The study of continuous-time difference equations pave the way for the discovery of new kinds of complexity. These equations appear as models of systems whose future depends not only on the current state but also on part of the past history. Even the simplest nonlinear continuous-time difference equation

gives a wonderful example of how very complex behaviors can be described with the help of very simple models (for example, see [ 13,14] and references therein). Over last decades, the focus of our research has been on this equation. As a result, the qualitative theory of equation (1) was completed and published not so long ago in [ 8] (the key ideas were put forward in [ 7,11]).

Each solution of (1) is determined by its values on $[0,1)$ and can be written as

where $f^n(z)=f(f^{n-1}(z)),f^0(z)=z$. Typical solutions of (1), even solutions of any smoothness index, behave in bizarre ways, for example as inFigure 1.

To describe the long-term properties of the solutions and explain how and why they occur, it is necessary to examine both the asymptotic properties of the iterative sequence ${\{f^n(z)\}}_n$ and the internal structure of the maps $f^n(z)$ by themselves for large $n$. As a rule, the structure of $f^n(z)$ becomes more and more intricate with the increase of $n$. The complexities in the behaviour of trajectories of the one-dimensional dynamical system $z\mapsto f(z)$ are transformed into a very complicated (even chaotic) structure of $f^n(z)$ as $n\to \infty$. But, at the same time, the sequence ${\{f^n(z)\}}_n$ is in a certain way (explained below) asymptotically periodic for almost all $f$.

With the aim to present in a uniform manner the findings in this direction, we considered the maps $f^0\left(z\right),f\left(z\right),f^2\left(z\right),\dots \text{, }{f}^n\left(z\right)\dots$ as the semigroup $\langle f\rangle$ generated by the map $f:I\to I$ in the space $C(I,I)$ of continuous maps of $I$ into itself (with $I$ being a closed bounded interval of the real line). If $\langle f\rangle$ is finite, then the sequence ${\{f^n(z)\}}_n$ is periodic and hence the structure of $f^n(z)$ is simple and repetitive as $n$ increases. To study the general case where $\langle f\rangle$ is infinite, we developed the notion of a limit semigroup for $\langle f\rangle$ in a wider map space. We succeeded in constructing a limit semigroup in the space of upper semicontinuous maps (naturally, this space is the first thing that comes to mind because semicontinuous maps are pointwise limits of sequences of continuous maps). All this enabled us to achieve our aim as well as to discover some new insights.

The results just discussed was published only in Russian and, moreover, in their entirety only in [ 8]. Therefore, we thought it appropriate to cover them all in this article (with some unimportant simplifications and omissions for brevity).

In order to indicate conditions for $\langle f\rangle$ to be infinite, we introduce the set of maps

Clearly, the set $\Im$ consists of continuous maps $g:I\to I$ such that for each one there exists an interval $\left[z_1,z_2\right]\subseteq I,$ $z_1\le z_2,$ with the properties:

If $g\in \Im ,$ then $g^n\in \Im ,$ $n=2,3,\dots$. If in addition $g$ is smooth, then either $g(z)=z$ or $g(z)\equiv \text{const}$.

Theorem 1 The semigroup $\langle f\rangle$ is finite if and only if

Proof. The `if' part is obvious since $f^{2m}=f^m.$ Consider the `only if' part. If $\langle f\rangle$ is finite, then one can find $k,l\ge 1$ such that $f^{k+l}=f^k.$ This gives$f^{ik+jl}=f^{ik}$ for all $i,j\ge 1.$ Putting $i=l,j=k,$ we get$f^{2kl}=f^{kl},$ and consequently,$f^{kl}\in \Im .$

The dynamics of maps generating finite semigroups is very simple (seeFigure 2). Let $\text{Fix}(f)$, $\text{Per}(f)$, and $\mathcal{D}(f)$ be the sets of fixed, periodic, and unstable points of $f$, respectively.

Theorem 2 If $\langle f\rangle$ is finite, then $\mathcal{D}(f)$ is empty, $\text{Per}(f)$ is connected and hence $\text{Per}(f)=\text{Fix}(f^2)$.

Proof. As $f$ bing continuous, whatever $z\in I$, $\epsilon >0$, and $n>0$, there exists ${\delta }_n={\delta }_n(z,\epsilon )>0$ such that $|f^i(z)-f^i(z^{\prime })|<\epsilon ,i=1,\dots ,n$, if $|z-z^{\prime }|<{\delta }_n$. As $\langle f\rangle$ is finite, $f^{2m}=f^m$ for a certain $m>0$. Therefore $|f^i(z)-f^i(z^{\prime })|<\epsilon$ for all $i=1,2,\dots$, if $|z-z^{\prime }|<{\delta }_{2m}$, i.e., all points $z\in I$ are stable under $f$ and $\mathcal{D}(f)=\varnothing$.

