Convergence Results for a Common Solution of a Finite Family of Equilibrium Problems and Quasi-Bregman Nonexpansive Mappings in Banach Space

Lemma 1 (see [21].)If $$ is uniformly Fréchet differentiable and bounded on bounded subsets of E, then ∇f is uniformly continuous on bounded subsets of E from the strong topology of E to the strong topology of E^∗.

Definition 2 (see [22].)The function f is said to be

• (i)

  essentially smooth, if ∂f is both locally bounded and single-valued on its domain,

• (ii)

  essentially strictly convex, if (∂f) ⁻¹ is locally bounded on its domain and f is strictly convex on every convex subset of dom⁡∂f,

• (iii)

  Legendre, if it is both essentially smooth and essentially strictly convex.

Remark 3. Let E be a reflexive Banach space. Then we have the following:

• (i)

  f is essentially smooth if and only if f^∗ is essentially strictly convex (see [22], Theorem 5.4).

• (ii)

  (∂f) ⁻¹ = ∂f^∗ (see [20]).

• (iii)

  f is Legendre if and only if f^∗ is Legendre (see [22], Corollary 5.5).

• (iv)

  If f is Legendre, then ∇f is a bijection satisfying ∇f = (∇f^∗) ⁻¹, ran ∇f = dom ∇f^∗ = int dom f^∗, and ran ∇f^∗ = dom f = int dom f (see [22], Theorem 5.10).

The following result was proved in [23] (see also [24]).

Lemma 4. Let E be a Banach space, let r > 0 be a constant, let ρ_r be the gauge of uniform convexity of g, and let $$ be a convex function which is uniformly convex on bounded subsets of E. Then,

• (i)

  for any x, y ∈ B_r and α ∈ (0,1),

  $$ ()

• (ii)

  for any x, y ∈ B_r,

  $$ ()

• (iii)

  if, in addition, g is bounded on bounded subsets and uniformly convex on bounded subsets of E then, for any x ∈ E, y^∗, z^∗ ∈ B_r, and α ∈ (0,1),

  $$ ()

Lemma 5 (see [25].)Let E be a Banach space, let r > 0 be a constant, and let $$ be a continuous and convex function which is uniformly convex on bounded subsets of E. Then

for all $$, and $$ with $$, where ρ_r is the gauge of uniform convexity of f.

Theorem 6. Let E be a reflexive Banach space and let $$ be a convex function which is bounded on bounded subsets of E. Then the following assertions are equivalent:

• (1)

  f is strongly coercive and uniformly convex on bounded subsets of E.

• (2)

  dom⁡f^∗ = E^∗, f^∗ is bounded on bounded subsets and uniformly smooth on bounded subsets of E^∗.

• (3)

  dom⁡f^∗ = E^∗, f^∗ is Fréchet differentiable and ∇f is uniformly norm-to-norm continuous on bounded subsets of E^∗.

Theorem 7. Let E be a reflexive Banach space and let $$ be a continuous convex function which is strongly coercive. Then the following assertions are equivalent:

• (1)

  f is bounded on bounded subsets and uniformly smooth on bounded subsets of E.

• (2)

  f^∗ is Fréchet differentiable and f^∗ is uniformly norm-to-norm continuous on bounded subsets of E^∗.

• (3)

  dom⁡f^∗ = E^∗, f^∗ is strongly coercive and uniformly convex on bounded subsets of E^∗.

Lemma 8. Let E be a reflexive Banach space, let $$ be a strongly coercive Bregman function, and let V be the function defined by

Then the following assertions hold:

• (1)

  D_f(x, ∇f(x^∗)) = V(x, x^∗) for all x ∈ E and x^∗ ∈ E^∗.

• (2)

  V(x, x^∗)+〈∇f^∗(x^∗) − x, y^∗〉≤V(x, x^∗ + y^∗) for all x ∈ E and x^∗, y^∗ ∈ E^∗.

Lemma 9 (see [26].)Let C be a nonempty, closed, and convex subset of a reflexive Banach space E. Let $$ be a Gâteaux differentiable and totally convex function and let x ∈ E. Then

• (a)

  $$ if and only if 〈∇f(x)−∇f(z), y − z〉≤0,   ∀ y ∈ C.

• (b)

  $$

Lemma 10 (see [29].)If x ∈ dom⁡f, then the following statements are equivalent:

• (i)

  The function f is totally convex at x.

