You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
author:
- |
    FERRUCCIO COLOMBINI\
    \
    \
    \
    FRANCESCO FANELLI\
    \
    \
date: 'February 16, 2010'
---

### Abstract {#abstract .unnumbered}

In this paper we obtain an energy estimate for a complete strictly hyperbolic operator with second order coefficients satisfying a log-Zygmund-continuity condition with respect to $t$, uniformly with respect to $x$, and a log-Lipschitz-continuity condition with respect to $x$, uniformly with respect to $t$.

### Keywords {#keywords .unnumbered}

Hyperbolic equations, log-Zygmund regularity, loss of derivatives.

### MS Classification 2000 {#ms-classification-2000 .unnumbered}

35L15

Introduction
============

Let us consider the second order operator $$\label{eq:P}
 P\,=\,\partial^2_t\,-\,\sum_{i,j=1}^n\partial_{x_i}(a_{ij}(t)\partial_{x_j})$$ and suppose that $P$ is strictly hyperbolic, i.e. there exist two positive constants $\lambda_0\leq\Lambda_0$ such that $$\label{eq:P_elliptic}
 \lambda_0\,|\xi|^2\,\leq\,\sum_{i,j=1}^n a_{ij}(t)\,\xi_i\xi_j\,\leq\,\Lambda_0\,|\xi|^2$$ for all $t\in{\mathbb}{R}$ and all $\xi\in{\mathbb}{R}^n$.

It is well-known (see e.g. [@Horm] and [@Treves]) that, if the coefficients $a_{ij}$ are Lipschitz-continuous, then the following energy estimate holds for the operator $P$: for all $s\in{\mathbb}{R}$, there exists $C_s>0$ such that $$\begin{aligned}
 \label{eq:en-estim_t_Lip}
& & \sup_{t\in[0,T]}\left(\|u(t,\cdot)\|_{H^{s+1}}\,+\,\|\partial_tu(t,\cdot)\|_{H^s}\right)\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\leq\,C_s\,\left(\|u(0,\cdot)\|_{H^{s+1}}\,+\,\|\partial_tu(0,\cdot)\|_{H^s}\,+\,
\int^T_0\|Pu(t,\cdot)\|_{H^s}\,dt\right) \nonumber\end{aligned}$$ for every function $u\in{\mathcal}{C}^2([0,T];H^\infty({\mathbb}{R}^n))$.\
In particular, the previous energy estimate implies that the Cauchy problem for (\[eq:P\]) is well-posed in the space $H^\infty$, with no loss of derivatives.

On the contrary, if the coefficients $a_{ij}$ are not Lipschitz-continuous, then (\[eq:en-estim\_t\_Lip\]) is no more true in general, as it is shown by an example given by Colombini, De Giorgi and Spagnolo in the paper [@Colomb-DeG-S]. Nevertheless, under suitable weaker regularity assumptions on the coefficients, one can recover the $H^\infty$-well-posedness again, but this time from an energy estimate with loss of derivatives.

A first result of this type was obtained in the quoted paper [@Colomb-DeG-S]. The authors supposed that there was a constant $C>0$ such that, for all ${\varepsilon}\in]0,T]$, $$\label{eq:cont_t_Log-Lip_int}
 \int^{T-{\varepsilon}}_0|a_{ij}(t+{\varepsilon})-a_{ij}(t)|\,dt\,\leq\,C\,{\varepsilon}\,\log\left(1+\frac{1}{{\varepsilon}}\right)\,.$$ The Fourier trasform with respect to $x$ of the equation, togheter with the new “approximate energy technique” (i.e. the approximation of the coefficients is different in different zones of the phase space), enabled them to obtain the following energy estimate: there exist strictly positive constants $K$ (indipendent of $s$) and $C_s$ such that $$\begin{aligned}
 \label{eq:en-estim_t_Log-Lip}
& & \sup_{t\in[0,T]}\left(\|u(t,\cdot)\|_{H^{s+1-K}}\,+\,\|\partial_tu(t,\cdot)\|_{H^{s-K}}\right)\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\leq\,C_s\,\left(\|u(0,\cdot)\|_{H^{s+1}}\,+\,\|\partial_tu(0,\cdot)\|_{H^s}\,+\,
\int^T_0\|Pu(t,\cdot)\|_{H^s}\,dt\right) \nonumber\end{aligned}$$ for all $u\in{\mathcal}{C}^2([0,T];H^\infty({\mathbb}{R}^n))$.

Considering again the case that the coefficients of $P$ depend only on the time variable, in the recent paper [@Tarama] (see also [@Yama]) Tarama has weakened the regularity hypotesis further, supposing a log-Zygmund type integral condition, i.e. that there exists a constant $C>0$ such that, for all ${\varepsilon}\in]0,T/2]$, $$\label{eq:cont_t_Log-Zyg_int}
 \int^{T-{\varepsilon}}_{{\varepsilon}}|a_{ij}(t+{\varepsilon})+a_{ij}(t-{\varepsilon})-2a_{ij}(t)|\,dt\,\leq\,C\,{\varepsilon}\,\log\left(1+\frac{1}{{\varepsilon}}\right)\,.$$ Nevertheless, he has been able to prove the well-posedness to the Cauchy problem for (\[eq:P\]) in the space $H^\infty$: the improvement with respect to [@Colomb-DeG-S] was obtained introducing a new type of approximate energy, which involves the second derivatives of the approximating coefficients.

Much more difficulties arise if the operator $P$ has coefficients depending both on the time variable $t$ and on the space variables $x$. This case was considered by Colombini and Lerner in the paper [@Colomb-Lern]. They supposed a pointwise log-Lipschitz regularity condition, i.e. that there exists $C>0$ such that, for all ${\varepsilon}\in]0,T]$, $$\label{eq:cont_t-x_Log-Lip_punt}
 \sup_{\begin{array}{c}
        \scriptstyle y,z\in[0,T]\times{\mathbb}{R}^n \\
    \scriptstyle |z|={\varepsilon}\end{array}} |a_{ij}(y+z)-a_{ij}(y)|\,\leq\,C\,{\varepsilon}\,\log\left(1+\frac{1}{{\varepsilon}}\right)\,.$$ Because the coefficients of the operator $P$ depend also on the space variables, here the Littlewood-Paley dyadic decomposition with respect to $x$ takes the place of the Fourier trasform, and it is, togheter with the approximate energy technique, the key tool to obtain the energy estimate: for all fixed $\theta\in]0,1/4]$, there exist $\beta$, $C>0$ and $T^*\in]0,T]$ such that $$\begin{aligned}
 \label{eq:en-estim_t-x_Log-Lip}
& & \sup_{t\in[0,T^*]}\left(\|u(t,\cdot)\|_{H^{-\theta+1-\beta t}}\,+\,\|\partial_tu(t,\cdot)\|_{H^{-\theta-\beta t}}\right)\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\leq\,C\,\left(\|u(0,\cdot)\|_{H^{-\theta+1}}\,+\,\|\partial_tu(0,\cdot)\|_{H^{-\theta}}\,+\,
\int^T_0\|Pu(t,\cdot)\|_{H^{-\theta-\beta t}}\,dt\right) \nonumber\end{aligned}$$ for all $u\in{\mathcal}{C}^2([0,T^*];H^\infty({\mathbb}{R}^n))$.\
In this case, the loss of derivatives gets worse with the increasing of time.

In a recent paper ( [@Colomb-DelSant_2]), Colombini and Del Santo considered the case of one space variable (i.e. $n=1$) and studied again the case of the coefficient $a$ depending both on $t$ and $x$, but under a special regularity condition: they mixed condition (\[eq:cont\_t\_Log-Zyg\_int\]) togheter with (\[eq:cont\_t-x\_Log-Lip\_punt\]). In particular, they supposed $a$ to be log-Zygmund-continuous with respect to $t$, uniformly with respect to $x$, and log-Lipschitz-continuous with respect to $x$, uniformly with respect to $t$. The dyadic decomposition technique and the Tarama’s approximate energy enabled them to obtain an estimate similar to (\[eq:en-estim\_t-x\_Log-Lip\]).

The reason why they focused on the case $n=1$ is that the case of several space variables needs some different and new ideas in the definition of the microlocal energy: this point still remains as an open problem.

In the present note, we will consider the case of the non-homogeneous operator $$\label{eq:L}
Lu\,=\,\partial_{t}^{2}u\,-\,\partial_x(a(t,x)\partial_x u)\,+\,b_0(t,x)\partial_tu\,+\,b_1(t,x)\partial_x u\,+\,c(t,x)u\;,$$ where the coefficient $a$ satisfy the same regularity assumptions as in [@Colomb-DelSant_2]. We will also suppose that $b_0$, $b_1\in L^\infty({\mathbb}{R}_t;{\mathcal}{C}^{\omega}({\mathbb}{R}_x))$, $\omega>0$, and $c$ bounded on ${\mathbb}{R}_t\times{\mathbb}{R}_x$. We will apply the Littlewood-Paley decomposition and the Tarama’s approximate energy again to obtain an energy estimate with a loss of derivatives that depends on $t$, as in (\[eq:en-estim\_t-x\_Log-Lip\]).

One can find the estimate of the second order coefficient $a$ in the paper [@Colomb-DelSant_2], however, for reader’s convenience, we will give here all the details.

