For any given z ∈ S, we get (3.7).
For any given z ∈ E ( b ) and any given positive integer k ≥ N 1, there are the following two different cases.
In fact, for any given z ∈ Γ, it follows from Lemma 3.1 that G ( z ) = 0, K i z = z and z = K i ( I − λ n, i G ∗ G ) z for each i ≥ 1.
Note that, for any given z ∈ S, z = T z and 〈 Proj C y s − y s, Proj C y s − z 〉 ≤ 0. By (3.3), it follows that (3.9).
Then for any given z τ ∈ E 0, there is a unique solution z = ( u, u t ) for the problem (2.8) in E 0. Furthermore, for i = 1, 2, let { z τ i, h i } ( z τ i ∈ E 0 and h i ∈ L b 2 ( R ; H ) ) be two initial conditions, and denote by z i the corresponding solutions to the problem (2.8).
In fact, for any given z ∈ Ω, it follows from Lemma 3.1, Lemma 3.2, and condition (iv) that z = J μ ( U i, K i ) ( I − λ n, i G ∗ G ) z, for each i ≥ 1, and ( I − λ n, i G ∗ G ) : H 1 × H 2 → H 1 × H 2 is a nonexpansive mapping.
The ex-post π can be obtained recursively via (11) for a given z, because any unit cost function is strictly concave with respect to its argument (in this case, π).
Of course, for a given z with a.e.
For a given z 0 ∈ H, compute z n + 1 by the iterative schemes u n = P K r z n, (25).
For a given z ∈ H and u ∈ C, u = P C z ⇔ 〈 u − z, v − u 〉 ≥ 0, ∀ v ∈ C. Lemma 2.4 Let H be a real Hilbert space.
For a given Z ∈ P ∗, R o ( Z ) ⊆ R i r ( Z ) if and only if I ( Y p ; X i | WQ ) ≤ I ( Y s ; X i | UWQ ).
