Suboxone, methadone and Vivitrol are popular medications often used to treat opioid addictions, and are often used in opioid replacement therapies (ORT). An addiction occurs when a person develops a psychological or physical dependence. Once this happens, the body goes through withdrawal whenever the drug is not in the system. If you're looking to get sober, most rehab centers will recommend using one of those drugs in your medical detox treatment plan. Here's a side-by-side comparison on all three of these drugs. Get a better idea of the characteristics and features of each medication. This will help you make a more informed decision that will benefit your recovery.
Opioid Replacement Therapy (ORT) uses weaker opioids as a substitute for stronger drugs. This means that Suboxone, methadone and Vivitrol are essentially weaker drugs with the same effects as stronger drugs, like oxycodone. ORT reduces the intensity of any withdrawal symptoms you may experience. You're less likely to experience life-threatening withdrawal symptoms. This treatment also curbs cravings. Because of all these factors, patients are less likely to relapse. ORT is one of the most effective treatments available, which is why it is extremely popular. 40% to 65% of patients who have gone through ORT are less likely to use other opioids again.
What Are Some of the Main Differences Between Suboxone, Methadone and Vivitrol?
Suboxone, methadone and Vivitrol all hope to achieve the same thing; however, they work in different ways. The differences are minor, but have a profound overall impact. Here's a quick look at the chemical differences between each type of medication.
Suboxone is made from buprenorphine and naloxone. Both of these drugs work in harmony to treat opioid addictions. Buprenorphine is a partial opioid agonist. This means that it stimulates the same areas in the brain as opioids do. In particular, it creates a sense of euphoria and satisfaction. Although it does the same task, buprenorphine does not have the same signal strength. It causes a much weaker reaction. This reaction prevents withdrawal symptoms. Naloxone is an opioid antagonist. It's mixed in with the buprenorphine to block receptors in the brain that respond to the opioids. This helps get rid of any lingering effects that may remain. In a way, naloxone helps reverse the effects of opioid usage.
Unlike Suboxone, methadone is a full opioid agonist. The brand name for this drug is Dolophine, and it consists of only one chemical compound. This medication is mostly used to treat opioids like heroin. In comparison to illicit street drugs, methadone has a slow onset of action. This gives it a smaller addictive potential than heroin. It also makes the withdrawal symptoms of methadone less intense. This makes it a lot easier for patients to taper off of the drug. Since methadone is still fairly strong, it is usually administered during inpatient detoxification. Most professionals find methadone to be a riskier option for patients who choose outpatient detoxification, as the dosage of the methadone administered must be exact.
Vivitrol is one of the newest options available for ORT. It's also known as naltrexone. While it's been used since the 1980s, this chemical was repackaged and altered in the late 2000s for treating alcohol dependence. Vivitrol is an opioid antagonist. This means that it blocks receptors in the brain, so that they cannot be stimulated by opioids. This prevents patients from feeling 'high' or euphoric when they take prescription opioids or illicit street drugs.
Since Suboxone, methadone and Vivitrol all work because they function similarly to opioids, they do come with some side effects. These side effects are much less intense in comparison to both opioids and opiates. This is why they are a popular treatment remedy. Take a look at some of the short-term and long-term side effects of each drug below.
The type of side effects associated with each drug should be carefully explored. Doctors should determine which drug is most suitable for each patient, based on their biological makeup. At times, some doctors may even switch patients to a different medication if the current one causes too many side effects.
Due to the differing chemical structure of each drug, they have varying addiction potentials. Out of all three medications, methadone has the highest risk of dependence. Suboxone contains buprenorphine. This drug has a "ceiling effect", which basically means that its effects will plateau after a certain dosage. Due to this plateau, Suboxone has a low risk of dependence. It is impossible for patients to develop any type of tolerance on the drug. Even if they were to take more and more of the drug, they are not likely to get high from it. The same can be said for Vivitrol. Much like Suboxone, it has a fairly low risk of dependence. This is because this drug is usually administered only once a month. On top of that, a medical professional will supervise the administration of the drug. Due to the way that Vivitrol is given to patients, it is also quite difficult to get addicted to it. Methadone, on the other hand, has a higher addiction potential than the rest of the drugs. In comparison to the other two, methadone is a stronger drug. It does not have a "ceiling effect", which is why it is often used to treat heroin addictions. This means that it is possible to build tolerance to methadone. A higher dosage results in a greater high.
While Suboxone, methadone and Vivitrol all play the same role in ORT, they have varying success rates. For optimal success, pair ORT with counselling and various types of therapies. The length of stay in addiction therapy will also have a profound impact. In general, patients have a higher chance at recovery if they opt for a longer stay at the rehab.
With that said, Suboxone treatments are much more effective if they are used for at least 55 days. Unfortunately, once completing the Suboxone treatments, 40% to 60% of patients relapsed. A higher relapse rate is associated with a shorter length of treatment. The success rate for Suboxone will also depend on how well the patients tapered off of the drug. Those who wean off of Suboxone too quickly have a higher risk of relapsing. On the other hand, those who wean off Suboxone gradually have a higher success rate.
Vivitrol is administered once a month via an injection. It is only administered once all alcohol and opioid molecules have left the body. Due to this reason, a naloxone challenge test must be performed before Vivitrol is prescribed. Those who develop symptoms of opioid withdrawals after the test should not start treatment. This medication is just as effective as Suboxone. Different study groups have seen different success rates. Study groups in UCLA have seen success rates of about 1 in 9. These individuals successfully recover from their addiction, and never relapse. It's important to note that not all patients will respond to Vivitrol. In fact, studies show that an average of only 40% of patients respond to this medication. As a result, your doctor may need to switch you over to a different medication if you're not getting any results.
Methadone is the standard care for opiate addictions. It has a success rate between 60% to 90%. The longer that patients use this treatment, the greater their success rate. While methadone has one of the highest success rates out of all the medications used in ORT, it needs to be administered carefully. As mentioned above, it has the highest risk of dependence. It's also easy for patients to overdose on this medication. As a result, it's imperative that patients gradually taper off of this drug. This means that a longer addiction treatment length is necessary when using methadone.
Patients who wean off of the medication gradually are less likely to experience physical symptoms. However, withdrawals are quite common still, and can last anywhere from hours to days. Here's a look at what to expect.
Other symptoms are similar to the flu. Those who are withdrawing from Suboxone may feel down under the weather.
Those addicted to many substances will go through a lengthier withdrawal timeline. In addition, those who attempt to quit cold turkey will experience more severe withdrawal symptoms.
Basically, Vivitrol withdrawal symptoms are non-existent or so mild that they shouldn't count. It is almost impossible to experience withdrawals from Vivitrol.
Since all three medications work in similar ways, the taper method is also similar in most cases. The taper schedule will be personalized to the needs of each patient. While some patients may be able to wean off of the drugs within a shorter time frame, others need more time. It's important to pay attention to the body, and how it's reacting to the taper.
It takes anywhere from 2 to 6 months to completely wean off of Suboxone use. The taper schedule will vary from patient to patient based on their body. With that said, once a dosage of 2 mg is reached, most doctors will follow the same schedule. They will reduce dosage by 0.5 mg every 2 weeks. Then, they will prescribe Suboxone every other day once a dosage of 0.5 mg is reached. Patients should not experience any physical withdrawal symptoms at this rate.
Decrease methadone dosage by 5 mg every 3 to 14 days. At this rate, patients should experience very little physical and psychological withdrawal symptoms. Once the dosage is lowered to about 20 mg, the decrease in dosage should be even more gradual.
The taper schedule for Vivitrol is a lot simpler. In most cases, doctors will simply stop administering any injections. Some doctors may administer a lower dosage each month before stopping. It all depends on the patient, and how they are reacting to the drug.
When weaning off of the drug, take this opportunity to focus on other therapies and counselling. If you're an artist, focus on your art. If you love to cook, focus on that. It's important that you don't obsess over the medication and the tapering schedule.
While each ORT medication can be used for months, and even years, the goal is to slowly wean off of them. Patients should try to live life without having to take any medications whatsoever. With that said, not all of these medications are suitable for long-term use. Long-term use of Suboxone can lead to severe side effects. As a result, most physicians only recommend using it for several months at most. Methadone, on the other hand, can be used for years; however, it is easy to build tolerance to this drug. Long-term use needs to be carefully monitored and supervised to prevent addiction. Supervision is also necessary to prevent misuse or abuse. Those who misuse methadone may get addicted to the drug. Vivitrol can also be used for long periods of time with ease. This drug is considered to be fairly safe. As a result, patients won't get addicted to it nor will they experience many intense side effects from long-term use. Unlike with Suboxone and methadone, Vivitrol treatments can only be started once all opioids are cleared from the system. Otherwise, the drug is not safe, and can cause harsh side effects.
If you're addicted to an opioid, ORT may be the way to go. It will reduce any withdrawal symptoms you may experience. It also makes it much easier to wean off of the opioids. With that said, the type of ORT that will be most effective for each patient will differ. The type of medication that the doctors will choose to use will depend on the amount of drugs taken, the length of the drug use and more. To get a better idea of what is most recommended for your situation, get a free phone assessment today. This will give you a better idea of what to expect on the road to sobriety. Our staff will walk you through the different medications and treatments we offer. We'll help you figure out what will work best for you, and what you should try. We'll also discuss the risks and benefits of each medication used in ORT, so you can make an informed decision.


text:
\section{Introduction}
The Schr\"odinger operator
 \begin{equation}
 \label{1}
  L y= -y''+v(x)y, \quad \quad \quad x \in \mathbb{R},
 \end{equation}
considered with a real-valued $\pi$-periodic potential $v \in
L^2([0,\pi]),$ is self-adjoint and its spectrum has a band-gap
structure, i.e., it consists of intervals $[\lambda^+_{n-1},
\lambda^-_{n}]$ separated by spectral gaps (instability zones)
$(\lambda^-_{n}, \lambda^+_{n})$, $n \in \mathbb{N}$. The Floquet
theory (e.g., see \cite{E}) shows that the endpoints $\lambda^-_{n}$,
$\lambda^+_{n}$ of these gaps are eigenvalues of the same
differential operator (\ref{1}) but considered on the interval
$[0,\pi]$ with periodic boundary conditions for even $n$ and
antiperiodic boundary conditions for odd $n$.

Hochstadt \cite{Ho1, Ho2} discovered that there is a close relation
between the rate of decay of the {\em spectral gap}
$\gamma_n=\lambda^+_n-\lambda^-_n$ and the smoothness of the
potential $v$. He proved that every finite zone potential is a
$C^\infty $-function, and moreover, \emph{if v is infinitely
differentiable then $\gamma_n$ decays faster than any power of
$1/n.$} Later several authors \cite{LP}-\cite{MT} studied this
phenomenon and showed that \emph{if $\gamma_n$ decays faster than any
power of $1/n,$ then v is infinitely differentiable}. Moreover,
Trubowitz \cite{Tr} proved that \emph{$v$ is analytic if and only if
$\gamma_n$ decays exponentially fast}.

If $v$ is a complex-valued function then the operator (\ref{1}) is
not self-adjoint and we cannot talk about spectral gaps. But
$\lambda^\pm_n $ are well defined for sufficiently large $n$ as
eigenvalues of (\ref{1}) considered on the interval $[0,\pi]$ with
periodic or antiperiodic boundary conditions, so we set again
$\gamma_n = \lambda_n^+ - \lambda_n^-$ and call it spectral gap.
Again the potential smoothness determines the decay rate of
$\gamma_n,$ but in general the opposite is not true. The decay rate
of $\gamma_n$ has no control on the smoothness of a complex valued
potential $v$ by itself as the Gasymov paper \cite{Gas} shows.

Tkachenko \cite{Tk92}--\cite{Tk94} discovered that the smoothness of
complex potentials could be controlled if one consider, together
with the spectral gap $\gamma_n,$ the deviation
$\delta^{Dir}_n=\lambda^+_n-\mu_n$, where $\mu_n$ is the closest to
$n^2$ Dirichlet eigenvalue of $L.$ He characterized in these terms
the $C^{\infty}$-smoothness and analyticity of complex valued
potentials $v.$  Moreover, Sansuc and Tkachenko \cite{ST} showed
that $v$ is in the Sobolev space $H^a$, $a\in \mathbb{N}$ if and
only if $\gamma_n $ and $\delta^{Dir}_n$ are in the weighted
sequence space $\ell_a^2=\ell^2((1+n^2)^{a/2})$.

The above results have been obtain by using Inverse Spectral Theory.
Kappeler and Mityagin \cite{KaMi01} suggested another approach based
on Fourier Analysis. To formulate their results, let us recall that
the smoothness of functions could be characterized by weights $\Omega
= (\Omega (k)),$  and the corresponding weighted Sobolev spaces are
defined by
$$
H(\Omega) = \{v(x) = \sum_{k \in \mathbb{Z}} v_k e^{2ikx}, \quad
\sum_{k \in \mathbb{Z}} |v_k|^2 (\Omega(k))^2 < \infty \}.
$$
A weight $\Omega $ is called sub-multiplicative, if $ \Omega (-k) =
\Omega (k) $ and $ \Omega (k+m) \leq \Omega (k)\Omega (m) $ for $ k,m
\geq 0. $ In these terms the main result in \cite{KaMi01} says that
if $\Omega $ is a sub-multiplicative weight, then
 \begin{equation}
  \label{i-ra}
 (A) \quad v\in H(\Omega) \quad \Longrightarrow \quad  (B) \quad
   (\gamma_n) ,  \left (\delta^{Dir}_n \right ) \in
  \ell^2(\Omega).
 \end{equation}
Djakov and Mityagin \cite{DM3,DM5, DM15} proved the inverse
implication $(B) \Rightarrow (A)$  under some additional mild
restrictions on the weight $\Omega. $  Similar results were obtained
for 1D Dirac operators (see \cite{DM7,DM6,DM15}).

