In this section, we introduce the concept of a weakly (*α, ψ, ξ*)-contractive mapping and give fixed point result for such mapping.

*Let* (*X, d*) *be a metric space*.

(1)

*A multi-valued mapping T* :*X* → *CL*(*X*) is called a weakly (*α, ψ, ξ*)-*contractive mapping if there exist three functions ψ* ∈ Φ, *ξ* ∈ Ξ *and α* :*X* × *X* → [0, ∞)*such that the following condition holds*:
$$x\in \text{X},y\in Tx\text{\hspace{0.17em}}\text{\hspace{0.17em}}with\text{\hspace{0.17em}}\text{\hspace{0.17em}}\alpha (x,y)\ge 1\u27f9\xi (d(y,Ty))\le \psi (\xi (M(x,y))),$$(2)

*where* $M(x,\text{\hspace{0.17em}}y)=\text{m}\text{a}\text{x}\left\{d(x,\text{\hspace{0.17em}}y),\text{\hspace{0.17em}}d(x,\text{\hspace{0.17em}}Tx),\text{\hspace{0.17em}}d(y,\text{\hspace{0.17em}}Ty),\text{\hspace{0.17em}}\frac{d(x,\text{\hspace{0.17em}}Ty)+d(y,\text{\hspace{0.17em}}Tx)}{2}\right\}$.

(2)

*If ψ is strictly increasing, then the weakly* (*α, ψ, ξ*)-*contractive mapping is called a strictly weakly* (*α, ψ, ξ*)-*contractive mapping*.

Next, we give first main result in this paper.

*Let* (*X*, *d*) *be a complete metric space and T* :*X* → *CL*(*X*) *be a strictly weakly* (*α*, *ψ*, *ξ*)-*contractive mapping satisfying the following conditions*:

(*S*1) *T is an α-admissible multi-valued mapping;*

(*S*2) *there exist* *x*_{0} ∊ *X* *and x*_{1} ∊ *Tx*_{0} *such that* *α*(*x*_{0},*x*_{1}) ≥ 1;

(*S*3) *T is a continuous multi-valued mapping*.

*Then T has a fixed point in X*.

For *x*_{0}, *x*_{1} in (*S*_{2}), if *x*_{0} = *x*_{1} or *x*_{1} ∊ *T* *x*_{1}, then *x*_{1} is a fixed point of *T*. We have nothing to prove.

So, we assume that *x*_{0} ≠ *x*_{1} and *x*_{1} ∉ *T**x*_{1}. Since *x*_{1} ∊ *T**x*_{0} and α(*x*_{0},*x*_{1}) ≥ 1, by the strictly weakly (*α*, *ψ*, *ξ*)-contractive condition of *T*, we get
$$\begin{array}{l}0<\xi \left(d\left({x}_{1},T{x}_{1}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \psi \left(\xi \left(\text{max}\left\{d\left({x}_{0},{x}_{1}\right),d\left({x}_{0},T{x}_{0}\right)d\left({x}_{1},T{x}_{1}\right),\frac{d\left({x}_{0},T{x}_{1}\right)d\left({x}_{1},T{x}_{0}\right)}{2}\right\}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\psi \left(\xi \left(\text{max}\left\{d\left({x}_{0},{x}_{0}\right),d\left({x}_{1},T{x}_{1}\right),\frac{d\left({x}_{0},T{x}_{1}\right)}{2}\right\}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \psi \left(\xi \left(\text{max}\left\{d\left({x}_{0},{x}_{1}\right),d\left({x}_{1},T{x}_{1}\right),\frac{d\left({x}_{0},T{x}_{1}\right)d\left({x}_{1},T{x}_{1}\right)}{2}\right\}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\psi \left(\xi \left(\text{max}\left\{d\left({x}_{0},{x}_{1}\right),d\left({x}_{1},T{x}_{1}\right)\right\}\right)\right).\end{array}$$(3)

