In this section, we present new fixed point theorems on the setting of complete metric spaces. Our first result is the following.

**Theorem 2.1** *Let* (X,d) *be a complete metric space*, \phi :X\to [0,\mathrm{\infty}) *be a lower semi*-*continuous function and* T:X\to X *be a given mapping*. *Suppose that for any* 0<a<b<\mathrm{\infty}, *there exists* 0<\gamma (a,b)<1 *such that for all* x,y\in X,

*Then* *T* *has a unique fixed point* {x}^{\ast}\in X. *Moreover*, *we have* \phi ({x}^{\ast})=0.

*Proof* Let {x}_{0} be an arbitrary point in *X*. Consider the sequence \{{x}_{n}\}\subset X defined by {x}_{n+1}=T{x}_{n} for all n\ge 0.

*Step I*. We prove that

d({x}_{n-1},{x}_{n})+\phi ({x}_{n-1})+\phi ({x}_{n})\to 0\phantom{\rule{1em}{0ex}}\text{as}n\to \mathrm{\infty}.

(3)

If for some *n* we have d({x}_{n-1},{x}_{n})+\phi ({x}_{n-1})+\phi ({x}_{n})=0, then {x}_{n-1} will be a fixed point of *T*. So, we can suppose that

0<d({x}_{n-1},{x}_{n})+\phi ({x}_{n-1})+\phi ({x}_{n}),\phantom{\rule{1em}{0ex}}\mathrm{\forall}n\ge 1.

(4)

Suppose that for some {n}_{0}\ge 1,

d({x}_{{n}_{0}-1},{x}_{{n}_{0}})+\phi ({x}_{{n}_{0}-1})+\phi ({x}_{{n}_{0}})<d({x}_{{n}_{0}},{x}_{{n}_{0}+1})+\phi ({x}_{{n}_{0}})+\phi ({x}_{{n}_{0}+1}).

(5)

From (4) and (5), we have

\begin{array}{rcl}0<a& :=& d({x}_{{n}_{0}-1},{x}_{{n}_{0}})+\phi ({x}_{{n}_{0}-1})+\phi ({x}_{{n}_{0}})\\ \le & d({x}_{{n}_{0}-1},{x}_{{n}_{0}})+\phi ({x}_{{n}_{0}-1})+\phi ({x}_{{n}_{0}})\\ <& d({x}_{{n}_{0}},{x}_{{n}_{0}+1})+\phi ({x}_{{n}_{0}})+\phi ({x}_{{n}_{0}+1}):=b.\end{array}

From (2), there exists \gamma (a,b)\in (0,1) such that

d({x}_{{n}_{0}},{x}_{{n}_{0}+1})+\phi ({x}_{{n}_{0}})+\phi ({x}_{{n}_{0}+1})\le \gamma (a,b)[d({x}_{{n}_{0}-1},{x}_{{n}_{0}})+\phi ({x}_{{n}_{0}-1})+\phi ({x}_{{n}_{0}})].

From (4) and (5), we get

0<d({x}_{{n}_{0}-1},{x}_{{n}_{0}})+\phi ({x}_{{n}_{0}-1})+\phi ({x}_{{n}_{0}})<\gamma (a,b)[d({x}_{{n}_{0}-1},{x}_{{n}_{0}})+\phi ({x}_{{n}_{0}-1})+\phi ({x}_{{n}_{0}})],

which implies that \gamma (a,b)>1, a contradiction. Thus, our assumption (5) is wrong. We deduce that

{\mathcal{U}}_{n}:=d({x}_{n},{x}_{n+1})+\phi ({x}_{n})+\phi ({x}_{n+1})\le {\mathcal{U}}_{n-1}:=d({x}_{n-1},{x}_{n})+\phi ({x}_{n-1})+\phi ({x}_{n}),\phantom{\rule{1em}{0ex}}\mathrm{\forall}n\ge 1.

