text:
\subsection*{0. Introduction }
\par\noindent
 Given a lineal $H_0$ and $x_0\in H_0$ and a linear injective
 $U_0: H_0 \to H_0$
 such that all ${U_0}^N \,,\quad N \in {\bf Z}$  exist,
 and all
 $\{{U_0}^N x_0 \,|\,N \in {\bf Z} \}$
 are linearly independent,
 anyone can define
 on
 $L_{x_0}:=span\{{U_0}^{\bf Z} x_0\}
           := span\{{U_0}^N x_0 \,|\,N \in {\bf Z} \}$
 a (pre)hilbert scalar product such that $U_0$ becomes a unitary operator.
 But what about the case where $H_0$ has an additional structure and some
 properties of $H_0$ are expressed in terms of that structure?
 Namely, what about the case where on $H_0$ is defined an indefinite inner
 product, say $\{\cdot , \cdot \}_0$, and $U_0$ is $\{,\}_0$-unitary at least
 on $span\{{U_0}^{\bf Z} x_0\}$ :
$$
    \{U_0 x, U_0 y \}_0 = \{x, y \}_0  \,, \quad
    \{U_0^{-1} x, U_0^{-1} y \}_0 = \{ x, y \}_0
$$
for all $x,y \in span\{{U_0}^{\bf Z} x_0\}$, what about such a case?
 If $\{x_0, U_0^N x_0 \}_0$ as a function of $N \in {\bf Z}$ can be presented
 as linear combination of positive defined functions of $N$, then one can
 define on $span\{{U_0}^{\bf Z} x_0\}$ a (pre)hilbert scalar product
 (may be degenerated),
 say $(,)$, such that {\bf $U_0$ becomes $(,)$-unitary} and
 {\bf simultaneously $\{,\}$ becomes continuous} with respect to the
 corresponding (pre)hilbert norm (may be seminorm).
 If not, if the presentation, mentioned above, is impossible,
 then such $(,)$ is impossible. But what is possible?
\par\addvspace{2\medskipamount}\par\noindent
 In this paper we will show that at least the following is possible:
\par\addvspace{2\medskipamount}\par\noindent
 Under some natural conditions, one can define a prehilbert scalar product,
 say $(,)$, on $span\{{U_0}^{\bf Z} x_0\}$, so that $\{,\}_0$ turned out to be
 continuous and after the corresponding
 completion of  $span\{{U_0}^{\bf Z} x_0\}$ , denote it here by
 ${L_{x_0}}_\sim$, this space
 ${L_{x_0}}_\sim$, indefinite inner product $\{,\}$ and $U_0$
 can be extended to the Hilbert
 space $H_{ext}$, continuous indefinite inner product $\{,\}_{ext}$
 on $H_{ext}$  and
 a $\{,\}_{ext}$-unitary operator $U_{ext}: H_{ext}\to H_{ext} $
 with the properties:
\par\addvspace{2\medskipamount}\par\noindent
 there exist two lineals
$$
 L_{+} \,,\, L_{-} \subset H_{ext}
$$
such that
\par\addvspace{\medskipamount}\par\noindent
(a) $U_{ext}^{\pm 1}L_{+} = L_{+} $
\par\addvspace{\medskipamount}\par\noindent
(b) $U_{ext}^{\pm 1}L_{-} = L_{-} $
\par\addvspace{\medskipamount}\par\noindent
(c) $\{x,x\}_{ext} > 0 \quad ( x \in L_{+}\backslash \{0\} )$
\par\addvspace{\medskipamount}\par\noindent
(d) $\{x,x\}_{ext} < 0 \quad ( x \in L_{-}\backslash \{0\} )$
\par\addvspace{\medskipamount}\par\noindent
(e)  $\{ L_{+} , L_{-}\}_{ext}=\{0\} $
\par\addvspace{\medskipamount}\par\noindent
(f)  $L_{+} \cap  L_{-} = \{0\}$
\par\addvspace{\medskipamount}\par\noindent
(g) if $\{ L_{+} + L_{-}, x \}_{ext}=0$  then  $x = 0$
\par\addvspace{\medskipamount}\par\noindent
(h)  $\overline{ L_{+} + L_{-} }= H $
\par\addvspace{\medskipamount}\par\noindent
 Moreover,
\par\addvspace{\medskipamount}\par\noindent
(c') $ \{x,x\}_{ext} > 0 \quad ( x \in \overline{L_{+}}\backslash \{0\} ) $
\par\addvspace{\medskipamount}\par\noindent
(d') $ \{x,x\}_{ext} < 0 \quad ( x \in \overline{L_{-}}\backslash \{0\} ) $
\par\addvspace{\medskipamount}\par\noindent
(e') $ \{ \overline{L_{+}} , \overline{L_{-}}\}_{ext}=\{0\} $
\par\addvspace{\medskipamount}\par\noindent
(f') $ \overline{L_{+}} \cap  \overline{L_{-}} = \{0\} $
\par\addvspace{\medskipamount}\par\noindent
 Indeed, if in addition $U$ and $U^{-1}$ are bounded, then
\par\addvspace{\medskipamount}\par\noindent
(a')  ${U_{ext}}^{\pm 1}\overline{L_{+}} = \overline{L_{+}}$
\par\addvspace{\medskipamount}\par\noindent
(b')  ${U_{ext}}^{\pm 1}\overline{L_{-}} = \overline{L_{-}}$
\par\addvspace{\medskipamount}\par\noindent
(overline stands for "closure of").
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 Of course, there exists $x_{0 \, ext} \in H_{ext}$ such that
\begin{eqnarray*}
 \{ x_{0 \, ext} , U_{ext}^N x_{0 \, ext}\}_{ext}
& = &
 \{ x_0 , U_0^N x_0\}_0  \,, \quad   N \in {\bf Z}
\end{eqnarray*}
 and one of methods of this paper consists in to take a Hilbert space $H_1$,
 indefinite inner product $\{,\}_1$ with desirable properties (a), (b), and so on,
 and then look for $x_1\in H_1$ such that
\begin{eqnarray*}
 \{ x_1 , U_1^N x_1\}_1
& = &
 \{ x_0 , U_0^N x_0\}_0  \,, \quad   N \in {\bf Z}
\end{eqnarray*}
 If we have success, we can identify "$1$"-objects with "ext"-objects.
 This is why the paper is called "Models of {\bf Z}-orbits ...".
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
  Logically we cosider the chain of transformaitions
$$
  \mbox{ restrict } \to
  \mbox{ change topology } \to
  \mbox{ make completion } \to
  \mbox{ extend }
  \quad
$$
$$
  \quad\quad
  \to\mbox{ restrict } \to
  \mbox{ change topology } \to
  \mbox{ make completion } \to
  \mbox{ extend }
$$
$$
  \to\mbox{ restrict } \to
  \cdots
$$
 The order of the components of the chain may be changed.
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 One can say that we want to develop a specific version
 of the technique of modelling operators (model operators)
 with the aim to construct the models with desired properties,
 not to prove something predesigned about the original abstract operator.
 The latter is secondary: the central object is model, not abstract operator.
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 Now about the "some natural conditions".
\\
 First, we will mostly suppose that the restriction $U_0$ on $L_{x_0}$
 has no non-zero so called neutral invariant subspace.
\\
 Second, to avoid problems with the definition of invariant subspace of
 unbounded operator, we will in some constructions suppose that
$$
    | \{x_0, U_0^N x_0\}_0 | \leq M\,e^{a\, |N| }
$$
 for some numbers $M, a$ and all integers $N$ .
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 Of course we suppose that the reader is familiar with the basics
 of the theory of operators acting in abstract indefinite inner product spaces.
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 In this paper the modelling operators are  combinations of bilateral
 weighted shifts. We only need of the basics of the basics of the theory of
 bilateral
 weighted shifts.
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
We should warn about some peculiarities of the system of notations
(in this paper).
For some reasons we prefer to define sesquilinear form to be
linear in the {\bf second } argument, not in the first.
Another, given an element $f$ of a lineal and a number $\lambda$,
we consider $f\,\lambda$ as a synonimous of $\lambda \, f$.
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
Now the details.

\newpage

\subsection*{1. Notes on Bilateral Weighted Shifts }
\par\noindent
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
Let
$\{u_n\}_{n\in {\bf Z}}$
 be a bilateral number sequence; we will suppose that
$u_n \not= 0$
 for all
$n\in {\bf Z}$.
Let
 $H$ be a separable Hilbert space, complex or real, with a fixed basis
$\{b_n\}_n$ indexed with $n \in {\bf Z}$, (bilateral basis).

 In this case let
$U$
 denote the {\bf shift }
\footnote{
 the full name is: the {\bf bilateral weighted shift of
$\{b_n\}_n$,
 to the right}.
}
 that is generated by the formula
$$
 U\,:\,b_n\,\mapsto\,\frac{u_{n+1}}{u_n}\,b_{n+1} \qquad . \eqno(*)
$$

 The general facts we need are these:

{\bf Observation 1.1}

 One constructs the
$U$ as follows:

 One starts extending the instruction
$(*)$
 on the linear span of the
$\{b_n\}_{n\in {\bf Z}}$
 so that the resulting operator becomes linear.
 That extension is unique and defines a linear densely defined
 operator, which is here denoted by
$U_{min}$,
 and which is closable.
 The closure of
$U_{min}$
 is just the
$U$.

 Now then, this
$U$
 is closed and at least densely defined and injective;
 it has dense range and the action of
$U^N$,
$U^{*-N}$,
$U^{*N}U^N$,
$U^{-N}U^{*-N}$
 (for any integer $N$) is generated by
$$
 U^N:b_n\mapsto \frac{u_{n+N}}{u_n}b_{n+N}\,;\qquad
 U^{*-N}:b_n\mapsto \frac{u_n^*}{u_{n+N}^*}b_{n+N}\,;
$$
$$
 U^{*N}U^N:b_n\mapsto {|\frac{u_{n+N}}{u_n}|}^2 b_n\,;\qquad
 U^{-N}U^{*-N}:b_n\mapsto {|\frac{u_n}{u_{n+N}}|}^2 b_n\,.
$$

 In particular,
$U^N$
 is bounded just when the number sequence
$\{|u_{n+N}/u_n|\}_n$
 is bounded.
 Moreover,
$$  \| U^N \| = sup\{ \; |u_{n+N}/u_n| \;  | n \in {\bf Z} \}    $$
\\
$\Box$

\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
This is all what we need from the theory of  bilateral weighted shifts.

\newpage

\par\addvspace{2\medskipamount}\par\noindent
\subsection*{2. Some properties of $\widehat{U}=U \oplus {U^*}^{-1}$
where $U$ is a bilateral weighted shift }
First of all recall that $\widehat{U}=U \oplus {U^*}^{-1}$
(acting in $H\oplus H$) is unitary
with respect to the indefinite inner product $\{,\}$, given by
\begin{eqnarray*}
 \{ x\oplus y , x_1\oplus y_1   \}
& := &
 (x, y_1) + (y, x_1) ;
\end{eqnarray*}
 where $(,)$ stands for Hilbert scalar product (in $H$)
\par\addvspace{2\medskipamount}\par\noindent
{\bf Remark 2.1.}
\\
 The reader maybe prefer to deal with the complex Hilbert space and
\begin{eqnarray*}
 \{ x\oplus y , x_1\oplus y_1   \}_s
& := &
 i( (x, y_1) - (y, x_1)  ;
\end{eqnarray*}
Recall
\begin{eqnarray*}
 \{ (I\oplus -iI)(x\oplus y) , (I\oplus iI) (x_1\oplus y_1)   \}
& := &
 i( (x, y_1) - (y, x_1)
\\&& =  \{ x\oplus y , x_1\oplus y_1 \}_s ;
\end{eqnarray*}
 and
\begin{eqnarray*}
 (I\oplus -iI)(U \oplus {U^*}^{-1}) (I\oplus iI)
& = &
 (I\oplus iI) (U \oplus {U^*}^{-1}) (I\oplus -iI)
\\
& = &
 U \oplus {U^*}^{-1} ;
\end{eqnarray*}
 So, in the complex space case both constructions are logically equivalent.
 On the other hand the former is algebraically more simple than latter,
 and we prefer to deal with $\{,\}$.
\\
$\Box$

\par\addvspace{2\medskipamount}\par\noindent
{\bf Observation 2.1.}
\begin{eqnarray*}
\{ \widehat{U}^N(b_0\oplus b_0), \widehat{U}^M(b_0\oplus b_0) \}
&=&
0 , \quad (M \not = N)
\\
\{ \widehat{U}^N(b_0\oplus b_0), \widehat{U}^N(b_0\oplus b_0) \}
&=&
2(b_0,b_0) = 2 > 0
\\
\{ \widehat{U}^N(b_0\oplus -b_0), \widehat{U}^M(b_0\oplus -b_0) \}
&=&
0 , \quad (M \not = N)
\\
\{ \widehat{U}^N(b_0\oplus -b_0), \widehat{U}^N(b_0\oplus -b_0) \}
&=&
-2(b_0,b_0) = -2 < 0
\\
\{ \widehat{U}^N(b_0\oplus b_0), \widehat{U}^M(b_0\oplus -b_0) \}
&=&
0 .
\\
\end{eqnarray*}

\par\addvspace{2\medskipamount}\par\noindent
{\bf Definition 2.1.}
\begin{eqnarray*}
 L_{+}
& := &
 span\{\widehat{U}^N(b_0\oplus b_0)   | N \in {\bf Z}\}
 \equiv span\{(U^N b_0\oplus {U^*}^{-N} b_0)   | N \in {\bf Z}\}
\\
 L_{-}
& := &
 span\{\widehat{U}^N(b_0\oplus -b_0)   | N \in {\bf Z}\}
 \equiv span\{(U^N b_0\oplus -{U^*}^{-N} b_0)   | N \in {\bf Z}\}
\\
\end{eqnarray*}
 {\bf Theorem 2.1.}
\begin{eqnarray*}
(1)
&&
  \widehat{U}^{\pm 1}L_{+} = L_{+}
\\
(2)
&&
  \widehat{U}^{\pm 1}L_{-} = L_{-}
\\
(3)
&&
b_N \oplus 0 \in L_{+} + L_{-}   \quad ( N \in {\bf Z} )
\\
&&
0 \oplus b_N \in L_{+} + L_{-}  \quad ( N \in {\bf Z} )
\\
(4)
&&
  \mbox{if } \{ L_{+} + L_{-}, x \}=0 \mbox{ then } x = 0
\\
(5)
&&
  \overline{ L_{+} + L_{-} }= \widehat{ H  }
\\
(6)
&&
  \{ L_{+} , L_{-}\}=\{0\}
\\
(7)
&&
 \{x,x\} > 0 \quad ( x \in L_{+}\backslash \{0\} )
\\
(8)
&&
 \{x,x\} < 0 \quad ( x \in L_{-}\backslash \{0\} )
\\
(9)
&&
  L_{+} \cap  L_{-} = \{0\}
\\
\end{eqnarray*}
 Moreover,
\begin{eqnarray*}
(6')
&&
 \{ \overline{L_{+}} , \overline{L_{-}}\}=\{0\}
\\
(7')
&&
 \{x,x\} > 0 \quad ( x \in \overline{L_{+}}\backslash \{0\} )
\\
(8')
&&
 \{x,x\} < 0 \quad ( x \in \overline{L_{-}}\backslash \{0\} )
\\
(9')
&&
 \overline{L_{+}} \cap  \overline{L_{-}} = \{0\}
\\
\end{eqnarray*}
 If in addition $U$ and $U^{-1}$ are bounded, then
\begin{eqnarray*}
(1')
&&
  \widehat{U}^{\pm 1}\overline{L_{+}} = \overline{L_{+}}
\\
(2')
&&
  \widehat{U}^{\pm 1}\overline{L_{-}} = \overline{L_{-}}
\end{eqnarray*}
\par\addvspace{2\medskipamount}\par\noindent
{\bf Proof} is straightforward.

