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\begin{document}
\begin{center}{\Large Law of large numbers for random walk with unbounded jumps and BDP with bounded jumps in random environment}\footnote{Supported by National
Nature Science Foundation of China (Grant No. 11226199) and Nature Science Foundation of Anhui Educational Committee (Grant No. KJ2014A085)}\end{center}

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\centerline {Hua-Ming \uppercase{Wang}$^a$ }

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      {\footnotesize$^a$Department of Mathematics, Anhui Normal University, Wuhu 241003, China

    E-mail\,$:$ hmking@mail.ahnu.edu.cn}
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\begin{center}\textbf{Abstract}\quad \end{center}

We study random walk with unbounded jumps in random environment. The environment is stationary and ergodic, uniformly elliptic and decays polynomially with speed $Dj^{-(3+\varepsilon_0)}$ for some small $\varepsilon_0>0$ and proper $D>0.$ We prove a law of large number with positive velocity under the condition that the annealed mean of the hitting time of the positive half lattice is finite. Secondly, we consider birth and death process with bounded jumps in stationary and ergodic environment. Under the uniformly elliptic condition, we prove a law of large number and give the explicit formula of its velocity.

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\textbf{Keywords:}\ random walk; random environment; unbounded jumps; birth and death process; skeleton process; environment viewed from particle

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\textbf{MSC 2010:}\
 60K37;  60J80
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\section{Introduction}

Our prime concern is to prove the Law of Large Number (LLN hereafter) for Birth and Death Process in Random Environment (BDPRE hereafter) with bounded jumps , say $\{N_t\}_{t\ge0}.$ We assume that at each discontinuity, the particle jumps at most a distance $R$ to the right or at most a distance $L$ to the left. For the nearest neighbour setting $L=R=1,$ if one defines $T_n$ the hitting time of $n,$ then $N_{T_n}=n$ and $\{T_n-T_{n-1}\}_{n\ge1}$ forms a stationary and mixing sequence under the annealed probability whenever the environment is i.i.d.. Therefore the LLN of $\{N_t\}_{t\ge0}$ follows from that of $\{T_n\}_{n\ge0}.$ For details, see Ritter \cite{rit}. For bounded-jump setting, the above approach does not work. Similarly, define the ladder times $T_0=0$ and $T_n=\inf\{t>T_{n-1}:N_{T_n}>N_{T_{n-1}}\},\ n\ge1.$ Two problems arise: (1) It is hard to tell the exact value of $N_{T_n},$ though we know that $n\le N_{T_n}\le nR;$ (2) $\{T_n-T_{n-1}\}_{n\ge1}$ is not a stationary sequence under the annealed probability even if the environment is i.i.d..

We turn to consider the $h$-skeleton process $\{N_{nh}\}_{n\ge0},\ h>0$ of $\{N_t\}_{t\ge0}.$ If the LLN of the $h$-skeleton process is proved, then the LLN of $\{N_t\}_{t\ge0}$ follows from some standard procedures. However, $\{N_{nh}\}_{n\ge0}$ is indeed a discrete time {\it random walk in random environment with unbounded  jumps,} since theoretically speaking, $\{N_t\}_{t\ge0}$ may have many jumps in a time interval of length $h.$

Although the developments of Random Walk in Random Environment (RWRE hereafter) with bounded jumps  were almost satisfying, RWRE with unbounded jumps was very seldom considered.  Andjel \cite{adj} proved the 0-1 law. Comets and Popov \cite{cp} proved  an LLN. In \cite{cp}, two main conditions were required essentially: a) the jumping probabilities of the walk have an exponential tail; b)   `` the random
walk is `uniformly' transient to the right (i.e., there are no `traps')".
We mention also that in Gallesco and Popov \cite{gpa,gpb}, the author studied the central limit theorem of random walks with unbounded jumps among random conductances.

The BDPRE with bounded jumps considered in this paper is the continuous time analogue of RWRE with bounded jumps.
The situation of RWRE with bounded jumps on $\mathbb Z$ is almost satisfying. We review here only the known results closely related to the transient RWRE. For the nearest neighbour setting, the 0-1 law and the LLN with explicit velocity in the case of i.i.d. environment were studied in Solomon \cite{sol}; the extension to stationary and ergodic environment
was given in Alili \cite{ali}. The RWRE with bounded jumps (jumps are non-nearest neighbor) was introduced in Key \cite{key}, where the middle Lyapunov exponents of a sequence of positive random matrices were used to give the recurrence criteria.
Letchikov \cite{letc} simplified the recurrence criteria and proved a $\log^2n$ law for the recurrent case as Sinai's walk.
The
regime where the LLN holds with a positive velocity was characterized
in Br\'emont \cite{brb,brc}  by ``the environment viewed from particle",
but no explicit formula is  available any longer. In \cite{hwa,hwb,hz}, the authors set up the branching structure for RWRE with bounded jumps and gave  the explicit formulae for the velocities of LLNs. At last we point out that Bolthausen and Goldsheid \cite{bg} studied the recurrence and transience of RWRE on a strip, which is a generalization of RWRE with bounded jumps.
The LLN for RWRE on a strip could be found in Roitershtein \cite{roi}.

In this paper, firstly we  prove an LLN for RWRE with unbounded jumps. In our setup, we only need to require that the jumping probabilities have an polynomial tail. Moreover  we do not need the condition `` the random walk is `uniformly' transient to the right" used in \cite{cp}. Our approach is based on ``the environment viewed from particle" which dates back to Kozlov \cite{K85}. But in Kozlov \cite{K85} and some later literatures, this approach is only used to treat the bounded-jump setting. We use the large deviation for martingales (with unbounded martingale differences) derived in Lesigne and Voln\'y \cite{lv} to estimate the tail of a martingale related to RWRE with unbounded jumps. By constructing an invariant measure for the auxiliary Markov chain of the environment viewed from particle, we could show the LLN for the summation of the local drift of RWRE with unbounded jumps.  In this way, we  prove the LLN for  RWRE with unbounded jumps.
Secondly, using the LLN of RWRE with unbounded jumps, we prove the LLN of BDPRE with bounded jumps. We assume a uniform ellipticity  condition on the environment, which is crucial for our development. Under such  ellipticity condition, we could show that the tail probability of the jumps for the $h$-skeleton process is exponentially bounded. Therefore we could use the LLN for RWRE with unbounded jumps to get the LLN for the $h$-skeleton process. Then the LLN for BDPRE with bounded jumps follows by some standard procedures.

The paper is organized as follows. In Section \ref{mr}, we define strictly the models  and state the main results. The LLN for RWRE with unbounded jumps is proved in Section \ref{pr}  while
the  LLNs for the  $h$-skeleton process $\{N_{nh}\}_{n\ge 0}$ and $\{N_t\}$ are proved in Section \ref{pec}.
We devote Section \ref{dvp} to give the explicit formula of the velocity $v_{\tilde\mbp}$ for the case $L=R=2$ by using the branching structure in the embedded process constructed in \cite{hwb}. An appendix section is also given at the end of the paper to discuss the existence of the process $\{N_t\}.$

\section{The models and results}\label{mr}

\subsection{RWRE with unbounded jumps}
Let $\Omega$ be the collection of $\omega=(\omega_{x})_{x\in\mathbb Z}$ where for $x\in \mathbb Z,$ $\omega_x=(\omega_{xy})_{y\in\mathbb Z}$ is a probability measure on $\mathbb Z,$ that is, $\omega_{xy}\ge 0$ for all $y\in\mathbb Z$ and $\sum_{y\in\mathbb Z}\omega_{xy}=1.$   Let $\theta$ be the shift operator on $\Omega$ defined by $(\theta\omega)_x:=\omega_{x+1}.$ Equip $\Omega$ with Borel $\sigma$-algebra $\mathcal F$ and let $\mbp$ be a probability measure on $(\Omega,\mathcal F).$ For a typical realization of $\omega,$ we consider a Markov chain $\{S_n\}_{n\ge 0}$ with transitional probabilities $$P_{\omega}^{x_0}(S_{n+1}=x+y\big |S_{n}=x)=\omega_{xy}\text{ for all }n\ge0,\ P_\omega^{x_0}(S_0=x_0)=1,$$
so that $P_\omega^{x_0}$ is the quenched law of the Markov chain starting from $x_0$ in the
environment $\omega.$
Define a new probability measure $P^{x_0}$ by $$P^{x_0}(\cdot)=\int P^{x_0}_\omega(\cdot)\mbp(d\omega),$$ which is called the annealed probability. We use $E_\omega^{x_0},$ $E^{x_0}$ and $\mbe$ to denote the expectation operator for $P_\omega^{x_0},$ $P^{x_0}$ and $\mbp$ respectively. The superscript $x_0$ will be omitted whenever it is $0.$

\noindent{\bf Condition B}

\noindent\textbf{(B1)} $(\Omega,\mathcal F, \mathbb P, \theta)$ forms a stationary and ergodic system.

\noindent\textbf{(B2)} There exists $\varepsilon>0$ such that $\mbp(\omega_{01}>\varepsilon)=1.$

\noindent\textbf{(B3)} There exist small $\varepsilon_0>0$ and proper $D>0,$ such that $\mbp$-a.s., $$\omega_{0j}<D|j|^{-(3+\varepsilon_0)}.$$

Define $T=\inf\{n>0:S_n>0\},$ which is the time $\{S_n\}_{n\ge 0}$ hits $[1,\infty)$  and let  $$U_k=\#\{0\le n<T:S_n=k\}.$$
Here and throughout, $``\#\{\ \}"$ denotes the number of elements  in set $\{\ \}.$
  We have the following ballistic LLN for $\{S_n\}_{n\ge0}.$
\begin{theorem}\label{unb}
  Suppose that condition B holds and $E(T)<\infty.$ Then  $$P\text{-a.s.,}\ \lim_{n\rto}\frac{S_n}{n}=v_\mbp>0$$
  where
   $$
  v_\mbp=\frac{\mbe\z(\sum_{i=1}^\infty\sum_{k\le 0}E_{\theta^{-k}\omega}\big(U_k|S_{T=i}\big)\sum_{j\in\mbz} j\omega_{0j}\y)}{\sum_{i=1}^\infty E\big(T|S_{T}=i\big)}.
$$
\end{theorem}
\begin{remark}
(1) Our proof is based on an approach known as ``the environment viewed from particle" introduced in Kozlov \cite{K85}. In \cite{K85} and the later literatures, this approach was only used to  treat the bounded-jump setting. In our setup, we construct an invariant measure $Q$ which is equivalent to $\mbp.$  Under $Q,$ $\overline{\omega}(n):=\theta^{S_n}\omega,\ n\ge0$ form a stationary and ergodic sequence.
Let $d(x,\omega)=E_\omega^x(S_1-S_0)$  be the local drift and set $$M_n=S_n-S_0-\sum_{k=0}^{n-1}d(S_n,\omega).$$ $\{M_n\}_{n\ge0}$ is a martingale with unbounded differences. But under (B3), we could use the large deviation for martingale to estimate the tail of $M_n.$ Therefore, using the auxiliary Markov chain $\{\overline{\omega}(n)\}_{n\ge0}$ we prove the LLN of $\{S_n\}_{n\ge0}.$

(2) For RWRE with bounded jumps. The velocity of the LLN could be written in terms of the environment $\omega$ explicitly because one could calculate the quenched mean $E_\omega (T)$ by using the branching structure constructed from the path of the random walk. See \cite{hwb} for details. However for unbounded-jump setting, we do not know how to calculate the quenched mean $E_\omega( T).$ Therefore, the condition $``E(T)<\infty"$ looks not so satisfying and consequently the velocity could not be given explicitly. For some special case, we could give the explicit formula for the velocity $v_\mbp.$ See the discussion below.
\end{remark}

If $\mbp(\omega_{0j}=0,j\ge 2)=1,$  define $$\Phi=\left(
        \begin{array}{ccccccc}
          0 & 1 & 0 & 0 &0 & 0&\cdots \\
          \omega_{0,1} & \omega_{0,0} &  \omega_{0,-1}& \omega_{0,-2} &\omega_{0,-3}  &\omega_{0,-4}& \cdots \\
          0 & \omega_{-1,1} & \omega_{-1,0} & \omega_{-1,-1} & \omega_{-1,-2} & \omega_{-1,-3}&\cdots \\
          0 & 0 & \omega_{-2,1} & \omega_{-2,0} & \omega_{-2,-1} &\omega_{-2,-2}&\cdots \\
          0 & 0 & 0& \omega_{-3,1} & \omega_{-3,0} &\omega_{-3,-1} &\cdots \\
          0 & 0 & 0&  0&\omega_{-4,1}&\omega_{-4,0}&\cdots\\
          \vdots & \vdots & \vdots & \vdots & \vdots & \vdots&\ddots \\
        \end{array}
      \right).
$$
Let $\pi=(\pi_1,\pi_0,\pi_{-1},\pi_{-2},...)$ be a solution of the equation $$\pi\Phi=\pi,$$ with $\pi_1=1.$
If $\mbe(\sum_{i\le0}\pi_i)<\infty,$ then it follows from the classical ergodic theory for Markov chain that $P$-a.s., $T<\infty$ and $E_\omega(T)=\sum_{i\le 0}\pi_i.$ We have the following corollary of Theorem \ref{unb}.

\begin{corollary}
  Suppose that condition B holds and $\mbe (\sum_{i\le 0}\pi_i)<\infty.$ Then $$P\text{-a.s.,}\ \lim_{n\rto}\frac{S_n}{n}=\frac{1}{\mbe\z(\sum_{i\le0}\pi_i\y)}.$$
\end{corollary}

\subsection{BDPRE with bounded jumps}
Next we define the birth and death process in random environment with bounded jumps. To construct the environment, fix $1\le L,R\in \mathbb Z$ and let $\tilde\Omega$ be the collection of $\tilde\omega=(\tilde\omega_i)_{i\in\mathbb Z}=(\mu^{L}_i,...,\mu^{1}_i,\lambda^1_i,...,\lambda^R_i)_{i\in \mathbb Z},$ where $\mu_i^l,\lambda_i^r\ge 0$ for all $i\in\mathbb Z,\ l=1,..,L$ and $r=1,...,R. $ Equip $\tilde\Omega$ with the Borel $\sigma$-algebra $\tilde{\mathcal{F}}$ and let $\tilde{\mathbb P}$ be a probability measure on $(\tilde\Omega,\tilde{\mathcal F}).$ Then the so-called  random environment $\tilde\omega$ is a random element of $\tilde\Omega$ chosen according to $\tilde{\mathbb P}.$
Given a realization of $\tilde\omega,$ let $\{N_t\}_{t\ge0}$ be a continuous time Markov chain, which waits at a state $n$ an exponentially distributed time with parameter $\sum_{l=1}^L\mu^{l}_n+\sum_{r=1}^R\lambda^{r}_n$ and then jumps to $n-i$ with probability ${\mu^i_n}/(\sum_{l=1}^L\mu^{l}_n+\sum_{r=1}^R\lambda^{r}_n),$ $i=1,...,L$ or to $n+j$ with probability ${\lambda^j_n}/(\sum_{l=1}^L\mu^{l}_n+\sum_{r=1}^R\lambda^{r}_n),$ $j=1,...,R.$ In this paper, we always assume that the paths of $\{N_t\}$ are right continuous. We call the process $\{N_t\}_{t\ge 0}$ a {\it birth and death process in random environment with bounded jumps}.

For a typical realization of $\tilde\omega,$ $\tilde P^x_{\tilde\omega}$ denotes the law induced by the process $\{N_t\}$ starting from $x.$ The measure $\tilde P^x_{\tilde\omega}$ is usually related as the {\it quenched} probability. Define the annealed probability measure $\tilde P^x$ by $\tilde P^x(\cdot)=\int_{\Omega}\tilde P^x_{\tilde\omega}(\cdot)\tilde{\mathbb P}(d\tilde\omega).$
The notations $\tilde E^x_{\tilde\omega},$ $\tilde E^x$ and $\tilde{\mathbb E}$ will be used to denote the expectation operators for $\tilde P^x_{\tilde\omega},$ $\tilde P^x$ and $\tilde{\mathbb P}$ respectively. The superscript $x$  will be omitted if it is $0.$ Let operator $\theta$ be the canonical shift on $\tilde\Omega$ defined by $(\theta\tilde\omega)_i=\tilde\omega_{i+1}.$

\noindent {\bf Condition C}

\noindent\textbf{(C1)} $(\tilde\Omega,\tilde{\mathcal F}, \tilde{\mathbb P}, \theta)$ forms a stationary and ergodic system.