• (ii)

  For any sequence {y_n} ⊂ dom⁡f,

  $$ ()

Lemma 11 (see [30].)The function f is totally convex on bounded sets if and only if the function f is sequentially consistent.

Lemma 12 (see [31].)Let $$ be a Gâteaux differentiable and totally convex function. If x₀ ∈ E and the sequence {D_f(x_n, x₀)} is bounded, then the sequence {x_n} is bounded too.

Lemma 13 (see [31].)Let $$ be a Gâteaux differentiable and totally convex function, x₀ ∈ E, and let C be a nonempty, closed, and convex subset of E. Suppose that the sequence {x_n} is bounded and any weak subsequential limit of {x_n} belongs to C. If $$ for any $$, then {x_n} converges strongly to $$.

Lemma 14 (see [32].)Let E be a real reflexive Banach space, let f : E → (−∞, +∞] be a proper lower semicontinuous function, and then f^∗ : E^∗ → (−∞, +∞] is a proper weak^∗ lower semicontinuous and convex function. Thus, for all z ∈ E, one has

Lemma 15 (see [33].)Let E be a real reflexive Banach space and let C be a nonempty closed convex subset of E. Let f : E → (−∞, +∞] be a Legendre function. If the bifunction $$ satisfies the conditions (A1)–(A4), then, the following hold:

• (i)

  $$ is single-valued.

• (ii)

  $$ is a Bregman firmly nonexpansive operator.

• (iii)

  $$.

• (iv)

  EP(g) is closed and convex subset of C.

• (v)

  For all x ∈ E and for all $$, one has

  $$ ()

Lemma 16 (see [34].)Let {a_n} be a sequence of nonnegative real numbers satisfying the following relation:

where {α_n}⊂(0,1) and {δ_n} is a real sequence satisfying the following conditions: Then, lim_n→∞a_n = 0.

Lemma 17 (see [35].)Let {a_n} be a sequence of real numbers such that there exists a subsequence {n_i} of {n} such that $$ for all $$. Then there exists a nondecreasing sequence $$ such that m_k → ∞ and the following properties are satisfied by all (sufficiently large) numbers $$:

In fact, m_k = max {j ≤ k : a_j < a_j+1}.

Theorem 18. Let C be a nonempty, closed, and convex subset of a real reflexive Banach space E and $$ a strongly coercive Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subset of E. For each j = 1,2, …, m, let g_j be a bifunction from C × C to $$ satisfying (A1)–(A4) and let $$ be a finite family of quasi-Bregman nonexpansive self-mapping of C such that $$, where F = F(T_NT_N−1T_N−2 ⋯ T₂T₁) = F(T₁T_NT_N−1T_N−2 ⋯ T₂) = ⋯ = F(T_N−1T_N−2 ⋯ T₂T₁T_N) ≠ ∅ and $$ Let $$ be a sequence generated by x₁ = x ∈ C,   C₁ = C, and

where T_[n] = T_n(mod N) and $$ and $$ satisfying lim_n→∞α_n = 0 and $$. Then $$ converges strongly to $$, where $$ is the Bregman projection of C onto Ω.

Proof. Let $$ from Lemma 15; we obtain

Now from (32), we obtain Also from (32), (26), and (34), we have Thus, by induction we obtain which implies that {x_n} is bounded and hence {y_n}, {T_[n]y_n}, {T_[n]x_n}, and {u_j,n} are all bounded for each j = 1,2, …, m. Now from (32) let z_n≔∇f^∗((1 − α_n)∇f(u_j,n)). Furthermore since α_n → 0 as n → ∞, we obtain Since f is strongly coercive and uniformly convex on bounded subsets of E, f^∗ is uniformly Fréchet differentiable on bounded sets. Moreover, f^∗ is bounded on bounded sets; from (37), we obtain On the other hand, in view of (3) in Theorem 6, we know that dom⁡f^∗ = E^∗ and f^∗ is strongly coercive and uniformly convex on bounded subsets. Let s = sup {‖∇f(y_n)‖, ‖∇f(T_[n]y_n)‖} and $$ be the gauge of uniform convexity of the conjugate function f^∗. Now from (32) and Lemmas 4 and 8, we obtain Now, we consider two cases.

