ibonacci representations of sequences in Hilbert space

Dynamical sampling deals with frames of the form $\{T^n\varphi\}_{n=0}^\infty$, where $T \in B({\mathcal{H}})$ belongs to certain classes of linear operators and $\varphi\in{\mathcal{H}}$. The purpose of this paper is to investigate a new representation, namely, Fibonacci representation of sequences $\{f_n\}_{n=1}^\infty$ in a Hilbert space ${\mathcal{H}}$; having the form $f_{n+2}=T(f_n+f_{n+1})$ for all $n\geqslant 1$ and a linear operator $T :\operatorname{span}\{f_n\}_{n=1}^\infty\to\operatorname{span}\{f_n\}_{n=1}^\infty$. We apply this kind of representations for complete sequences and frames. Finally, we present some properties of Fibonacci representation operators.

<span>**I**ntroduction</span>
=============================

We recall some definitions and standard results from frame theory.

1.  $F$ is called a *frame* for ${\mathcal{H}}$, if there exist two constants $A_F, B_F > 0$ such that $$\begin{aligned}
    \label{abc}
            A_F \|f\|^2\leqslant \sum_{i=1}^\infty |\langle f,f_i\rangle |^2 \leqslant  B_F \|f\|^2,\quad f\in {\mathcal{H}}.
            \end{aligned}$$

2.  $F$ is called a *Bessel sequence* with Bessel bound $B_F$ if at least the upper frame condition holds.

3.  $F$ is called *complete* in ${\mathcal{H}}$ if $\overline{\operatorname{span}}\{f_i\}_{i=1}^\infty={\mathcal{H}},$ i.e., $\operatorname{span}\{f_i\}_{i=1}^\infty$ is dense in ${\mathcal{H}}$.

4.  $F$ is called *linearly independent* if $\sum_{k=1}^m c_kf_k=0$ for some scalar coefficients $\{c_k\}_{k=1}^m$, then $c_k =0$ for all $k = 1,\cdots, m$. We say $F$ is *linearly dependent* if $F$ is not linearly independent.

@Ole_Book [Theorem 5.5.1]\[onto\] A sequence $F=\{f_i\}_{i=1}^\infty\subseteq {\mathcal{H}}$ is a frame for ${\mathcal{H}}$ if and only if $$T_F:\ell^2 \rightarrow{\mathcal{H}},\quad T_F(\{c_i\}_{i=1}^\infty )=\sum_{i=1}^\infty  c_if_i,$$ is a well-defined mapping from $\ell^2$ onto ${\mathcal{H}}$. Moreover, the adjoint of $T_F$ is given by $$\begin{aligned}
    T^*_F:{\mathcal{H}}\rightarrow \ell^2 ,\quad T_F^* f=\{\langle f,f_i \rangle\}_{i=1}^\infty .\end{aligned}$$

It is well known that if $F=\{f_i\}_{i=1}^\infty\subseteq {\mathcal{H}}$ is a Bessel sequence for ${\mathcal{H}}$, then $T_F$ is well-defined and bounded.

In @Ald_Acha, Aldroubi et al. introduced dynamic sampling which it deals with frame properties of sequences of the form $\{T^n\varphi\}_{n=0}^\infty$, where $T \in B({\mathcal{H}})$ belongs to certain classes of linear operators (such as diagonalizable normal operators) and $\varphi\in{\mathcal{H}}$. Various characterizations of frames having the form $ \{f_{k}\}_{k\in I}=\{T^k\varphi\}_{k\in I},$ where $T$ is a linear (not necessarily bounded) operator can be found in @O.M_Int [@O.M_Acha; @O.M.E_Jmaa; @RNO].

@O.M_Acha [Proposition 2.3]\[mnref\] Consider a frame sequence $F=\{f_i\}_{i=1}^\infty$ in a Hilbert space ${\mathcal{H}}$ which spans an infinite-dimensional subspace. The following are equivalent:

1.  $F$ is linearly independent.

2.  There exists a linear operator $T:\operatorname{span}\{f_i\}_{i=1}^\infty\rightarrow {\mathcal{H}}$ such that $Tf_i:=f_{i+1}$.

@O.M.E_Jmaa [Theorem 2.1]\[exbnd\] Consider a frame $F=\{f_i\}_{i=1}^\infty$ in ${\mathcal{H}}$. Then the following are equivalent:

1.  $F=\{T^{i-1} f_1\}_{i=1}^\infty$ for some $T\in B({\mathcal{H}})$.

2.  For some dual frame $G=\{g_i\}_{i=1}^\infty$ (and hence all) $$\begin{aligned}
            f_{j+1}=\sum_{i=1}^\infty\langle f_j,g_i\rangle f_{i+1},\forall j\in\mathbb{N}.
            \end{aligned}$$

3.  The $\ker T_F$ is invariant under the right-shift operator $\mathcal{T}:\ell^2\to\ell^2$ defined by $$\mathcal{T}(c_1, c_2,\cdots)=(0, c_1, c_2,\cdots).$$

Special sequences 
==================

It is well known, cf. @Ole_Book [Example 5.4.6] that if $\{e_n\}_{n=1}^\infty$ is an orthonormal basis $\{e_n\}_{n=1}^\infty$ for ${\mathcal{H}}$, then $\{e_n+e_{n+1}\}_{n=1}^\infty$ is complete and a Bessel sequence but not a frame. This motivates us to investigate some results concerning the sequences $F=\{f_n\}_{n=1}^\infty$, $M=\{f_n+f_{n+1}\}_{n=1}^\infty$ and $N=\{f_n-f_{n-1}\}_{n=1}^\infty$ in a Hilbert space ${\mathcal{H}}$.

\[lem1\] Let $\alpha$ and $\beta$ be nonzero scalers and $F=\{f_n\}_{n=1}^\infty \subseteq {\mathcal{H}}$. Then

1.  $F$ is a Bessel sequence for ${\mathcal{H}}$, if and only if $M=\{\alpha f_n +\beta f_{n+1}\}_{n=1}^\infty$ and $N=\{\alpha f_n -\beta f_{n+1}\}_{n=1}^\infty$ are Bessel sequences for ${\mathcal{H}}$.

2.  Suppose that $F$ is a Bessel sequence for ${\mathcal{H}}$. Then $F$ is complete, if and only if $M=\{\alpha f_n +\beta f_{n+1}\}_{n=1}^\infty$ is complete, whenever $|\alpha|\geqslant  |\beta|$.

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1.  Assume that $\{f_n\}_{n=1}^\infty$ is a Bessel sequence with Bessel bound $B_F$ and $\mu= \max\{|\alpha|^2,|\beta|^2\}$. Then for $f\in{\mathcal{H}}$, we have $$\begin{aligned}
            \sum_{n=1}^\infty |\langle f, \alpha f_n+\beta f_{n+1}\rangle|^2&+\sum_{n=1}^\infty |\langle f, \alpha f_n-\beta f_{n+1}\rangle|^2\\
            &= 2|\alpha|^2 \sum_{n=1}^\infty |\langle f, f_n\rangle|^2 +2|\beta|^2\sum_{n=1}^\infty |\langle f, f_{n+1}\rangle|^2\\
            & \leqslant  4 \mu \sum_{n=1}^\infty |\langle f, f_n\rangle|^2 \leqslant  4\mu B_F \|f\|^2.
            \end{aligned}$$ Then $M$ and $N$ are Bessel sequences. For the opposite implication, let $B_M$ and $B_N$ be Bessel bounds for sequences $M$ and $N$, respectively. Then $$\begin{aligned}
            2|\alpha|^2\sum_{n=1}^\infty |\langle f, f_n\rangle|^2&\leqslant  2  \sum_{n=1}^\infty |\langle f, \alpha f_n\rangle|^2+2\sum_{n=1}^\infty |\langle f, \beta f_{n+1}\rangle|^2\\
            &=\sum_{n=1}^\infty |\langle f, \alpha f_n +\beta f_{n+1} \rangle|^2 +\sum_{n=1}^\infty |\langle f, \alpha f_n -\beta f_{n+1}\rangle|^2\\
            &\leqslant  (B_M+B_N)\|f\|^2, \quad f\in {\mathcal{H}}.
            \end{aligned}$$

2.  Suppose that $F$ is complete and $f\in {\mathcal{H}}$ such that $\langle f, \alpha f_n +\beta f_{n+1}\rangle =0$ for all $n\in \mathbb{N}$. Then $\overline{\alpha}\langle f, f_n\rangle=-\overline{\beta} \langle f, f_{n+1}\rangle$ for all $n\in \mathbb{N}$. Since $|\alpha|\geqslant |\beta|$ and $$|\langle f,f_1 \rangle|^2 \sum_{n=0}^\infty \big|\dfrac{\alpha}{\beta}\big|^n =\sum_{n=1}^\infty |\langle f, f_n\rangle|^2 \leqslant  B_F \|f\|^2,$$ we get $\langle f, f_1\rangle=0$ and consequently $\langle f, f_n\rangle=0$ for $n\in\mathbb{N}$. Hence $f=0$ and this shows that $\{\alpha f_n +\beta f_{n+1}\}_{n=1}^\infty$ is complete. In order to show the other implication, assume that $M$ is complete and $f\in {\mathcal{H}}$ such that $\langle f,  f_n \rangle =0$ for all $n\in \mathbb{N}$. Since $$\langle f, \alpha f_n +\beta f_{n+1}\rangle =\overline{\alpha}\langle f, f_n\rangle+\overline{\beta} \langle f, f_{n+1}\rangle=0,\quad n\in\mathbb{N},$$ we conclude that $f=0$ and therefore $F$ is complete.