\noindent\textbf{(C2)} the measure $\tilde{\mathbb P}$ is uniformly elliptic, that is, $$\tilde{\mathbb P}\Big(\varepsilon<\mu_0^l, \lambda_0^r<M , 1\le l\le L, 1\le r\le R \Big)=1$$ for some small $\varepsilon>0$ and large $M>0.$

Under condition (C2), the process $\{N_t\}$ exists for $\tilde \mbp$-a.a. $\tilde\omega.$  For details, see the appendix section below.

Given $\tilde\omega,$ define for $i\in \mathbb Z,$
$$b_i(k)=\left\{\begin{array}{ll}
   \frac{\sum_{j=R-k+1}^R\lambda_i^j}{\mu_i^L}&\text{if } 1\le k\le R,\\
   - \frac{\sum_{j=k-R}^L\mu_i^j}{\mu_i^L}&\text{if } R +1\le k\le R+L-1,
  \end{array}\right.$$
and let \begin{equation*}A_i=\left(
        \begin{array}{cccc}
          0 & 1 & \cdots & 0 \\
           \vdots& \vdots & \ddots &\vdots  \\
          0 & 0 & \cdots & 1 \\
          b_{i}(1) & b_i(2) & \cdots & b_{i}(L+R-1) \\
        \end{array}\right)
\end{equation*} be an $(L+R-1)\times(L+R-1)$ matrix.

Since $A_i$ depends only on $\tilde\omega_i,$ $\{A_i\}_{i\in \mathbb Z}$ is an ergodic sequence of random matrices under $\tilde{\mathbb P}.$ Moreover, under condition (C2), $\tilde{\mathbb E} |\ln \|A_0^{-1}\||+\tilde{\mathbb E} |\ln \|A_0\||<\infty.$ Hence one could use Oseledec's multiplicative ergodic theorem (see \cite{osel}) to the sequence $\{A_i\}_{i\in \mathbb Z}.$ Consequently, we get the Lyapunov
exponents of the sequence $\{A_i\}_{i\in\mathbb Z}$ which we write in increasing order as
$$-\infty<\gamma_1\le \gamma_2\le...\le\gamma_{R+L-1}<\infty.$$
\begin{proposition}[Recurrence/transience criteria]\label{rtcrit} Suppose that condition C holds. Let $\gamma_1\le \gamma_2\le...\le\gamma_{R+L-1}$ be the Lyapunov exponents of the sequence $\{A_i\}_{i\in\mathbb Z}$ under the probability measure $\tilde\mbp.$ Then

\noindent(1) $\gamma_R>0\Rightarrow \tilde P(\lim_{t\rto}N_t=\infty)=1;$

\noindent(2) $\gamma_R=0\Rightarrow  \tilde P(-\infty=\liminf_{t\rto}N_t<\limsup_{t\rto}N_t=\infty)=1;$

\noindent(3) $\gamma_R<0\Rightarrow \tilde P(\lim_{t\rto}N_t=-\infty)=1.$
\end{proposition}
\proof Since the recurrence criteria for $\{N_t\}$ is the same as the embedded process $\{\chi_n\}$ defined below. Proposition \ref{rtcrit} is just a corollary of Theorem A in Letchikov \cite{letc}.\qed

Let $\tau_0=0,$ and for $n\ge 1$ define
$\tau_{n}=\inf\{t>\tau_{n-1}:N_t\neq N_{\tau_{n-1}}\}.$ Since the process $\{N_t\}$ exists, $\tilde P$-a.s., $\tau_n<\infty$ for all $n.$ Note that $\tau_n,\ n\ge 0$ are the consecutive discontinuities of $\{N_t\}.$ Let $\chi_n=N_{\tau_n},\ n\ge 0.$ Then $\{\chi_n\}$ is called the {\it embedded process} of $\{N_t\}.$
Next we study the LLN of  $\{N_t\}.$ Let $T_0=0$ and for $n\ge 1,$ define recursively $$T_{n}=\inf\{t>T_{n-1}:N_t>N_{T_{n-1}}\}.$$  We call $T_n,\ n\ge0$ the ladder times of the process $\{N_t\}.$
Define
\begin{equation}\label{vp}v_{\tilde{\mathbb P}}=\frac{\tilde\mbe \Big( \sum_{r=1}^R\sum_{k\le0}\tilde E_{\theta^{-k}\tilde\omega}\big(\sum_{j=1}^{U_k}\xi_{kj}|N_{T_1}=r\big)\big(\sum_{l=1}^L(-l)\mu_0^l+\sum_{r=1}^Rr\lambda_0^r\big)\Big)}{\sum_{r=1}^R\tilde E(T_1|N_{T_1}=r)}\end{equation}
where $$U_k:=\#\{n:N_{\tau_n}=k,\tau_n<T_1\}$$ is the number of times the embedded process $\{\chi_n\}$ has ever visited $k$ before it hits $[1,\infty),$ and given $\tilde\omega,$ $\xi_{kj},\ k\le 0,\ j\ge 0$ are independent random variables which are also all independent of $U_k$ such that $\tilde P_\omega(\xi_{kj}>t)=e^{-(\sum_{l=1}^L\mu_k^l+\sum_{r=1}^R\lambda_k^r)t},\ t\ge0.$

\begin{theorem}[LLN of $\{N_t\}$]\label{lln}
  Suppose that Condition C holds and $\gamma_R\ge 0.$  Then

\noindent(a) $ \tilde ET_1<\infty\Rightarrow \lim_{t\rto}\frac{N_t}{t}=v_{\tilde\mbp}>0,$ $\tilde P$-a.s.;

\noindent(b) $\tilde ET_1=\infty\Rightarrow \lim_{t\rto}\frac{N_t}{t}=0,$ $\tilde P$-a.s..
\end{theorem}

\begin{remark}\label{rc}
\noindent(i) $\tilde ET_1$ and the velocity $v_{\tilde\mbp}$ of LLN is not given directly in terms of $\tilde\omega.$ They could be calculated  by using the branching structure constructed in \cite{hwb}. Therefore  the formulae of both of $\tilde ET_1$ and $v_{\tilde\mbp}$ could be given explicitly. We treat the case $L=R=2$ to explain the idea in Section \ref{dvp}. See Theorem \ref{vpll} below. The special case $R=1, L> 1$ is discussed in Wang \cite{w}.

\noindent(ii). The case $\gamma_R\le 0$ could be treated in a similar way. We omit this part in this paper.

\noindent(iii). In the proof of part (a) of Theorem \ref{lln}, we only use the  condition
  ``\textbf{(C2')} for some small $\ka>0$ and large $K>0,$ $\tilde\mbp(\lambda_0^1>\ka, \sum_{l=1}^L\mu_0^l+\sum_{r=1}^R\lambda_0^r<K)=1.$"
Under condition (C2'), $\tilde\mbp(\lambda_0^r, \mu_0^l =0 \text{ for certain } 1\le r\le R, 1\le l\le L)>0$ is permitted. So (C2') is weaker than (C2). When proving  part (b) of Theorem \ref{lln} we borrow some results from \cite{brb} where  the uniform ellipticity in (C2) is used.
\end{remark}

To prove Theorem \ref{lln}, we need to prove the LLN of the $h$-skeleton process of $\{N_t\}.$
Fix a number $h>0,$ which will be assumed to be  small enough. For $n\ge 0,$ define $X_n=N_{nh}.$ Then $\{X_n\}$ is a discrete time RWRE with unbounded jumps, which is called the {\it $h$-skeleton process} of $\{N_t\}_{n\ge0}.$ For $i,j\in \mbz$ let $$p_{\tilde\omega}(h,i,j):=\tilde P_{\tilde\omega}(N_h=i+j|N_0=i)$$ be the transition probabilities of $\{X_n\}.$
Define $T_1^h=\inf\{k:X_k> 0\},$ and let
 $$U_k^h=\#\{0\le n<T_1^h:X_n=k\}.$$
\begin{theorem}[LLN of skeleton process]\label{llh}
  Suppose that condition C holds.  Then $\tilde P$-a.s., $\{X_n\}$ is transient to the right, recurrent or transient to the left according as $\gamma_R\ge 0,$ $\gamma_R=0$ or $\gamma_R\le 0.$   Moreover, if $\gamma_R\ge 0,$ then

 \noindent$\tilde ET_1^h=\infty\Rightarrow $ $\tilde P$-a.s., $\lim_{n\rto}\frac{X_n}{n}=0;$

 \noindent$\tilde ET_1^h<\infty\Rightarrow$ $\tilde P$-a.s., $\lim_{n\rto}\frac{X_n}{n}=v_{\tilde\mbp}^h>0,$

\noindent where
$
  v_{\tilde{\mbp}}^h=\frac{\tilde\mbe\z(\sum_{i=1}^\infty\sum_{k\le 0}\tilde E_{\theta^{-k}\tilde\omega}\big(U_k^h|X_{T_1^h=i}\big)\sum_{j\in\mbz} jp_{\tilde\omega}(h, 0,j)\y)}{\sum_{i=1}^\infty \tilde E\big(T_1^h|X_{T_1^h}=i\big)}.
$
\end{theorem}
\begin{remark}
 (i) The $h$-skeleton process $\{X_n\}$ is an RWRE with {\it unbounded jumps}. Under  (C2), we could show that,  for some proper constants $c_0, c_1>0,$ $\tilde\mbp$-a.s., $$p_{\tilde\omega}(h,i,j)<e^{c_0h}e^{-c_1|j|}.$$ Hence, the positive regime of the LLN in Theorem \ref{llh} follows from Theorem \ref{unb}.

 \noindent(ii) One sees easily from Theorem \ref{lln} that $v_{\tilde\mbp}^h$ is indeed independent of $h$ and $v_{\tilde\mbp}^h=v_{\tilde\mbp}.$ For details, see the proof of Theorem \ref{lln} below.
\end{remark}

\section{LLN for RWRE -Proof of Theorem \ref{unb}}\label{pr}

For $n\ge 0$, define $\overline\omega(n)=\theta^{S_n}\omega.$ $\{\overline\omega(n)\}_{n\ge0}$ is an $\Omega^\mathbb N$-valued process. It is usually called ``the environment viewed from particle". Let $$K(\omega,d\omega')=\sum_{j\in\mbz}\omega_{0j}\delta_{\omega'=\theta^j\omega}.$$
\begin{lemma} Under either $P$ or $P_\omega,$ $\{\overline\omega(n)\}_{n\ge0}$ is a Markov chain with transition kernel $K(\omega,\omega').$
\end{lemma}
\proof For test functions $f_1,...,f_n, f_{n+1}$ we have that \begin{eqnarray*}
  &&E_\omega\Big(\prod_{k=1}^{n+1}f_k(\overline\omega(k))\Big)=E_\omega\Big(\prod_{k=1}^{n}f_k(\overline\omega(k))E_\omega^{S_{n}}f_{n+1}(\theta^{S_1}\omega)\Big)\\
&&\quad\quad=E_\omega\Big(\prod_{k=1}^{n}f_k(\overline\omega(k))\sum_{j\in\mbz}\omega_{S_nj}f_{n+1}(\theta^{S_n+j}\omega)\Big)\\
&&\quad\quad=E_\omega\Big(\prod_{k=1}^{n}f_k(\overline\omega(k))\sum_{j\in\mbz}Kf_{n+1}(\overline\omega(n))\Big).
\end{eqnarray*}
Consequently, $\{\overline\omega(n)\}_{n\ge0}$ is a Markov chain under $P_\omega.$ Taking expectation, the above equations also yield the Markov property of  $\{\overline\omega(n)\}_{n\ge0}$ under $P.$ \qed

Whenever $E(T)<\infty,$
define the measures $$ Q(d\omega):=E\z(\displaystyle\sum_{i\ge1}\frac{\mathbf{1}_{S_{T}=i}}{P_\omega(S_{T}=i)}\sum_{k=0}^{T-1}1_{\overline\omega(k)\in d\omega}\y),\ \overline{Q}(d\omega)=\frac{Q(d\omega)}{E(T)}.$$

\begin{lemma}\label{sden}
  Suppose that Condition C holds and $E(T)<\infty.$ Then $Q$ is invariant under the kernel $K,$ that is $$Q(B)=\iint\mathbf1_{\omega'\in B}K(\omega,d\omega')Q(d\omega).$$
Moreover, $Q\sim\mathbb P$ and
$$\frac{dQ}{d\mbp}=\sum_{k\le0}\sum_{i\ge 1}E_{\theta^{-k}\omega}(U_k|S_{T}=i)=:\pi(\omega),$$ where $U_k=\#\{n\le T:S_n=k\}.$
\end{lemma}
\proof By the definition of $Q,$ we have that \begin{eqnarray*}
&&\iint\mathbf1_{\omega'\in B}K(\omega,d\omega')Q(d\omega)=\int P_\omega(\theta^{S_1}\omega\in B)Q(d\omega)\\
&&\quad\quad\quad = E\Big(\sum_{i\ge1}\frac{\mathbf{1}_{S_{T}=i}}{P_\omega(S_{T}=i)}\sum_{k=0}^{T-1}P_{\overline\omega(k)}(\theta^{S_1}\overline\omega(k)\in B)\Big)\\
&&\quad\quad\quad = E\Big(\sum_{i\ge1}\frac{\mathbf{1}_{S_{T}=i}}{P_\omega(S_{T}=i)}\sum_{k=0}^{T-1}P_{\overline\omega(k)}(\overline\omega(k+1)\in B)\Big)\\
&&\quad\quad\quad = E\Big(\sum_{i\ge1}\sum_{k\ge0}P_\omega\z(T\ge k+1,\overline\omega(k+1)\in B\big|S_{T}=i\y) \Big)\\
&&\quad\quad\quad = E\Big(\sum_{i\ge1}\sum_{k\ge1}P_\omega\z(T> k,\overline\omega(k)\in B\big|S_{T}=i\y) \Big)\\
&&\quad\quad\quad \quad\quad\quad+ E\Big(\sum_{i\ge1}P_\omega\z(T<\infty,\theta^i\omega\in B\big|S_{T}=i\y) \Big).\\
\end{eqnarray*}
Since $E(T)<\infty,$ then $P(T<\infty)=P(T> 0)=1.$ This fact together with the stationarity implies that the right-most hand of the last equations equals to
\begin{eqnarray*}
&& E\Big(\sum_{i\ge1}\sum_{k\ge1}P_\omega\z(T> k,\overline\omega(k)\in B\big|S_{T}=i\y) \Big)\\
&&\quad\quad\quad\quad\quad+ E\Big(\sum_{i\ge1}P_\omega\z(T> 0,\overline\omega(0)\in B\big|S_{T}=i\y) \Big)\\
&&\quad\quad=E\Big(\sum_{i\ge0}\sum_{k\ge0}P_\omega\z(T> k,\overline\omega(k)\in B\big|S_{T}=i\y) \Big)=Q(B).
\end{eqnarray*}
The first part of the lemma follows.
To prove the second part, for testing function $f(\omega),$ we have that
\begin{eqnarray}
 && \int f dQ=E\Big(\sum_{i\ge1}\frac{\mathbf{1}_{S_{T}=i}}{P_\omega(S_{T}=i)}\sum_{k=0}^{T-1}f(\overline\omega(k))\Big)\no\\
&&\quad\quad\quad\ =E\Big(\sum_{i\ge1}\frac{\mathbf{1}_{S_{T}=i}}{P_\omega(S_{T}=i)}\sum_{k=0}^{T-1}f(\theta^{S_k}\omega)\Big)\nonumber\\
&&\quad\quad\quad\  =E\Big(\sum_{i\ge1}\frac{\mathbf{1}_{S_{T}=i}}{P_\omega(S_{T}=i)}\sum_{k\le 0}U_kf(\theta^{k}\omega)\Big)\no\\
&&\quad\quad\quad\ =\mbe\Big(\sum_{i\ge1}E_\omega\Big(\sum_{k\le 0}U_kf(\theta^{k}\omega)\big|S_{T}=i\Big)\Big)\nonumber\\
&&\quad\quad\quad\ =\mbe\Big(f(\omega)\sum_{i\ge1}\sum_{k\le 0}E_{\theta^{-k}\omega}\Big(U_k\big|S_{T}=i\Big)\Big).\label{eqf}
\end{eqnarray}
By stationarity, we have that
\begin{eqnarray*}
\mbe\Big(\sum_{i\ge1}\sum_{k\le 0}E_{\theta^{-k}\omega}\Big(U_k\big|S_{T}=i\Big)\Big)=\sum_{i\ge1}\sum_{k\le 0}E\Big(U_k\big|S_{T}=i\Big)\le E(T)<\infty.
\end{eqnarray*}
Then it follows from  (\ref{eqf})  that $Q\sim \mbp$ and $$\frac{dQ}{d\mbp}=\sum_{k\le0}\sum_{i\ge 1}E_{\theta^{-k}\omega}(U_k|S_{T}=i).$$
\qed

\begin{lemma}\label{sbo}
  Under the conditions of Lemma \ref{sden}, $\{\overline\omega(n)\}$ is stationary and ergodic under the probability measure $\overline Q\times P_\omega.$
\end{lemma}
With Lemma \ref{sden} in hand, Lemma \ref{sbo} follows similarly as  Sznitman \cite{S02}, Theorem 1.2 or  Zeitouni \cite{ze04}, Corollary 2.1.25.

