# A New Generalization of Rhoades’ Condition

NİHAL TAŞ  Balıkesir University
Department of Mathematics
10145 Balıkesir, TURKEY<EMAIL_ADDRESS>and  NİHAL ÖZGÜR  Balıkesir
University
Department of Mathematics
10145 Balıkesir, TURKEY<EMAIL_ADDRESS>
###### Abstract.

In this paper, our aim is to obtain a new generalization of the well-kown
Rhoades’ contractive condition. To do this, we introduce the notion of an
$S$-normed space. We extend the Rhoades’ contractive condition to $S$-normed
spaces and define a new type of contractive conditions. We support our
theoretical results with necessary illustrative examples.

###### Key words and phrases:

$S$-metric space, $S$-normed space, fixed point theorem, Rhoades’ condition.

###### 2010 Mathematics Subject Classification:

54E35, 54E40, 54E45, 54E50.

##  1\. Introduction

Metric fixed point theory is important to find some applications in many areas
such as topology, analysis, differential equations etc. So different
generalizations of metric spaces were studied (see [3], [4], [5], [6], [9],
[10] and [15]). For example, Mustafa and Sims introduced a new notion of
“$G$-metric space” [10]. Mohanta proved some fixed point theorems for self-
mappings satisfying some kind of contractive type conditions on complete
$G$-metric spaces [9].

Recently Sedghi, Shobe and Aliouche have defined the concept of $S$-metric
spaces in [15] as follows:

######  Definition 1.1.

[15] Let $X$ be a nonempty set and $\mathcal{S}:X\times X\times
X\rightarrow[0,\infty)$ be a function satisfying the following conditions for
all $x,y,z,a\in X:$

$\mathbf{(S1)}$ $\mathcal{S}(x,y,z)=0$ if and only if $x=y=z$,

$\mathbf{(S2)}$
$\mathcal{S}(x,y,z)\leq\mathcal{S}(x,x,a)+\mathcal{S}(y,y,a)+\mathcal{S}(z,z,a)$.

Then $\mathcal{S}$ is called an $S$-metric on $X$ and the pair
$(X,\mathcal{S})$ is called an $S$-metric space.

Let $(X,d)$ be a complete metric space and $T$ be a self-mapping of $X$. In
[14], $T$ is called a Rhoades’ mapping if the following condition is
satisfied:

$\mathbf{(R25)}\text{ \ \ \ \
}d(Tx,Ty)<\max\\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\\},$

for each $x,y\in X$, $x\neq y.$ It was not given any fixed point result for a
Rhoades’ mapping in [14]. Since then, many fixed point theorems were obtained
by several authors for a Rhoades’ mapping (see [1], [2] and [8]). Furthermore,
the Rhoades’ condition was extended on $S$-metric spaces and it was presented
new fixed point results (see [12], [13] and [17]). Now we recall the Rhoades’
condition on an $S$-metric space.

Let $(X,\mathcal{S})$ be an $S$-metric space and $T$ be a self-mapping of $X$.
In [12] and [17], the present authors defined Rhoades’ condition
$\mathbf{(S25)}$ on $(X,\mathcal{S})$ as follows$:$

$\displaystyle\mathbf{(S25)}\text{ \ \ \ \ }\mathcal{S}(Tx,Tx,Ty)$
$\displaystyle<$ $\displaystyle
max\\{\mathcal{S}(x,x,y),\mathcal{S}(Tx,Tx,x),\mathcal{S}(Ty,Ty,y),$
$\displaystyle\mathcal{S}(Ty,Ty,x),\mathcal{S}(Tx,Tx,y)\\},$

for each $x,y\in X$, $x\neq y$.

In this paper, to obtain a new generalization of the Rhoades’ condition, we
introduce the notion of an $S$-normed space. We give some basic concepts and
topological definitions related to an $S$-norm. Then, we study a new form of
Rhoades’ condition $\mathbf{(R25)}$ on $S$-normed spaces and obtain a fixed
point theorem. In Section 2, we introduce the definition of an $S$-norm on $X$
and investigate some basic properties which are needed in the sequel. We
investigate the relationships among an $S$-norm and other known concepts by
counter examples. In Section 3, we define Rhoades’ condition $(\mathbf{NS25})$
on an $S$-normed space. We study a fixed point theorem using the condition
$(\mathbf{NS25})$ and the notions of reflexive $S$-Banach space,
$S$-normality, closure property and convexity. In Section 4, we investigate
some comparisons on $S$-normed spaces such as the relationships between the
conditions $(\mathbf{NR25})$ and $(\mathbf{NS25})$.

##  2\. $S$-Normed Spaces

In this section, we introduce the notion of an $S$-normed space and
investigate some basic concepts related to an $S$-norm. We study the
relationships between an $S$-metric and an $S$-norm (resp. an $S$-norm and a
norm).

######  Definition 2.1.

Let $X$ be a real vector space. A real valued function $\|.,.,.\|:X\times
X\times X\rightarrow\mathbb{R}$ is called an $S$-norm on $X$ if the following
conditions hold:

$\mathbf{(NS1)}$ $\|x,y,z\|\geq 0$ and $\|x,y,z\|=0$ if and only if $x=y=z=0$,

$\mathbf{(NS2)}$ $\|\lambda x,\lambda y,\lambda z\|=\mid\lambda\mid\|x,y,z\|$
for all $\lambda\in\mathbb{R}$ and $x,y,z\in X$,

$\mathbf{(NS3)}$
$\|x+x^{\prime},y+y^{\prime},z+z^{\prime}\|\leq\|0,x,z^{\prime}\|+\|0,y,x^{\prime}\|+\|0,z,y^{\prime}\|$
for all $x,y,z,x^{\prime},y^{\prime},z^{\prime}\in X$.

The pair $(X,\|.,.,.\|)$ is called an $S$-normed space.

######  Example 2.1.

Let $X=\mathbb{R}$ and $\|.,.,.\|:X\times X\times X\rightarrow\mathbb{R}$ be a
function defined by

$\|x,y,z\|=\mid x\mid+\mid y\mid+\mid z\mid,$

for all $x,y,z\in X$. Then $(X,\|.,.,.\|)$ is an $S$-normed space. Indeed, we
show that the function $\|x,y,z\|=\mid x\mid+\mid y\mid+\mid z\mid$ satisfies
the conditions $(\mathbf{NS1})$, $(\mathbf{NS2})$ and $(\mathbf{NS3})$.

$(\mathbf{NS1})$ By the definition, clearly we have $\|x,y,z\|\geq 0$ for all
$x,y,z\in X$. If $\|x,y,z\|=\mid x\mid+\mid y\mid+\mid z\mid=0$, we obtain
$x=y=z=0$.

$(\mathbf{NS2})$ Let $x,y,z\in X$ and $\lambda\in\mathbb{R}$. Then we have

$\displaystyle\|\lambda x,\lambda y,\lambda z\|$ $\displaystyle=$
$\displaystyle\mid\lambda x\mid+\mid\lambda y\mid+\mid\lambda
z\mid=\mid\lambda\mid\mid x\mid+\mid\lambda\mid\mid y\mid+\mid\lambda\mid\mid
z\mid$ $\displaystyle=$ $\displaystyle\mid\lambda\mid(\mid x\mid+\mid
y\mid+\mid z\mid)=\mid\lambda\mid\|x,y,z\|.$

$(\mathbf{NS3})$ Let $x,y,z,x^{\prime},y^{\prime},z^{\prime}\in X$. Then we
obtain

$\displaystyle\|x+x^{\prime},y+y^{\prime},z+z^{\prime}\|$ $\displaystyle=$
$\displaystyle\mid x+x^{\prime}\mid+\mid y+y^{\prime}\mid+\mid
z+z^{\prime}\mid$ $\displaystyle\leq$ $\displaystyle\mid x\mid+\mid
x^{\prime}\mid+\mid y\mid+\mid y^{\prime}\mid+\mid z\mid+\mid z^{\prime}\mid$
$\displaystyle\leq$ $\displaystyle\mid 0\mid+\mid x\mid+\mid
z^{\prime}\mid+\mid 0\mid+\mid y\mid+\mid x^{\prime}\mid+\mid 0\mid+\mid
z\mid+\mid y^{\prime}\mid$ $\displaystyle=$
$\displaystyle\|0,x,z^{\prime}\|+\|0,y,x^{\prime}\|+\|0,z,y^{\prime}\|.$

Consequently, the function $\|x,y,z\|=\mid x\mid+\mid y\mid+\mid z\mid$
satisfies the conditions $(\mathbf{NS1})$, $(\mathbf{NS2})$, $(\mathbf{NS3})$
and so $(X,\|.,.,.\|)$ is an $S$-normed space.

Now, we show that every $S$-norm generates an $S$-metric.

######  Proposition 2.1.

Let $(X,\|.,.,.\|)$ be an $S$-normed space. Then the function
$\mathcal{S}:X\times X\times X\rightarrow[0,\infty)$ defined by

$S(x,y,z)=\|x-y,y-z,z-x\|$ (2.1)

is an $S$-metric on $X$.

###### Proof.

Using the condition $(\mathbf{NS1})$, it can be easily seen that the condition
$(\mathbf{S1})$ is satisfied. We show that the condition $(\mathbf{S2})$ is
satisfied. By $(\mathbf{NS3})$, we have

$\displaystyle S(x,y,z)$ $\displaystyle=$ $\displaystyle\|x-y,y-z,z-x\|$
$\displaystyle=$ $\displaystyle\|x-a+a-y,y-a+a-z,z-a+a-x\|$
$\displaystyle\leq$ $\displaystyle\|0,x-a,a-x\|+\|0,y-a,a-y\|+\|0,z-a,a-z\|$
$\displaystyle=$ $\displaystyle S(x,x,a)+S(y,y,a)+S(z,z,a),$

for all $x,y,z,a\in X$.

Then, the function $\mathcal{S}$ is an $S$-metric and the pair $(X,S)$ is an
$S$-metric space. ∎

We call the $S$-metric defined in (2.1) as the $S$-metric generated by the
$S$-norm $\|.,.,.\|$ and denoted by $S_{\|.\|}$.

######  Corollary 2.1.

Every $S$-normed space is an $S$-metric space.

######  Example 2.2.

Let $X$ be a nonempty set, $(X,d)$ be a metric space and $\mathcal{S}:X\times
X\times X\rightarrow\left[0,\infty\right)$ be the function defined by

$\mathcal{S}(x,y,z)=d(x,y)+d(x,z)+d(y,z),$

for all $x,y,z\in X$. Then the function is an $S$-metric on $X$ [15].

