The first idea of the present article is that in the third section we want to prove a theorem based on rational $\alpha-\beta$-contractions, so we remind the necessary concepts for this type of operators. For more informations, we let the reader follow [13]. We recall the following crucial concepts.

Definition 1.1. Let $X$ be a nonempty set and $\alpha:X\times X\rightarrow[0,\infty)$ be a mapping. Then $f:X\rightarrow X$ is called $\alpha$-admissible, if it satisfies the following condition for each $x,y\in X$, with $\alpha(x,y)\geq 1\Longrightarrow\alpha(fx,fy)\geq 1$.

Definition 1.2. The mapping $\alpha:X\times X\rightarrow[0,\infty)$ is called transitive, if for each $x,y,z\in X$, with $\alpha(x,y)\geq 1$ and $\alpha(y,z)\geq 1$, we have $\alpha(x,z)\geq 1$.

Moreover, let’s denote by $Y$ the set of all functions $\beta:[0,\infty)\rightarrow[0,1)$, satisfying $\lim_{n\rightarrow\infty}\beta\left(t_{n}\right)=1\Longrightarrow\lim_{n\rightarrow\infty}t_{n}=0$.
Also, we recall the definition of $\alpha-\beta-$ contractions, given by Sintunavarat in [13].
Definition 1.3. Let ( $X,d$ ) be a metric space. A mapping $f:X\rightarrow X$ is called an $-\alpha-\beta$-contraction, if there exists $\alpha:X\times X\rightarrow[0,\infty)$ and $\beta\in Y$, such that $\left[\alpha(x,y)-1+\delta_{*}\right]^{d(fx,fy)}\leq\delta^{\beta(d(x,y))d(x,y)}$, for each $x,y\in X$, with $1<\delta\leq\delta_{*}$.

In [13], the author proved a series of theorems such as : existence of a fixed point assuming that the mapping $f$ is continuous, a theorem in which the continuity is dropped and a theorem for the uniqueness of fixed points. Also, this was based on the result of Geraghty [3] from 1973. Moreover, [8] Paunović et. al. extended the result of W. Sintunavarat in the framework of b-metric spaces. Additionally, they have studied fixed points for a given mapping $F:X\rightarrow X$, such that $[\alpha(x,y)-1+\delta]^{d(Fx,Fy)}\leq\delta^{\lambda M(x,y)}$, for each $x,y\in X$, with $1<\delta$, where $\lambda\in\left[0,\frac{1}{s}\right]$ and $M(x,y)=\left\{d(x,y),d(x,Fx),d(y,Fy),\frac{d(x,Fy)+d(y,Fx)}{2s}\right\}$, where $s$ was the coefficient of the b-metric space ( $X,d$ ).

Furthermore, Zabihi and Razani [14] considered rational type operators and developed some fixed point results in the framework of complete b -metric spaces. In the context of a metric space ( $X,d$ ), a self mapping $f$ on $X$ was considered, satisfying
$d(fx,fy)\leq\beta(d(x,y))M(x,y)+L\cdot N(x,y)$, where $L\geq 0$,
$M(x,y)=$ max $\left\{d(x,y),\frac{d(x,fx)d(y,fy)}{1+d(fx,fy)}\right\}$ and $N(x,y)=\min\{d(x,fx),d(x,fy),d(y,fx),d(y,fy)\}$.

Moreover, since we want to define some other type of $\alpha-\beta-$ contractions, we shall recall that the authors in [12] developed new fixed point theorems involving a new type of rational contractive Geraghty mapping in b-metric spaces. This rational Geraghty mapping is introduced as follows, in the case of metric spaces.

Definition 1.4. Let ( $X,d$ ) be a metric space. A mapping $f:X\rightarrow X$ is called a rational Geraghty of type $I$, if $d(fx,fy)\leq\beta(M(x,y))M(x,y)$, for each $x,y\in X$, where $\beta\in Y$ and $M(x,y)=\max\left\{d(x,y),\frac{d(x,fx)d(y,fy)}{1+d(x,y)},\frac{d(x,fx)d(y,fy)}{1+d(fx,fy)}\right\}$.