Define the local drift $d(x,\omega)=E_\omega^x(S_1-S_0)$ and set $$M_n=S_n-S_0-\sum_{k=0}^{n-1}d(S_n,\omega).$$
\begin{lemma}\label{coy}
  Under $P_\omega,$ $\{M_n\}$ is a martingale and $P$-a.s., $\lim_{n\rto}\frac{M_n}{n}=0.$
\end{lemma}
\proof Note that \begin{equation*}\begin{split}E_\omega&(M_n-M_{n-1}|M_{n-1},...,M_{0})\\
&=E_\omega(S_n-S_{n-1}-d(S_{n-1},\omega)|S_{n-1},...,S_0)=0.\end{split}\end{equation*} Then $\{M_n\}$ is a martingale under $P_\omega.$ By (B3), there exist some constants $c_2>0$ and $0<\varepsilon_1<\varepsilon_0$ such that $E_\omega(|M_n-M_{n-1}|^{2+\varepsilon_1})<c_2.$   Then one follows from Theorem 3.2 in \cite{lv} that there exists constant $c_3>0$ such that for $\lambda>0,$ and $n$ large enough,
\begin{equation}\label{azm}P(|M_n|>\sqrt{n}\lambda)\le \frac{c_3}{\lambda^{2+\varepsilon_1}}.\end{equation}
Choosing  $0<\varepsilon_2<\frac{1}{2}$ properly, we have $(2+\varepsilon_{1})\varepsilon_2>1.$
Setting $\lambda=n^{\varepsilon_2}$ in (\ref{azm}), it follows that
  $$P(|M_n|>n^{\frac{1}{2}+\varepsilon_2})\le c_3n^{-(2+\varepsilon_1)\varepsilon_2}.$$
An application of Borel-Cantelli's lemma yields that $$P\text{-a.s., } \lim_{n\rto}\frac{M_n}{n}=0.$$ \qed

\noindent{\it Proof of Theorem \ref{unb}:}
By Lemma \ref{sbo}, $\{\overline\omega(n)\}$ is a stationary and ergodic sequence under the measure $\overline Q\times P_\omega.$ Using Birkhoff's ergodic theorem, we have that for $\overline Q$-a.a. or $\mbp$-a.a. $\omega,$ $P_\omega$-a.s.,
\begin{equation}\label{cdo}
  \lim_{n\rto}\frac{1}{n}\sum_{k=0}^{n-1}d(S_k,\omega)=\lim_{n\rto}\frac{1}{n}\sum_{k=0}^{n-1}d(0,\overline\omega(k))=\int d(0,\omega)d\overline Q.
\end{equation}
We conclude from Lemma \ref{coy} and (\ref{cdo}) that $P$-a.s.,
$$\lim_{n\rto}\frac{S_n}{n}=\int d(0,\omega)d\overline Q=:v_\mbp.$$
It follows from Lemma \ref{sden} that
\begin{equation*}
  \begin{split}
  v_\mbp=\int d(0,\omega)d\overline Q=\frac{\mbe\z(\sum_{i=1}^\infty\sum_{k\le 0}E_{\theta^{-k}\omega}(U_k|S_{T=i})\sum_{j\in\mbz} j\omega_{0j}\y)}{\sum_{i=1}^\infty E(T|S_{T}=i)}.
\end{split}
\end{equation*}
Theorem \ref{unb} is proved. \qed

\section{LLN for BDPRE with bounded jumps}\label{pec}
\subsection{LLN for the $h$-skeleton process-Proof of Theorem \ref{llh}}\label{sllh}
The recurrence criteria for $\{X_n\}$ in Theorem \ref{llh} follows directly from Proposition \ref{rtcrit}. The zero speed regime of the LLN for $\{X_n\}$ follows from the counterpart of Theorem \ref{lln}. Next we prove
the non-zero speed regime of the LLN in Theorem \ref{llh}. To begin with, we estimate the tail probability of the transition probability
$p_{\tilde\omega}(h,i,j):=\tilde P_{\tilde\omega}(N_h=i+j|N_0=i)$
of $\{X_n\}.$
The following lemma shows that for fixed $i\in \mathbb Z,$ $\tilde{\mathbb P}$-a.s.,  $p_{\tilde\omega}(h,i,j)$ decays exponentially to $0$ as $|j|\rto.$
\begin{lemma}\label{esp}
  Suppose that Condition (C2) is satisfied. Then for $\tilde\mbp$-a.a. $\tilde\omega,$ there exist $0<c_0<\infty$ and $0<c_1<\infty,$ which are independent of $h$ and $\tilde\omega,$ such that for $|j|> \max\{L,R\},$ \begin{equation}\label{epij}
  p_{\tilde\omega}(h,i,j)<e^{c_0h}e^{-c_1|j|},
\end{equation} where $c_1$ could be made arbitrarily large by adjusting the value of $c_0.$
\end{lemma}
\proof We prove only the case $j>\max\{L,R\}.$ The case $j<-\max\{L,R\}$ follows similarly. Let $m=[\frac{j}{R}].$ Since at each discontinuity, the process $\{N_t\}$ jumps at most a distance $R$ to the right, then starting from $i,$ in order to reach $i+j,$  $\{N_t\}$ has at least $m$ discontinuities in a time period of length $h.$ Let $\eta_k,\ k=1,...,m$ be these discontinuities and $\tau_k$ be the waiting time after $\eta_k$ until the process $\{N_t\}$ leaves $N_{\eta_k}.$ Then we have that
\begin{equation}\label{ptau}
  p_{\tilde\omega}(h,i,j)= \tilde P_{\tilde\omega}(N_h=i+j|N_0=i)\le \tilde P^i_{\tilde\omega}(\tau_1+...+\tau_m\le h).
\end{equation}
Note that  $\tilde P_{\tilde\omega}^i(\tau_k>t|N_{\eta_k}=i_k)=e^{-s_kt}$ for $t>0$ where $s_k=\big(\sum_{l=1}^L\mu_{i_k}^l+\sum_{r=1}^R\lambda_{i_k}^r\big).$ Moreover under $\tilde P_{\tilde\omega}^i,$ $\tau_k,\ k=1,...,m$ are mutually independent. Then by Chebycheff's bound, it follows that  for $\lambda<0,$ $K=(L+R)M,$ $\ka=(L+R)\varepsilon,$
 \begin{eqnarray*}
&&\tilde P^i_{\tilde\omega}(\tau_1+...+\tau_m\le h)\\
&&\quad=\sum_{i_1,...,i_k}\tilde P^i_{\tilde\omega}\Big(\sum_{k=1}^m\tau_k\le h\Big|N_{\eta_k}=i_k,1\le k\le m\Big)\tilde P^i_{\tilde\omega}(N_{\eta_k}=i_k,1\le k\le m)\\
&&\quad\le\sum_{i_1,...,i_k}e^{-\lambda h}\tilde E_{\tilde\omega}^i\Big(e^{\lambda\sum_{k=1}^m\tau_k}\Big|N_{\eta_k}=i_k,1\le k\le m\Big)\\
&&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\tilde P_{\tilde\omega}^i(N_{\eta_k}=i_k,1\le k\le m)\\
&&\quad= e^{-\lambda h}\sum_{i_1,...,i_k}\prod_{k=1}^m \frac{s_k}{s_k-\lambda}\tilde P^i_{\tilde\omega}(N_{\eta_k}=i_k,1\le k\le m)\\
&&\quad\le  e^{-\lambda h}\Big(\frac{K}{\ka-\lambda}\Big)^m, \ \tilde\mbp\text{-a.s.,}
\end{eqnarray*}
where the last inequality follows from condition (C2). Substituting the above estimation to (\ref{ptau}), we have that $\mbp$-a.s.,
$$p_{\tilde\omega}(h,i,j)\le e^{-\lambda h}\Big(\frac{K}{\ka-\lambda}\Big)^m\le e^{-\lambda h}e^{\frac{j}{R}(\log K-\log(\ka-\lambda))}.$$
By choosing $\overline{\lambda}<0$ properly and letting  $c_0=-\overline{\lambda},$  $c_1=(\log(\ka-\overline\lambda)-\log K)/R,$ (\ref{epij}) is proved. Of course, we could make $c_1$ arbitrarily large by adjusting the value of $\overline\lambda.$ \qed

\begin{lemma}\label{sep} Under the measure $\tilde\mbp,$
  $\{p_{\tilde\omega}(h,i,j)\}_{i\in\mbz}$ is a stationary and ergodic sequence.
\end{lemma}
For the proof of the lemma, refer to Durrett \cite{dur}.
\qed

One follows from condition (C2) that for some small constant $c_4>0$ \begin{equation}\label{ell}p_{\tilde\omega}(h,0,1)\ge c_4.\end{equation}
Taking Lemma \ref{esp}, Lemma \ref{sep} and (\ref{ell}) together,  we could use Theorem \ref{unb} to conclude that
  $$\tilde E(T_1^h)<\infty\Rightarrow \tilde{P}\text{-a.s., } \lim_{n\rto}\frac{X_n}{n}=v_{\tilde\mbp}^h.$$
  We complete the proof of Theorem \ref{llh}. \qed

\subsection{The  LLN of $\{N_t\}$-Proof of Theorem \ref{lln}}
In this subsection,  using the LLN (nonzero speed regime) of the $h$-skeleton process proved in Subsection  \ref{sllh}, we prove the LLN of the process $\{N_t\}.$ Firstly, we show that for $h$ small enough, $\tilde E(T_1^h)<\infty$ whenever $\tilde E(T_1)<\infty.$ Therefore, under the condition of part (a) of Theorem \ref{lln}, the condition of the nonzero speed regime of LLN in Theorem \ref{llh} is also satisfied for $h$ small enough.
\begin{lemma}\label{tht}
 Fix $s>1.$ Suppose that condition (C2) holds.   Then, for $n$ large enough, $\tilde P$-a.s., $T_1^{\frac{1}{n^{s}}}\frac{1}{n^s}\le T_1+\frac{1}{n^s}.$
Consequently, we have that for $n$ large, if $\tilde E(T_1)<\infty,$ then $\tilde E\Big(T_1^{\frac{1}{n^{s}}}\Big)<\infty$ and $\tilde P$-a.s., $\lim_{n\rto}T_1^{\frac{1}{n^{s}}}\frac{1}{n^s}=T_1.$
\end{lemma}
\proof Given ${\tilde\omega},$ we have that
\begin{eqnarray}\label{tns}
  &&\tilde P_{\tilde\omega}\Big(T_1^{\frac{1}{n^{s}}}\frac{1}{n^s}> T_1+\frac{1}{n^s}\Big)\no\\
  &&\quad\quad\quad\le \tilde P_{\tilde\omega}\Big(\{N_t\} \text{ has at least one jump in } [T_1,T_1+\frac{1}{n^s})\Big)\nonumber\\
&&\quad\quad\quad \le \tilde P_{\tilde\omega}\Big(\{N_t\} \text{ leaves  }N_{T_1} \text{ within time } \frac{1}{n^s}\Big)\nonumber\\
&&\quad\quad\quad =\sum_{r=1}^R\tilde P_{\tilde\omega}(N_{T_1}=r)\tilde P_{\tilde\omega}\Big(\nu_r<\frac{1}{n^s}\big|N_{T_1}=r\Big)
\end{eqnarray}
where $\nu_r$ is the waiting time at state $r$ until the next jump of $\{N_t\}$ happens.
Since $\nu_r$ is exponentially distributed with parameter $\zeta_r:=\sum_{l=1}^L\mu_r^l+\sum_{k=1}^R\lambda_r^k.$
Using condition (C2), with $K=(L+R)M,$ it follows that $\tilde\mbp$-a.s., the right-most hand of  (\ref{tns}) equals to
$$\sum_{r=1}^R\tilde P_{\tilde\omega}(N_{T_1}=r)(1-e^{-\zeta_r \frac{1}{n^s}})\le 1-e^{-K \frac{1}{n^s}}. $$
Noting that $s>1,$ then we have that
$$\sum_{n=1}^\infty \tilde P\Big(T_1^{\frac{1}{n^{s}}}\frac{1}{n^s}> T_1+\frac{1}{n^s}\Big)<\infty.$$
An application of Borel Cantelli's lemma yields that for $n$ large, $\tilde P$-a.s.,
\begin{equation}\label{tst}
  T_1^{\frac{1}{n^{s}}}\frac{1}{n^s}\le T_1+\frac{1}{n^s}.
\end{equation}
Since $\tilde P$-a.s., $T_1^hh>T_1,$ then one follows from (\ref{tst}) that $\tilde P$-a.s.,
$\lim_{n\rto}T_1^{\frac{1}{n^{s}}}\frac{1}{n^s}= T_1.$ \qed

