(i) f is strongly convex on C, if ∃ α > 0 s.t. ∀ x, y ∈ C, f (y) − f (x) ≥ hy − x, ∇f (x)i + αky − xk

§2 A Generalized Proximal Point Algorithm

In this section, we introduce a generalized proximal point algorithm for solving vari- ational inequalities involving general set-valued operators. We first recall the concept of the strongly convex functions introduced by [28] and the some definitions of continuous property as follows.

Definition 2.1. Let C be a closed convex subset of X, f be differentiable on a neighborhood of C, and let ∇f be the gradient of f .

(i) f is strongly convex on C, if ∃ α > 0 s.t. ∀ x, y ∈ C, f (y) − f (x) ≥ hy − x, ∇f (x)i + αky − xk

;

4

(ii) f is convex on C, if ∀ x, y ∈ C, f (y) − f (x) ≥ hy − x, ∇f (x)i.

Also, for a set-valued operator T : X −→ X

, we shall say that

(iii) T is upper semicontinuous (u.s.c.) at x ∈ X, if for any open set G containing T (x), there is some neighborhood V (x) of x, such that T (y) ⊂ G for all y ∈ V (x);

(iv) T is (l,w)-u.s.c., if T is u.s.c. from line segments in X to the weak topology of X

; (v) T is (w,s)-u.s.c., if T is u.s.c. from the weak topology of X to the norm topology of

X

;

(vi) T is Lipschitz continuous, if there exists a constant m > 0 such that kw

− w

k ≤ mku − vk, ∀ w

∈ T (u), w

∈ T (v).

To establish our proximal point algorithm, we consider a differentiable auxiliary func- tion M : C −→ R, and denoted ∇M by M

. Define B : C × C × C −→ R by

B(x, y, z) = hM

(x) − M

(y), z − xi.

For some x ∈ C, y ∈ T (x), and a positive number ε, we introduce the auxiliary problem (P 2) as follows.

(P2) : find ˜ x ∈ C such that

εhy, z − xi + B(˜ x, x, z) ≥ 0, ∀z ∈ C. (2) In fact, if ˜ x exists and equal to x, then B(˜ x, x, z) = 0, which implies that ˜ x is a solution of the original problem (P 1) (i.e. (P 2) reduces to (P 1) ).

Algorithm 1 to VI(T,C) :

Step 1: Take any x

∈ C and y

∈ T (x

).

Step 2: Knowing (x

, y

) and ε

, compute x

∈ C such that

ε

hy

, z − x

i + B(x

, x

, z) ≥ 0, ∀z ∈ C. (3) Step 3: Take any y

∈ T (x

) and return to Step 2, until kx

− x

k is below

some threshold.

Remark. In particular, if M (x) =

kxk

, then M

(x) =

(

x, in a Hilbert space J (x), in a Banach space . In this case, our algorithm 1 reduces to the classical proximal point algorithm.

5

We also need the following well-known proposition, due to Karamardian [14, 17], see also Ortoga [20].

Proposition 2.2 [14, Proposition 6.1]. Let f be a differentiable function on an open convex set C of X, then f is strongly convex on C with constant α ⇐⇒ ∇f is strongly monotone with constant b on C, where b = 2α.

We give hereafter a lemma that will be used to prove the well-definedness of the sequence {x

} generated by our Algorithm 1.

Lemma 2.3. Let x ∈ C and T be weakly monotone with constant L. If M

is strongly monotone with constant b on C, and ε <

, then the operator F (y) = εT (y) + M

(y) − M

(x) is strongly monotone on C with constant b − εL. Moreover if T is (l,w)-u.s.c. and M

is (l,w)-continuous, then F is also (l,w)-u.s.c..

Proof : For all (y

, w

), (y

, w

) ∈ G(F ), we have some z

∈ T (y

) and z

∈ T (y

) such that

w

= εz

+ M

(y

) − M

(x), and

w

= εz

+ M

(y

) − M

(x).

Since T is weakly monotone,

hz

− z

, y

− y

i ≥ −Lky

− y

k

, and M

is strongly monotone,

hM

(y

) − M

(y

), y

− y

i ≥ bky

− y

k

. Thus,

hw

− w

, y

− y

i = εh(z

− z

) + M

(y

) − M

(y

), y

− y

i

= εhz

− z

, y

− y

i + hM

(y

) − M

(y

), y

− y

i

≥ ε(−L)ky

− y

k

+ bky

− y

k

= (b − εL)ky

− y

k

,

Hence F is strongly monotone. On the other hand, let G be open in the weak topology of X

and F (y) ⊂ G, where y ∈ L, and L is a line segment in X. This implies that

εT (y) ⊂ G + M

(x) − M

(y).

Since T and hence εT is (l, w)-u.s.c. at y, there is an open neighborhood V

(x) of X such that

εT (z) ⊂ G + M

(x) − M

(y), ∀z ∈ V

(x) ∩ L.

6

Hence, for all w ∈ εT (z),

w ∈ G + M

(x) − M

(y), which implies that

M

(y) ∈ G + M

(x) − w.

Since M

is (l, w)-continuous, there is an open neighborhood V

(x) of X such that M

(u) ∈ G + M

(x) − w, ∀u ∈ V

(x) ∩ L.

Thus above results imply

w + M

(u) − M

(x) ∈ G, ∀w ∈ εT (z), u ∈ V

(x) ∩ L, z ∈ V

(x) ∩ L.

Let V (x) = V

(x) ∩ V

(x). Then for all z ∈ V (x) ∩ L, we have F (y) = εT (z) + M

(z) − M

(x) ⊂ G.

Thus, we complete the proof. 2

Remark. Under different assumptions on T , the strongly monotone constants of F are different. For example,

(i) if T is monotone, then F is strongly monotone with constant b;

(ii) if T is strongly monotone with constant a, then F is strongly monotone with constant εa + b;

(iii) if T has the Dunn property, then F is strongly monotone with constant b.

7