Main result
===========

Let $a:{\mathbb}{R}^2\rightarrow{\mathbb}{R}$ be a function such that, for positive constants $\lambda_0\leq\Lambda_0$ and $C_0$, one has, for all $(t,x)\in{\mathbb}{R}^2$ and all $\tau>0$, $y>0$, $$\begin{aligned}
 & \lambda_0\leq a(t,x)\leq\Lambda_0   \label{(N)eq_condiz_1.1} \\
 & \sup_{(t,x)}|a(t+\tau,x)+a(t-\tau,x)-2a(t,x)|\leq C_0\,\tau\log\left(\frac{1}{\tau}+1\right) \label{(N)eq_condiz_1.2} \\
 & \sup_{(t,x)}|a(t,x+y)-a(t,x)|\leq C_0\,y\log\left(\frac{1}{y}+1\right) \label{(N)eq_condiz_1.3}\end{aligned}$$ Moreover, let $$\label{(N)eq_condiz_B}
b_0\,,\,b_1\;\in\,L^\infty({\mathbb}{R}_t;{\mathcal}{C}^{\omega}({\mathbb}{R}_x))\,,$$ where $\omega>0$, and $$\label{(N)eq_condiz_c}
c\;\in\,L^{\infty}({\mathbb}{R}_t\times{\mathbb}{R}_x)\;.$$

\[(N)teor\_teorema1\] Let us consider, on the whole space ${\mathbb}{R}^2$, the operator $$\label{(N)operatore}
Lu\,=\,\partial_{t}^{2}u\,-\,\partial_x(a(t,x)\partial_x u)\,+\,b_0(t,x)\partial_tu\,+\,b_1(t,x)\partial_x u\,+\,c(t,x)u\;,$$ where the coefficients $a$, $b_0$, $b_1$ and $c$ satisfy the hypothesis (\[(N)eq\_condiz\_1.1\])-(\[(N)eq\_condiz\_c\]).\
Then, for all fixed $$\theta\,\in\,\left]0,\min\left\{\frac{1}{2},\frac{\omega}{1+\log2}\right\}\right[\;,$$ there exist $\beta^*>0$, a time $T\in{\mathbb}{R}$ and a constant $C>0$ such that $$\begin{aligned}
 \label{(N)eq_tesi}
 & & \sup_{[0,T]}\left(\|u(t,\cdot)\|_{H^{1-\theta-\beta^*t}}+
\|\partial_t u(t,\cdot)\|_{H^{-\theta-\beta^*t}}\right)\,\leq \\
 & & \qquad\qquad\qquad\qquad\leq\,C\,\left(\|u(0,\cdot)\|_{H^{1-\theta}}+
\|\partial_t u(0,\cdot)\|_{H^{-\theta}}+\int_{0}^{T}\|Lu(t,\cdot)\|_{H^{-\theta-\beta^*t}}dt\right) \nonumber
$$ for all $u\in{\mathcal}{C}^2([0,T];H^\infty({\mathbb}{R}_x))$.

Proof of Theorem \[(N)teor\_teorema1\]
======================================

Approximation of the coefficient $a(t,x)$
-----------------------------------------

Let $\rho\in{\mathcal}{C}_{0}^{\infty}({\mathbb}{R})$ be an even function such that:

1.  $0\leq\rho\leq1$

2.  $supp\,\rho\,\subset\,[-1,1]$

3.  $\int\rho(s)ds=1$

4.  $|\rho'(s)|\leq2$;

for all $0<{\varepsilon}\leq1$, we set $\rho_{{\varepsilon}}(s)=\frac{1}{{\varepsilon}}\,\rho\left(\frac{s}{{\varepsilon}}\right)$.

Then, for all $0<{\varepsilon}\leq1$, we define $$\label{(N)eq_a_eps}
a_{\varepsilon}(t,x)\,:=\,\int_{{\mathbb}{R}_t\times{\mathbb}{R}_x}\rho_{\varepsilon}(t-s)\,\rho_{\varepsilon}(x-y)\,a(s,y)\,dsdy\;.$$

\[(N)lemma\_approx\] The following inequalities hold true:

1.  for all ${\varepsilon}\in]0,1]$, for all $(t,x)\in{\mathbb}{R}^2$, one has $$\label{(N)eq_condiz_1.1_eps}
    \lambda_0\,\leq\,a_{\varepsilon}(t,x)\,\leq\,\Lambda_0\;;$$

2.  for all ${\varepsilon}\in]0,1]$, one has $$\label{(N)eq_a_eps-a}
    \sup_{(t,x)}|a_{\varepsilon}(t,x)-a(t,x)|\,\leq\,
    \frac{3}{2}C_0\,{\varepsilon}\,\log\left(\frac{1}{{\varepsilon}}+1\right)\;;$$

3.  for all $\sigma\in]0,1[$, a constant $c_\sigma>0$ exists such that, for all ${\varepsilon}\in]0,1]$, $$\label{(N)eq_d_t_a_eps}
    \sup_{(t,x)}|\partial_t a_{\varepsilon}(t,x)|\,\leq\,c_\sigma (\Lambda_0+C_0)\,{\varepsilon}^{\sigma-1}\;;$$

4.  for all ${\varepsilon}\in]0,1]$, one has $$\begin{aligned}
    \sup_{(t,x)}|\partial_x a_{\varepsilon}(t,x)| & \leq &
    C_0\|\rho'\|_{L^1}\log\left(\frac{1}{{\varepsilon}}+1\right) \label{(N)eq_d_x_a_eps} \\
    \sup_{(t,x)}\left|\partial_{t}^{2} a_{\varepsilon}(t,x)\right| & \leq &
    \frac{C_0}{2}\|\rho''\|_{L^1}\frac{1}{{\varepsilon}}\log\left(\frac{1}{{\varepsilon}}+1\right) \label{(N)eq_d_t^2_a_eps} \\
    \sup_{(t,x)}|\partial_t\partial_x a_{\varepsilon}(t,x)| & \leq &
    C_0\|\rho'\|_{L^1}^{2}\frac{1}{{\varepsilon}}\log\left(\frac{1}{{\varepsilon}}+1\right) \label{(N)eq_d_t_d_x_a_eps}\;.\end{aligned}$$

Inequalities in (\[(N)eq\_condiz\_1.1\_eps\]) immediately follow from the fact that $|\rho|\leq1$.

Relation (\[(N)eq\_a\_eps-a\]), instead, follows from (\[(N)eq\_condiz\_1.2\]), after one has observed that $$a_{\varepsilon}(t,x)-a(t,x)=\frac{1}{2}\int_{{\mathbb}{R}_t}\rho_{\varepsilon}(s)\int_{{\mathbb}{R}_y}\rho_{\varepsilon}(x-y)(a(t+s,y)+a(t-s,y)-2a(t,y))\,dy\,ds\;,$$ where we have used the fact that $\rho$ is an even function. Moreover, one has $$\partial_t^2a_{\varepsilon}(t,x)=\frac{1}{2}\int\rho_{\varepsilon}''(s)\int\rho_{\varepsilon}(x-y)(a(t+s,y)+a(t-s,y)-2a(t,y))\,dy\,ds\;,$$ from which one can deduce (\[(N)eq\_d\_t\^2\_a\_eps\]).

Inequalities (\[(N)eq\_d\_x\_a\_eps\]) and (\[(N)eq\_d\_t\_d\_x\_a\_eps\]) derive from (\[(N)eq\_condiz\_1.3\]) in a very similar way.

Finally, relation (\[(N)eq\_d\_t\_a\_eps\]) is a consequence of the fact that (\[(N)eq\_condiz\_1.1\]) and (\[(N)eq\_condiz\_1.2\]) imply that for all $\sigma\in]0,1[$, a constant $c_\sigma'>0$ exists such that, for all $\tau>0$, one has $$\label{(N)eq_2.6}
\sup_{(t,x)}|a(t+\tau,x)-a(t,x)|\,\leq\,c_\sigma'\,(\Lambda_0+C_0)\,\tau^\sigma\;.$$

Littlewood-Paley decomposition {#sub-sec:L-P_dec}
------------------------------

We collect here some well-known facts about dyadic decomposition, referring to [@Bony], [@Colomb-Lern] and [@Metivier] for the details.

Let ${\varphi}_0\in{\mathcal}{C}_{0}^{\infty}({\mathbb}{R}_\xi)$ be an even function, decreasing on $[0,+\infty[$, such that $0\leq{\varphi}_0\leq1$ and $${\varphi}_0(\xi)=1\,\mbox{ if }\,|\xi|\leq1\quad\mbox{,}\quad{\varphi}_0(\xi)=0\,\mbox{ if }\,|\xi|\geq2.$$ We set ${\varphi}(\xi)\,=\,{\varphi}_0(\xi)-{\varphi}_0(2\xi)$ and, for $\nu\in{\mathbb}{N}\backslash\{0\}$, ${\varphi}_{\nu}(\xi):={\varphi}(2^{-\nu}\xi)$.

For a tempered distribution $u\in H^{-\infty}({\mathbb}{R})$, we define $$u_\nu(x)\,:=\,{\varphi}_\nu(D_x)u(x)\,=\,\frac{1}{2\pi}\int e^{ix\xi}{\varphi}_\nu(\xi)\,{\widehat}{u}(\xi)\,d\xi\,=\,
\frac{1}{2\pi}\int{\widehat}{{\varphi}_\nu}(y)u(x-y)\,dy\,;$$ for all $\nu$, $u_\nu$ is an entire analytic function belonging to $L^2$.\
Moreover, for all $s\in{\mathbb}{R}$ there exists a constant $C_s>0$ such that $$\label{eq:dyadic-dec_Sobolev-norm}
 \frac{1}{C_s}\,\sum^{+\infty}_{\nu=0}2^{2\nu s}\|u_\nu\|^2_{L^2}\,\leq\,\|u\|^2_{H^s}\,\leq\,
C_s\,\sum^{+\infty}_{\nu=0}2^{2\nu s}\|u_\nu\|^2_{L^2}$$ and the following inequalities (called “Bernstein’s inequalities”) hold: $$\begin{aligned}
 \|\partial_x u_\nu\|_{L^2} & \leq & 2^{\nu+1}\,\|u_\nu\|_{L^2}\qquad\qquad\;\mbox{  for all } \nu\geq0 \label{eq:Bern-ineq_deriv<} \\
 \|u_\nu\|_{L^2} & \leq & 2^{-(\nu-1)}\,\|\partial_x u_\nu\|_{L^2}\qquad\mbox{ for all } \nu\geq1 \label{eq:Bern-ineq_<deriv}\,.\end{aligned}$$

