A characterization of Smyth complete quasi-metric spaces via Caristi’s fixed point theorem

Abstract

We obtain a quasi-metric generalization of Caristi’s fixed point theorem for a kind of complete quasi-metric spaces. With the help of a suitable modification of its proof, we deduce a characterization of Smyth complete quasi-metric spaces which provides a quasi-metric generalization of the well-known characterization of metric completeness due to Kirk. Some illustrative examples are also given. As an application, we deduce a procedure which allows to easily show the existence of solution for the recurrence equation of certain algorithms.

1 Introduction and preliminaries

We start by recalling several notions and properties of the theory of quasi-metric spaces. Our basic references are [1] and [2].

By a quasi-metric on set X we mean a function $d:X\times X\rightarrow [0,\infty)$ such that for all $x,y,z\in X$: (i) $x=y\Leftrightarrow d(x,y)=d(y,x)=0$; (ii) $d(x,z)\leq d(x,y)+d(y,z)$.

A quasi-metric space is a pair $(X,d)$ such that X is a set and d is a quasi-metric on X.

Given a quasi-metric d on X, the function $d^{-1}$ defined by $d^{-1}(x,y)=d(y,x)$ is also a quasi-metric on X, called the conjugate of d, and the function $d^{s}$ defined by $d^{s}(x,y)=\max \{d(x,y),d^{-1}(x,y)\}$ is a metric on X.

Each quasi-metric d on X induces a $T_{0}$ topology $\tau_{d}$ on X which has as a base the family of open balls $\{B_{d}(x,r):x\in X, \varepsilon>0\}$, where $B_{d}(x,\varepsilon )=\{y\in X:d(x,y)<\varepsilon\}$ for all $x\in X$ and $\varepsilon>0$.

If $\tau_{d}$ is a $T_{1}$ topology on X, we say that $(X,d)$ is a $T_{1}$ quasi-metric space.

Note that a quasi-metric space $(X,d)$ is $T_{1}$ if and only if for each $x,y\in X$, condition $d(x,y)=0$ implies $x=y$.

There exist many different notions of Cauchy net, Cauchy sequence and quasi-metric completeness in the literature (see, e.g., [1–3]). For our purposes, here we will consider the following ones.

A net $(x_{\alpha})_{\alpha\in\Lambda}$ in a quasi-metric space $(X,d)$ is called left K-Cauchy if for each $\varepsilon>0$ there is $\alpha _{\varepsilon}\in\Lambda$ such that $d(x_{\alpha},x_{\beta })<\varepsilon$ whenever $\alpha_{\varepsilon}\leq\alpha\leq\beta$. The notion of a left K-Cauchy sequence is defined in the obvious manner.

We say that a quasi-metric space $(X,d)$ is complete if every left K-Cauchy net is convergent for $\tau_{d^{-1}}$, and say that it is sequentially complete if every left K-Cauchy sequence is convergent for $\tau_{d^{-1}}$. (Note that our notion of (sequential) completeness of $(X,d)$ coincides with the usual notion of right K-(sequential) completeness of $(X,d^{-1})$.)

A quasi-metric space $(X,d)$ is Smyth complete provided that every left K-Cauchy net in $(X,d)$ is convergent for $\tau_{d^{s}}$ (compare Definition 8 in [4], [5], p.454, etc.).

The following well-known result is a consequence of Definition 8 and Theorem 9 in [4] (see also [6], p.323, [7], p.347).

Proposition 1

A quasi-metric space $(X,d)$ is Smyth complete if and only if every left K-Cauchy sequence in $(X,d)$ is convergent for $\tau_{d^{s}}$.

The following implications are also known and easy to check:

$$\mbox{Smyth complete}\quad \Rightarrow \quad \mbox{complete}\quad \Rightarrow\quad \mbox{sequentially complete}. $$

However, the converse implications do not hold, in general. For instance, the Sorgenfrey quasi-metric space (see, e.g., [5], p.463 or Example 1.1.6 in [1]) provides a distinguished example of a complete $T_{1}$ quasi-metric space which is not Smyth complete, while Stoltenberg presented in Example 2.4 of [8] an example of a sequentially complete $T_{1}$ quasi-metric space which is not complete.

On the other hand, Caristi proved in 1976 the following important and well-known generalization of the Banach contraction principle.

Theorem 1

([9])

Let T be a self-mapping of a complete metric space $(X,d)$. If there is a lower semicontinuous function $\varphi:X\rightarrow[0,\infty)$ satisfying

$$d(x,Tx)\leq\varphi(x)-\varphi(Tx) $$

for all $x\in X$, then T has a fixed point in X.

Kirk showed in [10] that the validity of Caristi’s fixed point theorem in a metric space characterizes its completeness. More exactly, he proved the following.

Theorem 2

([10])

For a metric space $(X,d)$, the following conditions are equivalent:

• (1) $(X,d)$ is complete.

• (2) If T is a self-mapping of X such that there is a lower semicontinuous function $\varphi:X\rightarrow[ 0,\infty)$ satisfying $d(x,Tx)\leq\varphi(x)-\varphi(Tx)$ for all $x\in X$, then T has a fixed point in X.

Extensions and generalizations of Theorems 1 and 2 to partial metric spaces, cone metric spaces, quasi-metric spaces and probabilistic metric spaces have been obtained by several authors (see, e.g., [11–20]). In particular, Cobzaş ([15], Theorem 2.3) proved, among other interesting results, the following quasi-metric generalization of Caristi’s fixed point theorem.

Theorem 3

([15])

Let T be a self-mapping of a sequentially complete $T_{1}$ quasi-metric space $(X,d)$. If there is a function $\varphi:X\rightarrow[ 0,\infty)$ which is lower semicontinuous for $\tau_{d^{-1}}$ and satisfies

$$d(x,Tx)\leq\varphi(x)-\varphi(Tx) $$

for all $x\in X$, then T has a fixed point in X.

