Strong b-Suprametric Spaces and Fixed Point Principles

Abstract

In this paper, we introduce the strong b-suprametric spaces in which we prove the fixed point principles of Banach and Edelstein. Moreover, we prove a variational principle of Ekeland and deduce a Caristi fixed point theorem. Furthermore, we introduce the strong b-supranormed linear spaces in which we establish the fixed point principles of Brouwer and Schauder. As applications, we study the existence of solutions to an integral equation and to a third-order boundary value problem.

1 Introduction

Let X be a nonempty set and ${\mathbb {R}}_{+}$ be the set of all nonnegative real numbers. A semimetric is a distance function $d:X\,{ \times }\, X \,{ \rightarrow }\, {\mathbb {R}}_{+}$ that satisfies two axioms: $(d_{1})$: $d(x,y)\,{ = }\,0$ if and only if $x\,{ = }\,y$; $(d_{2})$: $d(x,y)\,{ = }\, d(y,x)$ for all $x,y\,{ \in }\, X$. It is well known that by adding the triangle inequality to the axioms of d it becomes continuous. In 1993, Czerwik [8] investigated a semimetric called b-metric, which satisfies the inequality: $d(x,y) \le b(d(x,z)+d(z,y))$, where b is a constant in $[1,+\infty )$ and $x,y,z \,{\in }\, X$. This notion has been studied previously by different authors, for the latest and rather complete bibliography, we refer the reader to the surveys of Berinde and Păcurar [2] and Karapınar [14]. Despite the b-metric is very useful in applications [7, 15, 27], it has a major drawback due to its lack of continuity [26]. In order to overcome this limitation, Kirk and Shahzad proposed a slight modification in the third axiom, see [16, 17].

The suprametric, which was introduced by the author in [3], is a semimetric that fulfill $d(x,y) \le d(x,z) + d(z,y) + \rho d(x,z)d(z,y)$, where $\rho$ is a constant in ${\mathbb {R}}_{+}$ and $x,y,z \,{\in }\, X$. This distance function is very useful to construct projective metrics of Thompson’s type [25], and to prove the existence of solutions to various classes of integral and matrix equations. Very recently, the suprametric has been utilized by Panda et al. [21, 22] to analyze complex valued fractional order neural networks and the existence of a solution of stochastic integral equations.

It is known that the space of p-integrable functions for $p\in (0,1)$ is a b-metric space, but it is unclear whether it is a suprametric space. The author introduced the b-suprametric [4], which subsume such functional space. Note that the distance function of the b-suprametric is not necessarily continuous [4, Example 2.10], although the continuity is very useful. In order to overcome this drawback here we introduce the strong b-suprametric distance function, a subfamily of the b-suprametric, and shows its continuity.

The objectives of this work are fourfold: (1) To introduce the strong b-suprametric space in which we establish fixed point theorems of Banach and Edelstein types. (2) To prove a variational principle through the Cantor’s intersection theorem, then to derive a Caristi fixed point result via this variational principle. (3) To introduce the strong b-supranormed linear space and to provide the fixed point principles of Brouwer and Schauder in such linear space. (4) To provide new sufficient conditions for the existence of a solution to an integral equation, via a Chebyshev type inequality, where the integral operator involved is not necessarily Lipschitzian with respect to a metric. Then, we show the existence of a unique solution to a third-order boundary value problem.

2 Strong b-Suprametric Spaces

Here and below, the symbols ${\mathbb {R}}$ and ${\mathbb {N}}$ will denote respectively the set of all real numbers and all nonnegative natural numbers. The symbol $\textrm{cl}(A)$ stands for the closure of a set A. We first need to recall the b-suprametric spaces from [4].

Definition 1.1

Let (X, d) be a semimetric space and $b\,{\ge }\,1$, $\rho \,{\ge }\,0$ be two real constants. The function d is called b-suprametric if:

($d_{3}$) $d(x,y)\le b \,(d(x,z)+d(z,y))+\rho d(x,z)d(z,y)$ for all $x,y,z\in X$.

A pair (X, d) is called b-suprametric space if X is a nonempty set and d is a b-suprametric.

In the previous definition, if $\rho \,{ = }\,0$ we obtain the b-metric [8] and if $b\,{ = }\,1$ we obtain the suprametric [3]. In the sequel, we focus on the following subclass of b-suprametric space.

Definition 1.2

Let (X, d) be a semimetric space and $b\,{\ge }\,1$, $\rho \,{\ge }\,0$ be two real constants. The function d is called strong b-suprametric (sb-suprametric space) if:

($d'_{3}$):

$d(x,y)\le b \, d(x,z)+d(z,y)+\rho d(x,z)d(z,y)$ for all $x, y, z \in X$.

A pair (X, d) is called sb-suprametric space if X is a nonempty set and d is an sb-suprametric.

Remark 1.3

From $(d_{2})$, it follows that we also have

$d''_{3}$:

$d(x,y)\le d(x,z)+b \, d(z,y)+\rho d(x,z)d(z,y)$, for all $x, y, z \in X$.

Examples 1.4

• All suprametric spaces of [3] are sb-suprametric spaces.

• Let $X=\{1,2,3\}$ and let $d:X\times X \rightarrow {\mathbb {R}}_{+}$ be a function defined by:

  $$\begin{aligned} d(x,y)=\left\{ \begin{array}{ll} 0,& \quad x=y,\\ 3,& \quad (x,y)\in \big \{(1,2),(2,1)\big \},\\ \frac{1}{2},& \quad \text {otherwise}. \end{array}\right. \end{aligned}$$

  Then (X, d) is an sb-suprametric space with coefficient $b=\textstyle \frac{3}{2}$ and $\rho =8$.

• Let $X=C_{+}[0,1]$ of continuous nonnegative functions endowed with

  $$\begin{aligned} \delta (x,y)=\sup _{t\in [0,1]}|x(t)-y(t)|(|x(t)-y(t)|+\textstyle \frac{1}{2}),\;\text { for all }x,y\in X. \end{aligned}$$

  Then $(X, \delta )$ is an sb-suprametric space for $b=\rho =2$.

Proposition 1.5

Let (X, d) be an sb-suprametric space, then for all $p, q, s, t\,{\in }\,X$

$$\begin{aligned} \frac{\rho (d(p,q)-d(s,t))^{2}}{(b+\rho d(p,q)+\rho d(s,t))^{2}}\le 2\big (d(p,t)+d(s,q)+\rho \, d(p,t)\, d(s,q)\big ). \end{aligned}$$

(1)

Proof

Let (X, d) be an $sb$-suprametric space with $\rho \,{>}\,0$ (the case $\rho \,{=}\,0$ is trivial). Then,

$$\begin{aligned} d(p,q)&\le d(p,s)+b\, d(s,q)+\rho \, d(p,s)\, d(s,q)\\&\le d(s,t)+b\, d(p,t)+\rho \, d(s,t) \, d(p,t)+ b \, d(s,q)\\&\quad +\rho \, \big (d(s,t)+b \, d(p,t)+\rho \, d(s,t) \, d(p,t)\big )\, d(s,q)\\&\le d(s,t)+\big (b+\rho \, d(s,t)\big ) \big (d(p,t)+ d(s,q)+\rho \, d(p,t)\, d(s,q)\big ), \end{aligned}$$

which implies

$$\begin{aligned} \frac{d(p,q)-d(s,t)}{(b+\rho \, d(s,t))}\le d(p,t)+ d(s,q)+\rho \, d(p,t)\, d(s,q). \end{aligned}$$

(2)

A similar argument shows that

$$\begin{aligned} \frac{d(s,t)-d(p,q)}{(b+\rho \, d(p,q))}\le d(p,t)+ d(s,q)+\rho \, d(p,t)\, d(s,q). \end{aligned}$$

(3)

Adding (2) to (3), we obtain

$$\begin{aligned} \frac{\rho (d(p,q)-d(s,t))^{2}}{(b+\rho d(p,q))(b+\rho d(s,t))}\le 2\big (d(p,t)+d(s,q)+\rho \, d(p,t)\, d(s,q)\big ), \end{aligned}$$

which implies (1). $\square$

Remark 1.6

Let $u\,{ \ge }\,0$ and $\rho \,{ > }\,0$. Assume that a sequence $\{u_{n}\}\subset {\mathbb {R}}_{+}$ satisfies

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{\rho \,(u_{n}-u)^{2}}{(b+\rho \,u_{n}+\rho \,u)^{2}}=0, \end{aligned}$$

then $u_{n}$ tends to u as $n\rightarrow \infty$. Otherwise, if $u_{n}$ does not tends to u there exists $\varepsilon >0$ such that for all integer $k>0$, $n(k)>k$ and $|u_{n(k)}-u|>\varepsilon$. Then,

$$\begin{aligned} \frac{\sqrt{\rho }\,\varepsilon }{b+\rho \,u_{n(k)}+\rho \,u} <\frac{\sqrt{\rho }\,|u_{n(k)}-u|}{b+\rho \,u_{n(k)}+\rho \,u}\rightarrow 0\text { as } k\rightarrow \infty , \end{aligned}$$

which implies that $u_{n(k)}$ tends to infinity as $k\rightarrow \infty$, and hence

$$\begin{aligned} \lim _{k\rightarrow \infty } \frac{{\rho }\,(u_{n(k)}-u)^{2}}{(b+\rho \,u_{n(k)}+\rho \,u)^{2}}=\frac{1}{\rho }, \end{aligned}$$

yields a contradiction.

Remark 1.7

Let $\{p_{n}\}$ and $\{q_{n}\}$ be sequences in X such that $\lim _{n\rightarrow \infty } d (p_{n}, t) \,{=}\,0$ and $\lim _{n\rightarrow \infty } d (q_{n}, s)\, {=}\, 0$, then by (1) and Remark 1.6, $\lim \limits _{n\rightarrow \infty } d (p_{n}, q_{n}) = d (s, t),$ and this means that d is continuous.

Let (X, d) be an sb-suprametric space. An open ball and a closed ball centered at $a\in X$ and of radius $r>0$, are respectively given by

$$\begin{aligned} B(a,r)&{:}{=}&\{x\in X: d(a,x)<r\} \text { and } B[a,r]{:}{=}\{x\in X: d(a,x)\le r\}. \end{aligned}$$

Proposition 1.8

Let (X, d) be an sb-suprametric space. Then

• (i) every open ball is an open set.

• (ii) every closed ball is a closed set.

Proof

To see (i), let $r>0$ and $a\in X$. For $y\in B(a,r)$ let

$$\begin{aligned} r_{1}&{:}{=}&\frac{r-d(y,a)}{b+\rho d(y,a)}, \end{aligned}$$

then if $x\in B(y,r_{1})$,

$$\begin{aligned} d(x,a)&\le b \, d(x,y)+ d(y,a)+\rho d(x,y)\, d(y,a)\\&< b \, r_{1}+ d(y,a)+\rho r_{1}\, d(y,a)=r. \end{aligned}$$

Thus, $B(y,r_{1})\subseteq B(a,r)$, and B(a, r) is open.

