American Journal of Mathematics and Statistics
p-ISSN: 2162-948X e-ISSN: 2162-8475
2012; 2(6): 213-216
doi: 10.5923/j.ajms.20120206.08
Amalendu Choudhury 1, T. Som 2
1Department of Mathematics and Statistics Haflong Govt. College, Haflong, Dima Hasao Assam. Pin, 788819, India
2Department of Applied Mathematics Indian Institute of Technology (BHU), Varanasi, 221005, India
Correspondence to: Amalendu Choudhury , Department of Mathematics and Statistics Haflong Govt. College, Haflong, Dima Hasao Assam. Pin, 788819, India.
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The present paper deals with few fixed point results for mappings satisfying some generalized contractive type inequality condition in 2-metric spaces, which generalize the results of Rhoades (1979) and Gahler (1963), Iseki (1975), Iseki et.al (1976) in turn. The inqualities involve a rational type of terms given by
,
, [ρ(x,fjmj(y),a) + ρ(y,fimi(x),a)] /2 etc under max composition.
Keywords: 2- Metric Space, Self Mapping, Generalized Contraction, Fixed Point
Cite this paper: Amalendu Choudhury , T. Som , "Some Fixed Point Results in 2-Metric Spaces", American Journal of Mathematics and Statistics, Vol. 2 No. 6, 2012, pp. 213-216. doi: 10.5923/j.ajms.20120206.08.
], [
] , [
] , [ρ(x,fjmj(y),a) + ρ(y,fimi(x),a)]/2 etc in the inequality condition under max composition, which is more general than the inequality condition used by previous authors.
], [
] } (i) Then f possesses a unique fixed point z.Proof : Let x0 ∈ X and define {xn} by xn+1 = f(xn), n = 0,1,2.. . From (i) we have ,ρ(xn+1, xm+1,a) = ρ(f(xn), f(xm),a) ≤ h max { ρ(xn,xm,a), ρ(xn,xn+1,a), ρ(xm,xm+1,a), [
], [
]}≤ h ρ(xn,xm,a)It cannot be that, ρ(xn,xm+1,a)/2 ≥ ρ(xn, xm,a) , ρ(xn+1, xm+1, a)Again ,ρ(xn,xm+1,a) ≤ ρ(xn,xm+1,xn+1) + ρ(xn,xn+1,a) + ρ(xn+1,xm+1,a)and we have , ρ(xn,xm+1,xn+1) = 0. Thus, ρ(xn+1,xm+1,a) ≤ h ρ(xn, xm, a)Similarly, ρ(xn,xm,a) ≤ h ρ(xn-1, xm-1,a)≤ h2 ρ(xn-2, xm-2,a)…………… ≤ hn ρ(x0, xk,a)Therefore for integers n, m where n > m ≥ 0, ρ(xn, xm, a) ≤ hn ρ(x0, xk,a) (ii)where k is a suitable integer satisfying 0 < k ≤ m. using property 4 of definition 1 and (ii),we get ρ(x0, xk,a) ≤ ρ(x0, xk,x1) + ρ(x0, x1,a) + ρ(x1, xk, a) ≤ ρ(x0, xk, x1) + ρ(x0, x1, a) + h ρ(x0, xk, a) ≤ ……………… ≤
ρ(x0, x1, a)Therefore, ρ(xn, xm, a) ≤
ρ(x0, x1, a)So it can be easily shown that ρ(x0, xk, x1) = 0 , (see[7]) where k/ is a suitable integer satisfying 0≤ k/≤ k. Therefore {xn} is Cauchy sequence, hence convergent. Let us consider the limit z, using (i), for any a ε X.ρ( xn+1, f(z), a) = ρ ( f(z),a, xn+1) = ρ( f(z), xn+1, a) ≤ h max{ ρ( z, xn, a), ρ(z, f(z), a), ρ(xn, xn+1, a), [
], [
] }Taking the limit of both sides as n → ∞, we have, ρ(z, f(z),a) ≤ h ρ(z, f(z),a)which implies z = f(z). Suppose z, w are fixed points of f. Then from (i), each a ε X, we haveρ(z, w, a) ≤ h ρ(z, w, a). Since 0 ≤ h < 1, and using ρ(a, b, c) ≠ 0.So we get, z = w. Therefore f has a unique fixed point z. Theorem 4.2 Let X be a complete 2- metric space, {fn} a sequence of mappings of X into X with fixed points zn, and f a mappings of X into X satisfying ρ(f(x),f(y),a) ≤ h max {ρ(x,y,a), ρ(x,f(x),a), ρ(y,f(y),a), ρ(x,f(y),a), ρ(y,f(x),a)} (iii) with fixed point z, such that fn → f uniformly on { zn : n = 1,2, ….}. Then zn → z.Proof: Let ε > 0. From the uniform convergence of {fn} on {zn : n = 1,2,..} there exists an integer N such that for all n ≥ N, ρ( f(zn), fn(zn), a) <
