In the first part of this section, we prove the well-definedness of the sequence {x

§3 Convergence Results With Strong Convexity

In the first part of this section, we prove the well-definedness of the sequence {x

} generated by Algorithm 1 under some weak monotonicity assumptions on the operator T . Then we give the convergence results obtained under an additional continuous assumption.

The following existence result is due to Shih and Tan [23].

Proposition 3.1 [23, Theorem 2]. Let C be a nonempty closed convex subset of X and let F be a (l, w)-u.s.c. and monotone set-valued operator from C into X

such that F (x) is weakly compact and convex for each x ∈ C. If C is either compact or

kzk→∞

lim

z∈C

inf

y∈F (z)

hy, z − x

i

kzk = ∞,

for some x

∈ C, then there exists a solution to the problem V I(F, C).

Using this, we can establish an existence result for strongly monotone operators.

Proposition 3.2. Let C be a nonempty closed convex subset of X and let F be a (l, w)-u.s.c. and strongly monotone set-valued operator from C into X

such that F (x) is weakly compact and convex for each x ∈ C, then there exists a unique solution to the problem V I(F, C).

Proof : Let z, x ∈ C and y

∈ F (z), y

∈ F (x). Since F is strongly monotone, there exists a > 0 such that

hy

, z − xi ≥ hy

, z − xi + akz − xk

≥ −ky

kkz − xk + akz − xk

= (akz − xk − ky

k)kz − xk.

Thus,

inf

y1∈F (z)

hy

, z − xi

kzk ≥ (akz − xk − ky

k) kz − xk kzk

→ ∞, as kzk −→ ∞.

By Proposition 3.1, V I(F, C) has a solution, and it is trivial that the solution is unique,

since F is strongly monotone. 2

In [25], Yao obtained the existence theorem for pseudomonotone operator as follows.

Proposition 3.3 [25, Corollary 2.1]. Let C be a nonempty closed convex subset of X and let F be a u.s.c. pseudomonotone set-valued operator from C into X

such that F (x) is weakly compact and convex for each x ∈ C. If C is either compact or

lim inf

kzk→∞

z∈C

inf

y∈F (z)

hy, z − x

i > 0,

for some x

∈ C, then there exists a solution to the problem V I(F, C).

Now we make some basic assumptions on our Algorithm 1.

Assumption 1.

(i) T is (l, w)-u.s.c. and weakly montone with constant L on C.

(ii) M is differentiable and strongly convex with constant α =

, and M

is (l, w)- continuous.

(iii) The sequence {ε

} satisfies the following condition : ε

> 0 ,

∞

X

k=0

ε

= +∞ ,

∞

X

k=0

ε

< +∞.

The following result analyzes the convergence of the Algorithm 1 involving pseudo- Dunn property.

Theorem 3.4. Suppose that the problem (P 1) has a solution x

. Under Assumption 1, for each step k, there exists a unique solution x

generated by Algorithm 1 to the auxiliary problem (P 2). Moreover, if T has the pseudo-Dunn property with constant t on C, and if

∀ k ∈ N, ε

≤ ε

and α < ε

<

, where α > 0, β > 0,

then the sequence {x

} is bounded. In addition, if M

is Lipschitz continuous with con- stant m, and if T is (w, s)-u.s.c. on C, then every weakly cluster point of the sequence {x

} is a solution of the problem (P 1).

Proof : Under Assumption 1, the operator F defined by F (y) = εT (y) + M

(y) − M

(x)

is (l, w)-u.s.c. and strongly monotone with constant b − εL on C , by Lemma 2.3. Ac- cording to Proposition 3.2, we get that there exists a unique solution to the auxiliary problem (P 2).

Consider the function Λ

defined by

Λ

(x, ε) := Φ

(x) + Ω

(x, ε), ∀ x ∈ C, where

Φ

(x) := M (x

) − M (x) − hM

(x), x

− xi, Ω

(x, ε) := εhy

, x − x

i.