\par\addvspace{2\medskipamount}\par\noindent
{\bf Definition 2.2.}
\begin{eqnarray*}
 L_{g\oplus h}
& = &
span \{ U^N g \oplus {U^*}^{-N} h  |
\quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
 \}
\end{eqnarray*}
\\
$\Box$

\par\addvspace{2\medskipamount}\par\noindent
{\bf Definition 2.1.}
Given $f \in H $ we write
$$
  f(n) := (b_n, f) \,;
$$
 Conversely, given $\{f(n)\}_n \in l_2(\bf Z)$ we put
$$
     f := \Sigma_n b_n(b_n, f) \in H  \,;
$$
\\
$\Box$

\par\addvspace{2\medskipamount}\par\noindent
{\bf Theorem 2.2.}
If $ \overline {f_{0}(-N) } =  f_{0}(N) $
then $ \{ L_{b_0 \oplus f_0} , L_{b_0 \oplus -f_0} \} =  \{0\} $
\par\addvspace{\medskipamount}\par\noindent
{\bf Proof.}
\\
\begin{eqnarray*}
\{b_0\oplus -f_{0} , \widehat{ U }^N ( b_0\oplus f_{0} ) \}
& = & \{b_0\oplus -f_{0}, U^N b_0\oplus {U^*}^{-N} f_{0} \}
\\
& = & (b_0,{U^*}^{-N} f_{0} ) - (f_{0} , U^N b_0 )
\\
& = & ({U}^{-N} b_0, f_{0} ) - (f_{0} , U^N b_0 )
\\
& = & \frac{\overline u_{-N}}{\overline u_0} f_{0}(-N)
       - \overline{f_{0}(N)}\frac{u_{N}}{u_0}
\\
& = &  0
\quad ,
\\\quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\\
$\Box$

\par\addvspace{2\medskipamount}\par\noindent
 {\bf Theorem 2.3.}
 Suppose
\begin{eqnarray*}
 \{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} , g_1\oplus h_1  \}
& = &
 \{0\}
\end{eqnarray*}
 then

\begin{eqnarray*}
h_1 & = &
 0
\end{eqnarray*}
\begin{eqnarray*}
( f_0 , {U}^{N} g_1 )
& = &
 0
 \,\quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}

\par\addvspace{2\medskipamount}\par\noindent
{\bf Proof}
\begin{eqnarray*}
\{ \widehat{ U }^N ( b_0\oplus \pm f_{0} ) , g_1\oplus h_1  \}
& = &
 0
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  Then
\begin{eqnarray*}
\{ \widehat{ U }^N ( b_0\oplus 0 ) , g_1\oplus h_1 \}
& = &
 0
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  and
\begin{eqnarray*}
\{ \widehat{ U }^N ( 0 \oplus f_0 ) , g_1\oplus h_1 \}
& = &
 0
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 Now obtain
\begin{eqnarray*}
 0
& = &
\{ \widehat{ U }^N ( b_0\oplus 0 ) , g_1\oplus h_1 \}
\\
& = &
\{  ( { U }^N b_0\oplus 0 ) , g_1\oplus h_1 \}
\\
& = &
( { U }^N b_0 ,  h_1 )
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
 0
& = &
\{ \widehat{ U }^N ( b_0\oplus 0 ) , g_1\oplus h_1 \}
=
\{  ( { U }^N b_0\oplus 0 ) , g_1\oplus h_1 \}
=
( { U }^N b_0 ,  h_1 )
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 Hence $h_1 = 0$.
Now
\begin{eqnarray*}
 0
& = &
\{ \widehat{ U }^N ( 0 \oplus f_0 ) , g_1\oplus h_1 \}
\\
& = &
\{  ( 0 \oplus {U^*}^{-N}f_0 ) , g_1\oplus h_1 \}
\\
& = &
({U^*}^{-N}f_0  ,  h_1 )
\\
& = &
(f_0  , {U}^{-N} h_1 )
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
Hence
\begin{eqnarray*}
 0
& = &
(f_0  , {U}^{N} h_1 )
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\\
$\Box$

\newpage

\par\addvspace{2\medskipamount}\par\noindent
\subsection*{3. Models of Orbits. Existence and Properties. }
\par\addvspace{2\medskipamount}\par\noindent
\par\noindent
 {\bf Definition 3.1.}
 Let $H_1, \{, \}_1$ and  $H_0, \{,\}_0$ be indefinite inner product spaces,
 $U_1$ and $U_0$ be respectively  $\{, \}_1$- and $\{, \}_0$-unitary operators
 (bounded or not), and let $x_1 \in H_1$,  $x_0 \in H_0$

 We say that the {\bf Z}-orbit
\begin{eqnarray*}
&&
 \{U_1^N x_1 |  N \in {\bf Z}\}
\end{eqnarray*}
 is a model of the orbit
\begin{eqnarray*}
&&
 \{U_0^N x_0 |  N \in {\bf Z}\}
\end{eqnarray*}
 iff
\begin{eqnarray*}
 \{ x_1 , U_1^N x_1\}_1
& = &
 \{ x_0 , U_0^N x_0\}_0  \,, \quad   N \in {\bf Z}
\end{eqnarray*}
\\
In this case we write
$$
 span \{U_1^N x_1 |  N \in {\bf Z}\} \approx_{\{,\}_1,\{,\}_0}
 span \{U_0^N x_0 |  N \in {\bf Z}\}
$$
 and/or simply
$$
  \{U_1^N x_1 |  N \in {\bf Z}\} \approx_{\{,\}_1, \{,\}_0}  \{U_0^N x_0 |  N \in {\bf Z}\}
$$
$\Box$
\par\addvspace{2\medskipamount}\par\noindent
 Throughout this section
\begin{eqnarray*}
 H_1
& := &
 H\oplus H
\end{eqnarray*}
\begin{eqnarray*}
 \{ x\oplus y , x_1\oplus y_1   \}_1
& := &
 \{ x\oplus y , x_1\oplus y_1   \}
\\
& := &
 (x, y_1) + (y, x_1) ;
\end{eqnarray*}
 where Hilbert space $H, (,)$ is the same as in the previous sections
 and let
\begin{eqnarray*}
 U_1
& := &
 U\oplus {U^*}^{-1}
\end{eqnarray*}
 for a suitable bilateral shift operator $U$ (bounded or not).
\par\addvspace{2\medskipamount}\par\noindent
 In addition suppose (mostly thechnical suppositions) that
$u_n$ is choosen so that
\begin{eqnarray*}
 \overline u_{-n}
& = &
 u_n \,, \qquad ( n \in {\bf Z} )
\end{eqnarray*}
 and
\begin{eqnarray*}
\Bigl\{\frac{1}{u_N}{ \{x_0, U_0^N\ x_0 \}_0 }\Bigr\}_{N \in {\bf Z}}
& \in  &
 l_2({\bf Z})
\end{eqnarray*}

\par\addvspace{2\medskipamount}\par\noindent
 First of all we try
\begin{eqnarray*}
 x_1
& := &
 b_0 \oplus f_0
\end{eqnarray*}
 for a suitable $f_0$
\par\noindent
\par\noindent
\par\noindent
\par\noindent
\par\noindent
 So, we consider the next equation, where $f_0$ is to be found.
\begin{eqnarray*}
\{b_0\oplus f_{0}, \widehat{ U }^N ( b_0\oplus f_{0} ) \}
=\{b_0\oplus f_{0}, U^N b_0\oplus {U^*}^{-N} f_{0} \}
& = &
\{x_0, U_0^N\ x_0 \}_0
\\
&&\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 Now we obtain
\begin{eqnarray*}
(b_0, {U^*}^{-N} f_{0}) + (f_{0}, {U}^{N} b_0)
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
( {U}^{-N} b_0, f_{0}) + (f_{0}, {U}^{N} b_0)
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
(\frac{u_{-N}}{u_0}b_{-N}, f_{0}) + (f_{0},\frac{u_N}{u_0} b_N)
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
\frac{\overline u_{-N}}{\overline  u_0}(b_{-N}, f_{0})
+ \frac{\overline u_N}{\overline u_0} (f_{0}, b_N)
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
\frac{\overline u_{-N}}{\overline  u_0}f_{0}(-N)
+ \frac{\overline u_N}{\overline u_0} \overline{f_{0}(N)}
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 Note
\begin{eqnarray*}
\{x_0, U_0^N\ x_0 \}_0
& = &
\overline{\{x_0, U_0^{-N}\ x_0 \}_0}
  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 Recall,
\begin{eqnarray*}
  u_N
& = &
\overline{u_{-N}}
  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  so that
\begin{eqnarray*}
 ( f_{0}(-N)
+ \overline{f_0(N)}) \frac{u_N}{u_0}
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  The formal solution is
\begin{eqnarray*}
  f_{0}(N)
& = &
\frac{1}{2}\frac{\overline u_0}{\overline u_N}\overline { \{x_0, U_0^N\ x_0 \}_0 }
+f_{00}(N)
\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 where
\begin{eqnarray*}
  \overline f_{00}(-N)
& = &
-f_{00}(N)
\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
Fix $f_{00} := 0 $ so that
\begin{eqnarray*}
  f_{0}(N)
& = &
\overline{f_{0}(-N)}
  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
In this case we have
\begin{eqnarray*}
  \{f_{0}(N)\}_{N \in {\bf Z}}
& \in &
 l_2({\bf Z })
\end{eqnarray*}
 and hence we really have obtained a solution.
\par\addvspace{2\medskipamount}\par\noindent
Thus we have obtained
\par\addvspace{2\medskipamount}\par\noindent
{\bf Theorem 3.1.}
 Choose
$u_n$ so that
\begin{eqnarray*}
 \overline u_{-n}
& = &
 u_n \,, \qquad ( n \in {\bf Z} )
\end{eqnarray*}
 and
\begin{eqnarray*}
\Bigl\{\frac{1}{u_N}{ \{x_0, U_0^N\ x_0 \}_0 }\Bigr\}_{N \in {\bf Z}}
& \in  &
 l_2({\bf Z})
\end{eqnarray*}
(it is always possible) and put
\begin{eqnarray*}
  f_{0}(N)
& = &
\frac{1}{2}\frac{\overline u_0}{\overline u_N}\overline { \{x_0, U_0^N\ x_0 \}_0 }
\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 In this case
\begin{eqnarray*}
  \{f_{0}(N)\}_{N \in {\bf Z}}
& \in &
 l_2({\bf Z })
\end{eqnarray*}
 and if one defines
$$
     x_1 := b_0 \oplus f_0
$$
  then
\begin{eqnarray*}
 \{ x_1 , U_1^N x_1\}_1
& = &
 \{ x_0 , U_0^N x_0\}_0  \,, \quad   N \in {\bf Z}
\end{eqnarray*}
\\
$\Box$
\par\addvspace{4\medskipamount}\par\noindent
  Let us discuss some properties of the obtained solution
\par\addvspace{4\medskipamount}\par\noindent
Recall the Definition 2.2:
\begin{eqnarray*}
 L_{g\oplus h}
& = &
span \{ U^N g \oplus {U^*}^{-N} h  |
\quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
 \}
\end{eqnarray*}
 and recall that by the {\bf Theorem 2.2} we have
\begin{eqnarray*}
 \{ L_{b_0 \oplus f_0} , L_{b_0 \oplus -f_0} \}
& = &
 \{0\}
\end{eqnarray*}
\par\addvspace{2\medskipamount}\par\noindent
 {\bf Theorem 3.2.}
 Suppose
$U, U^{-1}$ are bounded,
 $H_0$ is a complet normed space with the norm $||\cdot ||_0$
such that $|\{x,y\}_0| \leq ||x||_0||y||_0 \,, \quad (x,y \in H_0)$
and $u_n$ is choosen
so that
$$ \Sigma_n |\frac{u_0}{u_n}|^2 ||U_0^{n}\ x_0||_0^2 < \infty $$
(it is always possible if $U_0, U_0^{-1}$ are bounded)
\quad   If in addition
\begin{eqnarray*}
 \{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} , g_1\oplus h_1  \}
& = &
 \{0\}
\end{eqnarray*}
 then
$$
 \widetilde {g_1} := \Sigma_n \frac{u_0}{u_n} g_1(n) U_0^{n} x_0
  \in \overline {span\{U_0^{n} x_0 | n\in {\bf Z} \}}
  \subset H_0
$$
  and
\begin{eqnarray*}
 \{U_0^{N} x_0, \widetilde{g_1} \}_0
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
 \{U_0^{N} \widetilde{g_1}, \widetilde{g_1} \}_0
& = &
 0
\\ \quad N  = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 In particular,
$\overline {span \{U_0^{N} \widetilde{g_1}| n \in {\bf Z}  \} }$
is a $||\cdot||_0$ closed $U_0^{\pm 1}$-invariant $\{,\}_0$-neutral
subspace of $\overline {span \{U_0^{N} x_0 | n \in {\bf Z}  \} } \subset H_0$
(may be $\{0\}$).
\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 {\bf Proof }
\\
By the Theorem 2.3,
\begin{eqnarray*}
h_1 & = &
 0
\end{eqnarray*}
\begin{eqnarray*}
( f_0 , {U}^{N} g_1 )
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 On the other hand
\begin{eqnarray*}
 g_1
& = &
\Sigma_n g_1(n) b_n
\end{eqnarray*}
 Hence
\begin{eqnarray*}
 U^{-N}g_1
& = &
 \Sigma_n \frac{u_{n-N}}{u_n} g_1(n)b_{n-N}
\\
&&
 \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
( f_0 , \Sigma_n \frac{u_{n-N}}{u_n} g_1(n)b_{n-N} )
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
  \Sigma_n \frac{u_{n-N}}{u_n}  g_1(n) \overline{f_0(n-N)}
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
Recall,
\begin{eqnarray*}
  f_{0}(N) = (b_N , f_0)
& = &
\frac{1}{2}\frac{\overline u_0}{\overline u_N}\overline { \{x_0, U_0^N\ x_0 \}_0 }
\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
Hence
\begin{eqnarray*}
  f_{0}(n-N) = (b_{n-N} , f_0)
& = &
\frac{1}{2}\frac{\overline u_0}{\overline u_{n-N}}
\overline { \{x_0, U_0^{n-N}\ x_0 \}_0 }
\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  Hence
\begin{eqnarray*}
\Sigma_n
\frac{u_{n-N}}{u_n} g_1(n)
\frac{1}{2}\frac{ u_0 }{ u_{n-N} } \{x_0, U_0^{n-N}\ x_0 \}_0
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
\Sigma_n
\frac{u_0}{u_n} g_1(n)
\frac{1}{2}  \{x_0, U_0^{n-N}\ x_0 \}_0
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
\Sigma_n
\frac{u_0}{u_n} g_1(n)
\frac{1}{2}  \{U_0^{N} x_0,  U_0^{n}\ x_0 \}_0
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
 \{U_0^{N} x_0, \widetilde{g_1} \}_0
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
 \{U_0^{N}U_0^{n} x_0, \widetilde{g_1} \}_0
& = &
 0
\\ \quad N,n  = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
 \{U_0^{N} \widetilde{g_1}, \widetilde{g_1} \}_0
& = &
 0
\\ \quad N  = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  Quod erat demonstrandum.
\\
$\Box$
\par\addvspace{2\medskipamount}\par\noindent
 {\bf Corollary 3.1.}
\\
 Suppose
 $H_0 |\overline {span \{U_0^{N} \widetilde{g_1}| n \in {\bf Z}  \} }$
(the restriction of $H_0$ onto
        $\overline {span \{U_0^{N} \widetilde{g_1}| n \in {\bf Z}  \} }$)
has no non-zero neutral invariant subspace.
\\
Then
$$ \overline{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} } = H_1 $$
and
$$  L_{b_0 \oplus f_0} \cap L_{b_0 \oplus -f_0} = \{0\} $$