, for all zn , where M =
Now, ρ(zn, z, a) = ρ( fn(zn), f(z),a) ≤ ρ(fn(zn),f(z),f(zn) +ρ(fn(zn), f(zn),a) + ρ(f(zn),f(z),a) (iv)Again from (iii), ρ(f(zn),f(z),a) ≤ h max{ρ(zn,z,a),ρ(zn,f(zn),a),ρ(z,f(z),a),ρ(zn,f(z),a)ρ(z,f(zn),a)}≤ h max{ρ(zn, z, a), ρ(zn, f(zn),a)}so thatρ(fn(zn), f(z), f(zn)) = ρ(f(z), f(zn),zn) ≤ h max {ρ(zn, z, zn), ρ(zn, f(zn),zn)} = 0.Now (iv) becomes ρ(zn, z, a) ≤ ρ(fn(zn),f(zn),a) + h max{ρ( zn, z, a), ρ(zn, f(zn),a)}which implies ρ( zn, z, a) ≤
< ε. Thus zn → z.Our next result generalizes theorem 1.2 of Rhoades[7].Theorem 4.3 Let f and g be mappings of a complete 2–metric space X into itself satisfyingρ(f(x),g(y),a) ≤ h max {ρ(x,y,a), ρ(x,f(x),a), ρ(y,g(y),a), ρ(y,f(x),a), ρ(x,g(y),a),[
]} (v)for all x,y ε X, h a fixed constant satisfying 0 ≤ h <1 . Then f and g have a common fixed point z and (fg)n(x0) → z and (gf)n(x0) → z for each x0 ε X.Proof: Let x0 ε X and we define {xn} by x2n+1 = f(x2n) and x2n+2 = g(x2n+1)From (vii), we get ρ(x2n+1, x2n+2, a) = ρ(f(x2n), g(x2n+1), a) ≤ h max {ρ(x2n,x2n+1,a), ρ(x2n,x2n+1,a), ρ(x2n+1,x2n+2,a), ρ(x2n+1,x2n+1,a), ρ(x2n,x2n+2,a), [
] } ≤ h max { ρ(x2n,x2n+1,a), ρ(x2n, x2n+2,a)}Again , ρ(x2n,x2n+2,a) ≤ ρ(x2n,x2n+2,x2n+1) + ρ(x2n,x2n+1,a) + ρ(x2n+1,x2n+2,a)and we have ,ρ(x2n,x2n+2,x2n+1) = 0. Thus, ρ(x2n+1,x2n+2,a) ≤ h ρ(x2n, x2n+1, a)Similarly, ρ(x2n,x2n+1,a) ≤ h ρ (x2n-1, x2n,a)For arbitrary n, we have ρ(xn, xn+1, a) ≤ hn ρ(x0, x1, a) (vi) For any m > n and using property 4 of definition 1 and (vi)ρ(xm, xn,a) ≤
= hn(1-h)-1.[ρ(x0, x1, xm) + ρ(x0, x1, a)]we can easily shown that, ρ(x0, x1, xm) = 0. ([7] )So that {xn} is a Cauchy sequence and hence convergent.Let us consider the limit z.Now, ρ(f(z), z, a) ≤ ρ(f(z), z, x2n+2) + ρ(f(z), x2n+2, a) + ρ(x2n+2, z, a) (vii)From (vii), we get ρ(f(z), x2n+2, a) = ρ(f(z), g(x2n+1), a) ≤ h max{ ρ(z, x2n+1, a), ρ(z, f(z), a), ρ(x2n+1, x2n+2, a), ρ(x2n+1,f(z),a), ρ(z,x2n+2,a), [
]} (viii)Substituting (vii) into (viii) and taking limit as n → ∞, we have,ρ(f(z), z, a) ≤ h ρ(z, f(z), a) as 0 ≤ h < 1 , we get z = f(z).Therefore, z is a fixed point of f. Similarly, we can show that z is also a fixed point of g. For uniqueness, suppose z and w are common fixed points of f and g.Now from (v) ρ(z, w, a) = ρ(f(z), g(w), a) ≤ h max {ρ(z, w, a), ρ(z,f(z),a), ρ(w,g(w),a),ρ(z,g(w),a),ρ(w,f(z),a),[
} ≤ h max { ρ(z,w,a), 0 ,0 ,ρ(z,w,a), ρ(w,z,a),ρ(z,w,a)}or, ρ(z,w,a) ≤ h ρ(z,w,a), which implies z = w. Thus z is a unique common fixed point of f and g . An extension of theorem 1.3 of Rhoades[7] goes as follows:Theorem 4.4 Let X be a complete 2–metric space , {fn}, n = 1,2,…. a sequence of mapping fn: X → X, suppose there exists a sequence of non negative integers {mn} and a number h, 0 ≤ h < 1 such that, for all x,y ε X and every pair i ,j , i ≠ j and satisfyingρ(fimi(x),fjmj(y),a) ≤ h max {ρ(x,y,a), ρ(x,fimi(x),a),ρ(y,fjmj(y),a), ρ(y,fimi(x),a), ρ(x,fjmj(y), [ρ(x,fjmj(y),a) + ρ(y,fimi(x),a)] /2 } (ix) Then the mappings {fn} have a unique common fixed point.Proof : Taking gi = fimi, i = 1,2,3,…. in (ix), we have ρ(gi(x), gj(y),a) ≤ h max {ρ(x,y,a), ρ(x,gi(x),a), ρ(y,gj(y),a), ρ(y,gi(x),a), ρ(x,gj(y),a), [