We end this subsection quoting a result which will be useful in the following; for its proof, see [@Colomb-Lern].\
We denote with $[A,B]$ the commutator between two linear operators $A$ and $B$ and with ${\mathcal}{L}(L^2)$ the space of bounded linear operators from $L^2$ to $L^2$.

\[lemma:comm\]

1.  There exist $C>0$, $\nu_0\in{\mathbb}{N}$ such that, for all $a\in L^\infty({\mathbb}{R})$ satisfying $$\sup_{x\in{\mathbb}{R}}|a(x+y)-a(x)|\,\leq\,C_0\,y\,\log\left(1+\frac{1}{y}\right)$$ for all $y>0$, one has, for all $\nu\geq\nu_0$, $$\label{eq:comm_Log-Lip}
     \|\,[{\varphi}_\nu(D_x),a(x)]\,\|_{{\mathcal}{L}(L^2)}\,\leq\,C\,(\|a\|_{L^\infty}+C_0)\,2^{-\nu}\,\nu\,.$$

2.  There exist $C>0$, $\nu_0\in{\mathbb}{N}$ such that, for all $b\in{\mathcal}{C}^\omega({\mathbb}{R})$ and all $\nu\geq\nu_0$, one has $$\label{eq:comm_Holder}
     \|\,[{\varphi}_\nu(D_x),b(x)]\,\|_{{\mathcal}{L}(L^2)}\,\leq\,C\,\|b\|_{{\mathcal}{C}^\omega}\,2^{-\nu}\,.$$

Approximate and total energy
----------------------------

Let $T_0>0$ and $u\in{\mathcal}{C}^2([0,T_0];H^\infty({\mathbb}{R}_x))$. If we set $u_\nu(t,x)={\varphi}_\nu(D_x)u(t,x)$, we obtain $$\begin{aligned}
 \label{(N)eq_ni}
(Lu)_{\nu} & = & \partial_{t}^{2}u_{\nu}\,-\,\partial_x(a(t,x)\partial_x u_{\nu}) - \partial_x([{\varphi}_{\nu}(D_x),a]\partial_x u) \\
           & + &  b_0(t,x)\partial_t u_\nu + [{\varphi}_\nu(D_x),b_0]\partial_t u\,+\,
b_1(t,x)\partial_x u_\nu + [{\varphi}_\nu(D_x),b_1]\partial_x u \nonumber\\
           & + &  c(t,x)u_\nu + [{\varphi}_\nu(D_x),c]u \nonumber\;.\end{aligned}$$

Now, we introduce the approximate energy of $u_\nu$ (see [@Colomb-DelSant_2] and [@Tarama]), setting $$\label{(N)eq_e_nu-eps}
e_{\nu,\varepsilon}(t):=\int_{{\mathbb}{R}}\left(\frac{1}{\sqrt{a_\varepsilon}}\left|\partial_t u_\nu+
\frac{\partial_t\sqrt{a_\varepsilon}}{2\sqrt{a_\varepsilon}}u_\nu\right|^2+\sqrt{a_\varepsilon}|\partial_x u_\nu|^2+|u_\nu|^2\right)\,dx\;,$$ and, taken $\theta$ as in the hypothesis of Theorem \[(N)teor\_teorema1\], we define the total energy of $u$: $$\label{(N)eq_E_tot}
E(t)\,:=\,\sum_{\nu=0}^{+\infty}e^{-2\beta(\nu+1)t}\,2^{-2\nu\theta}\,e_{\nu,2^{-\nu}}(t)\;,$$ where $\beta>0$ will be fixed later on.

\[remark:E(0)-E(t)\] From (\[eq:dyadic-dec\_Sobolev-norm\]) and Bernstein’s inequalities (\[eq:Bern-ineq\_deriv&lt;\])-(\[eq:Bern-ineq\_&lt;deriv\]), it’s easy to see that there exist two positive constants $C_\theta$ and $C_\theta'$ such that $$\begin{aligned}
E(0) & \leq & C_\theta\,(\|\partial_tu(0,\cdot)\|_{H^{-\theta}}\,+\,\|u(0,\cdot)\|_{H^{1-\theta}}) \\ E(t) & \geq & C_\theta'\,(\|\partial_tu(t,\cdot)\|_{H^{-\theta-\beta^*t}}\,+\,
\|u(t,\cdot)\|_{H^{1-\theta-\beta^*t}}) $$ where we have set $\beta^*=\beta(\log2)^{-1}$.

First, we derive $e_{\nu,{\varepsilon}}$, defined by (\[(N)eq\_e\_nu-eps\]), with respect to the time variable, and we obtain $$\begin{aligned}
\frac{d}{dt}e_{\nu,{\varepsilon}}(t) & = & \int\frac{2}{\sqrt{a_{\varepsilon}}}\,\,\Re\left(\partial_t^2u_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \\
 & + & \int\frac{2}{\sqrt{a_{\varepsilon}}}\,\,\Re\left(R_{\varepsilon}u_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \\
 & + & \int\partial_t\sqrt{a_{\varepsilon}}\,\,|\partial_xu_\nu|^2\,dx \\
 & + & \int2\sqrt{a_{\varepsilon}}\,\,\Re(\partial_xu_\nu\,\cdot\,\overline{\partial_x\partial_tu_\nu})\,dx \\
 & + & \int2\,\,\Re(u_\nu\,\cdot\,\overline{\partial_tu_\nu})\,dx\;,\end{aligned}$$ where $R_{\varepsilon}v\,=\,\partial_t\left(\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}\right)v-
\left(\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}\right)^2v$. Now, we can put in the previous relation the value of $\partial_t^2u_\nu$, given by (\[(N)eq\_ni\]).

Integrating by parts and taking advantage of the spectral localisation of $u_\nu$, we have $$\begin{aligned}
 & & \int\frac{2}{\sqrt{a_{\varepsilon}}}\,\,\Re\left(\partial_x(a\partial_xu_\nu)\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx\,= \\
 & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\,\int2\frac{\partial_x\sqrt{a_{\varepsilon}}}{a_{\varepsilon}}\,a\,\,\Re\left(\partial_xu_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \\
 & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad-\,\int\frac{\partial_t\sqrt{a_{\varepsilon}}}{a_{\varepsilon}}\,a\,|\partial_xu_\nu|^2\,dx \\
 & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad-\,\int2\frac{a}{\sqrt{a_{\varepsilon}}}\,\,\Re\left(\partial_xu_\nu\,\cdot\,
\overline{\partial_x\partial_tu_\nu}\right)\,dx \\
 & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad-\,\int\frac{a}{\sqrt{a_{\varepsilon}}}\,\,
\partial_x\left(\frac{\partial_t\sqrt{a_{\varepsilon}}}{\sqrt{a_{\varepsilon}}}\right)\,
\Re(\partial_xu_\nu\,\cdot\,\overline{u_\nu})\,dx\,;\end{aligned}$$ taken care of the fact that $b_0$, $b_1$ and $c$ are real-valued, finally we obtain the complete expression for the time derivative of the approximate energy: $$\begin{aligned}
 \label{(N)eq_dt_e_nu,eps}
 & & \frac{d}{dt}e_{\nu,{\varepsilon}}(t)\,=\,\int\frac{2}{\sqrt{a_{\varepsilon}}}\,\,\Re
\left((Lu)_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \\
 & & \qquad\qquad\quad\,+\,\int\frac{2}{\sqrt{a_{\varepsilon}}}\,\,\Re\left(R_{\varepsilon}u_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \nonumber \\
 & & \qquad\qquad\quad\,+\,\int\partial_t\sqrt{a_{\varepsilon}}\left(1-\frac{a}{a_{\varepsilon}}\right)|\partial_xu_\nu|^2\,dx \nonumber \\
 & & \quad\qquad\qquad\,+\,\int2\left(\sqrt{a_{\varepsilon}}-\frac{a}{\sqrt{a_{\varepsilon}}}\right)\Re(\partial_xu_\nu\,\cdot\,
\overline{\partial_x\partial_tu_\nu})\,dx \nonumber \\
 & & \qquad\qquad\quad\,+\,\int2\frac{\partial_x\sqrt{a_{\varepsilon}}}{a_{\varepsilon}}\,a\,\,\Re\left(\partial_xu_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \nonumber \\
 & & \qquad\qquad\quad\,-\,
\int\frac{a}{\sqrt{a_{\varepsilon}}}\,\,\partial_x\left(\frac{\partial_t\sqrt{a_{\varepsilon}}}{\sqrt{a_{\varepsilon}}}\right)\,
\Re(\partial_xu_\nu\,\cdot\,\overline{u_\nu})\,dx \nonumber \\
 & & \quad\qquad\qquad\,+\,\int2\,\Re(u_\nu\,\cdot\,\overline{\partial_tu_\nu})\,dx \nonumber \\
 & & \quad\qquad\qquad\,+\,\int\frac{2}{\sqrt{a_{\varepsilon}}}\,\,\Re\left((\partial_x([{\varphi}_\nu(D_x),a]\partial_xu))\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \nonumber \\
 & & \qquad\qquad\quad\,-\,\int\frac{2}{\sqrt{a_{\varepsilon}}}\,b_0(t,x)\,\,\Re\left(\partial_tu_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \nonumber \\
 & & \qquad\qquad\quad\,-\,\int\frac{2}{\sqrt{a_{\varepsilon}}}\,\,\Re\left(([{\varphi}_\nu(D_x),b_0]\partial_tu)\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \nonumber \\
 & & \qquad\qquad\quad\,-\,\int\frac{2}{\sqrt{a_{\varepsilon}}}\,b_1(t,x)\,\,\Re\left(\partial_xu_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \nonumber \\
 & & \qquad\qquad\quad\,-\,\int\frac{2}{\sqrt{a_{\varepsilon}}}\,\,\Re\left(([{\varphi}_\nu(D_x),b_1]\partial_xu)\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \nonumber \\
 & & \qquad\qquad\quad\,-\,\int\frac{2}{\sqrt{a_{\varepsilon}}}\,c(t,x)\,\,\Re\left(u_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \nonumber \\
 & & \qquad\qquad\quad\,-\,\int\frac{2}{\sqrt{a_{\varepsilon}}}\,\,\Re\left(([{\varphi}_\nu(D_x),c]u)\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx \nonumber\;.\end{aligned}$$

Estimate for the approximate energy
-----------------------------------

We want to obtain an energy estimate; so, let us start to control each term of (\[(N)eq\_dt\_e\_nu,eps\]).\
Through the rest of the proof, we will denote with $C$, $C'$, $C''$ and ${\widehat}{C}$ constants depending only on $\lambda_0$, $\Lambda_0$, $C_0$ and on the norms of the coefficients of the lower order terms of the operator $L$ in their respective functional spaces, and which are allowed to vary from line to line.

