The theorems of the present section look like some theorems from [2], but condition (3) matters a lot. Our first theorem extends [2], Theorem 3.3.

### Theorem 3.1

*Let*
\((X,p)\)
*be a* 0-*complete d*-*metric space*, *and let*
\(f: X \rightarrow X\)
*be a mapping satisfying condition* (6) *or* (7), *for all*
\(x,y \in X\)
*and a*
\(\varphi\in \Phi\)
*such that* (3) *holds*. *Then*
*f*
*has a unique fixed point*; *if*
\(x = f(x)\), *then*
\(\lim_{n \rightarrow\infty }p(x,f^{n}(x_{0})) = p(x,x)= 0\) (*i*.*e*. \(x \in\operatorname{Ker}p\)), \(x_{0}\in X\).

### Proof

It is sufficient to prove that \(\lim_{m,n \rightarrow\infty }p(x_{n},x_{m})= 0\) holds for \(x_{n}= f^{n} (x_{0})\), \(n \in\mathbb{N}\) (see [2], Lemma 3.2). From the fact that \(\varphi\in\Psi_{P}\) (Lemma 2.3) it follows that \(\lim_{n \rightarrow\infty}p(x_{n+1},x_{n}) = 0\) (see [2], Lemma 3.1). Suppose that there exists an infinite set \(K \subset\mathbb{N}\) such that for each \(k \in K\) there exists an \(n \in\mathbb{N}\) for which \(p(x_{n+1+k},x_{k}) > \alpha> 0\) holds. Let \(n = n(k) > 0\) be the smallest numbers satisfying this inequality for \(k \in K\). For simplicity let us adopt \(x = f^{k}(x_{0})\) (\(x_{-1}= f^{k-1}(x_{0})\)), and \(x_{m} = f^{m}(x)\), \(m \in\mathbb{N}\). From

$$p(x_{n+1},x) \leq\varphi \bigl(m_{f}(x_{n},x_{-1}) \bigr) = \varphi \bigl(\max \bigl\{ p(x_{n}, x_{-1}),p(x_{n+1},x_{n}),p(x,x_{-1}) \bigr\} \bigr) $$

(see (7)) we get \(p(x_{n+1},x) \leq\varphi (p(x_{n},x_{-1}))\), for large *k* (or from (6) directly), as

$$\alpha< p(x_{n+1},x) \leq p(x_{n+1},x_{n}) + p(x_{n},x_{-1}) + p(x_{-1},x). $$

The inequality

$$p(x_{n},x_{-1}) \leq p(x_{n},x) + p(x,x_{-1}) \leq\alpha+ p(x,x_{-1}) $$

yields \(p(x_{n},x_{-1}) < \alpha+\epsilon\), for large *k*. Consequently, from (3) and \(\varphi(\beta) < \alpha\), \(\beta\leq\alpha\), we obtain

$$\alpha< p(x_{n+1},x) \leq\varphi \bigl(p(x_{n},x_{-1}) \bigr) \leq \alpha, $$

for large *k*, a contradiction, *i.e.*
\(\lim_{m,n \rightarrow \infty}p(x_{n},x_{m})= 0\). □

Now, Theorem 3.1, and [6], Lemma 29 yield the following extension of [2], Theorem 3.5.

### Theorem 3.2

*Let*
\((X,p)\)
*be a* 0-*complete d*-*metric space*, *and let*
\(f: X \rightarrow X\)
*be a mapping satisfying condition* (6) *or* (7), *for all*
\(x,y \in X\)
*with*
*f*
*replaced by*
\(f^{s} \)
*for an*
\(s \in\mathbb{N}\), *and a*
\(\varphi\in\Phi\)
*having property* (3). *Then*
*f*
*has a unique fixed point*; *if*
\(x = f(x)\), *then*
\(\lim_{n \rightarrow\infty}p(x,f^{n}(x_{0})) = p(x,x)= 0\), \(x_{0}\in X\).

A refinement of the proof of Theorem 3.1, yields the following extension of [2], Theorem 3.9.

### Theorem 3.3

*Let*
\((X,p)\)
*be a* 0-*complete d*-*metric space*, *and let*
\(f: X \rightarrow X\)
*be cyclic on*
\(X_{1},\ldots,X_{t}\). *Assume that* (6) *or* (7) *is satisfied for all*
\(x \in X_{j}\), \(y \in X_{j++}\), \(j=1,\ldots,t\)
*and a*
\(\varphi\in\Phi\)
*having property* (3). *Then*
*f*
*has a unique fixed point*; *if*
\(x = f(x)\), *then*
\(\lim_{n \rightarrow\infty}p(x,f^{n}(x_{0})) = p(x,x)= 0\), \(x_{0}\in X\).