This implies that \{{\mathcal{U}}_{n}\} is a non-increasing sequence of positive real numbers. Hence, there is r\ge 0 such that

d({x}_{n-1},{x}_{n})+\phi ({x}_{n-1})+\phi ({x}_{n})\to {r}^{+}\phantom{\rule{1em}{0ex}}\text{as}n\to \mathrm{\infty}.

(6)

Suppose that r>0. Then 0<r\le d({x}_{n-1},{x}_{n})+\phi ({x}_{n-1})+\phi ({x}_{n})\le {\mathcal{U}}_{0} for all n\ge 1. From (2), there exists \gamma (r,{\mathcal{U}}_{0})\in (0,1) such that

d({x}_{n},{x}_{n+1})+\phi ({x}_{n})+\phi ({x}_{n+1})\le \gamma (r,{\mathcal{U}}_{0})[d({x}_{n-1},{x}_{n})+\phi ({x}_{n-1})+\phi ({x}_{n})],\phantom{\rule{1em}{0ex}}\mathrm{\forall}n\ge 1.

Letting n\to \mathrm{\infty} in the above inequality and using (6), we get r\le \gamma (r,{\mathcal{U}}_{0})r, which implies that \gamma (r,{\mathcal{U}}_{0})\ge 1, a contradiction. Thus, r=0 and (3) holds.

*Step II*. We prove that \{{x}_{n}\} is a Cauchy sequence in (X,d).

Suppose that \{{x}_{n}\} is not a Cauchy sequence. Then there exists \epsilon >0 for which we can find two sequences of positive integers \{m(k)\} and \{n(k)\} such that for all positive integer *k*,

n(k)>m(k)>k,\phantom{\rule{2em}{0ex}}d({x}_{m(k)},{x}_{n(k)})\ge \epsilon ,\phantom{\rule{2em}{0ex}}d({x}_{m(k)},{x}_{n(k)-1})<\epsilon .

(7)

From (7), we have

\epsilon \le d({x}_{n(k)},{x}_{m(k)})\le d({x}_{m(k)},{x}_{n(k)-1})+d({x}_{n(k)-1},{x}_{n(k)})<\epsilon +d({x}_{n(k)},{x}_{n(k)-1}).

Thus, for all *k*, we have

\epsilon \le d({x}_{n(k)},{x}_{m(k)})<\epsilon +d({x}_{n(k)},{x}_{n(k)-1}).

Letting k\to \mathrm{\infty} in the above inequality and using (3), we obtain

d({x}_{n(k)},{x}_{m(k)})\to {\epsilon}^{+}\phantom{\rule{1em}{0ex}}\text{as}k\to \mathrm{\infty}.

(8)

On the other hand, we have

d({x}_{n(k)},{x}_{m(k)})\le d({x}_{n(k)},{x}_{n(k)+1})+d({x}_{n(k)+1},{x}_{m(k)+1})+d({x}_{m(k)+1},{x}_{m(k)})

and

d({x}_{n(k)+1},{x}_{m(k)+1})\le d({x}_{n(k)+1},{x}_{n(k)})+d({x}_{n(k)},{x}_{m(k)})+d({x}_{m(k)},{x}_{m(k)+1}).

Thus, we have

|d({x}_{n(k)+1},{x}_{m(k)+1})-d({x}_{n(k)},{x}_{m(k)})|\le d({x}_{n(k)+1},{x}_{n(k)})+d({x}_{m(k)},{x}_{m(k)+1}).

Letting k\to \mathrm{\infty} in the above inequality, using (3) and (8), we obtain that

d({x}_{n(k)+1},{x}_{m(k)+1})\to \epsilon \phantom{\rule{1em}{0ex}}\text{as}k\to \mathrm{\infty}.

