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Analysis of nonsmooth vector-valued functions associated with infinite-dimensional second-order cones

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to appear in Nonlinear Analysis, Theory, Methods and Applications, 2011

Analysis of nonsmooth vector-valued functions associated with infinite-dimensional second-order cones

Ching-Yu Yang 1 Department of Mathematics National Taiwan Normal University

Taipei 11677, Taiwan

Yu-Lin Chang2 Department of Mathematics National Taiwan Normal University

Taipei 11677, Taiwan

Jein-Shan Chen 3 Department of Mathematics National Taiwan Normal University

Taipei, Taiwan 11677

July 10, 2010

(1st revised on March 1, 2011) (2nd revised on May 12, 2011)

Abstract Given a Hilbert space H, the infinite-dimensional Lorentz/second-order cone K is introduced. For any x ∈ H, a spectral decomposition is introduced, and for any func- tion f : IR → IR, we define a corresponding vector-valued function fH(x) on Hilbert space H by applying f to the spectral values of the spectral decomposition of x ∈ H with respect to K. We show that this vector-valued function inherits from f the properties of con- tinuity, Lipschitz continuity, differentiability, smoothness, as well as s-semismoothness.

These results can be helpful for designing and analyzing solutions methods for solving infinite-dimensional second-order cone programs and complementarity problems.

1E-mail: yangcy@math.ntnu.edu.tw

2E-mail: ylchang@math.ntnu.edu.tw

3Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office.

The author’s work is partially supported by National Science Council of Taiwan. E-mail:

jschen@math.ntnu.edu.tw.

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Key words. Hilbert space, infinite-dimensional second-order cone, strong semismooth- ness.

1 Introduction

Let H be a real Hilbert space endowed with an inner product h·, ·i, and we write the norm induced by h·, ·i as k · k. For any given closed convex cone K ⊆ H,

K := {x ∈ H | hx, yi ≥ 0, ∀y ∈ K}

is the dual cone of K. A closed convex cone K in H is called self-dual if K coincides with its dual cone K; for example, the non-negative orthant cone IRn+ and the second-order cone (also called Lorentz cone) IKn := {(r, x0) ∈ IR × IRn−1 | r ≥ kx0k}. As discussed in [14], this Lorentz cone IKn can be rewritten as

IKn:=

(

x ∈ IRn

hx, ei ≥ 1

√2kxk

)

with e = (1, 0) ∈ IR × IRn−1.

This motivates us to consider the following closed convex cone in the Hilbert space H:

K(e, r) := {x ∈ H | hx, ei ≥ rkxk}

where e ∈ H with kek = 1 and r ∈ IR with 0 < r < 1. It can be seen that K(e, r) is pointed, i.e., 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 − r2 kx0k

)

. When H = IRn and e = (1, 0) ∈ IR × IRn−1, K(e,1

2) coincides with IKn. In view of 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

IK = K e, 1

√2

!

since two infinite-dimensional second-order cones IK(e1) and IK(e2) associated with dif- ferent unit elements e1 and e2 in H are isometric. This means there exists a bijective isometry P which maps IK(e1) onto IK(e2) such that kP xk = kxk for any x ∈ IK(e1). For

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example, let e1 = (1, 0, 0) and e2 = (0, 0, 1). Then, for any x ∈ IK(e1) and y ∈ IK(e2), we have the following relation:

y = P x =

0 0 1 0 1 0 1 0 0

x.

Moreover, this mapping preserves the Jordan algebra structure, i.e., P (x ◦ y) = P x ◦ P y.

In the infinite-dimensional Hilbert space, P is indeed a unitary operator. In light of this fact, we can consider the infinite-dimensional second-order cone associated with a fixed arbitrary unit element in H.

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 IK y (respectively, x IK y) if x − y ∈ intIK (respectively, x − y ∈ IK). 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), (1) where α1(x), α2(x) and vx(1), vx(2)are the spectral values and the associated spectral vectors of x, with respect to K, given by

αi(x) = λ + (−1)ikx0k , (2)

vx(i) =

1

2 e + (−1)i x0 kx0k

!

, x0 6= 0 1

2

e + (−1)iw, x0 = 0

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for i = 1, 2 with w being any vector in H satisfying kwk = 1. With this spectral decom- position, for any function f : R → R, the following vector-valued function associated with K is defined:

fH(x) = f (α1(x))vx(1)+ f (α2(x))v(2)x ∀x ∈ H. (4) The above definition is analogous to the one in finite-dimensional second-order cone case [22, 6].

The motivation of studying fH defined as in (4) is from concerning with the comple- mentarity problem associated with infinite-dimensional second-order cone IK, i.e., to find an x ∈ H such that

x ∈ IK, T (x) ∈ IK and hx, T (x)i = 0, (5) where T is a mapping from H to H. We denote this problem (5) as CP(IK, T ). More specifically, when dealing with such complementarity problem by nonsmooth function ap- proach, i.e., recasting it as a nonsmooth system of equations, we need to check what kind

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of properties of f can be inherited by fHso that we can know to what extent the conver- gence analysis of solutions methods based on such nonsmooth system can be obtained.

