [12] and was extended to the SDCP case by Tseng [14], then to the SOCCP case by Chen [2]. 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 [9] and a variant of cf_LT was also studied by Chen [2] 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 f_LT and cf_LT 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 f_LT and cf_LT 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 f_LT and cf_LT 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

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 lim_n→∞kx_n− xk = 0. A sequence of operators {T_n} → T means lim_n→∞kT_n− 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 [5], 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

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)v_x⁽¹⁾+ α2(x)v_x⁽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

α_i(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 x^1/2 ∈ K such that (x^1/2)² = (x^1/2)•(x^1/2) =

x, where

x^1/2 =p

α₁(x) v_x⁽¹⁾+p

α₂(x) v_x⁽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)⁻¹v_x⁽¹⁾+ α₂(x)⁻¹v_x⁽²⁾ = −x⁰+ λe

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

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

L_xy := x • y for any y ∈ H. (12)

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

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

Then, we have

(a) x _K0 ⇐⇒ L_x  0 and x _K0 ⇐⇒ L_x  0;

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

L⁻¹_x 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 =⇒ L²_z− L²_y− L²_x  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 x⁻_K 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+ x⁻_K and kxk² = kx⁺_Kk²+ kx⁻_Kk². (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.

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 l_x ∈ L(X , Y) such that

h→0lim

f (x + h) − f (x) − l_xh

khk = 0,

and l_x 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 [12] for the NCP case originally which was already extended to the SDP and SOCCP by Tseng [14] and Chen [2], respectively. We introduce them as below. Let f_LT be given as (5), i.e.,

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

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:

 hD_xψ(x, y), D_yψ(x, y)i ≥ 0, ∀(x, y) ∈ H × H.

hx, D_xψ(x, y)i + hy, D_yψ(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 [17] 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 [17]. 2

Lemma 3.3 Let φ_FB and ψ_FB be given as in (3) and (2), respectively. Then, (a) φ_FB(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

4ψ_FB(x, y) ≥ 2kφ_FB(x, y)₊k² ≥ k(−x)₊k²+ k(−y)₊k².

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 D_xψ₁(x, y) = (x)−, D_yψ₁(x, y) = (y)−.

(c) For every (x, y) ∈ H × H, we have

hD_xψ₁(x, y), D_yψ₁(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 hD_xψ(x, y), D_yψ(x, y)i = 0. (15) Two examples of such ψ are given in next two propositions.

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

(a) ψ_FB satisfies (7).

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

D_xψ_FB(x, y) = L_xL⁻¹_(x2+y2)1/2φ_FB(x, y) − φ_FB(x, y), D_yψ_FB(x, y) = L_yL⁻¹_(x2+y2)1/2φ_FB(x, y) − φ_FB(x, y).

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

Dxψ_FB(x, y) = λ

pλ²+ µ² − 1

!

φ_FB(x, y),

D_yψ_FB(x, y) = µ

pλ²+ µ² − 1

!

φ_FB(x, y)).

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

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

hx, D_xψ_FB(x, y)i + hy, D_yψ_FB(x, y)i = kφ_FB(x, y)k².

(e) ψ_FB belongs to Ψ₊₊.

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

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

ψ_YF(x, y) := ψ₀(hx, yi) + ψ_FB(x, y), and studied by Yamashita and Fukushima [16].

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

2kφ_FB(x, y)₊k², (16)

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

(a) ψ₂ satisfies (7).

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

D_xψ₂(x, y) = L_xL⁻¹_(x2+y2)1/2φ_FB(x, y)₊− φ_FB(x, y)₊, D_yψ₂(x, y) = L_yL⁻¹_(x2+y2)1/2φ_FB(x, y)₊− φ_FB(x, y)₊. If (x, y) 6= (0, 0) and x²+ y² ∈ int(K), then λ/ ²+ µ² 6= 0 and

D_xψ₂(x, y) = λ

pλ²+ µ² − 1

!

φ_FB(x, y)₊,

D_yψ₂(x, y) = µ

pλ²+ µ² − 1

!

φ_FB(x, y)₊.