Since complete and Smyth complete non-$T_{1}$ quasi-metric spaces provide efficient tools in several areas as asymmetric functional analysis, domain theory, theoretical computer science, complexity analysis of algorithms defined by recurrence equations, etc. (see, e.g., [1, 4, 5, 7, 21, 22] and their references), it seems natural to discuss the question of generalizing Theorem 3 to (non-necessarily $T_{1}$) quasi-metric spaces. In this direction, we shall give an example of a sequentially complete quasi-metric space for which Theorem 3 does not hold. We shall show that, nevertheless, Theorem 3 remains valid for complete quasi-metric spaces. A suitable and slight modification of the proof of that result will be used to deduce a characterization of Smyth complete quasi-metric spaces which provides a generalization to the quasi-metric framework of Kirk’s characterization of metric completeness. As an application, we obtain a procedure which allows to easily deduce the existence of solution for the recurrence equation of certain algorithms.

2 Results and examples

In order to simplify the terminology and the statements of our results, we shall use the following notions.

A self-mapping T of a quasi-metric space $(X,d)$ will be called a d-Caristi mapping (resp. a $d^{s}$-Caristi mapping) on $(X,d)$ if there is a function $\varphi:X\rightarrow[0,\infty)$ which is lower semicontinuous for $\tau_{d^{-1}}$ (resp. for $\tau_{d^{s}}$) and satisfies $d(x,Tx)\leq \varphi(x)-\varphi(Tx)$ for all $x\in X$.

Clearly, every d-Caristi mapping is a $d^{s}$-Caristi mapping. The following example shows that the converse is not true in general.

Example 1

Let d be the quasi-metric on the set $\mathbb{N}$ of all positive integer numbers, given by $d(x,x)=0$ for all $x\in \mathbb{N}$ and $d(x,y)=1/x$ for all $x,y\in\mathbb{N}$ with $x\neq y$. Clearly $(\mathbb {N},d)$ is a $T_{1}$ quasi-metric space such that $\tau_{d}$, and hence $\tau_{d^{s}}$ is the discrete topology on $\mathbb{N}$. Define $T:\mathbb{N}\rightarrow\mathbb{N}$ as $Tx=2x$ for all $x\in\mathbb{N}$. Then $d(x,Tx)=1/x=\varphi(x)-\varphi(Tx)$, where $\varphi:\mathbb {N}\rightarrow [0,\infty)$ is defined as $\varphi(x)=2/x$ for all $x\in\mathbb{N}$. Since $\tau_{d^{s}}$ is the discrete topology on $\mathbb{N}$, φ is lower semicontinuous for $\tau_{d^{s}}$ and thus T is a $d^{s}$-Caristi mapping on $(\mathbb{N},d)$. Finally, suppose that T is also a d-Caristi mapping. Then there exists a function $\varphi:\mathbb{N}\rightarrow [0,\infty)$ which is lower semicontinuous for $\tau_{d^{-1}}$ and satisfies $d(x,2x)=1/x\leq\varphi(x)-\varphi(2x)$ for all $x\in\mathbb{N}$. We easily deduce that $\varphi(1)\geq1+\varphi(2^{x})$ for all $x\in \mathbb{N}$, which contradicts that φ is a lower semicontinuous function for $\tau_{d^{-1}}$ because the sequence $(2^{n})_{n\in\mathbb{N}}$ converges to 1 for $\tau_{d^{-1}}$.

Our next example, based on Example 2.1 in [5], shows that condition $T_{1}$ cannot be removed in Theorem 3.

Example 2

Let $(\mathcal{A},d)$ be the non-$T_{1}$ quasi-metric space such that $\mathcal{A}$ is the family of all nonempty countable subsets of the set $\mathbb{R}$ of all real numbers, and d is the quasi-metric on $\mathcal{A}$ defined as $d(A,B)=0$ if $A\subseteq B$, and $d(A,B)=1$ otherwise. Let $(A_{n})_{n\in\mathbb{N}}$ be a left K-Cauchy sequence in $(\mathcal{A},d)$. Assume, without loss of generality, that $d(A_{n},A_{m})=0$ whenever $n\leq m$, i.e., $A_{n}\subseteq A_{m}$ whenever $n\leq m$. Since $\bigcup_{n\in\mathbb{N}}A_{n}\in\mathcal{A}$ and $d(A_{n},\bigcup_{n\in\mathbb{N}}A_{n})=0$ for all $n\in\mathbb{N}$, we deduce that $(\mathcal{A},d)$ is sequentially complete. Now let

$$\Lambda=\{A\in\mathcal{A}: A \mbox{ is a nonempty finite subset of }\mathbb{R} \mbox{ consisting of irrational numbers}\} $$

ordered by inclusion. Then the net $(A)_{A\in\Lambda}$ is left K-Cauchy in $(\mathcal{A},d)$ (see Example 2.1 in [5]) but it does not converge for $\tau_{d^{-1}}$ because the elements of $\mathcal{A}$ are countable subsets of $\mathbb{R}$. We conclude that $(\mathcal{A},d)$ is not complete.

However, we have the following extension of Theorem 3 whose proof is based on a classical technique used by Kirk [10], which is inspired in the partial order of Brøndsted [23, 24].

Theorem 4

Every d-Caristi mapping on a complete quasi-metric space $(X,d)$ has a fixed point in X.

