APREUTESEI Received 13 October 2003 and in revised form 10 December 2003 To the memory of my mother, Liliana Several existence theorems are given for some second-order difference equation
Trang 1MONOTONE OPERATORS
N C APREUTESEI
Received 13 October 2003 and in revised form 10 December 2003
To the memory of my mother, Liliana
Several existence theorems are given for some second-order difference equations associ-ated with maximal monotone operators in Hilbert spaces Boundary conditions of mono-tone type are attached The main tool used here is the theory of maximal monomono-tone operators
1 Introduction
In [1,2], the authors proved the existence of the solution of the boundary value problem
p(t)u (t) + r(t)u (t) ∈ Au(t) + f (t), a.e on [0,T], T > 0, (1.1)
u (0)∈ αu(0) − a, u (T) ∈ − βu(T) − b, (1.2) whereA : D(A) ⊆ H → H, α : D(α) ⊆ H → H, and β : D(β) ⊆ H → H are maximal
mono-tone operators in the real Hilbert spaceH (satisfying some specific properties), a, b are
given elements in the domainD(A) of A, f ∈ L2(0,T;H), and p,r : [0,T] →Rare contin-uous functions,p(t) ≥ k > 0 for all t ∈[0,T].
Particular cases of this problem were considered before in [9,10,12,15,16] Ifp ≡
1,r ≡0, f ≡0,T = ∞, and the boundary conditions areu(0) = a and sup { u(t) ,t ≥
0} < ∞instead of (1.2), the solutionu(t) of (1.1), (1.2) defines a semigroup of nonlinear contractions{ S1/2( t), t ≥0}on the closureD(A) of D(A) (see [9,10]) This semigroup and its infinitesimal generatorA1/2have some important properties (see [9,10,11,12])
A discretization of (1.1) is p i( u i+1 −2u i+u i−1) +r i( u i+1 − u i) ∈ k i Au i+g i, i =1,N,
whereN is a given natural number, p i, r i, k i > 0, g i ∈ H This leads to the finite difference
scheme
p i+r i
u i+1 −2p i+r i
u i+p i u i−1∈ k i Au i+g i, i =1,N, (1.3)
u1− u0∈ αu0− a, u N+1 − u N ∈ − βu N+1 − b, (1.4) wherea,b ∈ H are given, (p i)i=1,N, ( r i)i=1,N, and (k i)i=1,Nare sequences of positive num-bers, and (g i) i=1,N ∈ H N
Copyright©2004 Hindawi Publishing Corporation
Advances in Di fference Equations 2004:1 (2004) 11–22
2000 Mathematics Subject Classification: 39A12, 39A70, 47H05
URL: http://dx.doi.org/10.1155/S1687183904310046
Trang 2In this paper, we study the existence and uniqueness of the solution of problem (1.3), (1.4) under various conditions onA, α, and β.
The casep i ≡1,r i ≡0,g i ≡0 was discussed in [14] for the boundary conditionsu0= a
andu N+1 = b These boundary conditions can be seen as a particular case of (1.4) with
α = β = ∂j (the subdifferential of j), where j : H →Ris the lower-semicontinuous, con-vex, and proper function:
j(x) =
0, x =0,
In [6,8,13,14], one studies the existence, uniqueness, and asymptotic behavior of the solution of the difference equation
p i+r i
u i+1 −2p i+r i
u i+p i u i−1∈ k i Au i+g i, i ≥1, (1.6) (p i ≡1,r i ≡0 in [13,14] and the general case in [6,8]), subject to the boundary conditions
u0= a, sup
i≥0
Here · is the norm ofH In [7], the author establishes the existence for problem (1.3), (1.4) under the hypothesis thatA is also strongly monotone.
Other classes of difference or differential inclusions in abstract spaces are presented in [3,4,5]
InSection 2, we recall some notions and results that we need to show our main exis-tence theorems They are stated inSection 3and represent the discrete version of some results obtained in [1,2] for the continuous case
2 Preliminary results
In this section, we recall some fundamental elements on nonlinear analysis we need in this paper
IfH is a real Hilbert space with the scalar product ( ·,·) and the norm · , then the operatorA ⊆ H × H (with the domain D(A) and the range R(A)) is called a monotone operator if (x − x ,y − y )≥0 for allx,x ∈ D(A), y ∈ Ax, and y ∈ Ax The monotone operatorA ⊆ H × H is said to be maximal monotone if it is not properly enclosed in a
monotone operator A basic result of Minty (see [11, Theorem 1.2, page 9]) asserts that
A is maximal monotone if and only if A is monotone and the range of A + λI is the whole
spaceH for all λ > 0 (or equivalently, for only one λ0> 0) It is also known that a maximal
monotone and coercive operatorA is surjective, that is, its range R(A) is H.