(c) For every (x, y) = (x⁰+ λe, y⁰+ µe) ∈ H × H, we have hD_xψ₂(x, y), D_yψ₂(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, D_xψ₂(x, y)i + hy, D_yψ₂(x, y)i = kφ_FB(x, y)₊k².

(e) ψ₂ belongs to Ψ₊₊.

Proof. (a) Suppose ψ₂(x, y) = 0 and hx, yi ≤ 0. Let z := −φ_FB(x, y). Then (−z)₊ = φ_FB(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 φ_FB(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 φ_FB(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.

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

2kφ_FB(h, k)₊k²

≤ kφ_FB(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

D_xψ₂(0, 0) = D_yψ₂(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⁾z +α₂(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 [17], 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) = g₁(φ_FB(x, y)) and φ_FB is Fr´echet differentiable at (x, y) 6= (0, 0) with x²+ y² ∈ int(K) (see [5]). Hence, the chain rule yields

D_xψ₂(x, y) = D_xφ_FB(x, y)Dg₁(φ_FB(x, y)) = L_xL⁻¹

(x²+y²)^1/2φ_FB(x, y)₊− φ_FB(x, y)₊,

D_yψ₂(x, y) = D_yφ_FB(x, y)Dg₁(φ_FB(x, y)) = L_yL⁻¹_(x2+y2)1/2φ_FB(x, y)₊− φ_FB(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 φ_FB is no longer differentiable in this case. By the spectral factorization, we observe that

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

(18)

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

φ_FB(x, y) = λx⁰+ µy⁰

pλ²+ 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

pλ²+ 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 φ_FB(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

2kφ_FB(x, y)₊k² = 1 4α²₂

= 1

4 h

(p

λ² + µ²− (λ + µ))² +2p

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

have

[D_xw₂]h

= 1

pλ²+ µ² − λ²

(λ²+ µ²)pλ²+ µ²

!

hh, eix⁰+ λ

pλ²+ µ² − 1

!

h − hh, eie

− λµy⁰

(λ²+ µ²)pλ²+ µ²hh, ei

= 1

pλ² + µ²3



(λ²+ µ²)x⁰− λ²x⁰− λµy⁰



hh, ei + λ pλ²+ µ²

!

h − hh, eie

= λ

pλ²+ µ² − 1

!

h − hh, eie
,

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

[D_xψ₂(x, y)]h = 1

2α₂[D_xα₂]h

= 1 2α2

"

λ

pλ²+ µ² − 1

!

hh, ei + hw₂, [D_xw₂]hi kw₂k

#

= 1 2α₂

"

λ

pλ²+ µ² − 1

!

hh, ei + λ

pλ²+ µ² − 1

!hw₂, hi − hw₂, eihh, ei kw2k

#

= 1

2α₂ λ

pλ²+ µ² − 1

! w₂

kw₂k + e, h

 . It then follows that

D_xψ₂(x, y) = λ

pλ²+ µ² − 1

!

α₂u⁽²⁾ = λ

pλ²+ µ² − 1

!

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

D_yψ₂(x, y) = µ

pλ²+ µ² − 1

!

φ_FB(x, y)₊.

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

1

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

D_xψ₂(x, y) = λ

pλ²+ µ² − 1

!

φ_FB(x, y) = λ

pλ²+ µ² − 1

!

φ_FB(x, y)₊,

D_yψ₂(x, y) = µ

pλ²+ µ² − 1

!

φ_FB(x, y) = µ

pλ²+ µ² − 1

!

φ_FB(x, y)₊.

(22)

If there is (x⁰, y⁰) such that φ_FB(x⁰, y⁰) /∈ K ∪ −K and φFB(x⁰, y⁰) → φ_FB(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 φ_FB(x, y) ∈ −K then φFB(x, y)+ = 0. Thus, ψ2(x, y) = ¹₂kφ_FB(x, y)+k² = 0 and it is clear that its derivative under this subcase is

D_xψ₂(x, y) = 0 = λ

pλ²+ µ² − 1

!

