Dini’s Theorem for Fuzzy Number-Valued Continuous Functions

Abstract

This work aims to provide several versions of Dini’s theorem for fuzzy number-valued continuous functions defined on a compact set K. In this context, there is a wide variety of possibilities since, unlike the real line, we can consider different topologies and orders on the set of fuzzy numbers. For example, we will show that the fuzzy Dini’s theorem holds for the usual partial orders and the most commonly used topologies but does not hold for all orders in general.

1. Introduction

Dini’s theorem states that if a monotonic sequence of real-valued continuous functions defined on a compact set converges pointwise to a continuous function, then the convergence is uniform. It is one of the few results that guarantee pointwise and uniform convergence are equivalent.

Suitable versions of Dini’s theorem have been stated in other contexts, such as frames [1], computable analysis [2,3], operator theory [4], and set-valued functions [5,6], but there is a shortage of literature in the fuzzy setting. In fact, we can find some Dini-type theorems (see [7], Theorem 3.5, and [8]) for continuous functions defined on a compact set and taking values in the set of fuzzy numbers endowed with the supremum metric $d_{\infty }$, but the proofs are unclear since they use a difference between fuzzy numbers that is not properly defined.

In this paper, we prove several versions of Dini’s theorem for fuzzy number-valued continuous functions defined on a compact set K. In this fuzzy context, there is a wide variety of possibilities since, unlike the real line, we can consider different topologies (the level convergence topology and the topologies induced by the usual metrics) and orders on the set of fuzzy numbers. In fact, we will show that the fuzzy Dini’s theorem is true for the usual partial orders and most commonly used topologies, but we provide a counterexample that confirms it is not true, in general, for all orders.

We would like to remark that Dini’s theorem for the level convergence topology, the sendograph, and the endograph metrics has not yet been addressed in the literature. Such results can be framed within fuzzy analysis, which has important implications in several fields, such as fuzzy automata, neutrosophic topologies, fuzzy integrals, fuzzy decision making, and integro-differential equations (see [9,10,11,12]).

2. Preliminaries

Throughout this paper, the word space refers to a topological Tychonoff space. In particular, K stands for a compact (Hausdorff) space, unless otherwise stated. $\mathbb{R}$ represents the real numbers and $\mathbb{N}$ the natural numbers. As topological spaces, they will be endowed with their usual topologies. Let $F\left(\mathbb{R}\right)$ denote the family of all fuzzy subsets, that is, mappings $u:\mathbb{R}\to [0,1]$. For $u\in F\left(\mathbb{R}\right)$ and $\lambda \in [0,1]$, the $\lambda$-level set of u is defined by

$${\left[u\right]}^{\lambda }:=\{x\in \mathbb{R}:u\left(x\right)\ge \lambda \},\lambda \in [0,1],{\left[u\right]}^0:={\mathrm{cl}}_{\mathbb{R}}\{x\in \mathbb{R}:u\left(x\right)>0\},$$

where ${\mathrm{cl}}_{\mathbb{R}}A$ stands for the closure of A in $\mathbb{R}$. For example, consider the fuzzy set

$$u\left(x\right)=\left\{\begin{array}{cc}x\hfill & \mathrm{if}x\in [0,1]\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Then, the $\lambda$-level sets are

$$\begin{array}{cc}{\left[u\right]}^{\lambda }\hfill & =\{x\in \mathbb{R}:u(x)\ge \lambda \}=\{x\in \mathbb{R}:x\ge \lambda \}=[\lambda ,1],(\lambda >0),\hfill \\ {\left[u\right]}^0\hfill & ={\mathrm{cl}}_{\mathbb{R}}\{x\in \mathbb{R}:u\left(x\right)>0\}={\mathrm{cl}}_{\mathbb{R}}\left((0,1]\right)=[0,1].\hfill \end{array}$$

The fuzzy number space ${\mathbb{E}}^1$ is the set of elements $u\in F\left(\mathbb{R}\right)$ that satisfy the following properties:

• u is normal, i.e., there exists an $x_0\in \mathbb{R}$ such that $u\left(x_0\right)=1$;
• u is convex, i.e., $u(\alpha x+(1-\alpha )y)\ge min\left\{u\left(x\right),u\left(y\right)\right\}$ for all $x,y\in \mathbb{R},\alpha \in [0,1]$;
• u is upper-semicontinuous;
• ${\left[u\right]}^0$, the support of u, is a compact set in $\mathbb{R}$.

Note that if $u\in {\mathbb{E}}^1$, then the $\lambda$-level set ${\left[u\right]}^{\lambda }$ of u is a compact interval for each $\lambda \in [0,1]$. We denote by $u^{-}\left(\lambda \right)$ and $u^{+}\left(\lambda \right)$ the endpoints of such an interval, that is, ${\left[u\right]}^{\lambda }=[u^{-}\left(\lambda \right),u^{+}\left(\lambda \right)]$.

The well-known characterization of fuzzy numbers provided by Goetschel and Voxman enables us to determine a fuzzy number by knowing its $\lambda$-levels.

Theorem 1 

([13]). Let $u\in {\mathbb{E}}^1$ and ${\left[u\right]}^{\lambda }=[u^{-}\left(\lambda \right),u^{+}\left(\lambda \right)]$, $\lambda \in [0,1]$. Then, the pair of functions $u^{-}\left(\lambda \right)$ and $u^{+}\left(\lambda \right)$ has the following properties:

(i)

$u^{-}\left(\lambda \right)$ is a bounded, left-continuous, nondecreasing function on $(0,1]$;

(ii)

$u^{+}\left(\lambda \right)$ is a bounded, left-continuous, nonincreasing function on $(0,1]$;

(iii)

$u^{-}\left(\lambda \right)$ and $u^{+}\left(\lambda \right)$ are right-continuous at $\lambda =0$;

(iv)

$u^{-}\left(1\right)\le u^{+}\left(1\right)$.

Conversely, if a pair of functions $\alpha \left(\lambda \right)$ and $\beta \left(\lambda \right)$ satisfies the above conditions (i)–(iv), then there exists a unique $u\in {\mathbb{E}}^1$ such that ${\left[u\right]}^{\lambda }=[\alpha \left(\lambda \right),\beta \left(\lambda \right)]$ for each $\lambda \in [0,1]$.

Given $u,v\in {\mathbb{E}}^1$ and $k\in \mathbb{R}$, we can define $u+v$ and $ku$ using interval arithmetic (see, e.g., [14]), and it is well-known that ${\mathbb{E}}^1$ endowed with these two natural operations is not a vector space but a cone.

Every real number r can be considered a fuzzy number since r can be identified with the fuzzy number $\tilde{r}$ defined as

$$\tilde{r}\left(x\right):=\left\{\begin{array}{cc}1\hfill & \mathrm{if}x=r,\\ 0\hfill & \mathrm{if}x\ne r\end{array}\right.$$

and its $\lambda$-level sets are the singleton $\left\{r\right\}$.