Since M

is strongly monotone with constant b, by Proposition 2.2, M is strongly convex with constant α =

, and hence

Φ

(x

) ≥ b

2 kx

− x

k

≥ 0.

Since x

is a solution of the problem (P 1), Ω

(x, ε) is nonnegative, hence Λ

(x, ε) ≥ 0.

Now we study the variation of Λ

for Algorithm 1 :

Γ

= Λ

(x

, ε

) − Λ

(x

, ε

) = s

+ s

+ s

, where

s

:= M (x

) − M (x

) − hM

(x

), x

− x

i, s

:= hM

(x

) − M

(x

), x

− x

i,

s

:= ε

hy

, x

− x

i − ε

hy

, x

− x

i.

Since M

is strongly monotone,

s

≤ − b

2 kx

− x

k

.

Using (3) with z = x

, ε

hy

, x

− x

i + B(x

, x

, x

) ≥ 0, and therefore s

≤ ε

hy

, x

− x

i = ε

hy

, x

− x

i + ε

hy

, x

− x

i.

Using (1) with z = x

, we have some y

∈ T (x

) such that hy

, x

− x

i ≥ 0.

Since T has the pseudo-Dunn property with constant t and for some y

∈ T (x

), we have hy

, x

− x

i ≥ 1

t ky

− y

k

. Thus,

s

≤ − ε

t ky

− y

k

+ ε

hy

, x

− x

i.

Since ε

≤ ε

for each k, we obtain s

+ s

≤ − ε

t ky

− y

k

+ ε

hy

− y

, x

− x

i.

Therefore,

Γ

≤ − b

2 kx

− x

k

− ε

t ky

− y

k

+ε

hy

− y

, x

− x

i.

Since

ε

ky

− y

k · kx

− x

k ≤ ε

2λ ky

− y

k

+ λ

2 kx

− x

k

,

where λ > 0, we get Γ

≤ −( b

2 − λ

2 )kx

− x

k

− ε

( 1 t − ε

2λ )ky

− y

k

. Thus, this condition α < ε

<

, where α > 0, β > 0 yields

Γ

≤ −( b 2 − λ

2 )kx

− x

k

− αβ

t(t + β) ky

− y

k

,

and for λ < b, Γ

is negative unless x

= x

and y

= y

. Then, according to (3), {x

} is a solution to the problem (P 1). Note that the sequence {Λ

(x

, ε

)} is strictly decreasing, and since it is positive, it must be convergent and the difference between two consecutive terms tends to zero. Therefore, kx

− x

k and ky

− y

k are convergent to zero. Moreover, since the sequence {Λ

(x

, ε

)} converges, it is bounded, and so is {x

}.

Let ¯ x be a weakly cluster point of the sequence {x

}, and let {x

} be a subsequence weakly converging to ¯ x. By (3), since M

is Lipschitz continuous with constant m and ε

> α, we have some y

∈ T (x

),

hy

, z − x

i ≥ − m

α kx

− x

k · kz − x

k, ∀z ∈ C.

Since kx

− x

k converges to zero and y

strongly converges to y

, taking the limit for the subsequence {k

} in the last inequality yields

hy

, z − ¯ xi ≥ 0, ∀z ∈ C. (4)

Moreover, h¯ y, x

− ¯ xi converges to 0, for each ¯ y ∈ T (¯ x). If 0 ∈ T (¯ x), then ¯ x is clearly a solution of the problem (P 1). Otherwise, we let ¯ y ∈ T (¯ x) with ¯ y 6= 0, and let

w

= x

− h¯ y, x

− ¯ xi k¯ yk

y. ¯ Then,

h¯ y, w

− ¯ xi = 0, (5)

and kw

−x

k converges to 0. Thus, w

weakly converges to ¯ x. On the other hand, since T is (w, s)-u.s.c., for u

∈ T (w

) , u

strongly converges to y

, due to ky

− y

k −→ 0 and ku

− y

k −→ 0. By (5) and the pseudo-Dunn property of T , we get

hu

, w

− ¯ xi ≥ 1

t ku

− ¯ yk

.