By the property of *ψ*, the above relation is impossible if max{*d*(*x*_{0},*x*_{1}),*d*(*x*_{1},*Tx*_{1})} = *d*(*x*_{1},*Tx*_{1}). Hence, from (3), it follows that
$$0<\xi \left(d\left({x}_{1},T{x}_{1}\right)\right)\le \psi \left(\xi \left(d\left({x}_{0},{x}_{1}\right)\right)\right).$$(4)

Using Lemma 1.6, there exists *x*_{2} ∈ *Tx*_{1} such that
$$\xi \left(d\left({x}_{1},{x}_{2}\right)\right)<q\xi \left(d\left({x}_{1},T{x}_{1}\right)\right),$$(5)

where *q* is a fixed real number such that *q* > 1. If *x*_{1} = *x*_{2} or *x*_{2} ∈ *Tx*_{2}, then *x*_{2} is a fixed point of *T* and hence we have noting to prove. Now, we may assume that *x*_{1} ≠ *x*_{2} and *x*_{2} ∉ *Tx*_{2}. From (4) and (5), we obtain that
$$0<\xi \left(d\left({x}_{1},{x}_{2}\right)\right)<q\psi \left(\xi \left(d\left({x}_{0},{x}_{1}\right)\right)\right).$$(6)

Applying *ψ* in the above inequality, we get
$$0\text{\hspace{0.17em}}\text{\hspace{0.17em}}<\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\psi (\xi (d({x}_{1},{x}_{2})))\text{\hspace{0.17em}}\text{\hspace{0.17em}}<\text{\hspace{0.17em}}\psi (q\psi (\xi (d({x}_{0},{x}_{1}))))$$(7)

and so
$${q}_{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}:=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\psi (q\psi (\xi (d({x}_{0},{x}_{1}))))}{\psi (\xi (d({x}_{1},{x}_{2})))}>1.$$(8)

Since *T* is an *α*-admissible multi-valued mapping, we get *α*(*x*_{1}, *x*_{2}) ≥ 1. From the weakly (*α*, *ψ*, *ξ*)-contractive condition of *T*, we have
$$\begin{array}{l}0<\xi (d({x}_{2},T{x}_{2}))\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \psi \left(\xi \left(\text{max}\left\{d\left({x}_{1},{x}_{2}\right),d\left({x}_{1},T{x}_{1}\right)d\left({x}_{2},T{x}_{2}\right),\frac{d\left({x}_{1},T{x}_{2}\right)d\left({x}_{2},T{x}_{1}\right)}{2}\right\}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\psi \left(\xi \left(\text{max}\left\{d\left({x}_{1},{x}_{2}\right),d\left({x}_{2},T{x}_{2}\right),\frac{d\left({x}_{1},T{x}_{2}\right)}{2}\right\}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \psi \left(\xi \left(\text{max}\left\{d\left({x}_{1},{x}_{2}\right),d\left({x}_{2},T{x}_{2}\right),\frac{d\left({x}_{1},T{x}_{2}\right)d\left({x}_{2},T{x}_{2}\right)}{2}\right\}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\psi (\xi (\mathrm{max}\{d({x}_{1},{x}_{2}),\text{\hspace{0.17em}}d({x}_{2},T{x}_{2})\})).\end{array}$$(9)

From the above inequality, it follows that and thus
$$\mathrm{max}\{d({x}_{1},{x}_{2}),\text{\hspace{0.17em}}d({x}_{2},T{x}_{2})\}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}d({x}_{1},{x}_{2})$$(10)

and thus
$$0<\xi (d({x}_{2},T{x}_{2}))\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \text{\hspace{0.17em}}\psi (\xi (d({x}_{1},{x}_{2}))).$$(11)