Let $X=\mathbb{R}$. If we consider the usual metric $d$ on $X$, we obtain the
$S$-metric

$\mathcal{S}(x,y,z)=\mid x-y\mid+\mid x-z\mid+\mid y-z\mid,$

for all $x,y,z\in\mathbb{R}$. Using Proposition 2.1, we see that $\mathcal{S}$
is generated by the $S$-norm defined in Example 2.1. Indeed we have

$\displaystyle S(x,y,z)$ $\displaystyle=$ $\displaystyle\|x-y,y-z,z-x\|$
$\displaystyle=$ $\displaystyle\mid x-y\mid+\mid y-z\mid+\mid z-x\mid$
$\displaystyle=$ $\displaystyle\mid x-y\mid+\mid x-z\mid+\mid y-z\mid$
$\displaystyle=$ $\displaystyle d(x,y)+d(x,z)+d(y,z),$

for all $x,y,z\in X$.

######  Lemma 2.1.

An $S$-metric $\mathcal{S}$ generated by an $S$-norm on an $S$-normed space
$X$ satisfies the following conditions

  1. (1)

$\mathcal{S}(x+a,y+a,z+a)=\mathcal{S}(x,y,z)$,

  2. (2)

$\mathcal{S}(\lambda x,\lambda y,\lambda
z)=\left|\lambda\right|\mathcal{S}(x,y,z)$,

for $x,y,z,a\in X$ and every scalar $\lambda$.

###### Proof.

The proof follows easily from the Proposition 2.1. ∎

We note that every $S$-metric can not be generated by an $S$-norm as we have
seen in the following example:

######  Example 2.3.

Let $X$ be a nonempty set and a function $S:X\times X\times
X\rightarrow\left[0,\infty\right)$ be defined by

$S(x,y,z)=\left\\{\begin{array}[]{ccc}0&;&\text{if }x=y=z\\\
1&;&\text{otherwise}\end{array},\right.$

for all $x,y,z\in X$. Then the function $\mathcal{S}$ is an $S$-metric on $X$.
We call this $S$-metric is the discrete $S$-metric on $X$. The pair $(X,S)$ is
called discrete $S$-metric space. Now, we prove that this $S$-metric can not
be generated by an $S$-norm. On the contrary, we assume that this $S$-metric
is generated by an $S$-norm. Then the following equation should be satisfied
$:$

$S(x,y,z)=\|x-y,y-z,z-x\|,$

for all $x,y,z\in X$.

If we consider the case $x=y\neq z$ and $\mid\lambda\mid\neq 0,1$ then we
obtain

$\displaystyle S(\lambda x,\lambda y,\lambda z)$ $\displaystyle=$
$\displaystyle\|0,\lambda(y-z),\lambda(z-x)\|=1$ $\displaystyle\neq$
$\displaystyle\mid\lambda\mid
S(x,y,z)=\mid\lambda\mid\|0,y-z,z-x\|=\mid\lambda\mid\text{,}$

which is a contradiction with $(\mathbf{NS2})$. Consequently, this $S$-metric
can not be generated by an $S$-norm.

We use the following result in the next section.

######  Lemma 2.2.

Let $(X,\|.,.,.\|)$ be an $S$-normed space. We have

$\|0,x-y,y-x\|=\|0,y-x,x-y\|,$

for each $x,y\in X$.

###### Proof.

By the condition $(\mathbf{NS3})$, we get

$\|0,x-y,y-x\|\leq\|0,0,0\|+\|0,0,0\|+\|0,y-x,x-y\|=\|0,y-x,x-y\|$ (2.2)

and

$\|0,y-x,x-y\|\leq\|0,0,0\|+\|0,0,0\|+\|0,x-y,y-x\|=\|0,x-y,y-x\|.$ (2.3)

Using (2.2) and (2.3) we obtain $\|0,x-y,y-x\|=\|0,y-x,x-y\|$. ∎

We recall the definition of a norm on $X$ as follows.

Let $X$ be a real vector space. A real valued function
$\left\|.\right\|:X\rightarrow\mathbb{R}$ is called a norm on $X$ if the
following conditions hold:

$(\mathbf{N1})$ $\|x\|\geq 0$ for all $x\in X$.

$(\mathbf{N2})$ $\|x\|=0$ if and only if $x=0$ for all $x\in X$.

$(\mathbf{N3})$ $\|\lambda x\|=\mid\lambda\mid\|x\|$ for all
$\lambda\in\mathbb{R}$ and $x\in X$.

$(\mathbf{N4})$ $\|x+y\|\leq\|x\|+\|y\|$ for all $x,y\in X$.

The pair $(X,\|.\|)$ is called a normed space.

We show that every norm generates an $S$-norm. We give the following
proposition.

######  Proposition 2.2.

Let $(X,\|.\|)$ be a normed space and a function $\|.,.,.\|:X\times X\times
X\rightarrow\mathbb{R}$ be defined by

$\|x,y,z\|=\|x\|+\|y\|+\|z\|,$ (2.4)

for all $x,y,z\in X$. Then $(X,\|.,.,.\|)$ is an $S$-normed space.

###### Proof.

We show that the function $\|x,y,z\|=\|x\|+\|y\|+\|z\|$ satisfies the
conditions $(\mathbf{NS1})$, $(\mathbf{NS2})$ and $(\mathbf{NS3})$.

$(\mathbf{NS1})$ It is clear that $\|x,y,z\|\geq 0$ and $\|x,y,z\|=0$ if and
only if $x=y=z=0$.

$(\mathbf{NS2})$ Let $\lambda\in\mathbb{R}$ and $x,y,z\in X$. Then we obtain

$\displaystyle\|\lambda x,\lambda y,\lambda z\|$ $\displaystyle=$
$\displaystyle\|\lambda x\|+\|\lambda y\|+\|\lambda z\|$ $\displaystyle=$
$\displaystyle\mid\lambda\mid\|x\|+\mid\lambda\mid\|y\|+\mid\lambda\mid\|z\|$
$\displaystyle=$ $\displaystyle\mid\lambda\mid(\|x\|+\|y\|+\|z\|)$
$\displaystyle=$ $\displaystyle\mid\lambda\mid\|x,y,z\|.$

$(\mathbf{NS3})$ Let $x,y,z,x^{\prime},y^{\prime},z^{\prime}\in X$. Then we
obtain

$\displaystyle\|x+x^{\prime},y+y^{\prime},z+z^{\prime}\|$ $\displaystyle=$
$\displaystyle\|x+x^{\prime}\|+\|y+y^{\prime}\|+\|z+z^{\prime}\|$
$\displaystyle\leq$
$\displaystyle\|x\|+\|x^{\prime}\|+\|y\|+\|y^{\prime}\|+\|z\|+\|z^{\prime}\|$
$\displaystyle=$
$\displaystyle\|0\|+\|x\|+\|z^{\prime}\|+\|0\|+\|y\|+\|x^{\prime}\|+\|0\|+\|z\|+\|y^{\prime}\|$
$\displaystyle=$
$\displaystyle\|0,x,z^{\prime}\|+\|0,y,x^{\prime}\|+\|0,z,y^{\prime}\|.$

Consequently, the function $\|x,y,z\|=\|x\|+\|y\|+\|z\|$ satisfies the
conditions $(\mathbf{NS1})$, $(\mathbf{NS2})$, $(\mathbf{NS3})$ and so
$(X,\|.,.,.\|)$ is an $S$-normed space. ∎

We have proved that every norm on $X$ defines an $S$-norm on $X$. We call the
$S$-norm defined in (2.4) as the $S$-norm generated by the norm $\|.\|$. For
example, the $S$-norm defined in Example 2.1 is the $S$-norm generated by the
usual norm on $\mathbb{R}$.

There exists an $S$-norm which is not generated by a norm as we have seen in
the following example.

######  Example 2.4.

Let $X$ be a nonempty set and the function $\|.,.,.\|:X\times X\times
X\rightarrow\mathbb{R}$ be defined by

$\|x,y,z\|=\mid x-2y-2z\mid+\mid y-2x-2z\mid+\mid z-2y-2x\mid\text{,}$

for all $x,y,z\in X$. Then, the function $\|x,y,z\|$ is an $S$-norm on $X$,
but it is not generated by a norm.

Now, we show that the conditions $(\mathbf{NS1})$, $(\mathbf{NS2})$ and
$(\mathbf{NS3})$ are satisfied.

$(\mathbf{NS1})$ By the definition, clearly we obtain $\|x,y,z\|\geq 0$ and
$\|x,y,z\|=0$ if and only if $x=y=z=0$ for all $x,y,z\in X$.

$(\mathbf{NS2})$ We have

$\displaystyle\|\lambda x,\lambda y,\lambda z\|$ $\displaystyle=$
$\displaystyle\mid\lambda x-2\lambda y-2\lambda z\mid+\mid\lambda y-2\lambda
x-2\lambda z\mid+\mid\lambda z-2\lambda y-2\lambda x\mid$ $\displaystyle=$
$\displaystyle\mid\lambda\mid(\mid x-2y-2z\mid+\mid y-2x-2z\mid+\mid
z-2y-2x\mid)$ $\displaystyle=$ $\displaystyle\mid\lambda\mid\|x,y,z\|\text{,}$

for all $\lambda\in\mathbb{R}$ and $x,y,z\in X$.

$(\mathbf{NS3})$ Let $x,y,z,x^{\prime},y^{\prime},z^{\prime}\in X$. Then we
obtain

$\displaystyle\|x+x^{\prime},y+y^{\prime},z+z^{\prime}\|$ $\displaystyle=$
$\displaystyle\mid x+x^{\prime}-2y-2y^{\prime}-2z-2z^{\prime}\mid$
$\displaystyle+$ $\displaystyle\mid$ $\displaystyle
y+y^{\prime}-2x-2x^{\prime}-2z-2z^{\prime}\mid$ $\displaystyle+$
$\displaystyle\mid$ $\displaystyle
z+z^{\prime}-2y-2y^{\prime}-2x-2x^{\prime}\mid$ $\displaystyle\leq$
$\displaystyle\mid 2x+2z^{\prime}\mid+\mid x-2z^{\prime}\mid+\mid
z^{\prime}-2x\mid$ $\displaystyle+$ $\displaystyle\mid$ $\displaystyle
2y+2x^{\prime}\mid+\mid y-2x^{\prime}\mid+\mid x^{\prime}-2y\mid$
$\displaystyle+$ $\displaystyle\mid$ $\displaystyle 2z+2y^{\prime}\mid+\mid
z-2y^{\prime}\mid+\mid y^{\prime}-2z\mid$ $\displaystyle=$
$\displaystyle\|0,x,z^{\prime}\|+\|0,y,x^{\prime}\|+\|0,z,y^{\prime}\|\text{.}$

Consequently, the function $\|x,y,z\|=\mid x-2y-2z\mid+\mid y-2x-2z\mid+\mid
z-2y-2x\mid$ is an $S$-norm on $X$.

