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 R^{n}+ := {(x_{1}, · · · , x_{n}) ∈ R^{n}| x_{j} ≥ 0, j = 1, 2, . . . , n} and the
second-order cone (also called Lorentz cone) K^{n} := {(x_{1}, x_{2}) ∈ R × R^{n−1}| x_{1} ≥ kx_{2}k}.

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)k^{2}, (2)

which is induced by the Fischer-Burmeister (FB) function φ_{FB} : H × H → H defined as
φ_{FB}(x, y) := (x^{2}+ y^{2})^{1/2}− (x + y). (3)
Here x^{2} 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 R^{n} and is associated with the Lorentz cone K^{n},
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 ψ_{0} : R → R+ is any smooth function satisfying

ψ_{0}(t) = 0 ∀t ≤ 0 and ψ_{0}^{0}(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) := ψ_{0}^{∗}(F (ζ) • G(ζ)) + ψ(F (ζ), G(ζ)), (8)
where ψ_{0}^{∗} : H → R+ is given by

ψ^{∗}_{0}(w) = 1

2k(w)_{+}k^{2} (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 K^{n} := {(r, x^{0}) ∈ R × R^{n−1}| r ≥ kx^{0}k}. As discussed in [5], this Lorentz cone
K^{n} can be rewritten as

K^{n} :=

x ∈ R^{n}

hx, ei ≥ 1

√2kxk

with e = (1, 0) ∈ R × R^{n−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) =

x^{0}+ λe ∈ H

x^{0} ∈ hei^{⊥} and λ ≥ r

√1 − r^{2}kx^{0}k

.
When H = R^{n} and e = (1, 0) ∈ R × R^{n−1}, K(e,^{√}^{1}

2) coincides with K^{n}. By this, we shall
call K(e,^{√}^{1}_{2}) 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(e_{1},^{√}^{1}

2) and K(e_{2},^{√}^{1}

2) associated
with different unit elements e_{1} and e_{2} in H are isometric.

Unless specifically stated otherwise, we shall alternatively write any point x ∈ H as
x = x^{0}+ λe with x^{0} ∈ 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 = x^{0}+ 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)^{i}kx^{0}k + λ,
v^{(i)}x =

1 2

(−1)^{i} x^{0}
kx^{0}k + e

, if x^{0} 6= 0,

1

2((−1)^{i}w + e), if x^{0} = 0,

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

Next, we come to the Jordan product associated with the infinite-dimensional Lorentz
cone K. For any x = x^{0}+ λe ∈ H and y = y^{0}+ µe ∈ H, we define the Jordan product of
x and y by

x • y := (µx^{0}+ λy^{0}) + hx, yie. (11)
Clearly, when H = R^{n} and e = (1, 0) ∈ R × R^{n−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 = x^{0}+ λe ∈ H, x^{2} = x • x = 2λx^{0}+ kxk^{2}e ∈ K and hx^{2}, ei = kxk^{2}.
(e) If x = x^{0}+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

α_{1}(x) v_{x}^{(1)}+p

α_{2}(x) v_{x}^{(2)} =

0, if x = 0

x^{0}

2τ + τ e, otherwise with τ =

qλ+

√

λ^{2}−kx^{0}k^{2}

2 .

(f ) Every x = x^{0}+ λe ∈ H with λ^{2} − kx^{0}k^{2} 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} = α_{1}(x)^{−1}v_{x}^{(1)}+ α_{2}(x)^{−1}v_{x}^{(2)} = −x^{0}+ λe

det(x) = −x^{0} + λe
λ^{2}− kx^{0}k^{2}.
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_{x}y := 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 _{K}0 ⇐⇒ L_{x} 0 and x _{K}0 ⇐⇒ L_{x} 0;

(b) if x = x^{0} + λe with λ 6= 0 and |λ| 6= kx^{0}k, then L_{x} ∈ GL(H) with the inverse given
by

L^{−1}_{x} y = 1

λ y^{0}− hx^{−1}, yix^{0} + hx^{−1}, yie for any y = y^{0} + 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 _{K}0, the following implications hold:

z^{2} _{K} x^{2} + y^{2} =⇒ L^{2}_{z}− L^{2}_{y}− L^{2}_{x} 0,
z^{2} _{K}x^{2} =⇒ z Kx.

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

The following describes some important relations when x^{2}+ y^{2} lies on the boundary
of K.