\par\addvspace{2\medskipamount}\par\noindent
 {\bf Proof.}
By supposition and the Theorem 3.2 $\widetilde {g_1} =0 $.
Hence
$$ \{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} , g_1\oplus h_1  \} =0 $$
implies
$$ g_1\oplus h_1  = 0 \oplus 0 $$
Hence
$$ \overline{  L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} } = H_1 $$
 Now, by the Theorem 2.2,
$$\{ L_{b_0 \oplus f_0} , L_{b_0 \oplus -f_0}  \} =0$$
 and again
$$ \{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} , g_1\oplus h_1  \} =0 $$
implies
$$ g_1\oplus h_1  = 0 \oplus 0 $$
Hence
$$  L_{b_0 \oplus f_0} \cap L_{b_0 \oplus -f_0} = \{0\} $$
\\
$\Box$
\par\addvspace{2\medskipamount}\par\noindent
Thus, we have constructed the following chain of restrictions-extensions:
$$
 H_0 \supset L_{x_0} \approx_{\{,\}_0, \{,\}} L_{b_0 \oplus f_0} \subset
 L_{b_0 \oplus f_0} \dot{+}_{\{,\}} L_{b_0 \oplus -f_0}
 \subset_{dense} \overline{ L_{b_0 \oplus f_0} \dot{+}_{\{,\}} L_{b_0 \oplus -f_0}}
$$
$$
= \overline{L_{+} \dot{+}_{\{,\}} L_{-}}  \qquad (chain \; A )
$$
Here and in following $A\dot{+}B$ stands for $A+B$ with indicating
$A\cap B =\{0\}$
and  $A{+}_{\{,\}}B$ stands for $A+B$ with indicating $\{A,B\} =\{0\}$.
\par\addvspace{2\medskipamount}\par\noindent
\subsection*{4. Operator $U_0 \oplus U_0$ }
\par\addvspace{2\medskipamount}\par\noindent
{\bf Observation 4.1.}
\begin{eqnarray*}
\{b_0\oplus -f_{0}, \widehat{ U }^N ( b_0\oplus -f_{0} ) \}
&=&\{b_0\oplus -f_{0}, U^N b_0\oplus -{U^*}^{-N} f_{0} \}
\\
& = &
-\{x_0, U_0^N\ x_0 \}_0  \quad ,
\\
&&\quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
$\Box$
\\
This motivates the following considerations.

\par\addvspace{2\medskipamount}\par\noindent
{\bf Definition 4.1.}
\\
Put
\begin{eqnarray*}
 \widetilde H_0
& := &
 H_0\oplus H_0
\\
\|x \oplus y \|_{\sim}^2
& := &
\|x\|_{0}^2 + \|y \|_{0}^2 \, \quad ( x,y \in H_0 )
\\
 \widetilde U_0
& := &
 U_0\oplus U_0
\\
 \{ x\oplus y , x_1\oplus y_1     \}_{-}
& := &
 \{ x , x_1 \}_0 - \{ y , y_1 \}_0
\\
 L_{x_0}
& := &
 span\{ U^n x_0 | n \in {\bf Z} \}
\\
\overline{ L_{x_0} } &:=& \mbox{ closure of } L_{x_0}
                      \mbox{ with respect to } \|\cdot\|_0
\\
\widetilde L_{x_0}
& := &
L_{x_0}\oplus L_{x_0}
\\
\end{eqnarray*}
and define a linear map
\begin{eqnarray*}
 \Omega
& : &
 L_{x_0} \oplus L_{x_0} \subset \widetilde H_0
  \to
 L_{b_0\oplus f_0} + L_{b_0\oplus -f_0} \subset \widehat H
\\
\end{eqnarray*}
 as follows:
Firstly put
\begin{eqnarray*}
 \Omega ( \, U_0^N x_0 \oplus U_0^M x_0 \,)
& :=  &
 \widehat U^N (b_0\oplus f_0) + \widehat U^M  (b_0\oplus -f_0 )
\\
&&
   N,M \in {\bf Z}
\end{eqnarray*}
Then extend this map onto the span of $U_0^N x_0 \oplus U_0^M x_0$ \quad
$(N,M \in {\bf Z})$ as a linear map.
\par\addvspace{2\medskipamount}\par\noindent
{\bf Observation 4.2.}
\begin{eqnarray*}
 \{
 \Omega ( \, x\oplus y,) ,
 \Omega ( \, x_1\oplus y_1 \,)
 \}
 & =  &
 \{
   x\oplus y, ,
   x_1\oplus y_1
 \}_{-}
\\
&&
   x, y, x_1, y_1 \in L_{x_0}
\end{eqnarray*}
\\
Thus, if we take into account indefinite inner products, we may say that
$L_{x_0}\oplus L_{x_0}$ is embedded in
$L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}$,
we will write it symbolically as
$$
L_{x_0}\oplus L_{x_0} \subset_{\Omega, \{,\}_0 , \{,\} }
L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}
$$
If in addition
$U_0$ is bounded, has no nonzero $\{,\}_0$ invariant subspace,
then by the  Corollary 3.1
\\
$$ \overline{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} } = \widehat H \, \quad (a) $$
and
$$  L_{b_0 \oplus f_0} \cap L_{b_0 \oplus -f_0} = \{0\}\,; \quad (b)$$
As a result,
\begin{eqnarray*}
 \Omega
& : &
L_{x_0} \oplus L_{x_0} \subset \widetilde H_0
  \to
 L_{b_0\oplus f_0} + L_{b_0\oplus -f_0} \subset \widehat H
\\
\end{eqnarray*}
is  injective (note, surjective by the definition). Hence,
\begin{eqnarray*}
        \Omega^{-1}
& : &
 L_{b_0\oplus f_0} + L_{b_0\oplus -f_0} \subset \widehat H
  \to
 L_{x_0} \oplus L_{x_0} \subset \widetilde H_0
\\
\end{eqnarray*}
exists, and we may identify
$L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}$
and
$L_{x_0} \oplus L_{x_0}$
as indefinite inner product lineals, write it symbolically:
$$
L_{x_0}\oplus L_{x_0} \approx_{\Omega, \{,\}_0 , \{,\} }
L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}
$$
Thus, if take into account the relation $(b)$, we may write finally:
$$
L_{x_0}\oplus L_{x_0} \approx_{\Omega, \{,\}_0 , \{,\} }
L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}
\qquad\qquad\qquad\qquad
$$
$$
\qquad\qquad
\subset
\overline{L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}}
=\widehat{H}
=\overline{L_{+} \dot{+}_{\{,\}} L_{-}}
\qquad (chain \; B) ;
$$

\par\noindent
$\Box$
\par\addvspace{2\medskipamount}\par\noindent
As a result we have obtained
\par\addvspace{2\medskipamount}\par\noindent
{\bf Theorem 4.1.}
\\
Suppose $H_0$ is a Hilbert space.
Suppose
$U_0$ is bounded, has no nonzero $\{,\}_0$ invariant subspace, and $x_0$ is
a {\bf Z}-cyclic vector of $U_0$, i.e.,
$$
H_0=\overline{ L_{x_0} }
$$
Then
\par\addvspace{2\medskipamount}\par\noindent
$(a)$
there exists a (pre)hilbert norm on $L_{x_0}\oplus L_{x_0}$ such that
$\{,\}_{-}$ and $U_0\oplus U_0$ are continuous and
can be extended to continuous $\{,\}_{ext}$ and $U_{ext}$
onto $H_{ext} = \mbox{ the completion of } L_{x_0}\oplus L_{x_0}$ ;
\\

\par\addvspace{\medskipamount}\par\noindent
$(b)$ there are two $U_{ext}^{\pm 1}$-nvarint (closed) subspaces
${\cal L}_{+},{\cal L}_{-} \subset \widetilde{ H_0 }$ such that:
\par\addvspace{\medskipamount}\par\noindent
(1) ${\cal L}_{+}$ is strictly positive;
\par\addvspace{2\medskipamount}\par\noindent
(2) ${\cal L}_{-}$ is strictly negative;
\par\addvspace{2\medskipamount}\par\noindent
(3) ${\cal L}_{+}\cap {\cal L}_{-} = \{0\}$ ,
\par\addvspace{2\medskipamount}\par\noindent
(4) ${\cal L}_{+}+{\cal L}_{-} $ is dense in $\widetilde{H_0}$,
\par\addvspace{2\medskipamount}\par\noindent
(5) $\{{\cal L}_{+},{\cal L}_{-}\}_{-}=\{0\} $.
\par\addvspace{2\medskipamount}\par\noindent
(6) as semidefined subspaces, ${\cal L}_{+}$, ${\cal L}_{-}$ both are maximal.
\par\addvspace{2\medskipamount}\par\noindent
$\Box$
\par\addvspace{2\medskipamount}\par\noindent
Finally,
\par\addvspace{\medskipamount}\par\noindent
{\bf Remark 4.1.}
\\
Note, that, in contrast with $(chain\; A)$ in the previous subsection,
in $(chain\; B)$ all embeddings are dense.
$\Box$
\par\addvspace{2\medskipamount}\par\noindent

