Fixed and Best Proximity Points of Cyclic Jointly Accretive and Contractive Self-Mappings

Definition 2.1. T : Dom (T) ⊂ X → X is an accretive mapping if

for any λ ∈ R₀₊.

Note that, since X is also a vector space, x + λ  Tx    is in X for all x in X and all real λ. This fact facilitates also the motivation of the subsequent definitions as well as the presentation and the various proofs of the mathematical results in this paper. A strong convergence theorem for resolvent accretive operators in Banach spaces has been proved in [17].Two more restrictive (and also of more general applicability) definitions than Definition 2.1 to be then used are now introduced as follows:

Definition 2.2. T : Dom (T) ⊂ X → X  is a λ^*-accretive mapping, some λ^* ∈ R₀₊ if

for some λ^* ∈ R₀₊. A generalization is as follows:    T : Dom (T) ⊂ X → X  is $$-accretive for some $$ if

Definition 2.3. T : Dom (T) ⊂ X → X  is a weighted λ-accretive mapping, for some function   λ : X × X → R₀₊, if

The above concepts of accretive mapping generalize that of a nondecreasing function. Contractive and nonexpansive 2-cyclic self-mappings are defined as follows on unions of subsets of X.

Definition 2.4. T : A ∪ B → A ∪ B is a 2-cyclic k-contractive (resp., nonexpansive) self-mapping if

for some real k ∈ [0, 1) (resp. k = 1), [12, 13].

The concepts of Kannan-self mapping and 2-cyclic (α, β)-Kannan self-mapping which can be also a contractive mapping, and conversely if k <   1/3, [16], are defined below.

Definition 2.5. T : X → X  is a α-Kannan self-mapping if

for some real α ∈ [0, 1/2), [12, 13].

Definition 2.6. T : A ∪ B → A ∪ B  is an 2-cyclic (α, β)-Kannan self-mapping for some real α ∈ [0, 1/2) if it satisfies, for some β ∈ R₊.

The relevant concepts concerning 2-cyclic self-mappings are extended to p(≥2)-cyclic self-mappings in Section 3. Some simple explanation examples follow.

Example 2.7. Consider the scalar linear mapping from X ≡ A ≡ R to X as Tx = γx + γ₀    with γ, γ₀ ∈ R    endowed with the Euclidean distance d(x, y) = |x − y|; for all x, y ∈ X. Then,

for all x, y ∈ R for any λ ∈ R₀₊ provided that γ ∈ R₀₊. In this case, T : A ∪ B → X is accretive. It is also k-contractive if since d(Tx, Ty) = |Tx − Ty| = γ,   d(z, y) ≤ kd(x, y); for all x, y ∈ R. Also, if γ ∈ R₋, then d(x + λTx, y + λTy) ≥ |λ|γ| − 1|d(x, y) ≥ d(x, y); for all x, y ∈ R  if λ|γ| ≥ 2, that is, if $$. Then, T : R → R is $$-accretive and k-contractive if |γ| ≤ k < 1.

Example 2.8. Consider the metric space (R, d) with the distance being homogeneous and translation-invariant and a self-mapping T : R → R defined by Tx = −t|x|^psgn_e x = −t|x|^p−1x with   t ∈ R₀₊, p ∈ R₀₊,     and sgn_ex   = sgn   x if x ≠ 0    and sgn_e0 = 0. If pt = 0,     then T : R → R is accretive since

Furthermore, if t = 0, then 0 ∈ R is the unique fixed point with T^jx = 0; for all j ∈ Z₊. If p = 0 then, T^jx = t^j →   z = 0    as j →   ∞ if |  t| < 1 and then z = 0  is again the unique fixed point of T. In the general case, Tx = t|x|^psgn_ex implies holds if λ^*|t|  |x|^p−1 ≤ 1 that is, T : R → R is weighted λ^*(x, y)-accretive with λ^*(x, y) : = t  ⁻¹  min   (  |x|^1−p,   |y|^1−p). The restricted self-mapping T : [−1, 1] ⊂ X → [−1, 1]    is λ^*(≡t⁻¹)-accretive. Furthermore, if     p ≥ 1,     then    T : [−1, 1] ⊂   X → [−1, 1] is |t|-contractive if |t| <   1 and the iteration T^jx → 0    as j → ∞    with z = 0 being the unique fixed point since Note from the definition of the self-mapping Tx = −t|x|^p−1x on [−1,1] that it is also a 2-cyclic self-mapping from [−1,0] ∪ [0,1] to itself with the property T([−1,0]) = [0,1]    and T([0,   1]) = [−1,0].

Assertions 1. (1) If T : D(T) ⊂ X → X    is an accretive mapping, then it is λ^*-accretive, for all λ^* ∈ R₀₊. (2) If T : D(T) ⊂ X → X    is λ^*-accretive, then it is $$-accretive; for all $$. (3) Any nonexpansive self-mapping T : D(T) ⊂ X → X  is 0^*-accretive and conversely.

Theorem 2.9. Let (X, ∥ ∥)  be a Banach vector space with  (X, d)  being the associated complete metric space endowed with a norm-induced translation-invariant and homogeneous metric  d : X × X → R₀₊. Consider a self-mapping T : X → X    which restricted to    T : A ∪ B → A ∪ B    is a 2-cyclic k-contractive self-mapping where A and B are nonempty subsets of X. Then, the following properties hold.

• (i)

  Assume that the self-mapping T : X → X satisfies the constraint:

  $$ (2.12)
  with k, λ ∈ R₀₊ satisfying the constraint k(1 + kλ) < 1. Then, the restricted self-mapping T : A ∪ B → A ∪ B    satisfies
  $$ (2.13)
  irrespective of A and B being bounded or not.