The set of fuzzy numbers ${\mathbb{E}}^1$ can be endowed with several topologies. Most of them are provided by a metric based on the Hausdorff metric.

We recall here the definition of the Hausdorff metric. If $A,B$ are closed sets of a metric space $(X,d)$, then

$$d_H(A,B)=max\{H^{*}(A,B),H^{*}(B,A)\}$$

where

$$H^{*}(A,B)=\underset{a\in A}{sup}d(a,B)=\underset{a\in A}{sup}\{\underset{b\in B}{inf}d(a,b)\}.$$

The Hausdorff distance between compact intervals of the real line has a more simplified expression:

$$d_H([a^{-},a^{+}],[b^{-},b^{+}])=max\left\{\right|a^{-}-b^{-}|,|a^{+}-b^{+}\left|\right\}.$$

Some examples of metrics in ${\mathbb{E}}^1$ are as follows:

• ${\mathbb{E}}_{\infty }^1=({\mathbb{E}}^1,d_{\infty })$, where

  $$d_{\infty }(u,v)=\underset{\lambda \in [0,1]}{sup}\left\{d_H({\left[u\right]}^{\lambda },{\left[v\right]}^{\lambda })\right\}.$$
• ${\mathbb{E}}_p^1=({\mathbb{E}}^1,d_p)$, where

  $$d_p(u,v)={\left({\int }_0^1{\left(d_H({\left[u\right]}^{\lambda },{\left[v\right]}^{\lambda })\right)}^{p}d\lambda \right)}^{1/p}.$$
• ${\mathbb{E}}_{end}^1=({\mathbb{E}}^1,d_{end})$, where

  $$d_{end}(u,v)=d_H(end\left(u\right),end\left(v\right)),$$

  and

  $$end\left(u\right)=\left\{\right(x,\alpha )\in \mathbb{R}\times [0,1]:u(x)\ge \alpha \}(\mathrm{endograph}\mathrm{of}u).$$
• ${\mathbb{E}}_s^1=({\mathbb{E}}^1,d_s)$, where

  $$d_s(u,v)=d_H(send\left(u\right),send\left(v\right)),$$

  and $send\left(u\right)$ is the supported endograph or sendograph of u defined by

  $$send\left(u\right)=\{(x,\alpha )\in {\left[u\right]}^0\times [0,1]:u\left(x\right)\ge \alpha \}.$$
• ${\mathbb{E}}_{\ell }^1=({\mathbb{E}}^1,{\tau }_{\ell })$, where ${\tau }_{\ell }$ is the level convergence topology.

The latter does not come from a metric. We say that a net of fuzzy numbers ${\left\{u_{\alpha }\right\}}_{\alpha \in I}$ levelly converges to a fuzzy number u if, for each $\lambda \in [0,1]$, it holds that

$$\underset{\alpha }{lim}{u}_{\alpha }^{+}\left(\lambda \right)=u^{+}\left(\lambda \right)\mathrm{and}\underset{\alpha }{lim}{u}_{\alpha }^{-}\left(\lambda \right)=u^{-}\left(\lambda \right).$$

The level convergence provides ${\mathbb{E}}^1$ with a topology ${\tau }_{\ell }$, and $({\mathbb{E}}^1,{\tau }_{\ell })$ can be identified as a subset of $({\mathbb{R}}^{[0,1]}\times {\mathbb{R}}^{[0,1]},{\tau }_p)$, where ${\tau }_p$ is the pointwise topology (see [15]). The basic open sets in ${\tau }_{\ell }$ are given by

$$V(u,\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n\},\epsilon )=\{v\in {\mathbb{E}}^1:\underset{1\le k\le n}{max}\left\{\right|v^{-}\left({\lambda }_K\right)-u^{-}\left({\lambda }_k\right)|,|v^{+}\left({\lambda }_k\right)-u^{+}\left({\lambda }_k\right)\left|\right\}<\epsilon \}$$

where $u\in {\mathbb{E}}^1$, $\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n\}\subset [0,1]$ and $\epsilon >0$.

Let X be a topological space. A continuous function $f:X\to ({\mathbb{E}}^1,{\tau }_{\ell })$ is called a level continuous function, and $C(X,{\mathbb{E}}_{\ell }^1)$ denotes the set of all level continuous functions from X to ${\mathbb{E}}^1$.

In this paper, we focus on the following partial orders in ${\mathbb{E}}^1$. For $u,v\in {\mathbb{E}}^1$, define the following:

• $u\le v$⇔$u^{-}\left(\lambda \right)\le v^{-}\left(\lambda \right)\mathrm{and}{u}^{+}\left(\lambda \right)\le v^{+}\left(\lambda \right),$ for all $\lambda \in [0,1]$.
• $u⪯v$⇔${\left[u\right]}^{\lambda }\subseteq {\left[v\right]}^{\lambda },$ for all $\lambda \in [0,1]$.

The first one is compatible with the order in the real line and it is the most commonly used order in the set of fuzzy numbers ${\mathbb{E}}^1$.

3. Dini’s Theorem for Level Continuous Fuzzy Number-Valued Functions

We begin by studying a Dini-type result for fuzzy-valued level continuous functions. Indeed, we can relax the monotonicity in Dini’s theorem to pointwise monotonicity (see [16], Definition 3).

Definition 1. 

Given an order in ${\mathbb{E}}^1$, a sequence of functions ${\left\{f_n\right\}}_{n\in \mathbb{N}}\subset C(K,{\mathbb{E}}^1)$ is called pointwise monotonic if ${\left\{f_n\left(t\right)\right\}}_{n\in \mathbb{N}}$ is monotone for every $t\in K$.

Let us recall that every function $f:K\to {\mathbb{E}}^1$ gives, for each $\lambda \in [0,1]$, two real-valued functions

$$f{(·)}^{-}\left(\lambda \right):K\to \mathbb{R}\mathrm{and}f{(·)}^{+}\left(\lambda \right):K\to \mathbb{R}$$

defined by $t\mapsto f{\left(t\right)}^{-}\left(\lambda \right)$ and $t\mapsto f{\left(t\right)}^{+}\left(\lambda \right)$, respectively.

Theorem 2. 

Consider the partially ordered cone $({\mathbb{E}}^1,\le )$ or $({\mathbb{E}}^1,⪯)$. Let ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ be a pointwise monotonic sequence in $C(K,{\mathbb{E}}_{\ell }^1)$, which converges pointwise to $f\in C(K,{\mathbb{E}}_{\ell }^1)$. Then, ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ converges uniformly to f.

Proof. 

Let ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ be such a sequence. We need to show that for every $\epsilon >0$ and $\lambda \in [0,1]$, there exists $n_0\in \mathbb{N}$ such that

$$\begin{array}{cc}\hfill |f_n{\left(t\right)}^{-}\left(\lambda \right)-f{\left(t\right)}^{-}\left(\lambda \right)|& <\epsilon \mathrm{and}|f_n{\left(t\right)}^{+}\left(\lambda \right)-f{\left(t\right)}^{+}\left(\lambda \right)|<\epsilon \hfill \end{array}$$

(1)

holds for all $n\ge n_0$ and every $t\in K$.

