# b-Metric Generalization of Some Fixed Point Theorems

Authors: Mumtaz Ali; Muhammad Arshad
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We prove common fixed point theorems in terms ofb-metric spaces with new contractions. The results presented in this paper include b-metric generalizations of some fixed point theorems of Fisher, Pachpatte, and Sahu and Sharma.


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## Body

## 1. Introduction

Fixed point theory is a fascinating topic for research in both analysis and topology. In this direction the Banach contraction mapping theorem of 1922 popularly known as Banach contraction mapping principle is a rewarding result in fixed point theory. It has widespread applications in both pure and applied mathematics. The well-known Banach [1] contraction mapping principle states that if “Xis a complete metric space andT:X→Xis a contraction mapping ofXinto itself thenThas unique fixed point inX.” This celebrated principle has been generalized by several authors. In 1989, Bakhtin [2] introduced the concept of b-metric space which is generalization of renowned Banach contraction mapping principle. Czerwik [3, 4] extended the concept of b-metric space in 1993. Bakhtin’s concept of b-metric spaces has been extensively generalized and improved by several mathematicians for fixed points in several different ways, namely, Boriceanu [5], Bota et al. [6], Chen et al. [7], Hussain and Shah [8], Kutbi et al. [9], and Shukla [10] to name a few. In this paper, our main concern is to study common fixed point theorems in complete b-metric spaces for three self-mappings. The obtained results are generalizations of b-metric variant of fixed point theorems of Fisher, Pachpatte, and Sahu and Sharma.The following fixed point theorems were proved in [11–13].Theorem 1 (see [11]).
LetTbe a mapping of the complete metric space X into itself satisfying the inequality (1)dTx,Ty2≤α1dx,Txdy,Ty+α2dx,Tydy,Tx,∀x,y∈X,0≤α1<1,0≤α2thenThas  a  fixed  point  inX.Theorem 2 (see [12]).
IfTis a mapping of the complete metric space X into itself satisfying the inequality (2)dTx,Ty2≤α1dx,Txdy,Ty+dx,Tydy,Tx+α2dx,Txdy,Tx+dx,Tydy,Ty,for all x,y∈X, where α1,α2≥0 and α1+2α2<1 then T has a unique fixed point.Theorem 3 (see [13]).
IfT is a mapping of the complete metric space X into itself satisfying the inequality (3)dTx,Ty2≤α1dx,Txdy,Ty+dx,Tydy,Tx+α2dx,Txdy,Tx+dx,Tydy,Ty+α3dy,Tx2+dy,Ty2,for all x,y inX, where α1,α2,α3≥0 and α1+2α2+α3<1, then T has a unique fixed point.


## 2. Preliminaries

In this section we recall some basic definitions and necessary results from existing literature that will be used in the sequel.Definition 4 (see [3]).
LetX be a nonempty set and s≥1 be a given real number. A function d:X×X→R+ is said to be a b-metric on X if the following conditions hold:
(i)dx,y=0 if and only if x=y.
(ii)dx,y=dy,x for all x,y∈X.
(iii)dx,y≤sdx,z+dz,y for all x,y,z∈X.
The pairX,d is called b-metric space.It is clear from the definition ofb-metric that every metric space is b-metric for s=1, but the converse need not be true. The following example illustrates the above remarks.Example 5 (see [5]).
LetX=0,1,2. Define d:X×X→R+ by d0,0=d1,1=d2,2=0,d1,2=d2,1=d1,0=d0,1=1, and d0,2=d2,0=m≥2 for s=m/2. The function defined above is a b-metric space but is not a metric space for m>2.Proposition 6 (see [14]).
LetX be a nonempty set and the mappings f,g,h:X→X have a unique point of coincidence in X. If f,h and g,h are weakly compatible self-maps of X, then f,g,h have a unique common fixed point.Definition 7.
A sequencexn in a b-metric space X,d is called Cauchy sequence if and only if limm,n→∞dxn,xm=0.Definition 8.
A sequencexn in a b-metric space X,d is said to converge to a point x∈X if and only if limn→∞dxn,x=0. We denote this by limn→∞xn=x.Definition 9.
Ab-metric space X,d is said to be complete if and only if every Cauchy sequence in X converges to a point of X.Definition 10 (see [15]).