Proof

Let $(X,d)$ be a complete quasi-metric space and let $T:X\rightarrow X$ be a d-Caristi mapping on $(X,d)$. Then there exists a function $\varphi:X\rightarrow[0,\infty)$ which is lower semicontinuous for $\tau_{d^{-1}}$ and satisfies

$$d(x,Tx)\leq\varphi(x)-\varphi(Tx) $$

for all $x\in X$. As in the classical metric case, define a binary relation ⪯ on X by

$$x\preceq y\quad \Longleftrightarrow \quad d(x,y)\leq\varphi(x)-\varphi(y) $$

for all $x,y\in X$. Clearly ⪯ is a partial order on X. Note also that $x\preceq Tx$ for all $x\in X$.

We shall prove that every (nonempty) linearly ordered subset of the partially ordered set $(X,\preceq)$ has an upper bound. Indeed, let A be a (nonempty) linearly ordered subset of X. We show that the net $(x_{x})_{x\in A}$ is a left K-Cauchy net in $(X,d)$ where we have defined $x_{x}:=x$ for all $x\in A$. To this end, put $r=\inf_{x\in A}\varphi(x)$. Given an arbitrary $\varepsilon>0$, choose $x\in A$ such that $\varphi (x)< r+\varepsilon$. Thus, for any $y,z\in A$ with $x\preceq y\preceq z$, we obtain

$$d(y,z)\leq\varphi(y)-\varphi(z)\leq\varphi(x)-\varphi (z)< r+\varepsilon -r= \varepsilon. $$

Consequently, $(x_{x})_{x\in A}$ is a left K-Cauchy net in $(X,d)$, and hence it converges, for $\tau_{d^{-1}}$, to some $p\in X$. Fix $x\in A$ and let $\varepsilon>0$ be arbitrary. Then there is $y\in A$ such that $d(z,p)<\varepsilon$ and $\varphi(p)-\varphi(z)<\varepsilon$ whenever $z\in A$ and $y\preceq z$. Choose $z_{0}\in A$ with $x\preceq z_{0}$ and $y\preceq z_{0}$. Hence

$$\begin{aligned} d(x,p) \leq&d(x,z_{0})+d(z_{0},p)< \varphi(x)-\varphi (z_{0})+\varepsilon \\ < &\varphi(x)-\varphi(p)+2\varepsilon. \end{aligned}$$

Since ε is arbitrary, we deduce that $d(x,p)\leq\varphi (x)-\varphi(p)$, i.e., $x\preceq p$, so p is an upper bound of A. It follows from Zorn’s lemma that $(X,\preceq)$ has a maximal element, say a. Since $a\preceq Ta$, we conclude that $a=Ta$, so a is a fixed point of T. The proof is finished. □

Of course, Caristi’s fixed point theorem is a consequence of Theorem 4 when $(X,d)$ is a metric space. Next we present two examples of complete quasi-metric spaces $(X,d)$ with appropriate d-Caristi mappings, for which Caristi’s fixed point theorem cannot be applied to the metric space $(X,d^{s})$.

Example 3

Let $X=\mathbb{N}\cup\{\infty\}$. Define a nonnegative real-valued function d on $X\times X$ by $d(\infty ,\infty)=0$, $d(x,y)=\vert 1/x-1/y\vert$ if $x,y\in\mathbb{N}$, $d(x,\infty)=1/x$ and $d(\infty,x)=1$ for all $x\in\mathbb{N}$. It is easily seen that $(X,d)$ is a complete $T_{1}$ quasi-metric space (in fact, note $(X,\tau_{d^{-1}})$ is a compact topological space). Define $T:X\rightarrow X$ as $T\infty =\infty$, and $Tx=x^{2}$ for all $x\in\mathbb{N}$. Now define $\varphi:X\rightarrow [0,\infty)$ as $\varphi(\infty)=0$, and $\varphi(x)=1/x$ for all $x\in \mathbb{N}$. Then φ is clearly a lower semicontinuous function for $\tau_{d^{-1}}$. Since $d(\infty,T\infty)=d(1,T1)=0$, and for every $x\in X\backslash\{1,\infty\}$,

$$d(x,Tx)=\frac{1}{x}-\frac{1}{x^{2}}=\varphi(x)-\varphi(Tx), $$

we conclude that T is a d-Caristi mapping on $(X,d)$. Hence, we can apply Theorem 4 to this case. In fact, T has 1 and ∞ as fixed points. However, we cannot apply Caristi’s fixed point theorem to the metric space $(X,d^{s})$ because it is not complete. Indeed, $(x)_{x\in \mathbb{N}}$ is a Cauchy sequence in $(X,d^{s})$ that does not converge for $\tau_{d^{s}}$.

In the above example the metric space $(X,d^{s})$ is not complete. Now, we give an example of a complete quasi-metric space $(X,d)$ where the metric space $(X,d^{s})$ is complete and there is a d-Caristi mapping on $(X,d)$ which is not a Caristi mapping for the metric space $(X,d^{s})$.

Example 4

As in Example 3, let $X=\mathbb{N}\cup\{ \infty\}$. Define a nonnegative real-valued function d on $X\times X$ by $d(x,y)=0$ if $x\leq y$, and $d(x,y)=y$ if $y< x$ (here, ≤ denotes the usual order on X). It is routine to check that $(X,d)$ is a complete quasi-metric space (note that every net in X converges to ∞ for $\tau_{d^{-1}}$). Define $T:X\rightarrow X$ as $Tx=x+1$ for all $x\in \mathbb{N}$ and $T\infty=\infty$. Then $d(x,Tx)=0$ for all $x\in X$, so that T is trivially a d-Caristi mapping on $(X,d)$. Hence, we can apply Theorem 4. Finally, suppose that there exists a lower semicontinuous function, for $\tau_{d^{s}}$, $\varphi:X\rightarrow[0,\infty)$, such that $d^{s}(x,Tx)\leq\varphi(x)-\varphi(Tx)$ for all $x\in X$. Then

$$d^{s}(x,x+1)=x+1\leq\varphi(x)-\varphi(x+1) $$

for all $x\in\mathbb{N}$. We deduce that $\varphi(1)=\infty$, a contradiction. Hence, we cannot apply the classical Caristi fixed point theorem in this case.