Now, to see (ii), let $r\,{ > }\,0$ and $a\,{ \in }\, X$ and take a sequence $\{x_{n}\}$ in B[a, r] convergent to some x with respect to d. Then

$$\begin{aligned} d(a,x)&\le d(a,x_{n})+ b\,d(x_{n},x)+\rho d(a,x_{n})\, d(x_{n},x)\\&\le r+ (b+\rho r)\, d(x_{n},x), \end{aligned}$$

and as $n\rightarrow \infty$, we get $x\in B[a,r]$ which proves that B[a, r] is closed. $\square$

As a consequence, we obtain the following propositions.

Proposition 1.9

Let (X, d) be an sb-suprametric space. The family of open balls form a base of a topology on X.

Proof

Let $u\in B(x,\varepsilon ), B(y,\varepsilon ')$ and choose $r>0$ so that $(b+\rho r)d(x,u)+r<\varepsilon$ and $(b+\rho r)d(y,u)+r<\varepsilon '$. Then by taking a point $v\in B(u,r)$, we obtain

$$\begin{aligned} d(x,v)\le b d(x,u)+d(u,v)+\rho d(x,u)d(u,v)< (b+\rho r)d(x,u)+r<\varepsilon ,\\ d(y,v)\le b d(y,u)+d(x,u)+\rho d(y,u)d(u,v)<(b+\rho r)d(y,u)+r<\varepsilon ', \end{aligned}$$

which implies that $B(u,r)\subset B(x,\varepsilon )\cap B(y,\varepsilon ')$. Finally, we conclude by using [10, Lemma I.4.7] and the fact that every $x\,{ \in }\, X$ is also in $B(x,\tau )$ for some $\tau \,{ > }\,0$. $\square$

Proposition 1.10

An sb-suprametric space is normal.

Proof

Let (X, d) be an sb-suprametric space. If $x,y\,{ \in }\, X$ such $x\,{ \ne }\, y$, then $U{:}{=}B(x,\frac{d(x,y)}{2})$ and $V{:}{=}B(y,\frac{d(x,y)}{2\,b+\rho \,d(x,y)})$ are disjoint neighborhoods of x and y respectively. Otherwise, assume that $U\cap V\ne \varnothing$, so there exists $z\in U\cap V$. Thus, using that $d(x,z)<\frac{r}{2}$ and $d(y,z)<\frac{r}{2\,b+\rho \,r}$ where $r=d(x,y)$, we obtain

$$\begin{aligned} r=d(x,y)&\le d(x,z)+d(z,y)+\rho \, d(x,z)d(z,y)\\&<\frac{r}{2}+\frac{r}{2\,b+\rho \,r}+\rho \,\frac{r}{2}\,\frac{r}{2\,b+\rho \,r}=r, \end{aligned}$$

a contradiction, so our claim holds. We conclude therefore that X is Hausdorff.

Let now U and V be disjoint closed sets and let

$$\begin{aligned} d(x,U){{:}{=}}\inf \limits _{u\in U}d(x,u) \text { and } d(x,V){{:}{=}}\inf \limits _{v\in V}d(x,v). \end{aligned}$$

Define the sets

$$\begin{aligned} U'{:}{=}\big \{x\in X: d(x,U)<d(x,V)\big \}\text { and } V'{:}{=}\big \{x\in X: d(x,V)<d(x,U)\big \}. \end{aligned}$$

Then $U'$ and $V'$ are disjoint neighborhoods of U and V respectively. $\square$

Proposition 1.11

In an sb-suprametric space, if a sequence has a limit it is unique.

Proposition 1.12

The strong b-metric space [16] is normal.

Definition 1.13

Let (X, d) be an sb-suprametric space.

• (i) The sequence $\{x_{n}\}_{n\in {\mathbb {N}}}$ converges to $x \in X$ iff $\lim \limits _{n\rightarrow \infty }d(x_{n},x)\,{ = }\,0$.

• (ii) The sequence $\{x_{n}\}_{n\in {\mathbb {N}}}$ is Cauchy iff $\lim \limits _{n,m\rightarrow \infty }d(x_{n},x_{m})\,{ = }\,0$.

• (iii) (X, d) is complete iff any Cauchy sequence in X is convergent.

Remark 1.14

Let $X{=}C_{+}[0,1]$ be the set of continuous functions $x:[0,1]{\rightarrow } {\mathbb {R}}_{+}$ endowed with $\delta$ the sb-suprametric of Examples 1.4. The completeness of $(X,\delta )$ follows from that of (X, d) with $d(x,y){=}\sup \limits _{t\in [0,1]}|x(t){-}y(t)|$ for $x,y\in X$.

The next lemma is a direct consequence of Definition 1.13.

Lemma 1.15

In an sb-suprametric space, we have:

• (i) A convergent sequence is a Cauchy sequence.

• (ii) A Cauchy sequence converges iff it has a convergent subsequence.

• (iii) A point $u\in \textrm{cl}(U)$ iff there is a sequence $\{u_{n}\}\subset U$ converging to u.

Remark 1.16

If a sequence $\{x_{n}\}_{n\in {\mathbb {N}}}$ is Cauchy in a complete sb-suprametric (X, d), then there exists $x_{*}\in X$ such that $\lim \limits _{n\rightarrow \infty }d(x_{n},x_{*})=0$. By $d'_{3}$ follows that every subsequence $\{x_{n(k)}\} _{k\in {\mathbb {N}}}$ converges to $x_{*}$.

Remark 1.17

Let (X, d) be an $sb$-suprametric space. By $d'_{3}$, we have:

$$\begin{aligned} d(x_{0},x_{n}) \le b\max \{1,\rho ^{n}\} \textstyle \sum \limits _{i=1}^{n}{\textbf{e}}_{i}(d_{0},\ldots ,d_{n-1}), \end{aligned}$$

for all $n \in {\mathbb {N}}$, $x_{0},\ldots , x_{n} \in X$, where $d_{i-1}{:}{=}d(x_{i-1},x_{i})$ and ${\textbf{e}}_{i}$ is the i^th elementary symmetric polynomial in n variables, that is,

$$\begin{aligned} {\textbf{e}}_{i}(d_{0},\ldots ,d_{n-1})=\textstyle \sum \limits _{0\le j_{1}<j_{2}<\cdots <j_{i}\le n-1}\;\prod \limits _{k=1}^{i}d_{j_{k}}. \end{aligned}$$

It is easy to see that these polynomials possess the following properties:

Proposition 1.18

Let $n\in {\mathbb {N}}$ and ${\textbf{e}}_{i}(x_{0},\ldots ,x_{n-1})$ be an elementary symmetric polynomial of index $0\le i\,{\le }\,n$. Then,

• (i) $x_{k}\mapsto {\textbf{e}}_{i}(x_{0},\ldots ,x_{k},\ldots ,x_{n-1})$ is a nondecreasing function for $0\,{\le }\,k\,{\le }\,n$.

• (ii) ${\textbf{e}}_{i}(ax_{0},\ldots ,ax_{n-1}) = a^{i} {\textbf{e}}_{i}(x_{0},\ldots ,x_{n-1})$ for all $a\in {\mathbb {R}}_{+}$.

A covering of a set U in X is a family of open sets whose union contains U. A set $U \subseteq X$ is called compact if and only if every covering of U by open sets in X contains a finite sub-covering. A subset U of a topological space X is sequentially compact, if every sequence of points in U has a subsequence converging to a point of X. A set $N \subset X$ is called an $\varepsilon$-net for a set $U \subseteq X$ ($\varepsilon >0$), if there exists $x_{\varepsilon }\in N$ for every $x \in U$ such that $d(x, x_{\varepsilon }) < \varepsilon$. Next, we provide an extreme value theorem in sb-suprametric spaces.

Theorem 1.19

Let (X, d) be an sb-suprametric space. Let U be a compact subset of X and $f:U\rightarrow {\mathbb {R}}$ be a continuous function. Then,

• (i) f is bounded on U,

• (ii) f attains its supremum and its infimum.

Proof

The proof is similar to that of [18, Chap. 5. Theorem 1] (see also [10, Lemma I.5.8]). $\square$

The compactness is discussed in the rest of this section.

Theorem 1.20

For a set U in an sb-suprametric space X to be compact, it is necessary, and in the case of completeness of X, sufficient that there is a finite $\varepsilon$-net for the set U for every $\varepsilon > 0$.

Proof

The proof is similar to that of [18, Chap. 5. Theorem 3], except for the value of the distance between any two points, which does not exceed $\varepsilon _{n}\left( 1+b+\rho \varepsilon _{n}\right)$. $\square$

Corollary 1.21

A subset U of an sb-suprametric space is compact if and only if it is closed and sequentially compact.

Proof

The proof is exactly similar to that of [10, Theorem I.6.13]. $\square$

Corollary 1.22

Let (X, d) be a sb-suprametric space and $U\subseteq X$. If U is compact, then it is bounded.

Proof

Let $S_{n}=\{x_{1},\ldots ,x_{n}\}$ be a 1-net for U. Let $a\in X$, $x\in X$ and $x_{i}\in S_{n}$ for $i=1,\ldots ,n$. Then,

$$\begin{aligned} d(x,a)&\le b d(x,x_{i})+d(x_{i},a)+\rho d(x,x_{i})\, d(x_{i},a) \\&\le b \big (d(x,x_{i})+d(x_{i},a)\big )+\textstyle \frac{\rho }{2} \big (d(x,x_{i})+ d(x_{i},a)\big )^{2}\\&\le (b+\textstyle \frac{\rho }{2}) \big (1+\max _{i}d(x_{i},a)\big )^{2}<\infty . \end{aligned}$$

$\square$

3 Banach and Edelstein Fixed Point Theorems

We start by presenting a Banach fixed point result in sb-suprametric spaces.

Theorem 2.1

Let (X, d) be a complete sb-suprametric space and $f:X\rightarrow X$ be a given mapping. Assume that there exists $c\in [0,b^{-1})$ such that for all $x,y\in X$,

$$\begin{aligned} d(fx,fy)\le c\, d(x,y). \end{aligned}$$

(4)

Then f has a unique fixed point and $\big \{f^{n}x\big \}_{n\in {\mathbb {N}}}$ converges to it for all $x\in X$.