The analysis in \cite{KaMi01, DM3, DM5, DM15} is carried out under
the assumption $v\in L^2([0,\pi]).$  Using quasi-derivatives, Djakov
and Mityagin \cite{DM16} developed a Fourier method for studying the
spectra of $L_{Per^\pm}$ and $L_{Dir}$ in the case of periodic
singular potentials and  extended the above results. They proved that
if  $v \in H^{-1}_{per} (\mathbb{R})$ and $\Omega$ is a weight of the
form $\Omega(m)=\omega(m)/|m|$ for  $m \neq 0,$ with $\omega$ being a
sub-multiplicative weight, then  $(A) \Rightarrow (B),  $ and
conversely, if in addition $(\log \omega (n))/n$ decreases to zero,
then $(B) \Rightarrow (A)$ (see Theorem 37 in \cite{DMH}).

A crucial step in proving the implications $(A) \Rightarrow (B)$ and
$(B) \Rightarrow (A)$ is the following statement (which comes from
Lyapunov-Schmidt projection method, e.g., see Lemma 21 in
\cite{DM15}): {\em For large enough $n,$ the number $\lambda = n^2 +
z, \; |z| <n/4 $ is a periodic or antiperiodic eigenvalue if and only
if $z$ is an eigenvalue of the matrix} $
\begin{pmatrix} \alpha_n(z) & \beta^+_n(z) \\ \beta^-_n(z) &
\alpha_n(z)\end{pmatrix}.$ The entrees of this matrix
$\alpha_n(z)=\alpha_n(z;v)$ and $\beta^{\pm}_n(z)=\beta^{\pm}_n(z;v)$
are given by explicit expressions in terms of the Fourier
coefficients of the potential $v$ and depend analytically on $z$ and
$v.$  The functionals $\beta^\pm_n $ give lower and  upper bounds for
the gaps and deviations (e.g., see Theorem 29 in  \cite{DMH}): {If
$v \in H^{-1}_{per} (\mathbb{R})$ then, for sufficiently large} $n,$
\begin{equation}
\label{A1} \frac{1}{72}(|\beta_n^+(z_n^*)|+|\beta_n^-(z_n^*)|) \leq
|\gamma_n|+|\delta^{Dir}_n| \leq 58
 (|\beta_n^+(z_n^*)|+|\beta_n^-(z_n^*)|),
\end{equation}
where $z_n^*=  \frac{1}{2}(\lambda^+_n + \lambda^-_n) - n^2.$ Thus,
the implications $(A) \Rightarrow (B)$ and $(B) \Rightarrow (A)$ are
equivalent, respectively, to
\begin{equation}
\label{A2} (\tilde{A}):\quad v\in H(\Omega) \quad \Longrightarrow
\quad (\tilde{B}): \quad (|\beta_n^+(z_n^*)|+|\beta_n^-(z_n^*)|)\in
\ell^2 (\Omega),
\end{equation}
and $(\tilde{B}) \Rightarrow (\tilde{A}).$ In this way the problem of
analyzing the relationship between potential smoothness and decay
rate of the sequence $(|\gamma_n| + |\delta_n^{Dir}|)$ is reduced to
analysis of the functionals $\beta^{\pm}_n (z).$

The asymptotic behavior  $\beta^{\pm}_n (z)$ (or $\gamma_n$ and
$\delta^{Dir}_n$) plays also a crucial role in studying the Riesz
basis property of the system of root functions of the operators
$L_{Per^\pm}.$ In \cite[Section 5.2]{DM15}, it is shown (for
potentials $v \in L^2 ([0,\pi])$) that if the ratio $\beta^+_n
(z_n^*)/\beta^-_n (z_n^*) $ is not separated from $0$ or $\infty$
then the system of root functions of $L_{Per^\pm}$ does not contain a
Riesz basis (see Theorem 71 and its proof therein). Theorem 1 in
\cite{DM25} (or Theorem 2 in \cite{DM25a}) gives, for wide classes of
$L^2$-potentials,  a criterion for Riesz basis property in the same
terms. In its most general form, for singular potentials, this
criterion reads as follows (see Theorem 19 in \cite{DM28}):

{\em Criterion 1. Suppose $v\in H^{-1}_{per}(\mathbb{R});$ then the
set of root functions of $L_{Per^\pm}(v)$ contains Riesz bases if
and only if
\begin{equation}
\label{DM1} 0 < \inf_{\gamma_n\neq 0} |\beta^-_n (z_n^*)|/|\beta^+_n
(z_n^*)|, \quad \sup_{\gamma_n\neq 0} |\beta^-_n (z_n^*)|/|\beta^+_n
(z_n^*)|< \infty,
\end{equation}
where $n$ is even (respectively odd) in the case of periodic
(antiperiodic) boundary conditions.}

In \cite{GT11}  Gesztesy and Tkachenko obtained the following result.

{\em Criterion 2.   If $v \in L^2 ([0,\pi]),$ then there is a Riesz
basis consisting of root functions of the operator $L_{Per^\pm}(v)$
if and only if
\begin{equation}
\label{GT1} \sup_{\gamma_n\neq 0}|\delta^{Dir}_n|/|\gamma_n|<\infty,
\end{equation}
where $n$ is even (respectively odd) in the case of periodic
(antiperiodic) boundary conditions.}

They also noted that a similar criterion holds if (\ref{GT1}) is
replaced by
\begin{equation}
\label{GT2} \sup_{\gamma_n\neq 0}|\delta^{Neu}_n|/|\gamma_n|<\infty,
\end{equation}
where $\delta^{Neu}_n=\lambda^+_n - \nu_n $ and $\nu_n $ is the $n$th
Neumann eigenvalue.

Djakov and Mityagin \cite[Theorem 24]{DM28} proved, for potentials
$v\in H^{-1}_{per}(\mathbb{R}),$ that the conditions ({\ref{DM1}) and
(\ref{GT1}) are equivalent, so (\ref{GT1}) gives necessary and
sufficient conditions for Riesz basis property for singular
potentials as well.

However, if $v \not \in L^2 ([0,\pi])$ (say, for potentials $v \in
L^1 ([0,\pi])$ or  $v\in H^{-1}_{per}(\mathbb{R})$) \emph{it is not
known whether (\ref{GT2}) is equivalent to Riesz basis property.}

In this paper, we show for potentials $v\in L^p([0,\pi]), $ $ p>1,$
that the Neumann version of Criterion 2 holds, and the potential
smoothness could be characterized by the decay rate of $ |\gamma_n|
+ |\delta_n^{Neu}|.$ More precisely, the following two theorems
hold.

\begin{Theorem}
\label{ThmNeu}  Suppose $v\in L^p([0,\pi]), \; p>1$ and $\Omega$  is
a weight of the form $\Omega(m)=\omega(m)/m$ for  $m \neq 0,$ where
$\omega$ is a sub-multiplicative weight. Then
\begin{equation}
\label{B10} v \in H(\Omega) \Longrightarrow (|\gamma_n|),
(|\delta^{Neu}_n|) \in \ell^2 (\Omega);
\end{equation}
conversely, if in addition $(\log \omega (n))/n$ decreases to zero,
then
\begin{equation}
\label{B20}  (|\gamma_n|), (|\delta^{Neu}_n|) \in \ell^2
(\Omega)\Longrightarrow v \in H(\Omega).
\end{equation}
If $ \lim \frac{\log \omega (n)}{n} >0,$ (i.e. $\omega $ is of
exponential type), then
\begin{equation}
\label{B30} (\gamma_n), (\delta^{Neu}_n) \in \ell^2 (\Omega) \;
\Rightarrow \; \exists  \varepsilon >0: \; v \in H(e^{\varepsilon
|n|} ).
\end{equation}
\end{Theorem}

 \begin{Theorem}
 \label{Theorem}
  If $v \in L^p ([0,\pi]), \; p>1,$ then there is a Riesz
basis consisting of root functions of the operator $L_{Per^\pm}(v)$
if and only if
\begin{equation}
\label{GT10} \sup_{\gamma_n\neq 0}|\delta^{Neu}_n|/|\gamma_n|<\infty,
\end{equation}
where $n$ is respectively even (odd) for periodic (antiperiodic)
boundary conditions.
 \end{Theorem}

 We do not prove
Theorem~\ref{ThmNeu} and Theorem~\ref{Theorem} directly, but show
that they are valid by reducing their proofs to Theorem~37 in
\cite{DMH} and Theorem~19 in \cite{DM28}, respectively. For this end
we prove the following.

 \begin{Theorem}
 \label{Asymp}
If $v \in L^p ([0,\pi]), \; p>1,$ then, for sufficiently large $n,$
\begin{equation}
\label{A10} \frac{1}{80} (|\beta_n^+(z_n^*)|+|\beta_n^-(z_n^*)|)
\leq |\gamma_n|+|\delta^{Neu}_n| \leq 19
 (|\beta_n^+(z_n^*)|+|\beta_n^-(z_n^*)|).
\end{equation}
\end{Theorem}

Next we show that Theorem \ref{Asymp} implies Theorem~\ref{ThmNeu}
and Theorem~\ref{Theorem}}. Fix $v \in L^1 ([0,\pi]).$ We may assume
without loss of generality that $ V(0):=\int_0^\pi v(t) dt = 0 $
(otherwise one may consider $\tilde{v}(t) = v(t) -\frac{1}{\pi}
V(0)$ since a constant shift of the potential results in a shift the
spectra but root functions, spectral gaps and deviations  do not
change). Then the function $Q(x)= \int_0^x v(t) dt, $ extended
periodically on $\mathbb{R},$ has the property that $v= Q^\prime $
so we may think that $v\in H^{-1}_{per}(\mathbb{R})$  (see
\cite{DM16}) for details). By Theorem~29 in \cite{DMH} and
Theorem~\ref{Asymp}, (\ref{A1}) and (\ref{A10}) hold simultaneously,
so the sequences $(|\gamma_n|+|\delta^{Dir}_n|)$ are
$(|\gamma_n|+|\delta^{Neu}_n|)$ are asymptotically equivalent.
Therefore, every claim in Theorem~\ref{ThmNeu} follows from the
corresponding assertion in
 \cite[Theorem 37]{DMH}.

On the other hand the asymptotic equivalence of
$|\gamma_n|+|\delta^{Dir}_n|$  and $|\gamma_n|+|\delta^{Neu}_n|$
imply the asymptotic equivalence of
$\frac{|\delta_n^{Neu}|}{|\gamma_n|}$ and
$\frac{|\delta^{Dir}|}{|\gamma_n|},$ so (\ref{GT1}) and (\ref{GT10})
hold simultaneously if $v \in L^p ([0,\pi]).$ By Theorem 24 in
\cite{DM28}, (\ref{GT1}) gives necessary and sufficient conditions
for the Riesz basis property if $v\in H^{-1}_{per} (\mathbb{R}).$
Hence, Theorem~\ref{Theorem} is proved.

\section{Preliminary Results}
\vspace{3mm}

We consider the Hill--Schr\"odinger operator
\begin{equation}
\label{11.1} Ly = -y^{\prime \prime} + v(x) y
\end{equation}
with a (complex--valued) potential $v\in L^1([0,\pi])$, subject to
the following boundary conditions ($bc$):

($a$) {\em periodic} $(Per^+): \quad y(0) = y (\pi), \;\; y^\prime
(0) = y^\prime (\pi) $;

($b$) {\em anti--periodic}  $(Per^-): \quad  y(0) = - y (\pi), \;\;
y^\prime (0) = - y^\prime (\pi) $;

($c$) {\em Dirichlet} $ (Dir): \quad y(0) = 0, \;\; y(\pi) = 0; $

($d$) {\em Neumann} $ (Neu): \quad y'(0) = 0, \;\; y'(\pi) = 0. $

The closed operator generated by $L$  in the domain $$ Dom(L_{bc}) =
\{f' \text{ is absolutely continuous} : \; \; f \in (bc), \; Lf\in
L^2([0,\pi]) \} $$ will be denoted by $L_{bc},$ or $L_{Per^+},
L_{Per^-}, L_{Dir}, L_{Neu}$ correspondingly. $Dom(L_{bc})$ is dense
in $L^2([0,\pi])$ and  $L_{bc}=L_{bc}(v)$ satisfies
 \begin{equation}
 \label{L-L}
 (L_{bc}(v))^*=L_{bc}(\bar{v})\;\;\;\;\;bc=Per^\pm,\;Dir,\;Neu,
 \end{equation}
where $(L_{bc}(v))^*$ is the adjoint operator and  $\bar{v}$ is the
conjugate of $v$, i.e., $\bar{v}(x)=\overline{v(x)}$. In the
classical case where $v\in L^2([0,\pi])$, (\ref{L-L}) is a well known
fact. In the case where $v\in H^{-1}_{per}(\mathbb{R})$ it is
explicitly stated and proved for $bc=Per^\pm,\;Dir$ in \cite{DM16},
see Theorem~6 and Theorem~13 there. Following the same argument as in
\cite{DM16} one can easily see that it holds for $bc=Neu$ as well.

If $v=0$ we write $L^0_{bc},$ (or simply $L^0$). The spectra and
eigenfunctions of $L^0_{bc}$ are as follows:

($\tilde{a}$) $ Sp (L^0_{Per^+}) = \{n^2, \; n = 0,2,4, \ldots \};$
its eigenspaces are  $\mathcal{E}^0_n = Span \{e^{\pm inx} \} $ for
$n>0 $ and $E^0_0 = \{ const\}, \; \; \dim \mathcal{E}^0_n = 2 $ for
$n>0, $ and $\dim \mathcal{E}^0_0 = 1. $

($\tilde{b}$) $ Sp (L^0_{Per^-}) = \{n^2, \; n = 1,3,5, \ldots \};$
its eigenspaces are  $\mathcal{E}^0_n = Span \{e^{\pm inx} \}, $ and
$ \dim \mathcal{E}^0_n = 2. $

($\tilde{c}$) $ Sp (L^0_{Dir}) = \{n^2, \; n \in \mathbb{N} \};$ each
eigenvalue $n^2 $ is simple; its eigenspaces are $\mathcal{S}^0_n =
Span \{ s_n (x)\}, $ where $s_n (x)$ is the corresponding normalized
eigenfunction $ s_n (x) = \sqrt{2} \sin nx.$

($\tilde{d}$) $ Sp (L^0_{Neu}) = \{n^2, \; n \in \{0\}\cup\mathbb{N}
\};$ each eigenvalue $n^2 $ is simple; its eigenspaces are
$\mathcal{C}^0_n = Span \{ c_n (x)\}, $ where $c_n (x)$ is the
corresponding normalized eigenfunction $ c_0 (x) = 1$, $c_n (x) =
\sqrt{2} \cos nx $ for $n>0.$

 The sets of indices $2\mathbb{Z}$,
 $2\mathbb{Z}+1$,  $\mathbb{N}$, and $\{0\}\cup\mathbb{N}$ will be denoted by
 $\Gamma_{Per^+}$, $\Gamma_{Per^-}$, $\Gamma_{Dir}$ and
 $\Gamma_{Neu}$, respectively. For each $bc$
 we consider the corresponding canonical orthonormal basis of $L^0_{bc}$, namely
 $\mathcal{B}_{Per^+}=\{e^{inx}\}_{n\in\Gamma_{Per^+}}$,
  $\mathcal{B}_{Per^-}=\{e^{inx}\}_{n\in\Gamma_{Per^-}}$,
 $\mathcal{B}_{Dir}=\{s_n(x)\}_{n\in\Gamma_{Dir}}$, $\mathcal{B}_{Neu}=\{c_n(x)\}_{n\in\Gamma_{Neu}}.$

 The matrix elements of an operator $A$ with respect to the basis $\mathcal{B}_{bc}$ will
 be denoted by $A^{bc}_{nm}$, where $n,m \in \Gamma_{bc}$. The norm of an
 operator $A$ from $L^a([0,\pi])$ to $L^b([0,\pi])$ for $a,b \in [1,\infty]$ will be denoted
 by $\|A\|_{a\rightarrow b}$. We may also write $\|A\|$ instead of $\|A\|_{2\rightarrow
 2}$.

