In this section, we shall state and prove our main results.
Theorem 3.1
Let
\((X,q)\)
be a complete quasi
bmetric space and
\(T: X\to X\)
be an
αλcontraction. Suppose that

(i)
T
is an
αadmissible mapping;

(ii)
there exists
\(x_{0}\in X\)
such that
\(\alpha (x_{0},Tx_{0})\geq1\)
and
\(\alpha(Tx_{0},x_{0})\geq1\);

(iii)
T
is continuous on
\((X,q)\).
Then
T
has a fixed point.
Proof
By assumption (ii), there exists a point \(x_{0}\in X\) such that \(\alpha (x_{0},Tx_{0})\geq1\) and \(\alpha(Tx_{0},x_{0})\geq1\). Take \(x_{n}=T^{n} x_{0}\) for all \(n\geq0\). From (i), we have by induction
$$ \alpha(x_{n},x_{n+1})\geq1\quad \mbox{and} \quad \alpha (x_{n+1},x_{n})\geq1 \quad\mbox{for all } n=0,1,\ldots. $$
(7)
Applying (4) with \(x=x_{0}\) and \(y=x_{1}\) and using (7), we get
$$q(x_{1},x_{2})=q(Tx_{0},Tx_{1})\leq \alpha(x_{0},x_{1})q(Tx_{0},Tx_{1}) \leq\lambda(x_{0})q(x_{0},x_{1}). $$
We apply again (4) with \(x=x_{1}\) and \(y=x_{2}\) and using (7) together with the propriety of λ, we get
$$\begin{aligned} q(x_{2},x_{3})&=q(Tx_{1},Tx_{2})\leq \alpha(x_{1},x_{2})q(Tx_{1},Tx_{2}) \\ &\leq\lambda(x_{1})q(x_{1},Tx_{2}) \\ &=\lambda(fx_{0})q(x_{1},x_{2})\leq \lambda(x_{0})q(x_{1},x_{2}) \\ &\leq\bigl[\lambda(x_{0})\bigr]^{2} q(x_{0},x_{1}). \end{aligned}$$
A similar argument leads to
$$ q(x_{n},x_{n+1})\leq\bigl[\lambda(x_{0}) \bigr]^{n} q(x_{0},x_{1}). $$
(8)
The same procedure allows us to write
$$ q(x_{n+1},x_{n})\leq\bigl[\lambda(x_{1}) \bigr]^{n} q(x_{1},x_{0}). $$
(9)
Since \(\lambda(x_{0}) \) and \(\lambda(x_{1})\) are in \([0,1)\),
$$ \lim_{n\rightarrow\infty} q(x_{n+1},x_{n})= \lim_{n\rightarrow\infty} q(x_{n},x_{n+1})=0. $$
(10)
We shall prove that \(\{x_{n}\}\) is a Cauchy sequence in \((X,q)\).
First, we claim that \(\{x_{n}\}\) is a rightCauchy sequence in the quasi bmetric space \((X,q)\). Using (q2) and (8), we have for all \(n,k\in\mathbb{N}\)
$$\begin{aligned} q(x_{n},x_{n+k})&\leq sq(x_{n},x_{n+1})+s^{2} q(x_{n+1},x_{n+2})+\cdots+s^{k1} q(x_{n+k1},x_{n+k}) \\ &\leq\bigl(s \bigl[\lambda(x_{0})\bigr]^{n}+ s^{2} \bigl[\lambda(x_{0})\bigr]^{n+1}+ \cdots+s^{k1} \bigl[\lambda(x_{0})\bigr]^{n+k1}\bigr) \bigl[q(x_{0},x_{1})\bigr] \\ &\leq \sum_{i=n}^{n+k1} s^{i} \bigl[\lambda (x_{0})\bigr]^{i}\bigl[q(x_{0},x_{1}) \bigr] \\ &\leq \sum_{i=n}^{\infty} s^{i} \bigl[\lambda(x_{0})\bigr]^{i}\bigl[q(x_{0},x_{1}) \bigr]. \end{aligned}$$
(11)
Since \(s\lambda(x_{0})<1\),
$$ q(x_{n},x_{n+k})\rightarrow0 \quad\mbox{as } n \rightarrow\infty, \mbox{ for all } k. $$
(12)
It follows that \(\{x_{n}\}\) is a rightCauchy sequence in the quasi bmetric space \((X,q)\). Similarly, using (9), we see that \(\{ x_{n}\}\) is a leftCauchy sequence in the quasi bmetric space \((X,q)\). We deduce that \(\{x_{n}\}\) is a Cauchy sequence in the quasi bmetric space \((X,q)\).
Since \((X,q)\) is complete, the sequence \(\{x_{n}\}\) converges to some \(u\in X\), that is,
$$\lim_{n\rightarrow\infty} q(x_{n},u) = \lim_{n\rightarrow\infty} q(u,x_{n})=0. $$
The continuity of T yields
$$ \lim_{n\rightarrow\infty}q(Tx_{n},Tu)= \lim _{n\rightarrow\infty}q(x_{n+1},Tu)=0. $$
(13)
By uniqueness of the limit, we get \(Tu=u\). Therefore, u is a fixed point of T. □
Using the same techniques we obtain the following result.
Theorem 3.2
Let
\((X,q)\)
be a complete
bmetric space and
\(T: X\to X\)
be an
αλcontraction. Suppose that

