3.1 Sufficient condition for unconditional convergence
After giving Definitions 1 and 2 we are going to prove a result guaranteeing that a general second order difference equation as in (4) has an unconditional stable attractor.
Let $\mathrm{\infty}<c\le a<b\le d\le +\mathrm{\infty}$ and consider in the sequel a continuous function $\phi :(a,b)\times (c,d)\to (c,d)$, satisfying the following conditions:
(H1) $\phi ({x}_{1},y)<\phi ({x}_{2},y)$, whenever $a<{x}_{2}<{x}_{1}<b$ and $c<y<d$.
(H2) There exists
${F}_{\phi}:[a,b]\to [a,b]$ such that
$\frac{{F}_{\phi}(x)y}{\phi (x,y)y}\ge 1$
whenever $y\in (c,d)\setminus \{{F}_{\phi}(x)\}$.
The functions $\phi (\cdot ,y):[a,b]\to [c,d]$ and $\phi (x,\cdot ):(c,d)\to (c,d)$ are defined in the obvious way. Notice that $\phi (a,\cdot )$ is the limit of a monotone increasing sequence of continuous functions, thus it is lowersemicontinuous, likewise $\phi (b,\cdot )$ is an uppersemicontinuous function. Remember that we denote both φ and $\stackrel{\u02c6}{\phi}$ by φ.
The next lemma, which we prove at the end of this section, shows that if (H1) and (H2) holds we can get some information about the behavior and properties of φ and ${F}_{\phi}$.
Lemma 2 Let $\phi :(a,b)\times (c,d)\to (c,d)$,
where $\mathrm{\infty}<c\le a<b\le d\le +\mathrm{\infty}$,
be a continuous function satisfying (H1)
and (H2).
Then the function ${F}_{\phi}$ in (H2)
is unique and it is a continuous nonincreasing map,
thus it has a unique fixed point ${\mu}_{\phi}$.
Furthermore,
 (i)
$a<\phi (x,y)<b$ for all $x,y\in (a,b)$ and $a\le \phi (x,y)\le b$ for all $x,y\in [a,b]$.
 (ii)
If $x\in [a,b]$, $y\in (c,d)$ and $\phi (x,y)=y$, then $y={F}_{\phi}(x)$.
 (iii)
$\phi (x,{F}_{\phi}(x))={F}_{\phi}(x)$ for all $x\in [a,b]$.
 (iv)
${F}_{\phi}$ is decreasing in ${F}_{\phi}^{1}((a,b))$.
We are in conditions of presenting and proving our main result.
Theorem 1 Let $\phi :(a,b)\times (c,d)\to (c,d)$, where $\mathrm{\infty}<c\le a<b\le d\le +\mathrm{\infty}$, be a continuous function satisfying (H1) and (H2). If ${\mu}_{\phi}\in (a,b)$ and $Fix({F}_{\phi}\circ {F}_{\phi})=Fix({F}_{\phi})$, then ${\mu}_{\phi}$ is an unconditional stable attractor of φ in $(a,b)$.
Proof of Theorem 1 Consider
$\lambda \in {\mathrm{\Lambda}}_{m}^{2}$ and denote
${y}_{n}=\phi \circ \lambda ({y}_{n1},\dots ,{y}_{nm})$
for some ${y}_{1},\dots ,{y}_{m}\in (a,b)$. Notice that ${y}_{n}\in (a,b)$ for all n, as a consequence of (i) in Lemma 2.
We are going to prove first that
${\mu}_{\phi}$ is a stable equilibrium of
$\phi \circ \lambda $. By (iii) in Lemma 2, as
$\lambda ({\mu}_{\phi},\dots ,{\mu}_{\phi})=({\mu}_{\phi},{\mu}_{\phi}),$
we see that ${\mu}_{\phi}$ is an equilibrium.
Let
$\u03f5\in (0,min\{{\mu}_{\phi}a,b{\mu}_{\phi}\})$. Because of the continuity of
${F}_{\phi}$, there is
${a}^{\prime}\in ({\mu}_{\phi}\u03f5,{\mu}_{\phi})$ such that
${b}^{\prime}\equiv {F}_{\phi}\left({a}^{\prime}\right)\in ({\mu}_{\phi},{\mu}_{\phi}+\u03f5).$
As
$Fix({F}_{\phi}\circ {F}_{\phi})=Fix({F}_{\phi})$ and
${F}_{\phi}({F}_{\phi}(a))\ge a$, we have
${F}_{\phi}({F}_{\phi}(x))>x\phantom{\rule{1em}{0ex}}\text{for all}x\in [a,{\mu}_{\phi}).$
If
$x\in [{a}^{\prime},{b}^{\prime}]$, then
${F}_{\phi}(x)\le {F}_{\phi}\left({a}^{\prime}\right)={b}^{\prime}$
and
${F}_{\phi}(x)\ge {F}_{\phi}\left({b}^{\prime}\right)={F}_{\phi}\left({F}_{\phi}\left({a}^{\prime}\right)\right)>{a}^{\prime}.$