Proof

Assume that $\rho >0$, since the case $\rho =0$ is treated in [8] (see also [13]). Let $x_{0}\in X$ and define the sequence $\{x_{n}\}$ by $x_{n} = f^{n}x_{0}$ for all $n\in {\mathbb {N}}$, where $f^{n}$ is n^th iterates of f. For simplification let us introduce the notation: $d_{i,j}{:}{=}d(x_{i},x_{j})$, where $i,j\in {\mathbb {N}}$. Now, from (4), we get

$$\begin{aligned} d_{n,n+1}\le c d_{n-1,n}<d_{n-1,n}. \end{aligned}$$

Hence, $\{d_{n,n+1}\}$ is decreasing sequence and for all $k\in {\mathbb {N}}$, we have

$$\begin{aligned} d_{n,n+1}\le c^{n-k}\, d_{k,k+1},\;\text { for all } n>k. \end{aligned}$$

(5)

So $\lim \limits _{n\rightarrow \infty }d_{n,n+1}=0$, and therefore there exits $k\in {\mathbb {N}}$ such that for all $n\ge k$,

$$\begin{aligned} d_{n,n+1}\le 1. \end{aligned}$$

(6)

Next, we shall prove that the sequence $\{x_{n}\}$ is Cauchy. Using $d'_{3}$ and (6), and for sufficiently large integers p, q such that $q>p> k$ it follow that

$$\begin{aligned} d_{p,q}\le & d_{p,p+1}+b\,d_{p+1,q}+\rho \,d_{p,p+1}\,d_{p+1,q}\\\le & c^{p-k} d_{k,k+1} +b\, d_{p+1,q}+\rho \, c^{p-k} d_{k+1,k}\, d_{p+1,q}\\\le & c^{p-k} +(b+\rho \, c^{p-k} ) d_{p+1,q}, \end{aligned}$$

where

$$\begin{aligned} d_{p+1,q}\le & d_{p+1,p+2}+b\,d_{p+2,q}+\rho \,d_{p+1,p+2}\,d_{p+2,q}\\\le & c^{p-k+1} d_{k+1,k} +b\,d_{p+2,q}+\rho \, c^{p-k+1} d_{k+1,k} \,d_{p+2,q}\\\le & c^{p-k+1} +(b+\rho \, c^{p-k+1} ) d_{p+2,q}. \end{aligned}$$

By combining the previous inequalities, we obtain

$$\begin{aligned} d_{p,q}\le c^{p-k} +c^{p-k+1} (b+\rho \, c^{p-k} ) +(b+\rho \,c^{p-k}) (b+ \rho \,c^{p-k+1} ) d_{p+2,q}. \end{aligned}$$

Using (6) in all terms of the sum, we obtain by induction

$$\begin{aligned} d_{p,q} \le c^{p-k} \textstyle \sum \limits _{i=0}^{q-p-1}c^{i}\prod \limits _{j=0}^{i-1}(b+\rho \,c^{p-k+j}). \end{aligned}$$

Now, since $c\in [0,b^{-1})$, then

$$\begin{aligned} d_{p,q} \le c^{p-k} \textstyle \sum \limits _{i=0}^{q-p-1}c^{i}\prod \limits _{j=0}^{i-1}(b+\rho \, c^{j}). \end{aligned}$$

Using d’Alembert’s criterion of convergence of real series, we deduce that $\sum \limits _{i=0}^{\infty } u_{i}$ converges, where

$$\begin{aligned} u_{i}&{:}{=}&c^{i}\textstyle \prod \limits _{j=0}^{i-1}(b+\rho \,c^{j}). \end{aligned}$$

We conclude $d_{p,q}\rightarrow 0$ as $p,q \rightarrow \infty$, so the sequence $\{x_{n}\}$ is Cauchy. Thus, it follows that $\{x_{n}\}$ converges to some $x_{*}\in X$, say, since X is sb-complete, which proves that $\omega _{f}(x_{0})$ is nonempty. We now shall show that $x_{*}$ is a fixed point of f. By using (4), we get

$$\begin{aligned} d(fx_{n(k)},fx_{*})\le c\, d(x_{n(k)},x_{*}). \end{aligned}$$

By letting $k\rightarrow \infty$, we obtain by Proposition 1.11 and Remark 1.16 that $x_{*}=fx_{*}$. Finally, the uniqueness of the fixed point follows immediately from (4). $\square$

Remark 2.2

Theorem 2.1 generalizes [3, Theorem 2.1]. Note also that for the extended suprametric, introduced by Panda et al. [22], an additional continuity assumption was added to obtain the main fixed point theorem.

Proposition 2.3

Let (X, d) be an sb-suprametric space and let $f:X\rightarrow X$ be a Lipschitz mapping, that is, there is a constant $\lambda \in [0,\infty )$ such that for all $x,y\in X$,

$$\begin{aligned} d(fx,fy)\le \lambda \, d(x,y). \end{aligned}$$

Then f is continuous.

Proof

The proof is exactly the same as that of [3, Proposition 1.8]. $\square$

Let X be a topological space and $f:X\rightarrow X$ be a mapping. For $x_{0}\in X$ the $\omega$-limit set is given by

$$\begin{aligned} \omega _{f}(x_{0}){:}{=}\textstyle \bigcap \limits _{n\in {\mathbb {N}}}\textrm{cl}\left( \left\{ f^{k}x_{0}:k\ge n \right\} \right) . \end{aligned}$$

The next result follows immediately from Remark 1.7, Proposition 2.3 and [19, Theorem 1].

Theorem 2.4

Let (X, d) be an sb-suprametric space and let $f:X\rightarrow X$ be a contractive mapping, that is, for all $x,y\in X$ with $x\ne y$,

$$\begin{aligned} d(fx,fy)< d(x,y). \end{aligned}$$

If there exists $x_{0}\in X$ such that $\omega _{f}(x_{0})$ is nonempty, then f has a unique fixed point and $\big \{f^{n}x\big \}_{n\in {\mathbb {N}}}$ converges to this fixed point for all $x\in X$.

Remark 2.5

Theorem 2.4 generalize [11, Theorem 1] and [3, Theorem 2.3]. In this connection, see also [12, Chapter 1, Theorem 1.2] and [5, Section 6].

4 Ekeland Variational Principle and Caristi Fixed Point Theorem

We first present a Cantor’s intersection theorem in sb-suprametric spaces.

Theorem 3.1

Let (X, d) be a complete sb-suprametric space, and let $\left\{ C_{n}\right\} _{n\in {\mathbb {N}}}$ be a decreasing nested sequence of nonempty closed sets of X with

$$\begin{aligned} \textrm{diam}(C_{n})\mathrm{{:}{=}}\sup \left\{ d(x,y):x,y\in C_{n}\right\} \rightarrow 0 \text { as }n \rightarrow \infty . \end{aligned}$$

Then $\bigcap _{n\in {\mathbb {N}}} C_{n}=\{z\}$ for some $z\in X$.

Proof

Since $C_{n}$ is nonempty for all n, we take $z_{n}$ in every $C_{n}$. We then construct a Cauchy sequence $\{z_{n}\}$ because $d(z_{m},z_{n})\le \textrm{diam}(C_{N})$ for all m, n greater than some integer N and $\textrm{diam}(C_{N})\rightarrow 0$ as $N\rightarrow \infty$. Now, by completeness of (X, d), we deduce that $\{z_{n}\}$ converges to some z. Next, since $z_{n}\in C_{N}$ for all $n\ge N$ and $C_{N}$ is closed, $z\in \bigcap _{n\ge {N}} C_{n}$, which implies by the nestedness property that $z\in \bigcap _{n\in {\mathbb {N}}} C_{n}$. Assume now that there exists $z'\in \bigcap _{n\in {\mathbb {N}}} C_{n}$ such that $z'\ne z$, then $d(z,z')>0$, which implies that there exists $m\in {\mathbb {N}}$ such that $\textrm{diam}(C_{n})<d(z,z')$ for all $n\ge m$. Consequently, $z'\not \in C_{n}$ for all $n\ge m$ and therefore $z'\not \in \bigcap _{n\in {\mathbb {N}}} C_{n}$. This proves that $\bigcap _{n\in {\mathbb {N}}} C_{n}=\{z\}$. $\square$

We next present an Ekeland’s variational principle in the new spaces.

Theorem 3.2

Let (X, d) be a complete sb-suprametric space $(b \,{ > }\,1)$ and let $\phi : X \rightarrow {\mathbb {R}}\cup \{\pm \infty \}$ be a lower semicontinuous function which is proper and lower bounded. Then, for every $x_{0} \in X$ and $\varepsilon > 0$ with

$$\begin{aligned} \phi (x_{0})\le \inf _{x\in X}\phi (x)+\varepsilon , \end{aligned}$$

there exist $x_{\varepsilon }\in X$ and a sequence $\{x_{i}\}_{i\in {\mathbb {N}}}$ in X such that:

• (i) $\lim _{i\rightarrow \infty }d(x_{i},x_{\varepsilon })=0$.

• (ii) $d(x_{i},x_{\varepsilon })\le {2^{-n}}{\varepsilon }$, for all $i\in {\mathbb {N}}$.

• (iii) $\phi (x_{\varepsilon })+\sum \limits _{i=0}^{\infty }{b^{-i}}d(x_{\varepsilon },x_{i})\le \phi (x_{0})$.

• (iv) $\phi (x_{\varepsilon })+\sum \limits _{i=0}^{\infty }{b^{-i}}d(x_{\varepsilon },x_{i})< \phi (x)+\sum \limits _{i=0}^{\infty }{b^{-i}}d(x,x_{i})$ for all $x\ne x_{\varepsilon }$.

