Posts by Mira Anisiu

"BABEŞ -- BOLYAI" UNIVERSITY
Faculty of Mathematics
Re search Seminars
Seminar on Mathematical Analysis
Preprint Nr.7, 1991, pp.95 - 100

In the last years, the interest in metric fixed point theorems appeared again. New proofs were given [1, 2] for theorems of Hardy-Rogers, Ćirić - Reich - Rus.

In this note we study the case of mappings defined only on spheres, not on the entire metric space, following the well-known theorem:

THEOREM 1. Let ( $\mathrm{X},\mathrm{d}$$\mathrm{X},\mathrm{d}$X,d\mathrm{X}, \mathrm{d}$\mathrm{X},\mathrm{d}$ ) be a complete metric space, $\mathrm{f}:\bar{\mathrm{B}}(2,\mathrm{r})\to X$$\mathrm{f}:\overline{\mathrm{B}}(2,\mathrm{r})\to X$f: bar(B)(2,r)rarr X\mathrm{f}: \overline{\mathrm{B}}(2, \mathrm{r}) \rightarrow X$\mathrm{f}:\bar{\mathrm{B}}(2,\mathrm{r})\to X$ a $k$$k$kk$k$-contraction, $k\in [0,1)$$k\in [0,1)$k in[0,1)k \in[0,1)$k\in [0,1)$, i.e.

If

then $f$$f$ff$f$ has a unique fixed point $u\in \overline{B}(z,r)$$u\in \overline{B}(z,r)$u in bar(B)(z,r)u \in \bar{B}(z, r)$u\in \overline{B}(z,r)$.
Proof. The condition on $fz$$fz$fzf z$fz$ implies the fact that $f(\overline{B}(z,r))=\overline{B}(z,r)$$f(\overline{B}(z,r))=\overline{B}(z,r)$f( bar(B)(z,r))= bar(B)(z,r)f(\bar{B}(z, r)) =\bar{B}(z, r)$f(\overline{B}(z,r))=\overline{B}(z,r)$. Indeed, let $x\in \overline{B}(z,r)$$x\in \overline{B}(z,r)$x in bar(B)(z,r)x \in \bar{B}(z, r)$x\in \overline{B}(z,r)$. We estimate $d(z,fx)$$d(z,fx)$d(z,fx)d(z, f x)$d(z,fx)$ :

It follows that $f(\overline{B}(z,x))\subseteq \overline{B}(z,r)$$f(\overline{B}(z,x))\subseteq \overline{B}(z,r)$f( bar(B)(z,x))sube bar(B)(z,r)f(\bar{B}(z, x)) \subseteq \bar{B}(z, r)$f(\overline{B}(z,x))\subseteq \overline{B}(z,r)$ and Banach theorem applies in this complete metric space.