Indeed, the format of the aforementioned complementarity problem CP(IK, T ) indeed follows the direction of complementarity problems associated with symmetric cones in Euclidean Jordan algebra. Recently, nonlinear symmetric cone optimization and comple- mentarity problems in finite-dimensional spaces such as semidefinite cone optimization and complementarity problems, second-order cone optimization and complementarity problems, and general symmetric cone optimization and complementarity problems, be- come an active research field of mathematical programming. Taking second-order cone optimization and complementarity problems for example, there have proposed many ef- fective solution methods, including the interior point methods [1, 2, 3, 4], the smoothing Newton methods [5, 6, 7], the semismooth Newton methods [8, 9], and the merit function method [10, 11]. However, there are very limited works about nonlinear symmetric cone optimization and complementarity problems in infinite-dimensional spaces, for instance [12], in which with the JB algebras of finite rank primal-dual interior-point methods are presented for some special type of infinite-dimensional cone optimization problems.

It is our intention to extend the above methods for infinite-dimensional complemen- tarity problem CP(IK, T ), in which the vector-valued function fH will play a key role.

In this paper, we study the continuity and differential properties of the vector-valued function fH in general. In particular, we show that the properties of continuity, strict continuity (locally Lipschitz continuity), Lipschitz continuity, directional differentiabil- ity, differentiability, continuous differentiability, and s-semismoothness are each inherited by fH from f . These results can give some concept in designing solutions methods for solving infinite-dimensional second-order cone programs and infinite-dimensional second- order cone complementarity problems.

2 Preliminaries

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, (6) and write x2 = x ◦ x. Clearly, when H = Rn and e = (1, 0) ∈ R × Rn−1, this definition coincides with the one given in [13, Chatper II] which is the case of finite-dimensional second-order cone associated with Euclidean Jordan algebra. The following technical lemmas will be frequently used in the subsequent analysis.

Lemma 2.1 Let α1(x), α2(x) be the spectral values of x ∈ H and α1(y), α2(y) be the spectral values of y ∈ H. Then we have

1(x) − α1(y)|2+ |α2(x) − α2(y)|2 ≤ 2kx − yk2, (7)

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and hence, |αi(x) − αi(y)| ≤√

2kx − yk, ∀i = 1, 2.

Proof. The proof can be obtained by direct computation like in [22, Lemma 2]. 2

Lemma 2.2 Let x = x0+ λe ∈ H and y = y0+ µe ∈ H.

(a) If x0 6= 0 and y0 6= 0, then we have

v(i)x − v(i)y ≤ 1

kx0kkx − yk ∀i = 1, 2, (8)

where vx(i), vy(i) are the spectral vectors of x and y, respectively.

(b) If either x0 = 0 or y0 = 0, then we can choose vx(i), v(i)y such that the left hand side of inequality (8) is zero.

Proof. The proof is similar to [23, Lemma 3.2], so we omit it here. 2

Lemma 2.3 For any x 6= 0 ∈ H, the following hold.

(a) If g(x) = kxk, we have g0(x)h = hx, hi kxk . (b) If g(x) = x

kxk, we have g0(x)h = h

kxk − hx, hi kxk3 x.

Proof. (a) See Example 3.1 (V) of [14].

(b) First, we compute that g(x + h) − g(x)

= x + h

kx + hk− x kxk

= h

kx + hk− 1

kxk− 1

kx + hk

!

· x

= h

kx + hk−

qhx + h, x + hi −qhx, xi

qhx, xi ·qhx + h, x + hi · x

= h

kx + hk− 2hx, hi + hh, hi

qhx, xi ·qhx + h, x + hiqhx + h, x + hi +qhx, xi · x

= h

kxk− hx, hi

kxk3 x + o(khk).

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From the above, it is clear that g0(x)h = h

kxk− hx, hi

kxk3 x. 2

Semismooth function, as introduced by Mifflin [15] for functionals and further ex- tended by Qi and Sun [16] for vector-valued functions, is of particular interest due to the central role it plays in the superlinear convergence analysis of certain generalized Newton methods, see [16, 17] and references therein. Given a mapping F : IRn → IRm, it is well-known that if F is strictly continuous (locally Lipschitz continuous), then F is almost everywhere differentiable by Rademacher’s Theorem–see [18] and [19, Sec. 9J]. In this case, the generalized Jacobian ∂F (x) of F at x (in the Clarke sense) can be defined as the convex hull of the B-subdifferential ∂BF (x), where

BF (x) :=



lim

xj→x∇F (xj)|F is differentiable at xj ∈ IRn



.