Proof

Let $x_{0} \in X$ and $\varepsilon > 0$ and define the set

$$\begin{aligned} C_{0}{:}{=}\big \{x\in X: \phi (x)+d(x,x_{0})\le \phi (x_{0})\big \}. \end{aligned}$$

Clearly, $C_{0}$ is nonempty and closed since it contains $x_{0}$, d is continuous and $\phi$ is lower semi-continuous. Now, for all $y\in C_{0}$, we have

$$\begin{aligned} d(x,x_{0})\le \phi (x_{0})-\phi (y)\le \phi (x_{0})-\inf _{x\in X}\phi (x)\le \varepsilon . \end{aligned}$$

(7)

Choose $x_{1}\in C_{0}$ such that

$$\begin{aligned} \phi (x_{1})+d(x_{1},x_{0})\le \inf _{x\in C_{0}}\{\phi (x)+d(x,x_{0})\}+{(2b)^{-1}}{\varepsilon }, \end{aligned}$$

and consider the set

$$\begin{aligned} C_{1}&{:}{=}&\big \{x\in C_{0}: \phi (x)+d(x,x_{0})+b^{-1}d(x,x_{1})\le \phi (x_{1})+d(x_{1},x_{0})\big \}. \end{aligned}$$

By induction, we choose $x_{n-1}\in C_{n-2}$ $(n\ge 2)$ and consider

$$\begin{aligned} C_{n-1}&{:}{=}&\big \{x\in C_{n-2}: \phi (x)+\textstyle \sum \limits _{i=0}^{n-1}b^{-i}d(x,x_{i})\le \phi (x_{n-1})+\sum \limits _{i=0}^{n-2}b^{-i}d(x_{n-1},x_{i})\big \}. \end{aligned}$$

Then, we choose $x_{n}\in C_{n-1}$ such that

$$\begin{aligned} \phi (x_{n})+\textstyle \sum \limits _{i=0}^{n-1}b^{-i}d(x_{n},x_{i})\le \inf \limits _{x\in C_{n-1}}\left\{ \phi (x)+\sum \limits _{i=0}^{n-1}b^{-i}d(x_{n-1},x_{i})\right\} +(2b)^{-n}\varepsilon . \end{aligned}$$

(8)

Define again a set

$$\begin{aligned} C_{n}&{:}{=}&\left\{ x\in C_{n-1}: \phi (x)+\textstyle \sum \limits _{i=0}^{n}b^{-i}d(x,x_{i})\le \phi (x_{n})+\sum \limits _{i=0}^{n-1}b^{-i}d(x_{n},x_{i})\right\} . \end{aligned}$$

(9)

Clearly, $C_{n}$ is nonempty and closed since it contains $x_{n}$, d is continuous and $\phi$ is lower semicontinuous. Next, for all $y\in C_{n}$, we deduce from (8) and (9) that

$$\begin{aligned} b^{-n}d(y,x_{n})&\le \Bigg [ \phi (x_{n})+\textstyle \sum \limits _{i=0}^{n-1}b^{-i}d(x_{n},x_{i})\Bigg ]- \Bigg [\phi (y)+\sum \limits _{i=0}^{n-1}b^{-i}d(y,x_{i}) \Bigg ]\\&\le \Bigg [\phi (x_{n})+\textstyle \sum \limits _{i=0}^{n-1}b^{-i}d(x_{n},x_{i}) \Bigg ]{-}\inf \limits _{x\in C_{n-1}} \Bigg [ \phi (x)+\sum \limits _{i=0}^{n-1}b^{-i}d(x_{n},x_{i}) \Bigg ]\\&\le (2b)^{-n}\varepsilon . \end{aligned}$$

Hence, for all $y\in C_{n}$, we have

$$\begin{aligned} d(y,x_{n})\le 2^{-n}\varepsilon . \end{aligned}$$

(10)

this implies that (i) holds. Consequently, $\lim _{n\rightarrow \infty }\textrm{diam}(C_{n})= 0$ and the sequence $\{C_{n}\}_{n\in {\mathbb {N}}}$ is decreasing nested sequence of nonempty closed sets of X. So, by Theorem 3.1 it follows that $\bigcap _{n\in {\mathbb {N}}} C_{n}=\{x_{\varepsilon }\}$ for some $x_{\varepsilon }\in X$. Note that (ii) follows from (7) and (10). Now, since for all $x\ne x_{\varepsilon }$, $x\not \in \bigcap _{n\in {\mathbb {N}}} C_{n}$, thus there exists $m\in {\mathbb {N}}$ such that $x\not \in C_{m}$, then

$$\begin{aligned} \phi (x_{m})+\textstyle \sum \limits _{i=0}^{m-1}{b^{-i}}d(x_{m},x_{i}) < \phi (x)+\textstyle \sum \limits _{i=0}^{m}{b^{-i}}d(x,x_{i}). \end{aligned}$$

But $x\not \in C_{m}$ means that $x\not \in C_{k}$ for all $k\ge m$, so from the previous inequalities we conclude that for all $k\ge m$, we have

$$\begin{aligned} \phi (x_{\varepsilon })+\textstyle \sum \limits _{i=0}^{k}{b^{-i}}d(x_{\varepsilon },x_{i})&\le \phi (x_{k})+\textstyle \sum \limits _{i=0}^{k-1}{b^{-i}}d(x_{k},x_{i})\\&\le \phi (x_{m})+\textstyle \sum \limits _{i=0}^{m-1}{b^{-i}}d(x_{m},x_{i})\le \phi (x_{0}). \end{aligned}$$

Consequently (iii) and (iv) hold. $\square$

Remark 3.3

Theorem 3.2 generalizes [6, Theorem 2.2].

Corollary 3.4

Let (X, d) be a complete sb-suprametric space $(b \,{ > }\,1)$ and let $\phi : X \rightarrow {\mathbb {R}}\cup \{\pm \infty \}$ be a lower semi-continuous function which is proper and lower bounded. Then, for every $\varepsilon > 0$ there exist $x_{\varepsilon }\in X$ and a sequence $\{x_{i}\}_{i\in {\mathbb {N}}}$ in X such that:

• (i) $\lim _{i\rightarrow \infty }d(x_{i},x_{\varepsilon })=0$.

• (ii) $\phi (x_{\varepsilon })+\textstyle \sum \limits _{i=0}^{\infty }{b^{-i}}d(x_{\varepsilon },x_{i})\le \inf _{x\in X}\phi (x)+\varepsilon$.

• (iii) $\phi (x_{\varepsilon })+\textstyle \sum \limits _{i=0}^{\infty }{b^{-i}}d(x_{\varepsilon },x_{i}) \le \phi (x)+\textstyle \sum \limits _{i=0}^{\infty }{b^{-i}}d(x,x_{i})$ for all $x\in X$.

We next present a fixed point theorem in sb-suprametric spaces.

Theorem 3.5

Let (X, d) be a complete sb-suprametric space $(b\,{ > }\,1)$. Let $f:X\rightarrow X$ be a mapping for which there exists a proper, lower semicontinuous and lower bounded function $\phi :X\rightarrow {\mathbb {R}}\cup \{\pm \infty \}$ such that for all $x\in X$,

$$\begin{aligned} \frac{b^2+\rho }{b-1} d(x, f(x)) \le \phi (x) - \phi (f(x)). \end{aligned}$$

(11)

Then f has a fixed point.

Proof

Assume that for all $x \in X$, $f(x)\ne x$. By applying Corollary 3.4, we deduce that for every $\varepsilon > 0$ there exist $x_{\varepsilon }\in X$ and a sequence $\{x_{i}\}_{i\in {\mathbb {N}}}$ in X such that:

$$\begin{aligned} \phi (x_{\varepsilon })+\textstyle \sum \limits _{i=0}^{\infty }{b^{-i}}d(x_{\varepsilon },x_{i}) \le \phi (x)+\textstyle \sum \limits _{i=0}^{\infty }{b^{-i}}d(x,x_{i}), \end{aligned}$$

for all $x\in X$. By taking $x=f(x_{\varepsilon })$, where here $x\ne x_{\varepsilon }$, we get

$$\begin{aligned} \phi (x_{\varepsilon })-\phi (f(x_{\varepsilon }))< \textstyle \sum \limits _{i=0}^{\infty }{b^{-i}}d(f(x_{\varepsilon }),x_{i})-\textstyle \sum \limits _{i=0}^{\infty }{b^{-i}}d(x_{\varepsilon },x_{i}). \end{aligned}$$

Now, by $d'_{3}$ it follows that

$$\begin{aligned} \phi (x_{\varepsilon })-\phi (f(x_{\varepsilon }))< \textstyle \sum \limits _{i=0}^{\infty }b^{1-i}d(f(x_{\varepsilon }),x_{\varepsilon })+{\rho }\textstyle \sum \limits _{i=0}^{\infty }{b^{-i}}d(f(x_{\varepsilon }),x_{\varepsilon })d(x_{\varepsilon },x_{i}). \end{aligned}$$

Since $\lim _{i\rightarrow \infty }d(x_{\varepsilon },x_{i})=0$, then there exists an integer $N>0$ such that for all $i\ge N$, we have $d(x_{\varepsilon },x_{i})\le 1$. Hence,

$$\begin{aligned} \phi (x_{\varepsilon })-\phi (f(x_{\varepsilon }))< \Big (\textstyle \frac{b^2+\rho }{b-1} +\max _{0\le i\le N}\big \{b^{-i}d(x_{\varepsilon },x_{i})\big \}\Big )d(f(x_{\varepsilon }),x_{\varepsilon }). \end{aligned}$$

Next, we take $x=x_{\varepsilon }$ in (11), we obtain

$$\begin{aligned} \frac{b^2+\rho }{b-1} d(x_{\varepsilon }, f(x_{\varepsilon })) \le \phi (x_{\varepsilon }) - \phi (f(x_{\varepsilon })), \end{aligned}$$

and this inequality combined with the previous one yield a contradiction. We conclude that f has a fixed point. $\square$

The Caristi’s fixed point theorem in sb-suprametric spaces follows immediately by taking $\psi =\frac{b-1}{b^2+\rho } \phi$ in the previous theorem.

Corollary 3.6

Let (X, d) be a complete sb-suprametric space $(b\,{ > }\,1)$. Let $f:X\rightarrow X$ be a mapping for which there exists a proper, lower semicontinuous and lower bounded function $\psi :X\rightarrow {\mathbb {R}}\cup \{\pm \infty \}$ such that for all $x\in X$,

$$\begin{aligned} d(x, f(x)) \le \psi (x) - \psi (f(x)). \end{aligned}$$

Then f has a fixed point.

Proposition 3.7

Corollary 3.6 generalizes [17, Theorem 2.14].

5 Strong b-Supranormed Spaces

In this section, we introduce the concept of strong b-supranormed spaces and derive some of its properties.

Definition 4.1

Let X be a nonempty linear space and $b\,{\ge }\,1$, $\rho \,{\ge }\,0$ are two real constants. A function $\Vert \cdot \Vert :X\rightarrow {\mathbb {R}}_{+}$ is called $b$-supranorm if the following conditions hold:

(${n}_{1}$):

$\Vert x\Vert =0$ if and only if $x=0$,

(${n}_{2}$):

$\Vert \lambda x\Vert =|\lambda | \Vert x\Vert$, for all $x\in X$ and $\lambda \in {\mathbb {R}}$

(${n}_{3}$):

$\Vert x+y\Vert \le b(\Vert x\Vert +\Vert y\Vert )+\rho \, \Vert x\Vert \,\Vert y\Vert$ for all $x,y\in X$.

A pair $(X,\Vert \cdot \Vert )$ is called a b-supranorm space if X is a nonempty set and $\Vert \cdot \Vert$ is a b-supranorm. The pair $(X,\Vert \cdot \Vert )$ is called a supranorm space if $b=1$.

Definition 4.2

Let X be a nonempty linear space and $b\,{\ge }\,1$, $\rho \,{\ge }\,0$ are two real constants. A function $\Vert \cdot \Vert :X\rightarrow {\mathbb {R}}_{+}$ is called strong b-supranorm (sb-supranorm) if it satisfies (${n}_{1}$), (${n}_{2}$) and

($n'_{3}$):

$\Vert x+y\Vert \le b \,\Vert x\Vert +\Vert y\Vert +\rho \Vert x\Vert \Vert y\Vert$ for all $x, y\in X$.