\newpage
\bibliographystyle{unsrt}

\subsection*{0. Introduction }
\par\noindent
 Given a lineal $H_0$ and $x_0\in H_0$ and a linear injective
 $U_0: H_0 \to H_0$
 such that all ${U_0}^N \,,\quad N \in {\bf Z}$  exist,
 and all
 $\{{U_0}^N x_0 \,|\,N \in {\bf Z} \}$
 are linearly independent,
 anyone can define
 on
 $L_{x_0}:=span\{{U_0}^{\bf Z} x_0\}
           := span\{{U_0}^N x_0 \,|\,N \in {\bf Z} \}$
 a (pre)hilbert scalar product such that $U_0$ becomes a unitary operator.
 But what about the case where $H_0$ has an additional structure and some
 properties of $H_0$ are expressed in terms of that structure?
 Namely, what about the case where on $H_0$ is defined an indefinite inner
 product, say $\{\cdot , \cdot \}_0$, and $U_0$ is $\{,\}_0$-unitary at least
 on $span\{{U_0}^{\bf Z} x_0\}$ :
$$
    \{U_0 x, U_0 y \}_0 = \{x, y \}_0  \,, \quad
    \{U_0^{-1} x, U_0^{-1} y \}_0 = \{ x, y \}_0
$$
for all $x,y \in span\{{U_0}^{\bf Z} x_0\}$, what about such a case?
 If $\{x_0, U_0^N x_0 \}_0$ as a function of $N \in {\bf Z}$ can be presented
 as linear combination of positive defined functions of $N$, then one can
 define on $span\{{U_0}^{\bf Z} x_0\}$ a (pre)hilbert scalar product
 (may be degenerated),
 say $(,)$, such that {\bf $U_0$ becomes $(,)$-unitary} and
 {\bf simultaneously $\{,\}$ becomes continuous} with respect to the
 corresponding (pre)hilbert norm (may be seminorm).
 If not, if the presentation, mentioned above, is impossible,
 then such $(,)$ is impossible. But what is possible?
\par\addvspace{2\medskipamount}\par\noindent
 In this paper we will show that at least the following is possible:
\par\addvspace{2\medskipamount}\par\noindent
 Under some natural conditions, one can define a prehilbert scalar product,
 say $(,)$, on $span\{{U_0}^{\bf Z} x_0\}$, so that $\{,\}_0$ turned out to be
 continuous and after the corresponding
 completion of  $span\{{U_0}^{\bf Z} x_0\}$ , denote it here by
 ${L_{x_0}}_\sim$, this space
 ${L_{x_0}}_\sim$, indefinite inner product $\{,\}$ and $U_0$
 can be extended to the Hilbert
 space $H_{ext}$, continuous indefinite inner product $\{,\}_{ext}$
 on $H_{ext}$  and
 a $\{,\}_{ext}$-unitary operator $U_{ext}: H_{ext}\to H_{ext} $
 with the properties:
\par\addvspace{2\medskipamount}\par\noindent
 there exist two lineals
$$
 L_{+} \,,\, L_{-} \subset H_{ext}
$$
such that
\par\addvspace{\medskipamount}\par\noindent
(a) $U_{ext}^{\pm 1}L_{+} = L_{+} $
\par\addvspace{\medskipamount}\par\noindent
(b) $U_{ext}^{\pm 1}L_{-} = L_{-} $
\par\addvspace{\medskipamount}\par\noindent
(c) $\{x,x\}_{ext} > 0 \quad ( x \in L_{+}\backslash \{0\} )$
\par\addvspace{\medskipamount}\par\noindent
(d) $\{x,x\}_{ext} < 0 \quad ( x \in L_{-}\backslash \{0\} )$
\par\addvspace{\medskipamount}\par\noindent
(e)  $\{ L_{+} , L_{-}\}_{ext}=\{0\} $
\par\addvspace{\medskipamount}\par\noindent
(f)  $L_{+} \cap  L_{-} = \{0\}$
\par\addvspace{\medskipamount}\par\noindent
(g) if $\{ L_{+} + L_{-}, x \}_{ext}=0$  then  $x = 0$
\par\addvspace{\medskipamount}\par\noindent
(h)  $\overline{ L_{+} + L_{-} }= H $
\par\addvspace{\medskipamount}\par\noindent
 Moreover,
\par\addvspace{\medskipamount}\par\noindent
(c') $ \{x,x\}_{ext} > 0 \quad ( x \in \overline{L_{+}}\backslash \{0\} ) $
\par\addvspace{\medskipamount}\par\noindent
(d') $ \{x,x\}_{ext} < 0 \quad ( x \in \overline{L_{-}}\backslash \{0\} ) $
\par\addvspace{\medskipamount}\par\noindent
(e') $ \{ \overline{L_{+}} , \overline{L_{-}}\}_{ext}=\{0\} $
\par\addvspace{\medskipamount}\par\noindent
(f') $ \overline{L_{+}} \cap  \overline{L_{-}} = \{0\} $
\par\addvspace{\medskipamount}\par\noindent
 Indeed, if in addition $U$ and $U^{-1}$ are bounded, then
\par\addvspace{\medskipamount}\par\noindent
(a')  ${U_{ext}}^{\pm 1}\overline{L_{+}} = \overline{L_{+}}$
\par\addvspace{\medskipamount}\par\noindent
(b')  ${U_{ext}}^{\pm 1}\overline{L_{-}} = \overline{L_{-}}$
\par\addvspace{\medskipamount}\par\noindent
(overline stands for "closure of").
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 Of course, there exists $x_{0 \, ext} \in H_{ext}$ such that
\begin{eqnarray*}
 \{ x_{0 \, ext} , U_{ext}^N x_{0 \, ext}\}_{ext}
& = &
 \{ x_0 , U_0^N x_0\}_0  \,, \quad   N \in {\bf Z}
\end{eqnarray*}
 and one of methods of this paper consists in to take a Hilbert space $H_1$,
 indefinite inner product $\{,\}_1$ with desirable properties (a), (b), and so on,
 and then look for $x_1\in H_1$ such that
\begin{eqnarray*}
 \{ x_1 , U_1^N x_1\}_1
& = &
 \{ x_0 , U_0^N x_0\}_0  \,, \quad   N \in {\bf Z}
\end{eqnarray*}
 If we have success, we can identify "$1$"-objects with "ext"-objects.
 This is why the paper is called "Models of {\bf Z}-orbits ...".
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
  Logically we cosider the chain of transformaitions
$$
  \mbox{ restrict } \to
  \mbox{ change topology } \to
  \mbox{ make completion } \to
  \mbox{ extend }
  \quad
$$
$$
  \quad\quad
  \to\mbox{ restrict } \to
  \mbox{ change topology } \to
  \mbox{ make completion } \to
  \mbox{ extend }
$$
$$
  \to\mbox{ restrict } \to
  \cdots
$$
 The order of the components of the chain may be changed.
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 One can say that we want to develop a specific version
 of the technique of modelling operators (model operators)
 with the aim to construct the models with desired properties,
 not to prove something predesigned about the original abstract operator.
 The latter is secondary: the central object is model, not abstract operator.
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 Now about the "some natural conditions".
\\
 First, we will mostly suppose that the restriction $U_0$ on $L_{x_0}$
 has no non-zero so called neutral invariant subspace.
\\
 Second, to avoid problems with the definition of invariant subspace of
 unbounded operator, we will in some constructions suppose that
$$
    | \{x_0, U_0^N x_0\}_0 | \leq M\,e^{a\, |N| }
$$
 for some numbers $M, a$ and all integers $N$ .
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 Of course we suppose that the reader is familiar with the basics
 of the theory of operators acting in abstract indefinite inner product spaces.
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 In this paper the modelling operators are  combinations of bilateral
 weighted shifts. We only need of the basics of the basics of the theory of
 bilateral
 weighted shifts.
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
We should warn about some peculiarities of the system of notations
(in this paper).
For some reasons we prefer to define sesquilinear form to be
linear in the {\bf second } argument, not in the first.
Another, given an element $f$ of a lineal and a number $\lambda$,
we consider $f\,\lambda$ as a synonimous of $\lambda \, f$.
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
Now the details.

\newpage

\subsection*{1. Notes on Bilateral Weighted Shifts }
\par\noindent
\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
Let
$\{u_n\}_{n\in {\bf Z}}$
 be a bilateral number sequence; we will suppose that
$u_n \not= 0$
 for all
$n\in {\bf Z}$.
Let
 $H$ be a separable Hilbert space, complex or real, with a fixed basis
$\{b_n\}_n$ indexed with $n \in {\bf Z}$, (bilateral basis).

 In this case let
$U$
 denote the {\bf shift }
\footnote{
 the full name is: the {\bf bilateral weighted shift of
$\{b_n\}_n$,
 to the right}.
}
 that is generated by the formula
$$
 U\,:\,b_n\,\mapsto\,\frac{u_{n+1}}{u_n}\,b_{n+1} \qquad . \eqno(*)
$$

 The general facts we need are these:

{\bf Observation 1.1}

 One constructs the
$U$ as follows:

 One starts extending the instruction
$(*)$
 on the linear span of the
$\{b_n\}_{n\in {\bf Z}}$
 so that the resulting operator becomes linear.
 That extension is unique and defines a linear densely defined
 operator, which is here denoted by
$U_{min}$,
 and which is closable.
 The closure of
$U_{min}$
 is just the
$U$.

 Now then, this
$U$
 is closed and at least densely defined and injective;
 it has dense range and the action of
$U^N$,
$U^{*-N}$,
$U^{*N}U^N$,
$U^{-N}U^{*-N}$
 (for any integer $N$) is generated by
$$
 U^N:b_n\mapsto \frac{u_{n+N}}{u_n}b_{n+N}\,;\qquad
 U^{*-N}:b_n\mapsto \frac{u_n^*}{u_{n+N}^*}b_{n+N}\,;
$$
$$
 U^{*N}U^N:b_n\mapsto {|\frac{u_{n+N}}{u_n}|}^2 b_n\,;\qquad
 U^{-N}U^{*-N}:b_n\mapsto {|\frac{u_n}{u_{n+N}}|}^2 b_n\,.
$$

 In particular,
$U^N$
 is bounded just when the number sequence
$\{|u_{n+N}/u_n|\}_n$
 is bounded.
 Moreover,
$$  \| U^N \| = sup\{ \; |u_{n+N}/u_n| \;  | n \in {\bf Z} \}    $$
\\
$\Box$

\par\addvspace{\medskipamount}\par\noindent
\par\addvspace{\medskipamount}\par\noindent
This is all what we need from the theory of  bilateral weighted shifts.

\newpage

\par\addvspace{2\medskipamount}\par\noindent
\subsection*{2. Some properties of $\widehat{U}=U \oplus {U^*}^{-1}$
where $U$ is a bilateral weighted shift }
First of all recall that $\widehat{U}=U \oplus {U^*}^{-1}$
(acting in $H\oplus H$) is unitary
with respect to the indefinite inner product $\{,\}$, given by
\begin{eqnarray*}
 \{ x\oplus y , x_1\oplus y_1   \}
& := &
 (x, y_1) + (y, x_1) ;
\end{eqnarray*}
 where $(,)$ stands for Hilbert scalar product (in $H$)
\par\addvspace{2\medskipamount}\par\noindent
{\bf Remark 2.1.}
\\
 The reader maybe prefer to deal with the complex Hilbert space and
\begin{eqnarray*}
 \{ x\oplus y , x_1\oplus y_1   \}_s
& := &
 i( (x, y_1) - (y, x_1)  ;
\end{eqnarray*}
Recall
\begin{eqnarray*}
 \{ (I\oplus -iI)(x\oplus y) , (I\oplus iI) (x_1\oplus y_1)   \}
& := &
 i( (x, y_1) - (y, x_1)
\\&& =  \{ x\oplus y , x_1\oplus y_1 \}_s ;
\end{eqnarray*}
 and
\begin{eqnarray*}
 (I\oplus -iI)(U \oplus {U^*}^{-1}) (I\oplus iI)
& = &
 (I\oplus iI) (U \oplus {U^*}^{-1}) (I\oplus -iI)
\\
& = &
 U \oplus {U^*}^{-1} ;
\end{eqnarray*}
 So, in the complex space case both constructions are logically equivalent.
 On the other hand the former is algebraically more simple than latter,
 and we prefer to deal with $\{,\}$.
\\
$\Box$

\par\addvspace{2\medskipamount}\par\noindent
{\bf Observation 2.1.}
\begin{eqnarray*}
\{ \widehat{U}^N(b_0\oplus b_0), \widehat{U}^M(b_0\oplus b_0) \}
&=&
0 , \quad (M \not = N)
\\
\{ \widehat{U}^N(b_0\oplus b_0), \widehat{U}^N(b_0\oplus b_0) \}
&=&
2(b_0,b_0) = 2 > 0
\\
\{ \widehat{U}^N(b_0\oplus -b_0), \widehat{U}^M(b_0\oplus -b_0) \}
&=&
0 , \quad (M \not = N)
\\
\{ \widehat{U}^N(b_0\oplus -b_0), \widehat{U}^N(b_0\oplus -b_0) \}
&=&
-2(b_0,b_0) = -2 < 0
\\
\{ \widehat{U}^N(b_0\oplus b_0), \widehat{U}^M(b_0\oplus -b_0) \}
&=&
0 .
\\
\end{eqnarray*}

\par\addvspace{2\medskipamount}\par\noindent
{\bf Definition 2.1.}
\begin{eqnarray*}
 L_{+}
& := &
 span\{\widehat{U}^N(b_0\oplus b_0)   | N \in {\bf Z}\}
 \equiv span\{(U^N b_0\oplus {U^*}^{-N} b_0)   | N \in {\bf Z}\}
\\
 L_{-}
& := &
 span\{\widehat{U}^N(b_0\oplus -b_0)   | N \in {\bf Z}\}
 \equiv span\{(U^N b_0\oplus -{U^*}^{-N} b_0)   | N \in {\bf Z}\}
\\
\end{eqnarray*}
\par\addvspace{2\medskipamount}\par\noindent
 {\bf Theorem 2.1.}
\par\addvspace{\medskipamount}\par\noindent
\begin{eqnarray*}
(1)
&&
  \widehat{U}^{\pm 1}L_{+} = L_{+}
\\
(2)
&&
  \widehat{U}^{\pm 1}L_{-} = L_{-}
\\
(3)
&&
b_N \oplus 0 \in L_{+} + L_{-}   \quad ( N \in {\bf Z} )
\\
&&
0 \oplus b_N \in L_{+} + L_{-}  \quad ( N \in {\bf Z} )
\\
(4)
&&
  \mbox{if } \{ L_{+} + L_{-}, x \}=0 \mbox{ then } x = 0
\\
(5)
&&
  \overline{ L_{+} + L_{-} }= \widehat{ H  }
\\
(6)
&&
  \{ L_{+} , L_{-}\}=\{0\}
\\
(7)
&&
 \{x,x\} > 0 \quad ( x \in L_{+}\backslash \{0\} )
\\
(8)
&&
 \{x,x\} < 0 \quad ( x \in L_{-}\backslash \{0\} )
\\
(9)
&&
  L_{+} \cap  L_{-} = \{0\}
\\
\end{eqnarray*}
 Moreover,
\begin{eqnarray*}
(6')
&&
 \{ \overline{L_{+}} , \overline{L_{-}}\}=\{0\}
\\
(7')
&&
 \{x,x\} > 0 \quad ( x \in \overline{L_{+}}\backslash \{0\} )
\\
(8')
&&
 \{x,x\} < 0 \quad ( x \in \overline{L_{-}}\backslash \{0\} )
\\
(9')
&&
 \overline{L_{+}} \cap  \overline{L_{-}} = \{0\}
\\
\end{eqnarray*}
 If in addition $U$ and $U^{-1}$ are bounded, then
\begin{eqnarray*}
(1')
&&
  \widehat{U}^{\pm 1}\overline{L_{+}} = \overline{L_{+}}
\\
(2')
&&
  \widehat{U}^{\pm 1}\overline{L_{-}} = \overline{L_{-}}
\end{eqnarray*}
\par\addvspace{2\medskipamount}\par\noindent
{\bf Proof} is straightforward.

