In this section, we extend Definition 1.1 to introduce the notion b-metric space in the setting of \(C^{*}\)-algebras as follows.
Definition 2.1
Let \(\mathbb{A}\) be a \(C^{*}\)-algebra, and X be a nonempty set. Let \(b \in\mathbb{A}\) be such that \(\|b \| \geq1\). A mapping \(d_{b}\colon X \times X \rightarrow\mathbb{A}_{+} \) is said to be a \(C^{*}\)-algebra-valued b-metric on X if the following conditions hold for all \(x_{1},x_{2},x_{3} \in\mathbb{A}\):
-
(BM1)
\(d_{b}(x_{1},x_{2})=0_{\mathbb{A}} \Leftrightarrow x_{1}=x_{2} \).
-
(BM2)
\(d_{b}\) is symmetric, that is, \(d_{b}(x_{1},x_{2})=d_{b}(x_{2},x_{1})\).
-
(BM3)
\(d_{b}(x_{1},x_{2})\preceq b [d_{b}(x_{1},x_{3})+d_{b}(x_{3},x_{2})] \).
The triplet \((X,\mathbb{A}, d_{b})\) is called a \(C^{*}\)-algebra-valued b-metric space with coefficient b.
Remark 2.1
Note that:
-
(1)
If we take \(\mathbb{A}=\mathbb{R}\), then the new notion of \(C^{*}\)-algebra-valued b-metric space becomes equivalent to Definition 1.1 of the real b-metric space.
-
(2)
If we take \(b=1_{\mathbb{A}}\) in Definition 2.1, then \(d_{b}\) becomes the usual \(C^{*}\)-algebra-valued metric as defined in [11].
Thus, the class of ordinary \(C^{*}\)-algebra-valued metric spaces is clearly smaller than the class of \(C^{*}\)-algebra-valued b-metric spaces. In fact, there are \(C^{*}\)-algebra-valued b-metric spaces that are not \(C^{*}\)-algebra-valued metric spaces, as illustrated by the following example.
Example 2.1
Let \(X=\ell_{p}\) be the set of sequences \(\{x_{n}\}\) in \(\mathbb{R}\) such that \(\sum_{n=1}^{\infty}|x_{n}|^{p} < \infty\) and \(0< p<1\). Let \(\mathbb {A}=M_{2}(\mathbb{R})\). For \(x=x_{n}, y=y_{n} \in\ell_{p}\), define \(d_{b}:X \times X \rightarrow \mathbb{A}\) as follows:
$$d_{b}(x,y) = \begin{pmatrix} (\sum_{n=1}^{\infty}|x_{n}-y_{n}|^{p} )^{\frac{1}{p}} & 0 \\ 0 & (\sum_{n=1}^{\infty}|x_{n}-y_{n}|^{p} )^{\frac{1}{p}} \end{pmatrix}. $$
Then one can show that \(d_{b}\) is a \(C^{*}\)-algebra-valued b-metric space with coefficient \(b =\bigl( {\scriptsize\begin{matrix}{} 2^{\frac{1}{p}} & 0 \cr 0 & 2^{\frac{1}{p}} \end{matrix}}\bigr) \) such that \(\|b\|=2^{\frac{1}{p}}\). The claim follows from the following observation in [4]:
$$ \Biggl(\sum_{n=1}^{\infty}|x_{n}-z_{n}|^{p} \Biggr)^{\frac{1}{p}} \le 2^{\frac{1}{p}} \Biggl[ \Biggl(\sum _{n=1}^{\infty}|x_{n}-y_{n}|^{p} \Biggr)^{\frac {1}{p}} + \Biggl(\sum_{n=1}^{\infty}|y_{n}-z_{n}|^{p} \Biggr)^{\frac{1}{p}} \Biggr]. $$
Note that here \(d_{b}\) is not a usual \(C^{*}\)-algebra-valued metric on X.
From now on, we call a \(C^{*}\)-algebra-valued b-metric space simply a \(C^{*}\)-valued b-metric, and the triplet \((X,\mathbb{A},d_{b})\) is then called a \(C^{*}\)-valued b-metric space. Given \((X,\mathbb{A},d_{b})\), the following are natural deductions from the corresponding notions in \(C^{*}\)-valued metric spaces.