A pair $(X,\Vert \cdot \Vert )$ is called a strong b-supranormed (sb-supranormed) linear space if X is a nonempty set and $\Vert \cdot \Vert$ is a string b-supranorm. The pair $(X,\Vert \cdot \Vert )$ is called a strong supranormed linear space if $b=1$.

Remark 4.3

Using (${n}_{2}$), it follows that

($n''_{3}$):

$n''_{3}$ $\Vert x+y\Vert \le \Vert x\Vert +b \,\Vert y\Vert +\rho \Vert x\Vert \Vert y\Vert$ for all $x, y \in X$.

Examples 4.4

• Clearly, strong b-normed spaces of [16] are sb-supranormed spaces.

• If $\Vert \cdot \Vert$ is an sb-supranorm linear space X, then the function $d:X\times X\rightarrow {\mathbb {R}}_{+}$ given by $d(x,y)=\Vert x-y\Vert$ is an sb-suprametric.

• Consider the set $X={\mathbb {R}}^{2}$ endowed with a function $\Vert \cdot \Vert :X \rightarrow {\mathbb {R}}$ defined by

  $$\begin{aligned} \Vert (x,y) \Vert =\left| x-y \right| +\min (|x|,|y|). \end{aligned}$$

  It is not difficult to see that $(X,\Vert \cdot \Vert )$ is an sb-supranormed space for $b\,{=}\,\rho \,{=}\,2$.

Remark 4.5

Let $(X, \Vert \cdot \Vert )$ be an sb-supranormed linear space. If a sequence $\{x_{n}\}$ converges simultaneously to x and y, that is, $\lim \limits _{n\rightarrow \infty }\Vert x_{n}-x\Vert =\lim \limits _{n\rightarrow \infty }\Vert x_{n}-y\Vert =0$, then $x=y$, and this follows from (${n}_{1}$) and ($n'_{3}$), since we have

$$\begin{aligned} \Vert x-y\Vert \le \Vert x-x_{n}\Vert +b \Vert x_{n}-y\Vert +\rho \Vert x-x_{n}\Vert \Vert x_{n}-y\Vert . \end{aligned}$$

Moreover, we have the following inequality:

$$\begin{aligned} \textstyle \left\| \sum \limits _{i=0}^{n}x_{i}\right\| \le b\max \{1,\rho ^{n}\} \sum \limits _{i=1}^{n}{\textbf{e}}_{i}(\Vert x_{0}\Vert ,\ldots ,\Vert x_{n}\Vert ), \end{aligned}$$

for all $n \in {\mathbb {N}}$ and $x_{0},\ldots , x_{n} \in X$.

Lemma 4.6

Let $(X, \Vert \cdot \Vert )$ be an sb-supranormed linear space. Then, $\Vert \cdot \Vert$ is a continuous function.

Proof

Assume that $\rho >0$ and let $x,y \in X$, then

$$\begin{aligned} \Vert x\Vert =\Vert y+(x-y)\Vert \le \Vert y\Vert +b\,\Vert x-y\Vert +\rho \Vert y\Vert \,\Vert x-y\Vert , \end{aligned}$$

and consequently,

$$\begin{aligned} \frac{\Vert x\Vert -\Vert y\Vert }{b+\rho \Vert y\Vert }\le \Vert x-y\Vert . \end{aligned}$$

(12)

Similarly,

$$\begin{aligned} \frac{\Vert y\Vert -\Vert x\Vert }{b+\rho \Vert x\Vert }\le \Vert x-y\Vert . \end{aligned}$$

(13)

Hence, from (12) and (13), one gets

$$\begin{aligned} \frac{\rho (\Vert x\Vert -\Vert y\Vert )^{2}}{(b+\rho \Vert x\Vert +\rho \Vert y\Vert )^{2}}\le 2\Vert x-y\Vert . \end{aligned}$$

Therefore, if $\Vert x_{n}-x\Vert \rightarrow 0$ as $n\rightarrow \infty$, then

$$\begin{aligned} \frac{\rho (\Vert x_{n}\Vert -\Vert x\Vert )^{2}}{(b+\rho \Vert x_{n}\Vert +\rho \Vert x\Vert )^{2}}\le 2\Vert x_{n}-x\Vert \rightarrow 0\;\text { as }\, n\rightarrow \infty , \end{aligned}$$

and using Remark 1.6 it follows that $\Vert x_{n}\Vert \rightarrow \Vert x\Vert$ as $n\rightarrow \infty$, which implies that $\Vert \cdot \Vert$ is a continuous function. $\square$

Let $(X_{n}, d)$ be an n-dimensional sb-supranormed linear space and let $\{u_{1},\ldots , u_{n}\}$ be a base of $X_{n}$. For every $x\in X$ there exist unique coefficients $\alpha _{1},\ldots ,\alpha _{n}\in {\mathbb {R}}$ such that

$$\begin{aligned} x=\textstyle \sum \limits _{i=1}^{n}\alpha _{i} u_{i}, \end{aligned}$$

and define $\Vert \cdot \Vert _{0}:X \rightarrow {\mathbb {R}}_{+}$ by

$$\begin{aligned} \Vert x\Vert _{0}=\textstyle \sum \limits _{i=1}^{n} {\textbf {e}}_{i}(|\alpha _{1}|,\ldots ,|\alpha _{n}|). \end{aligned}$$

Theorem 4.7

Let $(X_{n}, \Vert \cdot \Vert )$ be an n-dimensional $sb$-supranormed linear space. Then, there exists $\beta > 0$ such that

$$\begin{aligned} \Vert x\Vert \le \beta \Vert x\Vert _{0},\;\text { for all } x \in X. \end{aligned}$$

(14)

Proof

Let $x\in X_{n}$ and $\{u_{1},\ldots , u_{n}\}$ be a base of $X_{n}$. Then, there exist unique coefficients $\alpha _{1},\ldots ,\alpha _{n}\in {\mathbb {R}}$ such that $x=\sum _{i=1}^{n}\alpha _{i} u_{i}$. Hence,

$$\begin{aligned} \Vert x\Vert&= \left\Vert \textstyle \sum \limits _{i=1}^{n}\alpha _{i} u_{i}\right\Vert \\&\le b\max \{1,\rho ^{n}\} \textstyle \sum \limits _{i=1}^{n} {\textbf {e}}_{i}(|\alpha _{1}| \, \Vert u_{1} \Vert ,\ldots ,|\alpha _{n}| \, \Vert u_{n} \Vert )\\&\le b\max \{1,\rho ^{n}\}\Big (\max _{_{1\le j_{1}<j_{2}<\cdots <j_{i}\le n}}\textstyle \prod \limits _{k=1}^{i}\Vert u_{j_{k}}\Vert \Big ) \textstyle \sum \limits _{i=1}^{n} {\textbf {e}}_{i}(|\alpha _{1}| ,\ldots ,|\alpha _{n}|)\\&= \beta \; \Vert x\Vert _{0}, \end{aligned}$$

where

$$\begin{aligned} \beta&{:}{=}&b\max \{1,\rho ^{n}\}\Big (\max _{_{1\le j_{1}<j_{2}<\cdots <j_{i}\le n}}\textstyle \prod \limits _{k=1}^{i}\Vert u_{j_{k}}\Vert \Big ). \end{aligned}$$

(15)

$\square$

Theorem 4.8

Let $(X_{n}, \Vert \cdot \Vert )$ be an n-dimensional $sb$-supranormed linear space. Then, $\Vert \cdot \Vert$ and $\Vert \cdot \Vert _{0}$ are equivalent, that is, there exists $\alpha ,\beta > 0$ such that for all $x \in X$

$$\begin{aligned} \alpha \Vert x\Vert _{0}\le \Vert x\Vert \le \beta \Vert x\Vert _{0}. \end{aligned}$$

Proof

Let $x\in X_{n}$ and $\{u_{1},\ldots , u_{n}\}$ be a base of $X_{n}$. Then, there exist unique coefficients $\alpha _{1},\ldots ,\alpha _{n}\in {\mathbb {R}}$ such that $x=\sum _{i=1}^{n}\alpha _{i} u_{i}$. Define a set U by

$$\begin{aligned} U&{:}{=}&\{x\in X: \Vert x\Vert _{0}=1\}. \end{aligned}$$

We first show that U is bounded. Let $\alpha ^{j}_{1},\ldots ,\alpha ^{j}_{n}\in {\mathbb {R}}$ for $j=1,2$ such that $x_{j}=\sum _{i=1}^{n}\alpha ^{j}_{i} u_{i}\in U$. Then

$$\begin{aligned} \Vert x_{1}-x_{2}\Vert&=\biggl \Vert \textstyle \sum \limits _{i=1}^{n}(\alpha ^{1}_{i}-\alpha ^{2}_{i}) e_{i}\biggr \Vert \\&\le b\max \{1,\rho ^{n}\} \textstyle \sum \limits _{i=1}^{n} {\textbf {e}}_{i}(|\alpha ^{1}_{1}-\alpha ^{2}_{1}| \, \Vert u_{1} \Vert ,\ldots ,|\alpha ^{1}_{n}-\alpha ^{2}_{n}| \, \Vert u_{n} \Vert )\\&\le \beta \textstyle \sum \limits _{i=1}^{n} {\textbf {e}}_{i}(|\alpha ^{1}_{1}-\alpha ^{2}_{1}| ,\ldots ,|\alpha ^{1}_{n}-\alpha ^{2}_{n}|)\\&\le \beta \textstyle \sum \limits _{i=1}^{n} {\textbf {e}}_{i}(|\alpha ^{1}_{1}|+|\alpha ^{2}_{1}| ,\ldots ,|\alpha ^{1}_{n}|+|\alpha ^{2}_{n}|) \\&\le \beta \, K, \end{aligned}$$

where $\beta$ is given by (15) and $K {:}{=}\max \big \{2^{k}n \left( ^{n}_{k}\right) :k=1,\ldots , n\big \}$, which proves that U is bounded.

Define now a function $\phi :X \rightarrow {\mathbb {R}}_{+}$ by $\phi (x)=\Vert x\Vert$. It follows by Lemma 4.6 that f is continuous. Note that U is strongly compact, since it is bounded and closed subset of ${\mathbb {R}}^{n}$. Using Theorem 1.19, we deduce that $\phi$ has an infimum $\alpha$ in U, which is different from zero because $\Vert x\Vert _{0}=1$ for every vector $x\in U$. Hence,

$$\begin{aligned} \alpha =\inf \{\phi (x): x\in U\}=\inf \{\Vert x\Vert : x\in U\}>0. \end{aligned}$$

Thus, from the fact $\frac{x}{\Vert x\Vert _{0}}\in U$ for all $x\in X$, it follows that

$$\begin{aligned} \left\| \frac{x}{\Vert x\Vert _{0}}\right\| \ge \alpha >0,\;\text { for all } x \in X, \end{aligned}$$

which implies that

$$\begin{aligned} \alpha \Vert x\Vert _{0} \le \Vert x\Vert ,\;\text { for all } x \in X. \end{aligned}$$

(16)

Finally, combine (14) and (16), we obtain the result. $\square$

Remark 4.9

As an immediate consequence of Theorem 4.8, any two sb-supranorms on a finite-dimensional space are equivalent.