Property 2.4 [5, Lemma 2.3] For any x = x^{0}+λe, y = y^{0}+µe ∈ H with x^{2}+y^{2} ∈ int(K),/
we have

λ^{2} = kx^{0}k^{2}, µ^{2} = ky^{0}k^{2}, λµ = hx^{0}, y^{0}i, λy^{0} = µx^{0}.

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 kxk^{2} = kx^{+}_{K}k^{2}+ kx^{−}_{K}k^{2}.
(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)^{+}_{K}k ≥ kx^{+}_{K}k.

(d) For any x ∈ K and y ∈ H with x^{2}− y^{2} ∈ 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 − yk^{2}, ∀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 − yk^{2}, ∀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_{x}h

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 x_{0} ∈ 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 x_{0}. 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) := ψ_{0}(hF (ζ), G(ζ)i) + ψ(F (ζ), G(ζ)),

where ψ_{0} : R → R+ satisfies (6) and ψ : H × H → R+ satisfies (7). We notice that
ψ_{0} 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_{x}ψ(x, y), D_{y}ψ(x, y)i ≥ 0, ∀(x, y) ∈ H × H.

hx, D_{x}ψ(x, y)i + hy, D_{y}ψ(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)−k^{2} and hx, (x)+i = k(x)+k^{2}
(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) = kxk^{2}, we have Dg(x)h = 2hx, hi.

(c) If g(x) = x

kxk, we have Dg(x)h = h

kxk − hx, hi
kxk^{3} 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)_{+}k^{2} ≥ k(−x)_{+}k^{2}+ k(−y)_{+}k^{2}.

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

(b) The first inequality follows from Property 2.5(a). Since (x^{2} + y^{2})^{1/2} − x ∈ K, by
Property 2.5(c), we can deduce that k((x^{2} + y^{2})^{1/2}− x − y)_{+}k^{2} ≥ k(−y)_{+}k^{2}. Similarly,
we can get k((x^{2}+ y^{2})^{1/2}− x − y)_{+}k^{2} ≥ k(−x)_{+}k^{2}. 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
ψ_{1}(x, y) := 1

2 k(−x)_{+}k^{2}+ k(−y)_{+}k^{2}

(14) Then, the following results hold.

(a) ψ_{1} satisfies (7).

(b) ψ_{1} is convex and Fr´echet differentiable at every (x, y) ∈ H × H with D_{x}ψ_{1}(x, y) =
(x)−, D_{y}ψ_{1}(x, y) = (y)−.

(c) For every (x, y) ∈ H × H, we have

hD_{x}ψ_{1}(x, y), D_{y}ψ_{1}(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)−k^{2}+ k(y)−k^{2}.

(e) ψ_{1} 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_{x}ψ(x, y), D_{y}ψ(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) = (x^{0}+ λe, y^{0}+ µe) ∈ H × H. Moreover,
D_{x}ψ_{FB}(0, 0) = D_{y}ψ_{FB}(0, 0) = 0. If (x, y) 6= (0, 0) and x^{2}+ y^{2} ∈ int(K), then

D_{x}ψ_{FB}(x, y) = L_{x}L^{−1}_{(x}_{2}_{+y}_{2}_{)}1/2φ_{FB}(x, y) − φ_{FB}(x, y),
D_{y}ψ_{FB}(x, y) = L_{y}L^{−1}_{(x}_{2}_{+y}_{2}_{)}1/2φ_{FB}(x, y) − φ_{FB}(x, y).

If (x, y) 6= (0, 0) and x^{2}+ y^{2} ∈ int(K), then λ/ ^{2}+ µ^{2} 6= 0 and

Dxψ_{FB}(x, y) = λ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y),

D_{y}ψ_{FB}(x, y) = µ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y)).

(c) For every (x, y) = (x^{0}+ λe, y^{0}+ 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) = (x^{0}+ λe, y^{0}+ µe) ∈ H × H, we have

hx, D_{x}ψ_{FB}(x, y)i + hy, D_{y}ψ_{FB}(x, y)i = kφ_{FB}(x, y)k^{2}.

(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) := ψ_{0}(hx, yi) + ψ_{FB}(x, y),
and studied by Yamashita and Fukushima [16].