-
(1)
A sequence \(\lbrace x_{n} \rbrace\) in X is said to be convergent to a point \(x \in X\) with respect to the algebra \(\mathbb{A}\) if and only if for any \(\epsilon>0\), there is an \(N \in\mathbb{N}\) such that \(\|d_{b}(x_{n},x) \| < \epsilon\) for all \(n> N\). Symbolically, we then write \(\lim_{n\rightarrow \infty} x_{n}=x\).
-
(2)
If for any \(\epsilon>0\), there exists \(N \in\mathbb{N}\) such that \(\|d_{b}(x_{n},x_{m}) \| < \epsilon\) for all \(n, m > N\), then the sequence \(\lbrace x_{n} \rbrace\) is called a Cauchy sequence with respect to \(\mathbb{A}\).
-
(3)
If every Cauchy sequence in X is convergent with respect to \(\mathbb{A}\), then the triplet \((X,\mathbb{A},d)\) is called a complete \(C^{*}\)-valued b-metric space.
Definition 2.2
Let \((X,\mathbb{A}, d_{b}) \) be a \(C^{*}\)-valued b-metric space. A contraction on X is a mapping \(T\colon X \rightarrow X \) if there exists an \(a\in\mathbb{A}\) with \(\| a \| < 1\) such that
$$ d_{b}(Tx,Ty)\preceq a^{*}d_{b}(x,y)a \quad\mbox{for all } x,y \in X. $$
(1)
Example 2.2
Let \(\mathbb{A}= \mathbb{R}^{2}\) and \(X=[0,\infty)\). Let ⪯ be the partial order on \(\mathbb{A}\) given by
$$\begin{aligned}& (a_{1},b_{1})\preceq(a_{2},b_{2}) \quad\Leftrightarrow\quad a_{1} \leq a_{2} \mbox{ and } b_{1} \leq b_{2}. \end{aligned}$$
Define
$$d_{b}\colon X \times X \rightarrow\mathbb{A},\qquad d_{b}(x,y)= \bigl((x-y)^{2},0\bigr). $$
Then \(d_{b}\) is \(C^{*}\)-valued b-metric with coefficient \((2,0)\), and with this \(d_{b}\), the triplet \((X,\mathbb{A},d_{b})\) becomes a \(C^{*}\)-valued b-metric. Consider \(T\colon X \rightarrow X\) given by \(Tx=\frac{x}{3}+5\); then T is a contraction on X with \(a=(\frac{1}{3},0)\):
$$\begin{aligned}[b] d_{b}(Tx,Ty)= \bigl((Tx-Ty)^{2},0 \bigr) = \biggl( \biggl(\frac{x}{3}-\frac{y}{3} \biggr)^{2},0 \biggr) = \biggl(\frac{1}{3},0 \biggr)d_{b}(x,y) \biggl( \frac{1}{3},0 \biggr). \end{aligned} $$
Theorem 2.1
Consider a complete
\(C^{*}\)-valued
b-metric space
\((X,\mathbb{A},d_{b})\)
with coefficient
b. Let
\(T\colon X \rightarrow X\)
be a contraction with the contraction constant
a
such that
\(\| b\| \|a \|^{2} < 1 \). Then
T
has a unique fixed point in
X.
Proof
If \(\mathbb{A} = \{0_{\mathbb{A}}\}\), then there is nothing to prove. Assume that \(\mathbb{A}\ne\{0_{\mathbb{A}}\}\).
Choose \(x_{0} \in X\) and define inductively a sequence \(\{x_{n}\}\) by the iterative scheme as
Then it follows that \(x_{n}=T^{n}x_{0}\) for \(n=0,1,2, \ldots\) . From the contraction condition (1) on T it follows that
$$\begin{aligned} d_{b}(x_{n},x_{n+1}) =& d_{b}(Tx_{n-1},Tx_{n}) \\ \preceq& a^{*}d_{b}(x_{n-1},x_{n})a \\ =& a^{*}d_{b}(Tx_{n-2},Tx_{n-1})a \\ \preceq& \bigl(a^{*}\bigr)^{2}d_{b}(x_{n-2},x_{n-1})a^{2} \\ \preceq& \bigl(a^{*}\bigr)^{3}d_{b}(x_{n-3},x_{n-2})a^{3} \preceq \bigl(a^{*}\bigr)^{n}d_{b}(x_{0},x_{1})a^{n}= \bigl(a^{*}\bigr)^{n}Da^{n}, \end{aligned}$$
where \(D=d_{b}(x_{0},x_{1})\).
