0$ denote the
square of such a bound.
Now, bearing these facts in mind, we have
\begin{eqnarray*}
\norm{\xi^{(k)}}^2
&\leq&
r_3 \sum_{m=m_1}^{\infty}\norm{\xi_{2m+2}^{(k)}}^2\\
&=&
r_3 \sum_{j=1}^{\infty}\sum_{p=0}^{m_{j+1}-m_j-1}
\norm{\xi_{2m_j+2p+2}^{(k)}}^2\\
&\leq&
r_3 \sum_{j=1}^{\infty}\norm{\xi_{2m_j+2}^{(k)}}^2
\left[1+r_2\sum_{j=1}^{m_{j+1}-m_j-1}
\left(\frac{m_j+p}{m_j+1}\right)^{v}\right]\\
&\leq&
r_1r_3\sum_{j=1}^{\infty}\frac{1}{m_j^\beta} +
r_1r_2r_3\sum_{j=1}^{\infty}\frac{1}{m_j^\beta}
\sum_{p=1}^{m_{j+1}-m_j-1}
\left(\frac{m_j+p}{m_j+1}\right)^{v}\,.
\end{eqnarray*}
The first term in the last inequality is clearly convergent,
since $m_j\geq j$ and $\beta>1$. The second term is also
convergent since (\ref{non-self}) is fulfilled.
Thus, we have proved that $\xi^{(1)}$ and $\xi^{(2)}$ belong
to $l_2(\mathbb{N},\mathbb{C}^2)$. It follows (by the
argumentation given in the proof of
Theorem~\ref{thm:self-adjointness}) that $x^{(1)}$ and
$x^{(2)}$ belong to $l_2(\mathbb{N})$. Moreover, they are
linearly independent as a consequence of the linear
independency of $\xi^{(1)}$ and $\xi^{(2)}$. The lack of
self-adjointness now follows immediately.
\end{proof}
\begin{remark}
Theorem \ref{thm:nonself-adjointness} does not
allow to obtain the converse of Theorem
\ref{thm:self-adjointness}.
\end{remark}
\begin{remark}
\label{all-of-them}
A fairly large set of sequences $\{m_j\}_{j\in{\mathbb N}}$ satisfies
condition (\ref{non-self}), ranging from those of linear growth, i.\ e.
$m_j\sim j$, up to those of very fast growth like
$m_j\sim j^j$.
\end{remark}
As we have mentioned previously, Jacobi operators of the form
proposed in the present work has been already studied for the
case of even-periodic sequences $\{c_n\}_{n\in{\mathbb N}}$. No
results are known (to the authors anyway) concerning the
odd-periodic case, although it has been suggested that, in
this case, these operators would not be self-adjoint
\cite{MR1924991}. Theorem~\ref{thm:nonself-adjointness}
allows us to prove that assertion easily.
\begin{corollary}
\label{corollary:non-self}
Let $\{c_n\}_{n\in{\mathbb N}}$ be an odd-periodic sequence.
Then the operator $J$,
given by the matrix representation (\ref{eq:jm}) with entries defined by
(\ref{eq:weights}) and (\ref{eq:diagonal}), is not self-adjoint.
\end{corollary}
\begin{proof}
Let $T$ be the period of the sequence. Define $m_j=jT$ for $j\in{\mathbb N}$.
Clearly, $\{m_j\}_{j\in{\mathbb N}}$ satisfies
(\ref{non-self}) for any $\beta>1$ and $v\geq 0$. Thus, it only remains
to verify (\ref{eq:nonself-adj-cond}) for $k=1,2$.
