# Characterization of∑-Semicompleteness via Caristi’s Fixed Point Theorem in Semimetric Spaces

Authors: Tomonari Suzuki
Journal: Journal of Function Spaces
 (2018)

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## Abstract

Introducing the concept of∑-semicompleteness in semimetric spaces, we extend Caristi’s fixed point theorem to ∑-semicomplete semimetric spaces. Via this extension, we characterize ∑-semicompleteness. We also give generalizations of the Banach contraction principle.


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## Body

## 1. Introduction

The following, very famous theorem is referred to as Caristi’s fixed point theorem. See also [1–10] and references therein.Theorem 1 (Theorem 1 in [11]).
Let(X,d) be a complete metric space and let T be a mapping on X. Let h be a lower semicontinuous function from X into [0,∞). Assume h(Tx)+d(x,Tx)≤h(x) for all x∈X. Then T has a fixed point.In 1976, Kirk proved that Caristi’s fixed point theorem characterizes the metric completeness.Theorem 2 (Theorem 2 in [12]).
Let(X,d) be a metric space. Then the following are equivalent:(i)
X is complete.(ii)
Every mappingT on X has a fixed point provided there exists a lower semicontinuous function h from X into [0,∞) satisfying h(Tx)+dx,Tx≤h(x) for all x∈X.Very recently, Theorem1 was extended to semimetric spaces.Theorem 3 (see [13]).
LetX,d be a (∑, ≠)-complete semimetric space and let T be a mapping on X. Let h be a function from X into -∞,+∞ which is proper, bounded from below, and sequentially lower semicontinuous from above in the sense of Definition 6. Assume hTx+dx,Tx≤hx for all x∈X. Then T has a fixed point.Remark 4.
See Definitions5 and 6 for the definitions of (∑,  ≠)-completeness and others.It is a very natural question of whether Theorem3 characterizes (∑, ≠)-completeness of the underlying space.In this paper, we give a negative answer to this question (see Example17). Motivated by this fact, we introduce the concept of ∑-semicompleteness and extend Theorem 3 to ∑-semicomplete semimetric spaces (see Corollary 12). And we characterize the ∑-semicompleteness via Corollary 12 (see Theorem 13). Also we give generalizations of the Banach contraction principle (see Section 4).