Assume, on the contrary, that there exists $\epsilon >0$ and ${\lambda }_0\in [0,1]$ such that for every $n_0\in \mathbb{N}$, we can find $n\ge n_0$ and $t_n\in K$ with

$$\begin{array}{cc}\hfill |f_n{\left(t_n\right)}^{-}\left({\lambda }_0\right)-f{\left(t_n\right)}^{-}\left({\lambda }_0\right)|& \ge \epsilon \mathrm{or}|f_n{\left(t_n\right)}^{+}\left({\lambda }_0\right)-f{\left(t_n\right)}^{+}\left({\lambda }_0\right)|\ge \epsilon .\hfill \end{array}$$

(2)

Define then, for every $n\in \mathbb{N}$,

$$B_n:=\{t\in K:max\{|f_n{\left(t\right)}^{+}\left({\lambda }_0\right)-f{\left(t\right)}^{+}\left({\lambda }_0\right)|,|f_n{\left(t\right)}^{-}\left({\lambda }_0\right)-f{\left(t\right)}^{-}\left({\lambda }_0\right)\left|\right\}\ge \epsilon \}.$$

Claim. $B_n$ is a closed set. Take a net ${\left\{t_d\right\}}_{d\in D}$ in $B_n$, which converges to a point $t\in K$. By the continuity of $f_n{(·)}^{\pm }\left({\lambda }_0\right)$ and $f{(·)}^{\pm }\left({\lambda }_0\right)$, we have that

$$\underset{d}{lim}{f}_n{\left(t_d\right)}^{\pm }\left({\lambda }_0\right)=f_n{\left(t\right)}^{\pm }\left({\lambda }_0\right)\mathrm{and}\underset{d}{lim}f{\left(t_d\right)}^{\pm }\left({\lambda }_0\right)=f{\left(t\right)}^{\pm }\left({\lambda }_0\right)$$

and, consequently,

$$max\left\{\right|f_n{\left(t\right)}^{\pm }\left({\lambda }_0\right)-f{\left(t\right)}^{\pm }\left({\lambda }_0\right)\left|\right\}=\underset{d}{lim}max\left\{\right|f_n{\left(t_d\right)}^{\pm }\left({\lambda }_0\right)-f{\left(t_d\right)}^{\pm }\left({\lambda }_0\right)|\ge \epsilon ,$$

which yields $t\in B_n$, so the claim is proved.

Since ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ is a pointwise monotonic sequence of functions that converges pointwise to f, then (see [17], Corollary 3)

$$\begin{array}{ccc}\hfill f_n\left(t\right)\le f_{n+1}\left(t\right)\le f\left(t\right)& \mathrm{or}& f_n\left(t\right)\ge f_{n+1}\left(t\right)\ge f\left(t\right),\hfill \\ \hfill (f_n\left(t\right)⪯f_{n+1}\left(t\right)⪯f\left(t\right)& \mathrm{or}& f_n\left(t\right)⪰f_{n+1}\left(t\right)⪰f\left(t\right),\mathrm{respectively})\hfill \end{array}$$

for every $t\in K$, yielding, in any case, $B_{n+1}\subseteq B_n$ for all $n\in \mathbb{N}$. We show just one case, the others being similar. Take $t\notin B_n$ and assume that $f_n\left(t\right)\le f_{n+1}\left(t\right)\le f\left(t\right)$. If $t\in B_{n+1}$, then

$$\begin{array}{cc}\hfill |f_n{\left(t\right)}^{-}\left({\lambda }_0\right)-f{\left(t\right)}^{-}\left({\lambda }_0\right)|& =f{\left(t\right)}^{-}\left({\lambda }_0\right)-f_{n+1}{\left(t\right)}^{-}\left({\lambda }_0\right)+(f_{n+1}{\left(t\right)}^{-}\left({\lambda }_0\right)-f_n{\left(t\right)}^{-}\left({\lambda }_0\right))\hfill \\ \hfill & \ge f{\left(t\right)}^{-}\left({\lambda }_0\right)-f_{n+1}{\left(t\right)}^{-}\left({\lambda }_0\right)\hfill \\ \hfill & =|f_{n+1}{\left(t\right)}^{-}\left({\lambda }_0\right)-f{\left(t\right)}^{-}\left({\lambda }_0\right)|\ge \epsilon ;\hfill \\ \hfill |f_n{\left(t\right)}^{+}\left({\lambda }_0\right)-f{\left(t\right)}^{+}\left({\lambda }_0\right)|& =f{\left(t\right)}^{+}\left({\lambda }_0\right)-f_{n+1}{\left(t\right)}^{+}\left({\lambda }_0\right)+(f_{n+1}{\left(t\right)}^{+}\left({\lambda }_0\right)-f_n{\left(t\right)}^{+}\left({\lambda }_0\right))\hfill \\ \hfill & \ge f{\left(t\right)}^{+}\left({\lambda }_0\right)-f_{n+1}{\left(t\right)}^{+}\left({\lambda }_0\right)\hfill \\ \hfill & =|f_{n+1}{\left(t\right)}^{+}\left({\lambda }_0\right)-f{\left(t\right)}^{+}\left({\lambda }_0\right)|\ge \epsilon .\hfill \end{array}$$

This contradicts that $t\notin B_n$. Hence, $t\notin B_{n+1}$, and it follows that $B_{n+1}\subseteq B_n$.

If $B_n\ne \varnothing$ for all $n\in \mathbb{N}$, with K being a compact set, we should obtain ${\bigcap }_{n\in \mathbb{N}}{B}_n\ne \varnothing$. Take $t_0\in {\bigcap }_{n\in \mathbb{N}}{B}_n$, which means that for all $n\in \mathbb{N}$,

$$max\left\{\right|f_n{\left(t_0\right)}^{-}\left({\lambda }_0\right)-f{\left(t_0\right)}^{-}\left({\lambda }_0\right)|,|f_n{\left(t\right)}^{+}\left({\lambda }_0\right)-f{\left(t\right)}^{+}\left({\lambda }_0\right)\left|\right\}\ge \epsilon$$

which is impossible, since ${\left\{f_n\left(t_0\right)\right\}}_{n\in \mathbb{N}}$ converges to $f\left(t_0\right)$ by hypothesis.

Thus, there exists $n_0\in \mathbb{N}$ such that $B_{n_0}=\varnothing$ and, consequently, $B_n=\varnothing$ for all $n\ge n_0$ as well. But this is a contradiction since, by our assumption, there must be $n\ge n_0$ and $t_n\in K$ satisfying Equation (2), which means that $t_n\in B_n$. Hence, Equation (1) holds, which finishes the proof. □

Now, from the above theorem, we can deduce the classical Dini’s theorem.