The notation ∂B is adopted from [17]. In [19, Chap. 9], the case of m = 1 is considered and the notations “ ¯∇” and “ ¯∂” are used instead of, respectively, “∂B” and “∂”. Assume F : IRn → IRm is strictly continuous, then F is said to be semismooth at x if F is directionally differentiable at x and, for any V ∈ ∂F (x + h) and h → 0, we have

F (x + h) − F (x) − V h = o(khk). (9)

Moreover, F is called ρ-order semismooth at x (0 < ρ < ∞) if F is semismooth at x and, for any V ∈ ∂F (x + h) and h → 0, we have

F (x + h) − F (x) − V h = O(khk1+ρ).

The Rademacher theorem does not hold in function spaces, see [20]. Hence, the aforementioned definitions of generalized Jacobian and semismoothness cannot be used in infinite-dimensional spaces. To overcome this difficulty, in the paper [20], so-called slanting functions and slant differentiability of operators in Banach spaces are proposed and used to formulate a concept of semismoothness in infinite-dimensional spaces. We shall introduce them as below. Let X, Y ⊂ H. A function F : X → Y is said to be directionally differentiable at x if the limit

δ+F (x; h) := lim

t→0+

F (x + th) − F (x)

t (10)

exists, where δ+F (x; h) is called the directional derivative of F at x with respect to the direction h. A function F : X → Y is said to be B-differentiable at x if it is directionally differentiable at x and

h→0lim

F (x + h) − F (x) − δ+F (x; h)

khk = 0 (11)

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in which we call δ+F (x; ·) the B-derivative of F at x. In finite dimensional Euclidean spaces, Shapiro [21] shows that a locally Lipschitz continuous function F is B-differentiable at x if and only if it is directionally differentiable at x. From (9) and (11) (also see [16]), it can be seen that F is semismooth at x if and only if F is B-differentiable (hence directionally differentiable) at x and, for each V ∈ ∂F (x + h), there has

δ+F (x; h) − V h = o(khk).

As mentioned earlier, these results do not hold in infinite-dimensional spaces. Therefore, the slant differentiability is introduced to circumvent this hurdle. In what follows, we state its definition.

Definition 2.1 Let D be an open domain in X and L(X, Y ) denote the set of all bounded linear operators from X onto Y .

(a) A function F : D ⊂ X → Y is said to be slantly differentiable at x ∈ D if there exists a mapping f : D → L(X, Y ) such that the family {f(x + h)} of bounded linear operators is uniformly bounded in the operator norm for h sufficiently small and

h→0lim

F (x + h) − F (x) − f(x + h)h

khk = 0. (12)

The function f is called a slanting function for F at x.

(b) A function F : D ⊂ X → Y is said to be slantly differentiable in an open domain D0 ⊂ D if there exists a mapping f : D → L(X, Y ) such that f is a slanting function for F at every x ∈ D0. In this case, f is called a slanting function for F in D0.

Definition 2.2 Suppose that f : D → L(X, Y ) is a slanting function for F at x ∈ D We denote the set

SF (x) :=



xlimk→xf(xk)



(13) and call it the slant derivative of F associated with f at x ∈ D. Note that f(x) ∈ ∂SF (x) which says ∂SF (x) is always nonempty.

A function F may be slantly differentiable at all points of D, but there is no common slanting function of F at all points of D. Moreover, a slantly differentiable function F at x can have infinitely many slanting functions at x. A slanting function f for F at x is a single-valued function, but not continuous in general. In addition, a continuous function is not necessarily slantly differentiable. For more details about slanting functions and slantly differentiability, please refer to [20].

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Definition 2.3 A mapping F : X → Y is said to be s-semismooth at x if there is a slanting function f for F in a neighborhood Nx of x such that f and the associated slant derivative satisfy the following two conditions.

(a) lim

t→0+f(x + th)h exists for every h ∈ X and

khk→0lim lim

t→0+f(x + th)h − f(x + h)h

khk = 0.

(b) f(x + h)h − V h = o(khk) for all V ∈ ∂SF (x + h).

We point it out that the function F defined in Definition 2.3 was called semismooth in [20]. However, we here rename it as “s-semismooth” because when X, Y are both finite-dimensional spaces it does not reduce to the original definition introduced by Qi and Sun [16] in finite-dimensional spaces. The main key causing this is the limits in

SF (x) and ∂BF (x) are approached by different ways. In order to distinguish such difference, we hence use the term “s-semismooth” to convey concept of semismoothness in infinite-dimensional spaces.

3 Continuous properties of f

H

In this section, we show properties of continuity and (local) Lipschitz continuity of fH. The arguments are straightforward by checking their definitions.

Proposition 3.1 Suppose x = x0+ λe ∈ H with spectral values α1(x), α2(x) and spectral vectors vx(1), vx(2). Let fH be defined as in (4). Then, fH is continuous at x ∈ H if and only if f is continuous at α1(x), α2(x).

Proof. (⇒) This part of proof is similar to the argument of [22, Proposition 2(a)].

(⇐) This direction of proof is also similar to [23, Proposition 2.2(a)], we omit it. 2

Proposition 3.2 Suppose x = x0+ λe ∈ H with spectral values α1(x), α2(x) and spectral vectors vx(1), vx(2). Let fH be defined as in (4). Then, the following hold.

