Solving Composite Fixed Point Problems with Block Updates

Problem 1.1

Let m be a strictly positive integer and let (ω_i)_1⩽i⩽m ∈ ]0, 1]^m be such that ∑i=1m ω_i = 1. For every i ∈ {0, …, m}, let T_i : 𝓗 → 𝓗 be α_i-averaged for some α_i ∈ ]0, 1[. The task is to find a fixed point of T₀ ∘ ∑i=1m ω_i T_i.

Definition 2.1

[15, Definition 2.1] A triangular array (μ_n,j)_{n∈ℕ,0⩽j⩽n} in [0, + ∞[ is concentrating if the following hold:

1. (∀ n ∈ ℕ) ∑j=0n μ_n,j = 1.

2. (∀ j ∈ ℕ) lim_n→+∞μ_n,j = 0.

3. Every sequence (ξ_n)_n∈ℕ in [0, + ∞[ that satisfies

   (∀n∈N)ξn+1⩽∑j=0nμn,jξj+εn, (2.1)

   for some summable sequence (ε_n)_n∈ℕ in [0, + ∞[, converges.

Lemma 2.2

Let C be a nonempty subset of 𝓗, let ϕ : [0, + ∞[ → [0, + ∞[ be strictly increasing and such that lim_t→+∞ ϕ(t) = + ∞, let (x_n)_n∈ℕ be a sequence in 𝓗, let (μ_n,j)_{n∈ℕ,0⩽j⩽n} be a concentrating array in [0, + ∞[, let (β_n)_n∈ℕ be a sequence in [0, + ∞[, and let (ε_n)_n∈ℕ be a summable sequence in [0, + ∞[ such that

Then the following hold:

1. (x_n)_n∈ℕ is bounded.

2. β_n → 0.

3. Suppose that every weak sequential cluster point of (x_n)_n∈ℕ belongs to C. Then (x_n)_n∈ℕ converges weakly to a point in C.

4. Suppose that (x_n)_n∈ℕ has a strong sequential cluster point in C. Then (x_n)_n∈ℕ converges strongly to a point in C.

Proof

Let x ∈ C. Let us first show that

It follows from (2.2) and Definition 2.1 that (ϕ(∥x_n − x∥))_n∈ℕ converges, say ϕ(∥x_n − x∥) → λ. However, since lim_t→+∞ ϕ(t) = + ∞, (∥x_n − x∥)_n∈ℕ is bounded and, to establish (2.3), it suffices to show that it does not have two distinct cluster points. Suppose to the contrary that there exist subsequences (∥x_kn − x∥)_n∈ℕ and (∥x_ln − x∥)_n∈ℕ such that ∥x_kn − x∥ → η and ∥x_ln − x∥ → ζ > η, and fix ε ∈ ]0, (ζ − η)/2[. Then, for n sufficiently large, ∥x_kn − x∥ ⩽ η + ε < ζ − ε ⩽ ∥x_ln − x∥ and, since ϕ is strictly increasing, ϕ(∥x_kn − x∥) ⩽ ϕ(η + ε) < ϕ(ζ − ε) ⩽ ϕ(∥x_ln − x∥). Taking the limit as n → + ∞ yields λ ⩽ ϕ(η + ε) < ϕ(ζ − ε) ⩽ λ, which is impossible.

1. and (iv): Clear in view of (2.3).

2. As shown above, there exists λ ∈ [0, + ∞[ such that ϕ(∥x_n − x∥) → λ. In turn, [28, Theorem 3.5.4] implies that ∑j=0n μ_n,jϕ(∥x_{j-x∥) → λ. We thus derive from (2.2) that 0 ⩽ βn ⩽ ∑j=0n μn,jϕ(∥xj − x∥) − ϕ(∥xn+1} − x∥) + ε_n → 0.

3. This follows from (2.3) and [6, Lemma 2.47]. □

Proposition 2.3

Let K be a strictly positive integer and let (μ_n,j)_{n∈ℕ,0⩽j⩽n} be a triangular array in [0, + ∞[ such that the following hold:

1. (∀ n ∈ ℕ) ∑j=0n μ_n,j = 1.

2. (∀ n ∈ ℕ)(∀ j ∈ ℕ) n − j ⩾ K ⇒ μ_n,j = 0.

