5 Stationary point conditions

From the previous discussions, we learn that the complementarity problem CP(IK, T) in the Hilbert space H with the continuously F-differentiable mapping T :H → Hcan be transformed into an unconstrained smooth minimization problem

min

x∈H f(x) := Ψ_t(x, T(x)) with 0< t <2. (37) However, when applying minimization algorithms for (37), we can only expect to obtain stationary points of (37). Thus, it is natural to ask under what conditions each stationary point of the minimization problem (37) is a solution to the problem CP(IK, T). To achieve this goal, we first establish some favorable properties for the differential Ψ^′_t(x, y).

This needs the following key lemma which generalizes the result of [9, Prop. 3.4] to H.

Lemma 5.1 For any x, y ∈ H and z ≻IK 0, the following implications hold

z² ≻IK x²+y² =⇒ L²_z−L²_x−L²_y ≻0, (38)

z² ≻IK x² =⇒ z ≻IK x. (39)

Moreover, the implications (38) and (39) remain true when “≻” is replaced by “≽”.

Proof. Similar to [9], we first prove (38) for the case where z = (x² +y² +δe)^1/2 for some δ >0. Fix any x, y ∈ H and any δ > 0. Let z = (x²+y² +δe)^1/2. It suffices to prove that for any nonzero vectorh in H,

0<⟨

h,(L²_z−L²_x−L²_y)h⟩

= ⟨L_zh, L_zh⟩ − ⟨L_zx, L_zx⟩ − ⟨L_zy, L_zy⟩

= ∥z•h∥²− ∥x•h∥²− ∥y•h∥². Letx=x^′+λe, y=y^′+µe,z =z^′+νeand h=h^′ +ξe. We calculate that

∥z•h∥²− ∥x•h∥²− ∥y•h∥² = ξ²∥z∥²+ν²∥h^′∥² + 4ξν⟨z^′, h^′⟩+⟨z^′, h^′⟩²

−[

ξ²∥x∥²+λ²∥h^′∥²+ 4ξλ⟨x^′, h^′⟩+⟨x^′, h^′⟩²]

−[

ξ²∥y∥²+µ²∥h^′∥²+ 4ξµ⟨y^′, h^′⟩+⟨y^′, h^′⟩²]

= ξ²[

∥z∥²− ∥x∥²− ∥y∥²] +[

ν²−λ²−µ²]

∥h^′∥² +(

⟨z^′, h^′⟩²− ⟨x^′, h^′⟩² − ⟨y^′, h^′⟩²)

+4 (ξν⟨z^′, h^′⟩ −ξλ⟨x^′, h^′⟩ −ξµ⟨y^′, h^′⟩). (40) From the expression of z = (x²+y²+δe)^1/2, it is not hard to obtain that

z^′ =τ⁻¹(λx^′ +µy^′) and ν=τ, where

τ =

√

∥x∥²+∥y∥²+δe+√

(∥x∥²+∥y∥²+δe)²−4∥λx^′+µy^′∥²

2 .

Substituting the expression ofz^′ above into ⟨z^′, h^′⟩ and using ν =τ yields that

ξν⟨z^′, h^′⟩ −ξλ⟨x^′, h^′⟩ −ξµ⟨y^′, h^′⟩ = 0 (41) and

[ν²−λ²−µ²]∥h^′∥²+⟨z^′, h^′⟩²− ⟨x^′, h^′⟩²− ⟨y^′, h^′⟩²

= [ν²−λ²−µ²]∥h^′∥²+(λ²+µ²−τ²)(⟨x^′, h^′⟩²+⟨y^′, h^′⟩²)

τ² − ⟨h^′, µx^′−λy^′⟩² τ²

= (τ²−λ² −µ²)(τ²∥h^′∥²− ⟨x^′, h^′⟩²− ⟨y^′, h^′⟩²)

τ² − ⟨h^′, µx^′−λy^′⟩²

τ² . (42)

Now combining equations (41)–(42) with equation (40) yields that

∥z•h∥²− ∥x•h∥² − ∥y•h∥² =ξ²[

∥z∥²− ∥x∥²− ∥y∥²]

− ⟨h^′, µx^′ −λy^′⟩² τ²

+(τ²−λ² −µ²)(τ²∥h^′∥²− ⟨x^′, h^′⟩²− ⟨y^′, h^′⟩²)

τ² . (43)

Notice that ∥z∥² =∥x∥²+∥y∥²+δ >∥x∥²+∥y∥² is equivalent to

∥λx^′ +µy^′∥²

τ² +τ² >∥x^′∥²+λ²+∥y^′∥²+µ².