Lemma 4.10

Let $(X_{n}, \Vert \cdot \Vert )$ be an n-dimensional $sb$-supranormed linear space, let U be a bounded set of X. Then, U is compact.

Proof

Let $x\in X_{n}$ and $\{u_{1},\ldots , u_{n}\}$ be a base of $X_{n}$. Then, there exist unique coefficients $\alpha _{1},\ldots ,\alpha _{n}\in {\mathbb {R}}$ such that $x=\sum _{i=1}^{n}\alpha _{i} u_{i}$. Let ${\overline{x}}=(\alpha _{1},\ldots ,\alpha _{n})\in {\mathbb {R}}^{n}$. Define the function $\phi :U \rightarrow {\mathbb {R}}^{n}$ by $\phi (x)={\overline{x}}$ for all $x\in U$ and let $V=\phi (U)$. Since the function $\Vert \cdot \Vert _{1}:X\rightarrow {\mathbb {R}}_{+}$ by $\Vert x\Vert _{1}{:}{=}\Vert \phi (x)\Vert _{n}$ is an sb-supranorm on $X_{n}$, where $\Vert \cdot \Vert _{n}$ is an sb-supranorm on ${\mathbb {R}}^{n}$. Hence, according to Remark 4.9, $\Vert \cdot \Vert _{1}$ is equivalent to the $sb$-supranorm $\Vert \cdot \Vert$. We conclude that there exist $\alpha ,\beta >0$ such that

$$\begin{aligned} \alpha \Vert {\overline{x}}\Vert _{n} \le \Vert x\Vert \le \beta \Vert {\overline{x}}\Vert _{n}. \end{aligned}$$

(17)

As consequences of (17), U bounded in X if and only if U bounded in ${\mathbb {R}}^{n}$, and a sequence $\{x_{n}\}$ is convergent in $(X, \Vert \cdot \Vert )$ if and only if the corresponding sequence $\{{\overline{x}}_{n}\}$ is convergent in ${\mathbb {R}}^{n}$. Consequently, the compactness of U bounded in X follows from the compactness of U bounded in ${\mathbb {R}}^{n}$. $\square$

6 Brouwer and Schauder Fixed Point Principles

We first recall the Brouwer fixed point principle in ${\mathbb {R}}^{n}$.

Theorem 5.1

(Brouwer) Let U be a bounded closed convex set of ${\mathbb {R}}^{n}$. If a mapping $f:U \rightarrow U$ is continuous, then it has a fixed point.

The Brouwer fixed point principle in sb-supranormed space is given next.

Theorem 5.2

Let $(X_{n}, \Vert \cdot \Vert )$ be an n-dimensional sb-supranormed linear space and let U be a bounded closed convex set of $X_{n}$. If a mapping $f:U \rightarrow U$ is continuous, then it has a fixed point.

Proof

Let $x\in X_{n}$ and $\{u_{1},\ldots , u_{n}\}$ be a base of $X_{n}$. Then, there exist unique coefficients $\alpha _{1},\ldots ,\alpha _{n}\in {\mathbb {R}}$ such that $x=\sum _{i=1}^{n}\alpha _{i} u_{i}$. Let ${\overline{x}}=(\alpha _{1},\ldots ,\alpha _{n})\in {\mathbb {R}}^{n}$. Define the function $\phi :U \rightarrow {\mathbb {R}}^{n}$ by $\phi (x)={\overline{x}}$ for all $x\in U$ and let $V=\phi (U)$. The function $\phi :U \rightarrow V$ is bijective, where the mapping $\phi ^{-1}:V\rightarrow U$ is given by $\phi ^{-1}({\overline{x}})=x$ for all ${\overline{x}}\in V$. Next, we will prove several claims: $\square$

Claim 1

$\phi :U \rightarrow V$ is an homeomorphism. Indeed $\phi$ is continuous in U, since by (16), we have

$$\begin{aligned} \Vert \phi (x)-\phi (x_{0})\Vert _{0} =\Vert {\overline{x}}-{\overline{x}}_{0}\Vert _{0} \le \alpha ^{-1}\Vert x-x_{0}\Vert ,\;\text { for all } x,x_{0} \in U, \end{aligned}$$

Similarly, $\phi ^{-1}$ is continuous in V, because by (14), we have

$$\begin{aligned} \Vert \phi ^{-1}({\overline{x}})-\phi ^{-1}({\overline{x}}_{0})\Vert = \Vert x-x_{0}\Vert \le \beta \Vert {\overline{x}}-{\overline{x}}_{0}\Vert _{0},\;\text { for all } {\overline{x}}, {\overline{x}}_{0} \in V. \end{aligned}$$

Hence, Claim 1 holds.

Claim 2

V is convex. Let ${\overline{x}}=(\alpha _{1},\ldots ,\alpha _{n}), {\overline{y}}=(\beta _{1},\ldots ,\beta _{n})\in V$, where $\alpha _{i},\beta _{i}\in {\mathbb {R}}$ for $i=1,\ldots ,n$. For all $\lambda \in [0,1]$, we obtain by convexity of U that

$$\begin{aligned} \lambda {\overline{x}}+(1-\lambda ){\overline{y}}&=\left( \lambda \alpha _{1}+(1-\lambda )\beta _{1}, \ldots , \lambda \alpha _{n}+(1-\lambda )\beta _{n} \right) \\&= \phi (\lambda {x}+(1-\lambda ){y})\in V, \end{aligned}$$

which implies that Claim 2 holds.

Claim 3

V is bounded. Let ${\overline{x}}, {\overline{y}} \in V$, then by the boundedness of U and Theorem 4.7 it follows that

$$\begin{aligned} \Vert {\overline{x}}-{\overline{y}}\Vert _{0}\le \alpha ^{-1}\Vert x-y\Vert \le \alpha ^{-1} d(U), \end{aligned}$$

where $d(U){:}{=}\max \{\Vert x-y\Vert : x,y\in U\}$, which proves Claim 3.

Claim 4

V is closed. Let $x=\sum _{i=1}^{n}\alpha _{i} u_{i}$, $x_{0}=\sum _{i=1}^{n}\beta _{i} u_{i}$, ${\overline{x}}=(\alpha _{1},\ldots ,\alpha _{n})\in V$, ${\overline{x}}_{0}=(\beta _{1},\ldots ,\beta _{n})$, where $\alpha _{i},\beta _{i}\in {\mathbb {R}}$ for $i=1,\ldots , n$. Assume that $\Vert {\overline{x}}-{\overline{x}}_{0}\Vert _{0}$ tends to zero. Now, by (16), we have

$$\begin{aligned} \Vert {x}-{x}_{0}\Vert \le \beta \Vert {\overline{x}}-{\overline{x}}_{0}\Vert _{0}, \end{aligned}$$

so it follows that by closedness of U that $x_{0}\in U$, which implies that ${\overline{x}}_{0}\in V$ and this prove the claim.

Claim 5

f has a fixed point. To show this, define the function $F:V\rightarrow V$ by $F=\phi {f}\phi ^{-1}$. By Theorem 5.1 and the previous claims, we deduce that there exists ${\overline{x}}\in V$ such that $F(x)=x$, that is,

$$\begin{aligned} \phi {f}\phi ^{-1}(x)=x, \end{aligned}$$

which is equivalent to $f(\phi ^{-1}(x))=\phi ^{-1}(x)$, and since $\phi ^{-1}(x)\in U$, then f has a fixed point in U. $\square$

Before establishing the fixed point principle of Schauder type in sb-supranormed spaces, we need to develop some auxiliary results. Let $(E,\Vert \cdot \Vert )$ be an sb-supranormed linear space and $N {:}{=}\{c_{1}, \ldots , c_{n}\}$ be a finite subset of E. For any fixed $\varepsilon > 0$, define the set

$$\begin{aligned} (N,{\varepsilon })&{:}{=}&\textstyle \bigcup \limits _{i=1}^{n} B(c_{i},\varepsilon ), \end{aligned}$$

where

$$\begin{aligned} B(c_{i},\varepsilon ){:}{=}\left\{ x\in E: \Vert x-c_{i}\Vert <\varepsilon \right\} ,\quad i=1,\ldots ,n. \end{aligned}$$

Define a mapping $\mu _{i}:(N,{\varepsilon })\rightarrow {\mathbb {R}}$ by

$$\begin{aligned} \mu _{i}(x){:}{=}\max \big [0, \varepsilon -\Vert x-c_{i}\Vert \big ],\quad i=1,\ldots ,n. \end{aligned}$$

Consider the Schauder projection $p_{\varepsilon }:(N,{\varepsilon }) \rightarrow \textrm{conv}(N)$ given by

$$\begin{aligned} p_{\varepsilon }(x)=\textstyle \left[ \,\sum \limits _{i=1}^{n}\mu _{i}(x)\right] ^{-1}\sum \limits _{i=1}^{n}\mu _{i}(x)c_{i}, \end{aligned}$$

Note that $p_{\varepsilon }((N,{\varepsilon }))\subset \textrm{conv}(N)$ as a convex combination of $\{c_{1}, \ldots , c_{n}\}$. Moreover, if $x\in (N,{\varepsilon })$, then there exists i such that $x\in B(C_{i},\varepsilon )$, so $\sum \nolimits _{i=1}^{n}\mu _{i}(x)\ne 0$, which means that $p_{\varepsilon }$ is well defined.

Lemma 5.3

Let $(E,\Vert \cdot \Vert )$ be an sb-supranormed linear space, U be a convex subset of E and $N=\{c_{1},\ldots ,c_{n}\}\subset U$. Then for a sufficiently small $\varepsilon > 0$, we have

• (i) $\Vert x-p_{\varepsilon }(x)\Vert \le n\,b\,\varepsilon \, \max \{1,\rho ^{n}\}$ for all $x\in (N,{\varepsilon })$,

• (ii) $p_{\varepsilon }:(N,{\varepsilon })\rightarrow \textrm{conv}(N)\subset U$ is a continuous compact mapping.

