Let
\(x_{0}\in X\) be such that
\(\alpha(gx_{0},Tx_{0})\geq1\) (using condition (ii)). Since
\(TX\subseteq gX\), we can choose a point
\(x_{1}\in X\) such that
\(Tx_{0}=gx_{1}\). Also, there exists
\(x_{2}\in X\) such that
\(Tx_{1}=gx_{2}\), this can be done through the reality
\(TX\subseteq gX\). Continuing this process having chosen
\(x_{1},x_{2},\ldots,x_{n} \in X\), we have
\(x_{n+1} \in X\) such that
$$ gx_{n+1}=Tx_{n},\quad n=0,1,2,\ldots. $$
(2.2)
By Lemma
2.1, we have
$$ \alpha(gx_{n},gx_{n+1})\geq1,\quad n=0,1,2, \ldots. $$
(2.3)
If
\(Tx_{n_{0}}=Tx_{n_{0}+1}\) for some
\(n_{0}\), then by (
2.2) we get
$$ gx_{n_{0}+1}=Tx_{n_{0}}=Tx_{n_{0}+1}, $$
that is,
T and
g have a coincidence point at
\(x=x_{n_{0}+1}\), and so the proof is completed. So, we suppose that for all
\(n \in\mathbb{N}\),
\(Tx_{n} \neq Tx_{n+1}\). Since the mapping
T is an almost generalized
\((\alpha\mbox{-}\psi\mbox{-}\varphi\mbox{-}\theta)\)-contractive mapping with respect to
g and using (
2.3), we obtain
$$\begin{aligned} \psi\bigl(d(gx_{n},gx_{n+1})\bigr)&\leq\psi \bigl(s^{3}d(gx_{n},gx_{n+1})\bigr)=\psi \bigl(s^{3}d(Tx_{n-1},Tx_{n})\bigr) \\ &\leq\alpha(gx_{n-1},gx_{n})\psi\bigl(s^{3}d(Tx_{n-1},Tx_{n}) \bigr) \\ &\leq\varphi\bigl(M(x_{n-1},x_{n})\bigr)+L\theta \bigl(N(x_{n-1},x_{n})\bigr) \end{aligned}$$
(2.4)
for all
\(n \in\mathbb{N}\), where
$$\begin{aligned} N(x_{n-1},x_{n})&=\min\bigl\{ d(gx_{n-1},Tx_{n-1}),d(gx_{n-1},Tx_{n}),d(gx_{n},Tx_{n-1}) \bigr\} \\ &=\min\bigl\{ d(gx_{n-1},gx_{n}),d(gx_{n-1},gx_{n+1}),d(gx_{n},gx_{n}) \bigr\} \\ &=0 \end{aligned}$$
and
$$\begin{aligned}& M(x_{n-1},x_{n}) \\& \quad =\max\biggl\{ d(gx_{n-1},gx_{n}),d(gx_{n-1},Tx_{n-1}),d(gx_{n},Tx_{n}), \frac {d(gx_{n-1},Tx_{n})+d(gx_{n},Tx_{n-1})}{2s}\biggr\} \\& \quad =\max\biggl\{ d(gx_{n-1},gx_{n}),d(gx_{n-1},gx_{n}),d(gx_{n},gx_{n+1}), \frac {d(gx_{n-1},gx_{n+1})+d(gx_{n},gx_{n})}{2s}\biggr\} \\& \quad =\max\biggl\{ d(gx_{n-1},gx_{n}),d(gx_{n},gx_{n+1}), \frac {1}{2s}d(gx_{n-1},gx_{n+1})\biggr\} . \end{aligned}$$
Since
$$ \frac{d(gx_{n-1},gx_{n+1})}{2s}\leq\frac {d(gx_{n-1},gx_{n})+d(gx_{n},gx_{n+1})}{2}\leq\max\bigl\{ d(gx_{n-1},gx_{n}),d(gx_{n},gx_{n+1}) \bigr\} , $$