Corollary 1 

(Dini’s theorem). Let ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ be a pointwise monotonic sequence in $C(K,\mathbb{R})$, which converges pointwise to $f\in C(K,\mathbb{R})$. Then, ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ converges uniformly to f.

Proof. 

Since $\mathbb{R}$ can be embedded in $({\mathbb{E}}^1,\le )$, we can consider, for every function $g\in C(K,\mathbb{R})$, the composition $\widehat{g}:K\to \mathbb{R}\hookrightarrow {\mathbb{E}}^1$, which is a level continuous function. Therefore, if ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ is such a sequence, we show that the fuzzy number-valued functions $\widehat{f_n}$ satisfy the conditions in Theorem 2. To this end, fix $t_0\in K$. Assume, without loss of generality, that $f_n\left(t_0\right){\le }_{\mathbb{R}}{f}_{n+1}\left(t_0\right)$. Then,

$$\widehat{f_n}\left(t_0\right)=\widehat{f_n\left(t_0\right)}\le \widehat{f_{n+1}\left(t_0\right)}=\widehat{f_{n+1}}\left(t_0\right)$$

and ${\left\{\widehat{f_n}\right\}}_{n\in \mathbb{N}}$ is pointwise monotonic. Moreover, since ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ pointwise converges to f, then

$$\underset{n}{lim}\widehat{f_n}\left(t_0\right)=\underset{n}{lim}\widehat{f_n\left(t_0\right)}=\underset{n}{lim}\widehat{f\left(t_0\right)}=\underset{n}{lim}\widehat{f}\left(t_0\right)$$

and ${\left\{\widehat{f_n}\right\}}_{n\in \mathbb{N}}$ pointwise converges to $\widehat{f}$. Therefore, by applying Theorem 2, we find that ${\left\{\widehat{f_n}\right\}}_{n\in \mathbb{N}}$ converges uniformly to $\widehat{f}$, which provides the conclusion. □

4. Dini’s Theorem for $D_{\infty }$ and $D_p$ Metrics

In this section, we endow ${\mathbb{E}}^1$ with the $d_{\infty }$ or $d_p$ metrics and use the well-known result that, in a metric space, $(X,d)$, the distance function $d:X\times X\to \mathbb{R}$ is a continuous function.

First, we need some technical lemmas, which state that the Hausdorff distance preserves the order ≤ and ⪯ in ${\mathbb{E}}^1$.

Lemma 1. 

If $I,J,L$ are closed sets of a metric space $(X,d)$ and $I\subseteq J\subseteq L$, then $d_H(I,L)\ge d_H(J,L)$ and $d_H(I,L)\ge d_H(I,J)$.

Proof. 

Since $I\subseteq J$, for every $i\in I$ and $l\in L$, we have

$$d(l,i)\ge \underset{j\in J}{inf}d(l,j)=d(l,J),$$

so

$$d(l,I)=\underset{i\in I}{inf}d(l,i)\ge d(l,J)$$

and then,

$$H^{*}(L,I)=\underset{l\in L}{sup}d(l,I)\ge \underset{l\in L}{sup}d(l,J)=H^{*}(L,J).$$

On the other hand, since $I\subseteq L$, it is clear that $d(i,L)=0$ for every $i\in I$. So,

$$H^{*}(I,L)=\underset{i\in I}{sup}d(i,L)=0$$

and since $J\subseteq L$, then $H^{*}(J,L)=0$. This yields

$$d_H(I,L)=H^{*}(L,I)\ge H^{*}(L,J)=d_H(J,L).$$

In a similar way, we can obtain the second property. □

As a consequence, we obtain the following result.

Lemma 2 

(Compare with [17], Lemma 3). If $u,v,w\in {\mathbb{E}}^1$ satisfy $u\le v\le w\mathrm{or}u⪯v⪯w,$ then, for every $\lambda \in [0,1]$,

$$d_H({\left[u\right]}^{\lambda },{\left[w\right]}^{\lambda })\ge d_H({\left[v\right]}^{\lambda },{\left[w\right]}^{\lambda })$$

(3)

and

$$d_H({\left[u\right]}^{\lambda },{\left[w\right]}^{\lambda })\ge d_H({\left[u\right]}^{\lambda },{\left[v\right]}^{\lambda }).$$

(4)

Proof. 

The proof is straightforward from Lemma 1. □

The following result states Dini’s theorem for $d_{\infty }$-continuous fuzzy number-valued functions, which corrects some of the proofs given in the literature (see [7,8]).

Theorem 3. 

Consider the partially ordered cone $({\mathbb{E}}^1,\le )$ or $({\mathbb{E}}^1,⪯)$. Let ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ be a pointwise monotonic sequence in $C(K,{\mathbb{E}}_{\infty }^1)$, which converges pointwise to $f\in C(K,{\mathbb{E}}_{\infty }^1)$. Then, ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ converges uniformly to f.

Proof. 

Let ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ be such a sequence. Fix $\epsilon >0$ and consider, for each $n\in \mathbb{N}$, the set

$$B_n=\{t\in K:d_{\infty }(f_n\left(t\right),f\left(t\right))\ge \epsilon \}.$$

$B_n$ is a closed set since $d_{\infty }(f_n\left(t\right),f\left(t\right))$ is a continuous function on K.

Claim. $B_{n+1}\subseteq B_n$ for all $n\in \mathbb{N}$. Consider $t_0\in B_{n+1}$ and assume that

$$f_n\left(t_0\right)\le f_{n+1}\left(t_0\right)\mathrm{or}{f}_n\left(t_0\right)⪯f_{n+1}\left(t_0\right).$$

Since ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ converges pointwise to f, [17] (Lemma 4), shows that we also have $f_n\left(t_0\right)\le f_{n+1}\left(t_0\right)\le f\left(t_0\right)$ (or $f_n\left(t_0\right)⪯f_{n+1}\left(t_0\right)⪯f\left(t_0\right)$, respectively). By using Equation (3) from Lemma 2, this results in

$$\begin{array}{cc}\hfill d_{\infty }(f_n\left(t_0\right),f\left(t_0\right))& =\underset{\lambda \in [0,1]}{sup}\left\{d_H({\left[f_n\left(t_0\right)\right]}^{\lambda },{\left[f\left(t_0\right)\right]}^{\lambda })\right\}\hfill \\ \hfill & \ge \underset{\lambda \in [0,1]}{sup}\left\{d_H({\left[f_{n+1}\left(t_0\right)\right]}^{\lambda },{\left[f\left(t_0\right)\right]}^{\lambda })\right\}\hfill \\ \hfill & =d_{\infty }(f_{n+1}\left(t_0\right),f\left(t_0\right)).\hfill \end{array}$$

