A NOTE ON FORA TYPE FIXED POINT THEOREMS

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A NOTE ON FORA TYPE FIXED POINT THEOREMS

Sanjay Kumar Gupta

Department of Mathematics, Rustamji Institute of Technology (India)

ABSTRACT

Singal and Lal generalized Nadler’s result to a larger class of spaces and a for larger class of mappings. Rhoades discussed a number of fixed point theorems dealing with rational type contractive conditions of some authors. Gupta generalized Nadler’s result to a larger class of spaces and a for larger class of mappings according to rational type

contractive conditions of Rhoades. In this paper we generalize Fora’s result according to some contractive conditions of Rhoades which involves single mappings.

Keywords: Complete Metric Space, Topological Space, Locally contraction in the first variable,

Continuous function, Fixed Point Property.

I. INTRODUCTION

Definition 1 [5]

Let (X, d) be a metric space and Z be any space. Let f be a mapping from X×Z into X×Z. Then f is said to be locally contraction mapping in the first variable if and only if for any zZ there exist an open set V(z)

containing z and a real number (z) [ 0,1) such that for each x, x* X and all uV(z)

d(



₁f(x, u),



₁f(x*,u) )≤ λ(z) d(x, x*), where



₁ is the projection of X×Z on X along Z.

or d(fu(x), fu(x*)) ≤ λ d(x, x*).

Theorem 2 [5]

Let (X, d) be a complete metric space. Let Z be a topological space with the fixed point property (f.p.p) and let f be a continuous function from X×Z into X×Z. If f is locally contraction mapping in the first variable,

then f has a fixed point.

In what follows X will denote a complete metric space, Z a topological space which has the f.p.p. and f is a mapping

from X×Z into X×Z. (m)' for 1≤m ≤10 will denote the condition (m)' from Rhoades [3] and also mentioned in Gupta

[4] with the modification that constants or functions that appear in (m)' depend on z. Thus f(4)' locally in the first

variable means the following:

for each z Z, there exist an open set V(z) containing z and numbers ,  0 with +  1 such that for each x, x*

 X, x* 0(x) and all u  V(z),

)

,

(

)

,

(

))

(

,

(

))

(

,

(

))

(

),

(

(

_*

*

* *

*

d

x

x

x

x

d

x

f

x

d

x

f

x

d

x

f

x

f

d

u u

u

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where 0(x) denotes the orbit of the function defined on X.

II. MAIN RESULTS

Now we prove the following results with the help of [4] and [5]:

Theorem 3

Let (X, d) be a complete metric space, Z a topological space which has the f.p.p. and let f: X×ZX×Z be a continuous mapping. If f satisfies (3)' locally in the first variable, then f has a fixed point.

Proof

: We define a sequence {tn} in X as follows:

For a fixed x0 in X, and any z Z,

1

);

),

(

(

)

(

)

(

,

)

(

0 1 1 0 0 0 0

0















n

z

x

f

f

x

f

t

z

t

x

t

x

f

zn

n z n n

z



.

Step I:

{tn} is a Cauchy sequence in X

Since f(3)' locally in the first variable i.e. for each z  Z, there exist an open set V(z) containing z and

numbers a, b, c ≥ 0 with a + b + c < 1 such that for each x, x*  X, x* 0(x) and all u  V(z),

)

,

(

.

)

,

(

))

(

,

(

)).

(

,

(

.

)

,

(

1

))

(

,

(

))].

(

,

(

1

[

))

(

),

(

(

_* * * * * * *

*

c

d

x

x

x

x

d

x

f

x

d

x

f

x

d

b

x

x

d

x

f

x

d

x

f

x

d

a

x

f

x

f

d

u u u u

u

u

_









(3.1)

Equation (3.1) implies that

)

,

(

.

)

,

(

))

(

,

(

)).

(

,

(

.

)

,

(

1

))

(

,

(

))].

(

,

(

1

[

))

(

),

(

(

_* * * * * * *

*

c

d

x

x

x

x

d

x

f

x

d

x

f

x

d

b

x

x

d

x

f

x

d

x

f

x

d

a

x

f

x

f

d

z z z z

z

z

_









(3.2)

Set x* = fz(x) in the above inequality to obtain

))

(

,

(

.

))

(

),

(

(

)

(

))

(

),

(

(

f

x

f

2

x

a

b

d

f

x

f

2

x

c

d

x

f

x

d

z z





z z



z

which implies that

))

(

,

(

.