Therefore, by taking the limit, we conclude that u

strongly converges to ¯ y. Then, y

=

¯

y ∈ T (¯ x) and hence ¯ x is a solution to the problem (P 1). 2

Corollary 3.5. Under all the conditions of Theorem 3.4, if M

is continuous from X equipped with the weak topology to X

equipped with the weak topology, then the sequence {x

} weakly converges to a unique solution of the problem (P 1).

Proof : Following the proof of Theorem 3.4, we need only to prove the uniqueness of

¯

x. Assume that the sequence {x

} has two weakly cluster points ¯ x and ˆ x. Then, both cluster points can be used as x

to defined the function Λ

. Consider the subsequence {k

} and {l

} such that x

and x

weakly converge to ¯ x and ˆ x, respectively. Then, by Lemma 1.4 (i) and the fact ¯ y = ˆ y, we get

Λ ˆ

(x

, ε

) = ¯ Λ

(x

, ε

) + R

(x

), where

R

(x

) = M (ˆ x) − M (¯ x) − hM

(x

), ˆ x − ¯ xi.

Note that ¯ Λ

(x

, ε

) and ˆ Λ

(x

, ε

) tend to some ¯ l and ˆ l, respectively. Since M

is (w, w)-continuous, R

(x

) converges to ˆ l − ¯ l. Since M is strongly convex with constant

b

2

, by Proposition 2.2, we obtain

ˆ l ≥ ¯ l + b

2 kˆ x − ¯ xk

. By interchangeing the role of ¯ x, the same calculations yield

¯ l ≥ ˆ l + b

2 kˆ x − ¯ xk

. Then

0 ≤ b

2 kˆ x − ¯ xk

≤ min{¯ l − ˆ l, ˆ l − ¯ l} ≤ 0.

Hence, ¯ x = ˆ x. 2

We note that, if T is strongly monotone with constant a and Lipschitz continuous with constant µ on C, then it has the Dunn property with constant

. Without monotonicity, we shall deal with the pseudomonotone case as follows.

Lemma 3.6. If T is strongly pseudomonotone with constant e and Lipschitz con- tinuous with constant µ on C, then it has the pseudo-Dunn property with constant

.

Proof : Given any (x

, y

) and (x

, y

) ∈ G(T ), since T is strongly pseudomonotone with constant e,

hy

, x

− x

i ≥ 0 =⇒ hy

, x

− x

i ≥ ekx

− x

k

. Since T is Lipschitz continuous with constant µ,

ky

− y

k ≤ µkx

− x

k,

which implies

kx

− x

k

≥ 1

µ

ky

− y

k

. So we have

hy

, x

− x

i ≥ e · 1

µ

ky

− y

k

.

We complete the proof. 2

We next analyze the convergence of Algorithm 1 invloving strongly monotone operator with Lipschitz property.

Theorem 3.7. Suppose that the problem (P 1) has a solution x

∈ C. Under As- sumption 1, for each step k, there exists a unique solution x

generated by Algorithm 1 to the auxiliary problem (P 2). If T is strongly pseudomonotone with constant e on C (x

is then unique) and Lipschitz continuous with constant µ on C, and if

∀ k ∈ N, α < ε

<

, where α > 0, β > 0,

then the sequence {x

} strongly converges to x

. Moreover, if M

is Lipschitz continuous with constant m on C and t =

+

< 1, then the error estimation of solution x

is

kx

− x

k ≤ t

1 − t kx

− x

k.

