Bregman Distance and Strong Convergence of Proximal-Type Algorithms

Proposition 1. Let X be a real Banach space.

• (i)

  J_φ is norm-to-weak* upper semicontinuous;

• (ii)

  for each x in X, the set J_φ(x) is convex and weakly closed in X^*;

• (iii)

  J_φ(−x) = −J_φ(x) and J_φ(λx) = (φ(∥λx∥)/φ(∥x∥))J_φ(x) for all nonzero x in X,    λ > 0;

• (iv)

  there holds

  $$ ()

• (v)

  J_φ is maximal monotone;

• (vi)

  if X is strictly convex, then J_φ is strictly monotone; that is,

  $$ ()

• (vii)

  if X is strictly convex and reflexive, then J_φ is single-valued monotone and demicontinuous.

Lemma 2. If a Banach space E has a uniformly Gâteaux differentiable norm, then J : E → E^* is uniformly norm-to-weak* continuous on nonempty bounded subsets of E to E^*.

Proof. Suppose not, and there exist norm one vectors x_n, y_n, z in E and a constant ϵ > 0 such that y_n − x_n → 0, and 〈z, J(y_n) − J(x_n)〉≥ϵ, for all n ∈ ℕ. For a fixed t > 0, define

Observe that Hence, Choose t = (2/ϵ)∥x_n − y_n∥. By the uniform Gâteaux differentiability of the norm, if n is large enough, both a_n and b_n are less than (1/2)ϵ, and so a_n + b_n < ϵ. We arrive at a contradiction.

Lemma 3. Let X be a smooth Banach space and J_φ : X → X^* the duality mapping with gauge φ. Then J_φ(y) = ∂Φ(∥y∥) for y ∈ X∖{0}; that is,

Lemma 4. Let X be a real Banach space and φ a gauge function.

• (a)

  Φ(∥·∥) is a convex and continuous function on X.

• (b)

  X is strictly convex if and only if Φ(∥·∥) is strictly convex.

Lemma 5. Let X be a real Banach space and φ a gauge function. Then the following assertions are equivalent:

• (a)

  X is uniformly convex;

• (b)

  Φ(∥·∥) is uniformly convex on the closed ball B_r : = {x ∈ X : ∥x∥ ≤ r}, where r > 0 is arbitrarily given. That is, there exists a strictly increasing convex function g_r : ℝ⁺ → ℝ⁺ with g_r(0) = 0 such that

  $$ ()

•   

  for all x, y in B_r and t in [0,1].

Proof. First we note that the general case can be reduced to the case r = 1. Suppose it holds

We will show that (20) holds for any r > 0. Set φ_r(t) = φ(rt) and g_r(t) = rg(t/r) for t ≥ 0. Then φ_r is still a gauge function, and let Φ_r be the function corresponding to φ_r as defined in (4), rΦ_r(t) = Φ(rt). Let x, y ∈ B_r and x_r = x/r and y_r = y/r. Then x_r, y_r ∈ B_X. Applying (21) to Φ_r, we get which is exactly (20). A similar argument also shows that the case r = 1 can be deduced from any other case r > 0.

Below, we assume r = 1 and g = g₁.

(b)⇒(a). Given x, y in X such that ∥x∥ = ∥y∥ = 1 and ∥x − y∥ = ε. Setting t = 1/2 in (20), we have

It turns out that This verifies that X is uniformly convex.

(a)⇒(b). Assume that X is uniformly convex, which implies that Φ(∥·∥) is strictly convex by Lemma 4(b). Define a function μ on [0,2] by setting μ(0) = 0, and for 0 < ε ≤ 2,

where B_X is the closed unit ball of X.

Claim. Consider μ(ε) > 0 for 0 < ε ≤ 2.

Suppose on the contrary that μ(ε) = 0 for some 0 < ε ≤ 2. Then we can find sequences {x_n}, {y_n} in B_X such that ∥x_n − y_n∥≥ε for all n, and

Without loss of generality, we may assume that It then follows from (26) that The strict convexity of Φ together with (28) implies that α + β < γ if α ≠ β. Since, on the other hand, by definition, γ ≤ α + β. We therefore must have α = β, which together with (28) implies that α = β = γ/2. If we set $$, $$, then $$ for all n; moreover, from (27), we get This contradicts the uniform convexity of X, and verifies that μ(ε) > 0 for all 0 < ε ≤ 2.