For *q*_{1} > 1, by using Lemma 1.6, there exists *x*_{3} ∈ *Tx*_{2} such that
$$\xi (d({x}_{2},{x}_{3}))<\text{\hspace{0.17em}}\text{\hspace{0.17em}}{q}_{1}\xi (d({x}_{2},T{x}_{2})).$$(12)

If *x*_{2} = *x*_{3} or *x*_{3} ∈ *Tx*_{3}, then *x*_{3} is a fixed point of *T* and hence we have noting to prove. Now, we may assume that *x*_{2} ≠ *x*_{3} and *x*_{3} ∉ *Tx*_{3}. From (11) and (12), we get
$$0\text{\hspace{0.17em}}<\text{\hspace{0.17em}}\xi (d({x}_{2},{x}_{3}))\text{\hspace{0.17em}}<\text{\hspace{0.17em}}{q}_{1}\psi (\xi (d({x}_{1},{x}_{2})))\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\psi (q\psi (\xi (d({x}_{0},{x}_{1})))).$$(13)

Since *ψ* is a strictly increasing function, we obtain
$$0\text{\hspace{0.17em}}<\text{\hspace{0.17em}}\psi \text{\hspace{0.17em}}(\xi (d({x}_{2},{x}_{3})))\text{\hspace{0.17em}}\text{\hspace{0.17em}}<\text{\hspace{0.17em}}{\psi}^{2}(q\psi (\xi (d({x}_{0},{x}_{1})))).$$(14)

By continuing this process, we can construct a sequence {*x*_{n}} in *X* such that *x*_{n} ≠ *x*_{n+1} ∈ *Tx*_{n},
$$a({x}_{n},{x}_{n+1})\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \text{\hspace{0.17em}}\text{\hspace{0.17em}}1$$(15)

and
$$0\text{\hspace{0.17em}}<\text{\hspace{0.17em}}\xi (d({x}_{n+1},\text{\hspace{0.17em}}{x}_{n+2}))\text{\hspace{0.17em}}<\text{\hspace{0.17em}}{\psi}^{n}(q\psi (\xi (d({x}_{0},{x}_{1}))))$$(16)

for all *n* ∈ ℕ∪{0}.

Next, we prove that {*x*_{n}} is a Cauchy sequence in *X*. Let *m, n* ∈ ℕ. Without loss of generality, we may assume that *m* > *n*. From the triangle inequality and (16), we obtain
$$\xi (d({x}_{m},{x}_{n}))\text{\hspace{0.17em}}\le \text{\hspace{0.17em}}{\displaystyle \sum}_{i=n}^{m-1}\xi (d({x}_{i},{x}_{i+1}))\text{\hspace{0.17em}}\text{\hspace{0.17em}}<{\displaystyle \sum}_{i=n}^{m-1}{\psi}^{i-1}\text{\hspace{0.17em}}(q\psi (\xi (d({x}_{0},{x}_{1}))))\text{\hspace{0.17em}}.$$

Using the property of *ψ*, we get $\underset{n,m\to \infty}{\mathrm{lim}}\xi (d({x}_{m},\text{\hspace{0.17em}}\text{}{x}_{n}))=\text{\hspace{0.17em}}\text{}0.$. By the continuity of *ξ* and (*ξ*3), we have
$$\underset{n,m\to \infty \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}{\mathrm{lim}}d({x}_{m}{x}_{n})=0$$(17)

This shows that {*x*_{n}} is a Cauchy sequence in (*X*, *d*). Since (*X*, *d*) is a complete metric space, there exists *x** ∈ *X* such that *x*_{n} → *x** as *n* → ∞, that is, $\underset{n\to \infty}{\mathrm{lim}}\text{\hspace{0.17em}}d({x}_{n,\text{\hspace{0.17em}}}{x}^{*})\text{\hspace{0.17em}}=\text{\hspace{0.17em}}0$. From the continuity of *T*, we obtain
$$\underset{n\to \infty}{\mathrm{lim}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}H(T{x}_{n},Tx*)=0.$$(18)

Further, we have
$$d(x*,Tx*)=\underset{n\to \infty}{\mathrm{lim}}d({x}_{n+1},Tx*)\le \text{\hspace{0.17em}}\underset{n\to \infty}{\mathrm{lim}}\text{\hspace{0.17em}}H(T{x}_{n},Tx*)=0.$$

By the closedness of *Tx**, we get *x** ∈ *Tx**. Therefore, *x** is a fixed point of *T*. This completes the proof. □

Next, we give second main result in this paper.