(i)
T
is an
αadmissible mapping;

(ii)
there exists
\(x_{0}\in X\)
such that
\(\alpha (x_{0},Tx_{0})\geq1\);

(iii)
T
is continuous on
\((X,q)\).
Then
T
has a fixed point.
Considering \(s=1\) in Theorem 3.1 (resp. Theorem 3.2), we have
Corollary 3.1
Let
\((X, q)\)
be a complete quasi metric space and
\(T: X\to X\)
be an
αλcontraction.
Suppose that:

(i)
T
is
αadmissible;

(ii)
there exists
\(x_{0}\in X\)
such that
\(\alpha (x_{0},Tx_{0})\geq1\)
and
\(\alpha(Tx_{0},x_{0})\geq1\);

(iii)
T
is continuous on
\((X,q)\).
Then there exists
\(u\in X\)
such that
\(u=Tu\).
Corollary 3.2
(Theorem 10, [1])
Let
\((X, d)\)
be a complete metric space and
\(T: X\to X\)
be an
αλcontraction satisfying the following conditions:

(i)
T
is
αadmissible;

(ii)
there exists
\(x_{0}\in X\)
such that
\(\alpha (x_{0},Tx_{0})\geq1\);

(iii)
T
is continuous on
\((X,d)\).
Then there exists
\(u\in X\)
such that
\(u=Tu\).
We may replace the continuity hypothesis of T in Theorem 3.1 (resp. Theorem 3.2) by one of the following hypotheses:
 (H):

If \(\{x_{n}\}\) is a sequence in X such that \(\alpha(x_{n},x_{n+1})\geq1\) and \(\alpha(x_{n+1},x_{n})\geq1\) for all n and \(x_{n} \rightarrow x\in X\) as \(n\rightarrow\infty\), then there exists a subsequence \(\{ x_{n(k)}\}\) of \(\{x_{n}\}\) such that \(\alpha(x_{n(k)},x)\geq1\), for all k.
 (R):

If \(\{x_{n}\}\) is a sequence in X such that \(\alpha(x_{n},x_{n+1})\geq 1\) for all n and \(x_{n} \rightarrow x\in X\) as \(n\rightarrow\infty\), then there exists a subsequence \(\{x_{n(k)}\}\) of \(\{x_{n}\}\) such that \(\alpha(x_{n(k)},x)\geq1\), for all k.
Theorem 3.3
Let
\((X,q)\)
be a complete quasi
bmetric space and
\(T: X\to X\)
be an
αλcontraction. Suppose that:

(i)
T
is
αadmissible;

(ii)
there exists
\(x_{0}\in X\)
such that
\(\alpha (x_{0},Tx_{0})\geq1\)
and
\(\alpha(Tx_{0},x_{0})\geq1\);