(a) fH is strictly continuous at x ∈ H if and only if f is strictly continuous at α1(x), α2(x).

(b) fH is Lipschitz continuous (with respect to k · k) if and only if f is Lipschitz contin- uous.

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Proof. (a) (⇐) Suppose f is strictly continuous at α1(x), α2(x). Then, there exist κi > 0 and δi > 0 for i = 1, 2, such that

|f (ξ) − f (ζ)| ≤ κi|ξ − ζ| ∀ξ, ζ ∈ [αi(x) − δi, αi(x) + δi] i = 1, 2.

Let δ = 1

2min{δ1, δ2} and for any y, z ∈ B(x, δ), we have fH(y) − fH(z)

= f (α1(y))vy(1)+ f (α2(y))v(2)y f (α1(z))v(1)z + f (α2(z))vz(2) (14)

= f (α1(y))vy(1)− vz(1)+



f (α1(y)) − f (α1(z))



vz(1) +f (α2(y))vy(2)− vz(2)+



f (α2(y)) − f (α2(z))



vz(2)

where y = α1(y)v(1)y + α2(y)vy(2) and z = α1(z)vz(1)+ α2(z)v(2)z . By Lemma 2.1, Lemma 2.2 and the similar argument in [22, Proposition 6(a)], the proof can be obtained.

(⇒) This part of proof is quite simple and similar to [22, Proposition 6(a)], we omit it here.

(b) The argument of proof is similar to [22, Proposition 6(c)]. 2

4 Differential properties of f

H

In this section, we show properties of directional differentiability, differentiability, con- tinuous differentiability and B-differentiability of fH. For simplicity, in the arguments we sometimes abbreviate αi(x) as αi when there is no ambiguity in the context. Note that, unlike in finite-dimensional second-order cone case [22], Prop. 4.1 and Prop. 4.2 are proved by different approaches since the chain rule for directional differentiability in infinite-dimensional space does not hold in general, see [21].

Proposition 4.1 Suppose x = x0+ λe ∈ H with spectral values α1(x), α2(x) and spectral vectors vx(1), vx(2). Let fH be defined as in (4). Then, fH is directionally differentiable at x ∈ H if and only if f is directionally differentiable at α1(x), α2(x).

Proof. (⇐) Suppose f is directionally differentiable at α1(x), α2(x). Fix x = x0+λe ∈ H and any direction h = h0+ le ∈ H, we discuss two cases as below.

Case (i): If x0 6= 0, then we have fH(x) = f (α1(x))vx(1) + f (α2(x))vx(2) where αi(x) = λ + (−1)ikx0k and vx(i)= 12e + (−1)i xkx00k

for i = 1, 2. Now x + th = (x0+ th0) + (λ + tl)e with spectral values αi(x + th) = λ + tl + (−1)ikx0+ th0k and spectral vectors v(i)x+th =

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1 2

e + (−1)i xkx00+th+th00k

 for i = 1, 2. We consider equation (14) again in which replacing y with x + th, then we have

fH(x + th) − fH(x)

= f (α1(x + th))vx+th(1) − vx(1)+



f (α1(x + th)) − f (α1(x))



vx(1) (15) +f (α2(x + th))v(2)x+th− v(2)x +



f (α2(x + th)) − f (α2(x))



v(2)x .

Because the process of checking argument is similar to [22, Proposition 3], we only present the result here.

By denoting

˜

a = f (α2(x)) − f (α1(x)) α2(x) − α1(x) ,

˜b = δ+f (α2(x); k2) + δ+f (α1(x); k1)

2 , (16)

˜

c = δ+f (α2(x); k2) − δ+f (α1(x); k1)

2 ,

where ki = hh, ei + (−1)i hxkx0,hi0k for i = 1, 2, we can write the expression of δ+fH(x; h) as

δ+fH(x; h) = ˜a h − hh, eie − hx0, hi kx0k2 x0

!

+ ˜be + ˜c x0

kx0k. (17)

Case (ii): If x0 = 0, we compute the directional derivative δ+fH(x; h) at x ∈ H for any direction h by definition. Let h = h0 + le ∈ H with h0 ∈ hei and l ∈ R. We discuss two subcases.

Subcase (a). If h0 6= 0, from the spectral decomposition, we choose v(i)x = 12e + (−1)i hkh00k



for i = 1, 2 such that

fH(x + th) = f (λ + th1)v(1)x + f (λ + th2)vx(2) fH(x) = f (λ)v(1)x + f (λ)v(2)x

where hi = l + (−1)ikh0k for i = 1, 2. Now, we compute

t→0lim+

fH(x + th) − fH(x) t

= lim

t→0+

f (λ + th1) − f (λ)

t v(1)x + lim

t→0+

f (λ + th2) − f (λ)

t vx(2) (18)

= δ+f (λ; l − kh0k)v(1)x + δ+f (λ; l + kh0k)vx(2). This shows that δ+fH(x; h) exists under this subcase.