\noindent{\it Proof of Theorem \ref{lln}:} We prove part (a) first. Suppose $\gamma_R\ge0$ and $\tilde E(T_1)<\infty.$ Fix $h>0$ small enough. For any $t>0,$ there is a unique number $n_t$ such that $n_th\le t<(n_{t}+1)h.$ Let $J_t$ be the number of jumps of $\{N_t\}$ in the time interval $[n_th, (n_{t}+1)h).$ Then, under condition (C2), a similar argument as the proof of Lemma \ref{esp} yields that there exist some positive constants $c_5$ and $c_6$ such that
\begin{equation}\label{bpk}\tilde P(J_t>n)\le e^{c_5h}e^{-nc_6}.\end{equation}
Note that the bound in (\ref{bpk}) is independent of $t.$ Applying Borel-Cantelli lemma, we have that $\tilde P$-a.s., \begin{equation}\label{jt0}\lim_{n\rto}\frac{J_t}{n}=0 \end{equation} uniformly in $t.$
Since at each discontinuity, $\{N_t\}$ jumps at most a distance $L$ to the left or at most a distance $R$ to the right, we have that
\begin{equation}\label{jbt}
  \frac{N_{n_th}-J_tL}{(n_t+1)h}\le \frac{N_t}{t}\le  \frac{N_{n_th}+J_tR}{n_th}.
\end{equation}
For $h$ small enough, by Lemma \ref{tht}, $\tilde E(T_1^h)<\infty$. Then we have from the nonzero speed regime of LLN in Theorem \ref{llh}, (\ref{jt0}) and (\ref{jbt}) that $\tilde P$-a.s.,
\begin{equation}\label{lmh}
  \lim_{t\rto}\frac{N_t}{t}=\frac{v_{\tilde\mbp}^h}{h}.
\end{equation}
To finish the proof of part (a) in Theorem \ref{lln}, it suffices to show that, for all $h>0,$
\begin{equation}\label{vph}\frac{v_{\tilde\mbp}^h}{h}=v_{\tilde\mbp}.\end{equation}
Indeed, since  $\frac{N_t}{t}$ is independent of $h$ and the limit in (\ref{lmh}) exists, then $\frac{v_{\tilde\mbp}^h}{h}$ is independent of $h.$
Therefore we have that
\begin{eqnarray}\label{lhd}
&&  \frac{v_{\tilde\mbp}^h}{h}=\lim_{h\rightarrow 0}\frac{v^h_{\tilde\mbp}}{h}=\lim_{h\rightarrow0}\frac{\tilde\mbe\z(\sum_{i=1}^\infty\sum_{k\le 0}\tilde E_{\theta^{-k}{\tilde\omega}}(U_k^h|X_{T_1^h=i})\sum_{j\in\mbz} jp_{\tilde\omega}(h, 0,j)\y)}{h\sum_{i=1}^\infty \tilde E(T_1^h|X_{T_1^h}=i)}\nonumber\\
&&\quad\quad\quad=\lim_{h\rightarrow0}\frac{\tilde\mbe\z(\sum_{i=1}^\infty\sum_{k\le 0}\tilde E_{\theta^{-k}{\tilde\omega}}(hU_k^h|N_{hT_1^h=i})\sum_{j\in\mbz} j\frac{p_{\tilde\omega}(h, 0,j)}{h}\y)}{\sum_{i=1}^\infty \tilde E(h T_1^h|N_{hT_1^h}=i)}.
\end{eqnarray}
Note that by Lemma \ref{esp}, under condition (C2), $\tilde P$-a.s., $\sum_{j\in\mbz} j\frac{p_{\tilde\omega}(h, 0,j)}{h}$ is uniformly bounded from above.
Moreover, by Lemma \ref{tht}, $\tilde P$-a.s., $hT^h_1$ is bounded by $T_1+h$ for $h$ small, and by stationarity
\begin{equation*}\begin{split}\tilde\mbe\Big(\sum_{i=1}^\infty\sum_{k\le 0}&\tilde E_{\theta^{-k}{\tilde\omega}}(hU_k^h|N_{hT_1^h=i})\Big)=\sum_{i=1}^\infty\sum_{k\le 0}\tilde\mbe\z(\tilde E_{\theta^{-k}{\tilde\omega}}(hU_k^h|N_{hT_1^h=i})\y)\\
&=\sum_{i=1}^\infty\sum_{k\le 0}\tilde E\z(hU_k^h|N_{hT_1^h=i}\y)=\sum_{i=1}^\infty \tilde E\z(hT_1^h|N_{hT_1^h=i}\y).\end{split}\end{equation*}
Since $\tilde P$-a.s., $\lim_{h\rto 0}hT^h_1=T_1,$ then by the above discussion, the condition of dominated convergence theorem is satisfied.
Note also that $\tilde P$-a.s.,
\begin{equation}\label{uhk}\lim_{h\rightarrow 0}hU^h_k=\lim_{h\rightarrow0}\sum_{j=0}^{T_1^h-1}h\mathbf1_{N_{jh}=k}=\int_{0}^{T_1}\mathbf1_{N_t=k}dt\overset{\mathcal D}{=}\sum_{i=1}^{U_k}\xi_{ki},\end{equation}
where $ ``A\overset{\mathcal D}{=}B"$ means that A equals to B in $\tilde P_{\tilde\omega}$ distribution.
Since $\tilde P$-a.s., $\lim_{h\rightarrow 0}hT^h_1=T_1,$  $hT^h_1\ge T_1,$ and the paths of $\{N_t\}$ are right continuous,  then by dominated convergence, it follows from (\ref{lhd}) and (\ref{uhk}) that
\begin{eqnarray*}
\frac{v_{\tilde\mbp}^h}{h}&=&\lim_{h\rightarrow0}\frac{\tilde\mbe\z(\sum_{i=1}^\infty\sum_{k\le 0}\tilde E_{\theta^{-k}{\tilde\omega}}(hU_k^h|N_{hT_1^h=i})\sum_{j\in\mbz} j\frac{p_{\tilde\omega}(h, 0,j)}{h}\y)}{\sum_{i=1}^\infty\tilde E(h T_1^h|N_{hT_1^h}=i)}\\
&=&\frac{\tilde\mbe \Big( \sum_{r=1}^R\sum_{k\le0}\tilde E_{\theta^{-k}{\tilde\omega}}\big(\sum_{j=1}^{U_k}\xi_{kj}|N_{T_1}=r\big)\big(\sum_{l=1}^L(-l)\mu_0^l+\sum_{r=1}^Rr\lambda_0^r\big)\Big)}{\sum_{r=1}^R\tilde E(T_1|N_{T_1}=r)}\\
&=&v_{\tilde\mbp}>0,
\end{eqnarray*}
where in the second equality, we use the facts $\tilde P(N_{T_1}=r)=0$ for $r>R$ and
$$\lim_{h\rightarrow 0}\frac{p_{\tilde\omega}(h,0,j)}{h}=\z\{\begin{array}{ll}
                                         \lambda_0^j & \text{if } j=1,...,R, \\
                                         \mu_0^j & \text{if } j=1,...,L, \\
                                         0 & \text{if }j<-L \text{ and } j>R.
                                       \end{array}\y.
$$
 Consequently, (\ref{vph}) is proved and part (a) of Theorem \ref{lln} follows.

Next, we turn to prove part (b) of Theorem \ref{lln}. Suppose $\gamma_R\ge 0$ and $\tilde E(T_1)=\infty.$ Recall that $\tau_n,\ n\ge 0$ are the consecutive discontinuities of $\{N_t\}$ and $\{\chi_n\}=\{N_{\tau_n}\}$ is the embedded process.  Let $\overline T_1:=\inf\{k:\chi_k>0\}.$
As the first step, we show that
\begin{equation}\label{tt}
 \tilde E(\overline T_1)=\infty.
\end{equation}
For this purpose, for $k\le 0$ let $\overline{U}_k=\#\{n<\overline T_1:\chi_n=k\}.$ Then Wald's equation implies that,  $$\tilde E_{\tilde\omega}(T_1)=\sum_{k\le 0}\tilde E_{\tilde\omega}\Big(\sum_{i=1}^{\overline U_k}\xi_{ki}\Big)=\sum_{k\le 0}\tilde E_{\tilde\omega}(\overline U_k)\tilde E_{\tilde\omega}(\xi_{k1}).$$
But condition (C2) implies that $\tilde\mbp$-a.s., $$\frac{1}{K}<\tilde E_{\tilde\omega}(\xi_{k1})=\frac{1}{\sum_{l=1}^L\mu_k^l+\sum_{r=1}^R\lambda_k^r}<\frac{1}{\ka}$$
with $K=(L+R)M$ and $\ka=(L+R)\varepsilon.$
Then it follows  that $\tilde\mbp$-a.s.,
$$\frac{1}{K}\sum_{k\le 0}\tilde E_{\tilde\omega}(\overline U_k)\le \tilde E_{\tilde\omega}(T_1)\le \frac{1}{\ka}\sum_{k\le 0}\tilde E_{\tilde\omega}(\overline U_k).$$
Taking expectation, we have that
$$\frac{\tilde E(\overline T_1)}{K}\le\tilde E(T_1)\le \frac{\tilde E(\overline T_1)}{\ka}.$$
Therefore (\ref{tt}) follows. Then it follows from Br\'emont \cite{brb} (See, Proposition 9.1, Theorem 9.2 and Corollary 9.3 therein.) that $\tilde P$-a.s., \begin{equation}\label{limnt}\lim_{n\rto}\frac{\chi_n}{n}=\lim_{n\rto}\frac{N_{\tau_n}}{n}=0.\end{equation}
As the second step, we show that for $h>0$ small enough and $n$ large enough
\begin{equation}\label{tnh}\tilde P\text{-a.s., } \frac{\tau_n}{n}>h.\end{equation}
For $k\ge 1,$ let $\nu_k=\tau_k-\tau_{k-1}.$ Then under $\tilde P_{\tilde\omega},$ $\nu_k,\ k\ge 1$ are mutually independent and all exponentially distributed.
By Chebycheff's bound, it follows that for $\lambda<0,$ \begin{eqnarray*}\label{lde}
  &&\tilde P_{\tilde\omega}(\tau_n\le nh)\le e^{-\lambda nh}\tilde E_{\tilde\omega}(e^{\lambda \sum_{k=1}^n\nu_k})=e^{-\lambda nh}\prod_{k=1}^n \tilde E_{\tilde\omega}(e^{\lambda \nu_k})\nonumber\\
&&\quad\quad\quad\quad\le e^{-n(\lambda h-\log K+\log(\ka-\lambda))}
\end{eqnarray*}
with $\ka=(L+R)\varepsilon$ and $K=(L+R)M.$
Choosing properly $\overline\lambda<0$ and $h$ small enough, we could make $c(h):=\overline\lambda h-\log K+\log(\ka-\overline\lambda)$ a strictly positive number.
Consequently we have $\sum_{n=1}^\infty \tilde P_{\tilde\omega}(\tau_n\le nh)<\infty.$ By applying Borel-Cantelli lemma, we have (\ref{tnh}).

Finally, (\ref{limnt}) and (\ref{tnh}) imply that $\tilde P$-a.s., \begin{equation}\label{lmntt}\lim_{n\rto}\frac{N_{\tau_n}}{\tau_n}=0.\end{equation}
For $t>0,$ there is a unique random number $n_t$ such that $\tau_{n_t}\le t<\tau_{n_t+1}.$
Since at each discontinuity, $\{N_t\}$ jumps at most a distance $L$ to the left or at most a distance $R$ to the right, we have  \begin{equation}\label{jbnt}\frac{N_{\tau_{n_t}}-L}{\tau_{n_t+1}}\le \frac{N_t}{t}\le \frac{N_{\tau_{n_t}}+R}{\tau_{n_t}}\end{equation}
It follows from (\ref{lmntt}) and (\ref{jbnt}) that $\tilde P$-a.s., $$\lim_{t\rto}\frac{N_t}{t}=0.$$
Part (b) of Theorem \ref{lln} is proved.
\qed

\section{Discussion of the velocity $v_{\tilde\mbp}$}\label{dvp}
In this section, we let $R=L=2$ and discuss the asymptotic velocity $v_{\tilde\mbp}.$ We have
\begin{theorem}\label{vpll}
  Let $\pi({\tilde\omega})$ and $D({\tilde\omega})$ be as in (\ref{do}) and (\ref{pio}) below. Suppose $L=R=2$ and $\tilde\mbe(\pi({\tilde\omega}))<\infty.$ Then
$\tilde P$-a.s., $$\lim_{t\rto}\frac{N_t}{t}=\frac{\tilde\mbe\z(\pi({\tilde\omega})(2\lambda_0^2+\lambda_0^1-\mu_0^1-2\mu_0^2)\y)}{\tilde\mbe(D({\tilde\omega}))}.$$
\end{theorem}
\proof Theorem \ref{vpll} is just a special case of Theorem \ref{lln}. We need only to calculate $v_{\tilde\mbp}.$
Recall that $\overline T_1=\inf\{n\ge 0:\chi_n > 0\}$ is the first ladder time
of the embedded process. Define $$\overline U_k=\#\{n\le \overline T_1:\chi_n=k\}$$ which is the occupation time at  state $k$ of the embedded process before $\overline{T}_1.$  In \cite{hwb}, the authors show that $\overline{U}_k$ could be written as the functional of a multitype branching process $\{Z_n\}_{n\le 1}$ whose offspring matrices are the function of the environment ${\tilde\omega}.$  To introduce those results, we  need to introduce some notations.

Fix $a<b.$ Let $\partial^+[a,b]=\{b,b+1\}$ and
$\partial^-[a,b]=\{a,a-1\}$ be the positive and negative boundaries
of $[a,b]$ correspondingly. For $k\in (a,b),$
$\zeta\in\partial^+[a,b]\cup\partial^-[a,b],$ define
\begin{equation}\label{epk}\no
\mcp_k(a,b,\zeta)=\tilde P_{\tilde\omega}^k(\{\chi_n\}\text{ exits the interval
}[a+1,b-1]\text{ at }\zeta).\end{equation}
For $j=1,2$ let $p_i^j=\frac{\lambda_i^j}{\mu_i^1+\mu_i^2+\lambda_i^1+\lambda_i^2}$ and $q_i^j=\frac{\mu_i^j}{\mu_i^1+\mu_i^2+\lambda_i^1+\lambda_i^2}.$