Let $F=\{f_n\}_{n=1}^\infty $, $M=\{\alpha f_n +\beta f_{n+1}\}_{n=1}^\infty$ and $N=\{\alpha f_n -\beta f_{n+1}\}_{n=1}^\infty $ be sequences in a Hilbert space ${\mathcal{H}}$ and $\alpha\neq0$. Then $F$ is a frame for ${\mathcal{H}}$, if and only if $M\cup N$ is a frame for ${\mathcal{H}}$.

Let $\mu= \max\{|\alpha|^2,|\beta|^2\}$. Then the result follows from $$\begin{aligned}
 |\alpha|^2\sum_{n=1}^\infty |\langle f, f_n\rangle|^2&\leqslant  2  \sum_{n=1}^\infty |\langle f, \alpha f_n\rangle|^2+2\sum_{n=1}^\infty |\langle f, \beta f_{n+1}\rangle|^2\\
    &=\sum_{n=1}^\infty |\langle f, \alpha f_n +\beta f_{n+1} \rangle|^2 +\sum_{n=1}^\infty |\langle f, \alpha f_n -\beta f_{n+1}\rangle|^2\\
    &= 2|\alpha|^2 \sum_{n=1}^\infty |\langle f, f_n\rangle|^2 +2|\beta|^2\sum_{n=1}^\infty |\langle f, f_{n+1}\rangle|^2\\
    & \leqslant  4 \mu \sum_{n=1}^\infty |\langle f, f_n\rangle|^2, \quad f\in{\mathcal{H}}.
    \end{aligned}$$

Let $M=\{f_n +f_{n+1}\}_{n=1}^\infty$ and $N=\{f_n-f_{n+1}\}_{n=1}^\infty$ be frames for ${\mathcal{H}}$. Then $F=\{f_n\}_{n=1}^\infty$ is a frame for ${\mathcal{H}}$ and $$\label{S_f=}
    4S_Ff=S_Mf+S_Nf+2\langle f, f_1\rangle f_1, \quad f\in {\mathcal{H}},$$ where $S_F, S_M$ and $S_N$ are frame operators for $F, M$ and $N$, respectively.

By Proposition \[lem1\], $F$ is a Bessel sequence for ${\mathcal{H}}$. Let $A_M$ and $A_N$ be lower frame bounds for $M$ and $N$, respectively. Then we have $$\begin{aligned}
    (A_M+A_N)\|f\|^2&\leqslant  \sum_{n=1}^\infty |\langle f, f_n+f_{n+1}\rangle|^2+\sum_{n=1}^\infty |\langle f, f_n-f_{n+1}\rangle|^2 \nonumber\\
    &= 2 \sum_{n=1}^\infty |\langle f, f_n\rangle|^2+ 2 \sum_{n=1}^\infty |\langle f, f_{n+1}\rangle|^2 \label{combes}\\
    &\leqslant  4 \sum_{n=1}^\infty |\langle f, f_n\rangle|^2,  \quad f\in {\mathcal{H}}\nonumber.
    \end{aligned}$$ Therefore, $F$ is a frame for ${\mathcal{H}}$. Furthermore, since $$\begin{aligned}
    &\sum_{n=1}^\infty |\langle f, f_n+f_{n+1}\rangle|^2+\sum_{n=1}^\infty |\langle f, f_n-f_{n+1}\rangle|^2 \nonumber\\
    &= 2 \sum_{n=1}^\infty |\langle f, f_n\rangle|^2+ 2 \sum_{n=1}^\infty |\langle f, f_{n+1}\rangle|^2,  \quad f\in {\mathcal{H}},
    \end{aligned}$$ we obtain $$\begin{aligned}
    \langle S_Mf ,f\rangle +\langle S_Nf ,f\rangle&=4\langle S_Ff ,f\rangle-2|\langle f ,f_1\rangle|^2\\
    &=4\langle S_Ff ,f\rangle -2\langle \langle f ,f_1\rangle f_1 ,f\rangle, \quad f\in {\mathcal{H}}.
    \end{aligned}$$ Then we obtain (\[S\_f=\]).

Let $M=\{f_n+f_{n+1}\}_{n=1}^\infty$, $N=\{f_n-f_{n+1}\}_{n=1}^\infty$ and $F=\{f_n\}_{n=1}^\infty$ be Bessel sequences for ${\mathcal{H}}$ and $\ker T_F$ be invariant under $\mathcal{T}_R$. Then $\ker T_F=\ker T_M\cap\ker T_N$.

Let $\{c_n\}_{n=1}^\infty  \in \ker T_F$. Since $\ker T_F$ is invariant under $\mathcal{T}_R$, we get $\{0,c_1,c_2,...\} \in \ker T_F$. Therefore $\{c_1,c_1+c_2,c_2+c_3,...\} , \{c_1,c_2-c_1,c_3-c_2,...\} \in \ker T_F$. Hence $$\begin{aligned}
    \sum_{n=1}^\infty c_n (f_n+f_{n+1})&=\sum_{n=1}^\infty (c_n+c_{n+1})f_{n+1} +c_1 f_1 =0,\\
\sum_{n=1}^\infty c_n (f_n-f_{n+1})&=\sum_{n=1}^\infty (c_{n+1}-c_{n})f_{n+1} +c_1 f_1=0.
    \end{aligned}$$ Then we conclude $\{c_n\}_{n=1}^\infty  \in \ker T_M\cap \ker T_N$. On the other hand, if $\{c_n\}_{n=1}^\infty  \in \ker T_M\cap \ker T_N$, then we have $$0=\sum_{n=1}^\infty c_n (f_n-f_{n+1})+\sum_{n=1}^\infty c_n (f_n+f_{n+1})=2\sum_{n=1}^\infty c_n f_n.$$ Therefore, $ \{c_n\}_{n=1}^\infty  \in \ker T_F$.

Fibonacci representation
========================

In this section we want to consider representation of a sequence $\{f_n\}_{n=1}^\infty\subseteq{\mathcal{H}}$ on the form $f_n=T(f_{n-1}+f_{n-2})$ for $n\geqslant 3$, where $T$ is a linear operator defined on an appropriate subspace of ${\mathcal{H}}$.

We say that a sequence $F=\{f_n\}_{n=1}^\infty$ has a *Fibonacci representation* if there is a linear operator $T:\operatorname{span}\{f_n\}_{n=1}^\infty\rightarrow \operatorname{span}\{f_n\}_{n=1}^\infty$ such that $f_n=T(f_{n-1}+f_{n-2})$ for $n\geqslant 3$. In the affirmative case, we say that $F$ is represented by $T$, and $T$ is called a *Fibonacci representation operator* with respect to $F$.

Throughout this segment, ${\mathcal{H}}$ denotes a Hilbert space and $\{e_n\}_{n=1}^\infty$ is an orthonormal basis for ${\mathcal{H}}$.

It is clear that $F=\{f_n\}_{n=1}^\infty=\{e_1, e_1, e_2,...\}$ is a frame for ${\mathcal{H}}$. We define the linear operator $T:\operatorname{span}\{e_n\}_{n=1}^\infty \rightarrow \operatorname{span}\{e_n\}_{n=1}^\infty$ by $$Te_1=\dfrac{e_2}{2},\quad Te_n=\sum_{i=0}^{n-2}(-1)^i e_{n-i+1}+(-1)^{n+1}\dfrac{e_2}{2}, \quad n\geqslant  2.$$ Then $F$ is represented by $T$. Indeed, $Te_1+Te_2=e_3$ and for $n\geqslant  4$ we have $$\begin{aligned}
    Te_{n-2}+Te_{n-1}&=\sum_{i=0}^{n-4}(-1)^i e_{n-i-1}+(-1)^{n-1}\dfrac{e_2}{2}+\sum_{i=0}^{n-3}(-1)^i e_{n-i}+(-1)^{n}\dfrac{e_2}{2}\\
    &=\sum_{i=0}^{n-4}(-1)^i e_{n-i-1}+e_{n}+\sum_{i=1}^{n-3}(-1)^i e_{n-i}\\
    &=\sum_{i=0}^{n-4}(-1)^i e_{n-i-1}+\sum_{i=0}^{n-4}(-1)^{i+1} e_{n-i-1}+e_{n}\\
    &=e_n.
    \end{aligned}$$ Note that $F$ is not linearly independent, and so by @O.M_Acha [Proposition 2.3], there does not exist a linear operator $S:\operatorname{span}\{e_n\}_{n=1}^\infty \rightarrow \operatorname{span}\{e_n\}_{n=1}^\infty $ such that $Se_1=e_1$ and $Se_{n-1}=e_n,\ n\geqslant  2 $.

\[exam\] The frame $F=\{f_n\}_{n=1}^\infty=\{e_1,e_2,e_3,e_1, e_4, e_5, e_6 ,...\}$ is represented by $T$, where $T:\operatorname{span}\{f_n\}_{n=1}^\infty \rightarrow \operatorname{span}\{f_n\}_{n=1}^\infty$ is defined by $$\begin{aligned}
        Te_1&=\frac{1}{2}(e_4+e_3-e_1),\quad Te_2=\frac{1}{2}(-e_4+e_3+e_1),\quad Te_3=\frac{1}{2}(e_4-e_3+e_1),\\
        Te_4&=e_5-Te_1,\quad Te_n=e_{n+1}-Te_{n-1},\quad n\geqslant  5.
        \end{aligned}$$

A sequence $F=\{f_n\}_{n=1}^\infty $ is represented by $T$, if and only if $M=\{f_n+f_{n+1}\}_{n=1}^\infty$ and $N=\{f_n-f_{n+1}\}_{n=1}^\infty $ are represented by $T$.