This theorem extends easily to Hardy - Rogers contractions.
THEOREM 2. Let ( $X,d$$X,d$X,dX, d$X,d$ ) be a complete metric space, $\mathbf{f}:\overline{B}(\mathbf{z},\mathbf{r})\to X$$\mathbf{f}:\overline{B}(\mathbf{z},\mathbf{r})\to X$f: bar(B)(z,r)rarr X\mathbf{f}: \bar{B}(\mathbf{z}, \mathbf{r}) \rightarrow X$\mathbf{f}:\overline{B}(\mathbf{z},\mathbf{r})\to X$ a Hardy - Rogers contraction, i.e. there exists $a_i\ge 0$$a_i\ge 0$a_(i) >= 0a_{i} \geq 0$a_i\ge 0$, $i=\overline{1,5},A=\sum _{i=1}^5{a}_i<1$$i=\overline{1,5},A=\sum _{i=1}^5 a_i<1$i= bar(1,5),A=sum_(i=1)^(5)a_(i) < 1i=\overline{1,5}, A=\sum_{i=1}^{5} a_{i}<1$i=\overline{1,5},A=\sum _{i=1}^5{a}_i<1$ such that
(3) $d(fx,fy)\le a_{1}d(x,y)+a_{2}d(x,fx)+a_{3}d(y,fy)+a_{4}d(x_y,fy)++a_{5}d(y,fx)$$d(fx,fy)\le a_{1}d(x,y)+a_{2}d(x,fx)+a_{3}d(y,fy)+a_{4}d\left(x_y,fy\right)++a_{5}d(y,fx)$d(fx,fy) <= a_(1)d(x,y)+a_(2)d(x,fx)+a_(3)d(y,fy)+a_(4)d(x_(y),fy)++a_(5)d(y,fx)d(f x, f y) \leq a_{1} d(x, y)+a_{2} d(x, f x)+a_{3} d(y, f y)+a_{4} d\left(x_{y}, f y\right)+ +a_{5} d(y, f x)$d(fx,fy)\le a_{1}d(x,y)+a_{2}d(x,fx)+a_{3}d(y,fy)+a_{4}d(x_y,fy)++a_{5}d(y,fx)$ for all $x,y$$x,y$x,yx, y$x,y$ in $\overline{B}(z,r)$$\overline{B}(z,r)$bar(B)(z,r)\bar{B}(z, r)$\overline{B}(z,r)$.
If
(4) $d(z,fz)\le \frac{2(1-A)}{2+A-a_1}r$$d(z,fz)\le \frac{2(1-A)}{2+A-a_1}r$d(z,fz) <= (2(1-A))/(2+A-a_(1))rd(z, f z) \leq \frac{2(1-A)}{2+A-a_{1}} r$d(z,fz)\le \frac{2(1-A)}{2+A-a_1}r$,
then $f$$f$ff$f$ has a unique fixed point $u\in \overline{B}(2,r)$$u\in \overline{B}(2,r)$u in bar(B)(2,r)u \in \bar{B}(2, r)$u\in \overline{B}(2,r)$.
Proof. To prove that $f(\overline{B}(z,r))\subset \overline{B}(z,r)$$f(\overline{B}(z,r))\subset \overline{B}(z,r)$f( bar(B)(z,r))sub bar(B)(z,r)f(\bar{B}(z, r)) \subset \bar{B}(z, r)$f(\overline{B}(z,r))\subset \overline{B}(z,r)$, we use a symmetric form of (3) which is obtained evaluating also $d(fy,fx)$$d(fy,fx)$d(fy,fx)d(f y, f x)$d(fy,fx)$ and adding,
(5) $d(fx,fy)\le a_{1}d(xy)+\frac{a_2+a_3}{2}[d(x,fx)+d(y,fy)]+$$d(fx,fy)\le a_{1}d(xy)+\frac{a_2+a_3}{2}[d(x,fx)+d(y,fy)]+$d(fx,fy) <= a_(1)d(xy)+(a_(2)+a_(3))/(2)[d(x,fx)+d(y,fy)]+d(f x, f y) \leq a_{1} d(x y)+\frac{a_{2}+a_{3}}{2}[d(x, f x)+d(y, f y)]+$d(fx,fy)\le a_{1}d(xy)+\frac{a_2+a_3}{2}[d(x,fx)+d(y,fy)]+$

$\overline{B}(z,I)$$\overline{B}(z,I)$bar(B)(z,I)\bar{B}(z, I)$\overline{B}(z,I)$.
Let $x$$x$xx$x$ be in $\overline{B}(z,r)$$\overline{B}(z,r)$bar(B)(z,r)\bar{B}(z, r)$\overline{B}(z,r)$, hence $d(x,z)\le r$$d(x,z)\le r$d(x,z) <= rd(x, z) \leq r$d(x,z)\le r$. We estimate