If $\mathcal{D}(f)=\varnothing$, then the set $\text{Fix}(f)$ is connected: otherwise one can find $z^{\prime },z^{″}\in \text{Fix}(f)$ such that $(z^{\prime },z^{″})\cap \text{Fix}(f)=\varnothing$; therefore the function $f(z)-z$ maintains its sign in $(z^{\prime },z^{″})$ and hence one of the points $z^{\prime },z^{″}$ is unstable under $f$, which gives a contradiction. Since $\mathcal{D}(f)=\varnothing$ implies $\mathcal{D}(f^n)=\varnothing$, all the sets $\text{Fix}(f^n),n=1,2,\dots ,$ are connected. Hence, if $z_1<z_2<\dots <z_n$ form a cycle of $f$, then $[z_1,z_n]\subset \text{Fix}(f^n),$ i.e. $f^n(z)=z$ for $z\in [z_1,z_n].$ As is known [ 1], this equality cannot hold for $n\ge 3.$ Consequently, $\text{Per}(f)$ coincides with $\text{Fix}(f^2)$ and is therefore connected.

From Theorem 2 it follows that the set of continuous maps generating finite semigroups is exhausted by the maps $f$ with the property: for $f$ there exists its own interval $\left[z_1,z_2\right]\subseteq I,$ $z_1\le z_2,$such that

Two simplest maps that satisfy (3) are shown inFigure 2. In the case where the semigroup$\langle f\rangle$ is finite, its structure is easily derived from (3) by elementary calculations.

Theorem 3 Let $k=\min \left\{m>0:f^m\in \Im \right\}.$ Then $\langle f\rangle$ is finite semigroup that has:

type $(k,1),$ if $\text{Fix}(f^2)=\text{Fix}(f),$

type $(k,2),$ if $\text{Fix}(f^2)\cancel{=}\text{Fix}(f)$ and $f^{k-1}(I)\cancel{=}\text{Fix}(f^2),$

type $(k-1,2),$ if $\text{Fix}(f^2)\cancel{=}\text{Fix}(f)$ and $f^{k-1}(I)=\text{Fix}(f^2),$

and $\langle f\rangle$ is a group if and only if $f^2$ is the identity map.

The map ofFigure 2(a) generates a semigroup of type (1,1), and the map ofFigure 2(b) generates a semigroup of type (1,2). Theorem 3 defines completely the finite semigroup $\langle f\rangle$. For example, if $\langle f\rangle$ has type $(k,2),$ then $\langle f\rangle$ consists of the maps$f^0,$ $f,\dots ,$ $f^{k+1},$ $f^{k+2}=f^k,$ $f^{k+3}=f^{k+1},$ $f^{k+4}=f^k,$ $\dots$

The above section shows that the semigroup $\langle f\rangle$ is typically infinite. In this case, we will try to describe its asymptotic properties using another, simpler semigroup.

Definition 1 Let the space $C(I,I)$ be embedded in a wider space $H_{\nu }(I,I)$ of maps $h:I\to I$, endowed with the metric $\nu .$ If there exists a periodic or almost periodic semigroup $\langle h\rangle$, $h\in H_{\nu }$, such that

then we say that $\langle h\rangle$ is the limit semigroup for $\langle f\rangle$ in $H_{\nu }.$

Definition 2 The semigroup $\langle h\rangle$ is called periodic if there exists a positive integer $N$ such that $h^{n+N}=h^n,n=0,1,\dots ,$ with $N$ referring to as the period of $\langle h\rangle$;

The semigroup $\langle h\rangle$ is called almost periodic if there are positive integers $N_1,N_2,\dots$ such that for any $\epsilon >0$ one can find $N_k$ satisfying $v\left(h^{n+j{N}_k},h^n\right)<\epsilon ,n,j=0,1,\dots ,$ with $N_1,N_2,N_3,\dots$ referring to as the almost-periods of $\langle h\rangle$.

Given $H_{\nu }$, the semigroup $\langle f\rangle$ have just one limit semigroup or none at all. Taking for $H_{\nu }$ this or that map space and using the corresponding limit semigroup (if exists), one can describe the asymptotic properties of $\langle f\rangle$ in less or more detail.

In this connection, the question arises whether $\langle f\rangle$ can have a limit semigroup in ``its own'' space $C(I,I)$. From the above section it follows that every finite $\langle f\rangle$ has a limit semigroup in $C(I,I)$ and this is its largest subgroup. Indeed, let $\langle f\rangle$ be finite. Then the map

generates the semigroup $\langle h_{\star }\rangle ,h\in C(I,I)$, which consists of the one element $f^{\star }$ if $\text{Per}\left(f\right)=\text{Fix}\left(f\right),$ and of the two elements $f^{\star }$ and $f\circ f^{\star }=f^{k+1}$ otherwise. Hence $\langle h_{\star }\rangle$ is a periodic group with period 1 or 2; moreover, $\langle h_{\star }\rangle$ is the largest subgroup of $\langle f\rangle$. By virtue of Theorem 3, $\Vert f^n-h_{\star }^n\Vert \to 0$ as $n\to \infty .$ Consequently: If $\langle f\rangle$ is finite, then its largest subgroup $\langle f\circ f^{\star }\rangle$ is the limit semigroup for $\langle f\rangle$ in $C(I,I).$

There is a representation for $f^{\star }$, in which $k$ does not appear explicitly. Let ${\omega }_f(z)$ be for the $\omega$-limit set of the trajectory of the point $z$ under $f$. The map $f^{\star }$ can be written as

The map $f^{\star }$ in the form (6) can also exist for certain continuous maps that generate infinite semigroup. These are maps $f$ with the property: for $f$ there is its own interval $\left[z_1,z_2\right]\subseteq I,$ $z_1\le z_2,$ such that

An illustration is inFigure 3: although $\langle f_1\rangle$ is finite and $\langle f_2\rangle$ is infinite, (6) yields $f_1^{\star }=f_2^{\star }$.