## 2. Preliminaries

Throughout this paper, we denote byN, Q, and R the sets of all positive integers, all rational numbers, and all real numbers, respectively. For an arbitrary set A, we also denote by #A the cardinal number of A.In this section, we give some preliminaries.Definition 5.
LetX be a nonempty set and let d be a function from X×X into [0,∞). Then X,d is said to be a semimetric space if the following hold: (D1)
d(x,x)=0.(D2)
d(x,y)=0⇒x=y.(D3)
d(x,y)=d(y,x) (symmetry).Definition 6.
LetX,d be a semimetric space, let xn be a sequence in X, and let x∈X. Let κ∈N and let h be a function from X into (-∞,+∞].(i)
{xn} is said toconverge to x if limn⁡d(xn,x)=0.(ii)
{xn} is said to beCauchy if limn⁡sup⁡dxn,xm:m>n=0.(iii)
{xn} is said to be ∑-Cauchy if ∑n=1∞d(xn,xn+1)<∞.(iv)
{xn} is said to be (∑, ≠)-Cauchy if xnn∈N are all different and ∑n=1∞d(xn,xn+1)<∞.(v)
X is said to beHausdorff if limn⁡d(xn,x)=0 and limn⁡d(xn,y)=0 imply x=y.(vi)
X is said to be κ-Hausdorff if (1)limn→∞⁡Dx,u1n,…,uκn,y=0implies x=y, where (2)Dx,u1n,…,uκn,y=dx,u1n+du1n,u2n+⋯+duκ-1n,uκn+duκn,y.(vii)
X is said to becomplete if every Cauchy sequence converges.(viii)
X is said to be ∑-complete if every ∑-Cauchy sequence converges.(ix)
X is said to be (∑, ≠)-complete if every (∑, ≠)-Cauchy sequence converges.(x)
X is said to besemicomplete if every Cauchy sequence has a convergent subsequence.(xi)
X is said to be ∑-semicomplete if every ∑-Cauchy sequence has a convergent subsequence.(xii)
X is said to be (∑, ≠)-semicomplete if every (∑, ≠)-Cauchy sequence has a convergent subsequence.(xiii)
d is said to besequentially lower semicontinuous if d(x,y)≤lim⁡infn⁡d(xn,yn) provided {xn} converges to x and {yn} converges to y.(xiv)
h is said to besequentially lower semicontinuous if h(x)≤lim⁡infn⁡h(xn) provided {xn} converges to x.(xv)
h is said to besequentially lower semicontinuous from above if h(x)≤limn⁡h(xn) provided {xn} converges to x and {hxn} is strictly decreasing.(xvi)
h is said to beproper if x∈X:hx∈R≠⌀.Remark 7.
(i)
The definitions ofκ-Hausdorffness and ∑-semicompleteness are new.(ii)
It is obvious thatX is Hausdorff ⇔X is 1-Hausdorff.(iii)
It is also obvious thatX is λ-Hausdorff ⇒X is κ-Hausdorff provided κ<λ.Proposition 8.
Let(X,d) be a semimetric space. Then the following implications hold: (I)
(i)⇒ (ii) ⇒ (iii) ⇔ (iv) ⇒ (vi).(II)
(i)⇒ (v) ⇒ (vi).(i)
X is ∑-complete.(ii)
X is (∑, ≠)-complete.(iii)
X is ∑-semicomplete.(iv)
X is (∑, ≠)-semicomplete.(v)
X is complete.(vi)
X is semicomplete.Proof.
(i)⇒ (ii) ⇒ (iv), (iii) ⇒ (iv), and (v) ⇒ (vi) obviously hold. We have already proved (i) ⇒ (v) in [13].
In order to prove (iv)⇒ (iii), we assume (iv). Let {xn} be a ∑-Cauchy sequence in X. We consider the following two cases: (a)
There existsz∈X satisfying #n∈N:xn=z=∞.(b)
For anyx∈X, #n∈N:xn=x<∞.
In the first case, some subsequence ofxn converges to z. In the second case, we define a subsequence fn of the sequence n in N as follows: f1=1. We assume that fn is defined. Then we define fn+1 by (3)fn+1=max⁡k∈N:xk=xfn+1.By induction, we have defined fn. We note that xf(n)(n∈N) are all different. We also have (4)∑n=1∞dxfn,xfn+1≤∑n=1∞dxn,xn+1<∞.Thus, xfn is (∑, ≠)-Cauchy. From (iv), there exists a subsequence of xfn which converges. It is obvious that the subsequence is also one of subsequences of xn. Therefore we have shown (iii).
Let us prove (iii)⇒ (vi). We assume (iii). Let xn be a Cauchy sequence in X. Choose a subsequence fn of n in N satisfying (5)sup⁡dxl,xm:m>l<2-nfor any l,n∈N with l≥f(n). Then xfn satisfies (6)∑n=1∞dxfn,xfn+1≤∑n=1∞2-n=1<∞,thus xfn is ∑-Cauchy. From (iii), there exist z∈X and a subsequence gn of n in N satisfying limn⁡d(xf∘gn,z)=0. Since {xf∘g(n)} is also a subsequence of {xn}, we obtain (vi).Proposition 9.
LetX,d be a semimetric space. Assume that d is sequentially lower semicontinuous. Then X is 2-Hausdorff.Proof.
Suppose(7)limn→∞⁡dx,un+dun,vn+dvn,y=0.Then un converges to x and vn converges to y. So we have (8)dx,y≤liminfn→∞⁡dun,vn=0.Thus, we obtain x=y.Proposition 10.
LetX,d be a ∑-complete semimetric space. Then X is κ-Hausdorff for any κ∈N.Proof.
Suppose(9)limn→∞⁡Dx,u1n,…,uκn,y=0.Choose a subsequence fn of n in N satisfying (10)Dx,u1fn,…,uκfn,y<2-nfor n∈N. Define a sequence {vn} in X as follows: (11)v2κ+2i+1=x,v2κ+2i+j+1=ujf2i+1,v2κ+2i+κ+2=y,v2κ+2i+2κ-j+3=ujf2i+2for i∈N∪0 and j∈1,…,κ. That is, vn is as follows: (12)x,u1f1,…,uκf1,y,uκf2,…,u1f2,x,u1f3,…,uκf3,y,uκf4,…,u1f4,x,….We have (13)∑i=1∞dvi,vi+1=∑i=0∞Dvκ+1i+1,…,vκ+1i+1+1<∑i=1∞2-i=1<∞.Thus, {vn} is ∑-Cauchy. Since X is ∑-complete, vn converges to some z. From the definition of vn, we have x=z=y. Thus, X is κ-Hausdorff.