On the contrary, we assume that this $S$-norm is generated by a norm. Then the
following equation should be satisfied

$\|x,y,z\|=\|x\|+\|y\|+\|z\|\text{,}$

for all $x,y,z\in X$.

If we consider $\|x,0,0\|$ and $\|x,x,0\|$ then we obtain

$\begin{array}[]{c}\|x,0,0\|=\|x\|=\mid x\mid+\mid 2x\mid+\mid 2x\mid=5\mid
x\mid,\\\ \|x,x,0\|=2\parallel x\parallel=\mid x\mid+\mid x\mid+\mid
4x\mid=6\mid x\mid\end{array}$

and so $\|x\|=5\mid x\mid$ and $\|x\|=3\mid x\mid$, which is a contradiction.
Hence this $S$-norm is not generated by a norm.

Now we prove that every $S$-norm generate a norm.

######  Proposition 2.3.

Let $X$ be a nonempty set, $(X,\|.,.,.\|)$ be an $S$-normed space and a
function $\|.\|:X\rightarrow\mathbb{R}$ be defined as follows:

$\|x\|=\|0,x,0\|+\|0,0,x\|,$

for all $x\in X$. Then the function $\|x\|=\|0,x,0\|+\|0,0,x\|$ is a norm on
$X$ and $(X,\|.\|)$ is a normed space.

###### Proof.

Using the conditions $(\mathbf{NS1})$ and $(\mathbf{NS2})$, it is clear that
we obtain the conditions $(\mathbf{N1})$, $(\mathbf{N2})$ and $(\mathbf{N3})$
are satisfied.

Now we show that the condition $(\mathbf{N4})$ is satisfied.

$(\mathbf{N4})$ Let $x,y\in X$. By the condition $(\mathbf{NS3})$, we have

$\displaystyle\|x+y\|$ $\displaystyle=$ $\displaystyle\|0,x+y,0\|+\|0,0,x+y\|$
$\displaystyle=$ $\displaystyle\|0,x+y,0\|+\|0,0,y+x\|$ $\displaystyle\leq$
$\displaystyle\|0,0,0\|+\|0,x,0\|+\|0,0,y\|+\|0,0,x\|+\|0,0,0\|+\|0,y,0\|$
$\displaystyle=$ $\displaystyle\|x\|+\|y\|.$

Consequently, the function $\|x\|=\|0,x,0\|+\|0,0,x\|$ is a norm on $X$ and
$(X,\|.\|)$ is a normed space. ∎

We call this norm as the norm generated by the $S$-norm $\|.,.,.\|$.

Let $X$ be a real vector space. New generalizations of normed spaces have been
studied in recent years. For example, Khan defined the notion of a $G$-norm
and studied some topological concepts in $G$-normed spaces [7]. Now we recall
the definition of a $G$-norm and give the relationship between a $G$-norm and
an $S$-norm.

######  Definition 2.2.

[7] Let $X$ be a real vector space. A real valued function $\|.,.,.\|:X\times
X\times X\rightarrow\mathbb{R}$ is called a $G$-norm on $X$ if the following
conditions hold$:$

$(\mathbf{NG1})$ $\|x,y,z\|\geq 0$ and $\|x,y,z\|=0$ if and only if $x=y=z=0$.

$(\mathbf{NG2})$ $\|x,y,z\|$ is invariant under permutations of $x,y,z$.

$(\mathbf{NG3})$ $\|\lambda x,\lambda y,\lambda z\|=\mid\lambda\mid\|x,y,z\|$
for all $\lambda\in\mathbb{R}$ and $x,y,z\in X$.

$(\mathbf{NG4})$
$\|x+x^{\prime},y+y^{\prime},z+z^{\prime}\|\leq\|x,y,z\|+\|x^{\prime},y^{\prime},z^{\prime}\|$
for all $x,y,z,x^{\prime},y^{\prime},z^{\prime}\in X$.

$(\mathbf{NG5})$ $\|x,y,z\|\geq\|x+y,0,z\|$ for all $x,y,z\in X$.

The pair $(X,\|.,.,.\|)$ is called a $G$-normed space.

######  Proposition 2.4.

Every $G$-normed space is an $S$-normed space.

###### Proof.

Using the conditions $(\mathbf{NG1})$ and $(\mathbf{NG3})$, we see that the
conditions $(\mathbf{NS1})$ and $(\mathbf{NS2})$ are satisfied. We only show
that the condition $(\mathbf{NS3})$ is satisfied.

$(\mathbf{NS3})$ Let $x,y,z,x^{\prime},y^{\prime},z^{\prime}\in X$. Using the
conditions $(\mathbf{NG2})$ and $(\mathbf{NG4})$, we obtain

$\displaystyle\|x+x^{\prime},y+y^{\prime},z+z^{\prime}\|$ $\displaystyle=$
$\displaystyle\|(x+0)+x^{\prime},0+(y+y^{\prime}),z^{\prime}+z\|$
$\displaystyle\leq$
$\displaystyle\|x+0,0,0+z^{\prime}\|+\|x^{\prime},y+y^{\prime},z\|$
$\displaystyle=$
$\displaystyle\|0,0+x,0+z^{\prime}\|+\|x^{\prime},y+y^{\prime},z\|$
$\displaystyle\leq$
$\displaystyle\|0,0,0\|+\|0,x,z^{\prime}\|+\|x^{\prime},y,0\|+\|0,y^{\prime},z\|$
$\displaystyle=$
$\displaystyle\|0,x,z^{\prime}\|+\|0,y,x^{\prime}\|+\|0,z,y^{\prime}\|\text{.}$

Consequently, the condition $(\mathbf{NS3})$ is satisfied. ∎

The converse of Proposition 2.4 can not be always true as we have seen in the
following example.

######  Example 2.5.

Let $X=\mathbb{R}$ and the $S$-norm be defined as in Example 2.4. If we put
$x=1$, $y=5$ and $z=0$, the condition $(\mathbf{NG5})$ is not satisfied.
Indeed, we have

$\|x,y,z\|=\mid x-2y-2z\mid+\mid y-2x-2z\mid+\mid z-2y-2x\mid=23$

and

$\|x+y,0,z\|=\mid x+y-2z\mid+\mid 2x+2y+2z\mid+\mid z-2y-2x\mid=30.$

Hence this $S$-norm is not a $G$-norm on $\mathbb{R}$.

Now we give the definitions of an open ball and a closed ball on an $S$-normed
space.

######  Definition 2.3.

Let $(X,\|.,.,.\|)$ be an $S$-normed space. For given $x_{0}$, $a_{1}$,
$a_{2}\in X$ and $r>0$, the open ball $B_{a_{1}}^{a_{2}}(x_{0},r)$ and the
closed ball $B_{a_{1}}^{a_{2}}[x_{0},r]$ are defined as follows:

$B_{a_{1}}^{a_{2}}(x_{0},r)=\\{y\in X:\|y-x_{0},y-a_{1},y-a_{2}\|<r\\}$

and

$B_{a_{1}}^{a_{2}}[x_{0},r]=\\{y\in X:\|y-x_{0},y-a_{1},y-a_{2}\|\leq
r\\}\text{.}$

######  Example 2.6.

Let us consider the $S$-normed space $(X,\|.,.,.\|)$ generated by the usual
norm on $X$, where $X=\mathbb{R}^{2}$ and

$\|x\|=\|(x_{1},x_{2})\|=\sqrt{x_{1}^{2}+x_{2}^{2}}\text{,}$

for all $x\in\mathbb{R}^{2}$. Then the open ball $B_{a_{1}}^{a_{2}}(x_{0},r)$
in $\mathbb{R}^{2}$ is a $3$-ellipse given by

$B_{a_{1}}^{a_{2}}(x_{0},r)=\\{y\in\mathbb{R}^{2}:\|y-x_{0}\|+\|y-a_{1}\|+\|y-a_{2}\|<r\\}\text{.}$

If we choose $y=(y_{1},y_{2})$, $x_{0}=(1,1)$, $a_{1}=(0,0)$, $a_{2}=(-1,-1)$
in $\mathbb{R}^{2}$ and $r=5$. Then we obtain

$B_{a_{1}}^{a_{2}}(x_{0},r)=\\{y\in\mathbb{R}^{2}:\sqrt{(y_{1}-1)^{2}+(y_{2}-1)^{2}}+\sqrt{y_{1}^{2}+y_{2}^{2}}+\sqrt{(y_{1}+1)^{2}+(y_{2}+1)^{2}}<5\\}\text{,}$
(2.5)

as shown in Figure 1(a).

(a) The open ball which is corresponding
to the $S$-norm defined in (2.5).

(b) The open ball which is corresponding
to the $S$-norm defined in (2.6).

Figure 1. Some open balls in $(\mathbb{R}^{2},\|.,.,.\|)$

Now we give the following example using the $S$-norm which is not generated by
a norm.

######  Example 2.7.

Let $X=\mathbb{R}^{2}$ and the function $\|.,.,.\|:X\times X\times
X\rightarrow\mathbb{R}$ be defined as in Example 2.4. Then we have

$\displaystyle\|x,y,z\|$ $\displaystyle=$ $\displaystyle\mid x-2y-2z\mid+\mid
y-2x-2z\mid+\mid z-2y-2x\mid$ $\displaystyle=$
$\displaystyle\sqrt{(x_{1}-2y_{1}-2z_{1})^{2}+(x_{2}-2y_{2}-2z_{2})^{2}}$
$\displaystyle+\sqrt{(y_{1}-2x_{1}-2z_{1})^{2}+(y_{2}-2x_{2}-2z_{2})^{2}}$
$\displaystyle+\sqrt{(z_{1}-2y_{1}-2x_{1})^{2}+(z_{2}-2y_{2}-2x_{2})^{2}},$

for all $x=(x_{1},x_{2})$, $y=(y_{1},y_{2})$,
$z=(z_{1},z_{2})\in\mathbb{R}^{2}$. Then $(\mathbb{R}^{2},\|.,.,.\|)$ is an
$S$-normed space. The open ball $B_{a_{1}}^{a_{2}}(x_{0},r)$ in
$\mathbb{R}^{2}$ is

$B_{a_{1}}^{a_{2}}(x_{0},r)=\\{y\in\mathbb{R}^{2}:\|y-x_{0},y-a_{1},y-a_{2}\|<r\\}\text{.}$

If we choose $y=(y_{1},y_{2})$, $x_{0}=(1,1)$, $a_{1}=(0,0)$, $a_{2}=(-1,-1)$
in $\mathbb{R}^{2}$ and $r=20$. Then we obtain

$\begin{array}[]{c}B_{a_{1}}^{a_{2}}(x_{0},r)=\\{y\in\mathbb{R}^{2}:\sqrt{(3y_{1}+3)^{2}+(3y_{2}+3)^{2}}+\sqrt{9y_{1}^{2}+9y_{2}^{2}}\\\
+\sqrt{(3-3y_{1})^{2}+(3-3y_{2})^{2}}<20\\}\text{,}\end{array}$ (2.6)

as shown in Figure 1(b).