Now suppose that \(m>n\); then the triangle inequality (BM3) for the b-metric \(d_{b}\) implies
$$\begin{aligned} d_{b}(x_{n},x_{m}) \preceq& b d(x_{n},x_{n+1}) + b^{2}d(x_{n+1},x_{n+2})+ \cdots+ b^{m-n-1}d(x_{m-2},x_{m-1}) \\ &{}+ b^{m-n-1}d(x_{m-1},x_{m}) \\ \preceq& b\bigl(a^{*}\bigr)^{n}Da^{n} +b^{2} \bigl(a^{*}\bigr)^{n+1}Da^{n+1} + \cdots+ b^{m-n-1} \bigl(a^{*}\bigr)^{m-2}Da^{m-2} \\ &{}+ s^{m-n-1}\bigl(a^{*}\bigr)^{m-1}Da^{m-1} \\ =& b\bigl[\bigl(a^{*}\bigr)^{n}Da^{n} +b\bigl(a^{*} \bigr)^{n+1}Da^{n+1} + \cdots+ b^{m-n-2}\bigl(a^{*} \bigr)^{m-2}Da^{m-2}\bigr] \\ &{}+ b^{m-n-1}\bigl(a^{*}\bigr)^{m-1}Da^{m-1} \\ =& b\sum_{k=n}^{m-2}b^{k-n} \bigl(a^{*}\bigr)^{k}Da^{k} + b^{m-n-1}\bigl(a^{*} \bigr)^{m-1}Da^{m-1} \\ =& b\sum_{k=n}^{m-1}b^{k-n} \bigl(a^{*}\bigr)^{k}D^{\frac{1}{2}}D^{\frac{1}{2}}a^{k} + b^{m-n-1}\bigl(a^{*}\bigr)^{m-1}D^{\frac{1}{2}}D^{\frac{1}{2}}a^{m-1} \\ =& b\sum_{k=n}^{m-1}b^{k-n} \bigl(D^{\frac{1}{2}}a^{k}\bigr)^{*} \bigl(D^{\frac{1}{2}}a^{k} \bigr) + b^{m-n-1}\bigl(D^{\frac{1}{2}}a^{m-1}\bigr)^{*} \bigl(D^{\frac{1}{2}}a^{m-1}\bigr) \\ =& b\sum_{k=n}^{m-1}b^{k-n}\bigl|D^{\frac{1}{2}}a^{k}\bigr|^{2} + b^{m-n-1}\bigl|D^{\frac {1}{2}}a^{m-1}\bigr|^{2} \\ \preceq& \Biggl\| b\sum_{k=n}^{m-1}b^{k-n}\bigl|D^{\frac{1}{2}}a^{k}\bigr|^{2} \Biggr\| 1_{\mathbb{A}} + \bigl\| b^{m-n-1}\bigl|D^{\frac{1}{2}}a^{m-1}\bigr|^{2} \bigr\| 1_{\mathbb{A}} \\ \preceq& \|b\|\sum_{k=n}^{m-1} \bigl\| b^{k-n}\bigr\| \bigl\| D^{\frac{1}{2}}\bigr\| ^{2} \bigl\| a^{k} \bigr\| ^{2} 1_{\mathbb{A}} + \bigl\| b^{m-n-1}\bigr\| \bigl\| D^{\frac{1}{2}} \bigr\| ^{2} \bigl\| a^{m-1}\bigr\| ^{2} 1_{\mathbb{A}} \\ \preceq& \|b\|\sum_{k=n}^{m-1} \|b\|^{k-n} \bigl\| D^{\frac{1}{2}}\bigr\| ^{2} \bigl\| a^{k} \bigr\| ^{2} 1_{\mathbb{A}} + \|b\|^{m-n-1} \bigl\| D^{\frac{1}{2}} \bigr\| ^{2} \bigl\| a^{m-1}\bigr\| ^{2} 1_{\mathbb{A}} \\ \preceq& \|b\|^{1-n} \bigl\| D^{\frac{1}{2}}\bigr\| ^{2}\sum _{k=n}^{m-1}\|b\|^{k} \bigl\| a^{2}\bigr\| ^{k} 1_{\mathbb{A}} + \|b\|^{-n}\|b \|^{m-1} \bigl\| D^{\frac{1}{2}}\bigr\| ^{2} \bigl\| a^{m-1} \bigr\| ^{2} 1_{\mathbb{A}} \\ \preceq& \|b\|^{1-n} \bigl\| D^{\frac{1}{2}}\bigr\| ^{2}\sum _{k=n}^{m-1}\bigl(\|b\| \bigl\| a^{2}\bigr\| \bigr)^{k} 1_{\mathbb{A}} + \|b\|^{-n}\bigl\| D^{\frac{1}{2}} \bigr\| ^{2}\bigl(\|b\| \bigl\| a^{2}\bigr\| \bigr)^{m-1} 1_{\mathbb{A}} \\ \longrightarrow& 0_{\mathbb{A}} \quad\mbox{as } m, n \rightarrow\infty, \end{aligned}$$