If (8) is not met, then $\langle f\rangle$ no longer has a limit semigroup in $C(I,I)$ and we need a wider map space. We choose the space $SC(I,2^I)$ of upper semicontinuous maps $h:I\to 2^I,$ endowed with the metric

where $\text{gr}h$ is for the graph of $h$ and ${\text{dist}}_H(\cdot ,\cdot )$ is for the Hausdorff distance between two sets. The space $SC(I,2^I)$ is compact in the topology induced by ${\rho }_{{\text{}}_{\text{}\text{}H}}$, and $C(I,I)$ is embedded in $SC(I,2^I)$.

Let $\text{Lim}$ denote the operation of passing to the limit in $SC(I,2^I)$. The meaning of the convergence in $SC(I,2^I)$ becomes clear from the equivalence of the relations

where $\text{Lt}$ is for the operation of passing to the topological limit. Thus, the convergence of maps in $SC(I,2^I)$ reduced to the existence of the topological limit of their graphs, which are treated as sets in the plane. It is therefore important to understand how the convergence of map graphs is related to the local behaviour of these maps. The answer is given by the following lemma.

Lemma 1 Let ${\{g_i(w)\}}_i$ be a sequence of maps from $C\left(J^{\prime },J^{″}\right)$ with $J^{\prime },J^{″}$ being closed bounded intervals, and let $V_{\epsilon }(w)$ stand for the $\epsilon$-neighborhood of $w\in J^{\prime }.$ The limit ${\text{Lt}}_{i\to \infty }\text{gr}{g}_i$ exists if and only if the limit $L{t}_{i\to \infty }{g}_i\left(V_{\epsilon }\left(w\right)\right)$ exists for any $w\in J^{\prime }$ and any $\epsilon >0$ less than a certain ${\epsilon }_w>0.$ Moreover, if so, then

Proof. The 'if' part is proved by verifying that ${\text{Ls}}_{i\to \infty }\text{gr}{g}_i\subseteq G\subseteq {\text{Li}}_{i\to \infty }\text{gr}{g}_i$ with $\text{Ls}$ and $\text{Li}$ being for the upper and lower topological limits, respectively. Take, say, the latter inclusion. If $(w,z)\in G,$ then $z\in L{t}_{i\to \infty }{g}_i\left(V_{\epsilon }\left(w\right)\right)$, $\epsilon <{\epsilon }_w$, i.e., for any $\epsilon <{\epsilon }_w$ there is a sequence of points $z_{i,\epsilon }\in J^{″}$ such that $z_{i,\epsilon }\in g_i(V_{\epsilon }(w))$ and $z_{i,\epsilon }\to z$ as $i\to \infty$. Let us consider the sequence ${\left\{z_{i,1/i}\right\}}_i,i>1/{\epsilon }_w.$ Since $z_{i,1/i}\in {\phi }_i(V_{1/i}(w))$, one can find $w_i\in V_{1/i}(w)$ such that $g(w_i)=z_{i,1/i}$ and $w_i\to w$ as $i\to \infty .$ Therefore $(w_i,g(w_i))\to (w,z)$ as $i\to \infty$, and hence $\left(w,z\right)\in {\text{Li}}_{i\to \infty }\text{gr}{g}_i$ as claimed.

The 'only if' part follows from the definition of topological limit. If ${\text{Lt}}_{i\to \infty }\text{gr}{g}_i$ exists, then ${\text{Lt}}_{i\to \infty }{g}_i\left(V_{\epsilon }\left(w\right)\right)$ also exists for every given $w\in J^{\prime }$ and sufficiently small $\epsilon >0$. Hence, the set $G_w$ is well defined and, by what was just shown, (10) holds.

The composition $h^{\prime }\circ h^{″}$ of $h^{\prime },h^{″}\in SC(I,2^I)$ is understood as the map $h^{\prime }\circ h^{″}={\displaystyle \underset{u\in h^{″}(z)}{\cup }{h}^{\prime }}(u),$ which obviously belongs to $SC(I,2^I).$ Let $h\in C(I,I)$ and $h_i\in SC(I,2^I),i=1,2,\dots$ From the topological sense of the convergence in $SC(I,2^I)$, it follows that, if ${\text{Lim}}_{i\to \infty }{h}_i$ exists, then $Li{m}_{i\to \infty }h\circ h_i,$ and $Li{m}_{i\to \infty }h\circ h$ also exist, moreover,

The equality $Li{m}_{i\to \infty }{h}_i\circ h=h_{\ast }\circ h$ is usually not satisfied.