φ_FB(x, y)₊,

Dyψ2(x, y) = 0 = µ

pλ²+ µ² − 1

!

φ_FB(x, y)+.

(23)

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

D_xψ₂(x⁰, y⁰) → D_xψ₂(x, y), D_yψ₂(x⁰, y⁰) → D_yψ₂(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 D_xψ₂(x, y) = D_yψ₂(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 hD_xψ₂(x, y), D_yψ₂(x, y)i

= h(LxL⁻¹_z − I)(φ_FB)+, (LyL⁻¹_z − I)(φ_FB)+i

= h(L_x− L_z)L⁻¹_z (φ_FB)₊, (L_y− L_z)L⁻¹_z (φ_FB)₊i

= h(L_y− L_z)(L_x− L_z)L⁻¹_z (φ_FB)₊, L⁻¹_z (φ_FB)₊i,

(24)

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

(L_z− L_x)(L_z − L_y)+(L_z− L_y)(L_z− L_x) = (L_z− L_x− L_y)²+(L²_z− L²_x− L²_y).

Since z ∈ K and z² = x²+ y², Property 2.3 implies L²_z − L²_x− L²_y  0. Then (24) yields hD_xψ₂(x, y), D_yψ₂(x, y)i ≥ 1

2k(L_z − L_x− L_y)L⁻¹_z (φ_FB)₊k²

= 1 2kL_φ

FBL⁻¹_z (φ_FB)₊k²,

where the equality uses L_z− L_x− L_y = L_z−x−y = L_φ

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

FBL⁻¹_z (φ_FB)₊k² = 0 and, by Property 2.1(d), L_φ

FBL⁻¹_z (φ_FB)₊= φ_FB • (L⁻¹_z (φ_FB)₊) = (L⁻¹_z (φ_FB)₊) • φ_FB = 0.

Since z = px²+ y² ∈ int(K) so that L⁻¹z  0 (see Property 2.1(d)), multiplying L⁻¹_z both side gives φ_FB• (φ_FB)₊= 0. From definition of Jordan product and Lemma 3.1(a), it implies (φ_FB)₊ = 0 and hence ψ₂ = 0. Conversely, if (φ_FB)₊ = 0, then it is clear that hD_xψ₂(x, y), D_yψ₂(x, y)i = 0.

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

hD_xψ₂(x, y), D_yψ₂(x, y)i = λ

pλ²+ µ² − 1

! µ

pλ²+ µ² − 1

!

kφ_FB(x, y)₊k² ≥ 0.

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

λ²+µ² = 1 or √ ^µ

λ²+µ² = 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 D_xψ₂(x, y) = D_xψ₂(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 D_xψ₂(x, y) = (L_xL⁻¹_z − I)φ_FB(x, y)₊, D_yψ₂(x, y) = (L_yL⁻¹_z − I)φ_FB(x, y)₊, where we let z =px² + y². Thus,

hx, D_xψ₂(x, y)i + hy, D_yψ₂(x, y)i

= hx, (L_xL⁻¹_z − I)φ_FB(x, y)₊i + hy, (L_yL⁻¹_z − I)φ_FB(x, y)₊i

= h(L_xL⁻¹_z − I)x, φ_FB(x, y)₊i + h(L_yL⁻¹_z − I)y, φ_FB(x, y)₊i

= hL⁻¹_z L_xx + L⁻¹_z L_yy − x − y, φ_FB(x, y)₊i

= hL⁻¹_z (x²+ y²) − x − y, φ_FB(x, y)₊i

= hL⁻¹_z z²− x − y, φ_FB(x, y)+i

= hz − x − y, φ_FB(x, y)₊i

= kφ_FB(x, y)₊k²,

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

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

D_xψ₂(x, y) = λ

pλ²+ µ² − 1

!

φ_FB(x, y)₊,

D_yψ₂(x, y) = µ

pλ²+ µ² − 1

!

φ_FB(x, y)₊.