Proposition 3.3 Let ψ_{2} : H × H → R+ be given by
ψ_{2}(x, y) = 1

2kφ_{FB}(x, y)_{+}k^{2}, (16)

where φ_{FB} is defined as (3). Then, the following results hold.

(a) ψ_{2} satisfies (7).

(b) ψ_{2} is Fr´echet differentiable at every (x, y) = (x^{0}+ λe, y^{0}+ µe) ∈ H × H. Moreover,
D_{x}ψ_{FB}(0, 0) = D_{y}ψ_{FB}(0, 0) = 0. If (x, y) 6= (0, 0) and x^{2}+ y^{2} ∈ int(K), then

D_{x}ψ_{2}(x, y) = L_{x}L^{−1}_{(x}_{2}_{+y}_{2}_{)}1/2φ_{FB}(x, y)_{+}− φ_{FB}(x, y)_{+},
D_{y}ψ_{2}(x, y) = L_{y}L^{−1}_{(x}_{2}_{+y}_{2}_{)}1/2φ_{FB}(x, y)_{+}− φ_{FB}(x, y)_{+}.
If (x, y) 6= (0, 0) and x^{2}+ y^{2} ∈ int(K), then λ/ ^{2}+ µ^{2} 6= 0 and

D_{x}ψ_{2}(x, y) = λ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y)_{+},

D_{y}ψ_{2}(x, y) = µ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y)_{+}.

(c) For every (x, y) = (x^{0}+ λe, y^{0}+ µe) ∈ H × H, we have
hD_{x}ψ_{2}(x, y), D_{y}ψ_{2}(x, y)i ≥ 0
and the equality holds whenever ψ_{2}(x, y) = 0.

(d) For every (x, y) = (x^{0}+ λe, y^{0}+ µe) ∈ H × H, we have

hx, D_{x}ψ_{2}(x, y)i + hy, D_{y}ψ_{2}(x, y)i = kφ_{FB}(x, y)_{+}k^{2}.

(e) ψ_{2} belongs to Ψ_{++}.

Proof. (a) Suppose ψ_{2}(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 = (x^{2}+ y^{2})^{1/2}+ z, squaring both sides and
simplifying yield

2x • y = 2((x^{2} + y^{2})^{1/2}• z) + z^{2}.

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

4hx, yi = 4h(x^{2}+ y^{2})^{1/2}, zi + 2kzk^{2}. (17)
Since (x^{2} + y^{2})^{1/2} ∈ K and z ∈ K, then we know h(x^{2} + y^{2})^{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, ψ_{2}(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 = h^{0}+ ¯λe, k = k^{0} + ¯µe ∈ H, let µ_{1} ≤ µ_{2} be the
spectral values and let v^{(1)}, v^{(2)} be the corresponding spectral vectors of h^{2}+ k^{2}. Hence,
by Property 2.1(e), we have

k(h^{2}+ k^{2})^{1/2}− h − kk = k√

µ_{1}v^{(1)}+√

µ_{2}v^{(2)}− h − kk

≤ √

µ_{1}kv^{(1)}k +√

µ_{2}kv^{(2)}k + khk + kkk

= (√

µ_{1}+√
µ_{2})/√

2 + khk + kkk.

Also

µ_{1} ≤ µ_{2} = khk^{2} + kkk^{2}+ 2k¯λh^{0}+ ¯µk^{0}k

≤ khk^{2} + kkk^{2}+ 2|¯λ|kh^{0}k + 2|¯µ|kk^{0}k

≤ 2(khk^{2}+ kkk^{2}).

Combining the above two inequalities yields
ψ_{2}(h, k) − ψ_{2}(0, 0) = 1

2kφ_{FB}(h, k)_{+}k^{2}

≤ kφ_{FB}(h, k)k^{2}

= k(h^{2}+ k^{2})^{1/2}− h − kk^{2}

≤ ((√

µ_{1}+√
µ_{2})/√

2 + khk + kkk)^{2}

≤ (2p

2khk^{2} + 2kkk^{2}/√

2 + khk + kkk)^{2}

= O(khk^{2}+ kkk^{2}),

where the first inequality is from Lemma 3.3. This shows that ψ_{2} is differentiable at
(0, 0) with

D_{x}ψ_{2}(0, 0) = D_{y}ψ_{2}(0, 0) = 0.