It follows
$(1-\frac{A-a_1}{2})d(z,fx)\le (1+\frac{A-a_1}{2})d(z,fz)+\frac{A+a_1}{2}d(x,z)\le$$\left(1-\frac{A-a_1}{2}\right)d(z,fx)\le \left(1+\frac{A-a_1}{2}\right)d(z,fz)+\frac{A+a_1}{2}d(x,z)\le$(1-(A-a_(1))/(2))d(z,fx) <= (1+(A-a_(1))/(2))d(z,fz)+(A+a_(1))/(2)d(x,z) <=\left(1-\frac{A-a_{1}}{2}\right) d(z, f x) \leq\left(1+\frac{A-a_{1}}{2}\right) d(z, f z)+\frac{A+a_{1}}{2} d(x, z) \leq$(1-\frac{A-a_1}{2})d(z,fx)\le (1+\frac{A-a_1}{2})d(z,fz)+\frac{A+a_1}{2}d(x,z)\le$
$\le (1-A)r+\frac{A+a_1}{2}r=(1-\frac{A-a_1}{2})r$$\le (1-A)r+\frac{A+a_1}{2}r=\left(1-\frac{A-a_1}{2}\right)r$<= (1-A)r+(A+a_(1))/(2)r=(1-(A-a_(1))/(2))r\leq(1-A) r+\frac{A+a_{1}}{2} r=\left(1-\frac{A-a_{1}}{2}\right) r$\le (1-A)r+\frac{A+a_1}{2}r=(1-\frac{A-a_1}{2})r$.
Dividing by $1-\frac{A-a_1}{2}>0$$1-\frac{A-a_1}{2}>0$1-(A-a_(1))/(2) > 01-\frac{A-a_{1}}{2}>0$1-\frac{A-a_1}{2}>0$ we obtain $d(z,fx)\le r$$d(z,fx)\le r$d(z,fx) <= rd(z, f x) \leq r$d(z,fx)\le r$ and Hardy

Ramark 1. For the Ciric - Reich - Rus contractions ( $a_1=a_1,a_2^{\prime }=a_3^{\prime }=b_1,a_4=a_5=0,a+2b<1$$a_1=a_1,a_2^{\prime }=a_3^{\prime }=b_1,a_4=a_5=0,a+2b<1$a_(1)=a_(1),a_(2)^(')=a_(3)^(')=b_(1),a_(4)=a_(5)=0,a+2b < 1a_{1}=a_{1}, a_{2}^{\prime}=a_{3}^{\prime}=b_{1}, a_{4}=a_{5}=0, a+2 b<1$a_1=a_1,a_2^{\prime }=a_3^{\prime }=b_1,a_4=a_5=0,a+2b<1$ ), condition (4) becomes:

For the Kannan contractions $(a_1=a_4=a_5=0,a_2=a_3=b<<\frac{1}{2}$$\left(a_1=a_4=a_5=0,a_2=a_3=b<\right.<\frac{1}{2}$(a_(1)=a_(4)=a_(5)=0,a_(2)=a_(3)=b < :} < (1)/(2)\left(a_{1}=a_{4}=a_{5}=0, a_{2}=a_{3}=b<\right. <\frac{1}{2}$(a_1=a_4=a_5=0,a_2=a_3=b<<\frac{1}{2}$ ), condition (4) writes

For the Banach contractions ( $a_1=k,a_i=0,i=\overline{2,5}$$a_1=k,a_i=0,i=\overline{2,5}$a_(1)=k,a_(i)=0,i= bar(2,5)a_{1}=k, a_{i}=0, i=\overline{2,5}$a_1=k,a_i=0,i=\overline{2,5}$ ),
condition (4) becomes (2).
Similar results hold for the more general contractions defined by Rus [3].

Let $:{\mathbf{R}}_{+}^5\to {\mathbf{R}}_{+}$$:{\mathbf{R}}_{+}^5\to {\mathbf{R}}_{+}$:R_(+)^(5)rarrR_(+): \mathbf{R}_{+}^{5} \rightarrow \mathbf{R}_{+}$:{\mathbf{R}}_{+}^5\to {\mathbf{R}}_{+}$be a continuous function such that
(a) if $r_i\le s_i,i=\overline{1,5}$$r_i\le s_i,i=\overline{1,5}$r_(i) <= s_(i),i= bar(1,5)r_{i} \leq s_{i}, i=\overline{1,5}$r_i\le s_i,i=\overline{1,5}$, then $\phi (r)\le \phi (s)$$\phi (r)\le \phi (s)$varphi(r) <= varphi(s)\varphi(r) \leq \varphi(s)$\phi (r)\le \phi (s)$;
(b) $\varphi (r)<r$$\varphi (r)<r$phi(r) < r\phi(r)<r$\varphi (r)<r$ for each $r>0$$r>0$r > 0r>0$r>0$, where $\varphi (r)=\phi (r,r,r,r,r)$$\varphi (r)=\phi (r,r,r,r,r)$phi(r)=varphi(r,r,r,r,r)\phi(r)=\varphi(r, r, r, r, r)$\varphi (r)=\phi (r,r,r,r,r)$;
(c) $r-\varphi (r)\to \mathrm{\infty }$$r-\varphi (r)\to \mathrm{\infty }$r-phi(r)rarr oor-\phi(r) \rightarrow \infty$r-\varphi (r)\to \mathrm{\infty }$ for $r\to \mathrm{\infty }$$r\to \mathrm{\infty }$r rarr oor \rightarrow \infty$r\to \mathrm{\infty }$.