Proof : By using Theorem 3.4 and Lemma 3.6, {x

} weakly converges to x

. Further- more, {x

} is strongly convergent. We illustrate as follows : As before, we consider Φ

(x), but ignore Ω

(x, ε). Note that Φ

(x) ≥ 0, and define ∆

= Φ

(x

)−Φ

(x

) = s

+s

, where the definitions of s

and s

are same as Theorem 3.4. Thus,

s

≤ − b

2 kx

− x

k

, and

s

≤ ε

hy

, x

− x

i

= ε

hy

− y

, x

− x

i + ε

hy

, x

− x

i.

By (1) with z = x

, and the strong pseudomonotonicity of T , we get s

≤ −eε

kx

− x

k

+ ε

hy

− y

, x

− x

i.

Since T is Lipschitz continuous with constant µ, it follows that

∆

≤ − b

2 kx

− x

k

− eε

kx

− x

k

+ ε

µkx

− x

k · kx

− x

k.

By the inequality:

ε

µkx

− x

k · kx

− x

k ≤ λ

2 kx

− x

k

+ ε

µ

2λ kx

− x

k

, with λ = b, we have

∆

≤ ε

( −e ε

+ µ

2b )kx

− x

k

≤ −αβ

2b kx

− x

k

≤ 0.

So the sequence {Φ

(x

)} is strictly decreasing (unless x

= x

) and bounded below by zero. Thus, {Φ

(x

)} converges. This yields ∆

tends to 0 and hence {x

} strongly converges to x

.

Now, by (3) with z = x

,

hε

y

+ M

(x

) − M

(x

), x

− x

i ≥ 0.

Since T is strongly pseudomonotone with constnat e, by (1) with z = x

, hy

, x

− x

i ≥ ekx

− x

k

.

It follows that

hM

(x

) − M

(x

), x

− x

i + ε

hy

− y

, x

− x

i ≥ eε

kx

− x

k

. (6) On the other hand, since M

is Lipschitz continuous with constant m, we have

hM

(x

) − M

(x

), x

− x

i + ε

hy

− y

, x

− x

i

≤ mkx

− x

k · kx

− x

k + ε

µkx

− x

k · kx

− x

k

= (m + ε

µ)kx

− x

k · kx

− x

k. (7) Thus, combining (6) with (7), we obtain

(m + ε

µ)kx

− x

k ≥ eε

kx

− x

k.

Hence,

kx

− x

k ≤ ( m eε

+ µ

e )kx

− x

k

≤ ( m eα + µ

e )kx

− x

k

≤ t(kx

− x

k + kx

− x

k).

It follows that kx

−x

k ≤

kx

−x

k. 2

The following two technical Lemmas will be used to the later theorem.

Lemma 3.8 [2, Lemma 5]. Let {x

}, {µ

} and {η

} be three sequences such that

∞

X

k=1

µ

< +∞,

x

≤ η

+

n−1

X

k=1

µ

sup

l≤k+1

x

, ∀ n = 1, 2, · · · .

If the sequence {η

} is bounded above, then the sequence {x

} is also bunded above.

Lemma 3.9 [2, Lemma 4]. Let f be a Lipschitz function on a Banach space X and consider the sequence {x

} ⊂ X and {ε

} are positive sequences such that

X

k∈N

ε

= +∞,

∃ ζ, ∀ k ∈ N , kx

− x

k ≤ ζε

,

∃ ξ,

k∈N

ε

|f (x

) − ξ| < +∞.

Then,

k→∞

lim f (x

) = ξ.

Under the condition that the norm of T does not increase faster than linearly with the norm of x; that is,

∃ p, q > 0 s.t. kx

k ≤ pkxk + q, ∀ (x, x

) ∈ G(T ), (?) the following result deals with the convergence of the Algorithm 1 involving α-strongly pseudomonotone operators.

Theorem 3.10. Suppose that the problem (P 1) has a solution x

∈ C. Assume T and M

satisfy Assumption 1. If T is α-strongly pseudomonotone with constant e on C, with the above property (?), then the sequence {x

} strongly converges to the unique solution x

.