(iii)
(H) holds.
Then there exists
\(u\in X\)
such that
\(u=Tu\).
Proof
Following the proof of Theorem 3.1, the sequence \(\{x_{n}\}\) is Cauchy and converges to some \(u\in X\) in \((X,q)\). Remember that (7) holds, so from condition (iii), there exists a subsequence \(\{x_{n(k)}\}\) of \(\{x_{n}\}\) such that \(\alpha(x_{n(k)},u)\geq1\), for all k. We shall show that \(u=Tu\).
We have, for all \(k\geq0\),
$$\begin{aligned} q(u,Tu)\leq sq(u,x_{n(k)+1})+sq(x_{n(k)+1},Tu). \end{aligned}$$
(14)
Taking \(x=x_{n(k)}\) and \(y=u\) in (4), we obtain
$$\begin{aligned} q(x_{n(k)+1},Tu)&=q(Tx_{n(k)},Tu)\leq\alpha(x_{n(k)},u) q(Tx_{n(k)},Tu) \\ &\leq\lambda(x_{n(k)})q(x_{n(k)},u)\leq\lambda (x_{0})q(x_{n(k)},u)< \frac{1}{s} q(x_{n(k)},u). \end{aligned}$$
Then we get for all \(k\geq0\)
$$\begin{aligned} q(u,Tu)\leq sq(u,x_{n(k)+1})+q(x_{n(k)},u). \end{aligned}$$
(15)
Letting \(k\rightarrow\infty\) in (15), we have
This yields \(Tu=u\). This completes the proof. □
We also state the following result. Its proof is very immediate.
Theorem 3.4
Let
\((X,q)\)
be a complete
bmetric space and
\(T: X\to X\)
be an
αλcontraction. Suppose that:

(i)
T
is
αadmissible;

(ii)
there exists
\(x_{0}\in X\)
such that
\(\alpha (x_{0},Tx_{0})\geq1\);

(iii)
(R) holds.
Then there exists
\(u\in X\)
such that
\(u=Tu\).
Considering \(s=1\) in Theorem 3.3 (resp. Theorem 3.4), we have
Corollary 3.3
Let
\((X, q)\)
be a complete quasi metric space and
\(T: X\to X\)
be an
αλcontraction.
Suppose that:

(i)
T
is
αadmissible;

(ii)
there exists
\(x_{0}\in X\)
such that
\(\alpha (x_{0},Tx_{0})\geq1\)
and
\(\alpha(Tx_{0},x_{0})\geq1\);

(iii)
(H) holds.
Then there exists
\(u\in X\)
such that
\(u=Tu\).
Corollary 3.4
(Theorem 12, [1])
Let
\((X, d)\)
be a complete metric space and
\(T: X\to X\)
be an
αλcontraction satisfying the following conditions:

(i)
T
is
αadmissible;

(ii)
there exists
\(x_{0}\in X\)
such that
\(\alpha (x_{0},Tx_{0})\geq1\);