First, let $F$ be represented by $T$. For every $n\in \mathbb{N}$, we have $$\begin{aligned}
T\big((f_n+f_{n+1})+(f_{n+1}+f_{n+2})\big)&=T(f_n+f_{n+1})+T(f_{n+1}+f_{n+2})\\
    &=f_{n+2}+f_{n+3},\\
    T\big((f_n-f_{n+1})+(f_{n+1}-f_{n+2})\big)&=T(f_n+f_{n+1})-T(f_{n+1}+f_{n+2})\\
    &=f_{n+2}-f_{n+3},
    \end{aligned}$$ Then $M$ and $N$ are represented by $T$. Conversely, if $M$ and $N$ are represented by $T$, then $$\begin{aligned}
    T(f_n&+f_{n+1})=\dfrac{1}{2}T(f_n+f_{n+1}+f_n-f_{n+1}+f_{n+1}+f_{n+2}+f_{n+1}-f_{n+2})\\
    &=\dfrac{1}{2}\Big(T(f_n+f_{n+1}+f_{n+1}+f_{n+2})+T(f_n-f_{n+1}+f_{n+1}-f_{n+2})\Big)\\
    &=\dfrac{1}{2}(f_{n+2}+f_{n+3}+f_{n+2}-f_{n+3})= f_{n+2}, \quad n\in \mathbb{N}.
    \end{aligned}$$ Hence $F$ is represented by $T$.

A frame may have more than one Fibonacci representation and a frame may not have any.

\[exam\] The frame $G=\{f_n\}_{n=1}^\infty=\{e_1,e_2,e_1,e_3,e_4,...\}$ does not have any Fibonacci representations. Indeed, if $G$ is represented by $T$, then $$\begin{aligned}
        Te_1+Te_2=e_1,\quad Te_2+Te_1=e_3,
        \end{aligned}$$ which is a contradiction. We note that $\{f_n+f_{n+1}\}_{n=1}^\infty$ is not linearly independent.

\[exam\] Consider the frame $E=\{e_n\}_{n=1}^\infty\subseteq{\mathcal{H}}$ and let $T,S:\operatorname{span}\{e_n\}_{n=1}^\infty\to\operatorname{span}\{e_n\}_{n=1}^\infty$ be linear operators defined by $$\begin{aligned}
    Te_1&=Te_2=\frac{1}{2}e_3,\quad Te_n=\frac{(-1)^n}{n}e_3-\sum_{i=4}^{n+1}(-1)^{n+i}e_i,\quad n\geqslant 4,\\
Se_1&=0,~ Se_2=e_3,~ Se_3=e_4-e_3,\quad Se_n=e_3-\sum_{i=4}^{n+1}(-1)^{n+i}e_i,\quad n\geqslant 4.
    \end{aligned}$$ Then it is easy to see that $E$ is represented by $T$ and $S$. We note that $\{e_n+e_{n+1}\}_{n=1}^\infty$ is linearly independent.

In general if $\{f_n\}_{n=1}^\infty\subseteq{\mathcal{H}}$ is linearly independent with a Fibonacci representation $T$, then for each $g\in\operatorname{span}\{f_n\}_{n=1}^\infty$ the linear operator $S:\operatorname{span}\{f_n\}_{n=1}^\infty\to\operatorname{span}\{f_n\}_{n=1}^\infty$ defined by $$S\Big(\sum_{i=1}^k c_i f_i\Big)=\sum_{i=1}^k c_i Tf_i+\sum_{i=1}^k (-1)^ic_i g$$ is a Fibonacci representation for $\{f_n\}_{n=1}^\infty$.

Now, we want to get a sufficient condition for a frame $F=\{f_n\}_{n=1}^\infty$ to have a Fibonacci representation. We need the following lemma.

\[lemf\] Consider a sequence $\{f_n\}_{n=1}^\infty$ in ${\mathcal{H}}$. Then the following hold:

1.  For $n\geqslant 2$, we have $$f_n=\sum_{i=0}^{m-1}(-1)^i (f_{n-i-1}+f_{n-i})+(-1)^{m} f_{n-m},\quad 1\leqslant  m \leqslant  n-1.$$

2.  $\operatorname{span}\{f_n\}_{n=1}^\infty=\operatorname{span}\big\{\{f_1\} \cup\{f_n+f_{n+1}\}_{n=1}^\infty\big\}$.

3.  If $\{f_n\}_{n=1}^\infty$ is linearly independent, then $\{f_n+f_{n+1}\}_{n=1}^\infty$ is linearly independent.

4.  If $\{f_1\} \cup\{f_n+f_{n+1}\}_{n=1}^\infty$ is linearly independent, then $\{f_n\}_{n=1}^\infty$ is linearly independent.

$(i)$ Let $n\geqslant 2$ and $1\leqslant  m \leqslant  n-1$. Then we have $$\begin{aligned}
    \sum_{i=0}^{m-1}(-1)^i (f_{n-i-1}+f_{n-i})
    &=\sum_{i=0}^{m-1}(-1)^i f_{n-i-1}+\sum_{i=0}^{m-1}(-1)^i f_{n-i}\\
    &=\sum_{i=1}^{m}(-1)^{i-1} f_{n-i}+\sum_{i=0}^{m-1}(-1)^i f_{n-i}\\
    &=\sum_{i=1}^{m-1}(-1)^{i-1} f_{n-i}+(-1)^{m-1} f_{n-m}+f_n+\sum_{i=1}^{m-1}(-1)^{i} f_{n-i}\\
    &=(-1)^{m-1} f_{n-m}+f_n.
    \end{aligned}$$ For the proof of $(ii)$, it is clear that $\operatorname{span}\{\{f_n+f_{n+1}\}_{n=1}^\infty \cup \{f_1\}\} \subseteq \operatorname{span}\{f_n\}_{n=1}^\infty$. On the other hand, by $(i)$ (for $m=n-1$) we infer $\operatorname{span}\{f_n\}_{n=1}^\infty\subseteq \operatorname{span}\{\{f_n+f_{n+1}\}_{n=1}^\infty \cup \{f_1\}\}$. This proves $(ii)$. To prove $(iii)$, let $\{c_n\}_{n=1}^k\subseteq \mathbb{C}$ such that $\sum_{n=1}^{k} c_n (f_n+f_{n+1})=0$. Then we have $$c_1f_1+\sum_{n=2}^{k}(c_{n-1}+c_n)f_n+c_k f_{k+1}=0.$$ Since $\{f_n\}_{n=1}^\infty$ is linearly independent, we get $c_1=c_k=0$ and $c_{n-1}+c_{n}=0$ for all $2\leqslant  n\leqslant  k$. Therefore, $c_n=0$ for all $1\leqslant  n\leqslant  k$. This completes the proof of $(iii)$.

To prove $(iv)$, let $\{c_n\}_{n=1}^N\subseteq \mathbb{C}$ such that $\sum_{n=1}^N c_n f_n=0$. Then by $(i)$, we have $$\begin{aligned}
        0=\sum_{n=1}^N c_n f_n
        &= c_1 f_1 +\sum_{n=2}^N c_nf_n\\
        &= c_1 f_1 +\sum_{n=2}^N c_n \Big( \sum_{i=0}^{n-2}(-1)^i (f_{n-i-1}+f_{n-i})+(-1)^{n-1} f_1\Big)\\
        &= \Big(c_1+\sum_{n=2}^N c_n(-1)^{n-1}\Big) f_1+\sum_{n=2}^N c_n \sum_{i=0}^{n-2} (-1)^i (f_{n-i-1}+f_{n-i})\\
        &= \Big(c_1+\sum_{n=2}^N c_n(-1)^{n-1}\Big) f_1+\sum_{i=0}^{N-2} \sum_{n=i+2}^{N} c_n (-1)^i (f_{n-i-1}+f_{n-i})\\
        &= \Big(c_1+\sum_{n=2}^N c_n(-1)^{n-1}\Big) f_1+\sum_{i=2}^{N} \sum_{n=i}^{N} c_n (-1)^{i-2} (f_{n-i+1}+f_{n-i+2})\\
        &= \Big(c_1+\sum_{n=2}^N c_n(-1)^{n-1}\Big) f_1+ \sum_{i=2}^N \sum_{n=0}^{N-i}c_{i+n} (-1)^{i}(f_{n+1}+f_{n+2})\\
        &= \Big(c_1+\sum_{n=2}^N c_n(-1)^{n-1}\Big) f_1+ \sum_{n=0}^{N-2} \Big(\sum_{i=2}^{N-n}c_{i+n} (-1)^{i}\Big)(f_{n+1}+f_{n+2}).
                        \end{aligned}$$ Since $\{f_1\} \cup\{f_n+f_{n+1}\}_{n=1}^\infty$ is linearly independent, we get $$\begin{aligned}
        c_1+\sum_{k=2}^N c_k(-1)^{k-1}=0,\quad
        \sum_{i=2}^{N-n}c_{i+n} (-1)^{i}=0, \quad 0\leqslant  n\leqslant  N-2.
        \end{aligned}$$ Hence we conclude that $c_n=0$ for all $ n=1,2,..,N.$ Then $\{f_n\}_{n=1}^\infty$ is linearly independent.