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Subcase (b). If h0 = 0, we choose v(i)x = 12(e + (−1)iw) for any w ∈ H with kwk = 1.

Analogous to (18), we have

t→0lim+

fH(x + th) − fH(x) t

= lim

t→0+

f (λ + tl) − f (λ)

t v(1)x + lim

t→0+

f (λ + tl) − f (λ)

t vx(2) (19)

= δ+f (λ; l)vx(1)+ δ+f (λ; l)v(2)x . Hence, δ+fH(x; h) exists under this subcase.

From all the above, we have proved that fH is directionally differentiable at x ∈ H when x0 = 0 and its directional derivative δ+fH(x; h) is either in form of (18) or (19).

(⇒) Suppose fHis directionally differentiable at x ∈ H, we will prove that f is direction- ally differentiable at α1, α2. For α1 ∈ R and any direction d1 ∈ R, let h = d1v(1)x + 0vx(2) where x = α1vx(1)+ α2vx(2). Then, x + th = (α1+ td1)vx(1)+ α2vx(2) and

fH(x + th) − fH(x)

t = f (α1+ td1) − f (α1) t vx(1).

Since fH is directionally differentiable at x, the above equation implies that δ+f (α1; d1) = lim

t→0+

f (α1+ td1) − f (α1)

t exists.

This means f is directionally differentiable at α1. Similarly, it can be verified that f is also directionally differentiable at α2. 2

Proposition 4.2 Suppose x = x0+ λe ∈ H with spectral values α1(x), α2(x) and spectral vectors vx(1), v(2)x . Let fH be defined as in (4). Then, fH is differentiable at x ∈ H if and only if f is differentiable at α1(x), α2(x).

Proof. (⇐) Suppose f is differentiable at α1, α2. Fix x = x0+λe ∈ H and h = h0+le ∈ H, we discuss two cases as below.

Case (i): If x0 6= 0, then we have fH(x) = f (α1)v(1)x + f (α2)vx(2) where αi = λ + (−1)ikx0k and v(i)x = 12e + (−1)i xkx00k

 for i = 1, 2. By using Lemma 2.3 and the chain rule and product rule for differentiation, the argument is similar to [22, Proposition 4] so we omit the process and present the result as following. Denoting

a = f (α2) − f (α1) α2− α1

, b = f02) + f01)

2 , c = f02) − f01)

2 . (20)

We can write the expression of (fH)0(x)h as (fH)0(x)h = ah + (b − a) hh, eie +hx0, hi

kx0k2 x0

!

+ c

kx0k(hx0, hie + hh, eix0). (21)

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Case (ii): The proof is identical to that of Case(ii) in Proposition 4.1, but with th replaced by h. We omit it and only present the formula of (fH)0(x)h as below. If x0 = 0, then

(fH)0(x)h = f0(λ)h. (22)

(⇒) This part of proof is similar to [23, Proposition 2.2(c)]. 2

Proposition 4.3 Suppose x = x0+ λe ∈ H with spectral values α1(x), α2(x) and spectral vectors v(1)x , v(2)x . Let fH be defined as in (4). Then, fH is continuously differentiable (smooth) at x ∈ H if and only if f is continuously differentiable at α1(x), α2(x).

Proof. (⇐) This part of proof is similar to [23, Proposition 2.2(d)], so we omit it.

(⇒) This direction of proof is some variant of argument in [22, Proposition 5], we also skip it here. 2

Proposition 4.4 Suppose x = x0+ λe ∈ H with spectral values α1(x), α2(x) and spectral vectors v(1)x , vx(2). Let fH be defined as in (4). Then, fH is B-differentiable at x ∈ H if and only if f is B-differentiable at α1(x), α2(x).

Proof. (⇐) If f is B-differentiable at α1(x), α2(x), f is directionally differentiable at α1(x), α2(x). By Proposition 4.1, fH is directionally differentiable at x. It remains to verify that

h→0lim

fH(x + h) − fH(x) − δ+fH(x; h)

khk = 0.

We write x = x0+ λe and h = h0+ le ∈ H. Again, two cases will be discussed.

Case (i): If x0 6= 0, considering equation (15) in which we replace x + th with x + h, it yields

fH(x + h) − fH(x)

= f (α1(x + h))vx+h(1) − vx(1)+ (f (α1(x + h)) − f (α1(x))) v(1)x (23) +f (α2(x + h))v(2)x+h− vx(2)+ (f (α2(x + h)) − f (α2(x))) vx(2).

Indeed, sum of the first and third can be simplified as

f (α1(x + h))v(1)x+h− vx(1)+ f (α2(x + h))v(2)x+h− vx(2)

= (f (α2(x + h)) − f (α1(x + h))) ·1

2 · x0+ h0

kx0+ h0k− x0 kx0k

!

= f (α2(x + h)) − f (α1(x + h))

2kx0k h0−hx0, h0i

kx0k2 x0+ o(kh0k)

!