That means that in the third section, we will present a generalized theorem for rational Geraghty mappings of type $I$.

In this section, we recall some general notions in the framework of multivalued analysis theory. Also, for the following preliminary notions and lemmas (such as: multivalued weakly Picard operators, data dependence of the fixed point set, Haussdorf metric properties) we refer the reader to [9], [10] and [11].

Let ( $X,d$ ) be a metric space and $P(X)$ be the family of all nonempty subsets of $X$.
We denote by $P_{cl}(X)$ the family of all nonempty subsets of $X$ which are closed, by $P_{b}(X)$ the family of all nonempty subsets of $X$ which are bounded and by $P_{cp}(X)$ the family of all nonempty subsets of $X$ which are compact.

Furthermore, we consider the following functionals
$D:P(X)\times P(X)\rightarrow\mathbb{R}_{+},D(A,B)=\inf\{d(a,b)/a\in A,b\in B\}$
$H:P_{b}(X)\times P_{b}(X)\rightarrow\mathbb{R}_{+},H(A,B)=\max\left\{\sup_{a\in A}D(a,B),\sup_{b\in B}D(b,A)\right\}$
$\rho:P_{b}(X)\times P_{b}(X)\rightarrow\mathbb{R}_{+},\rho(A,B)=\sup\{D(a,B)/a\in A\}$
We recall some useful results concerning the Haussdorf-Pompeiu generalized functional $H$.
Lemma 2.1. Let $q>1$ and $A,B\in P(X)$.
Then, for each $a\in A$, there exists $b\in B$ such that $d(a,b)\leq qH(A,B)$.
Lemma 2.2. Let $(X,d)$ be a metric space and $A,B\in P(X)$.
Suppose that there exists $\eta>0$ such that :
(i) for each $a\in A$, there exists $b\in B$ such that $d(a,b)\leq\eta$,
(ii) for each $b\in B$, there exists $a\in A$ such that $d(a,b)\leq\eta$.

Then $H(A,B)\leq\eta$.
Moreover, if $Y$ is a nonempty subset of $X$ and $T:Y\rightarrow P(X)$ a multivalued operator, then an element $x\in Y$ is
(a) a fixed point of $T$ if and only if $x\in Tx$;
(b) a strict fixed point of $T$ if and only if $\{x\}=Tx$;

Furthermore, we denote by $F_{T}$ the set of all fixed points of $T$ and by $(SF)_{T}$ the set of all strict fixed points of $T$.

We also remind the definition of the graphic of a multivalued operator, i.e.
$G_{T}:=\{(x,y)\in Y\times X/y\in Tx\}$.
Definition 2.3. Let ( $X,d$ ) be a metric space and $T:X\rightarrow P(X)$ a multivalued operator. We say that $T$ is a multivalued weakly Picard operator (briefly MWP) if for each $x\in X$ and for each $y\in Tx$, there exists a sequence $\left(x_{n}\right)\in X$, satisfying the following
(i) $x_{0}=x,x_{1}=y$;
(ii) $x_{n+1}\in Tx_{n}$, for each $n\in\mathbb{N}$;
(iii) the sequence $\left(x_{n}\right)$ is convergent to a fixed point of $T$.

Definition 2.4. Let ( $X,d$ ) be a metric space and $T:X\rightarrow P(X)$ an MWP.
Then $T$ is called a $c$-weakly Picard operator, with $c\in[0,\infty)$, if there exists a selection $t^{\infty}$ of $T^{\infty}$, such that
$d\left(x,t^{\infty}(x,y)\right)\leq cd(x,y)$, for each $(x,y)\in G_{T}$.
Now we focus our attention to the case of Hardy-Rogers type mappings. In [7], the basic notion of singlevalued Hardy-Rogers contraction appeared.

Definition 2.5. Let $(X,d)$ be a metric space and $T:X\rightarrow X$ be an operator such that there exists $\alpha,\beta\geq 0$ with $\alpha+\beta<1$, satisfying
$d(Tx,Ty)\leq\frac{\alpha d(y,Ty)[1+d(x,Tx)]}{1+d(x,y)}+\beta d(x,y)$, for each $x,y\in X$.
In [6], Oprea A. developed a theorem concerning multivalued rational contractions (of Hardy Rogers type).