Multiplying the two sides of the last inequality withτ², and then addingµ²∥x^′∥²+λ²∥y^′∥² to the both sides of the inequality, we obtain

(λ²+µ²)(∥x^′∥²+∥y^′∥²) +τ⁴−τ²(∥x^′∥²+λ²+∥y^′∥²+µ²)>∥µx^′−λy^′∥² or

(τ²− ∥x^′∥²− ∥y^′∥²)(τ²−λ²−µ²)>∥µx^′−λy^′∥². (44) This means that both τ²− ∥x^′∥²− ∥y^′∥² and τ² −λ²−µ² are positive or negative. If both are negative, then we would have∥x^′∥²+∥y^′∥² ≥τ² and λ²+µ² ≥τ², which leads to a contradiction that ∥x∥²+∥y∥² ≥2τ² ≥ ∥x∥²+∥y∥²+δ. Consequently,

τ²− ∥x^′∥²− ∥y^′∥² >0 and τ²−λ²−µ² >0.

Together with (43) and (44) and ∥z∥² >∥x∥²+∥y∥², it follows that

∥z•h∥²− ∥x•h∥²− ∥y•h∥²

≥ (τ²−λ²−µ²)(τ² − ∥x^′∥²− ∥y^′∥²)∥h^′∥²

τ² − ∥h^′∥²∥λx^′−µy^′∥² τ² >0.

So, (38) holds for any x, y ∈ H and z of the form z = (x²+y²+δe)^1/2 for some δ > 0.

In view of Lemma 2.2, the rest arguments are same as those of [9, Prop. 3.4]. 2

Lemma 5.2 Let Ψ_t be given in (15) with 0< t <2. Then, for any x, y ∈ H, (a) ⟨D₁Ψ_t(x, y), D₂Ψ_t(x, y)⟩ ≥0 with equality holding if and only if Φ_t(x, y) = 0;

(b) ⟨D₁Ψ_t(x, y), x⟩+⟨D₂Ψ_t(x, y), y⟩= 2Ψ_t(x, y);

(c) D₁Ψ_t(x, y) = 0⇐⇒D₂Ψ_t(x, y) = 0⇐⇒Ψ_t(x, y) = 0.

Proof. (a) We proceed the arguments by three cases shown as below. and λ and µ can not be zero simultaneously by Lemma 3.2, it follows that

(λ+ (t−1)µ

Since x•y= 0 implies ⟨x, y⟩= 0, the last equivalence means that

⟨D₁Ψ_t(x, y), D₂Ψ_t(x, y)⟩= 0 =⇒ ⟨Φ_t(x, y), L⁻_z¹Φ_t(x, y)⟩= 0 =⇒Φ_t(x, y) = 0

where the last implication is due toz ∈intIK and Lemma 2.2. Conversely, if Φ_t(x, y) = 0,

⟨D1Ψt(x, y), D2Ψt(x, y)⟩= 0 follows directly from (45). This proves part (a).

(b) It suffices to prove the assertion for (x, y)∈ H ×H\ {(0,0)}. If x²+y² ∈∂IK\ {0}, then using the formula (25), we obtain

⟨D₁Ψ_t(x, y), x⟩+⟨D₂Ψ_t(x, y), y⟩

= 2

⟨

Φ_t(x, y),

(λ+ (t−1)µ

τ −1

) x+

((t−1)λ+µ

τ −1

) y

⟩

= 2⟨Φt(x, y), z−(x+y)⟩= 2Ψt(x, y).

When x²+y² ∈intIK, from the formula (26), it follows that

⟨D₁Ψ_t(x, y), x⟩+⟨D₂Ψ_t(x, y), y⟩

= 2⟨L⁻_z¹Φ_t(x, y),(x+ (t−1)y)•x+ (y+ (t−1)x)•y⟩ −2⟨Φ_t(x, y), x+y⟩

= 2⟨L⁻_z¹Φ_t(x, y), z²⟩ −2⟨Φ_t(x, y), x+y⟩

= 2⟨Φ_t(x, y), z−(x+y)⟩= 2Ψ_t(x, y).

(c) The result is direct by part (a) and the expression of Ψ^′_tgiven by Theorem 4.1. 2 Now we are in a position to establish the main result of this section by Lemma 5.2.