Proof : Consider the Bregman distance

φ

(x) := M (x

) − M (x) − hM

(x), x

− xi, ∀x ∈ C. (8)

Since M

is strongly monotone with constant b, by Proposition 2.2, M is strongly convex

with constant α =

. Thus,

φ

(x

) = M (x

) − M (x

) − hM

(x

), x

− x

i

≥ αkx

− x

k

= b

2 kx

− x

k

.

Now we study the variation of φ

for Algorithm 1 : ∆

= φ

(x

) − φ

(x

) for each k. Notices that

∆

= M (x

) − M (x

) − hM

(x

), x

− x

i + hM

(x

) − M

(x

), x

− x

i

= S

+ S

, where

S

:= M (x

) − M (x

) − hM

(x

), x

− x

i

≤ − b

2 kx

− x

k

, (9)

and

S

:= hM

(x

) − M

(x

), x

− x

i

= −B(x

, x

, x

).

By (3) with z = x

,

ε

hy

, x

− x

i + B(x

, x

, x

) ≥ 0.

So

S

≤ ε

hy

, x

− x

i

= ε

hy

, x

− x

i + ε

hy

, x

− x

i.

By (1) with z = x

, we have some y

∈ T (x

) such that hy

, x

− x

i ≥ 0.

Since T is α-strongly pseudomonotone with constant e, we get hy

, x

−x

i ≤ −ekx

−x

k

. Thus,

S

≤ ε

hy

, x

− x

i − eε

kx

− x

k

≤ ε

ky

kkx

− x

k − eε

kx

− x

k

≤ (ε

)

2b ky

k

+ b

2 kx

− x

k

− eε

kx

− x

k

. (10) Therefore, combining (9) and (10), we obtain

∆

= S

+ S

≤ ε

2b ky

k

− eε

kx

− x

k

.

On the other hand, by the condition of assumption, we have ky

k ≤ pkx

k + q ≤ pkx

− x

k + pkx

k + q.

Then, by the simple inequaity :

(u + v)

≤ 2(u

+ v

), we get

ky

k

≤ 2p

kx

− x

k

+ 2(pkx

k + q)

. Thus,

∆

≤ ε

2b ky

k

− eε

kx

− x

k

≤ ε

(γkx

− x

k

+ δ) − eε

kx

− x

k

.

where γ =

and δ =

. Now, summing up this inequaity from k = 0 to n − 1, we have

b

2 kx

− x

k

≤ φ

(x

) =

n−1

X

k=0

∆

+ φ

(x

)

≤ φ

(x

) +

n−1

X

k=0

ε

(γkx

− x

k

+ δ) − e

n−1

X

k=0

ε

kx

− x

k

≤ φ

(x

) +

n−1

X

k=0

ε

(γkx

− x

k

+ δ). (11) Then,

kx

− x

k

≤ η

+

n−1

X

k=1

µ

kx

− x

k

, (12)

where

η

= 2

b φ

(x

) + 2γε

b kx

− x

k

+

n−1

X

k=0

2δε

b , µ

= 2γε

b , ∀ k = 1, 2, · · ·, n − 1.

Since

kx

− x

k

≤ sup

l≤k+1

kx

− x

k

, ∀ k = 1, 2, · · ·, it follows from (12) that

kx

− x

k

≤ η

+

n−1

X

k=1

µ

sup

l≤k+1

kx

− x

k

.

By Lemma 3.8, the sequence {kx

− x

k

} is bounded, and so is {x

}. Moreover, from Assumption 1 (iii) and (11), we get

X

k∈N

ε

kx

− x

k

< +∞.

Also by using (3) with z = x

and the strong monotonicity of M

, we get kx

− x

k ≤ ε

ky

k

b ≤ ε

(pkx

k + q)

b ≤ ζε

, ∀k = 1, 2, · · · ,

for some ζ > 0. Applying Lemma 3.9 to f (x) = kx−x

k

and ξ = 0, we conclude that {x

}

strongly converges to x

. 2