In the following theorem, the existence of coincidence points of a hybrid pair of single-valued and multi-valued mappings that satisfy Suzuki-Zamfirescu hybrid contraction condition in partial metric spaces is established.

**Theorem 2.1** *Let* (X,p) *be a partial metric space and* *Y* *be any non*-*empty set*. *Assume that a pair of mappings* f:Y\u27f6X *and* T:Y\u27f6C{B}^{p}(X) *satisfies Suzuki*-*Zamfirescu hybrid contraction condition with* T(Y)\subset f(Y). *If there exists* {u}_{0}\in Y *such that* f(Y) *is* (f,T)-*orbitally complete at* {u}_{0}, *then* C(f,T)\ne \varphi. *If* Y=X *and* (f,T) *is* *IT*-*commuting at coincidence points of* (f,T), *then* F(f,T)\ne \varphi *provided that* *fz* *is a fixed point of* *f* *for some* z\in C(f,T).

*Proof* Let h=1/\sqrt{r} and {u}_{0}\in Y be such that {y}_{0}=f{u}_{0}. By the given assumption, we have T{u}_{0}\subseteq f(Y). So, there exists a point {u}_{1}\in Y such that {y}_{1}=f{u}_{1}\in T{u}_{0}. As h>1, so by Lemma C, there exists a point {y}_{2}\in T{u}_{1} such that

p(f{u}_{1},{y}_{2})\le h{H}_{p}(T{u}_{0},T{u}_{1}).

Using the fact that T{u}_{1}\subseteq f(Y), we obtain a point {u}_{2}\in Y such that {y}_{2}=f{u}_{2}\in T{u}_{1}. Therefore,

p(f{u}_{1},f{u}_{2})\le h{H}_{p}(T{u}_{0},T{u}_{1}).

Since

\omega (r)p(f{u}_{0},T{u}_{0})\le \omega (r)p(f{u}_{0},f{u}_{1})\le p(f{u}_{0},f{u}_{1}),

we have

\begin{array}{rcl}p(f{u}_{1},f{u}_{2})& \le & h{H}_{p}(T{u}_{0},T{u}_{1})\\ \le & hrmax\{p(f{u}_{0},f{u}_{1}),\frac{p(f{u}_{0},T{u}_{0})+p(f{u}_{1},T{u}_{1})}{2},\frac{p(f{u}_{0},T{u}_{1})+p(f{u}_{1},T{u}_{0})}{2}\}\\ \le & \frac{1}{\sqrt{r}}rmax\{p({y}_{0},{y}_{1}),\frac{p({y}_{0},{y}_{1})+p({y}_{1},{y}_{2})}{2},\frac{p({y}_{0},{y}_{2})+p({y}_{1},{y}_{1})}{2}\}\\ \le & \sqrt{r}max\{p({y}_{0},{y}_{1}),\frac{p({y}_{0},{y}_{1})+p({y}_{1},{y}_{2})}{2}\}.\end{array}

If

max\{p({y}_{0},{y}_{1}),\frac{p({y}_{0},{y}_{1})+p({y}_{1},{y}_{2})}{2}\}=p({y}_{0},{y}_{1}),

then

p({y}_{1},{y}_{2})\le h{H}_{p}(T{u}_{0},T{u}_{1})\le \sqrt{r}p({y}_{0},{y}_{1}).

If

max\{p({y}_{0},{y}_{1}),\frac{p({y}_{0},{y}_{1})+p({y}_{1},{y}_{2})}{2}\}=\frac{p({y}_{0},{y}_{1})+p({y}_{1},{y}_{2})}{2},

then we obtain

p({y}_{1},{y}_{2})\le \frac{\sqrt{r}}{2-\sqrt{r}}p({y}_{0},{y}_{1})\le \sqrt{r}p({y}_{0},{y}_{1}).

As f{u}_{2}\in T{u}_{1}, we choose {y}_{3}\in T{u}_{2} such that p(f{u}_{2},{y}_{3})\le hH(T{u}_{1},T{u}_{2}). Using the fact that T{u}_{2}\subseteq f(Y), we obtain a point {u}_{3}\in Y such that {y}_{3}=f{u}_{3}\in T{u}_{2} and

p(f{u}_{2},f{u}_{3})\le h{H}_{p}(T{u}_{1},T{u}_{2}).