Theorem 5.1 Let T : H → H be a given continuously F-differentiable mapping and f(x) = Ψ_t(x, T(x)) with 0 < t <2. If T is monotone, then for every x∈ H, either (i) f(x) = 0 or (ii) f^′(x)̸= 0 and ⟨d(x), f^′(x)⟩<0 with d(x) =−D₂Ψ_t(x, T(x)).

Proof. Fix any x∈ H. From Theorem 4.2 and the continuous F-differentiability of T, it follows that f : H →IR₊ is continuously F-differentiable on H. By the chain rule of differential, we have for any v ∈ H,

f^′(x)v = Ψ^′_t(x, T(x))(v,0) + Ψ^′_t(x, T(x)) (0, T^′(x)v)

= ⟨D₁Ψ_t(x, T(x)), v⟩+⟨D₂Ψ_t(x, T(x)), T^′(x)v⟩, which means that

f^′(x) = D₁Ψ_t(x, T(x)) + (T^′(x))^TD₂Ψ_t(x, T(x)).

Suppose that f^′(x) = 0. Then the last equation implies that

⟨D₁Ψ_t(x, T(x)), D₂Ψ_t(x, T(x))⟩=−⟨D₂Ψ_t(x, T(x)), T^′(x)D₂Ψ_t(x, T(x))⟩.

SinceT is continuously F-differentiable and monotone, the right hand side is nonpositive, and consequently, ⟨D₁Ψ_t(x, T(x)), D₂Ψ_t(x, T(x))⟩ ≤ 0. Together with Lemma 5.2, it then follows that f(x) = Ψ_t(x, T(x)) = 0.

Suppose that f^′(x)̸= 0. Then, from the expression of d(x), it follows that

⟨d(x), f^′(x)⟩ = −⟨D₂Ψ_t(x, T(x)), D₁Ψ_t(x, T(x)) + (T^′(x))^TD₂Ψ_t(x, T(x))⟩

= − ⟨D₂Ψ_t(x, T(x)), D₁Ψ_t(x, T(x))⟩

−⟨

D₂Ψ_t(x, T(x)),(T^′(x))^TD₂Ψ_t(x, T(x))⟩

≤ − ⟨D₂Ψ_t(x, T(x)), D₁Ψ_t(x, T(x))⟩.

where the first inequality is using the monotonicity of T. By Lemma 5.2 (a), the right hand side is nonpositive and equals zero if and only if Ψ_t(x, T(x)) = 0, i.e.,xis a solution of the minimization problem (37). However, the latter can not be true since f^′(x)̸= 0, and consequently, ⟨d(x), f^′(x)⟩<0. The proof is completed. 2

Theorem 5.1 states that if x∈ H is not a solution of CP(IK, T), then we can always find a descent direction d(x) at this point. Based on this, an iterative descent algorithm can be designed for the self-dual conic complementarity problem CP(IK, T).

6 Conclusions

We have developed a merit function method for the infinitely dimensional SOCCP CP(IK, T) by extending Kanzow and Kleinmichel NCP-function to the Hilbert space.

We believe that the merit functions given in this paper will be useful in other contexts, and further research work will be given to the specific applications of the merit function method. Using the techniques in this paper, other well-known merit functions, for ex-ample, the Yamashita-Fukushima merit function [15] can be also extended analogously.

Specifically, we can define the Yamashita-Fukushima merit function in Hilbert space as ψ_YF(x, y) :=ψ₀(⟨x, y⟩) +ψ_FB(x, y) ∀x, y ∈ H × H,

where ψ0 : IR→IR+ is any smoothing function satisfying

ψ₀(t) = 0 ∀t≤0 and ψ^′₀(t)>0 ∀t >0.

In addition, although the spectral factorization of vectors is not used in the analysis of this paper, we want to point out that, by the Jordan product associated with IK, every x=x^′+λe ∈ H can be written as x=λ₁(x)u⁽¹⁾x +λ₂(x)u⁽²⁾x with

λ_j(x) =λ+ (−1)^j∥x^′∥ and u^(j)_x = 1 2

(e+ (−1)^jx¯^′)

, j= 1,2

where ¯x^′ = _∥x^x′_′∥ if x^′ ̸= 0, and otherwise ¯x^′ is an arbitrary unit vector in⟨e⟩^⊥.

Acknowledgments The authors would like to thank the referees for their careful reading of the paper and helpful suggestions.

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