### Terms with $a$ and $a_{\varepsilon}$

Thanks to relations (\[(N)eq\_condiz\_1.1\]), (\[(N)eq\_d\_t\_a\_eps\]) with $\sigma=1/2$, (\[(N)eq\_d\_t\^2\_a\_eps\]) and Bernstein’s inequalities, we deduce that there exists $C>0$, depending only on $\lambda_0$, $\Lambda_0$ and $C_0$, such that, for all $\nu\in{\mathbb}{N}$, $$\left|\int\frac{2}{\sqrt{a_{\varepsilon}}}\,\,\Re\left(R_{\varepsilon}u_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}
\right)\,dx\right|\,\leq\,C\,\frac{1}{{\varepsilon}}\,\log\left(\frac{1}{{\varepsilon}}+1\right)\,
2^{-\nu}\,e_{\nu,{\varepsilon}}(t)\;.$$ In the same way, from (\[(N)eq\_condiz\_1.1\]), (\[(N)eq\_a\_eps-a\]) and (\[(N)eq\_d\_t\_a\_eps\]), we have $$\left|\int\partial_t\sqrt{a_{\varepsilon}}\left(1-\frac{a}{a_{\varepsilon}}\right)|\partial_xu_\nu|^2\,dx\right|\,\leq\,
C\,\log\left(\frac{1}{{\varepsilon}}+1\right)\,e_{\nu,{\varepsilon}}(t)\;,$$ for a constant $C$ depending again only on $\lambda_0$, $\Lambda_0$ and $C_0$.

Moreover, again from (\[(N)eq\_condiz\_1.1\]) and (\[(N)eq\_a\_eps-a\]) and Bernstein’s inequalities, we obtain $$\begin{aligned}
\left|\int2\left(\sqrt{a_{\varepsilon}}-\frac{a}{\sqrt{a_{\varepsilon}}}\right)\Re(\partial_xu_\nu\,\cdot\,
\overline{\partial_x\partial_tu_\nu})\,dx\right| & \leq & C\,{\varepsilon}\,\log\left(\frac{1}{{\varepsilon}}+1\right)\,
\|\partial_xu_\nu\|_{L^2}\,\|\partial_x\partial_tu_\nu\|_{L^2} \\
 & \leq & C\,{\varepsilon}\,\log\left(\frac{1}{{\varepsilon}}+1\right)\,2^{\nu+1}\,
\|\partial_xu_\nu\|_{L^2}\,\|\partial_tu_\nu\|_{L^2}\;;\end{aligned}$$ but we have $$\|\partial_tu_\nu\|_{L^2}\,\leq\,\left\|\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu
\right\|_{L^2}\,+\,\left\|\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right\|_{L^2}$$ and $$\begin{aligned}
\left\|\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_0\right\|_{L^2} & \leq & C\,{\varepsilon}^{-1/2}\|u_0\|_{L^2} \\
\left\|\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right\|_{L^2} & \leq & C\,{\varepsilon}^{-1/2}\,2^{-\nu}\|u_\nu\|_{L^2}\qquad(\nu\geq1)\;,\end{aligned}$$ that give us the following: $$\left|\int2\left(\sqrt{a_{\varepsilon}}-\frac{a}{\sqrt{a_{\varepsilon}}}\right)\Re(\partial_xu_\nu\cdot
\overline{\partial_x\partial_tu_\nu})dx\right|\,\leq\,
C({\varepsilon}\,2^\nu+1)\log\left(\frac{1}{{\varepsilon}}+1\right)e_{\nu,{\varepsilon}}(t)\;.$$

In a very similar way, from (\[(N)eq\_d\_x\_a\_eps\]) one has $$\left|\int2\frac{\partial_x\sqrt{a_{\varepsilon}}}{a_{\varepsilon}}\,a\,\,\Re\left(\partial_xu_\nu\cdot
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)dx\right|
\,\leq\,C\log\left(\frac{1}{{\varepsilon}}+1\right)e_{\nu,{\varepsilon}}(t)\;;$$ moreover, from (\[(N)eq\_d\_t\_a\_eps\]) with $\sigma=1/2$, (\[(N)eq\_d\_x\_a\_eps\]) and (\[(N)eq\_d\_t\_d\_x\_a\_eps\]) we deduce $$\left|\int\frac{a}{\sqrt{a_{\varepsilon}}}\,\,\partial_x\left(\frac{\partial_t\sqrt{a_{\varepsilon}}}{\sqrt{a_{\varepsilon}}}\right)\,
\Re(\partial_xu_\nu\,\cdot\,\overline{u_\nu})\,dx\right|\,\leq\,C\,\frac{1}{{\varepsilon}}\,
\log\left(\frac{1}{{\varepsilon}}+1\right)\,2^{-\nu}\,e_{\nu,{\varepsilon}}(t)\;.$$

Finally, we have $$\left|\int2\,\Re(u_\nu\,\cdot\,\overline{\partial_tu_\nu})\,dx\right|\,\leq\,
C\,{\varepsilon}^{-1/2}\,2^{-\nu}\,e_{\nu,{\varepsilon}}(t)\;.$$

### Terms with $b_0$, $b_1$ and $c$

Thanks to the hypothesis (\[(N)eq\_condiz\_B\])-(\[(N)eq\_condiz\_c\]), one has that there exist suitable constants, depending only on $\lambda_0$, $\Lambda_0$, $C_0$ and on the norms of $b_0$ and $b_1$ in the space $L^\infty({\mathbb}{R}_t;{\mathcal}{C}^{\omega}({\mathbb}{R}_x))$ and of $c$ in $L^\infty({\mathbb}{R}_t\times{\mathbb}{R}_x)$, such that $$\begin{aligned}
& & \left|\int\frac{2}{\sqrt{a_{\varepsilon}}}\,b_0(t,x)\,\,\Re\left(\partial_tu_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx\right|\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,C\,\|\partial_t u_\nu\|_{L^2}\,
\left\|\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right\|_{L^2} \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,(C+C'\,2^{-\nu}{\varepsilon}^{-1/2})\,e_{\nu,{\varepsilon}}(t)\;; \\
& & \left|\int\frac{2}{\sqrt{a_{\varepsilon}}}\,b_1(t,x)\,\,\Re\left(\partial_xu_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx\right|
\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,C\,\int\sqrt[4]{a_{{\varepsilon}}}|\partial_x u_\nu|\,
\frac{1}{\sqrt[4]{a_{\varepsilon}}}\left|\partial_tu_\nu+
\frac{\partial_t\sqrt{a_{_{\varepsilon}}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right|\,dx \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,2C\,\int\sqrt{a_{\varepsilon}}|\partial_x u_\nu|^2\,+\,
\frac{1}{\sqrt{a_{\varepsilon}}}\left|\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu
\right|^2\,dx \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,2\,e_{\nu,{\varepsilon}}(t)\;; \\
& & \left|\int\frac{2}{\sqrt{a_{\varepsilon}}}\,c(t,x)\,\,\Re\left(u_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx\right|\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,C\,\int|u_\nu|\,\frac{1}{\sqrt[4]{a_{\varepsilon}}}
\left|\partial_tu_\nu+\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right|\,dx \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,2C\,e_{\nu,{\varepsilon}}(t)\;,\end{aligned}$$ where we have delt with $\|\partial_tu_\nu\|_{L^2}$ as before.

Now, we join the approximation parameter ${\varepsilon}$ with the dual variable $\xi$, setting $${\varepsilon}\,=\,2^{-\nu}\;;$$ so, from (\[(N)eq\_dt\_e\_nu,eps\]) and the previous inequalities, we obtain $$\begin{aligned}
& & \frac{d}{dt}e_{\nu,2^{-\nu}}(t)\,\leq\,{\widetilde}{C}\,(\nu+1)\,e_{\nu,2^{-\nu}} \label{(N)eq_dt_e_nu,eps_II}\\
& & \qquad\qquad\quad\quad\,+\,\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left((Lu)_\nu\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}
u_\nu\right)}\right)\,dx \nonumber \\
& & \qquad\qquad\quad\quad\,+\,\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left((\partial_x([{\varphi}_\nu(D_x),a]\partial_xu))\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}
u_\nu\right)}\right)\,dx \nonumber \\
& & \qquad\qquad\quad\quad\,-\,\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left(([{\varphi}_\nu(D_x),b_0]\partial_tu)\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}
u_\nu\right)}\right)\,dx \nonumber \\
& & \qquad\qquad\quad\quad\,-\,\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left(([{\varphi}_\nu(D_x),b_1]\partial_xu)\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}
u_\nu\right)}\right)\,dx \nonumber \\
& & \qquad\qquad\quad\quad\,-\,\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left(([{\varphi}_\nu(D_x),c]u)\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}
u_\nu\right)}\right)\,dx \nonumber\;,\end{aligned}$$ for a suitable constant ${\widetilde}{C}$, which depends only on $\lambda_0$, $\Lambda_0$, $C_0$ and on the norms of the coefficients of the operator $L$ in their respective functional spaces.

Estimates for the commutator terms
----------------------------------

Now, we have to deal with the commutator terms. As we will see, it’s useful to consider immediately the sum over $\nu\in{\mathbb}{N}$.