then we get
$$ \begin{aligned} &N(x_{n-1},x_{n})=0, \\ &M(x_{n-1},x_{n})= \max\bigl\{ d(gx_{n-1},gx_{n}),d(gx_{n},gx_{n+1}) \bigr\} . \end{aligned} $$
(2.5)
By (
2.4) and (
2.5), we have
$$ \psi\bigl(d(gx_{n},gx_{n+1})\bigr)\leq\varphi \bigl( \max\bigl\{ d(gx_{n-1},gx_{n}),d(gx_{n},gx_{n+1}) \bigr\} \bigr). $$
(2.6)
If for some
\(n \in\mathbb{N}\),
\(\max\{ d(gx_{n-1},gx_{n}),d(gx_{n},gx_{n+1})\}=d(gx_{n},gx_{n+1})\), then by (
2.6) and using the properties of the function
φ, we get
$$\begin{aligned} \psi\bigl(d(gx_{n},gx_{n+1})\bigr)&\leq\varphi\bigl( \max \bigl\{ d(gx_{n-1},gx_{n}),d(gx_{n},gx_{n+1}) \bigr\} \bigr)=\varphi\bigl( d(gx_{n},gx_{n+1})\bigr) \\ & <\psi\bigl(d(gx_{n},gx_{n+1})\bigr), \end{aligned}$$
which is a contradiction. So
$$ \psi\bigl(d(gx_{n},gx_{n+1})\bigr)\leq\varphi \bigl( d(gx_{n-1},gx_{n})\bigr) <\psi \bigl(d(gx_{n-1},gx_{n}) \bigr) \quad \mbox{for each }n\in\mathbb{N}. $$
(2.7)
From (
2.7), we deduce that
\(\{\psi(d(gx_{n},gx_{n+1}))\}\) is a nonnegative nonincreasing sequence. Since
ψ is increasing, the sequence
\(\{d(gx_{n},gx_{n+1})\}\) is nonincreasing, and consequently there exists
\(\delta\geq0\) such that
$$ \lim_{n \to\infty}d(gx_{n},gx_{n+1})=\delta. $$
We claim that
\(\delta=0\). On the contrary, assume that
$$ \lim_{n \to\infty}d(gx_{n},gx_{n+1})= \delta>0. $$
(2.8)
Since
ψ and
φ are continuous, then from (
2.7) and (
2.8) we have
$$ \psi(\delta)=\lim_{n \to\infty}\psi\bigl(d(gx_{n},gx_{n+1}) \bigr)=\lim_{n \to \infty}\varphi\bigl(d(gx_{n},gx_{n+1}) \bigr)=\varphi(\delta), $$
and so
\(\delta=0\), that is a contradiction. Thus
$$ \lim_{n \to\infty}d(gx_{n},gx_{n+1})=0. $$
(2.9)
Now, we claim that
$$ \lim_{n,m\rightarrow\infty} d(gx_{n},gx_{m})=0. $$
(2.10)
Assume, on the contrary, that there exist
\(\epsilon>0\) and subsequences
\(\{gx_{m(k)}\}\),
\(\{gx_{n(k)}\}\) of
\(\{gx_{n}\}\) with
\(n(k)>m(k)\geq k\) such that
$$ d(gx_{m(k)},gx_{n(k)})\geq\epsilon. $$
(2.11)
Additionally, corresponding to
\(m(k)\), we may choose
\(n(k)\) such that it is the smallest integer satisfying (
2.11) and
\(n(k)>m(k)\geq k\). Thus,
$$ d(gx_{m(k)},gx_{n(k)-1})< \epsilon. $$
(2.12)
Using the triangle inequality in a