For ( $X,d$$X,d$X,dX, d$X,d$ ) a metric space, a function $f:\overline{B}(z,r)\to X$$f:\overline{B}(z,r)\to X$f: bar(B)(z,r)rarr Xf: \bar{B}(z, r) \rightarrow X$f:\overline{B}(z,r)\to X$ is a $p$$p$pp$p$ -
contraction if
(6) $d(fx,fy)\le \phi (d(x,y)),d(x,fx),d(y,fy),d(x,fy),d(y,fx))$$d(fx,fy)\le \phi (d(x,y)),d(x,fx),d(y,fy),d(x,fy),d(y,fx))$d(fx,fy) <= varphi(d(x,y)),d(x,fx),d(y,fy),d(x,fy),d(y,fx))d(f x, f y) \leq \varphi(d(x, y)), d(x, f x), d(y, f y), d(x, f y), d(y, f x))$d(fx,fy)\le \phi (d(x,y)),d(x,fx),d(y,fy),d(x,fy),d(y,fx))$ for all $x,y\in \overline{B}(z,r)$$x,y\in \overline{B}(z,r)$x,y in bar(B)(z,r)x, y \in \bar{B}(z, r)$x,y\in \overline{B}(z,r)$.

Let $r_z$$r_z$r_(z)r_{z}$r_z$ be equal to sup $\{r\in R_{+}:r-\varphi (r)\le d(z,fz)\}$$\left\{r\in R_{+}:r-\varphi (r)\le d(z,fz)\right\}${r inR_(+):r-phi(r) <= d(z,fz)}\left\{r \in R_{+}: r-\phi(r) \leq d(z, f z)\right\}$\{r\in R_{+}:r-\varphi (r)\le d(z,fz)\}$ which is finite because $r=0$$r=0$r=0r=0$r=0$ is in the considered set (which is hence nonvoid) and $\phi$$\phi$varphi\varphi$\phi$ is continuous. We obtain a generalization of the fixed point theorem of Rus [3] which is given for. contractions defined on the entire space $X$$X$XX$X$.

THEOREM 3. Let $(X,d)$$(X,d)$(X,d)(X, d)$(X,d)$ be a complete metric space and $f$$f$ff$f$ : $\bar{\mathrm{B}}(\mathrm{Z},\mathrm{r})\to \mathrm{X}$$\overline{\mathrm{B}}(\mathrm{Z},\mathrm{r})\to \mathrm{X}$bar(B)(Z,r)rarrX\overline{\mathrm{B}}(\mathrm{Z}, \mathrm{r}) \rightarrow \mathrm{X}$\bar{\mathrm{B}}(\mathrm{Z},\mathrm{r})\to \mathrm{X}$ be a - contraction. If
(7) $d(z,fz)\le r-\varphi \left(r_z\right)$$d(z,fz)\le r-\varphi \left(r_z\right)$quad d(z,fz) <= r-phi(r_(z))\quad d(z, f z) \leq r-\phi\left(r_{z}\right)$d(z,fz)\le r-\varphi \left(r_z\right)$,
then $f$$f$ff$f$ has a unique fixed point $u\in \overline{B}(z,x),u=\underset{n\to \mathrm{\infty }}{lim}{f}^{n}z$$u\in \overline{B}(z,x),u=\underset{n\to \mathrm{\infty }}{lim} f^{n}z$u in bar(B)(z,x),u=lim_(n rarr oo)f^(n)zu \in \bar{B}(z, x), u=\lim _{n \rightarrow \infty} f^{n} z$u\in \overline{B}(z,x),u=\underset{n\to \mathrm{\infty }}{lim}{f}^{n}z$ and $d(f^{n}z,x)\le {\varphi }^n\left(r_z\right)$$d\left(f^{n}z,x\right)\le {\varphi }^n\left(r_z\right)$d(f^(n)z,x) <= phi^(n)(r_(z))d\left(f^{n} z, x\right) \leq \phi^{n}\left(r_{z}\right)$d(f^{n}z,x)\le {\varphi }^n\left(r_z\right)$.