Therefore, ${F}_{\phi}([{a}^{\prime},{b}^{\prime}])\subset [{a}^{\prime},{b}^{\prime}]$.
By replacing
a,
b by
${a}^{\prime}$,
${b}^{\prime}$ in Lemma 2(i), we see that
${y}_{n}\in ({a}^{\prime},{b}^{\prime})\subset ({\mu}_{\phi}\u03f5,{\mu}_{\phi}+\u03f5)\phantom{\rule{1em}{0ex}}\text{for all}n,$
whenever ${y}_{n}\in ({a}^{\prime},{b}^{\prime})$ for $n\le m$, thus ${\mu}_{\phi}$ is an unconditional stable equilibrium of φ.
Now, if we see that
$lim{F}_{\phi}({y}_{n})={\mu}_{\phi},$
we are done with the whole proof. Indeed, for each accumulation point
$\overline{y}$ of
$({y}_{n})$, one would have
${F}_{\phi}({\mu}_{\phi})={\mu}_{\phi}={F}_{\phi}(\overline{y}),$
because of the continuity of ${F}_{\phi}$. As ${\mu}_{\phi}\in (a,b)$, this implies $\overline{y}={\mu}_{\phi}$.
Therefore, as a consequence of Lemma 1, it suffices to find an increasing sequence
${n}_{k}$ of natural numbers such that
${a}_{k}\le {F}_{\phi}({y}_{n})\le {b}_{k}\phantom{\rule{1em}{0ex}}\text{for all}k\ge 0,n\ge {n}_{k}.$
Here,
${a}_{k}$ and
${b}_{k}$ are defined as in Lemma 1, with
${a}_{0}=a$ and
$\u03f5=1$,
${b}_{0}={F}_{\phi}(a);\phantom{\rule{2em}{0ex}}{a}_{k}={F}_{\phi}({b}_{k1}+\frac{1}{k});\phantom{\rule{2em}{0ex}}{b}_{k}={F}_{\phi}({a}_{k}\frac{1}{k})\phantom{\rule{1em}{0ex}}\text{for}k\ge 1.$
Let
${n}_{0}=m$, so that
${a}_{0}\le {F}_{\phi}({y}_{n})\le {F}_{\phi}(a)={b}_{0}\phantom{\rule{1em}{0ex}}\text{for}n\ge {n}_{0}.$
Having in mind that
$\lambda \in {\mathrm{\Lambda}}_{m}^{2}$ satisfies (3), we find
${n}_{k}$ from
${n}_{k1}$ as follows. Denote
${z}_{k}={F}_{\phi}^{1}({b}_{k1})$
and momentarily assume
$n>{n}_{k1}+m$ and
${b}_{k1}<{y}_{n}$ in such a way that
$\begin{array}{rcl}{b}_{k1}& <& {y}_{n}=\phi ({\lambda}_{1}({y}_{n1},\dots ,{y}_{nm}),{\lambda}_{2}({y}_{n1},\dots ,{y}_{nm}))\\ \le & \phi (min\{{y}_{n1},\dots ,{y}_{nm}\},{\lambda}_{2}({y}_{n1},\dots ,{y}_{nm}))\\ \le & \phi ({z}_{k},{\lambda}_{2}({y}_{n1},\dots ,{y}_{nm})),\end{array}$
which implies
$\phi ({z}_{k},{\lambda}_{2}({y}_{n1},\dots ,{y}_{nm}))\le {\lambda}_{2}({y}_{n1},\dots ,{y}_{nm})$
and then
${y}_{n}\le max\{{b}_{k1},{\lambda}_{2}({y}_{n1},\dots ,{y}_{nm})\}\le max\{{b}_{k1},{y}_{n1},\dots ,{y}_{nm}\}\equiv {w}_{n}$
for all $n>{n}_{k1}+m$.
As a consequence, the nonincreasing sequence ${w}_{n}$ is bounded below by ${b}_{k1}$. It cannot be the case that $lim{w}_{n}>{b}_{k1}$, because in such a case there is a subsequence ${y}_{{n}_{j}}>{b}_{k1}$ converging to $lim{w}_{n}$ and such that ${\lambda}_{2}({y}_{{n}_{j}1},\dots ,{y}_{{n}_{j}m})$ converges to a point $w\le lim{w}_{n}$.
Since
${y}_{{n}_{j}}\le \phi ({z}_{k},{\lambda}_{2}({y}_{{n}_{j}1},\dots ,{y}_{{n}_{j}m}))\le {w}_{{n}_{j}},$
one has
$\phi ({z}_{k},w)=lim{w}_{n}>{b}_{k1}=F({z}_{k})$
and then
$\phi ({z}_{k},w)w>F({z}_{k})w.$
By applying (H2), we see that
$lim{w}_{n}=\phi ({z}_{k},w)<w,$
a contradiction.
Therefore,
$lim{w}_{n}={b}_{k1}$ and there exists
${m}_{k}\ge {n}_{k1}$ such that
${y}_{n}<{b}_{k1}+\frac{1}{k}\phantom{\rule{1em}{0ex}}\text{for all}n\ge {m}_{k},$
that is,
${F}_{\phi}({y}_{n})\ge {a}_{k}\phantom{\rule{1em}{0ex}}\text{for all}n\ge {m}_{k}.$
Analogously, we see that there exists
${n}_{k}\ge {m}_{k}$ such that
${F}_{\phi}({y}_{n})\le {b}_{k}\phantom{\rule{1em}{0ex}}\text{for all}n\ge {n}_{k}.$
□
Proof of Lemma 2