Letf and g be self-mappings of a set X. If z=fx=gx for some x in X, then x is called a coincidence point of f and g and z is called a point of coincidence of f and g.Definition 11 (see [16]).
The mappingsf,g:X→X are weakly compatible, if, for every x∈X, the following holds: fgx=gfx whenever fx=gx.Definition 12.
A pointx∈X is said to be a fixed point of a self-map T:X→X if Tx=x.


## 3. Main Results

In this section we obtain coincidence points and common fixed point theorems for three maps in completeb-metric spaces. In order to start our main results we begin with a simple but useful Lemma.Lemma 13.
LetX,d be a complete b-metric space with the coefficient s≥1 and let f,g,h:X→X be self-mappings from X into itself satisfying the following conditions:
(i)fX∪gX⊆hX.
(ii)(4)dfx,gy2≤α1dhx,fxdhy,gy+dhx,gydhy,fx+α2dhy,gxdhy,fx+dhx,gydhy,gy+α3dhy,fx2+dhy,gy2,for all x,y∈X; α1,α2,α3≥0, such that(5)sα1+s2+sα2+α3<1.Then every sequence yn with initial point x0 is a Cauchy sequence in X.Proof.
Letx0∈X and choose a point x1∈X such that hx1=fx0 and for x1 there exists x2∈X such that hx2=gx1, and continuing this process we construct sequences xn and yn in X given by(6)y2n=hx2n+1=fx2n,y2n+1=hx2n+1=gx2n+1∀n≥0.Suppose that there existsh∈[0,1/s)such that (7)dyn,yn+1≤hdyn-1,yn∀n≥1.We show that yn is a Cauchy sequence in X. Using (4), we have(8)dy2n,y2n+12=dhx2n+1,hx2n+22=dfx2n,gx2n+12≤α1dhx2n,fx2ndhx2n+1,gx2n+1+dhx2n,gx2n+1dhx2n+1,fx2n+α2dhx2n+1,gx2ndhx2n+1,fx2n+dhx2n,gx2n+1dhx2n+1,gx2n+1+α3dhx2n+1,fx2n2+dhx2n+1,gx2n+12≤α1dhx2n,hx2n+1dhx2n+1,hx2n+2+dhx2n,hx2n+2dhx2n+1,hx2n+1+α2dhx2n,hx2n+!dhx2n+1,hx2n+1+dhx2nhx2n+2dhx2n+1,hx2n+2+α3dhx2n+1,hx2n+12+dhx2n+1,hx2n+22≤α1dy2n-1,y2ndy2n,y2n+1+dy2n-1,y2n+1dy2n,y2n+α2dy2n-1,y2ndy2n,y2n+dy2n-1,y2n+1dy2n,y2n+1+α3dy2n,y2n2+dy2n,y2n+12≤α1dy2n-1,y2n+α2dy2n-1,y2n+1+α3dy2n,y2n+1≤α1dy2n-1,y2n+sα2dy2n-1,y2n+dy2n,y2n+1+α3dy2n,y2n+1≤α1+sα2dy2n-1,y2n+sα2+α3dy2n,y2n+11-sα2+α3dy2n,y2n+1≤α1+sα2dy2n-1,y2n⟹dy2n,y2n+1≤α1+sα21-sα2+α3dy2n-1,y2n≤λdy2n-1,y2n,where (9)λ=α1+sα21-sα2+α3<1,dy2n,y2n+1≤λdy2n,y2n-1,dy2n,y2n+1≤λ2dy2n-1,y2n-2.Therefore, for all n∈N, we can write (10)dyn+1,yn+2≤λdyn,yn+1≤⋯≤λn+1dy0,y1.Now, for any m,n∈N,m>n, we have(11)dyn,ym≤sdyn,yn+1+sdyn+1,ym≤sdyn,yn+1+s2dyn+1,yn+2+s3dyn+2,ym≤sdyn,yn+1+s2dyn+1,yn+2+s3dyn+2,yn+3+⋯+sm-n-1dym-2,ym-1+sm-n-1dym-1,ym≤sdyn,yn+1+s2dyn+1,yn+2+s3dyn+2,yn+3+⋯+sm-n-1dym-2,ym-1+sm-ndym-1,ym≤sλn+s2λn+1+s3λn+2+⋯+sm-nλm-1dy1,y0≤sλn1-sλdy1,y0.Therefore, we have (12)dyn,ym≤sλn1-sλdy0,y1⟶0asn⟶∞.Thus, (13)dyn,ym⟶0asn⟶∞. Hence yn is a Cauchy sequence in b-metric space X.The next theorem isb-metric variant of Theorem 1.3 in [13].Theorem 14.