Case 2: If (x, y) 6= (0, 0) and x^{2}+y^{2} ∈ int(K), let z be factored as z = α1(z)u^{(1)}z +α_{2}(z)u^{(2)}z

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

2((z)_{+})^{2} = ˆg(α_{1}(z))u^{(1)}_{z} + ˆg(α_{2}(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 g_{1}(z) = 1

2k(z)_{+}k^{2} of g(z) is continuously Fr´echet differentiable as
well. By an easy computation, we have Dg_{1}(z) = (z)_{+}. Since ψ_{2}(x, y) = g_{1}(φ_{FB}(x, y))
and φ_{FB} is Fr´echet differentiable at (x, y) 6= (0, 0) with x^{2}+ y^{2} ∈ int(K) (see [5]). Hence,
the chain rule yields

D_{x}ψ_{2}(x, y) = D_{x}φ_{FB}(x, y)Dg_{1}(φ_{FB}(x, y)) = L_{x}L^{−1}

(x^{2}+y^{2})^{1/2}φ_{FB}(x, y)_{+}− φ_{FB}(x, y)_{+},

D_{y}ψ_{2}(x, y) = D_{y}φ_{FB}(x, y)Dg_{1}(φ_{FB}(x, y)) = L_{y}L^{−1}_{(x}_{2}_{+y}_{2}_{)}_{1/2}φ_{FB}(x, y)_{+}− φ_{FB}(x, y)_{+}.

Case 3: If (x, y) 6= (0, 0) and x^{2}+ y^{2} ∈ int(K), by direct computation, we know kxk/ ^{2}+
kyk^{2} = 2kλx^{0}+ µy^{0}k under this case. Since (x, y) 6= (0, 0), this also implies λx^{0}+ µy^{0} 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 α_{2} 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) = λx^{0}+ µy^{0}

pλ^{2}+ µ^{2} − (x^{0}+ y^{0}) +p

λ^{2}+ µ^{2}− (λ + µ)

e. (19)

Therefore, α_{2} and u^{(2)} are given as below:

α2 = pλ^{2}+ µ^{2}− (λ + µ) + kw2k,
u^{(2)} = 1

2

w_{2}
kw2k + e

, (20)

where w_{2} = λx^{0}+ µy^{0}

pλ^{2}+ µ^{2} − (x^{0}+ y^{0}).

To prove the differentiability of ψ_{2} 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)_{+} = α_{2}u^{(2)}, where α_{2} and u^{(2)} are given as in
(20). From the fact that ku^{(2)}k = 1/√

2, we obtain

ψ_{2}(x, y) = 1

2kφ_{FB}(x, y)_{+}k^{2} = 1
4α^{2}_{2}

= 1

4 h

(p

λ^{2} + µ^{2}− (λ + µ))^{2}
+2p

λ^{2}+ µ^{2}− (λ + µ)

kw_{2}k + kw_{2}k^{2}i
.

Since (x, y) 6= (0, 0) in this case, ψ_{2} is Fr´echet differentiable clearly. For all h ∈ H, we

have

[D_{x}w_{2}]h

= 1

pλ^{2}+ µ^{2} − λ^{2}

(λ^{2}+ µ^{2})pλ^{2}+ µ^{2}

!

hh, eix^{0}+ λ

pλ^{2}+ µ^{2} − 1

!

h − hh, eie

− λµy^{0}

(λ^{2}+ µ^{2})pλ^{2}+ µ^{2}hh, ei

= 1

pλ^{2} + µ^{2}3

(λ^{2}+ µ^{2})x^{0}− λ^{2}x^{0}− λµy^{0}

hh, ei + λ
pλ^{2}+ µ^{2}

!

h − hh, eie

= λ

pλ^{2}+ µ^{2} − 1

!

h − hh, eie,

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

[D_{x}ψ_{2}(x, y)]h = 1

2α_{2}[D_{x}α_{2}]h

= 1 2α2

"

λ

pλ^{2}+ µ^{2} − 1

!

hh, ei + hw_{2}, [D_{x}w_{2}]hi
kw_{2}k

#

= 1
2α_{2}

"

λ

pλ^{2}+ µ^{2} − 1

!

hh, ei + λ

pλ^{2}+ µ^{2} − 1

!hw_{2}, hi − hw_{2}, eihh, ei
kw2k

#

= 1

2α_{2} λ

pλ^{2}+ µ^{2} − 1

!
w_{2}

kw_{2}k + e, h

. It then follows that

D_{x}ψ_{2}(x, y) = λ

pλ^{2}+ µ^{2} − 1

!