### Proof

It is sufficient to prove that \(\lim_{m,n \rightarrow\infty }p(x_{n},x_{m})= 0\) holds for \(x_{n}= f^{n} (x_{0})\), \(n \in\mathbb{N}\) (see [2], Lemma 3.8). From the fact that \(\varphi\in\Psi_{P}\) (Lemma 2.3) it follows that \(\lim_{n \rightarrow\infty}p(x_{n+1},x_{n}) = 0\) (see [2], Lemma 3.7). Suppose that there exists an infinite set \(K \subset \mathbb{N}\) such that for each \(k \in K\) there exists an \(n \in\mathbb{N}\) for which \(p(x_{(n+1)t+k+1},x_{k}) > \alpha> 0\) holds. Let \(n = n(k) > 0\) be the smallest numbers satisfying this inequality for \(k \in K\). For simplicity let us adopt \(x = f^{k}(x_{0})\) (\(x_{-1}= f^{k-1}(x_{0})\)), and \(x_{m} = f^{m}(x)\), \(m \in\mathbb{N}\). Clearly, \(x \in X_{j}\) yields \(x_{nt+1},x_{(n+1)t+1}\in X_{j++}\). In view of (7) we have

$$ \begin{aligned} p(x_{(n+1)t+1},x) &\leq\varphi \bigl(m_{f}(x_{(n+1)t},x_{-1}) \bigr) \\ & =\varphi \bigl(\max \bigl\{ p(x_{(n+1)t},x_{-1}),p(x_{(n+1)t+1},x_{(n+1)t}),p(x,x_{-1}) \bigr\} \bigr), \end{aligned} $$

which, for large *k* (or from (6) directly) gives

$$p(x_{(n+1)t+1},x) \leq\varphi \bigl(p(x_{(n+1)t},x_{-1}) \bigr), $$

as

$$\alpha< p(x_{(n+1)t+1},x) \leq p(x_{(n+1)t+1},x_{(n+1)t}) + p(x_{(n+1)t},x_{-1}) + p(x_{-1},x). $$

Now,

$$ \begin{aligned} p(x_{(n+1)t},x_{-1}) &\leq p(x_{(n+1)t},x_{(n+1)t-1}) + \cdots+ p(x_{nt+2},x_{nt+1}) + p(x_{nt+1},x) +p(x,x_{-1}) \\ &\leq p(x_{(n+1)t},x_{(n+1)t-1}) + \cdots+ p(x_{nt+2},x_{nt+1})+ \alpha+ p(x,x_{-1}) \end{aligned} $$

yields \(p(x_{(n+1)t},x_{-1}) < \alpha+ \epsilon\), for large *k*. Consequently, from (3) and \(\varphi(\beta) < \alpha\), \(\beta\leq\alpha\), we obtain

$$\alpha< p(x_{(n+1)t+1},x) \leq\varphi \bigl(p(x_{(n+1)t},x_{-1}) \bigr) \leq\alpha, $$

for large *k*, a contradiction. Now, it is clear that \(\lim_{m,n \rightarrow\infty} p(x_{m+nt+1},x_{m}) = 0\). One step more is necessary for \(t > 1\). We have

$$ \begin{aligned} &\lim_{m,n \rightarrow\infty}p(x_{m+nt+s},x_{m}) \\ &\quad \leq\lim_{m,n \rightarrow\infty} \bigl[p(x_{m+nt+s},x_{m+nt+s-1}) + \cdots+ p(x_{m+nt+2},x_{m+nt+1}) + p(x_{m+nt+1},x_{m}) \bigr] = 0 \end{aligned} $$

for any \(s \in\{2,\ldots,t\}\), *i.e.*
\(\lim_{m,n \rightarrow \infty}p(x_{n},x_{m}) = 0\). □

Clearly Theorem 3.3 is more general than Theorem 3.1. The proof of Theorem 3.1 is easier, it helps to understand the idea of the proof of Theorem 3.3, and therefore, it is also presented.

Now, Theorem 3.3, and [6], Lemma 29 yield the following.

### Theorem 3.4

*Let*
\((X,p)\)
*be a* 0-*complete d*-*metric space*, *and let*
\(f: X \rightarrow X\)
*be a mapping such that*
\(f^{s}\)
*is cyclic on*
\(X_{1},\ldots,X_{t}\)
*for an*
\(s \in\mathbb{N}\). *Assume that* (6) *or* (7) *is satisfied for all*
\(x \in X_{j}\), \(y \in X_{j++}\), \(j=1,\ldots,t\)
*with*
*f*
*replaced by*
\(f^{s}\), *and a*
\(\varphi\in\Phi\)
*having property* (3). *Then*
*f*
*has a unique fixed point*; *if*
\(x = f(x)\), *then*
\(\lim_{n \rightarrow\infty}p(x,f^{n}(x_{0})) = p(x,x)= 0\), \(x_{0}\in X\).

### Remark 3.5

Let us note that [2], Lemmas 3.2, 3.8 stay valid if we assume that \((X,p)\) is 0-complete for orbits of *f*, *i.e.* (5) holds for \(x_{n}= f^{n}(x_{0} )\), \(x_{m} = f^{m}(x_{0})\), \(m,n \in\mathbb{N}\), \(x_{0}\in X\). Consequently, theorems of Section 3 stay valid if the assumption that \((X,p)\) is 0-complete is replaced by the requirement that \((X,p)\) is 0-complete for orbits of *f*.