LetX,d be a complete b-metric space with the coefficient s≥1. Suppose the self-maps f,g,h:X→X satisfy the condition(14)dfx,gy2≤α1dhx,fxdhy,gy+dhx,gydhy,fx+α2dhy,gxdhy,fx+dhx,gydhy,gy+α3dhy,fx2+dhy,gy2,where α1,α2,α3≥0 are nonnegative reals with(15)sα1+s2+sα2+α3<1.If fX∪gX⊆hX and hX is a complete subspace of X, then the maps f,g, and h have a coincidence point u in X. Moreover, if f,h and g,h are weakly compatible pairs. Then f,g, and h have a unique common fixed point in X.Proof.
Letx0be an arbitrary point in X and define the sequence yn in X such that y2n=hx2n+1=fx2n and y2n+1=hx2n+2=gx2n+1∀n=0,1,2,…. Now, we show that yn is a Cauchy sequence. So, by (4), we have(16)dy2n,y2n+12=dfx2n,gx2n+12≤α1dhx2n,fx2ndhx2n+1,gx2n+1+dhx2n,gx2n+1dhx2n+1,fx2n+α2dhx2n+1,gx2ndhx2n+1,fx2n+dhx2n,gx2n+1dhx2n+1,gx2n+1+α3dhx2n+1,fx2n2+dhx2n+1,gx2n+12≤α1dy2n-1,y2ndy2n,y2n+1+dy2n-1,y2ndy2n,y2n+α2dy2n,y2ndy2n,y2n+dy2n-1,y2n+1dy2n,y2n+1+α3dy2n,y2n2+dy2n,y2n+12≤α1dy2n-1,y2ndy2n,y2n+1+α2dy2n-1,y2n+1dy2n,y2n+1+α3dy2n,y2n+12⟹dy2n,y2n+1≤α1dy2n-1,y2n+α2dy2n-1,y2n+1+α3dy2n,y2n+1≤α1dy2n-1,y2n+sα2dy2n-1,y2n+dy2n,y2n+1+α3dy2n,y2n+1≤α1+sα2dy2n,y2n+1+sα2+α3dy2n-1,y2n⟹dy2n,y2n+1≤α1+sα31-sα2+α3dy2n-1,y2n≤λdy2n-1,y2n⟹dy2n,y2n+1≤λdy2n-1,y2n. Similarly, we can show that (17)dy2n+1,y2n+2≤λ2dy2n,y2n+1, where (18)λ=α1+sα21-sα2+α3<1s.Therefore, for all m,n∈N with m>n, we can get(19)yn+1,yn+2≤λdyn,yn+1≤⋯≤λn+1dy0,y1dyn,ym≤sdyn,yn+1+sdyn+1,ym≤sdyn,yn+1+s2dyn+1,yn+2+s3dyn+2,ym≤sdyn,yn+1+s2dyn+1,yn+2+s3dyn+2,yn+3+s4dyn+3,ym≤sdyn,yn+1+s2dyn+1,yn+2+s3dyn+2,yn+3+⋯+sm-n-2dym-3,ym-2+sm-n-1dym-2,ym-1+sm-ndym-1,ym⟹dyn,ym≤sλn1-sλdy0,y1.Thus, as n→∞,dyn,ym→0. It follows from Lemma 13 that yn is a Cauchy sequence and, by the completeness of X, yn converges to some y∈X. Therefore,(20)limn→∞⁡fx2n=limn→∞⁡gx2n+1=limn→∞⁡hx2n+1=limn→∞⁡hx2n+2=y.Thus, fX∪gX⊆hX implies either fX⊆hX or gX⊆hX.