If, on the contrary, $f_n\left(t_0\right)\ge f_{n+1}\left(t_0\right)$ (or $f_n\left(t_0\right)⪰f_{n+1}\left(t_0\right)$, respectively), [17] (Lemma 4) shows that we also have $f_n\left(t_0\right)\ge f_{n+1}\left(t_0\right)\ge f\left(t_0\right)$ ($f_n\left(t_0\right)⪰f_{n+1}\left(t_0\right)⪰f\left(t_0\right)$, respectively). By using Equation (4) from Lemma 2, this results in

$$\begin{array}{cc}\hfill d_{\infty }(f_n\left(t_0\right),f\left(t_0\right))& =\underset{\lambda \in [0,1]}{sup}\left\{d_H({\left[f_n\left(t_0\right)\right]}^{\lambda },{\left[f\left(t_0\right)\right]}^{\lambda })\right\}\hfill \\ \hfill & \ge \underset{\lambda \in [0,1]}{sup}\left\{d_H({\left[f_{n+1}\left(t_0\right)\right]}^{\lambda },{\left[f\left(t_0\right)\right]}^{\lambda })\right\}\hfill \\ \hfill & =d_{\infty }(f_{n+1}\left(t_0\right),f\left(t_0\right)).\hfill \end{array}$$

Therefore, we obtain in any case $d_{\infty }(f_n\left(t_0\right),f\left(t_0\right))\ge d_{\infty }(f_{n+1}\left(t_0\right),f\left(t_0\right))\ge \epsilon$. Thus, $t_0\in B_n$, which finishes the proof of the claim.

Reasoning again as in Theorem 2, we obtain $n_0\in \mathbb{N}$ such that $B_n=\varnothing$ for all $n\ge n_0$, which yields

$$d_{\infty }(f_n\left(t\right),f\left(t\right))<\epsilon$$

for all $n\ge n_0$ and every $t\in K$. This finishes the proof. □

The next proposition presents Dini’s theorem for $d_p$-continuous fuzzy number-valued functions with both orders.

Proposition 1. 

Consider the partially ordered cone $({\mathbb{E}}^1,\le )$ or $({\mathbb{E}}^1,⪯)$. Let ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ be a pointwise monotonic sequence in $C(K,{\mathbb{E}}_p^1)$, which converges pointwise to $f\in C(K,{\mathbb{E}}_p^1)$. Then, ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ converges uniformly to f.

Proof. 

Since $d_p$ also preserves the order, by using Lemma 2, the proof is similar to that of Theorem 3. □

5. Dini’s Theorem for Sendograph and Endograph Metrics

Recall that if $u\in {\mathbb{E}}^1$, then the sendograph of u is the compact set

$$send\left(u\right):=\{(x,\alpha )\in {\left[u\right]}^0\times [0,1]:u\left(x\right)\ge \alpha \}$$

and the sendograph metric in ${\mathbb{E}}^1$ is defined by

$$d_s(u,v)=d_H(send\left(u\right),send\left(v\right)).$$

Lemma 3 

([18] Theorem 4.5). Let $u,v,w\in ({\mathbb{E}}^1,\le )$. If $u\le v\le w$, then $d_s(u,w)\ge d_s(u,v)$.

Therefore, the technique in the previous theorems can be applied to the sendograph metric as well, providing the following Dini’s theorem for $d_s$-continuous fuzzy number-valued functions.

Theorem 4. 

Consider the partially ordered cone $({\mathbb{E}}^1,\le )$ or $({\mathbb{E}}^1,⪯)$. Let ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ be a pointwise monotonic sequence in $C(K,{\mathbb{E}}_s^1)$, which converges pointwise to $f\in C(K,{\mathbb{E}}_s^1)$. Then, ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ converges uniformly to f.

Proof. 

Let ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ be such a sequence. Fix $\epsilon >0$ and consider, for each $n\in \mathbb{N}$,

$$B_n=\{t\in K:d_s(f_n\left(t\right),f\left(t\right))\ge \epsilon \}.$$

As in the previous theorem, $B_n$ is a closed set, and we need to show that $B_{n+1}\subseteq B_n$.

Fix $t_0\in B_{n+1}$. First consider $({\mathbb{E}}^1,⪯)$. It is apparent that $(x,\alpha )\in send\left(f_n\left(t_0\right)\right)$ iff $x\in {\left[f_n\left(t_0\right)\right]}^{\alpha }$.

Since $f_n\left(t_0\right)⪯f_{n+1}\left(t_0\right)⪯f\left(t_0\right)$ or $f_n\left(t_0\right)⪰f_{n+1}\left(t_0\right)⪰f\left(t_0\right)$, we obtain $send\left(f_n\left(t_0\right)\right)\subseteq send\left(f_{n+1}\left(t_0\right)\right)\subseteq send\left(f\left(t_0\right)\right)$ (or $send\left(f_n\left(t_0\right)\right)\supseteq send\left(f_{n+1}\left(t_0\right)\right)\supseteq send\left(f\left(t_0\right)\right)$. Lemma 1 shows that

$$\begin{array}{cc}\hfill d_s(f_n(t{(}_0),f\left(t_0\right))& =d_H(send\left(f_n\left(t_0\right)\right),send\left(f\left(t_0\right)\right))\hfill \\ \hfill & \ge d_H(send\left(f_{n+1}\left(t_0\right)\right),send\left(f\left(t_0\right)\right))=d_s(f_{n+1}(t{(}_0),f\left(t_0\right)).\hfill \end{array}$$

Now consider $({\mathbb{E}}^1,\le )$. Since $f_n\left(t_0\right)\le f_{n+1}\left(t_0\right)\le f\left(t_0\right)$ or $f\left(t_0\right)\le f_{n+1}\left(t_0\right)\le f_n\left(t_0\right)$, by using Lemma 3, we also have

$$d_s(f_n\left(t_0\right),f\left(t_0\right))\ge d_s(f_{n+1}\left(t_0\right),f\left(t_0\right))$$

Therefore, in both orders, $d_s(f_n\left(t_0\right),f\left(t_0\right))\ge \epsilon$, which yields $t_0\in B_n$ and, consequently, $B_{n+1}\subseteq B_n$.

With K being compact, there is $n_0\in \mathbb{N}$ such that $B_n=\varnothing$ for all $n\ge n_0$. That is,

$$d_s(f_n\left(t\right),f\left(t\right))<\epsilon$$

for all $n\ge n_0$ and $t\in K$, which gives us the desired conclusion. □

Assume that $u,v\in {\mathbb{E}}^1$ satisfy $u⪯v$. Since $end\left(u\right)=send\left(u\right)\cup (\mathbb{R}\times \{0\left\}\right)$ and $end\left(v\right)=send\left(v\right)\cup (\mathbb{R}\times \{0\left\}\right)$, it is clear that $end\left(u\right)\subseteq end\left(v\right)$, yielding the following property: if $u⪯v⪯w$, then $d_{end}(u,v)\le d_{end}(u,w)$. On the other hand, [19] (Remark 6) points out that, if $u,v,w\in {\mathbb{E}}^1$ with $u\le v\le w$, then $d_{end}(u,w)\ge d_{end}(u,v)$ as well. Since no proof is given there, we show this property here for the sake of completeness.