\par\addvspace{2\medskipamount}\par\noindent
{\bf Definition 2.2.}
\begin{eqnarray*}
 L_{g\oplus h}
& = &
span \{ U^N g \oplus {U^*}^{-N} h  |
\quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
 \}
\end{eqnarray*}
\\
$\Box$

\par\addvspace{2\medskipamount}\par\noindent
{\bf Definition 2.3.}
\par\addvspace{\medskipamount}\par\noindent
Given $f \in H $ we write
$$
  f(n) := (b_n, f) \,;
$$
 Conversely, given $\{f(n)\}_n \in l_2(\bf Z)$ we put
$$
     f := \sum_{n\in {\bf Z}} b_n(b_n, f) \in H  \,;
$$
\\
$\Box$

\par\addvspace{2\medskipamount}\par\noindent
{\bf Theorem 2.2.}
\par\addvspace{\medskipamount}\par\noindent
If
$$
\frac{\overline u_{-N}}{\overline u_0} f_{0}(-N)
       = \overline{f_{0}(N)}\frac{u_{N}}{u_0}
$$
then $$ \{ L_{b_0 \oplus f_0} , L_{b_0 \oplus -f_0} \} =  \{0\} $$
\par\addvspace{\medskipamount}\par\noindent
{\bf Proof.}
\\
\begin{eqnarray*}
\{b_0\oplus -f_{0} , \widehat{ U }^N ( b_0\oplus f_{0} ) \}
& = & \{b_0\oplus -f_{0}, U^N b_0\oplus {U^*}^{-N} f_{0} \}
\\
& = & (b_0,{U^*}^{-N} f_{0} ) - (f_{0} , U^N b_0 )
\\
& = & ({U}^{-N} b_0, f_{0} ) - (f_{0} , U^N b_0 )
\\
& = & \frac{\overline u_{-N}}{\overline u_0} f_{0}(-N)
       - \overline{f_{0}(N)}\frac{u_{N}}{u_0}
\\
& = &  0
\quad ,
\\\quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\\
$\Box$

\par\addvspace{2\medskipamount}\par\noindent
 {\bf Theorem 2.3.}
\par\addvspace{\medskipamount}\par\noindent
 Suppose
\begin{eqnarray*}
 \{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} , g_1\oplus h_1  \}
& = &
 \{0\}
\end{eqnarray*}
Then

\begin{eqnarray*}
h_1 & = &
 0
\end{eqnarray*}
and
\begin{eqnarray*}
( f_0 , {U}^{N} g_1 )
& = &
 0
 \,\quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}

\par\addvspace{2\medskipamount}\par\noindent
{\bf Proof}
\begin{eqnarray*}
\{ \widehat{ U }^N ( b_0\oplus \pm f_{0} ) , g_1\oplus h_1  \}
& = &
 0
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  Then
\begin{eqnarray*}
\{ \widehat{ U }^N ( b_0\oplus 0 ) , g_1\oplus h_1 \}
& = &
 0
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  and
\begin{eqnarray*}
\{ \widehat{ U }^N ( 0 \oplus f_0 ) , g_1\oplus h_1 \}
& = &
 0
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 Now obtain
\begin{eqnarray*}
 0
& = &
\{ \widehat{ U }^N ( b_0\oplus 0 ) , g_1\oplus h_1 \}
\\
& = &
\{  ( { U }^N b_0\oplus 0 ) , g_1\oplus h_1 \}
\\
& = &
( { U }^N b_0 ,  h_1 )
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
 0
& = &
\{ \widehat{ U }^N ( b_0\oplus 0 ) , g_1\oplus h_1 \}
=
\{  ( { U }^N b_0\oplus 0 ) , g_1\oplus h_1 \}
=
( { U }^N b_0 ,  h_1 )
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 Hence $h_1 = 0$.
Now
\begin{eqnarray*}
 0
& = &
\{ \widehat{ U }^N ( 0 \oplus f_0 ) , g_1\oplus h_1 \}
\\
& = &
\{  ( 0 \oplus {U^*}^{-N}f_0 ) , g_1\oplus h_1 \}
\\
& = &
({U^*}^{-N}f_0  ,  g_1 )
\\
& = &
(f_0  , {U}^{-N} g_1 )
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
Hence
\begin{eqnarray*}
 0
& = &
(f_0  , {U}^{N} g_1 )
\\&& \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\\
$\Box$

\par\addvspace{2\medskipamount}\par\noindent
{\bf Corollary 2.1.}
\par\addvspace{\medskipamount}\par\noindent
Let $U^{\pm 1}$ be bounded, let $f_0 \not= 0$
and suppose that either
$\overline{span\{U^N g_1| N \in {\bf Z} \}} = \{0\}$
or
$\overline{span\{U^N g_1| N \in {\bf Z} \}} = H$
\\
Then
\begin{eqnarray*}
 \{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} , g_1\oplus h_1  \}
& = &
 \{0\}
\end{eqnarray*}
implies
\begin{eqnarray*}
g_1\oplus h_1 & = &
0 \oplus 0
\end{eqnarray*}
and as a result
\begin{eqnarray*}
 \overline{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} }
& = &
 H\oplus H  \,,
\end{eqnarray*}
\begin{eqnarray*}
  \overline{L_{b_0 \oplus f_0}} \cap  \overline{L_{b_0 \oplus -f_0}}
& = &
 \{0\} \,.
\end{eqnarray*}
$\Box$

\par\addvspace{2\medskipamount}\par\noindent
{\bf Remark 2.2}
\par\addvspace{\medskipamount}\par\noindent
If $U, U^{-1}$ are bounded then
$$
U\overline{span\{U^N g_1| N \in {\bf Z} \}}
=\overline{span\{U^N g_1| N \in {\bf Z} \}}
$$

\par\addvspace{2\medskipamount}\par\noindent
{\bf Definition 2.4.}
\par\addvspace{\medskipamount}\par\noindent
We say that $U$ has the {\bf Property A}, iff
from
$f_0\not =0 $
and
\begin{eqnarray*}
 \{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} , g_1\oplus h_1  \}
 & = &
 \{0\}
\end{eqnarray*}
one can deduce that
\begin{eqnarray*}
g_1\oplus h_1 & = &
0 \oplus 0
\end{eqnarray*}
\\
$\Box$

\par\addvspace{2\medskipamount}\par\noindent

\newpage

\par\addvspace{2\medskipamount}\par\noindent
\subsection*{3. Models of Orbits. Existence and Properties. }
\par\addvspace{2\medskipamount}\par\noindent
\par\noindent
 {\bf Definition 3.1.}
 Let $H_1, \{, \}_1$ and  $H_0, \{,\}_0$ be indefinite inner product spaces,
 $U_1$ and $U_0$ be respectively  $\{, \}_1$- and $\{, \}_0$-unitary operators
 (bounded or not), and let $x_1 \in H_1$,  $x_0 \in H_0$

 We say that the {\bf Z}-orbit
\begin{eqnarray*}
&&
 \{U_1^N x_1 |  N \in {\bf Z}\}
\end{eqnarray*}
 is a model of the orbit
\begin{eqnarray*}
&&
 \{U_0^N x_0 |  N \in {\bf Z}\}
\end{eqnarray*}
 iff
\begin{eqnarray*}
 \{ x_1 , U_1^N x_1\}_1
& = &
 \{ x_0 , U_0^N x_0\}_0  \,, \quad   N \in {\bf Z}
\end{eqnarray*}
\\
In this case we write
$$
 span \{U_1^N x_1 |  N \in {\bf Z}\} \approx_{\{,\}_1,\{,\}_0}
 span \{U_0^N x_0 |  N \in {\bf Z}\}
$$
 and/or simply
$$
  \{U_1^N x_1 |  N \in {\bf Z}\} \approx_{\{,\}_1, \{,\}_0}  \{U_0^N x_0 |  N \in {\bf Z}\}
$$
$\Box$
\par\addvspace{2\medskipamount}\par\noindent
 Throughout this section
\begin{eqnarray*}
 H_1
& := &
 H\oplus H
\end{eqnarray*}
\begin{eqnarray*}
 \{ x\oplus y , x_1\oplus y_1   \}_1
& := &
 \{ x\oplus y , x_1\oplus y_1   \}
\\
& := &
 (x, y_1) + (y, x_1) ;
\end{eqnarray*}
 where Hilbert space $H, (,)$ is the same as in the previous sections
 and let
\begin{eqnarray*}
 U_1
& := &
 U\oplus {U^*}^{-1}
\end{eqnarray*}
 for a suitable bilateral shift operator $U$ (bounded or not).
\par\addvspace{2\medskipamount}\par\noindent
 In addition suppose (mostly thechnical suppositions) that
$u_n$ is choosen so that
\begin{eqnarray*}
 \overline u_{-n}
& = &
 u_n \,, \qquad ( n \in {\bf Z} )
\end{eqnarray*}
 and
\begin{eqnarray*}
\Bigl\{\frac{1}{u_N}{ \{x_0, U_0^N\ x_0 \}_0 }\Bigr\}_{N \in {\bf Z}}
& \in  &
 l_2({\bf Z})
\end{eqnarray*}

\par\addvspace{2\medskipamount}\par\noindent
 First of all we try
\begin{eqnarray*}
 x_1
& := &
 b_0 \oplus f_0
\end{eqnarray*}
 for a suitable $f_0$
\par\noindent
\par\noindent
\par\noindent
\par\noindent
\par\noindent
 So, we consider the next equation, where $f_0$ is to be found.
\begin{eqnarray*}
\{b_0\oplus f_{0}, \widehat{ U }^N ( b_0\oplus f_{0} ) \}
=\{b_0\oplus f_{0}, U^N b_0\oplus {U^*}^{-N} f_{0} \}
& = &
\{x_0, U_0^N\ x_0 \}_0
\\
&&\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 Now we obtain
\begin{eqnarray*}
(b_0, {U^*}^{-N} f_{0}) + (f_{0}, {U}^{N} b_0)
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
( {U}^{-N} b_0, f_{0}) + (f_{0}, {U}^{N} b_0)
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
(\frac{u_{-N}}{u_0}b_{-N}, f_{0}) + (f_{0},\frac{u_N}{u_0} b_N)
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
\frac{\overline u_{-N}}{\overline  u_0}(b_{-N}, f_{0})
+ \frac{\overline u_N}{\overline u_0} (f_{0}, b_N)
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
\frac{\overline u_{-N}}{\overline  u_0}f_{0}(-N)
+ \frac{\overline u_N}{\overline u_0} \overline{f_{0}(N)}
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 Note
\begin{eqnarray*}
\{x_0, U_0^N\ x_0 \}_0
& = &
\overline{\{x_0, U_0^{-N}\ x_0 \}_0}
  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 Recall,
\begin{eqnarray*}
  u_N
& = &
\overline{u_{-N}}
  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  so that
\begin{eqnarray*}
 ( f_{0}(-N)
+ \overline{f_0(N)}) \frac{u_N}{u_0}
& = &
\{x_0, U_0^N\ x_0 \}_0  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  The formal solution is
\begin{eqnarray*}
  f_{0}(N)
& = &
\frac{1}{2}\frac{\overline u_0}{\overline u_N}\overline { \{x_0, U_0^N\ x_0 \}_0 }
+f_{00}(N)
\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 where
\begin{eqnarray*}
  \overline f_{00}(-N)
& = &
-f_{00}(N)
\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
Fix $f_{00} := 0 $ so that
\begin{eqnarray*}
  f_{0}(N)
& = &
\overline{f_{0}(-N)}
  \quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
In this case we have
\begin{eqnarray*}
  \{f_{0}(N)\}_{N \in {\bf Z}}
& \in &
 l_2({\bf Z })
\end{eqnarray*}
 and hence we really have obtained a solution.
\par\addvspace{2\medskipamount}\par\noindent
Thus we have obtained
\par\addvspace{2\medskipamount}\par\noindent
{\bf Theorem 3.1.}
\par\addvspace{\medskipamount}\par\noindent
 Choose
$u_n$ so that
\begin{eqnarray*}
 \overline u_{-n}
& = &
 u_n \,, \qquad ( n \in {\bf Z} )
\end{eqnarray*}
 and
\begin{eqnarray*}
\Bigl\{\frac{1}{u_N}{ \{x_0, U_0^N\ x_0 \}_0 }\Bigr\}_{N \in {\bf Z}}
& \in  &
 l_2({\bf Z})
\end{eqnarray*}
(it is always possible) and put
\begin{eqnarray*}
  f_{0}(N)
& = &
\frac{1}{2}\frac{\overline u_0}{\overline u_N}\overline { \{x_0, U_0^N\ x_0 \}_0 }
\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 In this case
\begin{eqnarray*}
  \{f_{0}(N)\}_{N \in {\bf Z}}
& \in &
 l_2({\bf Z })
\end{eqnarray*}
 and if one defines
$$
     x_1 := b_0 \oplus f_0
$$
  then
\begin{eqnarray*}
 \{ x_1 , U_1^N x_1\}_1
& = &
 \{ x_0 , U_0^N x_0\}_0  \,, \quad   N \in {\bf Z}
\end{eqnarray*}
\\
$\Box$
\par\addvspace{4\medskipamount}\par\noindent
  Let us discuss some properties of the obtained solution
\par\addvspace{4\medskipamount}\par\noindent
Recall the Definition 2.2:
\begin{eqnarray*}
 L_{g\oplus h}
& = &
span \{ U^N g \oplus {U^*}^{-N} h  |
\quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
 \}
\end{eqnarray*}
 and recall that by the {\bf Theorem 2.2} we have
\begin{eqnarray*}
 \{ L_{b_0 \oplus f_0} , L_{b_0 \oplus -f_0} \}
& = &
 \{0\}
\end{eqnarray*}
\par\addvspace{2\medskipamount}\par\noindent
 {\bf Theorem 3.2.}
\par\addvspace{\medskipamount}\par\noindent
 Suppose
 $H_0$ is a complet normed space with the norm $||\cdot ||_0$
such that $|\{x,y\}_0| \leq ||x||_0||y||_0 \,, \quad (x,y \in H_0)$
and $u_n$ is choosen
so that
$$ \sum_{n\in {\bf Z}} |\frac{u_0}{u_n}|^2 ||U_0^{n}\ x_0||_0^2 < \infty $$
and so that
$U, U^{-1}$ are bounded
(it is always possible if $U_0, U_0^{-1}$ are bounded)
\quad   If in addition
\begin{eqnarray*}
 \{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} , g_1\oplus h_1  \}
& = &
 \{0\}
\end{eqnarray*}
 then
$$
 \widetilde {g_1} := \sum_{n\in {\bf Z}} \frac{u_0}{u_n} g_1(n) U_0^{n} x_0
  \in \overline {span\{U_0^{n} x_0 | n\in {\bf Z} \}}
  \subset H_0
$$
  and
\begin{eqnarray*}
 \{U_0^{N} x_0, \widetilde{g_1} \}_0
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
 \{U_0^{N} \widetilde{g_1}, \widetilde{g_1} \}_0
& = &
 0
\\ \quad N  = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 In particular,
$\overline {span \{U_0^{N} \widetilde{g_1}| n \in {\bf Z}  \} }$
is a $||\cdot||_0$ closed $U_0^{\pm 1}$-invariant $\{,\}_0$-neutral
subspace of $\overline {span \{U_0^{N} x_0 | n \in {\bf Z}  \} } \subset H_0$
(may be $\{0\}$).
\par\noindent
\par\addvspace{\medskipamount}\par\noindent
 {\bf Proof }
\\
By the Theorem 2.3,
\begin{eqnarray*}
h_1 & = &
 0
\end{eqnarray*}
\begin{eqnarray*}
( f_0 , {U}^{N} g_1 )
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
 On the other hand
\begin{eqnarray*}
 g_1
& = &
\sum_{n\in {\bf Z}} g_1(n) b_n
\end{eqnarray*}
 Hence
\begin{eqnarray*}
 U^{-N}g_1
& = &
 \sum_{n\in {\bf Z}} \frac{u_{n-N}}{u_n} g_1(n)b_{n-N}
\\
&&
 \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
( f_0 , \sum_{n\in {\bf Z}} \frac{u_{n-N}}{u_n} g_1(n)b_{n-N} )
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
  \sum_{n\in {\bf Z}} \frac{u_{n-N}}{u_n}  g_1(n) \overline{f_0(n-N)}
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
Recall,
\begin{eqnarray*}
  f_{0}(N) = (b_N , f_0)
& = &
\frac{1}{2}\frac{\overline u_0}{\overline u_N}\overline { \{x_0, U_0^N\ x_0 \}_0 }
\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
Hence
\begin{eqnarray*}
  f_{0}(n-N) = (b_{n-N} , f_0)
& = &
\frac{1}{2}\frac{\overline u_0}{\overline u_{n-N}}
\overline { \{x_0, U_0^{n-N}\ x_0 \}_0 }
\quad , \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  Hence
\begin{eqnarray*}
\sum_{n\in {\bf Z}}
\frac{u_{n-N}}{u_n} g_1(n)
\frac{1}{2}\frac{ u_0 }{ u_{n-N} } \{x_0, U_0^{n-N}\ x_0 \}_0
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
\sum_{n\in {\bf Z}}
\frac{u_0}{u_n} g_1(n)
\frac{1}{2}  \{x_0, U_0^{n-N}\ x_0 \}_0
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
\sum_{n\in {\bf Z}}
\frac{u_0}{u_n} g_1(n)
\frac{1}{2}  \{U_0^{N} x_0,  U_0^{n}\ x_0 \}_0
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
 \{U_0^{N} x_0, \widetilde{g_1} \}_0
& = &
 0
\\ \quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
 \{U_0^{N}U_0^{n} x_0, \widetilde{g_1} \}_0
& = &
 0
\\ \quad N,n  = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
\begin{eqnarray*}
 \{U_0^{N} \widetilde{g_1}, \widetilde{g_1} \}_0
& = &
 0
\\ \quad N  = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
  Quod erat demonstrandum.
\\
$\Box$
\par\addvspace{2\medskipamount}\par\noindent