(iii)
(H) holds.
Then there exists
\(u\in X\)
such that
\(u=Tu\).
We provide the following examples.
Example 3.1
Let \(X=[0,\infty)\). Consider \(q(x,y)=(\max\{(xy),2(yx)\})^{2}\) for all \(x,y\in X\). We mention that \((X,q)\) is a complete quasi bmetric space with \(s=2\). Define \(T:X\to X\) and \(\alpha:X\times X\to[0,\infty)\) by
$$\begin{aligned} Tx= \textstyle\begin{cases} \ln(1+\frac{x}{2}) &\mbox{if } x\in[0,1],\\ 2(x1)+\ln\frac{3}{2}& \mbox{if } x>1, \end{cases}\displaystyle \quad \mbox{and}\quad \alpha(x,y)= \textstyle\begin{cases} 1 &\mbox{if } x,y\in[0,1],\\ 0 &\mbox{otherwise}. \end{cases}\displaystyle \end{aligned}$$
Now, we show that T is an αλcontraction where \(\lambda: X\rightarrow[0,\frac{1}{2})\) is defined by \(\lambda(x)=\frac{1}{4}\) for all \(x\in X\). To this aim, we distinguish the following cases:
Case 1: \(x,y\in X\) such that \(x\geq y\) and \(\alpha (x,y)=1\). We have
$$\begin{aligned} \alpha(x,y) q(Tx,Ty) =&q(Tx,Ty)=\biggl(\ln\biggl(1+\frac{x}{2}\biggr)\ln \biggl(1+\frac {y}{2}\biggr)\biggr)^{2} \\ \leq& \frac{1}{4}(xy)^{2}=\frac{1}{4}q(x,y)= \lambda(x)q(x,y). \end{aligned}$$
Case 2: \(x,y\in X\) such that \(x< y\) and \(\alpha (x,y)=1\). Similarly, we get
$$\begin{aligned} \alpha(x,y) q(Tx,Ty)\leq\lambda(x)q(x,y). \end{aligned}$$
Hence, (4) is verified and, since \(\lambda(Tx)=\lambda(x)\) for all \(x\in X\), the mapping T is an αλcontraction.
Note that T is αadmissible. Since T is continuous on \((X, \cdot )\) where \(\cdot \) is the standard metric on X, by Lemma 2.2, T is continuous on \((X,q)\). We mention that \(\alpha(1,T1)=\alpha (T1,1)=1\) and so condition (ii) of Theorem 3.1 is verified. Hence, all hypotheses of Theorem 3.1 hold. Note that \(u=0\) and \(v=2\ln(\frac{3}{2})\) are the two fixed points of T.
Example 3.2
Let \(X=[0,1]\). Consider \(q(x,y)=(\max\{xy,2(yx)\})^{2}\) for all \(x,y\in X\). We mention that \((X,q)\) is a quasi bmetric space with \(s=2\). Define \(T:X\to X\) and \(\alpha:X\times X\to[0,\infty)\) by
$$\begin{aligned} Tx= \textstyle\begin{cases} \frac{x^{2}}{2\sqrt{2}} &\mbox{if } 0\leq x< 1,\\ 1 &\mbox{if } x=1, \end{cases}\displaystyle \quad \mbox{and}\quad \alpha(x,y)= \textstyle\begin{cases} 1 &\mbox{if } x,y\in[0,1),\\ 0 &\mbox{otherwise}. \end{cases}\displaystyle \end{aligned}$$
We have
$$q\biggl(T1,T\frac{1}{2}\biggr)=q\biggl(\frac{3}{4}, \frac{1}{8\sqrt{2}}\biggr)=\biggl(\frac{3}{4}\frac {1}{8\sqrt{2}} \biggr)^{2}>\frac{1}{4}=q\biggl(1,\frac{1}{2}\biggr), $$
that is, T is not a Banach contraction on X. Now, we show that T is an αλcontraction where \(\lambda: X\rightarrow [0,\frac{1}{2})\) is defined by
$$\begin{aligned} \lambda(x)= \textstyle\begin{cases} \frac{(x+1)^{2}}{8} &\mbox{if } 0\leq x< 1,\\ \frac{2}{5} &\mbox{if } x=1. \end{cases}\displaystyle \end{aligned}$$
First of all, we show that \(\lambda(Tx)\leq\lambda(x)\) for all \(x\in X\). For \(x=1\), we have \(\lambda(T1)=\lambda(1)\). Also, for \(x\in[0,1)\), we have
$$\begin{aligned} \lambda(Tx)=\lambda\biggl(\frac{x^{2}}{2\sqrt{2}}\biggr)=\frac{1}{8}\biggl( \frac{x^{2}}{2\sqrt {2}}+1\biggr)^{2}\leq\frac{1}{8}(x+1)^{2}= \lambda(x). \end{aligned}$$
Again, we show that (4) is verified. To this aim, we distinguish the following cases:
Case 1: If \(x,y\in[0,1)\) such that \(x\leq y\), then we have
$$\begin{aligned} \alpha(x,y) q(Tx,Ty) =&q(Tx,Ty)=\frac{1}{2}\bigl(y^{2}x^{2} \bigr)^{2}=\frac {1}{2}(yx)^{2}(x+y)^{2}= \frac{1}{8}(x+y)^{2}q(x,y) \\ \leq& \frac{1}{8}(x+1)^{2}q(x,y)=\lambda(x)q(x,y). \end{aligned}$$
Case 2: If \(x,y\in[0,1)\) such that \(x>y\), then we obtain
$$\begin{aligned} \alpha(x,y) q(Tx,Ty)\leq\lambda(x)q(x,y). \end{aligned}$$
Case 3: If \((x,y)\notin[0,1)^{2}\), then we have \(\alpha(x,y)=0\), and so (4) is verified.
Thus, (4) is satisfied and the mapping T is an αλcontraction.
Note that T is αadmissible. By Lemma 2.2, T is not continuous on \((X,q)\), then Theorem 3.1 is not applicable. Also, it is easy to see that \(\alpha(\frac{1}{2},T\frac{1}{2})=\alpha(T\frac{1}{2},\frac{1}{2})=1\), and so condition (ii) of Theorem 3.3 is verified. Now, we show that condition (H) holds. Let \(\{x_{n}\}\) be a sequence in X such that \(\alpha(x_{n},x_{n+1})\geq 1\) and \(\alpha(x_{n+1},x_{n+})\geq1\) for all n and \(x_{n} \rightarrow u\) in \((X,q)\). Then \(\{x_{n}\}\subset[0,\frac{1}{2}]\) and \(x_{n} \rightarrow u\) in \((X,\cdot )\). Thus, \(u\in[0,\frac{1}{2}]\) and so \(\alpha(x_{n},u)=\alpha(u,x_{n})=1\) for all n.
Therefore, all hypotheses of Theorem 3.3 are satisfied. Here, \(\{ 0,1\}\) is the set of fixed points of T.
To prove uniqueness of the fixed point given in Theorem 3.1 (resp. Theorem 3.2, Theorem 3.3, Theorem 3.4), we need to take one of the following additional hypotheses:
 (U):

