Fixed Point Theorems of Quasicontractions on Cone Metric Spaces with Banach Algebras

Proposition 1 (see [7].)Let  A  be a Banach algebra with a unit  e, and let   x ∈ A. If the spectral radius  ρ(x)  of  x  is less than 1, that is,

then  e − x  is invertible. Actually,

Remark 2. In the literature on cone metric spaces, authors use  x < y  to mean  x ⩽ y  and  x ≠ y and  x ≪ y  to mean  y  −  x ∈ int P. To our knowledge, and from a topological point of view, the order relation  y − x ∈ int P  plays a very similar role in cone metric spaces as  x < y  does inℝ.

Definition 3 (see [8].)Let  X  be a nonempty set. Suppose the mapping  d :   X × X → A  satisfies

• (1)

  0 ⩽ d(x, y)  for all  x, y ∈ X  and  d(x, y) = 0  if and only if  x = y;

• (2)

  d(x, y) = d(y, x)  for all  x, y ∈ X;

• (3)

  d(x, y) ⩽ d(x, z) + d(z, x)  for all  x, y, z ∈ X.

Then, d is called a cone metric on  X, and  (X, d)  is called a cone metric space (with Banach algebra  A).

Definition 4 (see [8].)Let  (X, d)  be a cone metric space, and let   x ∈ X  and  {x_n}   be a sequence in  X. Then,

• (1)

  {x_n}  converges to  x  whenever for each c ∈ A with   0 < c   there is a natural number  N  such that   d(x_n, x) < c   for all   n⩾N. We denote this by   lim _n→∞x_n = x   or   x_n → x;

• (2)

  {x_n}  is a Cauchy sequence whenever for each   c ∈ A   with   0 < c   there is a natural number   N   such that   d(x_n, x_m) < c   for all   n, m⩾N;

• (3)

  (X, d)  is a complete cone metric space if every Cauchy sequence is convergent.

Proposition 5 (see [8].)Let  (X, d)  be a cone metric space, let   P  be a normal cone with normal constant  M, and let  {x_n}  be a sequence in  X. Then, {x_n}  converges to  x  if and only if  d(x_n, x) → 0  (n → ∞).

Proposition 6 (see [8].)Let  (X, d)  be a cone metric space, let   P  be a normal cone with normal constant  M, and let  {x_n}  be a sequence in  X. Then,   {x_n}  is a Cauchy sequence if and only if d(x_n,   x_m) → 0  (n,   m → ∞).

Definition 7. Let  (X, d)  be a cone metric space with Banach algebra  A. A mapping  T : X → X  is called a quasicontraction if for some  k ∈ P  with  ρ(k) < 1  and for all  x, y ∈ X, one has

where

Remark 8. In Definition 7, we only suppose the spectral radius of  k  is less than 1, while neither  k < e  nor  ∥k∥<1  is assumed. In fact, the condition  ρ(k) < 1  is weaker than that  ∥k∥<1. See the example in [6].

Theorem 9. Let  (X, d)  be a complete cone metric space with a Banach algebra  A, and let   P  be a normal cone with normal constant  M. If the mapping  T :   X → X  is a quasicontraction, then  T  has a unique fixed point in  X. And for any  x ∈ X, iterative sequence  {Tⁿx}  converges to the fixed point.

Lemma 10. Assume that the hypotheses in Theorem 9 are satisfied. Then, for each  n⩾1, and for all  i, j  such that  1 ⩽ i, j ⩽ n, one has

Proof. We present the proof by induction.

When  n = 1, which implies  i = j = 1, the conclusion is trivial.

Assume that the statement is true for  n = m; that is,

Now, we will prove that the statement is true for  n = m + 1. Note that in this case, if  1 ⩽ i,   j ⩽ m, then the statement is just (8). Thus, without loss of generality, we suppose that  j = m + 1  and  1 ⩽ i ⩽ m and denote  i = i₀.

By the definition of quasicontraction, we have

where

Firstly, we consider the case that  i₀ = 1; that is,

If  u = d(x₀, x_m), then

and the statement follows.

If  u = d(x₀, x₁), then

and the statement also follows.

If  u = d(x_m, x_m+1), then we set  i₁ = m  and we have

If  u = d(x₀, x_m+1), then

which implies Note that$$ and that  k  and  (e − k) ⁻¹  commute. Multiplying both sides by  (e − k) ⁻¹, we have and the statement also follows.

If$$, then

and the statement also follows.

Secondly, we consider the case that  2 ⩽ i₀ ⩽ m.

If$$ or$$ or$$, then, by (8), we have

and the statement follows.

If  u = d(x_m, x_m+1) or$$, then we set  i₁ = m  or  i₁ = i₀ − 1⩾1, respectively. And we have

In conclusion from discussions of both cases, it results that either the proof is complete, that is,

or there exists an integer  i₁  such that

As for the latter situation, we continue in a similar way, and come to the result that either

which implies that and the proof is complete, or there exists an integer  i₂  such that which implies that

Generally, if the procedure ends by the ℓ-th step with ℓ ⩽ m − 1, that is, there exist ℓ + 1  integers

such that and such that then Hence, the proof is complete.

Finally, if the procedure continues more than  m  steps, then there exist  m + 1  integers

such that Thus, there must exist two integers,  p  and  q, say, such that From (32), one sees that and therefore Note that which implies  e − k^q−p  is invertible. And since that we have So,

Therefore, by induction, the statement is proved.

Remark 11. Lemma 10 simply says that

Lemma 12. Assume that the hypotheses in Theorem 9 are satisfied. Then, {x_n}  is a Cauchy sequence.

Proof. For  1 < m < n, denote that

By the definition of quasicontraction, it follows that, for each  u ∈ C(m, n), there exists  v ∈ C(m − 1, n), such that Consequently, where and the last inequality comes from Lemma 10.

By the normality of  P, and noting that  ∥k^m∥→0  (m → ∞), we have

The proof is complete.

Proof. By Lemma 12 and the completeness of  (X, d), there is  x^* ∈ X  such that  x_n → x^*  (n → ∞). Then,

where

If  u = d(x_n−1, x^*) or  u = d(x_n−1, x_n) or  u = d(x^*, x_n), then  ∥u∥→0  (n → ∞). Hence,

If  u = d(x^*, Tx^*), then

Hence,

If  u = d(x_n−1, Tx^*), then

Hence, as  n → ∞.

In each case, we have  ∥d(x^*, Tx^*)∥ = 0. Thus, Tx^* = x^*.

Now, if  y^*  is another fixed point, then

where

If  u = d(x^*, Tx^*) = d(y^*, Ty^*) = 0, then  d(x^*, y^*) = 0.

If  u = d(x^*, y^*) = d(x^*, Ty^*) = d(y^*, Tx^*), then

which implies

Thus, the fixed point is unique. And we obtain Theorem 9.