Observe that the aforementioned example of Stoltenberg and Example 4 (or Example 3) above show that Theorems 3 and 4 are independent of each other.

Although we do not know whether the converse of Theorem 4 holds, i.e., if Kirk’s theorem can be generalized to complete quasi-metric spaces, we are going to show that it is possible to obtain such a generalization for Smyth complete quasi-metric spaces. To this end, the following essentially well-known fact (see, e.g., Proposition 1.2.4 in [1]) will be useful.

Proposition 2

Let $(x_{n})_{n\in\mathbb{N}}$ be a left K-Cauchy sequence in a quasi-metric space $(X,d)$. If $(x_{n})_{n\in\mathbb{N}}$ has a subsequence convergent to $x\in X$ for $\tau_{d^{s}}$, then $(x_{n})_{n\in\mathbb{N}}$ converges to x for $\tau_{d^{s}}$.

Theorem 5

A quasi-metric space $(X,d)$ is Smyth complete if and only if every $d^{s}$-Caristi mapping on $(X,d)$ has a fixed point in X.

Proof

Suppose that $(X,d)$ is a Smyth complete quasi-metric space, and let T be a $d^{s}$-Caristi mapping on $(X,d)$. Then there exists a function $\varphi:X\rightarrow[0,\infty)$ which is lower semicontinuous for $\tau_{d^{s}}$ and satisfies $d(x,Tx)\leq \varphi (x)-\varphi(Tx)$ for all $x\in X$. Exactly as in the proof of Theorem 4, we construct a left K-Cauchy net in $(X,d)$, which converges for $\tau_{d^{s}}$ to an element $p\in X$ by Smyth completeness of $(X,d)$. Finally, we deduce that p is a fixed point of T again as in the proof of Theorem 4 and taking into account that φ is now lower semicontinuous for $\tau _{d^{s}}$.

Conversely, it will be enough to prove, by Proposition 1, that every left K-Cauchy sequence in $(X,d)$ converges for $\tau_{d^{s}}$. Assume the contrary. Then there exists a left K-Cauchy sequence $(x_{n})_{n\in\mathbb {N}}$ in $(X,d)$ which is not convergent for $\tau_{d^{s}}$. For each $k\in \mathbb{N}$, there exists $n_{k}\geq k$ such that $d(x_{n_{k}},x_{n})<2^{-(k+1)}$ for all $n\geq n_{k}$. Therefore $d(x_{n_{k}},x_{n_{k+1}})<2^{-(k+1)}$ for all $k\in\mathbb{N}$. Put $y_{k}:=x_{n_{k}}$ for all $k\in\mathbb{N}$. Then, by Proposition 2, we can suppose, without loss of generality, that $y_{k}\neq y_{j}$ whenever $k\neq j$, and that the sequence $\{y_{k}:k\in\mathbb{N}\}$ does not have any convergent subsequence for $\tau_{d^{s}}$.

We want to show that the self-mapping T of X given by $Ty_{k}=y_{k+1}$ for all $k\in\mathbb{N}$, and $Tx=y_{1}$ for all $x\notin\{y_{k}:k\in \mathbb{N}\}$, is a $d^{s}$-Caristi mapping. To this end, construct a function $\varphi:X\rightarrow[0,\infty)$ as follows: $\varphi(y_{k})=2^{-k}$ for all $k\in\mathbb{N}$, and $\varphi(x)=d^{s}(x,y_{1})+1/2$ whenever $x\notin \{y_{k}:k\in\mathbb{N}\}$. Since, for each $k\in\mathbb{N}$, $\varphi (y_{k})<\varphi(x)$ whenever $x \notin\{y_{k}:k\in\mathbb{N}\}$, and the function $x\rightarrow d^{s}(x,y_{1})$ is continuous for $\tau _{d^{s}}$, we immediately deduce that φ is lower semicontinuous for $\tau _{d^{s}}$. Moreover, we have

$$d(y_{k},Ty_{k})=d(y_{k},y_{k+1})< 2^{-(k+1)}= \varphi(y_{k})-\varphi (Ty_{k}) $$

for all $k\in\mathbb{N}$, and

$$d(x,Tx)=d(x,y_{1})\leq d^{s}(x,y_{1})= \varphi(x)-\varphi(Tx) $$

for all $x \notin\{y_{k}:k\in\mathbb{N}\}$, so T is a $d^{s}$-Caristi mapping on $(X,d)$. However, T has no fixed point. This contradiction concludes the proof. □

As in the metric case, we are going to deduce a multivalued version of Theorem 5.

Given a quasi-metric space $(X,d)$, we denote by $\mathcal{P}_{0}(X)$ the collection of all nonempty subsets of X. A multivalued mapping $T:X\rightarrow\mathcal{P}_{0}(X)$ will be called $d^{s}$-Caristi on $(X,d)$ if there is a function $\varphi:X\rightarrow[0,\infty)$ which is lower semicontinuous for $\tau_{d^{s}}$ and satisfies the following condition: For each $x\in X$, there exists $y_{x}\in Tx$ such that $d(x,y_{x})\leq \varphi(x)-\varphi(y_{x})$.

As usual, we say that a point $z\in X$ is a fixed point of $T:X\rightarrow \mathcal{P}_{0}(X)$ if $z\in Tz$.