\par\addvspace{2\medskipamount}\par\noindent
 {\bf Theorem 3.3.}
\par\addvspace{\medskipamount}\par\noindent
 Suppose
 $H_0$ is a complet normed space with the norm $||\cdot ||_0$
such that $|\{x,y\}_0| \leq ||x||_0||y||_0 \,, \quad (x,y \in H_0)$
and $u_n$ is choosen
so that
$$P(\lambda)=\sum_{n\in {\bf Z}} \frac{u_0}{u_n} \lambda^{n}$$
as function of complex $\lambda$ is analytic on the spectrum of
$U_0$
(it is always possible if $U_0, U_0^{-1}$ are bounded)
\par\addvspace{\medskipamount}\par\noindent
Suppose
 $U_0 |\overline {span \{U_0^{N} \widetilde{g_1}| n \in {\bf Z}  \} }$
 (the restriction of $U_0$ onto
        $\overline {span \{U_0^{N} \widetilde{g_1}| n \in {\bf Z}  \} }$)
has no non-zero neutral invariant subspace.

\par\addvspace{\medskipamount}\par\noindent
Then
$span\{U_0^N x_0| N \in{\bf Z}\}$
is finite dimensional
or
$U$ has the Property A.
\par\addvspace{2\medskipamount}\par\noindent
 {\bf Proof.}
\par\addvspace{\medskipamount}\par\noindent
Let
\begin{eqnarray*}
 \{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} , g_1\oplus h_1  \}
& = &
 \{0\}  \quad\quad (*)
\end{eqnarray*}
 Then
$$
   \{g_1(n)\}_{n\in {bf Z}} \in l_2({\bf Z})
$$
 Then
$$P_1(\lambda)=\sum_{n\in {\bf Z}} g_1(n)\frac{u_0}{u_n} \lambda^{n}$$
as function of complex $\lambda$ is analytic on the spectrum of
$U_0$
\\
Then
$$P_1(U)=\sum_{n\in {\bf Z}} g_1(n)\frac{u_0}{u_n} U_0^{n}$$
 exists.
\\
Then
$$P_1(U)x_0=\sum_{n\in {\bf Z}} g_1(n)\frac{u_0}{u_n} U_0^{n}x_0$$
 exists.
\\
Then
$$ \sum_{n\in {\bf Z}} |\frac{u_0}{u_n}|^2 ||U_0^{n}\ x_0||_0^2 < \infty $$
Then
$$P_1(U)x_0=\widetilde {g_1}$$
\\
Supposition about neutral invariant subspaces implies $\widetilde {g_1}=0$
Thus $(*)$ implies
$$ P_1(U)x_0=0 $$
Suppose $(*)$ implies that $P_1(\lambda)$ is identically zero for any $g_1$.
In this case $(*)$ implies that $g_1=0$ for any $g_1$
and hence $U$ has the Property A.
Now suppose that there exists $g_1$ such that $P_1(\lambda)$ is not
identically zero. In this case, if for a $\lambda_0$ we have
$P_1(\lambda_0)=0$, then there exists a natural $n_0$ such that
$(U_0 -\lambda_0 I)^{n_0}x_0 =0$. Hence
$span\{U_0^N x_0| N \in{\bf Z}\}$
is finite dimensional.
Finally, if $P_1(\lambda)\not=0$ nowhere, then $P_1(U_0)$ is injective,
hence $x_0=0$ and $span\{U_0^N x_0| N \in{\bf Z}\}$
is again finite dimensional.
\\
$\Box$
\par\addvspace{2\medskipamount}\par\noindent
Thus, we have constructed the following chain of restrictions-extensions:
$$
 H_0 \supset L_{x_0} \approx_{\{,\}_0, \{,\}} L_{b_0 \oplus f_0} \subset
 L_{b_0 \oplus f_0} \dot{+}_{\{,\}} L_{b_0 \oplus -f_0}
 \subset_{dense} \overline{ L_{b_0 \oplus f_0} \dot{+}_{\{,\}} L_{b_0 \oplus -f_0}}
$$
$$
= \overline{L_{+} \dot{+}_{\{,\}} L_{-}}  \qquad (chain \; A )
$$
Here and in the following $A\dot{+}B$ stands for $A+B$ with indicating
$A\cap B =\{0\}$
and  $A{+}_{\{,\}}B$ stands for $A+B$ with indicating $\{A,B\} =\{0\}$.
\par\addvspace{2\medskipamount}\par\noindent
\subsection*{4. Operator $U_0 \oplus U_0$ }
\par\addvspace{2\medskipamount}\par\noindent
In this subsection we suppose that all ${U_0}^N x_0 \quad (N \in {\bf Z})$
are linearly independent and the sequence $\{u_n\}_n$ is choosen so that
$U$ has the Property A.
\par\addvspace{2\medskipamount}\par\noindent
{\bf Observation 4.1.}
\begin{eqnarray*}
\{b_0\oplus -f_{0}, \widehat{ U }^N ( b_0\oplus -f_{0} ) \}
&=&\{b_0\oplus -f_{0}, U^N b_0\oplus -{U^*}^{-N} f_{0} \}
\\
& = &
-\{x_0, U_0^N\ x_0 \}_0  \quad ,
\\
&&\quad N = 0, \pm 1, \pm 2, \pm 3, \ldots
\end{eqnarray*}
$\Box$
\\
This motivates the following considerations.

\par\addvspace{2\medskipamount}\par\noindent
{\bf Definition 4.1.}
\\
Put
\begin{eqnarray*}
 \widetilde H_0
& := &
 H_0\oplus H_0
\\
\|x \oplus y \|_{\sim}^2
& := &
\|x\|_{0}^2 + \|y \|_{0}^2 \, \quad ( x,y \in H_0 )
\\
 \widetilde U_0
& := &
 U_0\oplus U_0
\\
 \{ x\oplus y , x_1\oplus y_1     \}_{-}
& := &
 \{ x , x_1 \}_0 - \{ y , y_1 \}_0
\\
 L_{x_0}
& := &
 span\{ U^n x_0 | n \in {\bf Z} \}
\\
\overline{ L_{x_0} } &:=& \mbox{ closure of } L_{x_0}
                      \mbox{ with respect to } \|\cdot\|_0
\\
\widetilde L_{x_0}
& := &
L_{x_0}\oplus L_{x_0}
\\
\end{eqnarray*}
and define a linear map
\begin{eqnarray*}
 \Omega
& : &
 L_{x_0} \oplus L_{x_0} \subset \widetilde H_0
  \to
 L_{b_0\oplus f_0} + L_{b_0\oplus -f_0} \subset \widehat H
\\
\end{eqnarray*}
 as follows:
Firstly put
\begin{eqnarray*}
 \Omega ( \, U_0^N x_0 \oplus U_0^M x_0 \,)
& :=  &
 \widehat U^N (b_0\oplus f_0) + \widehat U^M  (b_0\oplus -f_0 )
\\
&&
   N,M \in {\bf Z}
\end{eqnarray*}
Then extend this map onto the span of $U_0^N x_0 \oplus U_0^M x_0$ \quad
$(N,M \in {\bf Z})$ as a linear map.
\par\addvspace{2\medskipamount}\par\noindent
{\bf Observation 4.2.}
\begin{eqnarray*}
 \{
 \Omega ( \, x\oplus y,) ,
 \Omega ( \, x_1\oplus y_1 \,)
 \}
 & =  &
 \{
   x\oplus y, ,
   x_1\oplus y_1
 \}_{-}
\\
&&
   x, y, x_1, y_1 \in L_{x_0}
\end{eqnarray*}
\\
Thus, if we take into account only indefinite inner products, we may say that
$L_{x_0}\oplus L_{x_0}$ is embedded in
$L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}$,
we will write it symbolically as
$$
L_{x_0}\oplus L_{x_0} \subset_{\Omega, \{,\}_0 , \{,\} }
L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}
$$
Recall, $U$ has the Property. As a consequece,
\\
$$ \overline{ L_{b_0 \oplus f_0} + L_{b_0 \oplus -f_0} } = \widehat H \, \quad (a) $$
and
$$  L_{b_0 \oplus f_0} \cap L_{b_0 \oplus -f_0} = \{0\}\,; \quad (b)$$
As a result,
\begin{eqnarray*}
 \Omega
& : &
L_{x_0} \oplus L_{x_0} \subset \widetilde H_0
  \to
 L_{b_0\oplus f_0} + L_{b_0\oplus -f_0} \subset \widehat H
\\
\end{eqnarray*}
is  injective (note, surjective by the definition). Hence,
\begin{eqnarray*}
        \Omega^{-1}
& : &
 L_{b_0\oplus f_0} + L_{b_0\oplus -f_0} \subset \widehat H
  \to
 L_{x_0} \oplus L_{x_0} \subset \widetilde H_0
\\
\end{eqnarray*}
exists, and we may identify
$L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}$
and
$L_{x_0} \oplus L_{x_0}$
as indefinite inner product lineals, write it symbolically:
$$
L_{x_0}\oplus L_{x_0} \approx_{\Omega, \{,\}_0 , \{,\} }
L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}
$$
Thus, if we take into account the relation $(b)$, we may write finally:
$$
L_{x_0}\oplus L_{x_0} \approx_{\Omega, \{,\}_0 , \{,\} }
L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}
\qquad\qquad\qquad\qquad
$$
$$
\qquad\qquad
\subset
\overline{L_{b_0\oplus f_0} + L_{b_0\oplus -f_0}}
=\widehat{H}
=\overline{L_{+} \dot{+}_{\{,\}} L_{-}}
\qquad (chain \; B) ;
$$

\par\noindent
$\Box$
\par\addvspace{2\medskipamount}\par\noindent
As a result we have obtained
\par\addvspace{2\medskipamount}\par\noindent
{\bf Theorem 4.1.}
\par\addvspace{\medskipamount}\par\noindent
Suppose $H_0$ is a Hilbert space.
Suppose
$U_0$ is bounded, has no nonzero $\{,\}_0$ invariant subspace, and $x_0$ is
a {\bf Z}-cyclic vector of $U_0$, i.e.,
$$
H_0=\overline{ L_{x_0} }
$$
Then
\par\addvspace{2\medskipamount}\par\noindent
$(a)$
there exists a (pre)hilbert norm on $L_{x_0}\oplus L_{x_0}$ such that
$\{,\}_{-}$ and $U_0\oplus U_0$ are continuous and
can be extended to continuous $\{,\}_{ext}$ and $U_{ext}$
onto $H_{ext} = \mbox{ the completion of } L_{x_0}\oplus L_{x_0}$ ;
\\

\par\addvspace{\medskipamount}\par\noindent
$(b)$ there are two $U_{ext}^{\pm 1}$-invarint (closed) subspaces
${\cal L}_{+},{\cal L}_{-} \subset \widetilde{ H_0 }$ such that:
\par\addvspace{\medskipamount}\par\noindent
(1) ${\cal L}_{+}$ is strictly positive;
\par\addvspace{2\medskipamount}\par\noindent
(2) ${\cal L}_{-}$ is strictly negative;
\par\addvspace{2\medskipamount}\par\noindent
(3) ${\cal L}_{+}\cap {\cal L}_{-} = \{0\}$ ,
\par\addvspace{2\medskipamount}\par\noindent
(4) ${\cal L}_{+}+{\cal L}_{-} $ is dense in $\widetilde{H_0}$,
\par\addvspace{2\medskipamount}\par\noindent
(5) $\{{\cal L}_{+},{\cal L}_{-}\}_{-}=\{0\} $.
\par\addvspace{2\medskipamount}\par\noindent
(6) as semidefined subspaces, ${\cal L}_{+}$, ${\cal L}_{-}$ both are maximal.
\par\addvspace{2\medskipamount}\par\noindent
$\Box$
\par\addvspace{2\medskipamount}\par\noindent
Finally,
\par\addvspace{\medskipamount}\par\noindent
{\bf Remark 4.1.}
\\
Note, that, in contrast with $(chain\; A)$ in the previous subsection,
in $(chain\; B)$ all embeddings are dense.
$\Box$
\par\addvspace{2\medskipamount}\par\noindent