If, furthermore, A and B are closed and convex and    A∩B ≠ ∅, then there exists a unique fixed point ω ∈ A∩  B of T : A ∪ B → A ∪ B such that there exists lim _j→∞ d(T^jx, T^jy) = 0; for all x ∈ A,   for all y ∈ B, implying that   lim _j→∞ T^jx = lim _j→∞ T^jy = ω. If, in addition,   dist (A, B) > 0  so that A∩B = ∅, then there exists lim _j→∞ d(T^jx, T^jy) = d  (z, Tz); for all x   ∈ A, for all y ∈ B for some best proximity points z ∈ A, Tz ∈ B which depend in general on x and y. Furthermore, if (X, ∥ ∥) is a uniformly convex Banach space, then T^2jx,   T^2j+1y → z₁ ∈ A and T^2jy,   T^2j+1x → Tz₁ ∈ B as →∞; for all  (x, y) ∈ A × B, where z₁ and z₂ are unique best proximity points in A and B of T : A ∪ B → A ∪ B.

• (ii)

  Assume that A and B are nondisjoint. Then, T : A ∪ B → X  is also k_c contractive and λ^*-accretive for any nonnegative λ^* ≤ k⁻²(k_c − k) and any k_c ∈ [k, 1). It is also nonexpansive and λ^*-accretive for any nonnegative λ^* ≤ k⁻²(1 − k).

• (iii)

  If k = 0 then T : A ∪ B → X is weighted λ-accretive for λ : X × X → R₀₊ for any λ^* ∈ R₊ and its restriction T : A ∪ B → A ∪ B is 2-cyclic 0-contractive.

• (iv)

  T : A ∪ B → X is weighted λ-accretive for λ : X × X → R₀₊ satisfying λ(x, y) ≤ k⁻²(k_c(x, y) − k)(d(x, y) − d_AB) for some k_c : X × X → [k, ∞). The restricted self-mapping T : A ∪ B → A ∪ B is also $$-contractive with $$ if $$ with $$. Also,  T : A ∪   B   → X is nonexpansive and weighted λ-accretive for λ : X ×   X   → R₀₊ satisfying λ  (x, y)   ≤ k⁻²(k_c(x, y) − k)(d(x, y) − d_AB) if k_c : X × X → [k, 1] which implies, furthermore, that λ : X × X → R₀₊  is bounded.

Proof. Let us denote d_AB : = dist (A, B).  Consider that the two following relations are verified simultaneously:

Since the distance d : X ×   X → R₀₊ is translation-invariant and homogeneous, then the substitution of (2.14) yields if A and B are disjoint sets, after using the subadditive property of distances, the following chained relationships since 0 ∈ X: with k_c : = k(1 + kλ^*) ≥ k. Note from (2.15) that and, if k_c < 1, then If d_AB = 0 then lim _j→∞ d(T^jx, T^jy) = 0. It is first proven that the existence of the limit of the distance implies that of the limit lim _j→∞ T^jz; for all z ∈ A ∪ B. Let be x_j =   T^jx, y_j = T^jy with    x_j, y_j ∈ A ∪ B. Then, since T : A ∪ B → A ∪ B being contractive is globally Lipschitz continuous. Then, lim _j→∞ d(T^jx, T^jy) = d(lim _j→∞ T^jx, lim _j→∞ T^jy) = 0  since, because the fact that the metric is translation-invariant, one gets As a result, lim _j→∞ d(T^jx, T^jy) = 0 if d_AB = 0 what implies which lim _j→∞ (T^jx − T^jy) = 0; for all x   ∈   A,     for all y   ∈   B, since T : A ∪   B → A ∪   B is globally Lipschitz continuous since it is contractive.

In addition, there exists   lim _j→∞ T^jx =   lim _j→∞ T^jy = ω ∈ A ∪ B; for all x   ∈   A, for all y   ∈   B. Assume not so that there exists x ∈   A such that ¬∃lim _j→∞ T^jx   and there exists a subsequence on nonnegative integers $$ such that $$. If so, one gets by taking y = Tx ∈ B    that $$ which contradicts lim _j→∞d(T^j(Tx),   T^jx) = 0. Then $$ is a Cauchy sequence for any x ∈ A ∪ B and then converges to a limit. Furthermore, ω ∈ A ∪ B since T^j(A ∪   B)⊆A ∪   B  for any j ∈   Z₀₊ and as j → ∞ since A and B are nonempty and closed. It has been proven that lim _j→∞ T^jx = lim _j→∞ T^jy = ω ∈ A ∪ B; for all x   ∈   A,   for all y   ∈   B.

It is now proven that  ω   = Tω ∈   Fix (T). Assume not, then, from triangle inequality,

which contradicts lim _j→∞  T^jω =   ω so that ω   = Tω ∈   Fix (T). It is now proven that ω ∈ Fix (T)∩(A∩B). Assume not, such that, for instance, T^jx ∈ A and $$. If so, since T(A)⊆B; T(B)⊆A, then the existing limit fulfils $$ which is impossible so that there would be no existing limit  lim _j→∞ T^jx in A ∪ B, contradicting the former result of its existence. Then, ω ∈ Fix (T)∩(A∩B) implying that Fix(T) ⊂ A∩B.

It is now proven by contradiction that ω = lim _j→∞ T^jx; for all x   ∈   A ∪   B is the unique fixed point of T : A ∪ B → A ∪ B. Assume that   ∃ ω₁(≠ω) ∈ Fix (T), then   lim _j→∞ T^jy₁ =   ω₁ for some y₁(≠y) ∈ B with no loss in generality and all x ∈   A. Thus, lim _j→∞d(  T^jx,   T^jy₁) = d  (ω,   ω₁) = 0⇒  ω   =   ω₁ which contradicts ω   ≠   ω₁   so that Fix(T) = {ω}.

Now, assume that A and B do not intersect so that dist  (A, B) = d_AB > 0. Then, one gets from the first inequality in (2.15) that for all x   ∈ A, y ∈   B, one gets

Note that since T(A)⊆B, T(B)⊆  A and dist  (A, B) = d_AB > 0, then    x ∈   A⇒T^jx ∈   A and T^jx ∉   B if j is even and T^jx ∈   B and T^jx ∉ A if j is odd    y ∈   B⇒T^jy ∈   B  and T^jy ∉ A if j is even and T^jy ∈   A and T^jy ∉ B if j is odd.