b-metric space and (
2.11) and (
2.12), we obtain that
$$\begin{aligned} \epsilon&\leq d(gx_{n(k)},gx_{m(k)})\leq sd(gx_{n(k)},gx_{n(k)-1})+sd(gx_{n(k)-1},gx_{m(k)}) \\ &< sd(gx_{n(k)},gx_{n(k)-1})+s\epsilon. \end{aligned}$$
Taking the upper limit as
\(k\rightarrow\infty\) and using (
2.9), we obtain
$$ \epsilon\leq\limsup_{k\rightarrow\infty} d(gx_{m(k)},gx_{n(k)}) \leq s\epsilon. $$
(2.13)
Also
$$\begin{aligned} \epsilon&\leq d(gx_{m(k)},gx_{n(k)})\leq sd(gx_{m(k)},gx_{n(k)+1})+sd(gx_{n(k)+1},gx_{n(k)}) \\ &\leq s^{2}d(gx_{m(k)},gx_{n(k)})+s^{2}d(gx_{n(k)},gx_{n(k)+1})+sd(gx_{n(k)+1},gx_{n(k)}) \\ &\leq s^{2}d(gx_{m(k)},gx_{n(k)})+ \bigl(s^{2}+s\bigr)d(gx_{n(k)},gx_{n(k)+1}). \end{aligned}$$
So, from (
2.9) and (
2.13), we have
$$ \frac{\epsilon}{s}\leq\limsup_{k\rightarrow\infty} d(gx_{m(k)},gx_{n(k)+1})\leq s^{2} \epsilon. $$
(2.14)
Also
$$\begin{aligned} \epsilon&\leq d(gx_{n(k)},gx_{m(k)})\leq sd(gx_{n(k)},gx_{m(k)+1})+sd(gx_{m(k)+1},gx_{m(k)}) \\ &\leq s^{2}d(gx_{n(k)},gx_{m(k)})+s^{2}d(gx_{m(k)},gx_{m(k)+1})+d(gx_{m(k)+1},gx_{m(k)}) \\ &\leq s^{2}d(gx_{n(k)},gx_{m(k)})+ \bigl(s^{2}+s\bigr)d(gx_{m(k)},gx_{m(k)+1}). \end{aligned}$$
So from (
2.9) and (
2.13), we get
$$ \frac{\epsilon}{s}\leq\limsup_{k\rightarrow\infty} d(gx_{n(k)},gx_{m(k)+1})\leq s^{2} \epsilon. $$
(2.15)
Also
$$d(gx_{m(k)+1},gx_{n(k)})\leq sd(gx_{m(k)+1},gx_{n(k)+1})+sd(gx_{n(k)+1},gx_{n(k)}), $$
so from (
2.9) and (
2.15), we have
$$ \frac{\epsilon}{s^{2}}\leq\limsup_{k\rightarrow\infty} d(gx_{n(k)+1},gx_{m(k)+1}). $$
(2.16)
Taking (
2.9), (
2.13), (
2.14) and (
2.15) into account, we get
$$\begin{aligned}& \limsup_{k\rightarrow\infty}M(x_{n(k)},x_{m(k)}) \\& \quad = \max \biggl\{ \limsup_{k\rightarrow\infty} d(gx_{n(k)},gx_{m(k)}), \limsup_{k\rightarrow\infty} d(gx_{n(k)},gx_{n(k)+1}),\limsup _{k\rightarrow\infty} d(gx_{m(k)},gx_{m(k)+1}), \\& \qquad \frac{\limsup_{k\rightarrow\infty} d(gx_{n(k)},gx_{m(k)+1})+\limsup_{k\rightarrow\infty} d(gx_{m(k)},gx_{n(k)+1})}{2s}\biggr\} \\& \quad \leq \max\biggl\{ s\epsilon,0,0,\frac{s^{2}\epsilon+s^{2}\epsilon}{2s}\biggr\} =s\epsilon. \end{aligned}$$
So,
$$ \limsup_{k\rightarrow\infty}M(x_{m(k)},x_{n(k)}) \leq \epsilon s. $$
