# Erratum to: Generalized metrics and Caristi’s theorem

The Original Article was published on 15 May 2013

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

$\begin{array}{rcl}{2}^{n+2}& =& 2\left({2}^{n+1}\right)\\ \ge & 2n\left(n+1\right)\\ =& \left(n+n\right)\left(n+1\right)\\ \ge & \left(n+1\right)\left(n+2\right).\end{array}$

## References

1. Kirk WA, Shahzad N: Generalized metrics and Caristi’s theorem. Fixed Point Theory Appl. 2013., 2013: Article ID 129

2. Branciari A: A fixed point theorem of Banach-Caccioppoli type on a class of generalizedmetric spaces. Publ. Math. (Debr.) 2000, 57: 31–37.

3. Jachymski J, Matkowski J, Świa̧tkowski T: Nonlinear contractions on semimetric spaces. J. Appl. Anal. 1995, 1(2):125–134.

## Author information

Authors

### Corresponding author 