Proof. If $fz=z$$fz=z$fz=zf z=z$fz=z$, the conclusion is obvious.
Let $fz\ne z$$fz\ne z$fz!=zf z \neq z$fz\ne z$; one has then $r_z>0$$r_z>0$r_(z) > 0r_{z}>0$r_z>0$. Let $z_0=z;z_1=fz$$z_0=z;z_1=fz$z_(0)=z;z_(1)=fzz_{0}=z ; z_{1}=f z$z_0=z;z_1=fz$ satisfies

hence $z_1\in \overline{B}(z,r)$$z_1\in \overline{B}(z,r)$z_(1)in bar(B)(z,r)z_{1} \in \bar{B}(z, r)$z_1\in \overline{B}(z,r)$. Suppose $z_n=f^{n}z\in \overline{B}(z,r)$$z_n=f^{n}z\in \overline{B}(z,r)$z_(n)=f^(n)z in bar(B)(z,r)z_{n}=f^{n} z \in \bar{B}(z, r)$z_n=f^{n}z\in \overline{B}(z,r)$. Then

$\le d(z,fz)+\phi (d(z,f^{n}z),d(z,fz),d(f^{n}z,f^{n+1}z),\overline{d}(z,f^{n+1}z),d(f^{n}z,fz))\le \le d(z,fz)+\varphi (\mathrm{diam}{O}_f(z,n+1)),O_f(z,n+1)=\{z,fz,\dots ,f^{n+1}z\}$$\le d(z,fz)+\phi \left(d\left(z,f^{n}z\right),d(z,fz),d\left(f^{n}z,f^{n+1}z\right),\overline{d}\left(z,f^{n+1}z\right),d\left(f^{n}z,fz\right)\right)\le \le d(z,fz)+\varphi \left(\mathrm{diam}{O}_f(z,n+1)\right),O_f(z,n+1)=\left\{z,fz,\dots ,f^{n+1}z\right\}$<= d(z,fz)+varphi(d(z,f^(n)z),d(z,fz),d(f^(n)z,f^(n+1)z),( bar(d))(z,f^(n+1)z),d(f^(n)z,fz))≤≤d(z,fz)+phi(diamO_(f)(z,n+1)),O_(f)(z,n+1)={z,fz,dots,f^(n+1)z}\leq d(z, f z)+\varphi\left(d\left(z, f^{n} z\right), d(z, f z), d\left(f^{n} z, f^{n+1} z\right), \bar{d}\left(z, f^{n+1} z\right), d\left(f^{n} z, f z\right)\right) \leq \leq d(z, f z)+\phi\left(\operatorname{diam} O_{f}(z, n+1)\right), O_{f}(z, n+1)=\left\{z, f z, \ldots, f^{n+1} z\right\}$\le d(z,fz)+\phi (d(z,f^{n}z),d(z,fz),d(f^{n}z,f^{n+1}z),\overline{d}(z,f^{n+1}z),d(f^{n}z,fz))\le \le d(z,fz)+\varphi (\mathrm{diam}{O}_f(z,n+1)),O_f(z,n+1)=\{z,fz,\dots ,f^{n+1}z\}$.

Because of (6), diam $o_f(z,n+1)$$o_f(z,n+1)$o_(f)(z,n+1)o_{f}(z, n+1)$o_f(z,n+1)$ is larger than $d(f^{i}z,f^{j}z)$$d\left(f^{i}z,f^{j}z\right)$d(f^(i)z,f^(j)z)d\left(f^{i} z, f^{j} z\right)$d(f^{i}z,f^{j}z)$, $i,j\ge 1$$i,j\ge 1$i,j >= 1i, j \geq 1$i,j\ge 1$, so there exists $p\le n+1$$p\le n+1$p <= n+1p \leq n+1$p\le n+1$ such that diam $O_f(z,n+1)==d(z,f^{P}z)$$O_f(z,n+1)==d\left(z,f^{P}z\right)$O_(f)(z,n+1)==d(z,f^(P)z){O_{f}}(z, n+1)= =d\left(z, f^{P} z\right)$O_f(z,n+1)==d(z,f^{P}z)$.