It turns out from (25) that

for all x, y in B_X, with g = 4μ.

By the dyadic rational argument used in the proof of [15, Theorem 2.2], we can extend the inequality (30) to the case of a general convex combination of x and y, namely,

for x, y in B_X and t in [0,1]. Note that g is increasing and continuous. By [16], the function g can also be assumed to be convex (the convexity of g is not needed in our argument throughout the rest of this paper however).

Lemma 6. Let X be a real uniformly convex Banach space. Then there exists a strictly increasing convex function g : ℝ⁺ → ℝ⁺ with g(0) = 0 such that

Proof. Since J_φ is the subdifferential of the functional Φ(∥·∥), we have for j_x in J_φ(x), x in X that

Let g be the function that satisfies (20) with r = 2 and assume that x, y ∈ B_X. Replacing y with x + λy,  0 < λ < 1, we obtain Taking limit as λ → 0, we get

Lemma 7 (see [17], Theorem 3.11, page 952.)Let X be a reflexive Banach space and $$ the duality map with gauge φ. Suppose that $$ is maximal monotone. Then the operator A + J_φ is surjective.

Proposition 8. Let X be a strictly convex smooth Banach space. Let x, y ∈ X. Then

Proof. See [18, Lemma 7.3(vi)]

Proposition 9. Let X be a smooth and uniformly convex Banach space. Then there exists a strictly increasing convex function g : ℝ⁺ → ℝ⁺ with g(0) = 0 such that

Proof. Let u, v ∈ B_X. By Lemma 6 we have

Proposition 10. Let X be a smooth and uniformly convex Banach space. Let {x_n} and {y_n} be two sequences in X such that D_φ(x_n, y_n) → 0. If {y_n} is bounded, then ∥x_n − y_n∥ → 0.

Proof. Assume {y_n} is bounded. From definition (5) we have

It follows that Since {y_n} is bounded, D_φ(x_n, y_n) → 0 and Φ(t)/t → ∞ as t → ∞ it follows from (41) that {∥x_n∥} is bounded, too.

We may now assume that {x_n} and {y_n} both lie in the closed unit ball B_X (otherwise consider the rescaled sequences {γx_n} and {γy_n} for a sufficiently small γ > 0). By Proposition 9, there exists a strictly increasing convex function g : ℝ⁺ → ℝ⁺ with g(0) = 0 such that

Since D_φ(x_n, y_n) → 0 and g is strictly increasing, we immediately conclude that ∥x_n − y_n∥→0.

Proposition 11 (see [20], Lemma 3.1.)Let X be a smooth Banach space. Let u ∈ X and {x_n} be a sequence in X such that {D_φ(x_n, u)} is bounded. Then {x_n} is bounded.

Proposition 12. Let X be a smooth Banach space. Then, for any fixed x in X, the scalar function D_φ(·, x) is continuous, weakly lower semicontinuous, and convex on X.

Proposition 13. Let C be a nonempty closed convex subset of a reflexive, strictly convex, and smooth Banach space X. Let x ∈ X and let f : X → [0, +∞] be a proper, convex, lower semicontinuous function with C ⊂ dom (f). Then there exists a unique element x₀ in C such that

Definition 14. In the setting of Proposition 13, we define the (φ, f)-generalized projection from X onto C by

In case φ(t) = t and f = 0, we notice that $$ coincides with Π_C. In case that X is a Hilbert space and φ(t) = t, denote $$ by $$.

Proposition 15. Let C be a nonempty closed convex subset of a reflexive, strictly convex, and smooth Banach space X and let f : X → [0, +∞] be a proper, convex, lower semicontinuous function with C ⊂ dom (f).

• (i)

  Let x₀ ∈ C and x ∈ X. Then the following assertions are equivalent:

  • (a)

    $$,

  • (b)

    〈z − x₀, J_φx₀ − J_φx〉 + f(z) − f(x₀) ≥ 0,

  •   

    ∀z ∈ C.