Let $V$ denote the operator of multiplication by $v$, i.e., $
(Vf)(x)=v(x)f(x).$ Then $L_{bc}=L^0_{bc}+V$ and we may use the
perturbation formula (see \cite{DM16}, equation (5.13))
 \begin{equation}
 \label{2-res}
  R_{\lambda}=R^0_{\lambda}+\sum_{m=1}^{\infty}K_{\lambda}(K_{\lambda}V
  K_{\lambda})^m K_{\lambda},
  \end{equation}
  where $R_{\lambda}=(\lambda-L_{bc})^{-1}$, $R^0_{\lambda}=(\lambda-L^0_{bc})^{-1}$ and $K_{\lambda}$ is
  a square root of $R^0_{\lambda}$, i.e.,  $K_{\lambda}^2=
  R^0_{\lambda}$. Of course, (\ref{2-res}) makes sense only if the
  series on the right converges.

  Since the matrix representation of $R^0_{\lambda}$ is
  \begin{equation}
    (R^0_{\lambda})^{bc}_{ij}=\frac{1}{\lambda-j^2}\delta_{ij},
    \quad i,j\in \Gamma_{bc}
    \end{equation}
we can define a square root $K=K_{\lambda}$ of $R^0_{\lambda}$ by
choosing its matrix representation as
    \begin{equation}
    (K_{\lambda})^{bc}_{ij}=\frac{1}{(\lambda-j^2)^{1/2}}\delta_{ij}, \quad i,j\in
    \Gamma_{bc},
    \end{equation}
where $z^{1/2}$ is the principal square root. If
 \begin{equation}
 \label{2-kvk}
 \|K_{\lambda}V K_{\lambda}\|_{2\rightarrow 2} < 1,
 \end{equation}
then $R_{\lambda}$ exists. Assuming only $v\in
H_{per}^{-1}(\mathbb{R})$, Djakov and Mityagin showed (see
\cite{DM16}, Lemmas 19 and 20) that there exists $N>0$, $N\in
\Gamma_{bc}$ such that (\ref{2-kvk}) holds for $\lambda \in H^N
\backslash R_N$ and also for all $n>N$, $n\in \Gamma_{bc}$
(\ref{2-kvk}) holds for $\lambda \in H_n \backslash D_n$ if
$bc=Per^{\pm}$ and for $\lambda \in G_n \backslash D_n$ if $bc=Dir$
where
 \begin{equation}
 H^N=\{\lambda \in \mathbb{C} : Re \; \lambda  \leq N^2+N\},
 \end{equation}
 \begin{equation}
 R_N=\{\lambda \in \mathbb{C} : -N\leq Re \;\lambda\leq N^2+N,\quad |Im \lambda|<N\},
 \end{equation}
 \begin{equation}
 H_n=\{\lambda \in \mathbb{C} : (n-1)^2\leq Re \;\lambda \leq (n+1)^2\},
 \end{equation}
 \begin{equation}
 G_n=\{\lambda \in \mathbb{C} : n^2-n\leq Re \;\lambda \leq n^2+n\},
 \end{equation}
 \begin{equation}
 D_n=\{ \lambda \in \mathbb{C} : |\lambda -n^2|< n/4\}.
 \end{equation}
Therefore, the following localization of the spectra holds:
  \begin{equation}
  \label{local}
   Sp(L_{bc}) \subset R_N \cup \bigcup_{n>N, n\in
\Gamma_{bc}}D_n, \;\;\; bc=Per^\pm,\;Dir.
 \end{equation}
Moreover, using the method of continuous parametrization of the
potential $v$, they showed that spectrum is discrete and
$$
 \sharp(Sp(L_{Per^+})\cap R_N)=2N+1, \quad  \sharp(Sp(L_{Per^+})\cap D_n)=2,\quad n>N, n\in
 \Gamma_{Per^+},
$$$$
 \sharp(Sp(L_{Per^-})\cap R_N)=2N, \quad  \sharp(Sp(L_{Per^-})\cap D_n)=2,\quad n>N, n\in
 \Gamma_{Per^-} ,
$$$$
 \sharp(Sp(L_{Dir})\cap R_N)=N, \quad  \sharp(Sp(L_{Dir})\cap D_n)=1,\quad n>N, n\in
 \Gamma_{Dir}.
$$

For Neumann $bc$, using the same argument as in \cite{DM16} one can
similarly localize and count the spectrum $Sp(L_{Neu})$ after
showing that (\ref{2-kvk}) holds for $ \lambda \not \in R_N \cup
\left \{\bigcup_{n>N, n\in \Gamma_{Neu}}D_n \right \}.$ However in
the case where $v\in L^1([0,\pi])$ we  estimate $\|K_{\lambda}V
K_{\lambda}\|$ explicitly for all $bc$ since we  need this estimate
later.
 \begin{Proposition}
 \label{2-p-kvk} If  $v\in L^1([0,\pi])$, there
 exist $C=C(v)$ and $N>0$, $N\in\Gamma_{bc}$  such that
\begin{equation}
 \|K_{\lambda}V K_{\lambda}\| \leq C \frac{\log N}{\sqrt{N}} \quad
 \text{for} \quad \lambda \in H^N \backslash R_N.
 \end{equation}
 Moreover for all $n>N$, $n\in\Gamma_{bc}$,
 \begin{equation}
 \label{KVK}
 \|K_{\lambda}V K_{\lambda}\| \leq C \frac{\log n}{n}
 \end{equation}
 for $\lambda \in H_n \backslash D_n$ if
 $bc=Per^{\pm},$ and for $\lambda \in G_n \backslash D_n$ if $bc=Dir,
 Neu$.
 \end{Proposition}
\begin{proof} Since the matrix
representation of $K_{\lambda}V K_{\lambda}$ is
 \begin{equation}
    (K_{\lambda}V
    K_{\lambda})^{bc}_{kj}=
    \sum_{k,j\in\Gamma_{bc}}
    \frac{V^{bc}_{kj}}{(\lambda-k^2)^{1/2}(\lambda-j^2)^{1/2}}
    \end{equation}
we can estimate its Hilbert-Schmidt norm as
 \begin{equation}
 \label{ineq}
 \|K_{\lambda}VK_{\lambda}\|_{HS}=
 \bigg{(}\sum_{k,j\in\Gamma_{bc}}\frac{|V^{bc}_{kj}|^2}
 {|\lambda-k^2||\lambda-j^2|}\bigg{)}^{1/2}\quad\quad\quad\quad
 \end{equation}
 $$\quad\quad\quad\quad \quad\quad\quad\leq \sup_{k,j \in
 \Gamma_{bc}}|V^{bc}_{kj}|\bigg{(}\sum_{k\in\Gamma_{bc}}
 \frac{1}{|\lambda-k^2|}\sum_{j\in\Gamma_{bc}}\frac{1}{|\lambda-j^2|}\bigg{)}^{1/2}.$$
From (5.27) and (5.28) in \cite{DM16} we know that there exists an
$N>0$ such that for all $n>N$ there exists an absolute constant
$C_1$ such that
 \begin{equation}
 \label{log}
 \sum_{k\in\Gamma_{bc}}\frac{1}{|\lambda-k^2|} \leq  C_1\frac{\log n}{n}
 \end{equation}
for $\lambda \in H_n\setminus D_n$ if $bc=Per^{\pm}$ and $\lambda
\in G_n\setminus D_n$ if $bc=Dir, Neu$. On the other hand to
estimate $\sup_{k,j \in
 \Gamma_{bc}}|V^{bc}_{kj}|$ let us denote the Fourier coefficients of $v$
with respect to the bases $\mathcal{B}_{Per^+}$ and
$\mathcal{B}_{Neu}$ by $V_+(k)$ and $V_c(k)$, respectively, i.e.,
 \begin{equation}
  V_+(k)=\frac{1}{\pi}\int_0^{\pi}v(x)e^{-ikx}dx, \quad k\in
  \Gamma_{Per^+},
  \end{equation}
     \begin{equation}
  V_c(k)=\frac{1}{\pi}\int_0^{\pi}v(x)\sqrt{2}\cos(kx)dx, \quad k\in \Gamma_{Neu}.
  \end{equation}
  Then one can easily see that
  \begin{equation} \label{r1} V^{Per^\pm}_{kj}=V_+(j-k),  \quad k,j \in
\Gamma_{Per^\pm}, \end{equation}
  \begin{equation}
  \label{r2}  V^{Dir}_{kj}=V_c(|j-k|)-V_c(j+k),  \quad
k,j \in \Gamma_{Dir},\end{equation}
 \begin{equation}
 \label{r3} V^{Neu}_{kj}=V_c(|j-k|)+V_c(j+k),
\quad k,j \in \Gamma_{Neu}.\end{equation}
 Hence
  \begin{equation}
  \label{in1}
  \sup_{k,j \in
 \Gamma_{Per^\pm}}|V^{Per^\pm}_{kj}| \leq \sup_{k \in \Gamma_{Per^+}}|V_+(k)| \leq
 \|v\|_1,
 \end{equation}
 and
 \begin{equation}
 \label{in2}
 \sup_{k,j \in
 \Gamma_{Dir}}|V^{Dir}_{kj}|,\;\; \sup_{k,j \in
 \Gamma_{Neu}}|V^{Neu}_{kj}| \leq 2\sup_{k \in \Gamma_{Neu}}|V_c(k)| \leq
 2\sqrt{2}\|v\|_1.
 \end{equation}
Now, together with (\ref{log}), (\ref{in1}) and (\ref{in2}),
(\ref{ineq}) implies that
 \begin{equation}
\|K_{\lambda}VK_{\lambda}\|_{HS} \leq 3\|v\|_p C_1\frac{\log n}{n}
 \end{equation}
for $\lambda \in H_n\setminus D_n$ if $bc=Per^{\pm}$ and $\lambda
\in G_n\setminus D_n$ if $bc=Dir, Neu$.

On the other hand, if $\lambda=x+it \in H^N\backslash R_N, \;$ with
$n^2 -n \leq x < n^2 +n$ in the case when $x\geq 0,$ then one can
see that
\begin{equation}
\label{lamda-kkare} |\lambda-k^2| \geq
\begin{cases}(k^2+N)/\sqrt{2} & \text{if} \quad x \leq 0, \\
(|n^2-k^2|+2N)/2\sqrt{2} & \text{if}
\quad x>0, \quad k \neq \pm n, \\
N & \text{if} \quad x >0, \quad k = \pm n.
\end{cases}
\end{equation}
By Lemma 79 in \cite{DM15}, for large enough $N$ we also have the
inequality
\begin{equation}
\label{101.11a} \sum_k \frac{1}{|n^2 -k^2|+N} \leq C_2 \frac{\log
N}{\sqrt{N}},
\end{equation}
where $C_2$ is an absolute constant. Combining (\ref{ineq}),
(\ref{in1}), (\ref{in2}), (\ref{lamda-kkare}), and (\ref{101.11a})
we obtain for large enough $N$ that
 \begin{equation}
\|K_{\lambda}VK_{\lambda}\|_{HS} \leq C_3 \|v\|_p \frac{\log
N}{\sqrt{N}} \quad \text{for} \quad \lambda \in H^N \backslash R_N,
 \end{equation}
where $C_3$ is an absolute constant. We complete the proof noting
that Hilbert-Schmidt norm of an operator dominates its $L^2$ norm.
\end{proof}

\begin{Proposition} For any potential $v\in L^1([0,\pi])$,
the spectrum of the operator $L_{Neu}(v)$ is discrete. Moreover
there exists an integer $N$ such that
  \begin{equation}
  \label{local2}
   Sp(L_{Neu}) \subset R_N \cup \bigcup_{n>N, n\in
\Gamma_{Neu}}D_n,
 \end{equation}
and \begin{equation}
 \sharp(Sp(L_{Neu})\cap R_N)=N+1, \quad
 \sharp(Sp(L_{Neu})\cap D_n)=1,\quad n>N, n\in
 \Gamma_{Neu}.
 \end{equation}
 \end{Proposition}

\begin{proof} Apply the proof of Theorem 21 in \cite{DM16} but use
Proposition \ref{2-p-kvk} instead of Lemmas 19 and 20 in
\cite{DM16}.
\end{proof}

\section{Estimates of $\|P_n-P^0_n\|_{{}_{{}_{2\rightarrow\infty}}}$}

For $bc=Per^\pm, Dir$ or $Neu$, we consider the Cauchy-Riesz
projections
\begin{equation}
\label{PP0} P_n = \frac{1}{2 \pi i} \int_{C_n} R_\lambda d\lambda,
\quad P^0_n = \frac{1}{2 \pi i} \int_{C_n} R^0_\lambda d\lambda,
\end{equation}
where $C_n=\partial D_n$. In this section we estimate the norms
$\|P_n-P^0_n\|_{{}_{{}_{2\rightarrow\infty}}}$ and
$\|D(P_n-P^0_n)\|_{{}_{{}_{2\rightarrow\infty}}}$, where
$D=\frac{d}{dx}$. First we consider two technical lemmas.