} (x)Let us consider x0 ε X and we define xn = gn(xn-1), n = 1,2,…Now from (x), we get ρ(xn, xn+1,a) = ρ( gn(xn-1), gn+1(xn),a)≤ h max { ρ(xn-1,xn,a), ρ(xn-1,xn,a), ρ(xn,xn+1,a),ρ(xn,xn,a), ρ(xn-1,xn+1,a), [
]}≤ h max { ρ(xn-1,xn,a), ρ(xn-1,xn,a), ρ(xn,xn+1,a),0, ρ(xn-1,xn+1,a),
}as in the proof of Theorem 2.3, we led to the conclusion that ρ(x2n,x2n+1,a) ≤ h ρ(x2n-1,x2n,a)and in general ρ(xn,xn+1,a) ≤ hn ρ(x0,x1,a).Therefore, {xn} is Cauchy sequence and converges to a limit z. Now we get from (x), ρ(gn(x),gm+1(xm),a) ≤ h max { ρ(z,xm,a), ρ(z,gn(z),a), ρ(xm,xm+1,a), ρ(xm,gn(z),a), ρ(z,xm+1,a),
}Taking limit as m → ∞, we obtain, ρ(gn(z),z,a) ≤ h ρ(gn(z),z,a) = h ρ(z,gn(z),a) which implies that gn(z) = z, as 0 ≤ h < 1.For each n, we have fn(z) = fn(gn(z)) = fn(fnmnz), which shows that fn(z) is a fixed point of gn. Uniqueness of this theorem follows easily from (ix) and by uniqueness, we have fn(z) = z .Hence z is a unique common fixed point of fn. Theorem 4.5 Let T and S be two self mappings of a complete metric space (X,d) satisfying ad(Tx,Sy,u) + bd(x,Tx,u) + cd(y,Sy,u) ≤ q max{d(x,y,u), d(x,Tx,u), d(y,Sy,u)} (xi) for x, y ε X and u ε X; a + c > q and a > q . Then T and S have a unique common fixed point.Proof: Let x0 ε X and we define {xn} by x2n+1 = Tx2n and x2n+2 = Sx2n+1. Now putting x = x2n and y = x2n+1 in (xi) we havead(Tx2n, Sx2n+1, u) + bd(x2n, Tx2n, u) + cd(x2n+1, Sx2n+1, u) ≤ q max{d(x2n, x2n+1, u), d(x2n, Tx2n, u), d(x2n+1,Sx2n+1,u)}or, ad(x2n+1, x2n+2, u) + bd(x2n, x2n+1, u) + cd(x2n+1, x2n+2, u) ≤ q max{d(x2n, x2n+1, u), d(x2n, x2n+1, u), d(x2n+1, x2n+2,u)}or, (a+c) d(x2n+1, x2n+2, u) + bd(x2n, x2n+1, u) ≤ qd(x2n, x2n+1, u)or, d(x2n+1, x2n+2, u) ≤
d(x2n, x2n+1, u) = h d(x2n, x2n+1, u), where h =
Similarly, d(x2n, x2n+1, u) ≤ h d(x2n-1, x2n, u) Therefore for any arbitrary nd(xn, xn+1, u) ≤ hn d(x0, x1, u) (xii)From (xii) using the property (4) of definition 1, we get, for any m > nd(xm, xn,u) ≤
≤ hn (1-h)-1[d(x0, x1, xm) + d(x0, x1, u)]We can easily shown that d(x0, x1, xm) = 0 ( Rhoades[7] )So that {xn} is a Cauchy sequence, hence convergent and {Tx2n},{Sx2n+1} also converge to z. From (xi), ad(Tx2n, Sz, u) + bd(x2n, Tx2n, u) + cd(z, Sz, u)≤ q max{d(x2n, z, u), d(x2n, Tx2n, u), d(z, Sz, u)}In the limiting case, we get ad(z, Sz, u) + cd(z, Sz, u) ≤ qd(z, Sz, u)or, (a + c -q) d(z, Sz, u) ≤ 0.Therefore, Sz = z since a+c > q.Thus z is a fixed point of S. Similarly we can show that z is also a fixed point of T. Hence z is a common fixed point of T and S.The uniqueness of the common fixed point can be easily shown by using (xi). This completes the proof of the theorem. Omitting the term bd(x,Tx,u) and cd(y,Sy,u) from the left hand side of theorem 4.5 we get the following result as a corollary of the above Theorem.Corollary 4.1 Let the self mappings T and S of a complete metric space (X, d) satisfyd(Tx, Sy, u) ≤ q max{d(x, y, u), d(x, Tx, u), d(y, Sy, u)} for x, y, u ε X. Then T and S have a unique common fixed point.