A multivalued operator $T:X\rightarrow P(X)$ is called a multivalued rational type contraction, if satisfies the following condition
$H(Tx,Ty)\leq\frac{\alpha D(y,Ty)[1+D(x,Tx)]}{1+d(x,y)}+\beta d(x,y)$, for each $x,y\in X$.
Oprea has showed that the multivalued rational contractions are MWP-operators and developed theorems for data dependence, fractal theory, Ulam-Hyers stability etc.

In [4], Kumari and Panthi introduced a new type of rational contractions, called modified HardyRogers contractions.

They introduced this as types of cyclic contractions for the case of families of dislocated metric spaces.

We recall the notion of singlevalued contractions in the context of metric spaces, i.e. singlevalued operator that satisfies

Also, regarding Hardy Rogers mappings, our purpose to define the concept of modified HardyRogers contractions under the multivalued case shall be presented in the last section, along with some fixed point results.

In this section we present a generalized theorem for rational Geraghty mappings or type $I$, using the $\alpha$-admissibility conditions by Sintunavarat.

Moreover, we will use the same terminology from the first section.
Definition 3.1. Let ( $X,d$ ) be a metric space.
A mapping $f:X\rightarrow X$ is called an $-\alpha-\beta$-rational Geraghty mapping of type $I$ if and only if there exists $\alpha:X\times X\rightarrow[0,\infty)$ and $\beta\in Y$, such that $\left[\alpha(x,y)-1+\delta_{*}\right]^{d(fx,fy)}\leq\delta^{\beta(M(x,y))M(x,y)}$, for each $x,y\in X$, with $1<\delta\leq\delta_{*}$, where $M(x,y)=\max\left\{d(x,y),\frac{d(x,fx)d(y,fy)}{1+d(x,y)},\frac{d(x,fx)d(y,fy)}{1+d(fx,fy)}\right\}$.

Our first main result of this section is the existence theorem for $\alpha-\beta$-rational Geraghty mappings of type $I$, using the assumption that $f$ is continuous. The techniques used in the theorem’s proof follow the same lines as in the theorems from [13].

Theorem 3.2. Let ( $X,d$ ) be a complete metric space and $f:X\rightarrow X$ an $\alpha-\beta$ rational Geraghty mapping of type I. Also, suppose that the following assumptions hold
(i) $f$ is $\alpha$-admissible,
(ii) $\alpha$ is transitive,
(iii) there exists $x_{0}\in X$, such that $\alpha\left(x_{0},fx_{0}\right)\geq 1$,
(iv) $f$ is continuous.

Then, there exists $x^{*}\in X$, such that $x^{*}=fx^{*}$.
Proof. - Let $x_{0}\in X$ satisfying $\alpha\left(x_{0},fx_{0}\right)\geq 1$.
Let’s consider the Picard sequence $x_{n+1}=fx_{n}$, for each $n\in\mathbb{N}$.
If there exists $n\in\mathbb{N}$ such that $x_{n}=x_{n-1}$, then $x_{n-1}$ is a fixed point and the conclusion holds.
Suppose that for each $n\in\mathbb{N},x_{n}\neq x_{n-1}$. So $d\left(x_{n-1},x_{n}\right)>0$, for each $n\in\mathbb{N}$.
From condition (i), we know that $f$ is $\alpha$-admissible. Since $\alpha\left(x_{0},x_{1}\right)=\alpha\left(x_{0},fx_{0}\right)\geq 1$, then we have that $\alpha\left(x_{1},x_{2}\right)=\alpha\left(x_{1},fx_{1}\right)=\alpha\left(fx_{0},fx_{1}\right)\geq 1$.