In the following, we give a sufficient condition for a sequence $F=\{f_n\}_{n=1}^\infty$ to have a Fibonacci representation.

\[mt\] Let $F=\{f_n\}_{n=1}^\infty$ be a sequence in ${\mathcal{H}}$. If $\{f_n+f_{n+1}\}_{n=1}^\infty $ is linearly independent, then $F$ has a Fibonacci representation.

First we assume that $f_1\in \operatorname{span}\{f_n+f_{n+1}\}_{n=1}^\infty$. Then by $(ii)$ of Lemma \[lemf\], we have $\operatorname{span}\{f_n+f_{n+1}\}_{n=1}^\infty=\operatorname{span}\{f_n\}_{n=1}^\infty.$ We define a linear operator $T:\operatorname{span}\{f_n\}_{n=1}^\infty \rightarrow \operatorname{span}\{f_n\}_{n=1}^\infty$ by $$T(f_n+f_{n+1})=f_{n+2}; \quad n\geqslant  2.$$ Since $\{f_n+f_{n+1}\}_{n=1}^\infty $ is linearly independent sequence, $T$ is well-defined and $F$ is represented by $T$. If $f_1\notin \operatorname{span}\{f_n+f_{n+1}\}_{n=1}^\infty$, then $\{f_1\}\cup \{f_{n}+f_{n+1}\}_{n=1}^\infty $ is linearly independent and so by Lemma \[lemf\] $(iv)$, $\{f_n\}_{n=1}^\infty$ is linearly independent. Hence we can define a linear operator $T:\operatorname{span}\{f_n\}_{n=1}^\infty \rightarrow\operatorname{span}\{f_n\}_{n=1}^\infty$ by $$Tf_n= \sum_{i=0}^{n} (-1)^{i} f_{n+1-i},\quad n\in\Bbb{N}.$$ We show that $F$ is represented by $T$. Indeed, $$\begin{aligned}
    Tf_n+Tf_{n+1}&=\sum_{i=0}^{n} (-1)^{i} f_{n+1-i}+\sum_{i=0}^{n+1} (-1)^{i} f_{n+2-i}\\
    &=\sum_{i=0}^{n} (-1)^{i} f_{n+1-i}+\sum_{i=1}^{n+1} (-1)^{i} f_{n+2-i} +f_{n+2}\\
    &=\sum_{i=0}^{n} (-1)^{i} f_{n+1-i}+\sum_{i=0}^{n} (-1)^{i+1} f_{n+1-i} +f_{n+2}=f_{n+2}.
    \end{aligned}$$

The following example shows that the converse of Theorem \[mt\] is not satisfied in general.

The frame $F=\{f_n\}_{n=1}^\infty=\{e_1,e_2,e_3,e_2,e_2,e_4,e_5,e_6,...\}$ is represented by the linear operator $T:\operatorname{span}\{f_n\}_{n=1}^\infty \rightarrow\operatorname{span}\{f_n\}_{n=1}^\infty$ given by $$\begin{aligned}
    Te_1&=e_3-\dfrac{e_4}{2},\quad
    Te_2=\dfrac{e_4}{2},\quad
    Te_3=e_2-\dfrac{e_4}{2},\\
    Te_n&=\sum_{i=0}^{n-4} (-1)^i e_{n-i+1}+(-1)^{n-3}\dfrac{e_4}{2}, \quad n\geqslant  4.
    \end{aligned}$$ But $\{f_n+f_{n+1}\}_{n=1}^\infty=\{e_1+e_2,e_2+e_3,e_3+e_2,2e_2,...\}$ is not linearly independent.

\[lis\] Let $F=\{f_n\}_{n=1}^\infty$ be a linear independent sequence in ${\mathcal{H}}$. Then $F$ has a Fibonachi representation.

It follows from Lemma \[lemf\] $(iii)$ and Theorem \[mt\].

Now, we provide sufficient conditions to make the converse of Theorem \[mt\] satisfy.

\[mt2\] Let $F=\{f_n\}_{n=1}^\infty$ be a complete sequence in an infinite dimensional Hilbert space ${\mathcal{H}}$ which is represented by $T$. If there exists $m\in \mathbb{N}$ such that $f_{m+1},Tf_1 \in \operatorname{span}\{f_n\}_{n=1}^{m}$, then $\{f_n+f_{n+1}\}_{n=1}^\infty $ is linearly independent.

Suppose that $\{f_n+f_{n+1}\}_{n=1}^\infty $ is not linearly independent. Then there exists $n_0\in \mathbb{N}$ such that $f_{n_0} +f_{n_0+1}=\sum_{n=1}^{n_0-1} c_n(f_n+f_{n+1}).$ Hence $$\label{14}
    f_{n_0+2}= T(f_{n_0} +f_{n_0+1})=\sum_{n=1}^{n_0-1} c_nf_{n+2} \in \operatorname{span}\{f_n\}_{n=1}^{n_0+1}.\\$$ Let $V=\operatorname{span}\{f_n\}_{n=1}^l$, where $l=\max\{n_0+1 , m\}$. By and $f_{m+1} \in \operatorname{span}\{f_n\}_{n=1}^{m}$, we get $f_{l+1} \in V$. We show $V$ is invariant under $T$. Suppose that $f=\sum_{n=1}^l c_nf_n\in V$. By using $(i)$ of Lemma \[lemf\], we have $$\begin{aligned}
    Tf=T\Big(\sum_{n=1}^l c_nf_n\Big)
    &= c_1 Tf_1 + \sum_{n=2}^l c_n Tf_n\\
    &=c_1Tf_1 + \sum_{n=2}^l c_n T\Big(\sum_{i=0}^{n}(-1)^i (f_{n-i-1}+f_{n-i})+(-1)^{n-1} f_1 \Big)\\
    &=c_1Tf_1 + \sum_{n=2}^l c_n \Big(\sum_{i=0}^{n}(-1)^i T(f_{n-i-1}+f_{n-i})+(-1)^{n-1} Tf_1\Big) \\
    &=\Big(c_1+\sum_{n=2}^l c_n(-1)^{n-1}\Big)Tf_1 + \sum_{n=2}^l c_n\sum_{i=0}^{n}(-1)^i f_{n-i+1}.
    \end{aligned}$$ Since $Tf_1 \in \operatorname{span}\{f_n\}_{n=1}^{m}\subseteq V$ and $f_{l+1}\in V$, the above argument proves that $V$ is invariant under $T$. Therefore $f_n \in V $ for all $n\geqslant  l+1$ and consequently $\operatorname{span}\{f_n\}_{n=1}^\infty=V$. Since $\{f_n\}_{n=1}^\infty$ is complete in ${\mathcal{H}}$, we have ${\mathcal{H}}=\overline{\operatorname{span}}\{f_n\}_{n=1}^\infty=\overline{V}=V$ which is in contradiction to $\dim {\mathcal{H}}=\infty$.

\[remcom\] Let $\{f_n\}_{n=1}^{\infty}$ be a complete and linearly dependent sequence with $f_1\neq0$ in an infinite dimensional Hilbert space ${\mathcal{H}}$. Then there exists $m\geqslant2$ such that $f_m\in \operatorname{span}\{f_n\}_{n=1}^{m-1}$ and $f_{m+1}\notin \operatorname{span}\{f_n\}_{n=1}^{m}$.