• (ii)

  Given x in X, one has

  $$ ()

Proposition 16. Let C be a nonempty closed convex subset of a smooth and uniformly convex Banach space X and f : X → [0, +∞] a proper, convex, lower semicontinuous function with C ⊂ dom (f). Let u ∈ X and let {x_n} be a bounded sequence in C such that lim _n→+∞f(x_n) = 0. If $$ for all n in ℕ, then $$.

Proof. Set $$. Then, by assumption, we have $$ for all n in ℕ. By the three-point identity (36), we have

Since {x_n} is a bounded sequence in C, there exists a subsequence $$ of {x_n} such that $$ converges weakly to some element z in C. Note that that is, f(z) = 0. Using Proposition 15, we have which implies that $$. It follows from Proposition 10 that $$. This implies that x_n converges strongly to $$.

Definition 17. Let X be a smooth Banach space, J_φ : X → X^* a duality mapping with gauge function φ, and C a nonempty subset of X. An operator T : C → X is called φ-firmly nonexpansive if

In the case of φ(t) = t, inequality (50) reduces to

If T satisfies condition (51), we call T of firmly nonexpansive type. The class of firmly nonexpansive type operators is studied by Kohsaka and Takahashi [28]. When X is a Hilbert space, inequality (50) reduces to the following inequality about firmly nonexpansive operators in the classical sense (see Goebel and Kirk [29]):

We now give useful characterizations of φ-firmly nonexpansive mappings which can be deduced from the Bregman distance (5).

Proposition 18. Let X be a smooth Banach space and C a nonempty closed convex subset of X. Let T : C → C be a φ-firmly nonexpansive mapping. Then

Proposition 19. Let X be a strictly convex smooth Banach space and C a nonempty closed convex subset of X. Let T : C → C be a φ-firmly nonexpansive mapping. Then the set F(T) of fixed points of T is closed and convex.

Proposition 20. Let X be a smooth Banach space, C a nonempty subset of X, and T : C → C a φ-firmly quasinonexpansive mapping. Then

for all x in C and p in F(T).

Proposition 21. Let X be a strictly convex Banach space with a uniformly Gâteaux differentiable norm. Let C be a nonempty closed convex subset of X and let T : C → C be a φ-firmly nonexpansive mapping. Then $$.

Proof. It is easy to see that $$. It remains to prove that $$. For this, suppose that $$. Then, there exists a sequence {x_n} in C such that x_n⇀u and ∥x_n − Tx_n∥ → 0. We need to prove that u ∈ F(T). Using (53), we get

It is not hard to find that (56) is reduced to the relation We note that Tx_n⇀u. Indeed, for any norm one linear functional ξ of X, we have Both terms in the right-hand side approach zero. Consequently,

Claim. Consider 〈Tx_n − Tu, J_φ(x_n) − J_φ(Tx_n)〉→0.

Since x_n⇀u and ∥x_n − Tx_n∥ → 0, there is a constant M > 0 such that all ∥x_n∥, ∥Tx_n∥ < M. It follows from the uniform norm to weak* continuity of J_φ on bounded subsets (Lemma 2); we have

Observe that Moreover, |∥x_n∥ − ∥Tx_n∥| ≤ ∥x_n − Tx_n∥ → 0. By the uniform continuity of φ on [0, M], we have The claim is thus verified.

We obtain from (57) and the claim that

This is equivalent to The strict monotonicity of the duality mapping J_φ implies that equality must hold. Namely, Tu = u or u ∈ F(T).

Definition 22. Let C be a nonempty closed convex subset of a smooth Banach space X and let J_φ : X → X^* be the duality mapping with gauge φ. Suppose that $$ is an operator satisfying the range condition

For each λ > 0, the φ-resolvent associated with operator A is the operator $$ defined by

For x in C and λ in (0, ∞), we have

If A is maximal monotone, then, by Lemma 7, we see that condition (65) holds for $$.

Remark 23. For smooth X and φ(t) = t^p−1 with p ∈ (1, +∞), we have J_φ = J_p and R^p,A = (J_p + A) ⁻¹J_p. For p = 2,  R^A : = R^2,A = (J + A) ⁻¹∘J and this kind of resolvent operators is studied in the literature (see [28, 33]).