 Writing
$\mcp_k(a,b,\zeta)$ as $\mcp_k(\zeta)$ temporarily, one follows from
Markov property that
\begin{equation*}
  \mcp_k(\zeta)=p_k^2\mcp_{k+2}(\zeta)+p_k^1\mcp_{k+1}(\zeta)+q_k^1\mcp_{k-1}(\zeta)+q_k^2\mcp_{k-2}(\zeta),
\end{equation*}
which leads to the following matrix form
$$  V_k(\zeta)=M_kV_{k+1}(\zeta)$$  where
$$  V_k(\zeta):=\left(\begin{array}{c}
             (\mcp_{k-1}-\mcp_{k-2})(\zeta) \\
             (\mcp_{k}-\mcp_{k-1})(\zeta) \\
             (\mcp_{k+1}-\mcp_{k})(\zeta) \\
           \end{array}
         \right), M_k=\left(
    \begin{array}{ccc}
      -\frac{\mu_k^1+\mu_k^2}{\mu_k^2} & \frac{\lambda_k^1+\lambda_k^2}{\mu_k^2} & \frac{\lambda_k^2}{\mu_k^2} \\
     1  &  0& 0\\
      0& 1 & 0 \\
    \end{array}
  \right).
$$
Then  $\mcp_k(a,b,b)$ and  $\mcp_k(a,b,b+1)$ follow from some standard procedure.
They could be expressed in terms of $\{M_i\}_{i\in\mathbb{Z}}.$
For $k\le i,$ let
$$f_k(i,i+1)=\mcp_k(-\infty,i+1,i+1)\text{ and } f_k(i,i+2)=\mcp_k(-\infty,i+1,i+2).$$
For $i\in\mbz,$ let
\begin{eqnarray*}
&&\alpha_{i,1}=\frac{q_i^1p_{i-1}^1}{1-q_{i-1}^1f_{i-2}(i-2,i-1)-q_{i-1}^2f_{i-3}(i-2,i-1)},\\
&&\alpha_{i,3}=\frac{q_i^1p_{i-1}^2}{1-q_{i-1}^1f_{i-2}(i-2,i-1)-q_{i-1}^2f_{i-3}(i-2,i-1)},\\
&&\beta_{i,1}=\frac{q_i^2f_{i-2}(i-2,i-1)p_{i-1}^1}{1-q_{i-1}^1f_{i-2}(i-2,i-1)-q_{i-1}^2f_{i-3}(i-2,i-1)},\\
&&\beta_{i,3}=\frac{q_i^2f_{i-2}(i-2,i-1)p_{i-1}^2}{1-q_{i-1}^1f_{i-2}(i-2,i-1)-q_{i-1}^2f_{i-3}(i-2,i-1)},\\
&&\gamma_{i,1}=\frac{q_{i+1}^2p_{i-1}^1}{1-q_{i-1}^1f_{i-2}(i-2,i-1)-q_{i-1}^2f_{i-3}(i-2,i-1)},\\
&&\gamma_{i,3}=\frac{q_{i+1}^2p_{i-1}^2}{1-q_{i-1}^1f_{i-2}(i-2,i-1)-q_{i-1}^2f_{i-3}(i-2,i-1)},\\
&&\ \alpha_{i,2}:=q_i^1-\alpha_{i,1}-\alpha_{i,3};\ \beta_{i,2}:=q_i^2-\beta_{i,1}-\beta_{i,3};\ \gamma_{i,2}:=q_{i+1}^2-\gamma_{i,1}-\gamma_{i,3}.
  \end{eqnarray*}
Set \begin{equation}\label{du}
u_1:=\z(\frac{\alpha_{1,1}}{\alpha_{1,1}+\alpha_{1,2}+\alpha_{1,3}},\frac{\alpha_{1,2}}{\alpha_{1,1}+\alpha_{1,2}+\alpha_{1,3}},\frac{\alpha_{1,3}}{\alpha_{1,1}+\alpha_{1,2}+\alpha_{1,3}},0,...,0\y)\in \mathbb R^9,
\end{equation}
and for $i\le 0$ define
$x_i=\frac{\alpha_{i,1}}{1-\alpha_{i,1}-\alpha_{i,2}-\beta_{i,1}-\beta_{i,2}},$
$y_i=\frac{\alpha_{i,2}}{1-\alpha_{i,1}-\alpha_{i,2}-\beta_{i,1}-\beta_{i,2}},$
$z_i=\frac{\beta_{i,1}}{1-\alpha_{i,1}-\alpha_{i,2}-\beta_{i,1}-\beta_{i,2}},$
$w_i=\frac{\beta_{i,2}}{1-\alpha_{i,1}-\alpha_{i,2}-\beta_{i,1}-\beta_{i,2}},$
 $1-v_i=\frac{\gamma_{i,3}}{\beta_{i+1,2}},$
 $s_i=\frac{\alpha_{i,3}}{\alpha_{i,3}+\beta_{i,3}}$ and
 $t_i=\frac{\gamma_{i,1}}{\gamma_{i,1}+\gamma_{i,2}}.$ Define also the matrices
\begin{equation}\label{q}
  Q_i=\left(
    \begin{array}{ccccccccc}
      x_i & y_i & 0 & z_i & w_i & 0 & 0 & 0 & 0 \\
       x_i & y_i & s_i & z_i & w_i & 1-s_i & 0 & 0 & 0 \\
       x_i & y_i & 0 & z_i & w_i & 0 & 0 & 0 & 0 \\
      x_i & y_i & 0 & z_i & w_i & 0 & t_i & 1-t_i  & 0 \\
      x_iv_i & y_iv_i & s_iv_i & z_iv_i & w_iv_i & (1-s_i)v_i & t_iv_i & (1-t_i) v_i & 1-v_i \\
    x_i & y_i & 0 & z_i & w_i & 0 & t_i & 1-t_i  & 0\\
      x_i & y_i & 0 & z_i & w_i & 0 & 0 & 0 & 0 \\
       x_i & y_i & s_i & z_i & w_i & 1-s_i & 0 & 0 & 0 \\
      x_i & y_i & 0 & z_i & w_i & 0 & 0 & 0 & 0 \\
    \end{array}
  \right).
\end{equation}
We have the following proposition which could be found in \cite{hwb}.
\begin{proposition} Suppose that $\gamma_R\ge 0.$ For $\tilde\mbp$-a.a. ${\tilde\omega},$ there exists a 9-type branching process $\{Z_n\},$ whose mean offspring matrices are as in (\ref{q}) and initial distribution mean is $u_1$ defined in (\ref{du}), such that in $\tilde P_{\tilde\omega}$-distribution, with $\mathbf v_1=(1,1,1,0,0,0,1,1,1)^T$ and $\mathbf v_2=(1,1,0,1,1,0,1,1,0)^T,$
  $$\overline U_k=Z_{k+1}\mathbf v_1+Z_k\mathbf v_2, \ k\le0.$$
Moreover, with the empty product being identity,
 \begin{equation}\label{enif}\begin{split}
&\tilde E_{\tilde\omega}(\overline U_k|\chi_{\overline T_1}=2)+\tilde E_{\tilde\omega}(\overline U_k|\chi_{\overline T_1}=1)\\
 &=\Big(\frac{\alpha_{1,1}}{\alpha_{1,1}+\alpha_{1,2}},\frac{\alpha_{1,2}}{\alpha_{1,1}+\alpha_{1,2}},1,0,...,0\Big)(Q_0Q_{-1}\cdots
Q_{k+1}\mathbf v_1+Q_0Q_{-1}\cdots Q_{k}\mathbf v_2).
\end{split}
  \end{equation}
\end{proposition}
Since after the $i$th visit of state $k,$  $\{N_t\}$  will wait here an exponentially distributed  time $\xi_{ki}$ with parameter $\mu_i^1+\mu_i^2+\lambda_i^1+\lambda_i^2,$ then given ${\tilde\omega},$ in $\tilde P_{\tilde\omega}$ distribution,
$$T_1=\sum_{k\le 0}\sum_{i=1}^{\overline U_k}\xi_{ki}.$$
Hence by Ward's equation, it follows from (\ref{enif}) that
\begin{eqnarray}\label{do}
 && \sum_{r=1}^2\tilde E_{\tilde\omega}(T_1|N_{T_1}=r)=\sum_{r=1}^2\sum_{k\le 0}\tilde E_{\tilde\omega}(\overline U_k|N_{T_1}=r)\tilde E_{\tilde\omega}(\xi_{k1})\no\\
&&= \sum_{k\le 0}\frac{1}{\mu_k^1+\mu_k^2+\lambda_k^1+\lambda_k^2}\no\\
&&\quad\quad\quad\times\Big(\frac{\alpha_{1,1}}{\alpha_{1,1}+\alpha_{1,2}},\frac{\alpha_{1,2}}{\alpha_{1,1}+\alpha_{1,2}},1,0,...,0\Big)\no\\
&&\quad\quad\quad\cdot(Q_0Q_{-1}\cdots
Q_{k+1}\mathbf v_1+Q_0Q_{-1}\cdots Q_{k}\mathbf v_2)\no\\
&&=:D({\tilde\omega})
\end{eqnarray}
and
\begin{equation}\label{pio}
  \begin{split}
  &\sum_{r=1}^2\sum_{k\le0}\tilde E_{\theta^{-k}{\tilde\omega}}\Big(\sum_{j=1}^{U_k}\xi_{kj}|N_{T_1}=r\Big)\\
&=\sum_{k= 0}^\infty\frac{1}{\mu_0^1+\mu_0^2+\lambda_0^1+\lambda_0^2}\\
&\quad\quad\quad\times\Big(\frac{\alpha_{k+1,1}}{\alpha_{k+1,1}+\alpha_{k+1,2}},\frac{\alpha_{k+1,2}}{\alpha_{k+1,1}+\alpha_{k+1,2}},1,0,...,0\Big)\\
&\quad\quad\quad\cdot(Q_kQ_{k-1}\cdots
Q_{1}\mathbf v_1+Q_kQ_{k-1}\cdots Q_{0}\mathbf v_2)\\
&=:\pi({\tilde\omega}).
\end{split}
\end{equation}
Substituting (\ref{do}) and (\ref{pio}) to (\ref{vp}),  we conclude that
\begin{equation*}
  \begin{split}
  v_{\tilde{\mathbb P}}&=\frac{\tilde\mbe \Big( \sum_{r=1}^2\sum_{k\le0}\tilde E_{\theta^{-k}{\tilde\omega}}\big(\sum_{j=1}^{U_k}\xi_{kj}|N_{T_1}=r\big)(2\lambda_0^2+\lambda_0^1-\mu_0^1-2\mu_0^2)\Big)}{\sum_{r=1}^R\tilde E(T_1|N_{T_1}=r)}\\
&=\frac{\tilde\mbe\z(\pi({\tilde\omega})(2\lambda_0^2+\lambda_0^1-\mu_0^1-2\mu_0^2)\y)}{\tilde\mbe(D({\tilde\omega}))}.
\end{split}
\end{equation*}
\qed

\section*{Appendix: On the existence of $\{N_t\}$}

Given $\tilde\omega,$ let $Q=(q_{ij})$ be a matrix with
$$q_{ij}=\left\{\begin{array}{ll}
                                       \lambda_i^r,&\text{ if } j=i+r,\ r=1,...,R; \\
                                       \mu_{i}^l,&\text{ if } j=i-l,\ l=1,...,L;\\
                                       -\big(\sum_{i=1}^L\mu_i^l+\sum_{r=1}^R\lambda_i^r), &\text{ if } j=i;\\
                                       0, &\text{ else.}
                                     \end{array}
\right.$$ Then  $Q$ is obviously a conservative Q-matrix. Note that under (C2), $Q$ is bounded from above. Hence the process $\{N_t\}$ exists (See for example Anderson \cite{ads}, Proposition 2.9, Chapter 2.). Next we give a condition which implies the existence of $\{N_t\}$ but is  weaker than (C2).
We have from classical argument that there exists at least one transition matrix $(\overline{p}_{\tilde\omega}(t,i,j))$ such that
\begin{equation}\label{uniq}
  \lim_{t\rightarrow0}\frac{\overline{p}_{{\tilde\omega}}(t,i,j)-\delta_{ij}}{t}=q_{ij},\ i,j\in\mathbb Z.
\end{equation}

Let $(p_{\tilde\omega}(h,i,j))$ be a standard transition matrix satisfying (\ref{uniq}). Let $\{N_t\}$ be a continuous time Markov chain with transition matrix  $(p_{\tilde\omega}(h,i,j)).$
Let  $\tau_0=0$ and
define $\tau_{n}:=\inf\{t>\tau_{n-1}:N_t\neq N_{\tau_{n-1}}\}$ recursively for $n\ge 1.$ Then $\tau_n,n\ge0$ are the consecutive discontinuities of the process $\{N_t\}.$
If \begin{equation*}\label{zz}\tilde\mbp\Big(\sum_{l=1}^L\mu_0^l+\sum_{r=1}^R\lambda_0^r>0\Big)=1,\end{equation*} then $\tilde P$-a.s., $\tau_n<\infty, n\ge0.$ Let $\chi_n=N_{\tau_n},$ for $n\ge0.$ The process $\{\chi_n\}_{n\ge0}$ is known as the embedded process of $\{N_t\}.$
\begin{proposition}
  Suppose  that $\tilde\mbp\big(\sum_{l=1}^L\mu_0^l+\sum_{r=1}^R\lambda_0^r>0\big)=1$ and
$$ \tilde{\mathbb P}\Big(\sum_{n=1}^\infty{\Big(\max_{1\le k\le R}\Big\{\sum_{r=1}^R\lambda_{nR-k}^r+\sum_{l=1}^L\mu_{nR-k}^l\Big\}\Big)^{-1}}=\infty\Big)=1,$$
 $$\tilde{\mathbb P}\Big(\sum_{n=-\infty}^0{\Big(\max_{1\le k\le L}\Big\{\sum_{r=1}^R\lambda_{nL-k}^r+\sum_{l=1}^L\mu_{nL-k}^l\Big\}\Big)^{-1}}=\infty\Big)=1.$$ Then for $\tilde\mbp$-a.a. ${\tilde\omega},$ there is a unique transition matrix $(p_{\tilde\omega}(h,i,j))$ which satisfies (\ref{uniq}).
\end{proposition}
\proof Considering the above defined $\{N_t\},$  since $\tilde\mbp\big(\sum_{l=1}^L\mu_0^l+\sum_{r=1}^R\lambda_0^r>0\big)=1,$ we have $\tilde P$-a.s., $\tau_n<\infty, n\ge 1.$ By the classical argument of the uniqueness of the $Q$-process, if $$\tilde P(\lim_{n\rto}\tau_n=\infty)=1,$$ then the minimal solution $p_{{\tilde\omega}}(t,i,j)$ is the unique $Q$-transition matrix. Let $q_{i}=-q_{ii},\ i\in\mbz.$
If \begin{equation}\label{tin}
  \tilde P\Big( \sum_{n=0}^\infty q^{-1}_{\chi_n}=\infty\Big)=1,
\end{equation} then we have (see Chung \cite{chung}, Theorem 1 in II.19) that $\tilde P(\lim_{n\rto}\tau_n=\infty)=1.$
Next we show (\ref{tin}). In fact, if the process $\{\chi_n\}_{n\ge 0}$ is recurrent or transient to the right, it must visit at least one state of each of the sets $B_n:=\{nR-k\}_{k=1}^R,n=1,2,....$ Then $\tilde P$-a.s.,
$$\sum_{n=0}^\infty q^{-1}_{\chi_n}\ge \sum_{n=1}^\infty{\Big(\max_{1\le k\le R}\Big\{\sum_{r=1}^R\lambda_{nR-k}^r+\sum_{l=1}^L\mu_{nR-k}^l\Big\}\Big)^{-1}}=\infty.$$
Else if the process $\{\chi_n\}_{n\ge 0}$ is transient to the left, it must visit at least one state of each of the sets $A_n:=\{nL-k\}_{k=1}^L,n=0,-1,-2,....$  It follows that $ P$-a.s., $$\sum_{n=0}^\infty q^{-1}_{\chi_n}\ge \sum_{n=-\infty}^0{\Big(\max_{1\le k\le L}\Big\{\sum_{r=1}^R\lambda_{nL-k}^r+\sum_{l=1}^L\mu_{nL-k}^l\Big\}\Big)^{-1}}=\infty.$$ Consequently (\ref{tin}) follows.\qed

\noindent{\large{\bf \large Acknowledgements:}} The author would like to thank Professor Wenming Hong for his useful comments on the paper.


TITLE: Proof inequality and series using Induction
QUESTION [1 upvotes]: How can I prove that
$$
\sum_{k=1}^n \frac{1}{k^2} ≤ 2 - \frac {1}{n}
$$
using induction?
I first tried to prove this for $P(1)$, so:
$$
\sum_{k=1}^n \frac{1}{1^2} ≤ 2 - \frac {1}1
$$
is proved. I then assume that $P(n)$ is proved.
Now I have to prove the same inequality for $P(n+1)$
$$
\sum_{k=1}^{n+1} \frac{1}{k^2} ≤ 2 - \frac {1}{n + 1}
$$
I concentrate on the argument on the left and make some changes:
$$
\sum_{k=1}^{n+1} \frac{1}{k^2}
$$
$$
\sum_{k=1}^{n} \frac{1}{k^2} + \frac {1}{n+1}
$$
And since $P(n)$ is proved I can change it like this
$$
2 - \frac {1}n + \frac {1}{n+1}
$$
But I can't prove it, cause in the end I get a wrong inequality.
Someone can please tell me what I'm doing wrong?

REPLY [1 votes]: This is a non-induction solution but included for reference only.
$$\sum_{k=1}^n\frac 1{k^2}\leq 1+\int_1^n \frac 1{k^2}=1+\left[-\frac 1k\right]_1^n=2-\frac 1n$$


TITLE: A possible disproof for the Descartes-Frenicle-Sorli Conjecture on odd perfect numbers
QUESTION [2 upvotes]: Let $\sigma(z)$ denote the sum of the divisors of $z \in \mathbb{N}$, the set of positive integers.  Denote the deficiency of $z$ by $D(z):=2z-\sigma(z)$, and the sum of the aliquot divisors by $s(z):=\sigma(z)-z$.  Finally, let the abundancy index of $z$ be denoted by $I(z):=\sigma(z)/z$.
If $n$ is odd and $\sigma(n)=2n$, then $n$ is said to be an odd perfect number.  Euler proved that an odd perfect number, if one exists, must have the form $n=p^k m^2$, where $p$ is the special / Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Starting from the fundamental equality
$$\frac{\sigma(m^2)}{p^k}=\frac{2m^2}{\sigma(p^k)}$$
one can derive
$$\frac{\sigma(m^2)}{p^k}=\frac{2m^2}{\sigma(p^k)}=\gcd(m^2,\sigma(m^2))$$
so that we ultimately have
$$\frac{D(m^2)}{s(p^k)}=\frac{2m^2 - \sigma(m^2)}{\sigma(p^k) - p^k}=\gcd(m^2,\sigma(m^2))$$
The following is copied verbatim from this answer to a closely related question:

It turns out that it is possible to express $\gcd(m^2, \sigma(m^2))$ as an integral linear combination of $m^2$ and $\sigma(m^2)$, in terms of $p$ alone.
To begin with, write
  $$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)}{\sigma(p^{k-1})}=\frac{(2m^2 - \sigma(m^2))(p-1)}{p^k - 1}.$$
Now, using the identity
  $$\frac{A}{B}=\frac{C}{D}=\frac{A-C}{B-D},$$
  where $B \neq 0$, $D \neq 0$, and $B \neq D$, we obtain
  $$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)-(2m^2 - \sigma(m^2))(p-1)}{p^k - (p^k - 1)},$$
  from which we get
  $$\gcd(m^2,\sigma(m^2))=\sigma(m^2)-(2m^2 - \sigma(m^2))(p-1)=2m^2 - p(2m^2 - \sigma(m^2)) = 2m^2 - pD(m^2),$$
  or equivalently,
  $$\gcd(m^2,\sigma(m^2))=2(1 - p)m^2  + p\sigma(m^2).$$

Here is my Question #1:

Is it possible to obtain the following equation?
$$2m^2 - \sigma(m^2)=\gcd(m^2,\sigma(m^2))=2(1 - p)m^2  + p\sigma(m^2)$$

MY ATTEMPT
Suppose that $2m^2 - \sigma(m^2)=\gcd(m^2,\sigma(m^2))$.  Then it follows that $m^2$ is deficient-perfect, which then implies that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.  But notice that the variable $k$ is not present in the last equation above, so that it is not really relevant to consider the value of $k$ here.
By assumption, we have
$$2m^2 - \sigma(m^2)=\gcd(m^2,\sigma(m^2))=2(1 - p)m^2  + p\sigma(m^2),$$
from which it follows (?) that
$$\bigg(2(1-p)=2\bigg) \land \bigg(-1 = p\bigg),$$
which results (?) in the contradiction
$$\bigg(0 = p\bigg) \land \bigg(-1 = p\bigg).$$
Note that either conjunct is already a contradiction, as $p$ being the special / Euler prime it ought to satisfy $p \equiv 1 \pmod 4$.
We therefore conclude that $m^2$ is not deficient-perfect.  This is equivalent to $k \neq 1$.
Here is my Question #2:

Is this proof correct?  Is it even logically sound?