Since $\{f_n\}_{n=1}^{\infty}$ is linearly dependent, there exists $k\geqslant2$ such that $f_k\in\operatorname{span}\{f_n\}_{n=1}^{k-1}$. We claim that there exists an integer $l>k$ such that $f_l\notin\operatorname{span}\{f_n\}_{n=1}^{l-1}$. If $f_l\in\operatorname{span}\{f_n\}_{n=1}^{l-1}$ for each $l>k$, then $f_{k+1}\in\operatorname{span}\{f_n\}_{n=1}^{k-1}$ because $f_k\in\operatorname{span}\{f_n\}_{n=1}^{k-1}$ and $f_{k+1}\in\operatorname{span}\{f_n\}_{n=1}^{k}$. Hence by induction we get $f_{l}\in\operatorname{span}\{f_n\}_{n=1}^{k-1}$ for each $l>k$. Therefore $\operatorname{span}\{f_n\}_{n=1}^{\infty}=\operatorname{span}\{f_n\}_{n=1}^{k-1}$. Since $\{f_n\}_{n=1}^{\infty}$ is complete and $\dim {\mathcal{H}}=\infty$, the contradiction is achieved. Now, let $i\in\Bbb{N}$ be the smallest number such that $f_{k+i}\notin\operatorname{span}\{f_n\}_{n=1}^{k+i-1}$. Putting $m=k+i-1$, we get $f_m\in \operatorname{span}\{f_n\}_{n=1}^{m-1}$ and $f_{m+1}\notin \operatorname{span}\{f_n\}_{n=1}^{m}$.

\[in\] Let $F=\{f_n\}_{n=1}^\infty $ be a sequence in ${\mathcal{H}}$ which is represented by $T$. Suppose that $f_m\in \operatorname{span}\{f_n\}_{n=1}^{m-1}$ and $f_{m+1} \notin \operatorname{span}\{f_n\}_{n=1}^{m-1}$ for some integer $m\geqslant 2$. Then $Tf_i\in \operatorname{span}\{f_n\}_{n=3}^{m+1}$ for $1\leqslant  i \leqslant  m$.

By the assumption, we have $f_m=\sum_{n=1}^{m-1}c_n f_n$, so $$\begin{aligned}
    \sum_{n=1}^{m-2}&\Big(\sum_{i=0}^{n-1}(-1)^i c_{n-i}\Big)(f_n+f_{n+1})\\
    &=\sum_{n=1}^{m-2}\Big(\sum_{i=0}^{n-1}(-1)^i c_{n-i}\Big)f_n + \sum_{n=1}^{m-2}\Big(\sum_{i=0}^{n-1}(-1)^i c_{n-i}\Big)f_{n+1}\\
    &=\sum_{n=1}^{m-2}\Big(\sum_{i=0}^{n-1}(-1)^i c_{n-i}\Big)f_n + \sum_{n=2}^{m-1}\Big(\sum_{i=0}^{n-2}(-1)^i c_{n-i-1}\Big)f_{n}\\
    &=c_1f_1+\sum_{n=2}^{m-2}\Big(\sum_{i=0}^{n-1}(-1)^i c_{n-i}\Big)f_n\\
    &\quad +\sum_{n=2}^{m-2}\Big(\sum_{i=0}^{n-2}(-1)^i c_{n-i-1}\Big)f_{n}+\Big(\sum_{i=0}^{m-3}(-1)^i c_{m-i-2}\Big)f_{m-1}\\
    &=c_1f_1+\sum_{n=2}^{m-2} \Big(\sum_{i=0}^{n-1}(-1)^i c_{n-i} + \sum_{i=0}^{n-2}(-1)^i c_{n-i-1}\Big) f_{n}\\
    &\quad +\Big(\sum_{i=0}^{m-3}(-1)^i c_{m-i-2}\Big)f_{m-1}\\
            &=c_1f_1+\sum_{n=2}^{m-2} \Big(c_n+\sum_{i=1}^{n-1}(-1)^i c_{n-i}+\sum_{i=1}^{n-1}(-1)^{i-1} c_{n-i}\Big)f_{n}\\
    &\quad+\Big(\sum_{i=0}^{m-3}(-1)^i c_{m-i-2}\Big)f_{m-1}\\
    &=c_1f_1+\sum_{n=2}^{m-2} c_nf_{n}+\Big(\sum_{i=0}^{m-3}(-1)^i c_{m-i-2}\Big)f_{m-1}\\
    &=f_{m}+\Big(-c_{m-1}+\sum_{i=0}^{m-3}(-1)^i c_{m-i-2}\Big)f_{m-1}\\
    &=f_{m}+\Big(\sum_{i=0}^{m-2}(-1)^{i-1} c_{m-i-1}\Big)f_{m-1},
    \end{aligned}$$ thus $$\begin{aligned}
\label{554}
    f_{m-1}+f_{m}=\sum_{n=1}^{m-2}\Big(\sum_{i=0}^{n-1}(-1)^i c_{n-i}\Big)(f_n+f_{n+1})
    +\Big(1-\sum_{i=0}^{m-2}(-1)^{i-1} c_{m-i-1}\Big)f_{m-1}.
    \end{aligned}$$ Since $F$ is represented by $T$, the equality implies that $$\label{12}
    f_{m+1}=\sum_{n=1}^{m-2}\Big(\sum_{i=0}^{n-1}(-1)^i c_{n-i}\Big)f_{n+2}+\Big(1-\sum_{i=0}^{m-2}(-1)^{i-1} c_{m-i-1}\Big)Tf_{m-1}.$$ If $1-\sum_{i=0}^{m-2}(-1)^{i-1} c_{m-i-1}=0$, then $f_{m+1}\in \operatorname{span}\{f_n\}_{n=3}^{m}\subseteq\operatorname{span}\{f_n\}_{n=1}^{m-1}$ which is a contradiction. Hence implies that $$\label{13}
    Tf_{m-1}=\dfrac{f_{m+1}-\sum_{n=1}^{m-2}\Big(\sum_{i=0}^{n-1}(-1)^i c_{n-i}\Big)f_{n+2}}{1-\sum_{i=0}^{m-2}(-1)^{i-1} c_{m-i-1}}\in\operatorname{span}\{f_n\}_{n=3}^{m+1}.$$ Also, by $(i)$ of Lemma \[lemf\], for $1\leqslant  j \leqslant  m-1$, we have $$\begin{aligned}
\label{15}
    f_{m+1}=Tf_{m}+ Tf_{m-1}&=T\Big(\sum_{i=0}^{j-1}(-1)^i (f_{m-i-1}+f_{m-i})+(-1)^{j} f_{m-j}\Big)+Tf_{m-1}\\
    &=\sum_{i=0}^{j-1}(-1)^i f_{m-i+1}+(-1)^{j} Tf_{m-j}+Tf_{m-1}.
    \end{aligned}$$ Therefore $$\label{116}
        Tf_{m-j}=(-1)^{j} \big(f_{m+1}-Tf_{m-1}-\sum_{i=0}^{j-1}(-1)^i f_{m-i+1}\big).$$ Hence it follows from and that $Tf_i \in \operatorname{span}\{f_n\}_{n=3}^{m+1}$ for each $ 1\leqslant  i \leqslant  m-1$.

\[in2\] Let $F=\{f_n\}_{n=1}^\infty $ be a sequence in ${\mathcal{H}}$ which is represented by $T$. Suppose that $f_m\in \operatorname{span}\{f_n\}_{n=1}^{m-1}$ and $f_{m+1} \notin \operatorname{span}\{f_n\}_{n=1}^{m-1}$ for some $m\in \mathbb{N}$. Then, $Tf_{m+i}\in \operatorname{span}\{f_n\}_{n=3}^{m+i+1}$ for each $i \in \mathbb{N}$.

Since $Tf_{m+i}=f_{m+i+1}-Tf_{m+i-1}$, the result follows by induction on $i$ and Proposition \[in\].

\[ranspn\] Let $F=\{f_n\}_{n=1}^\infty $ be a complete sequence in an infinite dimensional Hilbert space ${\mathcal{H}}$.

1.  If $F$ is linearly independent, then it has a Fibonacci representation $T$ such that $\operatorname{\mathcal{R}}(T)=\operatorname{span}\{f_n\}_{n=3}^\infty.$

2.  If $F$ is linearly dependent, then for any Fibonacci representation $T$ of $F$ we have $\operatorname{\mathcal{R}}(T)=\operatorname{span}\{f_n\}_{n=3}^\infty.$

First we note that if $F$ is represented by $T$, then $f_n=T(f_{n-1}+f_{n-2})\in \operatorname{\mathcal{R}}(T)$ for every $n\geqslant  3$, and consequently $\operatorname{span}\{f_n\}_{n=3}^\infty\subseteq \operatorname{\mathcal{R}}( T)$.

To prove $(i)$, consider the linear operator $T:\operatorname{span}\{f_n\}_{n=1}^\infty \rightarrow \operatorname{span}\{f_n\}_{n=1}^\infty$ defined by $$\begin{aligned}
    Tf_1=Tf_2=\frac{1}{2}f_3,\quad
    Tf_n= \sum_{i=0}^{n-3} (-1)^{i} f_{n+1-i}+ \frac{(-1)^n}{2}f_3, \quad n\geqslant  3.
    \end{aligned}$$ Then $$\begin{aligned}
    &Tf_1+Tf_2=f_3,\quad Tf_2+Tf_3=f_4,\\ &Tf_{n}+Tf_{n+1}=\sum_{i=0}^{n-3}(-1)^i f_{n-i+1}+\sum_{i=0}^{n-2}(-1)^i f_{n-i+2}=f_{n+2}, \quad n\geqslant  3.
    \end{aligned}$$ Hence $F$ is represented by $T$ and it is obvious that $\operatorname{\mathcal{R}}(T)\subseteq \operatorname{span}\{f_n\}_{n=3}^\infty$. In order to prove $(ii)$, by Proposition \[remcom\] there exists $m\geqslant 2$ such that $f_m\in \operatorname{span}\{f_n\}_{n=1}^{m-1}$ and $f_{m+1}\notin \operatorname{span}\{f_n\}_{n=1}^{m-1}$. If $F$ is represented by $T$, then by Proposition \[in\] and Corollary \[in2\] we have $\operatorname{\mathcal{R}}(T)\subseteq \operatorname{span}\{f_n\}_{n=3}^\infty.$

In Theorem \[mt2\], we showed that $\{f_{n}+f_{n+1}\}_{n=1}^\infty  $ is linearly independent under some conditions. In the following, we show that (under some conditions) by removing finitely many elements of $\{f_{n}+f_{n+1}\}_{n=1}^\infty  $ the remaining elements will be linearly independent.

\[mt3\] Let $F=\{f_n\}_{n=1}^\infty$ be a complete sequence in an infinite dimensional Hilbert space ${\mathcal{H}}$ which is represented by $T$. Then there exists $m\in \mathbb{N} $ such that $\{f_{m+n}+f_{m+n+1}\}_{n=1}^\infty  $ is linearly independent.