######  Definition 2.4.

Let $(X,\|.,.,.\|)$ be an $S$-normed space.

  1. (1)

A sequence $\\{x_{n}\\}$ in $X$ converges to $x$ if and only if

$\underset{n\rightarrow\infty}{\lim}\|0,x_{n}-x,x-x_{n}\|=0.$

That is, for each $\varepsilon>0$ there exists $n_{0}\in\mathbb{N}$ such that

$\|0,x_{n}-x,x-x_{n}\|<\varepsilon\text{,}$

for all $n\geq n_{0}$.

  2. (2)

A sequence $\\{x_{n}\\}$ in $X$ is called a Cauchy sequence if

$\underset{n,m,l\rightarrow\infty}{\lim}\|x_{n}-x_{m},x_{m}-x_{l},x_{l}-x_{n}\|=0\text{.}$

That is, for each $\varepsilon>0$ there exists $n_{0}\in\mathbb{N}$ such that

$\|x_{n}-x_{m},x_{m}-x_{l},x_{l}-x_{n}\|<\varepsilon\text{,}$

for all $n,m,l\geq n_{0}$.

  3. (3)

An $S$-normed space is called complete if each Cauchy sequence in $X$
converges in $X$.

  4. (4)

A complete $S$-normed space is called an $S$-Banach space.

######  Proposition 2.5.

Every convergent sequence in an $S$-normed space is a Cauchy sequence.

###### Proof.

Let the sequence $\\{x_{n}\\}$ in $X$ be convergent to $x$. For each
$\varepsilon>0$, there exists $n_{0}\in\mathbb{N}$ such that

$\|0,x_{n}-x,x-x_{n}\|<\frac{\varepsilon}{3}\text{,}$

for all $n\geq n_{0}$. We now show that for each $\varepsilon>0$ there exists
$n_{0}\in\mathbb{N}$ such that

$\|x_{n}-x_{m},x_{m}-x_{l},x_{l}-x_{n}\|<\varepsilon\text{,}$

for all $n,m,l\geq n_{0}$. Using the condition $(\mathbf{NS3})$, we obtain

$\displaystyle\|x_{n}-x_{m},x_{m}-x_{l},x_{l}-x_{n}\|$ $\displaystyle=$
$\displaystyle\|x_{n}-x+x-x_{m},x_{m}-x+x-x_{l},x_{l}-x+x-x_{n}\|$
$\displaystyle\leq$ $\displaystyle\|0,x_{n}-x,x-x_{n}\|+\|0,x_{m}-x,x-x_{m}\|$
$\displaystyle+\|0,x_{l}-x,x-x_{l}\|$ $\displaystyle<$
$\displaystyle\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon\text{.}$

Consequently, the sequence $\\{x_{n}\\}$ in $X$ is a Cauchy sequence. ∎

The converse of Proposition 2.5 can not be always true as we have seen in the
following example.

######  Example 2.8.

Let $X=(0,1)\subset\mathbb{R}$ and the function $\|.,.,.\|:X\times X\times
X\rightarrow\mathbb{R}$ be an $S$-norm generated by the usual norm on $X$. If
we consider the sequence $\\{x_{n}\\}=\left\\{\dfrac{1}{n}\right\\}$ on $X$,
then this sequence is a Cauchy sequence, but it is not a convergent sequence
on $X$.

Now we show that the sequence is a Cauchy sequence. For $x_{n}$, $x_{m}$,
$x_{l}\in X$, we obtain

$\displaystyle\underset{n,m,l\rightarrow\infty}{\lim}\|x_{n}-x_{m},x_{m}-x_{l},x_{l}-x_{n}\|$
$\displaystyle=$
$\displaystyle\underset{n,m,l\rightarrow\infty}{\lim}\left\|\frac{1}{n}-\frac{1}{m},\frac{1}{m}-\frac{1}{l},\frac{1}{l}-\frac{1}{n}\right\|$
$\displaystyle=$
$\displaystyle\underset{n,m,l\rightarrow\infty}{\lim}\left(\left|\frac{1}{n}-\frac{1}{m}\right|+\left|\frac{1}{m}-\frac{1}{l}\right|+\left|\frac{1}{l}-\frac{1}{n}\right|\right)=0\text{.}$

The sequence is convergent to $0$ as follows:

$\underset{n\rightarrow\infty}{\lim}\|0,x_{n}-x,x-x_{n}\|=\underset{n\rightarrow\infty}{\lim}\left\|0,\frac{1}{n}-0,0-\frac{1}{n}\right\|=0\text{,}$

for all $x_{n}\in X$. But $0\notin X$. Consequently, the sequence is not
convergent on $X$.

##  3\. A Fixed Point Theorem on $S$-Normed Spaces

In this section, we introduce the Rhoades’ condition on an $S$-normed space
and denote it by $(\mathbf{NS25})$. We prove a fixed point teorem using this
contractive condition.

At first, we give some definitions and a proposition which are needed in the
sequel.

######  Definition 3.1.

Let $(X,\|.,.,.\|)$ be an $S$-normed space and $E\subseteq X$. The closure of
$E$, denoted by $\overline{E}$, is the set of all $x\in X$ such that there
exists a sequence $\\{x_{n}\\}$ in $E$ converging to $x$. If $E=\overline{E}$,
then $E$ is called a closed set.

######  Definition 3.2.

Let $(X,\|.,.,.\|)$ be an $S$-normed space and $A\subseteq X$. The subset $A$
is called bounded if there exists $r>0$ such that

$\|0,x-y,y-x\|<r,$

for all $x,y\in A$.

######  Definition 3.3.

Let $(X,\|.,.,.\|)$ be an $S$-normed space and $A\subseteq X$. The
$S$-diameter of $A$ is defined by

$\delta^{s}(A)=\sup\\{\|0,x-y,y-x\|:x,y\in A\\}\text{.}$

If $A$ is bounded then we will write $\delta^{s}(A)<\infty$.

######  Definition 3.4.

Let $X$ be an $S$-Banach space, $A\subseteq X$ and $u\in X$.

  1. (1)

The $S$-radius of $A$ relative to a given $u\in X$ is defined by

$r_{u}^{s}(A)=\sup\\{\|0,u-x,x-u\|:x\in A\\}\text{.}$

  2. (2)

The $S$-Chebyshev radius of $A$ is defined by

$r^{s}(A)=\inf\\{r_{u}^{s}(A):u\in A\\}\text{.}$

  3. (3)

The $S$-Chebyshev centre of $A$ is defined by

$C^{s}(A)=\\{u\in A:r_{u}^{s}(A)=r^{s}(A)\\}\text{.}$

By Definition 3.3 and Definition 3.4, it can be easily seen the following
inequality:

$r^{s}(A)\leq r_{u}^{s}(A)\leq\delta^{s}(A)\text{.}$

######  Definition 3.5.

A point $u\in A$ is called $S$-diametral if $r_{u}^{s}(A)=\delta^{s}(A)$. If
$r_{u}^{s}(A)<\delta^{s}(A)$, then $u$ is called non-$S$-diametral.

######  Definition 3.6.

A convex subset of an $S$-Banach space $X$ has $S$-normal structure if every
$S$-bounded and convex subset of $A$ having $\delta^{s}(A)>0$ has at least one
non-$S$-diametral point.

######  Proposition 3.1.

If $X$ is a reflexive $S$-Banach space, $A$ is a nonempty, closed and convex
subset of $X$, then $C^{s}(A)$ is nonempty, closed and convex.

###### Proof.

It can be easily seen by definition of $C^{s}(A)$. ∎

Now we introduce the Rhoades’ condition $(\mathbf{NS25})$ on an $S$-Banach
space.

######  Definition 3.7.

Let $(X,\|.,.,.\|)$ be an $S$-Banach space and $T$ be a self-mapping of $X$.
We define

$\begin{array}[]{r}(\mathbf{NS25})\text{ \ \ }\|0,Tx-Ty,Ty-
Tx\|<\max\\{\|0,x-y,y-x\|,\|0,Tx-x,x-Tx\|,\\\
\|0,Ty-y,y-Ty\|,\|0,Ty-x,x-Ty\|,\|0,Tx-y,y-Tx\|\\}\text{,}\end{array}$

for each $x,y\in X$, $x\neq y$.

######  Lemma 3.1.

[16] Let $X$ be a Banach space. Then $X$ is reflexive if and only if for any
decreasing sequence $\\{K_{n}\\}$ of nonempty, bounded, closed and convex
subsets of $X$,

$\bigcap\limits_{n=1}^{\infty}K_{n}\neq\emptyset\text{.}$

######  Lemma 3.2.

Let $X$ be an $S$-Banach space. Then $X$ is reflexive if and only if for any
decreasing sequence $\\{K_{n}\\}$ of nonempty, bounded, closed and convex
subsets of $X$,

$\bigcap\limits_{n=1}^{\infty}K_{n}\neq\emptyset\text{.}$

###### Proof.

By the definition of reflexivity the proof follows easily. ∎

Recall that the convex hull of a set $A$ is denoted by $conv(A)$ and any
member of this set $conv(A)$ has the form

$\overset{n}{\sum\limits_{i=1}}\alpha_{i}x_{i}\text{,}$

where $x_{i}\in A_{i},\alpha_{i}\geq 0$ for all $i=1,...,n$ and
$\overset{n}{\sum\limits_{i=1}}\alpha_{i}=1$.

Now, we give the following fixed point theorem.

######  Theorem 3.1.

Let $X$ be a reflexive $S$-Banach space and $A$ be a nonempty, closed, bounded
and convex subset of $X$, having $S$-normal structure. If $T:A\rightarrow A$
is a continuous self-mapping satisfying the condition $(\mathbf{NS25})$ then
$T$ has a unique fixed point in $A$.

###### Proof.

At first, we show that the existence of the fixed point. Let $\mathcal{A}$ be
the family of every nonempty, closed and convex subsets of $A$. Also we assume
that if $F\in$ $\mathcal{A}$ then $TF\subseteq F$. The family $\mathcal{A}$ is
nonempty since $A\in$ $\mathcal{A}$. We can partially order $\mathcal{A}$ by
set inclusion, that is, if $F_{1}\subseteq F_{2}$ then $F_{1}\leq F_{2}$.

