The last equality in (
3.4) arises for any given
\(a, b \in X\), since
\(( X, \Vert \; \Vert )\) is a normed vector space, one has trivially
\(\Vert a + b - a \Vert = \Vert b \Vert \) and, since
\(0 \in X\), it follows that
\(F_{a + b, a} ( t ) = F_{b, 0} ( t )\),
\(\forall t \in \mathbf{R}_{ +}\), in the Menger PM space
\(( X,F,\Delta_{M} )\). Otherwise, the probability of
\(\Vert b \Vert \) to be less than some
\(t \in \mathbf{R}_{ +}\) would be distinct to the probability of
\(\Vert a + b - a \Vert \) to be less than such a positive real value
t, which is impossible. In the same way and, since
\(\Vert a + b + c - d - e - f \Vert = \Vert a - ( d + e + f - b - c ) \Vert \) for any given
\(a,b,c,d, e, f \in X\), it follows that
$$ \begin{aligned}[b] F_{a + b + c, d + e + f} ( t ) &= F_{a, d + e + f - b - c} ( t ) \\ &\ge \Delta _{M} \bigl( F_{a,d} ( t/2 ), F_{d, d + e + f - b - c} ( t/2 ) \bigr) \\ &\ge \Delta_{M} \bigl( F_{a,d} ( t/2 ), F_{b + c, e + f} ( t/2 ) \bigr) \\ &= \Delta_{M} \bigl( F_{a,d} ( t/2 ), \Delta_{M}F_{b, e + f - c} ( t/2 ) \bigr) \\ &\ge \Delta_{M} \bigl( F_{a,d} ( t/2 ), \Delta \bigl( F_{b, e} ( t/4 ) F_{e,e + f - c} ( t/4 ) \bigr) \bigr) \\ &= \Delta_{M} \bigl( F_{a,d} ( t/2 ), \Delta_{M} \bigl( F_{b, e} ( t/4 ) F_{c,f} ( t/4 ) \bigr) \bigr), \quad \forall t \in \mathbf{R}_{ +}, \end{aligned} $$
(3.5)
which becomes for
\(c = d = e = 0\) and
\(f = \sum_{i = 0}^{k} f_{i}\) and the symmetry of the probability density function (second property of (
2.1)), the associative property of the triangular norms (fourth property of (
2.3)) and the use of the minimum triangular norm:
$$\begin{aligned} &F_{a + f, b} ( t ) \\ &\quad \ge \Delta_{M} \bigl( F_{a,0} ( t/2 ), \Delta_{M} \bigl( F_{b, 0} ( t/4 ) , F_{f,0} ( t/4 ) \bigr) \bigr) \\ &\quad = \Delta _{M} \bigl( F_{\sum_{i = 0}^{k} f_{i},0} ( t/4 ), \Delta_{M} \bigl( F_{a,0} ( t/2 ),F_{b, 0} ( t/4 ) \bigr) \bigr) \\ &\quad \ge \Delta_{M} \bigl( \Delta_{M} \bigl( F_{f_{0}, 0} ( t/8 ),F_{\sum_{i = 1}^{k} f_{i},0} ( t/8 ) \bigr), \Delta_{M} \bigl( F_{a,0} ( t/2 ),F_{b, 0} ( t/4 ) \bigr) \bigr) \\ &\quad \cdots \\ &\quad \ge \Delta_{M} \Bigl( \Delta_{M} \bigl( F_{a,0} \bigl( t/ \bigl( 2^{k + 3} \bigr) \bigr), F_{b, 0} \bigl( t/ \bigl( 2^{k + 3} \bigr) \bigr) \bigr),\min_{0 \le i \le k} F_{f_{i}, 0} \bigl( t/ \bigl( 2^{k + 3} \bigr) \bigr) \Bigr) \\ &\quad \ge \min \Bigl( \min \bigl[ F_{a,0} \bigl( t/ \bigl( 2^{k + 3} \bigr) \bigr),F_{b, 0} \bigl( t/ \bigl( 2^{k + 3} \bigr) \bigr) \bigr],\min_{0 \le i \le k} \bigl[ F_{f_{i}, 0} \bigl( t/ \bigl( 2^{k + 3} \bigr) \bigr) \bigr] \Bigr) \\ &\quad \ge \min \Bigl( F_{a,0} \bigl( t/ \bigl( 2^{k + 3} \bigr) \bigr),F_{b, 0} \bigl( t/ \bigl( 2^{k + 3} \bigr) \bigr),\min _{0 \le i \le k} \bigl[ F_{f_{i}, 0} \bigl( t/ \bigl( 2^{k + 3} \bigr) \bigr) \bigr] \Bigr), \quad \forall t \in \mathbf{R}_{ +}, \end{aligned}$$
(3.6)
due to the fact that
\(F_{a,0} ( t/2 ) \ge F_{a,0} ( t/8 )\),
\(F_{b, 0} ( t/4 ) \ge F_{b, 0} ( t/8 )\) since the distribution function is non-decreasing. Now, define auxiliary sequences.