\newpage
\bibliographystyle{unsrt}

redpajama_set_name: RedPajamaArXiv


text:
Benjamin Batterham: Supporters call for release of accused NSW murderer
11:48am Mar 29, 2016
Thousands of Australians have signed online petitions demanding the immediate release of a NSW man accused of murdering an intruder to his home at the weekend.
Ricky Slater-Dickson, 34, died after allegedly being detained by Benjamin Batterham, 33, and another man at a Newcastle house following a fight in the early hours of Saturday morning.
Batterham was charged with recklessly inflicting grievous bodily harm but the charge was upgraded to murder after Slater-Dickson's life support was switched off on Sunday.
Three petitions on change.org which call for Batterham's release have amassed almost 8000 supporters as of midday Tuesday.
Ricky Slater, 34, died after his life support was switched off on Sunday.
The incident has kicked off a charged nationwide debate between Slater-Dickson's family and self-defence advocates.
"If you creep into someone's residence uninvited you deserve to die, it's that simple, good people don't invade peoples living space," wrote Michael Kinvig in comments on the change.org petition. Mr Kinvig's staunch stance was typical of many signatories on the petition.
"As a home owner I should be able to protect what it rightfully mine and as a mother I believe I should be able to protect my children from these low life scum," added Lea-Anne Dabron, in the comment section.
Tiara Keely told 9NEWS her brother, Mr Slater-Dickson a father-of-three, was a "loving, good person" who was "shy, quiet [and] placid".
"He didn't deserve what happened to him and we're fighting for justice," she said.
Police were told Mr Slater-Dickson had broken into Mr Batterham's home before an altercation broke out.
Few details of the incident have emerged, but it is believed Mr Batterham and another man allegedly chased Mr Slater-Dickson out onto the footpath, where a struggle ensued.
Mr Slater-Dickson's family claim he had not committed a break-and-enter but was in fact attending a party at Mr Batterham's residence.
Legal expert Sam Macedone said a person has the right to do whatever they think is necessary to protect themselves – as long as the response is reasonable given their fear at the time of the invasion.
Mr Batterham chose not to appear in court or make an application for bail on Monday and his case is due to be mentioned at Newcastle Local Court today.
With AAP
Auto news: Old Datsun sells for $1.1 million, breaks auction records in Japan - caradvice.com.au

redpajama_set_name: RedPajamaCommonCrawl


text:
Morrisons pay to break £10 an hour barrier for King's Lynn staff
By Greg Plummer
greg.plummer@iliffepublishing.co.uk
Morrisons has announced a new pay deal for its store colleagues, becoming the first UK supermarket group to guarantee pay of at least £10 an hour.
The new deal will start in April and will mean a significant pay increase for nearly 96,000 Morrisons colleagues.
For the majority of colleagues the pay increase is approximately nine per cent.
The Morrisons petrol station in King's Lynn.. (43943680)
The store's minimum hourly pay currently stands at £9.20 an hour.
Today's announcement means that since 2015, Morrisons pay has increased by more than 46 per cent.
Three-quarters of the costs of the increase will be met by direct payroll investment and a quarter by changing the discretionary annual colleague bonus scheme.
Having listened hard to colleagues, they said they would prefer to have a guaranteed amount in their hourly rate and receive it more regularly.
So, the annual budget for the discretionary bonus scheme is being rolled into colleagues' hourly rate, which is paid every four weeks.
David Potts, Morrisons CEO, said: "It's great to be able to say that in the UK from April this year, if you work at Morrisons supermarkets, you will earn at least £10 an hour.
"It's a symbolic and important milestone that represents another step in rewarding the incredibly important work that our colleagues do up and down the country.
"Over the last year we have seen renewed and widespread appreciation in the UK for our colleagues who have had an incredibly tough 2020, working tirelessly so Britain could eat well and shop safely throughout the pandemic.
"Morrisons colleagues have earned their status as key workers, and this pay increase, many times over."
Joanne McGuinness, Usdaw national officer, said: "The new consolidated hourly rate is now the leading rate of the major supermarkets, which is paid every hour and removes the uncertainty of a bonus payment.
"It is a big step forward, shows that Morrisons is prepared to invest in the staff to help grow the business and I hope that the deal is supported by our members in the ballot.
"It's been a tough time for food retail staff who have worked throughout the pandemic in difficult circumstances.
"They provide the essential service of keeping the nation fed and deserve our support, respect and appreciation. Most of all they deserve decent pay and this offer is a welcome boost."
The pay deal announced today is subject to a ballot opening on January 27.
In addition to the hourly pay increase, Morrisons will pay a London weighting. Rates for inner London will be 85p and for outer London 60p per hour.
Business Greg Plummer


The decorative plate PULCHRA SICUT LUNA (SERIES ASTRUM) shows an exact reproduction of the 16th century pianella, which can be admired in the church of San Donato (16th century) in Castelli (Abruzzo - Italy). In 1963 Carlo Levi described the church as "the Sistine Chapel of Italian majolica".
The decorative motif shows a double moon: a quarter of the moon inside a full moon surrounded by three stars in a dark blue rim. Moreover, the plate bears in the back the old Latin motto Pulchra Sicut Luna (i.e., beautiful as the moon) describing the virtues of the Virgin Mary.
A hook to hang the plate on the wall is included.
Curiosity: The great artistic value of these handmade products has been renowned in all the European continent since the 16th century. From this time on, the great request for items by the main European courts – makes the Castelli majolicas as the most popular and desired. Nowadays we can admire these antique pottery in the most important art museums around the world as the Hermitage Museum of Saint Petersburg, the Louvre Museum of Paris, the British Museum of London, the Castello Sforzesco Museum of Milan. Even today, the old Castelli ceramic tradition distinguishes itself because of the unaltered continuation of these secular and original production processes. A certificate guarantees the authenticity of every single artistic creation. Moreover, every item has an official seal – or the country of origin brand – impressed which assures the object is realized only and exclusively in the Municipality of Castelli.
GUARANTEE: This polished and high-valued item is handcrafted and hand painted in the old village of Castelli (Abruzzo – Italy). This artistic creation is entirely made according to the secular ceramic tradition. The back of the object bears the official seal – or mark of origin – ensuring the Customer that the product was manufactured only and exclusively in the municipality of Castelli. A certificate guarantees every creation attesting its authenticity.
Note: For decorative purposes only. Any slight imperfections in shape and colour should not be considered as a defect but forms part of the beauty and authenticity of the hand-crafted product. The images have the sole purpose of illustrating the product features. Actual colours may vary from the colour on the screen due to different types of lighting and with different monitor settings.


text:
Home » Lifestyle » San Francisco Boy Is Cancer-Free 5 Years After Saving City as 'Batkid'
San Francisco Boy Is Cancer-Free 5 Years After Saving City as 'Batkid'
11/16/2018 Lifestyle Comments Off on San Francisco Boy Is Cancer-Free 5 Years After Saving City as 'Batkid'
When Miles Scott was 5 years old and fighting Leukemia, all he wanted was to become Batman.
Back in November 2013, Make-A-Wish — known for granting "wishes" to children struggling from serious and life-threatening illnesses — worked with San Francisco's late mayor Ed Lee and the local police and fire departments (among many other groups) to transform the city into Batman's beloved city of Gotham.
Over 20,000 people flooded the streets — some holding posters of encouragement, others sporting Batman masks — to support Miles a.k.a. "Batkid" as he not only fought crime and villains, but also his serious illness.
Now, five years later, the boy, who captured the hearts of many, is cancer-free, Make-A-Wish confirmed.
"Miles has returned to being a typical kid—playing little league, going to school, helping his family farm, and even selling his first market goat in the local fair!" the non-profit organization said in a statement. "Now 10 years old and in fifth grade, Miles loves science and robotics."
Miles, who was diagnosed with leukemia when he was 1, still makes visits to an oncologist once a year, but fortunately has been cancer-free since his "Batday" in 2013.
RELATED: The Day SF Became Gotham City: How a Metropolis Is Making a 5-Year-Old's Wish Come True
In late 2013, Miles, wearing his Batsuit, paraded around San Francisco in a black Lamborgini — designed to resemble the Batmobile — and made stops at different scenes involving police and firefighters. He even had the opportunity to capture the Riddler.
To make the day more special, the San Francisco Chronicle published a special edition of the newspaper — the headline, "Batkid Saves City," was printed across the newly-titled Gotham City Chronical — and Miles was granted a key to the city.
The celebration caught the attention of many, even then-president Barack Obama and astronauts at the International Space Station, who sent words of encouragement.
Beyond a memorable day, his wish also became a medical milestone, as it marked the end of his treatments, Make-A-Wish said. His mom, Natalie, told the nonprofit at the time that it "meant closure for our family and an end to over three years of putting toxic drugs in our son's body."
RELATED: SNL's Vanessa Bayer on Her Teenage Battle with Cancer: I Was a Make-A-Wish Kid
And following his special day, Miles was convinced that he was Batkid.
"He just thought he was doing his job," Make-a-Wish director Jen Wilson, who put the citywide event together, told the San Francisco Chronical. "He took his work seriously. He thought Batkid might need to stick around."
Miles' special day was later turned into a documentary called Batkid Begins produced by Warner Bros. The film follows his leukemia journey from the time he was 18 months up until his Make-A-Wish day in 2013. It is available to rent or buy on Amazon Prime, Netflix, iTunes, Vudu, and Red Box.
RELATED VIDEO: 5-Year-Old With Terminal Brain Cancer Lives Out Dreams by Becoming Firefighter, Policeman and Pilot
The event also impacted more than just the Scott family.
In the days and years since "Batday," Make-A-Wish said they saw a major increase in donations, volunteers, referrals and other services. Their website also crashed for several hours on the day of Miles' wish because of the increase in traffic.
"It was an incredibly powerful boost to our organization," Wilson told the Chronical. "Batkid was responsible for that."
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Under the Madrid Protocol, areas of outstanding natural values, such as biodiversity, geology, or landscape, can be protected as Antarctic Specially Protected Areas (ASPAs). However, East Antarctica is the largest under-sampled region of continental shelf around Antarctica, despite more than 100 years of human activity in the region. As a result, secure protection through informed spatial management is not possible. But a strategy of adaptive management can be employed, using new information as it comes to hand, to strengthen and improve protective measures.
Most ASPAs in East Antarctica have been designated to protect biodiversity values, rather than non-living values such as geology or glaciology. However, our research focuses on the geological processes and characteristics that influence conservation values and vulnerability of Antarctic landscapes and geological features. It looks at landscape stability and vulnerability to disturbance – including rarity, fragility, sensitivity and ability to recover after disturbance. It also aims to develop methods to predict the distribution of rare or high value features, such as fossils or meteorites.
The research will provide the scientific foundation for spatial management and area protection in the East Antarctic, accounting for the special characteristics of Antarctic biodiversity and geodiversity, and the landscape and ecosystem processes that affect them.
How do spatial patterns and landscape influence vulnerability of the biodiversity of Antarctica and how must they be accommodated in spatial planning?
How do we select areas to include in the series of Antarctic Specially Protected Areas (ASPAs)?
How should the non-living values of Antarctica be included in spatial planning and area protection?
To answer these questions scientists are examining existing data using spatial analysis and modelling tools to identify the main scales of geology, biodiversity, habitat patchiness and connectedness. This will allow identification of the effects these factors have on the susceptibility and resilience of Antarctic ecosystems, and any gaps in the data that could be filled by future field surveys.
The research is also documenting landscape variability (through, for example, seabed mapping and satellite data of sea ice cover), which can be linked to biodiversity. Scientists are examining relationships between populations using taxonomy and molecular genetic techniques and developing models to explain the current distribution of the biodiversity of East Antarctica and inform predictions of how it might change.


The Bible hasn't been the source of inspiration for much of today's music. But Chris Lastovicka has drawn from it in his beautiful classical work, Fortune Has Turned. Drawing from The Book of Job specifically, his classical EP includes wonderful pieces written specifically for horns and piano. Chris has enlisted a powerhouse cast of musicians to create a wonderful, creative, sometimes downtrodden, but everlasting musical journey that is unlike anything heard today.
The musical arrangements on each track are complicated musical compositions, but if even if you are a classical music neophyte, it won't be difficult to understand that what you're hearing is musical wonderment. You'll come away appreciating the mastery in the music.
With these five, wondrous tracks ("The 7th Chapter of Job", "Abraxas", "The Tender Ones", "Shanti", "The End of Tyranny") you will hear the genius of Chris Lastovicka, and hear some of the most exquisite musical pieces recorded in this day and age.