In $\mathcal{A}$, if we define a decreasing net of subsets

$S=\\{F_{i}:F_{i}\in\text{$\mathcal{A}$},i\in I\\}\text{,}$

then by reflexivity, this net $S$ has nonempty intersection. Because it is a
decreasing net of nonempty, closed, bounded and convex subsets of $X$. If we
put $F_{0}=\bigcap\limits_{i\in I}F_{i}$ we have that $F_{0}$ is in
$\mathcal{A}$ and is a lower bound of $S$.

Using Zorn’s Lemma, there is a minimal element, denoted by $F$, in
$\mathcal{A}$ as $S$ is any arbitrary decreasing net in $\mathcal{A}$. We see
that this $F$ is a singleton.

Assume that $\delta^{s}(F)\neq\emptyset$. Since $F$ is nonempty, closed and
convex, $C^{s}(F)$ is a nonempty, closed and convex subset of $F$. We have
that

$r^{s}(F)<\delta^{s}(F)\text{,}$ $\delta^{s}(C^{s}(F))\leq
r^{s}(F)<\delta^{s}(F)$

and so $C^{s}(F)$ is a proper subset of $F$.

Let $(F_{m})_{m\in\mathbb{N}}$ be an increasing sequence of subsets of $F$,
defined by

$F_{1}=C^{s}(F)\text{ and }F_{m+1}=conv(F_{m}\cup TF_{m})\text{,}$

for all $m\in\mathbb{N}$. If we denote the $S$-diameters of these sets $F_{k}$
by $\delta_{k}^{s}=\delta^{s}(F_{k})$, we show that

$\delta_{k}^{s}\leq r^{s}(F)\text{,}$

for all $k\in\mathbb{N}$.

Using the $(PMI)$, we obtain

  1. (1)

For $k=1$,

$\delta_{1}^{s}=\delta^{s}(F_{1})=\delta^{s}(C^{s}(F))\leq r^{s}(F)\text{.}$

  2. (2)

If $\delta_{k}^{s}\leq r^{s}(F)$ for every $k=1,...,m$ then
$\delta_{m+1}^{s}\leq r^{s}(F)$.

We note that

$\delta_{m+1}^{s}=\delta^{s}(F_{m+1})=\delta^{s}(conv(F_{m}\cup
TF_{m}))=\delta^{s}(F_{m}\cup TF_{m})\text{.}$

By the definition of $S$-diameter, for any given $\varepsilon>0$ there are
$x^{\prime}$ and $y^{\prime}$ in $F_{m}\cup T(F_{m})$ satisfying

$\delta_{m+1}^{s}-\varepsilon<\|0,x^{\prime}-y^{\prime},y^{\prime}-x^{\prime}\|\leq\delta_{m+1}^{s}\text{.}$

We obtain the following three cases for $x^{\prime}$, $y^{\prime}$:

  1. (1)

$x^{\prime}$, $y^{\prime}\in F_{m}$ or

  2. (2)

$x^{\prime}\in F_{m}$ and $y^{\prime}\in TF_{m}$ or

  3. (3)

$x^{\prime}$, $y^{\prime}\in TF_{m}$.

Redefining $x^{\prime}$ and $y^{\prime}$ as follows:

  1. (1)

$x^{\prime}=x$ and $y^{\prime}=y$ with $x,y\in F_{m}$,

  2. (2)

$x^{\prime}=x$ and $y^{\prime}=Ty$ with $x,y\in F_{m}$,

  3. (3)

$x^{\prime}=Tx$ and $y^{\prime}=Ty$ with $x,y\in F_{m}$.

We show that in any case

$\delta_{m+1}^{s}-\varepsilon<r^{s}(F)\text{.}$

Case 1. By the definition of $\delta_{m}^{s}$ and the induction hypothesis, we
obtain

$\delta_{m+1}^{s}-\varepsilon<\|0,x-y,y-x\|\leq\delta_{m+1}^{s}\leq r^{s}(F)$
(3.1)

and so $\delta_{m+1}^{s}-\varepsilon<r^{s}(F)$.

Case 2. We obtain

$\delta_{m+1}^{s}-\varepsilon<\|0,x-Ty,Ty-x\|$

with $x$, $y\in F_{m}$. Then by the definition of $F_{m}$, we have $x,y\in
conv(F_{m-1}\cup TF_{m-1})$ and so there is a finite index set $I$ such that
$x=\sum\limits_{i\in I}\alpha_{i}x_{i}$, with $\sum\limits_{i\in
I}\alpha_{i}=1$, $\alpha_{i}\geq 0$ and $x_{i}\in F_{m-1}\cup TF_{m-1}$ for
any $i\in I$. We can separate the set $I$ in two disjoint subsets,
$I=I_{1}\cup I_{2}$, such that if $i\in I_{1}$ then $x_{i}\in F_{m-1}$ and if
$i\in I_{2}$ then $x_{i}\in TF_{m-1}$.

Now redefining $x_{i}$ as $x_{i}=Tx_{i}$ with $x_{i}\in F_{m-1}$, we obtain

$x=\sum\limits_{i\in I_{1}}\alpha_{i}x_{i}+\sum\limits_{i\in
I_{2}}\alpha_{i}Tx_{i}\text{.}$

Substituting in $\|0,x-Ty,Ty-x\|$, we get

$\|0,x-Ty,Ty-x\|\leq\sum\limits_{i\in I_{1}}\alpha_{i}\|0,x_{i}-Ty,Ty-
x_{i}\|+\sum\limits_{i\in I_{2}}\alpha_{i}\|0,Tx_{i}-Ty,Ty-Tx_{i}\|\text{.}$
(3.2)

Applying the condition $(\mathbf{NS25})$ to $\|0,Tx_{i}-Ty,Ty-Tx_{i}\|$, we
have

$\begin{array}[]{r}\|0,Tx_{i}-Ty,Ty-
Tx_{i}\|<\max\\{\|0,x_{i}-y,y-x_{i}\|,\|0,x_{i}-Tx_{i},Tx_{i}-x_{i}\|,\\\
\|0,y-Ty,Ty-y\|,\|0,x_{i}-Ty,Ty-
x_{i}\|,\|0,Tx_{i}-y,y-Tx_{i}\|\\}\text{.}\end{array}$ (3.3)

As $x_{i}\in F_{m-1}$, $Tx_{i},y\in F_{m}$, we have

$\begin{array}[]{l}\|0,x_{i}-y,y-x_{i}\|\leq r^{s}(F),\\\
\|0,x_{i}-Tx_{i},Tx_{i}-x_{i}\|\leq r^{s}(F),\\\ \|0,Tx_{i}-y,y-Tx_{i}\|\leq
r^{s}(F)\end{array}$

and replacing in (3.3), we obtain

$\|0,Tx_{i}-Ty,Ty-Tx_{i}\|<\max\\{r^{s}(F),\|0,y-Ty,Ty-y\|,\|0,x_{i}-Ty,Ty-
x_{i}\|\\}\text{.}$

Let us subdivide the index set $I_{2}$ in three disjoint subsets
$I_{2}=I_{2}^{1}\cup I_{2}^{2}\cup I_{2}^{3}$ such that

$\begin{array}[]{l}I_{2}^{1}=\\{i\in I_{2}:\|0,Tx_{i}-Ty,Ty-
Tx_{i}\|<r^{s}(F)\\},\\\ I_{2}^{2}=\\{i\in I_{2}:\|0,Tx_{i}-Ty,Ty-
Tx_{i}\|<\|0,x_{i}-Ty,Ty-x_{i}\|\\},\\\ I_{2}^{3}=\\{i\in
I_{2}:\|0,Tx_{i}-Ty,Ty-Tx_{i}\|<\|0,y-Ty,Ty-y\|\\}.\end{array}$

Then using (3.2), we have

$\begin{array}[]{c}\|0,x-Ty,Ty-x\|\leq\sum\limits_{i\in I_{1}\cup
I_{2}^{2}}\alpha_{i}\|0,x_{i}-Ty,Ty-x_{i}\|+\sum\limits_{i\in
I_{2}^{1}}\alpha_{i}r^{s}(F)\\\ +\sum\limits_{i\in
I_{2}^{3}}\alpha_{i}\|0,y-Ty,Ty-y\|\text{.}\end{array}$ (3.4)

Redefining $I_{1}$, $I_{2}$ and $I_{3}$

$\overline{I_{1}}=I_{1}\cup I_{2}^{2}\text{, }\overline{I_{2}}=I_{2}^{1}\text{
and }\overline{I_{3}}=I_{2}^{3}\text{.}$ (3.5)

Then we have $I=\overline{I_{1}}\cup\overline{I_{2}}\cup\overline{I_{3}}$,
with $\overline{I_{j}}\cap\overline{I_{k}}=\emptyset$. If $j\neq k$ and
$\sum\limits_{i\in I}\alpha_{i}=1$ then using (3.4), it becomes

$\begin{array}[]{c}\|0,x-Ty,Ty-x\|\leq\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}\|0,x_{i}-Ty,Ty-
x_{i}\|+\sum\limits_{i\in\overline{I_{2}}}\alpha_{i}r^{s}(F)\\\
+\sum\limits_{i\in\overline{I_{3}}}\alpha_{i}\|0,y-Ty,Ty-y\|\text{.}\end{array}$
(3.6)

If $A_{0}=\sum\limits_{i\in\overline{I_{2}}}\alpha_{i}$ and
$B_{0}=\sum\limits_{i\in\overline{I_{3}}}\alpha_{i}$ with
$\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}+A_{0}+B_{0}=1$. Using (3.6), we
have

$\|0,x-Ty,Ty-x\|\leq\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}\|0,x_{i}-Ty,Ty-
x_{i}\|+A_{0}r^{s}(F)+B_{0}\|0,y-Ty,Ty-y\|\text{.}$ (3.7)

For each $i\in\overline{I_{1}}$, $x_{i}\in F_{m-1}=conv(F_{m-2}\cup
TF_{m-2})$, there is a finite set $J_{i}$, such that

$x_{i}=\sum\limits_{j\in J_{i}}\beta_{i}^{j}x_{i}^{j}\text{,}$ (3.8)

with $x_{i}^{j}\in F_{m-2}\cup TF_{m-2}$, $\beta_{i}^{j}\geq 0$ and
$\sum\limits_{j\in J_{i}}\beta_{i}^{j}=1$ for any $j\in J_{i}$. Let
$J_{i}=J_{i}^{1}\cup J_{i}^{2}$, with $J_{i}^{1}\cap J_{i}^{2}=\emptyset$ such
that

$x_{i}=\sum\limits_{j\in J_{i}^{1}}\beta_{i}^{j}x_{i}^{j}+\sum\limits_{j\in
J_{i}^{2}}\beta_{i}^{j}Tx_{i}^{j}\text{.}$ (3.9)