Then, T^jx and T^jy are not both in either A or B if x and y are not both in either A or B for any j ∈   Z₀₊. As a result, lim _j→∞  sup d  (T^jx,   T^jy) < d_AB is impossible so that

for some best proximity points z   ∈   A and Tz   ∈   B or conversely. Then, where z_j = T^jx Thus, lim _j→∞ d  (z_j, Tz_j) = d_AB = d(z, Tz). It turns out that dist  (z_j,   Fr(A ∪ B)) → 0 and dist(Tz_j, Fr(A ∪ B)) → 0 as j → ∞. Otherwise, it would exist an infinite subsequence $$of $$ with $$being an infinite subset of Z₀₊  such that d(z_j, Tz_j) > d_AB for $$. On the other hand, since (X, ∥ ∥) is a normed space, then by taking the norm-translation invariant and homogeneous induced metric and since there exists lim _j→∞ d  (T^j+1x, T^j+1y) = d_AB, it follows that there exist j₁ ∈ Z₀₊ and δ   =   δ(ε, j₁) ∈ R₊ such that for any given ε   ∈   R₊; for all x ∈ A,   for all y ∈ B with T^jx   ∈   A,   T^j+1y ∈ A for any even j(≥j₁) ∈ Z₀₊ and T^jx   ∈   B,   T^j+1y   ∈   B, for any odd j(≥j₁) ∈ Z₀₊. As a result, by choosing the positive real constant arbitrarily small, one gets that T^2jx   →   T^2j+1y   →   z = z(x,   y) ∈ A (a best proximity point of A) and T^2j+1x → T^2jy → Tz ∈ B (a best proximity point of B), or vice-versa, as j → ∞ for any given x ∈ A and y ∈ B. A best proximity point z ∈ A ∪ B fulfils z = T²z. Best proximity points are unique in A and B as it is now proven by contradiction. Assume not, for instance, and with no loss in generality, assume that there exist two distinct best proximity points z₁ and z₂ in A. Then T²z₁ = z₁ and T²z₁ = z₂ contradict z₁ ≠ z₂ so that necessarily z₁ = T²z₁ ≠ z₂ = T²z₂. Since (X, ∥ ∥) is a uniformly convex Banach space, we take the norm-induced metric to consider such a space as the complete metric space (X, d) to obtain the following contradiction: since (X, ∥ ∥) is also a strictly convex Banach space and A and B are nonempty closed and convex sets. Then, z = T²z ∈ A is the unique best proximity point of T : A ∪ B → A ∪ B in A and Tz is its unique best proximity point in B. Then, Property (i) has been fully proven. Since A and B are not disjoint, then d_AB = 0, and T : A ∪ B → A ∪   B  is k_c-contractive and λ^*-accretive if λ^* = k⁻²(k_c − k) with k_c ∈ [k, 1). By taking k_c = 1, note that T : A ∪   B → X   is nonexpansive and k⁻²(1 − k)-accretive. Property (ii) has been proven.

To prove Property (iii), we now discuss if

is possible with 1 ≥ k_c and d_AB > 0. Note that  d_AB = dist (A, B) = d(z, Tz) for some z ∈ A. Define $$, if  d_AB ≠ 0 for some k_DA,   k_DB,   k_D ∈ R₊, where d_A : = diam A = k_DAd_AB and d_B : = diam B = k_DBd_AB. Three cases can occur in (2.26), namely,

• (a)

  If k = k_c then k²λd(x, y) ≤ 0⇔[kλ = 0∨d(x, y) ≤ 0] which is untrue if x ≠   y and kλ > 0 and it holds for either k = 0 or λ = 0,

• (b)

  k_c > k, then (2.26) is equivalent to

  $$ (2.27)
  Take x   ∈ A to be a best proximity point with so that d(x, Tx) = d_AB ≥ (k_c − k)/(k_c − k(1 + kλ))d_AB > d_AB which is untrue if kλ > 0 and true for kλ = 0,

• (c)

  1 ≥ k(1 + kλ) ≥ k_c < k, then (2.16) is equivalent to (k − k_c)d_AB ≥ [k(1 + k  λ) − k_c]  d(x,   y); for all x   ∈   A,   for all y   ∈   B, but $$. Thus, the above constraint is guaranteed to hold in the worst case if $$ which is a contradiction.

Property (iii) follows from the above three cases (a)–(c).

To prove Property (iv), consider again (2.26) by replacing the real constants λ and k_c with the real functions λ  : X × X → R₀₊ and k_c : X × X → [k, 1). Note that (2.26) holds through direct calculation if λ(x, y) ≤ k⁻²(k_c(x, y) − k)  (d(x, y)   − d_AB); for all x ∈ A,   for all y ∈ B for some k_c : X × X → [k, ∞). Thus, the self-mapping T : A ∪ B → X is weighted λ-accretive for λ  :  X × X → R₀₊ satisfying λ(x, y)   ≤ k⁻²(k_c(x, y) − k)(d(x, y) − d_AB) for some k_c : X × X → [k, ∞); and it is also $$-contractive with $$ if $$ with $$ and nonexpansive if k_c : X × X → [k, 1]. On the other hand, note that $$. If A and B are bounded and k_c : X × X → [k, 1], then

Property (iv) has been proven.

Remark 2.10. Note that Theorem 2.9 (iii) allows to overcome the weakness of Theorem 2.9 (ii) when A and B are disjoint by introducing the concept of weighted accretive mapping since for best proximity points z ∈ A ∪ B,   λ(z, Tz) = 0.

Remark 2.11. Note that the assumption that (X, ∥ ∥) is a uniformly convex Banach space could be replaced by a condition of strictly convex Banach space since uniformly convex Banach spaces are reflexive and strictly convex, [18]. In both cases, the existence and uniqueness of best proximity points of the 2-cyclic T : A ∪ B → A ∪ B in A and B are obtained provided that both sets are nonempty, convex, and closed.