REPLY [1 votes]: Here is a way to resolve the apparent contradiction:
By assumption, we have
$$2m^2 - \sigma(m^2) = 2(1 - p)m^2 + p\sigma(m^2) = 2m^2 - pD(m^2).$$
Cancelling $2m^2$ from both sides of the LHS and RHS of the last equation, we get
$$\frac{\sigma(m^2)}{p}=D(m^2)=2m^2 - \sigma(m^2)$$
from which it follows that
$$I(m^2)=\frac{\sigma(m^2)}{m^2}=\frac{2p}{p+1}.$$
The last equation then implies that $k=1$, which is expected from the equation
$$2m^2 - \sigma(m^2) = \gcd(m^2,\sigma(m^2)).$$


text:
\begin{document}

\title[Global weak solutions for a two-component Novikov system]
{Global weak solutions for a two-component Novikov system}

\author{Zhigang Li}
\address{College of sciences\\China University of Mining and Technology\\100083, Beijing, People's Republic of China}
\email{lzgcumtb@163.com}

\keywords{two-component Novikov system, weak solutions, global existence}

\begin{abstract}
In this paper, we mainly consider about the existence and uniqueness of global weak solutions for the two-component Novikov system. We first recall some results and definitions of strong solutions and weak solutions for the system, then by using the method of approximation of smooth solutions, we prove the existence and uniqueness of global weak solutions of the system.

Correspondence should be addressed to Zhigang Li; lzgcumtb@163.com
\end{abstract}

\maketitle

\section{Introduction}
In this paper, we consider the following two-component Novikov system (2NS) :
\begin{equation*}
\begin{cases}
m_t+3u_xvm+uvm_x=0,\\
n_t\ +3v_xun\ +uvn_x\ =0,\\
m=u-u_{xx}, n=v-v_{xx}.
\end{cases}
\end{equation*}
This system firstly appeared in reference [1] by Geng and Xue, who derived by arising the zero curvature equation
$$U_t-V_x+[U,V]=0$$
which equivalent to the compatibility condition of the $3 \times 3$ spectral system
$$\Phi_x=U\Phi,\ \ \ \Phi_t=V\Phi,$$
where the spacial part $U$ is
\begin{equation*}
U=\left(\begin{matrix}
0 & \lambda m & 1\\
0 & 0 & \lambda n\\
1 & 0 & 0
\end{matrix}\right),
\end{equation*}
and the time part $V$ is
\begin{equation*}
V=\left(\begin{matrix}
-u_xv+\frac{1}{3\lambda^2} & -\lambda uvm +\frac{u_x}{\lambda} & u_xv_x\\
-\frac{v}{\lambda} & u_xv-uv_x-\frac{2}{3\lambda^2} & -\lambda uvn-\frac{v_x}{\lambda}\\
-uv & \frac{u}{\lambda} & uv_x+\frac{1}{3\lambda^2}
\end{matrix}\right),
\end{equation*}
here $\lambda \in \mathbb{R}$ is spectral parameter.

The authors proved such system is integrable, and they also supplied many significant results, such as infinite many conserved quantities, Hamiltonian structure, and explicit multi-peakon traveling wave solutions. It should be noted that the bi-Hamiltonian structure was found by Li and Liu[2], which means 2NS is also integrable in Liouville sense. The single peakon solutions of 2NS are given by
$$u_c(t,x)=v_c(t,x)=\sqrt{c}e^{-|x-ct|},$$
and the periodic case are given by
$$u_p(t,x)=v_p(t,x)=\frac{\sqrt{c}}{\cosh \pi}\cosh\left(x-ct-2\pi[\frac{x-ct}{2\pi}]-\pi\right),$$
with periodic $2\pi$. More details on the derivation of peakons can be found in [3].

As 2NS is a multi-component system, we can exploit some reductions to reduce it into some single-component equations. A nature reduction is taking $u=v$, which is just the Novikov equation (NE) [4] with cubic nonlinearities,
\begin{equation*}
m_t+u^2m_x+3uu_xm=0,\ \ \ m=u-u_{xx},
\end{equation*}
it was firstly derived by Novikov via the symmetry classification method. The integrability of NE was shown by Hone and Wang, who proved that it is integrable both in Lax and Liouville sense. The Lax pair of NE can be obtained from one of 2NS by choosing $u=v$. More over, it also admits bi-Hamiltonian structure, infinitely many conserved quantities [5]. They also showed that NE is associated to the negative flow in the Sawada-Kotera hierarchy. The local well-posedness and ill-posedness of such equation can refer to [6-7], while for periodic case and for $s > \frac{5}{2}$, well-posedness had been proved by Tiglay [8], and the blow-up phenomena has studied in [9]. The global weak solutions of NE should be referred to [10].

Another important reduction of 2NS is Degasperis-Procesi equation (DPE) [11] if we take $v=1$,
$$m_t+um_x+3u_xm=0, \ \ \ m=u-u_{xx}.$$
It was proposed by Degasperis and Procesi, who considered the asymptotic integrability to the following dispersive PDE,
\begin{equation*}
u_t- \alpha^2u_{xxt} + \gamma u_{xxx} +c_0u_x =(c_1u^2+c_2u_x^2+c_3uu_{xx})_x.
\end{equation*}
In fact, if we change the coefficient of term $u_xm$ to 2 in DPE, it is just the famous Camassa-Holm equation (CHE) [12-14]. DPE is another integrable equation of b-family for $b=3$. As the same as CHE, DPE also arises bi-Hamiltonian structure, infinite many conserved quantities and peaked solutions, and it is connected with a negative flow in the Kaup-Kupershmidt hierarchy throw reciprocal transformation [15]. The well-posedness and stability of DPE have been shown in [16-17]. The global weak solutions of CHE has been studied in [18].

Himonas and Mantzavinos have illustrated 2NS is local well-posed if $u_0,v_0 \in H^s$ with $s>\frac{3}{2}$ in convolution form [19], which is equivalent to $m_0,n_0 \in H^s$ for $s>\frac{1}{2}$, and we have just proved if the initial data $m_0$ and $n_0$ do not change sign, then the corresponding strong solution $(m,n)$ exists globally in time [20]. While 2NS is ill-posed if $s<\frac{3}{2}$, which has been proven by Himonas, Holliman and Kenig in [21]. Thus, a nature question is how to modify the initial condition to guarantee there exists a unique weak solution globally in time.

The structure of our paper is organized as follows. In Section 2, we recall some basic definitions and lemmas which will be used in the sequel. We study the existence and uniqueness of the global weak solution of 2NS In section 3.

\textbf{Notations:} For $x$ variable, all the function space is over $\mathbb{R}$, and we drop $\mathbb{R}$ in our notations of function spaces if there is no ambiguity. $C_T$ represents a positive constant which depends only on $T$ and some norms of the initial data, and may be changed from line to line.

\section{Strong solutions and some priori estimates}
In this section, we recall some crucial lemmas and global existence of strong solutions to 2NS. For convenience, we also offer some priori estimates in this section. Note that the kernel of $\mathcal{S}^{-2}$ operator $(1-\partial_x^2)^{-1}$ is $g(x)=\frac{1}{2}e^{-|x|}$, which means $(1-\partial_x^2)^{-1}f=g*f$ for any function $f \in L^2$. Thus we can rewrite 2NS into the following convolution form,
\begin{equation}
\begin{cases}
u_t+uu_xv+g(x)*(3uu_xv+u_x^2v_x+uu_{xx}v_x)+\partial_x g(x)*uu_xv_x=0,\\
v_t+uvv_x+g(x)*(3uvv_x+u_xv_x^2+u_xvv_{xx})+\partial_x g(x)*u_xvv_x=0,
\end{cases}
\end{equation}

\begin{definition}
If $u(t,x),v(t,x) \in [C([0,T); H^s)\bigcap C^1([0,T); H^{s-1})]^2$ with $s>\frac{3}{2}$ is a solution to 2NS, then $u(t,x), v(t,x)$ is called a strong solution to 2NS.
\end{definition}

\begin{definition}
Assume $(u_0,v_0) \in H^s \times H^s$ with $s \in[0,\frac{3}{2}]$. If $(u(t,x),v(t,x))$ satisfies
\begin{equation*}
\begin{aligned}
&\int_0^T\int_\mathbb{R}\left(u\varphi_t-[uvu_x+g*(3uvu_x+u_x^2v_x+uu_{xx}v_x)+g_x*uu_xv_x]\varphi\right){\rm d}x{\rm d}t+\int_\mathbb{R}u_0\varphi(t,x){\rm d}x=0,\\
&\int_0^T\int_\mathbb{R}\left(v\varphi_t-[uvv_x+g*(3uvv_x+u_xv_x^2+u_xvv_{xx})+g_x*u_xvv_x]\varphi\right){\rm d}x{\rm d}t+\int_\mathbb{R}v_0\varphi(t,x){\rm d}x=0,\\
\end{aligned}
\end{equation*}
for any test function $\varphi \in C_c^\infty([0,T\times \mathbb{R})$, then $u(t,x), v(t,x)$ is called a weak solution tow 2NS.
\end{definition}

\begin{proposition}$[10]$
~

(1) Every strong solution is a weak solution.

(2) If $(u,v)$ is a weak solution and $u(t,x),v(t,x) \in [C([0,T); H^s)\bigcap C^1([0,T); H^{s-1})]^2$ with $s>\frac{3}{2}$, then it is a strong solution.
\end{proposition}

\begin{lemma}$[18]$
Let $T>0$. If
$$f,g \in L^2((0,T);H^1)\ \ \  and\ \ \ \frac{{\rm d}f}{{\rm d}t},\frac{{\rm d}g}{{\rm d}t} \in L^2((0,T);H^{-1}),$$
then $f$, $g$ are a.e. equal to a function continuous from $[0,T]$ into $L^2$ and
$$<f(t),g(t)>-<f(s),g(s)>=\int_s^t\Big\langle\frac{{\rm d}f(\tau)}{{\rm d}\tau},g(\tau)\Big\rangle {\rm d}\tau+\int_s^t\Big\langle\frac{{\rm d}g(\tau)}{{\rm d}\tau},f(\tau)\Big\rangle {\rm d}\tau$$
for all $s,t \in [0.T]$, where $< , >$ denote the duality paring between $H^1$ and $H^{-1}$.
\end{lemma}

Throughout this paper, let $\{\rho_n\}_{n=1}^\infty$ denote the mollifiers
$$\rho_n(x)=\left(\int_\mathbb{R}\rho(x){\rm d}x\right)^{-1}n\rho(nx),\ \ \ x \in \mathbb{R}, n \in \mathbb{N}^*,$$
where $\rho \in C_c^\infty$ is defined by
\begin{equation*}
\rho(x)=\begin{cases}e^{\frac{1}{x^2-1}},\ \ \ &|x|<1,\\
0,&|x| \ge 1.
\end{cases}
\end{equation*}

Next, we recall some crucial approximations results of convolution calculus.
\begin{lemma}$[18]$
Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ be uniformly continuous and bounded. If $g\in L^p$ with $1 \le p \le \infty$, then we have
$$[\rho_n *(fg)-(\rho_n*f)(\rho_n*g)]\rightarrow 0,$$
as $n\rightarrow \infty$ in $L^p$.
\end{lemma}

\begin{lemma}$[18]$
Suppose $f(t,\cdot) \in W^{1,1}$ is uniformly bounded in $W^{1,1}$ for all $t \in \mathbb{R}_+$. Then for a.e. $t \in \mathbb{R}_+$ and $1 \le p <\infty$, we have
$$\frac{1}{p}\frac{\rm d}{{\rm d}t}\int_\mathbb{R}|\rho_n*f|^p{\rm d}x=\int_\mathbb{R}|\rho_n*f|^{p-1}(\rho_n*f_t){\rm sgn}(\rho_n*f){\rm d}x,$$
and
$$\frac{1}{p}\frac{\rm d}{{\rm d}t}\int_\mathbb{R}|\rho_n*f_x|^p{\rm d}x=\int_\mathbb{R}|\rho_n*f_x|^{p-1}(\rho_n*f_{xt}){\rm sgn}(\rho_n*f_x){\rm d}x.$$
\end{lemma}

\begin{lemma}$[20]$
Suppose $u_0,v_0 \in H^s$, with $s \ge 3$, $m_0=(1-\partial_x^2)u_0$ and $n_0=(1-\partial_x^2)v_0$ are non-negative, then 2NS has a unique global strong solution
$$(u(t,x),v(t,x)) \in C(\mathbb{R}_+;H^s)\cap C^1(\mathbb{R}_+;H^{s-1}).$$
Moreover, $E(u,v)=\int_\mathbb{R}\Big(u(t,\cdot)v(t,\cdot)+u_x(t,\cdot)v_x(t,\cdot)\Big){\rm d}x$ is a conservation law.
\end{lemma}

At the end of this section, we offer some priori estimates for $u$, $v$ and potentials $m$, $n$.
\begin{lemma}$[20]$
Suppose $u_0,v_0 \in H^3$ and $m_0, n_0 \in L^p$ with $1 \le p \le \infty$ are non-negative, then there exists some constant $C>0$ only depends on the maximal existence time $T>0$ and initial data $(u_0,v_0)$, such that
\begin{equation*}
\begin{aligned}
(i)&\|u_x(t,\cdot)\|_{L^\infty} \le \|u(t,\cdot)\|_{L^\infty} \le \frac{\sqrt{2}}{2}\|u(t,\cdot)\|_{H^1} \le \frac{\sqrt{2}}{2}\|u_0\|_{H^1}e^{E(u_0,v_0)}t,\\
(ii)&\|v_x(t,\cdot)\|_{L^\infty} \le \|v(t,\cdot)\|_{L^\infty} \le \frac{\sqrt{2}}{2}\|v(t,\cdot)\|_{H^1} \le \frac{\sqrt{2}}{2}\|v_0\|_{H^1}e^{E(u_0,v_0)}t,\\
(iii)&\|u(t,\cdot)\|_{L^p},\|u_x(t,\cdot)\|_{L^p} \le \|m(t,\cdot)\|_{L^p} \le e^{Ct}\|m_0\|_{L^p},\\
(iv)&\|v(t,\cdot)\|_{L^p},\|v_x(t,\cdot)\|_{L^p} \le \|n(t,\cdot)\|_{L^p} \le e^{Ct}\|n_0\|_{L^p}.
\end{aligned}
\end{equation*}
\end{lemma}

\begin{proof}
For (i) and (ii), one can follow the proof of lemma 2.7 in [x]. Now we illustrate the third inequality. By virtue of the first equation of 2NS, we have
\begin{equation*}
\begin{aligned}
\frac{\rm d}{{\rm d}t}\int_\mathbb{R}m^p{\rm d}x&=\int_\mathbb{R}m^{p-1}m_t{\rm d}x=-\int_\mathbb{R}m^{p-1}(uvm_x+3u_xmv){\rm d}x=\int_\mathbb{R}(uv_x-2u_xv)m^p{\rm d}x\\
& \le (2\|u_x\|_{L^\infty}\|v\|_{L^\infty}+\|u\|_{L^\infty}\|v_x\|_{L^\infty})\int_\mathbb{R}m^p{\rm d}x\le C_T\int_\mathbb{R}m^p{\rm d}x,
\end{aligned}
\end{equation*}
and take advantage of Young's inequality and Gronwall's inequality, we obtain
\begin{equation*}
\begin{aligned}
\|u(t,\cdot)\|_{L^p}&=\|(g*m)(t,\cdot)\|_{L^p} \ \le \ \|g(t,\cdot)\|_{L^1}\|m(t,\cdot)\|_{L^p}\ \le e^{Ct}\|m_0\|_{L^p},\\
\|u_x(t,\cdot)\|_{L^p}&=\|(g_x*m)(t,\cdot)\|_{L^p} \le \|g_x(t,\cdot)\|_{L^1}\|m(t,\cdot)\|_{L^p} \le e^{Ct}\|m_0\|_{L^p}.
\end{aligned}
\end{equation*}
Similarly, one can check that (iv) is also holds.
\end{proof}

\section{Global weak solution}
Inspired by Zheng and Yin [22], now we state our main results as following theorem.
\begin{theorem}
Suppose $m_0,n_0 \in L^1\cap L^2$ and $m_0,n_0 \ge 0,\forall x\in\mathbb{R}.$ Then 2NS has a unique global weak solution $(u,v) \in C(\mathbb{R}_+;H^1)\cap C^1(\mathbb{R}_+;L^2)$, and the potential $(m,n) \in L_{loc}^\infty(\mathbb{R}_+;L^1\cap L^2)$.
\end{theorem}

\begin{proof}

\textbf{Step 1.} Let $u_0=(1-\partial_x^2)^{-1}m_0$, $v_0=(1-\partial_x^2)^{-1}n_0$. As $m_0,n_0 \in L^2$, by virtue of Lemma 2.8 with $p=2$, we have $u_0,v_0 \in H^1$. Define $u_0^k=\rho_k*u_0$ and $v_0^k=\rho_k*v_0$ belong to $H^\infty$ for $k \in \mathbb{N}^*$. By the convolution approximation, it is clearly that
$$u_0^k \rightarrow u_0\ \ \ {\rm and}\ \ \ v_0^k \rightarrow v_0\ \ \ {\rm in}\  H^1\ \ \ {\rm as}\ \ \ k\rightarrow \infty.$$
Noting that $m_0^k=u_0^k-u_{0,xx}^k=\rho_k*m_0 \ge 0$, and $n_0^k=v_0^k-v_{0,xx}^k=\rho_k*n_0 \ge 0$, for $k\in\mathbb{N}^*$, it is obvious that
$$m_0^k \rightarrow m_0\ \ \ {\rm and}\ \ \ n_0^k \rightarrow n_0\ \ \ {\rm in}\ L^1\cap L^2\ \ \ {\rm as}\ \ \ k\rightarrow \infty.$$
As $m_0$ and $n_0$ are smooth function in $H^\infty$, through Lemma 2.7 we have unique global strong solutions $m^k,n^k$ of 2NS with initial data $m_0^k,n_0^k$, and $u^k=g*m^k$, $v^k=g*n^k$.