For each $i\in\overline{I_{1}}$ we have

$\|0,x_{i}-Ty,Ty-x_{i}\|\leq\sum\limits_{j\in
J_{i}^{1}}\beta_{i}^{j}\|0,x_{i}^{j}-Ty,Ty-x_{i}^{j}\|+\sum\limits_{j\in
J_{i}^{2}}\beta_{i}^{j}\|0,Tx_{i}^{j}-Ty,Ty-Tx_{i}^{j}\|\text{.}$ (3.10)

Applying the condition $(\mathbf{NS25})$ to $\|0,Tx_{i}^{j}-Ty,Ty-
Tx_{i}^{j}\|$, we have

$\begin{array}[]{r}\|0,Tx_{i}^{j}-Ty,Ty-
Tx_{i}^{j}\|<\max\\{\|0,x_{i}^{j}-y,y-x_{i}^{j}\|,\|0,Tx_{i}^{j}-x_{i}^{j},x_{i}^{j}-Tx_{i}^{j}\|,\\\
\|0,y-Ty,Ty-y\|,\|0,Tx_{i}^{j}-y,y-Tx_{i}^{j}\|,\|0,x_{i}^{j}-Ty,Ty-
x_{i}^{j}\|\\}\text{.}\end{array}$ (3.11)

Since $x_{i}^{j}\in F_{m-2}$ and $y\in F_{m}$, we have

$\begin{array}[]{l}\|0,x_{i}^{j}-y,y-x_{i}^{j}\|\leq r^{s}(F),\\\
\|0,Tx_{i}^{j}-x_{i}^{j},x_{i}^{j}-Tx_{i}^{j}\|\leq r^{s}(F),\\\
\|0,Tx_{i}^{j}-y,y-Tx_{i}^{j}\|\leq r^{s}(F).\end{array}$

By (3.11), we obtain

$\|0,Tx_{i}^{j}-Ty,Ty-
Tx_{i}^{j}\|<\max\\{r^{s}(F),\|0,y-Ty,Ty-y\|,\|0,x_{i}^{j}-Ty,Ty-
x_{i}^{j}\|\\}\text{.}$

Let $J_{i}^{2}=J_{i}^{2_{1}}\cup J_{i}^{2_{2}}\cup J_{i}^{2_{3}}$ with
$J_{i}^{2_{k}}\cap J_{i}^{2_{p}}=\emptyset$ such that

$\begin{array}[]{l}J_{i}^{2_{1}}=\\{j\in J_{i}^{2}:\|0,Tx_{i}^{j}-Ty,Ty-
Tx_{i}^{j}\|<r^{s}(F)\\},\\\ J_{i}^{2_{2}}=\\{j\in
J_{i}^{2}:\|0,Tx_{i}^{j}-Ty,Ty-Tx_{i}^{j}\|<\|0,y-Ty,Ty-y\|\\},\\\
J_{i}^{2_{3}}=\\{j\in J_{i}^{2}:\|0,Tx_{i}^{j}-Ty,Ty-
Tx_{i}^{j}\|<\|0,x_{i}^{j}-Ty,Ty-x_{i}^{j}\|\\}.\end{array}$

Using (3.10), we obtain

$\begin{array}[]{c}\|0,x_{i}-Ty,Ty-x_{i}\|\leq\sum\limits_{j\in J_{i}^{1}\cup
J_{i}^{2_{3}}}\beta_{i}^{j}\|0,x_{i}^{j}-Ty,Ty-x_{i}^{j}\|+\sum\limits_{i\in
J_{i}^{2_{1}}}\beta_{i}^{j}r^{s}(F)\\\ +\sum\limits_{i\in
J_{i}^{2_{2}}}\beta_{i}^{j}\|0,y-Ty,Ty-y\|.\end{array}$ (3.12)

Let us denote by

$J_{1}^{i}=J_{i}^{1}\cup J_{i}^{2_{3}}\text{, }J_{2}^{i}=J_{i}^{2_{2}}\text{
and }J_{3}^{i}=J_{i}^{2_{1}}\text{.}$

Then using (3.12), we have

$\begin{array}[]{c}\|0,x_{i}-Ty,Ty-x_{i}\|\leq\sum\limits_{j\in
J_{1}^{i}}\beta_{i}^{j}\|0,x_{i}^{j}-Ty,Ty-x_{i}^{j}\|+\sum\limits_{i\in
J_{3}^{i}}\beta_{i}^{j}r^{s}(F)\\\ +\sum\limits_{i\in
J_{2}^{i}}\beta_{i}^{j}\|0,y-Ty,Ty-y\|.\end{array}$ (3.13)

If $A_{i}=\sum\limits_{i\in J_{3}^{i}}\beta_{i}^{j}$ and
$B_{i}=\sum\limits_{i\in J_{2}^{i}}\beta_{i}^{j}$ with $\sum\limits_{j\in
J_{1}^{i}}\beta_{i}^{j}+A_{i}+B_{i}=1$. Using (3.13), we obtain

$\|0,x_{i}-Ty,Ty-x_{i}\|\leq\sum\limits_{j\in
J_{1}^{i}}\beta_{i}^{j}\|0,x_{i}^{j}-Ty,Ty-
x_{i}^{j}\|+A_{i}r^{s}(F)+B_{i}\|0,y-Ty,Ty-y\|.$ (3.14)

Using (3.7) and $\|0,x_{i}-Ty,Ty-x_{i}\|$ by (3.14), we obtain

$\begin{array}[]{c}\|0,x-Ty,Ty-x\|\leq\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}\sum\limits_{j\in
J_{1}^{i}}\beta_{i}^{j}\|0,x_{i}^{j}-Ty,Ty-
x_{i}^{j}\|+\left[\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}A_{i}+A_{0}\right]r^{s}(F)\\\
+\left[\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}B_{i}+B_{0}\right]\|0,y-Ty,Ty-y\|.\end{array}$

Let $A_{1}=\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}A_{i}$ and
$B_{1}=\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}B_{i}$. Then we have

$\begin{array}[]{c}\|0,x-Ty,Ty-x\|\leq\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}\sum\limits_{j\in
J_{1}^{i}}\beta_{i}^{j}\|0,x_{i}^{j}-Ty,Ty-
x_{i}^{j}\|+(A_{1}+A_{0})r^{s}(F)\\\
+(B_{1}+B_{0})\|0,y-Ty,Ty-y\|.\end{array}$ (3.15)

We note that

$\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}\sum\limits_{j\in
J_{1}^{i}}\beta_{i}^{j}+\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}A_{i}+A_{0}+\sum\limits_{i\in\overline{I_{1}}}\alpha_{i}B_{i}+B_{0}=1.$

Let us take $K=\underset{i\in\overline{I_{1}}}{\bigcup}\left(\underset{j\in
J_{1}^{i}}{\bigcup}j\right)$ and denote the scalars by $\xi_{k}$. To each $k$
relative to the pair $(i,j)$, $x_{i}^{j}$ will be denoted by $x_{k}$.

Using (3.15), we obtain

$\begin{array}[]{c}\|0,x-Ty,Ty-x\|\leq\sum\limits_{k\in
K}\xi_{k}\|0,x_{k}-Ty,Ty-x_{k}\|+(A_{1}+A_{0})r^{s}(F)\\\
+(B_{1}+B_{0})\|0,y-Ty,Ty-y\|,\end{array}$

where $\sum\limits_{k\in K}\xi_{k}+A_{1}+A_{0}+B_{1}+B_{0}=1$ and $x_{k}\in
F_{m-2}$.

Repeating this process which is done for $x_{k}$, we get

$\begin{array}[]{c}\|0,x-Ty,Ty-x\|\leq\sum\limits_{p\in
P}\gamma_{p}\|0,x_{p}-Ty,Ty-x_{p}\|+\sum\limits_{k=0}^{m-1}A_{k}r^{s}(F)\\\
+\sum\limits_{k=0}^{m-1}B_{k}\|0,y-Ty,Ty-y\|,\end{array}$ (3.16)

where $\sum\limits_{p\in P}\gamma_{p}+\sum\limits_{i=0}^{m-1}(A_{i}+B_{i})=1$
and $x_{p}\in F_{1}=C^{s}(F)$.

Hence $\|0,x_{p}-Ty,Ty-x_{p}\|\leq r^{s}(F)$ and using (3.16), we obtain

$\|0,x-Ty,Ty-x\|\leq\sum\limits_{k=0}^{m-1}B_{k}\|0,y-Ty,Ty-y\|+\left(\sum\limits_{p\in
P}\gamma_{p}+\sum\limits_{k=0}^{m-1}A_{k}\right)r^{s}(F).$ (3.17)

Let us turn to $\|0,y-Ty,Ty-y\|$. Since $y\in conv(F_{m-1}\cup TF_{m-1})$, we
have $y=\sum\limits_{i\in I}\alpha_{i}y_{i}$ with $\sum\limits_{i\in
I}\alpha_{i}=1$, $y_{i}\in F_{m-1}\cup TF_{m-1}$ and $\alpha_{i}\geq 0$ for
all $i\in I$. Let $I=I_{1}\cup I_{2}$ such that $I_{1}\cap I_{2}=\emptyset$.
If $i\in I_{1}$ then $y_{i}\in F_{m-1}$ and if $i\in I_{2}$ then $y_{i}\in
TF_{m-1}$. Let $y_{i}=Ty_{i}$. Then we can write

$y=\sum\limits_{i\in I_{1}}\alpha_{i}y_{i}+\sum\limits_{i\in
I_{2}}\alpha_{i}Ty_{i},$

with $y_{i}\in F_{m-1}$.