Remark 2.12. Note that if either A or B is not closed, then its best proximity point of T : A ∪ B → A ∪ B is in its closure since T(A)⊆B⊆cl  B, T(B)⊆A⊆cl  A leads to T(A ∪ B)⊆A ∪ B⊆cl (A ∪ B) and T^k(A ∪ B)⊆cl (A ∪ B) for finitely many and for infinitely many iterations through the self-mapping T : A ∪ B → A ∪ B and Theorem 2.9 is still valid under this extension.

Theorem 2.13. The following properties hold.

• (i)

  Let (X, d) be a metric space endowed with a norm-induced translation-invariant and homogeneous metric d : X × X → R₀₊. Consider the λ^*-accretive mapping T : A ∪ B → Xfor some λ^* ∈ R₀₊ which restricted as T : A ∪ B → A ∪ B is 2-cyclic, where A and B are nonempty subsets of X subject to 0 ∈ A ∪ B. Then,

  $$ (2.29)
  If, furthermore, T : A ∪ B → A ∪ B  is k-contractive, then
  $$ (2.30)
    T : A ∪ B → A ∪ B  is guaranteed to be nonexpansive (resp., asymptotically nonexpansive) if
  $$ (2.31)
  respectively,
  $$ (2.32)

• (ii)

  Let (X, ∥ ∥) be a normed vector space. Consider a λ^*-accretive mapping T : A   ∪   B → X for some λ^* ∈   R₀₊ which restricted to T : A   ∪   B → A ∪   B is 2-cyclic, where A and B are nonempty subsets of X subject to 0 ∈ A ∪ B then

  $$ (2.33)
  If, furthermore, T : A ∪ B → A ∪ B is k-contractive, then
  $$ (2.34)
  T : A   ∪ B → A ∪ B  is nonexpansive (resp., asymptotically nonexpansive, [30]) if
  $$ (2.35)
  respectively,
  $$ (2.36)

Proof. To prove Property (i), define an induced by the metric norm as follows ∥x∥ = d(x,   0)since the metric is homogeneous and translation-invariant. Define the norm of T : A ∪ B → A ∪ B, that is, the norm of T on X restricted to A ∪ B as follows:

with the above set being closed, nonempty, and bounded from below. Since T : A ∪ B → A ∪ B is 2-cyclic and T : A ∪ B → X is λ^*-accretive (Definition 2.2), one gets by proceeding recursively since the metric is homogeneous and 0 ∈ A ∪ B, and I is the identity operator on X, where with the above set being closed, nonempty, and bounded from below. If  ∥(I+λT)^j∥ < 1 for some ∥(I+λT)^j∥ < 1, then we get the contradiction d(x, y) <   d(x, y); for all x ∈ A, for all y ∈ B in (2.38). Thus, ∥(I+λT)^j∥ = d((I  +λ  T)^jx,   0) ≥   1; for all j   ∈   Z₀₊, for  all  x(≠0) ∈ A ∪ B, for all λ ∈ [0, λ^*]. If now x and y are replaced with Tⁱx and Tⁱy for any i ∈ Z₀₊ in (2.30), one gets if T : A ∪ B → A ∪ B is a 2-cyclic k-contractive for some real k ∈ [0, 1) and λ^*-accretive mapping: for all x ∈ A, for all y(≠x) ∈ B, for all j(≥i) ∈ Z₊, for all i ∈ Z₊, for all λ ∈ [0, λ^*]. Then, 1 ≤ ∥(I+λT)^j∥   = d((I+λT)^jx,   0) < k⁻¹; for all j ∈   Z₊, for all x(≠0) ∈ A ∪ B, for all λ ∈ [0, λ^*]. If ∥(I+λT)^j∥ = d((I+λT)^jx, 0) = 1; for all j ∈ Z₊, for all x(≠0) ∈ A ∪ B, for all λ ∈ [0, λ^*], it turns out that T : A ∪ B → X is λ^*-accretive and T : A ∪ B → A ∪ B  is a 2-cyclic nonexpansive self-mapping. It is asymptotically nonexpansive if lim _j→∞  sup  d((I+λT)^jx,   0) = 1; for all x(≠0) ∈ A ∪ B, for  all  λ ∈ [0, λ^*]. Property (i) has been proven. The proof of Property (ii) for (X, ∥ ∥) being a normed vector space is identical to that of Property (i) without associating the norms to a metric.

Example 2.14. Assume that X =   R, A = R₋,  B =   R₊ and the 2-cyclic self-mapping $$ defined by the iteration rule A ∪ B∋x_j+1 = k_j  x_j ∈ A ∪ B with R∋k_j(∈[−k, k]) ≤   k ≤ 1, sgn  k_j+1 = −sgn k_j = sgn  x_j; for all j ∈   Z₀₊, and x₀ ∈ A ∪ B. Let d : R₀₊ → R₀₊ be the Euclidean metric.

• (a)

  If  k < 1, then $$ so that for any x₀ ∈ A ∪ B, x_j ∈ A ∪ B; for all j   ∈ Z₀₊x_j → z = 0 ∉ A ∪ B as j → ∞ with 0 ∈ cl (A ∪ B), Fix (t) = {0} ⊂ cl (A  ∩  B) but it is not in A∩B which is empty. If k = 1 and $$ (i.e., there are infinitely many values |k_i| being less than unity), then the conclusion is identical. If A and B are redefined as A = R₀₋, B = R₀₊, then Fix (t) = {0} ⊂ A∩B ≠ ∅.

• (b)

  If k_j = k = 1; for all j  ∈Z₀₊ the self-mapping $$is not expansive and there is no fixed point.

• (c)

  If k = 1 − σ for some σ(<1) ∈ R₊, then for    R₀∋λ ∈ [0, λ^*],

  $$ (2.41)
  so that $$is also λ^*-accretive and k₁ ∈ [k, 1)-contractive with λ^* = k₁k⁻¹ − 1.