(2.17)
Similarly, we have
$$ \limsup_{k\rightarrow\infty}N(x_{m(k)},x_{n(k)})=0. $$
(2.18)
Now, using inequality (
2.1) and Lemma
2.1, we have
$$\begin{aligned} \psi(s\epsilon)&=\psi\biggl(s^{3}\cdot\frac{\epsilon}{s^{2}}\biggr)\leq\psi \Bigl(s^{3}\limsup_{k \to\infty}d(gx_{m(k)+1},gx_{n(k)+1}) \Bigr) \\ &=\limsup_{k \to \infty}\psi\bigl(s^{3}d(gx_{m(k)+1},gx_{n(k)+1}) \bigr)=\limsup_{k \to\infty}\psi\bigl(s^{3} d(Tx_{m(k)},Tx_{n(k)}) \bigr) \\ &\leq\limsup_{k \to\infty} \alpha(gx_{m(k)},gx_{n(k)}) \psi\bigl(s^{3} d(Tx_{m(k)},Tx_{n(k)})\bigr) \\ &\leq\limsup_{k \to\infty}\bigl[\varphi\bigl(M(x_{m(k)},x_{n(k)}) \bigr)+L\theta \bigl(N(x_{m(k)},x_{n(k)})\bigr)\bigr] \\ &=\varphi\Bigl(\limsup_{k \to\infty}M(x_{m(k)},x_{n(k)}) \Bigr)+L\theta\Bigl(\limsup_{k \to\infty}N(x_{m(k)},x_{n(k)}) \Bigr) \\ &\leq\varphi(s\epsilon) \\ &<\psi(s\epsilon), \end{aligned}$$
which is a contradiction. So, we conclude that
\(\{gx_{n}\}\) is a Cauchy sequence in
\((X,d)\). By virtue of (
2.2) we get
\(\{Tx_{n}\}=\{gx_{n+1}\}\subseteq gX\) and
gX is closed, there exists
\(x \in X\) such that
$$ \lim_{n \to\infty}gx_{n}=gx. $$
(2.19)
Now, we claim that
x is a coincidence point of
T and
g. On the contrary, assume that
\(d(Tx,gx)>0\). Since
X is
α-regular with respect to
g and (
2.19), we have
$$ \alpha(gx_{n(k)+1},gx)\geq1 \quad \mbox{for all }k \in \mathbb{N}. $$
(2.20)
Also by the use of triangle inequality in a
b-metric space, we have
$$\begin{aligned} d(gx,Tx)&\leq sd(gx,gx_{n(k)+1})+sd(gx_{n(k)+1},Tx) \\ &=sd(gx,gx_{n(k)+1})+sd(Tx_{n(k)},Tx). \end{aligned}$$
In the above inequality, if
k tends to infinity, then we have
$$ d(gx,Tx) \leq\lim_{k \to\infty}sd(Tx_{n(k)},Tx). $$
(2.21)
By property of
ψ, (
2.20) and (
2.21), we have
$$\begin{aligned} \psi\bigl(s^{2}d(gx,Tx)\bigr)& \leq\lim_{k \to\infty} \psi \bigl(s^{3}d(Tx_{n(k)},Tx)\bigr)\leq\lim_{k \to\infty} \alpha(gx_{n(k)+1},gx) \psi\bigl(s^{3}d(Tx_{n(k)},Tx) \bigr) \\ &\leq\lim_{k \to\infty} \bigl[\varphi\bigl(M(x_{n(k)},x) \bigr)+L\theta \bigl(N(x_{n(k)},x)\bigr)\bigr] \\ &=\varphi\Bigl(\lim _{k \to\infty}M(x_{n(k)},x)\Bigr)+L\theta\Bigl(\lim _{k \to\infty}N(x_{n(k)},x)\Bigr) \\ &= \varphi\bigl(d(gx,Tx)\bigr) \\ &<\psi\bigl(d(gx,Tx)\bigr), \end{aligned}$$