3. inf_n∈ℕ μ_n,n > 0.

Then (μ_n,j)_{n∈ℕ,0⩽j⩽n} is a concentrating array.

Proof

Properties [a] and [b] in Definition 2.1 clearly hold. To verify [c], let (ξ_n)_n∈ℕ be a sequence in [0, + ∞[ and let (ε_n)_n∈ℕ be a summable sequence in [0, + ∞[ such that

Then, in view of (ii), for every integer n ⩾ K − 1,

Set μ = inf_n∈ℕμ_n,n. If μ = 1, then (i) and (2.5) imply that, for every integer n ⩾ K − 1,

and the convergence of (ξ_n)_n∈ℕ therefore follows from [6, Lemma 5.31]. We henceforth assume that μ < 1 and, without loss of generality, that K > 1. For every integer n ⩾ K − 1, define ξ̂_n = max_{0⩽k⩽K−1} ξ_n−k, and observe that (i) and (2.5) yield ξ_n+1 ⩽ ξ̂_n + ε_n. Hence,

and we deduce from [6, Lemma 5.31] that (ξ̂_n)_n∈ℕ converges to some number η ∈ [0, + ∞[. Therefore, if (ξ_n)_n∈ℕ converges, then its limit is η as well. Let us argue by contradiction by assuming that ξ_n ↛ η. Then there exists ν ∈ ]0, + ∞[ such that

Set

Since ξ̂_n → η and ∑_n∈ℕ ε_n < + ∞, let us fix an integer N ⩾ K − 1 such that

Then

while (2.5) and (i) imply that

It follows from (2.8) that there exists an integer n₀ ⩾ N such that |ξ_n0 − η| > ν, i.e.,

Suppose that ξ_n0 > η+ν. Then (2.9) and (2.10) imply that ν < ξ_n0 − η ⩽ ξ̂_n0 − η ⩽ ν^′ ⩽ ν/4, which is impossible. Therefore, 0 ⩽ ξ_n0 < η − ν and it follows from (2.12) that

Let us show by induction that, for every integer k ⩾ 1,

In view of (2.14), this inequality holds for k = 1. Now suppose that it holds for some integer k ⩾ 1. Then we deduce from (2.12) and (2.15) that

which completes the induction argument. Since μ ∈ ]0, 1[, we derive from (2.15), (2.11), and (2.9) that

Therefore, by (2.10),

We thus reach a contradiction and conclude that (ξ_n)_n∈ℕ converges. □

Example 2.4

Let I be a nonempty finite set, let (ω_i)_i∈I be a family in ]0, 1] such that ∑_i∈I ω_i = 1, let (I_n)_n∈ℕ be a sequence of nonempty subsets of I, and let K be a strictly positive integer such that (∀ n ∈ ℕ) ⋃_{0⩽k⩽K−1} I_n+k = I. Set

Then the following hold:

1. (μ_n,j)_{n∈ℕ,0⩽j⩽n} is a concentrating array.

2. Let ℕ ∋ n ⩾ K − 1, let (ξ_j)_0⩽j⩽n be in [0, + ∞[, and, for every i ∈ I, define ℓ_i(n) = max{k ∈ {n − K + 1, …, n} | i ∈ I_k. Then

   ∑j=0nμn,jξj=∑i∈Iωiξℓi(n). (2.20)

Proof

Let n ∈ ℕ. If n ⩾ K − 1, we have ⋃_{0⩽k⩽K−1} I_n−k = I and therefore

I is the union of the disjoint sets

1. It is clear from (2.19) that, for every integer j ∈ [0, n − K], μ_n,j = 0. In turn, we derive from (2.19) and (2.21) that

   ∑j=0nμn,j=μn,n=1,ifn<K;∑j=0nμn,j=∑j=n−K+1nμn,j=∑j=n−K+1n∑i∈Ij∖⋃k=j+1nIkωi=∑i∈Iωi=1,ifn⩾K. (2.22)

   Finally, inf_n∈ℕ μ_n,n = inf_n∈ℕ ∑_i∈In ω_i ⩾ min_i∈I ω_i > 0. All the properties of Proposition 2.3 are therefore satisfied.