First, we report an elementary lemma (see also [@Colomb-DelSant_2]), which we will use very often in the next calculations.

\[(N)lemma\_1\] There exist two continuous, decreasing functions $\alpha_{1}$, $\alpha_2:\,]0,1[\,\rightarrow\,]0,+\infty[$ such that $\lim_{c\rightarrow0^+}\alpha_j(c)=+\infty$ for $j=1\,,\,2$ and such that, for all $\delta\in]0,1]$ and all $n\geq1$, the following inequalities hold: $$\sum_{j=1}^{n}\,e^{\delta j}\,j^{-1/2}\,\leq\,\alpha_1(\delta)\,e^{\delta n}\,n^{-1/2}\,,\qquad\qquad
\sum_{j=n}^{+\infty}\,e^{-\delta j}\,j^{1/2}\,\leq\,\alpha_2(\delta)\,e^{-\delta n}\,n^{1/2}\,.$$

Before going on, we take $\beta>0$ and $T\in]0,T_0]$ such that $$\beta T\,=\,\frac{\theta}{2}\,\log2\;.$$

\[oss:beta\_T\] Notice that, thanks to the hypothesis of Theorem \[(N)teor\_teorema1\], this condition implies that $$\beta T\,\leq\,\frac{\omega-\theta}{2}\;.$$ Moreover, for all $t\in[0,T]$, we have: $$\begin{aligned}
\beta t+\frac{\theta}{2}\log2 & \geq & \frac{\theta}{2}\log2\;>0\\
\beta t+\frac{\theta}{2}\log2 & \leq & \theta\log2\;\leq\;\frac{1}{2}\log2\;<1\\
(1-\theta)\log2-\beta t & \geq & \left(1-\frac{3}{2}\theta\right)\log2\;\geq\;\left(1-\frac{3}{4}\right)\log2\;>0\\
(1-\theta)\log2-\beta t & \leq & (1-\theta)\log2\;\leq\;\log2\;<1\,.\end{aligned}$$

Finally, we set (with the same notations used in the subsection \[sub-sec:L-P\_dec\]) $$\psi_{\mu}\,=\,{\varphi}_{\mu-1}+{\varphi}_{\mu}+{\varphi}_{\mu+1}\qquad({\varphi}_{-1}\equiv0)\;.$$ As $\psi_\mu\equiv1$ on the support of ${\varphi}_\mu$, we can write $$\partial_xu_\mu\,=\,{\varphi}_\mu(D_x)\partial_xu\,=\,\Psi_\mu({\varphi}_\mu(D_x)\partial_xu)\,=\,\Psi_\mu\partial_xu_\mu\;,$$ where $\Psi_\mu$ is the operator related to $\psi_\mu$. So, given a generic function $f(t,x)$, one has $$\label{(N)eq_nu-mu}
[{\varphi}_\nu(D_x),f]\partial_xu\,=\,[{\varphi}_\nu(D_x),f]\left(\sum_{\mu\geq0}\partial_xu_\mu\right)\,=\,
\sum_{\mu\geq0}([{\varphi}_\nu(D_x),f]\Psi_\mu)\partial_xu_\mu\;.$$

After these preliminary remarks, we now can go on with commutators’ estimates.

### Term with $[{\varphi}_\nu(D_x),a]$

Due to Bernstein’s inequalities, we have $$\left\|\partial_x\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}
u_\nu\right)\right\|_{L^2}\,\leq\,C\,2^\nu\,\left(e_{\nu,2^{-\nu}}(t)\right)^{1/2}\;.$$ So, using (\[(N)eq\_nu-mu\]) and the fact that $a_{\varepsilon}$ is real-valued, one has $$\begin{aligned}
& & \left|\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left(\partial_x([{\varphi}_\nu(D_x),a]\partial_xu)\,\cdot\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}
u_\nu\right)}\right)\,dx\right|\,\leq\, \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\leq\,C\,\sum_\mu\|([{\varphi}_\nu(D_x),a]\Psi_\mu)\partial_xu_\mu\|_{L^2}\,
2^\nu\,\left(e_{\nu,2^{-\nu}}(t)\right)^{1/2} \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\leq\,C\,\sum_\mu\|[{\varphi}_\nu(D_x),a]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\,\left(e_{\mu,2^{-\mu}}(t)\right)^{1/2}\,
2^\nu\,\left(e_{\nu,2^{-\nu}}(t)\right)^{1/2}\;,\end{aligned}$$ with the constant $C$ which depends only on $\lambda_0$, $\Lambda_0$ and $C_0$.\
Then, $$\begin{aligned}
& & \left|\sum_{\nu\geq0}e^{-2\beta(\nu+1)t}2^{-2\nu\theta}
\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left(\partial_x([{\varphi}_\nu(D_x),a]\partial_xu)
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}
u_\nu\right)}\right)dx\right|\,\leq \\
& & \;\quad\qquad\leq\,C\,\sum_{\nu,\mu}k_{\nu\mu}\,(\nu+1)^{1/2}e^{-\beta(\nu+1)t}2^{-\nu\theta}
\left(e_{\nu,2^{-\nu}}\right)^{1/2}\,(\mu+1)^{1/2}e^{-\beta(\mu+1)t}2^{-\mu\theta}\left(e_{\mu,2^{-\mu}}\right)^{1/2}\;,\end{aligned}$$ where we have set $$\label{(N)eq_kernel_comm[a]}
k_{\nu\mu}\,=\,e^{-(\nu-\mu)\beta t}\,2^{-(\nu-\mu)\theta}\,2^{\nu}\,(\nu+1)^{-1/2}\,(\mu+1)^{-1/2}\,
\|[{\varphi}_\nu(D_x),a]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\;.$$ Observe that, if $|\nu-\mu|\geq3$, then ${\varphi}_\nu\psi_\mu\equiv0$, so $[{\varphi}_\nu(D_x),a]\Psi_\mu={\varphi}_\nu(D_x)[a,\Psi_\mu]$. Therefore, from lemma \[lemma:comm\], in particular from (\[eq:comm\_Log-Lip\]), we deduce that $$\|[{\varphi}_{\nu}(D_x),a(t,x)]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\leq
\left\{
\begin{array}{ll}
C\,2^{-\nu}(\nu+1) & \mbox{ if } |\nu-\mu|\leq2\;, \\
 & \\
C\,2^{-max\{\nu,\mu\}}(\max\{\nu,\mu\}+1) & \mbox{ if } |\nu-\mu|\geq3\;,
\end{array}
\right.$$ where the constant $C$ depends only on $\Lambda_0$ and $C_0$.

Now our aim is to apply Schur’s lemma, so to estimate the quantity $$\label{(N)eq_lemma_schur}
\sup_{\mu}\sum_{\nu}|k_{\nu\mu}|\,+\,\sup_{\nu}\sum_{\mu}|k_{\nu\mu}|\;.$$ To do this, we will use lemma \[(N)lemma\_1\] and the inequalities stated in remark (\[oss:beta\_T\]).

1.  Fix $\mu\leq2$. $$\begin{aligned}
    \sum_{\nu\geq0}|k_{\nu\mu}| & \leq & C\,e^{(\mu+1)\beta t}\,2^{(\mu+1)\theta}\,(\mu+1)^{-\frac{1}{2}}
    \sum_\nu e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta}\,(\nu+1)^{\frac{1}{2}} \\
    & = & C\,e^{(\mu+1)\beta t}\,2^{(\mu+1)\theta}\,(\mu+1)^{-\frac{1}{2}}
    \sum_\nu e^{-(\nu+1)(\beta t+\theta\log2)}\,(\nu+1)^{\frac{1}{2}} \\
    & \leq & C\,e^{3\beta t}\,2^{3\theta}\,\alpha_2(\beta t+\theta\log2) \\
    & \leq & C\,2^{\frac{9}{2}\theta}\,\alpha_2(\theta\log2)\;.\end{aligned}$$

2.  Now, take $\mu\geq3$ and first consider $$\begin{aligned}
    \sum_{\nu=0}^{\mu-3}|k_{\nu\mu}| & \leq & C\,e^{(\mu+1)\beta t}\,2^{-(\mu+1)(1-\theta)}\,(\mu+1)^{\frac{1}{2}}
    \sum_{\nu=0}^{\mu-3} e^{-(\nu+1)\beta t}\,2^{(\nu+1)(1-\theta)}\,(\nu+1)^{-\frac{1}{2}} \\
    & \leq & C\,e^{(\mu+1)\beta t}\,2^{-(\mu+1)(1-\theta)}\,(\mu+1)^{\frac{1}{2}}
    \sum_{\nu=0}^{\mu-3}e^{(\nu+1)(-\beta t+(1-\theta)\log2)}\,(\nu+1)^{-\frac{1}{2}} \\
    & \leq & C\,e^{(\mu+1)\beta t}\,2^{-(\mu+1)(1-\theta)}\,(\mu+1)^{\frac{1}{2}}\,
    \alpha_1(-\beta t+(1-\theta)\log2)\,\cdot \\
    & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\cdot\,
    e^{(-\beta t+(1-\theta)\log2)(\mu-2)}\,(\mu-2)^{-\frac{1}{2}} \\
    & \leq & C\,2^{\frac{9}{2}\theta}\,\alpha_1\left(\left(1-\frac{3}{2}\theta\right)\log2\right)\;.\end{aligned}$$

    For the second part of the sum, one has $$\begin{aligned}
    \sum_{\nu=\mu-2}^{+\infty}|k_{\nu\mu}| & \leq & C\,e^{(\mu+1)\beta t}\,2^{(\mu+1)\theta}\,(\mu+1)^{-1/2}
    \sum_{\nu=\mu-2}^{+\infty}e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta}\,(\nu+1)^{1/2} \\
    & \leq & C\,e^{(\mu+1)\beta t}\,2^{(\mu+1)\theta}\,(\mu+1)^{-1/2}\alpha_2(\beta t+\theta\log2)\,\cdot \\
    & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdot\,e^{-(\beta t+\theta\log2)(\mu-1)}\,(\mu-1)^{1/2} \\
    & \leq & C\,2^{\frac{7}{2}\theta}\,\alpha_2(\theta\log2)\;.\end{aligned}$$