Now the goal is to construct a special map $f^{\Delta }\in SC(I,2^I)$ so as $f\circ f^{\Delta }$ to generates a limit semigroup for $\langle f\rangle$ (of course, $f^{\Delta }$ must be identical to $f^{\star }$, if $\langle f\rangle$ is finite). We call $f^{\Delta }$ the resolvent map for $\langle f\rangle$ and accept that

So far, this is only a formal notation for $f^{\Delta }$, since the existence of the limit in (12) is not obvious. If $f^{\Delta }$ exists, then it belongs to the space $SC(I,\widehat{2^I})$, where $\widehat{2^I}$ is the subset of $2^I$, consisting of closed intervals (including degenerated ones).$$

Let $\text{Per}(f)=\text{Fix}(f)$. Then $L{t}_{i\to \infty }{f}^i\left(V\right)$ exists for any connected $V\subset \widehat{I}$ (see [ 6]) and, by Lemma 1, the sequences ${\left\{f^i\right\}}_i$ and ${\left\{f^{k!}\right\}}_k$ converge in $SC(\widehat{I},\widehat{I})$ to the same limit. Therefore $f^{\Delta }={\text{Lim}}_{i\to \infty }{f}^i$; an example is inFigure 4. If $\text{Per}(f)=\text{Fix}(f^s),s>0,$ similar arguments lead to the formula $f^{\Delta }={\text{Lim}}_{i\to \infty }{f}^{is}$ and $f^{\Delta }=f^{\star }$ for finite $\langle f\rangle$.

Where $\text{Per}(f)\cancel{=}\text{Fix}(f^s)$ for any $s>0,$ the resolvent map $f^{\Delta }$ may not exist. The simplest example occurs when some interval $[z^{\prime },z^{″}]\subset I$ collapses to a point, say, $z^{*}$ under $f$. Then $L{t}_{k\to \infty }\text{gr}{f}^{k!}|\left[z^{\prime },z^{″}\right]=\underset{k\to \infty }{\text{lim}}{f}^{k!}\left(z^{\ast }\right)$ when the limit on the right-hand side exists. If the trajectory of $z^{*}$ is neither asymptotically periodic nor asymptotically almost-periodic, then the sequence ${\left\{f^{k!}\right\}}_k$ diverges and $f^{\Delta }$ cannot exist. An example is delivered by the maps drawn inFigure 5. For each of $f_1$ and $f_2$, almost all points of [a,b] are not asymptotically periodic or asymptotically almost-periodic. Thereby, $f_1^{\Delta }$ is non-existent for almost all z* (because $\left[z^{\prime },z^{″}\right]$ collapses to $z^{*}$ under $f_1$) and $f_2^{\Delta }$ exists for all $z^{*}$ regardless of whether $z$ has the periodicity property or not, namely, $f_2^{\Delta }(z)=[a,b]$. Trajectories of almost all points of $[a,b]$ are dense in $[a,b]$ under both $f_1$ and $f_2$. This is what ensures the existence of $f_2^{\Delta }$, but for $f_1^{\Delta }$ this argument does not ``work'' whenever $z^{*}$ does not have periodicity properties under $f_1$.

Thus, in order for $f^{\Delta }$ to exists, $f$ must be subject to further restrictions. Before stating the theorem on the resolvent map, we need few more notions.

Definition 3 The set

is called the domain of influence of the point $z$ under the map $f$.

The set $Q_f(z)$ is nothing but the upper topological limit of the sets $f^i(V_{\delta }(z)),i=0,1,\dots ,$ as $i\to \infty$ and $\delta \to 0$. Consequently, $Q_f(z)$ shows how far from the trajectory of the point $z$ the trajectories of its nearby points go. For example, inFigure 4, the fixed point $z={\gamma }_1$ is attracting and the fixed point $z={\gamma }_2$ is repelling. Therefore, $Q_f({\gamma }_1)$ is the one-point set $\left\{{\gamma }_1\right\}$ (moreover, $Q_f(z)=\left\{{\gamma }_1\right\}$ for all $z$ that are attracted to $z={\gamma }_1$), and $Q_f({\gamma }_2)$ is the interval $[{\gamma }_1,b]$.

Let ${\omega }_f(z)$ stand for the $\omega$-limit set of the trajectory of $z$ under $f$. Clearly, $Q_f(z)$ contains ${\omega }_f(z)$. More precisely: if $z\cancel{\in }\text{D}(f)$, then $Q_f(z)={\omega }_f(z)$ and $Q_f(z)$ has no interior points, but if $z\in \text{D}(f)$, then at least one of its half-neighborhoods ``expands'' under $f$, therefore the interior of $Q_f(z)$ is nonempty and $Q_f(z)\supseteq {\omega }_f(z)$.