\textbf{Step 2.} For fixed $T>0$, by Lemma 2.8 and the fact $\|\rho_k*f\|_{W^{k,p}}\le\|\rho_k\|_{L^1}\|f\|_{W^{k,p}}$, we see $u^k,v^k$ is uniformly bounded in $L^\infty([0,T);H^1\cap W^{1,\infty})$, and $m^k,n^k$ is uniformly bounded in $L^\infty([0,T);L^1\cap L^2)$. In order to obtain the uniform boundedness of solution $u^k,v^k$ in $H^1((0,T)\times\mathbb{R})$ with initial data $u_0^k,v_0^k$, it remains to estimate $\|u_t^k(t,\cdot)\|_{L^2}$ and $\|v_t^k(t,\cdot)\|_{L^2}$. Rewriting the first equation of 2NS, we have
\begin{equation}
u_t^k+g_x*u^kv^km^k+2g*u_x^kv^km^k-g*u^kv_x^km^k=0.
\end{equation}
Taking advantage of Young's inequality yields
\begin{equation}
\begin{aligned}
&\|u_t^k(t,\cdot)\|_{L^2}\\
\le &\|(g_x*u^kv^km^k)(t,\cdot)\|_{L^2}+2\|(g*u_x^kv^km^k)(t,\cdot)\|_{L^2}+\|(g*u^kv_x^km^k)(t,\cdot)\|_{L^2}\\
\le &\|m^k(t,\cdot)\|_{L^2}\left(\|u^k(t,\cdot)\|_{L^\infty}\|v^k(t,\cdot)\|_{L^\infty}+2\|u_x^k(t,\cdot)\|_{L^\infty}\|v^k(t,\cdot)\|_{L^\infty}+\|u^k(t,\cdot)\|_{L^\infty}\|v_x^k(t,\cdot)\|_{L^\infty}\right)\\
\le &2\|m^k(t,\cdot)\|_{L^2}\|u^k(t,\cdot)\|_{W^{1,\infty}}\|v^k(t,\cdot)\|_{W^{1,\infty}},
\end{aligned}
\end{equation}
and the boundary of $\|v_t^k\|_{L^2}$ can be obtained in s similar way. Combining with the fact
$$\|u^k(t,\cdot)\|_{H^1} \le \|u(t,\cdot)\|_{H^1},\ \ \ \|v^k(t,\cdot)\|_{H^1} \le \|v(t,\cdot)\|_{H^1},$$
and Lemma 2.8, we see sequence $\{u^k,v^k\}_{k\in\mathbb{N}^*}$ is uniformly bounded in $H^1((0,T)\times\mathbb{R})$. Therefore, there exists a subsequence such that
\begin{equation}
\begin{aligned}
&(u^{k_j},v^{k_j}) \rightarrow (u,v)\ {\rm weakly\ in}\ H^1((0,T)\times\mathbb{R}),\ {\rm as}\ k_j\rightarrow \infty,\\
&(u^{k_j},v^{k_j}) \rightarrow (u,v)\ {\rm a.e.\ on}\ (0,T)\times\mathbb{R}, \ {\rm as}\ k_j\rightarrow \infty,
\end{aligned}
\end{equation}
for some $u,v\in H^1((0,T)\times\mathbb{R})$. By Lemma 2.7 and Lemma 2.8, we have that for fixed $t \in (0,T)$,
$$\mathbb{V}[u_x^{k_j}(t,\cdot)]=\|u_{xx}^{k_j}(t,\cdot)\|_{L^1}\le \|u^{k_j}(t,\cdot)\|_{L^1}+\|m^{k_j}(t,\cdot)\|_{L^1} \le 2\|m^{k_j}\|_{L^\infty((0,T);L^1)},$$
and
\begin{equation*}
\begin{aligned}
\|u_x^{k_j}(t,\cdot)\|_{L^\infty} &\le \frac{\sqrt{2}}{2}\|u^{k_j}(t,\cdot)\|_{H^1}\le \frac{\sqrt{2}}{2}\|u_0^{k_j}\|_{H^1}e^{E(u_0^{k_j},u_0^{k_j})t}\\
&\le \frac{\sqrt{2}}{2}\|u_0^{k_j}\|_{H^1}e^{\|u_0^{k_j}\|_{H^1}\|v_0^{k_j}\|_{H^1}t} \le \frac{\sqrt{2}}{2}\|u_0\|_{H^1}e^{\|u_0\|_{H^1}\|v_0\|_{H^1}t},
\end{aligned}
\end{equation*}
where $\mathbb{V}(f)$ is the total variation of function $f\in BV$. Applying Helly's theorem and the diagonal process, we obtain that there exists a subsequence, denoted again by $u_x^{k_j}(t, x)$, which converges pointwise for all $x \in \mathbb{R}$ and $t$ in a countable dense subset of $(0, T)$. In order to prove  $u_x^{k_j}(t, x)$ converges for a.e. $x$ and every $t \in (0, T)$, as $k_j \rightarrow \infty$, it will suffice to prove that $\|u_{xt}^{k_j}\|_{L^\infty((0,T);L^1)}$ is uniformly bounded. Differentiating equation (3) with respect to $x$, and together with the identity $\partial_x^2g*f=g*f-f$, we get
$$u_{xt}^k+g*u^kv^km^k-u^kv^km^k+2g_x*u_x^kv^km^k-g_x*u^kv_x^km^k=0.$$
Similar to the argument of (3), it is easy to check that there exists some constant $C_T>0$ such that
$$\|u_{xt}^{k_j}\|_{L^\infty((0,T);L^1)}\le C_T.$$
Thus we claim there exists some function such that $u_x^{k_j}(t,x)\rightarrow r(t,x)$ for a.e. $x\in \mathbb{R}$ and $\forall t\in (0,T)$ as $k_j\rightarrow \infty$. Since for almost all $(t,x)\in(0,T)\times\mathbb{R}$, $u_x^{k_j} \rightarrow u_x$ in $\mathcal{D}'((0,T)\times \mathbb{R})$, it follows that $r(t,x)=u_x(t,x)$ for a.e. $(t,x)\in(0,T)\times\mathbb{R}$. With a similar argument, we can also deduce that $v_x^{k_j} \rightarrow v_x$ a.e. on $(0,T)\times\mathbb{R}$. Therefore, we have
\begin{equation}
(u_x^{k_j}, v_x^{k_j})\rightarrow (u_x,v_x)\ \ \ {\rm a.e.\ on}\ (0,T)\times\mathbb{R}.
\end{equation}
Moreover, $u_x,v_x \in L^\infty((0,T)\times\mathbb{R})$.

Since $(m^k,n^k)$ is uniformly bounded in $L^\infty((0,T);L^1\cap L^2)$, there exists a subsequence such that
\begin{equation}
(m^{k_j},n^{k_j}) \rightarrow (m,n)\ \ \ {\rm weakly\  *\  in}\ L^\infty((0,T);L^1\cap L^2),\ \ \ {\rm as}\ k_j\rightarrow \infty.
\end{equation}

Next we show $m,n$ satisfy 2NS in distribution sense. We only deal with the term $u_xvm$ as an example, and the others are similar. By virtue of (5)-(7) and Lemma 2.5 , we see for any test function $\phi \in C_0^\infty((0,T)\times \mathbb{R})$,
\begin{equation*}
\begin{aligned}
&\int_0^T\int_\mathbb{R}\left(u_x^{k_j}v^{k_j}m^{k_j}-u_xvm\right)\phi{\rm d}x{\rm d}t\\
=&\int_0^T\int_\mathbb{R}m^{k_j}[u_x^{k_j}v^{k_j}-(u_xv)^{k_j}+(u_xv)^{k_j}-u_xv]\phi{\rm d}x{\rm d}t+\int_0^T\int_\mathbb{R}(m^{k_j}-m)u_xv\phi{\rm d}x{\rm d}t\\
\le&\|m^{k_j}\|_{L^\infty((0,T);L^2)}\|[u_x^{k_j}v^{k_j}-(u_xv)^{k_j}]\phi+[(u_xv)^{k_j}-u_xv]\phi\|_{L^1((0,T);L^2)}\\
&+\|m^{k_j}-m\|_{L^\infty((0,T);L^2)}\|u_xv\phi\|_{L^1((0,T);L^2)}\rightarrow 0,\ \ \ {\rm as}\ k_j\rightarrow \infty.
\end{aligned}
\end{equation*}

\textbf{Step 3.} Since for any fixed $T>0$, $(u^k,v^k)$ is bounded in $L^\infty((0,T);H^1)$, and $(u_t^k,v_t^k)$ is bounded in $L^\infty((0,T);L^2$ respectively, hence the map $t\rightarrow (u^k(t\,cdot), v^k(t,\cdot)) \in H^1$  is weakly equicontinuous on $[0,T]$. By virtue of Arzela-Ascoli theorem, we know there exists a subsequence of $(u^k(t,\cdot),v^k(t,\cdot))$ which converges weakly in $H^1$, uniformly in $t \in [0,T]$. The limit function is $(u,v)$. Since $T>0$ is arbitrary, we claim that
$$(u,v)\in C_w(\mathbb{R}_+;H^1).$$

In order to prove $u,v \in C(\mathbb{R}_+; H^1)$, it remains to show that $\|u(t,\cdot)\|_{H^1}$ and $\|v(t,\cdot)\|_{H^1}$ are continuous on $\mathbb{R}_+$. Since $(m,n)$ is a solution of 2NS in distributional sense, we see for a.e. $t\in \mathbb{R}_+$,
$$\rho_k*m_t+\rho_k*(uvm)_x-\rho_k*(uv_xm)+2\rho_k*(u_xvm)=0.$$
Multiplying with $\rho_k*u$ and integrating with respect to $x$ on $\mathbb{R}$, through Lemma 2.4, we have
\begin{equation}
\begin{aligned}
\frac{1}{2}\frac{\rm d}{{\rm d}t}\|\rho_k*u\|_{H^1}^2&=\int_\mathbb{R}\left[(\rho_k*u_x)(\rho_k*uvm)+(\rho_k*u)(\rho_kuv_xm)-2(\rho_k*u)(\rho_k*u_xvm)\right]{\rm d}x\\
&=G_k(t).
\end{aligned}
\end{equation}
Note that $u,v\in L^\infty((0,T);W^{1,\infty})$, $m,n\in L^\infty((0,T);L^1))$. Thus we also have $u,v\in L^\infty((0,T);W^{2,1}\hookrightarrow C_B^1)$, and for $t\in \mathbb{R}_+$, thanks to Lemma 2.5 we have
$$\lim_{k\rightarrow \infty}G_k(t)=\int_\mathbb{R}(u^2v_xm-uu_xvm){\rm d}\tau=G(t).$$
On the other hand, with H\"{o}lder inequality and Young's inequality, it is easy to check $|G_k(t)| \le C_T$ for $t\in (0,T)$, $k \in \mathbb{N}^*$. Thus integrating (8) with respect to $t$ and letting $k\rightarrow \infty$, we arrive at
$$\frac{1}{2}\|u(t,\cdot)\|_{H^1}^2-\frac{1}{2}\|u_0\|_{H^1}^2=\int_0^tG(\tau){\rm d}\tau.$$
This proves $u \in C(\mathbb{R}_+;H^1)$. With a similar argument, we also have $v \in C(\mathbb{R}_+;H^1)$. Because $u,v\in H^1((0,T)\times \mathbb{R})$, by Lemma 2.4 and (3), it is clear that $u,v\in C^1(\mathbb{R}_+;L^2)$.

\textbf{Step 4.} Finally, we prove the weak solution $(m,n)$ is unique. Suppose $(m_i,n_i)$ are two weak solutions of 2NS in $L_{loc}^\infty(\mathbb{R}_+;L^1\cap L^2)$, $i=1,2$, and the corresponding functions $(u_i,v_i)=(g*m_i,g*n_i) \in C(\mathbb{R}_+;H^1)\cap C^1(\mathbb{R}_+;L^2)$. Define the error of two solutions by $M=m_1-m_2$, $N=n_1-n_2$, $U=u_1-u_2$ and $V=v_1-v_2$, and we denote for any fixed $T>0$,
\begin{equation}
L=\sup_{t \in (0,T),i=1,2}\{\|m_i(t,\cdot)\|_{L^1\cap L^2}+\|n_i(t,\cdot)\|_{L^1\cap L^2}\},
\end{equation}
and with Young's inequality, we have
\begin{equation}
\|u_i(t,\cdot)\|_{W^{1,1}\cap W^{1,\infty}} \le \|m_i(t,\cdot)\|_{L^1} \le L,\ \ \ \|v_i(t,\cdot)\|_{W^{1,1}\cap W^{1,\infty}} \le \|n_i(t,\cdot)\|_{L^1} \le L.
\end{equation}

Substituting $u_i$, $v_i$, $m_i$ and $n_i$ into 2NS, it is easy to derive that the error terms $U$ and $U_x$ satisfy the following evolution equations
\begin{equation}
\begin{aligned}
&& U_t&+Uu_{1x}v_1+u_2U_xv_1+u_xu_{2x}V+g*\Big(3Uu_{1x}v_1+3u_2U_xv_1+3u_2u_{2x}V+(u_{1x}+u_{2x})U_xv_{1x}\\
&& &+u_{2x}^2V_x+\underline{U(u_1-m_1)v_{1x}}-u_{2x}U_xv_{1x}+u_2U_x(n_1-v_1)+u_2(u_2-m_2)V_x\Big)\\
&& &+g_x*\Big(Uu_{1x}v_{1x}+2\underline{u_2U_xv_{1x}}+u_xu_{2x}V_x\Big)=0,
\end{aligned}
\end{equation}

\begin{equation}
\begin{aligned}
&& U_{xt}&+U_xu_{1x}v_1+U(u_1-m_1)v_1+u_{2x}U_xv_1+\underline{u_2U_{xx}v_1}-u_2U_xv_{1x}+u_{2x}^2V+\underline{u_2(u_2-m_2)V}+g_x*\\
&& &\Big(3Uu_{1x}v_1+3u_2U_xv_1+3u_2u_{2x}V+(u_{1x}+u_{2x})U_xv_{1x}+u_{2x}^2V_x+U(u_1-m_1)v_{1x}-u_2U_xv_{1x}\\
&& &+u_2U_xv_{1x}+u_2U_x(n_1-v_1)+u_2(u_2-m_2)V_x\Big)+g*\Big(Uu_{1x}v_{1x}+2u_2U_xv_{1x}+u_2u_{2x}V_x\Big)=0.\\
\end{aligned}
\end{equation}

Our goal is to construct the following differential inequality
\begin{equation}
\begin{aligned}
&\frac{\rm d}{{\rm d}t}\|\rho_k*U\|_{L^2}+\frac{\rm d}{{\rm d}t}\|\rho_k*U_x\|_{L^2}+\frac{\rm d}{{\rm d}t}\|\rho_k*V\|_{L^2}+\frac{\rm d}{{\rm d}t}\|\rho_k*V_x\|_{L^2}\\
\le &C\left(\|\rho_k*U\|_{L^2}+\|\rho_k*U_x\|_{L^2}+\|\rho_k*V\|_{L^2}+\|\rho_k*V_x\|_{L^2}\right)+R_k(t),
\end{aligned}
\end{equation}
where
\begin{equation}
R_k(t)\rightarrow 0\ \ \  {\rm as}\ \ \  k\rightarrow \infty, \ \ \ {\rm  and}\ \ \  |R_k(t)| \le C_K,\ \ \  \forall t \in (0,T), k \in \mathbb{N}^*.
\end{equation}

In order to obtain (13), we should convolute (11) and (12) with $\rho_k$ and take inner product with $|\rho_k*U|{\rm sgn}(\rho_k*U)$ and $|\rho_k*U_x|{\rm sgn}(\rho_k*U_x)$ respectively. Due to the estimates are huge and cumbersome, here we only offer details for some special terms which are underlined.