Lemma 4. 

If $u,v,w\in {\mathbb{E}}^1$ with $u\le v\le w$, then $d_{end}(u,w)\ge d_{end}(u,v)$.

Proof. 

Assume that $u,v,w\in {\mathbb{E}}^1$ with $u\le v\le w$. In order to prove that $d_{end}(u,v)\le d_{end}(u,w)$, we need to show two conditions:

$$\forall (x,\lambda )\in end\left(u\right),\exists (y,\beta )\in end\left(v\right)\mathrm{such}\mathrm{that}d((x,\lambda ),(y,\beta ))\le d_{end}(u,w).$$

(5)

$$\forall (y,\beta )\in end\left(v\right),\exists (x,\lambda )\in end\left(u\right)\mathrm{such}\mathrm{that}d((x,\lambda ),(y,\beta ))\le d_{end}(u,w).$$

(6)

We start by checking condition (5). Consider $(x,\lambda )\in end\left(u\right)$. If $(x,\lambda )\in \mathbb{R}\times \left\{0\right\}\subset end\left(v\right)$, there is nothing to prove since $d((x,\lambda ),(x,\lambda ))=0\le d_{end}(u,w)$. Thus, we can assume that $\lambda >0$ and $x\in {\left[u\right]}^{\lambda }$.

Now, there exists $(z,\gamma )\in end\left(w\right)$ with

$$d((x,\lambda ),(z,\gamma ))=d((x,\lambda ),end\left(w\right))\le d_{end}(u,w).$$

Set $\rho =min\{\lambda ,\gamma \}$. If $\rho =0$, then $(z,\gamma )=(z,0)\in end\left(v\right)$, and we obtain

$$d((x,\lambda ),(z,0))\le d_{end}(u,w).$$

We assume that $\rho >0$. If $x\in {\left[v\right]}^{\rho }$, then $(x,\rho )\in end\left(v\right)$ and

$$d((x,\lambda ),(x,\rho ))=\lambda -\rho \le |\lambda -\gamma |\le d((x,\lambda ),(z,\gamma ))\le d_{end}(u,w).$$

If $x\notin {\left[v\right]}^{\rho }$, since $x\in {\left[u\right]}^{\lambda }\subseteq {\left[u\right]}^{\rho }$, we obtain

$$u^{-}\left(\rho \right)\le x<v^{-}\left(\rho \right)\le w^{-}\left(\rho \right)\le z,$$

yielding

$$\begin{array}{cc}\hfill d((x,\lambda ),(v^{-}\left(\rho \right),\rho ))& =\sqrt{|v^{-}{\left(\rho \right)-x|}^2+{|\lambda -\rho |}^2}\le \sqrt{{|z-x|}^2+{|\lambda -\gamma |}^2}\hfill \\ \hfill & =d((x,\lambda ),(z,\gamma ))\le d_{end}(u,w).\hfill \end{array}$$

Since $(v^{-}\left(\rho \right),\rho )\in end\left(v\right)$, we have proven this part.

Now, we will check condition (6). Consider $(y,\beta )\in end\left(v\right)$. As usual, if $(y,\beta )\in end\left(u\right)$, there is nothing to prove. If $(y,\beta )\in end\left(w\right)$, there exists $(x,\lambda )\in end\left(u\right)$ such that

$$d((x,\lambda ),(y,\beta ))=d(end\left(u\right),(y,\beta ))\le d_{end}(u,w)$$

and we obtain the conclusion.

Therefore, we can assume that $\beta >0$ and $y\in {\left[v\right]}^{\beta }$, and that the point $(y,\beta )$ does not lie in $end\left(u\right)$ or $end\left(w\right)$. Since $(u^{+}\left(\beta \right),\beta )\in end\left(u\right)$, there exists $(z,\gamma )\in end\left(w\right)$ such that

$$d((u^{+}\left(\beta \right),\beta ),(z,\gamma ))=d((u^{+}\left(\beta \right),\beta ),end\left(w\right))\le d_{end}(u,w).$$

Consider $\rho =min\{\beta ,\gamma \}$. If $\rho =0$, then $(z,\gamma )=(z,0)$. Consider $(y,0)\in end\left(u\right)$, and we have

$$d((y,\beta ),(y,0))=\beta \le d((u^{+}\left(\beta \right),\beta ),(z,0))\le d_{end}(u,w).$$

Assume now that $\rho >0$. Since $z\in {\left[w\right]}^{\gamma }\subseteq {\left[w\right]}^{\rho }$, $y\in {\left[v\right]}^{\rho }\setminus {\left[w\right]}^{\rho }$ and $y\in {\left[v\right]}^{\rho }\setminus {\left[u\right]}^{\rho }$, then

$$u^{+}\left(\rho \right)<y<w^{-}\left(\rho \right)\le z.$$

On the other hand, $u^{+}\left(\beta \right)\in {\left[u\right]}^{\beta }\subseteq {\left[u\right]}^{\rho }$, yielding $(u^{+}\left(\beta \right),\rho )\in end\left(u\right)$.

If $\rho =\gamma$, then

$$\begin{array}{cc}\hfill d((y,\beta ),(u^{+}\left(\beta \right),\rho ))& =\sqrt{{(y-u^{+}\left(\beta \right))}^2+{|\beta -\rho |}^2}<\sqrt{{(z-u^{+}\left(\beta \right))}^2+{|\beta -\gamma |}^2}\hfill \\ \hfill & =d((u^{+}\left(\beta \right),\beta ),(z,\gamma ))\le d_{end}(u,w).\hfill \end{array}$$

If $\rho =\beta$, we have

$$\begin{array}{cc}\hfill d((y,\beta ),(u^{+}\left(\beta \right),\rho ))& =|y-u^{+}\left(\beta \right)|<|z-u^{+}\left(\beta \right)|\hfill \\ \hfill & =d((u^{+}\left(\beta \right),\beta ),(z,\gamma ))\le d_{end}(u,w).\hfill \end{array}$$

Therefore, the condition (6) is also true and the proof is finished. □

Now, we can show the following theorem, which is a version of Dini’s theorem for $d_{end}$-continuous fuzzy number-valued functions.

Theorem 5. 

Consider the partially ordered cone $({\mathbb{E}}^1,\le )$ or $({\mathbb{E}}^1,⪯)$. Let ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ be a pointwise monotonic sequence in $C(K,{\mathbb{E}}_{end}^1)$, which converges pointwise to $f\in C(K,{\mathbb{E}}_{end}^1)$. Then, ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ converges uniformly to f.

Proof. 

Let ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ be such a sequence. Fix $\epsilon >0$ and consider, for each $n\in \mathbb{N}$,

$$B_n=\{t\in K:d_{end}(f_n\left(t\right),f\left(t\right))\ge \epsilon \}.$$

As in Theorem 4, $B_n$ is a closed set, and we need to show that $B_{n+1}\subseteq B_n$. To do this, fix $t_0\in B_{n+1}$.