If $F$ is linearly independent, then the result follows by $(iii)$ of Lemma \[lemf\]. Suppose that $F$ is linearly dependent. Then by Proposition \[remcom\] and Proposition \[in\], there exists $m\geqslant 2$ such that $f_m\in \operatorname{span}\{f_n\}_{n=1}^{m-1}$, $f_{m+1}\notin \operatorname{span}\{f_n\}_{n=1}^{m-1}$ and $Tf_1 \in \operatorname{span}\{f_n\}_{n=3}^{m+1}$. We prove $\{f_{m+n}+f_{m+n+1}\}_{n=1}^\infty  $ is linearly independent. Suppose by contradiction that $\{f_{m+n}+f_{m+n+1}\}_{n=1}^\infty  $ is not linearly independent. Then there exists $j\in \mathbb{N}$ such that $f_{m+j} +f_{m+j+1}=\sum_{n=1}^{j-1} c_n(f_{m+n}+f_{m+n+1}).$ Hence we have $$\label{17}
    f_{m+j+2}= T(f_{m+j} +f_{m+j+1})=\sum_{n=1}^{j-1} c_nf_{m+n+2} \in \operatorname{span}\{f_n\}_{n=1}^{m+j+1}.\\$$ Let $V=\operatorname{span}\{f_n\}_{n=1}^{m+j+1}$. We show that $V$ is invariant under $T$. Let $f=\sum_{n=1}^{m+j+1} c_nf_n\in V$. Then by $(i)$ of Lemma \[lemf\], we have $$\begin{aligned}
    Tf&= c_1 Tf_1 + \sum_{n=2}^{m+j+1} c_n Tf_n\\
    &=c_1Tf_1 + \sum_{n=2}^{m+j+1} c_nT\Big(\sum_{i=0}^{n-2}(-1)^i (f_{n-i-1}+f_{n-i})+(-1)^{n-1} f_1 \Big)\\
    &=c_1Tf_1 + \sum_{n=2}^{m+j+1} c_n \Big(\sum_{i=0}^{n-2}(-1)^i f_{n-i+1}+(-1)^{n-1} Tf_1\Big) \\
    &=\Big(c_1+\sum_{n=2}^{m+j+1} c_n(-1)^{n-1}\Big)Tf_1 + \sum_{n=2}^{j+m+1} c_n\sum_{i=0}^{n-2}(-1)^i f_{n-i+1}.
    \end{aligned}$$ Using $Tf_1 \in \operatorname{span}\{f_n\}_{n=3}^{m+1} \subseteq V$ and , we get $Tf \in V$. Then we conclude $f_n \in V $ for all $n\geqslant  m+j+2$. Thus, $\operatorname{span}\{f_n\}_{n=1}^\infty=V$ and since $\{f_n\}_{n=1}^\infty$ is complete in ${\mathcal{H}}$, we have ${\mathcal{H}}=\overline{\operatorname{span}}\{f_n\}_{n=1}^\infty=\overline{V}=V$ which is a contradiction.

Fibonacci Representation Operators
==================================

In a frame that indeed has the form $\{T^n\varphi\}_{n=0}^\infty$, where $T \in B({\mathcal{H}})$ and $\varphi\in{\mathcal{H}}$, all sequence members are represented by iterative actions of $T$ on $\varphi$. In the case where $\{f_n\}_{n=1}^\infty$ has a Fibonacci representation operator $T$, we expect (Theorem \[fib\]) all members of the sequence $\{f_n\}_{n=1}^\infty$ to be identified in terms of iterative actions of $T$ on elements $f_1$ and $f_2$. In this section, we present some results concerning Fibonacci representation operators. One of the results characterizes types of frame which can be represented in terms of a bounded operator $T$.

$[x]$ denotes the integer part of $x\in\Bbb{R}$ and ${\begin{pmatrix}        n  \\        k  \end{pmatrix} }:=\frac{n!}{k!(n-k)!}$ for integers $0\leqslant  k\leqslant  n$. We let ${\begin{pmatrix}        n  \\        k  \end{pmatrix} }:=0$ when $k>n$ or $k<0$.

\[fib\] Let $T:\operatorname{span}\{f_n\}_{n=1}^\infty \rightarrow \operatorname{span}\{f_n\}_{n=1}^\infty $ be a linear operator, then the following statements are equivalent:

1.  $F=\{f_n\}_{n=1}^\infty$ is represented by $T$.

2.  $Tf_1+Tf_2=f_3$ and $$\label{rep}
            f_n=\sum_{i=a_n}^{2a_n}\Big( {\begin{pmatrix}        i+b_n  \\        2i-2a_n+b_n  \end{pmatrix} } T^{i+b_n}f_2+ {\begin{pmatrix}        i+b_n  \\        2i-2a_n+b_n+1  \end{pmatrix} } T^{i+b_n+1}f_1\Big),\quad n\geqslant  4,$$ where $a_n=[\frac{n-1}{2}]$ , $b_n=n-2a_n-2$.