Inductively, one can show that $\alpha\left(x_{n-1},x_{n}\right)\geq 1$, for each $n\in\mathbb{N}$.
Now, we estimate

so $d\left(x_{n},x_{n+1}\right)\leq\beta\left(M\left(x_{n-1},x_{n}\right)\right)\cdot M\left(x_{n-1},x_{n}\right)$.
Moreover, we make the following computations :
$M\left(x_{n-1},x_{n}\right)=\max\left\{d\left(x_{n-1},x_{n}\right),\frac{d\left(x_{n-1},fx_{n-1}\right)d\left(x_{n},fx_{n}\right)}{1+d\left(x_{n-1},x_{n}\right)},\frac{d\left(x_{n-1},fx_{n-1}\right)d\left(x_{n},fx_{n}\right)}{1+d\left(fx_{n-1},fx_{n}\right)}\right\}$
$=\max\left\{d\left(x_{n-1},x_{n}\right),\frac{d\left(x_{n-1},x_{n}\right)d\left(x_{n},x_{n+1}\right)}{1+d\left(x_{n-1},x_{n}\right)},\frac{d\left(x_{n-1},x_{n}\right)d\left(x_{n},x_{n+1}\right)}{1+d\left(x_{n},x_{n+1}\right)}\right\}$
Since $\frac{d\left(x_{n-1},x_{n}\right)d\left(x_{n},x_{n+1}\right)}{1+d\left(x_{n-1},x_{n}\right)}=d\left(x_{n},x_{n+1}\right)\cdot\frac{d\left(x_{n-1},x_{n}\right)}{1+d\left(x_{n-1},x_{n}\right)}\leq d\left(x_{n},x_{n+1}\right)\cdot 1=d\left(x_{n},x_{n+1}\right)$ and
$\frac{d\left(x_{n-1},x_{n}\right)d\left(x_{n},x_{n+1}\right)}{1+d\left(x_{n},x_{n+1}\right)}=d\left(x_{n-1},x_{n}\right)\cdot\frac{d\left(x_{n},x_{n+1}\right)}{1+d\left(x_{n},x_{n+1}\right)}\leq d\left(x_{n-1},x_{n}\right)\cdot 1=d\left(x_{n-1},x_{n}\right)$,
we get that $M\left(x_{n-1},x_{n}\right)\leq\max\left\{d\left(x_{n-1},x_{n}\right),d\left(x_{n},x_{n+1}\right)\right\}$, for each $n\in\mathbb{N},n\geq 1$.

Now, we consider two cases.
(I) If $\max\left\{d\left(x_{n-1},x_{n}\right),d\left(x_{n},x_{n+1}\right)\right\}=d\left(x_{n},x_{n+1}\right)$, then we get
$d\left(x_{n},x_{n+1}\right)\leq\beta\left(M\left(x_{n-1},x_{n}\right)\right)\cdot d\left(x_{n},x_{n+1}\right)$.
Since $\beta\left(M\left(x_{n-1},x_{n}\right)\right)<1$, because $\beta\in Y$, we get the contradiction $d\left(x_{n},x_{n+1}\right)<d\left(x_{n},x_{n+1}\right)$.
(II) Then, only the second case is valid, i.e. $\max\left\{d\left(x_{n-1},x_{n}\right),d\left(x_{n},x_{n+1}\right)\right\}=d\left(x_{n-1},x_{n}\right)$, that is $d\left(x_{n},x_{n+1}\right)\leq\beta\left(M\left(x_{n-1},x_{n}\right)\right)d\left(x_{n-1},x_{n}\right)<d\left(x_{n-1},x_{n}\right)$, for each $n\in\mathbb{N}$.

So, the sequence $\left(d\left(x_{n},x_{n+1}\right)\right)$ is strictly decreasing and nonnegative. It implies that there exists $r\geq 0$, such that $d\left(x_{n},x_{n+1}\right)\rightarrow r$ as $n\rightarrow\infty$.