We have,
\begin{eqnarray}
\sum_{s=1}^{m_j}g_s^{(1)}
&=&
\frac{1}{2}\sum_{s=1}^{m_j}\frac{c_{2s-1}-c_{2s}-\alpha}{s}\nonumber\\
&=&
\frac{1}{2}\sum_{n=1}^{j}\sum_{s=m_{n-1}+1}^{m_n}\frac{c_{2s-1}-c_{2s}}{s}
- \frac{\alpha}{2}\sum_{s=1}^{m_j}\frac{1}{s}\nonumber\\
&=&
\frac{1}{2}\sum_{n=1}^{j}\sum_{s=1}^{T}\frac{c_{2s-1}-c_{2s}}{s+(j-1)T}
- \frac{\alpha}{2}\log(m_j) + O(1)\,,\label{per}
\end{eqnarray}
where the first term in the last equality follows from the periodicity of
$\{c_n\}_{n\in{\mathbb N}}$. Now, for $j\geq 2$,
\begin{eqnarray*}
\sum_{s=1}^{T}\frac{c_{2s-1}-c_{2s}}{s+(j-1)T}
&=&
\sum_{s=1}^{T}\frac{c_{2s-1}-c_{2s}}{(j-1)T}\\
& &
+\ \sum_{s=1}^{T}(c_{2s-1}-c_{2s})
\left[\frac{1}{s+(j-1)T}-\frac{1}{(j-1)T}\right].
\end{eqnarray*}
The first term above equals zero because of the
odd-periodicity of the sequence, while the second term is
$O(j^{-2})$. Therefore, the first term in (\ref{per}) is also
$O(1)$ thus yielding the expected inequality (with
$\beta=\alpha$) for $\sum_{s=1}^{h_m}g_s^{(1)}$. A similar
computation shows that also $\sum_{s=1}^{h_m}g_s^{(2)}$
fulfills the same inequality.
\end{proof}
The proof of Corollary~\ref{corollary:non-self} tells us that
an odd-periodic sequence $\{c_n\}_{n\in{\mathbb N}}$ produces a
non-self-adjoint operator because one can find a suitable
sequence $\{m_j\}_{j\in{\mathbb N}}$ for which, the contribution of
$c_{2s-1}-c_{2s}$
to the l.\ h.\ s.\ of (\ref{eq:nonself-adj-cond}) is nearly canceled
(and the same occurs for $c_{2s}-c_{2s+1}$). This idea may
be used to construct non-selfadjoint Jacobi operators out of
certain non-periodic sequences.
As a matter of fact, the following example is defined essentially in
the same way as the one discussed previously, having exactly one
modification.
\begin{example}
\label{ex:non-selfadjointness}
Consider a sequence $\{c_n\}_{n\in{\mathbb N}}$ defined by two given numbers
$c_1$ and $c_2$ arranged as follows:
\begin{equation}
\label{non-self-adjiont-model}
\underbrace{
\underbrace{
c_1\,c_2\,c_1\,c_2\cdots c_1\,c_2\,c_1\,c_2}_{2p_1}
c_2\,c_1\,c_2\,c_1\cdots c_2\,c_1\,c_2\,c_1}_{2p_2}
c_1\,c_2\,c_1\,c_2\cdots\,,
\end{equation}
where $\{p_j\}_{j\in{\mathbb Z}^+}$ is a sequence
with polynomial growth of the form $p_j=Cj^a$,
with $a\in{\mathbb N}\setminus\{1\}$ and $C>0$. That is, we
use the sequence defined by (\ref{eq:self-adjoint-example-g1})
with $c_3=c_1$. Alternatively, we may think
of this sequence as one obtained from a two-periodic sequence
by inserting dislocations at sites $p_j+1$. We claim that the
Jacobi operator defined by (\ref{non-self-adjiont-model}) is not
self-adjoint. To prove it, we first define
$m_j=p_{2j}$. This sequence satisfies
(\ref{non-self}) for any $\beta>1$, as a straightforward computation shows.
Next, we must verify (\ref{eq:nonself-adj-cond}) for $k=1,2$. For
$k=1$, (\ref{dislocation}) holds true. The computation done
in Example 1 shows that
\begin{eqnarray*}
\lefteqn{
\sum_{s=m_{n-1}+1}^{m_n}\!\frac{c_{2s-1}-c_{2s}}{s}}\\
&=&
\frac{c_2-c_1}{2}\left(\sum_{s=p_{2n-2}+1}^{p_{2n-1}}\frac{1}{s^2}
\left(\frac{p_{2n-2}\!\!-\!p_{2n-1}}
{1+\frac{p_{2n-1}-p_{2n-2}}{s}}\right) +
\sum_{s=2p_{2n-1}-p_{2n-2}+1}^{p_{2n}}\frac{1}{s}\right)\\
& &
+\ \frac{c_1-c_2}{2}\left( \sum_{s=p_{2n-1}+1}^{p_{2n}}\frac{1}{s^2}
\left(\frac{p_{2n-1}\!\!-\!p_{2n-2}}
{1+\frac{p_{2n-2}-p_{2n-1}}{s}}\right)+
\sum_{s=p_{2n-1}+1}^{p_{2n}-p_{2n-1}+p_{2n+2}}\frac{1}{s}\right).