The notion of the domain of influence of a point produces the notion of the domain of influence of a set:

which, of course, is equivalent to the formula $Q_f(A)={\cup }_{z\in \overline{A}}{Q}_f(z),A\subset I$.

Definition 4 A non-degenerated interval is called a $\xi$-interval of $f$, if the trajectories of all its points have the same $\omega$-limit set that is different from a cycle or the closure of an almost periodic trajectory.

The simplest variant of $\xi$-intervals is an interval that collapses (under $f$) into a point whose trajectory is neither asymptotically periodic nor asymptotically almost periodic, as inFigure 5(b).

Definition 5 An ordered set of intervals $J_0,J_1,...,j_{N-1}$ is said to be a cycle of intervals of the map $f$ with period $N$ if these intervals are cyclically permutated by $f$ and have mutually no common interior points.

Having introduced these definitions, we can pass to the resolvent map.

Theorem 4 If the map $f$ has no $\xi$-intervals, then the resolvent map $f^{\Delta }$ exists and it can be written in the form

Proof. The full proof is too long to include here, but the idea is that the theorem will be confirmed (in view of Lemma 1) if we will show that for any $z\in I$ there is ${\epsilon }_z>0$ such that $L{t}_{k\to \infty }{f}^{k!}\left(V_{\epsilon }\left(z\right)\right)$ exists for $0<\epsilon <{\epsilon }_z$ and

We consider only the simplest case where ${\omega }_f(z)$ is an attracting cycle. If so, then: $Q_{f^i}(z)={\omega }_{f^i}(z)$ for $i=1,2,\dots$; ${\text{Lt}}_{j\to \infty }{f}^{jN}\left(z\right)={\omega }_{f^N}\left(z\right)$ with $N$ being the period of the attracting cycle, and there exists ${\epsilon }_z>0$ such that ${\omega }_{f^N}(z^{\prime })={\omega }_{f^N}(z)$ for $z^{\prime }\in V_{\epsilon }(z)$ if $\epsilon <{\epsilon }_z$. Consequently,

As ${\{f^{k!}(z)\}}_{k\ge N}$ is a subsequence of ${\{f^{jN}(z)\}}_j$, we obtain (15) from the equalities

In case ${\omega }_f(z)$ is the closure of an almost-periodic trajectory, the proof is similar but more technically involved, it is presented in [ 11]. A proof that covers all cases can be found in [ 8], so far only in Russian.

Theorem 4 will hold if

${\omega }_f(z)$ is a cycle or the closure of an almost-periodic trajectory for all $z\in I$ . It is known [ 10,12] that (i) certainly holds if $\text{Per}(f)$ is closed (then ${\omega }_f(z)$ is a cycle whatever $z\in I$) and fails if $f$ has a cycle with a non-power-of-two period. Furthermore, (i) is no longer guaranteed if $f$ has cycles of all periods $1,2,2^2,\dots$ but no others.

Here is another sufficient condition for Theorem 4

$Q_f(z)$ is a cycle or a cycle of intervals for all $z\in I$.

If (ii) is modified as follows

(iii) $Q_f(z)$ is a cycle or a cycle of intervals for all $z\in I$ and their periods are uniformly bounded,

then $f^{\Delta }$ takes the very simple form

where $p<\infty$ is the least common multiple of the periods of all $Q_f(z),z\in I.$

So, the resolvent map $f^{\Delta }$ may or may not exist, and when it does, it is usually arranged complexly. The main properties of $f^{\Delta }$ are listed below.

The map $f^{\Delta }(z),$ thought of as a function form $I$ into $2^I,$ is upper semicontinuous. Its value is a non-degenerated closed interval for $z\in \text{D}(f)$ and a one-point set (singleton) for $z\in I\setminus \text{D}(f).$

The map $f^{\Delta }(z),$ thought of as a function form $I$ into $I$, is single-valued and continuous over $I\setminus \text{D}(f)$, and is multi-valued over $\text{D}(f).$

Definition 6 The set of values that $f^{\Delta }(z)$ takes on for $z\in \text{D}(f)$ is called its spectrum of jumps and denoted by $\text{J}(f^{\Delta })$.

Two simple illustrations are given by the maps ofFigure 4 andFigure 5(b): here $\text{J}(f^{\Delta })$ consists of the interval $[{\gamma }_1,b]$ and $\text{J}(f_2^{\Delta })$ consists of the intervals $\left[a,b\right]$ and $\left[c,b\right]$.

The spectrum of jumps $\text{J}(f^{\Delta })$ is composed from the connected component of domains of influence of the unstable points of $f$. Any two elements of $\text{J}(f^{\Delta })$ either do not have common interior points, or one of them contains the other.

In typical situations, $\text{J}(f^{\Delta })$ as well as the set of values of $f^{\Delta }(z)$ on $I\setminus \text{D}(f)$ are finite (in particular, this is the case for structurally stable $f$).

In many cases, the graph of $f^{\Delta }$ is locally self-similar and, furthermore, fractal: $\text{gr}{f}^{\Delta }$ is a plane curve whose fractal dimension is greater than 1.