For allx ∈ D(A), we denote by A0x the element of least norm in Ax:
A0x =inf
IfA is maximal monotone and A0x → ∞as x → ∞, thenA is surjective.
The operatorA ⊆ H × H (possibly multivalued) is said to be one to one if (Ax1)∩
(Ax2) Φ (with x1,x2∈ D(A)) implies x1= x2
Trang 3If A and B are maximal monotone in H and their domains satisfy the condition
(intD(A)) ∩ D(B) Φ, then A + B is maximal monotone (see [11, Theorem 1.7, page 46]) If A : D(A) ⊆ H → H is maximal monotone, then A is demiclosed, that is, from
[x n, y n] ∈ A, x n x and y n → y, then [x, y] ∈ A Here and everywhere below, we denote
by “” the weak convergence and by “ →” the strong convergence inH.
For every maximal monotone operatorA and the scalar λ > 0, we may consider the
single-valued and everywhere-defined operatorsJ λ andA λ, namely, J λ =(I + λA) −1and
A λ =(I − J λ) /λ They are called the resolvent and the Yosida approximation of A,
respec-tively Obviously, we haveJ λ x + λA λ x = x for all x ∈ H and for all λ > 0 Properties of these
operators can be found in, for example, [11, Proposition 1.1, page 42] or [11, Proposition 3.2, page 73]
Recall now another result concerning the sum of two maximal monotone operators (see [11, Theorem 3.6, page 82])
Theorem 2.1 If A : D(A) ⊆ H → H and B : D(B) ⊆ H → H are maximal monotone oper-ators in H such that D(A) ∩ D(B) Φ and (y,Aλ x) ≥ 0 for all [ x, y] ∈ B and for all λ > 0, then A + B is maximal monotone.
We end this section with some remarks on problem (1.3), (1.4) Denoting
θ i = p i
p i+r i, c i = k i
p i+r i, f i = g i
p i+r i, i =1,N, (2.2) problem (1.3), (1.4) becomes
u i+1 −1 +θ i
u i+θ i u i−1∈ c i Au i+f i, i =1,N,
u1− u0∈ αu0− a, u N+1 − u N ∈ − βu N+1 − b. (2.3)
Ifp i, r i, k i > 0, i =1,N, then θ i ∈(0, 1) andc i > 0 for all i =1,N.
Let (a i) i=1,Nbe the finite sequence given by
a0=1, a i = θ1···1 θ i, i =1,N, (2.4)
and letᏸ be the product space H N = H × ··· × H (N factors) endowed with the scalar
product
u i
i=1,N,
v i
i=1,N =
N
i=1
a i
u i, v i
It is clear thatH Nandᏸ coincide as sets and their norms are equivalent Observe that
a i θ i = a i−1, i =1,N. (2.6) Consider the operatorB in H N × H N:
B u
i
i=1,N =− u i+1+
1 +θ iu i − θ i u i−1
i=1,N,
D(B) =
u i
i=1,N ∈ H N,u1− u0∈ αu0− a,u N+1 − u N ∈ − βu N+1 − b. (2.7)
Trang 4This operator is not necessarily monotone inH N, but we have the following auxiliary result (see [7, Proposition 2.1])
Proposition 2.2 The operator B given above is maximal monotone in ᏸ.