(24)

= f (α2(x + h)) − f (α1(x + h))

α2(x) − α1(x) h − hh, eie − hx0, hi

kx0k2 x0+ o(kh0k)

!

,

(13)

where the second equality is due to Lemma 2.3(b) and the last equality uses the fact that α2(x) − α1(x) = 2kx0k. From (17), we know that

δ+fH(x; h)

= f (α2(x)) − f (α1(x))

α2(x) − α1(x) h − hh, eie − hx0, hi kx0k2 x0

!

(25) +δ+f (α1(x); k1)vx(1)+ δ+f (α2(x); k2)vx(2)

where ki = hh, ei + (−1)i hxkx0,hi0k for i = 1, 2. Since lim

h→0

h − hh, eie − hxkx00,hik2x0= 0, following almost the same arguments as in Proposition 4.1 gives

αi(x + h) − αi(x) = l + (−1)i(kx0+ h0k − kx0k)

= hh, ei + (−1)i hx0, h0i

kx0k + o(kh0k)

!

= ki+ (−1)io(kh0k) ∀i = 1, 2.

Let Ti := ki + (−1)io(kh0k) = αi(x + h) − αi(x) for i = 1, 2, we obtain

h→0lim

f (αi(x + h)) − f (αi(x)) khk

= lim

h→0

f (αi(x) + Ti· 1) − f (αi(x))

Ti · ki+ (−1)io(kh0k) khk

= δ+f (αi(x); 1) ·kei

= δ+f (αi(x);kei),

where the last equality uses the positive homogeneity property of δ+f (αi(x); ·) again and kei := lim

h→0 ki

khk. We notice that 0 < kkeik ≤ 2 and kei can be viewed as a directional vector here. By the above discussion, we have

h→0lim 1 khk

f (α1(x + h))v(1)x+h− vx(1)+ f (α2(x + h))v(2)x+h− vx(2)

− f (α2(x)) − f (α1(x))

α2(x) − α1(x) h − hh, eie − hx0, hi kx0k2 x0

!!

= lim

h→0

(f (α2(x + h)) − f (α2(x))) − (f (α1(x + h)) − f (α1(x)))

khk · (α2(x) − α1(x)) · h − hh, eie − hx0, hi kx0k2 x0

!

+ lim

h→0

f (α2(x + h)) − f (α1(x + h))

α2(x) − α1(x) · o(kh0k) khk

= δ+f (α2(x);fk2) − δ+f (α1(x);fk1)

α2(x) − α1(x) · 0 + 0 (26)

= 0.

(14)

By assumption, f is B-differentiable at α1(x), α2(x) and employ almost the same argu- ments, we compute

h→0lim

f (αi(x + h)) − f (αi(x)) − δ+f (αi(x); ki)

khk · vx(i)

= lim

h→0

f (αi(x) + Ti) − f (αi(x)) − δ+f (αi(x); Ti)

kTik · kki+ (−1)io(kh0k)k khk

+ δ+f (αi(x); Ti) − δ+f (αi(x); ki) khk

!

· v(i)x (27)

= 0 · lim

h→0kkeik + lim

h→0δ+f (αi(x); (−1)io(kh0k) khk )

!

· vx(i) = 0 ∀i = 1, 2.

Now from equations (23), (25), (26) and (27), we see that

h→0lim

fH(x + h) − fH(x) − δ+fH(x; h)

khk = 0

which says that fH is B-differentiable at x.

Case (ii): If x0 = 0, we need to further consider the following two subcases:

Subcase (a): If h0 6= 0, we choose v(i)x = 12e + (−1)i hkh00k

 for i = 1, 2 such that vx+h(i) = vx(i). Then,

fH(x + h) − fH(x)

= (f (α1(x + h)) − f (α1(x))) vx(1)+ (f (α2(x + h)) − f (α2(x))) v(2)x , and from Case (ii)(a) of Proposition 4.1, we have

δ+fH(x; h) = δ+f (λ; l − kh0k)vx(1)+ δ+f (λ; l + kh0k)v(2)x ,

where λ = α1(x) = α2(x). Again by the B-differentiability of f at α1(x) and α2(x), we have

h→0lim

fH(x + h) − fH(x) − δ+fH(x; h) khk

= lim

h→0

f (α1(x + h)) − f (α1(x)) − δ+f (α1(x); l − kh0k)

khk · v(1)x

+ lim

h→0

f (α2(x + h)) − f (α2(x)) − δ+f (α2(x); l + kh0k)

khk · vx(2)

= 0,

which implies the B-differentiability of fH at x.

Subcase (b): If h0 = 0, we choose vx(i) = 12(e + (−1)iw) with any w ∈ H with kwk = 1. With almost the same arguments as in Case (ii)-(b) of Proposition 4.1, the B- differentiability of fH can be verified, we omit the detail here.

(15)

(⇒) If fHis B-differentiable at x, then fHis directionally differentiable at x by definition.