Let C be a nonempty closed convex subset of a reflexive, strictly convex, and smooth Banach space X. Let $$ be a monotone operator satisfying the condition $$, where λ > 0. Using the smoothness and strict convexity of X, we obtain that $$ is single-valued. The conditions $$ ensure that $$ is the single-valued φ-resolvent operator form C into D(A). In other words,

Following [18, 28], we have the following proposition.

Proposition 24. Let C be a nonempty closed convex subset of a reflexive, strictly convex and smooth Banach space X and let J_φ : X → X^* be the duality mapping with gauge φ. Let $$ be a monotone operator satisfying the condition $$, where λ is a positive real number. Let $$ be a resolvent of A, where

• (a)

  $$ is φ-firmly nonexpansive mapping from C into C,

• (b)

  $$.

Remark 25. If 𝒯 is a singleton, that is, 𝒯 = {T}, or T(s) = T for all s > 0, then {T} always has property (𝒜).

Example 26. Let C be a nonempty closed convex subset of a Banach space X and T a nonexpansive mapping from C into C with F(T) ≠ ∅. Assume b_t in ℝ with 0 < a ≤ b_t ≤ b < 1 for all t > 0. Define G_t : C → C by G_tx = (1 − b_t)x + b_tTx for all x in C. Then T has property (𝒜) with respect to the family {G_t : t > 0}.

Proof. Let {x_t} _t>0 be a bounded net in C such that ∥x_t − G_t(x_t)∥→0 as t → ∞. Note that

and 0 < a ≤ b_t ≤ b < 1 for all t > 0. Therefore, ∥x_t − Tx_t∥→0 as t → +∞.

Example 27. Let C be a nonempty closed convex subset of a real Hilbert space H and let A ⊂ H × H be a monotone operator satisfying the following condition:

Let {z_t} _t>0 be a bounded net in C such that $$ as t → +∞. Then $$ as t → +∞ for each r > 0.

Proof. Let r, t > 0. By Takahashi [34], we have

Using (72), we have

Theorem 28. Let X be a uniformly convex Banach space with a uniformly Gâteaux differentiable norm. Let φ be a gauge function, C a nonempty closed convex subset of X, and f : X → [0, +∞] a proper, convex, lower semicontinuous function with C ⊂ dom (f). Let T : C → C be a φ-firmly nonexpansive mapping and 𝒯 : = {T_n} a sequence of φ-firmly nonexpansive self-mappings on C such that F(T) = ⋂_n∈ℕ F(T_n) ≠ ∅ and 𝒯 has property (𝒜) with respect to T. For u in C and C₁ = C with $$, define a sequence {x_n} in C as follows:

Then {x_n} converges strongly to $$.

Proof. We proceed the proof in the following steps:

Step 1. {x_n} is well defined.

Note that all C_n are closed and convex. For p in F(𝒯) and n in ℕ, we obtain from (54) that

It follows that p ∈ C_n and hence F(𝒯) ⊂ C_n. Therefore, {x_n} is well defined.

Step 2. {x_n} is bounded.

Let p ∈ F(𝒯). It follows from Proposition 15; we have

It follows that {D_φ(x_n, u)} is bounded and hence from Proposition 11, we obtain that {x_n} is bounded.

Step 3. Consider ∥x_n − Tx_n∥→0.

Note that x_n+1 ∈ C_n+1 ⊂ C_n. It follows from Proposition 15 that

This implies that Therefore, the sequence $$ (see (43)) is increasing. Note that $$ is bounded by (76). It follows that $$ exists. By (78), we obtain One can see that lim _n→+∞f(x_n) = 0. Using Proposition 10, we obtain that lim _n→+∞∥x_n − x_n+1∥ = 0. Since x_n+1 ∈ C_n+1, we have Using (79), we obtain that and hence lim _n→+∞D_φ(y_n, x_n) = 0. Note that {x_n} is bounded. Then, one can see from Proposition 10 that {y_n} is bounded and This implies that Since the family 𝒯 : = {T_n : n ∈ ℕ} has property (𝒜) with respect to T, it follows from (83) that lim _n→+∞∥x_n − Tx_n∥ = 0.

Step 4. The sequence {x_n} converges strongly to $$.