Recall here an existence theorem from [7], which we use in the sequel
Theorem 2.3 Assume that A, α, and β are maximal monotone operators in H with 0 ∈
D(A) ∩ D(α) ∩ D(β), A is also strongly monotone and
for all z ∈ α(x − y) (with x − y ∈ D(α)) and for all z ∈ β(x − y) (with x − y ∈ D(β)) If
θ i ∈ (0, 1), c i > 0, f i ∈ H, i =1,N, and a,b ∈ H, then problem (2.3) has a unique solution
(u i) i=1,N ∈ D(A) N
3 Existence theorems
LetH be a real Hilbert space with the norm · and the scalar product (·,·) Consider the maximal monotone operatorsA : D(A) ⊆ H → H, α : D(α) ⊆ H → H, and β : D(β) ⊆
H → H satisfying the properties
0∈ D(A) ∩ D(α) ∩ D(β), 0 ∈ α(0) ∩ β(0), (3.1)
A λ x − A λ y,z≥0 ∀ z ∈ α(x − y) with x − y ∈ D(α), (3.2)
A λ x − A λ y,z≤0 ∀ z ∈ − β(x − y) with x − y ∈ D(β). (3.3) Consider the difference inclusion (1.3), (1.4) As we have already discussed, problem (1.3), (1.4) has the equivalent form (2.3)
We first study the existence of the solution to problem (1.3), (1.4) in the casea = b =0, supposing that
A λ x,z≥0 ∀ z ∈ α(x) with x ∈ D(α), z ∈ β(x) with x ∈ D(β), (3.4)
and
R(α) is bounded, β0(x) −→ ∞ as x −→ ∞, (3.5) or
R(β) is bounded, α0(x) −→ ∞ as x −→ ∞ (3.6)
Theorem 3.1 Let A, α, and β be maximal monotone operators in the real Hilbert space H such that (3.1), (3.4), and (3.5) or (3.6) hold If p i, r i, k i > 0, i =1,N, and (g i) i=1,N ∈ H N , then the boundary value problem
p i+r iu i+1 −2p i+r iu i+p i u i−1∈ k i Au i+g i, i =1,N,
u1− u0∈ αu0
, u N+1 − u N ∈ − βu N+1
Trang 5has at least one solution (u i) i=1,N ∈ D(A) N The solution is unique up to an additive con-stant If A or α is one to one, then the solution is unique If A is, in addition, strongly mono-tone, then again uniqueness is obtained.
Proof We use the form (2.3) of the problem (3.7), wherea = b =0 ByProposition 2.2,
we know that the operator
B u
i
i=1,N =− u i+1+
1 +θ i
u i − θ i u i−1
i=1,N,
D(B) =
u i
i=1,N ∈ H N,u1− u0∈ αu0
,u N+1 − u N ∈ − βu N+1 (3.8)
is maximal monotone inᏸ Denote by| · |the norm inᏸ We show that
B u
i
i=1,N −→ ∞ as
u i
i=1,N −→ ∞ . (3.9)
Suppose by contradiction that (u n
i)i=1,N ∈ D(B) such that |(u n
i)i=1,N | → ∞asn → ∞
and| B((u n
i)i=1,N)| ≤ C1 If (a i) i=1,Nis the sequence given in (2.4), this means that
N
i=1
a iu n
i 2
−→ ∞,
N
i=1
a iu n i+1 − u n
i − θ i
u n
i − u n i−1 2
≤ C1. (3.10)
Assume that (3.5) holds By the boundary conditions in (3.7), we obtain thatu n
1− u n
0 is bounded, say u n
1− u n
0 ≤ C2, for alln ∈Nand
u n N+1 − u n
N −→ ∞ asn −→ ∞ifu n
N+1 −→ ∞ (3.11) The equalitya i( u n
i+1 − u n
i)= u n
1− u n
0+i k=1[a k( u n
k+1 − u n
k)− a k−1(u n
k − u n k−1)] implies that
a i( u n
i+1 − u n
i) ≤ C2+C3| B((u i)i=1,N) |and in view of (3.10), we get a i( u n
i+1 − u n
i) ≤ C4,
i =1,N, n ∈N In particular, a N( u n
N+1 − u n
N) ≤ C4 for alln ∈Nand from (3.11), we infer that u n
N+1 ≤ C5for alln ∈N
Using the boundedness ofu n
N+1anda k( u n
k+1 − u n
k) and the identity
u n
i = u n N+1 −
N
k=i
u n k+1 − u n
k
one arrives at u n
i ≤ C6, henceN
i=1a i u n
i 2≤ C7for alln ∈N But this is in contradic-tion with (3.10) and therefore (3.9) is true This shows thatB is coercive.
Next we show that
B
u i
i=1,N ,
A λ u i
i=1,N ≥0 ∀u i
i=1,N ∈ D(B), λ > 0. (3.13)
Trang 6
B
u i
i=1,N ,
A λ u i
i=1,N = −
N
i=1
a i
u i+1 − u i, A λ u i
− a i−1
u i − u i−1,A λ u i−1
+
N
i=1
a i−1
u i − u i−1,A λ u i − A λ u i−1
≥ − a N
u N+1 − u N, A λ u N
+
u1− u0,A λ u0
.
