Fixed Points of G-Type Quasi-Contractions on Graphs

Theorem 1. Let (X, d) be a complete metric space and let T be a self-map on X. Define the nonincreasing function θ from [0,1) onto (1/2, 1] by

Assume that there exists r ∈ [0,1), such that for all x, y ∈ X. Then, there exists a unique fixed point z of T. Moreover, lim _n→∞Tⁿx = z for all x ∈ X.

Definition 2. Let (X, d) be a metric space, T a self map on X, and G a graph with V(G) = X. We say that T is a G-type quasi-contraction whenever T preserves the edges of G and there exists r ∈ [0,1), such that

for all (x, y) ∈ E, where

Theorem 3. Let (X, d) be a complete metric space, T a G-type quasi-contraction map with r² + r < 1 such that (x, Tx), (x, T²x) ∈ E(G) for all x ∈ X. Then, T has a unique fixed point.

Proof. Take x₀ ∈ X. Since θ(r)d(x₀, Tx₀) ≤ d(x₀, Tx₀), we have

Since (1/2)d(x₀, T²x₀)≤(1/2)[d(x₀, Tx₀) + d(Tx₀, T²x₀)], we obtain d(T²x₀, Tx₀) ≤ rd(x₀, Tx₀). Hence, for all natural number n and so {Tⁿx₀} _n≥1 is a Cauchy sequence. Since X is complete, {Tⁿx₀} _n≥1 converges to some x^* ∈ X. Since G is a C-Graph, there is a subsequence $$ such that $$ for all k ≥ 1. Hence, $$ for all j ≥ 1. We claim that $$ for some natural number j₀. Arguing by contradiction, we assume that T^jx^* ≠ x^* for all j. Fix a natural number j and put x_n+1 = Tⁿx₀ for all n ≥ 1. Choose a natural number n₀ such that d(x_n, x^*) ≤ d(x^*, T^jx^*)/3 for all n ≥ n₀. If n_k ≥ n₀, then It follows that and so d(x^*, T^j+1x^*) ≤ rmax {d(x^*, T^jx^*), d(T^jx^*, T^j+1x^*)}. Since d(T^jx^*, T^j+1x^*) ≤ r^jd(x^*, Tx^*), we obtain for all j. Now, we assume that d(x^*, T²x^*) < d(T²x^*, T³x^*); then by (6), we have This is a contradiction, since r² + r < 1. So, we have and by (3), we obtain By considering the above inequality and (9), we deduce that that is a contradiction. Therefore, there exists j₀ ∈ ℕ such that $$. Since {Tⁿx^*} _n≥1 is a Cauchy sequence, we obtain x^* = Tx^*. In fact, if x^* ≠ Tx^*, from $$ for all n ≥ 1, it follows that {Tⁿx^*} _n≥1 is not a Cauchy sequence. Thus, x^* is a fixed point of T. The uniqueness of the fixed point follows easily.

Question 1. Does Theorem 3 hold for each r ∈ [0,1)?

Theorem 4. Let (X, d) be a complete metric space. Then, the following statements are equivalent

• (i)

  G  is weakly connected,

• (ii)

  for each G-type quasi-contraction map T : X_T → X_T and x, y ∈ X, the sequences {Tⁿx} _n≥1 and {Tⁿy} _n≥1 are Cauchy equivalent, where X_T = {x ∈ X : (x, Tx) ∈ E},

• (iii)

  for each G-type quasi-contraction map T : X → X, card(FixT) ≤ 1.

Proof. (i)⇒(ii) Let T : X → X be a G-type quasi-contraction map and x, y ∈ X. Since $$, there is a path {x₀ = x, …, x_N = y} in $$ from x to y. Since $$, $$ for all n and i = 1, …, N. Let 1 ≤ i ≤ N. Put x_i−1 = a and x_i = b. If one of the following inequalities holds

Then, we have If u_n ∈ {d(Tⁿ⁻¹a, Tⁿa), d(Tⁿ⁻¹b, Tⁿb)}, then If u_n = (1/2)[d(Tⁿ⁻¹a, Tⁿb) + d(Tⁿ⁻¹b, Tⁿa)], then Without loss of generality, suppose that d(Tⁿa, Tⁿb) ≤ rd(Tⁿ⁻¹a, Tⁿb). Then, and so d(Tⁿa, Tⁿb)≤(r/(1 − r))d(Tⁿ⁻¹a, Tⁿa)≤(rⁿ/(1 − r))d(a, Ta). Hence, Now, suppose that both of the inequalities (14) do not hold. If then and so If u_n = d(Tⁿ⁻¹a, Tⁿ⁻¹b), then we can continue in a similar process for n − 1. In the general case, we get d(Tⁿa, Tⁿb)≤(3rⁿ⁻¹/(1 − r))max {d(a, Ta), d(b, Tb)} and so d(Tⁿa, Tⁿb) → 0. Thus, Therefore, {Tⁿx} _n≥1 and {Tⁿy} _n≥1 are Cauchy equivalent.