Proof

Let $\varepsilon \in (0,1]$ be sufficiently small such that for every $1\,{\le }\, i\, {\le }\, n$ and any x in $(N,\varepsilon )$, $P_{i}(x)\le P_{1}(x)$, where $P_{i}(x){:}{=}{\textbf {e}}_{i}(\mu _{1}(x),\ldots ,\mu _{n}(x))$. Then,

$$\begin{aligned}&\Vert x-p_{\varepsilon }(x)\Vert = \textstyle \left[ \,\sum \limits _{i=1}^{n}\mu _{i}(x)\right] ^{-1}\left\| \sum \limits _{i=1}^{n}\mu _{i}(x) (x-c_{i})\right\| \\&\quad \le b \max \{1,\rho ^{n}\} \textstyle (P_{1}(x))^{-1} \sum \limits _{i=1}^{n} {\textbf {e}}_{i}(\mu _{1}(x)\Vert x-c_{1}\Vert ,\ldots ,\mu _{n}(x)\Vert x-c_{n}\Vert )\\&\quad \le b \max \{1,\rho ^{n}\}\textstyle (P_{1}(x))^{-1} \sum \limits _{i=1}^{n} {\textbf {e}}_{i}(\mu _{1}(x)\varepsilon ,\ldots ,\mu _{n}(x)\varepsilon )\\&\quad \le \textstyle b\,\varepsilon \,\max \{1,\rho ^{n}\} (P_{1}(x))^{-1} \sum \limits _{i=1}^{n}P_{i}(x)\\&\quad \le n\,b\,\varepsilon \,\max \{1,\rho ^{n}\}. \end{aligned}$$

Now, since $p_{\varepsilon }$ is a finite sum of continuous functions and $\Vert \cdot \Vert$ is continuous according to Lemma 4.6, then $p_{\varepsilon }$ is continuous. The compactness of $p_{\varepsilon }$ follows from Lemma 4.10, since its codomain is with finite-dimension. $\square$

Lemma 5.4

Let X be a topological space and E be an sb-supranormed linear space. Let U be a convex set of E and $f :X \rightarrow U$ be a compact mapping. Then for a sufficiently small $\varepsilon > 0$, there exists a finite set

$$\begin{aligned} N=\{c_{1},\ldots ,c_{n}\}\subset f(X)\subset U, \end{aligned}$$

and a finite-dimensional mapping $f_{\varepsilon } :X \rightarrow U$ such that:

• (i) $\Vert f_{\varepsilon } (x)-f(x)\Vert \le n\, b\, \varepsilon \, \max \{1,\rho ^{n}\}$ for all $x\in X$,

• (ii) $f_{\varepsilon } (X)\subset \textrm{conv}(N) \subset U$.

Proof

(i)::

By Theorem 1.20 and for sufficiently small $\varepsilon \in (0,1)$ there exists a finite $\varepsilon$-net $\{c_{1},\ldots ,c_{n}\} \subset f(X)$ because f(X) is compact in E. Now, if $y\in f(X)$, then $d(y,c_{i})<\varepsilon$ for some $i\in \{1,\ldots ,n\}$, thus $y\in B(c_{i},\varepsilon )$, so $y\in (N,\varepsilon )$ and this proves that $f(X)\subset (N,\varepsilon )$. Let $f_{\varepsilon }=p_{\varepsilon }f$. We deduce by Lemma 5.3 that

$$\begin{aligned} \Vert f_{\varepsilon }(x)-f(x)\Vert =\Vert p_{\varepsilon }y-y\Vert \le n\, b\, \varepsilon \, \max \{1,\rho ^{n}\}, \end{aligned}$$

where $y=f(x)\in (N,\varepsilon )$, for all $x\in X$.

(ii)::

Let $y\in f_{\varepsilon }(X)$. Thus, there is $z=f(x)\in (N,\varepsilon )$ for some $x\in X$ such that $y=p_{\varepsilon }(z)$. Consider

$$\begin{aligned} y=p_{\varepsilon }(z)=\textstyle \sum \limits _{i=1}^{n}\lambda _{i} c_{i},\;\; \textstyle \sum \limits _{i=1}^{n}\lambda _{i}=1, \;\;\lambda _{i}\in {\mathbb {R}},\;\; i=1,\ldots ,n. \end{aligned}$$

Thus, $y\in \textrm{conv}(N)\subset U$, and by convexity of U it follows that $f_{\varepsilon }\subset \textrm{conv}(N)\subset U$. $\square$

Let (X, d) be an sb-suprametric space, U be a nonempty set of X and $f :U\rightarrow X$ be a given mapping. If for a given $\varepsilon > 0$, there exists a point $x \in U$ such that $d(x,f(x)) < \varepsilon$, then we say that x is an $\varepsilon$-fixed point for f.

Theorem 5.5

Let (X, d) be an sb-suprametric space and U be a closed set of X. If a mapping $f :U \rightarrow X$ is compact, then f has a fixed point if and only if for each $\varepsilon > 0$ it has an $\varepsilon$-fixed point.

Proof

The necessary condition is trivial, so we only show the sufficient condition. Let $\varepsilon _{n}=\frac{1}{n}$, $n\in {\mathbb {N}}$. Assume there exists $u_{n}\in U$ for all $n\in {\mathbb {N}}$ such that $u_{n}$ are $\varepsilon _{n}$-fixed point, that is,

$$\begin{aligned} d(u_{n},f(u_{n}))< \textstyle \frac{1}{n},\;\text { for all }n\in {\mathbb {N}}. \end{aligned}$$

(18)

The mapping f is compact, so there exists a compact K such that $f(X)\subseteq K$. Thus, there exists a subsequence $\{u_{n_{k}}\}$ such that $f(u_{n_{k}})$ converges to some $u\in X$ as k tends to infinity. Now, using (18), it follows that for any $\varepsilon >0$ there exists $k_{0}\in {\mathbb {N}}$ such that for $k\ge k_{0}$, we have

$$\begin{aligned} d(u_{n_{k}}, u)&\le b d(u_{n_{k}}, f(u_{n_{k}})) +d(f(u_{n_{k}}), u) + \rho \,d(u_{n_{k}}, f(u_{n_{k}})) d(f(u_{n_{k}}), u) \\&\le \textstyle \frac{b}{n_{k}} + \varepsilon + \frac{\rho \, \varepsilon }{n_{k}} < \varepsilon (b+1+ \varepsilon ), \end{aligned}$$

which implies that $\{u_{n_{k}}\}$ converges to u in U because U is closed. Observe that $\{f(u_{n_{k}})\}$ converges to u and by continuity of f it converges also to f(u), which means by Proposition 1.11 that $u=f(u)$. $\square$

Remark 5.6

In Theorem 5.5, if $f :U \rightarrow U$ is compact, the assumption of closeness of U may be dropped, since the sequence $f(u_{n_{k}})$ converges to some $u\in \textrm{cl}\left( f(U)\right)$ which is a subset of U.

Finally, we present a Schauder fixed point principle.

Theorem 5.7

Let $(X, \Vert \cdot \Vert )$ be an sb-supranormed linear space and U be a convex set (not necessarily closed) of X. If a mapping $f:U \rightarrow U$ is compact, then it has a fixed point.

Proof

It suffice to show that f has an $\varepsilon$-fixed point. By Lemma 5.4 it follows that for a sufficiently small $\varepsilon >0$ there exists $f_{\varepsilon }: U\rightarrow U$ such that

• (i) $\Vert f_{\varepsilon } (x)-f(x)\Vert \le n\, b\, \varepsilon \, \max \{1,\rho ^{n}\}$ for all $x\in U$,

• (ii) $f_{\varepsilon } (U)\subset \textrm{conv}(N) \subset U$.

Since $\textrm{conv}(N) \subset U$, we get $f_{\varepsilon } (\textrm{conv}(N))\subset f_{\varepsilon }(U) \subset \textrm{conv}(N)$, which implies that $f_{\varepsilon }: \textrm{conv}(N)\rightarrow \textrm{conv}(N)$ is well defined. Since $\textrm{conv}(N)$ is bounded closed convex (see also [24, Propositions C.2 and C5]), we deduce by Theorem 5.2 that there exists $x_{\varepsilon }\in \textrm{conv}(N) \subset U$ such that $f_{\varepsilon }x_{\varepsilon }=x_{\varepsilon }$, so by (i), we obtain

$$\begin{aligned} \Vert f(x_{\varepsilon })-x_{\varepsilon }\Vert = \Vert f (x_{\varepsilon })-f_{\varepsilon }(x_{\varepsilon })\Vert \le n\, b\, \varepsilon \, \max \{1,\rho ^{n}\}, \end{aligned}$$

and by letting ${\varepsilon }$ tends to zero together with the continuity of f, we obtain the result by Theorem 5.5 and Remark 5.6. $\square$

Remark 5.8

Observe that U is not necessary closed, since Theorem 5.2 is applied to the selfmap $f_{\varepsilon }$ defined on the closed set $\textrm{conv}(N)$. Moreover and according to Remark 5.6, Theorem 5.5 can be applied without requiring the closeness of U. This answer the question in [9, Remark 13].

7 Applications

In this section, we study the existence of a unique solution to an integral equation as well as to a boundary value problem, as applications to the fixed point theorem proved in Section 2. We consider the integral equation:

$$\begin{aligned} x(t)=\lambda (t)+\int _{0}^{1} G(t,s)h(s,x(s))\textrm{d}s,\;\; t\in [0,1]. \end{aligned}$$

(19)

The problem of existence of a solution for the integral equation (19) will be discussed under the following assumptions:

($a_1$):

$\lambda :[0,1]\rightarrow {\mathbb {R}}_{+}$ is a continuous function.

($a_2$):

$h:[0,1]\,{\times }\,{\mathbb {R}}_{+} {\,\rightarrow \,} {\mathbb {R}}_{+}$ is a continuous function and there exists a continuous function $u:{\mathbb {R}}^{2}_{+} {\,\rightarrow \,} {\mathbb {R}}_{+}$ such that for all $(s,p,q)\,{\in }\, [0,1]\,{\times }\,{\mathbb {R}}_{+}^{2}$,

$$\begin{aligned} & u(p,p)=0, \end{aligned}$$

(20a)

$$\begin{aligned} & \left| h(s,p)-h(s,q)\right| \le u(p,q), \end{aligned}$$

(20b)

$$\begin{aligned} & u(p,q)^{2}+ {\textstyle \frac{1}{2}}u(p,q) \le |p-q|^{2}+ {\textstyle \frac{1}{2}}|p-q|. \end{aligned}$$

(20c)

($a_3$):

$G:[0,1]^{2} \rightarrow {\mathbb {R}}_{+}$ is a continuous function such that

$$\begin{aligned} c{:}{=}\max _{s,t\in [0,1]}G(s,t) <1. \end{aligned}$$

Before presenting the main result of this section, we derive an inequality of Chebyshev type. For more details on Chebyshev inequalities, we refer to Chapter IX of [20] and for more recent references, see for instance [1, 23].