(3.14)
Hypothesis (3.4) forx = u0andz = u1− u0gives us (u1− u0,A λ u0)≥0, while (3.4) for
x = u N+1andz = − u N+1+u N implies that−(u N+1 − u N, A λ u N) =(u N+1 − u N, A λ u N+1 −
A λ u N)−(u N+1 − u N,A λ u N+1) ≥0 Thus, by (3.14), inequality (3.13) follows
LetᏭ : D(A) N → H Nbe the operator
Ꮽ
u i
i=1,N =c1v1, ,c N v N
, v i ∈ Au i, u i ∈ D(A), i =1,N. (3.15) Since (0, ,0) ∈ D(Ꮽ) ∩ D(B) and (3.13) takes place, we deduce with the aid ofTheorem 2.1 and Proposition 2.2 the maximal monotonicity of B + Ꮽ in ᏸ Next, we can
eas-ily show that B((u i) i=1,N),Ꮽ((ui) i=1,N) ≥0, so|(B + Ꮽ)(u i) i=1,N | ≥ | B((u i) i=1,N)|, and from (3.9), one obtains the coercivity ofB + Ꮽ This shows that B + Ꮽ is surjective, that is,
for all (f i) i=1,N ∈ H N, there exists (u i) i=1,N ∈ D(Ꮽ) ∩ D(B) such that (B + Ꮽ)((u i) i=1,N)=
(− f i) i=1,N But this is the abstract form of (3.7) Thus the existence is proved
We show now that the difference of the two solutions (ui)i=1,Nand (v i) i=1,Nof (3.7) is
a constant Putw i = u i − v i, i =0,N + 1 Subtracting the corresponding equations of (2.3) foru iandv i, multiplying by a i w i, and summing from i =1 toi = N, one arrives with the
aid of the monotonicity ofA at
N
i=1
a iw i+1 − w i, w i
− a i θ iw i − w i−1,w i
or, in view of (2.6), at
N
i=1
a i
w i+1 − w i, w i
− a i−1
w i − w i−1,w i−1
≥
N
i=1
a i−1 w i − w i−1 2
By the boundary conditions in (2.3), we have
N
i=1
a i−1 w i − w i−1 2
≤ a N
w N+1 − w N, w N
−w1− w0,w0
≤0, (3.18)
sow0= w1= ··· = w N This implies that u i = v i+C, i =0,N, where C ∈ H is a constant.
IfA or α is one to one, then the uniqueness follows easily If A is maximal monotone and
strongly monotone, then we obtain
N
i=1
a iw i 2
+
N
i=1
a i−1 w i − w i−1 2
Trang 7Now we replace (3.4) by (3.2) and (3.3) and remove (3.5) and (3.6) Adding the bound-edness of the domain ofβ, we can state the following result.
Theorem 3.2 Let A, α, and β be maximal monotone operators in H such that D(β) is bounded and (3.1), (3.2), (3.3) hold If a,b ∈ H, (g i) i=1,N ∈ H N , and p i, r i, k i > 0, i =1,N, then problem (1.3), (1.4) admits at least one solution (u i) i=1,N ∈ D(A) N and the difference between two solutions is constant If A or α is one to one, then the solution is unique If A is also strongly monotone, then again uniqueness is obtained.
Proof We use again the equivalent form (2.3) of problem (1.3), (1.4) and the maximal monotone operatorᏭ given by (3.15) IfA λandᏭλare the Yosida approximations ofA
andᏭ, respectively, then Ꮽλ((u i) i=1,N)=(c1A λ u1, ,c N A λ u N) for all ( u i) i=1,N ∈ H N By
Proposition 2.2,B + Ꮽ λis maximal monotone inᏸ, therefore, R(B + Ꮽλ+λI) =ᏸ, that
is, for all (f i) i=1,N ∈ H N, for allλ > 0, the problem
u λ i+1 −1 +θ i
u λ
i+θ i u λ i−1= c i A λ u λ
i +λu λ
i+f i, i =1,N,
u λ
1− u λ
0∈ αu λ
0− a, u λ
N+1 − u λ
N ∈ − βu λ
N+1 − b, (3.20) has a unique solution (u λ
i)i=1,N ∈ H N (The uniqueness follows fromTheorem 2.3for the strongly monotone operatorᏭλ+λI.)
