For a map h:{\overline{\mathbb{R}}}^{k}\to {2}^{\overline{\mathbb{R}}}, the difference equation
{y}_{n}=h({y}_{n1},\dots ,{y}_{nk})
(4)
is always well defined whatever the initial points {y}_{1},\dots ,{y}_{k}\in \overline{\mathbb{R}} are, even though the {y}_{n} are subsets of \overline{\mathbb{R}}, rather than points.
A point \mu \in \overline{\mathbb{R}} is said to be an equilibrium for the map h if h(\mu ,\dots ,\mu )=\{\mu \}. The equilibrium μ is said to be stable if, for each neighborhood V of μ in \overline{\mathbb{R}}, there is a neighborhood W of (\mu ,\dots ,\mu ) in \mathcal{D}(h) such that {y}_{n}\in V for all n, whenever ({y}_{1},\dots ,{y}_{k})\in W.
The equilibrium μ is said to be an attractor in a neighborhood V of μ in \overline{\mathbb{R}}, if {y}_{n}\in \overline{\mathbb{R}} for all n and {y}_{n}\to \mu in \overline{\mathbb{R}}, whenever {y}_{n}\in V for n\le k.
Definition 1 The point μ is said to be an unconditional equilibrium of h (respectively, unconditional stable equilibrium, unconditional attractor in V) if it is an equilibrium (respectively, stable equilibrium, attractor in V) of h\circ \lambda for all \lambda \in {\mathrm{\Lambda}}^{k}.
Definition 2 We define the equilibria, stable equilibria, attractors, unconditional equilibria, unconditional stable equilibria and unconditional attractors of a continuous function \phi :U\subset {\overline{\mathbb{R}}}^{k}\to \overline{\mathbb{R}} to be those of \stackrel{\u02c6}{\phi}.
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
Then
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\}.
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.
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).
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. □