• (d)

  Now, define closed sets  R_ε+ : = {r(≥ε) ∈ R₊} and R_ε− : = {r(≤−ε) ∈ R₊} for any given ε ∈ R₀₊  so that d_AB = ε. The 2-cyclic self-mapping $$is re-defined by the iteration$$if |x_j+1  | ≥   ε  and x_j+1   =    −   ε sgn  x_j, for i = 1,2, otherwise, where$$for i = 1,2 with the real sequence $$being subject to k_j(∈[−k, k]) ≤ k ≤ 1, $$; i = 1,2, for all j∈Z₀₊  and x₀ ∈ A ∪ B.

  Then, for any ε   ∈ R₊  and any x₀ ∈ A ∪ B, there are two best proximity points z = −ε ∈ A and z₁ = ε ∈ B fulfilling −  ε = tε = −t²ε and d_AB = d(z, z₁) = d(z, tz) = d(tz₁, z₁).

• (e)

  Redefine X = R²  so that R²∋x =   (x⁽¹⁾,   x⁽²⁾)  ^T with x⁽¹⁾, x⁽²⁾ ∈   R; $$,$$. In the case that ε   = 0,   then A and B are open disjoint subsets (resp., $$,$$ are closed nondisjoint subsets with A∩  B = {  (0,   x)  ^T : x   ∈ R}).

  The 2-cyclic self-mapping $$is re-defined by the iteration rule:

  $$ (2.42)
  otherwise, where
  $$ (2.43)
  with the real sequence $$ being subject to    k_j  (∈[−k, k]) ≤ k ≤ 1, $$; for i = 1,2; for all j ∈   Z₀₊ and x₀ ∈ A ∪ B.

The same parallel conclusions to the above ones (a)–(c) follow related to the existence of the unique fixed point z = 0 in the closure of A and B but not in its empty intersection if either A or B is open, respectively, in the intersection of A and B (the vertical real line of zero abscissa) if they are closed. The same conclusion of (d) is valid for the best proximity points if ε > 0.

Corollary 2.15. The following properties hold.

• (i)

  Let (X, ∥ ∥) be a normed vector space with (X, d) being the associate metric space endowed with a norm-induced translation-invariant and homogeneous metric d : X × X → R  ₀₊ and consider the self-mapping T : X →   X so that the restricted T : A   ∪ B → X    is λ^*-accretive for some λ^* ∈ R₀₊, where A and B are nonempty subsets of X subject to 0 ∈   A ∪   B, and the restricted T : A ∪ B → A ∪ B  is 2-cyclic.Then,

  $$ (2.44)
  If, furthermore, T : A ∪ B →   A ∪ B  is k-contractive, then
  $$ (2.45)
  T : A ∪ B → A ∪ B  is guaranteed to be nonexpansive (resp., asymptotically nonexpansive) if

• (ii)

  Let (X, ∥ ∥) be a normed vector space. Then if T : A ∪ B → X  is a λ^*-accretive mapping and T : A ∪ B → A ∪ B  is 2-cyclic for some λ^* ∈   R₀₊ where A and B are nonempty subsets of X subject to 0 ∈ A ∪ B, then

  $$ (2.46)
  If, furthermore, T : A ∪ B → A ∪ B  is 2-cyclic k (∈[0, 1))-contractive, then
  $$ (2.47)

Outline of Proof It follows since the basic constraint of T : A ∪ B → X  being λ^*-accretive holds if

while it fails if

Remark 2.16. Theorem 2.13 and Corollary 2.15 are easily linked to Theorem 2.9 as follows. Assume that T : A ∪ B → A ∪ B is 2-cyclic k-contractive and T : A ∪ B →   X is a  λ^*-accretive mapping. Assume that there exists x ∈ A ∪ B such that ∥x∥ = d(x, 0) ≤ 1. Then, 1 ≤ ∥I   + λT∥ < k⁻¹; for all λ   ∈   [0, λ^*] from (2.47). This is guaranteed under sufficiency-type conditions with

with λ^* = k⁻²(k_c − k) for some real constants    k_c ∈ [k, 1), k ∈ [0,1). It is direct to see that Fix (T) = {0 ∈ Rⁿ} if 0 ∈ A∩B.

Example 2.17. Constraint (2.50) linking Theorem 2.13 and Corollary 2.15 to Theorem 2.9 is tested in a simple case as follows. Let A ≡ Dom (T) = B ≡ Im (T) ⊂ X ≡   Rⁿ. Rⁿ is a vector space endowed with the Euclidean norm induced by the homogeneous and translation-invariant Euclidean metricd : X × X → R₀₊. T is a linear self-mapping from Rⁿ to Rⁿ represented by a nonsingular constant matrix T in R^n×n. Then, ∥T∥  is the spectral (or ℓ₂-) norm of the k-contractive self-mapping T : X → X which is the matrix norm induced by the corresponding vector norm (the vector Euclidean norm being identical to the ℓ₂  vector norm as it is wellknown) fulfilling

with the symmetric matrix T^TT being a matrix having all its eigenvalues positive and less than one, since T is nonsingular, upper-bounded by a real constant k which is less than one. Thus, T : A ∪ B →   X is also λ^*-accretive for any real constant λ^* < k⁻²(1 − k) and k_c-contractive for any real k_c ∈ [k, 1). Assume now that for some integer 0 < p ≤ n with −k ≤ k_i(≠0) ≤   k < 1; for all $$. If p = n, then Fix (T) = {0 ∈ Rⁿ}. Also, Fix (T) = {0 ∈ Rⁿ}for any integer 0 < p < n    (then T is singular) but the last (n − p)-components of any   x ∈ A = X = Rⁿ are zeroed at the first iteration via T  so that if e_i    is the ith unit vector in Rⁿ    with its ith component being one, then Now, assume that the matrix T is of rank one with its first column being of the form with 0 < p < n, −k ≤ k_i(≠0) ≤ k < 1; for all $$. Then, (2.54) still holds by changing x ≠ 0 in the first equation to x₁ ≠ 0. Finally, assume that with 0 < p < n. Then, the self-mapping T : X → X is nonexpansive also noncontractive and Fix (T) = {0 ∈ R^p  } ⊕ R^n−p which is a vector subspace of Rⁿ, that is, there exist infinitely many fixed points each one being reached depending on the initial x in X with the property ∃  lim _j→∞  T^j  x   =   (0^T, y^T) ∈ Fix (T) for any given $$ with x ∈ R^p, y ∈ R^n−p.