We first prove that (u λ
i)i=1,Nis bounded inH with respect to λ To do this, we multiply
(3.20) bya i u λ
i and sum up fromi =1 toi = N Without any loss of generality, suppose that
0∈ A0 If not, we put A= A + A00 and f i = f i − c i A00 instead ofA and f i, respectively,
whereA0x denotes the element of least norm in Ax Since A λis monotone,A λ0 =0, and
a i θ i = a i−1, we derive
N
i=1
a i
u λ i+1 − u λ
i,u λ
i
−
N
i=1
a i−1
u λ
i − u λ i−1,u λ i−1
≥
N
i=1
a i−1 u λ
i − u λ i−1 2
+λ N i=1
a iu λ
i 2
+
N
i=1
a i
f i, u λ
i
,
(3.21)
hence
N
i=1
a i−1 u λ
i − u λ
i−1 2
≤ a N
u λ N+1 − u λ
N,u λ
N
−u λ
1− u λ
0,u λ
0
−
N
i=1
a i
f i, u λ
i
. (3.22)
Sinceu λ
1− u λ
0∈ α(u λ
0− a), 0 ∈ α(0), and α is monotone, we infer
−u λ
1− u λ
0,u λ
0
≤ −u λ
1− u λ
0,a≤ a ·u λ
1− u λ
and, similarly,
u λ
N+1 − u λ
N,u λ
N
≤ −u λ N+1 − u λ
N 2
+
u λ N+1 − u λ
N,u λ N+1 − b+
u λ N+1 − u λ
N,b, (3.24) so
u λ N+1 − u λ
N,u λ
N
≤ b ·u λ
N+1 − u λ
Trang 8Now (3.22), (3.23), and (3.25) yield
N
i=1
a i−1 u λ
i − u λ i−1 2
≤ a N b ·u λ
N+1 − u λ
N+ a ·u λ
1− u λ
0
+
N
i=1
a if i 2
1/2 N
i=1
a iu λ
i 2
1/2
(3.26)
The hypothesis thatD(β) is bounded and the boundary conditions imply the
bound-edness ofu λ
N+1with respect toλ Using this, together with the estimates
u λ
k ≤ N
i=1
a iu λ
i 2
1/2
,
u λ
k − u λ k−1 ≤ N
i=1
a i−1 u λ
i − u λ i−1 2
fork =1,N in (3.26), one deduces
N
i=1
a i−1 u λ
i − u λ
i−1 2
≤ C1+C2
N
i=1
a i−1 u λ
i − u λ i−1 2
1/2
+C3
N
i=1
a iu λ
i 2
1/2
, (3.28)
withC1,C2,C3> 0 independent of λ.
For eachi =1,N, we have u λ
i = u λ
0+i k=1(u λ
k − u λ k−1), so
u λ
i ≤ u λ
0 + N
k=1
1
a k−1
N
k=1
a k−1 u λ
k − u λ k−1 2
1/2
, i =1,N. (3.29)
From the boundary conditions, it follows that
u λ
0 2
≤ a ·u λ
0 + a N
i=1
a i−1 u λ
i − u λ i−1 2
1/2
and thus
u λ
0 ≤ a + a 1/2
N
i=1
a i−1 u λ
i − u λ i−1 2
1/4
Inequalities (3.29) and (3.31) imply that
u λ
i ≤ C4+C5
N
i=1
a i−1 u λ
i − u λ i−1 2
1/2
which, together with (3.28), leads to the boundedness
N
i=1
a i−1 u λ
i − u λ i−1 2
Trang 9Now (3.31) and (3.32) show that u λ
i ≤ C7,i =0,N and λ > 0, and therefore, N
i=1
a iu λ
i 2
All the constantsC j > 0 (j =1, ,13) here and below are independent of λ.
Multiplying (3.20) bya i A λ u λ
i and summing from 1 toN, we get via (2.6)
N
i=1
a i
u λ
i+1 − u λ
i,A λ u λ
i
− a i−1
u λ
i − u λ i−1,A λ u λ i−1
−
N
i=1
a i−1
u λ
i − u λ i−1,A λ u λ
i − A λ u λ i−1
=
N
i=1
a i c iA λ u λ
i 2
+λ N i=1
a iu λ
i,A λ u λ
i
+
N
i=1
a if i, A λ u λ
i.