\begin{Lemma}
\label{Alem} For every $p\in(1,2]$ there is a constant $T_p$ such
that
 \begin{equation} \label{Ax1}
 \bigg{(}\sum_{k\in \Gamma_{bc}}^{\infty}\frac{k^{p\varepsilon}}
 {|\lambda-k^2|^p}\bigg{)}^{1/p} \leq
 \frac{T_p}{n^{(1-\varepsilon)}} \quad \text{for} \quad \lambda \in C_n,
 \end{equation}
where $\varepsilon =0,1$ and $bc=Per^\pm, Dir, Neu$.
\end{Lemma}

\begin{proof} First note that for each $bc=Per^\pm, Dir, Neu$
 \begin{equation}
\label{Ap-sum}
 \sum_{k\in \Gamma_{bc}}^{\infty}\frac{|k|^{p\varepsilon}}{|\lambda-k^2|^p}
 \leq 2\sum_{k=0}^{\infty}\frac{|k|^{p\varepsilon}}{|\lambda-k^2|^p}.
 \end{equation}
If $\lambda \in C_n$ and $k \neq n$, then $|\lambda -n^2|=n/4,$ so
 \begin{equation}
 \label{Ap-sum22}
  |\lambda-k^2| \geq |n^2-k^2|-\frac{n}{4} \geq \frac{1}{2}|n^2-k^2|
 \end{equation}
because $|n^2-k^2| \geq 2n-1 \geq n.$ Hence
 \begin{equation}
 \label{AI1}
 \sum_{{\tiny{\begin{array}{c}k=0 \\(k \neq n)
\end{array}}}}^{\infty}\frac{k^{p\varepsilon}}{|\lambda-k^2|^p}
\leq 2^p \sum_{{\tiny{\begin{array}{c}k=0 \\(k \neq n)
\end{array}}}}^{\infty}\frac{k^{p\varepsilon}}{|n-k|^p|n+k|^p},
 \end{equation}
and the latter sum does not exceed
 \begin{equation}
 \label{ApI2}
 \frac{2^p}{n^{(1-\varepsilon)p}}\sum_{{\tiny{\begin{array}{c}k=0
\\(k \neq n)
\end{array}}}}^{\infty}\frac{1}{|n-k|^p}\leq
\frac{2^{p+1}}{n^{(1-\varepsilon)p}}\sum_{\tiny{k=1}}^{\infty}\frac{1}{k^p}.
\end{equation}
 Therefore, combining (\ref{Ap-sum}), (\ref{AI1}) and (\ref{ApI2}),
and taking into account that, for $\lambda\in C_n$, the $n^{th}$
summand of the right hand side of (\ref{Ap-sum}) is
$4^p/n^{(1-\varepsilon)p}$, we finally get
 $$ \label{Ax9}
 \sum_{k\in \Gamma_{bc}}^{\infty}\frac{k^{p\varepsilon}}{|\lambda-k^2|^p} \leq
 \frac{T^p_p}{n^{(1-\varepsilon)p}}
 $$
 with $T^p_p=2(4^p+
 2^{p+1}\sum_{\tiny{k=1}}^{\infty}\frac{1}{k^p})$. Hence (\ref{Ax1})
 holds.
\end{proof}

 \begin{Lemma}
\label{Av} Let $v\in L^p([0,\pi])$, $p\in (1,2]$, and let
$1/p+1/q=1$. Then, for $bc=Per^\pm,\; Dir,\; Neu$,  we have \begin{equation}
\label{Ag}
\bigg{(}\sum_{k\in\Gamma_{bc}}|V^{bc}_{kj}|^q\bigg{)}^{1/q} \leq
5\|v\|_p \quad \forall j\in\Gamma_{bc}. \end{equation}

\end{Lemma}
\begin{proof}
 From (\ref{r1}), (\ref{r2}) and (\ref{r3}) it follows
 \begin{equation} \label{Ac1}
\left(\sum_{k\in\Gamma_{bc}}|V^{bc}_{kj}|^q \right)^{1/q}=
 \bigg{(}\sum_{k\in\Gamma_{bc}}|V_+(j-k)|^q\bigg{)}^{1/q}=\|V_+\|_{\ell^q}
 \end{equation}
if $bc= Per^\pm,$ and
 \begin{equation}  \label{Ac2}
 \bigg{(}\sum_{k\in\Gamma_{bc}}|V^{bc}_{kj}|^q\bigg{)}^{1/q} \leq
\bigg{(}\sum_{k\in\Gamma_{bc}}|V_c(|j-k|)|^q\bigg{)}^{1/q}+
\bigg{(}\sum_{k\in\Gamma_{bc}}|V_c(j+k)|^q\bigg{)}^{1/q}
 \end{equation}
 $$ \leq 3\|V_c\|_{\ell^q} \quad \quad
 \quad\quad \quad \quad \quad \quad \quad \quad \quad$$
 if $bc= Dir$, $Neu$. On the other hand, by the Haussdorff-Young
(Hausdorff-Young Theorem, \cite[Theorem XII.2.3]{Zy90}),
 we have
 $\|V_+\|_{\ell^q} \leq \|v\|_p$ and $\|V_c\|_{\ell^q} \leq
 \sqrt{2}\|v\|_p$. These inequalities, together with (\ref{Ac1}) and
 (\ref{Ac2}), imply (\ref{Ag}).
\end{proof}

\begin{Proposition}
\label{Proposition} Let $D=\frac{d}{dx}$, $P_n$ and $P^0_n$ be
defined by (\ref{PP0}), and let $L=L_{bc}$ with $bc=Per^\pm, Dir,
Neu.$ If $v\in L^p([0,\pi])$, $p\in(1,2],$ then we have, for large
enough $n,$
 \begin{equation} \label{P-P0}
\|P_n-P^0_n\|_{{}_{{}_{2\rightarrow\infty}}} \leq \frac{M}{n} \end{equation}
and \begin{equation} \label{DP-DP0}
\|D(P_n-P^0_n)\|_{{}_{{}_{2\rightarrow\infty}}} \leq M, \end{equation} where
$M=M(v)$.
\end{Proposition}
\begin{proof} In view of (\ref{PP0}),
 \begin{equation}
 \label{3-P}
\|P_n-P^0_n\|_{{}_{{}_{2\rightarrow\infty}}} \leq \frac{1}{2 \pi }
\int_{C_n}\|R_\lambda-R^0_\lambda\|_{{}_{{}_{2\rightarrow\infty}}}d|\lambda|\leq
n\sup_{\lambda \in C_n}
\|R_\lambda-R^0_\lambda\|_{{}_{{}_{2\rightarrow\infty}}}.
 \end{equation}
In order to estimate
$\|D(P_n-P^0_n)\|_{{}_{{}_{2\rightarrow\infty}}}$, first we note that
 \begin{equation}
 \label{3-D-P}
D(P_n-P^0_n) =
\frac{1}{2\pi}\int_{C_n}D(R_\lambda-R^0_\lambda)d\lambda.
 \end{equation}
 Indeed, using integration by parts twice one can easily see that
 \begin{equation}
 \label{3-D-D-2}
\bigg{\langle} D\int_{C_n}(R_\lambda-R^0_\lambda)f d\lambda ,
g\bigg{\rangle}=\bigg{\langle} \int_{C_n}D(R_\lambda-R^0_\lambda) f
d\lambda,g \bigg{\rangle}
 \end{equation}
  for all $f\in L^2([0,\pi])$ and $g \in C^{\infty}_0([0,\pi])$. Since
  $C^{\infty}_0([0,\pi])$ is dense in
  $L^2([0,\pi]),$ (\ref{3-D-D-2}) implies (\ref{3-D-P}).
  Hence
   \begin{equation}
   \label{3-D-P-2}
\|D(P_n-P^0_n)\|_{{}_{{}_{2\rightarrow\infty}}} \leq
\frac{1}{2\pi}\int_{C_n}
\|D(R_\lambda-R^0_\lambda)\|_{{}_{{}_{2\rightarrow\infty}}}d|\lambda|
 \end{equation}
$$ \quad\quad\quad\quad\quad\quad\quad\leq n\sup_{\lambda \in C_n}
\|D(R_\lambda-R^0_\lambda)\|_{{}_{{}_{2\rightarrow\infty}}}.
$$
By (\ref{3-P}) and (\ref{3-D-P-2}) we see that in order to obtain
(\ref{P-P0}) and (\ref{DP-DP0}) it is enough to find good estimates
for $\|R_\lambda-R^0_\lambda\|_{{}_{{}_{2\rightarrow\infty}}}$ and
$\|D(R_\lambda-R^0_\lambda)\|_{{}_{{}_{2\rightarrow\infty}}}$ for
$\lambda\in C_n$. Next we will do that.

  Let $\varepsilon$ denote either $1$ or $0$, so
$D^\varepsilon=D$ for $\varepsilon=1$ and $D^\varepsilon=I$ for $\varepsilon=0$. Consider the operator
$D^{\varepsilon}(R_{\lambda}-R^0_{\lambda})$ for $\lambda\in C_n$. By
(\ref{2-res}) we can write
 \begin{equation}
 D^\varepsilon (R_{\lambda}-R^0_{\lambda})=D^\varepsilon
 R^0_{\lambda}VR^0_{\lambda}+\sum_{m=1}^{\infty}D^\varepsilon
 R^0_{\lambda}VK_{\lambda}(K_{\lambda}VK_{\lambda})^m K_{\lambda}.
 \end{equation}
Hence
 \begin{equation}
 \label{3-de}
 \|D^\varepsilon (R_{\lambda}-R^0_{\lambda})\|_{{}_{{}_{2\rightarrow\infty}}} \leq \|D^\varepsilon
 R^0_{\lambda}VR^0_{\lambda}\|_{{}_{{}_{2\rightarrow\infty}}}
 \quad\quad\quad\quad\quad\quad\quad\quad\quad\end{equation}
$$\quad\quad\quad\quad\quad\quad\quad\quad\quad+\sum_{m=1}^{\infty}\|D^\varepsilon
 R^0_{\lambda}VK_{\lambda}\|_{{}_{{}_{2\rightarrow\infty}}}
 \|K_{\lambda}VK_{\lambda}\|_{{}_{{}_{2\rightarrow2}}}^m
 \|K_{\lambda}\|_{{}_{{}_{2\rightarrow 2}}}.
$$
Since $\lambda\in C_n$, $\|K_{\lambda}VK_{\lambda}\|\leq 1/2$ for
sufficiently large $n$'s by Proposition \ref{2-p-kvk}. So
(\ref{3-de}) implies \begin{equation}
 \label{3-de2}
 \|D^\varepsilon (R_{\lambda}-R^0_{\lambda})\|_{{}_{{}_{2\rightarrow\infty}}} \leq \|D^\varepsilon
 R^0_{\lambda}VR^0_{\lambda}\|_{{}_{{}_{2\rightarrow\infty}}}
+2\|D^\varepsilon
R^0_{\lambda}VK_{\lambda}\|_{{}_{{}_{2\rightarrow\infty}}}
\|K_{\lambda}VK_{\lambda}\|
 \|K_{\lambda}\|.
 \end{equation}
It is easy to see that
 \begin{equation} \label{3-K}
\|K_{\lambda}\|=\sup_{k \in
\Gamma_{bc}}{\frac{1}{|\lambda-k^2|^{1/2}}}=
\frac{1}{|\lambda-n^2|^{1/2}}=\frac{2}{\sqrt{n}},\;\;\;
 \lambda\in C_n.
 \end{equation}
 Next we find estimates for $\|D^\varepsilon
 R^0_{\lambda}VR^0_{\lambda}\|_{{}_{{}_{2\rightarrow\infty}}}$ and $\|D^\varepsilon
 R^0_{\lambda}VK_{\lambda}\|_{{}_{{}_{2\rightarrow\infty}}}.$
 If we denote the Fourier coefficients of a function $f \in L^2([0,\pi])$
 with respect to the basis $\mathcal{B}_{bc}$ by $f^{bc}_j$, one can easily see that
 \begin{equation}
 (D^\varepsilon R^0_{\lambda}VR^0_{\lambda}f)(x)=\sum_{k,j\in\Gamma_{bc}}\frac{(i
k)^\varepsilon
V^{bc}_{kj}f^{bc}_j}{(\lambda-k^2)(\lambda-j^2)}e^{ikx},\;\;\;bc=Per^\pm.
 \end{equation}
 Noting also that $Ds_{k}=kc_k$ and $Dc_{k}=-ks_k$, we obtain for
 $bc=Dir,Neu$
  \begin{equation}
 (D^\varepsilon
 R^0_{\lambda}VR^0_{\lambda}f)(x)=\sum_{k,j\in\Gamma_{Dir}}\frac{k^\varepsilon
 V^{Dir}_{kj}f^{Dir}_j}{(\lambda-k^2)(\lambda-j^2)}\big{(}(1-\varepsilon)s_k(x)+\varepsilon
 c_k(x)\big{)},
 \end{equation}
 \begin{equation}
 (D^\varepsilon R^0_{\lambda}VR^0_{\lambda}f)(x)=\sum_{k,j\in\Gamma_{Neu}}\frac{k^\varepsilon
 V^{Neu}_{kj}f^{Neu}_j}{(\lambda-k^2)(\lambda-j^2)}\big{(}(1-\varepsilon)c_k(x)-\varepsilon
 s_k(x)\big{)}.
 \end{equation}
Since all basis functions are bounded above by $\sqrt{2}$ with
respect to supremum norm, the above expressions show that for each
$bc$ we have
 \begin{equation}\label{2I3}\|D^\varepsilon R^0_{\lambda}VR^0_{\lambda}f\|_{\infty} \leq
 2\sum_{k,j\in\Gamma_{bc}}\frac{|k|^\varepsilon
 |V^{bc}_{kj}||f^{bc}_j|}{|\lambda-k^2||\lambda-j^2|}.
 \end{equation}
We first apply the H\"older inequality about the index $k,$ and then
Lemma \ref{Alem} and Lemma \ref{Av} to get
 \begin{equation}\label{2I4}\|D^\varepsilon R^0_{\lambda}VR^0_{\lambda}f\|_{\infty} \leq 2
 \sum_{j\in\Gamma_{bc}}\frac{1}{|\lambda-j^2|}|f^{bc}_{j}|
 \bigg{(}\sum_{k\in\Gamma_{bc}}\frac{|k|^{p\varepsilon}}{|\lambda-k^2|^p}\bigg{)}^{1/p}
 \bigg{(}\sum_{k\in\Gamma_{bc}}|V^{bc}_{kj}|^q\bigg{)}^{1/q}
 \end{equation}
$$\leq 10 T_p \|v\|_p\frac{1}{n^{(1-\varepsilon)}}
 \sum_{j\in\Gamma_{bc}}\frac{1}{|\lambda-j^2|}|f^{bc}_{j}|.\quad\quad\quad$$
Applying the Cauchy inequality and Lemma \ref{Alem} one more time,
we finally get
 \begin{equation}\label{2I5}\|D^\varepsilon R^0_{\lambda}VR^0_{\lambda}f\|_{\infty}
\leq 10 T_2 T_p \|v\|_p\frac{1}{n^{(2-\varepsilon)}}\|f\|_2.
 \end{equation}
In a similar way it follows that
 \begin{equation}
 \label{2I5a}\|D^\varepsilon R^0_{\lambda}VK_{\lambda}f\|_{\infty}
 \leq  2\sum_{k,j\in\Gamma_{bc}}\frac{|k|^\varepsilon
 |V^{bc}_{kj}||f^{bc}_j|}{|\lambda-k^2||\lambda-j^2|^{1/2}},
 \end{equation}
so using the same argument as above but with (\ref{log}) at the last
step, we obtain
 \begin{equation}
 \label{2I6}
  \|D^\varepsilon
R^0_{\lambda}VK_{\lambda}f\|_{\infty} \leq 10 C_1^{1/2}T_p
\|v\|_p\frac{(\log n)^{1/2}}{n^{(3/2-\varepsilon)}}\|f\|_2.
 \end{equation}
Hence
 \begin{equation} \label{2I7}
 \|D^\varepsilon R^0_{\lambda}VR^0_{\lambda}\|_{{}_{{}_{2\rightarrow\infty}}}
 \leq \frac{\tilde{M}}{n^{(2-\varepsilon)}},
 \end{equation}
and
 \begin{equation} \label{2I8}
 \|D^\varepsilon R^0_{\lambda}VK_{\lambda}\|_{{}_{{}_{2\rightarrow\infty}}}
 \leq \tilde{M}\frac{(\log n)^{1/2}}{n^{(3/2-\varepsilon)}},
 \end{equation}
where $\tilde{M}=10 T_p \|v\|_p(T_2+ C_1^{1/2})$. Now we apply
(\ref{KVK}), (\ref{3-K}), (\ref{2I7}) and (\ref{2I8}) to the right
hand side of (\ref{3-de2}) and obtain
 \begin{equation}
\label{3-de3} \|D^\varepsilon
(R_{\lambda}-R^0_{\lambda})\|_{{}_{{}_{2\rightarrow\infty}}} \leq
\frac{\tilde{M}}{n^{(2-\varepsilon)}}\bigg{(}1+C_1\frac{(\log
n)^{3/2}}{n}\bigg{)} \leq \frac{2\tilde{M}}{n^{(2-\varepsilon)}}
 \end{equation}
for sufficiently large $n$'s. Finally, (\ref{3-P}), (\ref{3-D-P-2}),
and (\ref{3-de3}) imply (\ref{P-P0}) and (\ref{DP-DP0}), which
completes the proof.
 \end{proof}