For all \(x, y \in F(T)\), we have \(\alpha(x,y) \geq1\), where \(F(T)\) denotes the set of fixed points of T.
 (V):

For all \(x, y \in F(T)\), there exists \(z\in X\) such that \(\min\{\alpha(x,z),\alpha(z,y)\}\geq1\).
Theorem 3.5
Adding condition (U) to the hypotheses of Theorem
3.1 (resp. Theorem
3.2, Theorem
3.3, Theorem
3.4), we see that
u
is the unique fixed point of
T.
Proof
We argue by contradiction, that is, there exist \(u,v\in X\) such that \(u=Tu\) and \(v=Tv\) with \(u\neq v\). By (4) and the fact that \(\alpha(u,v)\geq1\), we get
$$0< q(u,v)\leq\alpha(u,v)q(u,v)=\alpha(u,v) q(Tu,Tv)\leq\lambda(u) q(u,v)< q(u,v), $$
which is a contradiction. Hence, \(u=v\). □
Theorem 3.6
Adding condition (V) to the hypotheses of Theorem
3.1 (resp. Theorem
3.2, Theorem
3.3, Theorem
3.4), we see that
u
is the unique fixed point of
T.
Proof
Suppose that there exist \(u,v\), two fixed points of T. By condition (V), there exists \(z\in X\) such that \(\min\{\alpha(u,z),\alpha(z,v)\} \geq1\). Since T is αadmissible, it follows that
$$\begin{aligned} \min\bigl\{ \alpha\bigl(u,T^{n}z\bigr),\alpha\bigl(T^{n}z,v \bigr)\bigr\} \geq1,\quad \forall n=0,1,\ldots. \end{aligned}$$
(16)
We have
$$\begin{aligned} q\bigl(u,T^{n+1}z\bigr) \leq&\alpha\bigl(u,T^{n}z\bigr)q \bigl(u,T^{n+1}z\bigr)=\alpha \bigl(u,T^{n}z\bigr)q\bigl(Tu,T \bigl(T^{n}z\bigr)\bigr) \\ \leq&\lambda(u) q\bigl(u,T^{n}z\bigr), \quad \forall n=0,1,\ldots. \end{aligned}$$
(17)
By induction, we obtain
$$\begin{aligned} q\bigl(u,T^{n}z\bigr)\leq\bigl[\lambda(u)\bigr]^{n} q(u,z), \quad \forall n=0,1,\ldots. \end{aligned}$$
(18)
A similar reasoning shows that
$$\begin{aligned} q\bigl(T^{n}z,v\bigr)\leq\bigl[\lambda(z)\bigr]^{n} q(z,v), \quad\forall n=0,1,\ldots. \end{aligned}$$
(19)
On the other side, we have
$$\begin{aligned} q(u,v)\leq sq\bigl(u,T^{n}z\bigr)+sq\bigl(T^{n}z,v\bigr), \quad \forall n=0,1,\ldots, \end{aligned}$$
(20)
which yields
$$\begin{aligned} q(u,v)\leq s\bigl[\lambda(u)\bigr]^{n} q(u,z)+s\bigl[\lambda(z) \bigr]^{n} q(z,v),\quad \forall n=0,1,\ldots. \end{aligned}$$
(21)
Passing to the limit as \(n\to\infty\), we obtain
$$\begin{aligned} q(u,v)\leq0, \end{aligned}$$
(22)
and so \(u=v\). □
The following examples illustrate Theorem 3.5.
Example 3.3
Let \(X=\{0,1,2,3\}\). Consider the function \(q:X\times X\to[0,\infty)\) defined by
$$ q(n,m)= \textstyle\begin{cases} 4& \mbox{if } n>m,\\ 0& \mbox{if } n=m,\\ 1& \mbox{if } n< m, \end{cases} $$
for all \(n,m\in X\) with \((n,m)\neq(0,1)\) and \(q(0,1)=9\). We mention that \((X,q)\) is a complete quasi bmetric space with \(s=2\).