If you do not wish for The Dude server to use your on-board flash memory you can specify data path in which server should keep all relevant data.
To change where the Dude servers stores it's stuff you need to use following command.
To move already existing dude data dir to new path.
2. Move existing data dir to new location via winbox or ftp/sftp.
Note: We do not suggest to use data stores that are formatted with FAT32 file-system, better before use re-format it as EXT3. For more details on how to do that consult [this ] wiki article.
This page was last edited on 23 February 2016, at 14:11.


text:
"The Greeks" Power Point
The Greeks
In 800-650 BC ,Greece consisted of Italy all the way to the edge of the
Black sea. This colonization affected Greek mathematics in that all early
Greek mathematics came from Asia minor, southern Italy, & Africa.
Coinage affected the Greeks - stimulated trade, gave rise to a
money economy based not only on agriculture but on movable
goods, created a leisure class.
Geography of Greece affected politics - no single ruler because
each state separated by mountains or sea. Alexander eventually
united Greece. (He died in 323 BC ruling over 2 million square miles
– gave rise to the Hellenistic age
Greek mathematics differ from the Babylonians and the Egyptians more profound, more abstract, and more rational. Notion of seeking
knowledge for its own sake.
Greek civilization begins in Iron Age, flourishing in the 4th and 5th
centuries BC. Most of the productive mathematics done in 350-200
Thales (625-547 BC)
Thales is credited with inventing the modern proof.
Thales is known as the first of the Seven Sages of Greece; coined the
maxim "Know thyself" and when asked what was the strangest thing he
had ever seen: "An aged tyrant."
Thales presumably got interested in geometry when he visited Egypt and
wanted to measure the pyramids by means of shadows. Thales also
determined how far a ship at sea is from the shore.
It is said that Thales predicted a solar eclipse but this would not have
been possible because astronomical records were not precise enough at
that time.
Pythagoras (580-500 BC) followed Thales. We know that he lived on the
Aegan island of Samos.
The distinctive feature of the school founded by Pythagoras was its aims
were political, philosophical, and religious.
The 300 or so students at the school studied the Quadrivium (number
theory, music, geometry, and astronomy).
What set the Pythagoreans apart from other sects was their belief that
knowledge is the greatest purification and to them knowledge meant
mathematics.
Mathematics played an essential part in life & religion. We saw last time
that music provided the Pythagoreans with the best instance of their
principle that "number was the cause of everything in nature."
Pythagoreans
Pythagoreans Celebrate Sunrise
by Fyodor Bronnikov
Nichomachus' Introductio Arithmeticae is the most complete
exposition of Pythagoras and his successors work.
The Pythagoreans did not have any symbols so it's likely that
they used pebbles to think of numbers geometrically – triangle
numbers, square numbers, etc.
The most important achievements of the Pythagorean
school in its influence of the number concept was the
discovery of the "irrational."
Commensurable
This idea caused great consternation among the Pythagoreans for it
challenged the adequacy of their philosophy that number was the
essence of all things. They called these numbers "unutterable.
Eventually Eudoxus (408-355BC) resolved this crisis. He revised
the theory of proportion applicable to incommensurable as well
commensurable quantities
Hippias(b. c. 460BC) is credited with discovering that the side of a
square and its diagonal are not commensurable.
i.e., no matter how small one goes, there is no magnitude dividing
evenly into both the square's side and its diagonal. This discovery
shattered all Pythagorean proofs using commensurability of all
(408-355
BC)
Eudoxus eventually gave proofs about similar triangles that did not rely
on this concept. Also it had an unsettling impact upon the
supremacy of whole numbers, for if not all quantities were
commensurable then, then whole numbers were somehow inadequate
to represent geometric lengths. This firmly established the supremacy
of geometry over arithmetic in all subsequent Greek mathematics.
For example, in a square of side length 1, with diagonal root 2, then root
2 arithmetic character is mysterious so better to think of magnitudes as
geometric entries!
Eudoxus is best known for 3 things: his general theory of proportion,
golden ratio, & method of exhaustion, which Archimedes refines.
Approximating
root 2
The Greeks were not the first to consider numerical values for root 2.
The Babylonians approximated the square root of 2.
Here the Babylonian clay tablet YBC 7289 (c. 1800–1600 BCE) gives an
approximation of in four sexagesimal figures, which is about six decimal
figures:1 + 24/60 + 51/(60)^2 + 10/(60)^3 = 1.41421296
Hippocrates(460-380 BC)
The mathematician that dominated the 2nd half of the 5th C BC was
Hippocrates began his life as a merchant but ended his life as a teacher
(like Thales) . He was robbed of his money (he was robbed by either pirates
or fraudulent customs officials) and went to Athens to prosecute the
offenders. He stayed and attended the lecturers of many philosophers. He
may have come in contact with some Pythagoreans.
The major accomplishment of Hippocrates is that he was the first to write a
systematically organized geometry textbook, called Stoicheia Elements,
that is, basic theorems, or building blocks of mathematical theory.
From then on, mathematicians from all over the ancient world could, at
least in principle, build on a common framework of basic concepts,
methods, and theorems, which stimulated the scientific progress of
Special Problems from Antiquity
Three problems that Hippocrates faced when he arrived in Athens:
Quadrature of the circle: Is it possible to construct a square whose
area is equal to that of a given circle? Equivalent to constructing a line
segment whose length is root pi times the radius of the circle.
Duplication of the cube: Find the edge of a cube, having twice the
volume of a given cube.
Trisection of an angle.
Add a 4th problem: Construction of a regular heptagon.
Plato insisted 3 things with these special problems: the task be
performed with straightedge and compass only.
With the straight edge, a line can be drawn through 2 points and
With the compass, a circle with a given center and radius can be
drawn.
Squaring a Circle
Surprisingly, the impossibility proofs for the three problems: doubling the
cube, trisecting an angle, and constructing a regular heptagon, all fall into
the same framework. Each depends on showing that roots of certain third
degree polynomial equations are not constructible.
The problem of squaring of a circle found itself in a completely different
category. Carl Louis Ferdinand von Lindemann (1852 – 1939) was a
German mathematician who was the first to prove that π is transcendental
in 1882. Since π is NOT algebraic, it is automatically not constructible.
The area of a circle with radius 1 is exactly π . If a square with such area
were constructible, so would be the number itself. The areas of this
square and this circle are equal. In 1882, it was proven that this figure
cannot be constructed in a finite number of steps with a compass and
straightedge.
Statue of
Euripides in
Louvre, Paris
Doubling the Cube
Doubling the Cube, the most famous of the collection, is often referred to
as the Delian problem due to a legend that the Delians had consulted
Plato on the subject.
In another form, the story asserts that the Athenians in 430 B.C.
consulted the oracle at Delos in the hope to stop the plague ravaging
their country. They were advised by Apollo to double his altar that had the
form of a cube. As a result of several failed attempts to satisfy the god,
the pestilence only worsened and at the end they turned to Plato for
According to a letter from the mathematician Eratosthenes(276-195 BC)
to King Ptolemy of Egypt, Euripides mentioned the Delian problem in
one of his (now lost) tragedies.
Euripides (480 BC – 406 BC) was the last of the three great tragedians of classical Athens
(the other two were Aeschylus and Sophocles).
To "double the cube" means to be given a
cube of some side length s and volume V, and
to construct a new cube, larger than the first,
with volume 2V and therefore side length
The problem is known to be impossible to
solve with only compass and straightedge,
because the cube root of 2≈ 1.25992105 is
not a constructible number.
Trisecting an Angle
That there may be a difficulty trisecting
an angle is rather surprising. Bisecting is
easy! The same easy approach applies
to bisecting both straight line segments
and circular arcs.
The problem of trisecting a line segment
is no more difficult than finding its n-th
part for an arbitrary n. However, the
general problem of trisecting an arbitrary
angle is not solvable in a finite number of
Two points must be made.
First, some angles are trisectable with a ruler and a compass. An obvious
example is supplied by the three quarters of the straight angle - 67.5o. (The
angle itself is constructible as it is obtained by two consecutive angle
bisections. Its third is obtained along the way.) Angles of 30o (draw a right
triangle with a side 1 and hypotenuse 2) and 45o (bisect the right angle) are
both constructible. Therefore, the latter also admits a classical trisection.
However, and this is where the difficulty lies, an arbitrary angle can't be
trisected with a ruler and a compass, as stated above.
Secondly, consider the geometric series 1/4 + 1/16 + 1/64 + ... that adds up to
1/3. Once we know how to bisect an angle, we may also find its 2-n-th part, for
any n. In particular, one may construct 1/4, 1/16, ... of any angle and, in
principle, find its third after an infinite number of steps. This solution is
universal but requires a forbidden (infinite) number of steps.
The problem was settled in 1837 by Pierre Laurent Wantzel (1814-1848) who
had proven that there was no way to trisect a 60o angle in the classical framework.
Constructing a Regular Heptagon
3-, 4-, 5-, and 6-gons (i.e., regular triangles, squares, pentagons, and
hexagons) are all easily constructible. Octagons are constructible on the
heels of squares with a single angle bisection. All polygons obtained
from the above four by doubling the number of sides are also
constructible. Not so for a heptagon, a 7-sided polygon.
In 1796, at the age of 19, Gauss showed that a regular heptadecagon (a
17-sided polygon) is constructible. Gauss showed that the heptadecagon
was only a particular case of a family of constructible regular polygons.
Any N-gon, where N is in the form 2np1...pm, where pi's are distinct
Fermat primes (i.e. primes in the form 22k + 1), is constructible. Wantzel
later proved that no other regular polygon has this property.
The Lune of Hippocrates
Some apparent partial solutions gave false hope in the "Squaring the
Circle" Problem for a long time. In this figure, the shaded figure is the
Lune of Hippocrates. Its area is equal to the area of the triangle ABC.
This is not a complete solution of the task (obviously since the
complete solution is proven to be impossible).
Explanation….
Data Warehouse to Cube
Math 258 - Webpages - Individual Webpages on Homepage
The Cube Activity - Florida International University
Chapter 10A-F Review
Thales and the Atomic Theory
Higher Tier WJEC GCSE Revision PPT(20 secs)
Applications involving right triangles


I've been on a ketogenic diet for a month now and for the most part, I'm feeling great and wishing I had discovered keto a long time ago. After years of constantly feeling tired and being frustrated with "healthy" diets that didn't work for me, I feel like I'm finally giving my body what it needs. I've had a few minor setbacks (due to eating ice cream and not hydrating enough), and I feel like I still have some improvements to make (like giving up dairy), but overall I feel like I'm continuing to make progress. Here's what happened during my first month of keto.
I stopped mindlessly snacking. Since I was tracking all of my macros and I knew any snacks would count towards my daily carb limit (which I set at 25 grams), I became more deliberate about everything I ate. This meant no more snacking out of stress or boredom — a big issue for me in the past.
My appetite decreased. After a few days, I simply couldn't eat as much as I used to. Sometimes I started feeling full after just a couple of minutes of eating. I had to remind myself to take smaller portions at mealtime.
I started eating less in general. Given the first two items above, I was eating a lot less food than I used to. I didn't feel heavy and bloated and completely stuffed after meals.
I started eating more greens. I used to get lazy about preparing meals and would sometimes just throw some gluten-free mac and cheese in the microwave for lunch. But because I stocked my fridge in advance with keto-friendly, low-carb veggies and leafy greens, I made sure to include lots of them in every meal.
My sinuses started clearing up. This is huge, and was completely unexpected. I've always had sinus issues. I'm always sneezing and blowing my nose, regardless of the time of year. I often get bronchitis. After less than a week on the keto diet, I realized I could actually breathe through my nose — and a little online research revealed I'm not the only one who's experienced these effects. It never occurred to me before that the sinus problems could be related to carbohydrate intake.
I got a touch of the keto diarrhea. This can happen if your digestive system isn't used to all the fat. For most people, this clears up after a few days.
One thing I did not experience, however, was the keto flu. I attribute this to the fact that my diet was already low in processed carbs and I was already taking a magnesium supplement. If you're even just curious about going keto, I would recommend cutting back on processed carbs immediately. Your body will thank you for it. You can always go back to your carb-loving life if you choose, but if you try to go from a high-carb diet straight to a keto diet, you're gonna have a bad time.
I started having more energy. I still wasn't exactly jumping out of bed in the morning, but once I was up and moving around, I felt pretty awake and less dependent on coffee.
I stopped experiencing the 3pm energy lull. I didn't even really notice when this happened, but I no longer felt like I needed a coffee break or a mid-afternoon nap.
I started feeling more productive and motivated to get things done in general. This likely goes along with the increased energy, plus I felt like the mental fog was lifting.
I lost 8 pounds. Losing water weight is common when you start a ketogenic diet, so this was not unexpected.
My skin cleared up and started looking amazing. I've always had oily skin and my face is usually a grease pit by afternoon. After a couple of weeks on keto, my skin was damn near glowing and felt much smoother. I'm still oily, but it doesn't feel as severe as before. Other people even commented that my skin looked great.
I lost 2 more pounds.
I ate a mini ice cream cone and felt extremely bloated the next day. Apparently that can happen with ice cream or too many carbs. I drank water all day and didn't eat until late afternoon when I finally started feeling hungry again.
I finally started experiencing a real increase in energy and mental clarity towards the end of the third week. I felt like I had more stamina when I went running, and I didn't feel completely mentally drained by the end of my work day.
I noticed I wasn't having any cravings. I had bought some dark chocolate so I could eat a small piece when I wanted something sweet, but I hadn't even touched it. I didn't really feel like I was missing my old snacks, because I just wasn't hungry.
Now I really started noticing the increased energy. I especially noticed it when we went rock climbing, and even though I hadn't climbed in a few months, I was able to get up more routes than usual and didn't even feel tired afterward.
I felt different in general, lighter and more upbeat. I didn't feel like stress was weighing me down. I felt more resilient.
I got super tired one day and I couldn't figure out why. This happened to be the day I was ovulating, so I'll have to look into this more and see if the same thing happens next month. I actually had to lie down and take a nap for a little while, but the rest of the week, I felt great.
I continued to feel more focused during the work day and more productive.
I got lazy with my water intake, drank half a glass of dry red wine one night, and woke up with a terrible hangover. I need to remember to keep drinking lots of water, especially if I'm going to drink any alcohol.
Did you experience any of the same things on your first month of keto? Are you considering keto and want to know what else to expect? Let me know what you think in the comments!
I am always so tired and groggy and have that 3pm energy lull. I want to feel the way it has made you feel!
You can do it! It might be tough at first but it's so worth it!


Pets are strictly forbidden in the forest, in watercraft, on beaches, on hiking trails, and in all public buildings. Elsewhere in the park they must be on a leash and under surveillance at all times. When on campgrounds, they must be kept inside the tent, trailer or other vehicle overnight.
Provincial parks in Quebec do not welcome dogs.


Could Bears roll the dice on Reggie Nelson to shore up a safety spot?
Everyone is pretty surprised that Nelson has not been signed to this point and he remains on the open market for one reason – the money he was expecting to get hasn't been there. Nelson turns 33 in September and is coming off his first Pro Bowl season with the Bengals as he had eight interceptions and 14 passes defended. Nelson has missed only three games in the last five seasons and played in all 16 last year. He's going to find a home eventually but I tend to doubt his destination will be Halas Hall given the Bears' push to sign younger players in free agency. They went with an older option at safety last March and that didn't turn out so well with Antrel Rolle. He'd been remarkably durable throughout the course of his career and wound up missing half the season. The Bears have to determine if that was an aberration or sign of things to come. It's worth noting there isn't any guaranteed money left in Rolle's deal so they could move on from him and take a shot with another aging safety but I don't know if general manager Ryan Pace is inclined to head down that path again or pay Nelson whatever it is he has been seeking.