Corollary

A quasi-metric space $(X,d)$ is Smyth complete if and only if every $d^{s}$-Caristi multivalued mapping on $(X,d)$ has a fixed point.

Proof

Suppose that $(X,d)$ is Smyth complete, and let $T:X\rightarrow\mathcal{P}_{0}(X)$ be a $d^{s}$-Caristi multivalued mapping. Then there is a function $\varphi:X\rightarrow[0,\infty)$ which is lower semicontinuous for $\tau_{d^{s}}$ and satisfies that for each $x\in X$ there exists $y_{x}\in Tx$ such that $d(x,y_{x})\leq\varphi(x)-\varphi (y_{x})$. Define a self-mapping f on X as follows: $fx=y_{x}$ for all $x\in X$. Obviously f is a $d^{s}$-Caristi mapping on $(X,d)$, so, by Theorem 5, there is $z\in X$ such that $z=fz$. Therefore $z=y_{z}$. Since $y_{z}\in Tz$, we conclude that z is a fixed point of T.

Conversely, suppose that every $d^{s}$-Caristi multivalued mapping on $(X,d)$ has a fixed point. Then every $d^{s}$-Caristi mapping on $(X,d)$ has a fixed point, so $(X,d)$ is Smyth complete by Theorem 5. □

Note that if $(X,d)$ is a quasi-metric space and T is a self-mapping of X such that $d(x,Tx)=0$ for all $x\in X$, then T is a $d^{s}$-Caristi mapping on $(X,d)$. If, in addition, $(X,d)$ is Smyth complete, then T has a fixed point by Theorem 5. Our next example illustrates this situation.

Example 5

Let Σ be a nonempty alphabet. Denote by $\Sigma^{\infty}$ the set of all finite and infinite words (sequences) over Σ, and denote by ϕ the empty word. For each $x,y\in \Sigma^{\infty}$, we define $x\sqcap y$ as the longest common prefix of x and y, and for each $x\in\Sigma^{\infty}$, we denote by $\ell(x)$ the length of x. Then $\ell(x)\in[1,\infty]$ whenever $x\neq\phi$ and $\ell(\phi)=0$. Now, for each $x,y\in\Sigma^{\infty}$, let $d(x,y)=0$ if x is a prefix of y, and $d(x,y)=2^{-\ell(x\sqcap y)}$ otherwise. Then d is a quasi-metric on $\Sigma^{\infty}$ [6, 25]. In fact, the quasi-metric space $(\Sigma^{\infty},d)$ is Smyth complete [5], Example 3.1. Define $T:\Sigma^{\infty}\rightarrow \Sigma ^{\infty}$ as follows: For each $x\in\Sigma^{\infty}$, Tx is an element of $\Sigma^{\infty}$ such that x is a prefix of Tx with $\ell (Tx)=\ell(x)+1$. Then $d(x,Tx)=0$ for all $x\in\Sigma^{\infty}$. By Theorem 5, T has a fixed point. In fact, $Tx=x$ if and only if $\ell (x)=\infty$.

Observe that if $(X,d)$ is a non-Smyth complete quasi-metric space such that $(X,d^{s})$ is complete, we can apply Caristi’s fixed point theorem to $(X,d^{s})$. However, by Theorem 5, there exists a $d^{s}$-Caristi mapping on $(X,d)$ without fixed point. We conclude this section with an example illustrating this fact.

Example 6

Let d be the quasi-metric on $\mathbb {R}$ given by $d(x,y)=y-x$ if $x\leq y$, and $d(x,y)=1$ if $x>y$. Then $(\mathbb{R},d)$ is the Sorgenfrey quasi-metric space. Since $d^{s}(x,y)\geq1$ for all $x,y\in\mathbb{R}$ with $x\neq y$, we deduce that the metric space $(\mathbb{R},d^{s})$ is complete and $\tau_{d^{s}}$ is the discrete topology on $\mathbb{R}$. As we indicated in Section 1, $(\mathbb{R},d)$ is not Smyth complete (indeed, note that the sequence $((n-1)/n)_{n\in\mathbb{N}}$ is left K-Cauchy but it does not converge for $\tau_{d^{s}}$). Define $T:\mathbb {R}\rightarrow\mathbb{R}$ as $Tx=0$ for all $x>0$, $T0=-1$, and $Tx=x/2$ for all $x<0$. Although T has no fixed point, we show that it is a $d^{s}$-Caristi mapping on $(\mathbb{R},d)$. To this end, define $\varphi:\mathbb{R} \rightarrow[0,\infty)$ as $\varphi(x)=3$ for all $x>0$, $\varphi (0)=2$, and $\varphi(x)=-x$ for all $x<0$. Obviously φ is lower semicontinuous for $\tau_{d^{s}}$. Moreover, for $x>0$, we obtain

$$d(x,Tx)=d(x,0)=1=\varphi(x)-\varphi(Tx). $$

For $x=0$, we obtain

$$d(x,Tx)=d(0,-1)=1=\varphi(x)-\varphi(Tx), $$

and for $x<0$,

$$d(x,Tx)=d\biggl(x,\frac{x}{2}\biggr)=-\frac{x}{2}=\varphi(x)- \varphi(Tx). $$

Hence T is a $d^{s}$-Caristi mapping on $(X,d)$ without fixed point. Finally, observe that for $x=-1$ one has

$$d^{s}(x,Tx)=1>\frac{1}{2}=\varphi(x)-\varphi(Tx). $$

3 An application

In this section we shall apply Theorem 5 to obtaining a general fixed point theorem in the setting of the complexity space, from which we shall deduce, in a unified and fast way, the existence of solution for a large class of algorithms defined by recurrence equations that includes Hanoi, Largetwo (average case), and Quicksort (worst case), (see, e.g., [26] for a detailed study of these algorithms).