(ii)⇒(iii) Let x, y ∈ Fix T. By using (ii) and the above process, we obtain easily that x = y.

(iii)⇒(i) If G is not weakly connected, then there exists x₀ such that $$ is not empty. Take $$ and define

Clearly, Fix T = {x₀, y₀}. Now, we show that T is a G-type quasi-contraction. For this reason, let (x, y) ∈ E. Since $$, either $$ or $$. In both cases, we get Tx = Ty. Thus, T is a G-type quasi-contraction which has two fixed points. This contradiction completes the proof.

Theorem 5. Let (X, d) be a complete metric space and let T be a G-type quasi-contraction and orbitally G-continuous self-map on X. Then,

• (i)

  for each x ∈ X_T, $$ is a Picard operator,

• (ii)

  $$.

Proof. Let x ∈ X_T. Then, $$. It is easy to check that {Tⁿ} _n≥1 is a Cauchy sequence. Let lim _n→∞Tⁿx = x^*. Since G is a C-Graph, there exists a subsequence $$ such that $$ for all k. Thus, $$ for all k. Since $$, $$. Since T is orbitally G-continuous, $$ which yields x^* = Tx^*. To prove (ii), define the mapping π by $$ for all x ∈ Fix T. It is sufficient to show that π is a bijection from Fix T onto $$. Since Δ⊆E(G), we get Fix T⊆X_T which yields π(Fix T)⊆𝒞. On the other hand, if x ∈ X_T, then $$ which implies that $$. Thus, π is a surjection from Fix T onto 𝒞. Now, if x₁, x₂ ∈ Fix T with π(x₁) = π(x₂), then $$ and so by using (i) we obtain

which implies that x₂ = x₁. Therefore, T is an injective and this completes the proof.

Lemma 6 (see [18].)Let X be a nonempty set and let T : X → X be a mapping. Then, there exists a subset Y⊆X such that TY = TX and T : Y → X is one-to-one.

Lemma 7 (see [8].)Let X be a nonempty set and that the mappings f, T : X → X have a unique point of coincidence v in X. If T and f are weakly compatible, then T and f have a unique common fixed point.

Theorem 8. Let (X, d) be a metric space, and let f and T be two self-maps on X such that TX⊆fX and fX is complete. Suppose that f and T satisfy the following conditions:

• (i)

  (fx, fy) ∈ E(G) implies that (Tx, Ty) ∈ E(G),

• (ii)

  if (fx, Tx) ∈ E(G) and Tx = fy for some y ∈ X, then (fx, Ty) ∈ E(G),

• (iii)

  there exists r ∈ [0,1) such that r² + r < 1 and θ(r)d(fx, Tx) ≤ d(fx, fy) implies that

  $$ ()

Then, T and f have a unique coincidence point. Moreover, if T and f are weakly compatible, then T and f have a unique fixed point.

Proof. By using Lemma 6, there exists Y ⊂ X such that f : Y → X is one-to-one and fY = fX. Define the self-map h : fY → fY by h(fx) = Tx. Clearly, h is well defined and h preserves the edges of G. In fact, (fx, fy) ∈ E(G) implies that (hfx, hfy) ∈ E(G). Note that θ(r)d(fx, hfx) ≤ d(fx, fy) implies that

Also, (fx, hfx) and (fx, h²fx) lie in E(G) for all x ∈ Y. To see this, take hfx = Tx. Then, Tx = fy for some y ∈ Y and so h²fx = Ty. By using (ii), (fx, Ty) ∈ E(G). Since fY is complete, by using Theorem 3, h has a unique fixed point in fY, namely, hfx^* = fx^*. Thus, x^* is a coincidence point of f and T. Note that the assumption (iii) shows the uniqueness of the coincidence point of f and T. Now, by using Lemma 7, it is easy to see that if f and T are weakly compatible, then f and T have a unique fixed point.