2. We have

   (∀j∈{n−K+1,…,n})(∀i∈Ij∖⋃k=j+1nIk)ℓi(n)=j. (2.23)

   Hence, in view of (2.19),

   (∀j∈{n−K+1,…,n})∑i∈Ij∖⋃k=j+1nIkωiξℓi(n)=∑i∈Ij∖⋃k=j+1nIkωiξj=μn,jξj. (2.24)

   Consequently, (2.21) yields

   ∑j=0nμn,jξj=∑j=n−K+1n∑i∈Ij∖⋃k=j+1nIkωiξℓi(n)=∑i∈Iωiξℓi(n), (2.25)

   which concludes the proof. □

Assumption 3.1

K is a strictly positive integer and (I_n)_n∈ℕ is a sequence of nonempty subsets of {1, …, m} such that

For every integer n ⩾ K − 1, define

The sequences (e_0,n)_n∈ℕ, (e_1,n)_n∈ℕ, …, (e_m,n)_n∈ℕ are in 𝓗 and satisfy

Theorem 3.2

Consider the setting of Problem 1.1 together with Assumption 3.1. Let ε ∈ ]0, 1[ and, for every n ∈ ℕ and every i ∈ {0} ∪ I_n, let α_i,n ∈ ]0, 1/(1 + ε)[ and let T_i,n : 𝓗 → 𝓗 be α_i,n-averaged. Suppose that, for every integer n ⩾ K − 1,

Let x₀ ∈ 𝓗, let (t_i,−1)_1⩽i⩽m ∈ 𝓗^m, and iterate

Let x be a solution to Problem 1.1. Then the following hold:

1. (x_n)_n∈ℕ is bounded.

2. Let i ∈ {1, …, m}. Then x_ℓi(n) − T_i,ℓi(n)x_ℓi(n) + T_i,ℓi(n) x − x → 0.

3. Let i ∈ {1, …, m} and j ∈ {1, …, m}. Then T_i,ℓi(n)x_ℓi(n) − T_j,ℓj(n) x_ℓj(n) − T_i,ℓi(n) x + T_j,ℓj(n) x → 0.

4. Let i ∈ {1, …, m}. Then x_ℓi(n) − x_n → 0.

5. x_n − T_0,n( ∑i=1m ω_i T_i,ℓi(n)x_n) → 0.

6. Suppose that every weak sequential cluster point of (x_n)_n∈ℕ solves Problem 1.1. Then the following hold:

   1. (x_n)_n∈ℕ converges weakly to a solution to Problem 1.1.

   2. Suppose that (x_n)_n∈ℕ has a strong sequential cluster point. Then (x_n)_n∈ℕ converges strongly to a solution to Problem 1.1.

7. For every n ⩾ K − 1 and every i ∈ {0} ∪ I_n, let ρ_i ∈ ]0, 1] be a Lipschitz constant of T_i,n. Suppose that (3.5) is implemented without errors and that, for some i ∈ {0, …, m}, ρ_i < 1. Then (x_n)_n∈ℕ converges linearly to the unique solution to Problem 1.1.

Proof

Let us fix temporarily an integer n ⩾ K − 1. We first observe that, by nonexpansiveness of the operators T_0,n and (T_i,ℓi(n))_1⩽i⩽m,

We also note that (3.2) and (3.5) yield

It follows from (3.5), (3.7), (3.4), and (3.6) that

Now define (μ_k,j)_{k∈ℕ,0⩽j⩽k} as in (2.19), with I = {1, …, m}, and set ε_n = ∥e_0,n∥ + ∑i=1m ∥e_i,ℓi(n)∥. Then we derive from Example 2.4(ii) that

and it follows from (3.8) and (3.3) that

Hence, Lemma 2.2(i) guarantees that

Consequently, using (3.3) and (3.6), we obtain

and

In addition, for every y ∈ 𝓗 and every z ∈ 𝓗, it follows from [6, Proposition 4.35] that

and, likewise, that

Hence, we deduce from (3.5), (3.7), (3.4), and [6, Lemma 2.14(ii)] that

It therefore follows from (3.9) that

Hence, Example 2.4(i), (3.3), and Lemma 2.2(ii) imply that

and that