\section{Proof of Theorem \ref{Asymp}}

In this section, we give a proof of Theorem \ref{Asymp}, i.e., we
show that the sequences $(|\gamma_n| +|\delta_n^{Neu}|)$  and
$(|\beta_n^- (z_n^*)|+|\beta_n^+ (z_n^*)|)$ are asymptotically
equivalent. The proof is based on the methods developed in
\cite{KaMi01,DM5,DM15}, but the technical details are different.

Let $L=L_{Per^{\pm}}$ and $L^0=L^0_{Per^{\pm}},$ and let $P_n$ and
$P_n^0$ be the corresponding projections defined by (\ref{PP0}).
Then $\mathcal{E}_n=Ran\,P_n  $  and $\mathcal{E}_n^0 =Ran\,P_n^0  $
are invariant subspaces of $L$ and $L^0,$ respectively. By Lemma 30
in \cite{DMH}, $\mathcal{E}_n$ has an orthonormal basis $\{f_n,
\varphi_n\}$ satisfying
 \begin{equation}
  \label{f}
 Lf_n=\lambda^+_n f_n
 \end{equation}
 \begin{equation}
 \label{phi} L\varphi_n=\lambda^+_n \varphi_n -\gamma_n \varphi_n +\xi_n
f_n.
 \end{equation}

\begin{Lemma}
\label{x+g-b+b} In the above notations, for large enough $n,$ \begin{equation}
\label{9.1} \frac{1}{5}(|\beta^+_n(z_n^*)|+|\beta^-_n(z_n^*)| ) \leq
|\xi_n|+|\gamma_n| \leq 9 (|\beta^+_n(z_n^*)|+|\beta^-_n(z_n^*)| )
 \end{equation}
\end{Lemma}

\begin{proof}  Indeed, combining (7.13) and (7.18) and (7.31)
in \cite{DMH} one can easily see that $|\xi_n| \leq
3(|\beta_n^+(z_n^*)|+|\beta_n^+(z_n^*)|)+4|\gamma_n|.$ This
inequality,  together with Lemma 20 in \cite{DMH}, implies that $
|\xi_n|+|\gamma_n| \leq 9 (|\beta_n^+(z_n^*)|+|\beta_n^+(z_n^*)|)$
for sufficiently large $n$'s. On the other hand by (7.31), (7.18),
and (7.14) in \cite{DMH} one gets  $
|\beta_n^+(z_n^*)|+|\beta_n^+(z_n^*)| \leq 5 (|\xi_n|+|\gamma_n|) $
for sufficiently large $n$'s.
\end{proof}

In the following, for simplicity, we suppress $n$ in all symbols
containing $n$. From now on, $P$  ($P^0$) denotes the Cauchy-Riesz
projection associated with $L$ ($L^0$) only. We denote the
projections associated with $L_{Neu}$  and $L^0_{Neu}$ by $P_{Neu}$
and $P^0_{Neu},$ respectively, and $\mathcal{C}=\mathcal{C}(v)$
denotes the one dimensional invariant subspace of
$L_{Neu}=L_{Neu}(v)$ corresponding to $P_{Neu}.$ We also set, for a
smooth function $u$ and a point $x_0 \in [0,\pi],$
$d_{x_0}(u)=\frac{du}{dx}(x_0).$

\begin{Lemma} Let $f, \varphi$ be an orthonormal basis in
$\mathcal{E}$ such that (\ref{f}) and (\ref{phi}) hold. Then there is
a unit vector $G=af+b\varphi$ in $\mathcal{E}$ satisfying
 \begin{equation}
 \label{bcH}
  d_0(G)=d_\pi(G)=0,
   \end{equation}
and there is a unit vector $g\in\mathcal{C}$ satisfying
 \begin{equation} \label{dere}
 \langle G,\bar{g}\rangle \delta^{Neu}=
  b \langle \varphi,\bar{g}\rangle \gamma - b\langle f,\bar{g}\rangle \xi
\end{equation} such that $\langle G,\bar{g}\rangle \in \mathbb{R}$ and
 \begin{equation}
 \label{4950}
  \langle G,\bar{g}\rangle \geq \frac{71}{72}
  \end{equation}
 for sufficiently large $n$.
\end{Lemma}
(Remark. (\ref{bcH}) means that $G$ is in the domain of $L_{Neu}.$)

\begin{proof} If $ d_0(f)=0 $ then
$d_\pi(f)=0$ since $f$ is either a periodic or antiperiodic
eigenfunction. Hence we can set $G=f$. Otherwise we set
$\tilde{G}(x)=d_0(\varphi)f(x)-d_0(f)\varphi(x)$. Then
$G=\tilde{G}/\|\tilde{G}\|$ satisfies (\ref{bcH}) because the
functions $f$ and $\varphi$ are simultaneously periodic or
antiperiodic.

By (\ref{bcH}), $G \in Dom(L)\cap Dom (L_{Neu}), $ so we have
$L_{Neu}G=LG.$ Hence it follows
 \begin{equation} \label{111111}
L_{Neu}G=aLf+bL\varphi=a\lambda^+f+b(\lambda^+\varphi-\gamma\ \varphi+\xi
f)\end{equation}
 $$ \quad \quad \quad \quad \quad \quad =\lambda^+(a
f+b\varphi)+b(\xi f-\gamma \varphi)=\lambda^+G+b(\xi f-\gamma \varphi).$$

Fix a unit vector $g\in \mathcal{C}$ so that
 \begin{equation}
  \label{accreditation}
   \langle G,\bar{g}\rangle= |\langle G,\bar{g}\rangle|,
 \end{equation}
Passing to conjugates in the equation $-g^{\prime \prime} +v(x) g=
\nu g $ one can see that
 \begin{equation}
 \label{de}
 L_{Neu}(\bar{v})\bar{g}=\bar{\nu}\bar{g}.
 \end{equation}
Taking inner product of both sides of (\ref{111111}) with $\bar{g}$
we get
 \begin{equation} \label{cln1} \langle L_{Neu}G,\bar{g}\rangle =
\lambda^+\langle G,\bar{g}\rangle+b(\xi\langle
f,\bar{g}\rangle-\gamma \langle \varphi,\bar{g}\rangle).
 \end{equation}
On the other hand, by (\ref{L-L}) and (\ref{de}),
we have
   \begin{equation} \label{cln2} \langle L_{Neu}(v)G,\bar{g}\rangle =
   \langle G,(L_{Neu}(v))^*\bar{g}\rangle=
\langle G,L_{Neu}(\bar{v})\bar{g}\rangle
  = \nu \langle G,\bar{g}\rangle.
 \end{equation}
  Now (\ref{cln1}) and (\ref{cln2}) imply (\ref{dere}).

Let $G^0=P^0G$ and  $\bar{g}^0=P^0_{Neu}\bar{g};$ then
$\|G^0\|,\|\bar{g}^0\| \leq 1$ since $P^0$ and $P^0_{Neu}$ are
orthogonal projections and $G$ and $\bar{g}$ are unit vectors.

We have
 $$
 \label{3IKA}
 \langle G,\bar{g}\rangle
=
 \langle G^0,\bar{g}^0\rangle +
\langle G^0,\bar{g}-\bar{g}^0\rangle + \langle G-G^0,
\bar{g}\rangle,
$$
so by the triangle and Cauchy inequalities it follows that
 $$
 \label{bodorey}
 |\langle G,\bar{g}\rangle| \geq |\langle G^0,\bar{g}^0\rangle|
 -\|\bar{g}-\bar{g}^0\|
- \|G-G^0\|.
 $$
By Proposition \ref{Proposition} we have
 \begin{equation} \label{G-G0}
\|G-G^0\|=\|(P-P^0)G\| \leq \|P-P^0\| \leq \frac{M}{n}
 \end{equation}
 and similarly
  \begin{equation}
  \label{g-g0}
  \|\bar{g}-\bar{g}^0\|=
  \|(P_{Neu}(\bar{v})-P^0_{Neu})\bar{g}\|
  \leq \|P_{Neu}(\bar{v})-P^0_{Neu}\| \leq
\frac{M}{n}.
 \end{equation}
  Hence, it follows that
 \begin{equation} \label{oyleboyle}
  |\langle G,\bar{g}\rangle| \geq |\langle G^0,\bar{g}^0\rangle |
-\frac{2M}{n}.
 \end{equation}
Next we estimate  $|\langle G^0,\bar{g}^0\rangle |$ from below in
order to get a lower bound for $|\langle G,\bar{g}\rangle |$. Since
$\mathcal{C}^0$ is spanned by $c_n(x)= \sqrt{2}\cos nx,$ $\bar{g}^0$
is of the form
 \begin{equation} \label{g0}
 \bar{g}^0 = e^{i\theta}\|\bar{g}^0\|\bigg{(}\frac{1}{\sqrt{2}}e^{inx}+
 \frac{1}{\sqrt{2}}e^{-inx}\bigg{)}
 \end{equation}
 for some $\theta \in [0,2\pi)$. Now let $G^0_1$ and $G^0_2$
 be the coefficients of $G^0$ in the basis
 $\{e^{inx},e^{-inx}\}$, i.e.,
 \begin{equation}
 \label{nelabel}
 G^0(x)=G^0_1e^{inx}+G^0_2e^{-inx}.
 \end{equation}
 Clearly $d_0(G^0)=in\,(G^0_1-G^0_2)$. Since
 $d_0(G)=0$, by Proposition \ref{Proposition} we also have
 \begin{equation} |d_0(G^0)|=|d_0(G)-d_0(G^0)| \leq\|(D
 P-DP^0)G\|_{\infty} \leq M.
 \end{equation}
 Hence we obtain
 \begin{equation}
 \label{ne}
 |G^0_1-G^0_2| \leq \frac{M}{n}
 \end{equation}
 and
 \begin{equation} \label{nelabelne} |G^0_2| \leq |G^0_1|+ |G^0_1-G^0_2| \leq
 |G^0_1|+\frac{M}{n}.
  \end{equation}
 From (\ref{G-G0}) it follows that
  $$
 \sqrt{|G^0_1|^2+|G^0_2|^2}=\|G^0\| \geq \|G\|-\|G-G^0\| \geq
 1-\frac{M}{n},
  $$
so by (\ref{nelabelne}) we get
 \begin{equation}
 \label{nelabelnelabel}
 |G^0_1| \geq \frac{1}{\sqrt{2}}-\frac{2M}{n}.
 \end{equation}
On the other hand (\ref{g0}) and (\ref{nelabel})  imply
 \begin{equation}
 \label{g0G0}
 |\langle G^0,\bar{g}^0 \rangle|=\frac{1}{\sqrt{2}}\|\bar{g}^0\||G^0_1+G^0_2|
 \geq \frac{1}{\sqrt{2}}\|\bar{g}^0\|\big{(}2|G^0_1|-|G^0_1-G^0_2|\big{)}.
 \end{equation}
 Combining (\ref{ne}), (\ref{nelabelnelabel}), (\ref{g0G0})
 and taking into account that
 $$\|\bar{g}^0\| \geq \|\bar{g}\|-\|\bar{g}-\bar{g}^0\| \geq
1-\frac{M}{n}$$ due to (\ref{g-g0}), we obtain
 \begin{equation}
 |\langle G^0,\bar{g}^0 \rangle| \geq 1-\frac{6M}{n}
 \end{equation}
 which, together with (\ref{oyleboyle}) and (\ref{accreditation}), implies
 \begin{equation}
 \langle G,\bar{g}\rangle \geq 1-\frac{8M}{n}.
 \end{equation}
Hence, for a sufficiently large $n$, $\langle G,\bar{g}\rangle \geq
71/72$.
 \end{proof}

 \begin{Corollary}
 \label{corollary}
 For sufficiently large $n$, we have
 \begin{equation}
 \label{corr2} |\gamma_n|+|\delta^{Neu}_n| \leq 19
 \big{(}|\beta^+_n(z_n^*)|+|\beta^-_n(z_n^*)|\big{)}.
 \end{equation}
 \end{Corollary}
\begin{proof}
Using (\ref{dere}), (\ref{4950}) and noting also that the absolute
values of $b$ and all inner products in the right-hand side of
(\ref{dere}) do not exceed $1$ we get $|\delta^{Neu}| \leq
72/71\big{(}|\xi|+|\gamma|\big{)}.$ This inequality, together with
Lemma \ref{x+g-b+b}, implies (\ref{corr2}).
 \end{proof}

Corollary \ref{corollary} proves the second inequality in
(\ref{A10}). In order to complete the proof of Theorem~\ref{Asymp} it
remains to prove the first inequality in (\ref{A10}).