The first four properties are easy to obtain from Theorem 4, and the last one needs some comments. Let ${\text{dim}}_{\text{box}}$ denote the box-counting dimension (one of the commonly used versions of fractal dimension). It is clear that

usually holds (at least when $\text{J}(f^{\Delta })$ is finite). The set $\text{D}(f)$ is known to be often fractal. Therefore, if $di{m}_{\text{box}}D\left(f\right)\ne 0,$ then the fractal dimension of $\text{gr}{f}^{\Delta }$ is greater than 1 and, moreover, it can even be equal to 2 (the latter is always the case where $\text{D}(f)$ contains an interval).

More details and arguments about $f^{\Delta }$ can be found in [ 8], but here the properties of $f^{\Delta }$ are illustrated inFigure 6, which displays the resolvent map for the logistics parabola

The dynamical properties of (17) are well known, so we use them without explanations.

If $\lambda =3.5$, then $\eta$ has an attracting period-4 cycle, which attracts all points of [ 1] except for the set $\text{D}(\eta )$, which consists of the repelling period-1 and period-2 periodic points of $\eta$ and their inverse images, and is therefore countable and nowhere dense. The resolvent map ${\eta }^{\Delta }(z)$, thought of as a function $[0,1]\to [0,1]$, is piecewise constant over $[0,1]\setminus \text{D}(\eta )$ and takes on four values that correspond to the points of the attracting cycle, but over $\text{D}(\eta )$, it becomes interval-valued with the spectrum of jumps consisting of three intervals (seeFigure 6(a)).

Where $\lambda =3.83$ and $\eta$ has an attractive period-3 cycle and repelling cycles of any period, the properties of ${\eta }^{\Delta }(z)$ are nearly the same as in the previous case (seeFigure 6(c)). The only difference is that $\text{D}(\eta )$, over which ${\eta }^{\Delta }(z)$ is interval-valued, is already uncountable but again nowhere dense (now $\text{D}(\eta )$ is a Cantor-like set, since it is the closure of the repelling periodic points of $\eta$ and their inverse images). In addition, $\text{D}(\eta )$ has positive fractal dimension and, hence, the graph of ${\eta }^{\Delta }(z)$ is a fractal.

Now a very different situation where $\lambda ={\lambda }_{\infty }(\approx 3.57)$ with ${\lambda }_{\infty }$ being the parameter value at which $\eta$ has only cycles with periods $1,2,2^2,\dots$ (seeFigure 6(b)). These cycles are repelling and their inverse images are dense on $\left[0,1\right]$. Hence, $\text{D}(\eta )$ is a countable dense set and all points of $[0,1]\setminus \text{D}(\eta )$ are attracted to the Cantor-like set $\mathcal{K}$ consisting of the condensation points of $\text{Per}(\eta )$. It follows that ${\eta }^{\Delta }(z)$ is singleton-valued over $[0,1]\setminus \text{D}(\eta )$ and its values run through the continual set $\mathcal{K}$; whereas ${\eta }^{\Delta }(z)$ is interval-valued over $\text{D}(\eta )$ and its spectrum of jumps is countable (the infimum of the lengths of intervals from $\text{J}({\eta }^{\Delta })$ equals $0$).

Graphics shown inFigure 6 have local self-similarity in the vicinity of each ``largest'' vertical segment in case (a), and each vertical segment in cases (b) and (c). Moreover, in the last case, there are an uncountable number of vertical segments, which leads to the fractality of the graph.

The existence of $f^{\Delta }$ given by (12) does not in itself guarantee that $f\circ f^{\Delta }$ generates the limit semigroup for $\langle f\rangle$. The last is automatically true if

(the former of these equalities entails the latter, but not the other way around).

The condition of Theorem 4, generally speaking, do not imply (18). A confirming example comes fromFigure 7. For $f$ depicted there, $f^{\Delta }$ exists, but $f$ and $f^{\Delta }$ do not commute: $f^{\Delta }(a)=[c,\gamma ]$, $f(f^{\Delta }(a))=[c,\gamma ]$, $f^{\Delta }(f(a))=f^{\Delta }(\gamma )=[c,{\gamma }_2]$, and hence $f^{\Delta }\circ f=f\circ f^{\Delta }$. Besides, $f^{\Delta }\circ f^{\Delta }\cancel{=}{f}^{\Delta }$. Here, either of the equalities in (18) fails because $f$ has an extremum at the unstable point $z_{°}$.