\end{eqnarray*}
By what has been shown before, the r.\ h.\ s.\ of this equality is
$O(n^{-2})$ as $n\to\infty$. This, together with (\ref{dislocation}), yields
\begin{equation*}
\sum_{s=1}^{m_j}g_s^{(1)}=
-\frac{\alpha}{2}\log(m_j) +O(1)\,.
\end{equation*}
In a similar fashion, we can show that the same equality
is true for $k=2$. Hence, we have verified that this example fulfills the
hypotheses of Theorem~\ref{thm:nonself-adjointness}.
We note that the case $p_m\sim m$ can not be included here
since the summability of
\begin{equation*}
\sum_{j=1}^\infty\sum_{s=m_{j-1}+1}^{m_j}\!\frac{c_{2s-1}-c_{2s}}{s}
\end{equation*}
is no longer ensured. This is to be expected because, for
$p_m\sim m$, the sequence $\{c_n\}_{n\in{\mathbb N}}$ is again
even-periodic so the associated Jacobi operator may be
self-adjoint (see Remark~\ref{614}).
\end{example}
Other examples of non-self-adjoint Jacobi operators (of the class discussed in
this work) could be devised along similar ideas. We think
that the example given above is of some interest because it is
derived from an operator defined by a even-periodic sequence
and therefore self-adjoint (provided that $|c_1-c_2|\geq
\alpha-1$, although this condition plays no role in our example)
\cite{MR1959871,MR1924991}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Discreteness of spectrum}
\label{sec:absence-cont-spectr}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In this section we show that, if the assumptions of
Theorem~\ref{thm:self-adjointness} are fulfilled and
$\{m_j\}_{j\in{\mathbb N}}$ obeys certain further conditions,
then $J$ has only discrete spectrum.
To simplify proofs, we will assume that the conditions of
Theorem~\ref{thm:self-adjointness} are satisfied for $k=1$ (so
$\xi^{(1)}$ is not in $l_2(\mathbb{N},\mathbb{C}^2)$).
We first prove an easy consequence of inequality
(\ref{cond-1-self-adjoint}).
\begin{lemma}
\label{lemma-bullshit}
If (\ref{cond-1-self-adjoint}) is true for $k=1$, then
\begin{equation}
\label{eq:silly-ineq}
\sum_{s=1}^{m_j}g_s^{(2)}
< -\frac{\alpha}{2}\log(m_j) + O(1)\,,
\quad\text{when }m\to\infty\,.
\end{equation}
An analogous assertion holds if the roles of $k=1$ and $k=2$
are interchanged.
\end{lemma}
\begin{proof}
We have
\begin{equation}
\label{bullshit}
\sum_{s=1}^{m_j}g_s^{(2)}
=
-\sum_{s=1}^{m_j}g_s^{(1)} + \sum_{s=1}^{m_j}\frac{c_{2s-1}-c_{2s+1}}{2s}
-\alpha\sum_{s=1}^{m_j}\frac{1}{s}\,.
\end{equation}
Furthermore,
\[
\sum_{s=1}^{m_j}\frac{c_{2s-1}-c_{2s+1}}{2s}
=
\sum_{s=2}^{m_j}\frac{c_{2s-1}}{2}\left(\frac{1}{s}-\frac{1}{s-1}\right)
+\frac{c_1}{2}-\frac{c_{2m_j+1}}{2m_j}\,.
\]
Thus, the second term in (\ref{bullshit}) is $O(1)$ because
the sequence $\{c_n\}_{n\in{\mathbb N}}$ is bounded and $(s^{-1}
- (s-1)^{-1})$ is $O(s^{-2})$. Since moreover the last term in
(\ref{bullshit}) equals $\alpha\log(m_j)$ up to a term $O(1)$, we
obtain
\begin{eqnarray*}
\sum_{s=1}^{m_j}g_s^{(2)}
&=&
-\sum_{s=1}^{m_j}g_s^{(1)} - \alpha\log(m_j) + O(1)\\
&\leq&
\frac{1}{2}\log(m_j) - \alpha\log(m_j) + O(1)\\
&<&
- \frac{\alpha}{2}\log(m_j) + O(1)\,,
\end{eqnarray*}
where the last inequality holds since $\alpha>1$.