By Proposition 34 in \cite{DMH}, if
 \begin{equation} \label{case1}
  \textit{Case 1}: \quad \quad \quad \frac{1}{4}|\beta^-(z^+)| \leq
 |\beta^+(z^+)| \leq 4 |\beta^-(z^+)|,\quad\quad\quad\quad\quad
 \end{equation}
 then we have
 \begin{equation}
 \label{case11}
 |\beta^+(z^*)|+|\beta^-(z^*)| \leq 2|\gamma|.\quad \quad
 \end{equation}
 Next we consider the complementary cases
$$
 \textit{Case 2(a)}:\quad 4|\beta^+(z^+)| < |\beta^-(z^+)|\quad
 \text{or} \quad \textit{Case 2(b)}:\quad 4|\beta^-(z^+)| <
 |\beta^+(z^+)|.$$
\begin{Lemma}
\label{LLL} If {\em Case 2(a)} or {\em Case 2(b)} holds, then we
have, for sufficiently large $n,$
 \begin{equation} \label{LLLL}
\frac{1}{4} \leq \frac{|d_0(f)|}{|d_0(\varphi)|} \leq 4.
 \end{equation}
 \end{Lemma}

\begin{proof} We consider only  \textit{Case 2(a)}, since the proof \textit{Case 2(b)} is
 similar. Let $f^0=P^0f$, $\varphi^0=P^0\varphi$ and let
$ f^0=f^0_1 e^{inx}+f^0_2 e^{-inx} $ and $ \varphi^0=\varphi^0_1
e^{inx}+\varphi^0_2 e^{-inx}. $
 In \textit{Case 2(a)}, if $v \in L^2([0,\pi])$
 it was shown in the proof of Lemma 64 in \cite{DM15} that the following
 inequalities hold (inequalities (4.51), (4.52), (4.54), and (4.55) in \cite{DM15}):
 \begin{equation}
 \label{6-f}
 |f^0_1| \geq \frac{2}{\sqrt{5}}-2\kappa_n,
 \quad|f^0_2| \leq \frac{1}{\sqrt{5}},\quad |\varphi^0_1| \leq \frac{1}{\sqrt{5}}+\kappa_n,
 \quad|\varphi^0_2|\geq
 \frac{2}{\sqrt{5}}-2\kappa_n,
 \end{equation}
where $\kappa_n$ is a sequence converging to zero. These
inequalities were derived using Lemma 21 and Proposition 11 in
\cite{DM15} which still hold in the case where $v \in
H^{-1}([0,\pi])$, including our case $v\in L^p([0,\pi])$ (See Lemma
6 and Proposition 44 in \cite{DMH}). Hence we can safely use them.

Note that $d_0(f^0)=in(f^0_1-f^0_2)$. Using (\ref{6-f}) we get
 \begin{equation}
 |d_0(f^0)| \geq n(|f^0_1|-|f^0_2|) \geq
 n\bigg{(}\frac{1}{\sqrt{5}}-2\kappa_n\bigg{)} \geq
 \frac{n}{\sqrt{6}}
 \end{equation}
for sufficiently large $n$. On the other hand we have
 \begin{equation}
 |d_0(f^0)| \leq n(|f^0_1|+|f^0_2|) \leq n\sqrt{2}\|f^0\| \leq
 \sqrt{2}n
 \end{equation}
Following the same argument for $d_0(\varphi^0)$, we have both
 \begin{equation}
  \frac{n}{\sqrt{6}} \leq |d_0(f^0)| \leq \sqrt{2}n\quad\text{and}\quad
  \frac{n}{\sqrt{6}} \leq |d_0(\varphi^0)| \leq \sqrt{2}n.
 \end{equation}
On the other hand by Proposition \ref{Proposition} we have
 \begin{equation}
 |d_0(f)-d_0(f^0)| \leq M \quad \text{and} \quad |d_0(\varphi)-d_0(\varphi^0)| \leq M.
 \end{equation}
 Hence, for sufficiently large $n$'s, we get
 \begin{equation}
 \frac{|d_0(f)|}{|d_0(\varphi)|} \leq
 \frac{|d_0(f^0)|+M}{|d_0(\varphi^0)|-M}\leq
 \frac{\sqrt{2}n+M}{n/\sqrt{6}-M}\leq 4
 \end{equation}
 and
 \begin{equation}
 \frac{|d_0(f)|}{|d_0(\varphi)|} \geq
 \frac{|d_0(f^0)|-M}{|d_0(\varphi^0)|+M}\geq
 \frac{n/\sqrt{6}-M}{\sqrt{2}n+M}\geq \frac{1}{4}.
 \end{equation}
 \end{proof}

\begin{Proposition}
 \label{prop2}
For sufficiently large $n$, we have
 \begin{equation}
 \label{corr} \big{(}|\beta^+_n(z_n^*)|+|\beta^-_n(z_n^*)|\big{)} \leq
 80 (|\gamma_n|+|\delta^{Neu}_n|)
 \end{equation}
 \end{Proposition}

\begin{proof} In view of
 (\ref{case11}), it remains to prove (\ref{corr}) if
 \textit{Case 2(a)} or \textit{Case 2(b)} holds.

 Now (\ref{dere}) implies that
 \begin{equation} \label{yokoyleyok}
 |b||\langle f,\bar{g}\rangle | |\xi| \leq |
 \delta^{Neu}|+|\gamma|.
 \end{equation}
 Thus, in order to estimate $|\xi|$ from above by
 $|\delta^{Neu}|+|\gamma|$ we need to
 find a lower bound to $|b||\langle f,\bar{g}\rangle |$. We have
 \begin{equation}
 \label{aydiaydi}
 |b||\langle f,\bar{g}\rangle |=|b|\big{|}\langle f,G \rangle+\langle f,\bar{g}-G
 \rangle\big{|}\geq |a||b|-\|\bar{g}-G\|
 \end{equation}
since $\|f\|=1$, $|b| \leq 1$ and $\langle f,G \rangle= \bar{a}.$ In
view of (\ref{4950})
 \begin{equation}
||\bar{g}-G||^2=||\bar{g}||^2+||G||^2-2 Re \,
\langle\bar{g},G\rangle=2-2\langle\bar{g},G\rangle \leq
\frac{1}{36},
 \end{equation}
  hence
 \begin{equation} \label{uwl} \|\bar{g}-G\| \leq\frac{1}{6}. \end{equation}
On the other hand, by the construction of $G$ we know $|b/a|=
|d_0(f)/d_0(\varphi)|,$ so Lemma \ref{LLL} implies that $1/4\leq |b/a|
\leq 4.$ Since $|a|^2+|b|^2=1$, a standard calculus argument shows
that
 \begin{equation} \label{sar1}
 |a||b| \geq\frac{4}{17}.
 \end{equation}
In view of (\ref{uwl}) and (\ref{sar1}), the right-hand side of
(\ref{aydiaydi}) is not less than $4/17-1/6 > 1/15,$ i.e.,
$|b||\langle f,\bar{g}\rangle |>1/15.$ Hence, by (\ref{yokoyleyok}),
it follows that
 \begin{equation} \label{bsbsb}
 |\xi| \leq 15 \big{(} |\delta^{Neu}|+|\gamma| \big{)}.
 \end{equation}
Now we complete the proof combining (\ref{bsbsb}) and Lemma
\ref{x+g-b+b}. \end{proof}

Corollary \ref{corollary} and Proposition \ref{prop2} show that
(\ref{A10}) holds, so Theorem~\ref{Asymp} is proved.

\section*{Acknowledgement}
The author wishes to express his gratitude to Prof. Plamen Djakov
who suggested the problem and offered invaluable assistance, support
and guidance.

redpajama_set_name: RedPajamaArXiv


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text:
Q: Should I use "that" after "whatever" Which is correct or better:

Anyways, I hope they do whatever that is for the benefit of them and the region.
Anyways, I hope they do whatever is for the benefit of them and the region.

redpajama_set_name: RedPajamaStackExchange


one goal was to honor Aaron Winborn and his battle with ALS by working on the Media module - one that he has been very involved with.
the other goal was to introduce people to core contributions and how the community at any level could get involved.
All of us at Mediacurrent had a blast throughout the day and hope to help organize and participate in more sprints in the near future.


If you follow this blog regularly, you'll remember that on Thursday, I told you that I was going to post a question about religion over at Yahoo Answers. It was a sort of follow-up to this post, but with a slightly bigger, more sensitive topic.
Is religion is a choice, in the same way that other beliefs are choices? For example, if I chose to subscribe to the moral belief that removing one's clothes was wrong, I wouldn't expect the Chippendales to make any special dispensation for me should I choose to work for them. Similarly, if religion is a choice, I wouldn't expect any special dispensation for wearing religious icons which do not conform to a company's uniform policy. I either have to give up my religion of choice, or work somewhere else.
Or is religion is more of a psychological and visceral calling? If, for example, I was blind, I would expect employers to do everything in their power to adapt to me. I could not stop being blind. Is religion more similar to this – You can't 'physically' stop believing?
Or is it the latter, but still fine to discriminate, just I would be discriminated against if I had an unshakeable belief that all women were evil?
Well, I asked, and you answered. In your droves. In fact, not only did you answer in your droves on the Yahoo site, when the post was picked up by Pickled Politics (an excellent blog, by the way) you answered there, too. And even though I closed the comments on the original post, you even went to the trouble of emailing me directly with your answers. I've read over 80 comments on the subject in less than a week – and I thank everyone for contributing.
Religion is the coward's version of hope.
Unless one is thoroughly brainwashed in a particular religion, they choose their belief. If it were not a choice, why would there be a need for so many different religions? They choose the one that makes them feel the most important, often at the expense of their fellow humans.
I'm interested by this comparison with other types of choice. When I come to pnder this political position or that one, there is usually one that is, to my mind, more persuasive than another. I don't feel I am "choosing" a political belief or position, in the same way that I choose whether to eat a banana or an apple. My conscience dicates only one option. The political position "chooses" me, just like the answer to 2 + 2 finds its way to me too.
Of course, politics is much harder, and the answer less obvious, than mathematics. New writing from new sources might weaken and eventually reverse that choice. Part of religion – at least the fundamentalist brand – seems to be the deliberate ignoring of new perspectives.
Either way, if 'finding' a political position can happen, I imagine it is even more likely to happen with religion. Although you can change your religion, I would be very surprised at people who say that it was a definite choice, not a calling. Indeed, when we see celebrities such as Madonna changing their religion like it is a diet fad, we criticise them for it!
In the case of Christianity, it is a gift. A gift of grace, pure unmerited favor from God. Even the faith we have is a gift of God.
If choice is defined broadly, then I guess religion defined narrowly is a choice.
Religion is, amongst other things, a way of programming(re-programming if you're a convert) your head in order to experience reality in a different way, and as such adherence to a fixed set of rules and rituals is essential.
Religions is a man-made placebo for humans.
Religion, like sexuality, is a constrained choice. Everyone has options, but that doesn't mean everyone has the same options.
For many it is a psychological need to believe in something bigger than oneself. It helps to make sense of things and to give one a sense of peace and ease.
It can also be a calling. If you are not brainwashed into believing a certain way and have formed your own opinion and beliefs…and one day you discover a path which has already been laid out, but follows the same beliefs that you already had….that's a calling. To choose to continue to follow that path or to follow a path that is not the same as the one you were raise with, is a choice.
Some form an opinion all their own which has nothing to do with the religion they were raised with and just go from there. It is based on reasoning, deduction and an open mind.
Religion SHOULD be a individual's OWN personal choice – but unfortunately all too often it is something that is "chosen for them" by members of their family or community – often through intimidation.
For example, how many so called followers of Islam are there who aren't "genuine" believers? (i.e, they would be prepared to sneak down to the pub for a quick pint if they thought they wouldn't get found out).
I bet there is a fair few such people – who claim to subscribe to a certain religion because "that is what is expected of them" rather than because that is what they genuinely believe.
A most unfortunate state of affairs, I think.
Religion is an expression of a deep yearning for a certainty which can never exist in the real Universe.
Religion is a man-made comforter, imperfectly describing what we are learning about the universe.
Religion is all about intimidation and power and Hollywood.
Go to any Spanish country for Good Friday and watch the street theatre with the sinners doing penance, walking the streets by torchlight, behind statues of Jesus and Mary, dressed in Klu-Klux-Klan outfits, bare-foot, the chains around their ankles dragging across the cobblestones, and you will see the frightening symbolism religion must have had in the days before Hollywood.
Religious symbolism is amateur dramatics and Hollywood.
Mix this up with our imperfect understanding of how our brains work – and that we are not well taught to question our perceptions – and we can see why we end up believing in fairytales and monsters rather than in the solidity of earth and the open space of the skies. In the west we live too much in our heads.
Inside our heads, depending on our mood at the time, we ricochet between our lower ancient reptilian brain (where we get fears of monsters from) and our more evolved, expansive, kinder, generous, socially-aware higher brain (heaven, angels, light, "enlightenment", our good side, our better nature).
Religious stories are a useful metaphorical soap opera to motivate us to do the real work of disciplining ourselves to be more socially aware and responsible for each other.
Greed and competitiveness puts the old defensive reptilian brain in charge.
Altruism brings the more evolved "other side" to come to the fore.
Metaphorical "stories" can sometimes help us to keep going.
But let the user beware. Here be dragons! Keep your feet on the ground.
I feel I should add a note at this point to say that if you're unhappy with your answer being featured here, please feel free to get in touch and I'll remove it.
Religion is a choice. Your job is a calling. And you are something else.
Please please go ahead and read all of the comments that have been made on this subject. You can read all of the Yahoo Answers comments here, and the debate is still rumbling over at Pickled Politics. If you want to leave your own thoughts, feel free to do so using the comments facility right here on sjhoward.co.uk.
So now I've received all of these answers to my question, you are perhaps wondering if I've managed to form my own opinion on the subject. I'm sorry to report that I'm just as confused as I was when I posed the question.
The exercise has shown, clear as day, that few agree on the nature of religion. How can we hope to have a sensible debate on religion, religious laws, and religious extremist if we can't even agree on the nature of religion itself?
For once, I'm not going to claim to have even an idea of the answer. For today at least, I'm happy to be the one posing the questions, in the hope that it will spark some consideration. I hope it has for you.
This 1,039th post was filed under: Headliner.
Trackback received at 09:42 on 31st December 2007.


