• Erratum
• Open Access

Erratum to: Generalized metrics and Caristi’s theorem

Fixed Point Theory and Applications20142014:177

https://doi.org/10.1186/1687-1812-2014-177

• Accepted: 1 August 2014
• Published:

The original article was published in Fixed Point Theory and Applications 2013 2013:129

The assertion in  that Caristi’s theorem holds in generalized metric spaces isbased, among other things, on the false assertion that if$\left\{{p}_{n}\right\}$ is a sequence in a generalized metric space$\left(X,d\right)$, and if $\left\{{p}_{n}\right\}$ satisfies ${\sum }_{i=1}^{\mathrm{\infty }}d\left({p}_{i},{p}_{i+1}\right)<\mathrm{\infty }$, then $\left\{{p}_{n}\right\}$ is a Cauchy sequence. In Example 1 below wegive a counter-example to this assertion, and in Example 2 we show that, infact, Caristi’s theorem fails in such spaces. We apologize for anyinconvenience.

For convenience we give the definition of a generalized metric space. The conceptis due to Branciari .

Definition 1 Let X be a nonempty set and $d:X×X\to \left[0,\mathrm{\infty }\right)$ a mapping such that for all $x,y\in X$ and all distinct points $u,v\in X$, each distinct from x and y:
1. (i)

$d\left(x,y\right)=0⇔x=y$,

2. (ii)

$d\left(x,y\right)=d\left(y,x\right)$,

3. (iii)

$d\left(x,y\right)\le d\left(x,u\right)+d\left(u,v\right)+d\left(v,y\right)$ (quadrilateral inequality).

Then X is called a generalized metric space.

The following example is a modification of Example 1 of .

Example 1 Let $X:=\mathbb{N}$, and define the function $d:\mathbb{N}×\mathbb{N}\to \mathbb{R}$ by putting, for all $m,n\in \mathbb{N}$ with $m>n$:
To see that $\left(X,d\right)$ is a generalized metric space, suppose$m,n\in \mathbb{N}$ with $m>n$ and suppose $p,q\in \mathbb{N}$ are distinct with each distinct from m andn. Also we assume $q>p$. We now show that
$d\left(n,m\right)\le d\left(n,p\right)+d\left(p,q\right)+d\left(q,m\right).$
(Q)
If one of the three numbers $|n-p|$, $q-p$ or $|q-m|$ is even, then, since
$d\left(n,m\right)\le 1,$
clearly (Q) holds. If all three numbers are odd, then, since $m-n=\left(m-q\right)+\left(q-p\right)+\left(p-n\right)$, $m-n$ is odd and
$d\left(n,m\right)=\sum _{i=n}^{m}d\left(i,i+1\right).$
In this instance there are four cases to consider:
1. (i)

$n,

2. (ii)

$p,

3. (iii)

$n,

4. (iv)

$p.

If (i) holds then
$\begin{array}{rcl}d\left(n,m\right)& =& \sum _{i=n}^{m}d\left(i,i+1\right)\\ =& \sum _{i=n}^{p}d\left(i,i+1\right)+\sum _{i=p}^{q}d\left(i,i+1\right)+\sum _{i=q}^{m}d\left(i,i+1\right)\\ =& d\left(n,p\right)+d\left(p,q\right)+d\left(q,m\right).\end{array}$
In the other three cases
$d\left(n,m\right)

Therefore $\left(X,d\right)$ is a generalized metric space. Now suppose$\left\{{n}_{k}\right\}$ is a Cauchy sequence in $\left(X,d\right)$. Then if ${n}_{i}\ne {n}_{k}$ and $d\left({n}_{i},{n}_{k}\right)<1$, $|{n}_{i}-{n}_{k}|$ must be odd. However, if $\left\{{n}_{k}\right\}$ is infinite, $|{n}_{i}-{n}_{k}|$ cannot be odd for all sufficiently large i, k. (Suppose ${n}_{i}>{n}_{j}>{n}_{k}$. If ${n}_{i}-{n}_{j}$ and ${n}_{j}-{n}_{k}$ are odd, then ${n}_{i}-{n}_{k}$ is even.) Thus any Cauchy sequence in$\left(X,d\right)$ must eventually be constant. It follows that$\left(X,d\right)$ is complete and that $\left\{n\right\}$ is not a Cauchy sequence in $\left(X,d\right)$. However, ${\sum }_{i=1}^{\mathrm{\infty }}d\left(i,i+1\right)<\mathrm{\infty }$.

Theorem 2 of  asserts that the analog of Caristi’s theorem holds in acomplete generalized metric space $\left(X,d\right)$. Thus a mapping $f:X\to X$ in such a space should always have a fixed pointif there exists a lower semicontinuous function $\phi :X\to {\mathbb{R}}^{+}$ such that

The following example shows this is not true in the space described inExample 1.

Example 2 Let $\left(X,d\right)$ be the space of Example 1, let$f\left(n\right)=n+1$ for $n\in \mathbb{N}$, and define $\phi :\mathbb{N}\to {\mathbb{R}}^{+}$ by setting $\phi \left(n\right)=\frac{2}{n}$. Obviously f has no fixed points and,because the space is discrete, φ is continuous. On the other hand, f satisfies Caristi’s condition:
$\frac{1}{{2}^{n}}=d\left(n,f\left(n\right)\right)\le \phi \left(n\right)-\phi \left(f\left(n\right)\right)=\frac{2}{n}-\frac{2}{n+1}.$
To see this, observe that
$\frac{1}{{2}^{n}}\le \frac{2}{n}-\frac{2}{n+1}=\frac{2}{n\left(n+1\right)}.$
This is equivalent to the assertion that
${2}^{n+1}\ge n\left(n+1\right).$
(C)
The proof is by induction. Clearly (C) holds if $n=1$ or $n=2$. Assume (C) holds for some $n\in \mathbb{N}$, $n\ge 2$. Then
$\begin{array}{rcl}{2}^{n+2}& =& 2\left({2}^{n+1}\right)\\ \ge & 2n\left(n+1\right)\\ =& \left(n+n\right)\left(n+1\right)\\ \ge & \left(n+1\right)\left(n+2\right).\end{array}$

Authors’ Affiliations

(1)
Department of Mathematics, University of Iowa, Iowa City, IA 52242, USA
(2)
Department of Mathematics, King Abdulaziz University, P.O. Box 80293, Jeddah, 21959, Saudi Arabia

References 