There are maps for which only the former equality in (18) breaks down. One example are

(the drawing is left to the reader). Here, $g$ has the semi-attracting fixed point $z=1$, which leads to that the domain of influence of a point $z\in [1,2]$ does not coincide with the domain of influence of its $\omega$-limit set, namely: $Q_g(z)={\omega }_g(z)=\left\{1\right\}$ and $Q_g({\omega }_g(z))=[0,1]$. Therefore, $g^{\Delta }(z)=1$ and $g^{\Delta }(g^{\Delta }(z))=[0,1]$ for $z\in [1,2].$ In order for the last equality in (18) to be also violated, we ``improve'' $g$ by replacing it with

which has much the same dynamical property as $g$. More precisely: $Q_{\overline{g}}(z)={\omega }_{\overline{g}}(z)=\left\{1\right\}$ and $Q_{\overline{g}}({\omega }_{\overline{g}}(z))=[0,1]$ for $z\in (1,2].$ As a result, ${\overline{g}}^{\Delta }=g^{\Delta }$, but ${\overline{g}}^{\Delta }\circ \overline{g}\cancel{=}\overline{g}\circ {\overline{g}}^{\Delta },$ whereas $g^{\Delta }\circ g=g\circ g^{\Delta }$. Indeed, ${\overline{g}}^{\Delta }(\overline{g}(z))=[0,1]$ and $\overline{g}\left({\overline{g}}^{\Delta }\left(z\right)\right)=\left\{1\right\}$ for $z\in [1,2]$.

There are smooth maps for which (18) is not satisfied, for example, $z\mapsto 4z^2(1-z).$

As the above examples demonstrate, violating (18) occurs when the set

is non-empty. It is not difficult to realize that this condition is necessary and sufficient.

Theorem 5 Let $f$ satisfy the condition of Theorem 4. The equalities (18) take place if and only if ${\mathcal{Q}}_{\cancel{=}}(f)=\varnothing .$

Proof. The 'only if' part is evident. The 'if' part follows from $f^n(Q_{f^m}(z))=Q_{f^m}(f^n(z))$ and $Q_{f^n}(Q_{f^m}(z))=Q_{f^{r(n,m)}}(z)$ with $r=\left(n,m\right)$ being the greatest common divisor of $n$ and $m$, because these formulas are valid only provided that ${\mathcal{Q}}_{\cancel{=}}(f)=\varnothing .$

The condition ${\mathcal{Q}}_{\cancel{=}}(f)=\varnothing$ does not seem very good for practical use, but it is much easier to check it where $f$ satisfies (iii) and, hence, $f^{\Delta }$ is given by (16) with some $p<\infty$. It suffices to analyse just $f^{2p}$ and not all the iterations of $f$.

Based on Theorem (5), from now on we will assume that

(v)the map $f$ has no $\xi$-intervals,

(vv)${\mathcal{Q}}_{\cancel{=}}(f)=\varnothing$,

and refer to (v),(vv) as the limit semigroup conditions or LSG-conditions for short.

These conditions cover most of continuous intervals maps, in particular, structurally stable maps. For instance, the logistic parabola (17) is structurally stable (and, hence, satisfies LSG-conditions) on a parameter set containing an open subset; this subset consists of those $\lambda$ for which there is an attracting cycle with multiplier less than 1 in absolute value. Of course, LSG-conditions can be met by maps that are not structurally stable. An example is again given by the logistic parabola. If $\lambda =3$, then $\eta$ has the attracting fixed point $z=2/3$ whose multiplier is equal to $-1$. In this case, $p=1$ and $Q_{{\eta }^2}\left(Z\right)=Q_{{\eta }^2}\left({\omega }_{{\eta }^2}\left(Z\right)\right)=\{\begin{array}{l}\left\{2/3\right\},z\in \left(0,1\right),\\ \left[0,3/4\right],z=0,z=1,\end{array}$(seeFigure 8), and thus the LSG-conditions are met.

Once LSG-conditions are valid, $f^{\Delta }$ exists and $f\circ f^{\Delta }$ generates in $SC(I,2^I)$ the semigroup $\langle f\circ f^{\Delta }\rangle$ that consists of just the maps~$h^n=f^n\circ f^{\Delta },$ $n=0,1,2,\dots$ Then from (11) and (18) it follows that

with the limit being uniform in $n$.

Theorem 6 Let LSG-conditions be met.

The semigroup $\langle f\circ f^{\Delta }\rangle$ is periodic or almost periodic.

The semigroup $\langle f\circ f^{\Delta }\rangle$ is the limit semigroup for $\langle f\rangle$ in the space $SC(I,2^I)$.

Proof. We consider the simple case where $f$ satisfies (iii). Then $f^{\Delta }$ is written in the form (16) with some $p<\infty$. Therefore $f^{2p}\circ f^{\Delta }=f^{\Delta }$ and $f^{n+2p}\circ f^{\Delta }=f^n\circ f^{\Delta },n=0,1,\dots$. Hence $\langle f\circ f^{\Delta }\rangle$ is periodic, as claimed in item 1. Item 2 can be considered true if we succeed in proving that

It is clear that $f^n\circ f^{\Delta }=f^s\circ f^{\Delta }$ with some $0\le s\le 2p-1$, $s=s(n)$. Since $f^s\circ f^{\Delta }={\text{Lim}}_{k\to \infty }{f}^{s+2pk},$ for any $\epsilon >0$ there exists $K(\epsilon ,s)>0$ such that ${\rho }_H\left(f^{s+2pk},f^s\circ f^{\Delta }\right)\le \epsilon$ if $k>K(\epsilon ,s)$. Hence,

which gives (20), and item 2 follows. If (iii) is not satisfied, the proof becomes more complicated and needs some additional background. It can be found in [ 8].