Case 1.Let fX⊆hX. Since hX is a complete subspace of fX∪gX and fX⊆hX implies hX is closed, hence, there exist v,y∈X such that hv=y. If gv≠y, then, by using (14), we get(21)dfx2n,gv2≤α1dhx2n,fx2ndhv,gv+dhx2n,gvdhv,fx2n+α2dhv,gx2ndhv,fx2n+dhx2n,gvdhv,gv+α3dhv,fx2n2+dhv,gv2. Taking limit as n→∞ yields (22)dfx2n,gv2≤α1dhx2n,fx2ndhv,gv+dhx2n,gvdhv,fx2n+α2dhv,gx2ndhv,fx2n+dhx2n,gvdhv,gv+α3dhv,fx2n2+dhv,gv2dy,gv2≤α1dy,ydy,gv+dy,gvdy,y+α2dy,ydy,y+dy,gvdy,gv+α3dy,y2+dy,gv2≤α2+α3dy,gv2⟹1-α2+α3dy,gv2≤0,and the above inequality is possible only if dy,gy2=0 which implies that y=gv. It follows that (23)hv=y=gv. Since g and h are weakly compatible, we have hgv=ghv and so(24)gy=hy.If y≠gy, then by (14) we have(25)dfx2n,gy2≤α1dhx2n,fx2ndhy,gy+dhx2n,gydhy,fx2n+α2dhy,gx2ndhy,fx2n+dhx2n,gydhy,gy+α3dhy,fx2n2+dhy,gy2.As n→∞, we have (26)dy,gy2≤α1dy,ydhy,gy+dy,gydhy,y+α2dhy,ydhy,y+dy,gydhy,gy+α3dhy,y2+dhy,gy2≤α1dy,ydgy,gy+dy,gydgy,y+α2dgy,ydgy,y+dy,gydgy,gy+α3dgy,y2+dgy,gy2≤α1+α2+α3dy,gy2⟹dy,gy2≤α1+α2+α3dy,gy2⟹1-α1+α2+α3dy,gy2≤0, and the inequality is possible only if (27)dy,gy=0⟹y=gy. By using (24), we have(28)hy=gy=y.
Case 2.If gX⊆hX, again there exists points v,y∈X such that y=hv if y≠fv, and then by using (14) we get (29)dfv,gy2≤α1dhv,fvdhy,gy+dhv,gydhy,fv+α2dhy,gvdhy,fv+dhv,gydhy,gy+α3dhy,fv2+dhy,gy2. It follows that (30)dfv,y2≤α1dy,fvdy,y+dy,ydy,fv+α2dy,ydy,fv+dy,ydy,y+α3dy,fv2+dy,y2⟹1-α3dy,fv2≤0, which is possible only if dy,fy=0⇒y=fv:(31)⟹fv=hv=y. Since f and h are weakly compatible and hence fhv=hfv and so fy=hy and by (28), we have (32)y=fy=gy=hy. Thus, y is the common fixed point of self-mappings f,g, and h. This completes the proof of the theorem.Uniqueness.In order to prove uniqueness, let y1≠y2 be two distinct common fixed points of the self-maps f,g, and h then we have by (14)(33)dy1,y22=dfy1,gy22≤α1dhy1,fy1dhy2,gy2+dhy1,gy2dhy2,fy1+α2dhy2,gy1dhy2,fy1+dhy1,gy2dhy2,gy2+α3dhy2,fy12+dhy2,gy22≤α1dy1,y1dy2,y2+dy1,y2dy1,y2+α2dy2,y1dy2,y1+dy1,y2dy2,y2+α3dy1,y22+dy2,y22≤α1+α2+α3dy1,y22⟹1-α1+α2+α3dy1,y22≤0,which is possible only if dy1,y2=0⇒y1=y2 which gives us uniqueness of y.Remark 15.
If we putα3=0, then we get Theorem 1.2 in [12].Now we present the modified form of Theorem3 in terms of b-metric spaces.Theorem 16.
LetX,d be a complete b-metric space with the coefficient s≥1 and suppose the self-maps f,g,h:X→X satisfy the condition(34)dfx,gy2≤α1dhx,fxdhy,gy+α2dhx,gydhy,gy1+dfx,hy+α3dhx,hydhy,gy1+dfx,hy+α4dhx,gydhy,gy+α5dhy,fx2+dhy,gy2,where α1,α2,α3,α4,α5≥0 are nonnegative reals with(35)sα1+s2+sα2+sα3+s2+sα4+α5<1.If fX∪gX⊆hX and hX is a complete subspace of X, then the maps f,g, and h have a coincidence point u in X. Moreover, if f,h and g,h are weakly compatible, then f,g, and h have a unique common fixed point in X.Proof.