*Let* (*X*, *d*) *be a complete metric space and T* :*X* → *CL*(*X*) *be a strictly weakly* (*α*, *ψ*, *ξ*)-*contractive mapping satisfying the following conditions*:

(*S*1) *T is an α-admissible multi-valued mapping;*

(*S*2) *there exist* *x*_{0} *and x*_{1} ∈ *Tx*_{0} such that *α*(*x*_{0}, *x*_{1}) ≥ 1;

$({S}_{3}^{\prime})$ *if* {*x*_{n}} *is a sequence in X with x*_{n+1} ∈ *Tx*_{n}, *x*_{n} → *x* ∈ *X* *as n* → ∞ and *α*(*x*_{n}, *x*_{n+1}) ≥ 1 *for all n* ∈ ℕ, *then we have* *ξ*(*d*(*x*_{n+1},*Tx*)) ≥ *ψ*(*ξ*(*M*(*x*_{n},*x*))) *for all n* ∈ ℕ.

*Then T has a fixed point in X*.

Following the proof of Theorem 2.2, we can construct a Cauchy sequence {*x*_{n}} in *X* such that *x*_{n} → *x** as *n* → ∞ and
$$\alpha ({x}_{n},{x}_{n+1})\ge 1$$(19)

for all *n* ∈ ℕ. From the condition $({S}_{3}^{\prime})$, we get
$$\xi \left(d\left({x}_{n+1},T{x}^{*}\right)\right)\le \psi \left(\xi \left(\mathrm{max}\left\{d\left({x}_{n},{x}^{*}\right),d\left({x}_{n},T{x}_{n}\right),d\left({x}^{*},T{x}^{*}\right),\frac{d\left({x}_{n},T{x}^{*}\right)+d\left({x}^{*},T{x}_{n}\right)}{2}\right\}\right)\right)$$(20)

for all *n* ∈ ℕ. Suppose that *d*(*x**, *Tx**) > 0 and let $\epsilon :\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{d({x}^{*},\text{\hspace{0.17em}}T\text{\hspace{0.17em}}{x}^{*})}{2}$. Since *x*_{n} → *x** as *n* → ∞, we can find *N*_{1} ∈ ℕ such that
$$d(x*,{x}_{n})<\frac{d(x*,Tx*)}{2}$$(21)

for all *n* ≥ *N*_{1}. Furthermore, we obtain
$$d(x*,T{x}_{n})\le d(x*,{x}_{n+1})<\frac{d(x*,Tx*)}{2}$$(22)

for all *n* ≥ *N*_{1}. Also, since {*x*_{n}} is a Cauchy sequence, there exists *N*_{2} ∈ ℕ such that
$$d({x}_{n},T{x}_{n})\le d({x}_{n},{x}_{n+1})<\frac{d(x*,Tx*)}{2}$$(23)

for all *n* ≥ *N*_{2}. Since *d*(*x*_{n}, *Tx**) → *d*(*x**, *Tx**) as *n* → ∞, it follows that there exists *N*_{3} ∈ ℕ such that
$$d({x}_{n},Tx*)<\frac{3d(x*,Tx*)}{2}$$(24)

for all *n* ≥ *N*_{3}. Using (21)-(24), it follows that
$$\mathrm{max}\left\{d({x}_{n},x*),d({x}_{n},T{x}_{n}),d(x*,Tx*),\frac{d({x}_{n},Tx*)+d(x*,T{x}_{n}}{2}\right\}=d(x*,Tx*)$$(25)