In this blog post we discuss renovation and restoration. Read on for our tips and advice.
Inexpensive old houses are available in some areas, but most have been snapped up and modernised years ago, and those that are left are generally no longer the bargains they once were. In rural areas it's still possible to buy old properties requiring total renovation or even total restoration, although they're rarer nowadays and are by no means cheap. There is a passion for rescuing old tumble-down houses and restoring them to their former glory, and 'ruins' sold at auction often far exceed their reserve price.
It's very expensive to restore an old property and can cost much more than building a new house, particularly if you do it properly using reclaimed materials.
Many people are lulled into a false sense of security and believe they're getting a wonderful bargain, without fully investigating the renovation costs, which are invariably higher than you imagined or planned! Some properties even lack basic services such as electricity, a reliable water supply and sanitation.
While you may get more for your money when buying an old home, the downside is that they require much more maintenance and upkeep than new homes, and heating costs can be high unless a property has good insulation.
Good Value Old properties can provide better value than new homes, although you must check their quality and condition carefully. As with most things in life, you generally get what you pay for, so you shouldn't expect a fully modernised property for a knock-down price. For those who can afford them, at the top end of the scale there's a wealth of beautiful mansions, castles and stately homes available with extensive grounds (some country homes even come with their own golf course!). Substantial period homes certainly don't come cheap, although this segment of the market has suffered in recent years and larger homes costing in excess of £750,000 are generally excellent value for money.
If you aspire to live the life of the landed gentry in your own stately home, bear in mind that the cost of their upkeep is usually astronomical. As a consequence many mansions have been converted into luxury apartments and townhouses in recent years.
If you're looking for something unusual try Pavilions of Splendour (22 Mount View Road, London N4 4HX, (020-8348 1234, www.heritage.co.uk) or SAVE Britain's Heritage (70 Cowcross Street, London EC1M 6EJ, ( 020-7253 3500, www.savebritainsheritage.org), a conservation charity that maintains a list of properties in need of restoration.


When it comes to Candida, this is a very common type of infection that can occur in the mouth, but also in other parts of the body. In general, it appears as a result of taking antibiotics for long periods of time and that is because these types of medications actually kill the beneficial bacteria in your system along with the harmful one. The good news is that treating Candida is actually simple and we're going to tell you more about that right now.
The easiest, fastest and most effective way of treating Candida is to start taking Candida supplements such as those at http://www.amazon.com/Purely-Holistic-Antifungals-Probiotics-Reoccurrence/dp/B0158VYYE8/. These supplements are made in the United States and given the fact that they contain probiotics, you will not have to worry about them destroying the good flora from your system. On top of that, these supplements will cleanse your body of Candida in four steps which makes them a lot more effective than anything out there on the market today. Given the fact that they come with a money back guarantee as well, you can easily get a refund if they haven't helped you treat your Candida.
One thing that you need to bear in mind is that whatever your diet consist of is very important and will undoubtedly affect your system and the chances of your Candida going away or not. For example, if you don't want to keep feeding your Candida, then you need o consider avoiding molds, mushrooms, vinegar and any type of fermented foods. That includes soy sauce and alcohol as well. Other types of foods you should certainly avoid include pork products, bacon and white flour, but also commercial soups and dry roasted nuts.
If you eat a lot of starchy foods, then now would be a great time to stop eating them. You shouldn't also limit your consumption of fruit juice and also fruit, since these contain sugar which feeds the Candida growth.
When you have Candida, there are also foods that you are allowed to eat and some of them include lamb, seafood, turkey, chicken and also eggs. Eating vegetables is also recommended, so you should also increase your consumption of kale, turnip, broccoli, but also onions, raw garlic and of course, cabbage. All in all, by keeping these tips in mind, getting rid of your Candida growth is going to be a matter of time.
There are many parents out there who want their children to start learning early and hence try to prepare their child for preschool with the help of lesson plans for toddlers. One of the key components of learning for a toddler is the development of reading skills. This article will specially focus on developing reading lesson plans for toddlers. It is very important for your toddler to get familiarized with the library and encourage him/her to check out the books. The library staff should take an initiative to introduce the child to the children book area. Many libraries schedule story time during different times of the year.
The child should be allowed to participate in library activities that involve reading and learning about new book releases. The trips to the library will help to imbibe a love for books in the child. Additionally you can also allow the child to choose a book to read once a week from the library. You must return home with that particular book and read it with the child. As soon as you finish reading the first chapter, discuss it with the child so that he grasps the story better.
Lesson plans for toddlers help to make the reading session more interactive in nature. You should also encourage conversation from the child at the time of reading the book. If you are reading a book on cats then you may ask the child "what sound does the cat makes?" You can also point out to the pictures in the book and ask to name the person, place or the animal in the picture. This type of interactive session helps the child to learn new concepts, spell and even read! Before you return the book to the library make sure the child is able to identify each and every picture in the book. Books are not just for reading; it is basically a tool to introduce the child to other subjects. The sheer number of books available for children in the market is totally unbelievable. There are books on ballet, football, fishing, painting and even camping ideas for toddlers!
If a child is not willing to read initially then think of incorporating small rewards in the lesson plans for toddlers. For example, you can set the child a particular target with respect to the number of pages to be read in a particular day. If he completes that target then you can reward the child with his favourite treat. There are many children who spend countless hours watching TV and playing video games. So it is very important for your child to get involved in mind stimulating activities. Moreover, lesson plans for toddlers are very effective in developing the reading IQ of the child.
If you heard about turmericcurcumin and you have no idea what it is, well, this is a special herbal medicine that has been used in Asian countries for thousands of years not only for cooking delicious dishes, but also for its medicinal properties. Given the fact that turmeric has anti-inflammatory properties, it can be easily used by anyone who wants to alleviate joint pain, muscle pains and other types of pains they may be suffering from. However, before you take turmeric, you should check with your doctor first.
Did you know that you can actually sue turmeric in order to treat various skin conditions, including cysts, cuts and also various wounds? In fact, turmeric is known to be a very powerful natural antibiotic and because of its health benefits it was extensively used in WW2 in order to treat gunshot wounds. To use it, you should carefully make a paste out of it and then apply it on your wounds. Be careful though, since turmeric can stain your clothes permanently.
You can easily use turmeric in order to soothe all types of pains, but most of the times it's going to help you a lot if you suffer from arthritis. Turmeric has the power of greatly reducing inflammation, especially in the bones so that is why you should use it in supplement or extract form, depending on which one you can find easier. Taking drugs is risky anyways, so why not focus on using a solution that is both healthy and affordable?
If you aren't aware of this yet, then you should know that turmeric has been used for a very long time now in order to help prevent cancer and that is because it can effectively destroy the cancerous cells in the body. Based on what doctors are saying, by taking a teaspoon of it per day, you'll easily be able to prevent cancer.
Last but not least, you can easily use turmeric for treating diabetes and it's best that you're going to take about 2 teaspoons of it a day. Better yet, it can help you lose weight and it's also going to help you prevent type two diabetes. All in all, just make sure you get it from a reputable online store, since there are many out there which sell fake versions of it that have nowhere near the aforementioned health benefits.


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Наді́я Олекса́ндрівна Жу́ркіна, за чоловіком — Кийок (; 28 серпня 1920 — 24 квітня 2002) — радянський військовик, в роки Другої світової війни — повітряний стрілець-радист 99-го окремого гвардійського розвідувального авіаційного полку 15-ї повітряної армії, гвардії старшина.

Одна з чотирьох жінок — повних кавалерів ордена Слави.

Життєпис
Народилася в місті Туринськ, нині Свердловської області Росії, в родині робітника. Росіянка. Після закінчення школи вступила до Московського юридичного інституту, одночасно навчалась в московському аероклубі.

До лав РСЧА призвана Туринським РВК 17 липня 1941 року. Закінчила курси радисток. Учасник німецько-радянської війни з липня 1943 року. Член ВКП(б) з 1943 року.

До квітня 1944 року повітряний стрілець-радист 99-го окремого гвардійського розвідувального авіаційного Забайкальського полку гвардії молодший сержант Н. О. Журкіна в складі екіпажу здійснила 30 успішних бойових вильотів на розвідку військ супротивника.

До квітня 1945 року здійснила 60 успішних бойових вильотів. За цей час з борту літака передала 84 радіограми з цінними розвідувальними даними про наявність руху шляхами військ і техніки супротивника, про залізничні перевезення і базування ворожої авіації на аеродромах. Під час виконання бойових завдань екіпажем, влучним і прицільним кулеметним вогнем відбивала всі атаки винищувачів супротивника.

У серпні 1945 року демобілізована. До виходу на пенсію працювала начальником відділу кадрів швейного об'єднання «Ригас айгербс» (Латвія).

З середини 1990-х років мешкала в московському пансіонаті для ветеранів № 17, де й померла. Похована на Ніколо-Архангельському кладовищі в Москві.

Нагороди і почесні звання
Нагороджена орденами Вітчизняної війни 1-го ступеня (11.03.1985), Червоної Зірки (30.04.1945), Дружби народів (01.10.1993), «Знак Пошани» (…), Слави 1-го (23.02.1948), 2-го (15.11.1944) та 3-го (30.04.1944) ступенів і медалями, в тому числі й «За відвагу» (29.07.1943).

Почесний громадянин міста Туринська.

Див. також
 Нечепорчукова Мотрона Семенівна
 Петрова Ніна Павлівна
 Станілієне Дануте Юргівна

Посилання

Уродженці Свердловської області
Жінки в Другій світовій війні
Радянські льотчики Другої світової війни
Члени КПРС
Померли в Москві
Поховані на Ніколо-Архангельському цвинтарі

redpajama_set_name: RedPajamaWikipedia


PHILIPSBURG--The Anti-Corruption Task Force TBO detained head of Princess Juliana International Airport (PJIA) Chief of Security J.S. (48) at his home on Wednesday.
The airport management was shocked to hear the news of the long-time employee being arrested.
J.S. is suspected of accepting bribes, money-laundering and complicity in violation of the duty to register foreign exchange. TBO officers searched his home and workplace on Wednesday.
J.S. will remain in custody and there was no comment from the airport management about their chief of security.


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redpajama_set_name: RedPajamaGithub


When French explorers Nicholas Baudin and François Péron visited Shark Bay in 1801 there were 23 species of mammals. Less than half of them remained in 1990.
These local extinctions were due to habitat destruction and competition for food by stock and rabbits and predation by introduced foxes and cats. Project Eden was launched to reverse this ecological destruction.
Vegetation and habitat recovered after cattle and sheep were removed. Acacia shrubs grew much more densely with seedlings able to grow and foods favoured by stock – delicate annuals, creepers and grasses – also returned.
Captive breeding pens were established to supply native animals for reintroduction to the peninsula. While a number of native animals were reintroduced, only two were successful – the bilby and mallefowl.
The Peron Captive Breeding Centre was established in 1996 to provide animals for reintroduction to Peron Peninsula. The centre bred more than 300 animals from five species – the western barred bandicoot, bilby, rufous hare-wallaby (mala), banded hare-wallaby and malleefowl.
The centre also improved understanding about reproduction, behaviour, diet and physiology of some arid environment species and has contributed to conservation of the hare-wallabies and other threatened species.
The Peron Captive Breeding Centre ceased operations in October 2013 after completing its translocation targets. At that stage it was too early to begin translocation preparations for the Dirk Hartog Island Return to 1616 project.
Find out about the status of Project Eden's translocations below.


The building is VACANT- Lots of value here for the savvy owner who wants to utilize the rich inherent community FAR(4.8) and Residential FAR(2.43) to create a robust building. Alternatively, the end user can create a dynamic home as this large building offers a massive amount of living space. This property rests in in the middle of transportation offering the flexibility of the M, J & L trains. The great growth opportunity can be seen in the 4.8 FAR for Facility; offering a wide range; educational, healthcare, religious and not-for-profit institutions. Moreover, the property has a 2.43 Residential FAR offering a wonderful residential component. This combination allows the buyer to create a robust building as the Zoning is R6, C2-3.Property Characteristics:?