## 3. Caristi’s Theorem

In this section, we first prove a Kirk-Saliga type fixed point theorem [14] in ∑-semicomplete semimetric spaces. See also [13].Letα be an ordinal number. We denote by α+ and α- the successor and the predecessor of α, respectively. α is said to beisolated if α- exists. On the other hand, α is said to belimit if α≠0 holds and α- does not exist. For κ∈N, we define α+κ by (14)α+κ=α+⋯+︷κ.Theorem 11.
LetX,d be a ∑-semicomplete semimetric space and let T be a mapping on X. Let h be a function from X into -∞,+∞ which is proper and bounded from below. Assume that h is sequentially lower semicontinuous from above in the sense of Definition 6. Assume also that there exists κ∈N satisfying the following: (i)
hTx≤h(x) for all x∈X.(ii)
hTκx+dx,Tx≤hx for all x∈X.
ThenT has a fixed point.Proof.
Define a functionH form X into (-∞,+∞] by (15)Hx=∑j=0κ-1hTjx,where T0 is the identity mapping on X. We have from (ii) (16)HTx+dx,Tx≤Hxforx∈X.
Arguing by contradiction, we assumeTx≠x for any x∈X. Let Ω be the first uncountable ordinal number. Using transfinite induction, we will define a net uα:α∈Ω satisfying the following: (P1 : α)
huα≤huβ and Huα<Huβ for any β∈Ω with β<α.(P2 : α)
huα<huβ for any β∈Ω with β+κ≤α.(P3 : α)
For anyε>0 and for any β∈Ω with β<α, there exists a finite sequence (γ0,…,γn)∈Ωn+1 satisfying (17)β=γ0<γ1<⋯<γn=α,Huα+∑j=0n-1duγj,uγj+1<Huβ+ε.
Fixu∈X with hu<∞. It follows from (i) that Hu<∞ holds. Put u0=u. Then (P1:0)–(P3:0) obviously hold.
Fixα∈Ω with 0<α and assume that (P1:β)–(P3:β) hold for β<α. We consider the following two cases: (a)
α is isolated.(b)
α is limit.
In the first case, we putuα=Tuα-. For any β<α, since β≤α- and uα-≠uα hold, we have by P1:α-, (i) and (16) (18)huα≤huα-≤huβ,Huα<Huα+duα-,uα≤Huα-≤Huβ.Thus, we have shown P1:α. For β∈Ω with β+κ≤α, we have by P1:α and (ii) (19)huα≤huβ+κ=hTκuβ<hTκuβ+duβ,Tuβ≤huβ.Thus, we have shown P2:α. Fix ε>0 and β∈Ω with β<α. In the case where β=α-, putting γ0=β and γ1=α, we have by (16) (20)Huα+∑j=0n-1duγj,uγj+1=HTuβ+duβ,Tuβ≤Huβ<Huβ+ε.In the other case, where β<α-, from P3:α-, there exists a finite sequence γ0,…,γn∈Ωn+1 satisfying (21)β=γ0<γ1<⋯<γn=α-,Huα-+∑j=0n-1duγj,uγj+1<Huβ+ε.Putting γn+1=α, we have by (16) (22)Huα+∑j=0nduγj,uγj+1≤Huα-+∑j=0n-1duγj,uγj+1<Huβ+ε.Thus, we have shown (P3:α). Therefore we have defined uα satisfying P1:α–P3:α in the first case.
In the second case, letβn be a strictly increasing sequence in Ω converging to α; that is, the following hold: (j)
βn<α for n∈N.(jj)
For anyβ<α, there exists n∈N satisfying β<βn.
For anyn∈N, from P3:βn+1, we can choose a finite sequence γn0,…,γnνn∈Ωνn+1 satisfying (23)βn=γn0<γn1<⋯<γnνn=βn+1,∑j=0νn-1duγnj,uγnj+1<Huβn-Huβn+1+2-n.Since h is bounded from below, H is also bounded from below. So we have (24)∑n=1∞∑j=0νn-1duγnj,uγnj+1<Huβ1-limn→∞⁡Huβn+1<∞.Since X is ∑-semicomplete, the sequence (25)β1=γ10,…,γ1ν1=β2=γ20,γ21,…,γ2ν2=β3=γ30,γ31,…has a subsequence δn such that uδn converges to some uα∈X. We note that δn is strictly increasing and it converges