for all *n* ≥ *N* := max{*N*_{1}, *N*_{2}, *N*_{3}}. For each *n* ≥ *N*, from (20) and the triangle inequality, it follows that
$$\begin{array}{l}\xi \left(d\left({x}^{*},T{x}^{*}\right)\right)\\ \le \xi \left(d\left({x}^{*},{x}_{n+1}\right)\right)+\xi \left(d\left({x}_{n+1},T{x}^{*}\right)\right)\\ \le \xi \left(d\left({x}^{*},{x}_{n+1}\right)\right)+\psi \left(\xi \left(\mathrm{max}\left\{d\left({x}_{n},{x}^{*}\right),d\left({x}_{n},T{x}_{n}\right),d\left({x}^{*},T{x}^{*}\right),\frac{d\left({x}_{n},T{x}^{*}\right)+d\left({x}^{*},T{x}_{n}\right)}{2}\right\}\right)\right)\\ =\xi \left(d\left({x}^{*},{x}_{n+1}\right)\right)+\psi \left(\xi \left(d\left({x}^{*},T{x}^{*}\right)\right)\right).\end{array}$$

Letting *n* → ∞ in the above inequality, we get
$$\xi (d(x*,Tx*))\le \psi (\xi (d(x*,Tx*))).$$

This implies that *ξ*(*d*(*x**,*Tx**)) = 0, which is a contradiction. Therefore, *d*(*x**,*Tx**)) = 0, that is, *x** ∈ *Tx**. This completes the proof. □

Next, we give an example to show that our result is more general than the results of Ali et al. [6] and many known results in literature.

*Let X* = [0, 100] *and the metric d : X* × *X* → ℝ *defined by d*(*x*, *y*) = |*x* – *y*| *for all x, y* ∈ *X*. *Define T* :*X* → *CL*(*X*) and *α* :*X* × *X* → [0,∞) *by*
$$Tx=\{\begin{array}{l}\left[0,\frac{x}{100}\right],\text{}x\in \left[0,1\right],\\ \left[{x}^{2}+5,50\right],\text{}x\in (1,5],\\ \left[\frac{x-1}{2},100\right],\text{}x\in (5,100],\end{array}$$

and
$$\alpha (x,y)=\text{\hspace{0.17em}}\{\begin{array}{cc}1,& x,y\in \text{\hspace{0.17em}}[0,5],\\ 0,& otherwise.\end{array}$$

*Here we show that the contractive condition (1) does not hold for all functions ψ* ∈ Ψ*and ξ* ∈ Ξ. *Let x* = 0 *and y = 2. We observe that*
$$\alpha (x,y)=\text{\hspace{0.17em}}\alpha (0,2)=1,$$

*but*
$$\xi (H(Tx,Ty))=\xi (H(T0,T2))=\xi (9)>\xi (7)>\psi (\xi (7))=\psi (\xi (M(0,2)))=\psi (\xi (M(x,y))).$$

*Therefore, the results of Ali et al. [6] are not applicable here*.

*Next, we show that Theorem 2.3 can be used to guarantee the existence of fixed point of T. Define the functions ψ, ξ* : [0, ∞) → [0, ∞) *by* $\text{\hspace{0.17em}}\psi \text{\hspace{0.17em}}(t)\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{t}{10}\text{\hspace{0.17em}}$ *and* $\text{\hspace{0.17em}}\xi (t)\text{\hspace{0.17em}}=\sqrt{t}\text{\hspace{0.17em}}$ *for all t* ∈ [0, ∞). *It is easy to see that ψ* ∈ Φ *and ξ* ∈ Ξ. *First, we show that T is a strictly weakly* (*α, ψ, ξ*)-*contractive mapping. Suppose that x* ∈ *X, y* ∈ *Tx* and *α*(*x, y*) > 1 and hence *x* ∈ [0,1] *and y* ∈ [0,0.01]. *Also, we obtain*
$$\begin{array}{c}\xi d(y,Ty)=\sqrt{d(y,Ty)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \text{\hspace{0.17em}}\sqrt{H(Tx,Ty)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\sqrt{\frac{\left|x-y\right|}{100}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{10}\sqrt{\left|x-y\right|}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \frac{1}{10}\text{\hspace{0.17em}}\sqrt{M(x,y)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\psi (\xi (M(x,y))).\\ \end{array}$$