Lemma 6.1

Let a and b be real numbers such that $a\, {< }\,b$, and let $w(t,\cdot )$ be a nonnegative measurable function for every $t\,{\in }\,[a,b]$. Let $x(s)=(x_{1}(s),x_{2}(s),\ldots ,x_{n}(s))$ such that $\{x_{i}\}_{1\le i \le n}$ are nonnegative functions defined on [a, b], and let u be a nonnegative function defined on $[a,b]\times {\mathbb {R}}^{n}_{+}$ such that $s\mapsto u(s,x(s))$ is integrable with respect to w(t, s) for every $t\in [a,b]$. Then

$$\begin{aligned} \left( \int _{a}^{b} u(s,x(s))w(t,s)\textrm{d}s \right) ^{2}\le \int _{a}^{b} u(s,x(s))^{2}w(t,s)\textrm{d}s \int _{a}^{b} w(t,s)\textrm{d}s,\;t\in [a,b]. \end{aligned}$$

Proof

We have

$$\begin{aligned} 0&\le \int _{a}^{b} \int _{a}^{b} \big (u(r,x(r))-u(s,x(s))\big )^{2}w(t,s)w(t,r)\textrm{d}s\textrm{d}r\\&= \int _{a}^{b} \int _{a}^{b}\big (u(r,x(r))^{2}-2u(r,x(r))u(s,x(s))+u(s,x(s))^{2}\big )w(t,s)w(t,r)\textrm{d}s\textrm{d}r\\&= \int _{a}^{b}\big ( u(r,x(r))^{2}\int _{a}^{b}w(t,s)\textrm{d}s-2u(r,x(r))\int _{a}^{b}u(s,x(s))w(t,s)\textrm{d}s\\ &\quad +\int _{a}^{b}u(s,x(s))^{2}w(t,s)\textrm{d}s\big )w(t,r)\textrm{d}r\\&= 2\int _{a}^{b} u(s,x(s))^{2}w(t,s)\textrm{d}s\int _{a}^{b}w(t,s)\textrm{d}s-2 \left( \int _{a}^{b}u(s,x(s))w(t,s)\textrm{d}s\right) ^{2}. \end{aligned}$$

$\square$

Theorem 6.2

Under assumptions ($a_1$)–($a_3$), the integral equation (19) has a unique solution in $C_{+}([0, 1])$.

Proof

Let $X=C_{+}([0, 1])$ be the set of continuous functions $x:[0, 1]\rightarrow {\mathbb {R}}_{+}$, endowed with the suprametric $\delta$ of Examples 1.4. First, by Remark 1.14, $(X,\delta )$ is a complete. Consider the operator $T:X\rightarrow X$ defined by

$$\begin{aligned} Tx(t)=\lambda (t)+\int _{0}^{1} G(t,s)h(s,x(s))\textrm{d}s,\;\; t\in [0,1]. \end{aligned}$$

Observe first that T is well defined. Let $x,y\in X$, then by using the assumptions ($a_1$)–($a_3$) and Lemma 6.1, we get

$$\begin{aligned}&|Tx(t)-Ty(t)|(|Tx(t)-Ty(t)|+{\textstyle \frac{1}{2}})\\&\quad = \left| \int _{0}^{1}\! G(t,s)\big (h(s,x(s)){-}h(s,y(s))\big )\textrm{d}s\right| \left( \left| \int _{0}^{1}\! G(t,s)\big (h(s,x(s)){-}h(s,y(s))\big )\textrm{d}s\right| {+}\textstyle \frac{1}{2}\right) \\&\quad \le \int _{0}^{1} G(t,s)\big |h(s,x(s))-h(s,y(s))\big |\textrm{d}s\left( \int _{0}^{1} G(t,s)\big |h(s,x(s))-h(s,y(s))\big |\textrm{d}s{+}{\textstyle \frac{1}{2}}\right) \\&\quad \le \int _{0}^{1} G(t,s)u(x(s),y(s))\textrm{d}s\left( \int _{0}^{1} G(t,s)u(x(s),y(s))\textrm{d}s{+}\textstyle \frac{1}{2}\right) \\&\quad \le \int _{0}^{1} G(t,s)\textrm{d}s\int _{0}^{1} G(t,s)u(x(s),y(s))^{2}\textrm{d}s{+}{\textstyle \frac{1}{2}}\int _{0}^{1} G(t,s)u(x(s),y(s))\textrm{d}s\\&\quad \le \int _{0}^{1} G(t,s)\big (u(x(s),y(s))^{2}+ {\textstyle \frac{1}{2}}u(x(s),y(s))\big )\textrm{d}s\\&\quad \le \int _{0}^{1} G(t,s)|x(s)-y(s)|\big (|x(s)-y(s)|+ {\textstyle \frac{1}{2}}\big )\textrm{d}s\\&\quad \le c\delta (x,y), \end{aligned}$$

and this implies

$$\begin{aligned} \delta (Tx,Ty)\le c\delta (x,y). \end{aligned}$$

By Theorem 2.1, we conclude that the integral equation (19) has a unique solution in X. $\square$

Next by Theorem 6.2, we show the existence of a unique solution in $C_{+}[0,1]$ to the following nonlinear third-order boundary value problem:

$$\begin{aligned}&x'''(t)+{\sqrt{t\,x(t)+1}}(1-e^{-t \, x(t)})=0,\quad t\in [0,1], \end{aligned}$$

(22a)

$$\begin{aligned}&x(0)=x'(1)=0\text { and }x(1)=1. \end{aligned}$$

(22b)

Proposition 6.3

The boundary value problem (22) has a unique solution in $C_{+}[0,1]$.

Proof

The boundary value problem (22) has a solution $x\in C_{+}[0,1]$ if and only if the operator $T:C_{+}[0,1]\rightarrow C_{+}[0,1]$ defined by

$$\begin{aligned} Tx(t)=\int _{0}^{1}G(t,s){\sqrt{ s\,x(s)+1}}(1-e^{-s \, x(s)})\textrm{d}s,\quad t\in [0,1], \end{aligned}$$

has a fixed point in $C_{+}[0,1]$, where the Green’s function associated to the homogeneous problem $x'''(t)=0$ that satisfies the boundary condition (22b) is given by

$$\begin{aligned} G(t,s)= \left\{ \begin{array}{ll} {\textstyle \frac{1}{2}}s^2(t-1)^2,& \quad 0 \le s\le t\le 1,\\ {\textstyle \frac{1}{2}}t(s-1)(s(t-2)+t)),& \quad 0 \le t\le s \le 1. \end{array} \right. \end{aligned}$$

Firstly, observe that T is well defined and ($a_1$) holds, where $\lambda =0$. Moreover, it is easy to see that G is continuous and satisfies ($a_3$), since we have

$$\begin{aligned} 0\le G(t,s) \le {\textstyle \frac{1}{2}},\text { for all }t,s\in [0,1]. \end{aligned}$$

Consider now the functions $h:[0,1]\times {\mathbb {R}}_{+}\rightarrow {\mathbb {R}}_{+}$ and $u:{\mathbb {R}}^{2}_{+}\rightarrow {\mathbb {R}}_{+}$ given by

$$\begin{aligned} h(s,p)=\sqrt{s\,p+1}(1-e^{-s \, p}) \;\text { and }\; u(p,q)=\sqrt{|p-q|+1}(1-e^{-|p-q|}). \end{aligned}$$

In order to use Theorem 6.2 and conclude that T has a unique solution in $C_{+}[0,1]$, we have to check ($a_2$). Note that h and u are continuous and it is not difficult to see that (20b) and (20c) follow from the next lemma. $\square$

Lemma 6.4

For all $(p,q,s)\in {\mathbb {R}}^{2}_{+}\times [0,1]$, we have $A\ge 0$ and $B\ge 0$, where

$$\begin{aligned} & A=\textstyle \sqrt{|p-q|+1}(1-e^{-|p-q|})-|\sqrt{p+1}(1-e^{-p})-\sqrt{q+1}(1-e^{-q})|,\\ & \quad B= p^2+\textstyle \frac{1}{2}p-(s\,p+1)(1-e^{-s\,p})^2-\frac{1}{2}\sqrt{s\,p+1}(1-e^{-s\,p}). \end{aligned}$$

Proof

Suppose, without loss of generality, that $p> q$. Then,

$$\begin{aligned} A=\textstyle \sqrt{p-q+1}(1-e^{q-p})-\sqrt{p+1}(1-e^{-p})+\sqrt{q+1}(1-e^{-q}) \end{aligned}$$

Using the mean value theorem twice, it follows that there exists $c\in (q, p)$ such that

$$\begin{aligned} A=\sqrt{p-q+1}(1-e^{q-p})-\textstyle \frac{1}{2}(p-q)\frac{1+e^{-c}+2ce^{-c}}{ \sqrt{c+1}}, \end{aligned}$$

and also there exists $c'\in (p-q, p)$ such that

$$\begin{aligned} A=\textstyle \sqrt{q+1}(1-e^{-q})-\frac{1}{2}q\frac{1+e^{-c'}+2c'e^{-c'}}{ \sqrt{c'+1}}. \end{aligned}$$

Now, since the function $h_{1}:{\mathbb {R}}_{+}\rightarrow {\mathbb {R}}$ given by $h_{1}(t)=\textstyle \frac{1+e^{-t}+2 t e^{-t}}{ \sqrt{t+1}}$ is decreasing on ${\mathbb {R}}_{+}$, we obtain

$$\begin{aligned} A\ge \textstyle \sqrt{p-q+1}(1-e^{q-p})-\textstyle \frac{1}{2}(p-q)\frac{1+e^{-q}+2 q e^{-q}}{ \sqrt{q+1}}, \end{aligned}$$

and

$$\begin{aligned} A\ge \textstyle \sqrt{q+1}(1-e^{-q})-\frac{1}{2}q\frac{1+e^{-(p-q)}+2(p-q)e^{-(p-q)}}{ \sqrt{p-q+1}}. \end{aligned}$$

Hence, it suffice to know the sign of $h_{2}(p-q)-h_{1}(q)$ and $h_{2}(q)-h_{1}(p-q)$, where the function $h_{2}:{\mathbb {R}}_{+}\rightarrow {\mathbb {R}}$ is given by $h_{2}(t)=\textstyle {2t^{-1}\sqrt{t+1}(1-e^{-t})}$. It is not difficult to see that $h_{2}$ is decreasing, so if $p-q\le q$, $h_{2}(p-q)-h_{1}(q)\ge h_{2}(p-q)-h_{1}(p-q)$ and if $p-q>q$, $h_{2}(q)-h_{1}(p-q)\ge h_{2}(q)-h_{1}(q)$. We conclude from the fact that $t\mapsto (h_{2}-h_{1})(t)$ is positive that $A\ge 0$. Finally, we have

$$\begin{aligned} B&\ge p^2+\textstyle \frac{1}{2}p-(s\,p+1)(1-e^{-s\,p})^2-\sqrt{s\,p+1}(1-e^{-s\,p})\\&\ge (s\,p)^2+\textstyle \frac{1}{2}s\,p-(s\,p+1-\sqrt{s\,p+1})(1-e^{-s\,p})\ge 0. \end{aligned}$$

$\square$