Then
$\mathrm{diam}{O}_f(z,n+1)=d(z,f^{P}z)\le d(z,fz)+d(fz,f^{P}z)\le$$\mathrm{diam}{O}_f(z,n+1)=d\left(z,f^{P}z\right)\le d(z,fz)+d\left(fz,f^{P}z\right)\le$diamO_(f)(z,n+1)=d(z,f^(P)z) <= d(z,fz)+d(fz,f^(P)z) <=\operatorname{diam} O_{f}(z, n+1)=d\left(z, f^{P} z\right) \leq d(z, f z)+d\left(f z, f^{P} z\right) \leq$\mathrm{diam}{O}_f(z,n+1)=d(z,f^{P}z)\le d(z,fz)+d(fz,f^{P}z)\le$
$\le d(z,fz)+\varphi (\mathrm{dian}{\rho }_f(z,n+1))$$\le d(z,fz)+\varphi \left(\mathrm{dian}{\rho }_f(z,n+1)\right)$<= d(z,fz)+phi(dianrho_(f)(z,n+1))quad\leq d(z, f z)+\phi\left(\operatorname{dian} \rho_{f}(z, n+1)\right) \quad$\le d(z,fz)+\varphi (\mathrm{dian}{\rho }_f(z,n+1))$ and
$\mathrm{diam}{O}_f(z,n+1)-\varphi (\mathrm{diam}{O}_f(z,n+1))\le d(z,fz)$$\mathrm{diam}{O}_f(z,n+1)-\varphi \left(\mathrm{diam}{O}_f(z,n+1)\right)\le d(z,fz)$diamO_(f)(z,n+1)-phi(diamO_(f)(z,n+1)) <= d(z,fz)\operatorname{diam} O_{f}(z, n+1)-\phi\left(\operatorname{diam} O_{f}(z, n+1)\right) \leq d(z, f z)$\mathrm{diam}{O}_f(z,n+1)-\varphi (\mathrm{diam}{O}_f(z,n+1))\le d(z,fz)$,
hence diam ${\rho }_f(z,n+1)\le r_z$${\rho }_f(z,n+1)\le r_z$rho_(f)(z,n+1) <= r_(z)\rho_{f}(z, n+1) \leq r_{z}${\rho }_f(z,n+1)\le r_z$.
It follows $d(z,z_{n+1})\le d(z,fz)+\varphi \left(x_z\right)\le r$$d\left(z,z_{n+1}\right)\le d(z,fz)+\varphi \left(x_z\right)\le r$d(z,z_(n+1)) <= d(z,fz)+phi(x_(z)) <= rd\left(z, z_{n+1}\right) \leq d(z, f z)+\phi\left(x_{z}\right) \leq r$d(z,z_{n+1})\le d(z,fz)+\varphi \left(x_z\right)\le r$.
The sequence ${\left\{z_n\right\}}_{n\in N}$${\left\{z_n\right\}}_{n\in N}${z_(n)}_(n in N)\left\{z_{n}\right\}_{n \in N}${\left\{z_n\right\}}_{n\in N}$ is well-defined for $f:\overline{B}(z,r)\to X$$f:\overline{B}(z,r)\to X$f: bar(B)(z,r)rarr Xf: \bar{B}(z, r) \rightarrow X$f:\overline{B}(z,r)\to X$.
We shall prove now that this sequence converges. Let $n\ge 1$$n\ge 1$n >= 1n \geq 1$n\ge 1$.