216 pages | 156 Color Illus.
Energy demand reduction is fast becoming a business activity for all companies and organisations because it can increase profits regardless of the nature of their core activity. The International Energy Agency believes that industry could improve its energy efficiency and reduce carbon dioxide emissions by almost a third using the best available practices and technologies.
compressed air and process controls.
Uniquely, it includes a whole chapter on greening data centres. It also looks at topics covered in greater detail in its companion volume, Energy Management in Buildings: insulation, lighting, renewable heating, cooling and HVAC systems. Further chapters examine minimising water use and how to make the financial case, both to prioritise measures for cost effectiveness, and to get management on board.
This title is aimed at all professional energy, industry and facilities managers, energy consultants, students, trainees and academics and can be read alongside training for ISO 50001 - Energy Management Systems. It takes the reader from basic concepts to the latest advanced thinking, with principles applicable anywhere in the world and in any climate.
Preface. Introduction. 1. Measuring Energy Consumption 2. Metering 3. Airtightness and Insulation 4. Lighting, Daylighting and Controls 5. Heating and Cooling 6. Heating, Ventilation and Air Conditioning Systems 7. Energy Reduction Technologies 8. Motors, Drives and Compressed Air 9. Refrigeration 10. Process Controls 11. Data Centres 12. Minimising Water Use 13. Making the Financial Case Conclusion. Appendix.
David Thorpe is community manager of Sustainable Cities Collective, an online community for leaders of major metropolitan areas, urban planning and sustainability professionals. Until 2013 he was for 13 years News Editor of Energy and Environment Management magazine, for which website he also wrote a weekly op-ed column. He is also the author of several books and hundreds of articles on related subjects. Formerly director of publications at the Centre for Alternative Technology, he has written two other books in the Earthscan Expert series, Sustainable Home Refurbishment and Solar Technology, and several B2B ebooks for the publisher DoSustainability. He runs his own sustainable media consultancy, Cyberium, manages the Green Deal Advice website, and blogs regularly as The Low Carbon Kid. Find him on Twitter @DavidKThorpe.


Laura Mercier's Loose Setting Translucent Powder is a staple in my collection, I love it! Ever since I discovered it was fantastic at keeping oil away and baking I've been using it religiously, so it was only right that I picked up the latest release – Laura Mercier Loose Setting Translucent Powder – Glow – £29.00.
I saw a few video reviews about this but wanted to try it out myself and see how it faired on deep/dark/black skin.
In terms of packaging, Translucent Glow is packaged similar to its predecessors, in a tub, with 29g of product. The only noticeable difference, packaging wise, is that the lid is mirrored so if you hate fingermarks, that's going to annoy you…a lot.
I really should have swatched this first because, at first glance, you would be forgiven for expecting Translucent Glow to produce a gorgeous warm sheen, but in reality, it has a white gold offset, meaning when blended out you get gold. Fantastic right? Not so much.
When I applied this all over to my face to lightly set my makeup, I could see these faint but weird shiny patches on my face. When I looked closer it was the subtle gold pigment from Translucent Glow. The gold pigment in this product doesn't actually blend very well, it's patchy…very patchy. Patchy on bare dry skin and patchy on moisturised skin.
Now, this may not be a problem on lighter skin tones but on my skin tone its obvious and I don't like it. It clings to the skin unevenly and with the contrast of the gold on my skin it just looks weird and sheeny.
I had to think of a new way of using this.
In the end, I decided to use it as I would a highlighter, applying sparingly to the tops of my cheeks and other typical highlighting zones which worked a lot better but if I wanted a highlighter I would have used the few I already own – I wanted a glowing setting powder!
If you have deep/dark skin this IS NOT going to work for setting your makeup. You'll be patchy and it'll be obvious up close which is such a shame as this powder had so much potential.
If you have any of the Hourglass Ambient Lighting Powders don't even think about adding LM Translucent Glow to your "buy" list, as I feel the Ambient Powders do the exact same thing but Translucent Glow is in a loose powder form and patchy.
I would honestly say save your £29 if you're melanin enriched and use something like a glow setting spray for now, it's cheaper and more effective.
EDIT: I've found out that Laura Mercier have a Medium/Deep version of Glow! I'm hoping that shade will redeem the cons of Translucent Glow. Stay tuned to the blog for the review of Medium/Deep Glow!


Watch & Download Betrayal (2012) : Full Movie Free On Putlocker HD While examining one of her patients, a doctor reveals that her husband is cheating with the patient's wife.
Who's The Cast on Film Betrayal 2012??
When Betrayal to release? ?


The night time experience for Ocean Independence at Monaco Yacht Show 2016, photographed by Big Dog Productions.
With 30,000 visitors over four days, 120 superyachts and 580 leading companies in yachting on show, we selected one of the most anticipated and exclusive superyacht events of the year – Monaco Yacht Show 2016 – to unveil a new-look exhibition stand and live experience for client, Ocean Independence.
An essential part of yacht marketing, Ocean Independence invests heavily in a strong presence ay the world's premier yacht shows; the brief was to create an engaging platform that would achieve industry standout and excite clients.
After creating a new identity for the Company earlier this year, we spearheaded and led the creative direction for a striking new exhibition stand. The design included: multi-media product display, creative lighting, an innovative light box installation and custom exterior furniture by Peter Glyn Machin.


Formular 1 super star, Lewis Hamilton has just won his third straight race victory after clinching the Russian Grand Prix ahead of Valtteri Bottas and Sebastian Vettel.
The British racing driver, 33, who races for Mercedes AMG Petronas clinched the win after his teammate, Valtteri Bottas, pulled over to let him through as the pair approached halfway in Sunday's Russian Grand Prix.
The team have avoided using team orders previously but clearly felt they had to maximise Hamilton's advantage over Vettel.
The victory means Mercedes continue their run of being unbeaten at Sochi since the race joined the calendar in 2014. Hamilton's eighth win of the season is ominous for Ferrari. Both teams brought aero upgrades to the race, largely centred on the front wing but for the first time this season, Mercedes seem to have gained a real advantage.
The Ferrari has struggled, particularly in the technical final sector for grip, where Hamilton has been strong through the last part of the lap all weekend. He exploited it but did not take great pleasure in the manner of his victory.
Hamilton who is rumoured to be dating rapper Nicki Minaj now leads Ferrari's Sebastian Vettel by 50 points in the standings with five races remaining as both look to win their fifth title.


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Douglas Walton Titles, Movies and Posters
The Bride of Frankenstein
CAST: James Whale, Boris Karloff, Elsa Lanchester, Ernest Thesiger, Colin Clive, Una O'Connor, Valerie Hobson, Dwight Frye, John Carradine, E.E. Clive, O.P. Heggie, Gavin Gordon, Douglas Walton, Billy Barty,
Kitten With a Whip
CAST: Heyes, Douglas, Ann-Margret, John Forsythe, Peter Brown, Patricia Barry, Richard Anderson, Diane Sayer, Ann Doran, Patrick Whyte, Audrey Dalton,
CAST: Peter Askin, Dalton Trumbo, Joan Allen, Brian Dennehy, Kirk Douglas, Michael Douglas, Paul Giamatt,
Charlie Chan in London
CAST: Eugene Forde, Warner Oland, Drue Leyton, Ray Milland Mona Barrie, Douglas Walton, Alan Mowbray, George Barraud, Paul England, Madge Bellamy, Walter Johnson, Murray Kinnell, E.E. Cliv,
Too Many Girls
CAST: George Abbott, Lucille Ball, Richard Carlson, Ann Miller, Eddie Bracken, Frances Langford, Desi Arnaz, Hal Le Roy, Libby Bennett, Harry Shannon, Douglas Walton, Chester Clute, Tiny Person, Ivy Scott,
Lone Texan
CAST: Paul Landres, Willard Parker, Grant Williams, Audrey Dalton, Douglas Kennedy,
Jesse James Jr.
CAST: George Sherman, Don 'Red' Barry, Lynn Merrick, Al St. John, Douglas Walton, Karl Hackett, Lee Shumway, Stanley Blystone,
Cole Younger, Gunfighter
CAST: R.G. Springsteen, Frank Lovejoy, James Best, Abby Dalton, Jan Merlin, Douglas Spencer, Ainslie Pryor, Frank Ferguson, Myron Healey, George Keymas, Dan Sheridan,
CAST: King Vidor, Hedy Lamarr, Robert Young, Ruth Hussey, Charles Coburn, Van Heflin, Fay Holden, Bonita Granville, Douglas Wood, Charles Halton, Leif Erickson, Phil Brown, David Clyde,
Mary of Scotland
CAST: John Ford, Katharine Hepburn, Fredric March, Florence Eldridge, Douglas Walton, John Carradine, Robert Barrat, Gavin Muir, Ian Keith, Moroni Olsen, William Stack, Ralph Forbes, Alan Mowbray, Frieda Inescort, Donald Crisp,
The Garden Murder Case
CAST: Edwin L. Marin, Edmund Lowe, Virginia Bruce, Benita Hume, Douglas Walton, Nat Pendleton, Gene Lockhart, H.B. Warner, Kent Smith, Grant Mitchell, Frieda Inescort, Henry B. Walthall, Jessie Ralph, Charles Trowbridge,
CAST: Willis Goldbeck, Wallace Beery, Marjorie Main, Donald Meek, Dorothy Morris, Howard Freeman, Connie Gilchrist, Tommy Batten, Gloria Dickson, Henry O'Neill, Richard Hall, Charles Halton, Morris Ankrum, Douglas Fowley, Chester Clute,
Nightmare County
CAST: Sean MacGregor, Sean MacGregor, Gayle Hemingway, Beau Gibson, R.N. Bullard, Woody Lee, Robert Reynolds, Michael Verona, Jody McCrea, Tod Fortner, Conchita Thornston, Duke Douglas, Ted Wilde, Robert Dalton, Conrad Hitchler,
CAST: John Reinhardt, Lee Tracy, Don Castle, Julie Bishop, Anabel Shaw, Douglas Walton, Regis Toomey, Francis Ford, Anthony Warde, Argentina Brunetti, Wilson Wood,
Fortunes of Captain Blood
CAST: Gordon Douglas, Louis Hayward, Patricia Medina, George Macready, Alfonso Bedoya, Dona Drake, Lowell Gilmore, Wilton Graff, Curt Bois, Lumsden Hare, Billy Bevan, Harry Cording, Duke York, Sven Hugo Borg, Martin Garralaga, James Fairfax, Charles Irwin, Terry Kilburn, Alberto Morin,
CAST: Philip Ford, Lloyd Bridges, Lorna Gray, Ray Collins, Sheila Ryan, Fletcher Chandler, Jeff Corey, Alan Carney, Emory Parnell, Don Beddoe, Nana Bryant, Charles Halton, Paul E. Burns, Douglas Evans,
The Dead Outside
CAST: Kerry Anne Mullaney, Sandra Louise Douglas, Alton Milne, Sharon Osdin, Vivienne Harvey, John Erskine, Phylis Douglas, Robin Morris, Max Adair, Ralph Berger, John Bruce, Linda Dickie, Cheryl Easton, Alan Fawcus, Leo Horsfield,
Babysitter Wanted
CAST: Jonas Barnes, Tina Houtz, Sarah Thompson, Nana Visitor, Jillian Schmitz, Matt Dallas, Linda Neal, Monty Bane, Douglas Rowe, Bruce Thomas, Kristen Dalton, Kai Caster, Brett Claywell, Bill Moseley, Jeff Markey, Cristie Schoen, Scott Spiegel,

redpajama_set_name: RedPajamaCommonCrawl


4th: Jay lined out to first.
6th: Jay grounded out to third.
8th: Jay singled to right, Happ scored, La Stella to second.
1st: Bryant singled to right.
2nd: Bryant homered to left (392 feet).
7th: Bryant singled to center.
8th: Bryant walked, La Stella to third, Jay to second.
1st: Rizzo walked, Bryant to second.
4th: Rizzo grounded out to shortstop.
7th: Rizzo grounded into double play, third to shortstop to first, Bryant out at second.
3rd: Contreras singled to center.
5th: Contreras grounded out to third.
7th: Contreras struck out looking.
1st: Zobrist reached on infield single to third, Bryant to third, Rizzo to second.
3rd: Zobrist grounded into fielder's choice to first, Contreras out at second.
8th: Zobrist safe at first on error by pitcher Alcántara.
9th: Almora Jr. grounded into double play, shortstop to second to first, Caratini out at second.
1st: Russell doubled to deep right, Bryant, Rizzo and Zobrist scored.
3rd: Russell lined out to left.
8th: Russell grounded into fielder's choice to second, Zobrist out at second.
1st: Heyward doubled to right center, Russell scored.
3rd: Heyward walked, Zobrist to second.
9th: Heyward popped out to second.
1st: Báez struck out looking.
3rd: Báez homered to left (422 feet), Zobrist and Heyward scored.
8th: Báez singled to shallow left, Russell to second.
3rd: Lester flied out to center.
6th: Lester struck out looking.
8th: La Stella singled to center, Russell scored, Happ to third.
1st: Pham grounded out to second.
5th: Pham flied out to right.
7th: Pham grounded out to third.
3rd: Fowler singled to right, Pham to third.
5th: Fowler grounded out to shortstop.
3rd: DeJong grounded into fielder's choice to shortstop, Fowler out at second.
6th: DeJong popped out to second.
8th: Mejia grounded out to third.
4th: Molina lined out to center.
8th: Kelly grounded out to third.
2nd: Gyorko homered to left (410 feet).
4th: Gyorko reached on infield single to shortstop.
6th: Gyorko singled to right.
9th: Garcia struck out looking.
2nd: Piscotty flied out to center.
4th: Piscotty walked, Gyorko to second.
2nd: Wong struck out looking.
4th: Wong grounded out to second, Gyorko to third, Piscotty to second.
9th: Rosario grounded into double play, shortstop to second to first, Piscotty out at second.
3rd: Bader flied out to left.
4th: Grichuk flied out to left.
7th: Voit homered to center (398 feet).
Addison Russell hit a three-run double in the first inning, then made a food run for fans in enemy territory while the Chicago Cubs beat the St. Louis Cardinals 10-2 Monday night to move within a win of cinching a second straight NL Central title.