to α. Taking a subsequence, without loss of generality, we may assume δn+κ≤δn+1 for n∈N. We have by P2:δn+1(26)huδn+1<huδnforn∈N.Thus, huδn is strictly decreasing. Fix ε>0 and β∈Ω with β<α. We can choose ν∈N satisfying (27)β<δν,duδν,uα<ε.Since h is sequentially lower semicontinuous from above, we have from (i) (28)huα≤limn→∞⁡huδn<huδν≤huβ,Huα≤κhuα≤κlimn→∞⁡huδn=κinf⁡huγ:γ<α=limn→∞⁡Huδn<Huδν<Huβ.We have shown P1:α and P2:α. We can choose a finite sequence (γ0,…,γn)∈Ωn+1 satisfying (29)β=γ0<γ1<⋯<γn=δν,Huδν+∑j=0n-1duγj,uγj+1<Huβ+ε.Putting γn+1=α, we have by (P1:α)(30)Huα+∑j=0nduγj,uγj+1<Huδν+∑j=0n-1duγj,uγj+1+ε<Huβ+2ε.Thus we have defined uα satisfying (P1:α)–(P3:α) in the second case.
Therefore by transfinite induction, we have defined the netuα:α∈Ω satisfying P1:α–P3:α for any α∈Ω. Since the net Huα:α∈Ω is strictly decreasing, we obtain (31)#Q=#N<#Ω≤#Q,which implies a contradiction. Therefore there exists a fixed point of T.Using Theorem11, we can generalize Theorem 3.Corollary 12.
Let(X,d) be a ∑-semicomplete semimetric space and let T be a mapping on X. Let h be a function from X into -∞,+∞ which is proper and bounded from below. Assume that h is sequentially lower semicontinuous from above in the sense of Definition 6. Assume also (32)hTx+dx,Tx≤hxfor all x∈X. Then T has a fixed point.Via Corollary12, we characterize the ∑-semicompleteness of X.Theorem 13.
LetX,d be a semimetric space. Then the following are equivalent: (i)
X is ∑-semicomplete.(ii)
Every mappingT on X has a fixed point provided there exists a function h from X into 0,∞ such that h is proper and sequentially lower semicontinuous and (32) holds for all x∈X.Proof.
By Corollary12, we obtain (i) ⇒ (ii).
In order to prove (ii)⇒ (i), we will show ¬ (i) ⇒¬ (ii). We assume that X is not ∑-semicomplete. Then by Proposition 8, X is not (∑, ≠)-semicomplete. So there exists a sequence xn in X such that xn(n∈N) are all different, ∑n=1∞d(xn,xn+1)<∞ holds, and there does not exist a subsequence which converges. Define a mapping T on X and a function h from X into 0,∞ by (33)Tx=xn+1ifx=xnfor somen∈Nx1ifx∉xn:n∈N,hx=∑j=n∞dxj,xj+1ifx=xnforsomen∈N∞ifx∉xn:n∈N.We note that T and h are well defined because xn(n∈N) are all different and ∑n=1∞dxn,xn+1<∞ holds. Then h is proper, (32) holds for any x∈X, and T does not have a fixed point. Let yn be a sequence in X converging to some y∈X. Arguing by contradiction, we assume (34)hy>liminfn→∞⁡hyn.Then from the definition of h, there exists a subsequence fn of n in N such that yf(n)∈xm:m>ν for any n∈N, where we put ν∈N∪0 by (35)ν=nify=xnforsomen∈N0ify∉xn:n∈N.Define a function g from N into m∈N:m>ν by yfn=xgn. We consider the following two cases: (a)
lim⁡supn⁡g(n)=∞.(b)
μ≔lim⁡supn⁡g(n)<∞.
In the first case,xn has a subsequence converging to y. This is a contradiction. In the second case, we have (36)∞=#n∈N:gn=μ=#n∈N:yfn=xμ≤#n∈N:yn=xμ,which implies y=xμ. This is also a contradiction. Therefore we obtain h(y)≤lim⁡infn⁡h(yn). Thus, h is sequentially lower semicontinuous.