Uniqueness of
${F}_{\phi}$: Let
${y}_{1}<{y}_{2}$ and
x in
$[a,b]$ such that
$\frac{{y}_{i}y}{\phi (x,y)y}\ge 1>0\phantom{\rule{1em}{0ex}}\text{for}i=1,2,y\in ({y}_{1},{y}_{2}).$
a contradiction.

(i): It suffices to prove the first assertion, because
$[a,b]$ is a closed set and, by definition,
$\phi (x,y)=\{lim\hspace{0.17em}sup\phi ({x}_{n},{y}_{n}):{x}_{n}\to x,{y}_{n}\to y,{x}_{n}\in (a,b),{y}_{n}\in (c,d)\}$
for all $(x,y)\in [a,b]\times [c,d]$. Assume now that $(x,y)\in (a,b)\times (a,b)$. We consider the following three possible situations. If $\phi (x,y)=y$, it is obvious that $\phi (x,y)\in (a,b)$.On the other hand, if $\phi (x,y)>y$ and ${x}^{\prime}\in (a,x)$, then
$a\le y<\phi (x,y)<\phi ({x}^{\prime},y)\le {F}_{\phi}\left({x}^{\prime}\right)\le b.$
Finally, if
$\phi (x,y)<y$ and
${x}^{\prime}\in (x,b)$, then
$b\ge y>\phi (x,y)>\phi ({x}^{\prime},y)\ge {F}_{\phi}\left({x}^{\prime}\right)\ge a.$

(ii): Suppose
$y\ne {F}_{\phi}(x)$. Since
$\phi (x,y)=y\in \overline{\mathbb{R}}$, then
$\phi (x,y)y=0$ or
$\phi (x,y)y=\overline{\mathbb{R}}$. In any event, it cannot be the case that
$\frac{{F}_{\phi}(x)y}{\phi (x,y)y}\ge 1,$
which contradicts hypothesis (H2), thus $y={F}_{\phi}(x)$.