Since

\omega (r)p(f{u}_{1},T{u}_{1})\le \omega (r)p(f{u}_{1},f{u}_{2})\le p(f{u}_{1},f{u}_{2}),

so we have

\begin{array}{rcl}p(f{u}_{2},f{u}_{3})& \le & h{H}_{p}(T{u}_{1},T{u}_{2})\\ \le & hrmax\{p(f{u}_{1},f{u}_{2}),\frac{p(f{u}_{1},T{u}_{1})+p(f{u}_{2},T{u}_{2})}{2},\frac{p(f{u}_{1},T{u}_{2})+p(f{u}_{2},T{u}_{1})}{2}\}\\ \le & \frac{1}{\sqrt{r}}rmax\{p({y}_{1},{y}_{2}),\frac{p({y}_{1},{y}_{2})+p({y}_{2},{y}_{3})}{2},\frac{p({y}_{1},{y}_{3})+p({y}_{2},{y}_{2})}{2}\}\\ \le & \sqrt{r}max\{p({y}_{1},{y}_{2}),\frac{p({y}_{1},{y}_{2})+p({y}_{2},{y}_{3})}{2}\}.\end{array}

Following the arguments similar to those given above, we obtain

p({y}_{2},{y}_{3})\le \sqrt{r}p({y}_{1},{y}_{2}),

which further implies that

p({y}_{2},{y}_{3})\le {(\sqrt{r})}^{2}p({y}_{0},{y}_{1}).

Continuing this process, we obtain a sequence \{{y}_{n}\}\subset Y such that for any integer n\ge 0, {y}_{n+1}=f{u}_{n+1}\in T{u}_{n} and

p({y}_{n},{y}_{n+1})\le {(\sqrt{r})}^{n}p({y}_{0},{y}_{1})

for every n\in \mathbb{N}. This shows that {lim}_{n\to \mathrm{\infty}}p({y}_{n},{y}_{n+1})=0. Since

p({y}_{n},{y}_{n})+p({y}_{n+1},{y}_{n+1})\le 2p({y}_{n},{y}_{n+1}),

so we obtain

\underset{n\to \mathrm{\infty}}{lim}p({y}_{n},{y}_{n})=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty}}{lim}p({y}_{n+1},{y}_{n+1})=0.

Now, for m>n\ge 1, we have

\begin{array}{rcl}{p}^{S}({y}_{n},{y}_{n+m})& =& 2p({y}_{n},{y}_{n+m})-p({y}_{n},{y}_{n})-p({y}_{n+m},{y}_{n+m})\\ \le & 2p({y}_{n},{y}_{n+1})+2p({y}_{n+1},{y}_{n+2})+\cdots +2p({y}_{n+m-1},{y}_{n+m})\\ \le & 2({(\sqrt{r})}^{n}+{(\sqrt{r})}^{n+1}+\cdots +{(\sqrt{r})}^{n+m-1})p({y}_{0},{y}_{1}).\end{array}

It follows that \{{y}_{n}\} is a Cauchy sequence in (f(Y),{p}^{S}). By Lemma A, we have \{{y}_{n}\} is a Cauchy sequence in (f(Y),p). Since (f(Y),p) is (f,T)-orbitally complete at {u}_{0}, so again by Lemma D, (f(Y),{p}^{S}) is (f,T)-orbitally complete at {u}_{0}. Hence, there exists an element u\in f(Y) such that {lim}_{n\to \mathrm{\infty}}{p}^{S}({y}_{n},y)=0. This implies that

\underset{n\to \mathrm{\infty}}{lim}p({y}_{n},u)=\underset{n\to \mathrm{\infty}}{lim}p({y}_{n},{y}_{n})=p(u,u)=0.

(2.1)

Let z\in {f}^{-1}u, then z\in Y and u=fz. Now,

give

\underset{n\to \mathrm{\infty}}{lim}p(f{u}_{n+1},Tx)=p(fz,Tx).

Similarly, we can show that

\underset{n\to \mathrm{\infty}}{lim}p(f{u}_{n},Tx)=p(fz,Tx).

Now, we will claim that

p(fz,Tx)\le rp(fz,fx)\phantom{\rule{1em}{0ex}}\text{for any}fx\in f(Y)-\{fz\}.