3.  Fix now $\nu\geq0$; we have $$\begin{aligned}
    \sum_{\mu=0}^{\nu+2}|k_{\nu\mu}| & \leq & C\,e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta}\,(\nu+1)^{1/2}
    \sum_{\mu=0}^{\nu+2}e^{(\mu+1)\beta t}\,2^{(\mu+1)\theta}\,(\mu+1)^{-1/2} \\
    & \leq & C\,e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta}\,(\nu+1)^{1/2}\,\alpha_1(\beta t+\theta\log2)\,\cdot \\
    & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdot\,e^{(\beta t+\theta\log2)(\nu+3)}\,(\nu+3)^{-1/2} \\
    & \leq & C\,2^{\frac{7}{2}\theta}\,\alpha_1(\theta\log2)\;.\end{aligned}$$ For the second part of the series, the following inequality holds: $$\begin{aligned}
    \sum_{\mu=\nu+3}^{+\infty}|k_{\nu\mu}| & \leq & C\,e^{-(\nu+1)\beta t}\,2^{(\nu+1)(1-\theta)}\,(\nu+1)^{-\frac{1}{2}}
    \sum_{\mu=\nu+3}^{+\infty}e^{(\mu+1)\beta t}\,2^{-(\mu+1)(1-\theta)}\,(\mu+1)^{\frac{1}{2}} \\
    & \leq & C\,e^{-(\nu+1)\beta t}\,2^{(\nu+1)(1-\theta)}\,(\nu+1)^{-\frac{1}{2}}\,
    \alpha_2(-\beta t+(1-\theta)\log2)\,\cdot \\
    & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\cdot\,
    e^{(-\beta t+(1-\theta)\log2)(\nu+4)}\,(\nu+4)^{\frac{1}{2}} \\
    & \leq & C\,2^{\frac{9}{2}\theta}\,\alpha_2\left(\left(1-\frac{3}{2}\theta\right)\log2\right)\;.\end{aligned}$$

In conclusion, there exists a positive function $\varPi$, with $\lim_{\theta\rightarrow0^+}\varPi(\theta)=+\infty$, such that $$\sup_{\mu}\sum_{\nu}|k_{\nu\mu}|\,+\,\sup_{\nu}\sum_{\mu}|k_{\nu\mu}|\,\leq\,C\,\varPi(\theta)\;,$$ and so $$\begin{aligned}
& & \hspace{-1cm}\left|\sum_{\nu\geq0}e^{-2\beta(\nu+1)t}\,2^{-2\nu\theta}
\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left(\partial_x([{\varphi}_\nu(D_x),a]\partial_xu)
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}
u_\nu\right)}\right)dx\right|\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad\qquad \leq\,C\,\varPi(\theta)\,
\sum_{\nu=0}^{+\infty}(\nu+1)\,e^{-2\beta(\nu+1)t}\,2^{-2\nu\theta}\,e_{\nu,2^{-\nu}}(t)\;.\end{aligned}$$

### Terms with $[{\varphi}_\nu(D_x),b_0]$ and $[{\varphi}_\nu(D_x),b_1]$

Now, let us consider $$\begin{aligned}
& & \left|\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left([{\varphi}_{\nu}(D_x),b_0(t,x)]\partial_t u\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}
u_\nu\right)}\right)\,dx\right|\,\leq \\
& & \quad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,2\,\|[{\varphi}_{\nu}(D_x),b_0(t,x)]\partial_t u\|_{L^2}\,
\left\|\frac{1}{\sqrt{a_{2^{-\nu}}}}\left|\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}
u_\nu\right|\;\right\|_{L^2} \\
& & \quad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,2\,\|[{\varphi}_{\nu}(D_x),b_0(t,x)]\partial_t u\|_{L^2}\,(e_{\nu,2^{-\nu}}(t))^{1/2}\;.\end{aligned}$$

Thanks to relation (\[(N)eq\_nu-mu\]), we have $$\begin{aligned}
\|[{\varphi}_{\nu}(D_x),b_0(t,x)]\partial_t u\|_{L^2} & = & 
\left\|[{\varphi}_{\nu}(D_x),b_0(t,x)]\sum_{\mu\geq0}\Psi_\mu\partial_tu_\mu\right\|_{L^2} \\
 & \leq & \sum_{\mu\geq0}\|[{\varphi}_{\nu}(D_x),b_0(t,x)]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\|\partial_tu_\mu\|_{L^2}\;.\end{aligned}$$ As we have done before, we have, for constants depending only on $\lambda_0$, $\Lambda_0$ and $C_0$, $$\begin{aligned}
\|\partial_tu_\mu\|_{L^2}\,=\,
\left(\int\frac{\sqrt{a_{2^{-\mu}}}}{\sqrt{a_{2^{-\mu}}}}|\partial_tu_\mu|^{2}dx\right)^{1/2} & \leq & \left(C\,e_{\mu,2^{-\mu}}(t)+
C'\,\left\|\frac{\partial_t\sqrt{a_{2^{-\mu}}}}{2\sqrt{a_{2^{-\mu}}}}u_\mu\right\|_{L^2}\right)^{1/2} \\
& \leq & \left(C e_{\mu,2^{-\mu}}(t)\,+\,C'\frac{1}{2^{-\mu}}2^{-2\mu}\|\partial_x u_\mu\|_{L^2}\right)^{1/2} \\
& \leq & (C+ C'\,2^{-\mu})^{1/2}\left(e_{\mu,2^{-\mu}}\right)^{1/2} \\
& \leq & C\,\left(e_{\mu,2^{-\mu}}\right)^{1/2}\;.\end{aligned}$$

Due to lemma \[lemma:comm\], we obtain $$\|[{\varphi}_{\nu}(D_x),b_0(t,x)]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\leq
\left\{
\begin{array}{ll}
C\,2^{-\nu\omega} & \quad\mbox{ if } |\nu-\mu|\leq2 \\
 & \\
C\,2^{-max\{\mu,\nu\}\omega} & \quad\mbox{ if } |\nu-\mu|\geq3
\end{array}
\right.$$ where $C$ is a constant depending only on $\|b_0\|_{L^\infty({\mathbb}{R}_t;\,{\mathcal}{C}^{\omega}({\mathbb}{R}_x))}$.\
As a matter of fact, the kernel of operator $[{\varphi}_\nu(D_x),b_0]$ is $$n(x,y)\,=\,{\widehat}{{\varphi}}(2^\nu(x-y))\,2^{\nu}\,(b_0(t,x)-b_0(t,y))\;;$$ so, to evaluate its norm we apply Schur’s lemma and, thanks to the fact that $b_0$ is $\omega$-hölder, we get the desired estimate.

Therefore, $$\begin{aligned}
& & \left|\sum_{\nu\geq0}e^{-2\beta t(\nu+1)}2^{-2\nu\theta} \int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left([\Delta_{\nu},b_0]\partial_tu\,
\overline{\left(\partial_t u_\nu+
\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}u_\nu\right)}\right)dx\right|\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\leq\,\sum_{\nu,\mu\geq0}e^{-\beta t(\nu+1)}\,2^{-\nu\theta}\,\left(e_{\nu,2^{-\nu}}\right)^{1/2}\,
e^{-\beta t(\mu+1)}\,2^{-\mu\theta}\,\left(e_{\mu,2^{-\mu}}\right)^{1/2}\,l_{\nu\mu}\;,\end{aligned}$$ where we have defined $$\label{(N)eq_kernel_comm[b_0]}
l_{\nu\mu}\,:=\,e^{-(\nu-\mu)\beta t}\,2^{-(\nu-\mu)\theta}\,\|[{\varphi}_{\nu}(D_x),b_0(t,x)]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\;.$$

As made before, we are going to estimate $l_{\nu\mu}$ applying Schur’s lemma.

1.  Let us fix $\mu\leq2$. Then $$\begin{aligned}
    \sum_{\nu\geq0}|l_{\nu\mu}| & \leq & C\,e^{(\mu+1)\beta t}\,2^{(\mu+1)\theta}\,
    \sum_{\nu\geq0}e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta}\,2^{-\nu\omega} \\
    & \leq & C\,e^{3\beta t}\,2^{3\theta}\,\sum_{\nu\geq0}e^{-(\nu+1)(\beta t+\theta\log2)}\,(\nu+1)^{1/2} \\
    & \leq & C\,e^{3\beta t}\,2^{3\theta}\,\alpha_2(\beta t+\theta\log2) \\
    & \leq & C\,2^{\frac{9}{2}\theta}\,\alpha_2(\theta\log2)\,.\end{aligned}$$

2.  Now, take $\mu\geq3$ and consider first $$\begin{aligned}
    \sum_{\nu=0}^{\mu-3}|l_{\nu\mu}| & \leq & C\,e^{\mu\beta t}\,2^{\mu\theta}\,2^{-\mu\omega}
    \sum_{\nu=0}^{\mu-3} e^{-\nu\beta t}\,2^{-\nu\theta} \\
    & \leq & C\,e^{\mu(\beta t-(\omega-\theta)\log2)}\,(\mu-2) \\
    & \leq & C\,e^{-\mu(\omega-\frac{\theta}{2})\log2}\,(\mu-2) \\
    & \leq & C\,M(\omega,\theta)\;,\end{aligned}$$ where $M(\omega,\theta)$ is the maximum of the function $z\mapsto e^{-\gamma z}\,(z-2)$, with $\gamma=\left(\omega-\frac{\theta}{2}\right)\log2$.\
    For the second part of the sum, we have $$\begin{aligned}
    \sum_{\nu=\mu-2}^{+\infty}|l_{\nu\mu}| & \leq & C\,e^{(\mu+1)\beta t}\,2^{(\mu+1)\theta}
    \sum_{\nu=\mu-2}^{+\infty}e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta}\,2^{-\nu\omega}
    \frac{(\nu+1)^{1/2}}{(\nu+1)^{1/2}} \\
    & \leq & C\,e^{(\mu+1)\beta t}\,2^{(\mu+1)\theta}\,(\mu-1)^{-1/2}\,\alpha_2(\beta t+\theta\log2)\,\cdot \\
    & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdot\,e^{-(\mu-1)\beta t}\,2^{-(\mu-1)\theta}\,(\mu-1)^{1/2} \\
    & \leq &  C\,e^{2(\beta t+\theta\log2)}\,\alpha_2(\theta\log2) \\
    & \leq & C\,2^{\frac{9}{2}\theta}\,\alpha_2(\theta\log2)\,.\end{aligned}$$