(9)

On the other hand, for all *k*, we have

\begin{array}{rcl}\epsilon & \le & d({x}_{n(k)},{x}_{m(k)})+\phi ({x}_{n(k)})+\phi ({x}_{m(k)})\\ \le & d({x}_{m(k)},{x}_{n(k)-1})+d({x}_{n(k)-1},{x}_{n(k)})+\mathrm{\Phi}\le \epsilon +{\mathcal{U}}_{0}+\mathrm{\Phi},\end{array}

where Φ is the upper bound of \{2\phi ({x}_{n})\} (note that from (3), \phi ({x}_{n})\to 0 as n\to \mathrm{\infty}). Applying (2), we obtain that there exists \gamma (\epsilon ,\epsilon +{\mathcal{U}}_{0}+\mathrm{\Phi})\in (0,1) such that

for all *k*. Letting k\to \mathrm{\infty} in the above inequality, using (3), (8) and (9), we obtain that \epsilon \le \gamma (\epsilon ,\epsilon +{\mathcal{U}}_{0}+\mathrm{\Phi})\epsilon, which implies that \gamma (\epsilon ,\epsilon +{\mathcal{U}}_{0}+\mathrm{\Phi})\ge 1, a contradiction with \gamma (\epsilon ,\epsilon +{\mathcal{U}}_{0}+\mathrm{\Phi})\in (0,1). Hence, we deduce that \{{x}_{n}\} is a Cauchy sequence in (X,d). Since (X,d) is complete, there exists {x}^{\ast}\in X such that {x}_{n}\to {x}^{\ast} as n\to \mathrm{\infty}.

*Step III*. We prove that

\phi \left({x}^{\ast}\right)=0.

(10)

From (3), we have \phi ({x}_{n})\to 0 as n\to \mathrm{\infty}. Since *φ* is lower semi-continuous, it follows that

0\le \phi \left({x}^{\ast}\right)\le \underset{n\to +\mathrm{\infty}}{lim\hspace{0.17em}inf}\phi ({x}_{n})=\underset{n\to +\mathrm{\infty}}{lim}\phi ({x}_{n})=0,

which implies (10).

*Step IV*. We prove that {x}^{\ast} is a fixed point of *T*. Consider the sets I,J\subseteq \mathbb{N} defined by

I:=\{n\in \mathbb{N}:{x}_{n}={x}^{\ast}\}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}J:=\{n\in \mathbb{N}:{x}_{n}\ne {x}^{\ast}\}.

As I\cup J=\mathbb{N}, at least one of these subsets is infinite. So, we consider two cases.

*Case 1*. *I* is an infinite subset. In this case, we can find a subsequence \{{x}_{n(p)}\} of \{{x}_{n}\} such that {x}_{n(p)}={x}^{\ast} for all *p*. Since {x}_{n(p)+1}=T{x}_{n(p)}=T{x}^{\ast} and {x}_{n(p)+1}\to {x}^{\ast} as p\to \mathrm{\infty}, we have necessarily {x}^{\ast}=T{x}^{\ast}, that is, {x}^{\ast} is a fixed point of *T*.

*Case 2*. *J* is an infinite subset. In this case, we can find a subsequence \{{x}_{n(p)}\} of \{{x}_{n}\} such that d({x}_{n(p)},{x}^{\ast})>0 for all *p*. Then, for every *p*, we can find 0<{A}_{p}<{B}_{p} such that

{A}_{p}\le d({x}_{n(p)},{x}^{\ast})+\phi ({x}_{n(p)})+\phi \left({x}^{\ast}\right)\le {B}_{p}.

From (2), we have

d({x}_{n(p)+1},T{x}^{\ast})+\phi ({x}_{n(p)+1})+\phi \left(T{x}^{\ast}\right)\le \gamma ({A}_{p},{B}_{p})[d({x}_{n(p)},{x}^{\ast})+\phi ({x}_{n(p)})+\phi \left({x}^{\ast}\right)].

Since \gamma ({A}_{p},{B}_{p})<1, for all *p* we deduce that

d({x}_{n(p)+1},T{x}^{\ast})+\phi ({x}_{n(p)+1})+\phi \left(T{x}^{\ast}\right)\le d({x}_{n(p)},{x}^{\ast})+\phi ({x}_{n(p)})+\phi \left({x}^{\ast}\right).