(2.2)

If x=z or fx=fz, then p(fx,Tx)=0. This gives {p}^{S}(fx,Tx)=0, which implies that fx\in Tx and we are done. Now from (2.1), there exists a positive integer {n}_{0} such that for all\phantom{\rule{0.25em}{0ex}}n\ge {n}_{0},

p(fz,f{u}_{n+1})\le \frac{1}{3}p(fz,fx)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}p(fz,f{u}_{n})\le \frac{1}{3}p(fz,fx).

So, for any n\ge {n}_{0}, we have

\begin{array}{rcl}\omega (r)p(f{u}_{n},T{u}_{n})& \le & p(f{u}_{n},T{u}_{n})\le p(f{u}_{n},f{u}_{n+1})\\ \le & p(f{u}_{n},fz)+p(fz,f{u}_{n+1})-p(fz,fz)\le \frac{2}{3}p(fz,fx)\\ \le & p(fz,fx)-\frac{1}{3}p(fz,fx)\le p(fz,fx)-p(fz,f{u}_{n})\\ \le & p(f{u}_{n},fx)-p(f{u}_{n},f{u}_{n})\le p(f{u}_{n},fx).\end{array}

Hence, for any n\ge {n}_{0}, we obtain

\omega (r)p(f{u}_{n},T{u}_{n})\le p(f{u}_{n},fx).

This implies

\begin{array}{rcl}p(f{u}_{n+1},Tx)& \le & {H}_{p}(T{u}_{n},Tx)\\ \le & rmax\{p(f{u}_{n},fx),\frac{p(f{u}_{n},T{u}_{n})+p(fx,Tx)}{2},\frac{p(f{u}_{n},Tx)+p(fx,T{u}_{n})}{2}\}\\ \le & rmax\{p({y}_{n},fx),\frac{p({y}_{n},{y}_{n+1})+p(fx,Tx)}{2},\frac{p({y}_{n},Tx)+p(fx,{y}_{n+1})}{2}\}\\ \le & rmax\{p({y}_{n},u)+p(u,fx)-p(u,u),\frac{p({y}_{n},{y}_{n+1})+p(fx,Tx)}{2},\\ \frac{p({y}_{n},u)+p(u,Tx)-p(u,u)+p(fx,u)+p(u,{y}_{n+1})-p(u,u)}{2}\}.\end{array}

On taking limit as *n* tends to ∞, we obtain

\begin{array}{rcl}p(fz,Tx)& \le & rmax\{p(u,fx),\frac{p(fx,Tx)}{2},\frac{p(u,Tx)+p(fx,u)}{2}\}\\ =& rmax\{p(fz,fx),\frac{p(fx,Tx)}{2},\frac{p(fz,Tx)+p(fx{,}_{,}fz)}{2}\}\\ \le & rmax\{p(fz,fx),\frac{p(fz,Tx)+p(fx{,}_{,}fz)}{2}\}.\end{array}

If

max\{p(fz,fx),\frac{p(fz,Tx)+p(fx{,}_{,}fz)}{2}\}=p(fz,fx),

then we are done. If

max\{p(fz,fx),\frac{p(fz,Tx)+p(fx{,}_{,}fz)}{2}\}=\frac{p(fz,Tx)+p(fx{,}_{,}fz)}{2},

then we obtain

p(fz,Tx)\le \frac{r}{2-r}p(fx,fz)\le rp(fx,fz)

and hence (2.2) holds. Next, we show that

{H}_{p}(Tz,Tx)\le rmax\{p(fz,fx),\frac{p(fx,Tx)+p(fz,Tz)}{2},\frac{p(fx,Tz)+p(fz,Tx)}{2}\}

(2.3)

for any x\in Y. If x=z, then fx=fz, and the claim follows from (2.2). Suppose that x\ne z, then fx\ne fz. As *f* is a non-constant single-valued mapping, we have

\begin{array}{rcl}p(fx,Tx)& \le & p(fx,fz)+p(fz,Tx)-p(fz,fz)\\ \le & p(fx,fz)+rp(fx,fz)\le (1+r)p(fx,fz).\end{array}

This implies

\omega (r)p(fx,Tx)\le p(fx,fz).

Therefore,

{H}_{p}(Tz,Tx)\le rmax\{p(fz,fx),\frac{p(fx,Tx)+p(fz,Tz)}{2},\frac{p(fx,Tz)+p(fz,Tx)}{2}\}.