If $\langle f\rangle$ is finite (in which case $f$ satisfies (iii) with $p=1$ or $p=2$), then its limit semigroup $\langle f\circ f^{\Delta }\rangle$ is periodic with period 1 or 2. At the same time, $\langle f\rangle$ is typically not periodic. It is periodic only if $f^2$ is the identity map, i.e., if either $f(z)=z,$ or $f(z)=-z+2c_{°}$ with $c_{°}$ being the midpoint of $I$. Indeed, in both the cases, $f^{-1}(z)=f(z)$ and therefore $\langle f\rangle$ is actually periodic with period 1 (in the former case) or 2 (in the latter one); moreover, $f^{\Delta }=f^{\star }$ and $f^{\star }$ is the identity map, i.e., $\langle f\circ f^{\Delta }\rangle$ coincides with $\langle f\rangle$, which means that $\langle f\rangle$ is the limit semigroup for itself in $C(I,I)$ as well as in $SC(I,2^I)$.

There is another approach to the study of transformation groups, which based on the notion of Ellis enveloping semigroup [ 2,3,4]. For $\langle f\rangle$, the enveloping semigroup $E(f)$ is defined as the closure of the set $\{f^n:I\to I,n=1,2,\mathrm{...}\}$ in the space $I^I$ of maps of $I$ into $I$, equipped with the pointwise convergence topology. The asymptotic properties of $\langle f\rangle$ are described by the following subset of $E(f)$:

Where $E^{\prime }\left(f\right)$ is a finite semigroup, it can be regarded in a certain sense as an analogue of the limit semigroup for $\langle f\rangle$ in $C(I,I)$. Indeed, $E^{\prime }\left(f\right)$ is finite if and only if

Then $E^{\prime }\left(f\right)$ is a periodic group of period $2^m$, namely,

This statement is a direct consequence of the following well-known fact: the trajectories of all $z\in I$ are asymptotically periodic under $f$ with their asymptotic periods being uniformly bounded, if and only if (21) holds (see [ 9]). In this case, it is easy to verify that $f_{\star }(z)=f^{\star }(z)$ and $f^{\Delta }(z)=f^{\star }(z)$ for $z\cancel{\in }\text{D}(f)$. Hence, $f_{\star }$ is identical to the resolvent map over $I\setminus \text{D}(f)$. In addition, $f_{\star }$ is continuous just if $\text{D}(f)=\varnothing$ (or, equivalently, $\text{Per}(f)$ is connected) and then $f_{\star }=f^{\star }$ over $I$. Otherwise, $f_{\star }(z)$ is a first Baire class function. Thus, the limit semigroup approach, compared with the enveloping semigroup approach, can give more information about the asymptotic properties of $\langle f\rangle$ and allows one to study a wider class of maps $f$.

LSG-conditions (v),(vv) are always satisfied by structurally stable maps, and then the limit semigroup $\langle f\circ f^{\Delta }\rangle$ exists and consists of the maps

When $f$ is not structurally stable, LSG-conditions may not be met. If this is due to the failure of only (vv), then $f^{\Delta }$ exists, but its associated maps (23) can no longer form a semigroup. Nevertheless, it may happen that the maps (23) constitute a periodic or almost periodic sequence and still retain the principal property of limit semigroups

If so, the maps (23) continue to describe the asymptotic properties of $\langle f\rangle$. In particular, such a situation occurs where $f$ satisfies (iii). Then (v) holds and the sequence (23) is periodic. Consequently, (24) holds true regardless of whether (vv) holds or not.

As an example, we again take the logistic parabola $\eta :z\mapsto {\lambda }_{*}z(1-z)$, where the parameter value ${\lambda }_{\ast }\left(\approx 3.678\right)$ is chosen so that the critical point $z=c=1/2$ is mapped onto the repelling fixed point $z=\gamma$ in three iterations, i.e., ${\eta }^3(c)=\gamma$ (seeFigure 9(b)). In this case, the domain of influence of each $z\in [0,1]$ is an interval. Namely, setting $c_1=\eta (c)$ and $c_2={\eta }^2(c),$ we get that $Q_{\eta }(0)=Q_{\eta }(1)=[0,c_1]$ and $Q_{\eta }\left(Z\right)=\left[c_z,c_1\right]$ for $z\cancel{=}0,1$, where $\left[c_2,c_1\right]$ absorbs all $z\in (0,1)$, each after finitely many iterations, and $\left[c_2,c_1\right]$ can be split into two adjoint intervals: $J_1=\left[\gamma ,c_1\right]$ and $J_2=\left[c_2,\gamma \right]$ that passe into each other under $\eta$ (seeFigure 9(a)). It follows that $\eta$, which is structurally unstable, satisfies (iii), and hence ${\eta }^{\Delta }$ exists (seeFigure 9(b)) and has the form (16) with $p=1$, namely,