*It is easy to observe that ψ is a strictly increasing function. Therefore, T is a strictly weakly* (*α, ψ, ξ*)-*contractive mapping. It is easy to see that T is an α-admissible multi-valued mapping. Moreover, there exists x*_{0} = 1 ∈ *X and* *x*_{1} = 0.0001 ∈ *Tx*_{0} *such that*
$$\alpha ({x}_{0},{x}_{1})=\alpha (1,0.0001)\ge 1.$$

*Finally, for each sequence* {*x*_{n}} *in X with x*_{n+1} ∈ *Tx*_{n}, *x*_{n} → *x* ∈ *X as n* → ∞ *and α*(*x*_{n}, *x*_{n+1} > 1 *for all n* ∈ ℕ, we get *x*_{n}, *x* ∈ [0,1]*for all n* ∈ ℕ *and so*
$$\begin{array}{c}\xi (d({x}_{n+1},Tx)=\sqrt{d({x}_{n+1},Tx)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \text{\hspace{0.17em}}\sqrt{H(T{x}_{n},Tx)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{\frac{\left|{x}_{n}-x\right|}{100}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{10}\sqrt{\left|{x}_{n}-x\right|}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\le \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{10}\sqrt{M({x}_{n},x)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\psi (\xi (M({x}_{n},x))).\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$$

*Therefore, the condition* $({{s}^{\prime}}_{3})$ *in Theorem 2.3 holds. By using Theorem 2.3, we can conclude that T has a fixed point in X. In this case, T has infinitely fixed points such as* 0, *6 and 7*.

From Remark 1.8, Theorems 2.2 and 2.3 apply particularly in the case when *T* is an α*-admissible multi-valued mapping.

It is easy to see that the contractive condition (1) implies the contractive condition (2). Therefore, we give the following results without the proof:

(Theorem 2.5 in [6]). *Let* (*X, d*) *be a complete metric space and T* :*X* → *CL*(*X*) *be a strictly* (*α, ψ, ξ*)-*contractive mapping satisfying the following conditions*:

(*S*_{1}) *T is an α-admissible (or a* -admissible) multi-valued mapping;*

(*S*_{2}) *there exist x*_{0} ∈ *X and x*_{1} ∈ *TX*_{0} *suchthat α*(*x*_{0}, *x*_{1}) > 1;

(*S*_{3}) *T is a continuous multi-valued mapping*.

*Then T has a fixed point in X*.

(Theorem 2.6 in [6]). *Let* (X, d) *be a complete metric space and T* :*X* → *CL*(*X*) *be a strictly* (*α, ψ, ξ*)-*contractive mapping satisfying the following conditions*:

(*S*_{1}) *T is an a-admissible* (*or α* -admissible*) *multi-valued mapping;*

(*S*_{2}) *there exist x*_{0} *and x*_{1} ∈ *Tx*_{0} *such that α*(*x*_{0}, *x*_{1}) ≥ 1;

${{S}^{\u2033}}_{3}$ *if* {*x*_{n}} *is a sequence in X with x*_{n} → *x* ∈ *X as n* → ∞ *and α*(*x*_{n}, *x*_{n} + 1) > 1 *for all n* ∈ ℕ, *then we have α*(*x*_{n}, *x*) > 1 *for all n* ∈ ℕ.

*Then T has a fixed point in X*.

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