Then, f is also directionally differentiable at αi(x), α2(x) by Proposition 4.1. In order to prove the B-differentiability of f at αi(x), α2(x), all we have to do is proving the following condition:

limt→0

f (αi(x) + t) − f (αi(x)) − δ+f (αi(x); t)

|t| = 0 ∀i = 1, 2.

Since fH is B-differentiable at x, the following condition is true:

h→0lim

fH(x + h) − fH(x) − δ+fH(x; h)

khk = 0.

Again, we write x = x0 + λe and h = h0 + le ∈ H and discuss two cases.

Case (i): If x0 6= 0, from the proof of first part, we know

h→0lim

fH(x + h) − fH(x) − δ+fH(x; h) khk

= lim

h→0

"

(f (α2(x + h)) − f (α2(x))) − (f (α1(x + h)) − f (α1(x)))

khk · (α2(x) − α1(x)) · h − hh, eie − hx0, hi kx0k2 x0

!

+f (α2(x + h)) − f (α1(x + h))

α2(x) − α1(x) · o(kh0k) khk +

2

X

i=1

f (αi(x + h)) − f (αi(x)) − δ+f (αi(x); ki)

khk · v(i)x

#

where ki = hh, ei + (−1)i hxkx0,hi0k for i = 1, 2.

Because lim

h→0

h − hh, eie − hxkx00,hik2x0= 0 and vx(1) ⊥ vx(2), we must have

h→0lim

f (αi(x + h)) − f (αi(x)) − δ+f (αi(x); ki)

khk = 0 ∀i = 1, 2. (28)

Note that h ∈ H is arbitrary, we can choose h = te where t ∈ R is also arbitrary. Then, we have

ki = αi(x + h) − αi(x) = t ∀i = 1, 2.

This together with the fact that t → 0 as h → 0 gives limt→0

f (αi(x) + t) − f (αi(x)) − δ+f (αi(x); t)

|t| = 0 ∀i = 1, 2,

which means that f is B-differentiable at αi(x) for i = 1, 2.

Case (ii): If x0 = 0, we consider the two subcases of h0 = 0 or h0 6= 0. The proof is routine check as earlier verifications, so we omit it. 2

(16)

5 S-semismooth properties of f

H

In this section, we show s-semismooth properties of fH. To this end, we first present some equivalent criteria for s-semismooth functions in infinite-dimensional spaces. In fact, we immediate obtain the following criteria from the very basic definition and combining some known results in [20].

Proposition 5.1 Suppose that F : X → Y is slantly differentiable on a neighborhood Nx of x. Let f be a slanting function for F in Nx and ∂SF be the slant derivative associated with f in Nx. Then, F is s-semismooth at x if and only if one of the following holds:

(a) lim

t→0+f(x + th)h exists for every h ∈ X,

khk→0lim lim

t→0+f(x + th)h − f(x + h)h

khk = 0, (29)

and

f(x + h)h − V h = o(khk) ∀ V ∈ ∂SF (x + h). (30) (b) F is B-differentiable at x, and

δ+F (x; h) − V h = o(khk) ∀ V ∈ ∂SF (x + h). (31) (c) F is B-differentiable at x, and

F (x + h) − F (x) − V h = o(khk) ∀ V ∈ ∂SF (x + h). (32) Proof. (a) This is clear from the original definition of s-semismooth function given as in Definition 2.3.

(b) This is result of [20, Theorem 3.3].

(c) Using part(a) and [20, Theorem 2.9] yield F being B-differentiable at x, and

δ+F (x; h) − f(x + h)h = o(khk). (33) Then, by definition of F being B-differentiable, condition (31) holds. 2

The conditions in Proposition 5.1 are indeed hard to be verified since it is difficult to write out the set ∂SF (x + h). Hence, we further establish some equivalent conditions which are useful in subsequent analysis regarding s-semismooth property which is the main contribution of this paper. We also want to point out the following observation.

Suppose that F : X → Y is slantly differentiable on a neighborhood Nx of x. Let f be a slanting function for F with uniform bound kfk ≤ L in Nx. It is easy to derive that kF (y) − F (z)k ≤ 2Lky − zk for any y, z ∈ Nx. However, we have no idea whether it is true or not for the opposite direction.

(17)

Proposition 5.2 Suppose that F : X → Y is slantly differentiable on a neighborhood Nx of x. Let f be a slanting function for F in Nx and ∂SF be the slant derivative associated with f in Nx. Then, the following hold.

(a) If F is s-semismooth at x, then F is B-differentiable at x, and

F (x + h) − F (x) − δ+F (x + h; h) = o(khk) (34) for all x + h at which F is B-differentiable.

(b) If F is B-differentiable on a neighborhood Nx of x and (34) holds for all x + h at which F is B-differentiable, then F is s-semismooth at x.