Letx0∈X and define a sequence of points yn in X as follows, y2n=hx2n+1=fx2nandy2n+1=hx2n+2=gx2n+1∀n=0,1,2,…. By using (34), we have (36)dy2n,y2n+12=dfx2n,gx2n+12≤α1dhx2n,fx2ndhx2n+1,gx2n+1+α2dhx2n,gx2n+1dhx2n+1,gx2n+11+dfx2n,hx2n+1+α3dhx2n,hx2n+1dhx2n+1,gx2n+11+dfx2n,hx2n+1+α4dhx2n,gx2n+1dhx2n+1,gx2n+1+α5dhx2n+1,fx2n2+dhx2n+1,gx2n+12≤α1dy2n-1,y2ndy2n,y2n+1+α2dy2n-1,y2n+1dy2n,y2n+11+dy2n,y2n+α3dy2n-1,y2ndy2n,y2n+11+dy2n,y2n+α4dy2n-1,y2n+1dy2n+1,y2n+α5dy2n,y2n2+dy2n,y2n+12. Or (37)dy2n,y2n+1≤α1dy2n-1,y2n+α2dy2n-1,y2n+1+α3dy2n-1,y2n+α4dy2n-1,y2n+1+α5dy2n,y2n+1≤α1dy2n-1,y2n+sα2dy2n-1,y2n+dy2n,y2n+1+α3dy2n-1,y2n+sα4dy2n-1,y2n+dy2n,y2n+1+α5dy2n,y2n+1≤sα2+sα4+α5dy2n,y2n+1+α1+sα2+α3+sα4dy2n,y2n-11-sα2+sα4+α5dy2n,y2n+1≤α1+sα2+α3+sα4dy2n,y2n-1⟹dy2n+1,y2n≤α1+sα2+α3+sα41-sα2+sα4+α5dy2n,y2n-1⟹dy2n+1,y2n≤λdy2n,y2n-1, where (38)λ=α1+sα2+α3+sα41-sα2+sα4+α5<1s. Similarly, it can be shown that (39)dy2n+1,y2n+2≤λdy2n,y2n+1. Therefore, for all n∈N we can get (40)dyn+1,yn+2≤λdyn,yn+1≤⋯≤λn+1dy0,y1. Now, for any m>n, we have (41)dyn,ym≤dyn,yn+1+dyn+1,yn+2+⋯+dym-1,ym≤λn+λn+1+⋯+λm-1dy1,y0≤λn1-λdy1,y0. Therefore, from Lemma 13, we have (42)dyn,ym≤sλn1-sλdy1,y0⟶0asm,n⟶∞, where sλ<1. It follows that the sequence yn is a Cauchy sequence and, by the completeness of X, ynconverges to some y∈X.
Therefore,(43)limn→∞⁡yn=limn→∞⁡fx2n=limn→∞⁡gx2n+1=limn→∞⁡hx2n+1=limn→∞⁡hx2n+2=y. Since hX is a complete subspace of X, there exists u,y∈X such that hu=y. If gu≠y, using (34), we get (44)dfx2n,gu2≤α1dhx2n,fx2ndhu,gu+α2dhx2n,gudhu,gu1+dfx2n,hu+α3dhx2n,hudhu,gu1+dfx2n,hu+α4dhx2n,gudhu,gu+α5dhu,fx2n2+dhu,gu2, and, on taking limit as n→∞, (45)dy,gu2≤α1dy,ydy,gu+α2dy,gudy,gu1+dy,y+α3dy,ydy,gu1+dy,y+α4dy,gudy,gu+α5dy,y2+dy,gu2≤α2+α4+α5dy,gu2⟹1-α2+α4+α5dy,gu2≤0,which is possible only if dy,gu2=0⇒y=gu. It follows that hu=y=gu. Since the pair of mappings g,hg is weakly compatible, we have ghu=hgu and so(46)gy=hy.If y≠gy, by (34), we get the following.