Since d({x}_{n(p)},{x}^{\ast})+\phi ({x}_{n(p)})+\phi ({x}^{\ast})\to \phi ({x}^{\ast})=0 as p\to +\mathrm{\infty} (see (3) and (10)), letting p\to +\mathrm{\infty} in the above inequality, we obtain that

\underset{p\to +\mathrm{\infty}}{lim}d({x}_{n(p)+1},T{x}^{\ast})+\phi \left(T{x}^{\ast}\right)\le 0

and so {x}_{n(p)+1}\to T{x}^{\ast} as p\to +\mathrm{\infty}. By the uniqueness of the limit, we have necessarily {x}^{\ast}=T{x}^{\ast}. Then {x}^{\ast} is a fixed point of *T*.

*Step V*. Uniqueness of the fixed point. Suppose that {y}^{\ast}\in X is another fixed point of *T*, that is, T{y}^{\ast}={y}^{\ast} and d({x}^{\ast},{y}^{\ast})>0. Since d({x}^{\ast},{y}^{\ast})>0, we can find 0<\alpha <\beta such that \alpha \le d({x}^{\ast},{y}^{\ast})+\phi ({x}^{\ast})+\phi ({y}^{\ast})\le \beta. From (2), we have

d(T{x}^{\ast},T{y}^{\ast})+\phi \left(T{x}^{\ast}\right)+\phi \left(T{y}^{\ast}\right)\le \gamma (\alpha ,\beta )[d({x}^{\ast},{y}^{\ast})+\phi \left({x}^{\ast}\right)+\phi \left({y}^{\ast}\right)],

that is,

d({x}^{\ast},{y}^{\ast})+\phi \left({x}^{\ast}\right)+\phi \left({y}^{\ast}\right)\le \gamma (\alpha ,\beta )[d({x}^{\ast},{y}^{\ast})+\phi \left({x}^{\ast}\right)+\phi \left({y}^{\ast}\right)],

which implies that \gamma (\alpha ,\beta )\ge 1, a contradiction with \gamma (\alpha ,\beta )\in (0,1). Then {x}^{\ast} is the unique fixed point of *T*. This makes end to the proof. □

An immediate consequence of Theorem 2.1 is the following.

**Theorem 2.2** *Let* (X,d) *be a complete metric space*, \phi :X\to [0,\mathrm{\infty}) *be a lower semi*-*continuous function and* T:X\to X *be a given mapping*. *Suppose that there exists a constant* \gamma \in (0,1) *such that for all* x,y\in X,

d(Tx,Ty)+\phi (Tx)+\phi (Ty)\le \gamma [d(x,y)+\phi (x)+\phi (y)].

*Then* *T* *has a unique fixed point* {x}^{\ast}\in X. *Moreover*, *we have* \phi ({x}^{\ast})=0.

**Remark 2.1** Taking \phi \equiv 0 in Theorem 2.2, we obtain the Banach contraction principle.

The following result is a generalization of Kannan’s fixed point theorem.

**Theorem 2.3** *Let* (X,d) *be a complete metric space*, \phi :X\to [0,\mathrm{\infty}) *be a lower semi*-*continuous function and* T:X\to X *be a given mapping*. *Suppose that there exists a constant* \gamma \in (0,1/2) *such that for all* x,y\in X,

d(Tx,Ty)+\phi (Tx)+\phi (Ty)\le \gamma [d(x,Tx)+d(y,Ty)+\phi (x)+\phi (y)+\phi (Tx)+\phi (Ty)].

(11)

*Then* *T* *has a unique fixed point* {x}^{\ast}\in X. *Moreover*, *we have* \phi ({x}^{\ast})=0.

*Proof* Let {x}_{0}\in X be an arbitrary point. Define the sequence \{{x}_{n}\} in *X* by {x}_{n+1}=T{x}_{n} for all n\ge 0. Applying (11) with x={x}_{1} and y={x}_{0}, we get immediately

d({x}_{2},{x}_{1})+\phi ({x}_{2})+\phi ({x}_{1})\le \left(\frac{\gamma}{1-\gamma}\right)[d({x}_{1},{x}_{0})+\phi ({x}_{1})+\phi ({x}_{0})].