Hence, (2.3) holds for any x\in Y. Note that

\begin{array}{rcl}p(Tz,f{u}_{n+2})& \le & {H}_{p}(Tz,T{u}_{n+1})\\ \le & rmax\{p(fz,f{u}_{n+1}),\frac{p(f{u}_{n+1},T{u}_{n+1})+p(fz,Tz)}{2},\\ \frac{p(f{u}_{n+1},Tz)+p(fz,T{u}_{n+1})}{2}\}\\ \le & rmax\{p(fz,{y}_{n+2}),\frac{p({y}_{n+2},{y}_{n+2})+p(fz,Tz)}{2},\\ \frac{p({y}_{n+2},fz)+p(fz,Tz)-p(fz,fz)+p(fz,{y}_{n+2})}{2}\}.\end{array}

On taking limit as n\to \mathrm{\infty}, we obtain

p(fz,Tz)\le \frac{r}{2}p(fz,Tz).

We obtain p(fz,Tz)=0, which further implies that {p}^{S}(fz,Tz)\le 2p(fz,Tz)=0. Hence, fz\in Tz. Further if Y=X and ffz=fz, then due to *IT*-commutativity of the pair (f,T), we have fz=ffz\in fTz\subseteq Tfz. This shows that *fz* is a common fixed point of the pair (f,T). □

**Corollary A** *Let* (X,p) *be a partial metric space and* *Y* *be any non*-*empty set*. *Assume that here exists* r\in [0,1) *such that the mappings* f:Y\u27f6X *and* T:Y\u27f6C{B}^{p}(X) *satisfy*

\omega (r)p(fx,Tx)\le p(fx,fy)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}{H}_{p}(Tx,Ty)\le rp(fx,fy)

*for all* x,y\in Y, *with* T(Y)\subset f(Y). *If there exists* {u}_{0}\in Y *such that* f(Y) *is* (f,T)-*orbitally complete at* {u}_{0}, *then* C(f,T)\ne \varphi. *If* Y=X *and* (f,T) *is* *IT*-*commuting at coincidence points of the pair* (f,T), *then* F(f,T)\ne \varphi *provided that* *fz* *is a fixed point of* *f* *for some* z\in C(f,T).

**Example 2.2** Let X=\{0,1,2\} and Y=\{0,1,2,3\}. Define a mapping p:X\times X\to {\mathbb{R}}^{+} as follows:

Then *p* is a partial metric on *X*. Let \omega (r) be as given in Theorem 2.1 and the mappings T:Y\u27f6C{B}^{p}(X) and f:Y\to X be given as

Tx=\{\begin{array}{cc}\{0\}\hfill & \text{when}x\ne 2,\hfill \\ \{0,1\}\hfill & \text{when}x=2,\hfill \end{array}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}fx=\{\begin{array}{cc}0,\hfill & \text{if}x\in \{0,1\},\hfill \\ 2,\hfill & \text{if}x=2,\hfill \\ 1,\hfill & \text{if}x=3.\hfill \end{array}

Note that

If we take r\ge \frac{3}{5} and \omega (r)\le \frac{5}{8}, then for all x,y\in Y,

\omega (r)p(fx,Tx)\le p(fx,fy)

holds. If we consider r=\frac{5}{6}, then \omega (r)=\frac{1}{6}. Then, for x,y\in \{0,1,3\}, we have {H}_{p}(Tx,Ty)=0, hence {H}_{p}(Tx,Ty)\le rp(fx,fy) is satisfied trivially. Now consider

Hence, for all x,y\in Y,

\omega (r)p(fx,Tx)\le p(fx,fy)

implies

{H}_{p}(Tx,Ty)\le rp(fx,fy).

Let {u}_{0}=1, {y}_{0}=f(1)=0. As T(0)\subseteq f(Y), there exists a point {u}_{1}=1 in *Y* such that {y}_{1}=f(1)=0\in T(1) and T(0)=\{0\}\subseteq f(Y), we obtain a point {u}_{2}=1 in *Y* such that {y}_{2}=0=f(1)\in T(1). Continuing this way, we construct an orbit \{{y}_{0}={y}_{1}={y}_{2}=\cdots =0\} for (f,T) at {u}_{0}=1. Also, f(Y) is (f,T)-orbitally complete at {u}_{0}=0. So, all the conditions of Corollary A are satisfied. Moreover, C(f,T)=\{0,1\}.

