Erratum to: Generalized metrics and Caristi’s theorem

The assertion in [1] that Caristi’s theorem holds in generalized metric spaces isbased, among other things, on the false assertion that if$\{p_n\}$ is a sequence in a generalized metric space$(X,d)$, and if $\{p_n\}$ satisfies ${\sum }_{i=1}^{\mathrm{\infty }}d(p_i,p_{i+1})<\mathrm{\infty }$, then $\{p_n\}$ 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 [2].

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

• (i) $d(x,y)=0\iff x=y$,

• (ii) $d(x,y)=d(y,x)$,

• (iii) $d(x,y)\le d(x,u)+d(u,v)+d(v,y)$ (quadrilateral inequality).

Then X is called a generalized metric space.

The following example is a modification of Example 1 of [3].

Example 1 Let $X:=\mathbb{N}$, and define the function $d:\mathbb{N}\times \mathbb{N}\to \mathbb{R}$ by putting, for all $m,n\in \mathbb{N}$ with $m>n$:

$$\begin{array}{c}d(n,n):=0;\hfill \\ d(m,n)=d(n,m):=\frac{1}{2^n}\text{if }m=n+1;\hfill \\ d(m,n)=d(n,m):=1\text{if }m-n\text{ is even};\hfill \\ d(m,n)=d(n,m):=\sum _{i=n}^{m}d(i,i+1)\text{if }m-n\text{ is odd}.\hfill \end{array}$$

To see that $(X,d)$ 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

$$d(n,m)\le d(n,p)+d(p,q)+d(q,m).$$

(Q)

If one of the three numbers $|n-p|$, $q-p$ or $|q-m|$ is even, then, since

$$d(n,m)\le 1,$$

clearly (Q) holds. If all three numbers are odd, then, since $m-n=(m-q)+(q-p)+(p-n)$, $m-n$ is odd and

$$d(n,m)=\sum _{i=n}^{m}d(i,i+1).$$

In this instance there are four cases to consider:

• (i) $n<p<q<m$,

• (ii) $p<n<q<m$,

• (iii) $n<p<m<q$,

• (iv) $p<n<m<q$.

If (i) holds then

$$\begin{array}{rcl}d(n,m)& =& \sum _{i=n}^{m}d(i,i+1)\\ =& \sum _{i=n}^{p}d(i,i+1)+\sum _{i=p}^{q}d(i,i+1)+\sum _{i=q}^{m}d(i,i+1)\\ =& d(n,p)+d(p,q)+d(q,m).\end{array}$$

In the other three cases

$$d(n,m)<d(n,p)+d(p,q)+d(q,m).$$

Therefore $(X,d)$ is a generalized metric space. Now suppose$\{n_k\}$ is a Cauchy sequence in $(X,d)$. Then if $n_i\ne n_k$ and $d(n_i,n_k)<1$, $|n_i-n_k|$ must be odd. However, if $\{n_k\}$ 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$(X,d)$ must eventually be constant. It follows that$(X,d)$ is complete and that $\{n\}$ is not a Cauchy sequence in $(X,d)$. However, ${\sum }_{i=1}^{\mathrm{\infty }}d(i,i+1)<\mathrm{\infty }$.

Theorem 2 of [1] asserts that the analog of Caristi’s theorem holds in acomplete generalized metric space $(X,d)$. 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

$$d(x,f(x))\le \phi (x)-\phi (f(x))\text{for each }x\in X.$$

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

Example 2 Let $(X,d)$ be the space of Example 1, let$f(n)=n+1$ for $n\in \mathbb{N}$, and define $\phi :\mathbb{N}\to {\mathbb{R}}^{+}$ by setting $\phi (n)=\frac{2}{n}$. Obviously f has no fixed points and,because the space is discrete, φ is continuous. On the other hand, f satisfies Caristi’s condition:

$$\frac{1}{2^n}=d(n,f(n))\le \phi (n)-\phi (f(n))=\frac{2}{n}-\frac{2}{n+1}.$$

To see this, observe that

$$\frac{1}{2^n}\le \frac{2}{n}-\frac{2}{n+1}=\frac{2}{n(n+1)}.$$

This is equivalent to the assertion that

$$2^{n+1}\ge n(n+1).$$

(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

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