Continuing this process, by induction, we obtain that

d({x}_{n+1},{x}_{n})+\phi ({x}_{n+1})+\phi ({x}_{n})\le {\left(\frac{\gamma}{1-\gamma}\right)}^{n}[d({x}_{1},{x}_{0})+\phi ({x}_{1})+\phi ({x}_{0})]

(12)

for all n\ge 0. Note that since \gamma \in (0,1/2), we have \gamma /(1-\gamma )\in (0,1), so from inequality (12), the sequence \{{x}_{n}\} is Cauchy in the metric space (X,d). Since (X,d) is a complete metric space, there exists {x}^{\ast}\in X such that {x}_{n}\to {x}^{\ast} as n\to \mathrm{\infty}. On the other hand, from (12), we have

\phi ({x}_{n})\le {\left(\frac{\gamma}{1-\gamma}\right)}^{n}[d({x}_{1},{x}_{0})+\phi ({x}_{1})+\phi ({x}_{0})],

for all n\ge 0. Letting n\to +\mathrm{\infty} in the above inequality, we get \phi ({x}_{n})\to 0 as n\to \mathrm{\infty}. Since *φ* is lower semi-continuous, it follows that

0\le \phi \left({x}^{\ast}\right)\le \underset{n\to +\mathrm{\infty}}{lim\hspace{0.17em}inf}\phi ({x}_{n})=\underset{n\to +\mathrm{\infty}}{lim}\phi ({x}_{n})=0,

that is, \phi ({x}^{\ast})=0. Now, we prove that {x}^{\ast} is a fixed point of *T*. Indeed, applying condition (11), we obtain that

for all n\ge 0. Letting n\to \mathrm{\infty}, we obtain

d({x}^{\ast},T{x}^{\ast})+\phi \left(T{x}^{\ast}\right)\le \gamma [d({x}^{\ast},T{x}^{\ast})+\phi \left(T{x}^{\ast}\right)],

which implies (since 0<\gamma <1/2) that

d({x}^{\ast},T{x}^{\ast})=\phi \left(T{x}^{\ast}\right)=0,

that is, {x}^{\ast}=T{x}^{\ast}.

Suppose now that {y}^{\ast} is another fixed point of *T*. Applying (11) with x=y={y}^{\ast}, we obtain

\phi \left({y}^{\ast}\right)\le 2\gamma \phi \left({y}^{\ast}\right),

which implies (since 0<\gamma <1/2) that \phi ({y}^{\ast})=0. Now, applying (11) with x={x}^{\ast} and y={y}^{\ast}, taking into consideration that \phi ({x}^{\ast})=\phi ({y}^{\ast})=0, we obtain d({x}^{\ast},{y}^{\ast})=0, which implies that {x}^{\ast}={y}^{\ast}. □

**Remark 2.2** Taking \phi \equiv 0 in Theorem 2.3, we obtain Kannan’s fixed point theorem [6].

Now, we state and prove a generalization of Reich’s fixed point theorem.

**Theorem 2.4** *Let* (X,d) *be a complete metric space*, \phi :X\to [0,\mathrm{\infty}) *be a lower semi*-*continuous function and* T:X\to X *be a given mapping*. *Suppose that there exist* \alpha ,\beta ,\gamma \in [0,\mathrm{\infty}), *with* \alpha +\beta +\gamma <1, *such that*

\begin{array}{rcl}d(Tx,Ty)+\phi (Tx)+\phi (Ty)& \le & \alpha [d(x,y)+\phi (x)+\phi (y)]+\beta [d(x,Tx)+\phi (x)+\phi (Tx)]\\ +\gamma [d(y,Ty)+\phi (y)+\phi (Ty)]\end{array}

(13)

*for all* x,y\in X. *Then* *T* *has a unique fixed point* {x}^{\ast}\in X. *Moreover*, *we have* \phi ({x}^{\ast})=0.

