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In this paper, in the proof of Theorem 2.1, the authors use the following result. We recall that, in this context, a "generalized metric space" is a metric for which the metric may assume infinite values.

Let $G$ be a non-empty set and $Y$ a Banach space. Define $S:=\{f:G\longrightarrow Y\}$ and the map $d:S\times S \longrightarrow [0,+\infty]$ as

$$ d(f,g):=\inf\{\alpha\geq 0: \|f(x)-g(x)\|\leq \alpha \psi(x) \quad \textrm{for all } x\in G\}, $$ where $\psi:G\longrightarrow [0,+\infty)$ is a given function.

Then, the authors state that $(S,d)$ is a complete generalized metric space. It is easy to show that $d$ satisfies all of the axioms of the metric definition. But, How can we prove that $d$ is complete?

Many thanks is advance for your comments and suggestions.

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Let $(f_n)_n$ be a Cauchy sequence in $S.$ Let $a_n=\sup \{d(f_m, f_{m'}):n\le m<m' \}.$ So $\lim_{n\to \infty}a_n=0.$

For any $x\in G$ we have $\|f_m(x)-f_{m'}(x)\|\le 2a_n\psi(x)$ whenever $n\le m<m',$ and $\lim_{n\to \infty}2a_n\psi(x)=0.$

So $(f_n(x))_n$ is a Cauchy sequence in $Y.$ Since $Y$ is complete, $f_n(x)$ converges to some $f(x)\in Y. $

Given $\epsilon\in \Bbb R^+, $ choose $n$ large enough that $\forall m\ge n\,(d(f_n,f_m)\le\epsilon).$ For each $x\in G$ the set $\{f_m(x):m\ge n\}$ is contained in the closed ball $C(x)=\overline {B(f_n(x),2\epsilon \cdot \psi(x))}$ of $Y,$ so $f_m(x)$ converges in $Y$ to a member of $C(x)$. That is, $\|f(x)-f_n(x)\|\le 2\epsilon \cdot \psi(x)\in C(x).$ Therefore $$\forall m\ge n\,\forall x\in G\,(\|f(x)-f_n(x)\|\le 2\epsilon \cdot \psi(x)\,).$$ That is,$ \forall m\ge n\,(\,d(f,f_n)\le 2\epsilon).$

So $\lim_{n\to \infty}d(f,f_n)=0.$

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  • $\begingroup$ Many thanks for your reply DanielWainfleet. I do not understand at all the number "2" in the proof....For instance, as $d(f_{m},f_{m'})\leq a_{n}$ for all $m'>m\geq n$, then $\|f_{m}(z)-f_{m'}(x)\|\leq a_{n}\psi(x)$, it is correct? $\endgroup$ – user123043 Apr 28 at 16:52
  • $\begingroup$ Just being cautious about the $\inf$ in the def'n of $d$, but that $\inf$ is actually a $\min$ so the $2$ could be dropped. $\endgroup$ – DanielWainfleet Apr 29 at 1:59
  • $\begingroup$ Ok, thank you very much. $\endgroup$ – user123043 Apr 29 at 16:09

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