(3.35)
Letc =inf{ c i, i =1,N } Then
c N
i=1
a iA λ u λ
i 2
≤ a N
u λ N+1 − u λ
N,A λ u λ
N
−u λ
1− u λ
0,A λ u λ
0
−
N
i=1
a i
f i, A λ u λ
i
. (3.36)
We observe that assumptions (3.2) and (3.3) and the boundary conditions yield
u λ N+1 − u λ
N,A λ u λ
N
≤A0b · u λ
N+1 − u λ
N,
−u λ
1− u λ
0,A λ u λ
0
≤A0a · u λ
1− u λ
therefore, (3.36) implies
c
N
i=1
a iA λ u λ
i 2
≤ a NA0b · u λ
N+1 − u λ
N+A0a · u λ
1− u λ
0
+
N
i=1
a if i 2
1/2 N
i=1
a iA λ u λ
i 2
1/2
(3.38)
In view of (3.29), (3.31), and the boundedness ofu λ
N+1, this means that
N
i=1
a iA λ u λ
i 2
≤ C9+C10
N
i=1
a iA λ u λ
i 2
1/2
+C11
N
i=1
a i−1 u λ
i − u λ i−1 2
1/2 (3.39)
According to (3.33), this leads to
N
i=1
a iA λ u λ
i 2
Trang 10We prove now thatu λ
i − u λ i−1is a Cauchy sequence with respect toλ Subtracting (3.20) withν in place of λ from the original equation (3.20), multiplying the result bya i( u λ
i −
u ν
i), and summing up fromi =1 toi = N, we find, with the aid of the equality x = J λ x +
λA λ x,
N
i=1
a i
u λ
i+1 − u ν
i+1 − u λ
i +u ν
i,u λ
i − u ν
i
−
N
i=1
a i−1
u λ
i − u ν
i − u λ i−1+u ν i−1,u λ i−1− u ν i−1
=
N
i=1
a i−1 u λ
i − u ν
i − u λ i−1+u ν i−1 2
+
N
i=1
a i c i
A λ u λ
i − A ν u ν
i,J λ u λ
i − J ν u ν
i
+
N
i=1
a i c i
A λ u λ
i − A ν u ν
i,λA λ u λ
i − νA ν u ν
i
+
N
i=1
a i
λu λ
i − νu ν
i,u λ
i − u ν
i
, (3.41)
hence
N
i=1
a i−1 u λ
i − u ν
i − u λ i−1+u ν i−1 2
≤ a Nu λ
N+1 − u ν N+1 − u λ
N+u ν
N,u λ
N − u ν
N
−u λ
1− u ν
1− u λ
0+u ν
0,u λ
0− u ν
0
+ (λ + ν) N
i=1
a i c i
A λ u λ
i,A ν u ν
i
+ (λ + ν) N
i=1
a i
u λ
i,u ν
i
.
(3.42)
The boundary conditions in (3.20) and the upper bounds (3.34) and (3.40) imply
N
i=1
a i−1 u λ
i − u ν
i − u λ i−1+u ν i−1 2
and therefore,u λ
i − u λ
i−1is a strongly convergent sequence inH.
Letu λ
i u i, i =1,N (on a subsequence denoted again by λ) Then u λ
i − u λ i−1→ u i −
u i−1, soB((u λ
i)i=1,N)→ B((u i)i=1,N) In addition, we have J λ u λ
i(= u λ
i − λA λ u λ
i) u iasλ →
0,i =1,N.
SinceA is demiclosed, this enables us to pass to the limit as λ →0 in (3.20) written under the form
− B u λ
i
i=1,N − λu λ
i
i=1,N −f i
i=1,N ∈Ꮽ J
λ u λ
i
i=1,N , (3.44) and one obtains that (u i) i=1,N verifies problem (2.3) The uniqueness follows like in
We now replace the boundedness ofD(β) by the conditions
inf
−(y,x)
inf
(y,x)
We get the following result
... maximal monotone in ᏸ.Recall here an existence theorem from [7], which we use in the sequel
Theorem 2.3 Assume that A, α, and β are maximal monotone operators in H with ∈... class="page_container" data-page="7">
Now we replace (3.4) by (3.2) and (3.3) and remove (3.5) and (3.6) Adding the bound-edness of the domain ofβ, we can state the following result.... =0,N + Subtracting the corresponding equations of (2.3) foru iand< i>v i, multiplying by a i w i, and summing from i