Asn→∞, we have (47)limn→∞⁡dfx2n,gy2≤limn→∞⁡α1d1hx2n,fx2ndhy,gy+α2dhx2n,gydhy,gy1+dfx2n,hy+α3dhx2n,hydhy,gy1+dy,y+α4dhx2n,gydhy,gy+α5dhy,fx2n2+dhy,gy2≤α1dy,ydgy,gy+α2dy,gydy,gy1+dy,y+α3dy,ydy,gy1+dy,y+α4dy,gydy,gy+α5dy,y2+dy,gy2≤α2+α4+α5dy,g2⟹1-α2+α4+α5dy,gy2, and the above inequality is possible only if dy,gy2=0⇒y=gy. Therefore, by (46) we get(48)y=gy=hy.Again, if y≠fu, by (34) we have the following.
Taking limit asn→∞, we have(49)dfu,gx2n+12=dy,fu2≤α1dhu,fudhx2n+1,gx2n+1+α2dhu,gx2n+1dhx2n+1,gx2n+11+dfu,hx2n+1+α3dhu,hx2n+1dhx2n+1,gx2n+11+dfu,hx2n+1+α4dhu,gx2n+1dhx2n+1,gx2n+1+α5dhx2n+1,fu2+dhx2n+1,gx2n+12≤α1dy,fudy,y+α2dy,ydy,y1+dfu,y+α3dy,ydy,y1+dfu,y+α4dy,ydy,y+α5dy,fu2+dy,y2≤α5dy,fu2⟹1-α5dy,fu2≤0,and this is possible only if dfu,y2=0⇒y=fu. Since f and h are weakly compatible, fhu=hfu:(50)⟹fy=gy.By (48) and (50) we have (51)fy=gy=hy=y.Thusyis the unique common fixed point of f,g, and h.
Uniqueness.In order to see the uniqueness of the common fixed point, let y1 and y2 be two distinct common fixed points of f,g, and h such that y1≠y2. Then by using (34) we get (52)dy1,y22=dfy1,gy22≤α1dhy1,fy1dhy2,gy2+α2dhy1,gy2dhy2,gy21+dfy1,hy2+α3dhy1,hy2dhy2,gy21+dfy1,hy2+α4dhy1,gy2dhy2,gy2+α5dhy2,fy12+dhy2,gy22≤α1dy1,y1dy2,y2+α2dy1,y2dy2,y21+dy1,y2+α3dy1,y2dy2,y21+dy1,y2+α4dy1,y2dy2,y2+α5dy2,y12+dy2,y22≤α5dy1,y22⟹1-α5dy1,y22≤0, and the inequality is possible only if dy1,y22=0 which implies that y1=y2 and the common fixed point is unique.Theorem16 yields the following corollaries.Corollary 17.
LetX,d be a complete b-metric space with coefficient s≥1 and f,g,h be a self-mapping of X into itself satisfying (53)dfx,gy2≤α1dhx,fxdhy,gy+dhx,gydhy,gy1+dfx,hy+α3dhx,hydhy,gy1+dfx,hy+dhx,gydhy,gy, for all x,y∈X; and α1,α3≥0 such that α1+s2+1α3<1 then f,g, and h have a unique common fixed point in X.Proof.
Puttingα5=0 and α2=α1, and α4=α3 in Theorem 16, we get the required result.Corollary 18.
LetX,d be a complete b-metric space and f,g,h:X→X be self-maps of X into itself satisfying the inequality (54)dfx,gy2≤α1dhx,fxdhy,gy+α2dhx,gydhy,gy1+dfx,hy, for all x,y∈X, such that α1+s2+sα2<1, and then f,g,h have a unique common fixed point in X.Proof.
Puttingα3=α4=α5=0 in Theorem 16 we get the required result.Remark 19.
Corollary17, is the result of Pachpatte [12].Remark 20.
Corollary18, is the result of Fisher [11].Example 21.
LetX=3,4,5, and d:X×X→R+ is defined as follows:(55)d3,4=d4,3=1,d5,4=d4,5=2025,d3,5=d5,3=125,d3,3=d4,4=d5,5=0. Then, it is easy to verify that X,d is a b-metric space with parameter s=4/5. Define the mappings f,g,h:X→X by (56)f3=f5=4,f4=5,g3=g4=g5=5,h3=4,h4=3,h5=5. Then, for α1=α2=α3=α4=α5=1/2, the self-maps f,g, and h satisfy all the conditions of Theorems 14 and 16 with 5 as the only common fixed point in X.



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*Source: 2658653-2018-04-15.xml*