On the other hand, the metric {p}^{S} induced by the partial metric *p* is given by

Now, we show that Corollary A is not applicable (in the case of a metric induced by a partial metric *p*) in this case. Since

\omega (r){p}^{S}(f1,T1)=\omega (r){p}^{S}(0,0)=0\le {p}^{S}(fx,fy)

is satisfied for any r\in [0,1), *x* and *y* in *X*, so it must imply {H}_{{p}^{S}}(T1,T2)\le rp(f1,f2). But

{H}_{{p}^{S}}(T1,T2)={H}_{{p}^{S}}(\{0\},\{0,1\})=\frac{1}{2}

and

{p}^{S}(f1,f2)={p}^{S}(0,2)=\frac{7}{15}<\frac{1}{2}.

Hence, for any r\in [0,1),

{H}_{{p}^{S}}(T1,T2)\nleqq rp(f1,f2).

**Corollary B** *Let* (X,p) *be a partial metric space*, *Y* *be any non*-*empty set and* f,T:Y\u27f6X *be such that* T(Y)\subset f(Y). *Suppose that there exists* {u}_{0}\in Y *such that* f(Y) *is* (f,T)-*orbitally complete at* {u}_{0}. *Assume further that there exists an* r\in [0,1) *such that*

\omega (r)p(fx,Tx)\le p(fx,fy)

*implies that*

p(Tx,Ty)\le rmax\{p(fx,fy),\frac{p(fx,Tx)+p(fy,Ty)}{2},\frac{p(fx,Ty)+p(fy,Tx)}{2}\}

*for all* x,y\in Y. *Then* C(f,T)\ne \varphi. *Further*, *if* Y=X *and the pair* (f,T) *is commuting at* *x* *where* x\in C(f,T), *then* F(f,T) *is a singleton*.

*Proof* It follows from Theorem 2.1, that C(f,T)\ne \varphi. If u\in C(f,T), then fu=Tu. Further, if Y=X and (f,T) is commuting at *u*, then ffu=fTu=Tfu. Now,

\omega (r)p(fu,Tfu)\le p(fu,Tfu)=p(fu,ffu)

implies that

\begin{array}{rcl}p(fu,ffu)& =& p(Tu,Tfu)\\ \le & r{M}_{p,f}(u,fu)\\ \le & rmax\{p(fu,ffu),\frac{p(ffu,Tfu)+p(fu,Tu)}{2},\frac{p(ffu,Tu)+p(fu,Tfu)}{2}\}\\ \le & rmax\{p(fu,ffu),\frac{p(ffu,ffu)+p(fu,Tu)}{2},\frac{p(ffu,fu)+p(fu,ffu)}{2}\}\\ \le & rmax\{p(fu,ffu),\frac{p(ffu,fu)+p(fu,ffu)}{2},\frac{p(ffu,fu)+p(fu,ffu)}{2}\}\\ \le & rp(fu,ffu).\end{array}

As r<1, we obtain p(fu,ffu)=0, which further implies that {p}^{S}(fu,ffu)\le 2p(fu,ffu)=0. Hence, *fu* is a common fixed point of *f* and *T*.

For uniqueness, assume there exist {z}_{1}\ne {z}_{2}, such that {z}_{1}=f{z}_{1}=T{z}_{1} and {z}_{2}=f{z}_{2}=T{z}_{2}. Then

\psi (r)p(f{z}_{1},T{z}_{1})\le p(f{z}_{1},T{z}_{1})=p(f{z}_{1},f{z}_{1})\le p(f{z}_{1},f{z}_{2}),

which implies

\begin{array}{rcl}p({z}_{1},{z}_{2})& =& p(T{z}_{1},T{z}_{2})\\ \le & rmax\{p(f{z}_{1},f{z}_{2}),\frac{p(f{z}_{1},T{z}_{1})+p(f{z}_{2},T{z}_{2})}{2},\frac{p(f{z}_{2},T{z}_{1})+p(f{z}_{1},T{z}_{2})}{2}\}\\ \le & rmax\{p({z}_{1},{z}_{2}),p({z}_{1},{z}_{2}),p({z}_{1},{z}_{2})\}\\ \le & rp({z}_{1},{z}_{2}).\end{array}

We obtain p({z}_{1},{z}_{2})=0, which further implies that {p}^{S}({z}_{1},{z}_{2})\le 2p({z}_{1},{z}_{2})=0. Hence, {z}_{1}={z}_{2}. □