which is a contradiction. Indeed,
$$\begin{aligned}& M(x_{n(k)},x) \\& \quad =\max\biggl\{ d(gx_{n(k)},gx),d(gx_{n(k)},Tx_{n(k)}),d(gx,Tx), \frac {d(gx_{n(k)},Tx)+d(gx,Tx_{n(k)})}{2s}\biggr\} \\& \quad =\max\biggl\{ d(gx_{n(k)},gx),d(gx_{n(k)},gx_{n(k)+1}),d(gx,Tx), \frac {d(gx_{n(k)},Tx)+d(gx,gx_{n(k)+1})}{2s}\biggr\} \end{aligned}$$
and
$$\begin{aligned} N(x_{n(k)},x)&=\min\bigl\{ d(gx_{n(k)},Tx_{n(k)}),d(gx_{n(k)},Tx),d(gx,Tx_{n(k)}) \bigr\} \\ &=\min\bigl\{ d(gx_{n(k)},gx_{n(k)+1}),d(gx_{n(k)},Tx),d(gx,gx_{n(k)+1}) \bigr\} . \end{aligned}$$
When
n tends to infinity, we deduce
$$\lim_{k \to\infty}M(x_{n(k)},x)=\max\biggl\{ d(gx,Tx), \frac{d(gx,Tx)}{2}\biggr\} =d(gx,Tx) $$
and
$$\lim_{k \to\infty}N(x_{n(k)},x)=0. $$
Hence,
\(d(gx,Tx)=0\), that is,
\(gx=Tx\) and
x is a coincidence point of
T and
g. We claim that if
\(Tu=gu\) and
\(Tv=gv\), then
\(gu=gv\). By hypotheses,
\(\alpha (u,v)\geq1\) or
\(\alpha(v,u)\geq1\). Suppose that
\(\alpha(u,v)\geq1\), then
$$ \psi\bigl(s^{3}d(gu,gv)\bigr)=\psi\bigl(s^{3}d(Tu,Tv)\bigr) \leq\alpha(u,v)\psi \bigl(s^{3}d(Tu,Tv)\bigr)\leq\varphi\bigl(M(u,v) \bigr)+L\theta\bigl(N(u,v)\bigr), $$
where
$$\begin{aligned} M(u,v)&=\max\biggl\{ d(gu,gv),d(gu,Tu),d(gv,Tv),\frac{d(gu,Tv)+d(gv,Tu)}{2s}\biggr\} \\ &=\max\biggl\{ d(gu,gv),d(gu,gu),d(gv,gv),\frac{d(gu,gv)+d(gv,gu)}{2s}\biggr\} \\ &=d(gu,gv) \end{aligned}$$
and
$$ N(u,v)=\min\bigl\{ d(gu,Tu),d(gu,Tv),d(gv,Tu)\bigr\} =\min\bigl\{ d(gu,gu),d(gu,gv),d(gv,gu)\bigr\} =0. $$
So,
$$ \psi\bigl(s^{3}d(gu,gv)\bigr)\leq\varphi\bigl(d(gu,gv)\bigr)<\psi \bigl(d(gu,gv)\bigr), $$
which is a contradiction. Thus we deduce that
\(gu=gv\). Similarly, if
\(\alpha(v,u)\geq1\), we can prove that
\(gu=gv\). Now, we show that
T and
g have a common fixed point. Indeed, if
\(w=Tu=gu\), owing to the weak compatibility of
T and
g, we get
\(Tw=T(gu)=g(Tu)=gw\). Thus
w is a coincidence point of
T and
g, then
\(gu=gw=w=Tw\). Therefore,
w is a common fixed point of
T and
g. The uniqueness of the common fixed point of
T and
g is a consequence of conditions (
2.1) and (b), and so we omit the details. □