$(i)\Rightarrow (ii)$ We prove by induction on $n$. For $n=4$, we have $a_4=1$ and $b_4=0 $. Then $$\begin{aligned}
    &{\begin{pmatrix}        1  \\        0  \end{pmatrix} } Tf_2+ {\begin{pmatrix}        1  \\        1  \end{pmatrix} } T^{2}f_1+ {\begin{pmatrix}        2  \\        2  \end{pmatrix} } T^{2}f_2+ {\begin{pmatrix}        2  \\        3  \end{pmatrix} } T^{3}f_1\\
    &=Tf_2+T(Tf_1+Tf_2)=Tf_2+Tf_3=f_4.
    \end{aligned}$$ Now, assume that $k>4$ and holds for all $n\leqslant  k$ and we prove for $n=k+1$. If $k+1$ is even, then $b_k=-1,~ b_{k-1}=b_{k+1}=0$ and $a_{k+1}=a_k=1+a_{k-1}$. Hence $$\begin{aligned}
    f_{k+1}&=Tf_k+Tf_{k-1}\\
    &=\sum_{i=a_k}^{2a_k} \Big( {\begin{pmatrix}        i-1  \\        2i-2a_k-1  \end{pmatrix} } T^{i}f_2+ {\begin{pmatrix}        i-1  \\        2i-2a_k  \end{pmatrix} } T^{i+1}f_1 \Big)\\
    &\quad+\sum_{i=a_{k-1}}^{2a_{k-1}} \Big({\begin{pmatrix}        i  \\        2i-2a_{k-1}  \end{pmatrix} } T^{i+1}f_2+ {\begin{pmatrix}        i  \\        2i-2a_{k-1}+1  \end{pmatrix} } T^{i+2}f_1\Big)\\
    &=\sum_{i=a_k}^{2a_{k}} \Big({\begin{pmatrix}        i-1  \\        2i-2a_{k}-1  \end{pmatrix} }T^{i}f_2+ {\begin{pmatrix}        i-1  \\        2i-2a_{k}  \end{pmatrix} } T^{i+1}f_1 \Big)\\
    &\quad+\sum_{i=a_k}^{2a_k-1} \Big({\begin{pmatrix}        i-1  \\        2i-2a_{k}  \end{pmatrix} } T^{i}f_2+ {\begin{pmatrix}        i-1  \\        2i-2a_{k}+1  \end{pmatrix} } T^{i+1}f_1\Big)\\
    &=\sum_{i=a_k}^{2a_{k}} \Big({\begin{pmatrix}        i-1  \\        2i-2a_{k}-1  \end{pmatrix} }T^{i}f_2+ {\begin{pmatrix}        i-1  \\        2i-2a_{k}  \end{pmatrix} } T^{i+1}f_1 \Big)\\
    &\quad+\sum_{i=a_k}^{2a_k} \Big({\begin{pmatrix}        i-1  \\        2i-2a_{k}  \end{pmatrix} } T^{i}f_2+ {\begin{pmatrix}        i-1  \\        2i-2a_{k}+1  \end{pmatrix} } T^{i+1}f_1\Big)\\
    &=\sum_{i=a_k}^{2a_k} \Big({\begin{pmatrix}        i  \\        2i-2a_{k}  \end{pmatrix} }  T^{i}f_2+ {\begin{pmatrix}        i  \\        2i-2a_{k}+1  \end{pmatrix} } T^{i+1}f_1\Big).
    \end{aligned}$$ Since $b_{k+1}=0$ and $a_{k+1}=a_k$, we obtain . If $k+1$ is odd, then $b_k=0,~ b_{k-1}=b_{k+1}=-1$ and $1+a_{k-1}=1+a_k=a_{k+1}$. Hence $$\begin{aligned}
    f_{k+1}&=Tf_k+Tf_{k-1}\\
    &=\sum_{i=a_{k}}^{2a_{k}} \Big( {\begin{pmatrix}        i  \\        2i-2a_{k}  \end{pmatrix} } T^{i+1}f_2+ {\begin{pmatrix}        i  \\        2i-2a_{k}+1  \end{pmatrix} } T^{i+2}f_1\Big)\\
    &\quad+\sum_{i=a_{k-1}}^{2a_{k-1}} \Big( {\begin{pmatrix}        i-1  \\        2i-2a_{k-1}-1  \end{pmatrix} }T^{i}f_2+ {\begin{pmatrix}        i-1  \\        2i-2a_{k-1}  \end{pmatrix} } T^{i+1}f_1\Big)\\
    &=\sum_{i=1+a_{k+1}}^{2a_{k+1}} \Big( {\begin{pmatrix}        i-2  \\        2i-2a_{k+1}-2  \end{pmatrix} } T^{i-1}f_2+ {\begin{pmatrix}        i-2  \\        2i-2a_{k+1}-1  \end{pmatrix} } T^{i}f_1\Big)\\
    &\quad+\sum_{i=a_{k+1}}^{2a_{k+1}-1} \Big( {\begin{pmatrix}        i-2  \\        2i-2a_{k+1}-1  \end{pmatrix} } T^{i-1}f_2+ {\begin{pmatrix}        i-2  \\        2i-2a_{k+1}  \end{pmatrix} } T^{i}f_1\Big)\\
    &=\sum_{i=a_{k+1}}^{2a_{k+1}} \Big( {\begin{pmatrix}        i-2  \\        2i-2a_{k+1}-2  \end{pmatrix} } T^{i-1}f_2+ {\begin{pmatrix}        i-2  \\        2i-2a_{k+1}-1  \end{pmatrix} } T^{i}f_1\Big)\\
    &\quad+\sum_{i=a_{k+1}}^{2a_{k+1}} \Big( {\begin{pmatrix}        i-2  \\        2i-2a_{k+1}-1  \end{pmatrix} } T^{i-1}f_2+ {\begin{pmatrix}        i-2  \\        2i-2a_{k+1}  \end{pmatrix} } T^{i}f_1\Big)\\
    &=\sum_{i=a_{k+1}}^{2a_{k+1}} \Big( {\begin{pmatrix}        i-1  \\        2i-2a_{k+1}-1  \end{pmatrix} } T^{i-1}f_2+ {\begin{pmatrix}        i-1  \\        2i-2a_{k+1}  \end{pmatrix} } T^{i}f_1\Big).
    \end{aligned}$$ Hence we get . To prove $(ii)\Rightarrow (i)$, there are two possibilities. If $n>4$ is odd, then $b_n=-1,~ b_{n-1}=b_{n+1}=0$ and $a_{n+1}=a_n=1+a_{n-1}$. Hence $$\begin{aligned}
    Tf_n&=\sum_{i=a_n}^{2a_n}\Big( {\begin{pmatrix}        i-1  \\        2i-2a_{n}-1  \end{pmatrix} } T^{i}f_2+{\begin{pmatrix}        i-1  \\        2i-2a_{n}  \end{pmatrix} }  T^{i+1}f_1\Big),\\
Tf_{n-1}&=\sum_{i=a_{n-1}}^{2a_{n-1}}\Big( {\begin{pmatrix}        i  \\        2i-2a_{n-1}  \end{pmatrix} } T^{i+1}f_2+ {\begin{pmatrix}        i  \\        2i-2a_{n-1}+1  \end{pmatrix} }  T^{i+2}f_1\Big)\\
    &=\sum_{i=a_n}^{2a_n-1} \Big({\begin{pmatrix}        i-1  \\        2i-2a_{n}  \end{pmatrix} } T^{i}f_2+ {\begin{pmatrix}        i-1  \\        2i-2a_{n}+1  \end{pmatrix} } T^{i+1}f_1\Big)\\
&=\sum_{i=a_n}^{2a_n} \Big({\begin{pmatrix}        i-1  \\        2i-2a_{n}  \end{pmatrix} } T^{i}f_2+ {\begin{pmatrix}        i-1  \\        2i-2a_{n}+1  \end{pmatrix} } T^{i+1}f_1\Big).
    \end{aligned}$$ Therefore, $$\begin{aligned}
    Tf_n+Tf_{n-1}=\sum_{i=a_n}^{2a_n} \Big({\begin{pmatrix}        i  \\        2i-2a_{n}  \end{pmatrix} } T^{i}f_2+ {\begin{pmatrix}        i  \\        2i-2a_{n}+1  \end{pmatrix} }  T^{i+1}f_1\Big)
    =f_{n+1}.
    \end{aligned}$$ If $n\geqslant 4$ is even, the argument is similar to the previous case.

We recall that $$\ell^2({\mathcal{H}}):=\Big\{\{f_n\}_{n=1}^\infty\subseteq {\mathcal{H}}: \sum_{n=1}^\infty \|f_n\|^2<\infty\Big\},$$ and $\mathcal{T}_L, \mathcal{T}_R:\ell^2({\mathcal{H}}) \rightarrow \ell^2({\mathcal{H}})$ are bounded linear operators defined by $$\mathcal{T}_L \{f_n\}_{n=1}^\infty= \{f_{n+1}\}_{n=1}^\infty ,\quad \mathcal{T}_R \{f_n\}_{n=1}^\infty= \{0,f_1,f_2,f_3,...\}.$$

\[bound\] Let $F=\{f_n\}_{n=1}^\infty $ be a Bessel sequence in ${\mathcal{H}}$ which is represented by $T_0$ and let $M=\{f_n+f_{n+1}\}_{n=1}^\infty$ be a frame for ${\mathcal{H}}$. Then $\ker T_M\subseteq \ker T_{\mathcal{T}_L^2F}$ if and only if $T:=T_0|_{\operatorname{span}\{f_n+f_{n+1}\}_{n=1}^\infty}$ is bounded with $\|T\|\leqslant  \sqrt{\frac{B_F}{A_M}}$, where $B_F$ is a Bessel bound for $F$ and $A_M$ is a lower frame bound for $M$.

Let $\ker T_M\subseteq \ker T_{\mathcal{T}_L^2F}$ and $f=\sum_{n=1}^{k} c_n (f_n+f_{n+1})$, where $\{c_n\}_{n=1}^\infty  \in\ell^2$ with $c_n=0$ for $n\geqslant  k+1$. Then $$\begin{aligned}
    Tf=T\Big(\sum_{n=1}^{k} c_n (f_n+f_{n+1})\Big)
    &=\sum_{n=1}^{k} c_n T(f_n+f_{n+1})\\
    &=\sum_{n=1}^{\infty} c_n f_{n+2}\\
&=\sum_{n=1}^{\infty} d_n f_{n+2} +\sum_{n=1}^{\infty} r_n f_{n+2},
    \end{aligned}$$ where $\{d_n\}_{n=1}^\infty  \in \ker T_M \subseteq \ker T_{\mathcal{T}_L^2F}$ and $\{r_n\}_{n=1}^\infty  \in (\ker T_M)^\perp$. Since $\{d_n\}_{n=1}^\infty  \in \ker T_{\mathcal{T}_L^2F}$, we have $\sum_{n=1}^{\infty} d_n f_{n+2}=0$ and consequently $Tf=\sum_{n=1}^{\infty} r_n f_{n+2}$. Therefore by applying @Ole_Book [Lemma 5.5.5], we have $$\begin{aligned}
    \|Tf\|^2&=\Big\|\sum_{n=1}^{\infty} r_n f_{n+2}\Big\|^2 \leqslant   B_F\sum_{n=1}^{\infty} |r_n|^2 \leqslant   \frac{B_F}{A_M}\Big\|\sum_{n=1}^{\infty} r_n (f_{n}+f_{n+1})\Big\|^2\\
    &= \frac{B_F}{A_M}\Big\|\sum_{n=1}^{\infty} r_n (f_{n}+f_{n+1}) + \sum_{n=1}^{\infty} d_n (f_{n}+f_{n+1})\Big\|^2\\
    &=\frac{B_F}{A_M}\Big\|\sum_{n=1}^{k} c_n (f_{n}+f_{n+1})\Big\|^2=\frac{B_F}{A_M}\|f\|^2.
    \end{aligned}$$ For the other implication, let $ \{c_n\}_{n=1}^\infty \in \ker T_{M}$. Since $\sum_{n=1}^\infty c_n(f_n+f_{n+1})=0$ and $T$ is bounded, we have $$0=T\Big(\sum_{n=1}^\infty c_n(f_n+f_{n+1})\Big)=\sum_{n=1}^\infty c_n f_{n+2}.$$ This means $\{c_n\}_{n=1}^\infty \in \ker T_{\mathcal{T}_L^2F}$.