Consider the order $({\mathbb{E}}^1,⪯)$. It is apparent that $(x,\alpha )\in end\left(f_n\left(t_0\right)\right)$ iff $x\in {\left[f_n\left(t_0\right)\right]}^{\alpha }$ or $(x,\alpha )\in \mathbb{R}\times \left\{0\right\}$.

Since $f_n\left(t_0\right)⪯f_{n+1}\left(t_0\right)⪯f\left(t_0\right)$ or $f_n\left(t_0\right)⪰f_{n+1}\left(t_0\right)⪰f\left(t_0\right)$, we obtain $end\left(f_n\left(t_0\right)\right)\subseteq end\left(f_{n+1}\left(t_0\right)\right)\subseteq end\left(f\left(t_0\right)\right)$ (or $end\left(f_n\left(t_0\right)\right)\supseteq end\left(f_{n+1}\left(t_0\right)\right)\supseteq end\left(f\left(t_0\right)\right)$. Lemma 1 shows that

$$\begin{array}{cc}\hfill d_{end}(f_n(t{(}_0),f\left(t_0\right))& =d_H(end\left(f_n\left(t_0\right)\right),end\left(f\left(t_0\right)\right))\hfill \\ \hfill & \ge d_H(end\left(f_{n+1}\left(t_0\right)\right),end\left(f\left(t_0\right)\right))=d_{end}(f_{n+1}(t{(}_0),f\left(t_0\right))\ge \epsilon .\hfill \end{array}$$

On the other hand, consider the order $({\mathbb{E}}^1,\le )$. Since $f_n\left(t_0\right)\le f_{n+1}\left(t_0\right)\le f\left(t_0\right)$ or $f\left(t_0\right)\le f_{n+1}\left(t_0\right)\le f_n\left(t_0\right)$, by using Lemma 4, we also have

$$d_{end}(f_n\left(t_0\right),f\left(t_0\right))\ge d_{end}(f_{n+1}\left(t_0\right),f\left(t_0\right))\ge \epsilon .$$

Therefore, in both orders, $d_{end}(f_n\left(t_0\right),f\left(t_0\right))\ge \epsilon$, which yields $t_0\in B_n$ and, consequently, $B_{n+1}\subseteq B_n$.

Since K is a compact set, there is $n_0\in \mathbb{N}$ such that $B_n=\varnothing$ for all $n\ge n_0$. That is,

$$d_{end}(f_n\left(t\right),f\left(t\right))<\epsilon$$

for all $n\ge n_0$ and $t\in K$, which means that ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ converges uniformly to f and the proof is finished. □

6. A Counterexample

We give an example showing that the fuzzy version of Dini’s theorem is not true for any order in ${\mathbb{E}}^1$. In [13] (Definition 2.5) a total (pre)order ≾ in the set of fuzzy numbers ${\mathbb{E}}^1$ is defined as

$$u\precsim v\iff {\int }_0^1\alpha \left[u^{-}\left(\alpha \right)+u^{+}\left(\alpha \right)\right]d\alpha \le {\int }_0^1\alpha \left[v^{-}\left(\alpha \right)+v^{+}\left(\alpha \right)\right]d\alpha .$$

Example 1. 

Consider the ordered cone $({\mathbb{E}}^1,\precsim )$. For every $x\in [0,1]$ and every $n\in \mathbb{N}$, consider the following real-valued functions:

$$\begin{array}{cc}\hfill f_n{\left(x\right)}^{-}\left(\lambda \right)& =0,\mathit{for}\mathit{all}\lambda \in [0,1],\hfill \\ \hfill f_n{\left(x\right)}^{+}\left(\lambda \right)& =\left\{\begin{array}{cc}{\displaystyle \frac{n}{\sqrt{e^n-1}}}{e}^{-{\displaystyle \frac{n}{e^n-1}}}{e}^{-n{\lambda }^2}\hfill & \mathit{if}0\le x\le {\displaystyle \frac{1}{\sqrt{e^n-1}}},\hfill \\ nx{e}^{-n{x}^2}{e}^{-n{\lambda }^2}\hfill & \mathit{if}{\displaystyle \frac{1}{\sqrt{e^n-1}}}<x\le 1,\hfill \end{array}\right.\mathit{for}\mathit{all}\lambda \in [0,1].\hfill \end{array}$$

Then, $f_n{\left(x\right)}^{-}\left(\lambda \right)$ and $f_n{\left(x\right)}^{+}\left(\lambda \right)$ satisfy the conditions in the Goetschel–Voxman characterization (Theorem 1), so there exists a unique fuzzy number $f_n\left(x\right)\in {\mathbb{E}}^1$ for all $x\in [0,1]$, such that

$${\left[f_n\left(x\right)\right]}^{\lambda }=[f_n{\left(x\right)}^{-}\left(\lambda \right),f_n{\left(x\right)}^{+}\left(\lambda \right)].$$

Claim: $f_n:[0,1]\to {\mathbb{E}}^1$ is a level continuous function for each $n\in \mathbb{N}$, and the sequence ${\left\{f_n\right\}}_{n\in \mathbb{N}}$ converges pointwise to 0.

To do this, fix $n_0$ and $x_0\in [0,1]$. Consider any fixed $\lambda \in [0,1]$. If $x_0\ne {\displaystyle \frac{1}{\sqrt{e^{n_0}-1}}}$, then

$$\underset{x\to x_0}{lim}{f}_{n_0}{\left(x\right)}^{+}\left(\lambda \right)=f_{n_0}{\left(x_0\right)}^{+}\left(\lambda \right).$$

If $x_0={\displaystyle \frac{1}{\sqrt{e^{n_0}-1}}}$, then

$$\begin{array}{cc}\hfill \underset{x\to x_0}{lim}{f}_{n_0}{\left(x\right)}^{+}\left(\lambda \right)& =\left\{\begin{array}{c}\underset{x\to x_0-}{lim}{\displaystyle \frac{n_0}{\sqrt{e^{n_0}-1}}}{e}^{-{\displaystyle \frac{n_0}{e^{n_0}-1}}}{e}^{-n_0{\lambda }^2}={\displaystyle \frac{n_0}{\sqrt{e^{n_0}-1}}}{e}^{-{\displaystyle \frac{n_0}{e^{n_0}-1}}}{e}^{-n_0{\lambda }^2}\hfill \\ \underset{x\to x_0+}{lim}{n}_{0}x{e}^{-n_0{x}^2}{e}^{-n_0{\lambda }^2}={\displaystyle \frac{n_0}{\sqrt{e^{n_0}-1}}}{e}^{-{\displaystyle \frac{n_0}{e^{n_0}-1}}}{e}^{-n_0{\lambda }^2}\hfill \end{array}\right.\hfill \\ \hfill & =f_{n_0}{\left(x_0\right)}^{+}\left(\lambda \right).\hfill \end{array}$$

Let us check the pointwise convergence to 0. If $x=0$, then

$$\underset{n\to +\infty }{lim}{f}_n{\left(0\right)}^{+}\left(\lambda \right)=\underset{n\to +\infty }{lim}{\displaystyle \frac{n}{\sqrt{e^n-1}}}{e}^{-{\displaystyle \frac{n}{e^n-1}}}{e}^{-n{\lambda }^2}=0$$

and if $x\ne 0$, there exists $n_0$ such that ${\displaystyle \frac{1}{\sqrt{e^n-1}}}<x$ for all $n\ge n_0$, so

$$\underset{n\to +\infty }{lim}{f}_n{\left(x\right)}^{+}\left(\lambda \right)=\underset{n\to +\infty }{lim}nx{e}^{-n{x}^2}{e}^{-n{\lambda }^2}=0.$$

Thus, the claim is proved.