3.  Fix now $\nu$. Initially, we have $$\begin{aligned}
    \sum_{\mu=0}^{\nu+2}|l_{\nu\mu}| & \leq & C\,e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta}\,2^{-\nu\omega}
    \sum_{\mu=0}^{\nu+2}e^{(\mu+1)(\beta t+\theta\log2)}\,\frac{(\mu+1)^{1/2}}{(\mu+1)^{1/2}} \\
    & \leq & C\,e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta}\,2^{-\nu\omega}\,(\nu+3)^{1/2}\,\alpha_1(\beta t+\theta\log2)\,\cdot \\
    & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdot\,e^{(\nu+3)\beta t}\,2^{(\nu+3)\theta}\,(\nu+3)^{-1/2} \\
    & \leq & C\,e^{2(\beta t+\theta\log2)}\,\alpha_1(\theta\log2) \\
    & \leq & C\,2^{3\theta}\,\alpha_1(\theta\log2)\;.\end{aligned}$$ Moreover, $$\begin{aligned}
    \sum_{\mu=\nu+3}^{+\infty}|l_{\nu\mu}| & \leq & C\,e^{-\nu\beta t}\,2^{-\nu\theta}
    \sum_{\mu=\nu+3}^{+\infty}e^{-\mu(-\beta t+(\omega-\theta)\log2)}\,\frac{\mu^{1/2}}{\mu^{1/2}} \\
    & \leq & C\,e^{-\nu\beta t}\,2^{-\nu\theta}\,(\nu+3)^{-1/2}\,\alpha_2((\omega-\theta)\log2-\beta t)\,\cdot \\
    & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdot\,e^{(\nu+3)\beta t}\,2^{-(\nu+3)(\omega-\theta)}\,(\nu+3)^{1/2}\\
    & \leq & C\,e^{3\beta t}\,2^{3\theta}\,2^{-(\nu+3)\omega}\,\alpha_2\left(\left(\omega-\frac{3}{2}\theta\right)\log2\right) \\
    & \leq & C\,2^{\frac{9}{2}\theta}\,\alpha_2\left(\left(\omega-\frac{3}{2}\theta\right)\log2\right)\,.\end{aligned}$$

From all these inequalities, thanks to Schur’s lemma, one has that there exists a constant ${\widetilde}{M}(\omega,\theta)$, depending only on $\|b_0\|_{L^\infty({\mathbb}{R}_t;\,{\mathcal}{C}^{\omega}({\mathbb}{R}_x))}$ and on the parameter $\theta$ (which we have fixed at the beginning of the calculations), such that $$\sup_{\mu}\sum_{\nu}|l_{\nu\mu}|\,+\,\sup_{\nu}\sum_{\mu}|l_{\nu\mu}|\,\leq\,C\,{\widetilde}{M}(\omega,\theta)\;;$$ from this relation, finally we get $$\begin{aligned}
& & \hspace{-0.5cm}\left|\sum_{\nu\geq0}e^{-2\beta t(\nu+1)}2^{-2\nu\theta} \int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,
\Re\left([{\varphi}_{\nu}(D_x),b_0]\partial_t u\,
\overline{\left(\partial_t u_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}u_\nu\right)}\right)dx\right|\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,C\,{\widetilde}{M}(\omega,\theta)
\sum_{\nu=0}^{+\infty}(\nu+1)\,e^{-2\beta(\nu+1)t}\,2^{-2\nu\theta}\,e_{\nu,2^{-\nu}}(t)\;.\end{aligned}$$

With regard to the term with the commutator $[{\varphi}_\nu(D_x),b_1(t,x)]$, notice that, as we made before, one has, for constants depending only on $\lambda_0$, $\Lambda_0$ and $C_0$, $$\begin{aligned}
& & \left|\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left([{\varphi}_{\nu}(D_x),b_1]\partial_xu\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}u_\nu\right)}
\right)\,dx\right|\,\leq \\
& & \qquad\qquad\leq\,C\,\sum_{\mu\geq0}\|[{\varphi}_\nu(D_x),b_1]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\,
\|\sqrt[4]{a_{2^{-\nu}}}\partial_xu_\mu\|_{L^2}\,\left\|\frac{1}{\sqrt[4]{a_{2^{-\nu}}}}
\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}u_\nu\right)\right\|_{L^2} \\
& & \qquad\qquad\leq\,C\,\sum_{\mu\geq0}\|[{\varphi}_\nu(D_x),b_1]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\,\left(e_{\nu,2^{-\nu}}\right)^{1/2}\left(e_{\mu,2^{-\mu}}\right)^{1/2}\;.\end{aligned}$$

Therefore, $$\begin{aligned}
& & \left|\sum_{\nu\geq0}e^{-2\beta t(\nu+1)}\,2^{-2\nu\theta}\,\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left([{\varphi}_{\nu}(D_x),b_1]\partial_xu\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}u_\nu\right)}
\right)dx\right|\,\leq\\
& & \qquad\qquad\qquad\qquad\qquad\leq\,C\,\sum_{\nu,\mu\geq0}e^{-\beta(\nu+1)t}\,2^{-\nu\theta}\,\left(e_{\nu,2^{-\nu}}\right)^{1/2}\,
e^{-\beta(\mu+1)t}\,2^{-\mu\theta}\,\left(e_{\mu,2^{-\mu}}\right)^{1/2}\,h_{\nu\mu}\;,\end{aligned}$$ where we have set again $$h_{\nu\mu}\,=\,e^{-(\nu-\mu)\beta t}\,2^{-(\nu-\mu)\theta}\,\|[{\varphi}_\nu(D_x),b_1]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\;.$$ As $b_0$ and $b_1$ satisfy to the same hypothesis, the commutator $[{\varphi}_\nu(D_x),b_1]$ verifies the same inequalities as $[{\varphi}_\nu(D_x),b_0]$; so, if we repeat the same calculations, we obtain $$\begin{aligned}
& & \left|\sum_{\nu\geq0}e^{-2\beta t(\nu+1)}\,2^{-2\nu\theta}\,\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left([{\varphi}_{\nu}(D_x),b_1]\partial_xu\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}u_\nu\right)}
\right)dx\right|\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,C\,{\widetilde}{M}(\omega,\theta)
\sum_{\nu=0}^{+\infty}(\nu+1)\,e^{-2\beta(\nu+1)t}\,2^{-2\nu\theta}\,e_{\nu,2^{-\nu}}(t)\;.\end{aligned}$$

### Term with $[{\varphi}_\nu(D_x),c]$

Finally, we have to deal with the commutator $[{\varphi}_\nu(D_x),c(t,x)]$.

First, observe that there exist constants, depending only on $\lambda_0$, $\Lambda_0$ and $C_0$ as usual, such that one has $$\begin{aligned}
& & \left|\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left([{\varphi}_{\nu}(D_x),c]\partial_xu\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}u_\nu\right)}
\right)\,dx\right|\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\leq\,C\,\|[{\varphi}_{\nu}(D_x),c]u\|_{L^2}\,\left\|\frac{1}{\sqrt[4]{a_{2^{-\nu}}}}\left(\partial_tu_\nu+
\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}u_\nu\right)\right\|_{L^2} \\
& & \qquad\qquad\qquad\qquad\qquad\leq\,C\,\sum_{\mu\geq0}\|[{\varphi}_\nu(D_x),c]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\,\|u_\mu\|_{L^2}\,
\left(e_{\nu,2^{-\nu}}(t)\right)^{1/2}\\
& & \qquad\qquad\qquad\qquad\qquad\leq\,2\,C\,\sum_{\mu\geq0}\|[{\varphi}_\nu(D_x),c]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\,
2^{-\mu}\,\|D_xu_\mu\|_{L^2}\,\left(e_{\nu,2^{-\nu}}(t)\right)^{1/2}\\
& & \qquad\qquad\qquad\qquad\qquad\leq\,2\,C\,\sum_{\mu\geq0}\|[{\varphi}_\nu(D_x),c]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\,2^{-\mu}\,
\left(e_{\mu,2^{-\mu}}(t)\right)^{1/2}\,\left(e_{\nu,2^{-\nu}}(t)\right)^{1/2}\;.\end{aligned}$$

Thereby, we get the estimate $$\begin{aligned}
& & \left|\sum_{\nu\geq0}e^{-2\beta t(\nu+1)}\,2^{-2\nu\theta}\,\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left([{\varphi}_{\nu}(D_x),c]\partial_xu
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}u_\nu\right)}
\right)dx\right|\,\leq \\
& & \qquad\qquad\qquad\leq\,2\,C\sum_{\nu,\mu\geq0}e^{-\beta t(\nu+1)}\,2^{-\nu\theta}\,\left(e_{\nu,2^{-\nu}}(t)\right)^{1/2}\,
e^{-\beta t(\mu+1)}\,2^{-\mu\theta}\,\left(e_{\mu,2^{-\mu}}(t)\right)^{1/2}\,m_{\nu\mu}\;,\end{aligned}$$ where we have defined $$m_{\nu\mu}\,=\,e^{-(\nu-\mu)\beta t}\,2^{-(\nu-\mu)\theta}\,2^{-\mu}\,\|[{\varphi}_\nu(D_x),c]\Psi_\mu\|_{{\mathcal}{L}(L^2)}\;.$$

The kernel of the operator $[{\varphi}_\nu(D_x),c]$ is $$n'(x,y)\,=\,{\widehat}{\psi}(2^\nu(x-y))\,2^{\nu}\,(c(t,y)-c(t,x))\,;$$ so, remembering that $c$ is bounded over ${\mathbb}{R}\times{\mathbb}{R}$, from Schur’s lemma one gets $$\|[{\varphi}_\nu(D_x),c]\|_{{\mathcal}{L}(L^2)}\,\leq\,C\qquad\qquad\forall\,\nu\geq0\,,$$ where the constant $C$ depends only on $\|c\|_{L^\infty({\mathbb}{R}_t\times{\mathbb}{R}_x)}$.