Substituting in $\|0,y-Ty,Ty-y\|$ we get

$\|0,y-Ty,Ty-y\|\leq\sum\limits_{i\in I_{1}}\alpha_{i}\|0,y_{i}-Ty,Ty-
y_{i}\|+\sum\limits_{i\in I_{2}}\alpha_{i}\|0,Ty_{i}-Ty,Ty-Ty_{i}\|.$ (3.18)

Using the condition $(\mathbf{NS}25)$, we obtain

$\begin{array}[]{r}\|0,Ty_{i}-Ty,Ty-
Ty_{i}\|<\max\\{\|0,y_{i}-y,y-y_{i}\|,\|0,y_{i}-Ty_{i},Ty_{i}-y_{i}\|,\\\
\|0,y-Ty,Ty-y\|,\|0,y_{i}-Ty,Ty-
y_{i}\|,\|0,Ty_{i}-y,y-Ty_{i}\|\\}\text{.}\end{array}$ (3.19)

Since $y_{i}\in F_{m-1}$, $y\in F_{m}$, we have

$\begin{array}[]{l}\|0,y_{i}-y,y-y_{i}\|\leq r^{s}(F),\\\
\|0,y_{i}-Ty_{i},Ty_{i}-y_{i}\|\leq r^{s}(F),\\\ \|0,Ty_{i}-y,y-Ty_{i}\|\leq
r^{s}(F).\end{array}$

Using (3.19), we obtain

$\|0,Ty_{i}-Ty,Ty-Ty_{i}\|<\max\\{r^{s}(F),\|0,y-Ty,Ty-y\|,\|0,y_{i}-Ty,Ty-
y_{i}\|\\}\text{.}$

Redefining the index set $I_{2}=I_{2}^{1}\cup I_{2}^{2}\cup I_{2}^{3}$ with

$\begin{array}[]{l}I_{2}^{1}=\\{i\in I_{2}:\|0,Ty_{i}-Ty,Ty-
Ty_{i}\|<r^{s}(F)\\},\\\ I_{2}^{2}=\\{i\in I_{2}:\|0,Ty_{i}-Ty,Ty-
Ty_{i}\|<\|0,y-Ty,Ty-y\|\\},\\\ I_{2}^{3}=\\{i\in I_{2}:\|0,Ty_{i}-Ty,Ty-
Ty_{i}\|<\|0,y_{i}-Ty,Ty-y_{i}\|\\}.\end{array}$

Now using (3.18), we get

$\begin{array}[]{c}\|0,y-Ty,Ty-y\|\leq\sum\limits_{i\in I_{1}\cup
I_{2}^{3}}\alpha_{i}\|0,y_{i}-Ty,Ty-y_{i}\|+\sum\limits_{i\in
I_{2}^{1}}\alpha_{i}r^{s}(F)\\\ +\sum\limits_{i\in
I_{2}^{2}}\alpha_{i}\|0,y-Ty,Ty-y\|.\end{array}$ (3.20)

We note that if $\sum\limits_{i\in I_{2}^{2}}\alpha_{i}=1$ then

$\|0,y-Ty,Ty-y\|\leq\|0,Ty_{i}-Ty,Ty-Ty_{i}\|<\|0,y-Ty,Ty-y\|,$

which is a contradiction.

Then $\sum\limits_{i\in I_{2}^{2}}\alpha_{i}<1$ and using (3.20), we obtain

$\begin{array}[]{c}\|0,y-Ty,Ty-y\|\leq\sum\limits_{i\in I_{1}\cup
I_{2}^{3}}\dfrac{\alpha_{i}}{1-\sum\limits_{i\in
I_{2}^{2}}\alpha_{i}}\|0,y_{i}-Ty,Ty-y_{i}\|\\\ +\sum\limits_{i\in
I_{2}^{1}}\dfrac{\alpha_{i}}{1-\sum\limits_{i\in
I_{2}^{2}}\alpha_{i}}r^{s}(F),\end{array}$ (3.21)

with $\sum\limits_{i\in I_{1}\cup
I_{2}^{3}}\dfrac{\alpha_{i}}{1-\sum\limits_{i\in
I_{2}^{2}}\alpha_{i}}+\sum\limits_{i\in
I_{2}^{1}}\dfrac{\alpha_{i}}{1-\sum\limits_{i\in I_{2}^{2}}\alpha_{i}}=1$.

Let $I_{1}=I_{1}\cup I_{2}^{3}$, $I_{2}=I_{2}^{1}$ and
$\beta_{i}=\dfrac{\alpha_{i}}{1-\sum\limits_{i\in I_{2}^{2}}\alpha_{i}}$.
Using (3.21), we obtain

$\|0,y-Ty,Ty-y\|\leq\sum\limits_{i\in I_{1}}\beta_{i}\|0,y_{i}-Ty,Ty-
y_{i}\|+\sum\limits_{i\in I_{2}}\beta_{i}r^{s}(F).$ (3.22)

If $A_{0}=\sum\limits_{i\in I_{2}}\beta_{i}$ then using (3.22), we have

$\|0,y-Ty,Ty-y\|\leq\sum\limits_{i\in I_{1}}\beta_{i}\|0,y_{i}-Ty,Ty-
y_{i}\|+A_{0}r^{s}(F),$ (3.23)

with $\sum\limits_{i\in I_{1}}\beta_{i}+A_{0}=1$ and $y_{i}\in
F_{m-1}=conv(F_{m-2}\cup TF_{m-2})$.

For each $i\in I_{1}$,

$y_{i}=\sum\limits_{j\in J_{i}^{1}}\gamma_{i}^{j}y_{i}^{j}+\sum\limits_{j\in
J_{i}^{2}}\gamma_{i}^{j}Ty_{i}^{j},$

with $y_{i}^{j}\in F_{m-2}$ and $\sum\limits_{j\in J_{i}^{1}\cup
J_{i}^{2}}\gamma_{i}^{j}=1$. So we obtain

$\|0,y_{i}-Ty,Ty-y_{i}\|\leq\sum\limits_{j\in
J_{i}^{1}}\gamma_{i}^{j}\|0,y_{i}^{j}-Ty,Ty-y_{i}^{j}\|+\sum\limits_{j\in
J_{i}^{2}}\gamma_{i}^{j}\|0,Ty_{i}^{j}-Ty,Ty-Ty_{i}^{j}\|.$ (3.24)

Using the condition $(\mathbf{NS25}),$ we have

$\begin{array}[]{r}\|0,Ty_{i}^{j}-Ty,Ty-
Ty_{i}^{j}\|<\max\\{\|0,y_{i}^{j}-y,y-y_{i}^{j}\|,\|0,Ty_{i}^{j}-y_{i}^{j},y_{i}^{j}-Ty_{i}^{j}\|,\\\
\|0,y-Ty,Ty-y\|,\|0,Ty_{i}^{j}-y,y-Ty_{i}^{j}\|,\|0,y_{i}^{j}-Ty,Ty-
y_{i}^{j}\|\\}\text{.}\end{array}$

Since $y_{i}^{j}\in F_{m-2}$, $Ty_{i}^{j}\in F_{m-1}$ and $y\in F_{m}$, we can
write

$\begin{array}[]{l}\|0,y_{i}^{j}-y,y-y_{i}^{j}\|\leq r^{s}(F),\\\
\|0,Ty_{i}^{j}-y_{i}^{j},y_{i}^{j}-Ty_{i}^{j}\|\leq r^{s}(F),\\\
\|0,Ty_{i}^{j}-y,y-Ty_{i}^{j}\|\leq r^{s}(F)\end{array}$

and

$\|0,Ty_{i}^{j}-Ty,Ty-
Ty_{i}^{j}\|<\max\\{r^{s}(F),\|0,y-Ty,Ty-y\|,\|0,y_{i}^{j}-Ty,Ty-
y_{i}^{j}\|\\}\text{.}$

Let $J_{i}^{2}$ be the union of the disjoint sets $J_{i}^{2}=J_{i}^{2_{1}}\cup
J_{i}^{2_{2}}\cup J_{i}^{2_{3}}$ such that

$\begin{array}[]{l}J_{i}^{2_{1}}=\\{j\in J_{i}^{2}:\|0,Ty_{i}^{j}-Ty,Ty-
Ty_{i}^{j}\|<\|0,y_{i}^{j}-Ty,Ty-y_{i}^{j}\|\\},\\\ J_{i}^{2_{2}}=\\{j\in
J_{i}^{2}:\|0,Ty_{i}^{j}-Ty,Ty-Ty_{i}^{j}\|<r^{s}(F)\\},\\\
J_{i}^{2_{3}}=\\{j\in J_{i}^{2}:\|0,Ty_{i}^{j}-Ty,Ty-
Ty_{i}^{j}\|<\|0,y-Ty,Ty-y\|\\}.\end{array}$

Using (3.24), we obtain

$\begin{array}[]{c}\|0,y_{i}-Ty,Ty-y_{i}\|\leq\sum\limits_{j\in J_{i}^{1}\cup
J_{i}^{2_{1}}}\gamma_{i}^{j}\|0,y_{i}^{j}-Ty,Ty-y_{i}^{j}\|+\sum\limits_{j\in
J_{i}^{2_{2}}}\gamma_{i}^{j}r^{s}(F)\\\ +\sum\limits_{j\in
J_{i}^{2_{3}}}\gamma_{i}^{j}\|0,y-Ty,Ty-y\|.\end{array}$ (3.25)

Now redefine the index sets $J_{i}^{1}=J_{i}^{1}\cup J_{i}^{2_{1}}$,
$J_{i}^{2}=J_{i}^{2_{2}},$ $J_{i}^{3}=J_{i}^{2_{3}}$ and using (3.25) we can
write

$\begin{array}[]{c}\|0,y_{i}-Ty,Ty-y_{i}\|\leq\sum\limits_{j\in
J_{i}^{1}}\gamma_{i}^{j}\|0,y_{i}^{j}-Ty,Ty-y_{i}^{j}\|+\sum\limits_{j\in
J_{i}^{2}}\gamma_{i}^{j}r^{s}(F)\\\ +\sum\limits_{j\in
J_{i}^{3}}\gamma_{i}^{j}\|0,y-Ty,Ty-y\|,\end{array}$

with $\sum\limits_{j\in J_{i}^{1}}\gamma_{i}^{j}+\sum\limits_{j\in
J_{i}^{2}}\gamma_{i}^{j}+\sum\limits_{j\in J_{i}^{3}}\gamma_{i}^{j}=1$.

Using the (3.23), we obtain

$\begin{array}[]{c}\|0,y-Ty,Ty-y\|\leq\sum\limits_{i\in
I_{1}}\beta_{i}\sum\limits_{j\in J_{i}^{1}}\gamma_{i}^{j}\|0,y_{i}^{j}-Ty,Ty-
y_{i}^{j}\|+\sum\limits_{i\in I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{3}}\gamma_{i}^{j}\|0,y-Ty,Ty-y\|\\\ +\left[\sum\limits_{i\in
I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{2}}\gamma_{i}^{j}+A_{0}\right]r^{s}(F).\end{array}$ (3.26)

If $\sum\limits_{i\in I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{3}}\gamma_{i}^{j}=1$ we have

$\|0,y-Ty,Ty-y\|<\|0,y-Ty,Ty-y\|,$

which is a contradiction.

Hence $\sum\limits_{i\in I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{3}}\gamma_{i}^{j}<1$ and using (3.26), we obtain

$\begin{array}[]{c}\|0,y-Ty,Ty-y\|\leq\dfrac{\sum\limits_{i\in
I_{1}}\beta_{i}\sum\limits_{j\in J_{i}^{1}}\gamma_{i}^{j}}{1-\sum\limits_{i\in
I_{1}}\beta_{i}\sum\limits_{j\in J_{i}^{3}}\gamma_{i}^{j}}\|0,y_{i}^{j}-Ty,Ty-
y_{i}^{j}\|\\\ +\dfrac{\sum\limits_{i\in I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{2}}\gamma_{i}^{j}+A_{0}}{1-\sum\limits_{i\in
I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{3}}\gamma_{i}^{j}}r^{s}(F),\end{array}$ (3.27)

with $\dfrac{\sum\limits_{i\in I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{1}}\gamma_{i}^{j}}{1-\sum\limits_{i\in I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{3}}\gamma_{i}^{j}}+\dfrac{\sum\limits_{i\in
I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{2}}\gamma_{i}^{j}+A_{0}}{1-\sum\limits_{i\in
I_{1}}\beta_{i}\sum\limits_{j\in J_{i}^{3}}\gamma_{i}^{j}}=1$.