Theorem 2.18. Let (X, ∥ ∥) be a normed vector space with (X, d) being the associated metric space endowed with a norm-induced translation-invariant and homogeneous metric d : X × X →   R  ₀₊. Let T : X∣A ∪ B → X∣A ∪ B  be a 2-cyclic k-contractive self-mapping so that T : A ∪ B → X is λ^*-accretive for some λ^* ∈   R₀₊ where A and B are nonempty subsets of X. Then,

for some finite real constants m₁ ∈ R₊, and m₂ ∈ R₀₊, which are independent of x and the jth power, and m₂  is zero if A   and B intersect. Furthermore, lim _j→∞sup  d(T^jx, T^j+1x) is finite irrespective of x ∈ A ∪ B.

Proof. One gets for λ   ∈   [0, λ^*], some λ^* ∈ R₀₊ and x ∈ A ∪ B that

so that one has for λ^* : = 1 − k⁻¹ε with ε   ∈   [ε₀, 1) for some real constant ε₀ ∈ [0,1) provided that k   ∈   (0, 1): and if k = 0 then Also, if k ∈ (0,   1), and if k = 0.

Theorem 2.19. Let (X, ∥ ∥) be a normed space with an associate metric space (X, d) endowed with a norm-induced translation-invariant and homogeneous metric d : X × X → R  ₀₊ and let T : X → X be a self-mapping on X which is k-contractive with k ∈ [0, 1/3) and 2-cyclic on A ∪ B, where A  and B are nonnecessarily disjoint nonempty subsets of X. If such sets A and B intersect then T : X →   X is also k_c-contractive with k_c : = k/(1 − 2k) = k/(1 − (2 + λ^*)k) ∈ [0,1) and λ^*-accretive with λ^* = ∞ if k_c = k = 0 and with $$ if k ∈ (0, 1/3). Irrespective of A and B being disjoint or not, T : A ∪ B → X is still λ^*-accretive and the following inequalities hold:

Proof. Direct calculations yield

which leads to the inequalities (2.65)–(2.67) with k_c : = k/(1 − (2 + λ^*)k) ∈ [0,1) and m : = (3 +   λ)(1 − k)/kk_c where k_c ∈ [0,1) with λ^* : = ∞ if k =   k_c = 0 and $$ if k_c : = k/(1 − 2k) ∈ (0,1) which holds if and only if k ∈ (0, 1/3). The proof is complete.

Remark 2.20. Compared to Theorem 2.9, Theorem 2.19 guarantees the simultaneous maintenance of the λ^*-accretive and contractive properties if the subsets of X intersect. Otherwise, the contractive property is not guaranteed if k > 0 to be λ^*-accretive for the nontrivial case of λ^* > 0 since m d_AB is larger than (1 − k_c)d_AB in general. However, the guaranteed value of λ^* is larger than that guaranteed in Theorem 2.9 to make compatible the accretive and contractive properties of the self-mapping. Also, the relevant properties (2.65)–(2.67) hold irrespective of the sets A and B being bounded or not. Note, in particular, that the uniformly bounded limit superior distance (2.67) is also independent of the boundedness or not of such subsets of X.

Theorem 2.21. Let (X, ∥ ∥) be a normed vector space with A and B being bounded nonempty subsets of X and 0 ∈ A ∪ B. Consider a 2-cyclic (k < 1/3)-contractive self-mapping T : A ∪ B → A ∪ B  with k ∈ [0, 1/3  ). Then, T : A ∪ B →   A ∪ B  is also a (k_c/(1 − k_c),   β)-Kannan self-mapping and T : A ∪ B → X is (1/3 − k)k⁻²-accretive for  k_c(∈R  ₊) = k(1 + k  λ^*), for all $$.

Proof. Since T : A ∪ B → A ∪ B is a 2-cyclic k(<1/3)-contractive self-mapping, then one gets for that the following relationships hold from the distance sub-additive property from the proof of Theorem 2.9(i), (2.15):

provided that since 1/3 > k_c : = k(1 + k  λ^*) ≥ k if λ^* : = (1/3 − k)k⁻² so that T : A ∪ B → X is (1/3 − k)k⁻²-accretive. Note that the function k = k(k_c) for a contractive self-mapping is the positive solution of λ^*k² + k − k_c = 0, that is, $$, which is wellposed since 0 ≤ k < 1 for 0 ≤ k_c < 1. Thus, T : A ∪ B → A ∪ B  is also a 2-cyclic (k_c/(1 − k_c),   β  )-Kannan self-mapping from Definition 2.6 since 0 ≤ k_c < 1/3    implies α : = k_c/(1 − k_c  ) < 1/2 with

Definition 3.1. $$is a p-cyclic weakly k-contractive (resp., weakly nonexpansive) self-mapping if

for some real constants k_i ∈ R₀₊  (resp., k_i ∈ R₊); for all $$ [12, 13] such that $$ (resp., k = 1).

Definition 3.2. $$ is a p-cyclic k-contractive (resp., nonexpansive) p-cyclic self-mapping if

for some real constants  k_i ∈ [0, 1) (resp., k_i = 1); for all $$ [12, 13].

Assertion 1. A p-cyclic weakly nonexpansive self-mapping $$ may be locally expansive for some (x,   y) ∈ A_i × A_i+1; for all $$ which cannot be best proximity points.