$$ \begin{aligned} &a = a ( n ) = \sum_{i = 0}^{k} \alpha_{i}^{ ( n ) } T^{i}x_{n + 1}, \\ &b = b ( n ) = \sum_{i = 0}^{k} \alpha_{i}^{ ( n )}T^{i}x_{n}, \\ &f_{i} = f_{i} ( n ) = \tilde{\alpha}_{i}^{ ( n ) } T^{i}x_{n + 1},\qquad f = f ( n ) = \sum_{i = 0}^{k} \tilde{\alpha}_{i}^{ ( n ) } T^{i}x_{n + 1} = \sum_{i = 0}^{k} f_{i}, \end{aligned} $$
(3.7)
\(\forall n \in \mathbf{Z}_{0 +}\), so that one gets from (
3.6)-(
3.7) into (
3.4) and since
\(0 \in X\):
$$\begin{aligned} &F_{x_{n + 2}, x_{n + 1}} \bigl( \phi ( t ) \bigr) \\ &\quad \ge \Delta_{M} \Bigl( \Delta_{M} \bigl( F_{\sum_{i = 0}^{k} \alpha_{i}^{ ( n ) } T^{i}x_{n + 1},0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 2} \bigr) \bigr) \bigr),F_{\sum_{i = 0}^{k} \alpha_{i}^{ ( n )}T^{i}x_{n}, 0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 2} \bigr) \bigr) \bigr) \bigr), \\ &\qquad {}\min_{0 \le i \le k}F_{\tilde{\alpha}_{i}^{ ( n ) } T^{i}x_{n + 1}, 0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 2} \bigr) \bigr) \bigr) \Bigr) \\ &\quad \ge \min \bigl( F_{x_{n + 1},0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 3}\rho \bigr) \bigr) \bigr),F_{x_{n}, 0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 3}\rho \bigr) \bigr) \bigr), \\ &\qquad {}F_{x_{n + 1}, 0} \bigl( \phi \bigl( ( k + 1 ) t/ \bigl( 2^{k + 3} ( 1 - \rho ) \bigr) \bigr) \bigr) \bigr),\quad \forall t \in \mathbf{R}_{ +}. \end{aligned}$$
(3.8)
From (
3.2)-(
3.3) and
\(\vert \tilde{\alpha}_{n} \vert \le 1 - \rho\), one gets
$$ \min_{0 \le i \le k}F_{\tilde{\alpha}_{i}^{ ( n ) } T^{i}x_{n + 1}, 0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 3} \bigr) \bigr) \bigr) \ge F_{x_{n + 1}, 0} \bigl( \phi \bigl( ( k + 1 ) t/ \bigl( 2^{k + 3} ( 1 - \rho ) \bigr) \bigr) \bigr),\quad \forall n \in \mathbf{Z}_{0 +}. $$
(3.9)
The use of the constraints (
3.2)-(
3.3), and the fact that
\(T_{n}0 = 0 \in X\),
\(\forall n \in \mathbf{Z}_{0 +}\), from (
3.1) and
\(0 \in X\), yield
$$\begin{aligned}& \begin{aligned}[b] F_{x_{n + 1}, 0} \bigl( \phi ( t ) \bigr) &= F_{\sum_{i = 0}^{k} \alpha_{i}^{ ( n ) } T^{i}x_{n}, 0} \bigl( \phi ( t ) \bigr) \\ &\ge F_{x_{n},0} \bigl( \phi \bigl( \rho^{ - 1}t \bigr) \bigr), \forall t \in \mathbf{R}_{ +},\quad \forall n \in \mathbf{Z}_{0 +}, \end{aligned} \end{aligned}$$
(3.10)