α_{2}u^{(2)} = λ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y)_{+}. (21)
Similarly, we can obtain that

D_{y}ψ_{2}(x, y) = µ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y)_{+}.

(ii) If φ_{FB}(x, y) ∈ K then φFB(x, y)_{+} = φ_{FB}(x, y), and hence ψ_{2}(x, y) = ^{1}_{2}kφ_{FB}(x, y)_{+}k^{2} =

1

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

D_{x}ψ_{2}(x, y) = λ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y) = λ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y)_{+},

D_{y}ψ_{2}(x, y) = µ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y) = µ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y)_{+}.

(22)

If there is (x^{0}, y^{0}) such that φ_{FB}(x^{0}, y^{0}) /∈ K ∪ −K and φFB(x^{0}, y^{0}) → φ_{FB}(x, y) ∈ K (the
neighborhood of point belonging to this subcase). From (21) and (22), it can be seen
that

Dxψ2(x^{0}, y^{0}) → Dxψ2(x, y), Dyψ2(x^{0}, y^{0}) → Dyψ2(x, y).

Thus, ψ_{2} is differentiable under this subcase.

(iii) If φ_{FB}(x, y) ∈ −K then φFB(x, y)+ = 0. Thus, ψ2(x, y) = ^{1}_{2}kφ_{FB}(x, y)+k^{2} = 0 and it
is clear that its derivative under this subcase is

D_{x}ψ_{2}(x, y) = 0 = λ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y)_{+},

Dyψ2(x, y) = 0 = µ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y)+.

(23)

Again, if there is (x^{0}, y^{0}) such that φ_{FB}(x^{0}, y^{0}) /∈ K ∪ −K and φFB(x^{0}, y^{0}) → φ_{FB}(x, y) ∈ K
(the neighborhood of point belonging to this subcase). From (21) and (23), it can be
seen that

D_{x}ψ_{2}(x^{0}, y^{0}) → D_{x}ψ_{2}(x, y), D_{y}ψ_{2}(x^{0}, y^{0}) → D_{y}ψ_{2}(x, y).

Thus, ψ_{2} 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 ψ_{2}(x, y) = 0. We follow the three cases as above.

Case 1: If (x, y) = (0, 0), by part (b), we know D_{x}ψ_{2}(x, y) = D_{y}ψ_{2}(x, y) = 0. Therefore,
the desired equality holds.

Case 2: If (x, y) 6= (0, 0) and x^{2} + y^{2} ∈ intK, by part (b), we have
hD_{x}ψ_{2}(x, y), D_{y}ψ_{2}(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 =px^{2}+ y^{2} 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}+(L^{2}_{z}− L^{2}_{x}− L^{2}_{y}).

Since z ∈ K and z^{2} = x^{2}+ y^{2}, Property 2.3 implies L^{2}_{z} − L^{2}_{x}− L^{2}_{y} 0. Then (24) yields
hD_{x}ψ_{2}(x, y), D_{y}ψ_{2}(x, y)i ≥ 1

2k(L_{z} − L_{x}− L_{y})L^{−1}_{z} (φ_{FB})_{+}k^{2}

= 1
2kL_{φ}

FBL^{−1}_{z} (φ_{FB})_{+}k^{2},

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})_{+}k^{2} = 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 = px^{2}+ y^{2} ∈ int(K) so that L^{−1}z 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 ψ_{2} = 0. Conversely, if (φ_{FB})_{+} = 0, then it is clear that
hD_{x}ψ_{2}(x, y), D_{y}ψ_{2}(x, y)i = 0.

Case 3: If (x, y) 6= (0, 0) and x^{2} + y^{2} ∈ int(K), by part (b), we have/

hD_{x}ψ_{2}(x, y), D_{y}ψ_{2}(x, y)i = λ

pλ^{2}+ µ^{2} − 1

! µ

pλ^{2}+ µ^{2} − 1

!

kφ_{FB}(x, y)_{+}k^{2} ≥ 0.