1

))

(

),

(

(

2

d

x

f

x

b

a

c

x

f

x

f

d

_{z z}



_z

















(3.3)

Now, set x = x*, then we have

))

(

,

(

.

1

))

(

),

(

(

_* 2 _*

d

x

_*

f

x

_*

b

a

c

x

f

x

f

d

_{z z}



_z

















(3.4)

Repeating above substitute we obtain

))

(

,

(

.

1

))

(

),

(

(

2 3 2

x

f

x

d

b

a

c

x

f

x

f

d

_{z z}



_z

















714 | P a g e

))

(

,

(

.

1

))

(

),

(

(

1

d

x

f

x

b

a

c

x

f

x

f

d

z

n n

z n

z





















Finally set x = x0, we get

1

1

),

,

(

.

)

,

(

_{1 0 1}

























b

a

c

h

where

t

t

d

h

t

t

d

_{n n} n

(3.5)

Using triangle inequality, we find for m>n

h

t

t

d

n

h

h

t

t

d

n

m

h

n

h

t

t

d

m

h

n

h

n

h

m

t

m

t

d

n

t

n

t

d

n

t

n

t

d

m

t

n

t

d









































1

)

1

,

0

(

.

1

)

1

,

0

(

).

1

(

)

1

,

0

(

).

1

....

1

(

)

,

1

(

....

)

2

,

1

(

)

1

,

(

)

,

(

Since hn0 as n, this inequality shows that {tn} is a Cauchy sequence. Since X is a complete metric space,

there exists a point tz . in X such that tn tz .

Step



I:



1f(tz, z)= tz

If possible let tz ≠ 1f(tz, z)=u*(say). Then d(u*, tz)=> 0. Since f is continuous, there exists an open set

U×V in X×Z such that

(tz, z)  U×V, U  S/4(tz) and f(U×V)  S/4(u*) ×Z

Since limn∞ tn= tz, there is a natural number k 1 such that tn U for all n k. But 1f(tk, z)=tk+1U.

Therefore f(tk, z)  S/4(u*)×Z, which contradicts the fact that f(U×V)  S/4(u*)×Z and d(u*, tz)=>o, so our

assumption is false, hence the required conclusion.

Step



I



: Let g: Z



Z defined by g(z)=



2f(tz, z). Then g is a continuous mapping

Let zZ and U be an open set containing g(z). Then f(tZ, z)  X×U. Since f is continuous at (tz, z), there exists an

open set G in Z and a positive real number >0 such that

(tz, z)  S(tz) ×G and f(S(tz) ×G) X×U.

Let W be an open set in Z containing z and h [0,1) be a real number. Then from above step I making decreasing sequence procedure we can write:

d(1 f(x, v),1 f(x*,v)) ≤ h. d(x, x*), for all x, x*  X, x* 0(x) and all v W.

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hn <

8





  



_















/8

t

,

t

d

h

1

1 0

andd(tz , tm) <

8



for allm  n

Since f(tn, z)  X × U and f is continuous at (tn , z) , there exists a basic open set Un ×Vn in X × Z such that

(tn , z) Un ×Vn, Un S/8 (tz), Vn GW and f(Un × Vn)  X×U.

Inductively, suppose that sets Un, Un-1,…Ui, Vn, Vn-1,…,Vi (1 ≤ i ≤ (n-1)) are choosen such that

(ti,z)  Ui × Vi, Ui S/8 (ti),Vi  Vi+1 andf(Ui × Vi) Ui+1×Z.

Since f is continuous at (ti-1, z) and f (ti-1, z)  Ui × Z, there exists a basic open set Ui-1 × Vi-1 in X×Z such that,

(ti-1, z)  Ui-1 × Vi-1, Ui-1 S/8 (ti-1), Vi-1 Vi andf(Ui-1 × Vi-1)  Ui ×Z.

In this way, we collect the open sets Un, Un-1,…,U0, Vn, Vn-1,…,V0 which are defined with the above mentioned

properties.

Now, we claim that g(V0)  U:

Let y  V0. Then from the above mention properties we have (t /

0, y)  U0 × V0 ,where t /

0 = xo.

Thus f(t/0, y)  U1 × Z ie. , t/1 = 1f(t/0, y)  U1,and consequently d(t/1, t1) <

8



.

Using the triangular inequality, we have

d(t/0 , t/1) = d(t0 , t1/ )  d(t0 , t1) + d(t1 , t/1) < d(t0 , t1) +

8



.

Since f(U1×V1)  U2×Z and (t/1, y)  U1×V1, therefore f(t/1, y) U2×Z ie t/2= 1 f(t/1 , y)  U2.