$$\begin{aligned}& \begin{aligned}[b] F_{x_{n + 2}, \sum_{i = 0}^{k} \alpha_{i}^{ ( n )}T^{i}x_{n + 1}} \bigl( \phi ( t ) \bigr) &= F_{\sum_{i = 0}^{k} \tilde{\alpha}_{i}^{ ( n ) } T^{i}x_{n + 1}, 0} \bigl( \phi ( t ) \bigr) \\ &\ge F_{x_{n + 1}, 0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 3} ( 1 - \rho ) \bigr) \bigr) \bigr) \\ &\ge F_{x_{n + 1}, 0} \bigl( \phi \bigl( \rho ^{ - 1} t \bigr) \bigr),\quad \forall t \in \mathbf{R}_{ +}, \forall n \in \mathbf{Z}_{0 +}, \end{aligned} \end{aligned}$$
(3.11)
for any sequence
\(\{ x_{n} \} \subset X\) generated from (
3.1) for any initial condition
\(x_{0} \in X\) since
ϕ is strictly increasing and the probability density function is non-decreasing. The use of (
3.10)-(
3.11) in (
3.8) yields by using recursive calculations:
$$\begin{aligned} &F_{x_{n + 1}, x_{n}} \bigl( \phi ( t ) \bigr) \\ &\quad \ge \min \bigl( F_{x_{n},0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 3}\rho \bigr) \bigr) \bigr),F_{x_{n - 1}, 0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 3}\rho \bigr) \bigr) \bigr) ,F_{x_{n}, 0} \bigl( \phi \bigl( ( k + 1 )t/ \bigl( 2^{k + 3} ( 1 - \rho ) \bigr) \bigr) \bigr) \bigr) \\ &\quad \cdots \\ &\quad \ge \min \bigl( F_{x_{0},0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 3}\rho^{n + 1} \bigr) \bigr) \bigr),F_{x_{0}, 0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 3}\rho^{n} \bigr) \bigr) \bigr) , \\ &\qquad F_{x_{0}, 0} \bigl( \phi \bigl( ( k + 1 ) t/ \bigl( 2^{k + 3} ( 1 - \rho )^{n + 1} \bigr) \bigr) \bigr) \bigr),\quad \forall t \in \mathbf{R}_{ +},\ \forall n \in \mathbf{Z}_{ +}. \end{aligned}$$
(3.12)
Since
\(\rho \in ( 0, 1 )\),
\(( 1 - \rho ) \in ( 0, 1 )\) and
\(\phi :\mathbf{R} \to \mathbf{R}_{0 +}\) is a Φ-function we have
\(\lim_{n \to \infty} \phi ( t/ ( 2^{k + 3}\rho^{n} ) ) = + \infty\) so that one finds from (
3.12) that the limit below exists:
$$\begin{aligned} &\lim_{n \to \infty} F_{x_{0},0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 3}\rho^{n + 1} \bigr) \bigr) \bigr) \\ &\quad = \lim _{n \to \infty} F_{x_{0}, 0} \bigl( \phi \bigl( t/ \bigl( 2^{k + 3}\rho^{n} \bigr) \bigr) \bigr) \\ &\quad = \lim_{n \to \infty} F_{x_{0}, 0} \bigl( \phi \bigl( ( k + 1 ) t/ \bigl( 2^{k + 3} ( 1 - \rho )^{n + 1} \bigr) \bigr) \bigr) = 1,\quad \forall t \in \mathbf{R}_{ +} \end{aligned}$$
(3.13)
since
\(F_{x,y}:\mathbf{R} \to [ 0, 1 ]\) is non-decreasing and left continuous with
\(F ( 0 ) = 0\) and
\(\sup_{t \in \mathbf{R}} F ( t ) = 1\). Then