Now, we show that $r=0$.
Let’s suppose that $r>0$.
We know that $d\left(x_{n+1},x_{n+2}\right)\leq\beta\left(M\left(x_{n},x_{n+1}\right)\right)d\left(x_{n},x_{n+1}\right)$. Taking the limit as $n\rightarrow\infty$, we get that $r\leq\lim_{n\rightarrow\infty}\beta\left(M\left(x_{n},x_{n+1}\right)\right)\cdot r$.
Because $r>0$,we get that $1\leq\lim_{n\rightarrow\infty}\beta\left(M\left(x_{n},x_{n+1}\right)\right)$.
But $\beta\left(M\left(x_{n},x_{n+1}\right)\right)<1$, so $\lim_{n\rightarrow\infty}\beta\left(M\left(x_{n},x_{n+1}\right)\right)\leq 1$. From all this, we find that $\lim_{n\rightarrow\infty}\beta\left(M\left(x_{n},x_{n+1}\right)\right)=1$. This implies that $\lim_{n\rightarrow\infty}M\left(x_{n},x_{n+1}\right)=0$.
Now, because $M\left(x_{n},x_{n+1}\right)$ is the maximum between three elements, if it’s limit is 0 , so all the elements have the limit 0 . This means that $d\left(x_{n},x_{n+1}\right)\rightarrow 0$. This is a contradiction!

• •

  We now show that $\left(x_{n}\right)$ is a Cauchy sequence.

  Report issue for preceding element

By reductio ad absurdum, let’s suppose that ( $x_{n}$ ) is not Cauchy. Then there exists $\varepsilon>0$ and there exists $n_{k}$ and $m_{k}$, such that $n_{k}>m_{k}\geq k$, with $d\left(x_{m_{k}},x_{n_{k}}\right)$ and $n_{k}$ being the smallest index satisfying the following

By triangular inequality, we have that $\varepsilon\leq d\left(x_{m_{k}},x_{n_{k}}\right)\leq d\left(x_{m_{k}},x_{n_{k}-1}\right)+d\left(x_{n_{k}-1},x_{n_{k}}\right)<\varepsilon+d\left(x_{n_{k}-1},x_{n_{k}}\right)$.

Since $d\left(x_{n_{k}-1},x_{n_{k}}\right)\rightarrow 0$, taking $k\rightarrow\infty$, it follows that $\lim_{k\rightarrow\infty}d\left(x_{m_{k}},x_{n_{k}}\right)=\varepsilon>0$.
Like in [13], since $\alpha$ is transitive, we observe that $\alpha\left(x_{m_{k}},x_{n_{k}}\right)\geq 1,k\in\mathbb{N}$.

Now, we make the follow estimation

So, we get that
$d\left(x_{m_{k}},x_{n_{k}}\right)\leq d\left(x_{m_{k}},x_{m_{k}+1}\right)+d\left(x_{n_{k}},x_{n_{k}+1}\right)+\beta\left(M\left(x_{m_{k}},x_{n_{k}}\right)\right)M\left(x_{m_{k}},x_{n_{k}}\right)$.
Furthermore

In the above inequality, taking the limit as $k\rightarrow\infty$, we have that
$\varepsilon\leq\lim_{k\rightarrow\infty}\beta\left(M\left(x_{m_{k}},x_{n_{k}}\right)\right)\cdot\varepsilon$.
Using the fact that $\varepsilon>0$, it follows that $\lim_{k\rightarrow\infty}\beta\left(M\left(x_{m_{k}},x_{n_{k}}\right)\right)=1$, i.e. $\leq\lim_{k\rightarrow\infty}M\left(x_{m_{k}},x_{n_{k}}\right)=0$.
Since $M\left(x_{m_{k}},x_{n_{k}}\right)$ is the maximum of three elements and it has the limit 0 , also because $M\left(x_{m_{k}},x_{n_{k}}\right)\geq d\left(x_{m_{k}},x_{n_{k}}\right)$, then all of the elements will have the limit 0 , so $d\left(x_{m_{k}},x_{n_{k}}\right)\rightarrow 0$, which is false; so ( $x_{n}$ ) is Cauchy.

• •

  Since $X$ is complete with respect to the metric $d$, there exists $x^{*}\in X$ such that $x^{*}=\lim_{n\rightarrow\infty}x_{n}$. Because $f$ is continuous, we infer that
  $x^{*}=\lim_{n\rightarrow\infty}x_{n}=\lim_{n\rightarrow\infty}fx_{n-1}=f\left(\lim_{n\rightarrow\infty}x_{n-1}\right)=fx^{*}$, so $x^{*}$ is a fixed point for the Geraghty-type mapping $f$.