Line 1:
\begin{equation*}
\begin{aligned}
&\int_\mathbb{R}|\rho_k*U|{\rm sgn}(\rho_k*U)\rho_k*g*(Um_1v_{1x}){\rm d}x\\
\le &\|\rho_k*U\|_{L^2}\|\rho_k*g*(Um_1v_{1x})\|_{L^2}\\
\le &\|\rho_k*U\|_{L^2}\|(\rho_k*U)(\rho_k*m_1)(\rho_k*v_{1x})\|_{L^1}+R_k(t)\\
\le &\|\rho_k*U\|_{L^2}\|\rho_k*U\|_{L^2}\|\rho_k*m_1\|_{L^2}\|\rho_k*v_{1x}\|_{L^\infty}+R^1_k(t)\\
\le &C\|\rho_k*U\|_{L^2}^2+R_k(t),
\end{aligned}
\end{equation*}
where
$$R^1_k(t)=\|\rho_k*U\|_{L^2}\left(\|\rho_k*(Um_1v_{1x})\|_{L^1}-\|(\rho_k*U)(\rho_k*m_1)(\rho_k*v_{1x})\|_{L^1}\right).$$
As $m_1 \in L^1$ and $U,v_1 \in W^{1,\infty}$, we see $R^1_k$ satisfies (14) via Lemma 2.5.

Line 2:
\begin{equation*}
\begin{aligned}
&\int_\mathbb{R}|\rho_k*U|{\rm sgn}(\rho_k*U)\rho_k*g_x*(u_2U_xv_{1x}){\rm d}x\\
\le &\|\rho_k*U\|_{L^2}\|\rho_k*(U_xu_2v_{1x})\|_{L^2}\\
=&\|\rho_k*U\|_{L^2}\|(\rho_k*U_x)(\rho_k*u_2v_{1x})\|_{L^2}+R^2_k(t)\\
\le &\|\rho_k*U\|_{L^2}\|\rho_k*U_x\|_{L^2}\|\rho_k*u_2\|_{L^\infty}\|\rho_k*v_{1x}\|_{L^\infty}+R^2_k(t)\\
\le &C\|\rho_k*U\|_{L^2}\|\rho_k*U_x\|_{L^2}+R^2_k(t),
\end{aligned}
\end{equation*}
with
$$R^2_k(t)=\|\rho_k*U\|_{L^2}\left(\|\rho_k*(U_xu_2v_{1x})\|_{L^2}-\|(\rho_k*U_x)(\rho_k*u_2)(\rho_k*v_{1x})\|_{L^2}\right),$$
which is also tends to zero as $k \rightarrow \infty$ by virtue of $U,u_1,v_1 \in H^1$.

Line 3:
\begin{equation*}
\begin{aligned}
&\int_\mathbb{R}|\rho_k*U_x|{\rm sgn}(\rho_k*U_x)\rho_k*(U_{xx}u_2v_1){\rm d}x\\
=&\int_\mathbb{R}|\rho_k*U_x|{\rm sgn}(\rho_k*U_x)(\rho_k*U_{xx}(\rho_k*u_2)(\rho_k*v_1){\rm d}x+R^3_k(t)\\
=&\frac{1}{2}\int_\mathbb{R}\frac{\partial}{\partial x}|\rho_k*U_x|^2(\rho_x*u_2)(\rho_k*v_1){\rm d}x+R^3_k(t)\\
=&-\frac{1}{2}\int_\mathbb{R}|\rho_k*U_x|^2\left[(\rho_k*u_{2x})(\rho_k*v_1)+(\rho_k*u_2)(\rho_k*v_{1x})\right]{\rm d}x+R^3_k(t)\\
\le &\frac{1}{2}\int_\mathbb{R}|\rho_k*U_x|^2\|\rho_k*u_2\|_{W^{1,\infty}}\|\rho_k*v_1\|_{W^{1,\infty}}+R^3_k(t),
\end{aligned}
\end{equation*}
with
$$R^3_k(t)=\int_\mathbb{R}|\rho_k*U_x|{\rm sgn}(\rho_k*U_x)\left[\rho_k*(U_{xx}u_2v_1)-(\rho_k*U_{xx})(\rho_k*u_2)(\rho_k*v_1)\right]{\rm d}x.$$

Line 4:
\begin{equation*}
\begin{aligned}
&\int_\mathbb{R}|\rho_k*U_x|{\rm sgn}(\rho_k*U_x)\rho_k*(u_2m_2V){\rm d}x\\
\le &\|\rho_k*U_x\|_{L^2}\|\rho_k*(u_2m_2V)\|_{L^2}\\
=& \|\rho_k*U_x\|_{L^2}\|(\rho_k*u_2)(\rho_k*m_2)(\rho_k*V)\|_{L^2}+R^4_k(t)\\
\le &\|\rho_k*U_x\|_{L^2}\|\rho_k*u_2\|_{L^\infty}\|\rho_k*m_2\|_{L^2}\|\rho_k*V\|_{L^\infty}+R^4_k(t)\\
\le &C\|\rho_k*U_x\|_{L^2}\|\rho_k*V\|_{H^1}+R^4_k(t),
\end{aligned}
\end{equation*}
with
$$R^4_k(t)=\|\rho_k*U_x\|_{L^2}\left(\|\rho_k*(u_2m_2V)\|_{L^2}-\|(\rho_k*u_2)(\rho_k*m_2)(\rho_k*V)\|_{L^2}\right).$$

For $V$ and $V_x$, one can following analogous estimates as above to check that both $\dfrac{\rm d}{{\rm d}x} \|\rho_k*V\|_{L^2}$ and
$\dfrac{\rm d}{{\rm d}x} \|\rho_k*V_x\|_{L^2}$ can be bounded by the right hand side of (13). Denote
$$A_k(t)=\|\rho_k*U\|_{L^2}+\|\rho_k*U_x\|_{L^2}+\|\rho_k*V\|_{L^2}+\|\rho_k*V_x\|_{L^2},$$
applying Young's inequality to (13) yields

\begin{equation*}
\begin{aligned}
A_k(t)=e^{Ct}\left(A_k(0)+\int_0^te^{-Cs}R_k(s){\rm d}s\right).
\end{aligned}
\end{equation*}
Since $R_k(t)$ satisfies (14) and $A_k(0)=0$, the uniqueness is obtained by letting $k \rightarrow \infty$.

\end{proof}

\begin{remark}
In the proof of Theorem 3.1, in order to obtain (5), we considered the total variation of $u_x^{k_j}$ and $v_x^{k_j}$, this is the reason we have to restrict $m_0,n_0 \in L^1$, and the uniqueness is guaranteed since our assumption is $m_0$ and $n_0$ also belong to $L^2$, and actually $p=2$ is the critical value of $L^p$ ($p \ge 2$) such that we can deduce the uniqueness of global weak solutions. More details can be found in the reference [22].
\end{remark}

\begin{remark}
As we have shown in introduction, 2NS admits peakon and periodic peakon solutions, and it is obvious $m_c(t,x)=n_c(t,x)=(1-\partial_x^2)\sqrt{c}e^{-|x-ct|}$ are not $L^p$ integrable function. Thus, how to weaken the conditions of $m_0$ and $n_0$ is still an unsolved problem.
\end{remark}

config: arxiv


TITLE: homogeneous first order differential equation
QUESTION [3 upvotes]: is there a method to solve $$\dfrac{dy}{dx} = f(x,y)$$, where $f(x,y)$ is a homogeneous function. I found some examples like $f(x,y)=(x+y)^2$ where it can be solved after converting it to Ricatti's equation.
thanks

REPLY [2 votes]: No as far as I know. There are certain particular cases that can be solved.

$f(x,y)$ homogeneous of degree $1$: change of variables $y=x\,u$.
$f(x,y) = g(a\,x+b\,y)$: change of variables $a\,x+b\,y=u$.
$f(x,y)=x^a\,y^b$: separeted variables.

But the equation $y'=x^a+y^a$, $a\ne0$, has an analytic solution (again afaik) only for $a=1$ and $a=2$ (in this last case in terms of Bessel functions.)


TITLE: Show that every closed subset of a metric space is the intersection of a countable number of open sets.
QUESTION [0 upvotes]: Show that every closed subset of a metric space is the intersection of a countable number of open sets.
This is Armstrong's Basic Topology Exercise 2.30 and it's following the section on Tietze extension theorem. I'm thinking about using the Tietze extension theorem, which states:

Any real-valued continuous function defined on a closed subset of a metric space can be extended over the whole space.

But I have no clue how to construct the open sets that we need to intersect...so maybe I'm not on the right track. Any hint please (to the question itself, not necessarily using Tietze extension theorem?)

REPLY [0 votes]: Hint: For each $x\in C$, the closed set, consider the coverings $\mathcal{A_n}=\{B(x,1/n)\}$.


TITLE: Present value of varying annuities
QUESTION [1 upvotes]: I need help with the following questions. In order to answer these questions, I am only allowed to use two of the tables of values for $\ddot a_n , \bar a_n$ (continuous annuity) and   $(I \ddot a)_n$ from n=1,...,30 and i=3%.
(h) Given an effective annual rate of 3%, calculate the present value at time 0 of a 15-year arithmetically increasing annuity immediate whereby the first annual payment is 2,452
and subsequent annual increment is 450.
$PV_0 = 2002 \ddot a_{15} + 450 (I \ddot a)_{15}$
$ = 2002*12.29607 + 450*91.60627$
=$65,839.55
(i) Consider a stream of 30-year continuous cash flows with a payment rate of 404t during year t. For instance, the payment rate during year 5 is $2,020. Calculate its present value
at time 0 given an effective annual rate of 3%.
$PV_0 = 404(I \bar a)_{30} + 404 \bar a_{30}$
$= 404(I \ddot a)_{30}* \frac{d}{\delta} + 404 \bar a_{30}$
$=404*268.7906* \frac {1-(1.03)^{-1}}{ln(1.03)} +404*19.893$
=$115,038.96
Any help is appreciated, especially for i) as I'm not too sure about that question.
Thanks

REPLY [1 votes]: You need to write out the cash flows and set up the equation of value.
For (h), an increasing annuity-immediate has the first payment of $2452$ at time $t = 1$, so the equation of value is $$PV = 2452 v + 2902 v^2 + \cdots + 8752 v^{15}$$ where $v = 1/(1+i)$ and $i = 0.03$.  Using your line of reasoning, this would be $$PV = 2002(v + v^2 + \cdots + v^{15}) + 450(v + 2v^2 + 3v^3 + \cdots 15v^{15}).$$  Therefore, the correct equation in actuarial notation would need to be $$PV = v\left(2002 \ddot a_{\overline{15}\rceil i} + 450 (I \ddot a)_{\overline{15}\rceil i}\right).$$
For (i), the equation of value is more difficult to express.  Consider the cash flow in year $t$ alone.  This corresponds to
$$404t {\bar a}_{\overline{1} \rceil i}$$
before discounting for the year in which the payment is made.  In other words, we are only taking the present value of the payments in year $t$ to time $t-1$.  Now if we discount for the year, this is an extra factor of $v^{t-1}$.  So the cash flow is
$$PV = 404 {\bar a}_{\overline{1}\rceil i} + 808 v {\bar a}_{\overline{1}\rceil i} + \cdots + 12120 v^{29} {\bar a}_{\overline{1}\rceil i}.$$  Factoring, we get $$PV = 404 {\bar a}_{\overline{1}\rceil i} \left( 1 + 2v + 3v^2 + \cdots + 30v^{29} \right) = 404 {\bar a}_{\overline{1}\rceil i} (I\ddot a)_{\overline{30}\rceil i}.$$


TITLE: Clarification for two different methods on showing that a surface is smooth
QUESTION [0 upvotes]: Let $S = \{(x,y,z) \in \mathbf{R}^3 \mid 4-4x^2-2y^2, x^2+y^2 < 1 \}$. Show that $S$ is a smooth surface.

I'm going to propose two methods that I was taught on how to show that a surface is smooth.
$(1)$ A surface is smooth if all the partials are $C^1$ and the normal vector is non zero at each point i.e there exists a tangent plane at each point on the surface for which the normal is non zero.
Parameterizing $S$ as $\varphi(x,y) =(x,y,4-4x^2-2y^2)$ I have that $$D\varphi(x,y) = \begin{bmatrix} 1 && 0 \\ 0 && 1 \\ -8x && -4y \end{bmatrix}$$
from which it can be seen that the partials exist and are continuous.
Now I was thinking that it would satisfy to check whether the Jacobian determinant is non-singular to satisfy the condition for the normals, but the Jacobian determinant cannot be computed for non-square matrices so I guess I have to compute the cross product of the partials.
Computing $\frac{\partial \varphi}{\partial x} \times \frac{\partial \varphi}{\partial y} = (8x,4y,1) \ne0$. Which would imply that $S$ is smooth.
$(2)$ The surface is a smooth $2$-dimensional surface if rank$(D\varphi(x,y)) = 2.$
This can be clearly seen from $D\varphi(x,y)$ for example if I reduce the matrix to echelon form I have that $$D\varphi(x,y) = \begin{bmatrix} 1 && 0 \\ 0 && 1 \\ 0 && 0 \end{bmatrix}$$
so the rank is clearly $2$ and $S$ is a smooth $2$-dimensional surface.
What I would like to understand is how do these two statements coincide. So how are they equivalent? And also If someone could give me an idea of why the idea of looking at the Jacobian went wrong that would be appreciated.

REPLY [1 votes]: $\frac{\partial \varphi}{\partial x} \times \frac{\partial \varphi}{\partial y}$ is non zero means that the  basis vectors are linearly independent so u get a tangent palne which is  $\mathbf{R}^2$.
And when u compute the Jacobian and find it is having rank 2 (where u calculated) it means that the linear transformation  : U -> $\mathbf{R}^2$.
So both are same.
additional ref. Bar page 94


TITLE: Understanding the definition of the fraction ring of a ring that is not a domain.
QUESTION [0 upvotes]: Let $R$ a ring and $S$ subset of $R$ that is closed under the multiplication. We denote $$S^{-1}R=(R\times S)/_\sim,$$
where $$(r,s)\sim(r',s')\iff \exists s''\in S: s''(rs'-r's)=0.$$
This equivalence relation $\sim$ looks weird to me. I tried to apply it an an example, for example $R=\mathbb Z/6\mathbb Z$ and $S=2\mathbb Z/6\mathbb Z$. How does it will work here ? I don't see it ! May be you have an easier example ?

REPLY [2 votes]: It's very simple: the multiplicative subset $S$ contains $0$,  since it's the ideal $2\mathbf Z/6\mathbf Z$. In this case, any $(r,s)\sim~(0,1)$, so $S^{-1}R$ is the null ring.
That's why, in the construction of rings of fractions, one usually requires the multiplicative subset does not contain $0$.
But you may localise at the prime ideal $2\mathbf Z/6\mathbf Z$. In this case, you can check $(\mathbf Z/6\mathbf Z)_{2\mathbf Z/6\mathbf Z}$ is isomorphic with the field $\mathbf Z/2\mathbf Z$, and the canonical map
\begin{align}
\mathbf Z/6\mathbf Z&\longrightarrow (\mathbf Z/6\mathbf Z)_{2\mathbf Z/6\mathbf Z}\\
\bar x&\longmapsto\frac{\bar x}{\bar 1}
\end{align}
is not injective: $\; \bar 0,\,\bar 2,\,\bar4$ map to $\dfrac{\bar 0}{\bar 1}$ and $\;\bar1,\,\bar 3,\,\bar 5$ map to $\;\dfrac{\bar1}{\bar1}$.