Moreover, ${\left\{f_n\right\}}_{n\ge 2}$ is a monotone decreasing sequence. In order to prove this, we define $T_n\left(x\right):={\int }_0^1\lambda f_n{\left(x\right)}^{+}\left(\lambda \right)d\lambda$, and we need to show that for each $x\in [0,1]$, ${\left\{T_n\left(x\right)\right\}}_{n\in \mathbb{N}}$ is a decreasing sequence of real numbers. We have

$$T_n\left(x\right)=\left\{\begin{array}{cc}{G}_n\left(x\right)\hfill & \mathit{if}0\le x\le {\displaystyle \frac{1}{\sqrt{e^n-1}}},\hfill \\ H_n\left(x\right)\hfill & \mathit{if}{\displaystyle \frac{1}{\sqrt{e^n-1}}}<x\le 1.\hfill \end{array}\right.$$

where

$$\begin{array}{cc}\hfill G_n\left(x\right)& :={\int }_0^1\lambda {\displaystyle \frac{n{e}^{-\frac{n}{e^n-1}}{e}^{-n{\lambda }^2}}{\sqrt{e^n-1}}}d\lambda ={\displaystyle \frac{e^{-\frac{n}{e^n-1}}}{2\sqrt{e^n-1}}}\left(1-e^{-n}\right),\hfill \\ \hfill H_n\left(x\right)& :={\int }_0^1\lambda nx{e}^{-n{x}^2}{e}^{-n{\lambda }^2}d\lambda ={\displaystyle \frac{x{e}^{-n{x}^2}}{2}}\left(1-e^{-n}\right).\hfill \end{array}$$

$G_n\left(x\right)$ is decreasing for $n\ge 2$ and $H_n\left(x\right)$ is decreasing for $n>log\left({\displaystyle \frac{1}{x^2}}+1\right)$, that is, ${\displaystyle \frac{1}{\sqrt{e^n-1}}}<x$. So, the only remaining case to show is

$$T_n\left(x\right)=G_n\left(x\right)>T_{n+1}\left(x\right)=H_{n+1}\left(x\right)\mathrm{if}{\displaystyle \frac{1}{\sqrt{e^{n+1}-1}}}<x\le {\displaystyle \frac{1}{\sqrt{e^n-1}}}.$$

(7)

We show that the function $F_n\left(x\right)=G_n\left(x\right)-H_{n+1}\left(x\right)>0$ for all $x\in J_n:=\left[{\displaystyle \frac{1}{\sqrt{e^{n+1}-1}}},{\displaystyle \frac{1}{\sqrt{e^n-1}}}\right]$.

First, the fact that $F_n\left({\displaystyle \frac{1}{\sqrt{e^n-1}}}\right)=G_n\left({\displaystyle \frac{1}{\sqrt{e^n-1}}}\right)-H_{n+1}\left({\displaystyle \frac{1}{\sqrt{e^n-1}}}\right)>0$ is equivalent to proving that ${\displaystyle \frac{e^n-\frac{1}{e}}{e^n-1}}<e^{{\displaystyle \frac{1}{e^n-1}}}$, which is true because

$$e^{{\displaystyle \frac{1}{e^n-1}}}>1+{\displaystyle \frac{1}{e^n-1}}={\displaystyle \frac{e^n}{e^n-1}}>{\displaystyle \frac{e^n-\frac{1}{e}}{e^n-1}}.$$

On the other hand, the derivative

$${\displaystyle \frac{d}{dx}}{H}_{n+1}\left(x\right)={\displaystyle \frac{e^{n+1}-1}{2e^{n+1}{e}^{(n+1)x^2}}}\left(1-2x^2(n+1)\right)$$

is always positive in $J_n$, since for $n\ge 2$, we have

$$1-2x^2(n+1)\ge 1-{\displaystyle \frac{2(n+1)}{e^n-1}}>0.$$

Therefore, ${\displaystyle \frac{d}{dx}}{F}_n\left(x\right)=-{\displaystyle \frac{d}{dx}}{H}_{n+1}\left(x\right)<0$, and it is a decreasing function in the interval $J_n$. The condition $F_n\left({\displaystyle \frac{1}{\sqrt{e^n-1}}}\right)>0$ implies that $F_n\left(x\right)>0$ for all $x\in J_n$, so Equation (7) holds.

We have just shown that the sequence ${\left\{T_n\left(x\right)\right\}}_{n\ge 2}$ is decreasing for each $x\in [0,1]$, which proves the claim.

Finally, we show that the convergence of ${\left\{f_n\right\}}_{n\ge 2}$ is not uniform. Assume, for the sake of contradiction, that for $\epsilon ={\displaystyle \frac{1}{4}}$ and $\lambda =0$, there exists $n_0$ such that $f_n{\left(x\right)}^{+}\left(0\right)<\epsilon$ holds for every $n\ge n_0$ and $x\in [0,1]$. This yields

$$nx{e}^{-n{x}^2}<{\displaystyle \frac{1}{4}}$$

for all $n\ge n_0$ and every $x>{\displaystyle \frac{1}{\sqrt{e^n-1}}}$, which means that

$${\displaystyle \frac{1}{2e^{\frac{n}{e^n-1}}}}-{\displaystyle \frac{1}{2n{e}^n}}={\int }_{{\displaystyle \frac{1}{\sqrt{e^n-1}}}}^{1}nx{e}^{-n{x}^2}dx\le {\displaystyle \frac{1}{4}}\left(1-{\displaystyle \frac{1}{\sqrt{e^n-1}}}\right).$$

Hence, by taking limits for n, we obtain ${\displaystyle \frac{1}{2}}\le {\displaystyle \frac{1}{4}}$, which is a contradiction.

7. Conclusions

We have studied several fuzzy versions of the celebrated Dini’s theorem for continuous fuzzy number-valued functions using different topologies and orders in the set of fuzzy numbers. Dini’s theorem is a key result that guarantees pointwise and uniform convergence are equivalent. Our approach clarifies some existing previous results that do not seem to be correctly proved. We have also provided a counterexample that shows that Dini’s theorem is not true for all orders.