Proof. Assume that   k_i > 1. Then, the following inequalities can occur for given    x ∈ A_i, y ∈ A_i+1:

• (1)
  $$ (3.3)
  In this case, and since d(x, y) < d_i is impossible, one concludes that
  $$ (3.4)
  so that (3.3) can only hold for best proximity points x ∈ A_i,  y = Tx ∈ A_i+1 for which $$ is nonexpansive. If d_i = d_i+1, then the last inequality of (3.4) becomes d  (Tx,   Ty) = d_i = d_i+1 so that Ty ∈ A_i+2 is also a best proximity point if A_i    are convex, for all $$,

• (2)
  $$ (3.5)
  and then $$is nonexpansive for (x,   y) ∈ A_i × A_i+1;

• (3)
  $$ (3.6)
  and then $$is expansive for (x,   y) ∈ A_i × A_i+1 which cannot be best proximity points since d  (x,   y) > d_i.

Remark 3.3. Note from Definitions 3.1 and 3.2 that a p-cyclic weakly contractive (resp., contractive) self-mapping is also weakly nonexpansive (resp., weakly contractive). Also, a nonexpansive (resp., contractive) self-mapping is also weakly nonexpansive (resp., weakly contractive). Note that if $$, is p-cyclic weakly nonexpansive and d_i = d₁; for all $$, then

where k_ℓp+i = k_i; for all ℓ ∈   Z₀₊, for all $$. Note that if d(  T^j−1x,   T^j−1y) > d₁; that is, T^j−1x,   T^j−1y are not best proximity points, then if k_i+j−1 > 1,     then d(T^jx, T^jy) > d(T^j−1x, T^j−1y) sincek_i+j−1d(T^j−1x, T^j−1y) + (1 − k_i+j−1)d₁ > d(T^j−1x, T^j−1y). Thus, a weakly nonexpansive self-mapping is not necessarily nonexpansive for each iteration. However, the composed self-mapping $$ defined as Tx = T(T^p−1x) = T^px; for all $$ is nonexpansive in the usual sense since if j = p, then $$, implies

Lemma 3.4. Assume that $$ is p-cyclic and nonexpansive. Then, d_i = d₁; for all $$.

Proof. If $$(i.e., the closures of the subsets intersect), then the proof is direct since d_i = 0; for all $$. Now, assume that 0 ≤ d_j < d₁ ≠   0 for some $$. Let z ∈ A₁  and z₁ = Tz ∈ A₂ best proximity points such that

since $$ is a p-cyclic nonexpansive self-mapping. Thus, any iterates T^jz and T^jz₁ are also best proximity points of some subset in$$; for all j ∈   Z₊. If d_j =   d₁; for all $$ does not hold, then from (3.9): Then d_i = 0;  for all $$which contradicts 0 ≤ d_j < d₁ ≠   0 for some $$ what is a contradiction or d_i = 0; for all $$, and $$.

Lemma 3.5. The following properties hold.

• (i)

  If $$ is a p-cyclic weakly nonexpansive self-mapping, then

  $$ (3.11)
  if$$satisfy
  $$ (3.12)

• (ii)

  If $$ is a p-cyclic nonexpansive self-mapping, then

• (iii)

  If $$ is a p-cyclic weakly contractive self-mapping, then

  $$ (3.14)
  for all x ∈ A_i, for all y ∈ A_i+1 if $$ satisfy the feasibility constraints $$ and $$. If d_i = d₁; for all $$, then
  $$ (3.15)
  for all x ∈ A_i, for all y ∈ A_i+1, for all $$ if $$ satisfy the feasibility constraints $$and $$

• (iv)

  If $$is a p-cyclic contractive self-mapping, then

• (v)

  If $$ and $$is a p-cyclic weakly contractive self-mapping, then

Proof. Property (i) follows from (3.7) for $$. Property (iii) follows from Property (i) since k < 1implies $$ as j → ∞. Property (ii) Follows from Property (i) for k_i = 1; for all $$ since d_i = d₁; for all $$ from Lemma 3.4. Property (iv) follows from Property (ii) for k_i < 1; for all $$since d_i = d₁; for all $$ from Lemma 3.4. Property (v) follows from Property (iii) since if all the subsets A_i; $$ intersect, then it follows necessarily d_i = d₁ = 0; for all $$ so that

Remark 3.6. Note that Lemma 3.5(v) also applies to contractive self-mappings since contractive self-mappings are weakly contractive.

Theorem 3.7. Assume that $$ is a p-cyclic weakly k-contractive self-mapping and the closures of the p subsets A_i; $$ of X intersect. Then, it exists a unique fixed point in $$ which is also in $$ if all such subsets    A_i; for all $$ of X, are closed.

Proof. The existence of a fixed point follows from Lemma 3.5(v). Its uniqueness follows by contradiction. Assume that there exist $$. Then, for some $$,  ∃ x ∈ A_i, y ∈ A_i+1 such that T^jx → z₁ and T^jy → z₂    as j → ∞. Then, by using triangle inequality for distances,

which implies by using Lemma 3.5(v) what contradicts z₁ ≠ z₂. Therefore, Fix (T) consists of a unique point in $$ which is also in $$ if the sets A_i; $$ are all closed.

Theorem 3.8. Let (X, ∥ ∥) be a uniformly convex Banach space endowed with the translation-invariant and homogeneous metric d : X × X → R₀₊    with nonempty convex subsets  A_i ⊂ X, for all $$ of pair-wise disjoint closures. Let $$ be a p-cyclic weakly k-contractive self-mapping so that the composed 2-cyclic self-mappings.    T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1, for all $$ are defined as T_ix = T  (T^p−1x); for all x ∈ A_i ∪ A_i+1; $$. Then, the following properties hold:

• (i)

  Any composed 2-cyclic self-mapping T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1, $$ is k-contractive provided that the constraint $$ holds. If, furthermore, it is assumed that A_i and A_i+1 are convex, then the 2-cyclic self-mapping T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1 self-mapping is extendable to T_i : cl (A_i ∪ A_i+1) → cl (A_i ∪ A_i+1), and that T(cl   A_i)⊆cl A_i+1; for all $$. Thus, the iterates $$ and $$; for all x ∈ A_i, for all y ∈ A_i+1 converge as j → ∞ to best proximity points in cl (A_i) and cl (A_i+1) which are also in A_i if A_i is closed, respectively, in A_i+1 if A_i+1 is closed.