Thus,

hx, Dxψ2(x, y)i + hy, Dyψ2(x, y)i

= λ

pλ²+ µ² − 1

!

hx, φ_FB(x, y)₊i + µ

pλ²+ µ² − 1

!

hy, φ_FB(x, y)₊i

=

* λx + µy

pλ²+ µ² − x − y, φ_FB(x, y)₊ +

= hφ_FB(x, y), φ_FB(x, y)+i

= kφ_FB(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 f_LT : 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 f_LT(ζ) ≥ 0 and f_LT(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 f_LT and

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

+(DF (ζ))^TD_xψ(F (ζ), G(ζ)) + (DG(ζ))^TD_yψ(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) f_LT(ζ) = 0 or (ii) f_LT(ζ) 6= 0 with hd(ζ), Df_LT(ζ)i < 0, where

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

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

(b) Fix any ζ ∈ H. From Theorem 4.2 in [5] and the Fr´echet differentiability of F and G, it follows that f_LT : H → R+ is Fr´echet differentiable on H. By the chain rule of differential, we have, for any v ∈ H,

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

+hD_xψ(F (ζ), G(ζ)), DF (ζ)vi + hD_yψ(F (ζ), G(ζ)), DG(ζ)vi, which means

Df_LT(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, Df_LTi = h−(DG)⁻¹(αG + D_xψ(F, G)), (DF )^T(αG + D_xψ(F, G)) +(DG)^T(αF + D_yψ(F, G))i

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

−hαG + D_xψ(F, G), αF + D_yψ(F, G)i

≤ −hαG + D_xψ(F, G), αF + D_yψ(F, G)i

= −α²hG, F i − α(hF, D_xψ(F, G)i) + hG, D_yψ(F, G)i)

−hDxψ(F, G), Dyψ(F, G)i

≤ −α²hG, F i − hD_xψ(F, G), D_yψ(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, Df_LT(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., f_LT(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., fc_LT(ζ) := ψ₀^∗(F (ζ) • G(ζ)) + ψ(F (ζ), G(ζ)),

where ψ^∗₀ : H → R+ is given as (9) and ψ : H × H → R+ satisfies (7). We notice that ψ₀^∗ possesses the following property:

ψ^∗₀(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 [2], 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, D_xψ₀^∗(x • y) = L_y(x • y)₊, and D_yψ^∗₀(x • y) = L_x(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 [17], 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 ψ^∗₀(x • y) = g₁(z(x, y)). Fix x = x⁰ + λe ∈ H, y = y⁰ + µe ∈ H and h = h⁰+ le ∈ H, then we have

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

= ly⁰+ µh⁰ + hh, yie

= y • h

= L_yh.

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

Proposition 3.5 Let cf_LT : H → R+ be given as (8) with ψ^∗₀ satisfying (9) and ψ satis- fying (7). Then, the following results hold.

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

(b) If ψ^∗₀, ψ and F, G are Fr´echet differentiable, then so is cf_LT and D cf_LT(ζ) = [(DF (ζ))^TL_G(ζ)+ (DG(ζ))^TL_{F (ζ)}](F (ζ) • G(ζ))₊

+(DF (ζ))^TD_xψ(F (ζ), G(ζ)) + (DG(ζ))^TD_yψ(F (ζ), G(ζ)).

4 Solution existence, error bound, and bounded level sets

In this section, using the above merit functions f_LT and cf_LT, 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

 .

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 ψ_FB, ψ₁, ψ₂ be given as (2), (14), (16), respectively. Then, ψ_FB, 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 ψ_FB and ψ₂, the inequality is still true due to Lemma 3.3. 2

By the definition of ψ^∗₀ given as (9), we can obtain the following lemma easily.

Lemma 4.3 Let ψ^∗₀ be given as (9). Then, ψ^∗₀ 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 f_LT be given by (5) with ψ satisfying (7). If there exists a constant τ such that

  τ − ρ

δ²  

≤ pρ²− δ²(γ²− 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 − ζ^∗k² ≤ max{1, hF (ζ), G(ζ)i} + k(−F (ζ))₊k + k(−G(ζ))₊k, ∀ζ ∈ H.

Moreover,

κkζ − ζ^∗k² ≤ ψ₀⁻¹(f_LT(ζ)) +

√2

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