  Report issue for preceding element

Now, also based on [13], we give a theorem where we dropped the continuity of the operator $f$.

Theorem 3.3. Let $(X,d)$ be a complete metric space and $f:X\rightarrow X$ an $\alpha-\beta$ rational Geraghty mappings of type I.

Let’s suppose that the following assumptions hold
(i) $f$ is $\alpha$-admissible,
(ii) $\alpha$ is transitive,
(iii) there exists $x_{0}\in X$, such that $\alpha\left(x_{0},fx_{0}\right)\geq 1$,
(iv) if ( $x_{n}$ ) is a sequence satisfying $\alpha\left(x_{n},x_{n+1}\right)\geq 1$ and $x_{n}\rightarrow x$ implies that $\alpha\left(x_{n},x\right)\geq 1$, for each $n\in\mathbb{N}$.

Then, there exists $x^{*}\in X$, such that $x^{*}=fx^{*}$.
Proof. In a similar manner like in the previous proof, we can show that $\left(x_{n}\right)$ is a Cauchy sequence and therefore there exists $x^{*}\in X$, such that $x_{n}\rightarrow x^{*}$ when $n\rightarrow\infty$.

From (iv), we have that $\alpha\left(x_{n},x^{*}\right)\geq 1$, for each $n\in\mathbb{N}$.
We make the following estimation

So, we have that $d\left(x^{*},fx^{*}\right)\leq d\left(x^{*},x_{n+1}\right)+\beta\left(M\left(x_{n},x^{*}\right)\right)M\left(x_{n},x^{*}\right)<d\left(x^{*},x_{n+1}\right)+M\left(x_{n},x^{*}\right)$.
Furthermore, we have that
$M\left(x_{n},x^{*}\right)=\max\left\{d\left(x_{n},x^{*}\right),\frac{d\left(x_{n},x_{n+1}\right)d\left(x^{*},fx^{*}\right)}{1+d\left(x_{n},x^{*}\right)},\frac{d\left(x_{n},x_{n+1}\right)d\left(x^{*},fx^{*}\right)}{1+d\left(x_{n+1},fx^{*}\right)}\right\}$.
Taking the limit as $n\rightarrow\infty$ and using the fact that $d\left(x_{n},x_{n+1}\right)\rightarrow 0$ and that $d\left(x_{n},x^{*}\right)\rightarrow 0$, we get that $M\left(x_{n-1},x_{n}\right)\rightarrow\max\left\{\lim_{n\rightarrow\infty}d\left(x_{n},x^{*}\right),0,0\right\}=\lim_{n\rightarrow\infty}d\left(x_{n},x^{*}\right)=0$.
Thus $d\left(x^{*},fx^{*}\right)\leq\lim_{n\rightarrow\infty}d\left(x_{n+1},x^{*}\right)+\lim_{n\rightarrow\infty}d\left(x_{n},x^{*}\right)=0$, so the conclusion holds properly.
Finally, we present the theorem for the uniqueness of the fixed point for the Geraghty type operator.

Theorem 3.4. Let’s suppose that all the assumptions from the last theorem are satisfied. Additionally, let’s suppose that one of the following assumptions are valid
(HO) if $a,b$ are two fixed points, then $\alpha(a,b)\geq 1$
(H1) for each $x,y\in X$, there exists $z\in X$ such that $\alpha(x,z)\geq 1$ and $\alpha(y,z)\geq 1$.
Then, $f$ admits a unique fixed point.
Proof. Let $x^{*},y^{*}$ two fixed points for the mapping $f$.
We consider two cases
(H0) We have that $\alpha\left(x^{*},y^{*}\right)\geq 1$.
From the Geraghty condition, it is easy to see that $d\left(x^{*},y^{*}\right)=0$, so the conclusion is true.
(H1) We have that there exists $z\in X$, with $\alpha\left(x^{*},z\right)\geq 1$ and $\alpha\left(y^{*},z\right)\geq 1$.
Since f is $\alpha$-admissible, by induction, we get that $\alpha\left(x^{*},f^{n}z\right)\geq 1$ and $\alpha\left(y^{*},f^{n}z\right)\geq 1$.
We make the following estimation