Let us recall that the so-called complexity space was introduced by Schellekens in [27] to the development of a topological foundation for the complexity analysis of algorithms and programs. Further contributions to the study of this space and its applications may be found in [7, 22, 28–30], etc.

The complexity space is the quasi-metric space $(\mathcal{C},d_{\mathcal {C}})$, where

$$\mathcal{C}=\Biggl\{ f:\mathbb{N}\rightarrow(0,\infty ]:\sum _{n=1}^{\infty }2^{-n}\frac{1}{f(n)}< \infty\Biggr\} , $$

and $d_{\mathcal{C}}$ is the quasi-metric on $\mathcal{C}$ given by

$$d_{\mathcal{C}}(f,g)=\sum_{n=1}^{\infty}2^{-n} \max \biggl( \frac {1}{g(n)}-\frac{1}{f(n)},0 \biggr) $$

for all $f,g\in\mathcal{C}$. (We adopt the convention that $1/\infty=0$.)

The set $\{f\in\mathcal{C}:f(n)<\infty\mbox{ for all }n\in\mathbb{N}\}$ is denoted by $\mathcal{C}_{0}$.

The elements of $\mathcal{C}$ are called complexity functions. According to Schellekens [27], p.540, given two complexity functions f and g, the numerical value $d_{\mathcal{C}}(f,g)$ (the complexity distance from f to g) can be interpreted as the relative progress made in lowering the complexity by replacing any program P with complexity function f by any program Q with complexity function g. Therefore, condition $d_{\mathcal{C}}(f,g)=0$, with $f\neq g$, can be read as the program P is at least as efficient as the program Q because $d_{C}(f,g)=0$ if and only if $f(n)\leq g(n)$ for all $n\in\mathbb{N}$. Obviously, the metric $(d_{\mathcal{C}})^{s}$ is not able to give this information since in the case that $d_{\mathcal {C}}(f,g)=0$, with $f\neq g$, we deduce that $d_{\mathcal {C}}(g,f)=(d_{\mathcal{C}})^{s}(f,g)$, and thus the last measure does not indicate that program is more efficient. However, we know that the program with complexity function f is more efficient than the one with complexity function g (see [27], p.541).

Now let c and a be positive real constants and $h\in\mathcal{C}_{0}$. Define

$$\mathcal{C}_{cah}=\bigl\{ f\in\mathcal{C}:f(1)=c\mbox{ and }f(n)\geq af(n-1)+h(n) \mbox{ for all }n\geq2\bigr\} . $$

Observe that $\mathcal{C}_{cah}\neq\emptyset$ since the complexity function $f_{1}$ defined by $f_{1}(1)=c$ and $f_{1}(n)=\infty$ for all $n\geq2$ clearly belongs to $\mathcal{C}_{cah}$.

The restriction of the quasi-metric $d_{\mathcal{C}}$ to $\mathcal{C}_{cah}$ will be denoted by $d_{\mathcal{C}_{cah}}$.

The following auxiliary results will be useful in the proof of the main result of this section (Theorem 6 below).

Lemma 1

Let $(f_{k})_{k\in\mathbb {N}}$ be a sequence in $\mathcal{C}$ such that $\lim_{k\rightarrow \infty}(d_{\mathcal{C}})^{s}(f,f_{k})=0$ for some $f\in \mathcal{C}$, and let $m\in\mathbb{N}$.

• (a) If $f(m)<\infty$, then $f_{k}(m)<\infty$ eventually, and $\lim_{k\rightarrow\infty}f_{k}(m)=f(m)$.

• (b) $f(m)=\infty$ if and only if $\lim_{k\rightarrow\infty }f_{k}(m)=\infty$.

Proof

Since $\lim_{k\rightarrow\infty}(d_{\mathcal {C}})^{s}(f,f_{k})=0$, for each $\varepsilon>0$, there is $k_{\varepsilon }\in \mathbb{N}$ such that

$$\sum_{n=1}^{\infty}2^{-n}\biggl\vert \frac{1}{f(n)}-\frac{1}{f_{k}(n)}\biggr\vert < \varepsilon $$

for all $k\geq k_{\varepsilon}$. In particular

$$ 2^{-m}\biggl\vert \frac{1}{f(m)}-\frac{1}{f_{k}(m)}\biggr\vert < \varepsilon $$

(1)

for all $k\geq k_{\varepsilon}$.

Suppose that $f(m)<\infty$. Taking $\varepsilon=2^{-m}/f(m)$, it follows from (1) that $f_{k}(m)<\infty$ for all $k\geq k_{\varepsilon}$. Hence $\lim_{k\rightarrow\infty}f_{k}(m)=f(m)$ by Proposition 2 of [7]. Thus, we have shown (a).

If $f(m)=\infty$, relation (1) gives $2^{-m}/\varepsilon< f_{k}(m)$ for all $k\geq k_{\varepsilon}$. Since $\varepsilon>0$ is chosen arbitrarily, we deduce that $\lim_{k\rightarrow\infty}f_{k}(m)=\infty$. Conversely, if $\lim_{k\rightarrow\infty}f_{k}(m)=\infty$, again it follows from (1) that $1/f(m)=0$, i.e., $f(m)=\infty$. Thus, we have shown (b). □

Lemma 2

([22])

The quasi-metric space $(\mathcal{C},d_{\mathcal{C}})$ is Smyth complete.

Lemma 3

Let c and a be positive real constants and $h\in\mathcal{C}_{0}$. Then the quasi-metric space $(\mathcal{C}_{cah},d_{\mathcal{C}_{cah}})$ is Smyth complete.