(iii): Since
${F}_{\phi}([a,b])\subset [a,b]\subset [c,d]$, it is worth considering the following three cases for each
$x\in [a,b]$: first,
$x\in (a,b)$,
${F}_{\phi}(x)\in (c,d)$ and then (after probing continuity, monotonicity and statement (iv)), we proceed with the case
$x\in \{a,b\}$,
${F}_{\phi}(x)\in (c,d)$ and finally with
$x\in [a,b]$,
${F}_{\phi}(x)\in \{c,d\}$.Case
$x\in (a,b)$ and
${F}_{\phi}(x)\in (c,d)$: Since
$\phi (x,y)>y$ when
$y<{F}_{\phi}(x)$ and
$\phi (x,y)<y$ when
$y>{F}_{\phi}(x)$, we see that
$\phi (x,{F}_{\phi}(x))={F}_{\phi}(x),$
because of the continuity of $\phi (x,\cdot )$.
${F}_{\phi}({x}_{1})\le {F}_{\phi}({x}_{2})\phantom{\rule{1em}{0ex}}\text{for}a\le {x}_{1}{x}_{2}\le b.$
If
$y\in [{F}_{\phi}({x}_{1}),{F}_{\phi}({x}_{2})]$, then
$y=a$,
$y=b$ or
$y\le \phi ({x}_{2},y)<\phi ({x}_{1},y)\le y$, thus
$[{F}_{\phi}({x}_{1}),{F}_{\phi}({x}_{2})]\subset \{a,b\}.$

Continuity: If
$x\in [a,b]$ and
$w\in I=(\underset{z\to x}{lim\hspace{0.17em}inf}{F}_{\phi}(z),\underset{y\to x}{lim\hspace{0.17em}sup}{F}_{\phi}(y)),$
then there exist two sequences
${y}_{n},{z}_{n}\to x$ with
${F}_{\phi}({z}_{n})<w<{F}_{\phi}({y}_{n}).$
Thus,
$\phi ({z}_{n},w)<w<\phi ({y}_{n},w)$
and, by (ii), one has $\phi (x,w)=w$. Since $w\in (c,d)$, this would imply ${F}_{\phi}(x)=w$ for all $w\in I$, which is impossible.

(iii) Case
$x\in \{a,b\}$ and
${F}_{\phi}(x)\in (c,d)$: Since
${F}_{\phi}(t)\le \phi (t,{F}_{\phi}(a))\le {F}_{\phi}(a)\phantom{\rule{1em}{0ex}}\text{for all}t\in (a,b),$
and because of the continuity of
${F}_{\phi}$, we have
${F}_{\phi}(a)=\underset{t}{sup}{F}_{\phi}(t)\le \underset{t}{sup}\phi (t,{F}_{\phi}(a))\le {F}_{\phi}(a),$
but
$\underset{t}{sup}\phi (t,{F}_{\phi}(a))=\phi (a,{F}_{\phi}(a)).$
Analogously, it can be seen that
$\phi (b,{F}_{\phi}(b))={F}_{\phi}(b)$.

(iii) Case
${F}_{\phi}(x)\in \{c,d\}$: First assume
${F}_{\phi}(x)=c$ and recall that, by definition,
$\phi (x,c)=\{lim\hspace{0.17em}sup\phi ({x}_{n},{y}_{n}):{x}_{n}\to x,{y}_{n}\to c,{x}_{n}\in (a,b),{y}_{n}\in (c,d)\}.$
Suppose
$\phi ({x}_{n},{y}_{n})\ge {c}^{\prime}>c$ for all
n. Then
$1\le \frac{{F}_{\phi}({x}_{n}){y}_{n}}{\phi ({x}_{n},{y}_{n}){y}_{n}}\le \frac{{F}_{\phi}({x}_{n}){y}_{n}}{{c}^{\prime}{y}_{n}},$
which implies ${F}_{\phi}({x}_{n})\ge {c}^{\prime}>c$, eventually for all n.Since ${F}_{\phi}({x}_{n})\to c$, we reach a contradiction. Therefore,
$\phi (x,{F}_{\phi}(x))=\{c\}=\{{F}_{\phi}(x)\}.$
Analogously, we see that
$\phi (x,{F}_{\phi}(x))=\{d\}$ when
${F}_{\phi}(x)=d$. □