In Theorem \[exbnd\], the invariance of $\ker T_F$ under the right-shift operator $\mathcal{T}$ is sufficient condition for the boundedness of $T$. It is obvious that the invariance of $\ker T_F$ under $\mathcal{T}$ is equivalent with $\ker T_F\subseteq \ker T_{\mathcal{T}_L F}$. In fact, for $\{c_n\}_{n=1}^\infty\in \ell^2$ we have $\mathcal{T}(\{c_n\}_{n=1}^\infty) \in \ker T_F$ if and only if $\{c_n\}_{n=1}^\infty\in \ker T_{\mathcal{T}_LF}.$

Let $F=\{f_n\}_{n=1}^\infty $ be a Bessel sequence in ${\mathcal{H}}$ which is represented by $T\in B({\mathcal{H}})$ and $M=\{f_n+f_{n+1}\}_{n=1}^\infty$ be a frame for ${\mathcal{H}}$. Then $T$ is injective if and only if $ \ker T_{\mathcal{T}_L^2F}\subseteq \ker T_M$

Let $T$ be injective and $\{c_n\}_{n=1}^\infty\in \ker T_{\mathcal{T}_L^2F}$. Then $$T\Big(\sum_{n=1}^\infty c_n(f_n+f_{n+1})\Big)= \sum_{n=1}^\infty c_nf_{n+2}=0.$$ Since $T$ is injective, we get $\sum_{n=1}^\infty c_n(f_n+f_{n+1})=0$ and consequently $\{c_n\}_{n=1}^\infty\in \ker T_M$. Conversely, assume that $f\in{\mathcal{H}}$ and $Tf=0$. Since $M=\{f_n+f_{n+1}\}_{n=1}^\infty$ is a frame for ${\mathcal{H}}$, we have $f=\sum_{n=1}^\infty c_n(f_n+f_{n+1})$ for some $\{c_n\}_{n=1}^\infty\in\ell^2$. Then $$\sum_{n=1}^\infty c_nf_{n+2}=T\Big(\sum_{n=1}^\infty c_n(f_n+f_{n+1})\Big)=Tf=0.$$ Hence $\{c_n\}_{n=1}^\infty\in \ker T_{\mathcal{T}_L^2F}$ and consequently $\{c_n\}_{n=1}^\infty\in \ker T_M$. This means $f=\sum_{n=1}^\infty c_n(f_n+f_{n+1})=0$ and the proof is completed.

Let $\{f_n\}_{n=1}^\infty $ be represented by $T$. Then the following hold:

-   If $K\in B({\mathcal{H}})$ is injective and has closed range, then $\{Kf_n\}_{n=1}^\infty$ has a Fibonacci representation.

-   If $K\in B({\mathcal{H}})$ is surjective, then $\{K^*f_n\}_{n=1}^\infty$ and $\{KK^*f_n\}_{n=1}^\infty$ have Fibonacci representations.

$(i)$ By Open Mapping Theorem, there exists a bonded linear operator $S:\operatorname{\mathcal{R}}(K)\rightarrow {\mathcal{H}}$ such that $SK=I_{{\mathcal{H}}}$. Therefore $$KTS(Kf_n+Kf_{n-1})=KT(f_n+f_{n-1})=Kf_{n+1},\quad n\geqslant 2.$$

To prove $(ii)$, by @Ole_Book [Lemma 2.4.1], $K^*$ is injective and has closed range. Also $KK^*$ is invertible. Then $(i)$ implies $(ii)$.

Let $\{f_n\}_{n=1}^\infty $ be a frame for ${\mathcal{H}}$ and represented by $T\in B({\mathcal{H}})$. If $Tf_1\in\operatorname{span}\{f_n\}_{n=3}^\infty$, then $\operatorname{\mathcal{R}}(T)$ is closed and $\operatorname{\mathcal{R}}(T)=\overline{\operatorname{span}}\{Tf_n\}_{n=1}^\infty=\overline{\operatorname{span}}\{f_n\}_{n=3}^\infty$.

For each $f\in{\mathcal{H}}$, there exists $\{c_n\}_{n=1}^\infty\in\ell^2$ such that $f=\sum_{n=1}^\infty c_nf_n$. Then $Tf=\sum_{n=1}^\infty c_nTf_n\in \overline{\operatorname{span}}\{Tf_n\}_{n=1}^\infty$, and therefore $\operatorname{\mathcal{R}}(T)\subseteq\overline{\operatorname{span}}\{Tf_n\}_{n=1}^\infty$. On the other hand, for $g\in\overline{\operatorname{span}}\{f_n\}_{n=3}^\infty$ there exists $\{c_n\}_{n=1}^\infty\in\ell^2$ such that $$g=\sum_{n=1}^\infty c_nf_{n+2}=\sum_{n=1}^\infty c_n T(f_n+f_{n+1})=T\Big(\sum_{n=1}^\infty c_n (f_n+f_{n+1})\Big)\in\operatorname{\mathcal{R}}(T).$$ Then $\overline{\operatorname{span}}\{f_n\}_{n=3}^\infty\subseteq\operatorname{\mathcal{R}}(T)$. Since $Tf_1\in\operatorname{span}\{f_n\}_{n=3}^\infty$, we can now apply $(ii)$ of Lemma \[lemf\] to conclude that $$\operatorname{span}\{Tf_n\}_{n=1}^\infty=\operatorname{span}\Big\{\{Tf_1\}\cup\{Tf_n+Tf_{n+1}\}_{n=1}^\infty\Big\}=\operatorname{span}\{f_n\}_{n=3}^\infty.$$ Therefore $\operatorname{\mathcal{R}}(T)=\overline{\operatorname{span}}\{Tf_n\}_{n=1}^\infty=\overline{\operatorname{span}}\{f_n\}_{n=3}^\infty$.

Let $\{f_n\}_{n=1}^\infty $ be a linearly dependent frame sequence represented by $T\in B({\mathcal{K}})$, where ${\mathcal{K}}=\overline{\operatorname{span}}\{f_n\}_{n=1}^\infty$ is an infinite dimensional Hilbert space. Then $\operatorname{\mathcal{R}}(T)$ is closed and $\operatorname{\mathcal{R}}(T)=\overline{\operatorname{span}}\{f_n\}_{n=3}^\infty$.

Let $T_0$ be the restriction of $T$ on $\operatorname{span}\{f_n\}_{n=1}^\infty$. Then by $(ii)$ of Corollary \[ranspn\], we have $\operatorname{\mathcal{R}}(T_0)=\operatorname{span}\{f_n\}_{n=3}^\infty$, and therefore $\operatorname{\mathcal{R}}(T)\subseteq\overline{\operatorname{span}}\{f_n\}_{n=3}^\infty$. On the other hand, Since $\{f_n\}_{n=1}^\infty $ is a frame sequence, for each $f\in\overline{\operatorname{span}}\{f_n\}_{n=3}^\infty$, there exists $\{c_n\}_{n=1}^\infty\in\ell^2$ such that $$f=\sum_{n=1}^\infty c_nf_{n+2}=\sum_{n=1}^\infty c_n(Tf_{n}+Tf_{n+1})=T\Big(\sum_{n=1}^\infty c_n(f_{n}+f_{n+1})\Big)\in\operatorname{\mathcal{R}}(T).$$ Hence $\overline{\operatorname{span}}\{f_n\}_{n=3}^\infty\subseteq \operatorname{\mathcal{R}}(T)$ and this completes the proof.

\[T=S\] Let $\{f_n\}_{n=1}^\infty $ be represented by $T$ and $S$. If $Tf_1=Sf_1$, then $T=S$.

Since $Tf_1=Sf_1$ and $T(f_n+f_{n+1})=f_{n+2}=S(f_n+f_{n+1})$ for all $n\in\Bbb{N}$, we get $Tf_n=Sf_n$ for all $n\in\Bbb{N}$ (we can use $(i)$ of Lemma \[lemf\]). This proves $T=S$ on $\operatorname{span}\{f_n\}_{n=1}^\infty$.

Let $\{f_n\}_{n=1}^\infty $ be represented by $T$ and $S$. If $f_1\in\operatorname{span}\{f_n+f_{n+1}\}_{n=k}^\infty$ for some $k\in\Bbb{N}$, then $T=S$.

Since $f_1\in\operatorname{span}\{f_n+f_{n+1}\}_{n=k}^\infty$, we have $f_1=\sum_{n=k}^m c_n(f_n+f_{n+1})$ for some scalars $c_k,\cdots,c_m$. Then $$Tf_1=T\Big(\sum_{n=k}^m c_n(f_n+f_{n+1})\Big)=\sum_{n=k}^m c_nT(f_n+f_{n+1})=\sum_{n=k}^m c_nf_{n+2}=\sum_{n=k}^m c_nS(f_n+f_{n+1})=Sf_1.$$ Therefore $T=S$ by Theorem \[T=S\].

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{'source': 'arXiv_out/2003'}