Proof

We first show that $\mathcal{C}_{cah}$ is a closed subset of the metric space $(\mathcal{C},(d_{\mathcal{C}})^{s})$. Indeed, let $(f_{k})_{k\in\mathbb{N}}$ be a sequence in $\mathcal{C}_{cah}$ and $f\in \mathcal{C}$ such that $\lim_{k\rightarrow\infty}(d_{\mathcal{C}})^{s}(f,f_{k})=0$. We shall show that $f(1)=c$ and $f(m)\geq af(m-1)+h(m)$ whenever $m\geq2$.

To this end, we distinguish the following cases.

Case 1. $m=1$. Then $f_{k}(1)=c$ for all $k\in\mathbb{N}$, so by Lemma 1(b), $f(1)<\infty$. Then $f(1)=c$ by Lemma 1(a).

Case 2. $m>1$ and $f(m)=\infty$. Then $f(m)\geq af(m-1)+h(m)$, obviously.

Case 3. $m>1$ and $f(m)<\infty$. Then, by Lemma 1(a), there is $k_{0}\in \mathbb{N}$ such that $f_{k}(m)<\infty$ for all $k\geq k_{0}$, and $\lim_{k\rightarrow\infty}f_{k}(m)=f(m)$. From this equality and the fact that $f_{k}\in\mathcal{C}_{cah}$, we deduce the existence of $k_{1}\geq k_{0}$ such that for each $k\geq k_{1}$,

$$ 1+f(m)\geq f_{k}(m)\geq af_{k}(m-1)+h(m). $$

(2)

Consequently, $f_{k}(m-1)<\infty$ for all $k\geq k_{1}$, and by Lemma 1(b), $f(m-1)<\infty$ (otherwise, $\lim_{k}f_{k}(m-1)=\infty$, which contradicts (2)). Therefore, we also have $\lim_{k\rightarrow\infty }f_{k}(m-1)=f(m-1)$, by Lemma 1(a).

Now choose an arbitrary $\varepsilon>0$. Then there exists $k_{\varepsilon }\in\mathbb{N}$ such that

$$\bigl\vert f_{k}(m-1)-f(m-1)\bigr\vert < \varepsilon \quad \mbox{and}\quad \bigl\vert f_{k}(m)-f(m)\bigr\vert < \varepsilon $$

for all $k\geq k_{\varepsilon}$. Hence

$$\varepsilon+f(m)>f_{k}(m)\geq af_{k}(m-1)+h(m)\geq a \bigl(f(m-1)+\varepsilon \bigr)+h(m) $$

for all $k\geq k_{\varepsilon}$. Thus $\varepsilon +f(m)>a(f(m-1)+\varepsilon)+h(m)$ for any $\varepsilon>0$, so $f(m)\geq af(m-1)+h(m)$. Consequently, $f\in C_{cah}$, and hence $\mathcal {C}_{cah}$ is closed in the metric space $(\mathcal{C},(d_{\mathcal{C}})^{s})$. Then $(\mathcal{C}_{cah},d_{\mathcal{C}_{cah}})$ is Smyth complete by Lemma 2. □

Theorem 6

Let c and a be positive real constants with $a\geq1$, let $h\in\mathcal{C}_{0}$, and let Ψ be the mapping on $\mathcal{C}_{cah}$ defined as

$$ \Psi(f) (n)=\left \{ \textstyle\begin{array}{l@{\quad}l} c & \textit{if }n=1, \\ f(n-1)+h(n) &\textit{if }n\geq2. \end{array}\displaystyle \right . $$

(3)

Then the following hold:

• (A) Ψ is a self-mapping on $\mathcal{C}_{cah}$.

• (B) For each $f\in\mathcal{C}_{cah}$,

  $$d_{\mathcal{C}_{cah}}(f,\Psi f)=\varphi(f)-\varphi(\Psi f), $$

  where $\varphi:\mathcal{C}_{cah}\rightarrow[0,\infty)$ is the lower semicontinuous function for $\tau_{(d_{\mathcal{C}_{cah}})^{s}}$ given by

  $$\varphi(f)=\frac{a+1}{2ac}-\sum_{n=1}^{\infty}2^{-n} \frac{1}{f(n)} $$

  for all $f\in\mathcal{C}_{cah}$.

• (C) Ψ has a fixed point in $\mathcal{C}_{cah}$.

Proof

(A) Let $f\in\mathcal{C}_{cah}$. Then $\Psi f(1)=c$ by definition of Ψ. We also have $\Psi f(2)=af(1)+h(2)=a\Psi f(1)+h(2)$.

Now let $n>2$. Then

$$\begin{aligned} \Psi f(n) =&af(n-1)+h(n) \\ \geq& a \bigl[ af(n-2)+h(n-1) \bigr] +h(n) \\ =&a\Psi f(n-1)+h(n). \end{aligned}$$

We conclude that $\Psi f\in\mathcal{C}_{cah}$.

(B) We first observe that, in fact, $\varphi(f)\geq0$ for all $f\in \mathcal{C}_{cah}$. Indeed, since $a\geq1$, we have $f(n)\geq f(n-1)$ for all $n\geq2$, and thus $f(n)\geq f(2)\geq ac$ for all $n\geq2$. Therefore

$$\begin{aligned} \sum_{n=1}^{\infty}2^{-n} \frac{1}{f(n)} =&\frac{1}{2c}+\sum_{n=2}^{\infty }2^{-n} \frac{1}{f(n)} \\ \leq&\frac{1}{2c}+\frac{1}{2ac}=\frac{a+1}{2ac}. \end{aligned}$$