Let $A_{1}=\dfrac{\sum\limits_{i\in I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{2}}\gamma_{i}^{j}+A_{0}}{1-\sum\limits_{i\in
I_{1}}\beta_{i}\sum\limits_{j\in J_{i}^{3}}\gamma_{i}^{j}}$ and denote the
index set by $K=\underset{i\in I_{1}}{\bigcup}\left(\underset{j\in
J_{i}^{1}}{\bigcup}j\right)$, write $\zeta_{k}$ for $k\in K$ relative to
$(i,j)$, that is

$\zeta_{k}=\dfrac{\sum\limits_{i\in I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{2}}\gamma_{i}^{j}}{1-\sum\limits_{i\in I_{1}}\beta_{i}\sum\limits_{j\in
J_{i}^{3}}\gamma_{i}^{j}}\text{.}$

Also we write $y_{k}$ for $y_{i}^{j}$. Then using (3.27), we obtain

$\|0,y-Ty,Ty-y\|\leq\sum\limits_{k\in K}\zeta_{k}\|0,y_{k}-Ty,Ty-
y_{k}\|+(A_{1}+A_{0})r^{s}(F),$

with $\sum\limits_{k\in K}\zeta_{k}+A_{1}+A_{0}=1$ and $y_{k}\in F_{m-2}$.

Repeating this process we get

$\|0,y-Ty,Ty-y\|\leq\sum\limits_{p\in P}\lambda_{p}\|0,y_{p}-Ty,Ty-
y_{p}\|+\sum\limits_{k=0}^{m-1}A_{k}r^{s}(F),$

where $y_{p}\in F_{1}$ and $\sum\limits_{p\in
P}\lambda_{p}+\sum\limits_{k=0}^{m-1}A_{k}=1$.

Then $\|0,y_{p}-Ty,Ty-y_{p}\|\leq r^{s}(F)$ and

$\|0,y-Ty,Ty-y\|\leq\left(\sum\limits_{p\in
P}\lambda_{p}+\sum\limits_{k=0}^{m-1}A_{k}\right)r^{s}(F)=r^{s}(F).$

Using (3.17), we get

$\|0,x-Ty,Ty-x\|\leq\left(\sum\limits_{k=0}^{m-1}B_{k}+\sum\limits_{p\in
P}\gamma_{p}+\sum\limits_{k=0}^{m-1}A_{k}\right)r^{s}(F),$

with $\sum\limits_{k=0}^{m-1}B_{k}+\sum\limits_{p\in
P}\gamma_{p}+\sum\limits_{k=0}^{m-1}A_{k}=1$.

Consequently, we obtain $\|0,x-Ty,Ty-x\|\leq r^{s}(F)$ and so

$\delta_{m+1}^{s}-\varepsilon<\|0,x-Ty,Ty-x\|\leq r^{s}(F).$

Case 3. For $x,y\in F_{m}$, we have

$\begin{array}[]{c}\delta_{m+1}^{s}-\varepsilon<\|0,Tx-Ty,Ty-
Tx\|<\max\\{\|0,x-y,y-x\|,\|0,Tx-x,x-Tx\|,\\\
\|0,y-Ty,Ty-y\|,\|0,x-Ty,Ty-x\|,\|0,Tx-y,y-Tx\|\\}\end{array}$ (3.28)

and repeating what has been done in Case 2, we get

$\delta_{m+1}^{s}-\varepsilon<\|0,Tx-Ty,Ty-Tx\|\leq r^{s}(F).$

In all three cases we have $\delta_{m+1}^{s}-\varepsilon<r^{s}(F)$. If
$\varepsilon$ tends to $0$ we get $\delta_{m+1}^{s}\leq r^{s}(F)$.

Let $F^{\infty}=\bigcup\limits_{n\in\mathbb{N}}F_{n}$. Then $F^{\infty}$ is
nonempty because $C^{s}(F)\neq\emptyset$. Since $F_{k}\subseteq F_{k+1}$, we
obtain

$\delta^{s}(F^{\infty})=\lim_{k\rightarrow\infty}\delta^{s}(F_{k})\leq
r^{s}(F).$

As $F_{k}\subset F$, $F^{\infty}\subseteq F$ and so
$\delta^{s}(F^{\infty})\leq r^{s}(F)$.

Using the $S$-normal structure of $F$ we have $r^{s}(F)<\delta^{s}(F)$ and
$\delta^{s}(F^{\infty})<\delta^{s}(F)$. So $F^{\infty}$ must be a proper
subset of $F$. We obtain that $F^{\infty}$ is convex and $TF^{\infty}\subseteq
F^{\infty}$.

Let $M=\overline{convF^{\infty}}=\overline{F^{\infty}}$, its diameter is the
same as $F^{\infty}$. So we have

$\delta^{s}(M)\leq r^{s}(F)<\delta^{s}(F)$

and $M$ is closed, nonempty and convex proper subset of $F$. Since $T$ is
continuous then $M$ is $T$-invariant and

$TM=TF^{\infty}\subseteq\overline{TF^{\infty}}\subseteq\overline{F^{\infty}}=M.$

So $M\in\mathcal{A}$ and $M\subsetneqq F$ contradicting the minimality of $F$.
Hence, it should be $\delta^{s}(F)=0$. Consequently, $F$ has a unique fixed
point under $T$. ∎

##  4\. Some Comparisons on $S$-Normed Spaces

In [12], the present authors defined Rhoades’ condition $(\mathbf{S25})$ using
the notion of an $S$-metric. Also, they investigated relationships between the
conditions $(\mathbf{S25})$ and $(\mathbf{R25})$ in [13].

In this section, we determine the relationships between the conditions
$(\mathbf{S25})$ (resp. $(\mathbf{NR25})$) and $(\mathbf{NS25})$.

At first, we recall the Rhoades’ condition on normed spaces as follows [11]:

Let $(X,\|.\|)$ be a Banach space and $T$ be a self-mapping of $X$.

$(\mathbf{NR25})\text{ \ \ \ \ }\|Tx-
Ty\|<\max\\{\|x-y\|,\|x-Tx\|,\|y-Ty\|,\|x-Ty\|,\|y-Tx\|\\},$

for each $x,y\in X$, $x\neq y$.

Now we give the relationship between $(\mathbf{S25})$ and $(\mathbf{NS25})$ in
the following proposition.

######  Proposition 4.1.

Let $(X,\|.,.,.\|)$ be an $S$-Banach space, $(X,S_{\|.\|})$ be the $S$-metric
space obtained by the $S$-metric generated by $\|.,.,.\|$ and $T$ be a self-
mapping of $X$. If $T$ satisfies the condition $(\mathbf{NS25})$ then $T$
satisfies the condition $(\mathbf{S25})$.

###### Proof.

Assume that $T$ satisfies the condition $(\mathbf{NS25})$. Using the condition
$(\mathbf{NS25})$, we have

$\begin{array}[]{l}S_{\|.\|}(Tx,Tx,Ty)=\|Tx-Tx,Tx-Ty,Ty-Tx\|=\|0,Tx-Ty,Ty-
Tx\|\\\ <\max\\{\|0,x-y,y-x\|,\|0,Tx-x,x-Tx\|,\|0,Ty-y,y-Ty\|,\\\
\|0,Ty-x,x-Ty\|,\|0,Tx-y,y-Tx\|\\}\\\
=\max\\{S_{\|.\|}(x,x,y),S_{\|.\|}(Tx,Tx,x),S_{\|.\|}(Ty,Ty,y),S_{\|.\|}(Ty,Ty,x),S_{\|.\|}(Tx,Tx,y)\\}\end{array}$

and so the condition $(\mathbf{S25})$ is satisfied by $T$ on $(X,S_{\|.\|})$.
∎

Now, we give the relationship between the conditions $(\mathbf{NR25})$ and
$(\mathbf{NS25})$ in the following proposition.

######  Proposition 4.2.

Let $(X,\|.\|)$ be a Banach space, $(X,\|.,.,.\|)$ be an $S$-normed space
obtained by the $S$-norm generated by $\|.\|$ and $T$ be a self-mapping of
$X$. If $T$ satisfies the condition $(\mathbf{NR25})$ then $T$ satisfies the
condition $(\mathbf{NS25})$.

###### Proof.

Let $T$ satisfies the condition $(\mathbf{NR25})$. Using the conditions
$(\mathbf{NR25})$ and $(\mathbf{N3})$, we have

$\begin{array}[]{l}\|0,Tx-Ty,Ty-Tx\|=\|0\|+\|Tx-Ty\|+\|Ty-Tx\|=2\|Tx-Ty\|\\\
<2\max\\{\|x-y\|,\|x-Tx\|,\|y-Ty\|,\|x-Ty\|,\|y-Tx\|\\}\\\
=\max\\{2\|x-y\|,2\|x-Tx\|,2\|y-Ty\|,2\|x-Ty\|,2\|y-Tx\|\\}\\\
=\max\\{\|x-y\|+\|y-x\|,\|x-Tx\|+\|Tx-x\|,\|y-Ty\|+\|Ty-y\|,\\\
\|x-Ty\|+\|Ty-x\|,\|y-Tx\|+\|Tx-y\|\\}\\\
=\max\\{\|0,x-y,y-x\|,\|0,Tx-x,x-Tx\|,\|0,Ty-y,y-Ty\|,\\\
\|0,Ty-x,x-Ty\|,\|0,Tx-y,y-Tx\|\\}\end{array}$

and so the condition $(\mathbf{NS25})$ is satisfied. ∎

Finally, we give the relationship between Theorem 3.1 and the following
theorem.

######  Theorem 4.1.

[11] Let $X$ be a reflexive Banach space and $A$ be a nonempty, closed,
bounded and convex subset of $X$, having normal structure. If $T:A\rightarrow
A$ is a continuous self-mapping satisfying the condition $(\mathbf{NR25})$
then $T$ has a unique fixed point in $A$.

Theorem 3.1 and Theorem 4.1 are coincide when $X$ is an $S$-Banach space
obtained by the $S$-norm generated by $\|.\|$. Clearly, Theorem 3.1 is a
generalization of Theorem 4.1 as we have seen in Section 2 that there are
$S$-norms which are not generated by any norm.

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