• (ii)

  If for some given $$, the sets A_i and A_i+1 are convex and closed, if any, then both best proximity points of T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1 of Property (i) are unique and belong, respectively, to A_i and A_i+1.

• (iii)

  Assume that the subsets A_i of X are convex, for all $$. If $$, then the best proximity points of Property (i) become a unique fixed point for all the composed 2-cyclic self-mappings T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1 which are k-contractive, $$. Such a fixed point is in $$ (and also in $$ if all the subsets A_i, $$, are closed).

Proof. Since    T(A_i)⊆A_i+1; for all $$, then for any $$, x ∈ A_i⇒T_ix ∈ A_i and x ∈ A_i+1⇒T_ix ∈ A_i+1    if p is even and x ∈ A_i⇒T_ix ∈ A_i+1 and x ∈ A_i+1⇒T_ix ∈ A_i if p is odd. Since $$ is p-cyclic weakly k-contractive then $$, then T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1  is 2-cyclic contractive provided that $$; $$. One has from Lemma 3.5(iv) that $$; for all x ∈ A_i, for  all  y ∈ A_i+1 for the given$$, where z = z  (i) ∈ cl (A_i)(z ∈ A_i if A_i is closed), z₁ = z₁(i) ∈ cl (A_i+1)(z ∈ cl (A_i)(z ∈ A_i+1 if A_i+1 is closed)) are best proximity points. Using Theorem 2.9(i) for 2-cyclic self-mappings in uniformly convex Banach spaces endowed with translation-invariant and homogeneous metric, one gets$$and $$as j → ∞; for all $$. Property (i) has been proven. Property (ii) was proven in Theorem  3.10, [12] for 2-cyclic k-contractive self- mappings in uniformly convex Banach spaces since they can be directly endowed with a norm-induced metric. The proof is valid here for a norm- induced distance in a uniformly convex Banach space since such distances are translation-invariant and homogeneous. It is also valid if the subsets are not closed with the fixed point then being in the nonempty intersection of their closures. Property (iii) follows directly from Lemma 3.5(v), which implies that T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1 is k-contractive for all $$, and the fact that all distances between the closures of all pairs of adjacent subsets are zero since (X, d) is a complete metric space since X is a Banach space.

Theorem 3.9. Let (X, ∥ ∥) be a uniformly convex Banach space endowed with the norm-induced translation-invariant and homogeneous metric d : X × X →   R₀₊ with nonempty subsets A_i ⊂ X, for all $$ of pair-wise disjoint closures. Let $$ be a p-cyclic k-contractive self-mapping so that the composed 2-cyclic self-mappings T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1, for all $$, are defined as T_ix = T  (T^p−1x); for all x ∈ A_i ∪ A_i+1; for all $$. Assume also that A_i is convex and T(cl  A_i)⊆cl  A_i+1; for all $$. Then, the following properties hold.

• (i)

  As j → ∞, the iterates T^jx and T^jy; for all x ∈ A_i, for all y ∈ A_i+1  converge to best proximity points in cl (A_i) and cl (A_i+1) which are also in A_i if A_i is closed, respectively, in A_i+1 if A_i+1 is closed for any$$. Also, for any given $$ such that the sets A_i and A_i+1 are convex and closed, if any, then both best proximity points of T : A_i ∪ A_i+1 → A_i ∪ A_i+1  of Property (i) are unique and belong, respectively, to A_i and A_i+1. If, furthermore, $$, then the best proximity points of Property (i) become a unique fixed point for the p-cyclic k-contractive self-mapping   T : A_i ∪ A_i+1 → A_i ∪ A_i+1. Such a fixed point is in $$ (and also in $$ if all the subsets A_i ⊂ X, $$ are closed).

• (ii)

  All the composed 2-cyclic self-mappings T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1, for all $$ are k-contractive. Thus, the iterates $$ and $$; for all x   ∈ A_i, for all y ∈ A_i+1converge as j → ∞ to best proximity points in cl (A_i)    and cl (A_i+1) which are also in A_i if A_i is closed, respectively in A_i+1 if A_i+1 is closed. For any given $$ such that the sets A_i and A_i+1 are closed and convex, if any, then both best proximity points of T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1 of Property (i) are unique and belong, respectively, to A_i and A_i+1. If, furthermore, $$, then the best proximity points of Property (i) become a unique fixed point for all the composed 2-cyclic self-mappings T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1 which are k-contractive; $$. Such a fixed point is in $$(and also in $$ if all the subsets A_i, $$ are closed).

Outline of Proof Property (ii) is the direct version of Theorem 3.8 applicable to the composed 2-cyclic self-mappings T_i : A_i ∪ A_i+1 → A_i ∪ A_i+1 which are all k-contractive since $$ is a p-cyclic k-contractive self-mapping. Since p-cyclic contractive self-mappings are nonexpansive, all the distances between adjacent subsets are identical (Lemma 3.4) so that there is no mutual constraint on distances contrarily to Theorem 3.8(i). Property (i) is close to Property (ii) by taking into account that $$ is also k-contractive.

Definition 3.10. $$ is a 2-cyclic (α, β)-Kannan self-mapping for some real α ∈ [0, 1/2) if it satisfies for some β   ∈   R₊:

Now, Theorem 2.9 and Theorems 2.18–2.21 for 2-cyclic accretive and Kannan self-mappings extend directly with direct replacements of their relevant parts as follows:

Theorem 3.11. Let (X, ∥ ∥) be a Banach space so that (X, d)is its associate complete metric space endowed with a norm-induced translation-invariant and homogeneous metric d : X × X → R₀₊. Consider a self-mapping T : X → X which is also a p-cyclic k-contractive self-mapping if restricted t$$, where A_i are nonempty convex subsets of X; for all $$. Then, Theorem 2.9 holds “mutatis-mutandis” by replacing the subsets A and B for pairs of adjacent subsets A_i and A_i+1, $$, $$, $$ and $$. In the same way, Theorems 2.18, 2.19, and 2.21 still hold.