Since {x_n} is bounded, there exists a subsequence $$ of {x_n} such that $$. Hence $$. From Proposition 19, F(T) is closed and convex. The nonemptiness of F(T) implies that the generalized projection $$ is well defined. Note that $$ and F(T) is contained in C_n; we have

Therefore, we conclude from Proposition 16 that {x_n} converges strongly to $$.

Theorem 29. Let X be a uniformly convex Banach space with a uniformly Gâteaux differentiable norm. Let φ be a gauge function, C a nonempty closed convex subset of X, and f : X → [0, +∞] a proper, convex, lower semicontinuous function with C ⊂ dom (f). Let $$ be a monotone operator with A⁻¹0 ≠ ∅ satisfying the following condition:

Let λ be a positive real number; for u in C and C₁ = C with $$, define a sequence {x_n} in C as follows: Then {x_n} converges strongly to $$.

Proof. Set T : = (J_φ + λA) ⁻¹J_φ. Note that T is φ-firmly nonexpansive mapping from C into C and F(T) = A⁻¹0. Further, every singleton family {T} enjoys property (𝒜). Therefore, Theorem 29 follows from Theorem 28.

Remark 30. Compared with other convergence theorems concerning proximal point algorithms in the literature (see, e.g., Agarwal et al. [35, Theorem 3.1]; Kamimura and Takahashi [26, Theorem 8]; Matsushita and Takahashi [32, Theorem 4.3]), Theorem 29 establishes a new proximal point algorithm for the problem of finding zeros of (not necessarily maximal) monotone operators in a uniformly convex Banach space with a uniformly Gâteaux differentiable norm.

Corollary 31. Let C be a nonempty closed convex subset of real Hilbert space H, and let f : X → [0, +∞] be a proper, convex, lower semicontinuous function with C ⊂ dom (f). Let A ⊂ H × H be a monotone operator satisfying condition (71) such that A⁻¹0 ≠ ∅. Let {λ_n} be a sequence in (0, ∞) such that lim _n→+∞λ_n = +∞, for u in C, let C₁ = C with $$, and define a sequence {x_n} in C as follows:

Then {x_n} converges strongly to $$.

Proof. Set $$ for λ > 0 and T_n : = (I + λ_nA) ⁻¹ for all n in ℕ. Note that T is a firmly nonexpansive mapping from C into C and F(T) = A⁻¹0. From (83), we have lim _n→+∞∥x_n − T_nx_n∥ = 0. Example 27 implies that 𝒯 : = {T_r : r > 0} has property (𝒜). It follows that lim _n→+∞∥x_n − Tx_n∥ = 0. Therefore, Corollary 31 follows from Theorem 28.

Lemma 32. Let C be a nonempty closed convex subset of a reflexive, strictly convex, and smooth Banach space X, and let φ be a gauge function. Let Θ : C × C → ℝ be a bi-function satisfying conditions (A1)–(A4), and let r > 0 and x ∈ X. Then there exists z in C such that

Lemma 33 (see [37].)One has the following.

• (1)

  $$ is single-valued.

• (2)

  $$ is a φ-firmly nonexpansive mapping; that is, for all x, y in X,

  $$ ()

• (4)

  $$.

• (5)

  GMEP(Θ, A, ψ) is closed and convex.

• (6)

  For all x in $$ and y in X, one has

  $$ ()

Theorem 34. Let C be a nonempty closed convex subset of a uniformly convex Banach space X with a uniformly Gâteaux differentiable norm. Let φ be a gauge function and let f : X → [0, ∞] be a proper, convex, lower semicontinuous function with C ⊂ dom (f). Let A : C → X^* be continuous and monotone, Θ : C × C → ℝ a bi-function satisfying conditions (A1)–(A4), and ψ : C → ℝ a lower semicontinuous and convex function. Assume that GMEP(Θ, A, ψ, φ) ≠ ∅. For u in C,  r > 0, and C₁ = C with $$, define a sequence {x_n} in C as follows:

Then {x_n} strongly converges to $$.

Proof. Note that $$ is a φ-firmly nonexpansive mapping from C into C and $$. Therefore, Theorem 34 follows from Theorem 28.