which follows from the observation that the summation in the first term is a geometric series, and \(\|b\|\|a^{2}\| < 1\) implies that both \((\|b\| \|a^{2}\|)^{m-1} \rightarrow0\) and \((\|b\| \|a^{2}\|)^{n-1} \rightarrow0\). This proves that \(\{x_{n}\} \) is a Cauchy sequence in X with respect to \(\mathbb{A,}\) and from the completeness of \((X, \mathbb{A}, d)\) it follows that \(x_{n} \rightarrow x \in X\), that is,
$$\lim_{n\rightarrow\infty} x_{n} = \lim_{n\rightarrow\infty} Tx_{n-1} = x . $$
We claim that x is a fixed point of T. In fact, from the triangle inequality (BM3) and the contraction condition (1) we have:
$$\begin{aligned} 0_{\mathbb{A}} \preceq& d(Tx,x) \\ \preceq& b\bigl[d(Tx,Tx_{n})+d(Tx_{n},x)\bigr] \\ \preceq& b a^{*}d(x,x_{n})a + d(x_{n-1},x) \longrightarrow 0_{\mathbb{A}} \quad\mbox{as } n\rightarrow\infty. \end{aligned}$$
This shows that \(Tx=x\).
To prove that x is the unique fixed point, we suppose that \(y\in X\) is another fixed point of T. Then again from the contraction condition (1) we have
$$\begin{aligned} 0_{\mathbb{A}} \preceq d(x,y) = d(Tx,Ty) \preceq a^{*} d(x,y) a. \end{aligned}$$
Using the norm of \(\mathbb{A}\), we have
$$\begin{aligned} 0\le\bigl\| d(x,y)\bigr\| \le\bigl\| a^{*} d(x,y) a\bigr\| \le\bigl\| a^{*}\bigr\| \bigl\| d(x,y)\bigr\| \|a\| =\|a\| ^{2} \bigl\| d(x,y)\bigr\| . \end{aligned}$$
The above inequality holds only when \(d(x,y) = 0_{\mathbb{A}}\). Hence, \(x=y\). □
Example 2.3
The mapping T of Example 2.2 satisfies the hypothesis of Theorem 2.1, and T has unique fixed point \(x=1.5\) in X.
Remark 2.2
Theorem 2.1 generalizes the following results.
-
(1)
By taking \(\mathbb{A} =\mathbb{R}\), the \(C^{*}\)-valued b-metric becomes simply the b-metric, and we immediately get the Banach contraction principle in b-metric spaces from Theorem 2.1.
-
(2)
Taking \(b=1\), [11], Theorem 2.1, becomes a special case of Theorem 2.1.