Proof. (a) The B-differentiability of F at x is clear by Proposition 5.1. It remains to claim that when F is B-differentiable at x + h, there has

kF (x + h) − F (x) − δ+F (x + h; h)k

khk → 0 as h → 0. (35)

If not, there exist a δ > 0 and a sequence hi → 0 such that F is B-differentiable at x + hi

for each i = 1, 2, . . . , and

kF (x + hi) − F (x) − δ+F (x + hi; hi)k

khik ≥ δ. (36)

By assumption, F is s-semismooth at x, then for each i ≥ 1 there exist Vi ∈ ∂SF (x + hi) and yi ∈ Nx+h such that

kVi− f(yi)k ≤ khik, kyi− (x + hi)k ≤ khik2 (37) and kF (x + hi) − F (x) − Vihik

khik → 0 as hi → 0. (38)

By [20, Proposition 2.8], for each hi there exist ti > 0 with 0 < ti ≤ khik such that kf(x + hi+ tihi)hi− δ+F (x + hi; hi)k ≤ khik2. (39) Now we compute

f(yi)hi − f(x + hi+ tihi)hi

= (F (x + hi+ tihi) − F (x) − f(x + hi+ tihi)(hi+ tihi))

−(F (yi) − F (x) − f(yi)(yi− x)) +(F (yi) − F (x + hi+ tihi)) +f(yi)(x + hi− yi)

+f(x + hi+ tihi)(tihi).

(40)

(18)

Because F is slantly differentiable at x, the first and second term of (40) implies F (x + hi+ tihi) − F (x) − f(x + hi+ tihi)(hi+ tihi)

khi+ tihik → 0 as i → ∞

and F (yi) − F (x) − f(yi)(yi− x)

kyi− xk → 0 as i → ∞

which lead to

F (x + hi+ tihi) − F (x) − f(x + hi+ tihi)(hi+ tihi) khik

= F (x + hi+ tihi) − F (x) − f(x + hi+ tihi)(hi+ tihi)

khi+ tihik ·khi+ tihik khik

→ 0 as i → ∞

(41)

and F (yi) − F (x) − f(yi)(yi− x) khik

= F (yi) − F (x) − f(yi)(yi− x)

kyi− xk · kyi− xk khik

→ 0 as i → ∞.

(42)

Here we use that fact that khi+ tihik = (1 + ti)khik and kyi− xk = kyi− x − hi+ hik ≤ kyi− x − hik + khik ≤ khik2+ khik. Besides, for the third, fourth and fifth term of (40), since F is slantly differentiable in a neighborhood Nx of x, kf(x)k is uniformly bounded in Nx, say kf(x)k ≤ M in Nx. Hence we have

kF (yi) − F (x + hi+ tihi)k ≤ M kyi− (x + hi+ tihi)k

≤ M (khik2 + tikhik), kf(yi)(x + hi− yi)k ≤ M kx + hi− yik ≤ M khik2 and

kf(x + hi+ tihi)(tihi)k ≤ M ktihik ≤ M khik2 which implies

kF (yi) − F (x + hi+ tihi)k

khik → 0 as i → ∞, (43)

kf(yi)(x + hi− yi)k

khik → 0 as i → ∞, (44)

kf(x + hi+ tihi)(tihi)k

khik → 0 as i → ∞. (45)

Combining (41)- (45) all together, we have

kf(yi)hi− f(x + hi+ tihi)hik

khik → 0 as i → ∞. (46)

(19)

Now consider

F (x + hi) − F (x) − δ+F (x + hi; hi)

= [F (x + hi) − F (x) − Vihi] +[Vihi− f(yi)hi]

+[f(yi)hi− f(x + hi+ tihi)hi]

+[f(x + hi+ tihi)hi− δ+F (x + hi; hi)].

From (38), (37), (46) and (39), we have

kF (x + hi) − F (x) − δ+F (x + hi; hi)k

khik → 0 as hi → 0

This is a contradiction to equation (36), hence (35) holds for all x + h at which F is B-differentiable.

(b) By Proposition 5.1(c), it suffice to show that for each V ∈ ∂SF (x + h), there has kF (x + h) − F (x) − V hk

khk → 0 as khk → 0.

If not, there exist δ > 0 and a sequence hi → 0, Vi ∈ ∂SF (x + hi) and yi ∈ Nx+hi such that kyi− (x + hi)k ≤ khik2, kVi− f(yi)k ≤ khik and

kF (x + hi) − F (x) − Vihik

khik ≥ δ.

By assumption, F is B-differentiable in a neighborhood of x and satisfies (34) which yields

kF (x + hi) − F (x) − δ+F (x + hi; hi)k

khik → 0 as khik → 0.

Then, we consider

F (x + hi) − F (x) − Vihi

= [F (x + hi) − F (x) − δ+F (x + hi; hi)]

+[f(yi)hi− Vihi]

+[f(x + hi+ tihi)hi− f(yi)hi]

+[δ+F (x + hi; hi) − f(x + hi+ tihi)hi].

With similar argument and choice of ti in part (a), we have kF (x + hi) − F (x) − Vihik

khik → 0 as i → ∞.

This leads to a contradiction. Thus, the proof is complete. 2

參考文獻

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