Let now $f\in\mathcal{C}_{cah}$ and $(f_{k})_{k\in\mathbb{N}}$ be a sequence in $\mathcal{C}_{cah}$ such that $\lim_{k\rightarrow\infty }(d_{\mathcal{C}_{cah}})^{s}(f,f_{k})=0$. Since

$$\begin{aligned} \varphi(f)-\varphi(f_{k}) =&\sum_{n=1}^{\infty}2^{-n} \frac {1}{f_{k}(n)}-\sum_{n=1}^{\infty}2^{-n} \frac{1}{f(n)} \\ \leq&\sum_{n=1}^{\infty}2^{-n}\biggl\vert \frac{1}{f_{k}(n)}-\frac {1}{f(n)}\biggr\vert \leq2(d_{\mathcal{C}_{cah}})^{s}(f,f_{k}), \end{aligned}$$

we deduce that $\varphi(f)\leq\lim\inf_{k\rightarrow\infty}\varphi (f_{k})$. Therefore φ is lower semicontinuous for $\tau_{(d_{\mathcal{C}_{cah}})^{s}}$.

Furthermore, for each $f\in\mathcal{C}_{cah}$, we have $f\geq\Psi f$, and hence

$$\begin{aligned} d_{\mathcal{C}_{cah}}(f,\Psi f) =&\sum_{n=1}^{\infty}2^{-n} \max \biggl( \frac{1}{\Psi f(n)}-\frac{1}{f(n)},0 \biggr) =\sum _{n=1}^{\infty }2^{-n} \biggl( \frac{1}{\Psi f(n)}-\frac{1}{f(n)} \biggr) \\ =&\sum_{n=1}^{\infty}2^{-n} \frac{1}{\Psi f(n)}-\sum_{n=1}^{\infty }2^{-n}\frac{1}{f(n)}=\varphi(f)-\varphi(\Psi f). \end{aligned}$$

(C) From (B) we deduce that Ψ is a $(d_{\mathcal {C}_{cah}})^{s}$-Caristi mapping on $(\mathcal{C}_{cah},d_{\mathcal{C}_{cah}})$. Then Ψ has a fixed point by Lemma 3 and Theorem 5. □

It follows from Theorem 6 that those algorithms defined by recurrence equations, whose associated functional is a mapping Ψ of type (3), admit a solution. We conclude the paper by applying this fact to deduce the existence of solution for the three algorithms mentioned at the beginning of this section.

Example 7

The algorithm Hanoi solves the celebrated Towers of Hanoi problem. The running time of computing of this algorithm is the solution of the recurrence equation $S:\mathbb{N}\rightarrow(0,\infty)$ given by

$$S(n)=\left \{ \textstyle\begin{array}{l@{\quad}l} c & \text{if }n=1, \\ 2S(n-1)+d & \text{if }n\geq2, \end{array}\displaystyle \right . $$

with $c,d>0$ (see, e.g., [26]). The functional $\Psi _{S}$ naturally associated to S is defined as

$$\Psi_{S}(f) (n)=\left \{ \textstyle\begin{array}{l@{\quad}l} c & \text{if }n=1, \\ 2f(n-1)+d & \text{if }n\geq2. \end{array}\displaystyle \right . $$

Clearly $\Psi_{S}$ is a mapping of type (3) for $a=2$, and $h\in \mathcal{C}_{0}$ defined as $h(n)=d$ for all $n\in\mathbb{N}$. By Theorem 6, there exists $f_{S}\in\mathcal{C}_{cah}$ such that $f_{S}=\Psi f_{S}$. Hence $f_{S}$ is a solution of the recurrence equation S.

Example 8

The algorithm Largetwo is a typical example of average case behavior whose running time of computing is the solution of the recurrence equation $S:\mathbb{N}\rightarrow(0,\infty)$ given by

$$S(n)=\left \{ \textstyle\begin{array}{l@{\quad}l} c & \text{if }n=1, \\ S(n-1)+2-1/n & \text{if }n\geq2, \end{array}\displaystyle \right . $$

with $c>0$ (see, e.g., [26]). The functional $\Psi _{S}$ naturally associated to S is defined as

$$\Psi_{S}(f) (n)=\left \{ \textstyle\begin{array}{l@{\quad}l} c & \text{if }n=1, \\ f(n-1)+2-1/n & \text{if }n\geq2. \end{array}\displaystyle \right . $$

Clearly $\Psi_{S}$ is a mapping of type (3) for $a=1$, and $h\in \mathcal{C}_{0}$ defined as $h(n)=2-1/n$ for all $n\in\mathbb{N}$. By Theorem 6, there exists $f_{S}\in\mathcal{C}_{cah}$ such that $f_{S}=\Psi f_{S}$. Hence $f_{S}$ is a solution of the recurrence equation S.

Example 9

The running time of computing of the well-known algorithm Quicksort is, for the worst case, the solution of the recurrence equation $S:\mathbb{N}\rightarrow(0,\infty)$ given by

$$S(n)=\left \{ \textstyle\begin{array}{l@{\quad}l} c & \text{if }n=1, \\ S(n-1)+bn & \text{if }n\geq2, \end{array}\displaystyle \right . $$

with $c,b>0$ (see, e.g., [26]). The functional $\Psi _{S}$ naturally associated to S is defined as

$$\Psi_{S}(f) (n)=\left \{ \textstyle\begin{array}{l@{\quad}l} c & \text{if }n=1, \\ f(n-1)+bn & \text{if }n\geq2. \end{array}\displaystyle \right . $$

Clearly $\Psi_{S}$ is a mapping of type (3) for $a=1$, and $h\in \mathcal{C}_{0}$ defined as $h(n)=bn$ for all $n\in\mathbb{N}$. By Theorem 6, there exists $f_{S}\in\mathcal{C}_{cah}$ such that $f_{S}=\Psi f_{S}$. Hence $f_{S}$ is a solution of the recurrence equation S.