## 4. Banach’s Theorem

The author has extended the Banach contraction principle [15, 16] to semicomplete semimetric spaces. Such a result will be published somewhere else. See also [17]. In this section, we give other generalizations.Theorem 14.
LetX,d be a 2-Hausdorff ∑-semicomplete semimetric space. Let T be a contraction on X. Then T has a unique fixed point z. Moreover, Tnx converges to z for all x∈X.Proof.
Letr∈0,1 satisfy (37)dTx,Ty≤rdx,yforx,y∈X.Fix u∈X. Then we have (38)∑n=1∞dTnu,Tn+1u≤∑n=1∞rndu,Tu<∞.Since X is ∑-semicomplete, there exists a subsequence fn of n in N such that Tfnu converges to some z∈X. We have (39)limn→∞dTfn+1u,Tz≤limn→∞rdTfnu,z=0.Thus, Tfn+1u converges to Tz. So we have (40)limn→∞⁡Dz,Tfnu,Tfn+1u,Tz=0.From 2-Hausdorffness of X, we obtain Tz=z.
For anyx∈X, we have (41)limn→∞⁡dTnx,z=limn→∞⁡dTnx,Tnz≤limn→∞⁡rndx,z=0.The uniqueness of the fixed point follows from (41).Theorem 15.
LetX,d be a (∑, ≠)-complete Hausdorff semimetric space. Let T be a contraction on X. Then T has a unique fixed point z. Moreover, Tnx converges to z for all x∈X.Proof.
Letr∈0,1 satisfy (37). Fix u∈X. We consider the following two cases: (a)
There existsν∈N satisfying Tν+1u=Tνu.(b)
Tnu≠Tn+1u for any n∈N.
In the first case,Tνu is a fixed point of T. In the second case, we have (42)dTn+1u,Tn+2u≤rdTnu,Tn+1u<dTnu,Tn+1ufor any n∈N. Hence Tnun∈N are all different. We also have (43)∑n=1∞dTnu,Tn+1u≤∑n=1∞rndu,Tu<∞.Since X is (∑, ≠)-complete, Tnu converges to some z∈X. We have (44)limn→∞⁡dTn+1u,Tz≤limn→∞⁡rdTnu,z=0.
Thus,Tn+1u converges to Tz. Since X is Hausdorff, we obtain Tz=z. Therefore in both cases, there exists a fixed point of T. As in the proof of Theorem 14, we can prove the remainder.By Propositions8–10 and Theorems 14 and 15, we obtain the following corollary.Corollary 16.
Let(X,d) be a semimetric space. Assume that either of the following holds: (i)
X is ∑-complete.(ii)
X is ∑-semicomplete and d is sequentially lower and semicontinuous.
LetT be a contraction on X. Then T has a unique fixed point z. Moreover, {Tnx} converges to z for all x∈X.


## 5. Example

We finally give an example which tells that Theorem3 does not characterize (∑, ≠)-completeness of the underlying space.Example 17.
PutX=0∪2-n:n∈N. Define a function d from X×X into 0,∞ as follows: (45)dx,x=0,d2-n,2-n+1=d2-n+1,2-n=2-n+1forn∈N,d2-2n,0=d0,2-2n=2-2nforn∈N,dx,y=1otherwise.Then the following assertions holds: (i)
(X,d) is a semimetric space.(ii)
X is not (∑, ≠)-semicomplete.(iii)
X is ∑-semicomplete.(iv)
Every mappingT on X has a fixed point provided there exists a function h from X into 0,∞ such that h is proper and sequentially lower semicontinuous and (32) holds for all x∈X.Proof.
(i) obviously holds. (iv) follows from (iii) and Corollary12.
We will show (ii). It is obvious that{2-n} is a (∑, ≠)-Cauchy sequence. However, we have lim⁡supn⁡d2-n,x=1 for all x∈X. Therefore we obtain (ii).
In order to prove (iii), we will show thatX is (∑, ≠)-semicomplete. Let xn be a (∑, ≠)-Cauchy sequence in X. We can choose μ,ν∈N satisfying (46)0∉xn:n≥ν,xν=2-μ.From the definition of d, we have xν+j=2-μ+j for j∈N. So {xn} has a subsequence converging to 0. We have shown that X is (∑, ≠)-semicomplete. By Proposition 8, X is ∑-semicomplete.



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