If the equality holds, then either φ_{FB}(x, y)_{+} = 0 or √ ^{λ}

λ^{2}+µ^{2} = 1 or √ ^{µ}

λ^{2}+µ^{2} = 1. In the
second case, we have µ = 0 and λ ≥ 0, so that Property 2.4 yields y^{0} = 0 and λ = kx^{0}k. In
the third case, we have λ = 0 and µ ≥ 0, so that Property 2.4 yields x^{0} = 0 and µ = ky^{0}k.

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_{x}ψ_{2}(x, y) = D_{x}ψ_{2}(x, y) = 0. Therefore,
the desired equality holds.

Case 2: If (x, y) 6= (0, 0) and x^{2} + y^{2} ∈ int(K), by part (b), we have
D_{x}ψ_{2}(x, y) = (L_{x}L^{−1}_{z} − I)φ_{FB}(x, y)_{+},
D_{y}ψ_{2}(x, y) = (L_{y}L^{−1}_{z} − I)φ_{FB}(x, y)_{+},
where we let z =px^{2} + y^{2}. Thus,

hx, D_{x}ψ_{2}(x, y)i + hy, D_{y}ψ_{2}(x, y)i

= hx, (L_{x}L^{−1}_{z} − I)φ_{FB}(x, y)_{+}i + hy, (L_{y}L^{−1}_{z} − I)φ_{FB}(x, y)_{+}i

= h(L_{x}L^{−1}_{z} − I)x, φ_{FB}(x, y)_{+}i + h(L_{y}L^{−1}_{z} − I)y, φ_{FB}(x, y)_{+}i

= hL^{−1}_{z} L_{x}x + L^{−1}_{z} L_{y}y − x − y, φ_{FB}(x, y)_{+}i

= hL^{−1}_{z} (x^{2}+ y^{2}) − x − y, φ_{FB}(x, y)_{+}i

= hL^{−1}_{z} z^{2}− x − y, φ_{FB}(x, y)+i

= hz − x − y, φ_{FB}(x, y)_{+}i

= kφ_{FB}(x, y)_{+}k^{2},

where the next-to-last equality follows from L_{z}z = z^{2}, so that L^{−1}_{z} z^{2} = z and the last
equality is from Lemma 3.1(a).

Case 3: If (x, y) 6= (0, 0) and x^{2} + y^{2} ∈ int(K), by part (b), we have/

D_{x}ψ_{2}(x, y) = λ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y)_{+},

D_{y}ψ_{2}(x, y) = µ

pλ^{2}+ µ^{2} − 1

!

φ_{FB}(x, y)_{+}.

Thus,

hx, Dxψ2(x, y)i + hy, Dyψ2(x, y)i

= λ

pλ^{2}+ µ^{2} − 1

!

hx, φ_{FB}(x, y)_{+}i + µ

pλ^{2}+ µ^{2} − 1

!

hy, φ_{FB}(x, y)_{+}i

=

* λx + µy

pλ^{2}+ µ^{2} − x − y, φ_{FB}(x, y)_{+}
+

= hφ_{FB}(x, y), φ_{FB}(x, y)+i

= kφ_{FB}(x, y)_{+}k^{2},

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 ψ_{0} 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) = Dψ_{0}(hF (ζ), G(ζ)i)[(DF (ζ))^{T}G(ζ) + (DG(ζ))^{T}F (ζ)]

+(DF (ζ))^{T}D_{x}ψ(F (ζ), G(ζ)) + (DG(ζ))^{T}D_{y}ψ(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})[Dψ_{0}(hF (ζ), G(ζ)i)G(ζ) + D_{x}ψ(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 = Dψ_{0}(hF (ζ), G(ζ)i)[hDF (ζ)v, G(ζ)i + hDG(ζ)v, F (ζ)i]

+hD_{x}ψ(F (ζ), G(ζ)), DF (ζ)vi + hD_{y}ψ(F (ζ), G(ζ)), DG(ζ)vi,
which means

Df_{LT}(ζ) = Dψ_{0}(hF (ζ), G(ζ)i)[(DF (ζ))^{T}G(ζ) + (DG(ζ))^{T}F (ζ)]

+(DF (ζ))^{T}Dxψ(F (ζ), G(ζ)) + (DG(ζ))^{T}Dyψ(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 := Dψ_{0}(hF (ζ), G(ζ)i) and drop the argument (ζ) for
simplicity. Then

hd, Df_{LT}i = h−(DG)^{−1}(αG + D_{x}ψ(F, G)), (DF )^{T}(αG + D_{x}ψ(F, G))
+(DG)^{T}(αF + D_{y}ψ(F, G))i