Again, we are going to estimate the kernel $m_{\nu\mu}$ to apply Schur’s lemma.

1.  First, we take $\mu\leq2$ and we have $$\begin{aligned}
    \sum_{\nu\geq0}|m_{\nu\mu}| & \leq &  C\,e^{(\mu+1)\beta t}\,2^{(\mu+1)\theta}\,2^{-\mu}
    \sum_{\nu\geq0}e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta} \\
    & \leq & C\,e^{3\beta t}\,2^{3\theta}\,\sum_{\nu\geq0}e^{-(\nu+1)(\beta t+\theta\log2)}\,(\nu+1)^{1/2} \\
    & \leq & C\,e^{3\beta t}\,2^{3\theta}\,\alpha_2(\beta t+\theta\log2) \\
    & \leq & C\,2^{\frac{9}{2}\theta}\,\alpha_2(\theta\log2)\;.\end{aligned}$$

2.  Now, we fix $\mu\geq3$ and we consider the first part of the series: $$\begin{aligned}
    \sum_{\nu=0}^{\mu-3}|m_{\nu\mu}| & \leq & C\,e^{(\mu+1)\beta t}\,2^{-(\mu+1)(1-\theta)}\,2^{-\mu}\,
    \sum_{\nu=0}^{\mu-3} e^{-(\nu+1)\beta t}\,2^{(\nu+1)(1-\theta)}\,2^{\mu-\nu}\frac{(\nu+1)^{1/2}}{(\nu+1)^{1/2}} \\
    & \leq & C\,e^{(\mu+1)\beta t}\,2^{-(\mu+1)(1-\theta)}\,(\mu-2)^{1/2}\,
    \sum_{\nu=0}^{\mu-3} e^{(\nu+1)((1-\theta)\log2-\beta t)}\,(\nu+1)^{-1/2} \\
    & \leq & C\,e^{(\mu+1)\beta t}\,2^{-(\mu+1)(1-\theta)}\,(\mu-2)^{1/2}\,
    \alpha_1((1-\theta)\log2-\beta t)\,\cdot \\
    & & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdot\,e^{-(\mu-2)\beta t}\,2^{(\mu-2)(1-\theta)}\,(\mu-2)^{-1/2} \\
    & \leq & C\,e^{3\beta t}\,2^{3\theta}\,\alpha_1\left(\left(1-\frac{3}{2}\theta\right)\log2\right) \\
    & \leq & C\,2^{\frac{9}{2}\theta}\,\alpha_1\left(\left(1-\frac{3}{2}\theta\right)\log2\right)\;.\end{aligned}$$ For the second part, one has: $$\begin{aligned}
    \sum_{\nu=\mu-2}^{+\infty}|m_{\nu\mu}| & \leq &  C\,e^{(\mu+1)\beta t}\,2^{(\mu+1)\theta}\,2^{-\mu}
    \sum_{\nu=\mu-2}^{+\infty}e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta}\frac{(\nu+1)^{1/2}}{(\nu+1)^{1/2}} \\
    & \leq & C\,e^{(\mu+1)\beta t}\,2^{(\mu+1)\theta}\,2^{-\mu}\,\alpha_2(\beta t+\theta\log2)\,
    e^{-(\mu-1)\beta t}\,2^{-(\mu-1)\theta} \\
    & \leq & C\,2^{3\theta}\,\alpha_2(\theta\log2)\;.\end{aligned}$$

3.  Now, we fix $\nu\geq0$. Initially, we consider $$\begin{aligned}
    \sum_{\mu=0}^{\nu+2}|m_{\nu\mu}| & \leq & C\,e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta}\,
    \sum_{\mu=0}^{\nu+2}e^{(\mu+1)(\beta t+\theta\log2)}\,2^{-\mu}\,\frac{(\mu+1)^{1/2}}{(\mu+1)^{1/2}} \\
    & \leq & C\,e^{-(\nu+1)\beta t}\,2^{-(\nu+1)\theta}\,(\nu+3)^{1/2}\,\alpha_1(\beta t+\theta\log2)\,\cdot \\
    & & \qquad\qquad\qquad\qquad\qquad\qquad\cdot\,e^{(\nu+3)\beta t}\,2^{(\nu+3)\theta}\,(\nu+3)^{-1/2} \\
    & \leq & C\,e^{2(\beta t+\theta\log2)}\,\alpha_1(\theta\log2) \\
    & \leq & C\,2^{3\theta}\,\alpha_1(\theta\log2)\;.\end{aligned}$$ The second part of the series, instead, can be treated as follow: $$\begin{aligned}
    \sum_{\mu=\nu+3}^{+\infty}|m_{\nu\mu}| & \leq & C\,e^{-(\nu+1)\beta t}\,2^{(\nu+1)(1-\theta)}\,
    \sum_{\mu=\nu+3}^{+\infty}e^{(\mu+1)\beta t}\,2^{-\mu}\,2^{-(\mu+1)(1-\theta)}\,2^{\mu-\nu}\,
    \frac{(\mu+1)^{\frac{1}{2}}}{(\mu+1)^{\frac{1}{2}}} \\
     & \leq & C\,e^{-(\nu+1)\beta t}\,2^{(\nu+1)(1-\theta)}\,(\nu+4)^{-\frac{1}{2}}\,
    \sum_{\mu=\nu+3}^{+\infty}e^{-(\mu+1)((1-\theta)\log2-\beta t)}\,(\mu+1)^{\frac{1}{2}} \\
    & \leq & C\,e^{-(\nu+1)\beta t}\,2^{(\nu+1)(1-\theta)}\,\alpha_2((1-\theta)\log2-\beta t)\,e^{(\nu+4)\beta t}\,2^{-(\nu+4)(1-\theta)} \\
    & \leq & C\,2^{\frac{9}{2}\theta}\,\alpha_2\left(\left(1-\frac{3}{2}\theta\right)\log2\right)\;.\end{aligned}$$

Finally, we obtain: $$\begin{aligned}
& & \left|\sum_{\nu\geq0}e^{-2\beta t(\nu+1)}\,2^{-2\nu\theta}\,\int\frac{2}{\sqrt{a_{2^{-\nu}}}}\,\,\Re\left([{\varphi}_{\nu}(D_x),c]\partial_xu\,
\overline{\left(\partial_tu_\nu+\frac{\partial_t\sqrt{a_{2^{-\nu}}}}{2\sqrt{a_{2^{-\nu}}}}u_\nu\right)}
\right)dx\right|\,\leq \\
& & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq\,C\,\varPi(\theta)\,
\sum_{\nu=0}^{+\infty}(\nu+1)\,e^{-2\beta(\nu+1)t}\,2^{-2\nu\theta}\,e_{\nu,2^{-\nu}}(t)\;,\end{aligned}$$ where the function $\varPi$ is the same used in the estimate of the term $[{\varphi}_\nu(D_x),a]$.

End of the proof of theorem \[(N)teor\_teorema1\]
-------------------------------------------------

Now we can complete the proof of theorem \[(N)teor\_teorema1\].

First, remembering the definition of total energy given by (\[(N)eq\_E\_tot\]), we have that there exists a constant $C>0$, depending only on $\theta$, such that $$\left|\sum_{\nu=0}^{+\infty}e^{-\beta(\nu+1)t}\,2^{-2\nu\theta}\,\int\frac{2}{\sqrt{a_{\varepsilon}}}\,\,\Re
\left((Lu)_\nu\,\cdot\,\overline{\left(\partial_tu_\nu+
\frac{\partial_t\sqrt{a_{\varepsilon}}}{2\sqrt{a_{\varepsilon}}}u_\nu\right)}\right)\,dx\right|\,\leq\,C\,(E(t))^{1/2}\,\|Lu\|_{H^{-\theta-\beta^*t}}\,.$$ Therefore, if we set ${\widetilde}{\varPi}(\omega,\theta)=\max\{{\widetilde}{M}(\omega,\theta),\varPi(\theta)\}$, from relation (\[(N)eq\_dt\_e\_nu,eps\_II\]) and from the estimates proved in the previous subsection, we have that, for suitable constants, depending only on $\lambda_0$, $\Lambda_0$, $C_0$ and on the norms of the coefficients of operator $L$ in their respective functional spaces, the following inequality holds: $$\begin{aligned}
\frac{d}{dt}E(t) & \leq & \left(C+C'\,{\widetilde}{\varPi}(\omega,\theta)-2\beta\right)\,
\sum_{\nu=0}^{+\infty}\,(\nu+1)\,e^{-2\beta(\nu+1)t}\,2^{-2\nu\theta}\,e_{\nu,2^{-\nu}}(t) \\
 & + & C''\,(E(t))^{1/2}\,\|Lu\|_{H^{-\theta-\beta^*t}}\,.\end{aligned}$$ Now, let us fix $\beta$ large enough, such that $C+C'\,{\widetilde}{\varPi}(\omega,\theta)-2\beta\leq0$: we can always do this, on condition that we take $T$ small enough. With this choice, we have $$\frac{d}{dt}E(t)\,\leq\,C''\,(E(t))^{1/2}\,\|Lu\|_{H^{-\theta-\beta^*t}}\;;$$ now, the thesis of the theorem follows from Gronwall’s lemma and remark \[remark:E(0)-E(t)\].

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