\(\lim_{n \to \infty} F_{x_{n + 1}, x_{n}} ( \phi ( t ) ) = 1\),
\(\forall t \in \mathbf{R}_{ +}\) from (
3.12)-(
3.13). Also, one sees for any given
\(\varepsilon \in \mathbf{R}_{ +}\),
\(\lambda \in ( 0, 1 ) \cap \mathbf{R}_{ +}\) that there exists
\(n_{0} = n_{0} ( \varepsilon,\lambda ) \in \mathbf{Z}_{0 +}\) such that, for all
\(n ( \in \mathbf{Z}_{ +} ) > n_{0}\),
$$\begin{aligned} F_{x_{n + 1}, x_{n}} ( \varepsilon ) ={}& F_{x_{n + 1}, x_{n}} \bigl( \phi \bigl( t_{1}^{ -} \bigr) \bigr) \\ \ge {}&\min \bigl( F_{x_{0},0} \bigl( \phi \bigl( t_{1}/ \bigl( 2^{k + 3}\rho^{n + 1} \bigr) \bigr) \bigr),F_{x_{0}, 0} \bigl( \phi \bigl( t_{1}/ \bigl( 2^{k + 3}\rho^{n} \bigr) \bigr) \bigr) , \\ &{} F_{x_{0}, 0} \bigl( \bigl( ( k + 1 ) t/ \bigl( 2^{k + 3} ( 1 - \rho )^{n + 1} \bigr) \bigr) \bigr) \bigr) \\ \ge {}&1 - \lambda \end{aligned}$$
(3.14)
from (
3.12) since
\(\lim_{n \to \infty} F_{x_{n + 1}, x_{n}} ( \phi ( t ) ) = 1\),
\(\phi :\mathbf{R} \to \mathbf{R}_{0 +}\) is left continuous and strictly increasing, continuous at
\(t = 0\) and
\(\phi ( t ) = 0\) if and only if
\(t = 0\), and then there exists
\(t_{1} = t_{1} ( \varepsilon ) \in \mathbf{R}_{ +}\) such that
\(\phi ( t_{1}^{ -} ) = \lim_{t \to t_{1}^{ -}} \phi ( t ) = \varepsilon\). Thus,
\(\{ x_{n} \} \subset X\) converges to some
\(a \in X\) (Definition
2.3) and it is bounded as a result. Also, one has from (
2.2) that, once having fixed any
\(\varepsilon \in \mathbf{R}_{ +}\),
\(\lambda \in ( 0, 1 ) \subset \mathbf{R}\) and for all integers
\(n, m > n_{0} = n_{0} ( \varepsilon, \lambda )\), there exists
\(\lambda_{1} ( \le \lambda ) \in ( 0, 1 ) \subset \mathbf{R}\) such that, since the distribution function is non-decreasing,
$$\begin{aligned} F_{x_{n} , x_{m}} ( \varepsilon ) &\ge \Delta_{M} \bigl( F_{x_{n} , a} ( \varepsilon /2 ), F_{a , x_{m}} ( \varepsilon /2 ) \bigr) \\ &\ge \Delta_{ M} ( 1 - \lambda_{1}, 1 - \lambda_{1} ) = 1 - \lambda_{1} \ge 1 - \lambda \end{aligned}$$
(3.15)
and
\(\{ x_{n} \} \subset X\) is a Cauchy sequence (Definition
2.4) and then it is bounded. It is clear from (
3.8) that
\(F_{x_{n}, 0} ( \phi ( t/ ( 2^{k + 3}\rho ) ) ) \to 1\),
\(\forall t \in \mathbf{R}_{ +}\) as
\(n \to \infty\) so that
\(\{ x_{n} \} \to 0\) from the first property in (
2.1) of the PM space
\(( X,F, \Delta )\). Since
\(T_{n}0 = 0\),
\(\forall n \in \mathbf{Z}_{0 +}\), from (
3.1), we find that