In this way we find the sequence t/n (y)= t/n, for which t/i = 1f(t/i-1, y)  Ui; i =1,2,..., n.

Moreover, t/n Unand Un S/8(tz), therefore d(t/n, tz) <

8



.

Using the triangular inequality, we find, for m ≥ n.

d(t/n , tz)  d(tz , t/n) + d(t/n , t/n+1) +...+ d(t/m-1, t/m)

<

8



+ hn . d(t/0 , t/1) +----+hm-1 d(t/0 , t/1)

=

h

h

n



1

(1-h

m-n_{). d(t}/ 0 , t/1) +

8



< (

h

h

n



1

).[d(t

/ 0 , t/1) +

8



] +

8



<

8



+

8



=

4



If ty= lim t/n , then the above inequality shows that d(ty , tz) /4.

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Therefore our claim is proved and hence g is continuous mapping.

Now as in step III of the theorem 3, g: ZZ is continuous mapping. Since Z has the f.p.p., there is a point z0Z

such that g(z0) = z0 and by step I above, we have, 1 f(tz0, z0) = tz0. But z0 =g(z0) = 2 f(tz0, z0). Hence f(tzo, z0)=(tzo,

zo) i.e. (tz0,z0) is a fixed point of f. This completes the proof.

Theorem 4:

Let (X, d) be a complete metric space, Z a topological space with the f.p.p and f: X×Z X×Z be a continuous mapping. If f satisfies any one of the conditions (2)', (5)', (6)', (7)', (8)', (9)', and (10)' locally in the first variable then f has a fixed point.

Proof:

We define a sequence tn(z)=tn in X as follows:

For a fixed x0 in X and any z  Z,

to= x0, tn= 1f(tn-1, z); n 1

Step-I: {tn} is a Cauchy sequence in X

If f satisfies any one of the conditions (2)', (5)', (6)', (7)', (8)', (9)', and (10)' locally in the first variable then we have find following equations respectively,

)

,

(

1

)

,

(

t

t

₁

d

t

₀

t

₁

d

n

n

n



















_



(4.1)

)

,

(

)

,

(

t

t

₁

k

d

t

₀

t

₁

d

n

n

n 



(4.2)

)

,

(

1

)

,

(

_{0 1}

4 3 1

4 2 1

1

d

t

t

t

t

d

n

n

n



























_

_

_







(4.3)

)

,

(

)

,

(

t

t

₁

d

t

₀

t

₁

d

_{n n}





n (4.4)

)

,

(

)

,

(

t

t

₁

q

₁

d

t

₀

t

₁

d

_{n n}



n (4.5)

)

,

(

)

,

(

t

t

₁

q

₂

d

t

₀

t

₁

d

n

n

n 



(4.6)

and

(

,

)

1

)

,

(

_{0 1}

5 4 2 1

5 4 3 2

1

d

t

t

t

t

d

n

n

n































_

_

_

_









(4.7)

In each of these cases we see that {tn} is a Cauchy sequence in X. However, by the completeness of X, there is a

point tz in X such that tn converges to tz.

Step



I:

1 f (tz, z) = tz

Step



I:

Let g : ZZ defined by g(z) = 2 f(tz, z). Then g is a continuous mapping:

(As by theorem 3)

Finally, since Z has the f.p.p., there is a point z0 Z such that g(z0) =z0 and by step I above, we have,1

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III. CONCLUSION

We observe from [4] that conditions (1)', (4)' are stronger than (3)'.Therefore the above theorem 3 has two corollaries corresponding to each of these two conditions. Also we observe that condition (4)' is stronger than conditions (6)'and (8)'. Hence the above theorem 4 have one corollary corresponding to condition (4)'. Clearly Nadler’s result [2] is also corollary of above theorem 3 and theorem 4.

REFERENCES

[1] M.K. Singal & Sunder Lal, A note on fixed point theorems for product spaces, The MSRI – Korea Publications, Number 1, (1986), 103-107.

[2] S.B. Nadler jr., Sequence of contractions and fixed point, Pacific J. Math 27, (1968), 579-585.

[3] B.E. Rhoades, Proving fixed point theorems using general principles, Indian J. Pure App. Math.,27(8), (1996),

741-770.

[4] Sanjay Kumar Gupta, Some fixed point theorems on product spaces by the setting of Rhoades, Int. J. of Advance Tech. in Engg. & Science, Vol. No. 5, No.01, (2017), 162-169.