Case  1. Suppose that there exists $$ such that {D_f(p, x_n)} is nonincreasing. In this situation {D_f(p, x_n)} is convergent. Then from (40) we obtain

which implies, by the property of ρ_s and since β_n ∈ [c, d]⊂(0,1), Since f is strongly coercive and uniformly convex on bounded subsets of E, f^∗ is uniformly Fréchet differentiable on bounded sets. Moreover, f^∗ is bounded on bounded sets; from (43), we obtain Now from (4), we obtain and therefore

Also, from (28) in Lemma 15, we have

Then, we have from Lemma 10 that

Also, from (b) of Lemma 9, we have

Then, we have from Lemma 10 that

From (38) and (48), we obtain

From (50) and (51), we obtain

Since f is strongly coercive and uniformly convex on bounded subsets of E, f^∗ is uniformly Fréchet differentiable on bounded sets. Moreover, f^∗ is bounded on bounded sets; from (52), we obtain

Also from (44) and (52)

Now from (4) and (34), we obtain

therefore, from (53), we obtain Also thus Also, from (56) Then, we have from Lemma 10 that Then from (32) and (44), we have

This implies

from (44), (52), and (60), we obtain This implies that Also from (52) and (62), we obtain But as n → ∞. Hence

From the uniformly continuous ∇f, we have from (66) that

From (4), (35), and (69), we obtain which implies Also from quasi-Bregman nonexpansivity of T_[n], we have which implies and from the uniform continuous ∇f, we obtain

Also from (4) and (64), we obtain

From (64), (66), and (73), we obtain which from uniform continuous ∇f implies and from (4) and (77), we obtain

From (4), (71), (77), and (78)

Also from (4), (71), and (79)

Using the quasi-Bregman nonexpansivity of T_(i) for each i, we obtain the following finite table:

Then, applying Lemma 10 on each line above, we obtain and adding up this table, we obtain Using this and (68), we obtain Also from quasi-Bregman nonexpansivity of T_(i), for each i, we have as  n → ∞. Then, we have from Lemma 10 that Since then, from (52), (84), and (86), we obtain Following the argument from (85), (86), and (88) by replacing y_n with z_n and using (51), we obtain

Let $$ be a subsequence of {x_n}. Since {x_n} is bounded and E is reflexive, without loss of generality, we may assume that $$ for some q ∈ F and since x_n − z_n → 0 as n → ∞, then $$. Since the pool of mappings of T_[n] is finite, passing to a further subsequence if necessary, we may further assume that, for some i ∈ {1,2, …, N}, from (89), we get

and also

Noticing that $$ for each j = 1,2, …, m, we obtain

Hence

From (A2), we note that, for each j = 1,2, …, m,

Taking the limit as i → ∞ in above inequality and from (A4) and $$, we have g_j(y, q) ≤ 0 for each j = 1,2, …, m. For 0 < t < 1 and y ∈ C, define y_t = ty + (1 − t)q. Noticing that y, q ∈ C, we obtain y_t ∈ C, which yield that g_j(y_t, q) ≤ 0. It follows from (A1) that That is, for each j = 1,2, …, m, we have g_j(y_t, y) ≥ 0.

Let t ↓ 0; from (A3), we obtain g_j(q, y) ≥ 0 for any y ∈ C, for each j = 1,2, …, m. This implies that $$ Hence q ∈ Ω. It follows from the definition of the Bregman projection that

It follows from Lemma 16 and (41) that D_f(p, x_n) → 0 as n → ∞. Consequently, from Lemma 10, we obtain x_n → p as n → ∞.

Case  2. Suppose D_f(p, x_n) is not monotone decreasing sequences; then set Φ_n≔D_f(p, x_n) and let $$ be a mapping defined for all n ≥ N₀ for some sufficiently large N₀ by

Then by Lemma 17  τ(n) is a nondecreasing sequence such that τ(n) → ∞ as n → ∞ and Φ_τ(n) ≤ Φ_τ(n)+1, for n ≥ N₀. Then from (40) and the fact that α_τ(n) → 0, we obtain that Following the same argument as in Case  1, we obtain and also we obtain

Then from (41), we obtain that

It follows from (101) and Φ_n ≤ Φ_τ(n)+1, α_τ(n) > 0 that

as τ(n) → ∞. Thus Furthermore, for n ≥ N₀, if n ≠ τ(n) (i.e., τ(n) < n), because Φ_j > Φ_j+1 for τ(n) + 1 ≤ j ≤ n. it then follows that for all n ≥ N₀ we have This implies that lim_n→∞Φ_n = 0, and hence D_f(p, x_n) → 0 as n → ∞. Consequently, from Lemma 10, we obtain x_n → p as n → ∞. Therefore from the above two cases, we conclude that {x_n} converges strongly to p ∈ Ω and this completes the proof.