\(x = 0\) is an equilibrium point of (
3.1) and a fixed point of all the members of the sequence of self-mappings
\(\{ T_{n} \}\). It is now proved that such an equilibrium is unique. Assume that the uniqueness fails and that there is
\(a \ne 0\) which is also an equilibrium point of (
3.1). Since
ϕ is strictly increasing and
\(\rho < 1\), choose
\(\delta \in \mathbf{R}_{ +}\), such that
\(\phi ( \rho^{ - 1}t ) - \delta > \phi ( t )\), and define
\(\lambda_{1} \in ( 0, 1 )\) such that
\(\lambda_{1} = \lambda_{1} ( t ) = 1 - F_{a,0} ( \rho^{ - 1}t )\). Since
\(a \ne 0\) according to the first property of the PM spaces in (
2.1), implies that
\(F_{a,0} ( \rho^{ - 1}t ) < 1\) from the first property of the probabilistic metric space
\(( X,F, \Delta )\) (
2.1). Now, fix any
\(\varepsilon \in \mathbf{R}_{ +}\) and
\(\lambda ( < \lambda_{1} ) \in ( 0, 1 ) \cap \mathbf{R}\) such that if some solution sequence of (
3.1)
\(\{ x_{n} \} ( \subset X ) \to a\) then there exists
\(n_{0} = n_{0} ( \varepsilon, \lambda ) \in \mathbf{Z}_{0 +}\) such that for all
\(n, m ( \in \mathbf{Z}_{ +} ) > n_{0}\), one has
$$\begin{aligned} 1 - \lambda_{1}& = F_{a,0} \bigl( \phi \bigl( \rho^{ - 1}t \bigr) \bigr) \ge \Delta_{M} \bigl( F_{a, x_{n}} ( \delta ), F_{x_{n}, x_{m}} \bigl( \phi \bigl( \rho^{ - 1}t \bigr) - \delta \bigr) \bigr) \\ &> \Delta_{M} ( 1 - \lambda, 1 - \lambda ) = 1 - \lambda, \end{aligned}$$
(3.16)
which contradicts the choice
\(\lambda < \lambda_{1}\) so that
\(x = 0\) is the unique equilibrium point of (
3.1) and trivially a fixed point of
\(T_{n}:X \to X\). This fixed point is unique. Assume that this is not the case. Then
\(T_{n}0 = 0\),
\(\forall n \in \mathbf{Z}_{0 +}\), and assume that for some
\(m \in \mathbf{Z}_{0 +}\),
\(T_{m}^{n}x = T_{m}x = x \ne 0\). Thus,
\(F_{x, 0} ( \varepsilon ) = F_{T_{m}^{n}x, T_{m}^{n}0} ( \varepsilon ) \ge F_{x, 0} ( \phi ( \rho^{ - n}t ) )\) for any given
\(\varepsilon \in \mathbf{R}_{ +}\) and a unique
\(t = t ( \varepsilon ) \in \mathbf{R}_{ +}\) satisfying
\(\phi ( t ) = \varepsilon\), which exists since
\(\phi : \mathbf{R}_{0 +} \to \mathbf{R}_{0 +}\) is a Φ-function. As a result,
$$ \mathop{\lim\inf}\limits_{n \to \infty} \bigl( F_{x, 0} ( \varepsilon ) - F_{x, 0} \bigl( \phi \bigl( \rho^{ - n}t \bigr) \bigr) \bigr) = F_{x, 0} ( \varepsilon ) - \lim_{t \to \infty} F_{x, 0} ( t ) = F_{x, 0} ( \varepsilon ) - 1 \ge 0. $$
(3.17)