Define \(T:X\to X\) and \(\alpha:X\times X\to[0,\infty)\) by
$$\begin{aligned} T0=3,\qquad T1=0,\qquad T2=T3=2 \quad\mbox{and}\quad \alpha(n,m)= \textstyle\begin{cases} 1 &\mbox{if } n,m\in\{2,3\},\\ 0 &\mbox{otherwise}. \end{cases}\displaystyle \end{aligned}$$
We have
$$q(T0,T2)=q(3,2)=4>1=q(0,2), $$
that is, T is not a Banach contraction on X. Now, we show that T is an αλcontraction where \(\lambda: X\rightarrow [0,\frac{1}{2})\) is defined by \(\lambda(n)=\frac{1}{4}\) for all \(n\in X\). To this aim, we distinguish the following cases:
Case 1: If \(n,m\in X\) such that \(\alpha(n,m)=1\), then \(n,m\in\{2,3\}\). So
$$\begin{aligned} \alpha(n,m) q(Tn,Tm)=q(2,2)=0\leq\lambda(n)q(n,m). \end{aligned}$$
Case 2: If \(n,m\in X\) such that \(\alpha(n,m)=0\), then (4) is satisfied.
Thus, (4) holds and since \(\lambda(Tn)=\lambda(n)\) for all \(n\in X\), so the mapping T is an αλcontraction.
Note that T is αadmissible. In fact, let \(n,m\in X\) such that \(\alpha(n,m)\geq1\), then \(n,m\in\{2,3\}\), and so \(\alpha(Tn,Tm)=\alpha(2,2)=1\). Moreover, T is continuous on \((X,q)\). In fact if \(\{x_{n}\}\) is a sequence in X such that \(x_{n} \rightarrow u\) in \((X,q)\), it easy to see that there exists \(N\in\mathbb{N}\) such that \(x_{n}=u\) for all \(n\geq N\) and so \(Tx_{n}=Tu\) for all \(n\geq N\). It follows that \(\lim_{n\to\infty }q(Tx_{n},Tu)=0\), that is, T is continuous on \((X,q)\). Also, since \(\alpha(3,T3)=\alpha(3,2)=1\), and \(\alpha(T3,3)=\alpha(2,3)=1\), condition (ii) of Theorem 3.3 is verified. Therefore, all hypotheses of Theorem 3.3 are satisfied. Here, 2 is the unique fixed point of T.
Example 3.4
Going back again to Example 3.2 where \(X=[0,1]\) is endowed with the quasi bmetric \(q(x,y)=(\max\{xy,2(yx)\})^{2}\) for all \(x,y\in X\). Define \(T:X\to X\) and \(\alpha:X\times X\to[0,\infty)\) by
$$\begin{aligned} Tx= \textstyle\begin{cases} \frac{x^{2}}{2\sqrt{2}} &\mbox{if } 0\leq x< 1,\\ \frac{3}{4} &\mbox{if } x=1, \end{cases}\displaystyle \quad \mbox{and}\quad \alpha(x,y)= \textstyle\begin{cases} 1 &\mbox{if } x,y\in[0,\frac{1}{2}],\\ 0 &\mbox{otherwise}. \end{cases}\displaystyle \end{aligned}$$
We know that T is an αλcontraction where \(\lambda: X\rightarrow[0,\frac{1}{2})\) is defined by
$$\begin{aligned} \lambda(x)= \textstyle\begin{cases} \frac{(x+1)^{2}}{8}& \mbox{if } 0\leq x< 1,\\ \frac{2}{5}& \mbox{if } x=1. \end{cases}\displaystyle \end{aligned}$$
Therefore, all hypotheses of Theorem 3.3 are satisfied. Here, 0 is the unique fixed point of T.