\end{proof}
\begin{theorem}
\label{arafat}
Suppose that the hypotheses of
Theorem~\ref{thm:self-adjointness} are satisfied for $k=1$
(then $x^{(1)}$ is not in $l_2(\mathbb{N})$). In addition, assume
that $\{m_j\}_{j\in{\mathbb N}}$ satisfies
\begin{equation}
\label{one-more}
\sum_{j=1}^{\infty}\frac{m_{j+1}-m_j}{m_j^\alpha}
\left(\frac{m_{j+1}}{m_j}\right)^{\check{v}}<\infty\,,
\end{equation}
with $\check{v}=\max\{0,\,\sup_s\{c_{2s}-c_{2s+1}-\alpha\}\}$.
Then $x^{(2)}$ is in $l_2(\mathbb{N})$. The assertion holds also if
we interchange $k=1$ and $k=2$ and take
$\check{v}=\max\{0,\,\sup_s\{c_{2s-1}-c_{2s}-\alpha\}\}$.
\end{theorem}
\begin{proof}
By Lemma~\ref{lemma-bullshit} we have (\ref{eq:silly-ineq})
and then, taking into account (\ref{one-more}) and reasoning
as in Theorem ~\ref{thm:nonself-adjointness}, we obtain that
$\xi^{(2)}$ is in $l_2(\mathbb{N},\mathbb{C}^2)$.
\end{proof}
\begin{remark}
\label{last}
In view of Remarks~\ref{divergent} and \ref{all-of-them}, the set of
sequences $\{m_j\}_{j\in{\mathbb N}}$ that fulfill both
(\ref{cond-2-self-adjoint}) and (\ref{one-more}) is far from being empty.
\end{remark}
\begin{corollary}
\label{pp}
$J$ has only pure point spectrum
whenever the assumptions of Theorem~\ref{arafat} are met.
\end{corollary}
\begin{proof}
Under the assumptions of the previous theorem we have that
$x^{(1)}\not\in l_2(\mathbb{N})$ and $x^{(2)}\in
l_2(\mathbb{N})$. This behavior of the basis
$\{x^{(1)},\,x^{(2)}\}$ in the space of solutions of
(\ref{eq:main-recurrence}) is the same for any
$\zeta\in\mathbb{C}$. Thus, we always have a subordinate
solution for the generalized eigenequation of the self-adjoint
operator $J$. By applying Subordinacy Theory
\cite[Theorem 3]{MR1179528}, we conclude that the spectrum is
pure point.
\end{proof}
\begin{remark}
Notice that Theorem~\ref{thm:self-adjointness} does not
necessarily imply Theorem~\ref{arafat}. However, in account of what is
said in Remark~\ref{last}, the implication is true for a large set
of Jacobi operators $J$.
\end{remark}
We notice that (\ref{cond-2-self-adjoint}) along with
(\ref{eq:silly-ineq}) imply a certain correlation between the
slow decay of $x^{(1)}$ and the fast decay of $x^{(2)}$ (or
vice-versa). We shall use this to show that the resolvent $(J-\zeta I)^{-1}$
is Hilbert-Schmidt. Unfortunately,
(\ref{cond-1-self-adjoint}) is not good enough for that purpose.
The next technical result accounts for what is needed.
\begin{lemma}
\label{life-saver}
Suppose that for a certain sequence
$\{m_j\}_{j\in{\mathbb{N}}}$ one of the following holds:
\begin{enumerate}[i.]
\item For any natural $i$ greater than some $i_0\in\mathbb{N}$,
\begin{equation*}
\sum_{s=m_i+1}^{m_{i+1}}g_s^{(1)}
\ge -\frac{1}{2}\log\left(\frac{m_{i+1}}{m_i}\right) +
f_i\,,
\end{equation*}
where $\sum_{i=j}^lf_i$ has an upper bound for any $j$ and $l$
($i_0\le j