*Proof* Let {x}_{0}\in X be an arbitrary point. Define the sequence \{{x}_{n}\} in *X* by {x}_{n+1}=T{x}_{n} for all n\ge 0. Applying (13) with x={x}_{1} and y={x}_{0}, using the triangular inequality, we have

\begin{array}{rcl}d({x}_{2},{x}_{1})+\phi ({x}_{2})+\phi ({x}_{1})& \le & \alpha [d({x}_{1},{x}_{0})+\phi ({x}_{1})+\phi ({x}_{0})]+\beta [d({x}_{2},{x}_{1})+\phi ({x}_{1})+\phi ({x}_{2})]\\ +\gamma [d({x}_{1},{x}_{0})+\phi ({x}_{0})+\phi ({x}_{1})].\end{array}

This implies that

d({x}_{2},{x}_{1})+\phi ({x}_{2})+\phi ({x}_{1})\le \left(\frac{\alpha +\gamma}{1-\beta}\right)[d({x}_{1},{x}_{0})+\phi ({x}_{1})+\phi ({x}_{0})],

where (\alpha +\gamma )/(1-\beta )\in (0,1). Continuing this process, by induction, we obtain that

d({x}_{n+1},{x}_{n})+\phi ({x}_{n+1})+\phi ({x}_{n})\le {\left(\frac{\alpha +\gamma}{1-\beta}\right)}^{n}[d({x}_{1},{x}_{0})+\phi ({x}_{1})+\phi ({x}_{0})]

for all n\ge 0. The rest of the proof is similar to the proof of Theorem 2.3. □

**Remark 2.3** Taking \phi \equiv 0 in Theorem 2.4, we obtain Reich’s fixed point theorem [10].

Now, we give a generalization of Chatterjea’s fixed point theorem.

**Theorem 2.5** *Let* (X,d) *be a complete metric space*, \phi :X\to [0,\mathrm{\infty}) *be a lower semi*-*continuous function and* T:X\to X *be a given mapping*. *Suppose that there exists a constant* k\in (0,1/2) *such that*

d(Tx,Ty)+\phi (Tx)+\phi (Ty)\le k[d(x,Ty)+d(y,Tx)+\phi (x)+\phi (y)+\phi (Tx)+\phi (Ty)]

(14)

*for all* x,y\in X. *Then* *T* *has a unique fixed point* {x}^{\ast}\in X. *Moreover*, *we have* \phi ({x}^{\ast})=0.

*Proof* Let {x}_{0}\in X be an arbitrary point. Define the sequence \{{x}_{n}\} in *X* by {x}_{n+1}=T{x}_{n} for all n\ge 0. Applying (14) with x={x}_{1} and y={x}_{0}, using the triangular inequality, we have

\begin{array}{rcl}d({x}_{2},{x}_{1})+\phi ({x}_{2})+\phi ({x}_{1})& \le & k[d({x}_{1},{x}_{1})+d({x}_{0},{x}_{2})+\phi ({x}_{1})+\phi ({x}_{0})+\phi ({x}_{2})+\phi ({x}_{1})]\\ \le & k[d({x}_{0},{x}_{1})+d({x}_{1},{x}_{2})+\phi ({x}_{1})+\phi ({x}_{0})+\phi ({x}_{2})+\phi ({x}_{1})].\end{array}

This implies that

d({x}_{2},{x}_{1})+\phi ({x}_{2})+\phi ({x}_{1})\le \left(\frac{k}{1-k}\right)[d({x}_{1},{x}_{0})+\phi ({x}_{1})+\phi ({x}_{0})].

Continuing this process, by induction, we obtain that

d({x}_{n+1},{x}_{n})+\phi ({x}_{n+1})+\phi ({x}_{n})\le {\left(\frac{k}{1-k}\right)}^{n}[d({x}_{1},{x}_{0})+\phi ({x}_{1})+\phi ({x}_{0})]

for all n\ge 0. The rest of the proof is similar to the proof of Theorem 2.3. □

**Remark 2.4** Taking \phi \equiv 0 in Theorem 2.5, we obtain Chatterjea’s fixed point theorem [3].