We estimate:

But $r_z>0$$r_z>0$r_(z) > 0r_{z}>0$r_z>0$, so ${\varphi }^n\left(r_z\right)\to 0(n\to \mathrm{\infty })$${\varphi }^n\left(r_z\right)\to 0(n\to \mathrm{\infty })$phi^(n)(r_(z))rarr0(n rarr oo)\phi^{n}\left(r_{z}\right) \rightarrow 0(n \rightarrow \infty)${\varphi }^n\left(r_z\right)\to 0(n\to \mathrm{\infty })$; indeed, the sequence is descreasing and bounded and if its limit were a * 0 , from the continuity of $\varphi$$\varphi$phi\phi$\varphi$ it would follow $a=\varphi (a)$$a=\varphi (a)$a=phi(a)a=\phi(a)$a=\varphi (a)$, contradicting the property (b).

It follows that $\left\{z_n\right\}⊏\overline{B}(z,r)$$\left\{z_n\right\}⊏\overline{B}(z,r)${z_(n)}⊏ bar(B)(z,r)\left\{z_{n}\right\} \sqsubset \bar{B}(z, r)$\left\{z_n\right\}⊏\overline{B}(z,r)$ is a cauchy sequence, hence it converges to a limit $u$$u$uu$u$ in $\overline{B}(z,r)$$\overline{B}(z,r)$bar(B)(z,r)\bar{B}(z, r)$\overline{B}(z,r)$, which is a fixed point for f. Indeed, suppose $u\ast$$u\ast$u**u *$u\ast$ fu. Then, for $n\in N$$n\in N$n in Nn \in N$n\in N$
$d(u,fu)\le d(u,f^{n}z)+d(f^{n}z,fu)\le$$d(u,fu)\le d\left(u,f^{n}z\right)+d\left(f^{n}z,fu\right)\le$d(u,fu) <= d(u,f^(n)z)+d(f^(n)z,fu) <=d(u, f u) \leq d\left(u, f^{n} z\right)+d\left(f^{n} z, f u\right) \leq$d(u,fu)\le d(u,f^{n}z)+d(f^{n}z,fu)\le$
$\le d(u,f^{n}z)+\phi (d(f^{n-1}z,u),d(f^{n-1}z,f^{n}z),d(u,fu),d(f^{n-1}z,fu)$$\le d\left(u,f^{n}z\right)+\phi \left(d\left(f^{n-1}z,u\right),d\left(f^{n-1}z,f^{n}z\right),d(u,fu),d\left(f^{n-1}z,fu\right)\right.$<= d(u,f^(n)z)+varphi(d(f^(n-1)z,u),d(f^(n-1)z,f^(n)z),d(u,fu),d(f^(n-1)z,fu):}\leq d\left(u, f^{n} z\right)+\varphi\left(d\left(f^{n-1} z, u\right), d\left(f^{n-1} z, f^{n} z\right), d(u, f u), d\left(f^{n-1} z, f u\right)\right.$\le d(u,f^{n}z)+\phi (d(f^{n-1}z,u),d(f^{n-1}z,f^{n}z),d(u,fu),d(f^{n-1}z,fu)$, $d(u,f^{n}z)\}$$\left.d\left(u,f^{n}z\right)\right\}${:d(u,f^(n)z)}\left.d\left(u, f^{n} z\right)\right\}$d(u,f^{n}z)\}$
and for $n\to \mathrm{\infty }$$n\to \mathrm{\infty }$n rarr oon \rightarrow \infty$n\to \mathrm{\infty }$
$d(u,fu)\le \phi (0,0,d(u,fu),d(u,fu),0)\le \varphi (d(u,fu))<d(u,fu)$$d(u,fu)\le \phi (0,0,d(u,fu),d(u,fu),0)\le \varphi (d(u,fu))<d(u,fu)$d(u,fu) <= varphi(0,0,d(u,fu),d(u,fu),0) <= phi(d(u,fu)) < d(u,fu)d(u, f u) \leq \varphi(0,0, d(u, f u), d(u, f u), 0) \leq \phi(d(u, f u))<d(u, f u)$d(u,fu)\le \phi (0,0,d(u,fu),d(u,fu),0)\le \varphi (d(u,fu))<d(u,fu)$,
which im a contradiction.
The uniquaness of the fixed point can be aasily established
in a similar way.
mamark 2. Condition (3) in Theorem 2 corresponds to a $\phi$$\phi$varphi\varphi$\phi$ contraction with