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

−hαG + D_{x}ψ(F, G), αF + D_{y}ψ(F, G)i

≤ −hαG + D_{x}ψ(F, G), αF + D_{y}ψ(F, G)i

= −α^{2}hG, F i − α(hF, D_{x}ψ(F, G)i) + hG, D_{y}ψ(F, G)i)

−hDxψ(F, G), Dyψ(F, G)i

≤ −α^{2}hG, F i − hD_{x}ψ(F, G), D_{y}ψ(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 tDψ_{0}(t) > 0 if
and only if t > 0 since ψ_{0} 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)^{−1}being positive
definite. 2

Next, we further consider another class of merit functions given as (8), i.e.,
fc_{LT}(ζ) := ψ_{0}^{∗}(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 ψ_{0}^{∗}
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 ψ_{0}^{∗}(x•y) := 1

2k(x•y)_{+}k^{2}is differentiable for all (x, y) ∈ H×H.

Moreover, D_{x}ψ_{0}^{∗}(x • y) = L_{y}(x • y)_{+}, and D_{y}ψ^{∗}_{0}(x • y) = L_{x}(x • y)_{+}.

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

g(z) := 1

2((z)_{+})^{2} = ˆg(α_{1}(z))u^{(1)}+ ˆg(α_{2}(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 g_{1}(z) = 1

2k(z)_{+}k^{2} of g(z) is continuously Fr´echet differentiable as
well. By an easy computation, we have Dg_{1}(z) = (z)_{+}. Now, let

z(x, y) := x • y = (λy^{0}+ µx^{0}) + hx, yie,

then we have ψ^{∗}_{0}(x • y) = g_{1}(z(x, y)). Fix x = x^{0} + λe ∈ H, y = y^{0} + µe ∈ H and
h = h^{0}+ le ∈ H, then we have

[D_{x}z(x, y)]h = hh, eiy^{0} + µ(h − hh, eie) + hh, yie

= ly^{0}+ µh^{0} + hh, yie

= y • h

= L_{y}h.

Similarly, we can obtain [D_{y}z(x, y)]h = L_{x}h. 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))^{T}L_{G(ζ)}+ (DG(ζ))^{T}L_{F (ζ)}](F (ζ) • G(ζ))_{+}

+(DF (ζ))^{T}D_{x}ψ(F (ζ), G(ζ)) + (DG(ζ))^{T}D_{y}ψ(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 = x^{0}+ λe ∈ H and y = y^{0}+ 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)+ = α_{2}u^{(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 := (µx^{0}+ λy^{0}) + 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)+= α_{2}u^{(2)}, where
α_{2} = hx, yi + kµx^{0} + λy^{0}k,
u^{(2)} = 1

2

µx^{0}+ λy^{0}
kµx^{0}+ λy^{0}k+ 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)_{+}k^{2} = 1
2α^{2}_{2}

= 1

2(hx, yi^{2}+ 2hx, yikµx^{0}+ λy^{0}k + kµx^{0}+ λy^{0}k^{2})

≥ 1

2hx, yi^{2}.
Then, we complete the proof. 2

Lemma 4.2 Let ψ_{FB}, ψ_{1}, ψ_{2} be given as (2), (14), (16), respectively. Then, ψ_{FB}, ψ_{1},
and ψ_{2} satisfy the following inequality.

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

Proof. For ψ_{1}, it is clear by definition (14) with α = ^{1}_{2}. For ψ_{FB} and ψ_{2}, 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
ψ_{0}^{∗}(ω) ≥ βk(ω)+k^{2}, ∀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

δ^{2}

≤ pρ^{2}− δ^{2}(γ^{2}− σ^{2})

δ^{2} , ρ > δp

γ^{2}− σ^{2}, (26)

where σ = 1 −p1 − 2η + γ^{2}. 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^{∗}k^{2} ≤ max{1, hF (ζ), G(ζ)i} + k(−F (ζ))_{+}k + k(−G(ζ))_{+}k, ∀ζ ∈ H.

Moreover,

κkζ − ζ^{∗}k^{2} ≤ ψ_{0}^{−1}(f_{LT}(ζ)) +

√2

√αf_{LT}(ζ)^{1/2}, ∀ζ ∈ H,
where α is a positive constant.